Timeline for Complex semisimple Lie algebra modules with non-semisimple Cartan action
Current License: CC BY-SA 4.0
13 events
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| Mar 18, 2023 at 13:29 | comment | added | მამუკა ჯიბლაძე | @LSpice Yes, I meant this paper and yes, I meant the adjoint representation. In one of its realizations the element corresponds to the infinite (in all directions) matrix with ones on the diagonal above the main one and zeroes elsewhere. And now I realize this case is in fact not relevant for the question since the corresponding Lie algebra is not finite-dimensional... | |
| Mar 18, 2023 at 12:00 | comment | added | LSpice | @მამუკაჯიბლაძე, re, could you say more where you see this usage? In "Fixed-point varieties on affine flag manifolds", semisimple elements, including rss, are diagonalizable in any algebraic representation over $\bar F$. Thus, for KL, an ss element is tn iff its eigenvalues (in any faithful algebraic representation/$\bar F$) are all strictly less than $1$ in absolute value. (But there's no requirement about eigenvalues in, say, an abstract-gp representation of $G(F)$; maybe that's what you mean.) | |
| Mar 18, 2023 at 8:46 | comment | added | Qixian Zhao | Certainly a lot of natural reps are not weight modules (weight module = module with a semisimple Cartan action). For example, extensions of highest weight modules may not be weight modules. Also, if you take a semisimple real group with no compact Cartan, then the Harish-Chandra modules of the irred reps of such a group may not be weight modules. | |
| Mar 18, 2023 at 6:48 | comment | added | მამუკა ჯიბლაძე | @LSpice No I did not have in mind the $p$-adic case as I know next to nothing about it. What I want to say is that some specialists call regular semisimple elements without any eigenvalues/nontrivial eigenspaces. | |
| Mar 17, 2023 at 17:59 | comment | added | LSpice | @მამუკაჯიბლაძე, re, by "that $x$" do you mean multiplication by $p$ from my comment? That eigenspace is not proper, but it is non-trivial, so I am still accustomed to call $p$ an eigenvalue. For KL's elements, it is possible that their eigenvalues live in some algebraic extension, but being semisimple means that they are diagonalizable in any algebraic representation over a separably closed field (which is enough to detect top.nilptce.). | |
| Mar 17, 2023 at 15:39 | comment | added | მამუკა ჯიბლაძე | @LSpice Actually if I am not mistaken that $x$ does not have any proper eigenspaces, hence also no eigenvalues at all. I believe also another $x$, from my first comment, is such. | |
| Mar 17, 2023 at 15:34 | comment | added | მამუკა ჯიბლაძე | Sorry for possible confusion: I should write $\operatorname{ad}(x)^n\to0$ rather than $\to\infty$. | |
| Mar 17, 2023 at 14:45 | comment | added | LSpice | @მამუკაჯიბლაძე, that is less exotic than it might sound, and can be more about the eigenvalues than about any weird eigenspace behaviour; for example, multiplication by $p$ on a 1-dimensional $p$-adic vector space does exactly that. | |
| Mar 17, 2023 at 14:21 | history | edited | László Szabados | CC BY-SA 4.0 |
added 164 characters in body
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| Mar 17, 2023 at 13:13 | comment | added | მამუკა ჯიბლაძე | I have no idea what is standard, I just meant that there are several possible meanings, some more restrictive, some less. For example, I encountered (in a paper by Kazhdan and Lusztig) elements $x$ they call regular semisimple which are topologically nilpotent, which means that $\operatorname{ad}(x)^n\to\infty$ when $n\to\infty$. | |
| Mar 17, 2023 at 12:24 | comment | added | László Szabados | I would naively suggest that the definition be that $M$ has a (Hamel) basis consisting of commoning eigenvectors for the elements of the Cartan. Is this too restrictive, or non-standard? | |
| Mar 17, 2023 at 12:08 | comment | added | მამუკა ჯიბლაძე | What is your definition of semisimple for an operator on an infinite-dimensional space? Is the operator of multiplication by $x$ in $\mathbb C[x,x^{-1}]$ semisimple, for example? | |
| Mar 17, 2023 at 11:48 | history | asked | László Szabados | CC BY-SA 4.0 |