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I was wondering if there is any obvious reason or quick proof that for every $g\in GL_n$ the centralizer $Z_{GL_n}(g)$ is connected. Also I wanted to see why for any semisimple $s\in Sp_{2n}$ the centralizer $Z_{Sp_{2n}}(s)$ is connected. Thanks.

flag	asked May 16 2011 at 13:14 HNuer 915●2●7

	Milne, ALA ( jmilne.org/math/CourseNotes/ala.html ), I, Exercise 18-1: "Use the criterion (18.1) to show that the centralizer of a torus in a connected algebraic group is connected." Now the question is whether you can find, for your element $g$ or $s$, a torus which has the same centralizer as the element. Probably something like the closure of the set of powers of $g$ (rsp. $s$)? – darij grinberg May 16 2011 at 15:09
	Thanks to Brian Conrad for explaining me why the procedure I proposed in the comment above doesn't work. The closure of the set of powers of $g$ needs not be a torus. – darij grinberg May 17 2011 at 18:52

Jim Humphreys · Answer 1 · 2011-05-16 17:13:02Z UTC

Centralizers of arbitrary elements in a general linear group (over an arbitrary algebraically closed field) are connected for an easy reason: the centralizer in the space of $n \times n$ matrices is just a subspace, while the centralizer in $GL_n$ is a principal open set in this affine space (nonzeros of the determinant polynomial) and hence also connected.

Symplectic or other semisimple groups require a much more subtle approach, though there may well be a fairly direct approach in the symplectic case (at least in characteristic 0). A basic theorem due to Springer and Steinberg states: In a connected and simply connected semisimple algebraic group, over an arbitrary algebraically closed field, the centralizer of every semisimple element is connected. This theorem is still waiting for a really transparent proof, but it's written up in Chapter 2 of my 1995 AMS book Conjugacy Classes in Semisimple Algebraic Groups following Steinberg's papers and Tata lecture notes. (For the general linear case above, see 1.2 in that book.)

P.S. Maybe I should emphasize that I'm using algebraic group language, in the spirit of the tag here, so that a closed connected subgroup corresponds to having an underlying irreducible affine variety structure: the regular functions form a domain. Over $\mathbb{C}$ this notion of "connected" agrees with the topological one (and similarly the algebraic notion of "simply connected" agrees with the topological one in the case of semisimple groups), as shown by Chevalley and Borel. But working over $\mathbb{R}$ is more complicated.

Neil Strickland · Answer 2 · 2011-05-16 13:54:45Z UTC

I'll assume you're working with the abstract group $GL_n(\mathbb{C})$. Probably you could do a similar thing with algebraic groups. Let $\lambda$ be a complex number whose argument is different from the arguments of any of the eigenvalues of $g$. Then the straight line from $g$ to $-\lambda I_n$ lies wholly in $Z_{GL_n}(g)$. It is then easy to connect $-\lambda I_n$ to $I_n$ in $ZGL_n\subseteq Z_{GL_n}(g)$.

Connectedness of Centralizers in $GL_n$

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