Elementary questions regarding characterisation of versors
Fred
Many geometries with which I am reasonably familiar ---
in particular spherical Cl(n+1,0,0), Euclidean Cl(n,0,1),
Moebius / conformal Cl(n+1,1,0), Laguerre Cl(n,1,1), Lie-sphere
Cl(n+1,2,0), Euclidean 3-space line geometry Cl(3,3,0) ---
follow the same broad structure. There is a quadratic metric;
a group of isometries preserving that metric; and a theorem
(Cartan-Dieudonne) that the isometries are generated by
reflections in primes (that is, subspaces of codimension unity).
The isometry group of a geometry with this structure is best
represented algebraically via a Clifford algebra with signature
corresponding to the canonical form of the metric, vectors
(grade 1) representing prime reflections homogeneously, Clifford
product corresponding to composition of isometries,
and Clifford conjugation to application of an isometry.
Question 1: just exactly why does Clifford product represent
composition --- in particular, why should the product of
generators be anti-commutative?
Continuing, by Cartan-Dieudonne every isometry is representable
by a Clifford algebra "versor", that is a (non-unique) product
of vectors; and every versor conjugates vectors only into vectors
(and so versors only into versors).
Question 2: is any multor (multivector element of the algebra)
which conjugates vectors into vectors necessarily a versor?
[[ Note that the versor representing a given isometry is not
necessarily unique, even modulo a scalar factor. Consider 2-space
Lie-sphere geometry with generators
x^2 = y^2 = u^2 = +1, v^2 = r^2 = -1,
where x,y represent reflection in the y,x-axes, u,v geometric
inversion in the real and imaginary unit circles, and r eversion
of orientation.
The pseudar (pseudo-scalar) x y u v r commutes with everything,
and so represents the same (identity) transformation as 1. This
has the unexpected consequence that there are only Spin
transformations: for example, geometric inversion in the unit circle,
represented by the odd versor u, is represented also by its even
dual x y v r; and the latter is kinematic, connected to the
identity via the continuous (Spin-up) isoclinic elliptic birotation
(cos t/2 + sin t/2 x y) (cos t/2 + sin t/2 v r) . ]]
Question 3: 1 (or any nonzero scalar) is obviously a versor,
representing the identity isometry, and an empty product of vectors.
But is 0 a versor?
An important part is played in the later geometries above by
isotropic (magnitude zero) versors --- for example, axes of
parabolic rotations, and Lie-sphere cycles (oriented circles or
lines). Normalising versors with nonzero magnitudes --- in order,
for example, to compare them --- is easy, at any rate in principle:
compare the signs of the magnitudes, and if they're equal, divide
by the square roots of the absolute magnitudes. However ...
Question 4: How might isotropic vectors or versors be normalised?
Finally, an elementary howler into which I tumble at regular
intervals: given versors A,B, it is a theorem that the magnitude
of their product should equal the product of their magnitudes.
Yet
||(x r + y v)(x v + y r)|| = 0,
||(x r + y v)|| = ||(x v + y r)|| = -2,
and (-2)(-2) = 4 > 0 ---
Question 5: Where did I go wrong (again)?
Fred Lunnon [19/01/10]
Lanco
Please see the uploaded file.
Lanco
Fred
Lanco has pointed out that Hestenes' definition of "versor" is
restricted to invertible products of vectors. I had overlooked
this completely, thereby establishing two relevant principles:
firstly, that I don't get out enough [a situation I had hoped
to remedy by joining this list]; secondly, that mathematical
nomenclature is too important to be left to mathematicians.
Why might "versor" be thus restricted? Simply because these are
the members of the "Clifford group" (usually denoted by uppercase
Gamma) introduced in traditional treatments of Clifford algebra,
which is a theoretically convenient concept because --- well, it's
a group, ain't it! But from an engineering point of view, excluding
non-invertibles is just plain daft --- like redefining "matrix" to
exclude singular matrices, or "vector" to exclude isotropic
(light-like, zero magnitude) vectors. Sure, it would give you a group
--- but it would no longer be not compact, which is in practice far
more inconvenient.
