Over an algebraically closed field $k$, every one-dimensional torus embedded (as a closed algebraic subgroup) into GL2 is diagonalisable, and the embedding is $t\mapsto (t^m,t^n)$ for some integers $m,n$ with $gcd(m,n)=1$.

What happens when the field $k$ is not algebraically closed? Can I find somewhere the classification of all subtori of GL2, up to conjugation?

By a torus I of course mean a closed algebraic subgroup of GL2, which is isomorphic to $K^{*}$ over the algebraical closure $K$ of $k$.

One example is, when $k=\mathbb{R}$, the group of rotations. This group is algebraically given by matrices of the form $[a,b,c,d]$ (sorry it seems that matrices do not appear correctly, but you should see what I mean) with $a=d$, $b=-c$ and $a^2+b^2=1$. The same example works over any field, and it is not diagonalisable if $-1$ is not a square in the field.

I would like to have a complete classification.

  • 5
    $\begingroup$ I think (though I'm not completely sure) that (at least over a field of characteristic not $2$) the nondiagonalizable tori of the type you mention are the subgroups of $SL2$ that stabilize a quadratic form that does not factor. These correspond to the nontrivial elements of the group $k^\ast/(k^\ast)^2$, i.e., such quadratic forms are, up to a change of basis (i.e., conjugation in $SL2$), of the form $x^2+\lambda y^2$ where $\lambda$ is not a square in $k$, and two such forms $x^2+\lambda_i y^2$ are equivalent (up to change of basis) if and only if $\lambda_1/\lambda_2$ is a square in $k$. $\endgroup$ – Robert Bryant Feb 16 '13 at 0:52
  • 4
    $\begingroup$ I believe Bryant's answer is correct; another way of saying this: let $l/k$ be a quadratic extension with $char k\neq 2$. Treat $l$ as a two dimensional vector space over $k$; then $l^*$ the multiplicative group of $l$ acts on $l$ by multiplication, and is $k$ linear. Hence $l^*$ lies in $GL_2(k)$. There is the norm map $l^*\rightarrow k^*$ and the kernel $N^1_{l/k}$ is a torus which over the algebraic closure, is $K^*$ but over $k$, it is not isomorphic to $k^*$. $\endgroup$ – Venkataramana Feb 16 '13 at 1:11
  • 2
    $\begingroup$ If $T \subset {\rm{GL}}_n$ is a maximal $k$-torus then by $k_s$-rational conjugacy of maximal $k_s$-tori we see $T(k_s)$ generates a Galois-stable etale $k_s$-subalgebra of ${\rm{Mat}}_n(k_s)$ that descends to an etale maximal commutative $k$-subalgebra $A \subset {\rm{Mat}}_n(k)$. Clearly $T$ is contained in the $k$-torus $\underline{A}^{\times}$ of units of $A$, and this is an equality for dimension reasons. Thus, $A \mapsto \underline{A}^{\times}$ is a bijection between the sets of maximal $k$-tori of ${\rm{GL}}_n$ and the etale maximal commutative $k$-subalgebras of ${\rm{Mat}}_n(k)$. $\endgroup$ – user30379 Feb 16 '13 at 8:27

In general, a group that is isomorphic to a $d$-dimensional torus over the algebraic closure of the ground field can be identified by the action of the Galois group on its character group / group of one-parameter subgroups, which is $\mathbb Z^d$, so you get a homomorphism $Gal(\bar{k}/k) \to GL_d(\mathbb Z)$. In the one-dimensional case, this is just a quadratic character, so tori correspond to quadratic fields, in the manner discussed in the comments. In particular, all $1$-dimensional tori in any group embed into $GL_2$.

The reason you can recover it from the Galois action on the character group is that the points of the torus are exactly the homomorphisms from the character group to $GL_1(\bar{k})$ that are Galois-equivariant. This is a version of Pontryagin duality.

This is part of the general theory that twists of an object $X$ over a field $k$ are classified by the Galois cohomology group $H^1(k,X)$, or, for twists over an arbitrary base, an etale cohomology group.

