# Torelli theorem over the complex numbers

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

Concretely if

where is a polarization on the Jacobian –– then this data suffices to produce an isomorphism . A seemingly stronger statement is

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

Remark: The above seems stronger to me because it doesn’t explicitly mention the curve, however it requires knowing that 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 determine a divisor and a map where is the Grassmanian of k-planes in .

The data determines a Hermitian metric and ultimately an ample line bundle , one description of is simply the zero locus of a section of .

Aside: note by the Riemann-Roch Theorem for Abelian Varieties [pg 54 jmilne abelian varieties], , and for different choices of sections we get linearly equivalent effective divisors, but there is a unique effective divisor giving the line bundle, so is well defined.

The map is the Gauss map of . In particular, because is a group, I can translate and naturally identify all the tangent spaces with the one at the identity . The map is .

Step 2: Understand the geometry of via the symmetric power

Riemann’s Theorem on the theta divisor. Denote by the symmetric power of a curve ; set theoretically this is the space of effective divisors of degree d. Fix a base point and a basis . There is a map defined by

(0)

or algebraically: think of as line bundles of degree 0 then .

Thm1: For a suitable constant , where .

Aside: no proof yet but at least heres why is a divisor.

Roughly, so it suffices to show is generically 1-1, this shows has the same dimension, i.e. codimension 1. Now

. (1)

where and . Remains to show iff by Riemann-Roch. The following is a standard result that I’ll add a reference to or a proof of later

Lemma: If then for all but finitely many points .

Apply the lemma to to get an open set with the property that for all . Let be the diagonal map on the ith and jth factors. Let and finally let ; it is an open dense set and for all .

In relation to the main theorem the significance of this result is that carries a lot of useful information about the geometry of the curve and so up to a constant this useful information is sitting in which only depends on .

This is where the need to break into cases arises. For , and there is nothing to prove.

Step 3: Describe the smooth locus for the case and is non hyperelliptic.

Using thm1 we identify points of with points of ; for write for .

Claims

a) is smooth iff ; such are sometimes called regular.

b) For regular, can identify with the span of :

The span of and Geometric Version of Riemann-Roch

For as stated the canonical divisor gives embedding into .

For a divisor, set span where ranges in . Can check (as projective spaces). Whence the geometric version of the Riemann-Roch Theorem

(2)

Proofs a,b:

Case 1: where are distinct. Let local coordinates near , then is a local chart near . In view of (0), if , are coordinates on the map on tangent spaces is

Where locally on the curve; 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) which has (uprojectivized) dimension when . In other words, the Jacobian of the map has maximal rank so its locally an isomorphism and is smooth, hence the target point is a smooth point.

Case 2: of the form . Geometrically, is now the smallest linear space containing the point of plus a tangent direction.

Same approach works but have to use symmetric polynomials for local chart i.e. . The upshot is the rows of the Jacobian look like .

Have shown implies is a smooth point. If is positive dimensional then there is positive dimensional family of points mapping to and the tangent space contains the union of for . This is a secant variety and not contained in a proper linear subspace, i.e. the tangent space at is too big, i.e. not a smooth point.

Step 4: Let be the branch locus of . Then ; is the set of hyperplanes that contain a tangent line of .

In light of step 3, points in the image of correspond to hyperplane sections of the canonical curves. A general hyperplane section has points of the curve and any of those points determine the hyperplane, i.e. divisors or points of whose image under map to the given hyperplane section. In other words is a finite map of degree .

If is not tangent at any point to its not hard to show that in a small nbd has pre images. So . There is a significant subtlety but can show on where is the union of hyperplane not intersecting in general position. Then use an incidence correspondence to show that is irreducible$, i.e. it is the image of

and maps to with irreducible fibers, which implies is irreducible, and ultimately .

Step 5: implies

I would prefer a statement of the form: knowing is the dual to some curve allows you to produce a curve such that .

The idea: For have meets a point with multiplicity at least 2. However, for a generic such that , Bertini’s theorem implies is smooth outside of . But must have multiplicity at least 2 somewhere, so it must occur in . Therefore is a tangent line of .

The rest of the argument breaks into cases showing

(2)

gives the desired iso.

For the case this follows from the fact that no line is tangent to at two points:

Suppose on the contrary . Consider . Then as a projective line. So geo Riemann-Roch says . But then . By Clifford’s theorem equality holds iff or is hyper elliptic and is a multiple of . For we can rule out and this contradicts that is not hyperelliptic.

For , the canonical embedding gives a quartic curve in 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 , .

Step 6: is hyperelliptic. In this case it is standard [Hartshorne IV.5.3] that the canonical map factors as where is the uple embedding. Consequently

a)

b) is the union of plus as ranges over the branch points of

c) The above gives the data of a double cover of branched along the branch locus of , and this suffices to determine the original curve.

For hyperelliptic curves the branch locus consists of plus a locus associated to the branch locus of the canonical embedding. In effect this determines a double cover of branched at points, and this is enough to recover the curve. See Griffiths and Harris for details.

## About this entry

You’re currently reading “Torelli theorem over the complex numbers,” an entry on Math Meandering

- Published:
- April 23, 2010 / 11:54 pm

- Category:
- alg. geo., Teleman, wall scribble

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