Facts about alternating bilinear maps on vector spaces

Let F be a field, let V be an n-dimensional F-vector space, and let f : V \times V \rightarrow W be a bilinear map, where W is an F-vector space.

  1. Prove that if f is an alternating F-bilinear map on V then f(x,y) = -f(y,x) for all x,y \in V.
  2. Suppose \mathsf{char}\ F \neq 2. Prove that f(x,y) is an alternating bilinear map on V if and only if f(x,y) = -f(y,x) for all x,y \in V.
  3. Suppose \mathsf{char}\ F = 2. Prove that every alternating bilinear form f(x,y) on V is symmetric. (I.e. f(x,y) = f(y,x) for all x,y \in V.) Prove that there exist symmetric bilinear maps which are not alternating.

Let x,y \in V, and suppose f is an alternating bilinear map. Now 0 = f(x-y,x-y) = f(x,x) - f(x,y) - f(y,x) + f(y,y), so that f(x,y) = -f(x,y).

Suppose \mathsf{char}\ F \neq 2; in particular, 2 is a unit in F. If f is bilinear such that f(x,y) = -f(y,x) for all x,y \in V, then in particular we have f(x,x) = -f(x,x), so that 2f(x,x) = 0. Thus f(x,x) = 0 for all x \in V, and so f is alternating. Conversely, if f is alternating then by part (1) above we have f(x,y) = -f(y,x) for all x, y \in V.

Now suppose \mathsf{char}\ F = 2. Note that (x+y) \otimes (x+y) = x \otimes x - x \otimes y + y \otimes x - y \otimes y. Mod \mathcal{A}^2(V), we have x \otimes y - y \otimes x = 0. In particular, the submodule \mathcal{C}^2(V) generated by all tensors of the form x \otimes y - y \otimes x is contained in \mathcal{A}^2(V).

We have already seen that \mathsf{dim}_F\ \mathcal{S}^2(V) = {{n+1} \choose {n-1}} = \frac{n(n+1)}{2} and \mathsf{dim}_F\ \bigwedge^2(V) = {n \choose 2} = \frac{n(n-1)}{2}. Thus the containment \mathcal{C}^2(V) \subsetneq \mathcal{A}^2(V) is proper.

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  • mcoulont  On October 18, 2011 at 6:28 am

    In 2, the bilinearity of f seems to be an hypothesis to prove the equivalence.

    • nbloomf  On October 19, 2011 at 9:57 am

      Thanks! I think I understand and fixed the problem.

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