EDIT: For a very explicit description, the Galois rep associated to the quadratic field extension $\mathbb Q(\sqrt{D})$ is the group of determinant $1$ matrices that preserve the quadratic form $x^2-Dy^2$, which means

$\left(\begin{array}{cc} a & Db \\ b & a \end{array}\right) $

for $a^2-D b^2 =1 $

for $D=-1$ this reduces to the classic $SO_2$, and for $D=1$ this is just a regular split torus.

| cite | improve this answer | |
  • $\begingroup$ Thx Will. So in my case I have just GL_2(Z), which is Z/2Z. Does this helps to get a precise description of the tori? $\endgroup$ – Jérémy Blanc Feb 16 '13 at 11:01
  • 1
    $\begingroup$ Yes, the description is in Aakumadula's comment above. $\endgroup$ – Ben Webster Feb 16 '13 at 14:26
  • $\begingroup$ Thx. By "precise description" I really meant something precise, like diagonal tori or matrices of the form [a,b*mu, b,a], with some conditions on a,b. $\endgroup$ – Jérémy Blanc Feb 17 '13 at 0:44
  • $\begingroup$ Does this suffice? $\endgroup$ – Will Sawin Feb 17 '13 at 6:04
  • 3
    $\begingroup$ The $H^1(k,X)$ in this answer should be $H^1(k,{\rm{Aut}}_{X/k})$ (and one should note the relevance of projectivity hypotheses to ensure effectivity of descent, unless "object" is understood in some sense like algebraic spaces). $\endgroup$ – user28172 Feb 17 '13 at 10:29

The $n$-dimensional tori up to isomorphism can be described by Galois cohomology, see for instance, Serre's Galois cohomology, or my thesis (http://arxiv.org/abs/math/0409453) for more details. These are given by (equivalence classes of) Galois representations taking values in $GL_n(\mathbb{Z})$.

For instance, for real numbers, the Galois group is $\mathbb{Z}/2\mathbb{Z}$ and has three indecomposable integral representations; the trivial 1-dimensional representation, the sign representation and the 2 dimensional permutation representation (where the Galois group operates by permuting a basis). Thus we get three 2-dimensional tori over $\mathbb{R}$; the split one, the restriction of scalars from $\mathbb{G}_m$ from $\mathbb{C}$ to $\mathbb{R}$, and the $SO_2 \times \mathbb{G}_m$. Since the last torus is not embeddable in $GL_2$, we get only two types of tori in $GL_2$ over $\mathbb{R}$.

Over $\mathbb{Q}$, the Galois group is much more complicated and hence has many more two-dimensional representations (one corresponding to each quadratic extension). Thus we get many non-conjugate maximal tori in $GL_2$ over $\mathbb{Q}$.

| cite | improve this answer | |
  • $\begingroup$ Thanks for the comment. I would like to use the fact that I am restricted to GL2 to really have all tori. For example, over $\mathbb{Q}$ the results should not be so complicated. $\endgroup$ – Jérémy Blanc Feb 16 '13 at 11:00
  • 1
    $\begingroup$ You can do this via the explicit description of all quadratic characters of $GL_2(\mathbb Q)$ coming from Kummer theory (or class field theory). In particular, every quadratic character trivializes over $\mathbb Q(\sqrt{D})$ for some $D$, and the corresponding torus is the group of determinant $1$ matrices that preserve the quadratic form $x^2-Dy^2$, as was first pointed out by Robert Bryant. $\endgroup$ – Will Sawin Feb 16 '13 at 15:44
  • $\begingroup$ Ok, thx. It seems thus that 2-dim tori are either diagonalisable or of the form [a,muc,c,a] with a,c \in k^{}. And 1-dimensional tori can be found in a similar way. $\endgroup$ – Jérémy Blanc Feb 17 '13 at 0:46
  • $\begingroup$ Thanks Will, your comment elaborates my answer very well. Jeremy, by complicated for $\mathbb{Q}$ I meant only in comparison with the case over $\mathbb{R}$. $\endgroup$ – Shripad Feb 18 '13 at 6:59
  • $\begingroup$ @ Shripad, yes I see that torus can be complicated in general over Q (more than over R); I just meant that in GL2 we can have an explicit description (given above in the answer of Will). Thx anyway for your answers. $\endgroup$ – Jérémy Blanc Feb 18 '13 at 21:01

In the special case $G= \mathrm{GL}_2(k)$, a more direct approach than those indicated in the answers and comments is certainly possible even though it doesn't do much to illuminate the general case.

