You have a quadratic map $Q:S^2V^*\to S^2\Lambda^2V^*$ which you can polarize to a linear map $L:S^2S^2V^*\to S^2\Lambda^2V^*$ and this will be $\mathrm{SL}(V)$-equivariant.

If $n=\dim V\geq4$, then as modules over $\mathrm{SL}(V)$ Magma tells me that we have $\Lambda^2S^2V^*\cong V_{4,\dots}\oplus V_{0,2,\dots}$ and $S^2\Lambda^2V^*\cong V_{0,0,0,1,\dots}\oplus V_{0,2,\dots}$ (the dots mean "complete with zeroes to form a partition of length $n$, and the $V_{\mathrm{something}}$ are highest-weight modules in their ‘usual’ notation). It follows that there is up to scalars *one* non-zero linear $\mathrm{SL}(V)$-linear map $\Lambda^2S^2V^*\to S^2\Lambda^2V^*$. Since $L$ is non-zero, $L$ is that map, and its image is the summand $V_{0,2,\dots}$ of $S^2\Lambda^2V^*$.

I would imagine (but I do not know if) the image of the quadratic map $Q$ is also that $V_{0,2,\dots}$. To answer your question, one would then need to characterise that summand. The other summand, $V_{0,0,0,1,\dots}$ is isomorphic to $\Lambda^4V^*$, so maybe the $V_{0,2,\dots}$ inside $S^2\Lambda^2V^*$ is just the kernel of the anti-symmetrization $S^2\Lambda^2V^*\to\Lambda^4V^*$.

very farfrom sufficient. – thomas Nov 17 '09 at 20:27