Torelli theorem over the complex numbers

Statement: $M_g$ is the moduli space for genus $g$ curves and $A_g$ is the moduli space for Abelian varieties of dimension $g$ with a principal polarization, then the map $M_g \to A_g$ associated to a curve its Jacobian is injective.

Concretely if

$\begin{array}{ccc} C_1 & \mbox{} & C_2 \\ \downarrow & \mbox{} & \downarrow \\ J(C_1), E_1 & \cong & J(C_2), E_2 \end{array}$

where $E_i$ is a polarization on the Jacobian –$E \in H^2(J,\mathbb{Z}) = \wedge^2H^1(J,\mathbb{Z})$– then this data suffices to produce an isomorphism $C_1 \cong C_2$.  A seemingly stronger statement is

Thm: if $J, E$ is a prinicipally polarized Abelian variety which is the Jacobian of some curve, then the curve can be recovered from the data $J,E$.

Remark: The above seems stronger to me because it doesn’t explicitly mention the curve, however it requires knowing that $J$ is a Jacobian; in general determining if a principally polarized Abelian variety is a Jacobian is very difficult.

The following outline follows the proof given in Griffiths and Harris.

Proof:

Step 1 The data $J, E$ determine a divisor $\Theta$ and a map $G \colon \Theta^{smooth} \to Gr(g-1,g) = (\mathbb{P}^{g-1})^*$ where $Gr(k,n)$ is the Grassmanian of k-planes in $\mathbb{C}^n$.

The data $E$ determines a Hermitian metric and ultimately an ample line bundle $O(E)$, one description of $\Theta$ is simply the zero locus of a section of $O(E)$.

Aside: note by the Riemann-Roch Theorem for Abelian Varieties [pg 54 jmilne abelian varieties], $h^0(O(E)) = 1$, and for different choices of sections we get linearly equivalent effective divisors, but there is a unique effective divisor giving the line bundle, so $\Theta$ is well defined.

The map $G \colon \Theta^{smooth} \to Gr(g-1, g)$ is the Gauss map of $\Theta \subset J$.  In particular, because $J$ is a group, I can translate and naturally identify all the tangent spaces with the one at the identity $T_eJ \cong \mathbb{C}^g$.  The map is $p \mapsto T_p\Theta \subset T_pJ \cong \mathbb{C}^g$.

Step 2: Understand the geometry of $\Theta$ via the symmetric power $S^{g-1}C$

Riemann’s Theorem on the theta divisor. Denote by $S^dC$ the symmetric power of a curve $C$; set theoretically this is the space of effective divisors of degree d.  Fix a base point $p_0 \in C$ and a basis $w_1,..., w_g \in H^0(\Omega_C)$.  There is a map $\phi^d\colon S^dC \to J$ defined by

$\sum_i p_i \mapsto \bigl(\sum_i \int_{p_0}^{p_i} w_1,..., \sum_i \int_{p_0}^{p_i}w_g \bigr)$ (0)

or algebraically: think of $J$ as line bundles of degree 0 then $\phi^d(\sum_i p_i) = O(\sum_i p_i - dp_0)$.

Thm1: For a suitable constant $\kappa \in J$, $\Theta = W_{g-1} + \kappa$ where $W_{g-1} = Im(\phi^{g-1})$.

Aside: no proof yet but at least heres why $W_{g-1}$ is a divisor.

Roughly, $\dim S^{g-1}C = g-1$ so it suffices to show $\phi^{g-1}$ is generically 1-1, this shows $W_{g-1}$ has the same dimension, i.e. codimension 1.  Now

$(\phi^d)^{-1}(q) = |D|$.  (1)

where $D \in S^dC$ and $q = \phi^d(D)$.  Remains to show $h^0(D) = 1$ iff $h^0(K-D) = 1$ by Riemann-Roch.  The following is a standard result that I’ll add a reference to or a proof of later

Lemma: If $h^0(D)>0$ then for all but finitely many points $h^o(D-p) = h^0(D)-1$.