Anyhow, the upshot is that, to avoid confusion, we need a word
for the actual useful concept: a product of vectors. Does anybody
out there already know of such a coinage? Failing that, I thought
up "convector" (for composition of vectors); "conversor" (for
compactified versor); or "jector" (since singular products act like
projections into subspaces). I've also lately felt the need for
a symbol analogous to "Cl(p,q)" for (the semigroup of) products of
vectors: off the cuff I coined "Vl(p,q)" in lurker.txt, but in view
of Lanco's remark, that's a dead duck.
Finally, in the elementary questions posted earlier, for "versor"
read "jector", or "conversor", or "convector", or whatever ...
and apologies for this confusion! I'll discuss Lanco's replies to
those questions shortly.
Fred Lunnon
eskki paramasivan
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Fred
As a result of the versor / jector confusion discussed earlier, much
of Lanco's replies in Versor+Questions.pdf unfortunately misses the
points I had intended to make.
For instance, to take Question 5 (which was anyway tongue-in-cheek),
magnitude of product equals product of magnitudes even if the
arguments
are non-invertible jectors.
But in fact, A = (x r + y v) and B = (x v + y r) are not jectors:
a quick way to establish this is to compute the reversion products
A^+ A = 2(x y v r - 1) etc, which would necessarily be scalar (and
equal the magnitude) for jectors, not just versors.
Similarly for Question 3. In a homogeneous setting (nonzero scalar
factors irrelevant), multor zero occupies a special place ---
it represents no geometric object, and furthermore its grade is
strictly speaking indeterminate (though I think of it as grade -1).
It's hard to see how it can be interpreted as a (possibly rescaled)
product of (some sequence of) vectors; on the other hand, compactness
seems to demand that it be included in the set of jectors, as the
limit of a sequence of (say) scalars shrinking indefinitely.
[But I'm not sure I've thought this out properly: perhaps I need
a topologist --- never thought I'd find myself saying that!]
Regarding Question 2, Lanco offers to show that a vector-preserving
(invertible!) multor which is entirely even-grade or entirely
odd-grade is necessarily a versor.
[Unfortunately, he has switched from using x,y for generators to
using them for vectors ... I'll use X,Y for vectors!
X*Y here must denote the bilinear product of vectors X,Y based on the
metric, which being scalar commutes with every multor.]
The proof depends in turn on the proof of his 6(b), with which I have
two issues. Firstly, an aesthetic quibble: it involves matrices,
which
ought surely to be unnecessary in establishing elementary properties
of GA [particularly when they involve the grisly Atiyah-Bott-Shapiro
twist!] Secondly, the deduction from
X C - parity(C) X = 0
that C must be scalar is surely false: consider for example
X = x, C = parity(C) = y z ,
where x,y,z are generators with x^2 = y^2 = z^2 = 1 .
Finally, regarding Question 1, I don't doubt for a moment that
Clifford algebra "works" --- which I think is more or less what
6(a) and 6(b) are concerned with establishing. What I'm asking for
here is rather some compelling intuitive insight illuminating why,
in the first place, we (or W.K.Clifford) should set up an algebra
to model algebraically perpendicular prime reflections, in which
x y + y x = 0, rather than (say) x y - y x = 0 ?
It's a particularly tantalising conundrum when --- as in my case
--- the representations in question are homogeneous, and the sign
of a multor eventually irrelevant!
Fred Lunnon
Lanco
Lanco
Fred
Well, don't be discouraged --- can you fix anything?
Speaking of fixing things, my sharp-eyed partner here has noticed
I wrote earlier
<< Sure, it would give you a group
--- but it would no longer be not compact, which is in practice far
more inconvenient. >>
when I meant to say
<< Sure, it would give you a group
--- but it would no longer be compact, which is in practice far
more inconvenient. >>
WFL
Fred
I earlier misinterpreted the term "versor" to include isotropic (zero
magnitude) products of vectors.
Thinking some more about nomenclature for these things, maybe
"proversor" is better coinage --- it emphasises the connection with
versors (as usually understood), and that these objects are a little
more primitive. So, a versor is just the same thing as an invertible
proverser.
