Q: Geometric Algebra instead of Vector Algebra??
John Baez
Matt McIrvin wrote:
>Anyway, whatever happened to differential forms? In this discussion they
>have been conspicuous by their absence (though the p-forms that John just
>mentioned might be same thing-- I'm not entirely up on the terminology).
>Some of the claimed advantages of using Clifford algebras also apply in
>exactly the same way to differential forms.
Differential forms, p-forms, totally antisymmetric tensors, the Grassmann
algebra, the exterior algebra --- these are all different names for more
or less the same thing. The proliferation of synonymous monickers
is a token of their ubiquity. I am always suspicious that some of fans
of "geometric algebra" (i.e., Clifford algebras) don't know enough about
differential forms.
Let me be more precise. If you give me any vector space V, I can
define a new vector space, usually denoted Lambda V, called the
"exterior algebra" or "Grassmann algebra" over V. It's defined as
roughly as follows. It consists of linear combinations of formal
products of elements of V, like
v_1 ^ v_2 ^ ... ^ v_p
The product is usually denoted with a "wedge" and called the "wedge
product" or "exterior product" or "Grassman product". (Of course,
when you write Lambda in Greek the way you should, it looks like a
big wedge, which is why they call the exterior algebra Lambda V.) The
wedge product is required to satisfy some rules: it's associative, it
distributes over addition, gets along nicely with multiplication by
scalars, and most importantly it's ANTICOMMUTATIVE:
v ^ w = - w ^ v.
It's easy to check that if V has dimension n, the exterior algebra
over V has dimension 2^n. (Note: above we include the case p = 0,
that is, the wedge product of zero vectors, which we write as 1.)
Now if V is the cotangent space at some point of a manifold,
we call elements of the exterior algebra over V "differential forms",
and those that are linear combinations of products of p guys are called
"p-forms". So you've got 1-forms, 2-forms, on up to n-forms if your
manifold is n-dimensional.
>For instance, there has been a lot of talk about the independence of the
>antisymmetric "wedge product" from the special properties of
>three-dimensional space; but the exterior product (also called a wedge
>product) of two differential forms is *also*
>an antisymmetric generalization of the cross product that works in
>arbitrary dimensions. There are also pseudoscalar-like objects (3-forms)
>and pseudovector-like objects (2-forms) in three-dimensional space.
(Hey, so you know about 2-forms and 3-forms, just not p-forms! Okay.
Sorry for the jargonizing.)
As Matt notes, differential forms and their wedge product are the most
common way folks generalize the cross product to arbitrary dimensions.
How does the exterior algebra compare to the Clifford algebra? Well,
the big difference is that the Clifford algebra depends on a *metric*.
The Clifford algebra CV is just like the exterior algebra except that the
product of v and w is usually just written vw, and more importantly,
the anticommutativity law is replaced by
vw + wv = <v,w>
where <v,w> is the inner product of v and w, which depends on a metric.
The Clifford algebra is also 2^n-dimensional when the vector space
you started with was n-dimensional. There is in fact a natural
isomorphism (as vector spaces, not as algebras) between the Clifford
algebra and the exterior algebra. So it's not surprising that fans
of one, and fans of the other, can argue endlessly. However, they
are not isomorphic as algebras, so they really perform quite different
jobs. In particular, as I noted before, geometric algebra fans seem
to use "p-blades" --- that is, p-fold products
v_1 ... v_p
in the Clifford algebra --- to represent little infinitesimal
p-dimensional "volume elements". I would prefer to use the p-form
v_1 ^ ... ^ v_p
It may seem there's no big difference, but they are really quite
different in some ways, because the exterior algebra
algebra has
v ^ w + w ^ v = 0
while the Clifford algebra has
vw + wv = <v,w>
When dealing with oriented area elements it is very natural to have
v ^ w = - w ^ v.
Note that the exterior algebra is just a special case of a Clifford
algebra where the metric is set to zero. Modern geometers use
both extensively, and do lots of cool stuff with the relations between
the two. It's not a case of "either you like exterior algebras,
or Clifford algebras".
As general relativists and some other field theorists know, differential
forms are a very natural, coordinate-independent way of talking about
field theories.