Apply the lemma to $K$ to get an open set $U$ with the property that $h^0(K-P) = h^0(K) - 1$ for all $P \in U$.  Let $\Delta_{ij} \colon C \to S^{g-1}$ be the diagonal map on the ith and jth factors.  Let $U_{ij} = S^{g-1} - \Delta_{ij}(C)$ and finally let $V = \cap_{i\ne j} U_{ij}$; it is an open dense set and $h^o(D) = 1$ for all $D \in V$.

In relation to the main theorem the significance of this result is that $S^{g-1}C \to J$ carries a lot of useful information about the geometry of the curve and so up to a constant this useful information is sitting in $\Theta$ which only depends on $E$.

This is where the need to break into cases arises.  For $g = 1$, $C \cong J(C)$ and there is nothing to prove.

Step 3: Describe the smooth locus $\Theta^{smooth}$ for the case $g(C) \ge 3$ and $C$ is non hyperelliptic.

Using thm1 we identify points of $\Theta$ with points of $W_{g-1}$; for $q \in \Theta$ write $q = \mu(D):= \phi^{g-1}(D)$ for $D \in S^{g-1}C$.

Claims

a) $\mu(D) \in \Theta$ is smooth iff $\dim |D| = 0$; such $D$ are sometimes called regular.

b) For $D$ regular, can identify $T_{\mu(D)} \Theta$ with the span of $D$:

The span of $D$ and Geometric Version of Riemann-Roch

For $C$ as stated the canonical divisor $K$ gives embedding into $\mathbb{P}^{g-1}$.

For $D$ a divisor, set span $D = \cap E$ where $E$ ranges in $\{E \in |K| : E \ge D\}$.  Can check $\dim span(D) + \dim |K-D| = g-2$ (as projective spaces). Whence the geometric version of the Riemann-Roch Theorem

$\dim |D| = \deg D - \dim span(D) -1$ (2)

Proofs a,b:

Case 1: $q = \mu(D)$ where $D = p_1 + ... + p_{g-1}$ are distinct.  Let $z_i$ local coordinates near $p_i$, then $(z_1, ..., z_{g-1})$ is a local chart near $D$.  In view of (0), if $x_i$, are coordinates on $J$ the map on tangent spaces is

$\frac{\partial}{\partial z_i} \mapsto W_1(p_i)\frac{\partial}{\partial x_1} + ... + W_g(p_i) \frac{\partial}{\partial x_g}$

Where $W_idz_i = w_i$ locally on the curve; $[W_1(p_i):...:W_g(p_i)]$ is a point on the canonical curve and b) follows.

The Jacobian of this map is a matrix whose columns (or rows) consists of a basis of (the unprojectivized) $span(D)$ which has (uprojectivized) dimension $g-1$ when $\dim |D| = 0$.  In other words, the Jacobian of the map has maximal rank so its locally an isomorphism and $S^{g-1}C$ is smooth, hence the target point is a smooth point.

Case 2: $D$ of the form $D = 2p_1 + ... + p_{g-2}$.  Geometrically, $span(D)$ is now the smallest linear space containing the point of $D$ plus a tangent direction.

Same approach works but have to use symmetric polynomials for local chart i.e. $(\frac{z_1+z_2}{2}, z_1z_2, z_3,..., z_{g-1})$.  The upshot is the rows of the Jacobian look like $\phi_K(p_1), \phi'_K(p_1),..., \phi_K(p_{g-1})$.

Have shown $\dim |D| = 0$ implies $\mu(D)$ is a smooth point.  If $|D|$ is positive dimensional then there is positive dimensional family of points mapping to $\mu(D)$ and the tangent space contains the union of $D\mu(T_ES^{g-1}C)$ for $E \in |D|$.  This is a secant variety and not contained in a proper linear subspace, i.e. the tangent space at $\mu(D)$ is too big, i.e. not a smooth point.

