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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~

Jack Schmidt
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ShinyaSakai
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1 Answers1

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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.

Philipp Hartwig
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  • 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. – ShinyaSakai Jul 05 '11 at 11:59
  • 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 $T'$. – Philipp Hartwig Jul 06 '11 at 08:57
  • @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? – M Turgeon May 02 '12 at 19:15
  • @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 $T$. – Philipp Hartwig May 03 '12 at 06:10
  • @PhilippHartwig Fair enough! Thank you! – M Turgeon May 03 '12 at 12:37
  • @PhilippHartwig sorry for late commenting this question, why we have that $gTg^{-1} \subset SL(n,K)$? – Riccardo Nov 27 '14 at 10:38
  • @Riccardo Because $SL(n,K)\subset GL(n,K)$ is a normal subgroup. – Philipp Hartwig Dec 14 '14 at 17:08