Exterior algebras, wedge products, and all that...
Edward Green
the body of "mathematical methods for physicists" which seems to have
been codified about 1905.
Sometimes I notice that authors writing about the wedge product talk
of forms. A "form", as far as I can tell, is a funny new-math name
for a co-vector (like a "Bra"), or linear function on the vectors into
the scalars. An n-form is a device like that which eats n vectors at
once and spits out a scalar. Obscure looking rules are given for
combining a p form and a q form -- if I have this correctly -- to make
a brand new p+q form, and that rule is the "wedge product".
However, sometimes I see the mischievous authors combining vectors via
the wedge product. Of course the multi-vectors and the n-forms are
pretty much dual, or something like that: maybe that's what allows us
to act like we didn't give a damn whether we wedge some old forms or
vectors.
"Category theory" would be another part of the new math, though, as
far as I can see, it's simply more of the old math. I mean, what is
math but abstracting common properties from a group of objects... I
should probably say "class" ... to from, say "groups". Then we can
reason about the abstraction instead of all the little exemplars. The
only thing vaguely new in category theory is possibly adding another
layer of abstraction. But that's what math does. It's not some
goddamn new world. The New Practitioners always want to act as if
they were doing something fundamentally new in type, but they seldom
are.
Anyway, if we are making an exterior algebra thingie, do we have to
choose at the outset whether we are going to eat vectors or forms, or
can we take some of each?
Tom Roberts
> Anyway, if we are making an exterior algebra thingie, do we have to
> choose at the outset whether we are going to eat vectors or forms, or
> can we take some of each?
Note, please, that forms are necessarily fields on a manifold, but
vectors and covectors need not be. So you're probably mixing types when
you say "vectors or forms" (but not necessarily, as "vector fields or
forms" does not mix types in that manner, and people often omit "fields"
in this context).
In the abstract, covectors are dual to vectors -- there is NOTHING to
tell you what vectors are, just that they are elements of a vector space
satisfying certain relationships in their definition. So you can
interchange vectors with covectors and all required relationships still
hold.
But in physics we normally use these objects in conjunction with a
manifold, using the tangent space at each point of the manifold as the
vector space which defines vectors (really vector fields on the
manifold, but we conventionally call these "vectors", potentially
confusing the neophyte). Now there is no ambiguity -- vectors are
members of the tangent spaces and covectors are their dual (members of
the cotangent spaces). This makes vectors behave under a diffeomorphism
of the manifold via push-forward, and covectors behave via a pull-back.
Since functions also behave via pull-back, the abstract notation for a
vector is v and for a covector is w(.) (the "." indicating an
unspecified argument, here of type vector) -- the dependence on point in
the manifold is generally omitted.
Derivatives also behave via pull-back under a diffeomorphism, and that
is what makes differential forms necessarily be covector fields on a
manifold (i.e. not vector fields). The antisymmetry of the wedge product
is essential -- that is what makes the connection terms cancel when one
takes the antisymmetrized covariant derivative of a form (this is the
exterior derivative).
BTW I have never seen wedge products of vectors; there are good reasons
why wedge products are natural for forms, but they don't hold for
vectors. In particular, when integrating on a manifold the integrand is
essentially always a form (or the integral is unlikely to make sense).
Tom Roberts
Uncle Ben
The only place I have seen this stuff used is in MTW's "Gravitation."
They seem to think it is the cat's pajamas. I'm glad to see that it
is in safe hands. Don't try this at home, kiddes!
Uncle Ben
Uncle Ben
Rock Brentwood
> This seems to be part of the "new math", which means it is not part of
> the body of "mathematical methods for physicists" which seems to have
> been codified about 1905.
Maxwell made use of Grassman algebra in his treatise and essentially
wrote everything in the language of differential forms. The reason
this isn't recognized clearly as such these days is because
(1) all the differential forms were kept under integral signs
(2) 19th century language was used: "2-form" was called a "quantity
referred to a surface", "1-form" a "quantity referred to a curve" or
line; "3-form" as a "quantity referred to a volume".
The hype about his use of vectors is largely a red-herring
distraction. Much greater use was made of differential forms.
By "used Grassman algebra" I mean that he explicitly cited and used
the rule dx dy = -dy dx in at least one point in the treatise.
There was also the extensive discussion in Chapter 1 of what is
essentially just the Stokes' theorem; plus a 19th-century-language
discussion of the related issues of (co)homology -- referred to as
"cyclosis".
Edward Green
> Note, please, that forms are necessarily fields on a manifold, but
> vectors and covectors need not be. So you're probably mixing types when
> you say "vectors or forms" (but not necessarily, as "vector fields or
> forms" does not mix types in that manner, and people often omit "fields"
> in this context).
I had reached about that conclusion, although I believe I have also
seen "form" in contexts where no field was necessary or implied. I'll
take it that "form" generally implies a field, while "covector" does
not, but this is not ironclad.
> In the abstract, covectors are dual to vectors -- there is NOTHING to
> tell you what vectors are, just that they are elements of a vector space
> satisfying certain relationships in their definition. So you can
> interchange vectors with covectors and all required relationships still
> hold.
So if we can make sense of a certain structure over covectors, we can
definitely make sense of an isomorphic structure over the vectors;
though it's an open question if we can make sense of mixed cases (a
wedge product of a vector with a covector).
> But in physics we normally use these objects in conjunction with a
> manifold, using the tangent space at each point of the manifold as the
> vector space which defines vectors (really vector fields on the
> manifold, but we conventionally call these "vectors", potentially
> confusing the neophyte). Now there is no ambiguity -- vectors are
> members of the tangent spaces and covectors are their dual (members of
> the cotangent spaces).
Ok.
> This makes vectors behave under a diffeomorphism
> of the manifold via push-forward, and covectors behave via a pull-back.
> Since functions also behave via pull-back, the abstract notation for a
> vector is v and for a covector is w(.) (the "." indicating an
> unspecified argument, here of type vector) -- the dependence on point in
> the manifold is generally omitted.
Grokking "push-forward" and "pull-back" is on my short list of things
to do, mathematically (when I've tried so far, I've gotten lost in the
mists of category theory). Perhaps it's as simple as noting that if a
vector transforms via Tv, then a covector must transform via wT^(-1)
to preserve the inner product invariant?
> Derivatives also behave via pull-back under a diffeomorphism, and that
> is what makes differential forms necessarily be covector fields on a
> manifold (i.e. not vector fields). The antisymmetry of the wedge product
> is essential -- that is what makes the connection terms cancel when one
> takes the antisymmetrized covariant derivative of a form (this is the
> exterior derivative).
Hmm... I'm not quite ready to understand that, but I'm getting closer.
> BTW I have never seen wedge products of vectors;...
If you happen to have a copy of "Lie Groups, Lie Algebras, and Some of
Their Applications" handy (Robert Gilmore, Dover reprint, 2005), you
will find on pages 30 and 36 what certainly look like wedge products
of basis vectors (not covectors)
Also, in
http://en.wikipedia.org/wiki/Exterior_algebra
one finds that wedge products live in spaces "containing V [the vector
space] as a subspace" -- N.B., not "the dual space to V". However
rereading down a little further, I also find:
"The exterior algebra is one example of a bialgebra, meaning that its
dual space also possesses a product, and this dual product is
compatible with the wedge product. This dual algebra is precisely the
algebra of alternating multilinear forms on V, and the pairing between
the exterior algebra and its dual is given by the interior product."
Well, aha! That exactly answers my question, even though I don't
follow the last assertion. (Attempts to grok down the "interior
product" are not immediately successful. It is a prickly and
apparently not widely used concept).
I also see in Gilmore that there is a symmetrized product of two
(co)vectors, as well as an anti-symmetrized product, written as a
downward pointing wedge: u \/ v vs. u /\ v (not apparently,
unfortunately, the "interior product").
> there are good reasons
> why wedge products are natural for forms, but they don't hold for
> vectors. In particular, when integrating on a manifold the integrand is
> essentially always a form (or the integral is unlikely to make sense).
Thank you for your response.
Pmb
"Edward Green" <spamsp...@netzero.com> wrote in message
news:edb76104-f438-4529...@n33g2000pri.googlegroups.com...
> On Jul 24, 8:29 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
>> Note, please, that forms are necessarily fields on a manifold, but
>> vectors and covectors need not be. So you're probably mixing types when
>> you say "vectors or forms" (but not necessarily, as "vector fields or
>> forms" does not mix types in that manner, and people often omit "fields"
>> in this context).
>
> I had reached about that conclusion, although I believe I have also
> seen "form" in contexts where no field was necessary or implied. I'll
> take it that "form" generally implies a field, while "covector" does
> not, but this is not ironclad.
I'm not sure what you mean here? Do you believe that all forms are fields
like Tom Roberts claims? If so then why? Consider the 1-form which is the
dual to 4-momentum. 4-momentum is not a field so in what sense do you
believe that the 1-form dual to 4-momentum is a field?
>> In the abstract, covectors are dual to vectors -- there is NOTHING to
>> tell you what vectors are, just that they are elements of a vector space
>> satisfying certain relationships in their definition.
That certainly is not true as evidenced by any GR text or differential
geometry/tensor analysis text.
Pete
Tom Roberts
> On Jul 24, 8:29 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> Note, please, that forms are necessarily fields on a manifold, but
>> vectors and covectors need not be. So you're probably mixing types when
>> you say "vectors or forms" (but not necessarily, as "vector fields or
>> forms" does not mix types in that manner, and people often omit "fields"
>> in this context).
>
> I had reached about that conclusion, although I believe I have also
> seen "form" in contexts where no field was necessary or implied. I'll
> take it that "form" generally implies a field, while "covector" does
> not, but this is not ironclad.
Be careful to avoid the PUN on "field" between some physicists and
mathematicians: to some physicists "field" is a physical quantity, while
to mathematicians the correct usage (here) is "field on a manifold", and
no relationship to any physical quantity is implied -- this just means a
function of position on the manifold (and ONLY of position). I used it
in the second sense; I'm not sure which sense you are using the word.
Mathematicians also use "field" in quite different way,
in the sense of a commutative division algebra (e.g. the
reals). And to illustrate the complex interrelationships,
puns, and potential confusions, I'll point out that a
vector space requires a field (comm. div. algebra) for
its definition -- so a vector field directly involves
"field" with two rather different meanings....
> Grokking "push-forward" and "pull-back" is on my short list of things
> to do, mathematically (when I've tried so far, I've gotten lost in the
> mists of category theory). Perhaps it's as simple as noting that if a
> vector transforms via Tv, then a covector must transform via wT^(-1)
> to preserve the inner product invariant?
These are basic operations, based on mappings. Given a mapping from A to
B [@], if you have a function on A you have no idea how to apply it to B
[#], but if you have a function on B it's clear how to apply it to A --
just map from A to B and then apply the function. That is a pull-back
because it goes backwards from the mapping (you "pulled" the function
from B to A via the mapping, which doesn't actually map functions at
all). If your mapping is continuous, then if you have a directed
interval on A, you know how to map the directed interval to B -- simply
map each endpoint and declare their images to be the endpoints of the
mapped interval (can have difficulties for finite intervals, but
infinitesimal ones are OK and more interesting as they are essentially
tangent vectors). That is a push-forward, as it goes the same direction
as the mapping (you "pushed" the directed interval from A to B via the
mapping, which doesn't actually map directed intervals at all). This
applies to the tangent space of a manifold, so vectors push-forward
naturally under such a continuous mapping (and a diffeomorphism is such
a mapping, of course).
[@] I don't say what A and B are; think of them as regions
of a manifold, but that can be relaxed and the concepts are
more general.
[#] unless you have an inverse mapping. The basic concepts
apply to non-invertible mappings, but the useful mappings
in physics (diffeomorphisms) are invertible.
The book that brought this all into focus for me is:
Baez and Muniain, _Gauge_Fields,_Knots,_and_Gravity_.
I recommend it highly.
Tom Roberts
schoenf...@gmail.com
[...]
> BTW I have never seen wedge products of vectors;
[1] The wedge product of two vectors is a bivector.
[2] Bivectors (and other blades) are used extensively in physics (see
Faraday bivector).
[3] The cross-product is the dual of a bivector -- A x B = dual( A /\
B)
> there are good reasons why wedge products are natural for forms,
The wedge product goes far beyond differential forms.
> but they don't hold for
> vectors.
That's incorrect - the wedge product is well-defined for vectors.
[...]
Daryl McCullough
>Grokking "push-forward" and "pull-back" is on my short list of things
>to do, mathematically (when I've tried so far, I've gotten lost in the
>mists of category theory). Perhaps it's as simple as noting that if a
>vector transforms via Tv, then a covector must transform via wT^(-1)
>to preserve the inner product invariant?
I think that's pretty much it. But here's a way to see this
"forward and back" business mathematically. If you have a
path P(s) through spacetime where s is proper time (for instance),
then P is a function from R (the reals) to M (the manifold, which
is spacetime in this case). A scalar field, such as "the temperature
at location X" is a function going the other way, from M to R.
A tangent vector is a linear approximation to a path (at a point),
while a cotangent vector is a linear approximation to a scalar
field.
A coordinate system is a map from one manifold (spacetime) to
another (R^4, the set of 4-tuples of reals). If P(s) is a
path on M, and m is a map from M to R^4, then you can compose
them to get a map on R^4: P'(s) = m(P(s)) is a function from R to R^4.
Scalar fields work just the opposite. If Phi'(X') is a scalar
field on R^4 (a function from R^4 to R), and m is a map
from M to R^4, then you can compose them to get a scalar
field on M (a function from M to R):
Phi(X) = Phi'(m(X))
So a mapping from M to R^4 can be used to convert a path on M
to a path on R^4, or it can be used to convert a scalar field
on R^4 to a scalar field on M. So scalar fields map in the
opposite direction from paths. (And since covectors and
tangent vectors are derivatives of scalar fields and paths,
it follows that covectors map in the opposite direction from
tangent vectors).
--
Daryl McCullough
Ithaca, NY
Edward Green
wrote:
> ... here's a way to see this
> "forward and back" business mathematically. If you have a
> path P(s) through spacetime where s is proper time (for instance),
> then P is a function from R (the reals) to M (the manifold, which
> is spacetime in this case). A scalar field, such as "the temperature
> at location X" is a function going the other way, from M to R.
> A tangent vector is a linear approximation to a path (at a point),
> while a cotangent vector is a linear approximation to a scalar
> field.
>
> A coordinate system is a map from one manifold (spacetime) to
> another (R^4, the set of 4-tuples of reals). If P(s) is a
> path on M, and m is a map from M to R^4, then you can compose
> them to get a map on R^4: P'(s) = m(P(s)) is a function from R to R^4.
>
> Scalar fields work just the opposite. If Phi'(X') is a scalar
> field on R^4 (a function from R^4 to R), and m is a map
> from M to R^4, then you can compose them to get a scalar
> field on M (a function from M to R):
>
> Phi(X) = Phi'(m(X))
>
> So a mapping from M to R^4 can be used to convert a path on M
> to a path on R^4, or it can be used to convert a scalar field
> on R^4 to a scalar field on M. So scalar fields map in the
> opposite direction from paths. (And since covectors and
> tangent vectors are derivatives of scalar fields and paths,
> it follows that covectors map in the opposite direction from
> tangent vectors).
Odd. I can follow every step in your argument, though it doesn't seem
to have an epiphanic result all the same. :-) I do see that a
function from M to R^4 (coordinate system) can either be applied after
a map from R to M (path), resulting in a path in R^4, or else _before_
a map from R^4 to R, thus resulting in a scalar field on M. So scalar
fields and paths move in opposite direction under the influence of the
same map -- and I notice in this case we didn't say anything about the
map being invertible.
So let me take a wild guess: the coordinate map m(X) "pushes forward"
the path, since the path travels in the same direction as the function
m (from M to R^4), or else "pulls back" the scalar field, which
travels in the opposite direction (from R^4 to M).
In general then, shall we say, if we have a map from a space X to a
space Y, if we apply the map _after_ a preliminary map into X:
W -> X -> Y == W -> Y
we are pushing forward the first mapping into Y, whereas if we apply
the given map _before_ a secondary map on Y:
X -> Y -> Z == X -> Z
we are pulling back the latter map onto X?
I notice that in either case we simply have a row of three symbols
contracted to two symbols, so that in describing, for example,
X -> Y -> Z == X -> Z
we can apparently either say that the function Y -> Z is pushing
forward the first function into Z, or else that the function X -> Y is
pulling back the second function onto X!
Edward Green
> On Jul 24, 6:38 pm, Edward Green <spamspamsp...@netzero.com> wrote:
>
> > This seems to be part of the "new math", which means it is not part of
> > the body of "mathematical methods for physicists" which seems to have
> > been codified about 1905.
>
> Maxwell made use of Grassman algebra in his treatise and essentially
> wrote everything in the language of differential forms. The reason
> this isn't recognized clearly as such these days is because
> (1) all the differential forms were kept under integral signs
> (2) 19th century language was used: "2-form" was called a "quantity
> referred to a surface", "1-form" a "quantity referred to a curve" or
> line; "3-form" as a "quantity referred to a volume".
>
> The hype about his use of vectors is largely a red-herring
> distraction. Much greater use was made of differential forms.
>
> By "used Grassman algebra" I mean that he explicitly cited and used
> the rule dx dy = -dy dx in at least one point in the treatise.
Well, thanks for the historical insight, but I still have the opinion
that when what we are wont to call "19th century mathematical methods"
had congealed around the early part of the last century, Grassmann
algebra had largely dropped out of sight, awaiting its rediscovery in
the latter part of the century with the aura of a brave new world.
It may be more powerful and general, but power comes at a price: I
never remember struggling so hard to understand the friendly old cross
and inner products, divergence, curl, and all that. And it seems to
be be possible to do quite a bit of practical physics, especially EM,
limiting ourselves to these specialized devices. Simplification has
power as does generalization.
I did find a nice web book by John Browne at
http://www.grassmannalgebra.info/grassmannalgebra/book/#Book
In it I am delighted to learn that we must not stop with the simple
exterior product. Oh no! We must of course graduate to the interior
product. But wait, there's more! There is also the all but forgotten
regressive product! Why, we can't even _begin_ to define the interior
product until we've seen the regressive product, and something called
the complement! What are these young guns thinking, touting their
brilliant exterior product, while giving short shrift to the interior
and regressive products? It's only a piece of the puzzle -- we must
grok all of Grassmann I think, or be dilettantes.
What a can of worms.
> There was also the extensive discussion in Chapter 1 of what is
> essentially just the Stokes' theorem; plus a 19th-century-language
> discussion of the related issues of (co)homology -- referred to as
> "cyclosis".
(co)homology remains a closed book to me for now. A project for
another day.
Edward Green
> Edward Green wrote:
> > On Jul 24, 8:29 pm, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> >> Note, please, that forms are necessarily fields on a manifold, but
> >> vectors and covectors need not be. So you're probably mixing types when
> >> you say "vectors or forms" (but not necessarily, as "vector fields or
> >> forms" does not mix types in that manner, and people often omit "fields"
> >> in this context).
>
> > I had reached about that conclusion, although I believe I have also
> > seen "form" in contexts where no field was necessary or implied. I'll
> > take it that "form" generally implies a field, while "covector" does
> > not, but this is not ironclad.
>
> Be careful to avoid the PUN on "field" between some physicists and
> mathematicians: to some physicists "field" is a physical quantity, while
> to mathematicians the correct usage (here) is "field on a manifold", and
> no relationship to any physical quantity is implied -- this just means a
> function of position on the manifold (and ONLY of position). I used it
> in the second sense; I'm not sure which sense you are using the word.
Have no fear... I meant it in the mathematical sense, and not...
> Mathematicians also use "field" in quite different way,
> in the sense of a commutative division algebra (e.g. the
> reals). And to illustrate the complex interrelationships,
> puns, and potential confusions, I'll point out that a
> vector space requires a field (comm. div. algebra) for
> its definition -- so a vector field directly involves
> "field" with two rather different meanings....
...either.
> > Grokking "push-forward" and "pull-back" is on my short list of things
> > to do, mathematically (when I've tried so far, I've gotten lost in the
> > mists of category theory). Perhaps it's as simple as noting that if a
> > vector transforms via Tv, then a covector must transform via wT^(-1)
> > to preserve the inner product invariant?
>
> These are basic operations, based on mappings. Given a mapping from A to
> B [@], if you have a function on A you have no idea how to apply it to B
> [#], but if you have a function on B it's clear how to apply it to A --
> just map from A to B and then apply the function. That is a pull-back
> because it goes backwards from the mapping (you "pulled" the function
> from B to A via the mapping, which doesn't actually map functions at
> all). If your mapping is continuous, then if you have a directed
> interval on A, you know how to map the directed interval to B -- simply
> map each endpoint and declare their images to be the endpoints of the
> mapped interval (can have difficulties for finite intervals, but
> infinitesimal ones are OK and more interesting as they are essentially
> tangent vectors). That is a push-forward, as it goes the same direction
> as the mapping (you "pushed" the directed interval from A to B via the
> mapping, which doesn't actually map directed intervals at all). This
> applies to the tangent space of a manifold, so vectors push-forward
> naturally under such a continuous mapping (and a diffeomorphism is such
> a mapping, of course).
Thank you. That nicely reenforces Daryl McCullough's description.
I've got it: at least in the simplest approximation.
> [@] I don't say what A and B are; think of them as regions
> of a manifold, but that can be relaxed and the concepts are
> more general.
>
> [#] unless you have an inverse mapping. The basic concepts
> apply to non-invertible mappings, but the useful mappings
> in physics (diffeomorphisms) are invertible.
>
> The book that brought this all into focus for me is:
>
> Baez and Muniain, _Gauge_Fields,_Knots,_and_Gravity_.
> I recommend it highly.
Ah... good old John Baez! Only $41 and change on Amazon.com for the
paperback. A bargain.
jmfbahciv
Paperback!? I have the hardcopy. IIRC, there was an errata sheet
that was very important for the hard copy (but I can't seem to
find it; I'm sure I printed it out). Double-check that the
paperback has the corrections.
/BAH
Tom Roberts
> On Jul 25, 10:29 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>> BTW I have never seen wedge products of vectors;
>
> [1] The wedge product of two vectors is a bivector.
Sure, for an appropriately defined wedge product.
> [2] Bivectors (and other blades) are used extensively in physics (see
> Faraday bivector).
