On the definition of "right-handed triple"
Benjamin J. Tilly
ba...@guitar.ucr.edu (john baez) writes:
> >I know another student who complained who was not a physics student,
> >but I agree that people who complain about it are fairly rare.
>
> Over on sci.physics we inveigh against the right-hand rule regularly!
> Ron Maimon claims to have refused to learn it. I learned it, but
> being left-handed I immediately recognized its artificiality and
> noted that one could do everything with a left-hand rule just as well.
> I think it's the sort of thing physicists get especially worked up over
> because the cross product appears in some laws of physics, which makes
> one wonder "HUH? So you mean god is right-handed??" - and then eventually
> one learns about polar versus axial vectors, or 1-forms and 2-forms, and
> is vastly relieved.
Not me! I learned the rule as a handy notation for a plane and an
orientation, recognized the fact that one could also use a left-hand
rule to sort things out so long as you are consistent, and then never
saw the point in worrying about it. Of course later I learned the other
notations, but while they are better for generalizations I am
unconvinced that for three dimensional space that there is anything
worse about the notation for cross-products. The choice of the right
hand is no more artificial to me than the choice of putting matrices of
the left of the vector that we are multiplying rather than switching
the order and taking transposes.
Ben Tilly
john baez
Benjamin...@dartmouth.edu (Benjamin J. Tilly) writes:
>Not me! I learned the rule as a handy notation for a plane and an
>orientation, recognized the fact that one could also use a left-hand
>rule to sort things out so long as you are consistent, and then never
>saw the point in worrying about it. Of course later I learned the other
>notations, but while they are better for generalizations I am
>unconvinced that for three dimensional space that there is anything
>worse about the notation for cross-products. The choice of the right
>hand is no more artificial to me than the choice of putting matrices of
>the left of the vector that we are multiplying rather than switching
>the order and taking transposes.
Well, the technical term for it is that the operation of cross product
V x V -> V, for V 3-dimensional, is not natural. (Dunno if you have
seen the category-theoretic definition of naturality.) The wedge product
V x V -> Lambda^2V is natural. Matrix multiplication End(V) x V -> V is
also natural. Certainly the way you WRITE matrix multiplication is
arbitrary but that's a bit different. That's why the cross product
bugged me and the rest didn't.
Edwin Clark
Why all the hostility toward unnatural things. They can be fun.
As for the cross product: doesn't the fact that the Lie algebra
of the rotation group on R^3 is R^3 under the cross product
give it some respectability?
--Edwin Clark
Matt Austern
> Well, the technical term for it is that the operation of cross product
> V x V -> V, for V 3-dimensional, is not natural. (Dunno if you have
> seen the category-theoretic definition of naturality.)
I, for one, haven't seen that definition. (I occasionally pretend to
know something about math, but I really don't.) I've seen phrases
like "natural isomorphism" every so often, but I actually hadn't
realized that this actually had a rigorous technical meaning.
If the definision is sufficiently non-technical that I can understand
it without really knowing any category theory, I'd be interested in
hearing it.
--
Matthew Austern Maybe we can eventually make language a
ma...@physics.berkeley.edu complete impediment to understanding.
john baez
(Edwin Clark) writes:
>Why all the hostility toward unnatural things. They can be fun.
Sure, but when you try to use them they cause problems because of their
unnaturality. Try defining cross products of vector fields on an
3-dimensional nonorientable manifold. You can't, whereas the wedge
product is perfectly fine. Also, it would be rather startling if
the laws of electromagnetism depended on a choice of orientation. They
don't, so it's nice to know there's a way to write them down that
doesn't involve operations that DO depend on this choice.
Mind you, I use cross products whenever I feel like it. This is the
sort of matter of principle that only certain people need to worry
about. I'm working on quantum gravity, which goes along with the fact
that I have always been extremely interested in the very essence of the
concept of space and nitpicky issues of this kind; also, general
relativity is all about naturality.
>As for the cross product: doesn't the fact that the Lie algebra
>of the rotation group on R^3 is R^3 under the cross product
>give it some respectability?
Interesting point there. Starting with a 3d vector space V with an
inner product we can naturally form the rotation group SO(V), and then
naturally form its Lie algebra so(V). This thing has a bracket so(V) x
so(V) -> so(V) which is perfectly natural and is ALMOST the same thing
as the cross product. However, if one wants to get a cross product map
V x V -> V one has to pick an isomorphism V <-> so(V). This cannot be
done naturally! One has to use a right-hand or left-hand rule at this
point.
Moral: if you like cross products and naturality work with the Lie
algebra. That's respectable. :-)
I hope y'all don't take me for an utterly stuffy pedant; when I teach
vector calculus I only make one joke about how being left-handed
I think it should be changed to a left-hand rule (LEFT HANDERS RULE!),
and then plunge. (Presumably this joke will serve as a pinprick of
puzzlement to some of the students.)
john baez
ma...@physics.berkeley.edu writes:
>In article <2b44ea$6...@galaxy.ucr.edu> ba...@guitar.ucr.edu (john baez) writes:
>
>> Well, the technical term for it is that the operation of cross product
>> V x V -> V, for V 3-dimensional, is not natural. (Dunno if you have
>> seen the category-theoretic definition of naturality.)
>I, for one, haven't seen that definition. (I occasionally pretend to
>know something about math, but I really don't.) I've seen phrases
>like "natural isomorphism" every so often, but I actually hadn't
>realized that this actually had a rigorous technical meaning.
Yes, the first great thing about category theory is that it really makes
this concept of "naturality" utterly precise! (And in a way that really
corresponds pretty well to the intuitive idea.)
>If the definision is sufficiently non-technical that I can understand
>it without really knowing any category theory, I'd be interested in
>hearing it.
Sure, I'll present it just in this special case without all the
trimmings. What would it mean to define the cross product of 3d vectors
naturally? It would mean that we could choose for ANY 3d vector space V
a bilinear map V x V -> V such that
1) When V was R^3 it'd be the usual cross product (v,w) -> v x w.
2) Naturality: given two different 3d vector spaces V and W and an
isomorphism f: V -> W, the way of taking cross products in V is
compatible with that in W, i.e.:
f(v x w) = f(v) x f(w).
To see that it is impossible to meet both conditions, just consider
f: R^3 -> R^3 to be the operation of "parity", sending
(x,y,z) -> -(x,y,z).
Then we have
f(v x w) = -f(v) x f(w).
For a physicist I might roughly describe that naturality condition as
being "coordinate-independence." You see, if someone hands us a 3d
vector space we have to *choose* coordinates to think of it as R^3.
Unnaturality is a sophisticated way of saying that the cross product must
depend on this choice. Sophisticated, because it doesn't even
introduce the notion of coordinates explicitly! The idea is that if I
have "my" 3d vector space V and you have "yours," W, to "translate"
between our vector spaces (like translating from French to German) we
need an isomorphism f: V -> W. And we'll be happiest if my way of doing
cross products is compatible with yours. The cross product would be
natural if everyone, when handed a 3d vector space, could define a cross
product on it in such a way that no matter what isomorphisms they used
to communicate to each other the cross products would be compatible.
Writing this I'm struck again at what an outrageous demand naturality
seems to be! But there are lots of examples of natural operations and
natural isomorphisms, some of which Matthew has bumped into. For
example, if we have two vector spaces V and W we can take their
Cartesian product in two orders: V x W or W x V. These are, however,
naturally isomorphic, a natural isomorphism being
(v,w) -> (w,v)
(Exercise: state this precisely as above and prove it.)
The essence of special and, even more so, general relativity is
naturality. This is why Louis Crane has gone on a bit of limb and is trying
to formulate a theory of quantum gravity in terms of sophisticated ideas
from category theory. Folks who are sufficiently daring might try:
Categorical Physics, by Louis Crane, 9 pages amstex, hep-th/9301061.
Pertti Lounesto
I have heard some people arguing about that but not making sense.
A non-trivial cross-product exists only in dimensions 3 and 7,
if we require the cross product to satisfy the properties:
1) the cross-product is a bilinear mapping R^n x R^n -> R^n,
(u,v) -> uxv
2) uxv is perpendicular to both u and v
3) (uxv)^2 = u^2 v^2 - (u.v)^2 (Lagrange's identity)
where u^2 = |u|^2 and u.v is the Euclidean scalar product.
Of course, in seven dimensions there is a multitude of orientations.
Pertti Lounesto
Pertti....@hut.fi Institute of Mathematics
voice: +358-0-4513020 Helsinki University of Technology
fax: +358-0-4513016 SF-02150 Espoo, Finland
john baez
>In article <MATT.93N...@physics2.berkeley.edu>
>ma...@physics.berkeley.edu writes:
>Sure, I'll present it just in this special case without all the
>trimmings. What would it mean to define the cross product of 3d vectors
>naturally? It would mean that we could choose for ANY 3d vector space V
>a bilinear map V x V -> V such that
>1) When V was R^3 it'd be the usual cross product (v,w) -> v x w.
>2) Naturality: given two different 3d vector spaces V and W and an
>isomorphism f: V -> W, the way of taking cross products in V is
>compatible with that in W, i.e.:
>f(v x w) = f(v) x f(w).
Unfortunately I can't cancel posts; here it occurred to me right after
posting that I was demanding more naturality than one really had any
right to expect. Since the cross product of two vectors is
perpendicular to each, etc., it's clear that one at least needs to fix a notion
of dot product to have a chance of defining the cross product. Thus I
am way too ambitious to expect "naturality in the category of 3d vector
spaces," but might possibly expect "naturality in the category of 3d
inner product spaces" - 3d vector spaces equipped with a dot product.
An isomorphism between such spaces is just a 1-1 and onto map of vector
spaces that preserves the dot product. So one could formulate
naturality in this category as:
...we could choose for ANY 3d inner product space V a bilinear map V x V
-> V such that
1) When V was R^3 it'd be the usual cross product (v,w) -> v x w.
