# Jim Humphreys

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bio website math.umass.edu/~jeh location U. Massachusetts, Amherst age 75 member for 5 years, 1 month seen 14 hours ago profile views 15,775
More-or-less retired professor at UMass Amherst. Basically an algebraist with interests in Lie theory, specifically representation theory and related algebraic geometry.

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 Apr17 awarded Nice Answer Apr16 awarded Nice Answer Apr11 comment One-Parameter Families of Indecomposable Representations of the Preprojective Algebra of type A5 It's probably useful to add the location for a free download of the 2005 paper: numdam.org/numdam-bin/fitem?id=ASENS_2005_4_38_2_193_0. Apr10 comment Rank of a special linear group over a finite field As Geoff says, Steinberg proved this statement in a much more general context for groups of Lie type, following a long history of developments which he refers to. His paper is here: cms.math.ca/10.4153/CJM-1962-018-0 Apr8 comment Quotient of Flag varieties @Dave: This seems to be most relevant to the question. It may be useful to cite the published version, though it's hard for most people to access and may not differ from the preprint: ams.org/mathscinet-getitem?mr=3011793 Apr7 answered Solvable Lie algebras: embedded in upper triangular matrices? Apr3 revised Bruhat order and the Robinson-Schensted correspondence added 38 characters in body Apr1 answered Character table of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$ Apr1 comment Equivalence of Lie subalgebras, within a (irreducible) representation Dietrich's comment is well-taken and points to a fundamental difficulty with this kind of broad question: every finite dimensional Lie algebra embeds in some general linear algebra by Ado/Iwasawa theorems, so a lot of Lie algebras embed in the trace zero subalgebra (which is simple in most characteristics and acts irreducibly on the underlying vector space). Mar31 revised Root in positive Weyl chamber edited body Mar31 comment Root in positive Weyl chamber @shu: Having gone back to the foundational material on root systems, I'm motivated to outline the full story more carefully. This is all fairly elementary, starting with the axioms, but it does need to be done systematically. Though some shortcuts are possible for the narrow question you've raised, the details in the outline are useful to know about. Mar31 revised Root in positive Weyl chamber added 1802 characters in body Mar28 revised Root in positive Weyl chamber added 343 characters in body Mar28 answered Root in positive Weyl chamber Mar27 comment Root in positive Weyl chamber @shu: This example is correct, since the highest root happens to coincide with $\rho$. But it is apparently the only such case. (I echoed Sasha's comment too quickly, since the highest root can lie inside the chamber $K$. But it doesn't usually coincide with $\rho$. So I don't quite understand what you asking.) Mar27 comment Subgroups generated by opposite root groups Have you consulted the classic 1965 IHES paper by Borel-Tits on the structure of (isotropic) reductive groups over arbitrary fields, or the later papers by Bruhat-Tits specializing to local fields? These are freely available online at numdam.org and could help to answer your questions. (Also, your boldface notation here is unhelpful. It's better to follow Borel-Tits.) Mar27 comment Root in positive Weyl chamber @shu: Your question isn't clear to me. The set $R \cap K$ is empty to begin with (as Sasha comments, only one or two roots can be dominant weights, and they don't lie in the interior of $K$). On the other hand, $\rho$ isn't a root but does lie in the (dominant) open Weyl chamber $K$. What are you trying to find a proof for without using classification? Mar27 comment Simply connected Lie groups homeomorphic to R^n are solvable @user61471: It would help to have more precise references for the "many proofs" you mention, for example in the short paper you mention in your comment: ams.org/mathscinet-getitem?mr=975639 Mar26 comment Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation? @Dave: Yes, I'm only considering simple Lie algebras. I've also tried to clarify my answer yet again but will give up edits at this point. Mar26 revised Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation? added 336 characters in body
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