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The moduli space of K3 surfaces forms a 20-dimensional family with countably many 19-dimensional components $M_d$ corresponding to the polarized K3s $(X,L)$ with $L^2=d$. The moduli space $M_d$ has a natural locus $M_d^W$ for each lattice $W$ with an embedding $W\subset H^{1,1}(X)\cap H^2(X,\mathbb Z)$. Is it known how these loci intersect?

For instance, is it true that $M_{d}^W\cap M_{e}\neq \emptyset$ for each $W,d,e$ provided $M_{d}^W\neq \emptyset$?

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I won't completely answer your question, but will try to just rephrase it in a certain way. You are asking when two given moduli spaces of lattice-polarized K3 surfaces $M_L$ and $M_{L'}$ intersect. This is equivalent to the existence of a lattice with $L,L'\hookrightarrow N$ such that $M_N$ is non-empty. Necessary is the existence of such an $N$ which is a sub-lattice of the K3 lattice of rank less than or equal to $20$ and hyperbolic signature.

One approach to the second question would be to consider a hyperbolic lattice $L$ whose quadratic form represents the integer $d$, and enlarge it to a hyperbolic lattice $N$ which also represents the integer $e$. For instance by adding in vectors of negative square in $L^{\perp}$. This may help resolve your question in the low-rank case. The only obstructions to a lattice of rank $5$ or more representing an integer are congruence obstructions. I think one can eliminate any such obstructions by increasing the rank of $L$ by two. So maybe $rk(L)\leq 18$ is fine?

The Picard lattice of any Kummer surface will represent every integer because it contains a rank $16$ lattice with intersection form $(-2)^{16}$. When the rank of $L$ is $19$ or $20$, your K3 is necessarily a Kummer surface, so perhaps that resolves your second question.

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  • $\begingroup$ Thanks for the answer - this looks like it might resolve the question yes! $\endgroup$
    – gsvr
    Commented Oct 5, 2015 at 11:32

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