Maths > Abelian varieties > Complex abelian varieties and the MumfordTate 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 adic representations attached to abelian varieties, and in particular the images
of these representations.
I will define algebraic groups approximating
, which are often more convenient to work with.
I will end by stating the MumfordTate conjecture, linking
to the MumfordTate group.
Let be an abelian variety of dimension
over a number field
.
Recall from last time that the Tate module
gives us a representation
We define
to be
, a compact subgroup of
.
The example of elliptic curves
If is an elliptic curve over a number field
without complex multiplication, then Serre's Open Image Theorem (proved in the 60s) says that
is an open subgroup of
.
Since
is compact, it follows that
has finite index in
.
Next suppose that is an elliptic curve over a number field
with complex multiplication by
.
Recall that
can be embedded in
as the invertible matrices of the form
.
This extends to an embedding of
in
.
Let
be the image of this embedding.
(The structure of
depends on whether
has a square root in
or not.
If yes, then
is conjugate to the diagonal subgroup of
.
If no, then
is isomorphic to the group of units of
.)
Now if , then it follows from the main theorem of complex multiplication that
is an open subgroup of
.
If
, then
has an index2 subgroup which is an open subgroup of
,
but
is not itself contained in
.
If you compare the above groups with the MumfordTate groups of elliptic curves,
you will notice that they are rather similar, as long as you ignore the complications about "finiteindex subgroups".
There is another reason to not care about finiteindex subgroups:
let
be an abelian variety defined over a number field
.
It is often convenient to be able to replace
by a finite extension
;
but if we do this, then
gets replaced by a finiteindex subgroup
(because
is a finiteindex subgroup of
).
Algebraic envelope of
Let be the smallest algebraic subgroup (over
) of
such that
contains
.
This is sometimes given the unwieldy name of
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 is sometimes more useful than
:

It is easier to compute
because of the structure theorems for algebraic groups.

Algebraic groups have few finiteindex algebraic subgroups, so the problem that the group changes when you enlarge the base field is reduced, although not eliminated.
Let be an elliptic curve over
.

If
does not have CM, then
.

If
has CM by the field
and
, then
is the
algebraicgroup version of
.

If
has CM by
and
, then
has the
algebraicgroup version of
as an index2 subgroup.
If is to be a good approximation to
,
it is important to know that
is not too much bigger than
.
Let
denote the elements of
such that both the matrix in question and its inverse have all their entries in
.
Bogomolov proved in 1980 that
is always an open subgroup of
in the
adic topology, and so is of finite index.
Connected components of
Any algebraic group has finitely many connected components in the Zariski topology.
The connected component containing the identity is a normal subgroup of
.
(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 be the connected component of
containing the identity.
Connected algebraic groups do not have any finiteindex algebraic subgroups (because they are also irreducible, so not a union of finitely many closed subsets).
So
is unchanged if we enlarge the base field
.
To summarise: neither nor
is necessarily contained in the other, but both are finiteindex subgroups of
.
The MumfordTate conjecture
is a connected algebraic group over
, and the MumfordTate group
is a connected algebraic group over
.
The MumfordTate conjecture states that
is equal to
(the latter just means
viewed as an algebraic group over
).
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, and
where
and
.
We saw last time that
is canonically isomorphic to
.
Tensoring this isomorphism with
gives a canonical isomorphism between
and
,
which we use to identify
and
.
From the examples, we see that the MumfordTate 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 in the CM case).
In general, Deligne showed using Hodgetheoretic techniques that
is always contained in
.
Various other cases are known when restrictions are placed on the dimension and endomorphism ring of
, for example whenever
has CM, or when
is odd and
.
Great Post Martin! Here are two questions just to ensure that I'm still with you:
Is
the smallest algebraic subgroup of
defined over
such that
? Therefore, is
?
Is the following true?
Here are some other questions/comments:
You briefly mentioned a property of a representation  "irreducible". I guess this means that the Tate module
, as a Galois module, does not decompose. Is irreducibility of an
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).
Is there a way to think about these
for lots of
s and lots of abelian varieties at once? I'm trying to say that, as I vary over all nonCM elliptic curves over a fixed field
, and all primes
, then
is almost always
(and it would be nice to find the exceptional
, or at least a bound for them in terms of
). For fixed
this is certainly true. (I had reason to believe that, if
was not all of
, then it must be contained in a Borel, but your answer last week dispels of this).
There seems to be one case of the conjecture open, that when
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?
Probably I was not as clear as I should have been about
 and
points. It is not true at all that
 this is why we introducted
and in fact
For example, in the nonCM elliptic curve case, you have
which is much bigger than
.
The
points matter because it is
which is a good approximation to
itself. The
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
which have entries in
and determinant in
, which is equivalent to saying that both
and
have entries in
.
Right...I think I had two problems. First, I assumed that
was the biggest thing in the picture. But more importantly, I didn't understand what it meant to say that
(which is a priori a subvariety of
) contains
(which is just a subgroup of
. Now I think it just means
. But if this is wrong, please put me right. Is the point of this envelope that,
isn't a geometric object?
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
 fixed above.
Yes,
is just a plain old group (well a profinite group) and the point of
is to have an algebraic group.
Irreducible means that
does not have a
submodule. Certainly surjectivity implies this, but the converse is false. For example,
being of finite index in
is enough, but this is still far from necessary.
Consider an elliptic curve
over
with CM by
, and suppose that
. Now
(working with
instead of
does not affect reducibility) is
. There are two cases, depending on whether
is a square in
or not.
If
is a square in
, then
. So
, and
is reducible: there is one line on which the first copy of
acts nontrivially, and one line on which the second copy acts nontrivially.
If
is not a square in
, then
. Now
is isomorphic to the field
and
is the multiplicative group
acting in the natural way. Since
is a field, there are no subrepresentations. So
is irreducible, but the representation is far from surjective.
For twodimensional representations, this is related to Borel subgroups:
is irreducible if and only if
is not contained in a Borel subgroup defined over
. This is because
is precisely the set of matrices which fix the line
.
From the second paragraph onwards in your latest comment above, every instance of
(excluding the first two) should be replaced with
.
I spent a long time thinking about why, in case 2 above,
is isomorphic to
. I think the following is why:
is a module over End
, so
is a vector space over
. And it must be of dimension one because also,
, so
as vector spaces. If this is wrong, please put me right.
In 1 above, is
saying that the image are the diagonal matrices?
I guess my claim
``
not surjectiveBorel" is equivalent (for twodimensional things) to saying "Irreducible implies surjective" which I now know is false. Indeed, for elliptic curves over
,
is always irreducible, but not always surjective.
So far
adic representations have the center stage. But what about mod
? Is there an analogous theory? Does there exist a "mod
" version of the Mumford Tate group/conjecture?
You are correct about
and
, and about why
is isomorphic to
.
says that you can choose a basis for
such that the image is diagonal matrices. Another natural choice of basis is the one that makes the image things of the form
, because this gives a group isomorphic to
whether
is a square or not. Exercise: write down a change of basis which diagonalises
and observe that you need
to do this.
I don't know anything about considering all nonCM 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
is sufficient for all nonCM curves over
, but you may have known that already.
Well, the stuff about
adic representations is already related to mod
because you just reduce
mod
. The MumfordTate group is something that works over the complex numbers, so I doubt that a mod
version would make sense.