Rephrasing one of my earlier queries, I should have asked
> Question 3: 1 (or any nonzero scalar) is obviously a proversor,
> representing the identity isometry, and an empty product of vectors.
> But is 0 a proversor?
and later added
> ... In a homogeneous setting (nonzero scalar
> factors irrelevant), multor zero occupies a special place ---
> it represents no geometric object, and furthermore its grade is
> strictly speaking indeterminate (though I think of it as grade -1).
> It's hard to see how it can be interpreted as a (possibly rescaled)
> product of (some sequence of) vectors; on the other hand, compactness
> seems to demand that it be included in the set of proversors, as the
> limit of a sequence of (say) scalars shrinking indefinitely.
I was looking at this in the wrong way. Provided we operate in Cl
(p,q)
where p,q > 0, zero is obviously a proversor, since 0 = (x - v)^2,
where
x,v are generators with x^2 = -v^2 = 1 --- duh!
But this reasoning fails in spherical space, either Cl(p,0) or Cl
(0,q);
furthermore, similarly -1 would not be a proversor in Cl(p,0) ...
The whole idea was to avoid complications of this sort; so the
final definition should probably just read instead:
"A proversor is either a scalar or of a Clifford product of vectors".
Fred Lunnon
Peeter Joot
I can't keep up with your rate of nomenclature invention, and have had trouble in the past even differentiating the existing names. I think it would be more productive to do a literature search and build a glossary of existing terms. Where the usage is overloaded or conflicting, cite the authors and the frequency of use of the various terms, or any specializations where one is preferred over the other. This should include at least the following to start
scalar
multivector
blade
versor
simple element
pseudoscalar
...
Peeter
Fred
I dislike having to coin new terms, but find it hard to avoid here.
One problem is that many people working in this area have restricted
their attention to Cl(p,0) or (almost the same thing in practice)
Cl(0,q) --- which may well explain for example why "versor" has
acquired its overly-restrictive meaning, since isotropic proversors
are absent from spherical algebras.
It might also explain why Hestenes seems to have failed to take into
account that his "magnitude" |X| might be imaginary; and I have just
realised that I have inadvertently instead been using "magnitude" for
its (real) square, which I denote by ||X||. Li appears to favour
superscript 2 to denote ||X|| when X is a vector --- I dislike this
usage, since X^2 /= ||X|| when the grade of X exceeds 1.
Accidental new terminology is a nuisance; but more irritating is
the introduction of apparently random variation of adequate existing
notation of which their author was already well aware. I still
have not seen Li's book, but it appears that he is responsible for
introducing a tilde (~ over expression) to denote reversion, instead
of Hestenes' superscript dagger; also a caret (^ over expression)
to denote parity transform, instead of Hestenes' superscript asterisk
(*). What on earth can be the benefit of such destructively confusing
permutations?
Neither is Hestenes entirely blameless in this respect --- using
tilde
in one paper to denote Clifford conjugate, in another to denote dual
(as I also prefer). See
@book{[Hes99b],
author = {D.~Hestenes},
title = {New Foundations for Classical Mechanics},
edition = 1,
publisher = {Kluwer},
year = 1998,
}
@incollection{[Hes99],
author = {D.~Hestenes and H.~Li and A.~Rockwood},
title = {New Algebraic Tools for Classical Geometry},
chapter = 1,
editor = {G.~Sommer},
booktitle = {Geometric Computing with {C}lifford Algebra},
publisher = {Springer},
year = 1999,
}
It also must be said that there are far too many superfluous
operators
in use, presenting another obstacle to communication: in my
experience
a perfectly practical minimalist set comprises generators, addition/
subtraction, Clifford and scalar product, grade extraction (and
creation,
for programming purposes), reversion, duality, odd and even split,
magnitude
(squared) and bilinear product; and I suppose parity transform, which
has occasional relevance to non-homogeneous representations ---
its only effect on a (pro)versor being possibly to change the sign.