Such are usually forms in disguise as vectors. Most likely in older
books that did not take a geometrical approach, and do not distinguish
vectors from covectors. In flat 3-d space the distinction can be glossed
over, unless one is rigorous (and physics books aren't).
> [3] The cross-product is the dual of a bivector -- A x B = dual( A /\
> B)
Hmmm. One generally finds that the vectors one would use in a cross
product (e.g. angular momentum) are really better modeled as covectors.
But again: in flat 3-space the distinction between vectors and covectors
can be glossed over (one must accept pseudo-vectors -- that kludge is a
GREAT BIG FLAG that something is not rigorously correct, and better
treatments have no need for them, using 2-forms instead).
>> there are good reasons why wedge products are natural for forms,
>
> The wedge product goes far beyond differential forms.
As I said, I have never seen it for vectors. But I looked in: Flanders,
_Differential_Forms_with_Applications_to_the_Physical_Sciences_. He does
not even list "wedge product" in the index. But his Chapter 2 introduces
"p-vectors" and he uses antisymmetrized products of them. Ch. 2 is
disconnected from ch 3 in which he introduces local forms. So this seems
to be one of the older books (1963) that glosses over vectors vs
covectors. In ch V where he introduces differential forms, he mentions
they transform as covectors (he writes in the older style of components
rather than tensors).
>> but they don't hold for
>> vectors.
>
> That's incorrect - the wedge product is well-defined for vectors.
Sure. But it's not "natural" for vectors in the same way it is natural
for forms (read what I wrote).
Pmb wrote:
> Do you believe that all forms are fields
> like Tom Roberts claims? If so then why?
Because being a field on the manifold is part of the definition of a
differential form, and "form" is a nearly universal shortcut for
"differential form" -- the local variety is rarely used, and AFAICT is
not useful except in constructing differential forms, which ARE fields
on the manifold.
> Consider the 1-form which is the
> dual to 4-momentum. 4-momentum is not a field so in what sense do you
> believe that the 1-form dual to 4-momentum is a field?
I think it is not useful to say "1-form" in this case, because that is a
LOCAL form, and such are not used in physics (or, I suspect, math)
except to define differential forms (and are not really needed for
that). Most people will consider your usage to refer to a differential
form, which is wrong.
The COVECTOR dual to the 4-momentum of an object is certainly not a
field on the manifold -- both are naturally functions of the object's
proper time, not of position in the manifold.
>> In the abstract, covectors are dual to vectors -- there is NOTHING to
>> tell you what vectors are, just that they are elements of a vector space
>> satisfying certain relationships in their definition.
>
> That certainly is not true as evidenced by any GR text or differential
> geometry/tensor analysis text.
You need to distinguish vectors and covectors IN THE ABSTRACT, and in a
particular application. As I said shortly afterward, related to a
manifold the definitions of tangent vector and covector are clear and
cannot be interchanged -- both GR and differential geometry are
instances of this. But IN THE ABSTRACT, the space of covectors dual to a
given vector space is itself a vector space, so vector and covector can
be interchanged and all required properties still hold.
Tom Roberts
Timothy Golden BandTechnology.com
> On Jul 25, 10:29 am, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> > BTW I have never seen wedge products of vectors;
>
> [1] The wedge product of two vectors is a bivector.
>
> Faraday bivector).
>
> B)
>
>
> The wedge product goes far beyond differential forms.
>
> > vectors.
>
> That's incorrect - the wedge product is well-defined for vectors.
>
Can anyone tell me how to do
(1,1,1,1) /\ ( 1,2,3,4) ?
We do accept that there is an easy answer for
(1,1,1) x (1,2,3)
or
(1,1,1) /\ (1,2,3)
right? It is
( 3 - 2, 3 - 1, 2 - 1 )
= ( 1,2,1) .
In all of the talk about the exterior product I have a great level of
confusion to the point of noninstantiability. Could someone please
instantiate an exterior product in 4D for me?
- Tim
As I try to get to this exterior product I am having a hard time
finding the 4D definition.
Even just a link that explains how to do this fully would be helpful.
schoenf...@gmail.com
Try this site:
http://www.euclideanspace.com/maths/algebra/clifford/d4/index.htm
schoenf...@gmail.com
>
> - Show quoted text -
You need to extend your notation and algebra.
A bivector is the 2D generalization of a vector - a directed plane
segment - a 2-blade.
A trivector is the 3D generalization of a vector - a directed volume
segment - a 3-blade.
A k-vector is the k-D generalization of a vector - a k-blade.
A scalar is a 0-blade.
A multivector is the sum of any number of blades.
multivector = scalar + vector + bivector + ....
EXAMPLE:
A complex number is a multivector as
z = x + y i
adds a scalar and a bivector (the imaginary number is a bivector).
Multivectors are the N-dimensional generalization of a "number". A 2-D
multivector is a complex number. Past dimension 2 though, the algebra
loses the "field" structure (in mathematician sense).
So in 4D
the basis vectors are e1, e2, e3, e4
the basis bivectors are e12, e23, e34, e14, e24, e31
the basis trivectors are e123, e234, e341, e124
the basis quadvector is e1234
wedge product of basis vectors multiplies as follows
e_n /\ k = e_n ;/ k is scalar, e_n arbitrary basis vector
e_n /\ e_n = 1 ;/ wedge of a basis vector by itself is the
dimensionless number 1
e1 /\ e2 = e12 = -e21
e2 /\ e3 = e23 = -e23
e3 /\ e1 = e31 = -e13
e12 /\ e3 = e123
e12 /\ e1 = e121 = -e112 = -e2 (every flip of indices introduces
negation symbol, two adjactent and equal indices cancel).
e12 /\ e2 = e122 = e1
... etc
e1234 /\ e5 = 0 ;/ for a 4D space.
NOTE: Only on a flat manifold does the antisymmetry of the wedge
product hold -- e12 = -e21.
So to answer your question
Z = (A1,B1,C1,D1) /\ ( A2,B2,C2,D2)
= (0,A1,A2,C3,D4,0,0,0,0,0,0,0) /\ (0,B1,B2,B3,B4,0,0,0,0,0,0,0)
Now multipy that out
= (A1 e1 + A2 e2 + A3 e3 + A4 e4) * (A1 e1 + A2 e2 + A3 e3 + A4 e4)
And take the bivector component of that multivector. That will be your
answer!
schoenf...@gmail.com
CLARIFICATION: the even subalgebra of the 2D multivectors are the
complex numbers. In other words
2D Multivector = scalar + vector + bivector
setting the vector component to null gives
Complex number = scalar + bivector
> So in 4D
> the basis vectors are e1, e2, e3, e4
> the basis bivectors are e12, e23, e34, e14, e24, e31
> the basis trivectors are e123, e234, e341, e124
> the basis quadvector is e1234
>
> wedge product of basis vectors multiplies as follows
>
> e_n /\ k = e_n ;/ k is scalar, e_n arbitrary basis vector
>
> e_n /\ e_n = 1 ;/ wedge of a basis vector by itself is the
> dimensionless number 1
Correction: that should be 0, not 1.
[...]
The direction you should look at is geometric algebra from a computer
graphics point of view. The usefulness of this math is lost with the
usual "clifford algebra" approach.
Timothy Golden BandTechnology.com
Yeah, right in here is the self reference that makes this definition
bullshit.
Can you in fact instantiate the exterior product? I don't ask that you
use my example.
Please, come up with your own. The comfort that is taken with the
generalization of the cross product may be lost into the meaningless
bliss of the mathematician who has built out an empty theory.
Really, so in 4D which of the blade forms do you require? Since the
unitary form seemed to work in 3D shall we extend to a two-blade for
4D? Then can we get an actual instantiated answer from this
construction?
Shall I even make out some values?
So we'll make an m-dimensional n-blade as
(a11,a12,...a1m),(a21,a22,...a2m),(an1,an2,...anm)
So now in 4D can we ask for an actual instantiated answer of
(1.11,1.12,1.13,1.14),(1.21,1.22,1.23,1.24) /\ (2.11,2.12,2.13,2.14)
(2.21,2.22,2.23,2.24) ?
I think not since the definition of
e1 /\ e2
is not actually given (see your own notes below). Its result is
undefined.
What does it take to instantiate a 4D cross (exterior) product?
> e_n /\ e_n = 1 ;/ wedge of a basis vector by itself is the
> dimensionless number 1
>
> e1 /\ e2 = e12 = -e21
This line above seems to me the failing point. Why are we needing a
new basis vector e12 ?
What is this and why then was the old cross product not needing this
new vector space?
Have you indeed generated a new space? In some ways this is
appropriate to product definitions yet the way that it is described
within this theory is as if to hoodwink us. I'm pretty sure you meant
to say
e12 = - e21 in your notation and I do not mean to quibble over
that. It is the instantiation of an entirely new and ill defined e12
which I am questioning. If e12 cannot be resolved in terms of its
sources e1 and e2 then how can we ever hope to instantiate an answer?
The quantity of these uniquely defined new basis vectors for a 4D are
e12 e13 e14 e23 e24 e34
right? Are we admitting an additional 6D in the resultant?
Stepping back to the 3D we the same attempt yields
e12 e13 e23
but these are not defined either are they? Nicely there are only three
of them for the 3D but still assigning back to the original e1, e2, e3
is not sensible.
> e2 /\ e3 = e23 = -e23
> e3 /\ e1 = e31 = -e13
>
> e12 /\ e3 = e123
> e12 /\ e1 = e121 = -e112 = -e2 (every flip of indices introduces
> negation symbol, two adjactent and equal indices cancel).
> e12 /\ e2 = e122 = e1
> ... etc
> e1234 /\ e5 = 0 ;/ for a 4D space.
>
> NOTE: Only on a flat manifold does the antisymmetry of the wedge
> product hold -- e12 = -e21.
>
> So to answer your question
>
> Z = (A1,B1,C1,D1) /\ ( A2,B2,C2,D2)
>
> = (0,A1,A2,C3,D4,0,0,0,0,0,0,0) /\ (0,B1,B2,B3,B4,0,0,0,0,0,0,0)
Err... is this line above accurate? isn't there a leading zero that
shouldn't be there?
also aren't your trailing zeros going to null out the blades? Also
wouldn't there be just four trailing zeros for a 2-blade in 4D? Really
all of your notation here is horrible. Here I'll attempt a clean up
and further generalization without nulling out the 2-blade (assuming
that is the correct interpretation of what you are trying here)
Z = A /\ B .
A = ( a11, a12, a13, a14 ), ( a21, a22, a23, a24 ) . (ie a two-
blade in 4D)
B = ... . (symmetric to A)
>
> Now multipy that out
>
> = (A1 e1 + A2 e2 + A3 e3 + A4 e4) * (A1 e1 + A2 e2 + A3 e3 + A4 e4)
Now here it looks as if you've addressed the ill defined
e1 /\ e2
and resolved it. How did you do that? Or did your zeros do that for
you?
>
> And take the bivector component of that multivector. That will be your
> answer!
The instantiation of actual values is what I seek. These values then
generalized would form a complete definition in 4D. For instance if we
were to use values such as 1.11 for the left value of the product's
first blade's first component then we'd effectively have generalized
and without collapsing the math we'd have a clean prototype of one
type of this class of n-blades. Any unique value will suffice so long
as the math is not reduced. If it actually existed don't you think
someone would have bothered to instantiate it as such? Would this then
be the generalized cross product? Instead the need to go over to
manifolds and tangents as an abstraction leads us away from
instantiable vectors. Too much of math is in the twilight zone. Let's
get a clean instantiation of a 4D exterior product and if there are
multiple classes of it lets try a simple class but not so simple that
it shrouds the full result. I am sincerely interested in this problem
and have spent some energy on it but am not convinced of the math. The
3D cross product is instantiable on the existing basis e1,e2,e3. If
this is not true of the 4D version then are we as physicists accepting
that the product creates space? This is not necessarily a stopper,
especially since products rarely actually indicate resultants on the
existing space. Hence the desire to head for n-blades or whatever.
Here these feedback loops are conspiring into confusion for me. As I
go back to the definition it states that the resultant exists in the
space of the operands doesn't it? But then it goes on to define new
unit vectors for the resultant e.g. e12 = e1 /\ e2. So I challenge the
integrity of this math for not coming clean with its own defficiency.
- Tim
Edward Green
<tttppp...@yahoo.com> wrote:
> Can anyone tell me how to do
> (1,1,1,1) /\ ( 1,2,3,4) ?
I was beginning to wonder about things like that myself. :-)
Let me try first grope.
First, the result has 4-choose-2 = 6 components
The "basis vectors" of this 6-d space are the 6 unique combinations
e1^e2, e1^e3, e1^e4, e2^e3, e2^e4, e3^e4
where e1,e2,e3,e4 are the bases of the 4-space, and, the product being
anti-symmetric, e1^e1 = 0, e3^e2 = -e2^e3, etc.
Now, expanding and collecting terms (keeping the 6-basis in the above
order)
(1,1,1,1) /\ ( 1,2,3,4)
= [(2-1),(3-1),(4-1),(3-2),(4-2),(4-3)] = [1,2,3,1,2,1]
Look ma! I instantiated a wedge product! I instantiated a wedge
product!
Perhaps we should write this as a 4x4 antisymmetric tensor, according
to taste.
No sir, I have no bloody idea what this means.
Timothy Golden BandTechnology.com
Nice work Mr. Green. Now how do you interperet these six dimensions?
I am not entirely convinced of your answer yet, but it seems simple
enough.
The unique form of the /\ basis vectors is satisfied.
Is it true that in the 3D cross product we impose them back onto the
original basis?
How is this exterior product then the generalization of the cross
product?
I suppose we might be admitting that the old practice is
inappropriate.
Can we make make a framework here in an open way that allows us to
pick out a decent path?
I guess this sends me back to electromagnetics versus angular
momentum; two operations using the cross product in completely
different ways whose resultants are differing interpretations on the
same operator.
- Tim
Edward Green
<tttppp...@yahoo.com> wrote:
> Nice work Mr. Green.
If it were not for your brilliant suggestion that we actually try to
write something explicit down, I would not have attempted it. Thanks.
> Now how do you interperet these six dimensions?
Ah, I don't know yet (or possibly ever), but I'm sure Misner, Thorne
and Wheeler do: they are full of pictures of stacks of planes and egg
crate structures. Their ilk is probably responsible for the prejudice
that we have to do with forms (linear functions on vectors), or even
differential forms -- since that seems to be the application in GR.
However, the basic definition of wedge product starts with a vector
space, which we expand into a variety of higher dimensional spaces,
and a dual to a vector space is as good as a vector space, of course.
Here is something else I gleaned: the object v^w can be viewed as
something living in its n-choose-2 space, or equivalently as an
antisymmetrized tensor product living in n^2 space. Hence in 3
dimensions (the story often told) the cross product can either be
viewed as an antisymmetric 3x3 matrix or else a 3-vector. What I
didn't think about up to now is, though, that this is deeper than just
noting that the antisymmetric 3x3 tensor has only 3 independent
components: these components really _work_ as components of a 3
vector! You can draw it in space, get sensible results contracting it
with other 3 vectors, and so forth. I don't get why this is so.
Now, in 4 space, our antisymmetric 4x4 tensor has 6 independent
coordinates, so can no longer be mapped back to 4 space. What is the
meaning of this new 6-space?
Beats the heck out of me, right now! It ought to express an oriented
area (2-d surface) in 4 space. Can you think of a reason an object
representing oriented 2-d surfaces in 4 space has 6 degrees of
freedom? Wait! It's coming to me! Each of the 6 basis vectors e1^e2,
etc., expresses an independent direction a normal to a 2-surface can
point in 4-space!
Whoa. Heavy, dude.
> I am not entirely convinced of your answer yet, but it seems simple
> enough.
> The unique form of the /\ basis vectors is satisfied.
> Is it true that in the 3D cross product we impose them back onto the
> original basis?
Apparently, but I don't understand why this works... unless... hmm.
Given that a surface normal perpendicular to e1,e2 can only lie along
e3. Sounds simple enough. Just don't think too hard about how this
works out in R^4. The "orientation" of an e1^e2 surface is both
somehow fixed (it's a basis vector), yet we can't associate it with a
fixed direction in 4-space. 2-surfaces don't have "normals" in 4-space
the way we understand the idea in 3-space. Apparently.
(So maybe I shouldn't have said "direction a normal can point", in
relation to 4-space, but rather "independent orientation"?).
> How is this exterior product then the generalization of the cross
> product?
It _is_ the cross product for all practical purposes, in R^3.
> I suppose we might be admitting that the old practice is
> inappropriate.
> Can we make make a framework here in an open way that allows us to
> pick out a decent path?
>
> I guess this sends me back to electromagnetics versus angular
> momentum; two operations using the cross product in completely
> different ways whose resultants are differing interpretations on the
> same operator.
Both ought to bear some interpretation as an oriented area?
BTW, I have a personal bugbear that tells me something about angular
momentum is in fact hiding behind the applications of the cross
product in EM, but that's another topic.
pmb
It would have been more polite if you just simply said that you don't
respond to probing questions by me. It could have saved me some time.
Plonk
Edward Green
> "Edward Green" <spamspamsp...@netzero.com> wrote in message
> > I had reached about that conclusion, although I believe I have also
> > seen "form" in contexts where no field was necessary or implied. I'll
> > take it that "form" generally implies a field, while "covector" does
> > not, but this is not ironclad.
>
> I'm not sure what you mean here? Do you believe that all forms are fields
> like Tom Roberts claims? If so then why? Consider the 1-form which is the
> dual to 4-momentum. 4-momentum is not a field so in what sense do you
> believe that the 1-form dual to 4-momentum is a field?
This response is bizarre. Just about every word of _my_ response
negated or qualified the proposition that "all forms are fields"!
Edward Green
> > (1,1,1,1) /\ ( 1,2,3,4)
>
> > = [(2-1),(3-1),(4-1),(3-2),(4-2),(4-3)] = [1,2,3,1,2,1]
> Now how do you interperet these six dimensions?
Possible short answer: the 6-vector has a magnitude equal to the area
of the parallelogram spanned by the given vectors in 4-space, and
components equal to the projection of this area onto each of the basis
planes, e1^e2, etc..
Edward Green
Thanks for the heads-up.
I've been working several weeks on a Dover reprint on Lie groups, and
I'm still struggling with the "introductory" material. Age, fatigue,
other demands on my time, and lack of adept people to talk to take
their toll. So it's probably a fantasy that I will get much out of
Baez and Muniain.
Matthew Johnson
Edward Green says...
[snip]
>I've been working several weeks on a Dover reprint on Lie groups,
I don't think any of the Dover reprints on Lie groups were really that good...
> and
>I'm still struggling with the "introductory" material.
so that could explain why you are still struggling;)
Now if only they would reprint Lipkin's classic "Lie Groups for Pedestrians"!
Or have they done it? Google is showing me a link to Dover Pubs for this book.
You can get a Preview of it at Google Books.
jmfbahciv
> On Jul 26, 7:36 am, jmfbahciv <jmfbahciv@aol> wrote:
> Thanks for the heads-up.
You are welcome. Now I've got that sheet on my list of things
to look for while I mosey around the house. I didn't know I'd
"lost" it.
>
> I've been working several weeks on a Dover reprint on Lie groups, and
> I'm still struggling with the "introductory" material.
I looked at one of those books and left it on the shelf. When my
eyes cross immediately, I know I'm looking at material way out of
my league.
> Age, fatigue,
> other demands on my time, and lack of adept people to talk to take
> their toll. So it's probably a fantasy that I will get much out of
> Baez and Muniain.
I found it readable. I don't think I understood any of it, but
I'm still essentially at Calculus 101 first semester levels
because I don't use it daily.
/BAH
Timothy Golden BandTechnology.com
It seems lame to me that they have to repeat their definition within
their result. But further also that their answer is so complicated in
the higher D. In effect it seems that
1.23 e1 /\ e2
means 1.23 D-2 dimensional units in directions not e1 and not e2.
so for the 4D case we'd have
r3 e3 , r4 e4, where r3 r4 = 1.23 .
Does it follow that in 5D there would be 3D results?
Then also if we want to get back to ordinary vectors in 4D we have the
option of doing
u ^ v ^ w
right?
This actually does seem interesting since classical physics attempts
to take dual relations integrated over the universe on 3D. If a 4D
force vector was to be instantiated by the exterior product then we
would want triple interactions wouldn't we? This is daunting as well.
I understand that higher dimensions are less straightforward than
their predecessors but here that complication is getting taken into
absurdity isn't it? Even if we were to validate this mathematical
construction what would be its physical correspondence? The cross
product is in use and if this is the general form then it certainly is
clear that clean behaviors end at 3D. Maybe this is how they come to
develop this so far. At least it helps explain why nobody bothers to
do out a 4D wedge product.
As I go back to Wolfram and wikipedia I'm not even convinced that your
interpretation is good. Yes, its consistent with the cross product and
I have just gone there, but if it is true couldn't they have gotten
some of that meaning into their definition? There are interesting
effects in associative algebra, but I'm not even willing to put this
product in the algebraic category where its definition leads to its
own reuse and such wild results. Simply put this is an ill defined
product and I am surprised that it passes as a clean math with such
success.
I guess the first breakdown for a simplistic minded person like myself
is that algebraic operators put their result back in the space of
their sources. This was true of the cross product but is not true of
this wedge product. Yet they claim bilinear behavior. I do hate to
scrutinize down to this extent because I start to feel uncertain about
whether what I think I'm reading is what I'm understanding. Anyhow
another way to consider this is to ask whether the results are
superposable? We can only truly superpose within each new basis and
not accross them, maintaining a triangular structure or with the
redundant dual sign an antisymmetric matrix. Doesn't this then
informationally imply a 6D resultant for the 4D basis?
I've got better math than this to work physics out on:
http://bandtechnology.com/PolySigned
While there are similarities and the antisymmetric form does ring a
bell, I step back to spacetime behavior and electromagnetism.
Structured spacetime is an acceptable construction especially given
unidirectional time. The electromagnetic constants which provide the
speed of light are already attributed to spacetime. The Minkowski
metric already imposes structure on a 4D basis. We should take
relativistic theory the rest of the way so that it does not merely
take c, but takes electricity and magnetism as well. The natural
structured form of spacetime is not Cartesian based. In a brute way I
would take this criticism and explain away the wierdness of the wedge
product results as the extension of an inappropriate (Cartesian)
basis. I don't mean to discourage you. I encourage you in another
direction.
