# Differntial of the Torelli morphism and the multiplication map

Consider the Torelli morphism $T$: $M_{g}$-----------> $A_{g}$. The differential of this morphism is the map

$dT$ : $T_{M_{g}}$---------> $T_{A_{g}}$ between the tangent bundles. Now at the point $[C]$ $\in$ $M_{g}$, this becomes the map

$H^{1}(T_{C})$ -----> $Sym^2 H^{1}(O_{C})$ .

On the other hand, we have the natural multiplication map:

$\mu$ : $Sym^2 H^{0}(\Omega^1_{C})$------------->$H^{0}(\Omega^2_{C})$

Serre duality says that $H^{1}(T_{C})$ is dual to $H^{0}(\Omega^2_{C})$ and likewise $Sym^2 H^{1}(O_{C})$ is dual to $Sym^2H^{0}(\Omega^1_{C})$. This propmts the following question:

Is it true that the differential of the Torelli morphism at the point $[C]$ is always the dual of the multiplication map?

-

$\DeclareMathOperator{\Jac}{\mathrm{Jac}}\DeclareMathOperator{\Sym}{\mathrm{Sym}}$ The answer is yes.

First of all let's see how to identify the tangent space of a point $[\Jac C] \in A_g$ with $\Sym^2 H^1(C,O_C)$.

The 1st order deformations of an abelian variety $A$ (not necessarily preserving the polarization) are given by $H^1(A,T_A) \cong H^1(A,O_A) \otimes H^0(A,\Omega_A)^\vee$ since $T_A$ is the trivial bundle $O_A \otimes H^0(A,\Omega_A)^\vee$. (You get a trivialization of $T_A$ by translation on the group $A$, identifying each fiber canonically with the tangent space at the origin.) The map $H^1(C,\mathbb C) \to H^1(\Jac C, \mathbb C)$ is an isomorphism of Hodge structures, so for a Jacobian the latter deformation space can be rewritten as $H^1(C,O_C) \otimes H^0(C,\Omega_C)^\vee$. Serre duality (on $C$!) identifies $H^0(C,\Omega_C)^\vee$ with $H^0(C,O_C)$, so for a Jacobian the deformations can be written as $H^1(C,O_C) \otimes H^1(C,O_C)$. Finally $\Sym^2 H^1(C,O_C)$ sits inside the deformation space as the deformations preserving the polarization on $\Jac C$ (and hence the isomorphism $H^1(A,O_A)\cong H^1(A,\Omega_A)^\vee$).

Now cup product defines a map $H^1(C,T_C) \otimes H^0(C,\Omega_C) \to H^1(C,O_C)$, which by dualizing gives $H^1(C,T_C) \to H^1(C,O_C) \otimes H^0(C,\Omega_C)^\vee$. The claim is that this is the differential of the Torelli map. A proof can be found in Arbarello, Cornalba, Griffiths "The geometry of algebraic curves vol. 2" in the chapter on deformation theory, section 8.

Transposing and applying Serre duality gives also a map $H^0(C,\Omega_C) \otimes H^0(C,\Omega_C) \to H^0(C, \Omega_C^{\otimes 2})$ -- the codifferential of Torelli -- and what we want to show is that this map, too, is given by cup product. So what you need to know is that Serre duality is compatible with cup product maps. But this is immediate if you think of Serre duality as defined by cup products taking values in $H^1(C,\Omega_C)$ with its canonical isomorphism with $\mathbb C$ defined via the trace map.

-
Thank you very much Dan for your complete answer. Is your argument only valid over C or it works in any characteristic? I think the proof in ACG book only applies over C. On the other hand I have some reasons to believe that this identification can not be made in characteristic p. Could you clarify this? Thank you in advance. – Cyrus Aug 9 '11 at 21:07
Although I don't know a reference, I can not imagine that the map $H^1(T_C) \otimes H^0(\Omega_C) \to H^1(O_C)$ is not given by cup product in positive characteristic. That would be madness. The only issue I can see is that you will not be able to identify symmetric tensors with the symmetric square in characteristic two. – Dan Petersen Aug 9 '11 at 22:04
Nowadays I suppose those letters refer to "stacks", which I understand to mean you don't care that the tangent space at a singular point of Mg is not the same as at a smooth point hence neither is the differential. ignorant comment by Old guy. – roy smith Aug 10 '11 at 4:46
Dan is completely right, the only thing you have to do (as Dan indicates) in positive characteristic is to note that first order deformations deformations of principally polarised abelian varieties is a subspace of the first order deformation space of abelian varieties. Hence we get a subspace of $\mathrm{Hom}(H^0(C,\Omega^1),H^1(C,\mathcal O_C))=H^1(C,\mathcal O_C)^{\oplus2}$ consisting of the symmetric tensors. This is $\Gamma^2H^1(C,\mathcal O_C)$ rather than $S^2H^1(C,\mathcal O_C)$. – Torsten Ekedahl Aug 10 '11 at 6:00
if you are really interested in this question as stated, and not just in the easier one of the induced map of deformations spaces, i.e. the differential of the map of functors, you might look at: Algebraic geometry: Angers 1979 ; variétés de petite dimension: the article of oort and steenbrink, on the local torelli problem. – roy smith Aug 31 '11 at 2:00