Martin Orr's Blog

Maths > Abelian varieties > Finiteness theorems and the Faltings height

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 K and an abelian variety A defined over K, there are only finitely many isomorphism classes of abelian varieties defined over K and isogenous to A.


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

Diagram of implications among conjectures proved by Faltings

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

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

The Shafarevich conjecture for curves (Shafarevich's original conjecture) is the same thing, but with "abelian varieties of dimension g" replaced by "nonsingular complete curves of genus g", with the condition that if g = 1 then we only consider curves having a K-rational point (elliptic curves). The Mordell conjecture, that a curve of genus g \geq 2 over a number field K has finitely many K-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 K and an abelian variety A defined over K, there are only finitely many isomorphism classes of abelian varieties defined over K and isogenous to A.

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

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 A be an abelian variety over a number field K. Then T_\ell A \otimes \mathbb{Q}_\ell is semisimple as a representation of \operatorname{Gal}(\bar{K}/K) (that is, it is a direct sum of irreducible representations).

Tate conjecture. Let A be an abelian variety over a number field K. Then \operatorname{End}_{\mathrm{Gal}(\bar{K}/K)} T_\ell A \cong \operatorname{End} A \otimes \mathbb{Z}_\ell.

Any endomorphism of A induces an endomorphism of T_\ell A, so there is a homomorphism of \mathbb{Z}_\ell-modules \operatorname{End} A \otimes \mathbb{Z}_\ell \to \operatorname{End}_{\mathrm{Gal}} T_\ell A. It is injective because an endomorphism of A which vanishes at all \ell^n-torsion points, for all n, must be 0. So Tate's conjecture is really the claim that this map is surjective.

To prove semisimplicity, we need to show that if V = T_\ell A \otimes \mathbb{Q}_\ell and W is a Galois-invariant subspace of V, then there exists an idempotent e \in \operatorname{End}_\mathrm{Gal} V such that eV = W. Then V splits as the direct sum eV \oplus (1-e)V.

The key step is the following lemma:

Lemma. Let W \subseteq V be a Galois-invariant subspace. Then there exists u \in \operatorname{End} A \otimes \mathbb{Q}_\ell such that uV = W.

In this lemma, u lies not just in \operatorname{End}_\mathrm{Gal} V but in the apparently smaller \operatorname{End} A \otimes \mathbb{Q}_\ell. This is important because:

  1. We know that \operatorname{End} A \otimes \mathbb{Q}_\ell is a product of matrix rings of division algebras. This allows us to upgrade our u \in \operatorname{End} A \otimes \mathbb{Q}_\ell to an idempotent e satisfying eV = W, as required to prove semisimplicity.

  2. The Tate conjecture is all about showing that things lie in \operatorname{End} A \otimes \mathbb{Q}_\ell.

To prove the lemma, we construct abelian varieties B_n and isogenies \phi_n : B_n \to A (defined over K) such that  \phi_n(T_\ell B_n) = (T_\ell A \cap W) + \ell^n T_\ell A.

By Finiteness I, the B_n belong to only finitely many isomorphism classes so there is an infinite set I \subseteq \mathbb{N} such that all the B_n (for n \in I) are isomorphic. We use the \phi_n (n \in I) to construct elements u_n \in \operatorname{End} A \otimes \mathbb{Q}_\ell such that some subsequence converges (in the \ell-adic topology) to an element u which has the desired property uV = W.

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

To prove the Tate conjecture, let \alpha \in \operatorname{End}_\mathrm{Gal} V. We use the double centraliser theorem from algebra and apply the lemma to the graph \{ (v, \alpha v) | v \in V \} of \alpha, a Galois-invariant subspace of T_\ell(A \times A) \otimes \mathbb{Q}_\ell.

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 A \times B, we deduce that if A and B are two abelian varieties over K, then \operatorname{Hom}_{\mathrm{Gal}} (T_\ell A, T_\ell B) \cong \operatorname{Hom} (A, B) \otimes \mathbb{Z}_\ell. We further deduce that if A and B are two abelian varieties such that T_\ell A \otimes \mathbb{Q}_\ell \cong T_\ell B \otimes \mathbb{Q}_\ell as Galois representations, then A and B are isogenous.

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

The idea here is that a semisimple representation \rho of a group G is determined by the traces of \rho(g) for all g \in G. In the case of a Galois representation, Frobenius elements are dense in \mathrm{Gal}(\bar{K}/K), 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 K and S and an integer d, there is a finite set T of primes of K such that a semisimple \mathbb{Q}_\ell-representation of \mathrm{Gal}(\bar{K}/K) of dimension d unramified outside S is determined by the traces of Frobenius elements of primes in T. This argument ultimately relies on Hermite's theorem that there are finitely many extensions of K of bounded degree unramified outside S.