This region of math is kaleidoscopic and probably you've come accross
other kaleidoscopic associative algebra work. Getting back to
spacetime is relevant to physics.
- Tim
Edward Green
wrote:
> In article <ef9ab5b3-183b-4a87-9818-751a301d1...@2g2000hsn.googlegroups.com>,
> Edward Green says...
>
> [snip]
>
> >I've been working several weeks on a Dover reprint on Lie groups,
>
> I don't think any of the Dover reprints on Lie groups were really that good...
>
> > and
> >I'm still struggling with the "introductory" material.
>
> so that could explain why you are still struggling;)
>
> Now if only they would reprint Lipkin's classic "Lie Groups for Pedestrians"!
Beautiful! I have just happened to have received this book from
Amazon. With your recommendation I will put aside Gilmore for now and
start fresh with Lipkin.
Edward Green
wrote:
> Now if only they would reprint Lipkin'sclassic "Lie Groups for Pedestrians"!
Those are some robust pedestrians, I tell you. Probably race-walkers!
After skimming about 1/2 the book, I get a vague picture like this:
(1) we can learn something about angular momentum by performing
algebraic manipulations of its related (quantum mechanical) operators,
without necessarily troubling much about representations of the wave
function
(2) we might not have known it, but we were playing with a Lie
algebra!
(3) other Lie algebras, in this context realized by operator algebras,
somehow suggest themselves to our baffled brains, enabling us to
deduce feature of other systems, without troubling much about
representations of their wave functions
(4) once we have made some insightful deductions from a Lie algebra,
naturally we do not have to repeat our insightfulness if the same
algebra appears again in some different context -- which might even
involve a different Lie group!
(5) Lie group? Well, theoretically, this is where the algebras came
from, but apparently, in physics, we put the cart before the horse,
and are more interested in the algebras than the groups, which reflect
some symmetry of the system we often have no ready interpretation for.
Comments? Enlightenments? Insults?
Edward Green
<tttppp...@yahoo.com> wrote:
> On Jul 29, 12:28 pm, Edward Green <spamspamsp...@netzero.com> wrote:
>
> > On Jul 28, 12:46 pm, "Timothy Golden BandTechnology.com"
>
> > <tttppp...@yahoo.com> wrote:
> > > On Jul 28, 12:30 am, Edward Green <spamspamsp...@netzero.com> wrote:
> > > > (1,1,1,1) /\ ( 1,2,3,4)
>
> > > > = [(2-1),(3-1),(4-1),(3-2),(4-2),(4-3)] = [1,2,3,1,2,1]
> > > Now how do you interperet these six dimensions?
>
> > Possible short answer: the 6-vector has a magnitude equal to the area
> > of the parallelogram spanned by the given vectors in 4-space, and
> > components equal to the projection of this area onto each of the basis
> > planes, e1^e2, etc..
>
> It seems lame to me that they have to repeat their definition within
> their result.
I'm not sure what you mean by this, but bear in mind this is _my_
speculation, not "their" result. All I'm tolerably sure of is the
numeric values.
> But further also that their answer is so complicated in
> the higher D. In effect it seems that
> 1.23 e1 /\ e2
> means 1.23 D-2 dimensional units in directions not e1 and not e2.
It might be better to think of this as (oriented) area.
> so for the 4D case we'd have
> r3 e3 , r4 e4, where r3 r4 = 1.23 .
>
> Does it follow that in 5D there would be 3D results?
The wedge product of two 5-d vectors would have 5-choose-2 = 10
components.
> Then also if we want to get back to ordinary vectors in 4D we have the
> option of doing
> u ^ v ^ w
> right?
About right, but this brings the "Hodge dual" thing into it (no, I'm
no pretending to be an exterior algebra savant, yet). Actually, we
already do this in 3-d, where the cross product is "really" the wedge
product, but its basis vectors (e1^e2, etc.) can be taken as "dual" to
the ordinary 1-d basis vectors (here, e3). So in 4-d your triple
wedge product would have basis elements of form e1^e2^e3, etc., which
we should believe, for some reason, can be usefully represented or
substituted for by e4.
> This actually does seem interesting since classical physics attempts
> to take dual relations integrated over the universe on 3D. If a 4D
> force vector was to be instantiated by the exterior product then we
> would want triple interactions wouldn't we? This is daunting as well.
>
> I understand that higher dimensions are less straightforward than
> their predecessors but here that complication is getting taken into
> absurdity isn't it? Even if we were to validate this mathematical
> construction what would be its physical correspondence? The cross
> product is in use and if this is the general form then it certainly is
> clear that clean behaviors end at 3D. Maybe this is how they come to
> develop this so far. At least it helps explain why nobody bothers to
> do out a 4D wedge product.
I don't know if that's true or not. Whether or not we understand it
or like it, I think it's a little bizarre to seriously doubt whether
it is self-consistent or doable. It exists.
> As I go back to Wolfram and wikipedia I'm not even convinced that your
> interpretation is good. Yes, its consistent with the cross product and
> I have just gone there, but if it is true couldn't they have gotten
> some of that meaning into their definition? There are interesting
> effects in associative algebra, but I'm not even willing to put this
> product in the algebraic category where its definition leads to its
> own reuse and such wild results. Simply put this is an ill defined
> product and I am surprised that it passes as a clean math with such
> success.
Ill-defined? That smacks of hubris. It's just hard, that's all.
> I guess the first breakdown for a simplistic minded person like myself
> is that algebraic operators put their result back in the space of
> their sources. This was true of the cross product but is not true of
> this wedge product. Yet they claim bilinear behavior. I do hate to
> scrutinize down to this extent because I start to feel uncertain about
> whether what I think I'm reading is what I'm understanding. Anyhow
> another way to consider this is to ask whether the results are
> superposable? We can only truly superpose within each new basis and
> not accross them, maintaining a triangular structure or with the
> redundant dual sign an antisymmetric matrix. Doesn't this then
> informationally imply a 6D resultant for the 4D basis?
The wedge product (I've learned to say from my betters) is not really
an "algebra" over the space of vector inputs: if this were true, it
would return outputs in the space of inputs, as you surmise. It's an
algebra (again, I'm not trying to ape expertise, but I've stumbled a
few steps further down this road, apparently) over a _direct sum_ of
(oh, hell, let's go for the terminological gold) _fully
antisymmetrized tensor products_.
Let me try to illustrate this for the 4-d case: we have 4-vectors, and
then we have 6 component objects, and then we have 6 again (they are
just the binomial coefficients), and finally we close with 4 again, as
you noted above. So our "exterior product space" really has 4 + 6 + 6
+ 4 dimensions, although (as far as I know, at least at first) we only
seem to be interested in objects which lie wholly in one of the
subspaces.
I think in fact we can form exterior products over mixed objects
(something in general in each subspace), make sense of the results (at
least computationally), and return a result within the same composite
space. In that sense its a real "algebra", over this extended space.
This guy Grassmann was evidently a brilliant dude, and while I share
some of your distaste for people who seem to view exterior algebra as
the new utopia, I think we have to accept that it is self-consistent,
and maybe even profound. Shit.
> I've got better math than this to work physics out on:
> http://bandtechnology.com/PolySigned
> While there are similarities and the antisymmetric form does ring a
> bell, I step back to spacetime behavior and electromagnetism.
> Structured spacetime is an acceptable construction especially given
> unidirectional time. The electromagnetic constants which provide the
> speed of light are already attributed to spacetime. The Minkowski
> metric already imposes structure on a 4D basis. We should take
> relativistic theory the rest of the way so that it does not merely
> take c, but takes electricity and magnetism as well. The natural
> structured form of spacetime is not Cartesian based. In a brute way I
> would take this criticism and explain away the wierdness of the wedge
> product results as the extension of an inappropriate (Cartesian)
> basis. I don't mean to discourage you. I encourage you in another
> direction.
> This region of math is kaleidoscopic and probably you've come accross
> other kaleidoscopic associative algebra work. Getting back to
> spacetime is relevant to physics.
Well, thanks for your oblique encouragement. Got to go now!
Matthew Johnson
Edward Green says...
>
>On Jul 29, 10:40=A0pm, Matthew Johnson <matthew_mem...@newsguy.org>
>wrote:
>
>> Now if only they would reprint Lipkin'sclassic "Lie Groups for Pedestrian=
>s"!
>
>Those are some robust pedestrians, I tell you. Probably race-walkers!
It was written back in the days when most graduate students in physics didn't
know representation theory that well; so the 'pedestrians' were people with
strong physics backgrounds, but not so strong in Lie theory.
>After skimming about 1/2 the book, I get a vague picture like this:
Within the limits of what can be achieved by skimming a physics book, you have
done pretty well. Just don't be too surprised if on further, more detailed
reading, you discover your initial impressions require some modifications.
>(1) we can learn something about angular momentum by performing
>algebraic manipulations of its related (quantum mechanical) operators,
>without necessarily troubling much about representations of the wave
>function
This is true not just for angular momentum, but for other physical quantities as
well.
There is one general principle I did not see you mention, one that is important:
the Lie algebra of a (Lie) group G determines G up to "local isomorphism". This
is what justifies solving problems in Lie groups by referring to their algebras,
where the problems are often easier to solve.
Timothy Golden BandTechnology.com
Thanks for filling this out a bit further. The 'ill-defined' nature of
this product hinges on whether the operator
/\
is used within its own definition. As we go to a fully instantiated
version e.g. the 4D wedge product that you've been so good to
participate in here then we do see the reuse of the operator within
the result. This is a fairly fundamental crux. Who would get away with
such an attemp here on usenet if they were providing this argument for
the first time? In terms of circular constructions we rarely even see
people attempt them because they are so clearly improper. For instance
if I were to attemp to provide some new math and I explained to you
that
6.787 & 5.67 = 1.23 & 3.42
you'd still be left asking for a definition of the new & operator
wouldn't you? Wouldn't you also yield this criticism back to the
provider if you were serious enough about the problem? How much of
higher mathematics is like the Grassman algebra? People have been
pushing the limits to get published and couldn't we consider that as
the permissivity of pushing limits goes on we could wind up in such a
system? It does bring me back to instantiation to ground out a
definition. Instantiation is not requisite to abstraction. This is a
tension that does not seem to exist for the mathematician but it does
for the physicist doesn't it? As a designer I could design a self
sufficient spaceship with all sorts of coherent doohickies that fit
together so tightly as to be brilliant yet there is no proof that the
spaceship can even be built. This next stage that we would enter of
actually building the spaceship is instantiation. Here even still on
paper I do cast my doubt on the wedge product as logically as I can. A
caveat is that if the definition is not self-referential then this
criticism is oblique. Your interpretation does actually reduce this
trouble though when the result is studied at its face value seems
contorted. Still, the connection to the antisymmetric tensor and hence
electromagnetism is a fine one to the cross product.
Dimensionally it is clear that raising and lowering in dimension
carries much more complexity than merely the length of the series of
numbers needed for the representation. At 4D topological surprises
which are not directly apparent do surface. To go in the other
direction (3D to 2D) is also a wise check on generality. Strangest of
all I can expose a 0D algebra and so in terms of dimensional reduction
can expose a unidirectional time displacement. Perhaps though part of
the trouble is that this operator seems to require two arguments
whereas logically it seems we could have
/\ x
come out too under your own interpretation. Maybe that will ding a
bell for you. I do see raw value in the antisymmetric representation.
But, the quantity of systems that are yielding into that space are
more than just this Grassman algebra perspective. The kaleidoscope
seems to be powered by associative algebra and under that constraint
alone there are some peculiar theorems. These theorems do relate to
what I call dimensional reduction. I am afraid that so long as people
overlook unidirectional time that the proper basis will not be worked
on. This is pertinent in that the signal within relativity theory of
the electromagnetic tensor that lines up with the wedge product is
blowing off unidirectional time which actualy be partly why the
redundant form exists. The non-redundant form
a11
a21 a22
a31 a32 a33
...
I call a 'tatrix' for triangular matrix and I believe that this is the
correct primitive format. Informationally it is consistent with the
antisymmetric tensor but is much simplified especially since its
values carry no sign; their positions are their signs. Sign is
dimension plus one, though I can argue that dimension is sign minus
one since the polysign numbers generate their own geometry without the
need for the Cartesian product. Still, this progressive tatrix form
does take a Cartesian style superposition, yet since they are
structurally unique entities even this usage is lightened. Sorry; that
is all semi-tangential to the wedgie.
Is the definition of the wedge product self-referential?
That in a line is all that I should have said.
- Tim
Timothy Golden BandTechnology.com
> On Jul 28, 12:46 pm, "Timothy Golden BandTechnology.com"
>
> <tttppp...@yahoo.com> wrote:
>
> If it were not for your brilliant suggestion that we actually try to
> write something explicit down, I would not have attempted it. Thanks.
>
>
> > enough.
> > The unique form of the /\ basis vectors is satisfied.
> > Is it true that in the 3D cross product we impose them back onto the
> > original basis?
>
> Given that a surface normal perpendicular to e1,e2 can only lie along
> e3. Sounds simple enough. Just don't think too hard about how this
> works out in R^4. The "orientation" of an e1^e2 surface is both
> somehow fixed (it's a basis vector), yet we can't associate it with a
> fixed direction in 4-space. 2-surfaces don't have "normals" in 4-space
> the way we understand the idea in 3-space. Apparently.
>
> (So maybe I shouldn't have said "direction a normal can point", in
> relation to 4-space, but rather "independent orientation"?).
>
> > product?
>
>
> > inappropriate.
> > Can we make make a framework here in an open way that allows us to
> > pick out a decent path?
>
> > I guess this sends me back to electromagnetics versus angular
> > momentum; two operations using the cross product in completely
> > different ways whose resultants are differing interpretations on the
> > same operator.
>
>
> BTW, I have a personal bugbear that tells me something about angular
> momentum is in fact hiding behind the applications of the cross
According to
http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node4.html
some of this interpretation may be backwards. Maybe they have it
backwards though.
Thanks to the ill-defined operator
/\
will we ever know who is right? Whereas you've offered the idea that
x /\ y
eliminates these components from the resultant their interpretation
claims them to be in the resultant.
This is probably best resolved by using the tangent space concept.
According to their interpretation of a/\b
"it is a directed area, or bivector, oriented in the plane
containing a and b."
whereas we have inverted that (in the 4D case). Why is this sort of
language is left out of their (Wolfram and Wikipedia) definition?
Here is another piece of criticism from this site:
"There is another language which has some claim to achieve useful
unifications. The use of `differential forms' became popular with
physicists, particularly as a result of its use in the excellent, and
deservedly influential, `Big Black Book' by Misner, Thorne &
Wheeler[33]. Differential forms are skew multilinear functions, so
that, like multivectors of grade k, they achieve the aim of coordinate
independence. By being scalar-valued, however, differential forms of
different grades cannot be combined in the way multivectors can in
geometric algebra. Consequently, rotors and spinors cannot be so
easily expressed in the language of differential forms. In addition,
the `inner product', which is necessary to a great deal of physics,
has to be grafted into this approach through the use of the duality
operation, and so the language of differential forms never unifies the
inner and outer products in the manner achieved by geometric algebra."
- http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node12.html
I don't fully understand this but it does seem pertinent.
Thanks Thomas Heger for pointing this link out to me.
- Tim
peeter
<tttppp...@yahoo.com> wrote:
...
> inner and outer products in the manner achieved by geometric algebra."
> -http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node12.html
>
> I don't fully understand this but it does seem pertinent.
There is some good explaination of clifford/geometric algebra concepts
here:
- http://www.science.uva.nl/ga/tutorials/
See the 2003 game developer lecture. It's a interactive GA tutorial/
presentation for a game programmers conference that provides a really
good intro and has a lot of examples that I found helpful to get an
intuitive feel for all the various product operations and object
types.
Even if you weren't trying to learn the algebra, if you have done any
traditonal vector algebra/calculus, IMO its worthwhile to download
this just to just to see the animation of how the old cross product
varies with changes to the vectors.
Other sources of info:
- The book "Geometric Algebra for Computer Science". Very clear and
easy to understand (much more so than either of the "New Foundations
for Classical Mechanics" or "Geometric Algebra for Physicists" books).
These actually contain a fair amount of the non-conformal parts of
that GASC book (the parts that were interesting to me) :
http://staff.science.uva.nl/~leo/clifford/dorst-mann-I.pdf
http://staff.science.uva.nl/~fontijne/phd.html
- lect1.pdf from the following is also a good starting point:
http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectures.zip
- http://en.wikipedia.org/wiki/Geometric_algebra
Also has some links at the end to other sources (like the GA primer).
- http://www.geocities.com/peeter_joot/
This has a collection of some personal notes of mine on various bits
of math/physics, many of which were written as I've learned geometric
algebra. With only an undergrad engineering education I'm no
physisist, but I'm having fun learning;)
Edward Green
<tttppp...@yahoo.com> wrote:
<small, I hope well-meant, snip>
> Is the definition of the wedge product self-referential?
> That in a line is all that I should have said.
I am not feeling very brilliant right now, but I do see what you are
getting at. I set out to evaluate something of the form "v /\ u", and
in the end I wind up with an expansion in terms of things like "e1 /\
e2", which still look very mysterious, and don't seem to have been
defined, so what's the point?
I could claim something weaselly, and say that all you really asked me
to do was evaluate a numeric example, and these intermediate terms in
the calculation were merely scaffolding, which eventually fell away to
give you what you asked for: a numeric answer.
I could also say something general about how some mathematics is
learned by the "jump in" paradigm, wherein we learn how to manipulate
the symbols, what use the symbols may be and what use we may make of
the output, and eventually, learning the game moves, we begin to
cultivate the illusion that we have, or could have, built the machine
ourselves, rather than just having learned to drive.
Since we are accepting the cross product as something nice and
friendly and intuitive, and I realize I have probably known its
essential properties and rules of manipulation for over 30 years, I
realize that I cannot recall ever having proved, or possibly even seen
proved, or skimmed over the proof, that the components we get by the
familiar manipulation have the advertised property: that they
represent a vector orthogonal to the two original vectors -- in a
sense determined by the right-hand rule -- and of length equal to the
spanned area of the inputs.
There were times and years I would have sweated this, and pored over
the terms until I found a way to understand, by direct intuitive
apprehension, that the output did have these properties: but evidently
when I first learned of the cross product it was not one of those
times and years, as I have no memory of having ever tried this.
Perhaps I ought to do this now, as we are trying to understand the
wedge product, and I see that the cross product is not, as we might
think, a simpler alternative to exterior algebra, but rather its
simplest non-trivial example: albeit a bit fudged up by this
additional Hodge dual thing, where we label its three independent
components with the original vector bases rather than these strange
"wedge bases".
And it's not quite true, I think -- or at least I will take the
contrarian position to the tyro's -- that "the cross product only
exists in three dimensions". The wedge product of two 4-dimensional
vectors looks a _lot_ like the component-wise cross product -- its
components are all of form (a_i)(b_j) - (a_j)(b_i) -- and only differs
in that it no longer lives in a space with the dimensions of the
inputs. So it seems like a natural generalization of the cross
product to me.
Much else occurs to me to write, but little fresh insight. I tried to
read Baez and Munian tonight, and when I reached a quote which ended
with the words "generally convariant", precisely, it was lights out.
After that clarity suffered. I should not eat, and read.
peeter
...
> getting at. I set out to evaluate something of the form "v /\ u", and
> in the end I wind up with an expansion in terms of things like "e1 /\
> e2", which still look very mysterious, and don't seem to have been
> defined, so what's the point?
Instead of focusing on the cross product, look at the simpler example
of the R2 case first. A way to look at this in a clifford algebra
context is by considering complex numbers to be a special
representation of two dimensional vectors.
In a nutshell one can consider a complex number to be a vector with
one of the unit vectors factored out, as in the following example:
r = x e_1 + y e_2 = e_1 (x + e_1 e_2 y) e_1 = e_1 Z
(one could left or right factor out either of the e_j unit vectors
resulting in similar (ie: congugate) "complex number"
representations).
Now that may look a bit strange at first since we are taught that we
can't multiply or divide vectors, but these can be considered
perfectly well defined operations with respect to the geometric
(clifford) vector product. In a nutshell, that product defines the
square of a vector as a scalar with magnitude equal to its squared
length. A consequence of this rule, plus associativity and linearity,
is the product of two perpendicular vectors is an entity that changes
sign with commutation. With these as rules for multiplication,
squaring e_1 e_2 (= e_1 \wedge e_2) gives:
(e_1 e_2)(e_1 e_2) = e_1 (e_2 e_1) e_2 = - e_1 e_1 e_2 e_2 = -1
ie: a product e_1 e_2 operates as a unit imaginary, and encodes a
representation for an oriented plane (that is true in higher
dimensions as well as the two of this specific example). All the
complex number operations (multiplication, conguation, inversion, ...)
have natural equivalents expressed in this fashion. I was quite
excited seeing this for the first time since the complex
multiplication "rule":
(a,b)(c,d) = (ab -bd, bc + ad)
always seemed like something that had been pulled out of a magic hat
as opposed to something that followed from a logical argument. I
similarily disliked the way the cross product appeared to also be
pulled out of a magic hat, not to mention the fact that it doesn't
work for >3D, or even 2D when you have quantities like torque that
ought to be perfectly well defined in a plane without a normal to
describe the orientation.
Simple Simon
> excited seeing this for the first time since the complex
> multiplication "rule":
>
> (a,b)(c,d) = (ab -bd, bc + ad)
Shouldn't this be (a,b)(c,d) = (ac - bd, bc + ad)?
peeter
> Shouldn't this be (a,b)(c,d) = (ac - bd, bc + ad)?
yes.
peeter
yes.
Timothy Golden BandTechnology.com
> On Aug 5, 3:58 pm, "Timothy Golden BandTechnology.com"
>
> <tttppp...@yahoo.com> wrote:
>
> ...
>
> > inner and outer products in the manner achieved by geometric algebra."
> > -http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node12.html
>
> > I don't fully understand this but it does seem pertinent.
