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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 $G$?

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  • $\begingroup$ 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. $\endgroup$
    – Jim Humphreys
    Commented May 6, 2011 at 22:45
  • $\begingroup$ 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). $\endgroup$
    – Jim Humphreys
    Commented May 20, 2011 at 17:25

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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.

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  • $\begingroup$ 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. $\endgroup$
    – asm
    Commented May 20, 2011 at 15:19
  • $\begingroup$ As @LaurentMoret-Bailly said, the tori are parameterised by $(G/N_G(T))(k)$. Your argument shows that the tori are usually not parameterised by $G(k)/N_G(T)(k)$, but that is usually only a proper subset of $(G/N_G(T))(k)$. The test for $k$-rationality of the coset of $N_G(T)$ containing $g \in G(\overline k)$ is exactly as you have described. $\endgroup$
    – LSpice
    Commented Jan 30, 2020 at 20:07

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