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

Tags abelian-varieties, alg-geom, faltings, maths, number-theory


  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 B_n and \phi_n satisfying that property? And how about using \phi_n to construct u_n 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 \alpha furnishes a u \in \mbox{End}(A \times A) \otimes \mathbb{Q}_l such that u \cdot T_l(A \times A) \otimes \mathbb{Q}_l = \left\{ (v,\alpha v) : v \in V \right\}. But where does the U \in \mbox{End}(A) \otimes \mathbb{Q}_l come from, whose graph on V is the same as the graph of \alpha?

    3. How do you upgrade your u to an idempotent e?

    4. Because this is an important point: how does V_l(A) \cong V_l(B) as Galois modules imply that A and B 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  \{ a \in \operatorname{End} A \otimes \mathbb{Q}_\ell | aV \subseteq W \}.

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

    Now I will answer question 1. Let W_n = (T_\ell A \cap W) + \ell^n T_\ell A.

    (In the original text of the blog, I made a typo in the statement of the lemma, putting (T_\ell A \cap W) + \ell^n W instead of (T_\ell A \cap W) + \ell^n T_\ell A. This is now fixed.)

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

    Lemma. Let U be a Galois-invariant sub-\mathbb{Z}_\ell-module of T_\ell A of finite index. Then there exists an abelian variety B and an isogeny \phi : B \to A (defined over K) such that \phi(T_\ell B) = U.

    Proof. Note that U being of finite index is equivalent to containing \ell^n T_\ell A for some n. Choose such an n, and let \tilde{U} be the reduction of U modulo \ell^n.

    Then \tilde{U} is a finite subgroup of A(\bar{K}) defined over K. We can take B to be the quotient A/\tilde{U}. Now \tilde{U} is contained in the kernel of [\ell^n] : A \to A so [\ell^n] factors through A/\tilde{U}. This gives us \phi : A/\tilde{U} \to A. You can check that \phi(T_\ell B) = U.

    Then to construct u_n, let n_0 be the smallest element of I. For each n \in I, we can choose an isomorphism  f_n : B_{n_0} \to B_n. Then \phi_n f_n is an isogeny B_{n_0} \to A satisfying \phi_n(T_\ell B_{n_0}) = W_n.

    To get an element of \operatorname{End} A \otimes \mathbb{Q}_\ell, we compose this with \phi_{n_0}^{-1}. Of course \phi_{n_0} is not usually invertible as an isogeny, but it has an inverse in \operatorname{Hom}(A, B_{n_0}) \otimes \mathbb{Q}. We set  u_n = \phi_n f_n \phi_{n_0}^{-1} \in \operatorname{End} A \otimes \mathbb{Q} and this satisfies u_n(W_{n_0}) = W_n.

    In particular W_n \subseteq W_{n_0} so u_n \in \operatorname{End}(W_{n_0}). Now \operatorname{End}(W_{n_0}) is compact (it is a finite rank \mathbb{Z}_\ell-module) so some subsequence of u_n \; (n \in I) converges, say to u \in \operatorname{End}(A) \otimes \mathbb{Q}_\ell.

    Taking the limit of u_n(W_{n_0}) = W_n gives u(W_{n_0}) = W \cap T_\ell A, and multiplying by \mathbb{Q}_\ell gives uV = W.

  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 U \in \operatorname{End} A \otimes \mathbb{Q}_\ell inducing \alpha. 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 V be a finite-dimensional vector space and R a subalgebra of \operatorname{End} V. Let R' = \operatorname{End}_R V (that is, the set of endomorphisms of V which commute with R) and let R'' = \operatorname{End}_{R'} V. If V is semisimple as an R-module, then R'' = R.

    We apply this theorem with V = V_\ell A and R = \operatorname{End} A \otimes \mathbb{Q}_\ell. By the theorem, it will suffice to show that \operatorname{End}
_\mathrm{Gal} V = R'', i.e. that all \alpha \in \operatorname{End}
_\mathrm{Gal} V commute with all c \in R'.

    Note that \operatorname{End}(A \times A) \otimes \mathbb{Q}_\ell = \mathrm{M}_2(R). Since c \in R', \bigl( \begin{smallmatrix}
 c & 0 \\ 0 & c
\end{smallmatrix} \bigr)
\in \mathrm{M}_2(\operatorname{End}_{\mathbb{Q}_\ell} V) commutes with all of \operatorname{End}(A \times A) \otimes \mathbb{Q}_\ell. In particular, it commutes with the u such that u(V \times V) = W. Hence \bigl( \begin{smallmatrix}
 c & 0 \\ 0 & c
\end{smallmatrix} \bigr) W
\subseteq W.