>
> There is some good explaination of clifford/geometric algebra concepts
> here:
>
> -http://www.science.uva.nl/ga/tutorials/
Hi Peeter. I've followed your link above here which eventually gets me
to a tutorial paper
http://www.science.uva.nl/ga/files/cga-1.pdf
which claims that
a /\ b
should be interpreted as an area in the plane of a and b. Maybe I've
abbreviated a bit though since they are using basis components
(e1,e2,e3). This is somewhat the problem; If the we compose a and b as
a1 e1 + a2 e2 + a3 e3
(likewise for b)
should we be any happier with the
r12 e1 /\ e2 + r23 e2 /\ e3 + r31 e3 /\e2
result as three areas? Are these three areas superposable into one?
Apparently not. What then does this say about tensor notions of space?
Are we supposed to accept this space as isotropic? Then why can't we
compute the original
a /\ b
as a singular instance as if a were an e1 component and b were an e2
component? If you say we can then clearly the three component answer
ought to superpose to the single component answer that I've just
constructed. All that should be needed is a single projection mapping
a to e1, b to e2, and the normal of a and b to e3. The odor here is of
a fine cheese, but who knows how to make this cheese from first
principles? This cheese is a self referential form and when we push
one way on it it bulges out in another way. I guess its a soft cheese
then. Is this American cheese?
Your link is concerned strictly with three dimensions. We attempt the
same work in four dimensions to verify the generality of the solution.
It is too easy to pull a hazy trick with the 3D case. At some
humanistic level perhaps we as readers should give ourselves more
credit; when we read something and it seems a bit hazy must we always
attribute that haze to our own inability? Or is it instead the
author's inability to instantiate their construction completely?
Unfortunately within this communication paradigm their is a channel
involved as well which is the existent language. This is a serious
problem for humans and the mathematician's claim to have escaped the
human condition cannot be true. As we attempt to interperet shouldn't
we also admit that we each have our own internal representation? The
deepest flaws are typically invalid assumptions which wind up being
transferred without ever being questioned. To get to the bottom of the
problem we simply need to seek such assumptions and verify them. Then
hopefully we come out with a solid understanding deeper than mere
acceptance of what someone else smarter than us has written. If in the
process we expose a misnomer then that ought to be pursued.
I haven't gotten to your other links yet but I'll try a few, hopefully
today. This math is diffused through many circles and I do like your
first link as an instance but it is stuck in 3D.
- Tim
>
> See the 2003 game developer lecture. It's a interactive GA tutorial/
> presentation for a game programmers conference that provides a really
> good intro and has a lot of examples that I found helpful to get an
> intuitive feel for all the various product operations and object
> types.
>
> Even if you weren't trying to learn the algebra, if you have done any
> traditonal vector algebra/calculus, IMO its worthwhile to download
> this just to just to see the animation of how the old cross product
> varies with changes to the vectors.
>
> Other sources of info:
>
> - The book "Geometric Algebra for Computer Science". Very clear and
> easy to understand (much more so than either of the "New Foundations
> for Classical Mechanics" or "Geometric Algebra for Physicists" books).
>
> These actually contain a fair amount of the non-conformal parts of
> that GASC book (the parts that were interesting to me) :
>
>
> - lect1.pdf from the following is also a good starting point:
>
> http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectu...
>
> -http://en.wikipedia.org/wiki/Geometric_algebra
>
> Also has some links at the end to other sources (like the GA primer).
>
> -http://www.geocities.com/peeter_joot/
Matthew Johnson
Timothy Golden BandTechnology.com says...
[snip]
>Your link is concerned strictly with three dimensions. We attempt the
>same work in four dimensions to verify the generality of the solution.
>It is too easy to pull a hazy trick with the 3D case.
But there are lots of things that work in 3D that do not in others. There really
is special about 3D.
[snip]
Timothy Golden BandTechnology.com
Unfortunately if we try to reinterperet this work we just wind up with
another rendition of it and there are already far too many reditions.
To me the concise form is an attempted refutation which I am open to
being either successful or unsuccessful. It seems that one attack may
be an actual instantiation whereby we question the quality of the
results and whether those results have any meaning.
Beneath here I've provided peeter with another puzzle that might
couple with the first. Are we going to allow the usual tensor forms to
exist on this product? If so then by choosing basis vectors which
match the problem instance a/\b we can compute a single component
solution. We can then hold this single component solution alongside
the multiple component solution and ask how do they match up? Where is
the transform in the new /\ space that allows this simple one
component occurrence to turn into a six component nightmare for the 4D
instance? Even within the hazily defined /\ space could this be a
sufficient attack?
- Tim
peeter
<tttppp...@yahoo.com> wrote:
> which claims that
> a /\ b
> should be interpreted as an area in the plane of a and b. Maybe I've
> abbreviated a bit though since they are using basis components
> (e1,e2,e3). This is somewhat the problem; If the we compose a and b as
> a1 e1 + a2 e2 + a3 e3
> (likewise for b)
> should we be any happier with the
> r12 e1 /\ e2 + r23 e2 /\ e3 + r31 e3 /\e2
> result as three areas? Are these three areas superposable into one?
Such a composition only works in two and three dimensions. Suppose
you have:
a = \sum a_i e_i
b = \sum b_i e_i
You can interpret a^b as a oriented area regardless of the number of
dimensions. If you expand this wedge product by components you'll
get:
a ^ b = \sum_{i<j} (a_i b_j - a_j b_i) e_i ^ e_j
Observe that the coefficients are all determinants. So, for your 4D
question, when you have basis bivectors e12, e13, e14, e23, e24, e34,
you will have a determinant coefficient associated with each of these,
but only when this sum is a bivector. You can't however arbitrary
form a linear combination of these basis bivectors that can
neccessarily be interpretted as an area. Such an object is a grade 2
multivector. You can do various operations on such a multivector
(example: project a vector onto it), but you won't find such an object
as a result in geometrical problems unless it is as a result of
something like numerical error. An example is:
e12 + e34
There isn't a way to form this using a wedge product from any two
vectors since there is no common vector. In 3D all planes through an
origin happen to intersect in at least a line, so an arbitrary sum of
bivectors can be also interpretted as a bivector, but that (and 2D)
are special cases.
If you are to pick the 4D bivector basis, and form arbitrary linear
combinations (ie: form a grade 2 multivector that may not be a
bivector) then a geometrical analogy breaks down. In such a case for
interpretation you have to look for where you would actually get such
objects. The most obvious real example of this is the faraday
bivector, where a general grade two multivector can be used to very
naturally express completely antisymmetric grade 2 dimension-4 tensor
of the electrodynamic field (F^{\mu\nu}). Here a possible geometric
interpretation is possible if you view the subspaces spanned by {e12,
e13, e14} and {e23, e24, e34} as separate orthogonal bivector
subspaces (ie: 3 components for the spacetime planes that you can
associate with the electric field and 3 components for the purely
spacial planes that can be associated with the magnetic field).
Peeter
Edward Green
<tttppp...@yahoo.com> wrote:
<...>
> According to
> http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node4.html
> some of this interpretation may be backwards. Maybe they have it
> backwards though.
I'm not sure what it was I may or may not have had backwards.
Your reference is a barbarity, though! :-) It is not enough to
struggle with exterior or Grassmann algebra -- no, we must have a
different algebra now: a geometric, or Clifford algebra! The two seem
to differ primarily on how we handle the generalization of the other
classic vector operation: the dot product. Grassmann generalizes it
to a different sort of product, the interior product, whereas Clifford
somehow gloms both generalized dot and cross together as aspects of a
single operation. So we are confronted with not one, but two closely
related Nineteenth Century Inventions Touted by the Faithful as the
Greatest Thing since Sliced Bread.
> Thanks to the ill-defined operator
> /\
> will we ever know who is right? Whereas you've offered the idea that
> x /\ y
> eliminates these components from the resultant their interpretation
> claims them to be in the resultant.
> This is probably best resolved by using the tangent space concept.
Not sure about that. We may find strong applications in manifolds,
but so far we haven't been discussing anything which can't live in a
single copy of a vector space (and its associated higher dimensional
cohorts). Much can be resolved before we ever set foot on a manifold.
"Vector space" or "linear space" is an exceedingly general
mathematical idea. Are all the concepts of exterior and geometric
algebras applicable to a general vector space? An exterior algebra on
Hilbert space? Think of the possibilities! And we've been worrying
about four piddling components of spacetime.
> According to their interpretation of a/\b
> "it is a directed area, or bivector, oriented in the plane
> containing a and b."
> whereas we have inverted that (in the 4D case).
Have we? Maybe my first confused groping sounded that way. In my
now, just slightly less confused state of groping, that statement
sounds about right.
> Why is this sort of
> language is left out of their (Wolfram and Wikipedia) definition?
Wikipedia is great, but it's probably a mistake to rely on it as a
sole source.
> Here is another piece of criticism from this site:
> "There is another language which has some claim to achieve useful
> unifications. The use of `differential forms' became popular with
> physicists, particularly as a result of its use in the excellent, and
> deservedly influential, `Big Black Book' by Misner, Thorne &
> Wheeler[33]. Differential forms are skew multilinear functions, so
> that, like multivectors of grade k, they achieve the aim of coordinate
> independence. By being scalar-valued, however, differential forms of
> different grades cannot be combined in the way multivectors can in
> geometric algebra. Consequently, rotors and spinors cannot be so
> easily expressed in the language of differential forms. In addition,
> the `inner product', which is necessary to a great deal of physics,
> has to be grafted into this approach through the use of the duality
> operation, and so the language of differential forms never unifies the
> inner and outer products in the manner achieved by geometric algebra."
> -http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node12.html
>
> I don't fully understand this but it does seem pertinent.
Cat fight! Cliffordists attack Grassmannites.
> Thanks Thomas Heger for pointing this link out to me.
Thanks for sharing your educated skepticism with me, with references.
In the absence of complete mastery, we must proceed on Bayesian
probabilities, and I think I've mentioned already, mine seem to
diverge from yours in one particular. You seem to assign some
positive probability to the idea that one or both formulations are
somehow "wrong". I'm willing to entertain unflattering adjectives
like "annoying", "overblown" and "trendy" in connection with some of
the advocacy, but as far as logical coherency of the structures, my
prior on the chance of serious deficiency here is about nil.
Matthew Johnson
Edward Green says...
[snip]
>So we are confronted with not one, but two closely
>related Nineteenth Century Inventions Touted by the Faithful as the
>Greatest Thing since Sliced Bread.
Didn't you know? That is the only way to get computer programmers to read it
before they run off to play with their spaghetti code like a kitten plays with a
ball of yarn;)
That is why "structured programming" was marketed in the same way decades ago,
and "object oriented design" after that. Now it is "Geometric ALgrebra" and
"design patterns"!
[snip]
Timothy Golden BandTechnology.com
Look, why is this operation even called a product? When we start out
with two clean 4D vectors
( a1, a2, a3, a4 ), ( b1, b2, b3, b4 )
and wind up with six resultant components it would seem that they have
decreased the number of components by two, yet the meaning of the six
components can only come from these source vectors which do form one
simple area. Somehow the six areas that they are generating are just
this one area. By making the area unspecific have we done any real
service? How is this a product? It is not; it is merely a
rerepresentation of the source data with some minor loss of
information. A traditional product loses half of the information and
yields an element in the source space
(c1, c2, c3, c4)
but this was not easy to do in general dimension so they wound up
trying like hell and this is where they wound up. Hamilton found the
quaternion and Clifford generalized that and I do agree with your
whiticism on the marriage of it all into one contorted pile and how
much it reeks, especially this wedge definition. If enough of us
question this maybe we'd get a real response. I'm willing to question
all products but this one in particular is already walking around with
a wedgie.
If only I could go back and talk with Hamilton. I could show him how
even the real numbers are tied directly to the complex numbers and
that right beneath them is unidirectional time with its notion of now
inherent. That above these one does not need to hopscotch about or
settle for noncommutative products, or for these whacky wedge devices
that send one outside of the basis.
Does the new math that I speak of come with a cost? Yes, but it is a
highly desirable cost in that support for spacetime is built into the
math. Even while all of the higher dimension algebras are perfectly
behaved in terms of commutative, associative, and distributive laws
the quotient is not so straightforward. Distance in these high
dimension spaces does not behave so well under product as it does in
the traditional spaces. This is a blessing in disguise, though the
casual encounter often ends there in dismissal for an even higher
math. I seriously doubt if it can go any further but even if it could
would we wish to work with such a math when what we see about us is
spacetime? The polysign numbers
http://bandtechnology.com/PolySigned
form the correct basis for physics and the data formats which send hot
signals through the corners of your mind lay there as well. The
antisymmetric tensor and this wedge product mimic electromagnetism
under the isotropic space model. Yet, if electromagnetic behavior is
an attribute of spacetime itself as we take the speed of light to be
then we ought to seek a structured form of spacetime. We have to give
up the isotropic space concept. As I study the words "isotropic space"
closely they make no sense. Such a space is devoid. Especially, the
alternative of choosing a structured space that matches the format
that is fueling those hot spots in your mind will enable
electromagnetism to be natural to the spacetime structure. Thus we can
pose the question
Did Maxwell unify electricity and magnetism?
The default answer in my upbringing would be yes to this question, yet
now I can pose a second round of question
Why then do we still need the words electricity and magnetism?
and we see that we do accept these as undeniably unique features. This
comes about because we have insisted upon isotropic space. Relativity
theory does break this, but noone other than myself seems willing to
admit it here. The Minkowski metric is the first successful structured
spacetime. Obviously there will be a mental blockage is assigning
structure to space itself since we've been tought an isotropic fairy
tale. The reality which I propose admits the relative reference frame
and it is via that construct that two structured spacetime entities
can coexist with unique and continuous reference to each other. There
may well be quantum reference frames as well whereby the structures
can only click out discrete forms but even these may be referenced
relatively. But in our continuous space if I were to propose to you
that your antisymmetric format informationally merely held
a11
a21 a22
a31 a32 a33
...
wouldn't you accept? Now if we interperet this format further we can
move on to
1D: a11
2D: a21 a22
3D: a31 a32 a33
...
and this we can label a 'progressive' form. It is structured and it is
by taking this structure literally that I believe we will make
progress. We do not need to worry so much about how we got here...
We've already seen about four different ways. I only want to get here
for the moment to address the relative reference frame concept. While
this is a structured geometry we are still free to couple two of these
entities relatively accross the structure. For instance going into
particle terminology particle a's a21 component could align with
U( b32 + 2.3 b33 ) as an instance where U forms a unit vector.
So what then of a structured spacetime so long as it is one that can
still take on such relative reference frames? Can't we see that we've
got our terminology all gunked up? The keywords that I would focus on
are
isotropic, relativity, tensor, structure
and admit that
isotropic <-> tensor
are nearly synonomous and further that there is a real break between
these and
relativity <-> structure
Upon assigning the metric
s s = - x1 x1 + x2 x2 + x3 x3 + x4 x4
can't we admit that the isotropic concept was broken? Einsteins math
was wrong to use the tensor in its fullest form and he did have a hard
time with that didn't he? That math is a pseudotensor form. Because
time is zero dimensional and we will work in a progressive form in the
future this issue will resolve itself. Electricity and Magnetism are
evidenced by force fields. They are forces peculiarly unique to each
other yet directly coupled all the way down to an electron's spin.
Maxwell built out of raw charge and isotropic space. Somehow we still
are tied to these by classical training and we are winding up in
quantum mechanics. Could it be that a semiclassical approach will
trade off some complexity in one way and take it back in another? If
the compression ratio is raised (informationally speaking) then a
cleaner theory could alight. Already the polysign number can build the
traditional number systems and more. It joins discrete and continuous
qualities together differently. In the final say is it merely
correspondence with reality that is the measure? What math constructs
unidirectional time? The polysign numbers do this. They stand on their
own without any need for all of this rhetoric.
- Tim
Timothy Golden BandTechnology.com
>
> <tttppp...@yahoo.com> wrote:
>
> ...
>
> > inner and outer products in the manner achieved by geometric algebra."
> > -http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node12.html
>
> > I don't fully understand this but it does seem pertinent.
>
> There is some good explaination of clifford/geometric algebra concepts
> here:
>
>
> See the 2003 game developer lecture. It's a interactive GA tutorial/
> presentation for a game programmers conference that provides a really
> good intro and has a lot of examples that I found helpful to get an
> intuitive feel for all the various product operations and object
> types.
>
> Even if you weren't trying to learn the algebra, if you have done any
> traditonal vector algebra/calculus, IMO its worthwhile to download
> this just to just to see the animation of how the old cross product
> varies with changes to the vectors.
>
> Other sources of info:
>
> - The book "Geometric Algebra for Computer Science". Very clear and
> easy to understand (much more so than either of the "New Foundations
> for Classical Mechanics" or "Geometric Algebra for Physicists" books).
>
> These actually contain a fair amount of the non-conformal parts of
> that GASC book (the parts that were interesting to me) :
>
I don't see how line 11 of the math in the following link came about.
> http://staff.science.uva.nl/~leo/clifford/dorst-mann-I.pdf
Perhaps it is just a notational issue in what he means by squaring the
expression ei/\ej. I think he meant the geometric product but
regardless the breakdown of the math doesn't make sense does it? I'm
pretty sure this is incorrect so could you take a look and let me know
what you think? I'm not asking for yet another rendition of what's
already been provided. We're trying to look a bit deeper than the
surface. Dorst and Mann starts out so nice I want to believe them but
what if they slipped up here? Again here
a/\b
is taken to simply imply a planar area in a and b. That seems pretty
well settled in terms of consensus though it you go to tangent spaces
I think it will wind up inverted to how Green put it on this thread.
"Even through the definition of the geometric product is simple,
proving that it actually is a sensible product is non-trivial."
- pg.27 of
> http://staff.science.uva.nl/~fontijne/phd.html
I haven't read this thoroughly but does seem to not be whitewashing
the topic.
>
> - lect1.pdf from the following is also a good starting point:
>
> http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectu...
Yes, this is pretty nice too. You've got some good stuff. Thanks for
sharing it here. I'd forgotten that the Grassman algebra relates to
string theory which exposes what a nice little history stream they put
in their introduction and how current it is. If the stringers are
working in this space as well then all the better. The braneish nature
of a progressive dimensional structure seems quite alive already.
>
> -http://en.wikipedia.org/wiki/Geometric_algebra
>
> Also has some links at the end to other sources (like the GA primer).
>
> -http://www.geocities.com/peeter_joot/
>
> This has a collection of some personal notes of mine on various bits
> of math/physics, many of which were written as I've learned geometric
> algebra. With only an undergrad engineering education I'm no
> physisist, but I'm having fun learning;)
It's pretty clear that the geometric product is the end goal in the
construction and so the wedge product is more like a stepping stone to
it. Still, if that stepping stone is flawed you can't get over to the
geometric product. On top of this I'm not getting what we are gaining
by entering this whacked out space of superposed interdimensional
quantities. I do see some signals but rather than resolve those
signals coherently it feels more like a heap of spaghetti. A lot of
people have their fork in it and can spin out a bite of the stuff but
is this all that you want to eat? What are you going to make out of
this? I'm pretty sure this is just an alternative map to straight
linear algebra. The last lecture(16) of
http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectures.zip
has some heavy hitting information, but was all of this math necessary
to develop those relations? I'm pretty sure most of those were done
with linear algebra and big brother tensor without all of this spag. I
do mean to get to your Aug 6 post but just haven't quite made it
through this one yet.
- Tim
peeter
<tttppp...@yahoo.com> wrote:
> I don't see how line 11 of the math in the following link came about.>http://staff.science.uva.nl/~leo/clifford/dorst-mann-I.pdf
>
> Perhaps it is just a notational issue in what he means by squaring the
> expression ei/\ej. I think he meant the geometric product but
Yes, this requires the geometric product (ie: you need to provide a
context where the square of a wedge product has meaning).
for two unit vectors that are perpendicular you have:
e_i e_j = e_i . e_j + e_i ^ e_j = e_i e_j = - e_j ^ e_i = - e_j e_i
> It's pretty clear that the geometric product is the end goal in the
> construction and so the wedge product is more like a stepping stone to
> it.
Yes, the wedge product in this context is a special case of the
geometric product. For the two vector case, the wedge product of a
with b can be conceptually thought of as the antisymmetric parts of
the geometric product, whereas the dot product is the symmetric parts:
a b = 1/2 (ab + ba) + 1/2 (ab - ba) = a.b + a^b
Axiomatically, this can be a natural way to introduce it without any a-
priori introduction of either the dot product or the wedge product.
Require the contraction rule as a fundamental property. For example,
in a euclidian context:
a a = \abs(a)^2
Then expand (a+b)^2:
(a+b)^2 = a^2 + b^2 + 2 ( 1/2(ab + ba) )
This last term can be then be identified with the dot product with a
comparision to the coordinate expansion of the same result. Next
observe from the above that for perpendicular vectors one has:
ab + ba = 0
or:
ab = -ba
This and the fact that 1/2 (ab - ba) has the other anitsymmetric
property expected from the wedge product. ie:
a = b =>
1/2(ab -ba) = 0
are good justifiers for _defining_ this as the wedge product.
Peeter
Timothy Golden BandTechnology.com
simply have
ab = 0 .
Anyhow nowhere in this math am I seeing a minus one as a result and
that is their result that I am either trying to understand or refute.
> This and the fact that 1/2 (ab - ba) has the other anitsymmetric
> property expected from the wedge product. ie:
>
> a = b =>
>
> 1/2(ab -ba) = 0
>
> are good justifiers for _defining_ this as the wedge product.
>
> Peeter
None of this seems to yield the minus one result that they've gotten
on line 11.
I'll just type it up here to see if that helps us get anywhere with
it. They have
( ei /\ ej )( ei /\ ej )
= ( ei ej )( ei ej )
= ei ej ei ej
= - ei ei ej ej
= - 1 .
Somehow we are supposed to wind up with minus one from doing the
geometric product of a unit area with itself.
Their math doesn't appear to be compressed either. It looks as if
they've done every step out. Yet they are taking advantage of the
commutative property which has no place in these products. What
product are they using when they write
ei ej ei ej ?
This is ill defined I think. Certainly by the text we should go back
to line 10 to see what they are doing. It looks to me as if
ei ej
is a geometric product so placing four terms side by side then
flipping the middle two around (commutative) does not appear to be
valid to me. This is poor notation and an illegal operation, or at
least this is my current claim. I am open to proof to the contrary. If
I try the geometric square of the unit two-blade I'd get
( ei /\ ej ) ( ei /\ ej )
= ( ei /\ ej ) /\ ( ei /\ ej ) + ( ei /\ ej ) . ( ei /\ ej ) .
