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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"?2011-07-13
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    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.2011-07-13
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    are these assumptions you tacitly make?2011-07-13
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    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 $\langle \cdot,\cdot \rangle=\lambda\kappa$.2011-07-13
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    @Qiaochu, yes, I mean a semisimple Lie algebra.2011-07-13
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    @Olivier, thank you very much.2011-07-13
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    @user9791 has any of this answered your question?2011-07-13
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    @Olivier, yes. But can you compute some example of $<\alpha_i, \alpha_i>$ for type $B$ or other non-simply-laced types?2011-07-13
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    @Olivier, originally I am confused about the ratio of lengths long roots and short roots. Now I think that the ratio of the lengths of a long root and a short root is the same as the number of edges joining the two roots in Dynkin diagram by using the table in Section 9.4 of the book of Humphreys. So I can compute $<\alpha_i, \alpha_i>$ now.2011-07-13
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    @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

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