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The intersection of 3 quadrics in $P^5$ is a K3 surface $S$. There is a natural map $S^{[2]} \to G(1,5)$ well defined everywhere, because a generic K3 doesn't contain any line and this family is maximal. This is moreover injective, because a line can meet $S$ in at most 2 points (otherwise it is contained in each of the 3 quadrics, and hence it lays on $S$).

Is it also an embedding?

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up vote 3 down vote accepted

The embedding of $S$ into $P^5$ induces an embedding $S^{[2]} \to (P^5)^{[2]}$. On the other hand, it is easy to check that $(P^5)^{[2]}$ is isomorphic to $P_{G(1,5)}(S^2U)$, the projectivization of the symmetric square of the tautological rank 2 bundle on the Grassmannian. This gives the map which you want.

EDIT. Since the map is injective, it remains to show that it is also injective on tangent spaces. For this note that if we consider just one quadric $Q$ then its Hilbert square $Q^{[2]}$ is the zero locus of a rank 2 vector bundle on $P_{G(1,5)}(S^2U)$, namely of $O(2h)\otimes p^*S^2U / O(-u)$, where $h$ is the pullback of the generator of $Pic G(1,5)$, $u$ is the relative $O(1)$ for the projectivization, and $p$ is the projection to the Grassmannian. It follows from this that $Q^{[2]}$ is isomorphic to the blowup of $G(1,5)$ in the subscheme of lines lying on $Q$. It follows, that the map $S^{[2]} \to G(1,5)$ is injective on tangent spaces over all points corresponding to lines which do not lie at least in one quadric passing through $S$. Since this property is satisfied by all lines in the image, the map is an embedding.

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It doesn't answer the question, imho. – Barbara Mar 14 '13 at 20:01
@Barbara: You are completely right, sorry. I added an explanation to the answer. – Sasha Mar 15 '13 at 6:39

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