# About root space decomposition of complex semisimple Lie algebra

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?

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.

• 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"? May 8, 2019 at 10:32
• 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. May 8, 2019 at 10:34
• Thank you very much. May 8, 2019 at 10:48