--
Matt McIrvin <http://world.std.com/~mmcirvin/>
John Baez
da...@cogentex.com (Daryl McCullough) writes:
>I guess since the Clifford algebra already has
>a metric built-in (for two basis vectors e_i and e_j, we have
>e_i e_j - e_j e_i = 2 g_ij) there is no particular need for covectors and
>mixed tensors?
Sort of. The way I'd put it is that the Clifford algebra CV can only be
defined when one starts with a vector space V equipped with a
metric. If this metric is nondegenerate, it gives an isomorphism
between V and its dual, so it doesn't make sense to worry terribly much
about the difference between vectors and covectors.
On the other hand, there are all sorts of tensors that don't live
in the Clifford algebra in any natural way, even if one charitably uses
the isomorphism between V and its dual provided by a nondegenerate metric.
For example, symmetric rank-2 tensors, like the metric itself! So
one cannot throw out tensors and work only with the Clifford algebra,
even in the presence of a nondegenerate metric.
>John Baez writes:
>>In particular, as I noted before, geometric algebra fans seem
>>to use "p-blades" --- that is, p-fold products v_1 ... v_p
>>in the Clifford algebra --- to represent little infinitesimal
>>p-dimensional "volume elements". I would prefer to use the p-form
>>v_1 ^ ... ^ v_p
>Are you sure?
That was my impression based on a recent cursory peek at Hestenes' book.
I may be wrong; I hope so.
>I would have guessed that in geometric algebras, the
>volume would not be p-blade, but a totally antisymmetric linear combination
>of p-plades. So, in three dimensions, the volume associated with vectors
>u,v,w would be 1/4 (uvw - vuw - wvu + wuv)
That'd be better. Here one is secretly using the vector space
isomorphism between the Clifford algebra and the exterior algebra
to transport the wedge product in the exterior algebra over to the
Clifford algebra.
>>Note that the exterior algebra is just a special case of a Clifford
>>algebra where the metric is set to zero.
>That's kind of interesting. Clifford algebra is the generalization of
>exterior algeba to allow nonzero anticommutators, just as (could there
>be some connection here?) quantum mechanics can be thought of as the
>generalization of classical mechanics to allow nonzero commutators.
Yes, there is a very big and interesting connection here. The
Clifford algebra is a "deformation quantization" of the exterior algebra,
just as the Weyl algebra (the usual noncommutative algebra of p's and q's)
is a quantization of the symmetric algebra (also known as the algebra
of polynomials in the p's and q's).
I went on a long rant about this in sci.physics last spring.
The basic set of analogies to understand is:
bosons fermions
commutators anticommutators
symplectic structure omega(.,.) metric g(.,.)
symmetric algebra SV exterior algebra Lambda V
Weyl algebra WV Clifford algebra CV
symplectic group Sp(n) orthogonal group O(n)
metaplectic group Mp(n) spin group Spin(n)
In one world you start with a vector space V with a nondegenerate
metric g satisfying
g(v,w) = g(w,v)
and cook up the Clifford algebra with
vw + wv = hbar g(v,w)
which when Planck's constant hbar goes to zero, becomes the
exterior algebra. Then you do all sorts of fermionic stuff.
In the other world you start with a vector space V with a
symplectic structure, a nondegenerate bilinear omega satisfying
omega(v,w) = -omega(w,v)
and cook of the Weyl algebra with
vw - wv = hbar omega(v,w)
which when hbar goes to zero becomes the symmetric algebra.
Then you do all sorts of bosonic stuff.
I went on and on about all the wonderful things that happen...
and then I posed the puzzle: what sort of vector space has
both a nondegenerate metric AND a symplectic structure? Such
a thing would let you study both bosons and fermions, in a
unified way. It would obviously be very important.
If you give up, read "Introduction to Algebraic and Constructive
Quantum Field Theory" by Baez, Segal, and Zhou, which goes into
more detail on this stuff than you'd ever want.
For other stuff on Clifford algebras as quantized exterior
algebras, try Berline, Getlzer and Vergne's book "Heat kernels
and Dirac operators". This stuff turns out to be crucial in
modern differential topology.
