Suppose $G$ is a reductive algebraic group defined over a field $k$.
Is this right?
If so, it means for each maximal $k$-split $k$-tori $T$ there are many maximal $k$-anisotropic ones $T'$ that commute with $T$? Finitely many?
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4$\begingroup$ In $G=GL(2)$ over $k=\mathbf R$, you have two kinds of maximal tori, those which are split and those which are not. The former are $\mathbf R$-isomorphic to the diagonal torus, while the latter are $\mathbf R$-isomorphic to the algebraic subgroup of matrices of the form $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$ with $(a,b)\neq (0,0)$. However, the former are not $\mathbf R$-isomorphic to the latter, because an $\mathbf R$-isomorphism won't change the fact for a torus to be split or non-split. $\endgroup$– ACLCommented Oct 1, 2015 at 22:14
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1$\begingroup$ As ACL points out, even the most familiar matrix groups over non-algebraically closed fields $k$ show what the problem is. The group $\mathrm{SL}_2$ over a finite field $k$ usually has both $k$-split and $k$-anistropic maximal tori. While all of those $k$-tori (of dimension 1) are conjugate over an algebraic closure of $k$, they need not be conjugate over $k$ itself: not every semisimple matrix can be diagonalized over $k$. $\endgroup$– Jim HumphreysCommented Oct 1, 2015 at 22:57
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2$\begingroup$ @Jerry: I don't understand your "If so ...." part, since a maximal $k$-torus will typically be its own centralizer in $G$. $\endgroup$– Jim HumphreysCommented Oct 1, 2015 at 23:02
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$\begingroup$ Thanks, now I understand the picture. Forget about the "if so.." part, I agree that it doesn't make sense. $\endgroup$– JerryCommented Oct 2, 2015 at 17:35
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1$\begingroup$ I'm not even sure what the question is any more but if the maximal $k$-split torus in $G$ is trivial then any maximal torus, of which there could certainly be infinitely many, will commute with it. $\endgroup$– ericCommented Oct 3, 2015 at 14:28
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