Diagonalizable subgroups of a connected linear algebraic group - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T12:21:12Z http://mathoverflow.net/feeds/question/60781 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60781/diagonalizable-subgroups-of-a-connected-linear-algebraic-group Diagonalizable subgroups of a connected linear algebraic group Mikhail Borovoi 2011-04-06T08:00:44Z 2011-04-11T21:52:20Z <p>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 $T\subset G$?</p> <p>This is so for $G=\mathrm{GL}_n$. Is this true for any connected linear $G$ (or any connected reductive $G$)?</p> <p>I am stuck with this simple question...</p> <p><strong>Edit.</strong> The answer to the original question is NO, see Angelo's answer. However, is it true that any <em>cyclic</em> finite diagonalizable subgroup $C$ of $G$ is contained in some torus $T\subset G$?</p> http://mathoverflow.net/questions/60781/diagonalizable-subgroups-of-a-connected-linear-algebraic-group/60789#60789 Answer by Angelo for Diagonalizable subgroups of a connected linear algebraic group Angelo 2011-04-06T09:53:09Z 2011-04-06T09:53:09Z <p>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.</p> http://mathoverflow.net/questions/60781/diagonalizable-subgroups-of-a-connected-linear-algebraic-group/60801#60801 Answer by Jim Humphreys for Diagonalizable subgroups of a connected linear algebraic group Jim Humphreys 2011-04-06T11:16:07Z 2011-04-07T13:02:46Z <p>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 <em>simply connected</em> or not; this shows up in the standard example where the adjoint group <code>$\mathrm{PGL}$</code> fails to be simply connected. Here you have the deep theorem: If <code>$G$</code> 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 <em>Advances</em> paper). </p> <p>Some of the results are written up in later textbooks and in the first two chapters of my 1990 AMS book <em>Conjugacy Classes in Semisimple Algebraic Groups</em> (with the relevant example for the question here given in 1.12). </p> <p>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 <em>Linear Algebraic Groups</em>), though it does take a while to get that far into the theory. </p> http://mathoverflow.net/questions/60781/diagonalizable-subgroups-of-a-connected-linear-algebraic-group/60991#60991 Answer by George McNinch for Diagonalizable subgroups of a connected linear algebraic group George McNinch 2011-04-07T20:01:46Z 2011-04-11T21:52:20Z <p>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 $0 &lt; \dim U &lt; \dim V$ with $\dim U$ <em>even</em>, such that the restriction of $\beta$ to $U$ and $W$ is non-degenerate.</p> <p>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. </p> <p>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.</p> <p>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).</p>