2) Naturality: given two different 3d inner product spaces V and W and an
isomorphism f: V -> W, the way of taking cross products in V is
compatible with that in W, i.e.:
f(v x w) = f(v) x f(w).
But again the cross product would NOT be natural in this category, for
the same reason: there's an orientation-reversing isomorphism.
My main point to Matt and the physicists here is that naturality is not
a cut-and-dried affair; there are some things you can do naturally *if*
one has some structure, and if one has more structure around there is
more one can do naturally. Indeed, the cross product IS natural in the
category of 3d vector spaces equipped with a dot product AND an
orientation (a notion of a "right-handed basis").
The point is that it can be worth keeping track of the minimal amount of
structure around necessary to do a certain construction naturally. In
general relativity, for example, there are a few important categories:
1) smooth manifolds
2) smooth manifolds with metric
3) oriented smooth manifolds
4) oriented smooth manifolds with metric
and some more subtle ones, like manifolds with "spin structure," which
is the structure needed to define a notion of spinor fields.
john baez
(Pertti Lounesto) writes:
>Is there a difference between the terms "natural" and "canonical"?
>I have heard some people arguing about that but not making sense.
I don't think MOST people know what the difference is, and it wasn't
until Jim Dolan, who knows categories better than most, explained it to
me that I really got it.
So, when most people use these terms, don't expect there to be a
difference. But when a categorist does, do!
Briefly, "natural" means compatible with all isomorphisms, in the
sense I described in my post (in an example). "Canonical" means
compatible with all morphisms. This is a more stringent requirement.
Jim enlightened me with the following puzzle: there is only one
CANONICAL way to pick a permutation of a finite set -- pick the identity
permutation. However, there are two NATURAL ways. One is to pick the
identity permutation. What's the other????
The exterior derivative, or "d" operator, which is so important in the
theory of differential forms, is not only natural but canonical. That
is, for any smooth mapping F between manifolds, NOT JUST DIFFEOMORPHISMS,
one has
F_\ast d \omega = d F_\ast \omega.
This equation really is the sort of commutative diagram that occurs in
the general definition of "natural" and "canonical".
Ron Maimon
|>
|> Not me! I learned the rule as a handy notation for a plane and an
|> orientation, recognized the fact that one could also use a left-hand
|> rule to sort things out so long as you are consistent, and then never
|> saw the point in worrying about it. Of course later I learned the other
|> notations, but while they are better for generalizations I am
|> unconvinced that for three dimensional space that there is anything
|> worse about the notation for cross-products. The choice of the right
|> hand is no more artificial to me than the choice of putting matrices of
|> the left of the vector that we are multiplying rather than switching
|> the order and taking transposes.
|>
|> Ben Tilly
Actually, I didn't mind cross products so much when I wrote them down in
component form. What I hated was having to know which way is the "right"
way to draw the z axis given the x and y axis. Even if you draw it wrong,
and use the same formula for the cross product, you get the same answer.
Of course, when you draw the picture you end up using your left hand instead
of youre right ....
so I just used the formula for the cross product, aware that I have to be
careful when I do coordinate inversions, and never learned the right hand
rule. It's not that I did it on purpose, but I learned physics on my own
and never had a need for it. I screwed up the direction of many magnetic
field problems in my introductory courses and AP tests because of it, but
I'm thankfully past that now.
As for wedge products, I thought those were almost as bad as the right hand
rule when I first learned them, because I learned them from a really formal
professor, who didn't bother explaining their interpretation as linear maps,
and this was after I knew index tensor notation really fluently. Now, I don't
hate them quite so much.
I keep on coming up with things that are horribly awkward to express with
wedges and d's and easy to express in indices. I still think there is no
need for a different notation for forms and tensors, and I would prefer
to put indices on them all, but I would get lynched if I said this too loud.
Ron Maimon
Ron Maimon
|>
|> Why all the hostility toward unnatural things. They can be fun.
|> As for the cross product: doesn't the fact that the Lie algebra
|> of the rotation group on R^3 is R^3 under the cross product
|> give it some respectability?
|>
|> --Edwin Clark
|>
I don't know about you, but I like to feel like every law of physics I
know is in a form in which I would have put it myself.
The laws involving cross products are not in a form I would ever put them.
In addition, do you remember this identity?
( A x B ) x C = B (A . C) - A ( B . C )
I never do. I remember the much more obvious identity
( AiBj - AjBi) Ci = AiCi Bj - Aj BiCi
where the summation over "i" is understood.
I like things when they are obvious. I hate things I am supposed to
memorize.
Ron Maimon
Ron Maimon
|>
|> and some more subtle ones, like manifolds with "spin structure," which
|> is the structure needed to define a notion of spinor fields.
|>
please explain this further!!
I had a lot of trouble understanding spinor fields because I couldn't see
how to make them Generally Covariant.
Ron Maimon
john baez
>Now why is it important? Well constructions which are natural are a lot
>nicer. For example constructions involving the tangent spaces and the
>cotangent spaces at a point on a smooth manifold can be turned into
>constructions on the manifold if the construction is natural, but
>otherwise there are problems. (Something which John Baez will be glad
>to enlighten you on... :-)
>Do I pass John?
A+.
I think that one needs to be doing a certain kind of mathematics to
fully appreciate the importance of naturality. Maybe one needs to be
regularly engaged in constructing new mathematical objects. Computer
programmers would probably understand this pretty well by analogy to the
constant problem of compatibility and "standards" in software. If a
construction is natural that automatically implies it will be compatible
with a whole lot of other constructions. E.g., in the example Tilly
mentions above, a natural construction of one vector space from another
will automatically give you a nice way to construct one vector *bundle*
from another. Unnatural constructions won't.
If people only wrote a few computer programs and then used them for the
rest of eternity, compatibility wouldn't be such a big issue. It's when
people are rapidly developing more and more new software that should
interact with existing and not-yet-existing software that compatibility
becomes important. Similarly, if one is just going to be using the cross
product in electromagnetism, fluid dynamics, or classical mechanics,
its unnaturality isn't a big problem. However, when one is developing
new mathematical frameworks a lot one starts to value naturality.
Benjamin J. Tilly
ma...@physics2.berkeley.edu (Matt Austern) writes:
> In article <2b44ea$6...@galaxy.ucr.edu> ba...@guitar.ucr.edu (john baez) writes:
>
> > Well, the technical term for it is that the operation of cross product
> > V x V -> V, for V 3-dimensional, is not natural. (Dunno if you have
> > seen the category-theoretic definition of naturality.)
>
> I, for one, haven't seen that definition. (I occasionally pretend to
> know something about math, but I really don't.) I've seen phrases
> like "natural isomorphism" every so often, but I actually hadn't
> realized that this actually had a rigorous technical meaning.
>
> If the definision is sufficiently non-technical that I can understand
> it without really knowing any category theory, I'd be interested in
> hearing it.
If you want the *very* informal definition, here goes. You have some
mathematical structure, which gives you some sort of information. If
you can define an operation on that structure using only that
information, then it is natural, otherwise it is not. For an example of
what I mean consider a vector space V and its dual V*, the set of
linear functionals on V. They are not naturally isomorphic, even though
they are isomorphic, because to get an isomorphism you would need more
information, such as a basis for V. However V *is* naturally isomorphic
to V**, and the natural isomorphism is to send a vector v to the
function on functionals of evaluating the functional f at v. That is we
get a functional on functionals v' with v'(f)=f(v). For the technical
definition I suggest that you read a category theory book.
Now why is it important? Well constructions which are natural are a lot
nicer. For example constructions involving the tangent spaces and the
cotangent spaces at a point on a smooth manifold can be turned into
constructions on the manifold if the construction is natural, but
otherwise there are problems. (Something which John Baez will be glad
to enlighten you on... :-)
Do I pass John?
Ben Tilly
john baez
>please explain this further!!
I will only explain this if you say 3 times, "differential forms are
wonderful." [Evil laughter.]
Better yet, why don't you do obeisance to those mathematicians you like
to make fun of and take some courses in differential geometry! [Cackles
of glee.]
Well....
Let me be brief and then expand upon it as needed. A Riemannian
manifold M gives rise to a bundle called the orthonormal frame bundle
O(M). For each point of x the space of orthonormal frames of the
tangent space at x is the fiber of O(M) at x; it looks just like the
group O(n) if M is n-dimensional - but not naturally so. If M is
orientable we can pick a global orientation on it at define the oriented
orthonormal frame bundle SO(M) consisting of oriented frame; the fiber
at x looks like SO(n).
The key to spinors is that SO(n) has a universal covering group Spin(n);
it's a 2-fold cover for n > 2. E.g. for SO(3) one has Spin(3) = SU(2).
Spinors are reps of Spin(n). Let's write f: Spin(n) -> SO(n) for this
2-1 homomorphism.
So to define spinors globally on M we need to cook up a bundle Spin(M)
whose fiber at any x looks like Spin(n) and which covers SO(M). More
precisely, we say a *spin structure* for an oriented Riemannian manifold
is a principal Spin(n)-bundle Spin(M) equipped with a bundle morphism
F: Spin(M) -> SO(M) which satisfies the obvious compatibility condition
F(p)f(g) = F(pg)
for all p in Spin(M), g in Spin(n). (The group Spin(n) acts on Spin(M)
on the right, ditto for SO(n) and SO(M).)
There may be NO choice of spin structure or more than one. Existence of
a spin structure is equivalent to vanishing of the second
Stiefel-Whitney class of M, which is the kind of topological condition
that topologists are pretty good at checking. If a spin structure
exists, spin structures are in 1-1 correspondence with elements of
H^1(M,Z_2), which is the kind of cohomology group topologists are pretty
good at computing.