This step is simply invoking the definition of the geometric product.
Since
v /\ v = 0
is an unconditional law of the wedge product I'd say we're down to the
question of what a dot product on a 2-blade is.
If there is any geometric sense to this we'll have something like the
vector dot product where
a . a = | a | | a |
where | a | is the magnitude of a. If we allow a 2-blade to have a
magnitude then that magnitude would be its area which in this case is
unity. So under this interpretation we'd get positive unity, not
negative unity. Now we have two paths of thought with contradictory
results.
Thanks for considering this. Perhaps we should call in Leo Dorst and
Stephen Mann to get their professional opinions on this. Something is
wrong either with me or with their paper. Like your own results above
here I keep seeing a reliance upon zeros in other writings as well. On
Wiki they state that when Q = 0 the Clifford algebra is achieved. This
is pretty sad. Getting back to the quaternion Hamilton was happy to
express his work in terms of unit vectors and had easily instantiable
quantities with a product that remained in the space of its sources. I
think Hamilton would like the polysign numbers
http://bandtechnology.com/PolySigned
They not only show a natural well behaved extension to the complex
numbers; they also expose that the complex numbers are merely an
extension from the real numbers with no additional rules. They show a
zero dimensional construction that matches time. All of these are true
algebras all the way up to any dimension you want and down to zero
dimensions. They are associative, commutative, and distributive.
Cartesian thought paralyzes most from getting here. The polysign
family is profound but with noone to appreciate them they sit idle.
- Tim
peeter
<tttppp...@yahoo.com> wrote:
> http://groups.google.com/group/sci.physics/browse_frm/thread/2c86c90e...
> On Aug 8, 12:37 pm, peeter <peeter.j...@gmail.com> wrote:
>
> No. For perpendicular vectors a and b where ab is the dot product we
> simply have
> ab = 0 .
Yes, but this isn't the dot product. The wedge product by itself
doesn't have a non-wedgie product (can't resist saying that;) with any
other wedge product. Differential forms avoid such a requirement via
use of the hodge dual and a dot product on wedged elements. The
product used in the papers, is called the geometric or clifford
product, can be (*) defined like so, using a "complex-number-
like" (**) composition of dot product and wedge product:
a b = a . b + a ^ b
(a dot b plus a wedge b)
So for perpendicular vectors, since a . b = 0 (a dot b = 0) one has:
a b = a ^ b
since the wedge product is antisymetric this is:
a b = - b ^ a
conversely:
b a = b . a + b ^ a
b . a == 0 (dot product is commutative). Combining results, for two
perpendicular vectors, one has:
b ^ a = -ab = ba
> commutative property which has no place in these products. What
> product are they using when they write
> ei ej ei ej ?
> This is ill defined I think.
This is the geometric product above, which for unit vectors will
therefore be:
e_i e_j = delta_{ij} + e_i ^ e_j
(ie: dot product of e_i with e_j, for unit vectors e_i, and e_j == 0
if perpendicular, and 1 if equal. If equal the wedge product term is
zero since a ^ a == 0).
Peeter
(*) I prefer an axiomatic definition, but this first form is probably
easier to introduce provided one has had exposure to the properties of
the dot and wedge products separately.
(**) I say that this generalized vector product is "complex number
like" above since the wedge product of two perpendicular unit vectors
plays an identical role to the i of complex algebra, both squaring to
produce a scalar -1 value. This allows one to essentially introduce
ad-hoc complex numbers using the geometric product via product
composition of any pair of non-colinear vectors in an arbitrary plane
of choice (and that plane does not have to be in a two or three
dimensional space).
Timothy Golden BandTechnology.com
Peeter, you seem to be engaged in a long rabbit run. This entire
subject suits you. Good luck. I stand by my previous post and you have
done nothing to address the issue except to side step it. The notion
that geometric algebra yields complex numbers may be a misnomer by the
work of your favorite authors in comparison to my few lines. I am
simply studying
( ei /\ ej )( ei /\ ej )
as a geometric product for which they get minus unity. I myself see an
easy path to yield positive unity. Not only does their answer look
wrong, but I only came to question it because their method looks
wrong. Humans are subject to poor behaviors as we are social animals.
So I understand that you may look askance at me. Still can't you see
that there ought to be a direct route to proving me wrong? Instead
what we see is that the math which is called geometric algebra is so
indirect that it steps on its own toes and has its waistband nearly up
to its shoulders. I try again:
( ei /\ ej )( ei /\ ej )
= (ei /\ ej) /\ (ei /\ ej) + (ei /\ ej) . (ei /\ ej)
where '.' represents the dot product and '/\' is the wedge product.
I've merely applied the definition of the geometric product. Now since
we are told that
v /\ v = 0
unconditionally we can cast the first part of the sum above away:
(ei /\ ej)/\(ei /\ ej) = 0
leaving
( ei /\ ej )( ei /\ ej )
= ( ei /\ ej ) . ( ei /\ ej )
to ponder. I have yet to see a formal explanation of dot products of
wedge products, but one poor rendition I've seen claims that the grade
difference of the sources is the grade of the result. This does jibe
with my own more primitive thinking especially for this case of
identical sources. So the result of
(ei /\ ej) . (ei /\ ej) = + 1
seems reasonable just as
ei . ei = + 1
though as I've explained elsewhere the wedge product itself may be ill
defined since it requires its own reuse within its definition. This is
a large pile of material which has self references within it holding
it together, hence the rabbit path. Be sure to scurry down the hole
before you make it all the way around for I'll be waiting to see you
come around again.
As I understand it here we are questioning the claim that this algebra
can sweep up and contain the complex plane. This is why their result
must come out to minus unity and so they've contorted the notation
badly. We see a similar contortion in lecture1 at page17 of Lasenby
and Doran from
http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectures.zip
When things get hazy please consider that it may not be merely the
haze in your own head; that some of it may be coming directly from
what you are reading. We are supposed to be capable of practicing
mathematics with rigor and so make an argument without the haze. Even
such arguments must still be grounded in a series of definitions or
axioms. These axioms can be freely constructed though their simplicity
and lack of tangling is paramount. If there were a tangle we might
hope that someone else could come along and untangle the mess. Yet if
noone is willing to admit the tangle in the first place because this
is mathematics and everything is presterilized and clean and it must
have been in order to get published then the problem becomes one of
human behavior. Can you really expect a group of people who got their
PhDs in this topic and have been working it into grant money that
ensures their family's well being for a few generations to confront
the tangle? Yes, that would be nice. But if there is no solution then
unfortunately they've self destructed the subject. It's probably best
to lead this group rather than follow in the leaders' wake, since your
own credibility can then be accented. The holdouts will be the real
losers still pushing their snake oil onto the youth generations from
now, perhaps with nobody even noticing the difference. How many topics
suffer this level of behavior? Academic institutions are under
pressure to be 'productive'. How much of modern productivity is
desperation in disguise?
Human reality suffers an unusual phenomenon in which belief becomes
reality merely via consensus. That can even be a sincere position, so
lets not put sincerity as a high value commodity. Hah. Openness seems
more the paradigm to consider. In structures which have their bishops
and priests and popes we will not hear what goes on behind closed
doors. In the technical subjects even when the door is left open if
the haze is thick enough what happened in the room even becomes
questionable between the leaders. Is this an accurate portrayal of the
human condition? What right does any human scientific thinker have to
remove themselves from it? It seems to me that the perfect scientific
stance is a fundamentally invalid assumption until problems are
accepted as open rather than as closed and done.
- Tim
Edward Green
<tttppp...@yahoo.com> wrote:
> http://bandtechnology.com/PolySigned
Hmm... Ok. Some interesting history:
http://en.wikipedia.org/wiki/Negative_and_non-negative_numbers
"In Hellenistic Egypt, Diophantus in the third century A.D. referred
to an equation that was equivalent to 4x + 20 = 0 (which has a
negative solution) in Arithmetica, saying that the equation was
absurd."
"In the 15th century, Nicolas Chuquet, a Frenchman, used negative
numbers as exponents and referred to them as 'absurd numbers'. "
http://en.wikipedia.org/wiki/Imaginary_number
"Imaginary numbers were defined in 1572 by Rafael Bombelli. At the
time, such numbers were thought not to exist, much as zero and the
negative numbers were regarded by some as fictitious or useless.
http://en.wikipedia.org/wiki/Quaternion
"In mathematics, quaternions are a non-commutative extension of
complex numbers. They were first described by the Irish mathematician
Sir William Rowan Hamilton".
http://en.wikipedia.org/wiki/Surreal_numbers
"In mathematics, surreal numbers are the elements of a field[1]
containing the real numbers as well as infinite and infinitesimal
numbers, respectively larger or smaller in absolute value than any
positive real number, and therefore the surreals are algebraically
similar to superreal numbers and hyperreal numbers".
So, once upon a time, we only had positive numbers, and then we had
negative numbers, and then... well, the rest is history (and
undoubtedly incomplete history in this post). Now, you are proposing
to add to the list with "polysign" numbers, by extending the list of
signs beyond "+" and "-". Bravo! It is the genius -- or the
pathology -- of mathematics to look at a given structure and ask what
would happen if we relaxed this, or added that.
Your disdain for the work of Grassmann and Clifford, which takes
something of the same flavor as the early skepticism for negative, and
later imaginary numbers (by the time we got the quaternions, I guess
people were inured to the fancy of mathematicians) is possibly ironic,
since you want to introduce yet another flavor of number system, or
algebraic system, to the menu. :-)
peeter
<tttppp...@yahoo.com> wrote:
> Peeter, you seem to be engaged in a long rabbit run. This entire
> subject suits you. Good luck. I stand by my previous post and you have
> done nothing to address the issue except to side step it. The notion
Hi Tim. I'm at a loss of how to reply to much of this. The only
specific question I see is "how to define the dot product on wedge
products". This can be done with or without geometric algebra. For
example, the Dover book by Flanders "differential forms with
applications to the physical sciences", does so with a specific
determinant, and then uses that to define the hodge dual. Geometric
algebra also has such a generalized dot product, but I don't think it
would be an effective use of my time trying to discuss either.
Other than the question of the generalized dot product, when I look at
my previous reply I believe it has the answers to the mathematics
parts of your last questions, and I cannot think of how to explain
those any more effectively.
Peeter
Timothy Golden BandTechnology.com
Good enough Peeter. Nobody else has stepped in to help clarify this
either. When we get two differing numerical answers to the same
expression and cannot resolve the discrepancy then we've got a strong
math problem. Is it your problem or mine or theirs? Maybe none of the
above. I have a pretty strong nose for piles of bullshit and I do
sincerely think that this is one. However sincerity is not enough is
it? Now I've instantiated a claim to dispute this geometric algebra,
or at least under this conversation with you one portion, but the
criticism of the wedge product goes to the core. I don't actually
believe that
v /\ v = 0
makes any sense, but then I also don't believe that the wedge product
makes any sense. If we simply accept the graphical meaning of the
wedge and especially in grade two as an area then how can we be happy
with a six component decomposition for a 3D wedge product? How do
those six areas cleanly represent what was two vectors forming a clean
area? The meaning of the wedge operation is unclear. Now, let's
consider a constant unit area and vary those two vectors that form it.
As the angle between them grows closer and closer we'll see the extent
of the area shoot out quite far but all the while maintaining unity
area and really both components are still variable in magnitude. We
can do this down to arbitrarily small angle and maintain the constant
area, yet when we get to zero angle we are supposed to accept a drop
out of this constant area. Is it illegal? To exclude zero angle as
illegal does not seem right to me. It would be wise to consider that
as we procede very close in angle to zero that the vectors forming the
area grew very large; that an infinite bi-ray could exist whose normal
is ill defined and whose area is ill-defined, yet whose wedge product
v1 /\ v2
is balanced. For instance if we force one of the vectors to be unity
length and assign the variable len to the other at an angle theta we
will see that for unity area
1 = len * sin( theta )
where '*' is multiplication of raw numbers. We now have a clean
relation between a length and an angle for a given area. We can now
approach theta = 0 by halving an initial nonzero value of theta and
produce corresponding len to maintain this unity area.
The cleanest description for
v /\ v
is that it is undefined. The zero stance I believe to be false. This
would be most consistent with the methods of real number arithmetic
where division by zero is undefined. At a philosophical level if one
were to choose the zero value for v/\v then it would follow that you
could eliminate the zero exception out of the field laws of real
arithmetic, therby forcing
1 / 0 = 0.
But we see the lie don't we? as we get close to zero the expression in
fact grows very large. So as we construct
1 / b
and drop b in half succesively who are we to claim that the result is
going to be zero when b zeros out? The analogy isn't perfect, but I do
believe it is general enough to apply back onto the alleged wedge
product. To place an exception into a math is highly undesirable and I
do see that as an open topic, so here is an argument for someone who
believes in the real number as it has been done in the past applied
onto the wedge product. To widen the view is to question the product
and the quotient in general. If I take three oranges and multiply them
by five oranges they say I will have fifteen oranges but as I try to
count them I only see eight oranges. Is any product sensible? The
space of the resultant does seem to be a new space. Should we create
space? Perhaps we are barking up a quotient natural tree. By assuming
the quotient to be natural perhaps then the zero exception need not
exist. Perhaps zero does not exist. Geometry has long been tied to the
real number in our thinking. The usage of a zero as the origin... I
think that zero ought to be a unity, but I haven't fully figured out
how to make a new physics of it.
- Tim
Timothy Golden BandTechnology.com
Very fine criticism Mr. Green. If you understand how three-signed
numbers are the complex numbers and how the rules are identical to the
real numbers, then I think you will see an instance of the quality of
the construction. The fundamental twists that a Cartesian thinker has
to take to unwind the polysign family are challenging but the final
result is consistent with much of existing math and physics. P1 is
especially troubling yet if one follows the general sign rules and
makes no exceptions then we see a unidirectional zero dimensional
definition consistent with time behavior in its geometrical quality.
In terms of disdain I could get a lot more filthy with my language.
I'll wait for Uncle Al to show up on this thread for that. I do mean
my criticism of the wedge product. It's not just to promote polysign
numbers. I'm amazed that the wedge product is so readily accepted.
The polysign rules redefine the real number. There are already an
obnoxious number of definitions. The catch is that out of this one the
complex numbers, time, and a myriad of higher dimension algebras
follow with no additional work. All are well behaved algebraically.
Polysign math deeply challenges the existent system yet does so
without conflict.
- Tim
peeter
<tttppp...@yahoo.com> wrote:
> ... I don't actually
> believe that
> v /\ v = 0
> makes any sense, but then I also don't believe that the wedge product
> makes any sense. If we simply accept the graphical meaning of the
If you want to calculate with it, then I would say your beliefs and
requirement to accept this is what is getting in the way. Math is not
a matter of belief. It is nice to have intuition, especially
geometrical, but it not required. When it comes down to it, the wedge
product is a definition. You take that definition and work with it.
The wedge product has a single fundamental defining properly, it is a
bilinear antisymmetric operator. The antisymmetry can be expressed in
symbols as:
v ^ w = - v ^ w
An implication of this _definition_ is that with w = v you have:
v ^ v = - v ^ v
so one writes v ^ v = 0 (a zero element in the space of /\^2).
Now, you can go on and fit this wedge product in other contexts,
geometrical and otherwise, but those again will be subject to
definitions. Since you have some geometrical intuitive requirements,
I suggest you discard them, at least until you are comfortable with
the calculations. Take a specific application of the wedge product,
as defined above, and do something concrete with it. For example, you
need nothing more than the definition above to prove Cramer's rule.
I'd suggest trying this for the simple real numbered cases of a two
variable R^2 linear equation. If you don't have any trouble with
that, then move on and try it for the three by three and general
cases.
Peeter
Androcles
"peeter" <peete...@gmail.com> wrote in message
news:4e280b58-f601-4716...@m45g2000hsb.googlegroups.com...
On Aug 11, 8:28 am, "Timothy Golden BandTechnology.com"
<tttppp...@yahoo.com> wrote:
> ... I don't actually
> believe that
> v /\ v = 0
> makes any sense, but then I also don't believe that the wedge product
> makes any sense. If we simply accept the graphical meaning of the
If you want to calculate with
You can't.
Spaceman
> If you want to calculate with it, then I would say your beliefs and
> requirement to accept this is what is getting in the way. Math is not
> a matter of belief. It is nice to have intuition, especially
> geometrical, but it not required. When it comes down to it, the wedge
> product is a definition. You take that definition and work with it.
> The wedge product has a single fundamental defining properly, it is a
> bilinear antisymmetric operator. The antisymmetry can be expressed in
> symbols as:
>
> v ^ w = - v ^ w
If in the above
Let v = -4
and w = 1
You would end up with
16 = -16
Hi peeter
Do you truly believe that
v = -v?
or that every negative number equals a positive number?
peeter
wrote:
> > v ^ w = - v ^ w
>
> If in the above
> Let v = -4
> and w = 1
> You would end up with
> 16 = -16
It looks like you are jumping in without context. v and w are not
numbers here but elements of a vector space and this product (wedge,
written here in plain text with ^, where it is also not clear that
this does not represent exponentiation) defines a mapping from pairs
of vector spaces to elements of a "product space". Adjusting your
question to use the simplest example of one dimensional vectors, with
a basis vector e_1
v = -4 e_1
w = 1 e_1
v ^ w = -4 e_1 ^ 1 e_1 = -4 (e_1 ^ e_1) = 0
You have to go to at least two dimensions to get a non-zero result:
v = a_1 e_1 + a_2 e_2
w = b_1 e_1 + b_2 e_2
v ^ w = a_1 b_2 e_1 ^ e_2 + a_2 b_1 e_2 ^ e_1 + a_1 b_1 e_1 ^ e_1 +
a_2 b_2 e_2 ^ e_2 = (a_1 b_2 - b_1 a_2) e_1 ^ e_2
Here e_1 ^ e_2 can be considered a basis element for the product space
in question (this is a linear space much like a vector space). All
the elements that are linear combinations of e_1 ^ e_1 and/or e_2 ^
e_2 are zeros of this product space.
Peeter
Timothy Golden BandTechnology.com
wrote:
I do think that Peeter is coherent however he is only taking the
puzzle one piece at a time. Within a circular thought process this
will not expose the circle. Again we see zeros entering the topic and
getting misused.
Peter also has made a choice to dodge the minus unity part of this
thread.
I'm waiting to see a direct address of the conflict exposed on
( ei /\ ej )( ei /\ ej ) ,
the geometric product of a unity wedge with itself which two of
Peeter's favorite authors claim is negative unity while I claim it is
positive unity.
So far I think I have about three semi-unique attacks on this topic:
1. The definition of the wedge product is self referential and so
cannot be a definition in the axiomatic sense.
2. The wedge product v/\v=0 is questionable in the calculus sense on
a continuum of 2-blade areas. Simply sweep a unit area 2-blade down to
small angle.
3. Results of Dorst and Mann yield
( ei /\ ej )( ei /\ ej ) = - 1
whereas I believe the correct answer is
( ei /\ ej )( ei /\ ej ) = + 1.
This result appears to link to the claim of geometric algebra's
compatibility with the complex plane. That claim may be false.
Have a little self respect Peeter. You need not be a pawn of this
system. We all are struggling with this information and its validity.
Yet to merely posit one small piece at a time is not going to be a
strong enough defense.
I am the one who has put myself out on a limb here and I am entirely
open to that limb falling to the ground. Please attack freely and
fully. Let's get this over with. If only it was this easy. But you see
my arguments on the human race do hold. If you concede then your own
sense of reality slips doesn't it? As the problems open up the world
feels a bit more overwhelming. The mathematician is happy to
construct a fairy land and has no requirement that reality be
consistent with their math. A math need only be self consistent. Here
though this math called geometric algebra does seem to have some
rotten spots even within this self consistency requirement. If I were
to tell you that the definition of the operator
&
was
x & y = z & w
wouldn't you care to criticize me? This is exactly what has been
perpetrated in the wedge product (but times (D^2-D)/2) as best I can
see. Please disarm me. So far you've merely empowered me by neglecting
my argument.
- Tim
peeter
<tttppp...@yahoo.com> wrote:
> Have a little self respect Peeter.
I have plenty of self respect, but I have now exausted my patience for
what I originally misinterpretted as honest questions that I could
answer. I'm not going to be baited into anger to write long diatribes
on philosophy in what I thought was a mathematics thread.
Peeter
Spaceman
Hi Timothy,
I have to ramble about something , forgive me.
A negative number multiplied by itself should not
actually produce a positive number.
Let me ask you a question....
If a negative number is multiplied, and because it is multiplied
it can jump the 0, shouldn't a positive number be allowed to do
the same?
Or should we only allow such "jumping of the 0" to one direction
and not the other?
Or better yet, should we not allow such jumping the 0 at all
and allow negative times negative to equal a negative?
> Have a little self respect Peeter. You need not be a pawn of this
> system. We all are struggling with this information and its validity.
> Yet to merely posit one small piece at a time is not going to be a
> strong enough defense.
> I am the one who has put myself out on a limb here and I am entirely
> open to that limb falling to the ground. Please attack freely and
> fully. Let's get this over with. If only it was this easy. But you see
> my arguments on the human race do hold. If you concede then your own
> sense of reality slips doesn't it? As the problems open up the world
> feels a bit more overwhelming. The mathematician is happy to
> construct a fairy land and has no requirement that reality be
> consistent with their math. A math need only be self consistent. Here
> though this math called geometric algebra does seem to have some
> rotten spots even within this self consistency requirement. If I were
> to tell you that the definition of the operator
> &
> was
> x & y = z & w
> wouldn't you care to criticize me? This is exactly what has been
> perpetrated in the wedge product (but times (D^2-D)/2) as best I can
> see. Please disarm me. So far you've merely empowered me by neglecting
> my argument.
I agree about mathematicians losing the reality.
But part of thier loss of reality stems from the allowance
of a postive number (in the negative dimension) to jump
the dimension if multiplied by itself.
In reality, there are no "real" negatives.
All negatives are simply positives headign in a different
direction.
If I asked you to move negative 5 meters aways from
where you are right now, wouldn't you actually have to move
a positive 5 meters away no matter what direction you called
negative?
