# Jacobi’s inversion theorem

In this post, I will discuss Jacobi’s inversion theorem. It is a follow up in a series of posts about Riemann surfaces, the first of which being this one. This theorem tells us in what sense the Abel-Jacobi map is surjective.

Theorem (Jacobi’s inversion):Let be a compact Riemann surface of genus and take any . Then for any , we can find points such that.

In other words, for a basis of , for all , there is and paths from to such that

for all .

With this theorem coupled with Abel’s theorem, we will have completely proved the exactness of

.

**Lemma 1: **The set of effective divisors of degree on is a compact complex manifold.

**Proof of lemma 1:** Consider the action of the permutation group on

( times).

The quotient, denoted , inherits the quotient topology making a continuous map. Clearly, is in bijection with . Suppose for a moment that . We will write an element of as a sum to indicate the unimportance of the ordering (note that the ‘s are not necessarily distinct). We consider the map

that takes to the d-tuple consisting of the coefficents of the monic polynomial having as its roots, i.e

if

.

In other words, where is the ith elementary symmetric polynomial. By the fundamental theorem of algbera, is a bijection, and in fact a homeomorphism.

The difficulty is in seeing that is also continuous. To see this, take an open set and consider . In other words,

with .

We need to show that there is an open set . Write

which, when defined, is the number of zeros of the polynomial inside . We can choose an open set containing small enough so that is defined for all . Then since is continuous and takes only integer values, we have

for all .

But this means that for all , all the roots of are in so that as we wanted.

This gives the structure of a complex manifold of dimension , in fact biholomorphic to .

Now we put a similar complex manifold structure on for arbitrary . Consider and take holomorphic charts around on such that if and if . We obviously have an injective mapping

where , by doing as above;

.

The verifications that this gives a homeomorphism onto an open set of is just as in the earlier case. These maps thus provide with a holomorphic atlas.

QED

Fixing a base point we get a (holomorphic) injection

and thus holomorphic mappings

(mod )

by composing and . Jacobi’s inversion theorem says that is surjective.

**Lemma 2: ** Let be a holomorphic map between two compact connected complex manifolds of the same dimension. If is not everywhere singular, i.e. the Jacobian matrix is not identically zero, then is necessarily surjective.

**Proof of lemma 2:** This is immediate from the proper mapping theorem which says that if is an analytic subvariety, then is an analytic subvariety of . In this case, would be a compact subvariety containing an open set which would mean . Griffiths-Harris presents a more elementary proof which does not use the rather deep proper mapping theorem:

Consider a volume form on . Since is not identically and since preserves the orientation (being holomorphic), we have

.

Since for any we have the volume form is exact in and

for some -form on . But then if , we have

,

which contradicts earlier considerations.

QED

To prove the theorem, we thus have to show that is not everywhere singular.

**Proof of the theorem:** At points such that the ‘s are distinct, the quotient map is locally a biholomorphism. So choosing disjoint charts in centered at , we get a chart

.

In such coordinates, for near , we have

mod .

So

.

But with near , this is

hence

The jacobian matrix of near is thus given by

.

It suffices to see that this matrix is of full rank for some choice of and basis . We simply do Gauss reduction: Choose such that . Then subtracting a multiple of to , we make it so that

.

The ‘s are still a basis of and we continue like this, choosing a such that etc… We eventually arrive at the form

which is of maximal rank since for all . This shows that is not everywhere singular, so it is surjective by lemma 2.

QED

Putting everything together, we showed that the set of (isomorphism classes) of topologically trivial holomorphic line bundles (i.e. with zero first Chern class) has the structure of a complex torus of dimension . Indeed, the group consisting of those (isomorphism classes of) holomorphic line bundles is isomorphic to , which by Abel’s and Jacobi’s theorem is isomorphic to .

Note that in fact the fibres of consist of projective spaces. Indeed, if , then by Abel’s theorem the fiber is the set of effective divisors linearly equivalent to , which corresponds to the projectivisation of . It can be shown that generically the fiber of is a point and that is a **birational map**.

## Trackbacks & Pingbacks