Let $G$ be a split symplectic group over a number field $K$, and let $T \subset G$ be a maximal torus defined over $K$ (not necessarily a split torus). The long roots of $T$ form a $Gal(\overline{K}/K)$ stable set. Let $H$ be the subgroup generated by $T$ and the root subgroups corresponding to the long roots.$H$ is isogenous to $SL_2^n$ over $\overline{K}$. Will $H$ be quasisplit over $K$?
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1$\begingroup$ The question needs more details (and more precision). For example, the context suggests that you require $T$ to be defined over $K$. And "long roots of $T$" doesn't quite make sense: try "long roots of $G$ relative to $T$""? Also, $H$ is generated not by the roots (along with $T$) but by the corresponding 1dimensional root subgroups of $G$ (along with $T$). Can you say more explicitly what you mean by "geometrically isogenous" and why $H$ should be related in this way to a product of rank 1 groups? What is $n$? And what if $G$ has rank 2? $\endgroup$ – Jim Humphreys Oct 19 '15 at 21:17

$\begingroup$ Thanks for pointing out the imprecisions  I've edited my question to make it more precise :) So  working inside the Lie algebra of G, the direct sum of the root spaces corresponding to the long roots and the lie algebra of $T$ is closed under the lie bracket. Picking a Borel of $H$ (not necessarily over $K$) containing $T$, it's true that the root spaces corresponding to different positive roots commute with each other, whence H has to be isogenous to a power of $SL_2$ over $\overline{K}$ $\endgroup$ – user81686 Oct 19 '15 at 21:26

$\begingroup$ The formulation is still unclear to me. In particular, I don't see where your description of $H$ comes from. Subgroups of maximal rank (such as $H$) in a symplectic group are wellunderstood and don't include a product of $n$ commuting rank 1 groups. (assuming $n$ is the rank). Also, "quasisplit" is not appropriate in a situation where the groups involved have split but not quasisplit forms. Is there a reason for $K$ to be a number field here? (You need different tags, such as 'algebraicgroups' and perhas 'nt.numbertheory'. The question doesn't involve representation theory.) $\endgroup$ – Jim Humphreys Oct 20 '15 at 14:00