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Jun 2, 2021 at 14:13 comment added LSpice Name of the particular paper: Friedman and Morgan - Automorphism sheaves, spectral covers, and the Kostant and Steinberg sections.
Nov 2, 2016 at 6:19 history edited Francois Ziegler
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Oct 31, 2016 at 18:56 comment added David Ben-Zvi @SHP: I don't think (Lie algebra) regular centralizers (which have the same dimension as a Cartan subalgebra) are contained in Cartan subalgebras unless $X$ is itself regular semisimple (so that its centralizer is a Cartan). For example the Lie algebra centralizer of a regular nilpotent element (such as the $X$ in your post) always consists entirely of nilpotent matrices.
Oct 31, 2016 at 14:17 answer added Jim Humphreys timeline score: 6
Oct 31, 2016 at 13:34 history edited Francois Ziegler CC BY-SA 3.0
Removed unnecessary simple connectedness assumption
Oct 31, 2016 at 9:16 vote accept SHP
Oct 31, 2016 at 9:15 comment added SHP @PeterMcNamara Right, thanks. The correct argument is that $Z_{\mathfrak{g}}(X)$ is contained in the generalized $0$-weight space of $\mathrm{ad}(X)$ which is a Cartan subalgebra since $X$ is regular.
Oct 31, 2016 at 5:26 answer added Francois Ziegler timeline score: 6
Oct 31, 2016 at 0:03 comment added Peter McNamara A small correction: The centraliser $Z_{\mathfrak{g}}(X)$ is not always a Cartan subalgebra when X is regular (it's not self-normalising when X isn't semisimple). But for regular X, this centraliser is still abelian, so the argument that the identity component of $Z_G(X)$ is abelian still holds, without assuming simply-connectedness.
Oct 30, 2016 at 17:11 history edited SHP CC BY-SA 3.0
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Oct 30, 2016 at 17:02 answer added Will Sawin timeline score: 2
Oct 30, 2016 at 17:02 history edited SHP CC BY-SA 3.0
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Oct 30, 2016 at 16:24 history asked SHP CC BY-SA 3.0