3
$\begingroup$

Let $\mathfrak{g}$ be a complex semisimple Lie algebra and $\mathfrak{h}$ be a Cartan subalgebra of $\mathfrak{g}$. We can use the Killing form to identify $\mathfrak{h}$ and $\mathfrak{h}^*$ ($\phi\in \mathfrak{h}^*$ corresponds to $t_{\phi}\in \mathfrak{h}$), where $t_{\phi}$ satisfies $\phi(h)=\kappa(t_{\phi}, h)$ for all $h\in \mathfrak{h}$ ($\kappa$ is the Killing form on $\mathfrak{g}$). Since $\kappa$ is non-degenerate, it is easy to show that the above correspondence is injective. But how to show that the above correspondence is surjective? Why each $h\in \mathfrak{h}$ is of the form $t_{\phi}$ for some $\phi \in \mathfrak{h}^*$?

I think there are other ways to identify $\mathfrak{h}$ and $\mathfrak{h}^*$. I am reading a paper path description of type B q-characters. On page 3, line&nbps;2 of section 2, it is said that $\mathfrak{h}$ and $\mathfrak{h}^*$ can be identified by using the invariant inner product $\langle\, , \rangle$ on $\mathfrak{g}$ normalized in such a way that the square length of the maximal root equals $2$. What is the relation of this form and the correspondence above?

Let $\langle\, , \rangle$ be the form defined as above. How to compute $\langle\alpha_i, \alpha_i\rangle$ explicitly for all types ($\alpha_i$ are simple roots)? Thank you very much.

  • 0
    @user9791: if you mean that the algebra be semisimple, please edit the question to add that information. It is usually best when questions are complete, and readers do not have to go through a whole thread of comments to find out what is being asked! :=)2011-07-13

0 Answers 0