MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be a finite field and $G$ a connected, reductive linear algebraic group defined over $k$. It is well-known that the union of the maximal tori of $G$ is dense in $G$ (more generally, if $G$ is not reductive, the union of its Cartan subgroups is dense).

It is also true, that $G$ is generated by the maximal tori defined over $k$ (this is proven in a paper by Borel and Springer from 1968).

However, is the union of the maximal tori which are defined over $k$ still dense in $G$?

share|cite|improve this question
The Tohoku paper by Borel and Springer (1968) is valuable for its treatment of finer points about rationality properties of linear algebraic groups and their Lie algebras which don't usually fit into textbooks. It's actually part II of a paper for which part I appears in the proceedings (1966) of the 1965 AMS Summer Institute in Boulder. Part II, including generation of a connected reductive group over a finite field by maximal tori defined over that field (a nontrivial result), is based partly on older work by Rosenlicht and then Grothendieck but doesn't use scheme language explicitly. – Jim Humphreys May 6 '11 at 22:45
Concerning the accepted answer and the newer proposed one, the key point is just the finiteness of the number of maximal tori defined over $k$ (so their union is closed and of dimension equal to the rank of $G$): in concrete terms, such a torus is the zero set in $G$ of finitely many polynomials with coefficients in $k$. For more detailed discussion see for instance the Springer-Steinberg article (section II.1) in the 1970 Lect. Notes in Math. 131 (Springer). – Jim Humphreys May 20 '11 at 17:25
up vote 5 down vote accepted

The set of maximal tori defined over $k$ is finite: if $T$ is one of them, they correspond to the $k$-rational points of the $k$-variety $G/\text{(normalizer of }T)$. So their union is not dense, unless $G$ is a torus.

share|cite|improve this answer

I don't see why they would correspond to the $k$-rational points of $G$/(normalizer of $T$). A conjugate of $T$ can be defined over $k$ even if the conjugating matrix $g$ is not $k$-rational. It is sufficient that $g \phi(g)^{-1}$ is contained in the normalizer of $T$, where $\phi$ denotes the Frobenius.

Nevertheless, I believe there are only finitely many since their $k$-conjugacy classes correspond to certain equivalence classes of the Weyl group.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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