# Martin Orr's Blog

## Finiteness theorems for abelian varieties

Posted by Martin Orr on Monday, 19 September 2011 at 16:34

Faltings famously proved the Mordell, Shafarevich and Tate conjectures in 1983. In this post I will discuss the relationships between the Tate and Shafarevich conjectures and some other finiteness theorems for abelian varieties.

Everything which I call a conjecture in this post is known to be true: they all follow from Finiteness Theorem I. Proving Finiteness Theorem I was the bulk of Faltings' work, but I am not going to talk about that today.

Finiteness Theorem I. Given a number field and an abelian variety defined over , there are only finitely many isomorphism classes of abelian varieties defined over and isogenous to .

### Overview

Here is a diagram of the standard route by which these conjectures are proved (a little simpler than the path taken by Faltings).

At the centre of these results is the Shafarevich conjecture for abelian varieties.

Shafarevich conjecture for abelian varieties. Given a number field , a finite set of primes of , and an integer , there are only finitely many isomorphism classes of abelian varieties of dimension , defined over and having good reduction outside .

The Shafarevich conjecture for curves (Shafarevich's original conjecture) is the same thing, but with "abelian varieties of dimension " replaced by "nonsingular complete curves of genus ", with the condition that if then we only consider curves having a -rational point (elliptic curves). The Mordell conjecture, that a curve of genus over a number field has finitely many -points, is outside the scope of this post.

The Shafarevich conjecture for abelian varieties is equivalent to the following pair of theorems.

Finiteness Theorem I. Given a number field and an abelian variety defined over , there are only finitely many isomorphism classes of abelian varieties defined over and isogenous to .

Finiteness Theorem II. Given a number field , a finite set of primes of , and an integer , there are only finitely many isogeny classes of abelian varieties of dimension , defined over and having good reduction outside .

The surprising thing, which I will sketch in a little more detail below, is that Finiteness Theorem I implies Finiteness Theorem II, passing through the Tate conjecture and semisimplicity theorem.

### The semisimplicity theorem and the Tate conjecture

I will now sketch the proof that Finiteness Theorem I implies the semisimplicity theorem and the Tate conjecture.

Semisimplicity theorem. Let be an abelian variety over a number field . Then is semisimple as a representation of (that is, it is a direct sum of irreducible representations).

Tate conjecture. Let be an abelian variety over a number field . Then .

Any endomorphism of induces an endomorphism of , so there is a homomorphism of -modules . It is injective because an endomorphism of which vanishes at all -torsion points, for all , must be . So Tate's conjecture is really the claim that this map is surjective.

To prove semisimplicity, we need to show that if and is a Galois-invariant subspace of , then there exists an idempotent such that . Then splits as the direct sum .

The key step is the following lemma:

Lemma. Let be a Galois-invariant subspace. Then there exists such that .

In this lemma, lies not just in but in the apparently smaller . This is important because:

1. We know that is a product of matrix rings of division algebras. This allows us to upgrade our to an idempotent satisfying , as required to prove semisimplicity.

2. The Tate conjecture is all about showing that things lie in .

To prove the lemma, we construct abelian varieties and isogenies (defined over ) such that

By Finiteness I, the belong to only finitely many isomorphism classes so there is an infinite set such that all the (for ) are isomorphic. We use the () to construct elements such that some subsequence converges (in the -adic topology) to an element which has the desired property .

Once the lemma is proved, the semisimplicity theorem follows as in point 1 above.

To prove the Tate conjecture, let . We use the double centraliser theorem from algebra and apply the lemma to the graph of , a Galois-invariant subspace of .

### Finiteness Theorem II

Finally I will sketch the proof that semisimplicity and the Tate conjecture imply Finiteness Theorem II.

The Tate conjecture has some important corollaries: by applying it to , we deduce that if and are two abelian varieties over , then . We further deduce that if and are two abelian varieties such that as Galois representations, then and are isogenous.

Hence to prove Finiteness Theorem II it suffices to show that the -adic Tate modules of abelian varieties of dimension over , with good reduction outside , fall into finitely many isomorphism classes.

The idea here is that a semisimple representation of a group is determined by the traces of for all . In the case of a Galois representation, Frobenius elements are dense in , so it suffices to use traces of Frobenius elements.

A clever argument shows that we only need to look at finitely many Frobenius elements. Precisely, given and and an integer , there is a finite set of primes of such that a semisimple -representation of of dimension unramified outside is determined by the traces of Frobenius elements of primes in . This argument ultimately relies on Hermite's theorem that there are finitely many extensions of of bounded degree unramified outside .

The Weil conjectures imply that for each prime , the trace of the Frobenius at can take only finitely many values. So we are done.

1. The Faltings height of an abelian variety over $\mathbb{Q}$ From Martin's Blog

The Faltings height is a real number attached to an abelian variety (defined over a number field), which is at the centre of Faltings’ proof of Finiteness Theorem I. In this post all I will do is define the Faltings height of an abelian vari...

2. The Masser-Wüstholz period theorem From Martin's Blog

I wanted to write a post about the Masser-Wüstholz isogeny theorem, which gives a quantitative version of Finiteness Theorem I. But it turned out to be too long so for today I will focus on the main ingredient in the proof of the isogeny theorem:...

3. The Masser-Wüstholz isogeny theorem From Martin's Blog

Let and be two isogenous abelian varieties over a number field . Can we be sure that there is an isogeny between them of small degree, where “small” is an explicit function of and ? In particular, our bound should not depend on ; this ...