For a quick intro to spin structures, try Analysis Manifolds and Physics
by Choquet-Bruhat et al. For more than you probably want to know, try Spin
Geometry by Lawson and Michelson.
john baez
difficulties:
john baez writes:
>In article <LOUNESTO.9...@dopey.hut.fi> loun...@dopey.hut.fi
>(Pertti Lounesto) writes:
>>Is there a difference between the terms "natural" and "canonical"?
>>I have heard some people arguing about that but not making sense.
>
>I don't think MOST people know what the difference is, and it wasn't
>until Jim Dolan, who knows categories better than most, explained it to
>me that I really got it.
>
>So, when most people use these terms, don't expect there to be a
>difference. But when a categorist does, do!
>
>Briefly, "natural" means compatible with all isomorphisms, in the
>sense I described in my post (in an example). "Canonical" means
>compatible with all morphisms. This is a more stringent requirement.
hmm, do i also get credit for your getting the distinction _precisely_
backwards?
[yes, it's all your fault - jb]
again, briefly: vague ideas about "naturalness" and "canonicalness" go
back a long time. then eilenberg and maclane invented the formalism
of category theory around the 1940's, and the centerpiece of this
formalism was a precise formal definition of the concept of "natural
map". the formalism eventually caught on, and so it is generally a
good idea nowadays to use the terminology "natural" in accord with the
eilenberg-maclane definition. the terminology "canonical", however,
is still somewhat up for grabs; sometimes you hear people who have
practically never even heard of category theory (or who wish that they
hadn't) using it in a very informal way. listening to this informal
usage of the term "canonical", and acting as a descriptive (that is,
non-prescriptive) linguist, i have proposed that what people usually
really tend to mean by "canonical" is "natural with respect to the
(implicitly relevant class of) _isomorphisms_".
thus canonicalness in the sense i'm talking about here is in fact a
special case of the concept of naturalness. but, in the situation
where you might actually want to use the terms "natural" and
"canonical" in a directly contrasting way (namely, when you have some
class of morphisms which are _not_ all isomorphisms), notice that it
is "naturalness" that is the more stringent requirement. this extra
stringency is what gives category theory its extra bite (as compared
to mere group theory, which is what you are doing when you are only
interested in naturalness with respect to isomorphisms).
>Jim enlightened me with the following puzzle: there is only one
>CANONICAL way to pick a permutation of a finite set -- pick the identity
>permutation. However, there are two NATURAL ways. One is to pick the
>identity permutation. What's the other????
so remember, in the above puzzle and elsewhere, that john always says
"canonical" for "natural", and vice versa. (also, the puzzle still
works even if the sets aren't required to be finite, by the way.)
-james dolan
[If there is enough demand I will try to get Jim to write up the answer.]
Ron Maimon
|> In article <2b6qd1$q...@scunix2.harvard.edu> rma...@husc9.Harvard.EDU (Ron Maimon) writes:
|> >In article <2b6e3l$e...@galaxy.ucr.edu>, ba...@guitar.ucr.edu (john baez) writes:
|>
|>
(I try to give a translation of what you said into english, but I am not so sure
I succeeded, please correct me if I am wrong)
|> Let me be brief and then expand upon it as needed. A Riemannian
|> manifold M gives rise to a bundle called the orthonormal frame bundle
|> O(M). For each point of x the space of orthonormal frames of the
|> tangent space at x is the fiber of O(M) at x; it looks just like the
|> group O(n) if M is n-dimensional - but not naturally so. If M is
|> orientable we can pick a global orientation on it at define the oriented
|> orthonormal frame bundle SO(M) consisting of oriented frame; the fiber
|> at x looks like SO(n).
Ok, so youre picking a basis for the tangent space at every point, and making
sure it's orthonormal at every point.
|>
|> The key to spinors is that SO(n) has a universal covering group Spin(n);
|> it's a 2-fold cover for n > 2. E.g. for SO(3) one has Spin(3) = SU(2).
|> Spinors are reps of Spin(n). Let's write f: Spin(n) -> SO(n) for this
|> 2-1 homomorphism.
then you notice that the set of isomorphisms of orthonormal basis to orthonormal
basis is O(n), but this O(n) can be different at different points, but it should
be different in a continuous differentiable way.
And then you say that if the manifold is orientable you can reduce this to SO(n)
and work with Spin(n), the covering group. I don't see why you can't work with
the original O(n), can't you reflect spinors?
nevermind
You can define any spin field by just defining it locally.
|>
|> So to define spinors globally on M we need to cook up a bundle Spin(M)
|> whose fiber at any x looks like Spin(n) and which covers SO(M). More
|> precisely, we say a *spin structure* for an oriented Riemannian manifold
|> is a principal Spin(n)-bundle Spin(M) equipped with a bundle morphism
|> F: Spin(M) -> SO(M) which satisfies the obvious compatibility condition
|>
|> F(p)f(g) = F(pg)
|>
|> for all p in Spin(M), g in Spin(n). (The group Spin(n) acts on Spin(M)
|> on the right, ditto for SO(n) and SO(M).)
This part slips into mathematical teminology so badly I can't really follow it,
but this is what I think you mean:
and then we define a quadruplet of numbers at each point such that changing the
orthogonal basis at that point transforms the components like a spinor would
change under the same basis change.
And we make sure that it's consistent, in that if I bring the spinor field
defined in one coordinate system + frame system to it's form in another
coordinate system + frame system it had better turn out to be a function of
the frame and coordinates only.
|>
|> There may be NO choice of spin structure or more than one. Existence of
|> a spin structure is equivalent to vanishing of the second
|> Stiefel-Whitney class of M, which is the kind of topological condition
|> that topologists are pretty good at checking. If a spin structure
|> exists, spin structures are in 1-1 correspondence with elements of
|> H^1(M,Z_2), which is the kind of cohomology group topologists are pretty
|> good at computing.
|>
does this mean that certain universes are incompatible with the existence
of electrons?
I don't see why this has to be so.
|> For a quick intro to spin structures, try Analysis Manifolds and Physics
|> by Choquet-Bruhat et al. For more than you probably want to know, try Spin
|> Geometry by Lawson and Michelson.
|>
|>
actually, I havn't looked these up, but they seem to mathematics oriented for
me.
I just want a way to formulate the field equations of a spinor field on a
curved space time.
I think that what you said in the post is enough to do this, but I have to
think about it.
Ron Maimon
john baez
>In article <2b75na$h...@galaxy.ucr.edu>, ba...@guitar.ucr.edu (john baez) writes:
>(I try to give a translation of what you said into english, but I am
>not so sure I succeeded, please correct me if I am wrong)
>|> Let me be brief and then expand upon it as needed. A Riemannian
>|> manifold M gives rise to a bundle called the orthonormal frame bundle
>|> O(M). For each point of x the space of orthonormal frames of the
>|> tangent space at x is the fiber of O(M) at x; it looks just like the
>|> group O(n) if M is n-dimensional - but not naturally so. If M is
>|> orientable we can pick a global orientation on it at define the oriented
>|> orthonormal frame bundle SO(M) consisting of oriented frame; the fiber
>|> at x looks like SO(n).
>Ok, so youre picking a basis for the tangent space at every point, and making
>sure it's orthonormal at every point.
Not quite, I'm letting O(M)_x be the set of ALL o.n. bases for the
tangent space at x. If we pick one o.n. basis, any other differs just
by a rotation, so O(M)_x is isomorphic (but not naturally, since we made
a choice) to O(n). SO(M)_x is the set of all oriented ("right-handed")
ones and similar remarks apply.
>|> The key to spinors is that SO(n) has a universal covering group Spin(n);
>|> it's a 2-fold cover for n > 2. E.g. for SO(3) one has Spin(3) = SU(2).
>|> Spinors are reps of Spin(n). Let's write f: Spin(n) -> SO(n) for this
>|> 2-1 homomorphism.
>then you notice that the set of isomorphisms of orthonormal basis to
>orthonormal basis is O(n), but this O(n) can be different at different
>points, but it should be different in a continuous differentiable way.
>And then you say that if the manifold is orientable you can reduce this
>to SO(n) and work with Spin(n), the covering group. I don't see why you
>can't work with the original O(n), can't you reflect spinors?
Oh, you can do that too, it's just a bit more subtle. The two-fold
cover of SO(n) is called Spin(n), and the two-fold cover of O(n) is
cutely called Pin(n). A manifold may have more Pin structures than Spin
structures because there may be more than one way to globally define a
notion of "reflecting spinors."
>nevermind
>You can define any spin field by just defining it locally.
>|> So to define spinors globally on M we need to cook up a bundle Spin(M)
>|> whose fiber at any x looks like Spin(n) and which covers SO(M). More
>|> precisely, we say a *spin structure* for an oriented Riemannian manifold
>|> is a principal Spin(n)-bundle Spin(M) equipped with a bundle morphism
>|> F: Spin(M) -> SO(M) which satisfies the obvious compatibility condition
>|>
>|> F(p)f(g) = F(pg)
>|>
>|> for all p in Spin(M), g in Spin(n). (The group Spin(n) acts on Spin(M)
>|> on the right, ditto for SO(n) and SO(M).)
>This part slips into mathematical teminology so badly I can't really
>follow it, but this is what I think you mean:
>and then we define a quadruplet of numbers at each point such that
>changing the orthogonal basis at that point transforms the components
>like a spinor would change under the same basis change.
>And we make sure that it's consistent, in that if I bring the spinor field
>defined in one coordinate system + frame system to it's form in another
>coordinate system + frame system it had better turn out to be a function of
>the frame and coordinates only.
Yup.
>|> There may be NO choice of spin structure or more than one. Existence of
>|> a spin structure is equivalent to vanishing of the second
>|> Stiefel-Whitney class of M, which is the kind of topological condition
>|> that topologists are pretty good at checking. If a spin structure
>|> exists, spin structures are in 1-1 correspondence with elements of
>|> H^1(M,Z_2), which is the kind of cohomology group topologists are pretty
>|> good at computing.