Your suggestion of a glossary is an excellent one --- I have already
sketched one for my own notes, though it grows at an alarming rate.
Keeping track of incompatible coinages, such as Li vs Hestenes,
looks to be something of a challenge though --- it is high time the
notation started to stabilise!
scalar --- element in some ground field, providing components for
multivectors; usually a real or complex number;
identified with multivectors of grade 0.
generator --- one of p+q variables x_i satisfying (Clifford)
constraints
x_i x_i = +1 for 0<i<=p; -1 for p<i<=p+q; 0 for p+q<i<=p+q+r;
x_i x_j + x_j x_i = 0 for i /= j .
multivector --- polynomial with coefficients in the scalar field,
and variables reduced modulo given generator constraints.
geometric (Clifford) algebra, Cl(p,q,r) --- set of multivectors, given
p,q,r.
vector --- purely linear (grade 1) multivector polynomial.
simple element --- product of vectors or scalar [PROBABLY].
versor --- product of vectors (or scalar) with nonzero magnitude.
blade --- versor with nonzero components all having the same grade.
pseudoscalar --- multivector with (a single) component of maximum
possible grade p+q+r.
I haven't come across a definition of "simple element" so far, but
from
the way the term is used it looks as if it must mean the same thing
as
my "provector". "Simple" in general describes some object which is
NOT
the product of smaller ones --- e.g. simple group, semi-simple Lie
algebra
--- so I don't think much of this misleading coinage!
Fred Lunnon
Fred
On Jan 29, 12:17 am, Fred <fred.lun...@gmail.com> wrote:
> ... and I have just
> realised that I have inadvertently instead been using "magnitude" for
> its (real) square, which I denote by ||X||. Li appears to favour
> superscript 2 to denote ||X|| when X is a vector --- I dislike this
> usage, since X^2 /= ||X|| when the grade of X exceeds 1.
It looks like "norm" is a reasonably safe and well-established word
for ||X|| = X^+ X (the product of the reversion of X with X),
scalar
when X is a "proversor" (scalar or product of vectors).
> I haven't come across a definition of "simple element" so far, but
> from the way the term is used it looks as if it must mean the same
> thing as my "provector". "Simple" in general describes some object
> which is NOT the product of smaller ones --- e.g. simple group,
> semi-simple Lie algebra --- so I don't think much of this misleading
> coinage!
I can't believe it --- I meant "proversor" --- groan ....
Another word frequently encountered in this connection is "spinor",
whose correct definiton has been eluding me for several years. I
could
fill a small book with the incompatible meanings I've struggled to
reconcile: proversor, (invertible) versor, even versor,
orthogonal matrix, ideal of Clifford algebra, matrix with versor
elements, member of (abstract) Pin group, of Spin group ...
The most reliable definition I can suggest is unfortunately also the
vaguest: "spinors" can apparently refer to any representation of
(meaning assignment of some sort of coordinates to) the elements
of a Lie group (essentially the isometries of the kind of geometric
space usually discussed in GA applications).
Examples are --- well, pretty much all of the above. The upshot is
that
this term demands regarding with extreme prejudice when encountered
in the literature: the writer is most likely as hazy about its exact
meaning
as is the reader, and just trying to impress. Deprecated!
There doesn't seem to be much in the way of a glossary available on
the
web. I'm considering uploading a suitably edited version of my
personal
version (TeX / PDF) to solicit comments, typos, obscurities,
corrections,
further suggestions, etc in the customary fashion. Comments?
Fred Lunnon
Fred
> the introduction of apparently random variation of adequate existing
> notation of which their author was already well aware. I still
> have not seen Li's book, but it appears that he is responsible for
> introducing a tilde (~ over expression) to denote reversion, instead
> of Hestenes' superscript dagger; also a caret (^ over expression)
> to denote parity transform, instead of Hestenes' superscript asterisk
> (*). What on earth can be the benefit of such destructively confusing
> permutations?
I was rushing to judgement here --- in fact, the "new" notation turns
out to be used by Ian Porteous in "Clifford Algebras and the
Classical Groups", and probably going back to 1981: a pedigree even
I hesitate to argue with. I can see I may have to admit defeat here,
and roll over to join the majority!
