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.


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.


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.


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