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I hope this is not too elementary.

Let $B: V\ \times V \to W$ be a skew-symmetric bilinear map where $V$, $W$ are finite dimensional real vector spaces. Assume that $B (u, v)$ is never zero for linearly independent $u$, $v\in V$.

What can we say about the dimensions of $V$ and $W$?

An elementary estimate is

$$ \dim W \geq \dim V - 1 $$

In fact, for $u\in V$, $u\neq0$, the linear map $B_u : V\to W$, $B_u(v)=B(u,v)$, is injective on a codimension $1$ subspace of $V$. Can the above estimate be improved? Recall that the image of a bilinear map does not need to be a subspace.

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A better bound is $\dim W\geq 2\dim V-3$. This is obtained as follows. In the projective space $\mathbb{P}(\wedge^2V)$, the set of decomposable bivectors is the Grassmannian $\mathbb{G}(2,V)$, of dimension $2\dim V-4$. The bilinear map $B$ induces a linear map $b:\wedge^2V\rightarrow W$, and the hypothesis is $\mathbb{P}(\mathrm{Ker}\, b)\cap \mathbb{G}(2,V)=\varnothing$. This implies $2\dim V-4<\mathrm{codim}\, \mathrm{Ker}\, b=\dim \mathrm{Im}\, b$, hence the result.

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  • $\begingroup$ Simple and smart, thanks! In fact, the better bound suffices for what I need. $\endgroup$
    – Claudio Gorodski
    Jan 9, 2015 at 16:00
  • $\begingroup$ This is way late, but I think the argument works over $\mathbb C$ but not over $\mathbb R$, which is what I wanted. The obvious idea of complexifying $B$ does not seem to preserve the assumption on $B$. $\endgroup$
    – Claudio Gorodski
    Feb 21, 2015 at 14:07
  • $\begingroup$ I think this is a counter-example: $\mathbb G(2,\mathbb R^4)$ has dim $4$ and embedds in $\mathbb P(\Lambda^2\mathbb R^4)$ which has dim $5$, defined $\endgroup$
    – Claudio Gorodski
    Feb 21, 2015 at 14:09
  • $\begingroup$ the equation $a_{12}a_{34}-a_{13}a_{24}+a_{14}a_{23}=0$. Construct a linear map $\endgroup$
    – Claudio Gorodski
    Feb 21, 2015 at 14:10
  • $\begingroup$ $b:\Lambda^2\mathbb R^4\to W^3$ whose kernel is $a_{12}=a_{34}$, $a_{14}=a_{23}$, $a_{13}=-a_{24}$, and consider its projectivization $U^2 \subset \mathbb P (\Lambda^2\mathbb R^4)$. Then $U^2$ does not meet $\mathbb G(2,\mathbb R^4)$. On the other hand, $2\dim V - 3 = 5 > 3 = \dim W$. $\endgroup$
    – Claudio Gorodski
    Feb 21, 2015 at 14:14

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