>does this mean that certain universes are incompatible with the existence
>of electrons?
Basically, yeah.
>I don't see why this has to be so.
Well, it's just like in certain universes there is no globally
well-defined notion of right and left. It's a topology kind of thing.
Of course there are sneaky ways out, but they require more work - check
out the remarks on "generalized spin structures" in Choquet-Bruhat. For
most universes *I'd* care to live in, spin structures suffice.
>|> For a quick intro to spin structures, try Analysis Manifolds and Physics
>|> by Choquet-Bruhat et al. For more than you probably want to know, try Spin
>|> Geometry by Lawson and Michelson.
>actually, I havn't looked these up, but they seem too mathematics oriented for
>me.
Choquet-Bruhat is the "bible" of mathematics a hot-shot theoretical
physicist like yourself should probably know. (E.g., Witten has it all
contained in the fingernail of his left pinkie.) But you're right that
reading it is probably not the most efficient way to figure out this
one question.
>I just want a way to formulate the field equations of a spinor field on a
>curved space time.
>I think that what you said in the post is enough to do this, but I have to
>think about it.
Yeah, what I said is enough, the main point is a *choice* of a globally
consistent way of double-covering the rotation groups of the tangent spaces
is required; such a choice is a spin structure, and there may be none or
lots. (On R^n there's just one.)
Pertti Lounesto
(axb)xc = b(a.c) - a(b.c)
ax(bxc) = b(a.c) - c(a.b)
However, if we use the wedge product a^b, representing the parallelogram
with a and b as edges and sense of rotation from from a to b, and
the left and right contractions a_|b and a|_b, the above formulas are
(a^b)|_c = a^(b|_c) - (a|_c)^b
a_|(b^c) = (a_|b)^c - b^(a_|c)
In the first formula the vector c contracts (= drops the degree of) the
bivector a^b, and in the second formula the conrtraction by the vector a
can be recognized as a derivation.
Pertti Lounesto
Ron Maimon, accidentally sent out in the middle of writing.
It is indeed a bit difficult to remember the formulas
ax(bxc) = b(a.c) - c(a.b)
(axb)xc = b(a.c) - a(b.c)
If one uses instead the wedge product a^b, which represents
the parallelogram with a and b as edges and has a sense of
rotation from a to b, and the left and right contractions
u_|v and u|_v, the above formulas become
a_|(b^c) = (a_|b)^c - b^(a_|c)
(a^b)|_c = a^(b|_c) - (a|_c)^b
The left contraction u_|v is defined in the exterior algebra
\bigwedge R^3 of R^3 by its characteristic properties
a_|b = a.b for a,b in R^3
a_|(u^v) = (a_|u)^v +- u^(a_|v)
(u^v)_|w = u_|(v_|w) for u,v,w in \bigwedge R^3
where the second rule tells that contraction by a vector is a
derivation (the element u is even or odd and in front
of it is the sign of its parity) and the third rule tells that
\bigwedge R^3 has a linear structure over \bigwedge R^3
(scalar multiplication being contraction).
The left contraction could also be defined by the non-degenerate
scalar product <u,v> on \bigwedge R^3 extending the scalar
product on R^3 by <a1^a2^...^ak,b1^b2^...^bk> = det(ai.bj) and
by linearity and orthogonality to all of \bigwedge R^3 and setting
<u_|v,w> = <v,u~^w> for all w in \bigwedge R^3
where u~ is the reversion of u.
The left and right contractions obey several nice rules like
u_|(v|_w) = (u_|v)|_w (= associativity when the middle factor is
contracted). This theorem was proved just a month ago by
Quintino Souza (Campinas, Brazil).
The unnaturality of the cross product can be circumvented by
the wedge product and properly used contraction (but the prise is
that in addition to vectors and scalars we have
also bivectors).
Torsten Ekedahl
> I'm posting this for Jim Dolan, who can't, due to technical
> difficulties:
> .
> .
> again, briefly: vague ideas about "naturalness" and "canonicalness" go
> back a long time. then eilenberg and maclane invented the formalism
> of category theory around the 1940's, and the centerpiece of this
> formalism was a precise formal definition of the concept of "natural
> map". the formalism eventually caught on, and so it is generally a
> good idea nowadays to use the terminology "natural" in accord with the
> eilenberg-maclane definition. the terminology "canonical", however,
> is still somewhat up for grabs; sometimes you hear people who have
> practically never even heard of category theory (or who wish that they
> hadn't) using it in a very informal way. listening to this informal
> usage of the term "canonical", and acting as a descriptive (that is,
> non-prescriptive) linguist, i have proposed that what people usually
> really tend to mean by "canonical" is "natural with respect to the
> (implicitly relevant class of) _isomorphisms_".
> .
> .
> >Jim enlightened me with the following puzzle: there is only one
> >CANONICAL way to pick a permutation of a finite set -- pick the identity
> >permutation. However, there are two NATURAL ways. One is to pick the
> >identity permutation. What's the other????
> .
> .
> -james dolan
One informal but still (!) very interesting use of canonical is in the
sense that somehow for some reason there is one abvious, natural etc
choice which almost always is a natural transformation in the
technical sense, One example is the map from a vector space to its
double dual given by evaluation. It certainly is a natural
transformation but more than that being the obvious map. What makes
the thing really interesting is that canonical maps are expected to
have certain properties. First they are normally stable under
compositions and inverses (if they exist). This is maybe not so
interesting but there are two other properties which when combined
with the first give some interesting things.
The first of those is that whereas it in general is a very bad
thing to identify isomorphic objects as long as they are canonically
isomorphic you will never run into trouble. This has led to the use
(introduced by Deligne I think) to use the equality sign for canonical
ismorphisms. The necessary adjunct to this first property is the
second:
The only canonical isomorphism from an object to itself is the
identity or possibly minus the identity. This last property can
sometimes be hard to verify in concrete situations but usually much
harder is to determine the sign.
Note that this last property is certainly not a provable result
but rather a meta principle which has to be verified in each
individual case.
Let me finish by an example. There is for a smooth and proper
n-dimensional algebraic variety X over a field k a canonical trace map
H^n(X,o_X) -> k, where o_X is the sheaf of n-forms. This is the
algebraic analogue of integration of C^oo 2n-forms when k is the
complex numbers. Now, when X=P^n one may compute H^n(X,o_X) using Cech
cohomology and the standard covering consisting of the open sets where
one coordinate is non-zero. Then dx_1/x_1 ^ ... ^ dx_n/x_n is an
n-form which is regular on the intersection of all elements of the
cover and hence a Cech cocycle which in fact is a basis element for
H^n(P^n,o_P^n). In this way the trace map becomes a canonical
ismorphism from k to k (using this cocycle as basis element).
According to the principle this should be 1 or -1, in other words, the
trace of dx_1/x_1 ^ ... ^ dx_n/x_n should be 1 or -1. This can
actually be proved by using naturality of the trace and working over
the base ring Z instead of over a field. As the only invertible
elements of Z are 1 and -1 we are finished. (This can be used in many
situations in algebraic geometry to verify the meta principle). One
would then be tempted to assume that the trace is actually always 1.
This is not true, the real answer is (-1)^(n(n+1)/2) (or
(-1)^(n(n-1)/2), my memory is a little bit unclear). This shows
without any doubt that one must admit -1 in the second principle....
--
Torsten Ekedahl
te...@matematik.su.se
Ron Maimon
way to make it convenient for numerical computation, like if I wanted to put the
whole thing on a computer and simulate a spinor field on a curved background.
Is there any choice of basis at every point that makes computations easier? like
choosing a coordinate basis for tensors does?
Ron Maimon
john baez
torsten ekedahl writes:
>In article <2b8utd$m...@galaxy.ucr.edu> ba...@guitar.ucr.edu (john baez) writes:
>> I'm posting this for Jim Dolan, who can't, due to technical
>> difficulties:
...
>> hadn't) using it in a very informal way. listening to this informal
>> usage of the term "canonical", and acting as a descriptive (that is,
>> non-prescriptive) linguist, i have proposed that what people usually
>> really tend to mean by "canonical" is "natural with respect to the
>> (implicitly relevant class of) _isomorphisms_".
...
>One informal but still (!) very interesting use of canonical is in the
>sense that somehow for some reason there is one abvious, natural etc
>choice which almost always is a natural transformation in the
>technical sense, One example is the map from a vector space to its
...
> Let me finish by an example. There is for a smooth and proper
>n-dimensional algebraic variety X over a field k a canonical trace map
>H^n(X,o_X) -> k, where o_X is the sheaf of n-forms. This is the
>algebraic analogue of integration of C^oo 2n-forms when k is the
>complex numbers. Now, when X=P^n one may compute H^n(X,o_X) using Cech
thanks for helping to prove my point. the canonical integration map
from c^infinity 2n-forms to k in the case where k:=the complex numbers
ain't natural, it's only canonical!!! if it were natural, it
would be compatible with _all_ morphisms, not just with the isomorphisms.
but it is conspicuously _not_ compatible with covering maps; the
pullback of the canonical 2n-form along an m-to-1 covering map is off by
a factor of m from being the canonical 2n-form on the covering space.
of course, you _can_ get away with saying that integration of the
2n-forms here is natural, _if_ you are careful to specify the class
of morphisms with respect to which naturality can legitimately be
claimed, namely in this case the isomorphisms.
the stick-out-like-a-sore-thumb obviousness that makes a map
intuitively canonical relies only upon the map being compatible
with isomorphisms. being compatible with non-isomorphisms
as well is generally a much stronger condition, very interesting
in its own way.
the e-mail and postings responding to my postings about naturalness
vs. canonicalness have convinced me quite strongly of the validity
of the claim that i at first made only rather tentatively: the de
facto standard meaning of "canonical" is "natural with respect to
the (implicitly relevant class of) isomorphisms".
