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Maths > Abelian varieties > Complex abelian varieties and the Mumford-Tate conjecture

Images of Galois representations

Posted by Martin Orr on Saturday, 27 November 2010 at 16:22

In this post, I will continue to talk about the \ell-adic representations attached to abelian varieties, and in particular the images G_\ell of these representations. I will define algebraic groups approximating G_\ell, which are often more convenient to work with. I will end by stating the Mumford-Tate conjecture, linking G_\ell to the Mumford-Tate group.

Let A be an abelian variety of dimension g over a number field K. Recall from last time that the Tate module T_\ell A gives us a representation  \rho_\ell : \operatorname{Gal}(\bar{K}/K) \to \operatorname{GL}_{2g}(\mathbb{Z}_\ell). We define G_\ell to be \rho_\ell(\operatorname{Gal}(\bar{K}/K)), a compact subgroup of \operatorname{GL}_{2g}(\mathbb{Z}_\ell).

The example of elliptic curves

If E is an elliptic curve over a number field K without complex multiplication, then Serre's Open Image Theorem (proved in the 60s) says that G_\ell is an open subgroup of \operatorname{GL}_{2g}(\mathbb{Z}_\ell). Since \operatorname{GL}_{2g}(\mathbb{Z}_\ell) is compact, it follows that G_\ell has finite index in \operatorname{GL}_{2g}(\mathbb{Z}_\ell).

Next suppose that E is an elliptic curve over a number field K with complex multiplication by F = \mathbb{Q}(\sqrt{-d}). Recall that \mathfrak{o}_F^\times can be embedded in \operatorname{GL}_2(\mathbb{Z}) as the invertible matrices of the form \bigl( \begin{smallmatrix} u & -dv \\ v & u \end{smallmatrix} \bigr). This extends to an embedding of (\mathfrak{o}_F \otimes_{\mathbb{Z}} \mathbb{Z}_\ell)^\times in \operatorname{GL}_2(\mathbb{Z}_\ell). Let R_\ell be the image of this embedding. (The structure of R_\ell depends on whether -d has a square root in \mathbb{Z}_\ell or not. If yes, then R_\ell is conjugate to the diagonal subgroup of \operatorname{GL}_2(\mathbb{Z}_\ell). If no, then R_\ell is isomorphic to the group of units of \mathbb{Z}_\ell[\sqrt{-d}].)

Now if K \supseteq F, then it follows from the main theorem of complex multiplication that G_\ell is an open subgroup of R_\ell. If K \not\supseteq F, then G_\ell has an index-2 subgroup which is an open subgroup of R_\ell, but G_\ell is not itself contained in R_\ell.

If you compare the above groups G_\ell with the Mumford-Tate groups of elliptic curves, you will notice that they are rather similar, as long as you ignore the complications about "finite-index subgroups". There is another reason to not care about finite-index subgroups: let A be an abelian variety defined over a number field K. It is often convenient to be able to replace K by a finite extension L; but if we do this, then G_\ell gets replaced by a finite-index subgroup (because \operatorname{Gal}(\bar{L}/L) is a finite-index subgroup of \operatorname{Gal}(\bar{K}/K)).

Algebraic envelope of G_\ell

Let G_\ell^{alg} be the smallest algebraic subgroup (over \mathbb{Q}_\ell) of \operatorname{GL}_{2g} such that G_\ell^{alg}(\mathbb{Q}_\ell) contains G_\ell. This is sometimes given the unwieldy name of \ell-adic algebraic monodromy group, but I am not sure whether anyone other than Pink and my supervisor uses this name.

A couple of reasons why G_\ell^{alg} is sometimes more useful than G_\ell:

  1. It is easier to compute G_\ell^{alg} because of the structure theorems for algebraic groups.

  2. Algebraic groups have few finite-index algebraic subgroups, so the problem that the group changes when you enlarge the base field is reduced, although not eliminated.

Let E be an elliptic curve over K.

  1. If E does not have CM, then G_\ell^{alg} = \operatorname{GL}_2.

  2. If E has CM by the field F and K \supseteq F, then G_\ell^{alg} is the \mathbb{Q}_\ell-algebraic-group version of (F \otimes_{\mathbb{Q}} \mathbb{Q}_\ell)^\times.

  3. If E has CM by F and K \not\supseteq F, then G_\ell^{alg} has the \mathbb{Q}_\ell-algebraic-group version of (F \otimes_{\mathbb{Q}} \mathbb{Q}_\ell)^\times as an index-2 subgroup.

