Timeline for Cyclic vectors in irreducible representations of simple Lie algebras
Current License: CC BY-SA 4.0
10 events
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| Jul 23, 2019 at 20:11 | comment | added | YCor | The definition in the Elashvili-Jibladze-Kac paper (ArXiv link) is that of a cyclic element with respect to a given nilpotent element, so I don't see an immediate confusion. Actually in general, there's a wide-spread confusion, namely calling cyclic both endomorphisms and vectors (an endomorphism is cyclic if it admits a cyclic vector...). This confusion is worse in the context of Lie algebras, where elements play both the role of endomorphisms and vectors. An advantage of "cyclic" on "special" is that the terminology gives a hint to the definition. | |
| Jul 23, 2019 at 19:19 | history | edited | Jim Humphreys | CC BY-SA 4.0 |
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| Jul 6, 2019 at 0:44 | comment | added | Jim Humphreys | @AThomas: Your formulation is definitely not optimal, starting with the notion of "cyclic" element of the simple Lie algebra: are there any such elements, and do they lead to interesting theorems? More importantly, in your basic definition "the whoe vector space" is ambiguous, and referring to a matrix reqjuires a choice of bssis. Maybe it's better, in view of Victor's answer, to refer to the elements of the Lie algebra as something like "special" rather than "cyclic"? Anyway, existence of such elements is the first goal. One tool is the Chevalley-Jordan decomposition. | |
| Jul 5, 2019 at 9:10 | comment | added | AThomas | Sorry about the misleading title. I'm interested in cyclic elements of the Lie algebra. | |
| Jul 4, 2019 at 23:12 | comment | added | YCor | I repeat, the question is whether an irreducible $\mathfrak{g}$-module $V$ is cyclic as module of the 1-dimensional Lie algebra generated by $\mathrm{ad}(x)$ for given $x\in\mathfrak{g}$ (and not whether it's cyclic as $\mathfrak{g}$-module). For instance when $V$ is the adjoint rep of $\mathfrak{sl}_d$, the answer is different for $d=2$ and $d\ge 3$. For $d=2$ it holds for every $x\neq 0$; for $d\ge 3$ it holds for no $x$. | |
| Jul 4, 2019 at 23:04 | comment | added | Jim Humphreys | @YCor: I don't understand your comment. Why does it matter that $x$ is diagonal (any element will do provided it's nonzero, to generate the simple Lie algebra under the irreducible adjoint rep)? | |
| Jul 4, 2019 at 13:35 | comment | added | YCor | Well, what I mean is that if you take any diagonal $x\in\mathfrak{g}=\mathfrak{sl}_d$ for $d\ge 3$, then $\mathfrak{ad}(x)$ does not make $\mathfrak{g}$ a cyclic module (over the single endomorphism $\mathfrak{ad}(x)$); actually for no $x$ at all by a simple additional argument. | |
| Jul 4, 2019 at 13:32 | comment | added | Jim Humphreys | @YCor: I'm not sure what is exactly the question being asked, but it's clear in a finite dimensional simpe Lie algebra that each nonzero vectror in the Lie algebra generates the entire Lie algebra under the (irreduible) adjoint representaton. | |
| Jul 4, 2019 at 13:24 | comment | added | YCor | If I understand correctly, the question is not which vector are cyclic (which is a trivial question), but which elements of the Lie algebra act cyclically. | |
| Jul 4, 2019 at 13:08 | history | answered | Jim Humphreys | CC BY-SA 4.0 |