linear algebra - Which quadratic forms on $Lambda^2 V$ come from quadratic forms on $V$?

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

If $n=dim Vgeq4$, 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 $S^2Lambda^2V^*cong V_{0,0,0,1,dots}oplus V_{0,2,dots}$ (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 $Lambda^2S^2V^*to S^2Lambda^2V^*$. Since $L$ is non-zero, $L$ is that map, and its image is the summand $V_{0,2,dots}$ of $S^2Lambda^2V^*$.

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 $S^2Lambda^2V^*$ is just the kernel of the anti-symmetrization $S^2Lambda^2V^*toLambda^4V^*$.