Lie Groups & Brackets

Lee Smooth Manifolds Hey look at that, Lie and Lee, heh.

(Lee 7-18, T7-35) Suppose $G$ is a Lie group, and $N,H\subseteq G$ are Lie subgroups such that $N$ is closed and normal, $N\cap H=\{e\}$, and $NH=G$. Then the map $(n,h)\mapsto nh$ is a Lie group isomorphism between $N{⋊}_{\theta }H$ and $G$, where $\theta :H\times N\to N$ is the action by conjugation: ${\theta }_h(n)=hn{h}^{-1}$.

Remark: The original problem additionally states $H$ is closed, but this is an unnecessary hypothesis, as confirmed by the errata.

Proof) As $N$ is a closed Lie subgroup, it is an embedded submanifold. This, in addition to $N$ being normal, ensures $\theta$ is a smooth action by automorphisms. Clearly $(n,h)\mapsto nh$ is a smooth bijection between $N{⋊}_{\theta }H$ and $G$, and thus it suffices to show it is a group homomorphism. This follows by the computation:

$$(n,h)(n^{\prime },h^{\prime })=(n{\theta }_h(n^{\prime }),h{h}^{\prime })=(nh{n}^{\prime }{h}^{-1},h{h}^{\prime }),nh{n}^{\prime }{h}^{-1}h{h}^{\prime }=nh{n}^{\prime }{h}^{\prime }$$

(Lee 7-19) Suppose $G$, $N$, and $H$ are Lie groups. Prove that $G$ is isomorphic to a semidirect product $N⋊H$ if and only if there are Lie group homomorphisms $\phi :G\to H$ and $\psi :H\to G$ such that $\phi \circ \psi =I{d}_H$ and $\text{Ker}\phi \cong N$.

Proof) Assume there is some $\theta :H\times N\to N$ such that $N{⋊}_{\theta }H\cong G$. Let $\Phi :N{⋊}_{\theta }H\to G$ be this isomorphism, and define $\phi =\pi \circ {\Phi }^{-1}$ where $\pi$ is the projection onto the second factor and $\varphi (h)=\Phi (e,h)$. These functions clearly satisfy the desired conditions.

We now prove the converse. $\psi$ is an injective Lie group homomorphism, and thus is a isomorphism between $H$ and the Lie subgroup $\psi (H)\subset G$. If we take $N=\text{Ker}\phi$, $N$ is a closed normal Lie subgroup of $G$. Clearly, $\psi (H)\cap N=\{e\}$, and thus it suffices to show $N\psi (H)=G$, as then the hypotheses of 7-18 apply. This follows from the fact that there is a group isomorphism $G/N\cong H\cong \psi (H)$ induced by $\phi$. For any $g\in G$, $k=\phi (\psi (g))$ satisfies $\psi (k)=\psi (g)$, and thus $g=nk$for some $n\in N$. As $k\in \psi (H)$ and $n\in N$, we are done.

(Lee E8.29) For $f,g\in C^{\infty }(M)$ and $X,Y\in \mathfrak{X}(M)$,

$$[fX,gY]=fg[X,Y]+(fXg)Y-(gYf)X$$

Proof) For any $h\in C^{\infty }(M)$,

$$[fX,gY]h=fX(gYh)-gY(fXh)=fgX(Yh)+fYhXg-gfY(Xh)-gXhYf$$ $$=fg[X,Y]h+(fXg)Yh-(gYf)Xh=(fg[X,Y]+(fXg)Y-(gYf)X)h$$