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edited Apr 11 2011 at 21:52

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

occurred Apr 11 2011 at 14:19

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edited Apr 11 2011 at 14:19

George McNinch
1,930●1●5●11

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).

deleted 1255 characters in body

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edited Apr 8 2011 at 13:07

George McNinch
1,930●1●5●11

diagonalizable subgroup of a reductive connected linear algebraic group $G$, then $s$ is ahence $D$ is contained in a maximal torus of $G$ (e.g. G$. Indeed, by [BorelLAG,11.10] $s$ is contained in a Borel subgroup , of $G$, and then the claimfollows from the connected solvable case [Borel LAG,10.6])LAG,10.6].

Second, to repent somewhat for my earlier goof, I note the following:

Claim: There are simply connected groups containing non-toral diagonalizable subgroups.

Let me construct earlier posted an example: let $k$ alg. closed of characteristic $\ne 2$ and let $M$ be theorthogonal group $SO(7) = SO(V,\beta)$ over $k$. Of course, $M$ is notsimply connected. I want to show first answer claiming that $M$ contains the centralizer of adiagonalizable subgroup $D_1$ of order $4$ contained in no maximal torus of $M$. To constructsuch a $D_1$, write $V$ as an orthogonal sum $V = V_1 \perp V_0$ with $\dim V_1 = 3$ and $\dim V_0 = 4$, wherethe restriction to each $V_i$ of the form $\beta$ is non-degenerate.

Now define an involution $s \in M$ acting as the identityon $V_1$ and as multiplication by $-1$ on $V_0$.Note that $C_M(s)$identifies with {$(g_0,g_1) \in O(V_0) \timesO(V_1) \mid \det(g_0) = \det(g_1)$}. In particular, $C_M(s)$is not G$ was connected ; it has component group of order 2. Choose aninvolution $t \in C_M(s) \setminus C_M(s)^0$ and set $D_1 =\langle s,t \rangle$. Then $D_1$ G$ is a commutative group of order 4simply connected. Since$C_M(s)$ contains a maximal torus of $M$, and since $t$ is notcontained in any maximal torus of $C_M(s)$, it follows that $D_1$ isnot contained The argument I gave was wrong, because while the centralizer in any maximal torus of $M$.

Now let $G$ be a (simply connected) simple group of type $G_2$ over$k$. Then $G$ contains asingle semisimple element $t$ of finite order whosecentralizer $M$ is isomorphic to $SO(7)$. always connected (One can see this subgroup for $M$ using the fact G$ simply connectedand reductive), in general that the extended Dynkin diagram of type $G_2$ contains $B_2$ as a "sub-diagram". That subdiagram corresponds centralizer -- when the char is not 2 while reductive -- to the Dynkin diagram of a semisimple subgroup which is not againsimply connected (so the centralizer of a finite-order semisimple elementinduction I described fails).And note

I think it is not just the case that this subgroup $M$ my proof was wrong, but indeedthere are diagonalizable subgroups of type $B_2$ can't be simply connected e.ggroups withdisconnected centralizers. because it acts faithfuly on the 7-dimensional representation of $G=G_2$. Probably there This is an explicit way to see this element $t$and its centralizer using the description of $G$ as automorphisms of suggested by the Cayley algebraresult[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 don't seem made a hash of it. I hope toknow how return to do it...)

Then evidently this when I have a bit more time... but I thought I'd post thediagonalizable subgroup $D=\langle t,D_1 \rangle$ is contained inno maximal torus of $G$.response for the "cyclic" case, anyhow.

Post Deleted by George McNinch

occurred Apr 7 2011 at 23:00

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answered Apr 7 2011 at 20:01

George McNinch
1,930●1●5●11

I have two comments to make:

First, the cyclic case: If $D = \langle s \rangle$ is a cyclic diagonalizable subgroup of a reductive group $G$, then $s$ is a semisimple element of $G$ (of finite order). In particular, $s$ and hence $D$ is contained in a maximal torus of $G$ (e.g. by [Borel LAG,11.10] $s$ is contained in a Borel subgroup, and then the claim follows from the connected solvable case [Borel LAG,10.6]).

Second, to repent somewhat for my earlier goof, I note the following:

Claim: There are simply connected groups containing non-toral diagonalizable subgroups.

Let me construct an example: let $k$ alg. closed of characteristic $\ne 2$ and let $M$ be the orthogonal group $SO(7) = SO(V,\beta)$ over $k$. Of course, $M$ is not simply connected. I want to show first that $M$ contains a diagonalizable subgroup $D_1$ of order $4$ contained in no maximal torus of $M$. To construct such a $D_1$, write $V$ as an orthogonal sum $V = V_1 \perp V_0$ with $\dim V_1 = 3$ and $\dim V_0 = 4$, where the restriction to each $V_i$ of the form $\beta$ is non-degenerate.

Now define an involution $s \in M$ acting as the identity on $V_1$ and as multiplication by $-1$ on $V_0$. Note that $C_M(s)$ identifies with {$(g_0,g_1) \in O(V_0) \times O(V_1) \mid \det(g_0) = \det(g_1)$}. In particular, $C_M(s)$ is not connected; it has component group of order 2. Choose an involution $t \in C_M(s) \setminus C_M(s)^0$ and set $D_1 = \langle s,t \rangle$. Then $D_1$ is a commutative group of order 4. Since $C_M(s)$ contains a maximal torus of $M$, and since $t$ is not contained in any maximal torus of $C_M(s)$, it follows that $D_1$ is not contained in any maximal torus of $M$.

Now let $G$ be a (simply connected) simple group of type $G_2$ over $k$. Then $G$ contains a semisimple element $t$ of finite order whose centralizer $M$ is isomorphic to $SO(7)$.

(One can see this subgroup $M$ using the fact that the extended Dynkin diagram of type $G_2$ contains $B_2$ as a "sub-diagram". That subdiagram corresponds -- when the char is not 2 -- to the Dynkin diagram of a semisimple subgroup which is the centralizer of a finite-order semisimple element. And note that this subgroup $M$ of type $B_2$ can't be simply connected e.g. because it acts faithfuly on the 7-dimensional representation of $G=G_2$. Probably there is an explicit way to see this element $t$ and its centralizer using the description of $G$ as automorphisms of the Cayley algebra, but I don't seem to know how to do it...)

Then evidently the diagonalizable subgroup $D=\langle t,D_1 \rangle$ is contained in no maximal torus of $G$.