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The following Lemma is in Beauville-Donagi, and I always took it for granted. Now I've tried to find a proof, but got stuck. They say it is a really simple lemma, so I may just be overlooking something easy.

Let $V$ be a vector space of dimension $6$, and let $W \subset \bigwedge^2 V^*$ be a subspace of dimension $2$. Assume that every form in $W$ is degenerate. Then there is a subspace $K \subset V$ of dimension $4$ such that each form in $W$ restricts to $0$ on $K$.

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First note that if a 2-form is degenerate, it is 0 on some 4-subspace (take a lagrangian subspace of the quotient by the kernel).

Now, assume not. Pick two elements that span $W$. If either of them has 4-d kernel, it is 0 on any 4-d subspace, and we can use whichever on the other vanishes on.

Thus, every element of $W$ has 2-d kernel. If two elements had different kernels, then one of their linear combinations would have no kernel. Thus, they all kill the same 2-d subspace. Thus, we've reduced to the statement that any two 2-forms on a 4-d space $Z$ have a common Lagrangian subspace. Pick any line $L$; this is isotropic for both, since all lines are. Consider the intersections of the symplectic orthogonals of $L$ for the two 2-forms. These are 3-d, so their intersection is a 2-space. Now you win.

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"If two elements had different kernels, then one of their linear combinations would have no kernel." This is the crucial point. Why is this so? By the way the claim "If either of them has 4-d kernel, it is 0 on any 4-d subspace" is false, but that case is easy to handle anyway. –  Andrea Ferretti Jan 30 '10 at 14:31
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