Lie Algebras

Lee Smooth Manifolds

(Lee E8-25) Prove that if $G$ is an abelian Lie group, $\text{Lie}(G)$ is abelian.

Proof) The inversion map $i:G\to G$ is a Lie group homomorphism as $(gh{)}^{-1}=h^{-1}{g}^{-1}=g^{-1}{h}^{-1}$. As it is also bijective, it is in fact a Lie group isomorphism and thus a Lie algebra isomorphism.

Note that if $m:G\times G\to G$ is the multiplication map, $d{m}_{(e,e)}:T_{(e,e)}(G\times G)\cong T_{e}G\oplus T_{e}G\to T_{e}G$ is given by $d{m}_{(e,e)}(X,Y)=X+Y$. This is because, when restricted to the embedded submanifold $G\times \{e\}\subset G\times G$, $m$ collapses into the identity map $\iota :G\to G$, and thus $d{m}_{(e,e)}(X,0)=d\iota X=X$. Analogously, $d{m}_{(e,e)}(0,Y)=Y$ for all $Y\in T_{e}G$, and as $dm$ is linear this forces $d{m}_{(e,e)}(X,Y)=X+Y$. In turn, as $m(g,i(g))=e$, if we let $\eta :g\mapsto (g,i(g))$, for any $g\in G$, $d{m}_{\eta (g)}\circ d{\eta }_g=0$. Thus, substituting $g=e$ and applying to any $v\in T_{e}G$,

$$0=d{m}_{\eta (e)}\circ d{\eta }_e(v)=v+d{i}_e(v)\Rightarrow d{i}_e(v)=-v$$

We now prove the theorem. Let $i_{\ast }:\text{Lie}(G)\to \text{Lie}(G)$ be the pushforward isomorphism. Let $X,Y\in \text{Lie}(G)$ be arbitrary and take $x=X_e$ for convenience. For any $v\in T_{e}G$, let $\xi (v)\in \text{Lie}(G)$ be the vector field such that $\xi (v{)}_e=v$. $i_{\ast }X$ must satisfy $i_{\ast }{X}_e=d{i}_e(x)=-x\Rightarrow i_{\ast }X=\xi (-x)=-X$. Thus,

$$-[X,Y]=i_{\ast }[X,Y]=[i_{\ast }X,i_{\ast }Y]=[-X,-Y]=[X,Y]\Rightarrow [X,Y]=0$$

$\text{Lie}(G)$ is therefore abelian.

(Lee E8-31) Let $\mathfrak{g}$ be a Lie algebra. A linear subspace $\mathfrak{h}\subset \mathfrak{g}$ is called an ideal in $\mathfrak{g}$ if $[X,Y]\in \mathfrak{h}$ whenever $X\in \mathfrak{h}$ and $Y\in \mathfrak{g}$. (a) Show that if $\mathfrak{h}$ is an ideal in $\mathfrak{g}$, the quotient space $\mathfrak{g}/\mathfrak{h}$ has a unique Lie algebra structure such that the projection $\pi :\mathfrak{g}\to \mathfrak{g}/\mathfrak{h}$ is a Lie algebra homomorphism. (b) Show that a subspace $\mathfrak{h}\subset \mathfrak{g}$ is an ideal iff it is the kernel of a Lie algebra homomorphism.

Proof) For (a) we first show that $\pi ([X,Y])=\pi ([X+H_1,Y+H_2])$ where $X,Y\in \mathfrak{g}$ and $H_1,H_2\in \mathfrak{h}$ are arbitrary. This follows from the fact that the Lie bracket is bilinear and alternating, as

$$[X+H_1,Y+H_2]=[X,Y]+[H_1,Y]+[X,H_2]+[H_1,H_2]\in [X,Y]+\mathfrak{h}$$

For any $Q_1,Q_2\in \mathfrak{g}/\mathfrak{h}$, we define $[Q_1,Q_2]$ as $\pi ([R_1,R_2])$ where $R_i$ is a representative of $Q_i$. By the computation above this is independent of the choice of $R_i$, and thus well-defined. The Lie algebra properties naturally descend from the fact that $\pi$ is linear. Uniqueness is similarly trivial. This completes (a). (b) easily follows as the kernel of Lie algebra homomorphism is trivially an ideal, whereas an arbitrary ideal $\mathfrak{h}\subset \mathfrak{g}$ is the kernel of its corresponding projection $\pi$ with the quotient space equipped with the Lie algebra structure from (a).