Step 4: Let $B$ be the branch locus of $G \colon \Theta^{smooth} \to Gr(g-1,g)$.  Then $\bar B = C^*$; $C^*$ is the set of hyperplanes that contain a tangent line of $C$.

In light of step 3, points in the image of $G$ correspond to hyperplane sections of the canonical curves.  A general hyperplane section has $2g-2$ points of the curve and any $g-1$ of those points determine the hyperplane, i.e. ${2g -2 \choose g-1}$ divisors or points of $\Theta$ whose image under $G$ map to the given hyperplane section.  In other words $G$ is a finite map of degree ${2g-2 \choose g-1}$.

If $E \in |K|$ is not tangent at any point to $C$ its not hard to show that in a small nbd $G$ has ${2g-2 \choose g-1}$ pre images.  So $B \subset C^*$.  There is a significant subtlety but can show $B = C^*$ on $Gr(g-1,g) - V$ where $V$ is the union of hyperplane not intersecting $C$ in general position.  Then use an incidence correspondence to show that $C^*$ is irreducible\$, i.e. it is the image of

$I = \{(p,H)| T_pC \subset H\} \subset C \times Gr(g-1,g)$

and $I$ maps to $C$ with irreducible fibers, which implies $I$ is irreducible, and ultimately $\bar B = C^*$.

Step 5: $C^* = C^{'*}$ implies $C \cong C'$

I would prefer a statement of the form: knowing $C^*$ is the dual to some curve allows you to produce a curve $C_1$ such that $C_1^* = C^*$.

The idea: For $H \in C^{'*} = C^*$ have $H\cap C'$ meets a point with multiplicity at least 2.  However, for a generic $H \in C^*$ such that $T_pC \subset H$, Bertini’s theorem implies $H \cap C'$ is smooth outside of $C' \cap T_pC$.  But $H\cap C'$ must have multiplicity at least 2 somewhere, so it must occur in $T_pC \cap C'$. Therefore $T_pC$ is a tangent line of $C'$.

The rest of the argument breaks into cases showing

$C \to C'$ (2)

$p\mapsto p'$

$T_pC = T_{p'}C'$

gives the desired iso.

For the case $g>3$ this follows from the fact that no line is tangent to $C$ at two points:

Suppose on the contrary $T_pC = T_qC$.  Consider $D = 2p + 2q$.  Then $\dim span(D) = 1$ as a projective line.  So geo Riemann-Roch says $h^0(D) = 4 - 1 = 3$.  But then $2 = |D| = 1/2 \deg D$.  By Clifford’s theorem equality holds iff $D = 0,K$ or $C$ is hyper elliptic and $D$ is a multiple of $g^1_2$.  For $g>3$ we can rule out $D = K$ and this contradicts that $C$ is not hyperelliptic.

For $g = 3$, the canonical embedding gives a quartic curve in $\mathbb{P}^2$ either I can argue there are only finitely many bitangents by, for example, appealing to equations and use a similar argument as above, or I can use that for curves in $\mathbb{P}^2$, $(C^*)^* = C$.

Step 6: $C$ is hyperelliptic.  In this case it is standard [Hartshorne IV.5.3] that the canonical map $\phi_K \colon C \to \mathbb{P}^{g-1}$ factors as $\phi_K = \psi\circ g^1_2$ where $\psi$ is the $(g-1)$ uple embedding.  Consequently

a) $\phi_K(C) \cong \mathbb{P}^1$

b) $B$ is the union of $C^*$ plus $\{H| p \in H \subset \mathbb{P}^2\}$ as $p$ ranges over the branch points of $\phi_K$

c) The above gives the data of a double cover of $\mathbb{P}^1$ branched along the branch locus of $\phi_K$, and this suffices to determine the original curve.

For hyperelliptic curves the branch locus $B$ consists of $C^*$ plus a locus associated to the branch locus of the canonical embedding.  In effect this determines a double cover of $\mathbb{P}^1$ branched at $2g-2$ points, and this is enough to recover the curve.  See Griffiths and Harris for details.