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Let $K$ be an algebraically closed field. Let $SL(n,K), GL(n,K)$ denote the special linear group and general group respectively, and $D(n,K)$ is the diagonal subgroup of $GL(n,K)$.

Then $SL(n,K) \cap D(n,K)$ must be a torus, i.e. a connected diagonalizable subgroup, of $SL(n,K)$. But, must it be maximal?

When char$K=0$, this seems to be the case. When $K=\mathbb{Z}/(2)$ and $n=2$, $SL(2,K) \cap D(2,K)=\{e\}$, here, $e$ denotes the identity element of the two groups. And $e$ is the only diagonalizable matrix in $SL(2,K)$, so this is verified.

But how to prove that this is true in general? And if is not true, what is its counterexample?

Many thanks~

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Let $T$ be a torus in $SL(n,K)$. Then $T$, considered as a subset of $M(n\times n,K)$, is a commuting set of diagonalizable matrices and as such is simultaneously diagonalizable, i.e. there is a $g\in GL(n,K)$ with $gTg^{-1}\subseteq SL(n,K)\cap D(n,K)$. Hence the dimension of any torus is bounded by the dimension of the intersection in question which shows that this intersection is indeed a maximal torus.

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  • $\begingroup$ Thank you very much. If I am not mistaken, any two maximal tori in a given algebraic group are conjugate. Must any conjugation of a maximal torus be a maximal torus? If the answer is yes, then the dimension consideration can be removed. $\endgroup$
    – ShinyaSakai
    Commented Jul 5, 2011 at 11:59
  • $\begingroup$ Sure, either again for dimension reasons, or if $T$ is a maximal torus and $T'$ is a torus containing $gTg^{-1}$, then $T$ is contained in $g^{-1}T'g$, hence they are equal and hence so are $gTg^{-1}$ and $T'$. $\endgroup$
    – Philipp Hartwig
    Commented Jul 6, 2011 at 8:57
  • $\begingroup$ @PhilippHartwig Can we generalize this? Namely, if $T$ is a maximal torus in $G$ and $H$ is a closed subgroup of $G$ such that the intersection $T\cap H$ is connected, does it follow that $T\cap H$ is a maximal torus in $H$? If not, what is missing? $\endgroup$
    – M Turgeon
    Commented May 2, 2012 at 19:15
  • $\begingroup$ @MTurgeon You need a closed normal(!) subgroup $H$ of $G$, then the proof carries over. If $H$ is not necessarily normal in $G$, the statement you are asking for fails: $H$ could be a maximal torus in $G$ different from $T$. $\endgroup$
    – Philipp Hartwig
    Commented May 3, 2012 at 6:10
  • $\begingroup$ @PhilippHartwig Fair enough! Thank you! $\endgroup$
    – M Turgeon
    Commented May 3, 2012 at 12:37

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