Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Let $W$ be its Weyl group.

I would like to know whether $W$ is always finite? If so, why?


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Yes, it is always finite. Let $\mathfrak h$ be a Cartan subalgebra of $\mathfrak g$, and let $\Phi$ be the root system of $(\mathfrak g,\mathfrak h)$. By definition the Weyl group of $(\mathfrak g,\mathfrak h)$ is generated by all reflection $s_\alpha$ ($\alpha\in\Phi$), where$$s_\alpha(v)=v-2\frac{\langle\alpha, v\rangle}{\langle\alpha,\alpha\rangle}\alpha$$and $\langle\cdot,\cdot\rangle$ is the inner product in $\mathfrak h^*$ induced by the Killing form. It turns out that each $s_\alpha$ preserves $\Phi$ and that therefore each element of the Weyl group preserves $\Phi$. But $\Phi$ generates $\mathfrak h^*$ and therefore the Weyl group can be seen as a subgroup of the group of permutations of $\Phi$, which is finite.

  • $\begingroup$ Thank you very much. $\endgroup$ – James Cheung Mar 16 at 14:36
  • $\begingroup$ If my answer was useful, perhaps that you could mark it as the accepted one. $\endgroup$ – José Carlos Santos Mar 16 at 14:39
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    $\begingroup$ That $\Phi$ generates $\mathfrak h^*$ is not really needed, right? The main point is that $\Phi$ is finite, hence so is its permutation group, and $W$ is contained therein. $\endgroup$ – Torsten Schoeneberg Mar 16 at 15:47
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    $\begingroup$ Suppose that our vector space is $\mathbb{R}^2$, that $\Phi=\{(1,0),(-1,0)\}$, and that $W$ is the group of endomorphisms of $\mathbb{R}^2$ spanned by $s(x,y)=(-x+y,-y)$. Then $W$ is infinite, in spite of the fact that $s(\Phi)\subset\Phi$. $\endgroup$ – José Carlos Santos Mar 16 at 17:39

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