Spaceman
Ok, let me rephrase my question then,
Peter,
Do you actually believe that a negative "vector space" equals
a positive "vector" space?
such as the v^w = -v^w that you have accepted?
Daryl McCullough
>Now I've instantiated a claim to dispute this geometric algebra,
>or at least under this conversation with you one portion, but the
>criticism of the wedge product goes to the core. I don't actually
>believe that
> v /\ v = 0
>makes any sense, but then I also don't believe that the wedge product
>makes any sense.
Okay, well here is an excercise that helps understand v /\ v = 0.
Suppose you have four points P1, P2, P3, P4 forming the vertices of
a parallelogram. What is its area? One way to figure it out is to let
vector A be the displacement from P1 to P2, and let vector B be the
displacement from P2 to P3. Then the area of the parallelogram is given
by |A /\ B|. We want this to be zero if A and B are parallel (because
then the parallelogram is flat, with zero area).
In Euclidean geometry, |A /\ B| = |A||B| sin(theta) where theta
is the angle between A and B. Again, if A and B are parallel, then
the angle theta is zero, and the wedge product gives zero.
Another way to see it is that the meaning of A /\ B is that it is a
directed area (a plane, or a 2D surface) determined from the vectors
A and B. A directed area is an area, together with an orientation in the
same way that a vector is a length together with an orientation.
If A and B are in the same direction, then it doesn't specify
a plane.
With three unit vectors e_1, e_2, e_3, you can make 3
different parallelograms: one with sides e_1 and e_2,
one with sides e_2 and e_3, and one with sides e_1 and e_3.
With four unit vectors e_0, e_1, e_2, e_3, you can make
6 different parallelograms.
>If we simply accept the graphical meaning of the
>wedge and especially in grade two as an area then how can we be happy
>with a six component decomposition for a 3D wedge product?
A 3D wedge product A /\ B /\ C is a directed volume, in the
same sense that a 2D wedge product is a directed area, and
a vector is a directed length. To specify an area, you give
two non-collinear vectors A and B, and the corresponding area
is the parallelogram with sides A and B. To specify
a volume, you give three non-coplanar vectors A, B, and C,
and the corresponding volume is the parallelepiped
(that's a 3D figure with 6 sides, each side being a 4-sided
parallelogram, see http://en.wikipedia.org/wiki/Parallelepiped)
with two faces being a parallelogram with sides A and B, two
more faces being a parallelogram with sides A and C, and two
more faces being a parallelogram with sides B and C.
With three unit vectors e_1, e_2 and e_3, you can make only
one parallelepiped: e_1 /\ e_2 /\ e_3. With 4 unit vectors, you
can make four different parallelepipeds: e_1 /\ e_2 /\ e_3
e_0 /\ e_2 /\ e_3, e_0 /\ e_1 /\ e_3, e_0 /\ e_1 /\ e_2.
>How do those six areas cleanly represent what was two vectors
>forming a clean area?
Well, that area (a parallelogram) has two edges. In 4-D
spacetime, (so there are 4 basis vectors), there are 6 different
ways to choose two vectors to serve as the two edges.
>The meaning of the wedge operation is unclear. Now, let's
>consider a constant unit area and vary those two vectors that form it.
>As the angle between them grows closer and closer we'll see the extent
>of the area shoot out quite far but all the while maintaining unity
>area and really both components are still variable in magnitude. We
>can do this down to arbitrarily small angle and maintain the constant
>area, yet when we get to zero angle we are supposed to accept a drop
>out of this constant area. Is it illegal?
When the angle between the edges of a parallelogram go to zero,
then the edges no longer specify a plane. You need non-collinear
edges to specify a plane.
A wedge product of two vectors is not just a number, like an
area. It also has an *orientation*, like a vector. You can't
determine an orientation for a plane determined by collinear
vectors.
>To exclude zero angle as illegal does not seem right to me.
>It would be wise to consider that as we procede very close
>in angle to zero that the vectors forming the area grew very
>large; that an infinite bi-ray could exist whose normal is
>ill defined and whose area is ill-defined, yet whose wedge product
> v1 /\ v2
>is balanced. For instance if we force one of the vectors to be unity
>length and assign the variable len to the other at an angle theta we
>will see that for unity area
> 1 = len * sin( theta )
>where '*' is multiplication of raw numbers. We now have a clean
>relation between a length and an angle for a given area. We can now
>approach theta = 0 by halving an initial nonzero value of theta and
>produce corresponding len to maintain this unity area.
>
>The cleanest description for
> v /\ v
>is that it is undefined. The zero stance I believe to be false.
The point is that we want a directed area to be defined by
only two things: (1) an area, and (2) an orientation. So
different parallelograms resulting in the same area and
orientation are to be considered the same wedge product.
Let A and B be two vectors, and consider the two wedge
products:
w1 = A /\ B
w2 = A /\ (A + B)
If you draw the corresponding parallelograms, you can
easily see that they have the same area. And they also
have the same orientation. So we should consider them
the same wedge product. So we have
A /\ B = A /\ (A + B)
But if we want /\ to be a bilinear operator, then it
must be that
A /\ (A + B) = A /\ A + A /\ B
So we have to make A /\ A = 0.
One way to think of it is that the only parallelogram that
*lacks* an orientation is the one with zero area. This is
very much like the fact that a vector with zero length
also lacks an orientation.
--
Daryl McCullough
Ithaca, NY
Timothy Golden BandTechnology.com
wrote:
> > If you want to calculate with it, then I would say your beliefs and
> > requirement to accept this is what is getting in the way. Math is not
> > a matter of belief. It is nice to have intuition, especially
> > geometrical, but it not required. When it comes down to it, the wedge
> > product is a definition. You take that definition and work with it.
> > The wedge product has a single fundamental defining properly, it is a
> > bilinear antisymmetric operator. The antisymmetry can be expressed in
> > symbols as:
>
This line above is a typo I think. It's actually
v ^ w = - w ^ v
If the sources are vectors v and w we are then supposed to accept that
their areas are signed. Still, maybe this is yet another valid attack
on the wedge product. I take a vector ( 1.2, 3.4, 5.6 ) and another
(1.3, 2.4, 3.5). Now using the continuum concept again I maintain
their unit vector directions but shift their magnitudes such that they
conserve area but eventually swap magnitudes. Now we see that they are
supposed to invert, yet they cannot since we've provided a constant
area under the entire transition. Here we see two fairly similarly
directed vectors. Oh, I see, this is not a complete argument but it
does run back over to tensor concepts. The space that geometric
algebra works over in not isotropic. It's the only way that you could
try to enforce the negative result. Still, just going down this path
of thought I think exposes the absurdity of the construction. These
operations are on a fixed reference frame. This is fine, but that
quality should be pointed out more clearly than the subject builds it
out. Especially because they start in a vector space and wind up in
the wierd space of wedge products they can blow off such concerns. I'm
not quite sure that this attack is a number four in the arsenal
against geometric algebra but it's close to one.
I do take interest in forms that are structured and especially of this
antisymmetric tensor representation but their construction seems
completely foul as they've built it. It's like they were reaching for
the antisymmetric tensor the whole time and wound up in this
contraption to get them there.
- Tim
>
> If in the above
> Let v = -4
> and w = 1
> You would end up with
> 16 = -16
>
peeter
wrote:
> Do you actually believe that a negative "vector space" equals
> a positive "vector" space?
A statement like "does a negative vector space equal a positive vector
space?" doesn't actually make sense. We are talking about elements of
a vector space, and not the vector space itself. Loosely, you would
say that an element A of a vector space that qualifies as a "zero" in
a vector space will also satisfy the identity:
A = -A
These are fairly fundamental concepts, so I would recommend that that
you pick up a good introductory book on linear or vector algebra.
There are thousands of such books (your public library probably has
five of them), any of which will explain in much more detail than I
would be inclined to.
Peeter
Timothy Golden BandTechnology.com
wrote:
I think you've made a reasonable argument here.
If you attempt to apply relativity on your 1D line then you have some
options to keep in mind. One of these is to choose whether or not each
particle position has its own reference directions. This would be the
fullest freedom. Maybe a bit beyond there is to give each one its own
scale as well; then they'd each have their own numerical distance to
each other. I haven't found any good use for that yet but already just
in a 1D relative frame there are a fair number of qualities to keep in
mind. Now have your products gained some freedom? How you construct
them determines their result. In effect we may have now made generic
particles with no qualities and put all of the quality into their
reference frames. Does particle 'a' communicate its impression of b
back to b? Ugh, you might be thinking. Me too. How much freedom is
there? I've messed around with lots of these and up in 2D (the complex
plane) it's tough to feel comfortable. We want clean motives, not
arbitrary ones. This is the best reason that I can provide to bother
working in the polysign basis. It provides natural support for
spacetime, including unidirectional time. One of the consequences of
polysign numbers is that sign is dimension, just off by one:
n = D + 1
where n is the signature of a number system and D is the geometrical
dimension of that number system. So for instance the four-signed
numbers are three dimensional. Under product we have to admit that
even the real line is anisotropic. Your doubts seem valid in terms of
symmetry. Under the polysign math it is appropriate to look upon your
real line manipulations as rotation. In a 1D space the rotation
options are binary.
The classical force equations do seem to use sign. Under coulomb
forces we do witness attraction and repulsion so a two-signed paradigm
seems fitting, though the resultant force is in terms of acceleration,
not raw position. Natural products seem not to make any sense in the
source space. Somehow they are coming out in the second derivative.
Sign is discrete and charge is discrete. Mass is likewise discrete and
seems one-signed, though there is an electron mass and so forth which
don't seem to mesh quite so cleanly as with charge.
Polysign math uses magnitude as fundamental, not as derived. So we
combine a sign s with a magnitude x to get to the rudimentary
s x
but still upon taking products in polysign I have to admit that I
don't have a deeply convincing argument to remedy your criticism.
Likewise if I take five apples and multiply them by two apples I still
only seem to wind up with seven apples. Superposition is native.
Somehow we've got to work on getting our products to the second
derivative cleanly. Going up in dimension points to torque forces even
at 2D. I don't have anything to say with finality, but I do think this
is a valid direction to expolore. As we look out and see cosmological
jets and we look in and claim electron spin and plenty of vortex
dynamics in between these scales it is easy to accept rotational
influences, though modern physics prefers a linear approach.
- Tim
Spaceman
> On Aug 11, 1:05 pm, "Spaceman" <space...@yourclockmalfunctioned.duh>
> wrote:
>> Do you actually believe that a negative "vector space" equals
>> a positive "vector" space?
>> such as the v^w = -v^w that you have accepted?- Hide quoted text -
>
> A statement like "does a negative vector space equal a positive vector
> space?" doesn't actually make sense. We are talking about elements of
> a vector space, and not the vector space itself. Loosely, you would
> say that an element A of a vector space that qualifies as a "zero" in
> a vector space will also satisfy the identity:
>
> A = -A
Actually the statement does make sense if you actually tried
to think about it.
Set A equaling a negative set A is a problem along with
element A equaling a negative element A would be,
if not 0.
Do you only wish to use elements or sets that are 0 or would you like
to work with other elements and sets also.
A = A
not
A = -A
This is fundamental algebra also, nevermind elements, sets, and so on..
The algebra must be ok in all situations, not just a 0
only situation.
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
Spaceman
The problems all stem with non-euclidian geometry.
Euclidian geometry and 3D is beautiful and works so well it is amazing
anyone would want to work with anything other than such.
But, for the "rubber rulers" and "malfunctioning clocks" of relativity
Someone needed to accept non euclidian geometry for the
math to have any structure and yet all it does is actually produce
paradox after paradox and more bullshit math such as
u+v for high speeds is wrong yet it is ok to use it in
u+v/1+(u*v)c^2
So basically they have
(supposedly wrong answer)/1 + (u*v)c^2
and think it is ok to use such "supposed wrong" math
in thier own math.
If you remove observational limit of lightspeed and use basic
good old fashion math along with good old fashion euclidian
3D and geometry and of course with good old fashion
single standards for time and distance, no paradoxes appear
in anything occuring from cause to effect.
Timothy Golden BandTechnology.com
> On 10 Aug, 23:21, amy666 <tommy1...@hotmail.com> wrote:
>
> > for any questions about the polysigned , you may ask me.
>
> Good to know.
>
> > nice to see someone intresting in polysigned again here on sci.math.
>
> I too am very interested. I have already some code in place with
> closed real interval arithmetic, but rather than extend to the complex
> numbers, I'm looking for a way to extend to the polysigned. The
> problem is, I'd have so many questions...
I'm happy to help you out either privately via email or here. I have
some code though it's not maintained for distribution. If you're
interested I'll post a link. It's C++ code but mostly straightforward.
The polysign numbers are straightforward, but modern humans are so
trained upon the real number as their reference that the leap is
difficult.
Tommy you've managed to split this thread by posting to just sci.math.
I'm posting back to the original groups so hopefully that will bring
us back into the main thread through my google interface. On
sci.physics your messages are not present.
- Tim
>
> -LV
>
> > regards
>
> > tommy1729
Timothy Golden BandTechnology.com
Going over to the Minkowski metric does get into negative distance.
Einsteins theory does yield experimental difference doesn't it?
Do you predict the angular difference in a star's light passing to us
near the sun? I tend to agree with the rubber ruler problem and that
all clocks are mechanical apparatus and so reliant upon space to
perform. If we apply time dilation to a photon we'll see that the
photon thinks that it is instantaneous. Zero seconds elapse from its
emission to its absorption in its own frame of reference. Yet in our
own frame delays can be measured. I do think that this feels absurd.
Is it consistent to simply grant the photon some small mass? That is
one way out isn't it? I haven't tried to do this math out and to do it
we'd wind up using some more of Einsteins equations.
I think Einstein was brilliant and you only need read a small portion
of his work (though I don't read German) to see that he was a man of
reason. But that doesn't mean that he was perfectly correct or that we
cannot still look at those problems as open problems and seek
alternate solutions.
The usage of time as a full dimension; that I can cast serious doubt
upon. Is electromagnetism the source of the velocity of light? Where
is electromagnetism in the foundation of relativity theory? Should we
anticipate that a new theory would contain electromagnetism within it
since some of those behaviors(electrin and magnetic constants) are
attributed to space itself?
If Einstein was free to curve space then what else can we do with it?
Could we really have free space? Like an ether that could be gobbled
up but you'd never run out of it. I don't know. If we are going to
play with spacetime should we consider that the fundamental question
Why spacetime?
has been left out of theory. Three dimensions of space are not a
trivial choice, yet the math that has been used does construct this
trivially. So much so that when string theory comes along they are
left out of the lurch on this simple minded question because those
that came before never bothered to answer it either. This particular
point I found to be troubling and so I attribute my own awareness of
this subject to the string theorists who were happy to build a ten
dimensional space and write in by hand three or four extended
dimensions. Theory thus far has affixed three dimensions to space by
choice. A more rigorous math system will produce spacetime as well as
the physics in spacetime.
There are actually a lot of people messing around with dimension and
trying to get spacetime. There is an article in Scientific American of
a group doing just that in this or last months issue. The quantum
gravity people insist on this as well. Brane theorists (who were
string theorists) are branching out in as many directions as they can
playing out variations and attempting a new mastery of dimension.
There is already so much to wade through out and about that you'll
have no problem swamping yourself.
I have found a fundamental math that is consistent with spacetime. It
is called polysign numbers. Anyone can produce this math if they sit
down and work it out themselves. It isn't obvious, but likewise it's
difficult to go astray since little else will make any sense as you
try to generalize sign. This math leads me to make quite a lot of
statements and in terms of relativity I point out that the Minkowski
metric is an instance of structured spacetime. By taking this further
we will arrive in a consistently structured spacetime; the 3D
isotropic space is a misnomer. Relativity itself is the isotropic
feature. That freedom to choose ones own reference frame alleviates
the math beneath from having any isotropic requirement. Now, lets ask
if Maxwell unified electricity and magnetism?
Why do we still need these two words for something which has been
unified? Is electricity unique from magnetism? This takes us right
back down to geometry and a rotational plane with a line orthogonal to
it and this is the structure that I propose space actually is. Space
itself is structured but it is structured per particle. Hence an
electrons spin becomes self consistent since we've now inserted
electromagnetic behaviors into space itself. Informationally we've
shifted some of the complexity into spacetime and alleviated the
particle of some of that complexity. Here again we see that Maxwell's
equations are for raw charge in explicitly isotropic space, yet
Maxwell himself described the magnetic field as circulation of space.
If he were exposed to electron spin he'd be revising his system
because the raw charge is not a valid assumption.
There are a lot of signals that line up nicely. One strong one is the
electromagnetic tensor whose format mimics the tatrix of the polysign
progression. I think it is hot flashes of this tensor that are keeping
algebraic geometry and its wedge product alive. This field of math,
especially the wedge product, is contorted. I have read one small
snippet that claims string theory is relying upon this. I haven't gone
into there yet. My study of this subject is only about a week old but
I have been meaning to get into it since I read an insinuation of a
generalized cross product in a linear algebra book by Zelinski:
"As a matter of cultural interest, perhaps we should remark that
there is an 'exterior product' in n-space that is essentially the
cross product when n = 3; but the exterior product of two vectors is a
kind of tensor, not another vector." -1968, of A First Course in
LINEAR ALGEBRA, Daniel Zelinski
This is the only paper reference that I've used so far and it's pretty
thin coverage. Still, it exposes that this math is fairly old. I wish
somebody would come and defend it better that what we've got so far.
- Tim
Spaceman
That is because it is absurd,
It is a direct violation of the science of measurement.
It has accepted and uses multiple standards for time and had to
include the multiple standards for distance to make up for the
wrotten math that actually violates basic math like there was no
reality at all.
The most simplest violation it uses is a transform.
They will scream that the answer for u+v is wrong when
used alone, yet they will use it in u+v/1 + (u*v)c^2.
This is like using
(wrong answer)/ 1 + (u*v)c^2
How much of a joke is that to begin with?
What they are ignoring is the limit of observation caused
by a limited speed of light to prop up this violation.
A simple mathematical problem with collision speeds
occurs.
Take two ships traveling towards point A at 0.9c wrt A.
Each ship is on opposite sides one coming from direction x
and one coming from direction -x.
The closing speed is 1.8c. and point A knows this well.
But according to the ships using the stupid ass transform,
The collision will not occur until after point A says it will.
So.
Spacetime is a complete joke to reality and to the science
of measurement.
It has faults that would end up making zillion dollar starships
crash into things that were supposedly not there yet all because
of some stupid ass use of a limitation of light for observation.
I know one thing, I will never let a relativist drive my staship
at close to light speeds, and if it could go FTL, I would nto even let
them see it just so they could stay stupid forever.
:)
(I know... I don't have a dang starship)
But if I did. a relativist would never be able to use it correctly.
> Is it consistent to simply grant the photon some small mass? That is
> one way out isn't it? I haven't tried to do this math out and to do it
> we'd wind up using some more of Einsteins equations.
> I think Einstein was brilliant and you only need read a small portion
> of his work (though I don't read German) to see that he was a man of
> reason. But that doesn't mean that he was perfectly correct or that we
> cannot still look at those problems as open problems and seek
> alternate solutions.
Einstein was brilliant, he was also smart enough to get the rubber rulers
and malfunctioning clocks accepted as scientific instead of finding
out the true cause for the clocks malfunctions.
He got all to accept the "rubber rulers" instead.
I am sorry, but I can not defend anything that props up spacetime,
Spacetime is a mixing of two seperate measurement systems.
It is against my scientific practice and also very much against
mechanical laws.
When using Euclidian 3D space and a seperate dimension for time.
Everything works out perfectally when you find each and
every Newtonian force that actually occurs and causes effects.
Spacetime will only make a starship of the future crash when
it should not accorging to the on board clock.
The first step is to find out what "physically" changes the clocks
tick rate from acceleration and g-potential differences.
In short what "truly causes gravity" instead of accepting a
(counting method merged with a distance measurement method)
(spacetime) as a physical cause.
Then science will again take bounds and leaps instead of small
steps.
Timothy Golden BandTechnology.com
I just found after equation 5 of
http://mathworld.wolfram.com/WedgeProduct.html
"while the formula v^v=0 holds when v has degree one, it does not
hold in general."
So I've got to withdraw my number 3 above here. Still I think that
v^v=0 is poor as described in number 2.
- Tim
Timothy Golden BandTechnology.com
> I am sorry, but I can not defend anything that props up spacetime,
> Spacetime is a mixing of two seperate measurement systems.
> It is against my scientific practice and also very much against
> mechanical laws.
> When using Euclidian 3D space and a seperate dimension for time.
> Everything works out perfectally when you find each and
> every Newtonian force that actually occurs and causes effects.
> Spacetime will only make a starship of the future crash when
> it should not accorging to the on board clock.
> The first step is to find out what "physically" changes the clocks
> tick rate from acceleration and g-potential differences.
> In short what "truly causes gravity" instead of accepting a
> (counting method merged with a distance measurement method)
> (spacetime) as a physical cause.
> Then science will again take bounds and leaps instead of small
> steps.
>
> --
> James M Driscoll Jr
> Creator of the Clock Malfunction Theory
> Spaceman
You haven't addressed the starlight bending as it reaches us by
passing near the sun's surface.
Could you please?
- Tim
Mariano Suárez-Alvarez
It appears so very odd to me, that someone may admit
such a total ignorance of what the wedge product is
and how it behaves in such an explicit way, and, concurrently,
present what's supposed to be a critique of the notion and,
to boot, offer a replacement.
Essentially most of the very florid and apparently deep
state of philosophical angst your writing on this matters
transpires through words like 'appears to be', 'conflict'
and so on, would be very efficiently helped by the simple
and straightforward procedure of studying a little bit of
multilinear algebra. While at the time of Grassmann and
Hamilton such tones are understandable, with the appropriate
dose of historical perspective, nowadays, to be honest, they
provide very little apart from a minor, poor entertainment.
Today one does not "withdraw claims regarding that
all elements of the exterior algebra square to zero",
as it is better to reserve such verbs to more worthy
contexts: today one simply gets a minimally good text,
sits down, and studies, preferrably leaving all
pontification on the matter at hand to a point in
time in which one really knows what one is talking about.
-- m
-- m
Spaceman
Heat effect,
It even occurs when light is seen passing over a hot tar road
but the asphault has massive currents of different heat flowing
from wind and air.