If G_\ell^{alg} is to be a good approximation to G_\ell, it is important to know that G_\ell^{alg} is not too much bigger than G_\ell. Let G_\ell^{alg}(\mathbb{Z}_\ell) denote the elements of G_\ell^{alg}(\mathbb{Q}_\ell) such that both the matrix in question and its inverse have all their entries in \mathbb{Z}_\ell. Bogomolov proved in 1980 that G_\ell is always an open subgroup of G_\ell^{alg}(\mathbb{Z}_\ell) in the \ell-adic topology, and so is of finite index.

Connected components of G_\ell^{alg}

Any algebraic group G has finitely many connected components in the Zariski topology. The connected component containing the identity is a normal subgroup of G. (In algebraic geometry, we usually work with irreducible components rather than connected components; but for algebraic groups irreducible components and connected components are the same, and we usually call them connected components, because "irreducible" has a different meaning in the context of representations.)

Let G_\ell^\circ be the connected component of G_\ell^{alg} containing the identity. Connected algebraic groups do not have any finite-index algebraic subgroups (because they are also irreducible, so not a union of finitely many closed subsets). So G_\ell^\circ is unchanged if we enlarge the base field K.

To summarise: neither G_\ell^\circ(\mathbb{Z}_\ell) nor G_\ell is necessarily contained in the other, but both are finite-index subgroups of G_\ell^{alg}(\mathbb{Z}_\ell).

The Mumford-Tate conjecture

G_\ell^\circ is a connected algebraic group over \mathbb{Q}_\ell, and the Mumford-Tate group M is a connected algebraic group over \mathbb{Q}. The Mumford-Tate conjecture states that G_\ell^\circ is equal to M \times_{\mathbb{Q}} \mathbb{Q}_\ell (the latter just means M viewed as an algebraic group over \mathbb{Q}_\ell).

In order for it to make sense to say that these groups are equal, and not merely isomorphic, we need them both to be subgroups of some common group. By definition, M \subseteq \operatorname{GL}(V) and G_\ell \subseteq \operatorname{GL}(V_\ell) where V = H_1(A, \mathbb{Q}) and V_\ell = T_\ell A \otimes_{\mathbb{Z}_\ell} \mathbb{Q}_\ell. We saw last time that T_\ell A is canonically isomorphic to \Lambda \otimes_{\mathbb{Z}} \mathbb{Z}_\ell. Tensoring this isomorphism with \mathbb{Q} gives a canonical isomorphism between V \otimes_{\mathbb{Q}} \mathbb{Q}_\ell and V_\ell, which we use to identify \operatorname{GL}(V) \times_{\mathbb{Q}} \mathbb{Q}_\ell and \operatorname{GL}(V_\ell).

From the examples, we see that the Mumford-Tate conjecture is true for elliptic curves (at least that the groups are isomorphic; you need a little more to check that they are equal as subgroups of \operatorname{GL}_2 in the CM case). In general, Deligne showed using Hodge-theoretic techniques that G_\ell^\circ is always contained in M \times \mathbb{Q}_\ell. Various other cases are known when restrictions are placed on the dimension and endomorphism ring of A, for example whenever A has CM, or when \dim A is odd and \operatorname{End} A = \mathbb{Z}.

Tags abelian-varieties, alg-geom, alg-groups, hodge, maths


  1. Deligne's Principle A and the Mumford-Tate conjecture From Martin's Blog

    In this post I will fill in a missing detail from two weeks ago, where I mentioned that the Mumford-Tate group is determined by the Hodge classes. More precisely, I will show that an element g in GL(H_1(A, Z)) is in the Mumford-Tate group if and ...


  1. Barinder said on Sunday, 28 November 2010 at 19:40 :

    Great Post Martin! Here are two questions just to ensure that I'm still with you:

    1. Is G_l^{alg} the smallest algebraic subgroup of GL_{2g} defined over \mathbb{Q} such that G_l \subset G_l^{alg}(\mathbb{Q}_l)? Therefore, is G_l^{alg}(\mathbb{Q}_l) \subset GL_{2g}(\mathbb{Z}_l)?

