# A very basic question about Abel-Jacobi map

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...

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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? –  Charles Siegel Jun 29 '11 at 20:43
Oops, I'm sorry. I shouldn't be taking symmetric product, just ordinary product. –  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.

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Thanks, Oscar! Indeed I've seen your paper already.. –  Kevin H. Lin Jul 6 '11 at 2:03

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.

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Thanks, Francesco! This is very nice. –  Kevin H. Lin Jul 6 '11 at 1:11