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Suppose $\mathfrak{g}$ is a semisimple Lie algebra, and denote $\mathfrak{h}$ one of its Cartan subalgebras. I would like to prove the set of roots of $\mathfrak{g}$ necessarily spans $\mathfrak{h}^\star$. A linear form $\alpha\in\mathfrak{h}^\star$ is a root of $\mathfrak{g}$, the set of which denoted $\Delta$, if the set {$g\in\mathfrak{g}|\exists n\in\mathbb{N}~s.t.~(ad(h)-\alpha(h))^ng=0\forall h\in\mathfrak{h}$} is not {$0$}.

A proof should be possible by contradiction. But I fail to see the logic behind the following statement: suppose $\Delta$ does not span $\mathfrak{h}^\star$, this is equivalent to saying there exists some $h\in\mathfrak{h}$ such that $\alpha(h)=0$ for all $\alpha\in\Delta$. Could someone offer an explanation why this statement is true?

Rescy_
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1 Answers1

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Assume that the root system $\Phi$ does not generate $\mathfrak{h}^{\ast}$. Then, by duality there exists an $h\in \mathfrak{h}$, $h\neq 0$ with $\alpha(h)=0$ for all $\alpha \in \Phi$. Why is this? Say, $(h_1^*,\ldots ,h_{\ell}^*)$ is a basis of $\mathfrak{h}^{\ast}$ extending the basis of $\Phi$ by, say, $h_{\ell}^*\not\in \Phi$. Then we consider the dual basis $(h_1,\ldots ,h_{\ell})$ of $\mathfrak{h}$ with $h_i^*(h_j)= \delta_{ij}$. We have $h_i^*(h_{\ell})= 0$ for all $i\neq \ell$ and hence $\alpha(h)=\alpha(h_{\ell})=0$ for all $\alpha\in \Phi$, since $(h_1^*,\ldots, h_{\ell-1}^*)$ is a basis of $\Phi$. But this implies $[h,\mathfrak{g}_{\alpha}]=0$ for all $\alpha \in \Phi$. Because of $[h,\mathfrak{g}_{0}]=0$ we have $[h,\mathfrak{g}]=0$, hence $h\in Z(\mathfrak{g})=0$, which is a contradiction. Hence we have ${\rm Span}(\Phi)=\mathfrak{h}^{\ast}$.

Dietrich Burde
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