    2. Is the following true?

    G_l^{alg}(\mathbb{Z}_l) = \{ M \in G_l^{alg}(\mathbb{Q}_l) : det(M) \in \mathbb{Z}_l^{\ast}\}

    Here are some other questions/comments:

    1. You briefly mentioned a property of a representation - "irreducible". I guess this means that the Tate module T_l(A), as a Galois module, does not decompose. Is irreducibility of an l-adic representation equivalent to it being surjective? If not, is there an easy example I can hold in my head? (I think surjective implies irreducible, but not sure about converse).

    2. Is there a way to think about these G_l for lots of ls and lots of abelian varieties at once? I'm trying to say that, as I vary over all non-CM elliptic curves over a fixed field K, and all primes l \geq 5, then G_l is almost always GL_2(\mathbb{Z}_l) (and it would be nice to find the exceptional l, or at least a bound for them in terms of K). For fixed E this is certainly true. (I had reason to believe that, if G_l was not all of GL_2(\mathbb{Z}_l), then it must be contained in a Borel, but your answer last week dispels of this).

    3. There seems to be one case of the conjecture open, that when A is not CM and dim(A) is even. Why is this case harder than odd dimension? Would it help to know that the representation was modular?

  2. Martin Orr said on Monday, 29 November 2010 at 17:46 :

    Probably I was not as clear as I should have been about \mathbb{Z}_\ell- and \mathbb{Q}_\ell-points. It is not true at all that G_\ell^{alg}(\mathbb{Q}_\ell) \subset \operatorname{GL}_{2g}(\mathbb{Z}_\ell) - this is why we introducted G_\ell^{alg}(\mathbb{Z}_\ell) and in fact  G_\ell^{alg}(\mathbb{Z}_\ell) = G_\ell^{alg}(\mathbb{Q}_\ell) \cap \operatorname{GL}_{2g}(\mathbb{Z}_\ell). For example, in the non-CM elliptic curve case, you have G_\ell^{alg}(\mathbb{Q}_\ell) = \operatorname{GL}_2(\mathbb{Q}_\ell) which is much bigger than G_\ell^{alg}(\mathbb{Z}_\ell) = \operatorname{GL}_2(\mathbb{Z}_\ell).

    The \mathbb{Z}_\ell-points matter because it is G_\ell^{alg}(\mathbb{Z}_\ell) which is a good approximation to G_\ell itself. The \mathbb{Q}_\ell-points come in when you want to use the theory of algebraic groups, because algebraic geometry is much easier when you are working over a field than when you are working over a ring.

    Your claim 2 is not quite true - you need to take those M \in \operatorname{GL}_{2g}(\mathbb{Q}_\ell) which have entries in \mathbb{Z}_\ell and determinant in \mathbb{Z}_\ell^\times, which is equivalent to saying that both M and M^{-1} have entries in \mathbb{Z}_\ell.

  3. Barinder said on Monday, 29 November 2010 at 19:55 :

    Right...I think I had two problems. First, I assumed that GL_{2g}(\mathbb{Z}_l) was the biggest thing in the picture. But more importantly, I didn't understand what it meant to say that G_l^{alg} (which is a priori a subvariety of GL_{2g}) contains G_l (which is just a subgroup of GL_{2g}(\mathbb{Z}_l). Now I think it just means G_l \subset G_l^{alg}(\mathbb{Q}_l). But if this is wrong, please put me right. Is the point of this envelope that, G_l isn't a geometric object?

  4. Martin Orr said on Monday, 29 November 2010 at 21:06 :

    Ah, sorry. I try to be careful about distinguishing between algebraic varieties and their sets of points over particular fields, but in this case I missed it. You are indeed right that I meant to say G_\ell \subset G_\ell^{alg}(\mathbb{Q}_\ell) - fixed above.

    Yes, G_\ell is just a plain old group (well a profinite group) and the point of G_\ell^{alg} is to have an algebraic group.

  5. Martin Orr said on Tuesday, 30 November 2010 at 09:11 :

    Irreducible means that V_\ell does not have a \operatorname{Gal}(\bar{K}/K)-submodule. Certainly surjectivity implies this, but the converse is false. For example, G_\ell being of finite index in \operatorname{GL}_{2g} is enough, but this is still far from necessary.

    Consider an elliptic curve E over K with CM by F, and suppose that K \supset F. Now G_\ell^{alg}(\mathbb{Q}_\ell) (working with G_\ell^{alg} instead of G_\ell does not affect reducibility) is (F \otimes_{\mathbb{Q}} \mathbb{Q}_\ell)^\times. There are two cases, depending on whether -d is a square in \mathbb{Q}_\ell or not.

