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 $\pi(\psi^{-1}(q))$?


1 Answer 1


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.

  • $\begingroup$ Thank you for the answer, I will check the details. By any chance, do you have any clue about Question 3? $\endgroup$ Commented Jul 15, 2022 at 6:17
  • $\begingroup$ @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 $\pi ( \psi^{-1} (q))$. $\endgroup$
    – Will Sawin
    Commented Jul 15, 2022 at 11:23
  • $\begingroup$ 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))$? $\endgroup$ Commented Jul 21, 2022 at 16:21
  • $\begingroup$ @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 $a,b,c$. $\endgroup$
    – Will Sawin
    Commented Jul 21, 2022 at 16:27
  • $\begingroup$ This can clearly happen even if $a\neq b, a\neq c$. $\endgroup$
    – Will Sawin
    Commented Jul 21, 2022 at 16:27

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