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$\mathfrak{g}$ is a complex semisimple lie algebra which is a subalgebra of some $\mathfrak{g}l(n,\mathbb{C})$, we have chosen a compact real form $\mathfrak{l}$ of $\mathfrak{g}$ and let $K$ be the compact subgroup of $GL(n,\mathbb{C})$ whose lie algebra is $\mathfrak{l}$, we have chosen a maximal commutative subalgebra $\mathfrak{t}$ of $\mathfrak{l}$ and we work with associated cartan subalgebra $\mathfrak{h}=\mathfrak{t}+i\mathfrak{t}$, we have chosen 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: $$Z(\mathfrak{t})=\{A\in K: Ad_A(H)=H,\forall H\in \mathfrak{t}\}$$ $$N(\mathfrak{t})=\{Ad_A(H)\in \mathfrak{t},\forall H\in\mathfrak{t}\}$$ Define Weyl group, $W=N(\mathfrak{t})/Z(\mathfrak{t})$, I do not understand the action of weyl group on $\mathfrak{t}$ here is that: We can define an action of $W$on $\mathfrak{t}$ as follows . For each element $w\in W$ choose an element $A$ corresponding to equivalence class in $N(\mathfrak{t})$. Then for $H\in\mathfrak{t}$ we define the action $w.H$ of $w$ on $H$ by $w.H=Ad_A(H)$. could any one help me what is going on here possibly by an example?

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I will update my answer to your other question with an example of how this can look. Maybe this question should be closed as a duplicate. – Tobias Kildetoft Oct 15 '12 at 11:50
@Tobias,are you telling me to ask that in a different question? – Un Chien Andalou Oct 15 '12 at 11:51
please answer at your convenient place, thank you! – Un Chien Andalou Oct 15 '12 at 11:54
I have edited my answer to your other question to include an example. – Tobias Kildetoft Oct 15 '12 at 12:01

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