-james dolan
john baez
This is in those books I mentioned, and I don't really have the energy
or intelligence or whatever to give brief clear description of how the
whole thing works, so let me just randomly say some more stuff that may
or may not help.
When we want to local computations it is nice to work in a coordinate
chart and pick a "section of SO(M) over the chart," which is a terse
way of saying a smoothly varying oriented orthonormal basis of the
tangent space e_i at each point in the chart. Because Spin(M) is a 2-fold
cover of SO(M) there are two "sections of Spin(M) lifting the section of
SO(M)"... each of which gives a way to identify the fiber of Spin(M) at
any point x in the chart with the group Spin(n). This in turn gives a
way to think of a spinor field as a function with values in some vector
space.
All of this can be done perfectly explicitly if one wants to do juicy
coordinate-ridden explicit calculations, or make your computer do 'em!
Say our o.n. basis at some point is e_1,....,e_n (vectors whose
components you can write out in coordinates if desired). Then there is
a gadget called the Clifford algebra which is generated by the dual
basis e^1,....,e^n and has relations e^ie^j + e^je^i = eta^{ij}. (Here
eta is the Dirac delta when we are in the Riemannian case, or something
like the diagonal matrix with ++++------- down the diagonal when we're
in the Lorentzian case.) This is an algebra of dimension 2^n, it's easy
to believe. It is isomorphic to the 2^{n/2} x 2^{n/2} complex matrices
when n is even (here I note for experts that I'm talking about Clifford
algebras over C), and the isomorphism is explicitly describable, e.g.
for n = 4 the e^i's are isomorphic to the Dirac gamma matrices. The
spinors then become isomorphic to the column vectors of length 2^{n/2}.
(Things work out differently when n is odd, there is an interesting
even/odd thing going on here.) One can use all this and some more stuff
to write down the Dirac operator in local coordinates and start grinding
away.
This was neither especially complete nor coherent....
Tal Kubo
>It is isomorphic to the 2^{n/2} x 2^{n/2} complex matrices
>when n is even (here I note for experts that I'm talking about Clifford
>algebras over C), and the isomorphism is explicitly describable, e.g.
>for n = 4 the e^i's are isomorphic to the Dirac gamma matrices. The
>spinors then become isomorphic to the column vectors of length 2^{n/2}.
>(Things work out differently when n is odd, there is an interesting
>even/odd thing going on here.) One can use all this and some more stuff
It's actually an interesting mod 8 thing, basically Bott periodicity.
john baez
Working with Clifford algebras over C one gets Bott periodicity of
period 2: the (n+2)nd complex Clifford algebra is isomorphic to the
nth times the 2x2 complex matrix algebra. This is why complex K-theory
is periodic with period 2. It's also why if one looks at U(infty), the
direct limit of the unitary groups U(n), one has pi_n(U) = pi_{n+2}(U).
Working with Clifford algebras over R one gets Bott periodicity of
period 8: the (n+8)th real Clifford algebra is isomorphic to the
nth times the 16x16 real matrix algebra. So real K-theory is periodic
with period 8, and it turns out that the homotopy groups of O(infty),
the limit of the orthogonal groups, have pi_n(O) = pi_{n+8}(O).
The complex case is also simpler because in the real case there are
nondegenerate quadratic forms of different signature, so that the above
story (where I was implicitly talking about Clifford algebras coming
from a positive definite quadratic form) is only part of the picture.
I.e., there is really one real Clifford algebra C(p,q) for each
signature (p,q) with p+q = n. The Dirac matrices, for example, generate
C(1,3) (or C(3,1) if you prefer - but beware, these are NOT isomorphic).
All of these real Clifford algebras become the same when you complexify
them. Physicists are sometimes pretty free about complexifying their Clifford
algebras, e.g. working with complex linear combinations of Dirac
matrices. But sometimes they are careful and us mathematicians have to
struggle to remember the difference between Weyl, Dirac, and Majorana
spinors (I rarely succeed).
Pertti Lounesto
have only periodicity of 2 (not 8).
John Baez (California) is right: Clifford algebras over R
have a periodicity of 8.
Real Clifford algebras have in fact a periodicity of (8x8)/2 = 32, see
P. Lounesto: Scalar products of spinors and an extension of
Brauer-Wall groups. Found. Phys. 11 (1981), 721-740.
P. Budinich, A. Trautman: The Spinorial Chessboard, Springer, 1988.
John Baez said that look for H.B. Lawson, M.-L. Michelsohn: Spin
Geometry, Univ. Ceara, Brazil, 1983, Princeton, 1989, if you want
to know more of spinor structures on manifolds. Here is a problem.
Lawson&Michelsohn claim on page 97 of the 1983 edition (Chapter II
Spin Geometry and the Dirac Operators, Section 2 Spin manifolds and
spin cobordism) that the n-dimensional projective space P^n(R)
has a spinor structure only if n = 3 (mod 4). On the other hand,
A. Crumeyrolle: Orthogonal and Symplectic Clifford Algebras, Spinor
Structures, Kluwer, 1990, claims on pages 202, 203 that P^n(R)
has a spinor structure when n = 4 (mod 4) or n = 3 (mod 4).
They seem to use both the same defintion as outlined by John Baez.
So one of them must be wrong. The question is: which one?
John Baez further said that he cannot always remember what are
Majorana, Weyl and Dirac spinors. These spinors are classified
for the first time by their bilinear covariants (= observables)
in a geometric way in the article P. Lounesto: Clifford algebras
and Hestenes spinors, Found. Phys. 23 (1993), 1203-1237. This
article also introduces a new class of spinors, sitting somewhere
between the Dirac spinors of the electron and the Majorana and Weyl
spinors of the massless neutrino.
Michael Weiss
Briefly, "natural" means compatible with all isomorphisms, in the
sense I described in my post (in an example). "Canonical" means
compatible with all morphisms. This is a more stringent requirement.
Jim enlightened me with the following puzzle: there is only one
CANONICAL way to pick a permutation of a finite set -- pick the identity
permutation. However, there are two NATURAL ways. One is to pick the
identity permutation. What's the other????
Off the cuff guess: pick the identity unless the set has two elements, in
which case exchange them.
This suggests that "canonical" is the "right" notion, and "natural" is just
a curiosity.
Now is there a natural/canonical definition of "right" :=?
Michael Weiss
Me:
But your description of Spin(M), nice as it was,
didn't answer Ron's original question, how to make spinor fields generally
covariant.
jb:
Actually I did... what you describe below is interesting but not really
the point (as I see it), which was: how to write the Dirac equation in a
generally covariant way, i.e., a way that manifestly doesn't depend on
any coordinates. Of course I didn't quite explain this either, but the
best way to do it is not to introduce coordinate charts in the first place.
One mainly needs to make sense of gamma^i partial_i where gamma^i are
"gamma matrices" (things in the Clifford algebra of the tangent space)
and partial_i are an o.n. basis of the tangent space (not necessarily
coming from a coordinate system!!!!)
me:
Or reinterpreted: suppose we have a vector field on a piece of M, i.e., a
section of the tangent bundle over the piece. Now put a chart on the
piece. This induces a section of the general frame bundle over the piece,
which determines a basis in each tangent plane, which enables us to give
coordinates for each vector in the vector field. If we choose a different
chart, then the two frames at each fiber are related by an element of
GL(n), which tells us how to transform coordinates. If each chart is
orthonormal with the right orientation, we can use SO(M) instead of the
general frame bundle, and SO(n) instead of GL(n).
How do you do the analogous thing for Spin(M)? Does a chart on the piece
induce a section of Spin(M)? A spinor field should be (by analogy) a
section of a bundle each of whose fibers is acted on by Spin(n); for 3D,
this means something that looks like C^2. Etc. etc.
jb:
Well, almost; Spin(M) is a two-fold cover of SO(M) so if you have a
section of SO(M) over a chart (or simply connected region) you
automatically get two sections Spin(M) lifting it.
me:
I read the question as, "How do spinor fields transform under
coordinate changes?" Then one could look at both sides of the Dirac
equation, and observe that they transform the same way, and conclude that
the equation was covariant.
Or trying to get away from coordinates: I wouldn't think that a spinor
field would be a section of Spin(M), any more than a vector field is a
section of the frame bundle.
Making sense of gamma^i partial_i is nice, but this addresses just the
Dirac operator, not what it's operating on.
john baez
>In article <2b6er0$e...@galaxy.ucr.edu> ba...@guitar.ucr.edu (john baez) writes:
> Briefly, "natural" means compatible with all isomorphisms, in the
> sense I described in my post (in an example). "Canonical" means
> compatible with all morphisms. This is a more stringent requirement.
> Jim enlightened me with the following puzzle: there is only one
> CANONICAL way to pick a permutation of a finite set -- pick the identity
> permutation. However, there are two NATURAL ways. One is to pick the
> identity permutation. What's the other????
>Off the cuff guess: pick the identity unless the set has two elements, in
>which case exchange them.
Nailed it.
>This suggests that "canonical" is the "right" notion, and "natural" is just
>a curiosity.
Well, first of all, as Jim Dolan noted, in the above I got "natural" and
"canonical" backwards! So let me repeat, for the benefit of all:
In article <COLUMBUS.9...@strident.think.com>
colu...@strident.think.com (Michael Weiss) would have written:
>In article <2b6er0$e...@galaxy.ucr.edu> ba...@guitar.ucr.edu (john baez)
should have written:
> Briefly, "canonical" means compatible with all isomorphisms, in the
> sense I described in my post (in an example). "natural" means
> compatible with all morphisms. This is a more stringent requirement.
