Is the Torelli map an immersion?

Question

The Torelli map $\tau\colon M_g \to A_g$ sends a curve C to its Jacobian (along with the canonical principal polarization associated to C); see this question for a description which works for families.

Theorem (Torelli): If $\tau(C) \cong \tau(C')$, then $C \cong C'$.

If one prefers to work with coarse spaces (instead of stacks) it is okay to just say that $\tau$ is injective.

Question: Is $\tau$ an immersion?

(One remark: $\tau$ isn't a closed immersion -- the closure of its image consists of products of Jacobians!)

Answer 1

Respectfully, I disagree with Tony's answer. The infinitesimal Torelli problem fails for $g>2$ at the points of $M_g$ corresponding to the hyperelliptic curves. And in general the situation is trickier than one would expect.

The tangent space to the deformation space of a curve $C$ is $H^1(T_C)$, and the tangent space to the deformation space of its Jacobian is $Sym^2(H^1(\mathcal O_C))$. The infinitesimal Torelli map is an immersion iff the map of these tangent spaces

$$ H^1(T_C) \to Sym^2( H^1(\mathcal O_C) )$$

is an injection. Dually, the following map should be a surjection:

$$ Sym^2 ( H^0(K_C) ) \to H^0( 2K_C ), $$

where $K_C$ denotes the canonical class of the curve $C$. This is a surjection iff $g=1,2$ or $g=3$ and $C$ is not hyperelliptic; by a result of Max Noether.

Therefore, for $g\ge 3$ the Torelli map OF STACKS $\tau:M_g\to A_g$ is not an immersion. It is an immersion outside of the hyperelliptic locus $H_g$. Also, the restriction $\tau_{H_g}:H_g\to A_g$ is an immersion.

On the other hand, the Torelli map between the coarse moduli spaces IS an immersion in char 0. This is a result of Oort and Steenbrink "The local Torelli problem for algebraic curves" (1979).

F. Catanese gave a nice overview of the various flavors of Torelli maps (infinitesimal, local, global, generic) in "Infinitesimal Torelli problems and counterexamples to Torelli problems" (chapter 8 in "Topics in transcendental algebraic geometry" edited by Griffiths).

P.S. "Stacks" can be replaced everywhere by the "moduli spaces with level structure of level $l\ge3$ (which are fine moduli spaces).

P.P.S. The space of the first-order deformations of an abelian variety $A$ is $H^1(T_A)$. Since $T_A$ is a trivial vector bundle of rank $g$, and the cotangent space at the origin is $H^0(\Omega^1_A)$, this space equals $H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee}$ and has dimension $g^2$.

A polarization is a homomorphism $\lambda:A\to A^t$ from $A$ to the dual abelian variety $A^t$. It induces an isomorphism (in char 0, or for a principal polarization) from the tangent space at the origin $T_{A,0}=H^0(\Omega_A^1)^{\vee}$ to the tangent space at the origin $T_{A^t,0}=H^1(\mathcal O_A)$. This gives an isomorphism $$ H^1(\mathcal O_A) \otimes H^0(\Omega^1_A)^{\vee} \to H^1(\mathcal O_A) \otimes H^1(\mathcal O_A). $$

The subspace of first-order deformations which preserve the polarization $\lambda$ can be identified with the tensors mapping to zero in $\wedge^2 H^1(\mathcal O_A)$, and so is isomorphic to $Sym^2 H^1(\mathcal O_A)$, outside of characteristic 2.

Answer 2

Yes the Torelli map is an immersion, at least over $\mathbb{C}$. The Torelli theorem says that the map separates points (on the stack level it also says tha the map is representable) and the infinitesimal Torelli theorem says that the map seprates tangent directions. Over $\mathbb{C}$ the infinitesimal Torelli theorem follows easily from Griffiths' infinitesimal period relations.