4 typo fix!

Here is another example similar to Angelo's construction of a non-toral diagonalizable subgroup of a reductive group. I'll suppose that the characteristic is not 2. Let $G = SO(V) = SO(V,\beta)$ for $\dim V > 2$, and write $V$ as an orthogonal sum $V = U \perp W$ for $0 < \dim U < \dim V$ with $\dim U$ even, such that the restriction of $\beta$ to $U$ and $W$ is non-degenerate.

Let $t \in G$ act as the identity on $W$ and as $-1$ on $U$. Then the centralizer $M=C_G(t)$ identifies with the subgroup {$(x,y) \in O(U) \times O(VO(W) \mid \det(x) = \det(y)$}. In particular, this centralizer is not connected: $M/M^0$ has order 2.

One can evidently choose an involution $s \in M \setminus M^0$, and then $D = \langle t,s\rangle$ is a diag. subgroup of $G$ which is contained in no maximal torus.

Part of this construction can be made in char. 2. Instead of $t$, you have to take a non-smooth subgroup $\mu \simeq \mu_2$, essentially given by the action of a semisimple element $X \in \operatorname{Lie}(G)$ ($X$ should act as $1$ on $U$ and $0$ on $W$). Then $M=C_G(\mu) = C_G(X)$ is again disconnected (well, now you can't argue by determinants) with component group of order $2$. But this doesn't seem to lead to a non-toral diagonalizable subgroup (any finite order element representating the non-trivial coset of $M/M^0$ has a non-trivial unipotent part).

Post Undeleted by George McNinch
3 deleted 31 characters in body; edited body

I have two comments to make:

First, the cyclic case: If $D = \langle s \rangle$

Here is another example similar to Angelo's construction of a cyclicnon-toral diagonalizable subgroup of a connected linear algebraic reductive group$G$, then $s$ . I'll suppose that the characteristic is asemisimple element of $G$ (of finite order)not 2.In particular, Let $s$ G = SO(V) = SO(V,\beta)$for$\dim V > 2$, and hence write$D$is contained in a maximal torus of V$ as an orthogonal sum$G$. IndeedV = U \perp W$for$0 < \dim U < \dim V$with$\dim U$even,by [BorelLAG,11.10] such that the restriction of$s$\beta$ to $U$ and $W$ is contained non-degenerate.

Let $t \in a Borel subgroup of G$ act as the identity on $G$, W$and then as$-1$on$U$. Then the follows from centralizer$M=C_G(t)$identifies with the subgroup {$(x,y) \in O(U) \times O(V) \mid \det(x) = \det(y)$}. In particular,this centralizer is not connectedsolvable case [Borel LAG,10.6]:$M/M^0$has order 2. Second, I earlier posted One can evidently choose an answer claiming that the centralizer of involution$s \in M \setminus M^0$, and then$D = \langle t,s\rangle$is a diagonalizable diag. subgroup of$G$was connected in$G$which is simply connectedcontainedin no maximal torus.The argument I gave was wrong, because while the centralizer Part of this construction can be made in char. 2. Instead of$G$t$, you haveto take a non-smooth subgroup $\mu \simeq \mu_2$, essentially given bythe action of a single semisimple element is always connected $X \in \operatorname{Lie}(G)$ (for $X$ shouldact as $G$ simply connected1$on$U$and reductive), in general that centralizer -- while reductive --$0$on$W$). Then$M=C_G(\mu) = C_G(X)$is not againsimply connected (so the induction I described fails). I think it is not just the case that my proof was wrong, but indeedthere are diagonalizable subgroups of simply connected groups withdisconnected centralizers. This is suggested (well, now you can't argue by the result[Springer-Steinberg, Theorem II, 5.8] that Jim mentioned in a commentfollowing my incorrect proof. Anyhow, I tried yesterday to write down anexample, but yesterday was a busy day and I made a hash determinants) with componentgroup of it. I hope order$2$. But this doesn't seem to return lead to this when I have a bit more time... but I thought I'd post theresponse for non-toral diagonalizablesubgroup (any finite order element representating the "cyclic" case, anyhownon-trivialcoset of$M/M^0\$ has a non-trivial unipotent part).

2 deleted 1255 characters in body
Post Deleted by George McNinch
1