I would like to ask about the setting of the book "Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$" by Humphreys. I would like to know whether the semisimple Lie algebra $\mathfrak{g}$ is finite dimensional or not. I search the book and did not find any assumption about the semisimple Lie algebra $\mathfrak{g}$ being finite dimensional. However, the book mentions root space decomposition of $\mathfrak{g}$, which depends on the finite dimensionality of $\mathfrak{g}$ by Humphreys' other book "Introduction to Lie Algebras and Representation Theory". Also a Cartan subalgebra $\mathfrak{h}$ of $\mathfrak{g}$ is mentioned to be finite dimensional on p.1 of "Representations of Semisimple Lie Algebras in the BGG Category $\mathcal{O}$".

My questions:

  1. Is the semisimple Lie algebra $\mathfrak{g}$ finite dimensional?
  2. If $\mathfrak{g}$ is not assumed to be finite dimensional, do we still have the root space decomposition of complex semisimple Lie algebra with respect to a Cartan subalgebra?
  3. If $\mathfrak{g}$ is not assumed to be finite dimensional, does the root space decomposition imply that any complex semisimple Lie algebra with a Cartan subalgebra is finite dimensional?
  4. If $\mathfrak{g}$ is not assumed to be finite dimensional, how do we prove $\mathfrak{h}$ is finite dimensional?
  • $\begingroup$ Yes, the Lie algebra $\mathfrak{g}$ is finite dimensional. I'm sorry if this traditional setting isn't clear enough, but beyond this is mainly the Kac-Moody case which is not usually viewed as "semisimple". $\endgroup$ May 8, 2019 at 22:17
  • $\begingroup$ Thank you for your reply. $\endgroup$ May 9, 2019 at 6:42
  • $\begingroup$ By the way, I would like to know does $\mathfrak{g}$ is complex semisimple Lie algebra imply $\mathfrak{g}$ is finite dimensional by the root space decomposition or do people just omit the words "finite dimensional" when assuming $\mathfrak{g}$ is finite dimensional complex semisimple since it is a traditional setting? $\endgroup$ May 9, 2019 at 6:51

1 Answer 1

  1. In the book, $\mathfrak{g}$ is always assumed to be finite-dimensional. (I believe that this is explicitly stated in the first chapter of the book.)
  2. Depending on $\mathfrak{g}$ and on the definition of Cartan subalgebras in the infinite-dimensional setting, you may have an infinite-dimensional semisimple Lie algebra $\mathfrak{g}$ and a Cartan subalgebra $\mathfrak{h}$ such that, with respect to $\mathfrak{h}$, the root-space decomposition of $\mathfrak{g}$ exists. You can look at something like root-reductive Lie algebras.
  3. See my answer to 2.
  4. Also see my answer to 2.
  • $\begingroup$ Thank you for your answers. But I really cannot find out the assumption $\mathfrak{g}$ is finite dimensional throughout the book by manually searching the word "finite". I just find out some sentences implicitly suggest that $\mathfrak{g}$ is finite dimensional. e.g. p.36 (the last sentence) and p.259 (the last sentence in Section 13.6). If you can find it out, please let me know. $\endgroup$ May 8, 2019 at 12:15
  • $\begingroup$ Well, isn't a semisimple lie algebra just a (finite) direct sum of simple Lie algebras, each of which is finite dimensional. $\endgroup$
    – Filip
    Jul 27, 2021 at 23:14

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