> Jim enlightened me with the following puzzle: there is only one
> NATURAL way to pick a permutation of a finite set -- pick the identity
> permutation. However, there are two CANONICAL ways. One is to pick the
> identity permutation. What's the other????
>Off the cuff guess: pick the identity unless the set has two elements, in
>which case exchange them.
Nailed it.
>This suggests that "natural" is the "right" notion, and "canonical" is just
>a curiosity.
Well, I wouldn't go that far. Jim pointed out a nice example: the top
homology class of an oriented n-manifold is canonical but not natural.
They are both reasonably useful notions, but naturality is much better.
Thank god the exterior derivative operator is not merely
"coordinate-independent" (canonical) but natural, for example - this is
what relates DeRham cohomology so nicely to topologically defined
cohomology.
I can now keep the notions of "natural" and "canonical" straight now by
remembering the example above and noting that, while canonical, it is very
unnatural.
Ron Maimon
|>
|> me:
|> I read the question as, "How do spinor fields transform under
|> coordinate changes?" Then one could look at both sides of the Dirac
|> equation, and observe that they transform the same way, and conclude that
|> the equation was covariant.
|>
unfortunately this can only be done for tensor fields, since these are
the only representations of GL(n). You can't really define a "spinor field
relative to this coordinate system" you have to define a spinor field
relative to both a coordinate system and a system of bases for the tangent
space.
what I mean is that there is no analog to the tensor transformation law for
spinors. If you are given four functions psi(i,x) defined on a maniford there
is no way to give them a spinor transformation law and define them with respect
to each coordinate change.
I understand this (I think- I don't know the proof that the spinor transformations
can't be extended to work for general coordinate transformations - John may know
it though)
What I don't understand is how to do a computation with a spinor field on a
computer- it seems an awful waste to pick four arbitrary vectors at each point
in order to describe the spinor. This is way too many numbers! I "should" need
only four numbers at each point, but instead I need 20!
This is highly inefficient. I think there might be a way out of this by choosing
the orthonormal basis as the orthonormalization of the coordinate basis. This
will allow me to get away with only four numbers at each point. Not that I'm doing
any spinor field computations, but I do want to know how to be able
to do them in principle.
|> Or trying to get away from coordinates: I wouldn't think that a spinor
|> field would be a section of Spin(M), any more than a vector field is a
|> section of the frame bundle.
well, the nice thing about a vector field is that you can write it down relative
to the coordinates only, since these determine a basis for the tangent space.
Unfortunately this basis is not necessarily orthonormal, and so doesn't determine
the spinor components in any way that makes sense.
|>
|> Making sense of gamma^i partial_i is nice, but this addresses just the
|> Dirac operator, not what it's operating on.
I think this is the question of how to take a covariant derivative of a spinor
field. This is not so obvious in the formalism as I understand it.
I need to look up those references though, since I don't want to waste John's
time with this explanation.
Ron Maimon
Greg Watson
>...
>There may be NO choice of spin structure or more than one. Existence of
>a spin structure is equivalent to vanishing of the second
>Stiefel-Whitney class of M, which is the kind of topological condition
>that topologists are pretty good at checking. If a spin structure
>exists, spin structures are in 1-1 correspondence with elements of
>H^1(M,Z_2), which is the kind of cohomology group topologists are pretty
>good at computing.
Is there a way of understanding in physical terms what is the meaning
and/or consequences of a manifold (a) not admitting a spin structure;
(b) admitting more than one spin structure? Examples?
Greg Watson
(Still trying to understand characteristic classes)
john baez
Well, in really "physical" terms about all I could say is that if there
is no spin structure we can't formulate the Dirac equation (well, unless
we do extra tricks), while if there are many, there are many ways.
DeWitt-Morette (one of the 3 authors of the marvelous tome, Analysis
Manifolds and Physics) has done some calculations with spinor fields on
manifolds showing how different spin structures can give different
vacuum energy densities for a theory of free fermions. Roughly
speaking, if one chooses a "twisted" spin structure the field might
never be able to settle down as much, so the energy would be bigger.
But in more mathematical terms let me just say this... cover your
Riemannian manifold with charts that look like R^n. For any two charts
U_i and U_j there is a transition function g_{ij} which is defined on
U_i intersect U_j and has values in SO(n), the rotation group. This is
just the matrix-valued function of the form
dx'_{alpha} / dx_{beta}
where x'_{alpha} denotes the coordinates in U_i and x_{beta} denotes the
coordinates in U_j. The transition functions satisfy the usual
consistency condition (fancy term: cocycle condition)
g_{ij} g_{jk} g_{ki} = 1
Now say we want to construct a spin structure. This means we need to
define transition functions h_{ij} with values not in SO(n) but the
2-fold cover Spin(n) satisfying the consistency condition and such that
F(h_{ij}) = g_{ij}
where F: Spin(n) -> SO(n) is the 2-1 homomorphism. The point is this:
assuming the overlaps U_i intersect U_j are contractible, there are two
ways to pick a continuous function h_{ij} with F(h_{ij}) = g_{ij} since
F is 2-1; if h_{ij} works so does -h_{ij}. So we can just randomly pick
any h_{ij} with F(h_{ij}) = g_{ij} and we'll get
h_{ij} h_{jk} h_{ki} = plus or minus 1.
But we may not be able to pick the h_{ij}'s in ANY way that makes
h_{ij} h_{jk} h_{ki} = 1
for all triple intersections! Or there may be several "inequivalent"
ways (I won't define this here). Thus there may be no spin structure or
more than one.
It's very much like how a nonorientable manifold has no globally
consistent way of defining a right-handed triple. In fact all the above
is what mathematicians call Cech cohomology, and the condition for
orientability is that the first Stiefel-Whitney class (which can easily
be described as a Cech cohomology class) vanish, while the condition for
having a spin structure is that the second one vanish.
Matthew MacIntyre
: I can see how you define the spin structure on a manifold, but I don't see any
I don't know the answer to this, but if there is some spinor- knowledgeable
person out there, I have another question for you: Diffeomorphisms of
spacetime affect tangent vectors, covectors etc in the obvious way. I can
see how diffeomorphisms [to be concrete, let's take the parity mapping,ie
a particular orientation-reversing map] can have a well-defined effect on
spinors if the spin structure is unique. But is parity well-defined
[assuming that the space-time has the right kind of orientation-reversing
map] if the spin structure is not unique? How?
Dale Hall
>John Baez said that look for H.B. Lawson, M.-L. Michelsohn: Spin
>Geometry, Univ. Ceara, Brazil, 1983, Princeton, 1989, if you want
>to know more of spinor structures on manifolds. Here is a problem.
>Lawson&Michelsohn claim on page 97 of the 1983 edition (Chapter II
>Spin Geometry and the Dirac Operators, Section 2 Spin manifolds and
>spin cobordism) that the n-dimensional projective space P^n(R)
>has a spinor structure only if n = 3 (mod 4). On the other hand,
>A. Crumeyrolle: Orthogonal and Symplectic Clifford Algebras, Spinor
>Structures, Kluwer, 1990, claims on pages 202, 203 that P^n(R)
>has a spinor structure when n = 4 (mod 4) or n = 3 (mod 4).
>They seem to use both the same defintion as outlined by John Baez.
>So one of them must be wrong. The question is: which one?
>
Lawson and Michelsohn are correct: a Spin structure on the manifold M is a
lifting of the structure group of the tangent bundle TM to Spin(n), the
(non-trivial) double cover of SO(n) (which is the universal cover for n>2).
To do this, one requires (1) a reduction of the structure group from O(n) to
the identity component SO(n) [my people call this an orientation], and (2)
another feature, essentially a compatibility condition, that ensures that the
various locally-defined liftings can be patched together compatibly.
What is required for (1) is that the first Stiefel-Whitney class w_1(TM) = 0.
The condition (2) is that w_2(TM), the second Stiefel-Whitney class, be 0 as
well. Considering n-dimensional real projective space, RP^n, the tangent
bundle of RP^n is known to be (stably) the Whitney sum of (n+1) copies of the
canonical line bundle \lambda over RP^n. Thus the jth Stiefel-Whitney class
w_j(RP^n) = C(n+1,j) w_1^j
where w_1 = the nonzero element of H^1(RP^n;Z_2). Here, C(n+1,j) is the
binomial coefficient. Note that w_1^k is nonzero for k <= n.
To wrap it all up, for the space RP^n to admit a Spin structure, we need
C(n+1,1) w_1^1 = (n+1) w_1 = 0, so n+1 = 0 mod 2, or, n even
C(n+1,2) w_1^2 = ((n+1)n / 2) w_1^2 = 0, so (n+1)*n = 0 mod 4
the second condition gives n=3 or 4 mod 4. However, for n=0 mod 4, we don't
have orientability (i.e., w_1 != 0). Thus, only n=3 mod 4 works.
HOWEVER...
Some people talk of the union of two copies of Spin(n), mapped componentwise
to two copies of SO(n) as a pair of covering maps. Since the union of two
copies of SO(n) might as well be O(n), people who do this will call the result
Pin(n) by analogy with O(n) & SO(n) (Pin and S Pin, get it? Haar, Haar) This
construction would then allow one to produce a Pin(n) structure for RP^n, if
n = 0 mod 4.
>>>>>extended technical discussion for the needy. shoes for the dead. <<<<
There is a fairly elementary description of the process of changing the
structure group of a bundle, together with the implications in cohomology,
if you're willing to believe in the notion of a "classifying space" for fibre
bundles over nice spaces (I think of CW complexes as nice, you may prefer to
lie back and think of England, er, um, manifolds):
First, you should know that, given an effective action (i.e., only the identity
acts trivially) of the group G on the vector space V, the set of isomorphism
classes of principal G-bundles over a suitable space X is in a natural (there's
that word again) 1-1 correspondence with the set of vector bundles over X with
typical fibre V and structure group G. Further, the set of principal G-bundles
over X is in a natural 1-1 correspondence with the set of homotopy classes of
maps from X into BG, the classifying space of G. BG can be constructed as
(for the orthogonal and unitary groups) a limit of Grassmannian manifolds, or
(in general) as an infinite join (G*G*G*G*G*...) modulo the diagonal G-action.
The operation of reducing the structure group from G to H < G looks like this:
BH
|
V
X --> BG
where the intent is to create the diagonal arrow from X to BH. In the case
of the reduction of the structure group from O(n) to SO(n), the above diagram
has extensions up and down, corresponding to the exact sequence
SO(n) --> O(n) --> Z_2.
Upon use of the "B - construction", one obtains:
Z_2 (the group with 2 elements)
|
V
BSO(n)
|
V
X --> BO(n)
|
V
BZ_2 = K(Z_2,1) = RP^\infty
The vertical portion of this diagram has the property that, to lift the
horizontal map one stage, it is neccessary and sufficient to show that
the composition of the horizontal map with the following vertical map is null-
homotopic (i.e., compresses to a point in the space below). Thus, to lift
the map X --> BO(n) to X --> BSO(n) one needs the composition
X --> BO(n) --> K(Z_2,1)
to be null-homotopic. The composition defines an element of H^1(X; Z_2),
which we call the first Stiefel-Whitney class w_1(the bundle). If the bundle
is the tangent bundle TM, it is called w_1(M). The first Stiefel-Whitney
class maps H_1(M;Z) --> Z_2 or \pi_1(M) --> Z_2 (by using the fact that
H_1 is the abelianization of the fundamental group), by assigning a +1 if
the cycle (or loop) preserves orientation, and -1 if the loop reverses orien-
tation. Then w_1 = 0 iff the bundle is orientable.
The next situation regards a similar, but different, situation. Here, we
have the diagram:
BSpin(n)
|
V
X ---> BSO(n)
The extension of this diagram requires knowing how it was arrived at:
Z_2 --> Spin(n) --> SO(n)
is an exact sequence of groups, and as such leads to a fibration
B(Z_2) --> BSpin(n) --> BSO(n).
Extending the sequence of spaces to the right as a fibre sequence:
BSpin(n) --> BSO(n) --> B(B(Z_2))
which makes sense only when you know that ^^^this exists, and is the
space K(Z_2,2), and one obtains:
BSpin(n)
|
V
X ---> BSO(n)
|
V
K(Z_2,2)
In order to lift the horizontal map now, one needs the composition
X --> BSO(n) --> K(Z_2,2)
to be null-homotopic, or, identifying maps into K(Z_2,2) with elements
of H^2(*,Z_2), the element w_2(bundle) = 0, where w_2 is the second Stiefel-
Whitney class.
The business of Pin(n) works similarly, but you have only the second diagram
above in modified form. Here, we have the exact sequence:
Z_2 ---> Pin(n) ---> O(n)
leading to a fibre sequence
K(Z_2,1) --> BZ_2 --> BPin(n) --> BO(n) --> B(B(Z_2)) = K(Z_2,2)
and the diagram:
BPin(n)
|
V
X ---> BO(n)
|
V
K(Z_2,2)
The mapping X --> K(Z_2,2) is, just as before, the second Stiefel-Whitney
class, without having to worry about w_1 vanishing.
Lee Rudolph
>I can now keep the notions of "natural" and "canonical" straight now by
>remembering the example above and noting that, while canonical, it is very
>unnatural.
Apparently Mackey was the referee of Eilenberg and MacLane's first paper
on "Natural Transformations of Functors", and tried to dissuade them
from the use of the word "natural", without success.
Lee Rudolph
Allen Knutson
}}Is there a way of understanding in physical terms what is the meaning
}}and/or consequences of a manifold (a) not admitting a spin structure;
}}(b) admitting more than one spin structure? Examples?
...
}vacuum energy densities for a theory of free fermions. Roughly
}speaking, if one chooses a "twisted" spin structure the field might
}never be able to settle down as much, so the energy would be bigger.
Huh? How can we get a preferred "untwisted" spin structure? Certainly I
don't prefer one orientation to the other, when orienting a manifold.
Aren't all spin structures created equal?
To add something new and maybe useful to people new to this; why do we
try and put on orientations and spin structures but not anything else?
Given our naked manifold, the structure group of the tangent bundle is GL(n).
We can always put on a Riemannian metric, reducing it to O(n). (One
high-level reason this is true is that GL(n)/O(n) is contractible, and a
high-level reason that's true is that O(n) is the maximal compact subgroup.
But I digress.) First, let's ask for our group to be (0-)connected,
which takes us to SO(n). Now let's ask for our group to be 1-connected,
i.e. pi_1=0, too. So we lift to Spin(n). Now let's ask for it to be
2-connected. Hurray! All compact Lie groups have pi_2=0 already.
Now let's ask for it to be 3-connected. Bzzt: the only connected simply
connected compact Lie group with pi_3=0 is the 1-element group. So this
amounts to picking a trivialization of the bundle. And that's the last word.
Allen K.
john baez
>To wrap it all up, for the space RP^n to admit a Spin structure, we need
>
> C(n+1,1) w_1^1 = (n+1) w_1 = 0, so n+1 = 0 mod 2, or, n even
You meant n odd, of course. (And used that below.)
Your description of characteristic classes in terms of classifying
spaces was great. I need to be repeatedly exposed to this stuff to
reduce my fear of it. I have a feeling that to describe why w_1 and w_2
are obstructions to orientability and existence of spin structures,
physicists might find Cech cohomology simpler, particularly if one
doesn't call it "Cech cohomology". Transition functions are, I think,
something lots of them have a handle on. I tried to sketch the
rudiments of that approach a while back...
john baez
>ba...@guitar.ucr.edu (john baez) writes:
>}}Is there a way of understanding in physical terms what is the meaning
>}}and/or consequences of a manifold (a) not admitting a spin structure;
>}}(b) admitting more than one spin structure? Examples?
>}vacuum energy densities for a theory of free fermions. Roughly
>}speaking, if one chooses a "twisted" spin structure the field might
>}never be able to settle down as much, so the energy would be bigger.
>
>Huh? How can we get a preferred "untwisted" spin structure? Certainly I
>don't prefer one orientation to the other, when orienting a manifold.
>Aren't all spin structures created equal?
I didn't mean to imply the existence of a preferred "untwisted" one, but
I think that the eigenvalues of the Dirac operator may depend on which
spin structure you pick. The eigenvalue nearest zero is some sort of
measure of how "twisted" the least twisted section of the spinor bundle
is. There are after all typically lots of spin structures, in 1-1
correspondence with H^1(M,Z_2). By the way if this correspondence were
NATURAL there would be a "best" spin structure, corresponding to zero.
I'm in a bit of a rush, so I'll just say that a peek at p. 81 of
Michelson and Lawson makes it clear that this'd only be true if the
sequence 1.4 split naturally when it was short exact, and it doesn't
look like it.
However this sequence DOES have a "best" splitting when the tangent
bundle of our manifold M is trivial, say. Consider e.g. the torus T^n
This has trivial tangent bundle and a bunch of spin structures, each of
which gives rise to a Spin(n)-bundle over the torus, only one of which
is trivial - the "best" one corresponding to the zero element of
H^1(T^n,Z_2). I have a feeling that only for this "best" spin structure
does the Dirac operator have a zero eigenvalue - corresponding the
constant section of the spinor bundle.
>To add something new and maybe useful to people new to this....
Nice.
Pertti Lounesto
> Physicists are sometimes pretty free about complexifying their Clifford
> algebras, e.g. working with complex linear combinations of Dirac
> matrices. But sometimes they are careful and us mathematicians have to
> struggle to remember the difference between Weyl, Dirac, and Majorana
> spinors (I rarely succeed).
Indeed, the complixification of Clifford algebras and working with complex
linear combinations of Dirac's gamma matrices has been a source of long
confusion and many misunderstandings -- and has resulted in inadequate
understanding of Weyl, Majorana and Dirac spinors -- even by physicists.
Let me try to clear up this confusion / lack of understanding. In fact,
I have found new spinors residing in between the Dirac spinors of the
electron and the Majorana and Weyl spinors of the neutrino. My study is
completely algebraic, but I have revealed the geometric properties of
the new spinors which I will call here the flag-dipole spinors. It is
now the task of experimental physicists to make tests -- after some
physical interpretation of my new flag-dipole spinors have been set.
The main idea is to classify spinors by their bilinear covariants
(= observables). Below I will give a "short" introduction to my
invention, but somebody who wants to know more is directed to my
two articles:
P. Lounesto: Clifford algebras and Hestenes spinors, Found. Phys.
23 (1993), 1203-1237.
P. Lounesto: Clifford algebras, relativity and quantum mechanics.
In Proceedings of SILARG-8, July 25-30, Brazil (eds. P. Letelier,
W.A. Rodrigues), World Scientific (in press).
Let \psi be a solution of the Dirac equation. Then we have the
bilinear covariants (=observables)
\sigma = \psi\dagger\gamma_0 \psi
J_\mu = \psi\dagger\gamma_0 \gamma_\mu \psi
S_{\mu\nu} = \psi\dagger\gamma_0 i\gamma_{\mu\nu} \psi
K_\mu = \psi\dagger\gamma_0 i\gamma_{0123}\gamma_\mu \psi
\omega = -\psi\dagger\gamma_0 \gamma_{0123} \psi
where \gamma_{0123} = \gamma_0\gamma_1\gamma_2\gamma_3.
The vector J = J_\mu\gamma^\mu is the Dirac current,
and v = J/\rho is the velocity for non-zero \rho
= sqrt(\sigma^2+\omega^2).
