# Lately in Lie Groups

Rmk: Here $\mathfrak{g}$ is a complex semisimple lie algebra.  Making it complex is just so that you always have eigenvalues when you want them.

Most of this material is coming from Appendix C and Appendix D of Fulton and Harris.  We proved the absolute Jordan decomposition for a semisimple lie algebra.  That is, for $X \in \mathfrak{g}$ with $\mathfrak{g}$ semisimple and you have a faithful representation $\mathfrak{g} \to \mathfrak{gl}(V)$ then then you can use normal linear algebra Jordan decomposition to get $X = X_s + X_n$ etc.  In fact $X_s, X_n$ come from elements of $\mathfrak{g}$.  These are absolute in the sense that for any other representation, $\rho(X_s) = \rho(X)_s$.

We started talking about Cartan Subalgebras.  : maximal abelian subalgebras of a semisimple lie algebra that consist only of semisimple elements.  In fact we proved that in such a situation, its sufficient that the subalgebra just be maximal with respect to only having semisimple elements, i.e. it will then be abelian.  In fact any subalgebra of a semisimple $\mathfrak{g}$ consisting of only semisimple elements will be abelian (rough idea: if this wasn’t the case there would be an element with a nonzero eigenvalue and eigenvector, argue by contradiction …).

(Prop D.3) For a semisiple lie algebra, a regular element $H$ is one such that the dimension of

$C(H) = \{X|[H,X] = 0\}$

Is minimal.  The prop. says that for a minimal element, $C(H)$ is a cartan subalgebra.

The proof uses at one point that is $X \in C(H)$ then $X_s,X_n \in C(H)$ (this is ok), but also claims that $X_s$ acts nilpotently, its not said in the proof, but it seems this claims is appealing to exercise D.8 in the appendix.

So now the story is you have $\mathfrak{g}$ and you decompose it into root spaces

$\mathfrak{g} = \mathfrak{h} \oplus \mathfrak{g}_\alpha$

Where the spaces $\mathfrak{g}_\alpha$ have the following properties (the $\alpha$‘s are called roots)

1. They are 1-dimensional
2. they consist of eigenvector for $ad_H$ for all $H \in \mathfrak{h}$
3. they are parametrized by functionals $\alpha \in \mathfrak{h}^*$, i.e. $\mathfrak{g}_\alpha = \{X|ad_H (X) = \alpha(H) X$.
4. if $\alpha$ is a root so is $-\alpha$.
5. $[\mathfrak{g}_\alpha, \mathfrak{g}_\beta] \subset \mathfrak{g}_{\alpha + \beta}$
6. from 5 it follows that for the killing form $B$ and for $\alpha + \beta \ne 0$ then $B(\mathfrak{g}_\alpha, \mathfrak{g}_\beta) = 0$.

For the proof in Fulton and Harris of 1, its given that $\alpha$ is a root, and then its showed that $2 \alpha$ cannot be a root.  So it $1/2 \alpha$ was a root then $\alpha$ could not be root, so $1/2 \alpha$ cannot be a root.