It's also the case that superscript asterisk X* is traditionally used
to denote the Hodge duality operator, corresponding to duality in
Clifford algebra (rather than to parity transform, prescribed by
Hestenes).
The issue remains that it should be perfectly possible to devise
a practical text-encoding for a working set of Clifford operations,
as I was attempting to do in lurker.txt --- it's not acceptable that
people have to resort to putting up a pdf file in order to
communicate
a mathematical message.
Has anybody thought about this? Fred Lunnon
Fred
On Jan 28, 3:34 pm, Fred <fred.lun...@gmail.com> wrote:
> I earlier misinterpreted the term "versor" to include isotropic (zero
> magnitude) products of vectors.
> the final definition should probably just read instead:
>
> "A proversor is either a scalar or of a Clifford product of vectors".
Following up on Peeter Joot's suggestion of a literature survey
reveals endemic confusion over whether "versor" and "blade"
should constrained to be invertible, or allowed possibly to be
isotropic.
Most of the time, it seems assumed that a "versor" is invertible,
while a "blade" may be isotropic; but Leo Dorst in particular,
who has published extensively on these matters, seems quite
unable to make up his mind about either term, even in the course
of a single book or paper.
This illustrates the confusion which follows inevitably on a poor
notational decision: in this case, the original definition of
"versor"
as implicitly invertible. It's obviously too late to repair the
situation:
but at the very least, any author making use of either term would
be well advised to state his preferred definitions explicitly.
Perhaps these matters are not all that important to people who
have already acquired a good grasp of the fundamentals ---
once the expert reader is aware of the problem, it's not hard to
spot which assumptions are in force. But apparent inconsistency
can present a severe obstacle to a beginner --- and beginners
find GA ideas quite difficult enough without having to cope with
this as well!
Fred Lunnon
Fred
In a private communication, Lanco asks
<< You should tell why you are interested in non zero products of
vectors giving a non invertible result. >>
Before I discuss this --- no doubt to Peeter's disgust --- I want
to
share my latest inspiration on the nomenclature front, having become
disillusioned with "proversor" after twice in succession discovering
myself(!) mis-typing it as "provector". Ruminating on the knock-on
effect of "versor" confusion on "blade", I noticed that the teeth of
a comb are blades, alternating with gaps in between; likewise the
(possibly) nonzero graded subvectors of a product of vectors, having
all odd-grade or all even-grade, alternate with identically empty
grades.
So let's hear it for the definition:
A "comb" is either a scalar or of a Clifford product of vectors.
And while I'm at it,
A "subvector" <X>_k --- or k-subvector, 1-subvector
(vector), 2-subvector (bivector), etc. --- is a subset
comprising all terms of a multivector X having the same
grade (k, where specified).
Now a "blade" is defined as a comb which is also a subvector;
a "versor" as an invertible comb.
Hestenes calls a subvector a "k-vector" --- a gross misnomer,
since elsewhere routinely used to denote a vector of _length_ k,
rather than _grade_ k; furthermore, how is the user supposed to
avoid specifying a value for the grade when unknown? Deprecated!
"Isotropic" combs, with norm zero so non-invertible, occur
naturally as soon as we move outside the algebras Cl(p,0,0) and
Cl(0,q,0) clung to by neophytes. The most immediately applicable
example is probably motor geometry using algebra Cl(3,0,1),
explored by robotics specialists, but apparently unknown to the
many other people who should be using it for Euclidean 3-space
[instead of employing the many strange contraptions proposed
elsewhere for this purpose]. Isotropic combs in this algebra
represent points and lines at projective infinity, i.e.
directions of (pencils of parallel) lines, and "normal vectors"
of planes.
Other examples are 3-space conformal geometry (Moebius group)
using Cl(4,1,0), where all points are isotropic; and 3-space
contact geometry (Lie-sphere group) using Cl(4,2,0), where all
cycles --- points, oriented spheres and planes, etc --- are
isotropic.
