Let $C$ be a compact Riemann surface, let $C^2$ be the cartesian square of $C$, let $J(C)$ be the degree zero Jacobian of $C$, and let $\delta : C^2 \to J(C)$ be the map $(x,y) \mapsto [\mathcal{O}(x-y)]$.

In this paper http://arxiv.org/abs/math/9810054 of Hain and Reed, page 9, they say that it is an elementary exercise in algebraic topology to show that $\delta^\ast(\phi) = \Delta + (\psi_1 + \psi_2)/2$.

Explanation of notation:

  • $\phi$ denotes the symplectic form $\sum dx_i \wedge dy_i$ on $J(C)$, where $x_i, y_i$ are coordinates on the torus $J(C) = H^1(C;\mathcal{O}) / H^1(C;\mathbb{Z})$ corresponding to a symplectic basis of $H_1(C;\mathbb{Z}) \cong H^1(C;\mathbb{Z})$. This is also the class of the theta divisor $\Theta$.

  • $\psi_i$ is the first Chern class of the relative cotangent bundle of the $i$th projection $C^{2} \to C$.

  • $\Delta$ is the class of the diagonal $C \to C^{2}$, $x \mapsto (x,x)$.

My question is: How to do this "elementary exercise"? It ought to be easy but I'm just not seeing it...

  • 1
    $\begingroup$ Am I missing something, or is your map actually to the Kummer of the Jacobian? How do you send the set $\{x,y\}$ to $x-y$ without a sign ambiguity? $\endgroup$ – Charles Siegel Jun 29 '11 at 20:43
  • $\begingroup$ Oops, I'm sorry. I shouldn't be taking symmetric product, just ordinary product. $\endgroup$ – Kevin H. Lin Jun 29 '11 at 20:49

Let me write $x_i$ and $y_i$ for a symplectic basis of cohomology of $C$, and $a_i$, $b_i$ for the linear dual basis of the first homology of $C$. It is enough to find $\delta^*(dx_i)$ and $\delta^*(dy_i)$ in terms of $x_i$ and $y_i$.

But to do this we just evaluate $\langle \delta^*(dx_i), a_i \otimes 1 \rangle$, $\langle \delta^*(dx_i), b_i \otimes 1 \rangle$ and so on, which is easy as $\delta_*(a_i \otimes 1) = a_i$, $\delta_*(1 \otimes a_i) = -a_i$ and so on.

Hence $\delta^*(dx_i) = x_i \otimes 1 - 1 \otimes x_i$ and $\delta^*(dy_i) = y_i \otimes 1 - 1 \otimes y_i$. Hence $$\delta^*(\phi) = \sum (x_i \otimes 1 - 1 \otimes x_i)(y_i \otimes 1 - 1 \otimes y_i)$$ and multiplying the whole mess out and using relations like $x_iy_i = [C] = \tfrac{1}{\chi(C)} \psi$ (which perhaps has a sign, depending on your conventions), you get the required expression.

Addendum: You may also be interested in the following. In "Relations among tautological classes revisited" I gave the following alternative description of the class $\delta^*(\phi)$ on $\mathcal{C}_g^2$, the square of the universal curve, which parametrises curves with two ordered non-necessarily distinct points. The paper uses different notation, but I will stick to yours.

Let $\pi : \overline{\mathcal{C}}_g^2 \to \mathcal{C}_g^2$ be the universal curve, and $\Delta_i \in H^2(\overline{\mathcal{C}}_g^2)$ for $i=1,2$ denote the class of the locus of the first and second points respectively. Then $$\delta^*(\phi) = -\pi_!((\Delta_1 - \Delta_2)^2).$$ Again, depending on your conventions there may be a sign and a power of 2 somewhere.

| cite | improve this answer | |

Let me give another proof in the spirit of algebraic geometry.

By Riemann's theorem, for any symmetric theta divisor $\Theta$ there exists a theta characteristic $\kappa$ (i.e., a line bundle such that $\kappa^{\otimes 2}=\omega_C$) on $C$ such that $$W_{g-1}=\Theta + \kappa,$$ where $W_{g-1}$ is the image of the abelian sum mapping $$u \colon \textrm{Sym}^{g-1}(C) \to J(C).$$

$\kappa$ is called the Riemann's constant, and one has

$$\delta^* \Theta = q_1^* (\kappa) \otimes q_2^*(\kappa) \otimes \mathcal{O}_{C^2}(\Delta), \quad \quad (\star)$$

where $q_i \colon C^2 \to C$ are the natural projections.

The proof of such a formula is easy, and it is based on the Seesaw Principle: in other words, one shows that the restrictions to $C \times \{ p \}$ and $\{p \} \times C$ of both sides of $(\star)$ coincide for all $p \in C$.

For the details, see [Birkenhake-Lange, Complex Abelian Varieties, Proposition 11.10.2 $(a)$], putting $\eta=\kappa$ into the statement.

Since $\kappa$ is a theta characteristic it follows that the cohomology class of $q_1^* (\kappa) \otimes q_2^*(\kappa)$ is exactly $\frac{1}{2}(\psi_1 + \psi_2)$. On the other hand, as you noticed, the cohomology class of $\Theta$ is $\phi$, so we are done.

| cite | improve this answer | |

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.