Subject: Re: Cohomology question
Date: Mon, 3 Sep 2001 05:05:13 +0000
From: Tom Goodwillie <tomg@math.brown.edu>
To: Don Davis <dmd1@Lehigh.EDU>

>
>In other words, for what n and m, for every skew-symmetric bilinear form
>
>F: Z^n\wedge Z^n -> Z^m there exist linearly independent v,w such that
>F(v,w)=0 ?
>
>For example, this statement holds for m<n-1.
>(In this situation the kernel of F(v,*): Z^n-> Z^m has dim>1. Hence
>there is w in the kernel which is linearly independent from v).
>Does the statement hold for bigger m?
>

Denote by m(n) the largest number for which it holds.

m(2)=0
m(3)=2
m(4)=2
m(5)=6

After that I don't know, but here are some general comments.

Sikora remarks that m(n) is at least n-2.

If n is odd then m(n) is at least n-1.
(Given n-1 forms (f_1,...,f_{n-1}), consider the first form f_1.
There is a nonzero vector v such that f_1(v,w)=0 for all w,
because n is odd. Apply Sikora's argument to that v and the
remaining forms (f_2,...,f_{n-1}).)

The question has "coefficients" in Z. You could just as well say Q;
it's the same question.
But you can ask the same question with coefficients in any field k.
Let m_k(n) be the greatest m such that for any skew-symmetric pairing
of k^n into k_m there exist independent v and w such that F(v,w)=0.

Consider the Grassmannian of 2-planes in k^n as being embedded in the
projective space of lines in the second exterior power of k^n.
Then we are talking about the greatest m such that every linear
subspace of that projective space of codimension m must intersect
the Grassmannian.

For k algebraically closed, then, m_k(n)=2n-4, the dimension of the
Grassmannian, since a d-dimensional closed projective variety will
intersect every linear space of codimension d but not every one
of codimension d+1.

For a general field (or at least an infinite field) k this argument gives
that 2n-4 is an upper bound for m_k(n).

m_k(4) can be 2, 3, or 4 depending on k. It's 2 for the rationals,
reals, or p-adics but 3 for finite fields.
This is a pleasant exercise
using the Pfaffian (square root of the determinant)
of 4 by 4 skew-symmetric matrices.

m_k(5)=6 for all k, I believe, but the proof is not very interesting.

So I don't see a pattern.

Tom Goodwillie