bio | website | math.umass.edu/~jeh |
---|---|---|
location | U. Massachusetts, Amherst | |
age | 75 | |
visits | member for | 5 years, 1 month |
seen | 14 hours ago | |
stats | 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.
Apr 17 |
awarded | Nice Answer |
Apr 16 |
awarded | Nice Answer |
Apr 11 |
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. |
Apr 10 |
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 |
Apr 8 |
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 |
Apr 7 |
answered | Solvable Lie algebras: embedded in upper triangular matrices? |
Apr 3 |
revised |
Bruhat order and the Robinson-Schensted correspondence
added 38 characters in body |
Apr 1 |
answered | Character table of $\mathrm{SL}_2(\mathbb{Z}/p^n\mathbb{Z})$ |
Apr 1 |
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). |
Mar 31 |
revised |
Root in positive Weyl chamber
edited body |
Mar 31 |
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. |
Mar 31 |
revised |
Root in positive Weyl chamber
added 1802 characters in body |
Mar 28 |
revised |
Root in positive Weyl chamber
added 343 characters in body |
Mar 28 |
answered | Root in positive Weyl chamber |
Mar 27 |
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.) |
Mar 27 |
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.) |
Mar 27 |
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? |
Mar 27 |
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 |
Mar 26 |
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. |
Mar 26 |
revised |
Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
added 336 characters in body |