Timeline for Centralizers of regular elements are abelian
Current License: CC BY-SA 3.0
9 events
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| Nov 11, 2016 at 16:56 | comment | added | Jim Humphreys | Sorry to have overlooked this question. My edited answer may be more precise, since I am indeed thinking about regular nilpotent elements as the crucial case. | |
| Nov 2, 2016 at 23:52 | comment | added | Francois Ziegler | @JimHumphreys : Oh, $(∗)$ does not split over $Z_H(X)$ when $G=SL_2(\mathbf C)$ and $X=\operatorname{diag}(1,−1)$. For then $Z_G(X)=\{\operatorname{diag}(z,1/z):z\in\mathbf C^\times\}\simeq\mathbf C^\times$ covers $Z_H(X)\simeq\mathbf C^\times/\{\pm1\}$ nontrivially. So I guess you only meant the splitting for $X$ nilpotent. Where does one find algebraic proofs of 1) this, and 2) how commutativity follows in general? | |
| Nov 2, 2016 at 13:44 | history | edited | Francois Ziegler | CC BY-SA 3.0 |
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| Nov 2, 2016 at 13:29 | comment | added | Jim Humphreys | Yes, I misread Kostant's symbol. However, in this paper he is getting closer to the algebraic group viewpoint soon developed by Steinberg and Springer. He starts here with a reductive Lie algebra, though in the question it's semisimple, along with its adjoint group $G$. To get to an arbitrary semisimple group over $\mathbb{C}$ is then easy, since the center of such a group is finite. No need for a "discrete center" or other steps here. Note that Kostant's argument is essentially algebraic. | |
| Nov 2, 2016 at 6:19 | history | edited | Francois Ziegler | CC BY-SA 3.0 |
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| Nov 2, 2016 at 6:18 | comment | added | Francois Ziegler | @JimHumphreys : Kostant's cited and linked Prop. 14 (p. 362) deals with all (regular) $X$, not just nilpotent! See his definition of $\mathfrak r$ (3.4.3, p. 358). As to "necessity" of the above argument: I was happy with sufficient. If, as the last § of your answer suggests, there is an algebraic proof that $(*)$ actually splits over $Z_H(X)$, so that $Z_G(X)=C\times Z_H(X)$, I haven't seen it -- but would love to. | |
| Nov 1, 2016 at 20:48 | comment | added | Jim Humphreys | A couple of other belated comments. Kostant's 1959 and 1963 papers are pioneering but also somewhat improvised. His notion of "regular" shifted from (semisimple) elements having a Cartan subalgebra as centralizer to the more general notion which Steinberg then generalized to an algebraic setting. Since only semisimple Lie groups are mentioned in the question, Ad has finite kernel equal to the center, so your analytic arguments are unnecessary. On the other hand, the theorem of Kostant you cite deals just with "nilpotent" elements, but the question includes all $X$. | |
| Oct 31, 2016 at 9:16 | vote | accept | SHP | ||
| Oct 31, 2016 at 5:26 | history | answered | Francois Ziegler | CC BY-SA 3.0 |