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Consider two simple Lie groups $G_1$ and $G_2$. Let $G_1$ have $W_1$ as a Weyl group and $G_2$ have $W_2$ as a Weyl group. Is it true that the Weyl group of $G_1 \times G_2$ is $W_1 \times W_2$?

Consider in particular a case in which $G_1$ is abelian, and therefore does not have roots, nor a Weyl group.

Is it true that the Weyl chambers in which the weight lattice of $G_1\times G_2$ is partitioned are of the form $\Lambda_1 \times C_i$, where $\Lambda_1$ is the whole weight lattice of $G_1$ and $C_i$ is a Weyl chamber of the weight lattice of $G_2$?

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  • $\begingroup$ An Abelian (connected) Lie group will not be simple. However an Abelian (and reductive) Lie group, even though it has no roots, does have a Weyl group; however, the Weyl group is trivial (it has one element). $\endgroup$ Feb 1, 2014 at 15:50

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The answer is yes. In more detail, suppose that $G_1$ and $G_2$ are reductive complex algebraic groups. Suppose that $T_1$ and $T_2$ are maximal tori of $G_1$ and $G_2$, respectively. Note that $T_1\times T_2$ is a maximal torus of $G_1\times G_2$. So, the Weyl group $W$ of $G_1\times G_2$ is $$W=N_{G_1\times G_2}(T_1\times T_2)/(T_1\times T_2)$$ $$\cong N_{G_1}(T_1)/T_1\times N_{G_2}(T_2)/T_2$$ $$=W_1\times W_2.$$

For your second question, let $\mathfrak{g}_1$ and $\mathfrak{g}_2$ denote the Lie algebras of $G_1$ and $G_2$, respectively. The Lie algebra of $G_1\times G_2$ is naturally isomorphic to $\mathfrak{g}_1\oplus\mathfrak{g}_2$. The adjoint representation of $G_1\times G_2$ is given by the property that $G_i$ acts via its adjoint representation on $\mathfrak{g}_i$ and trivially on the other summand. So, the roots of $G_1\times G_2$ are obtained by taking those of $G_1$ and $G_2$ and extending them trivially on $T_2$ and $T_1$, respectively. Hence, the root hyperplanes are of the form $H_1\times\mathfrak{t}_2$ and $\mathfrak{t}_1\times H_2$, where $H_1$ and $H_2$ are the root hyperplanes for $G_1$ and $G_2$, respectively, and $\mathfrak{t}_1$ and $\mathfrak{t}_2$ are the Lie algebras of $T_1$ and $T_2$, respectively. I think this should allow you to answer your second question.

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  • $\begingroup$ Yes, thank you. This has been really helpful. $\endgroup$ Feb 1, 2014 at 16:55

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