Alan Katz
Matt McIrvin wrote:
>Anyway, whatever happened to differential forms? In this discussion they
>have been conspicuous by their absence (though the p-forms that John just
>mentioned might be same thing-- I'm not entirely up on the terminology).
>Some of the claimed advantages of using Clifford algebras also apply in
>exactly the same way to differential forms.
and John Baez said:
> Differential forms, p-forms, totally antisymmetric tensors, the Grassmann
> algebra, the exterior algebra --- these are all different names for more
> or less the same thing. The proliferation of synonymous monickers
> is a token of their ubiquity. I am always suspicious that some of fans
> of "geometric algebra" (i.e., Clifford algebras) don't know enough about
> differential forms.
I've always wondered if there was a simple way to see WHY p-forms are
defined to be totally antisymmetric (other then, it works). From what
I've read, the geometric algebra approach is to define a (general)
vector product which is associative, and break it up into a symmetric
and antisymmetric part. The wedge-product is the antisymmetric one,
which I can sort-of visualize.
The differential forms approach I have trouble visualizing. What is
it about geometry (or differential geometry) that requires these
antisymmetric entities? (orientablility relative to two vectors in
dimensions > 2??)
(I had the same trouble, as I think many people did, of understanding
exactly WHAT the vector cross product is (as opposed to how to use it)).
Alan (Ka...@Halcyon.com)
Matt McIrvin
In article <53jq4a$7...@agate.berkeley.edu>, ka...@coho.halcyon.com (Alan
Katz) wrote:
> I've always wondered if there was a simple way to see WHY p-forms are
> defined to be totally antisymmetric (other then, it works).
Here's one way that I suppose might help. One way of thinking of a 2-form
that is the wedge product of two 1-forms is to imagine the 1-forms locally
as arrows in space, and the 2-form as defining the plane and area of the
parallelogram spanned by them. If the arrows are pointing in the same
direction, the parallelogram shrinks to nothing; if you shift one of the
arrowheads in a direction parallel to the other arrow, it doesn't change
the area of the parallelogram. So the symmetric part of the tensor product
doesn't contribute to the 2-form.
In general, the wedge product of p 1-forms defines the p-volume and
orientation of a parallelepiped in p dimensions spanned by p little arrows.
If you shift one of the vectors in the direction of any of the other
vectors, it just changes the shape of the box, not its volume, so if this
is a linear product, it can't depend on the bits of the tensor product that
involve components of vectors pointing in the same direction. If any two
of the sides point in the same direction, the whole thing collapses to
zero volume, like a sat-on cardboard box.
> (I had the same trouble, as I think many people did, of understanding
> exactly WHAT the vector cross product is (as opposed to how to use it)).
The reason you had the same trouble is that the cross product is just a
modified special case of the same thing.
james dolan
allen katz writes:
>I've always wondered if there was a simple way to see WHY p-forms are
>defined to be totally antisymmetric (other then, it works). From what
>I've read, the geometric algebra approach is to define a (general)
>vector product which is associative, and break it up into a symmetric
>and antisymmetric part. The wedge-product is the antisymmetric one,
>which I can sort-of visualize.
>
>The differential forms approach I have trouble visualizing. What is
>it about geometry (or differential geometry) that requires these
>antisymmetric entities? (orientablility relative to two vectors in
>dimensions > 2??)
one way of thinking about it is that a p-linear form f is
anti-symmetric if and only if f(x1,...,xp) = f(y1,...,yp) whenever the
matrix whose row vectors are y1,...,yp is obtained from the matrix
whose row vectors are x1,...,xp by multiplying by a p-by-p matrix of
determinant one.
this is a pretty easy calculation to check- think for example about the
case p=2 and the determinant one matrix:
0 -1
1 0 .
then think about the geometric significance of determinants, and you
might see why p-forms on a manifold m are just what you want when you
want a kind of geometric object that can be integrated in an invariant
way over arbitrary p-dimensional surfaces in m.
(some people prefer to take anti-symmetric p-linear forms as the
fundamental concept, and then derive from that the importance of
determinants. i much prefer the opposite approach.)