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Take a smooth complex projective variety $X$, consider $H^k(X,\mathbb Z)$, and take the global period domain as described, for example, in Voisin's Hodge theory book, 10.1.3: it's a subset of a flag variety, modulo the action of $\operatorname{GL}(H^k(X,\mathbb Z))$. We can also take the global period domain for $H^k(X,\mathbb Q)$, which is the same subset modulo $\operatorname{GL}(H^k(X,\mathbb Q))$, which is no longer a discrete group.

When can we expect the map from the period domain for integral Hodge structures to the period domain for rational Hodge structures to be generically injective? What about the same map for polarized Hodge structures on $H^k_\text{prim}$, or the map from the polarized period domain to the unpolarized one? It seems like this should be in a textbook, but I haven't found it.

Some remarks in the introduction to Voisin's paper "Schiffer variations..." suggest that this forgetful map is generically injective for hypersurfaces of most degrees and dimensions, and for curves of genus $g \ge 4$ but not genus 3. I'm not even sure what happens for K3 surfaces.

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If the image of the usual period map is an open subset of the period domain, this map is certainly not generically injective: For a typical point in the image, its translate by any element of $GL_n(\mathbb Q)$ that does not lie in $GL_n(\mathbb Z)$ will also lie in the image, and then the preimages of these two points will be different but map to the same point in the period domain modulo $GL_n(\mathbb Q)$.

The same argument works if you consider a period domain for a subgroup of $GL_n$: For example the moduli space of genus $g$ curves has open image in thee moduli space of dimension $g$ abelian varieties for $g \leq 3$, and this moduli space is the quotient of a suitable period domain by $Sp_{2g}(\mathbb Z)$. The same thing works for K3 surfaces using an orthogonal group preserving the class of the ample line bundle on the K3 surface (which needs to be fixed to define a moduli space of 3 surfaces).

On the other hand when the image of the period map has positive codimension, we should expect this map to be generically injective: This is equivalent to saying that for a typical point in the image, its $GL_n(\mathbb Q)$-orbit avoids the image. This is a countable intersection of conditions whose complement has positive codimension (unless the image is preserved by some element of $GL_n(\mathbb Q) \setminus GL_n(\mathbb Z)$, which possibly one can check doesn't happen and certainly is unlikely) and so occurs generically.

If you're interested in injectivity and not generic injectivity, this shouldn't happen unless the image of the period map is less than half the dimension of the period domain (since we want this condition to have codimension greater than the dimension of the space and not just positive codimension) and even then is hard to rule out (since a single unlikely intersection would prevent injectivity).

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  • $\begingroup$ This is great, thanks! So it's important to focus on the map from a moduli space of varieties to either period domain, and not about the map between the two period domains. $\endgroup$ Commented May 9 at 1:44
  • $\begingroup$ By the way, you can make the moduli space of unpolarized K3s: take the moduli space of marked K3s, discussed in Huybrechts' book section 7.2, and then divide by automorphisms of the lattice. But neither the marked moduli space nor the quotient is Hausdorff. $\endgroup$ Commented May 9 at 1:44
  • $\begingroup$ @NickAddington Well, you said smooth projective complex variety, and the K3s in the moduli space of unpolarized K3s are neither projective nor varieties. $\endgroup$
    – Will Sawin
    Commented May 9 at 6:58
  • $\begingroup$ Ah ha, so I did. $\endgroup$ Commented May 9 at 15:24
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    $\begingroup$ @NickAddington One further observation: One can study the failure of injectivity in general using results about the Hodge locus, since an isomorphism of rational Hodge structures between $H^k(X)$ and $H^k(Y)$ gives a Hodge class on $X \times Y$ (or more precisely in $H^k(X)^\vee \otimes H^k(Y)$). There is some subtlety since the converse is not always true but some results in this direction should be helpful. $\endgroup$
    – Will Sawin
    Commented May 9 at 15:32

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