Martin's Blog

Tate modules

Posted by martin on Sunday, 21 November 2010 at 17:32

I said after my last post that I would write something about $\ell$-adic representations coming from abelian varieties. I have finally got around to doing so: here I will tell the story of how these representations are defined, and show that the Tate module is canonically isomorphic to $H_1(A, \mathbb{Z}) \otimes \mathbb{Z}_\ell$. Next time I will relate this to Mumford-Tate groups.

Torsion points

Let $A$ be an abelian variety over $\mathbb{C}$ of dimension $g$. We shall look at the $N$-torsion points of $A$, written $A[N]$. The group $A[N]$ is isomorphic to $(\mathbb{Z}/N\mathbb{Z})^{2g}$.

(This fact holds over any algebraically closed field, as long as $N$ is not divisible by the characteristic, but I shall give a proof valid only over $\mathbb{C}$ because this proof will show the relationship with singular homology.)

For one proof of this, recall that $A$ is isomorphic to $\mathbb{C}^g/\Lambda$ via the exponential map, for some lattice $\Lambda \subseteq \mathbb{C}^g$. This isomorphism takes $A[N]$ to $\tfrac{1}{N}\Lambda/\Lambda$.

Now a fundamental domain for $\mathbb{C}^g/\Lambda$ is a parallelotope in $\mathbb{C}^g \cong \mathbb{R}^{2g}$, and the points of $\tfrac{1}{N}\Lambda$ inside this form an $N \times N \times \dotsb \times N$ ($2g$ times) grid. For example here is a picture of the 3-torsion points of an elliptic curve:

9 points in a parallelogram

At this point we have proved that $A[N] \cong (\mathbb{Z}/N\mathbb{Z})^{2g}$, but in fact we have proved more: there is a canonical isomorphism between $A[N]$ and $H_1(A, \mathbb{Z}) \otimes_\mathbb{Z} \mathbb{Z}/N\mathbb{Z}$. This is because $\Lambda$ is canonically isomorphic to $H_1(A, \mathbb{Z})$ and multiplication by $N$ gives an isomorphism $\tfrac{1}{N}\Lambda/\Lambda \to \Lambda \otimes_\mathbb{Z} \mathbb{Z}/N\mathbb{Z}$.

Tate modules

Now fix a rational prime $\ell$. We want to look at $\ell^n$-torsion points for all $n$ at once.

One way of doing this would be to take the union of all the $A[\ell^n]$. This gives an abelian group, but there is no interesting module structure.

Instead consider the sequence of groups $A[\ell^n]$ with the multiplication-by-$\ell$ morphisms (from the group law on the abelian variety): $ \usepackage{xypic} \xymatrix{ \dotsc \ar[r]^{[\ell]} & A[\ell^{n+1}] \ar[r]^{[\ell]} & A[\ell^n] \ar[r]^{[\ell]} & \dotsc \ar[r] & A[\ell] \ar[r] & 0 } $

Under the isomorphisms $A[\ell^] \cong \Lambda \otimes \mathbb{Z}/\ell^n\mathbb{Z}$, the multiplication-by-$\ell$ morphism $A[\ell^{n+1}] \to A[\ell^n]$ becomes the reduction-mod-$\ell^n$ map $\Lambda \otimes \mathbb{Z}/\ell^{n+1}\mathbb{Z} \to \Lambda \otimes \mathbb{Z}/\ell^n\mathbb{Z}$.

So when we take the inverse limit $\lim\limits_\leftarrow A[\ell^n]$, we get something isomorphic to $\Lambda \otimes_\mathbb{Z} \mathbb{Z}_\ell$. This is called the $\ell$-adic Tate module, written $T_\ell A$.

Galois action

Now suppose that $A$ is defined over a number field $K$. In this case the Tate modules carry extra structure, namely an action of $\mathop{\mathrm{Gal}}(\bar{K}/K)$.

When I talk about the “coordinates” of a point in $A(\mathbb{C})$, I shall mean the homogeneous coordinates under some embedding in projective space. These of course are completely different from the coordinates of a corresponding point in $\mathbb{C}^g/\Lambda$.

Since the group law on $A$ is defined by polynomials with coefficients in $K$, there exists a set of polynomials with coefficients in $K$ whose complex solutions are precisely the $\ell^n$-torsion points. (Writing down these polynomials is pretty horrible but we only care that they exist.)

It follows that if we let $\mathop{\mathrm{Aut}}(\mathbb{C}/K)$ act on the coordinates of points in $A(\mathbb{C})$, it takes $\ell^n$-torsion points to $\ell^n$-torsion points. Since there are only finitely many $\ell^n$-torsion points, their coordinates must be algebraic numbers.

So the group $\mathop{\mathrm{Gal}}(\bar{K}/K)$ acts on $A[\ell^n]$. Furthermore, the action is compatible with the multiplication-by-$\ell$ morphisms, so we can pass to the limit and get an action of $\mathop{\mathrm{Gal}}(\bar{K}/K)$ on $T_\ell A$.

Image of the Galois representation

We can interpret the image action of $\mathop{\mathrm{Gal}}(\bar{K}/K)$ on $T_\ell A$ as a group homomorphism $ \rho_\ell : \mathop{\mathrm{Gal}}(\bar{K}/K) \to \mathop{\mathrm{GL}}\nolimits_{2g}(\mathbb{Z}_\ell). $

Both sides of this homomorphism are profinite groups, so carry a natural topology. For any $n$, the preimage under $\rho_\ell$ of those matrices which reduce to $1$ modulo $\ell^n$ is the set of elements of the Galois group which act trivially on $A[\ell^n]$. Since $K(A[\ell^n])$ is a finite extension of $K$, this subgroup is open in $\mathop{\mathrm{Gal}}(\bar{K}/K)$.

It follows that $\rho_\ell$ is continuous, and so its image $G_\ell$ is a compact subgroup of $\mathop{\mathrm{GL}}_{2g}(\mathbb{Z}_\ell)$. This image is interesting because it tells us about the Galois groups of the fields generated by torsion points: $\mathop{\mathrm{Gal}}(K(A[\ell^n])/K)$ is isomorphic to the reduction mod $\ell^n$ of $G_\ell$. I shall write more about $G_\ell$ and its conjectural relation to the Mumford-Tate group next time.

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  1. Images of Galois representations From Martin's Blog
    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...

Comments

  1. Barinder said about 4 hours later:

    Nice Post. I’ve been thinking about this recently; in particular, I wanted to say that for only finitely many primes $l$ is the image $G_l$ contained in a Borel subgroup of $GL_{2g}(\mathbb{Z}_l)$. Is this true in general? What about for the mod-$l$ representations that one can get from the $l$-adic one? In my case, I had a bunch of $l$-adic representations coming from weight 2 newforms (of any level), and wanted to say something like, almost always, the image is inside a Borel subgroup.

    I’m looking forward to your next post.

  2. Martin Orr said about 17 hours later:

    Are your abelian varieties defined over number fields (as opposed to say local fields)? From what I remember about the varieties associated to modular forms, the answer is yes.

    Then I think that, as long as your variety is not CM, the image is never contained in a Borel subgroup. A quick sketch: let $G_\ell^{alg}$ be the Zariski closure of the image of $\rho_\ell$ in $\operatorname{GL}_{2g}(\mathbb{Q}_\ell)$. By Faltings the connected component of this is reductive. If it is contained in a Borel then it is solvable. A group which is both reductive and solvable must be commutative. If the connected component of $G_\ell^{alg}$ is commutative, then the abelian variety is CM.

    In the CM case it matters what you take as your base field. If the base field is large enough (something like if it contains the Galois closure of the CM field) then $G_\ell$ is commutative, so is contained in a Borel.

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