2
$\begingroup$

Let $\frak{g}$ be a complex semisimple Lie algebra. I would like to know about infinite-dimensional representations $M$ of $\frak{g}$ for which the Cartan $\frak{h} \subseteq \frak{g}$ does not act semisimply. Are such representations interesting, and if they are, what are some motivating examples. Also, what can we say about the "number" of such representations? Are they more common "in nature" than representations with a semisimple Cartan action, or are they more rare?

Edit: To clarify, by the Cartan acting semisimply I mean that $M$ has a (Hamel) basis consisting of commoning eigenvectors for the elements of the Cartan.

Improve this question
$\endgroup$
11
  • 4
    $\begingroup$ 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? $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 17, 2023 at 12:08
  • $\begingroup$ 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? $\endgroup$
    – László Szabados
    Commented Mar 17, 2023 at 12:24
  • $\begingroup$ 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 $n\to\infty$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 17, 2023 at 13:13
  • 1
    $\begingroup$ @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. $\endgroup$
    – LSpice
    Commented Mar 17, 2023 at 14:45
  • 1
    $\begingroup$ @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... $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 18, 2023 at 13:29

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .