# Some questions about the (projectivized cotangent bundle of the) symmetric square of a genus $3$ curve

Let $$C$$ be a smooth, non-hyperelliptic curve of genus $$3$$ and $$X:= \mathrm{Sym}^2{C}$$ its symmetric square. Then $$X$$ is a smooth, minimal surface of general type with $$p_g=q=3, \, K^2=6$$.

Calling $$\Omega_X$$ the cotangent bundle of $$X$$, we see that $$\Omega_X$$ is globally generated and $$h^0(X, \, \Omega_X)=3$$; in fact, since $$C$$ is non-hyperelliptic, the Albanese map of $$X$$ is an embedding into a principally polarized abelian threefold.

For all $$a \in C$$, denote by $$C_a$$ the curve in $$X$$ given by $$C_a:=\{ a+x \; | \; x \in C \}.$$ Then $$\mathscr{C}:=\{C_a \}$$ is an irreducible, $$1$$-dimensional family of smooth curves on $$X$$, parametrized by $$C$$; every $$C_a \in \mathscr{C}$$ is isomorphic to $$C$$ and satisfies $$(C_a)^2=1$$.

Question 1. For some undirect reason, I know that the restriction of $$\Omega_X$$ to $$C_a$$ splits and contains a trivial summand. What is a direct proof of this fact? Is there any reference?

Consider now the following diagram: $$\require{AMScd}$$ $$\begin{CD} \mathbb{P}(\Omega_X) @>{\psi}>> \mathbb{P}H^0(X, \, \Omega_X) \simeq \mathbb{P}^2\\ @V{\pi}VV\\ X {} \end{CD}$$

where $$\pi \colon \mathbb{P}(\Omega_X) \to X$$ is the standard projection and $$\psi \colon \mathbb{P}(\Omega_X) \to \mathbb{P}^2$$ is induced by the complete linear system $$|\mathcal{O}_{\mathbb{P}(\Omega_X)}(1)|$$.

By the fact mentioned in Question 1, the map $$\psi$$, when restricted to the ruled surface $$\pi^{-1}(C_a)$$, contracts a section (corresponding to the trivial summand of the restriction of $$\Omega_X$$ to $$C_a$$) to a point $$p_a \in \mathbb{P}^2$$. Call $$\Delta$$ the locus described in $$\mathbb{P}^2$$ by the points $$p_a$$, when $$C_a$$ varies in $$\mathscr{C}$$.

Question 2. Does $$\Delta$$ coincide with the canonical image of $$C$$?

Finally note that, if $$p_a \in \Delta$$, then $$\require{enclose} \enclose{horizontalstrike}{C_a = \pi(\psi^{-1}(p_a))}$$ $$C_a$$ is a component of $$\pi(\psi^{-1}(p_a))$$.

Question 3. Take any point $$q \in \mathbb{P}^2$$ such that $$q \notin \Delta$$. What is $$\pi(\psi^{-1}(q))$$?

Question 1: This has nothing to do with the genus.

$$C_a$$ is the image of a map $$C \to C \times C \to X$$ so $$\Omega_X |_{C_a}$$ maps to the restriction of $$\Omega_{C\times C}$$ to $$a \times C$$, which is $$\mathcal O_C \oplus \omega_C$$.

This map of vector bundles is an isomorphism away from $$a$$, so it is an injection of sheaves with cokernel supported at $$a$$. The cokernel has length $$1$$ since $$C_a$$ intersects the branch divisor of $$C \times C \to X$$, where the ramification has order $$2$$, transversely. Furthermore, we can check that each factor $$\mathcal O_C$$ and $$\omega_C$$ has nontrivial map to the cokernel. A quick way to see this is that sections of $$\Omega_X$$ come from 1-forms on $$C$$, with the map to $$\omega_C$$ given by taking the fiber of the $$1$$-form at $$x$$ and the map to $$\mathcal O_C$$ given by taking the fiber at $$a$$, so in the $$x=a$$ case both these $$1$$-forms are nonzero.

We have an exact sequence $$0 \to \Omega_X|_{C_a} \to \mathcal O_C \oplus \omega_C \to \mathcal O_{\{a\}} \to 0.$$

This gives a natural map $$\Omega_X|_{C_a} \to \mathcal O_C$$. To split that map, consider the long exact sequence on global sections

$$0 \to H^0(X, \Omega_X|_{C_a}) \to H^0(C,\mathcal O_C) \oplus H^0(C, \omega_C) \to \kappa$$ where $$\kappa$$ is the base field. The last arrow can be written concretely as $$(s_1, s_2) \to \alpha_1 s_1(a) + \alpha_2 s_2(a)$$ where $$s_1, s_2$$ are sections of $$\mathcal O_C$$ and $$\omega_C$$ respectively and $$\alpha_1,\alpha_2$$ are nonzero by the earlier nontriviality. Then because $$\omega_C$$ is globally generated, we can choose a section $$s_2$$ with $$s_2(a) = - \alpha_1/\alpha_2$$, and then $$(1,s_2)$$ is a section of $$\Omega_X|_{C_a}$$ that maps to the unit section of $$\mathcal O_C$$ and thus gives a splitting. Note that the splitting depends on a choice.

Question 2: Yes.

Luckily, our choice doesn't matter for the definition of $$\Delta$$. Points of the projectivization correspond to one-dimensional quotients of the underlying vector space, so the points corresponding to the $$\mathcal O_C$$ summand you are looking for are the points corresponding to the $$\mathcal O_C$$ quotient, which is canonically defined by pulling back a differential on $$X$$ to $$C \times C$$ and evaluating in the direction of varying $$a$$. So $$p_a$$ is the point corresponding to the linear form on $$H^0(X, \Omega_X) = H^0(C,\omega_C)$$ obtained by taking a global $$1$$-form, restricting to the divisor $$[x]+ [a]$$, and then evaluating at $$a$$. This is the same as just evaluating at $$a$$, which of course is the point corresponding to $$a$$ under the canonical embedding.

Claim between Question 2 and Question 3: I don't believe this is true. I think $$\pi ( \psi^{-1} ( p_a))$$ has another component, consisting of pairs $$b,c\in C$$ such that $$\omega_C ( -a-b-c)$$ has a global section.

• Thank you for the answer, I will check the details. By any chance, do you have any clue about Question 3? Commented Jul 15, 2022 at 6:17
• @FrancescoPolizzi Here is the best description I found: Any point of $\mathbb P^2 \setminus \Delta$ defines a map $C \to \mathbb P^1$ of degree $4$ by projecting from that point. From any degree four covering of curves we can make a degree six covering, points of whose fibers, correspond to size two subsets of the fiber of the original covering, and the total space of the associated degree $6$ covering is $\pi ( \psi^{-1} (q))$. Commented Jul 15, 2022 at 11:23
• I am checking the details, and everything seems ok, so far. May I ask you how you detected the other component of $\pi ( \psi^{-1} ( p_a))$? Commented Jul 21, 2022 at 16:21
• @FrancescoPolizzi I think of $\mathbb P H^0(X, \Omega_X))$ as the space of nontrivial linear forms on 1-forms, up to scaling, and $\pi^{-1} (b+c)$ as the space of linear forms that depend only on the values of the 1-form at $b$ and $c$, i.e. linear forms vanishing on the unique $1$-form that vanishes on $b$ and $c$. Then $p_a$ is the linear form evaluation at $a$, so $\psi^{-1}(p_a)$ consists of all pairs $b+c$ such that the linear form evaluation at $a$ vanishes on the unique $1$-form vanishing at $b$ and $c$, or in other words such that there exists a $1$-form vanishing at $a,b,c$. Commented Jul 21, 2022 at 16:27
• This can clearly happen even if $a\neq b, a\neq c$. Commented Jul 21, 2022 at 16:27