    So for any x \in V, (cx, c\alpha x) \in W. But W contains only one element of the form (cx, -), so this implies that c\alpha x = \alpha cx. So \alpha commutes with c 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  \operatorname{Hom}_\mathrm{Gal}(V_\ell A, V_\ell B) \cong \operatorname{Hom}(A, B) \otimes \mathbb{Q}_\ell from the Tate conjecture applied to A \times B. So an isomorphism V_\ell A \to V_\ell B comes from something in \operatorname{Hom}(A, B) \otimes \mathbb{Q}_\ell. The only problem is the tensoring with \mathbb{Q}_\ell.

    The injections in \operatorname{Hom}_\mathrm{Gal}(V_\ell A, V_\ell B) form an open set, and by hypothesis this set is non-empty. By the Tate conjecture, \operatorname{Hom}(A, B) \otimes \mathbb{Q} is dense in \operatorname{Hom}_\mathrm{Gal}(V_\ell A, V_\ell B). Hence there is an element \phi \in \operatorname{Hom}(A, B) \otimes \mathbb{Q} which induces an injection \phi_\ell : V_\ell A \to V_\ell B.

    By multiplying by an integer, we may assume that \phi \in \operatorname{Hom}(A, B). Now the injectivity of \phi_\ell implies that \phi has finite kernel, and the surjectivity of \phi_\ell implies that \phi 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 \ell representation, which is finite, instead of the \ell-adic representation.

    More precisely: let \rho be a semisimple \mathbb{Z}_\ell-representation of \operatorname{Gal}(\bar{K}/K) of dimension d, with V being the \mathbb{Z}_\ell-module on which it acts. Let M be the sub-\mathbb{Z}_\ell-module of \operatorname{End} V generated by the image of \rho.

    Let \bar\rho be the representation obtained by reducing \rho mod \ell. The image of \bar\rho is finite, so we can take a finite set \Sigma \subseteq \operatorname{Gal}(\bar{K}/K) which hits all of the image of \bar\rho. Then \bar\rho(\Sigma) generates M/\ell M as a \mathbb{F}_\ell-module, so by Nakayama's lemma \rho(\Sigma) generates M as a \mathbb{Z}_\ell-module.

    However the set \Sigma may vary from one representation to another. This is not a problem because the representation \bar\rho factors through \operatorname{Gal}(L/K) for some field L of degree at most \ell^{d^2} over K, and by Hermite's theorem there are only finitely many such fields unramified outside S. Let L_0 be the composite of all these fields, and let \Sigma_0 be a finite subset of \operatorname{Gal}(\bar{K}/K) which surjects onto \operatorname{Gal}(L_0/K). (Using Chebotarev we can find primes whose Frobenii give us \Sigma_0.)

    Then for any \sigma \in \operatorname{Gal}(\bar{K}/K), \rho(\sigma) is a \mathbb{Z}_\ell-linear combination of \rho(\Sigma_0). This is still not enough to prove the theorem, because the coefficients of this combination may depend on \rho.

    To solve this we work with two representations \rho and \rho' simultaneously, and apply the above argument to their direct sum \rho^* = \rho \oplus \rho'. Let M^* be the sub-\mathbb{Z}_\ell-module of \operatorname{End}(V \oplus V') generated by the image of \rho^*. The mod \ell representation \bar\rho^* factors through \operatorname{Gal}(L_0/K) so by the previous argument, \rho^*(\Sigma_0) generates M^* as a \mathbb{Z}_\ell-module. This implies that for every \sigma \in \operatorname{Gal}(\bar{K}/K), we can write \rho(\sigma) as a linear combination of \rho(\Sigma_0) and \rho'(\sigma) as a linear combination of \rho'(\Sigma_0) with the same coefficients. So if the traces of \rho and \rho' agree on \Sigma_0, 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 e such that eV \subseteq W, and I couldn't see why it had to be equality, until I realised that this follows from the existence of the u in the right ideal with uV = W.

    Also, there is a very small typo in Comment number 3, where you construct u_n; it should read ``\phi_nf_n is an isogeny such that \phi_nf_n(T_lB_{n_0}) = W_n."

Post a comment

Markdown syntax with embedded LaTeX.
Type LaTeX between dollar signs, and enclose them between backticks to protect it from Markdown.
All comments are subject to moderation before they appear on the blog.