Even if the user insists on drawing an (artificial) distinction
between subspaces and isometries, isotropic combs still intrude
as limiting cases of non-isotropic (versors). For example, a
(continuous) hyperbolic rotation in conformal geometry is
represented by a 2-comb of the form
cosh t/2 + sinh t/2 R
where R is a 2-blade with ||R|| = -1, and t represents time.
After infinite time, the rotation becomes isotropic, projecting
onto one of a pair of eigen-points. To construct a rotation with
given source and sink eigen-points, the first step is to multiply
these together, giving the isotropic comb 1 + R .
Isotropic rotations also appear when a higher-grade (Spin)
isometry is factored into rotations, their extents (time, angle,
distance) requiring subsequent attachment via a separate stage.
Fred Lunnon
Lanco
New glossary may be more acceptable, when linked to new main results.
To get to the essential let us consider conformal geometric algebra,
CGA,
or more generally Cl(n,1,r).
It seems that the basic objects in the CGA are blades as in the
classification at
http://www.geometricalgebra.net/euclidtable.pdf
of all blades. Here the singular blades, i.e. blades with square 0,
are directions and tangents (including points) and their dual
versions.
A classification of versors with the same clarity seems not to have
been published.
But focus seems to be on maps generated with an isotropic vector a:
x -- > a x a
If a is in the kernel of the metric the map is zero, and otherwise it
can be proved,
that the range has dimension 1.
To generate composite maps as in the versor case with at least one
isotropic factor
still gives a range of maximal dimension 1.
Blades of degree greater than 1 are mapped onto zero.
It is correct that the source and sink for a hyperbolic rotation are
related as you write.
But these points can also be found as eigenvectors.
If you have a new main conjecture, that could justify a special word
for these
versorlike object, I would suggest the word pseudo-versor.
Lanco
Fred
On Feb 3, 4:21 pm, Lanco <a....@c.dk> wrote:
> Hi Fred
>
> New glossary may be more acceptable, when linked to new main results.
That's a fair point --- though it's also true that clumsy notation can
prevent
people from finding new results. For example, much elementary
theory relating to versors does not depend on them being invertible,
so actually applies to combs as well.
> To get to the essential let us consider conformal geometric algebra,
> CGA,
> or more generally Cl(n,1,r).
>
> It seems that the basic objects in the CGA are blades as in the
> classification at
> http://www.geometricalgebra.net/euclidtable.pdf
> of all blades. Here the singular blades, i.e. blades with square 0,
> are directions and tangents (including points) and their dual
> versions.
I did look at this table --- unfortunately, the chapter explaining it
has
been censored from googlebooks, so it remains rather mysterious.
Even so, it does strike me as rather complicated --- some of the
reason for this may be that Cl(4,1,0) is not entirely appropriate for
Euclidean geometry, which is more naturally (and efficiently)
modelled
using Cl(3,0,1).
> A classification of versors with the same clarity seems not to have
> been published.
Agreed --- in fact, a lot of space in Dorst's book seems to be taken
up by rather superficial manipulations, while fundamental results
concerning isometries remain unaddressed.
> But focus seems to be on maps generated with an isotropic vector a:
> x -- > a x a
> If a is in the kernel of the metric the map is zero, and otherwise it
> can be proved,
> that the range has dimension 1.
I don't understand this --- any isotropic multivector lies in the
"kernel of the metric" by definition! If vector X is orthogonal to
isotropic vector A, then A~ X A = 0; otherwise, A~ X A :=: A
[meaning the two are scalar multiples of one another].
> To generate composite maps as in the versor case with at least one
> isotropic factor
> still gives a range of maximal dimension 1.
> Blades of degree greater than 1 are mapped onto zero.
Once again, this doesn't look quite right.
Taking generators x x = y y = +1, v v = -1,
let A = 1 + y v, X = y x; then for example,
A~ X A = (1 - y v) y x (1 + y v) = -2(x y + x v) :=: x(y + v).
> It is correct that the source and sink for a hyperbolic rotation are
> related as you write.
> But these points can also be found as eigenvectors.
