4
$\begingroup$

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

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$
12
  • $\begingroup$ Thank you for the answer, I will check the details. By any chance, do you have any clue about Question 3? $\endgroup$ Jul 15 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
    Jul 15 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))$? $\endgroup$ Jul 21 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
    Jul 21 at 16:27
  • $\begingroup$ This can clearly happen even if $a\neq b, a\neq c$. $\endgroup$
    – Will Sawin
    Jul 21 at 16:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.