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Dec 16, 2015 at 11:24 comment added question @Paul: Do you mean that the torus of $\Omega_{2l}(K,f)$ is $Diag(t_1,\cdots, t_l,t_1^{-1},\cdots , t_l^{-1})$ when $K$ is not algebraically closed and the torus of $SO_{2l}(K,f)$ is in general bigger than it?
Dec 8, 2015 at 19:11 answer added Jim Humphreys timeline score: 3
Dec 8, 2015 at 16:49 comment added Paul Levy In answer to your question, the commutator subgroup $\Omega_{2l}(K,f)$ is clearly contained in ${\rm SO}_{2l}(K,f)$ (take determinants). They are equal for an algebraically closed field, but otherwise need not be, e.g. for the case $l=1$ we have $\Omega_{2l}(K,f)=\{ {\rm Diag}(t,t^{-1}):t\in (K^*)^2\}\subsetneq {\rm SO}_{2l}(K,f)$.
Dec 8, 2015 at 16:32 comment added Paul Levy Although Carter uses the same symbol $\overline{G}$ for the universal Chevalley group and the subgroup of ${\rm GL}_{2l}(K)$ generated by the subgroups ${\rm exp} (te_r)$, they aren't actually the same in type $D_l$: the centre of the former group is a Klein four-group, while the centre of the latter group is cyclic of order 2.
Dec 8, 2015 at 12:43 review First posts
Dec 8, 2015 at 12:56
Dec 8, 2015 at 12:41 history asked question CC BY-SA 3.0