But the whole idea of GA is to avoid doing things (more clumsily)
using
matrices!
For example, suppose in conformal geometry you are given some
hyperbolic
rotation R = S + L, where S is scalar and L a 2-blade with ||L|| =
-1.
Let the isotropic comb / provector / pseudovector Q = 1 + L, and
choose a random vector X. Provided you're not so unlucky that X turns
out orthogonal to them, the eigenpoints --- or more generally
eigencycles,
in contact / Lie sphere geometry --- are simply
Q~ X Q and Q X Q~
> If you have a new main conjecture, that could justify a special word
> for these
> versorlike object, I would suggest the word pseudo-versor.
> Lanco
Another example of a result using isotropic combs is the algorithm
for
finding principal angles which I (mis)stated elsewhere; but I think
that's probably enough for one post!
Fred
Lanco
" ...much elementary theory relating to versors does not
depend on them being invertible,..."
No, everything about versors has to do with their invertibility.
Take e.g. Hestenes http://geocalc.clas.asu.edu/pdf/CompGeom-ch1.pdf
where versors nearly always is used in connection
with orthogonal mappings expressed as
x --> Par(U) x Inverse(U)
" > http://www.geometricalgebra.net/euclidtable.pdf
... it does strike me as rather complicated --- some of the
reason for this may be that Cl(4,1,0) is not entirely appropriate for
Euclidean geometry, which is more naturally (and efficiently)
modelled..."
May be it is a study worth.
"> But focus seems to be on maps generated with an isotropic vector a:
> x -- > a x a
> If a is in the kernel of the metric the map is zero, and otherwise it
> can be proved,
> that the range has dimension 1.
I don't understand this --- any isotropic multivector lies in the
"kernel of the metric" by definition! If vector X is orthogonal to
isotropic vector A, then A~ X A = 0; otherwise, A~ X A :=: A
[meaning the two are scalar multiples of one another]."
I wrote vector a, and it is in the kernel, if a . x = 0 for all
vectors x.
I use capital for multivectors.
We may prove this:
If x and a vectors and aa=0, then
the range for x -- > f(x)= a x a is either {0} or all
multiples of a.
Sketch of proof, if a<>0:
Select a basis a,b, e1, ...ek, such that bb=0, a . b=1, and a,b are
orthogonal to the rest of the vectors. Then evidently
f(a)=0, f(e1)=0,..., f(ek)=0 and f(b) =(ab+ba-ba)a=2(a . b)a=2 a
Hence, if you have a composite mapping
with x, a, b vectors and aa=bb=0, then the range for
x -- > f(x)= ba x ab is either {0} or all multipla of b
- and likewise for several factors.
I had in mind the extension of f by outermorphism to a mapping F.
For orthogonal x and y, therefore
F(x y)=F(x) (wedge) F(y) ,
which is zero, if the range of f is at most 1-dimensional.
"; then for example,
A~ X A = (1 - y v) y x (1 + y v) = -2(x y + x v) :=: x(y + v)."
Your example proves, that the formula A~ X A doesn't extend as a
outermorphism.
So here is the divergens.
"> But these points can also be found as eigenvectors.
But the whole idea of GA is to avoid doing things (more clumsily)
using matrices!"
It is not at all about matrices, when e.g. Hestenes works with
eigenvectors
or eigenblades.
"For example, ...the eigenpoints ...--- are simply
Q~ X Q and Q X Q~"
Agreed. - And this has to do with the range being at most 1-
dimensional,
when X is a vector and Q contains an isotropic factor.
Generalizations to Q with multiple factors seems problematic:
If Q has e.g. 4 isotropic factors with the first pair orthogonal to
the second,
and X is a vector the mapping is zero.
"Another example of a result using isotropic combs is the algorithm
for finding principal angles ..."
It seems that your are testing some of your ideas on the group.
So let us see this isotropic application about principal angles.
Lanco
Fred
On Feb 4, 11:49 pm, Lanco <a....@c.dk> wrote
> " ...much elementary theory relating to versors does not
> depend on them being invertible,..."