The bivector S = 1/2 S_{\mu\nu}\gamma^{\mu\nu} is called
in the literature the electromagnetic moment bivector.
The vector s = K/\rho is called the spin vector, when
K = K_\mu\gamma^\mu and \rho is non-zero.
These bilinear covariants satisfy certain quadratic equations
called the Fierz identities (discovered by Pauli and Kofink)
J^2 = \sigma^2+\omega^2 K^2 = -J^2
J.K = 0 J\wedge K = -(\omega+\gamma_{0123}\sigma)S
which enable us to recover the original spinor \psi up to a
phase factor, in case of a non-zero \rho.
The vector J cannot be zero, because J^0 > 0.
For non-zero \rho we must have also non-zero K, since
K^2 = -\rho^2. Similarly, computing the square of the bivector
S, we find S^2 = (\omega - \gamma_{0123}\sigma)^2, which
implies that S cannot be zero for a non-zero \rho.
Therefore, in case of a non-zero \rho we are left with three
alternatives:
1) Both \sigma and \omega are non-zero
2) \sigma is non-zero but \omega vanishes
3) \sigma vanishes but \omega is non-zero
All these three alternatives describe the Dirac spinors of
the electron -- and their properties are fairly well known.
We can form a multivector, more precisely, a sum of a scalar,
a vector, a bivector, a three-vector and a volume element
Z = \sigma + J +iS + iK\gamma_{0123} + \omega\gamma_{0123}
(This multivector is nilpotent in the class 3 above.)
This multivector was the starting point for the studies of
J. Crawford: On the algebra of Dirac bispinor densities:
Factorization and inversion theorems, J. Math. Phys. 26
(1985), 1439-1441, who found the decomposition
Z = (\sigma+J+\omega\gamma_{0123})
(1-i(\sigma+\omega\gamma_{0123})^{-1}K\gamma_{0123})
With this factorization Crawford was able to reconstruct the
original Dirac spinor of the electron from its bilinear
covariants. However, Crawford's second factor did not
commute with the first factor unless \omega = 0, and finding
such commuting factors was an open question from 1985 till this
year 1993. Such commuting factors are (for a non-zero \rho)
P = \sigma + J + \omega\gamma_{0123}
\Sigma = 1 - i\gamma_{0123}JK^{-1}
so that Z = P \Sigma = \Sigma P. Here P is a multiple of the
energy projection operator, and \Sigma is twice the spin
projection operator (at least in the class 2).
However, everything goes quite differently in case \rho = 0.
This concerns the case of Weyl spinors, where S = 0, and
Majorana spinors for which K = 0. In this null case there also
exist new kind of spinors which not have not been discussed
before, namely the case when both S and K are non-zero.
Weyl spinors are eigenspinors of the helicity projection
operators
\psi+ = 1/2(1-i\gamma_{0123})\psi+
\psi- = 1/2(1+i\gamma_{0123})\psi-
and Majorana spinors are eigenspinors of the charge conjugation
operator
\psi+ = -i\gamma_2 (\psi+)*
\psi- = +i\gamma_2 (\psi-)*
where (\psi)* is the complex conjugate of \psi (be careful
-- complex conjugate in the matrix representation -- that is,
all entries of the column spinor are complex conjugated).
The above definitions of the Weyl and Majorana spinors are
the usual ones which you can find for instance in the classic
Bjorken-Drell. The Majorana and Weyl spinors cannot be
studied by Crawford's reconstruction theorem (of spinors from
their bilinear covariants), because for them \rho = 0.
However, without going to detail, in order to keep this poster
short, let me just give the end-result to the problem of finding
also Majorana and Weyl spinors from their bilinear covariants:
In these cases we have Z = J + iS + iK\gamma_{0123}. The
multivector Z must be nilpotent, which implies
Z = J(1 + is + ih\gamma_{0123})
where s is a space-like vector, s <= 0, orthogonal to J,
J.s = 0, and (s + h\gamma_{0123})^2 = -1. This implies that
Z cannot reduce to J, whereby either K or S must be
non-zero. Thereby, we can have the following three more
alternatives for spinors (in addition to classes 1,2,3 above)
4) Both S and K are non-zero. These spinors have a flag
S and a dipole consisting of J and K, and I will call
them flag-dipole spinors.
5) S is non-zero but K vanishes. These spinors have a flag
S and a pole J. I will call them flag-pole spinors.
These spinors contain as a special case the Majorana spinors.
A flag-pole spinor is a Majorana spinor multiplied by a
unit complex number. These flag-poles are stable under the
U(1)-gauge and are not the Penrose flags, which turn 2\pi
under a half-turn.
6) S vanishes but K is non-zero. These spinors have a
dipole of J and K. They correspond to the Weyl spinors.
If J = K, then we have positive helicity, and if J = -K,
then we have negative helicity.
The number of parameters in the sets of bilinear covariants
(or spinors without the U(1)-gauge) is
class number 1 2 3 4 5 6
parameters 7 6 6 5 4 3
If the U(1)-gauge is taken into consideration, then the number
of parameters is raised by one unit in all classes except the
class number 5 of Majorana spinors (Weyl spinors with U(1)-
gauge and Majorana spinors both have 4 parameters and can be
mapped onto each other -- which enables attaching Penrose flags
also to the Weyl spinors -- read my article to see how Penrose
flags are attached to Majorana spinors).
Clearly, there are no more than the above 6 classes of possible
spinors. The class 4 of the flag-dipole spinors does not
represent the Dirac spinors of the electron, because in this
class both \sigma = 0 and \omega = 0. The new class 4 does
not represent the Weyl or Majorana spinors either. However,
a purely algebraic study shows that these new spinors truely
exist, and have slipped the theorists up till today.
Pertti Lounesto
projective spaces that "Lawson and Michelsohn are correct" about the
controversy between Lawson&Michelsohn and Crumeyrolle.
Here is another difference in opinion -- the same question as before:
who is right?
Lawson&Michelsohn write in their book on Spin Geometry (1989 p. 56
and 1983 p. 78) that Spin+(3,3) is isomorphic to the two-fold
cover of SL(4,R). On the other hand Ian Porteous claims in his
article entitled "Clifford algebra tables" in the Proceedings of
the 3rd Workshop in Clifford Algebras (edited by R. Delanghe and
H. Serras), Kluwer, 1993, that Spin+(3,3) is isomorphic to SL(4,R).
They cannot evidently be both right. So who is wrong?
Torsten Ekedahl
> I'm posting this for Jim Dolan, who can't, due to technical
> difficulties:
> .
> .
> again, briefly: vague ideas about "naturalness" and "canonicalness" go
> back a long time. then eilenberg and maclane invented the formalism
> of category theory around the 1940's, and the centerpiece of this
> formalism was a precise formal definition of the concept of "natural
> map". the formalism eventually caught on, and so it is generally a
> good idea nowadays to use the terminology "natural" in accord with the
> eilenberg-maclane definition. the terminology "canonical", however,
> is still somewhat up for grabs; sometimes you hear people who have
> practically never even heard of category theory (or who wish that they
> hadn't) using it in a very informal way. listening to this informal
> usage of the term "canonical", and acting as a descriptive (that is,
> non-prescriptive) linguist, i have proposed that what people usually
> really tend to mean by "canonical" is "natural with respect to the
> (implicitly relevant class of) _isomorphisms_".
> .
> .
> >Jim enlightened me with the following puzzle: there is only one
> >CANONICAL way to pick a permutation of a finite set -- pick the identity
> >permutation. However, there are two NATURAL ways. One is to pick the
> >identity permutation. What's the other????
> .
> .
> -james dolan
One informal but still (!) very interesting use of canonical is in the
sense that somehow for some reason there is one abvious, natural etc
choice (which almost always is a natural transformation in the
technical sense). One example is the map from a vector space to its
double dual given by evaluation. It certainly is a natural
transformation but more than that being the obvious map. What makes
the thing really interesting is that canonical maps are expected to
have certain properties. First they are normally stable under
compositions and inverses (if they exist). This is maybe not so
interesting but there are two other properties which when combined
with the first give some interesting things.
The first of those is that whereas it in general is a very bad
thing to identify isomorphic objects as long as they are canonically
isomorphic you will never run into trouble. This has led to the use
(introduced by Deligne I think) to use the equality sign for canonical
ismorphisms. The necessary adjunct to this first property is the
second:
The only canonical isomorphism from an object to itself is the
identity or possibly minus the identity. This last property can
sometimes be hard to verify in concrete situations but usually much
harder is to determine the sign.
Note that this last property is certainly not a provable result
but rather a meta principle which has to be verified in each
individual case.
Let me finish by an example. There is for a smooth and proper
n-dimensional algebraic variety X over a field k a canonical trace map
H^n(X,o_X) -> k, where o_X is the sheaf of n-forms. This is the
algebraic analogue of integration of C^oo 2n-forms when k is the
complex numbers. Now, when X=P^n one may compute H^n(X,o_X) using Cech
john baez
>Here is another difference in opinion -- the same question as before:
>who is right?
>Lawson&Michelsohn write in their book on Spin Geometry (1989 p. 56
>and 1983 p. 78) that Spin+(3,3) is isomorphic to the two-fold
>cover of SL(4,R). On the other hand Ian Porteous claims in his
>article entitled "Clifford algebra tables" in the Proceedings of
>the 3rd Workshop in Clifford Algebras (edited by R. Delanghe and
>H. Serras), Kluwer, 1993, that Spin+(3,3) is isomorphic to SL(4,R).
>They cannot evidently be both right. So who is wrong?
Well, spin groups are supposed to be simply connected by definition,
right? And SL(4,R) is not; it has pi_1 = Z/2. So it sounds like Lawson
and Michelson have a better chance on this one.
Torsten Ekedahl
choice which almost always is a natural transformation in the
technical sense, One example is the map from a vector space to its