Martin's Blog

Character groups of algebraic tori

Posted by Martin Orr on Sunday, 24 January 2010 at 18:10

In this post I will return to the subject of algebraic tori. Just as Pontryagin duality classifies locally compact abelian groups through their characters, so algebraic tori are also classified by their characters.

In order to account for the arithmetic phenomenon of non-split tori, we need to include a Galois action on the character group. The primary result of this post is that there is an anti-equivalence of categories between {k-tori} and {finitely generated free abelian groups with a continuous action of \mathop{\mathrm{Gal}}(k^s/k)}.

Characters of split tori

A character of an algebraic group G is a homomorphism G \to \mathbb{G}_m.

The characters of G form an abelian group under pointwise multiplication, denoted X^*(G). The map sending G to its group of characters is a contravariant functor {algebraic groups over k} to {abelian groups}.

For G = \mathbb{G}_m, the characters are the maps \mathbb{G}_m \to \mathbb{G}_m of the form t \mapsto t^n (for n \in \mathbb{Z}).

Any character of \mathbb{G}_m^r is determined by its values on each of the factors, so is  (t_1, t_2, \ldots, t_r) \mapsto t_1^{n_1} t_2^{n_2} \ldots t_r^{n_r} for some n_1, n_2, \ldots, n_r \in \mathbb{Z}.

Hence the X^*(\mathbb{G}_m^r) is the finitely generated free abelian group \mathbb{Z}^r.

The character group functor {split k-tori} to {f.g. free abelian groups} is essentially surjective because isomorphism classes on both sides are indexed by nonnegative integers.

Hopf algebra of a split torus

Let G and H be algebraic groups and \phi : X^*(H) \to X^*(G) a homomorphism. To prove that the above functor is full and faithful, we need to prove that \phi corresponds to a unique homomorphism G \to H.

It is quite simple to write this homomorphism down in coordinates, but proving it using Hopf algebras gives us practice for later when using coordinates is not so easy.

In particular the following lemma is important:

Lemma. If G is a split torus, then its characters form a k-vector space basis for the Hopf algebra.

Proof. Note first that by composing with the natural inclusion \mathbb{G}_m \hookrightarrow \mathbb{A}^1, characters can be viewed as regular functions on G.

The ring of regular functions of \mathbb{G}_m^r is k[T_1, T_1^{-1}, \ldots, T_r, T_r^{-1}].

This has basis \{ T_1^{n_1} \ldots T_r^{n_r} : n_1, \ldots, n_r \in \mathbb{Z} \}; but this is precisely the set of characters of G.

End of proof.

Thanks to the lemma, \phi induces a linear map (Hopf algebra of H) to (Hopf algebra of G).

This linear map always preserves the coalgebra structure, and because \phi is a homomorphism it also preserves the algebra structure. And so the anti-equivalence between algebraic groups and Hopf algebras gives a unique homomorphism G \to H.

Characters of non-split tori

Now let G be an arbitrary torus over a field k.

If G is non-split, the characters of G defined over k no longer contain enough information. Instead we work with characters defined over the separable closure: X_{k^s}^*(G) = \mathop{\mathrm{Hom}}(G_{k^s}, \mathbb{G}_{m,k^s}).

Because all tori over a separably closed field are split, X_{k^s}^*(G) is a finitely-generated free abelian group.

Galois action on the characters

X_{k^s}^*(G) comes with a continuous action of \mathop{\mathrm{Gal}}(k^s/k).

To construct the action we use Hopf algebras: let A be the Hopf algebra of G (over k).

Then the Hopf algebra of G_{k^s} is A \otimes_k k^s and \mathop{\mathrm{Gal}}(k^s/k) acts on this via the second factor, and this action takes characters to characters.

(The coordinate description of this action is much more complicated: first embed G in affine space over k. With respect to this embedding, regular functions defined over k^s are polynomials with coefficients in k^s. Let \mathop{\mathrm{Gal}}(k^s/k) act on these coefficients.)

A morphism of tori G_{k^s} \to H_{k^s} is defined over k iff the corresponding homomorphism of character groups commutes with the Galois actions.

So X_{k^s}^* is a full and faithful contravariant functor {k-tori} to {f.g. free abelian groups with a continuous action of \mathop{\mathrm{Gal}}(k^s/k)}.

Obtaining a torus from a character group

All that remains to complete the anti-equivalence of categories is to construct a k-torus for any given character group (with Galois action). This Galois descent construction is very awkward to do rigorously without using Hopf algebras.

So suppose we are given X, a f.g. free abelian group of rank r with a Galois action.

Let A be the Hopf algebra of \mathbb{G}_m^r over k^s.

We know that characters of \mathbb{G}_m^r form a basis for A; write e_\chi for the basis vector corresponding to \chi \in X.

Let \mathop{\mathrm{Gal}}(k^s/k) act on A by

 \sigma(\sum_\chi \lambda_\chi e_\chi) = \sum_\chi \sigma(\lambda_\chi) e_{\sigma(\chi)}.

This is a \mathop{\mathrm{Gal}}(k^s/k)-semilinear action and commutes with the Hopf algebra structure.

Let B be the subset of A fixed by \mathop{\mathrm{Gal}}(k^s/k). By a standard Galois descent argument, B is a k-Hopf algebra and B \otimes_k k^s = A.

Then the algebraic group over k with Hopf algebra B is the required torus.

Tags alg-geom, alg-groups, maths, tori


  1. Representations of algebraic tori From Martin's Blog

    I am finally ready to finish my series on algebraic tori, by talking about their representations. I shall show that these representations can be classified by a grading on the vector space of the representation, after extending scalars to the sepa...


  1. Alex Youcis said on Wednesday, 01 October 2014 at 07:55 :

    Both of the things you say are awkward to do without Hopf algebras are not if you think scheme theoretically.

    For the action of G_k on T_{\overline{k}}, you can just think about this as T\times_{\text{Spec}(k)}\text{Spec}(\overline{k}). Then G_k acts on the right factor.

    Also, thinking scheme theoretically, the descent argument in the last paragraph follows immediately from standard descent of affine morphisms.

    I don't think anything I've said is easier than what you wrote, I thought I would just comment. :)

  2. Martin Orr said on Tuesday, 07 October 2014 at 23:02 :

    Thank you Alex.

    I think that when I wrote this, I was probably not comfortable with things like descent for affine morphisms. Given the dates, I asked this question while working on this post, and Pete Clark's answer convinced me that I should use coordinate rings for the proof.

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