> No, everything about versors has to do with their invertibility.
Let X,Y be combs/proversors/pseudo-versors, i.e. products of vectors.
Then
||X Y|| = ||X|| ||Y||;
X~ Y X is a comb having the same grade as Y;
X may be factorised into vectors, essentially via the proof of
the Cartan-Dieudonne theorem: choose a random vector F, compute
G = X~ F X, H = |G| F + |F| G, X <- H X (H is a factor).
Iterate until X becomes vector or scalar.
[Above, |Y| denotes the square root of the absolute value of Y~ Y.
If G = 0, the "random" F was unfortunate (orthogonal to X), so choose
another F; similarly if |F| = 0.]
None of these depends on invertibility!
[Of course, you could argue that factorising a comb X of grade k
is equivalent to factorising its maximum-grade tooth/subvector/
k-vector part <X>_k, which is necessarily a blade and therefore is
"allowed" to be isotropic anyway.
But why should you have to? Indeed, throwing away the information
conatined in the remainder of X is almost certainly a bad idea
where approximate numerical computations is concerned.]
Fred Lunnon
Fred
My apologies to those who receive this twice --- I goofed over
starting
a new thread! WFL
On Feb 4, 11:49 pm, Lanco <a....@c.dk> wrote:
> " --- some of the
> reason for this may be that Cl(4,1,0) is not entirely appropriate for
> Euclidean geometry, which is more naturally (and efficiently)
> modelled..."
>
> May be it is a study worth.
I've only recently become aware that Cl(3,0,1) --- or at any
rate, its grade-2 rotors --- have long been studied in robotics
under the name "motors". Unfortunately, robotics has little use
for subspaces (as opposed to isometries); with the result that
the potential for elementary geometrical computations as required
by CAD, graphics and vision have not been developed there.
Jon M. Selig goes some way towards remedying this in his textbook
"Geometric Fundamentals of Robotics", Springer (2005)
[though he insists quixotically on using Cl(0,3,1) instead ---
both algebras are for practical purposes essentially the same].
Many other authors have preferred prefer to ignore Cl(p,q,r)
for r > 0, which is strange since (1) Grassman algebra Cl(0,0,r)
is well established for differential forms; and (2) Clifford's
original geometry for Euclidean space was also degenerate
[grade 3 only, since he excluded improper (odd grade) isometries].
Because the algebra is "degenerate" (with a generator squaring to
zero), duality becomes a nontrivial operation in its own right,
prompting in turn the introduction of a separate dual product
operator (sorry, Peeter!).
I will upload an introductory account DCQ_teabag.pdf illustrating
the application of Cl(3,0,1) --- called "DCQ's" or Dual Complex
Quaternions, reflecting their algebraic ring structure ---
to a toy problem in Euclidean 3-space geometry.
The notation differs somewhat from Porteous/Dorst:
||X|| denotes norm X~ X ;
bullet denotes Clifford product X Y ;
circle denotes dual product (X* Y*)* .
Notice too that Dorst's convention of conjugating Z -> X Z /X ---
meaning presumably (X Z X~) / ||X|| --- involves unnecessarily
dividing every component of the product by the norm, besides
making it impossible to apply an isotropic transformation.
By the way, why the slightly perverse right-to-left convention is
so popular with other GA authors is something of a mystery to me:
it might be reasonable in a mathematical context, where isometries
are regarded as functions acting on points thus Y(X(Z));
however, in a computational context, it makes no sense at all.
Notice too that Dorst's convention of conjugating Z -> X Z /X ---
meaning presumably (X Z X~) / ||X|| --- involves unnecessarily
dividing every component of the product by the norm, besides making
it impossible to apply an isotropic transformation.
> "Another example of a result using isotropic combs is the algorithm
> for finding principal angles ..."
>
> It seems that your are testing some of your ideas on the group.
> So let us see this isotropic application about principal angles.
I intend to return to this topic in another thread. WFL
Lanco
Of course it is an exaggeration to write everything.
A little correction to the proof:
if a<>0... should be "if a is not in the kernel of the
metric..."
Lanco