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Let $\mathfrak{g}=\mathfrak{h} \oplus \bigoplus_{\alpha \in \Phi}\, \mathfrak{g}_\alpha$ be a complex semisimple Lie algebra, where $\mathfrak{g}_{\alpha}=\mathbb{C} \cdot e_\alpha$ is 1-dimensional. Moreover, let $\Delta \subseteq \Phi^+$ be a set of simple roots. It is known, that $\mathfrak{g}=\langle \, h_i,e_i,f_i \,| \, \alpha_i \in \Delta \, \rangle$, where $e_i=e_{\alpha_i}$, $f_i=e_{-\alpha_i}$ and $h_i=[e_i,f_i]$.

Let $\alpha = \sum_{\alpha_i \in \Delta} \,n_i \alpha_i \in \Phi^+$. How can we express $e_\alpha$ in terms of $e_i$'s?

Bipolar Minds
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1 Answers1

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One has $\mathfrak g _{\alpha+\beta} = [\mathfrak g_\alpha , \mathfrak g_\beta]$, hence the root spaces $\mathfrak g_\alpha$ for $\alpha =\sum n_i \alpha_i$ can easily be expressed in terms of nested commutators of the root spaces $\mathfrak g_{\alpha_i}$. Cf. What is the basis of the root space $\mathfrak g_\theta$, where $\theta = \alpha_1+\cdots+\alpha_n$?

Then if for $\alpha \notin \Delta$ there is no restriction on $e_\alpha$ except that it is a vector space basis (i.e. non-zero element) of the (one-dimensional) space $\mathfrak g_\alpha$, obviously you can choose any in there. Be careful though that according to a specific definition of a Cartan-Weyl or Chevalley basis, there are further restrictions / relations, which might force you to carefully choose a correct scaling for each $e_\alpha$.

Note: One should be careful though to order the nested commutators in the right way. Compare ladder operators of simple Lie algebra and simple roots.

Torsten Schoeneberg
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