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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.

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Do you mean "let $\mathfrak{g}$ be a semisimple Lie algebra"? –  Qiaochu Yuan Jul 13 '11 at 17:13
    
It seems to me there are several wrong statements in your question. I assume all Lie algebras to be finite dimensional, maybe that's where I differ from you. You talk about $\mathfrak{g}$'s Cartan subalgebra, but there may be many of them, although they are all conjugate when (I believe) the ground field is algebraically closed. Also, $\kappa$ non degenerate is equivalent (again modulo some condition on the ground field) to the Lie algebra being semisimple. Finally (and why I suspect you might be interested in the infinite dimensional case), injectivity implies bijectivity in finite dimension. –  Olivier Bégassat Jul 13 '11 at 17:21
    
are these assumptions you tacitly make? –  Olivier Bégassat Jul 13 '11 at 17:22
2  
I think that when you are dealing with a simple Lie algebra, there is essentially only one invariant bilinear form, which is the Killing form. So, it may simply be that the author considers the Killing form multiplied by some constant such that $\langle \alpha, \alpha \rangle=2$ for the maximal root, where $\langle \cdot,\cdot \rangle=\lambda\kappa$. –  Olivier Bégassat Jul 13 '11 at 17:26
    
@Qiaochu, yes, I mean a semisimple Lie algebra. –  LJR Jul 13 '11 at 17:31

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