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Let $C$ be a smooth curve of genus $g\geq 3$, embedded in its Jacobian $X=\textrm{Jac } C$ via an Abel map. Let $\textrm{Hilb}_1(X)$ be the Hilbert scheme of curves in $X$, and let $[C]\in\textrm{Hilb}_1(X)$ be the corresponding point. In Some Remarks and Examples on Continuous Systems and Moduli, Griffiths computed the dimension of the tangent space to $\textrm{Hilb}_1(X)$ at $[C]$. The answer is

  • $g$, if $C$ is non-hyperelliptic, and
  • $2g-2$ if $C$ is hyperelliptic.

Say $g=3$ and $C$ is hyperelliptic. In On the module of intermediate Jacobians, Lieberman showed $$\textrm{Hilb}_1(X,[C])_{\textrm{red}}\cong X=\textrm{Jac } C,$$ so Griffiths' result says the Hilbert scheme component $\textrm{Hilb}_1(X,[C])\subset \textrm{Hilb}_1(X)$ is a thickening of the Jacobian (tangent spaces are $4$-dimensional everywhere).

Question. What is the scheme structure on $\textrm{Hilb}_1(X,[C])$, when $C$ is hyperelliptic of genus $3$? Is it known to be something like $X\times \textrm{Spec }\mathbb C[\epsilon]$, or anything more complicated?

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    $\begingroup$ For the Abel map $u:C\to \text{Pic}^1_{C/k}$ with conormal sheaf $N^\vee_u$, the long exact sequence of the short exact sequence $$0\to N^\vee_u \to H^0(C,\Omega_{C/k})\otimes_k \mathcal{O}_C \to \Omega_C \to 0,$$ identifies the fiber at $[u]$ of the conormal sheaf of the reduced locus $X\hookrightarrow \text{Hilb}_{X/k}$ with the fiber at $[C]$ of the conormal sheaf $N^\vee_{\mathcal{H}_g/\mathcal{M}_g}$ of the hyperelliptic locus inside $\mathcal{M}_g$. This isomorphism seems to be canonical. So at least the conormal sheaf of the reduced locus is canonically trivialized . . . $\endgroup$
    – Jason Starr
    Aug 14, 2017 at 12:53
  • $\begingroup$ . . . So the extension of $\Omega_{X/k}$ by $N^\vee_{\mathcal{H}_g/\mathcal{M}_g}$ coming from the scheme structure on $\text{Hilb}_{X/k}$ is the same as a $k$-linear map $$N_{\mathcal{H}_g/\mathcal{M}_g,[C]}\to H^1(X,T_X).$$ Your question is whether or not this is the zero map. That seems likely. Of course there is the nonzero Kodaira-Spencer map $T_{\mathcal{M}_g,[C]}\to H^1(X,T_X)$, but this is nonzero on $T_{\mathcal{H}_g,[C]}$. So it does not factor through $N_{\mathcal{H}_g/\mathcal{M}_g,[C]}$ . . . $\endgroup$
    – Jason Starr
    Aug 14, 2017 at 12:58
  • $\begingroup$ . . . Over $\mathbb{C}$, you may be able to use curvature results on these bundles to prove that there is no nonzero map over $\mathcal{H}_g$ from the normal bundle $N_{\mathcal{H}_g/\mathcal{M_g}}$ to the pullback via the Torelli map of the tangent bundle of the moduli space of Abelian varieties. $\endgroup$
    – Jason Starr
    Aug 14, 2017 at 13:01
  • $\begingroup$ Thanks! can you say more on the curvature results you were mentioning in your last comment? do you have a reference in mind? $\endgroup$
    – Andrea Ricolfi
    Aug 15, 2017 at 6:22
  • $\begingroup$ I do not know any specific curvature result, but the curvatures of the "natural" metrics on "natural" vector bundles on $\mathcal{M}_g$ have been studied quite intensively. $\endgroup$
    – Jason Starr
    Aug 15, 2017 at 13:26

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