4. Tate classes From Martin's Blog

In my last post I talked about Hodge classes on abelian varieties. Today I will talk about the analogue in l-adic cohomology, called Tate classes. These are defined to be those classes on which the action of the Galois group is given by multiplying ...

1. Barinder Banwait said on Saturday, 24 September 2011 at 20:18 :

Great Post Martin! I especially like that diagram, it looks like it took you a long time to TeX/HTML.

1. The proof of your lemma is rather sketchy. How does one construct the and satisfying that property? And how about using to construct satisfying the subsequence property? Neither of these two steps is clear to me.

2. I was also confused by your "To prove the Tate conjecture" paragraph. Applying the lemma to the graph of furnishes a such that But where does the come from, whose graph on is the same as the graph of ?

4. Because this is an important point: how does as Galois modules imply that and are isogenous?

5. In your 'Clever Argument' paragraph, does Chebotarev's theorem come in? Does Weil Restriction of Scalars come in?

2. Martin Orr said on Tuesday, 27 September 2011 at 09:31 :

I made the diagram in LibreOffice Draw. I agree that my proofs are rather sketchy, as I thought it would be too long if I wrote more. For this morning I will just deal with your question 3.

We use that every right ideal of a product of matrix rings of division algebras is generated by an idempotent. This is because any right ideal of a matrix ring of a division algebra consists of those matrices which are zero in some specified rows and anything in the other rows. Such an ideal is generated by the matrix with 1s in the appropriate places on the diagonal and zeroes elsewhere. Apply this to the right ideal

3. Martin Orr said on Wednesday, 28 September 2011 at 09:52 :

Now I will answer question 1. Let .

(In the original text of the blog, I made a typo in the statement of the lemma, putting instead of . This is now fixed.)

For the construction of and , we use the following general lemma.

Lemma. Let be a Galois-invariant sub--module of of finite index. Then there exists an abelian variety and an isogeny (defined over ) such that .

Proof. Note that being of finite index is equivalent to containing for some . Choose such an and let be the reduction of modulo .

Then is a finite subgroup of defined over . We can take to be the quotient . Now is contained in the kernel of so factors through . This gives us . You can check that .

Then to construct , let be the smallest element of . For each , we can choose an isomorphism Then is an isogeny satisfying .

To get an element of , we compose this with . Of course is not usually invertible as an isogeny, but it has an inverse in . We set and this satisfies .

In particular so . Now is compact (it is a finite rank -module) so some subsequence of converges, say to .

Taking the limit of gives , and multiplying by gives .

4. Martin Orr said on Thursday, 29 September 2011 at 11:32 :

For your question 2: in proving the Tate conjecture, we do not actually construct a inducing . Instead we use the double centraliser theorem. I tend to remember the theorem like this, though this is not the most general possible statement:

Theorem. Let be a finite-dimensional vector space and a subalgebra of . Let (that is, the set of endomorphisms of which commute with ) and let . If is semisimple as an -module, then .

We apply this theorem with and . By the theorem, it will suffice to show that , i.e. that all commute with all .

Note that . Since , commutes with all of . In particular, it commutes with the such that . Hence .

So for any , . But contains only one element of the form , so this implies that . So commutes with as required.

5. Martin Orr said on Friday, 30 September 2011 at 16:50 :

Your question 4 is indeed important, and it looks like it should be quite easy. You can deduce from the Tate conjecture applied to . So an isomorphism comes from something in . The only problem is the tensoring with .

The injections in form an open set, and by hypothesis this set is non-empty. By the Tate conjecture, is dense in . Hence there is an element which induces an injection .

By multiplying by an integer, we may assume that . Now the injectivity of implies that has finite kernel, and the surjectivity of implies that is surjective.

6. Martin Orr said on Saturday, 01 October 2011 at 15:20 :

And for question 5, yes the clever argument uses the Chebotarev density theorem. But I think that the clever bit is the use of Nakayama’s lemma to allow you to look at the mod representation, which is finite, instead of the -adic representation.

More precisely: let be a semisimple -representation of of dimension , with being the -module on which it acts. Let be the sub--module of generated by the image of .

Let be the representation obtained by reducing mod . The image of is finite, so we can take a finite set which hits all of the image of . Then generates as a -module, so by Nakayama's lemma generates as a -module.

However the set may vary from one representation to another. This is not a problem because the representation factors through for some field of degree at most over , and by Hermite's theorem there are only finitely many such fields unramified outside . Let be the composite of all these fields, and let be a finite subset of which surjects onto . (Using Chebotarev we can find primes whose Frobenii give us .)

Then for any , is a -linear combination of . This is still not enough to prove the theorem, because the coefficients of this combination may depend on .

To solve this we work with two representations and simultaneously, and apply the above argument to their direct sum . Let be the sub--module of generated by the image of . The mod representation factors through so by the previous argument, generates as a -module. This implies that for every , we can write as a linear combination of and as a linear combination of with the same coefficients. So if the traces of and agree on , then they agree everywhere, as required.

7. Barinder Banwait said on Monday, 10 October 2011 at 21:51 :

Thanks a lot for those clarifications Martin!

Regarding your first rebuttal, about my question 3, your argument gives us an such that , and I couldn't see why it had to be equality, until I realised that this follows from the existence of the in the right ideal with .

Also, there is a very small typo in Comment number 3, where you construct ; it should read  is an isogeny such that ."