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.
• 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. – Torsten Schoeneberg Mar 16 at 15:47
• 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 $s(\Phi)\subset\Phi$. – José Carlos Santos Mar 16 at 17:39