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Let $G$ be a connected linear algebraic group over an algebraically closed field $k$ of characteristic 0. Let $D\subset G$ be a closed diagonalizable subgroup of $G$ (a subgroup of multiplicative type). Is it true that $D$ is contained in some torus $T\subset G$?

This is so for $G=\mathrm{GL}_n$. Is this true for any connected linear $G$ (or any connected reductive $G$)?

I am stuck with this simple question...

Edit. The answer to the original question is NO, see Angelo's answer. However, is it true that any cyclic finite diagonalizable subgroup $C$ of $G$ is contained in some torus $T\subset G$?

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For the cyclic case: If $D = \langle s \rangle$ is a cyclic diagonalizable subgroup of a connected linear algebraic 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 $G$. Indeed, by [Borel LAG,11.10] $s$ is contained in a Borel subgroup of $G$, and then the claim follows from the connected solvable case [Borel LAG,10.6]. – George McNinch Apr 8 '11 at 13:15
The answer for cyclic finite diagonalizable groups is affirmative for connected reductive $G$; this is Lemma 7.1 in the Appendix of (where $\mu_n$ is written, but the initial reduction to algebraically closed ground field does not use that the cyclic group is split -- i.e., constant Cartier dual -- and so it gives the result in general). – nfdc23 Nov 2 '15 at 15:33
up vote 13 down vote accepted

No. For example, $\mathrm{PGL}_n$ contains a subgroup $G$ isomorphic to the product of two cyclic subgroups of order $n$, generated by the classes of the diagonal matrix whose entries are the powers of a fixed primitive $n^{\rm th}$ root of 1, and the permutation matrix corresponding to a cycle of length $n$. The inverse image of this subgroup in $\mathrm{GL}_n$ is not commutative, while the inverse image of a maximal torus in $\mathrm{PGL}_n$ is a maximal torus in $\mathrm{GL}_n$, so $G$ is not contained in a torus.

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Thank you, Angelo! – Mikhail Borovoi Apr 6 '11 at 12:58

To reinforce Angelo's example, it's worthwhile to point out the broader setting for this kind of question: the study of centralizers and connectedness properties in a semisimple (or more generally reductive) algebraic group. An older but very useful source is part II of the extensive notes by T.A. Springer and R. Steinberg on conjugacy classes, part of an IAS seminar (Lect. Notes in Math. 131, Springer, 1970). A crucial question is whether a given connected semisimple group is simply connected or not; this shows up in the standard example where the adjoint group $\mathrm{PGL}$ fails to be simply connected. Here you have the deep theorem: If $G$ is a connected, simply connected algebraic group over an algebraically closed field, then all centralizers of semisimple elements are connected. (It's elementary on the other hand to prove that all centralizers in a general linear group are connected.) The role of the characteristic of the field is also discussed in depth by Springer and Steinberg, as well as the role of "torsion primes" (treated mpre fully in Steinberg's 1975 Advances paper).

Some of the results are written up in later textbooks and in the first two chapters of my 1990 AMS book Conjugacy Classes in Semisimple Algebraic Groups (with the relevant example for the question here given in 1.12).

ADDED: To answer the added question, in any connected algebraic group it's true that an arbitrary semisimple element and hence the cyclic subgroup it generates lies in some maximal torus. This is part of the standard development of Borel-Chevalley structure theory (see for example Section 22.3 of my book Linear Algebraic Groups), though it does take a while to get that far into the theory.

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The connectedness theorem you quote shows that if G is simply connected and $D$ is a smooth diagonalizable subgroup scheme, it is contained in a maximal torus. I believe it is still true for any -- possibly non-reduced -- diagonalizable subgroup scheme $D$ that the centralizer of $D$ in $G$ is connected when $G$ is simply connected; I'm unaware of an existing reference, though. (In fact, I have some notes on related matters that are waiting to be written up carefully...) – George McNinch Apr 6 '11 at 11:36
Thank you, Jim! – Mikhail Borovoi Apr 6 '11 at 12:59
@George McNinch: Could you please explain, HOW the connectedness theorem quoted by Jim Humphreys shows that if $G$ is simply connected and $D$ is a smooth diagonalizable subgroup scheme, then $D$ is contained in a maximal torus. – Mikhail Borovoi Apr 7 '11 at 9:07
Sorry - my argument in support of that remark wasn't correct. – George McNinch Apr 7 '11 at 13:18
@George Also the assertion of your remark is not correct, see Jim's answer to my question – Mikhail Borovoi Apr 8 '11 at 6:41

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 {$(x,y) \in O(U) \times O(W) \mid \det(x) = \det(y)$}. 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).

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Thank you, George! – Mikhail Borovoi Apr 11 '11 at 20:19

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