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Let $G$ be be a connected real algebraic reductive Lie group. Is it always possible to find finitely many maximal algebraic $\mathbf{R}$-tori $\{T_i\}_{i=1}^n$ such that the group generated by the $T_i$'s is $G$.

P.S. The assumption for $G$ to be reductive is essential since $\mathbb{G}_a$ does not contain any real torus.

added: The algebraicity assumption was added in order to avoid one-parameter subgroups like $\gamma:\mathbf{R}^{\times}\rightarrow GL_2(\mathbf{R})$: $$ t\mapsto \left( \begin{array}{cc} 1 & \log|t| \newline 0 & 1 \end{array} \right) $$

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Hugo -- what kind of tori do you have in mind? I.e., do you mean compact tori or products of $\mathbb{R}^*$'s or something else? –  algori Jan 24 '12 at 19:10
By an $\mathbf{R}$-torus I mean the $\mathbf{R}$-valued points of an algebraic group $H$ defined over $\mathbf{R}$ such that $H\otimes_{\mathbf{R}}\mathbf{C}\simeq (\mathbb{G}_m/\mathbf{C})^n$. So for example, with a rank $1$ torus (case $n=1$), if $T$ is split then $H(\mathbf{R})\simeq \mathbf{R}^{\times}$ and if it is non-split (and non-empty) it is isomorphic to $SO(2)$. –  Hugo Chapdelaine Jan 24 '12 at 19:22
So my maximal $\mathbf{R}$-tori will look like $(\mathbf{R}^{\times})^n\times (S^1)^m$ where $S^1$ stands for the unit circle. May be also I should have restricted my question to real algebraic reductive group –  Hugo Chapdelaine Jan 24 '12 at 19:25
You also need $\mathbb C^*$ factors –  Jeffrey Adams Jan 25 '12 at 18:16

3 Answers 3

up vote 3 down vote accepted

At least for affine algebraic groups this is true. Indeed, let $G$ be a connected affine real algebraic group. Assume $G$ contains at least one diagonalizable element. Then diagonalizable elements are dense in $G$. Let $U$ be a neighborhood of 0 in the Lie algebra $g$ of $G$ such that $exp|U$ is a diffeomorphism. Choose a basis $e_1,\ldots,e_k\in U$ of $g$ that consists of diagonalizable elements.

Set $G_k$ to be the set of the real points of the Zariski closure of $\{exp(te_k)\mid t\in\mathbb{R}\}$. The map $\mathbb{R}^k\to G$ defined as $(t_1,\ldots, t_k)\mapsto exp(te_1)\cdot\cdots\cdot exp(te_k)$ is a diffeomorphism in a neighborhood of 0. So every element of $G$ that is sufficiently close to the unit can be written as a product of elements of $G_i$'s. Since $G$ is connected, so can be any element of $G$. On the other hand, each $G_i$ is a torus and as such is included in a maximal torus.

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Thanks algori, I did not think of taking the exponential of diagonalizable element! So for this density result on diagonalizable elements, are you assuming that $G$ is reductive? –  Hugo Chapdelaine Jan 24 '12 at 21:13
Hugo -- no, I'm just assuming that $G$ is algebraic and contains a non-trivial diagonalizable element. In this case non-diagonalizable elements are contained in an algebraic subset $\neq G$. –  algori Jan 24 '12 at 21:20
So I guess that the "simplest" example of a non-reductive real algebraic that contains a semi-simple element is the set of invertible matrices of form $\left( \begin{array}{cc} a & b \newline 0 & 1 \end{array} \right) $ –  Hugo Chapdelaine Jan 24 '12 at 21:33
@algori: It's safest to stay with reductive groups, since you might have a direct product of a reductive and a unipotent group in which semisimple elements fail to be dense. –  Jim Humphreys Jan 24 '12 at 21:40
But then I don't quite see how you can generate this group with algebraic one-parameter subgroups "generated" by diagonalizable elements... –  Hugo Chapdelaine Jan 24 '12 at 21:41

For a more algebraic version of algori's approach, I'd suggest a conceptually simple method based on the Borel-Tits structure theory. Of course, some tweaking would be needed to deal with real Lie groups which aren't algebraic, like the universal covering group of $SL_2(\mathbb{R})$. Without loss of generality, take $G$ to be connected and semisimple as well as defined over $\mathbb{R}$. The split case is straightforward, since here the group is generated by finitely many copies of $SL_2$ or $PGL_2$ (corresponding to roots). A maximal torus in each such subgroup lies in some split maximal torus of $G$, so it's enough to see what happens in rank 1. The diagonal group is a maximal torus in $SL_2$ and lies in two opposite Borel subgroups (upper and lower triangular). In each Borel subgroup the fixed maximal torus and an arbitrary distinct conjugate of it generate this 2-dimensional connected group over $\mathbb{R}$, so three maximal tori here generate the entire $SL_2$.

When $G$ is defined but not split over $\mathbb{R}$, more careful use of the Borel-Tits theory (allowing even anisotropic = compact tori) is needed but the proof should follow a similar outline. I won't try to make this more explicit but want to emphasize that there is an underlying idea applicable to algebraic groups defined over other fields. Lie group methods are certainly most appropriate for the question asked.

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So this is a comment to my question. In general if $G/k$ ($k$ is any field) is a connected linear algebraic $k$-group then one can show that it is generated over $\bar{k}$ by its Cartan subgroups defined over $k$. (See the remark on the top of p. 220 of Borel's book on linear algebraic groups). For dimension reason we may take only finitely many of them. Finally, since I was assuming that $G$ is reductive it follows that every Cartan subgroup is a maximal torus and thus it is possible to generate $G$ with only finitely many $k$-tori.

Here is Brian Conrad's proof that was communicated to me through email in the case where $G$ is reductive and $k$ is infinite:

Consider a smooth connected affine $k$-group $G$. For every $X$ in $Lie(G)$ that is semisimple (relative to $G$, which is to say in the sense defined in Borel's book on algebraic groups), it is an important fact (proved in Borel's book on algebraic groups (see (**) in 9.1 and the Proposition just after it) that the scheme-theoretic centralizer $Z_G(X)$ of $X$ under the adjoint action of $G$ on $Lie(G)$ is smooth due to semisimplicity of $X$.

We proceed by induction on $dim(G)$, using the reductivity, and we work over infinite fields. (The case of finite fields is true, but requires other ideas.) We may assume $G$ isn't a torus, so it is not commutative and over $\bar{k}$ there are (by the classical structure theory over alg. closed fields) subgroups of type $SL_2$ or $PGL_2$ whose Lie algebras are not commutative (in any characteristic). Hence, $Lie(G)$ is not commutative. In the proof of 18.2(ii) in Borel's book he shows that $G$ is generated by finitely many of the centralizers $Z_G(X_i)^{\circ}$ for finitely many $X_i$ in $Lie(G)$ that are regular semisimple in a sense defined in 18.1. In particular, the discussion there implies (using that $Lie(G)$ is not commutative) that if $X\in Lie(G)$ is regular ss then $ad_G(X)$ is not zero and hence $Z_G(X)$ is not equal to $G$. Thus, the finitely many smooth connected $k$-subgroups $H_i = Z_G(X_i)^{\circ}$ that generate $G$ have strictly smaller dimension. But $Z_G(X)^{\circ}$ is always reductive for semisimple $X\in Lie(G)$ (13.19 in Borel's book), so by dimension induction we win. QED

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