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.

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