Let T be a non-trivial torus in a semisimple group G of adjoint type defined over Q. I was wondering whether there exists a reductive subgroup H of G (defined over Q !) such that T is a maximal torus of H and such that T is not in the centre of H.
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$\begingroup$ At least in the split case, I think the answer to your question is sometimes no (but would have to analyze it using the standard structure theory). It's useful to distinguish the two possible cases, where your given torus is regular or singular. In the singular case I don't see how a subgrooup $H$ can exist, even in the first nontrivial case of type $A_2$. $\endgroup$– Jim HumphreysCommented Oct 28, 2013 at 23:21
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Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H_0$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, whose maximal subgroup is $T \times T'$. Taking $H$ the derived group of $H_0$ will kill the center of $H_0$. $T' \times T$ meets $H$ against a maximal torus of $H$, which is the intersection of the roots of $G$ whose kernel contains $T$.
Using this process, you do not really get $T$ as maximal torus, but rather a kind of “arithmetic closure” of it. Is this close enough?
answered Oct 28, 2013 at 14:50
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$\begingroup$ Many thanks for this reply! I just do not quite see why T is a maximal torus of this H; isn't T X T' in H? $\endgroup$ Commented Oct 28, 2013 at 15:08
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$\begingroup$ My bad, you need to take the derived group! $\endgroup$ Commented Oct 28, 2013 at 15:10
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$\begingroup$ This construction does not seem to work. Let $S$ be a maximal torus in $G$ and let $\lambda:{\rm{GL}}_1 \rightarrow S$ be a "generic" cocharacter. Then $Z_G(\lambda) = S$ and its derived group is 1. So if one begins with $T = \lambda({\rm{GL}}_1)$ and chooses $T'$ an isogeny complement to $T$ in $S$ then $T' \cdot T = S$, etc. Am I misunderstanding the proposed answer? $\endgroup$– MarguaxCommented Oct 28, 2013 at 15:18
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$\begingroup$ Maybe I should add that I am interested in the case where T is anisotropic over Q... $\endgroup$ Commented Oct 28, 2013 at 15:27
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$\begingroup$ @Marguax You are right, the construction I proposed only works in the case of where $T$ is degenerated. $\endgroup$ Commented Oct 28, 2013 at 15:27