The Sun is constantly generating massive amounts of sub particles
and these particles are generated in a basic outward spherical direction.
It is a lens effect basically also.
It could even be caused by a massive static field of free electrons.
There are many "causes" that could be explored but have not been
attempted because all the money has been wasted on "relativity and
spacetime"
alone, sadly.
:(
--
James M Driscoll Jr
Creator of the Clock Malfunction Theory
Spaceman
> - Tim
Timothy Golden BandTechnology.com
Mariano, why don't you explain to me how
( ei ^ ej )( ei ^ ej) = - 1 .
When you sit down with a lot of books that tell you how to do this you
are sure to get their answer aren't you? If the math is invalid, and
all maths must be open to such challenge, then you would be the one
who is lacking.
The left hand expression above expands by definition to
( ei ^ ej ) ^ ( ei ^ ej ) + ( ei ^ ej ) . ( ei ^ ej ) .
From Wolfram we see that higher grade wedge products like the left
component of the sum above do not necessarily have to be zero. What
else gets conceded? The old rule that
a ^ b = - b ^ a
must also be invalid for higher grades, otherwise this statement from
Wolfram cannot make any sense. Now could we go back to seeking some
reduced result? Is it valid to believe that
( ei ^ ej ) . ( ei ^ ej ) = + 1 ?
Does this then mean that to get out the result of minus one that
( ei ^ ej ) ^ ( ei ^ ej ) = - 2 ?
Are these things even properly defined in this math? No, they are not.
The wedge product reuses itself within its own operational definition.
What of my other points? You have come out of the woodwork, now why
not substantiate your own position by correcting me on something so
simple as the idea that as one considers a unity area wedge product
and sweeps it to small angle that suddenly it turns to zero even while
it held its unity area right up till then. This is more like the
division by zero exception of the real numbers. Yes, its more
complicated, but not much more. We can have arbitrarily small angle
and maintain unity area. It follows that
v ^ v
is undefined. When we approach division by zero we cannot simply plop
a zero in for something that cannot be computed. In the division by
zero case the result gets very large whereas in this case the result
remains constant. This difference does not alter the outcome.
My lines of attack are crossed and that is because even if I argue
about
a ^ b = - b ^ a
I'm already dismissive of the very definition of '^'.
Hamilton did not use these contraptions to develop the quaternion did
he?
I think in terms of rhetoric we should use it in this environment to
get poeple like you to speak up. I do try to write carefully. I guess
it may get a bit flamboyant at times. Does this help a reader
reconsider the situation?
What about that situation? Are academic institutions under pressure to
be productive even while they may not actually have much product? Does
this effect have an influence on the product? After the first round of
variations have been played out the stretch to the second round grows
fictional. I do think that the wedge product is a second round sort of
thing because the first round didn't go far enough. It's tripped up,
tricked out, and has its waistband up to its chest.
You Mariano have preached the mimic. You have pledged a closed mind.
The book must be right. Preserve the book. Are you religious? That's
not very scientific.
Or maybe this subject just isn't your strong suit so you hesitate to
straighten out a beginner. Straighten me out please. I think that at
the junction of the geometric product to the complex plane is a good
place to start, but really the attention needs to drive down to the
definition of the wedge product. Is the wedge product definition self
referential? Informationally isn't this important? I've already
clearly stated this elsewhere on this thread. Mariano has criticism of
my usage of rhetoric but he has thus far no critical content on the
math beneath the rhetoric.
- Tim
p.s. Mariano, I know you are a high caliber mathematician. Could you
please actually apply a bit of your ability to this rather than parrot
the book?
I've layed my arguments pretty cleanly I think though this thread has
gotten so long I'm a bit lost. At the title "Is Geometric Algebra a
valid subject?" is a fine starting position. the points that I make
are brief there and may be filled out a little bit wider elsewhere,
but they are an easy read and bulleted too.
Daryl McCullough
>We can have arbitrarily small angle
>and maintain unity area. It follows that
> v ^ v
>is undefined.
No, it doesn't. |v ^ v| is the area of a parallelogram
with angle zero between adjacent edges. That is perfectly
well-defined; it is zero.
The only way for A^B to approach a nonzero value as
the angle between A and B goes to zero is if |A| goes
to infinity, or |B| goes to infinity. So you could
say that v^v in the case where v has infinite length
is undefined. That's the case analogous to 0/0 in
arithmetic. If v has a finite length, then v^v is 0.
There is no ambiguity.
>When we approach division by zero we cannot simply plop
>a zero in for something that cannot be computed.
There is nothing corresponding to division by zero here.
v^v is a perfectly meaningful bivector. It is the zero
bivector. There is no other interpretation possible.
>My lines of attack are crossed and that is because even if I argue
>about
> a ^ b = - b ^ a
>I'm already dismissive of the very definition of '^'.
Yes, it's true by definition. The meaning of a bivector A^B
is that it is an oriented area. B^A has the opposite orientation.
>Hamilton did not use these contraptions to develop the quaternion did
>he?
They are basically the same thing.
Hamilton's quaternions I, J, K represent the
bivectors I = e_y ^ e_z, J = e_z ^ e_x,
K = e_x ^ e_y.
For orthonormal vectors e_x, e_y, e_z, the
multiplication table according to geometric algebra is:
e_x e_x = 1
e_x e_y = e_x ^ e_y = - e^y ^ e_x
e_x e_z = - e_z ^ e_x = e_x ^ e_z
e_y e_x = - e_x ^ e_y = e^y ^ e_x
e_y e_y = 1
e_y e_z = e_y ^ e_z = - e_z ^ e_y
e_z e_x = e_z ^ e_x = - e_x ^ e_z
e_z e_y = - e_y ^ e_z = e_z ^ e_y
e_z e_z = 1
From these multiplication rules, we deduce the multiplication
rules for I, J, and K:
I I = -1
I J = K
I K = -J
J I = -K
J J = -1
J K = I
K I = J
K J = -I
K K = -1
So Hamilton's quaternions are the special case of
geometric algebra in which you only allow even multivectors
(scalars and bivectors).
Timothy Golden BandTechnology.com
wrote:
> Timothy Golden BandTechnology.com says...
>
> >We can have arbitrarily small angle
> >and maintain unity area. It follows that
> > v ^ v
> >is undefined.
>
> No, it doesn't. |v ^ v| is the area of a parallelogram
> with angle zero between adjacent edges. That is perfectly
> well-defined; it is zero.
>
> The only way for A^B to approach a nonzero value as
> the angle between A and B goes to zero is if |A| goes
> to infinity, or |B| goes to infinity. So you could
> say that v^v in the case where v has infinite length
arbitrarily close.
> is undefined. That's the case analogous to 0/0 in
> arithmetic. If v has a finite length, then v^v is 0.
> There is no ambiguity.
>
> >When we approach division by zero we cannot simply plop
> >a zero in for something that cannot be computed.
>
> There is nothing corresponding to division by zero here.
Sorry but you just said there was above here. I'm not trying to be an
ass. I might be one, but I'm trying to apply some logic to the
situation and here you are inverting. Is it because you used 0/0 that
you feel OK here?
I guess maybe rather than worry about this point let's just to go back
to getting arbitrarily close to zero angle and observe that constant
area can be maintained. Keeping one arm of the bivector at unity we'll
see that
L sin( theta )
is the area, where L is the length of the other arm and theta is the
angle between them.
> v^v is a perfectly meaningful bivector. It is the zero
> bivector. There is no other interpretation possible.
That depends upon what v is according to Wolfram. If v is grade 2 then
the interpretation of v^v is not tied to zero. I find this troubling
but rather than get all troubled out I'm trying to just hit one point
and one point alone. Here that point is the zero claim on v^v can be
challenged by building it continuously. By getting arbitrarily close
to theta=0 we can maintain constant area, yet that area is supposed to
suddenly drop to zero when we hit zero. I'm proposing that this is
actually an undefined position and that it is like division by 0 ( 1/0
if you like) which is likewise undefined and uncomputable but can be
gotten arbitrarily close to. In the division by zero case we know that
the result grows very large whereas here the result remains constant.
So using your math above we cannot simply declare that
I I = - 1
can we? As you say we deduce it. Let's assume that
ex ^ ey = I .
Now we need to get
( ex ^ ey )( ex ^ ey ) = - 1
right? We know that by the definition of geometric product we have
( ex ^ ey ) ^ ( ex ^ ey ) + ( ex ^ ey ) . ( ex ^ ey )
So now how do these two values evaluate to a scalar? I'm pretty
comfortable with the last part (the dot product) being positive unity.
Somehow then the
( ex ^ ey ) ^ ( ex ^ ey )
part must be minus two right?
Can you show me how this comes to be? Really the sources that I've
seen do it differently:
( ex ^ ey )( ex ^ ey ) = e1e2e1e2 = - e1e1e2e2 = -1 .
This is from Lasenby and Doran from lecture 1 pg.15 of
http://www.mrao.cam.ac.uk/~clifford/ptIIIcourse/GeometricAlgebraLectures.zip
but I've seen it in at least one other pub to. I do not understand
how those steps work. How have they managed to eliminate the wedge
product? There are a small series of steps involved. I understand that
e1e2 = e1 ^ e2 = e1 ^ e2 + e1 . e2
but when they go to a form like
e1e2e1e2
I think a mistake might have been made. How do we say this in general?
(AB)(CD)
is the geometric product of two geometric products. This does not give
us the freedom to say
(AB)(CD) = ABCD
I don't believe that this step is valid in their logic. Is this
supposed to stem from
A(BC) = (AB)C = ABC ?
The application of this form is on triples, not doubles. The form we
are looking at does not fit this associative law. Please here is a
simple place to prove me so wrong. Why doesn't someone slap me around
a bit here? The form
(AB)(CD)
is a hierarchy of identical operations that do not flatten out as far
as I can tell. Especially given the complex nature of units involved
one should have a system of wedge areas capable of canceling each
other via wedge products, unless somehow the dot product provides this
strange ability. I see forms close to done when we substitute E for CD
to get closer to the form we want but I still resist accepting.
Furthermore we've merely assumed the associative law. This likewise
has to be proven. We cannot just take it on blind faith. I have not
seen this derived yet but still I use it here. Maybe I just have to
admit that I hate this math. Who can have any sense of what a
geometric product of a geometric product even is? How will we
attribute this minus one to its dot and wedge components? I've smoked
some grass man but I can't smoke this Grassman stuff. It's too skunky.
- Tim
Timothy Golden BandTechnology.com
wrote:
I am open to what you are describing, though the math will have to
work out. The simplest way to help the math work out would be to grant
the photon a mass. But massive things aren't supposed to travel the
speed of light. Anyway if we use e=mcc then the e=hv of a photon does
translate to a variable mass hv/c/c.
Variable mass is not really a problem since orbital motion is more a
function of velocity, the more massive photons carrying more momentum
while they receive more force. Would this force then be bidirectional?
In effect the light given off by a galaxy could then be a binding
factor for that galaxy.
Then maybe some of the dark matter problem is actually bright matter
being ignored.
Einstein's black box experiment of a photon absorption process assumes
that the photon is like a cannonball, yet here we could consider the
photon being sucked in via affinity. It's strange that repulsion and
attraction can be both modeled via attraction. This can be taken over
to Rutherford's gold foil experiment as criticism. When a particle
returns back toward its source at some sharp angle who are we to claim
that the interaction was repulsive? We know that gravity can slingshot
and slingshot pretty strongly an object back where it came from. If a
radiant particle path comes close to the far side of the object then
the angle can be sharp. This all without even exiting the particle
paradigm. At close distance we are a bit free to put in additional
dynamics too.
- Tim
Spaceman
So the first thing we do is ignore the "statement" about massive things
not being able to travel faster than light since no "physical" proof
of such a limit has ever been proven anyway.
> Variable mass is not really a problem since orbital motion is more a
> function of velocity, the more massive photons carrying more momentum
> while they receive more force. Would this force then be bidirectional?
> In effect the light given off by a galaxy could then be a binding
> factor for that galaxy.
> Then maybe some of the dark matter problem is actually bright matter
> being ignored.
It sure could be "real" mass that is just too small to actually detect
using limited detection methods of today.
> Einstein's black box experiment of a photon absorption process assumes
> that the photon is like a cannonball, yet here we could consider the
> photon being sucked in via affinity. It's strange that repulsion and
> attraction can be both modeled via attraction. This can be taken over
> to Rutherford's gold foil experiment as criticism. When a particle
> returns back toward its source at some sharp angle who are we to claim
> that the interaction was repulsive? We know that gravity can slingshot
> and slingshot pretty strongly an object back where it came from. If a
> radiant particle path comes close to the far side of the object then
> the angle can be sharp. This all without even exiting the particle
> paradigm. At close distance we are a bit free to put in additional
> dynamics too.
The only people afraid to come back to the "physicals" of physics
are those who are "pushing" that "non physicals crap" such as 0 mass
and pure energy only.
Such "massless crap" is not even physics, it is more like "religion".
A "force" that is not made of matter.
:)
Timothy Golden BandTechnology.com
I am sad that nobody is taking this thread seriously. Upon granting
the bivector an adjustable angle as the wedge product is claimed to do
in its graphical sense we see that there is no normal defined for
v ^ v
nor can there be any area defined. We can create differential areas
with normals in other orientations but at this particular bivector no
such normal is possible. This is just another way of looking at this
problem. Beneath it is another trouble: the definition of the wedge
operator uses itself within itself. This detail is being completely
neglected on this thread. Preach the books if you like. I thought we
were here to discuss such things.
- Tim
Mariano Suárez-Alvarez
It is impossible to discuss anything until you
bother reading up on how, precisely, the exterior
algebra is constructed. There is no "using itself
within itself" going on, anywhere, and you will
notice that immediately upon considering how it
is done.
There is very little, in fact, to "discuss" about
the whole thing. The construction is one of those
boring, uneventful constructions: you can start,
for example, with your vector space, you construct
the tensor algebra on it, you divide out by an
appropriate ideal and lo, you've got yourself an
exterior algebra. The details have been written up
already many times painstakingly (Bourbaki's Chapter
3 in his Algebra comes to mind), and I doubt anyone
feels very inclined to repeat them here---I know
I don't.
-- m
Edward Green
<mariano.suarezalva...@gmail.com> wrote:
That, in itself, is a useful compass.
> The details have been written up
> already many times painstakingly (Bourbaki's Chapter
> 3 in his Algebra comes to mind), and I doubt anyone
> feels very inclined to repeat them here---I know
> I don't.
Just what subset of the world has access to Bourbaki, though? It
seems to be out of print, and if you could find it, it would be
hideously expensive. For Average Man, it may as well be one of the
illuminated manuscripts passed among the nobility, before the
invention of movable type.
Edward Green
<mariano.suarezalva...@gmail.com> wrote:
<...>
> Essentially most of the very florid and apparently deep
> state of philosophical angst your writing on this matters
> transpires through words like 'appears to be', 'conflict'
> and so on, would be very efficiently helped by the simple
> and straightforward procedure of studying a little bit of
> multilinear algebra. While at the time of Grassmann and
> Hamilton such tones are understandable, with the appropriate
> dose of historical perspective, nowadays, to be honest, they
> provide very little apart from a minor, poor entertainment.
Leaving aside your deep tone of learned sarcasm, I essentially said
the same thing. I admit I know next to nothing about exterior
algebra, but the weight of Bayesian priors is certainly in favor of
its self-consistency and existence. There may be poor and circular
expositions of exterior algebra though, just as they may be poor
expositions of special relativity. There too the faction that doubts
self consistency, even given their own lack of understanding, seem to
be making a very dubious Bayesian judgment on likelihood of
generations of scientists deluding themselves regarding something so
relatively (no pun) simple.
Then again I have my own crackpot judgments, where I think I know
better than everybody else. I have a strong Bayesian prior that that
entire quantum optics cottage industry is in some sense deluding
themselves on the significance of their continued confirmation of
things related to the Bell inequalities. Why do I think it's possible
I know better than everybody else in a by now well established
specialty, and am reasonable in my prior, while I think that Timothy
is unreasonable in his apparently similar prior?
Well, of course, I have an explanation!
Because I am not conjecturing a failure of mathematical consistency,
which is relatively easy -- or at least sure -- of being checked, but
a failure of physical inference, which is more delicate. You
naturally will think it is I who am deluded, but that's the nature of
the beast.
Timothy Golden BandTechnology.com
I have read enough to make my statements. Please now, let's go into
it.
When you go to perform the operation
a ^ b
on vectors a and b you wind up with results in terms of an
e1,e2,e3,... basis
e1 ^ e2, e1 ^ e3, etc.
These factors are using the ^ operator so the operation is self
referential.
Is this operation the definition? Alternatively there is the graphical
interpretation of an area whose normal and scalar area form a
sufficient representation. Never do the two get related graphically.
How the simple area is supposed to decompose into a series of
primitive areas in terms of the basis is somewhat at the crux. This is
never provided graphically from any of the sources I have read. It is
up to the operator. The operator is self referential. Look, if I were
to define the cross product and reuse the cross product in my
definition wouldn't you have a hard time with that? You simply cannot
define in this way, especially without addressing the logical
conflict.
>
> There is very little, in fact, to "discuss" about
> the whole thing. The construction is one of those
> boring, uneventful constructions: you can start,
> for example, with your vector space, you construct
> the tensor algebra on it, you divide out by an
> appropriate ideal and lo, you've got yourself an
> exterior algebra.
Well, here again is an opportunity to attack. Tensor algebra is
reference frame independent. Geometric algebra is reference frame
specific. Actually maybe it is the usage of real value out of thin
area that then mimics the quaternion or complex number that is where I
am getting this from. It's another sloppy point. The equivalence of
the geometric algebra form to the complex number or quaternion is
complicated due to the units involved. What once was straight vector
math is in this system highly contorted. Was the old complex plane
actually wrong? Must we use the new units of e1^e2 and a real
component that washes up out of thin air? What then of the other two
dimensions? Are the complex numbers now three dimensional? Here it is
said by someone more professional than I:
"By virtue of the last two properties the bivector e1^e2 becomes
our first candidate for the role of the unit imaginary i, and in 2-
dimensional applications it fulfills this role admirably. Indeed, we
see that the even-grade elements z=x+ye1e2 form a natural subalgebra,
equivalent to the complex numbers."
- http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node6.html
Where here is the discussion of dimension? This interpretation is a
compromised form that merely has added a heap of complication and that
to achieve a sloppy result.
Again, rather than attack my math thought directly Mariano chooses to
stay above and preach the book. People like this make up the system.
To shoot me down fully should be an easy little thing. I must have
made some fundamental mistake along the way to get to such a
misleading interpretation right? All that you need to do is point it
out. Firstly is the self referential definition. You deny it, but it
seems so balatantly apparent to me that it surprises me that you wipe
it over with no detail found.
> The details have been written up
> already many times painstakingly (Bourbaki's Chapter
> 3 in his Algebra comes to mind), and I doubt anyone
> feels very inclined to repeat them here---I know
> I don't.
>
> -- m
As E.G. says I do not have easy access to Bourbaki. If you do think
that the web resources are incorrect then that is perhaps fueling my
reaction to this math. But do you think that Dorst and Mann are of
questionable integrity?
What then is wrong with using these resources? Really, preaching the
book is such a weak position. By doing so you've dropped integrity on
the floor. I'm not asking you to burn the book, just to question it.
Let's see now, what sort of snipe might actually get Mariano to bight?
You are a booker! Meant to rhyme with hooker, as in a whore to
academia. You booker! Ooh, its like booger too. I reccomend you take
the pages on wedge product and seal them together with boogers so that
you will not have to worry about them ever again. They'll just sit
there in your book and you'll flip through it and the little carat
symbol will just carry on elsewhere without conflict. You'll never
need to worry about carats within carats.
- Tim
Edward Green
<...>
> > There is very little, in fact, to "discuss" about
> > the whole thing. The construction is one of those
> > boring, uneventful constructions: you can start,
> > for example, with your vector space, you construct
> > the tensor algebra on it, you divide out by an
> > appropriate ideal and lo, you've got yourself an
> > exterior algebra.
>
> Well, here again is an opportunity to attack. Tensor algebra is
> reference frame independent. Geometric algebra is reference frame
> specific.
I doubt that very much. Some amateurish expositions we may have seen
may have resorted directly to components, but I have little doubt that
exterior algebra is a purely "geometric" construction. (Suarez wrote
"exterior" algebra while you wrote "geometric", but I'll take it we
are talking about exterior algebras, albeit that geometric algebras
are closely related).
As to why components make their appearance so quickly in expositions
by amateurs or would-be pedants, it may be related to why we first
learned about vectors as triples of real numbers -- whether or not
it's strictly necessary, it's a pedagogic habit to introduce geometric
concepts as identical with their components.
By the way, I take it that you, like I, are completely comfortable
with the idea of dividing a tensor algebra by an ideal (deep
sarcasm)? I have it on my list of things to do mathematically to
understand this generalized sense of division -- I've seen it enough
times, in the context of quotient groups and so forth -- and I don't
know what an ideal is, beyond having looked it up once or twice; but I
wasn't ready for the definition, and it didn't stick. Shall we try
again?
<...>
Mariano Suárez-Alvarez
No it is not. What is true, and what any sensible textbook
will tell you, is that if {e1, ..., en} is a basis for
a vector space V, then the set of products
B2 = { ei ^ ej : 1 <= e < j <= n }
is a basis for the exterior square of V, which I'll write
L^2V. This implies, in particular, that every element of
L^V can be uniquely written as a linear combination of
elements of B2.
It follow that
(*) if v, w are two elements of V, then v ^ w, which is an
element of L^2V by definition, and just as every other
element of L^2V, can be written as a linear
combination of elements of B2.
The wedge product of elements in V, producing
elements in L^2V is defined *absolutely* independently
of the observation (*).
This can be observed, as I said, in any minimally
good text.
> [...]
> > There is very little, in fact, to "discuss" about
> > the whole thing. The construction is one of those
> > boring, uneventful constructions: you can start,
> > for example, with your vector space, you construct
> > the tensor algebra on it, you divide out by an
> > appropriate ideal and lo, you've got yourself an
> > exterior algebra.
