# Commutator of Lie sub-algebra

I have a problem understanding the proof of Proposition 8.20, page 211, in Besse's Einstein Manifolds. He considers a semi-simple Lie algebra $\mathfrak{g}$ with $\operatorname{Ad}$-invariant scalar product, its adjoint group $G$, and an orbit $M$ in $\mathfrak{g}$ under the adjoint representation. By $\operatorname{Ad}$-invariance the Lie algebra decomposes into an orthogonal direct sum $$\mathfrak{g}=\ker\operatorname{ad}_w\oplus\operatorname{im}\operatorname{ad}_w\equiv L_w\oplus M_w\quad\forall w\in M.$$ $L_w$ is the Lie algebra of the stabilizer group $G_w$.

Proposition. The stabilizer group $G_w$ is the commutator group $C(S_w)$ of its connected center $S_w$.

Proof. The Lie algebra $\mathfrak{s}_w$ of the connected center $S_w$ of $G_w$ is the center of $L_w$. Now, since $\operatorname{ad}_w$ has no kernel on $M_w$, we infer that $L_w$ is exactly the commutator of $\mathfrak{s}_w$ in $\mathfrak{g}$, so that the commutator group of $S_w$ is exactly the connected component of the identity of $G_w$, that is $G_w$ itself.

What is meant by "$L_w$ is exactly the commutator of $\mathfrak{s}_w$ in $\mathfrak{g}$"? (Usually the commutator of $\mathfrak{s}_w$ is the Lie sub-algebra $[\mathfrak{s}_w,\mathfrak{s}_w]$ generated by all commutators in $\mathfrak{s}_w$, but this definition does not make sense here.)

• What is meant by "commutator group of $S_{\omega}$" in the first place ? This should give us the commutator subalgebra of $\mathfrak{s}_{\omega}=Lie(S_{\omega})$. – Dietrich Burde Jan 8 '16 at 13:25
• @DietrichBurde Thank you for pointing out that also "commutator group of $S_w$" is not clear to me (again the usual definition does not make sense). So maybe $C(S_w)=\{g\in G_w\mid gs=sg~\forall s\in S_w\}$ is the commutant of $S_w$ in $G_w$? An explanation would be appreciated. – gofvonx Jan 8 '16 at 14:34
• The commutator group $C(S)$ is described in section $8.19$ in Besse's book as the union of all maximal tori containing $S$. – Dietrich Burde Jan 9 '16 at 10:17