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Let $G=U(n)$ a compact Lie group and $T$ a maximal torus in $G$ (subgroup of diagonal matrix). We define $W=N(T)/T$ the Weyl group where $N(T)$ is the normalizer of $T \in G$. I have to prove that $W \rightarrow G/T \rightarrow G/N(T)$ is a covering space. So I think I have to prove that the action of $W$ on $G/T$ is free. Is there a simple proof of this?

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up vote 2 down vote accepted

You just have to write down the action of $W$ on $G/T$ :

$N(T)$ acts on $G$ by multiplication from the right : $x\cdot g = gx^{-1}$ and as you can easily see, the action descends to a well defined action of $W$ on $G/T$ by : $(xT)\cdot (gT) = (gx^{-1})T$.

Since the Weyl group $W$ of $G$ is finite, you only have to check that the action is in fact free which is straightforward (every stabilizer subgroup is trivial).

Hence, the map that you were looking for is a covering : $G/T\rightarrow (G/T)/(N(T)/T)\simeq G/N(T)$. This is exactly the space of maximal tori of $G$.

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