>
> Well, here again is an opportunity to attack. Tensor algebra is
> reference frame independent. Geometric algebra is reference frame
> specific. Actually maybe it is the usage of real value out of thin
> area that then mimics the quaternion or complex number that is where I
> am getting this from. It's another sloppy point. The equivalence of
> the geometric algebra form to the complex number or quaternion is
> complicated due to the units involved. What once was straight vector
> math is in this system highly contorted. Was the old complex plane
> actually wrong? Must we use the new units of e1^e2 and a real
> component that washes up out of thin air? What then of the other two
> dimensions? Are the complex numbers now three dimensional? Here it is
> said by someone more professional than I:
> "By virtue of the last two properties the bivector e1^e2 becomes
> our first candidate for the role of the unit imaginary i, and in 2-
> dimensional applications it fulfills this role admirably. Indeed, we
> see that the even-grade elements z=x+ye1e2 form a natural subalgebra,
> equivalent to the complex numbers."
> -http://www.mrao.cam.ac.uk/~clifford/introduction/intro/node6.html
>
> Where here is the discussion of dimension? This interpretation is a
> compromised form that merely has added a heap of complication and that
> to achieve a sloppy result.
Honestly, I have no idea of what you are trying to say here.
As it apparently is an "attack" on the definition of exterior
algebra, I can but imagine the loss incurred to be small.
> Again, rather than attack my math thought directly Mariano chooses to
> stay above and preach the book. People like this make up the system.
> To shoot me down fully should be an easy little thing. I must have
> made some fundamental mistake along the way to get to such a
> misleading interpretation right? All that you need to do is point it
> out. Firstly is the self referential definition. You deny it, but it
> seems so balatantly apparent to me that it surprises me that you wipe
> it over with no detail found.
I already told you what the problem with what
you say is: you do not seem to be aware of how
the exterior algebra and friends are really
defined. Your objections fall, in consequence,
very much into the "not even wrong" field.
> > The details have been written up
> > already many times painstakingly (Bourbaki's Chapter
> > 3 in his Algebra comes to mind), and I doubt anyone
> > feels very inclined to repeat them here---I know
> > I don't.
>
> > -- m
>
> As E.G. says I do not have easy access to Bourbaki. If you do think
> that the web resources are incorrect then that is perhaps fueling my
> reaction to this math. But do you think that Dorst and Mann are of
> questionable integrity?
You keep using terms like "attack" and "questionable
integrity" in the context of such boring things as
the definition of exterior algebra. The effect is not
very different from what one gets from some of the
characters of Moliere...
> What then is wrong with using these resources?
I don't know. I am not familiar at all with Dorst
and Mann.
> Really, preaching the
> book is such a weak position. By doing so you've dropped integrity on
> the floor. I'm not asking you to burn the book, just to question it.
> Let's see now, what sort of snipe might actually get Mariano to bight?
> You are a booker! Meant to rhyme with hooker, as in a whore to
> academia. You booker! Ooh, its like booger too. I reccomend you take
> the pages on wedge product and seal them together with boogers so that
> you will not have to worry about them ever again. They'll just sit
> there in your book and you'll flip through it and the little carat
> symbol will just carry on elsewhere without conflict. You'll never
> need to worry about carats within carats.
I cannot really do much about this, apart from
wondering in what way you think it is conducent
to rational discussion and, again without much
loss, ignoring it.
-- m
peeter
<mariano.suarezalva...@gmail.com> wrote:
It's been a while since I looked at this thread, and there are a
couple points worth further clarification.
> I don't know. I am not familiar at all with Dorst
> and Mann.
These authors, and others such as Hestenes, Lasenby, and Doran,
describe Clifford algebra, re-branded as "geometric algebra" in a
physics biased context. Elements of the algebra are built up by grade
selection. A product of two vectors has a grade 0 and grade 2 part ;
product of a grade 2 element with a vector has a grade 1 and grade 3
part, ... The grade two part of the vector product can be interpreted
as a wedge product (or bivector) as it has the required antisymmetric
properties. A consequence of building higher grade elements in terms
of a multiplicative process instead of additive is that all the
elements are what a text such as:
http://www.grassmannalgebra.info/grassmannalgebra/book/index.htm
call "simple elements". You won't find an additive element such as
e_1 ^ e_2 + e_3 ^ e_4 that can't be expressed as the wedge product of
two other vectors, so the property of general additive exterior
algebras of v^v != 0 that is mentioned in the wolfram text isn't
typically of interest. You have to be careful when you introduce
elements such as the faraday bivector for electromagnetism since it
isn't a bivector in the strict sense. It is instead built up in an
additive fashion from pairs of electric-field and magnetic-field
bivectors.
Axiomatically geometric algebra requires the following definitions:
- linearity
- contraction (a^a = |a|^2 = length of Euclidean vector).
- associativity. a (bc) = a(bc)
In terms of these, the dot product (definition) of two vectors is:
a . b = 1/2(ab + ba)
and the wedge product (definition) of two vectors is:
a ^ b = 1/2(ab - ba)
It can be shown that both of these definitions are consistent with
standard definitions of the dot and wedge product of two vectors.
For the product of two vectors, one can perform a split into symmetric
and antisymmetric components:
ab = 1/2(ab + ba) + 1/2(ab - ba)
based on the definitions above, this is a split into a dot product
(grade 0) term, and a wedge product (grade 2) term respectively.
A question that was asked is how for two orthogonal unit vectors, e_1,
and e_2 the product:
(e_1 e_2)(e_1 e_2) = -1
This does follow from the axioms above. A possible missing ingredient
contributing to confusion here is perhaps the fact that associativity
is one of the axioms.
Peeter
Timothy Golden BandTechnology.com
<mariano.suarezalva...@gmail.com> wrote:
> On Aug 23, 10:37 am, "Timothy Golden BandTechnology.com"
>
> <tttppp...@yahoo.com> wrote:
> > On Aug 22, 1:16 pm, Mariano Suárez-Alvarez
>
> > <mariano.suarezalva...@gmail.com> wrote:
> > I have read enough to make my statements. Please now, let's go into
> > it.
> > When you go to perform the operation
> > a ^ b
> > on vectors a and b you wind up with results in terms of an
> > e1,e2,e3,... basis
> > e1 ^ e2, e1 ^ e3, etc.
> > These factors are using the ^ operator so the operation is self
> > referential.
>
> No it is not. What is true, and what any sensible textbook
> will tell you, is that if {e1, ..., en} is a basis for
> a vector space V, then the set of products
>
> B2 = { ei ^ ej : 1 <= e < j <= n }
Here we see a first usage of the carat symbol to mean a wedge product.
>
> is a basis for the exterior square of V, which I'll write
> L^2V. This implies, in particular, that every element of
> L^V can be uniquely written as a linear combination of
> elements of B2.
>
> It follow that
>
> (*) if v, w are two elements of V, then v ^ w, which is an
> element of L^2V by definition, and just as every other
> element of L^2V, can be written as a linear
> combination of elements of B2.
Since the wedge symbol was already used by you above here this is a
following definition. It is important to now repeat whay you've
written above (building L^2v) and consider that when you say that v^w
is an element that you are doing so in a circular form. I have yet to
see this circle broken and you've clearly repeated it here.
Particularly you make the statement:
"v ^ w, which is an element of L^2V by definition"
This statement is in conflict. You built L^2v out of elements of ei^ej
which are using the operator that you claim to be constructing. This
text in quotes above here needs to be weakened. These few words are
tying the circle. I would be happy to banter on these few words alone
and will be disappointed if you choose to dodge.
Again, my own rendition of what you've just written is that you've
defined the ^ operator with the ^ operator, these ^ operators being
the same thing, namely the wedge product. This is a logical conflict.
We can step out to the graphical sense but in that case the area was
good enough on its own without all of this additional gook. Also, this
is just one sliver of my criticism though it is the most basic part so
I thank you for trying to resolve it. Unfortunately I remain
unconvinced by what you've written. Obviously we cannot take much
freedom in the renditions of this math. There are a thousand ways to
say it but they all will have to conform with the math. If you did
have a noncircular definition then I do think it would be wise to
promote it. A scalar with a normal vector are a fine general
definition of a planar area and that fomat is informationally much
less controversial. Since the issue ov v^v will never come up I would
reccomend stepping back to the normal vector with a scalar area
interpretation and see what else fits structurally. This method will
not need the wedge product at all. The fruits which form the
attachment to this wedge product are still ten steps away and by
pushing toward those features it seems that they've overlooked this
weakness in the fundamentals. The qualities of the fruit may be
compromised.
- Tim
Ahh, but it is not the discussion that I've been trying to have!
You are the one who has chosen this route and I am merely following
through with feedback on your criticism and pointing out your own
criticism's weaknesses. Those who will preach the book are fronting a
closed mind to an open problem. You tell me to just follow the book.
Follow the book. Preserve the book...
Future work does not lay in that direction. Here we have a format
where those books can be challenged without having the conversation
shut down by a professor in a closet sized office whisking us off,
that professor likely being a fine mimic. How many students have taken
this very complaint to their professor and been rejected? Plenty I
would guess. Anyhow, I hope you can at least get a few chuckles now
and then rather than take it all as toxic. The booker booger stuff was
supposed to be funny.
>
> -- m
Mariano Suárez-Alvarez
Oh well. That's fine by me, really.
-- m
Timothy Golden BandTechnology.com
So it seems you accept my criticism but claim that the construction is
consistent. Now I freely construct a spaceship with regenerative
breaking. The regenerative breaking decelerates the spaceship and with
it I am able to generate energy. By pointing my spaceship backwards I
can accelerate forwards while building my energy supply ad infinitum.
It's a beautiful system and it relies entirely on regenerative
breaking. I build my regenerative breaks out of regenerative breaks.
Can't you see the logical discrepancy in this construction? I am free
to construct this on paper. Yet any who rely or even comply with
regenerative breaking may be travelling a sorry path.
You've verified a fundamental problem with mathematics as a distinct
subject. By ignoring physics and philosophy; reality really; the
mathematician takes freedoms as they see fit and if these freedoms do
not comply with physics or philosophy then what? Shall we call that
pure mathematics? Not at all. Pure mathematics complies with physics
and philosophy and logic as well. It may even form them. We are free
to construct upon axioms granted by choice. Yet we should not accept a
circular definition, especially if it leads into grungy territory.
This thinking then pleads that we consider the other details of the
system. Having accepted the circular definition which Mariano has
previously denied shall we go on to consider its consequences? They
are indirectly related, yet when we consider the whole bundle not as a
series of splinters but as one broken stick then a judgement can be
clearly made as to whether we accept this math or not. That certainly
is an individual decision and we do need people to consider many
variations, including failed variations. What we do not need is denial
of failure. Especially not in such a way that covers it up and then
imposes it on the next generation. Where religion begins and math ends
are open problems. We are caught as humans with mimicry responses that
allow such complex informational systems to arise. Some like this.
They are the best mimics. The straight A's: the teacher's pet. Woof,
woof.
- Tim
Timothy Golden BandTechnology.com
I like your logic here. Yet if mathematical consistency is broken then
you will concede that this is a deeper issue right? It is so good to
step out to this level of the unknown. We know that we grow attached
to ideas, especially those that we've committed alot of energy to.
Pythagoras is a fine instance. Just as I am fallible so is the
establishment. I admit that the odds are stacked in their favor. Still
I am practiced in constructing math. Upon entering the issues of
dimensionality we are entering a kaleidoscopic realm. Somebody should
have come along and challenged a detail in my area format above here.
We know that the normal definition is more complicated than a single
vector in high (4D+) dimensions. Yet the informationally compact form
I am fairly sure will take on the antisymmetric tensor format, or the
tatrix format as I like to call the unsigned nonredundant version.
This is the kaleidoscope again. I am pretty sure of it though I
haven't worked through it fully. Anyhow clearly going to two full 4D
vectors to assign a normal in 4D is redundant so there is a more
compact form, essentially declaring the second vector as a remainder
off of the first full vector. That is maybe a start of where to go.
This is just a dimensional reduction with a cutoff since the object is
an area. Beneath the area then is a segment which mimics the vector
and then beneath that a point. This point is the strangest of all. It
still has a little bit of algebra even though it sits at the origin.
In Cartesian math the 1X1 antisymmetric tensor is just a zero right?
The polysign fits nicely here since it collapses just that way, but
collapsed it becomes necessary to carry around the signs with the
numbers. So that string of diagonal zeros that the wedge somehow calls
zero (ei^ei) but not undonditionally
(ei^ej)^(ei^ej) = - 2 ?
is sort of like polysign but twisted by carats within carats within
carats. The best way over to proper geometric algebra is to challenge
the existing geometric algebra. Since they've taken that name I do not
use it. In all the time that I've studied I've only come to the wedge
product recently even while I've been seeking a generalized cross
product for years. The high level connections which evade
quantification are strong. I think there is a bit more to construct.
The construction of the wedge brings an area to life. Beneath this we
should bring the segment to life. Does it have two endpoints or just
one? Should we really be toying with such a generic area that has no
position? If we want geometric algebra shouldn't we be able to get
away from the origin? Somehow we've settled into vectors without
tails. These are data formats but a natural structure should be
available. Perhaps the natural format is the tatrix or its complicated
cousin the antisymmetric tensor, but interpereting it cleanly is as
much a physics puzzle as it is a math puzzle.
- Tim
Timothy Golden BandTechnology.com
> On Aug 23, 9:37 am, "Timothy Golden BandTechnology.com"
>
> <tttppp...@yahoo.com> wrote:
> > On Aug 22, 1:16 pm, Mariano Suárez-Alvarez
>
> <...>
>
> > > There is very little, in fact, to "discuss" about
> > > the whole thing. The construction is one of those
> > > boring, uneventful constructions: you can start,
> > > for example, with your vector space, you construct
> > > the tensor algebra on it, you divide out by an
> > > appropriate ideal and lo, you've got yourself an
> > > exterior algebra.
>
> > Well, here again is an opportunity to attack. Tensor algebra is
> > reference frame independent. Geometric algebra is reference frame
> > specific.
>
> I doubt that very much. Some amateurish expositions we may have seen
> may have resorted directly to components, but I have little doubt that
> exterior algebra is a purely "geometric" construction. (Suarez wrote
> "exterior" algebra while you wrote "geometric", but I'll take it we
> are talking about exterior algebras, albeit that geometric algebras
> are closely related).
The complex plane is claimed to be mimiced by geometric algebra. The
complex plane is not isotropically behaved under product. If we happen
to be near the real axis we'll see small rotations and if we are near
the imaginary axis we'll see lots of flipping about. Once product is
thrown in as a behavior to maintain isotropic space over I'm pretty
sure there is a need to consider how we'll even invoke reference
frames. There are choices to be made. The original tensor algebra as I
understand it was about not caring what frame of reference the math
was put in. It did not have any product to consider. Rotations in that
true tensor (isotropic) realm are projections. I do see this as more
sleight of hand since they do not go into any detail of this leaning
on tensor math. Again too, there is no graphical interpretation on the
gooky complex representation in terms of a real axis resulting from
the dot products and the e1,e2 bases of areas. Whacking out a new line
orthogonal to e1 and e2 won't cut it will it? This is just down at 2D.
What a mess.
>
> As to why components make their appearance so quickly in expositions
> by amateurs or would-be pedants, it may be related to why we first
> learned about vectors as triples of real numbers -- whether or not
> it's strictly necessary, it's a pedagogic habit to introduce geometric
> concepts as identical with their components.
>
> By the way, I take it that you, like I, are completely comfortable
> with the idea of dividing a tensor algebra by an ideal (deep
> sarcasm)? I have it on my list of things to do mathematically to
> understand this generalized sense of division -- I've seen it enough
> times, in the context of quotient groups and so forth -- and I don't
> know what an ideal is, beyond having looked it up once or twice; but I
> wasn't ready for the definition, and it didn't stick. Shall we try
> again?
>
> <...>
I am curious about the quotient math. Somehow I've missed this post
and got sidetracked with Mariano. The trouble is if I don't really
believe in this product then its inverse should be equally
meaningless. I likewise do not know what the term ideal is within this
math. Still I have my own quotient issues to try to deal with in
polysign and this could be a helpful cross study. I don't know if I
can contribute much though.
As you talk about components above it seems like without them we would
be caught in a graphical interpretation. That can work in 3D but
beyond it's too much. Anyhow there don't seem to be graphical
renditions of the wedged area being broken down graphically into a
series of pieces, even in 3D. Then too as I pointed out above the
usage of a scalar does not map to any usual graphical representation.
They've added a dimension to the basis. Ordinary dot products as
projections do remain in the space that they were declared in and are
projection components of the original vector/object in its new frame
of reference. Everything has been turned on end in this subject with a
maze of holes left behind.
If I start calling polysign "geometric algebra" maybe I could get some
traction. At the very least by perturbing the algebraic geometrists
they might take a look.
- Tim
- Tim
Timothy Golden BandTechnology.com
geometric algebra:
1. The definition of the wedge product is self referential and so
cannot be a definition in the axiomatic sense.
a continuum of 2-blade areas. Simply sweep a unit area 2-blade down to
undefined.
3. Results of Dorst and Mann yield
whereas by applying the definition of geometric product we get
( ei ^ ej ) . ( ei ^ ej ) + ( ei ^ ej ) ^ ( ei ^ ej )
which requires that we now define these operations which need to
resolve to a scalar to yield consistency.
The most straightforward solution is to put
( ei ^ ej ) . ( ei ^ ej ) = + 1
( ei ^ ej ) ^ ( ei ^ ej ) = - 2
straightforward.
This result appears to link to the claim of geometric algebra's
compatibility with the complex plane. That claim may be false.
I am putting myself out on a limb here and I am entirely open to
falling to the ground. Please attack freely and fully. So far Peeter
and Mariano have tried but it seems that I've gotten the better of
them. That or the rhetoric was just too much. There is plenty of math
to consider here and I would appreciate a sincere attempt.
- Tim
Timothy Golden BandTechnology.com
galathaea
there is a hidden story here
the quantity of good books on geometric intuitions
in algebraic characterisations
of grassmannian spatial relations
is surprisingly small
and there are _many_ books that claim to cover the topic
the problem is that
so many of them give no good process or motivation
many physics texts will start defining a dynamics
over some poisson manifold
and suddenly lapse into deep theorem proving abstractness
a number of mathematical texts will simply
explore grassmannians qua grassmannians
an algebraic type
but the origin is very geometric
joe harris has a book
"algebraic geometry: a first course"
that actual does a decent job of going over the structure
grassmannian varieties G(k, V) constructed
from the k-dimensional linear subspaces
of the n-dimensional vector space V
right away he points out the ever crucial
"plucker embedding"
into a subvariety of the projective space of k-multivectors
P(/\^k V)
it's not an incredibly easier path
but phi: G(k, V) -> P(/\^k V)
does provide a convenient place to calculate
harris then sketches how to calculate
the matrix relations for the polynomials
and when he gets to describing which polynomials
generate this embedded subvariety of P(/\^k V)
he describes really simply why introducing dual spaces
through the natural identification
of /\^k V and /\^(n-k) V*
(this is motivated by the standard visualisation
of a line in an n-dimensional space
parametrising leafs of subspaces
with it's points)
he then derives the plucker relations
and begins giving geometric descriptions of these varieties
(cut out by intersecting quadrics from the construction)
but
of course
that doesn't give the motivation
that a physically oriented person would desire
what does that have to do with fermions then?
well this same fundamental algebra
underlies the fermionic pauli exclusion principle
as some of the better quantum field theory books describe
bosonic propagators are constructed over complex variables
fermionic propagators are built from grassmanians
fundamentally
again
this is a relationship between V and a derived space
but here the space is the differential space TV
(the tangent space)
and the grassmanian arises
because of the anticommutativity of fermions
through the fundamental fourier relationship
between x and d/dx
the differential structure of anticommuting variables obeys
* d^2f/(dx)^2 = 0 (ie. exact, e.g. nilpotent)
* / d
| f dx = -- f
/ dx
and path integrals and all that fun follows
all of this is still fundamentally the same geometric construction
with the plucker embedding giving the fundamental
creation and annihilation operators
good texts like jean zinn-justin's
"quantum field theory and critical phenomena"
go over qft with this algebraic approach from early on
<..>
so
if i were to try to answer your questions:
1) the construction of wedge products
can be presented in a non self-referential way
using the space G(k, V) of k-dim linear subspaces of V
or the embedding in projective k-multivectors
this isn't saying that some aren't defining it circularly
i have seen really sloppy treatments of the theory
2) the construction of the wedge product
over G(k, V) or P(/\^k V)
might be confusing here
since the scalar 0 is easily confused with the zero vector
it might be easier to try to understand
why the quotient of the tensor algebra T(V) of V
with the two-sided ideal x (x) x
gives the same grassmannian construction
this quotient relationship shows an algebraic construction
that immediately explains x /\ x = 0
(it's basically defining the algebra
from the free tensor structure
by this relation)
3) exterior products do not produce scalars
this seems like another poor explanation
probably trying to base the intuitions
on the cross product construction
i know it can be frustrating
often having to make additional effort
due these poor explanations in certain areas
in 3D real euclidean space
/\^2 V is isomorphic to /\ V*
(1 = 3 - 2)
so the 3D cross product may be seen as a covector
this may again be due to the source sloppily mixing
the scalar 0 and 0 vectors and multivectors
the equation
(x ^ y) ^ (x ^ y) = -2 should be looked at more closely
even if this were only about 3D cross products
and covectors identified vectors
(x ^ y) should be the same form both times
and so the same associated vector
which cross producted with itself is already defined
(0)
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galathaea: prankster, fablist, magician, liar
galathaea
my point was
that i think you shouldn't build aversion
to the grassmannians
i have mentioned barycentric coordinates in the past
in relation to your polysign numbers
and i think in general
if you look at the works of mobius or plucker
you will find some great historical motivation for your tools
so i would hate to learn that you had formed negative opinions
about this tool that has some interesting connections
with your own
you see
i'm pretty convinced that your polysign
is actually a very natural additive homogeneous coordinate
and you can take your invention many deep ways
both to help you understand these deeper theories
and to likely discover some new things along the way
and i know you have the intelligence to pursue it
so i really want to stress to you
the importance of grassmanians
V x TV is a very central physical space
the phase space on which interactions are formed
and i know you are looking for
redefinition of the space of interactions
edward green has asked a lot recently
about these exterior products lately
and i understand so much
just how poor the available texts are
and how i struggled quite a bit to get a reasonable picture
and i just think
both of you are right there
if only you guys could find the description that "clicks"