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John Baez science forum Guru Wannabe
Joined: 01 May 2005
Posts: 220

Posted: Tue Jun 27, 2006 5:23 am Post subject:
Classifying nondegenerate bilinear forms



Suppose K^n is an ndimensional vector space over some nice
field K, like R or C. Do people know how to classify
nondegenerate bilinear forms on V, where we regard two as
the same if we can get from one to the other by the action
of GL(n,K)?
I know how it works for R or C when the bilinear form is
symmetric or antisymmetric, but tonight I'm curious about
the general case, where it could be neither symmetric nor
antisymmetric.
We can always take our bilinear form and write it as a sum
of a symmetric and antisymmetric part  does that help?
It's possible for both these parts to be degenerate, even
if the whole form is nondegenerate. 

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William C Waterhouse science forum beginner
Joined: 04 May 2005
Posts: 17

Posted: Tue Jun 27, 2006 9:45 pm Post subject:
Re: Classifying nondegenerate bilinear forms



In article <e7qfdd$62s$1@glue.ucr.edu>,
baez@math.removethis.ucr.andthis.edu (John Baez) writes:
Quote:  Suppose K^n is an ndimensional vector space over some nice
field K, like R or C. Do people know how to classify
nondegenerate bilinear forms on V, where we regard two as
the same if we can get from one to the other by the action
of GL(n,K)?
I know how it works for R or C when the bilinear form is
symmetric or antisymmetric, but tonight I'm curious about
the general case, where it could be neither symmetric nor
antisymmetric.
We can always take our bilinear form and write it as a sum
of a symmetric and antisymmetric part  does that help?
It's possible for both these parts to be degenerate, even
if the whole form is nondegenerate.

Yes, it's worked out (though somewhat messy in characteristic 2).
The basic reference is
Carl Riehm, The equivalence of bilinear forms.
J. Algebra 31 (1974), 4566.
The split into symmetric and antisymmetric is reasonable, but it
does encounter the problem mentioned. The solution is to look
at the "antisymmetry": that is, if B is the matrix (and nonsingular),
then B^t = BS for some S. Changing B to some H^t B H will change S to
a similar matrix. (Abstractly, the form b is an isomorphism
from V to V*; its dual b^t then maps V (== V**) to V* also, and thus
we get a unique s with b^t = bs).
So the first invariant is the rational canonical form of S. Over
the complex numbers, that's it. In any case, you can split V into
biorthogonal subspaces where S involves only one (or two related)
irreducible polynomials.
What happens in general there is that you get invariants that are
nondegenerate symmetric or skew or hermitian forms over finite
extensions of K. (In characteristic 2 you also need some quadratic forms).
Finally, a singular B splits into a nonsingular part and one or more
"basic" singular parts; the parts are unique up to isomorphism. See
Peter Gabriel, Appendix: degenerate bilinear forms.
J. Algebra 31 (1974), 6772.
William C. Waterhouse
Penn State 

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Ilya Zakharevich science forum beginner
Joined: 20 May 2005
Posts: 27

Posted: Fri Jun 30, 2006 10:01 am Post subject:
Re: Classifying nondegenerate bilinear forms



[A complimentary Cc of this posting was sent to
John Baez
<baez@math.removethis.ucr.andthis.edu>], who wrote in article <e7qfdd$62s$1@glue.ucr.edu>:
Quote:  We can always take our bilinear form and write it as a sum
of a symmetric and antisymmetric part  does that help?
It's possible for both these parts to be degenerate, even
if the whole form is nondegenerate.

The classification follows from those in
@article {Thom91Pen,
old = MR91k:15031,
AUTHOR = {Thompson, Robert C.},
TITLE = {Pencils of complex and real symmetric and skew matrices},
JOURNAL = {Linear Algebra Appl.},
FJOURNAL = {Linear Algebra and its Applications},
VOLUME = {147},
YEAR = {1991},
PAGES = {323371},
ISSN = {00243795},
CODEN = {LAAPAW},
}
Below is my short summary (of the complex case, I have one for real
one somewhere too).
Hope this helps,
Ilya
====================================================================
Splittable matrix: block matrix of the form
0 A and 0 A
A^t 0 A^t 0
are symmetric and skewsymmetric (correspondingly) for any matrix A.
If A is m x n, then the total matrix is (n+m) x (n+m). For any pair
of m x n matrices A, B one can get a pair of splittable (skew)symmetric
matrices; call such pairs spliced. There are 3 flavors for such a pair:
[symm,symm], [symm,skew], [skew,skew] pairs; each flavor gets its own
spliced pair.
If the pair (A, B) is decomposable, the corresponding spliced pair
is decomposable. Thus if we want to find splittable examples of
indecomposable pairs, one must splice indecomposable pair of matrices.
Note that indecomposable pairs (A,B) are isomorphic (Kronecker
classification) to one of
Jordan lambdapair ( J(lambda,n), Id(n) ) n x n
Jordan infinitypair ( Id(n), J(0,n) ) n x n
decreasing Kronecker pair ( k(n), K(n) ) (n1) x n
inreasing Kronecker pair ( k^t(n), K^t(n) ) (n+1) x n.
Here k(n) and K(n) are two "almostdiagonal" (n1) x n matrices with
matrix elements delta(i,j) and delta(i+1,j) correspondingly. For n = 4:
1 0 0 0 0 1 0 0
0 1 0 0 and 0 0 1 0
0 0 1 0 0 0 0 1.
Since pairs (A, B) and (A^t, B^t) lead to isomorphic spliced pairs, it is
enough to consider only decreasing Kronecker pairs. It turns out that
spliced pair for a Kronecker pair is indecomposable in all 3 flavors of
pairs. Call the resulting splittable pair a spliced Kronecker pair.
Blocks for complex pencils: either spliced Kronecker, or spliced Jordan,
or pure Jordan; can never be isomorphic, so it is enough to describe which
spliced Jordan blocks are indecomposable, and provide examples of pure
Jordan blocks.
[skew, skew]: always evendimensional, always splittable. Jordan
blocks and the increasing Kronecker blocks lead to nonisomorphic splittable
pairs. Thus any block is spliced, and all spliced blocks are different.
[symm, symm]: spliced Jordan pair is decomposable. Example of pure Jordan
pair: reflect matrices for "usual" Jordan pair in vertical axis.
[symm, skew]: spliced Jordan pair is indecomposable if lambda is not0 and
notinfinity; or if lambda=0 and n is odd; or if lambda=infinity and n
is even. Pure Jordan pairs exist for:
lambda=infinity and odd n: reflect "usual" Jordan pair in vertical axis,
and make subdiagonal elements skewsymmetric
(1s in the first half, 1s in the other).
lambda=0 and even n: reflect "usual" Jordan pair in vertical axis,
and make diagonal elements skewsymmetric
(1s in the first half, 1s in the other). 

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