31,794 reputation
366140
bio website math.umass.edu/~jeh
location U. Massachusetts, Amherst
age 75
visits member for 5 years, 1 month
seen 27 mins ago
More-or-less retired professor at UMass Amherst. Basically an algebraist with interests in Lie theory, specifically representation theory and related algebraic geometry.

16h
revised Root in positive Weyl chamber
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16h
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.
16h
revised Root in positive Weyl chamber
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revised Root in positive Weyl chamber
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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?
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Mar
26
revised Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
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Mar
26
revised Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
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Mar
25
comment Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
@Dave: I was just having some second thoughts, so I edited my answer. I may still be oversimplifying.
Mar
25
revised Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
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Mar
25
comment Automorphisms of B_n
This question should probably be asked on stackexchange.com rather than here. Assuming you are dealing with simple Lie algebras over a field such as $\mathbb{C}$, the automorphisms are well known and are all inner if there is no graph automorphism (as in type $B_n$).
Mar
25
answered Is an $\mathfrak{sl}_2$-triple determined up to Lie algebra automorphism by the adjoint representation?
Mar
25
revised Prescribed spherical representations, symplectic group $Sp(n)$
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Mar
25
answered Prescribed spherical representations, symplectic group $Sp(n)$
Mar
25
comment Generators of invariant polynomials of semisimple Lie algebra
This looks like a nice higher-level way to realize the algebraic results. By the way, the fact that the algebra of invariant polynomial functions is generated in all cases by powers of trace functions is quite classical. But Bourbaki's exercise makes a subtle refinement in the choice of a minimal set of generators for type $D_\ell$.