How are Jacobians of genus $3$ curves different from one another?

Question

There are two types of smooth projective (complex) curves of genus $3$: plane quartics, and hyperelliptic curves. The Torelli morphism $M_3\to A_3$, assigning a curve to its (principally polarized) Jacobian, is bijective on points. Thus $A_3$ consists of Jacobians of genus $3$ curves. Because of the "partition" of $M_3$ as {plane quartics} $\amalg$ {hyperelliptics}, I would expect this to be reflected somehow in $A_3$. Whence my questions:

Question 1. What are the main differences between Jacobians of plane quartics and of hyperelliptic curves?

(I am intentionally vague here, as I am curious about any sort of difference).

It would be also interesting to know if the deformation theory of the curve inside its Jacobian is sensibly different according to the curve itself; this leads to a second question:

Question 2. (Deformations inside the Jacobian) What do the normal bundles $\mathcal N_{C/Jac(C)}$ look like, when $C$ is a plane quartic or a hyperelliptic curve, respectively?

Thanks for any clue on this.

Answer 1

Let $L$ be a principal polarization on the Jacobian of the genus $3$ curve. Then $L^2$ gives an embedding of the Kummer variety $X$ of this Jacobian (i.e., quotient by the action $x \mapsto -x$). Khaled showed that this embedding is projectively normal (i.e., the restriction map

$Sym^n(H^0(X, L^2)) \to H^0(X, L^{2n})$

is surjective for all $n$) if and only if $C$ is not hyperelliptic. This was actually useful in a paper I wrote (and I will also take the opportunity to advertise): see Proposition 6.9 of http://arxiv.org/abs/1203.2575 and also the reference for Khaled's paper.

Answer 2

For Question 2. From the conormal bundle sequence $$0\to\mathcal N^{\vee}\to \Omega_{Jac(C)}\to \omega_C\to 0$$ we see that we can identify $\mathcal N^{\vee}$ with the kernel/syzygy bundle $M_{\omega_C}$ given by the exact sequence $$0\to M_{\omega_C}\to H^0(C,\omega_C)\to\omega_C.$$ This bundle is semistable of slope $-2$ and if $g\geq 3$ it fails to be stable precisely for hyperelliptic curves. See (3.4)Corollary in https://www.imsc.res.in/~kapil/papers/chap1/index.html

To see the instability appear for hyperelliptic curves, it is enough to observe sections of $M_{\omega_C}\otimes{\mathcal O}_C(H)$, where $H$ is the hyperelliptic pencil giving the $2:1$ map to $\mathbb P^1$, such that $h^0(C,H)=2$. Twisting in the sequence above and taking sections, we have $$0\to H^0(C,M_{\omega_C}\otimes{\cal O}_C(H))\to\mathbb C^{2g}\to\mathbb C^{g+1}.$$

Answer 3

For Question 2, a quick description of the normal bundle $N_{C|J(C)}$ is (See Lemma 3.8 of [1]):

$$h^0(C,N_{C|J(C)})=h^1(C,N_{C|J(C)})= \begin{cases} 3 ~\text{for} ~C~\text{non-hyperelliptic}\\ 4 ~\text{for} ~C~\text{hyperelliptic}.\tag{1}\label{1} \end{cases}$$

This tells us some information of the deformation theory of the pair $(J(C),C)$. Let $\mathcal{D}$ be the deformation space of such pairs, for $C\in M_3$ and $J(C)$ the Jacobian of $C$, then the tangent space of $\mathcal{D}$ at $(J(C),C)$ is the hypercohomology $\mathbb{H}^1(J(C),C)$, which fits into the exact sequence

$$0\to H^0(J(C),T_{J(C)})\to H^0(C,N_{C|J(C)})\to \mathbb{H}^1(J(C),C)\xrightarrow{d} H^1(J(C),T_{J(C)})\to H^1(C,N_{C|J(C)})\to 0.$$

Using $T_{J(C)}\cong \mathcal{O}^{\oplus 3}_{J(C)}$, one has $h^0(J(C),T_{J(C)})=3$ and $h^1(J(C),T_{J(C)})=9$, together with $(\ref{1})$, one has $\dim \mathbb{H}^1(J(C),C)$ is always equal to $6$, so $\mathcal{D}$ is smooth, no matter $C$ is hyperelliptic or not.

However, the distinction comes from the rank of $d$, which is the answer to your Question 2:

If $C$ is non-hyperelliptic, $d$ is of maximal rank, so when $J(C)$ deforms in the 6-dimensional p.p.a.v. locus, the curve $C$ deforms with $J(C)$; If $C$ is hyperelliptic, $d$ has a one-dimensional kernel, so $C$ will be obstructed when $J(C)$ deforms normal to the 5-dimensional hyperelliptic locus.

The proof of $(\ref{1})$ is essentially a classical fact known as Noether's theorem, which states that $$\text{Sym}^2H^0(C,K_C)\to H^0(C,2K_C)$$ is an isomorphism when $C$ is non-hyperellptic curve and has a one-dimensional kernel when $C$ is hyperellptic.

Question 1 is essentially answered in rfauffar's comment, namely $\tau(C)$ coincides with $C$ up to a translation in torus if and only if $C$ is hyperelliptic, where $\tau:J(C)\to J(C)$ is the involution given by $(-1)$-map and we identify $C$ with its Abel-Jacobi image in $J(C)$.

I'd like to mention a theorem related to this fact. Note that $\tau$ always induces identity map on $H_2(J(C),\mathbb Z)$, so $\tau(C)$ is homologous to $C$ in $J(C)$, however, they are in general not algebraically equivalent:

Theorem: (G. Ceresa [2], 1983) When $C$ is general in $M_3$, $C$ is not algebraically equivalent to $\tau(C)$.

Based on this result, F. Bardelli and M. Nori showed

Theorem: (F. Bardelli [1], 1989, M. Nori [3], 1989) Let $A$ be a general abelian threefold, the Griffiths group $\mathcal{G}_1(A)$ (the group of algebraically trivial 1-cycles modulo homological equivalence) is not finitely generated.

[1] Bardelli, Fabio. Curves of genus three on a general abelian threefold and the nonfinite generation of the Griffiths group. Arithmetic of complex manifolds (Erlangen, 1988), 10–26, Lecture Notes in Math., 1399, Springer, Berlin, 1989.

[2] Ceresa, G. $C$ is not algebraically equivalent to $C^−$ in its Jacobian. Ann. of Math. (2) 117 (1983), no. 2, 285–291.

[3] Nori, Madhav V. Cycles on the generic abelian threefold. Proc. Indian Acad. Sci. Math. Sci. 99 (1989), no. 3, 191–196.