Martin's Blog

Images of Galois representations

Posted by martin 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 $\operatorname{End} A = \mathbb{Z}$.

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Comments

  1. Barinder said 1 day later:

    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 $l$s 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 2 days later:

    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 2 days later:

    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 2 days later:

    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 3 days later:

    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 3 days later:

    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 4 days later:

    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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