    1. If -d is a square in \mathbb{Q}_\ell, then F \otimes_{\mathbb{Q}} \mathbb{Q}_\ell \cong \mathbb{Q}_\ell \times \mathbb{Q}_\ell. So G_\ell^{alg}(\mathbb{Q}_\ell) \cong \mathbb{Q}_\ell^\times \times \mathbb{Q}_\ell^\times, and V_\ell is reducible: there is one line on which the first copy of \mathbb{Q}_\ell^\times acts non-trivially, and one line on which the second copy acts non-trivially.

    2. If -d is not a square in \mathbb{Q}_\ell, then F \otimes_{\mathbb{Q}} \mathbb{Q}_\ell \cong \mathbb{Q}_\ell(\sqrt{-d}). Now V_\ell is isomorphic to the field \mathbb{Q}_\ell(\sqrt{-d}) and G_\ell^{alg}(\mathbb{Q}_\ell) is the multiplicative group \mathbb{Q}_\ell(\sqrt{-d})^\times acting in the natural way. Since \mathbb{Q}_\ell(\sqrt{-d}) is a field, there are no subrepresentations. So V_\ell is irreducible, but the representation is far from surjective.

    For two-dimensional representations, this is related to Borel subgroups: V_\ell is irreducible if and only if G_\ell is not contained in a Borel subgroup defined over \mathbb{Q}_\ell. This is because \{ \bigl(\begin{smallmatrix} * & * \\ 0 & * \end{smallmatrix}\bigr) \} is precisely the set of matrices which fix the line \{ \bigl( \begin{smallmatrix} * \\ 0 \end{smallmatrix} \bigr) \}.

  6. Barinder said on Tuesday, 30 November 2010 at 23:37 :

    From the second paragraph onwards in your latest comment above, every instance of K (excluding the first two) should be replaced with F.

    I spent a long time thinking about why, in case 2 above, V_l is isomorphic to \mathbb{Q}_l(\sqrt{-d}). I think the following is why: T_l(E) is a module over End(E), so V_l is a vector space over \mathbb{Q}_l(\sqrt{-d}). And it must be of dimension one because also, T_l(E) \cong \mathbb{Z}_l^2, so V_l \cong \mathbb{Q}_l^2 as vector spaces. If this is wrong, please put me right.

    In 1 above, is G_l^{alg}(\mathbb{Q}_l) \cong \mathbb{Q}_l^{\times} \times \mathbb{Q}_l^{\times} saying that the image are the diagonal matrices?

    I guess my claim ``G_l not surjective \Rightarrow G_l \subset Borel" is equivalent (for two-dimensional things) to saying "Irreducible implies surjective" which I now know is false. Indeed, for elliptic curves over \mathbb{Q}, V_l is always irreducible, but not always surjective.

    So far l-adic representations have the center stage. But what about mod l? Is there an analogous theory? Does there exist a "mod l" version of the Mumford Tate group/conjecture?

  7. Martin Orr said on Wednesday, 01 December 2010 at 10:02 :

    You are correct about K and F, and about why V_\ell is isomorphic to \mathbb{Q}_\ell(\sqrt{-d}).

    G_\ell^{alg}(\mathbb{Q}_\ell) \cong \mathbb{Q}_\ell^\times \times \mathbb{Q}_\ell^\times says that you can choose a basis for V_\ell such that the image is diagonal matrices. Another natural choice of basis is the one that makes the image things of the form \bigl( \begin{smallmatrix} a & -db \\ b & a \end{smallmatrix} \bigr), because this gives a group isomorphic to (F \otimes \mathbb{Q}_\ell)^\times whether -d is a square or not. Exercise: write down a change of basis which diagonalises \bigl( \begin{smallmatrix} a & -db \\ b & a \end{smallmatrix} \bigr) and observe that you need \sqrt{-d} to do this.

    I don't know anything about considering all non-CM elliptic curves at once. I also tend to ignore the difference between surjective and "almost surjective" (i.e. image of finite index). At the end of Serre's article 'Quelques applications du théorème de densité de Chebotarev' (Pub. Math. IHES 1981 or Volume III of his Oeuvres), he asks whether \ell \geq 41 is sufficient for all non-CM curves over \mathbb{Q}, but you may have known that already.

    Well, the stuff about \ell-adic representations is already related to mod \ell because you just reduce G_\ell mod \ell. The Mumford-Tate group is something that works over the complex numbers, so I doubt that a mod \ell version would make sense.

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