It is well-known that for a complex semisimple Lie algebra $\mathfrak{g}$ with Cartan subalgebra $\mathfrak{h}$ and root system $\Phi$, there is a root space decomposition $\mathfrak{g}=\mathfrak{h}\oplus\bigoplus_{\alpha\in \Phi}\mathfrak{g}_\alpha$, where $\mathfrak{g}_\alpha:=\{x\in \mathfrak{g} : h\cdot x=\alpha(h)x, \forall h\in\mathfrak{h}\}$.

Meanwhile, $\Phi$ is a finite set of vectors and $\mathfrak{g}_\alpha$ is 1-dimensional.

My question: Does this implies any complex semisimple Lie algebra is finite dimensional? Or do I miss something?


1 Answer 1


You are missing the fact that the theorem about the existence of a root space decomposition is stated from the start as a theorem about finite-dimensional semisimple Lie algebras.

  • $\begingroup$ However, I read two books of Humphreys: Introduction to Lie Algebras and Representation Theory and Representations of Semisimple Lie Algebras in the BGG Category O. The sections about semisimple Lie algebra $\mathfrak{g}$ has not assume $\mathfrak{g}$ is finite dimensional but just assume $\mathfrak{g}$ is a complex semisimple Lie algebra with Cartan subalgebra $\mathfrak{h}$ . Results about root space decomposition are also included in these two books. Why is that? Do you have any references about the assumption "Being finite dimensional semisimple Lie algebra"? $\endgroup$ May 8, 2019 at 10:32
  • 1
    $\begingroup$ In Humphrey's textbook, in iw written, on page 1, “In this book we shall be concerned almost exclusively with Lie algebras $L$ whose underlying vector space is finite dimensional over $F$. This will always be assumed, unless otherwise stated. $\endgroup$ May 8, 2019 at 10:34
  • $\begingroup$ Thank you very much. $\endgroup$ May 8, 2019 at 10:48

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