To simplify a bit, note that the center (consisting of nonzero scalar matrices) is itself a 1-dimensional torus not conjugate to others; so it's safe to consider just the derived group $\mathrm{SL}_2(k)$. Here the 1-dimensional tori are just the maximal tori in the algebraic group setting, usually called Cartan subgroups in the more specialized Lie group version when $k= \mathbb{C}$ or $\mathbb{R}$.

In the Lie group case, you are asking for all conjugacy classes of Cartan subgroups in the real group. This is an old problem, solved in general by Kostant for semisimple Lie groups (and independently by Sugiura, available online here). It's equivalent to finding the conjugacy classes of Cartan subalgebras. In your special case there are two classes, represented by the diagonal torus and a compact version.

In the algebraic group case, you are similarly asking for all conjugacy classes of maximal $k$-tori in the case when $k$ fails to be algebraically closed. Here again there is a lot of general theory, organized by Borel-Tits, with the machinery of Galois cohomology then being invoked to discuss $k$-forms. But some short-cuts are likely in your rank 1 situation. What approach you take depends a lot on what your ultimate interest is and which fields are most important.

For higher rank groups, the whole problem has a different flavor. In the algebraic group setting, you'd be looking at the dual of the character group of a maximal torus consisting on co-characters (or 1-parameter subgroups) relative to a splitting field and its subfield.

| cite | improve this answer | |

A finite dimensional commutative $k$-algebra is étale if it is isomorphic to a product of separable field extensions of $k$. For any $n \ge 1$, the conjugacy classes of maximal $k$-tori in $\operatorname{GL}_n = \operatorname{GL}(V)$ are in 1-1 correspondence with isomorphism classes of étale $k$-algebras $E$ of dimension $n$ over $k$.

Here is how the correspondence goes: given a maximal torus $T$ of $\operatorname{GL}(V)$, the Lie algebra $\operatorname{Lie}(T)$ is an $n$-dimensional etale subalgebra of $\operatorname{End}_k(V)$.

Conversely, given $E$ an étale algebra of dimension $n$ over $k$, view $E$ as a left $E$-module. The module structure determines a $k$-algebra embedding $E \to \operatorname{End}_k(E)$ and hence an injective homomorphism of algebraic groups $E^\times \to \operatorname{GL}(E) = \operatorname{GL}_n$, where $E^\times$ is the "unit group of $E$ viewed as an algebraic group". The image of this homomorphism is the desired maximal torus of $\operatorname{GL}_n$.

It remains to see that étale subalgebras $E$ and $F$ of $\operatorname{End}_k(V)$ are conjugate by an element of $\operatorname{GL}_n(k)$ if and only if they are isomorphic $k$-algebras. This follows from the observation that if $E$ is an étale subalgebra of $\operatorname{End}_k(V)$ with $\dim_k E = \dim_k V$, then viewed as $E$-module, $V$ is isomorphic to the regular representation of $E$.

From the point of view of Galois cohomology, here is what is going on. Write $G = \operatorname{GL}_n$, let $T_0$ be a split maximal $k$-torus, and let $N = N_G(T_0)$. The $k$-variety $\mathcal{T}$ of all maximal tori of $G$ is $\mathcal{T} = G/N$, and there is an exact sequence of pointed sets $$1 \to N(k) \to G(k) \to \mathcal{T}(k) \to H^1(k,N) \to 1$$ since $H^1(k,G)$ is trivial (Hilbert 90).

Now, $N$ is the semidirect product of the split torus $T_0$ and the symmetric group $S_n$. Since $H^1(k,T_0)$ is trivial (Hilbert 90 again), and since the quotient homomorphism $N \to S_n$ has a section, $H^1(k,N)$ identifies with $H^1(k,S_n)$, which in turn identifies with the set $\operatorname{Et}_n(k)$ of iso. classes of étale $k$-algebras of dimension $n$.

Thus the mapping $\mathcal{T}(k) \to H^1(k,N) = \operatorname{Et}_n(k)$ is onto, and by "twisting" one identifies the fibers of this mapping with the $G(k)$-orbits on $\mathcal{T}(k)$; i.e. with the conjugacy classes of maximal $k$-tori in $\operatorname{GL}_n$.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.