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The Hom scheme of two projective varieties over some field is constructed as an open subfunctor of the Hilbert scheme of the product of the two schemes by Grothendieck. So it is a countable union of quasi-projective varieties. For example even in the simplest case $\mathcal{Hom}(\mathbb{P}_k^1, \mathbb{P}_k^1)$ there seems to be infinitely many components. So my question is threefold:

  1. Is there some general condition that ensures it is a finite union of quasi-projective varieties?
  2. Is there some general condition that ensures each component in the countable union is a projective variety?
  3. For what type of morphisms $f: X\rightarrow Y$ can one say the map $\mathcal{Hom}(Z,X)\rightarrow \mathcal{Hom}(Z,Y)$ is a proper morphism on the connected components? I know that if $f$ is a closed immersion, the induced map is also closed. I was speculating that if $f$ is finite then the induced map is going to be proper, is something like this true?
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    $\begingroup$ A sufficient condition for 1 and 2 is if the source is finite étale (then the Hom is étale-locally just a finite product). In 3, $f$ finite does not suffice for properness. For example, take $Y$ to be a point and $Z=Spec(k[x]/(x^2))$. Then $Hom(Z,Y)$ is a point but $Hom(Z,X)$ is the tangent bundle of $X$, which can have positive dimension. $\endgroup$ Mar 14, 2022 at 17:29
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    $\begingroup$ That's the finite étale case. If $X$ is not reduced then $Hom(Z,X)$ will not be finite. $\endgroup$ Mar 14, 2022 at 17:51
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    $\begingroup$ You can find answers to your first two questions in Section 3 of arxiv.org/pdf/1807.03665.pdf $\endgroup$ Mar 14, 2022 at 18:02
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    $\begingroup$ It seems that you can check the valuative criterion when $Z$ is normal and $X \to Y$ is finite, in order to show properness. $\endgroup$ Mar 14, 2022 at 20:38
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    $\begingroup$ @user127776 I'm not sure if normality is necessary. Let $R$ be a valuation ring with fraction field $K$. Basically, the point is that you have an integral extension $O_Y \to O_X$ and a homomorphism $O_X \to O_Z \otimes K$ such that the image of $O_Y$ is contained in $O_Z \otimes R$. If $O_Z \otimes R$ is integrally closed in its field of fractions, then it follows that the image of $O_X$ is also contained in $O_Z \otimes R$. $\endgroup$ Mar 14, 2022 at 23:54

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Let $k$ be an algebraically closed field of characteristic zero and let $X$ be a projective variety over $k$. To try and answer your first two questions, let us try formalize the property that Hom-schemes $\mathrm{Hom}(Y,X)$ are finite unions of quasi-projective schemes.

Definition. We say that $X$ is bounded over $k$ if, for every normal projective variety $Y$ over $k$, the Hom-scheme $\mathrm{Hom}(Y,X)$ is of finite type over $k$.

We can then show the following result; see [1] and [2].

(I think that your question is only concerned with 1 and 6 in the following result, but the rest might also be useful to you.)

Theorem. The following are equivalent.

  1. $X$ is bounded over $k$.
  2. For every algebraically closed field $L$ containing $k$, the projective variety $X_L$ is bounded over $L$.
  3. For every smooth projective curve $C$ over $k$, the Hom-scheme $Hom(C,X)$ is of finite type.
  4. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ has only finitely many connected components.
  5. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is quasi-projective.
  6. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$.
  7. For every normal projective variety $Y$ over $k$, the Hom-scheme $Hom(Y,X)$ is a projective scheme over $k$. The subscheme of non-connected morphisms $Hom^{nc}(Y,X)$ is a bounded projective scheme of dimension $<\dim X$ which maps finitely to $X$.

Boundedness is most likely equivalent to "hyperbolicity". In fact, it is implied by hyperbolicity (Kobayashi, Brody, ...). More precisely:

Theorem. Assume $k=\mathbb{C}$. If $X$ is hyperbolic (i.e., every holomorphic map $\mathbb{C}\to X^{an}$ is constant), then $X$ is bounded.

Conjecture. Assume $k=\mathbb{C}$. If $X$ is bounded over $\mathbb{C}$, then $X$ is hyperbolic.

This should answer your first question. For your second question: if $X$ has no rational curves, then every Hom-scheme $Hom(Y,X)$ has projective components (but it can have infinitely many components). Conversely, if all Hom-schemes $Hom(Y,X)$ have (only) projective components, then $X$ has no rational curves. (Use that the components of $Hom(\mathbb{P}^1,\mathbb{P}^1)$ are affine varieties of increasing dimension.)

Abelian varieties have no rational curves and give examples of non-finite type Hom-schemes with each component projective. Note that non-trivial abelian varieties are far from being hyperbolic, so there's no contradiction to the above conjecture.

References.

[1] R. van Bommel, A. Javanpeykar, L. Kamenova. Boundedness in families with applications to arithmetic hyperbolicity https://arxiv.org/abs/1907.11225

[2] A. Javanpeykar and L. Kamenova Demailly's notion of algebraic hyperbolicity: geometricity, boundedness, moduli of maps https://arxiv.org/abs/1807.03665

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    $\begingroup$ Thanks for your fantastic answer. $\endgroup$
    – user127776
    Mar 16, 2022 at 0:19
  • $\begingroup$ You're wecome. Feel free to email me if you have more questions. $\endgroup$ Mar 17, 2022 at 1:38

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