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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: $$Z(\mathfrak{t})=\{A\in K: Ad_A(H)=H,\forall H\in \mathfrak{t}\}$$ $$N(\mathfrak{t})=\{Ad_A(H)\in \mathfrak{t},\forall H\in\mathfrak{t}\}$$ I am not getting why $Z(\mathfrak{t})$ is a Normal subgroup of $N(\mathfrak{t})$, could any one help me to show this? and I must say I am not getting how the group look like I mean what type of element will be there in the group $W=N(\mathfrak{t})/Z(\mathfrak{t})$, please help in detail. thank you

this is a doubt from Brian C Hall page 174.

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2 Answers 2

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Consider the group homomorphism from $N(\mathfrak{t})$ to $\rm{Aut}(\mathfrak{t})$ given by $A\mapsto \rm{Ad}_A$. The kernel of this map is exactly $Z(\mathfrak{t})$, which is then a normal subgroup.

This also tells you one way to think of the elements of the quotient, namely as certain automorphisms of $\mathfrak{t}$.

Let us consider the easiest example, namely $GL_n(\mathbb{C})$. Here the Lie algebra is just $gl_n(\mathbb{C})$, and the adjoint action is given by conjugation. Our $\mathfrak{t}$ is simply the subalgebra consisting of diagonal matrices.

The normalizer of this subalgebra turns out to be matrices that have exactly one non-zero element on each row and column (so like permutation matrices, but instead of having exactly one $1$ on each row and column, that $1$ can be any non-zero element from $\mathbb{C}$). That this is the case is a nice exercise.

We also see that the centralizer of $\mathfrak{t}$ is exactly the subgroup consisting of diagonal matrices (again a nice exercise).

Finally, this shows that when we mod out by the centralizer, we can "ignore" those arbitrary non-zero elements and assume they are in fact $1$'s, so we just get that our quotient can be identified with the subgroup of permutation matrices, which then has an obvious action on the subalgebra of diagonal matrices, by simply permuting the entries.

To really get a good understanding of this, I now urge you to try going through the same example with $SL_n(\mathbb{C})$ instead.

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Thank you for your reply –  Une Femme Douce Oct 15 '12 at 11:47

Since $Z(t)$ is the centralizer and $N(t)$ a (normal) subgroup of $K$ you have $$ Z \triangleleft N\,\,\Leftrightarrow\,\forall\,z\in{Z},\forall\,n\in{N}\, nzn^{-1}=znn^{-1}=z\in{Z}. $$

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Thank you for your reply –  Une Femme Douce Oct 15 '12 at 11:49

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