# Weyl group of $\mathfrak{sl}(2,\mathbb{C})$

$\mathfrak{g}$ is a complex semisimple lie algebra which is a subalgebra of some $\mathfrak{gl}(n,\mathbb{C})$, we have chosen a compact real form $\mathfrak{l}$ of $\mathfrak{g}$ and let $K$ be the compact subgroup of $\operatorname{GL}(n,\mathbb{C})$ whose lie algebra is $\mathfrak{l}$, we have chosen a maximal commutative subalgebra $\mathfrak{t}$ of $\mathfrak{l}$ and we work with the associated cartan subalgebra $\mathfrak{h}=\mathfrak{t}+i\mathfrak{t}$, we have chosen an inner product on $\mathfrak{g}$ that is invariant under the adjoint action of $K$ and that takes real values on $\mathfrak{l}$. Consider the subgroups: $$Z(\mathfrak{t})=\{A\in K: \operatorname{Ad}_A(H)=H,\forall H\in \mathfrak{t}\}$$ $$N(\mathfrak{t})=\{\operatorname{Ad}_A(H)\in \mathfrak{t},\forall H\in\mathfrak{t}\}$$ Weyl group is defined as $W=N(\mathfrak{t})/Z(\mathfrak{t})$, for $\mathfrak{sl}(2,\mathbb{C})$ we take $\mathfrak{t}=\left\{\left(\begin{smallmatrix}ia&0\\0&-ia\end{smallmatrix}\right):a\in\mathbb{R}\right\}$

I understand $Z(\mathfrak{t})=\left\{\left(\begin{smallmatrix}e^{ia}&0\\0&e^{-ia}\end{smallmatrix}\right):a\in\mathbb{R}\right\}$ and $N(\mathfrak{t})=\left\{\left(\begin{smallmatrix}e^{ia}&0\\0&e^{-ia}\end{smallmatrix}\right):a\in\mathbb{R}\right\}$ or $N(\mathfrak{t})=\left\{\left(\begin{smallmatrix}0&e^{ia}\\-e^{-ia}&0\end{smallmatrix}\right):a\in\mathbb{R}\right\}$ but I dont understand why the quotient group $N(\mathfrak{t})/Z(\mathfrak{t})$ has $2$ elements? Could it not be only one element?

And I do not understand the action of the Weyl group on $\mathfrak{t}$, please help.

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