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Let $G$ and $H$ be matrix Lie groups and $\mathfrak g$, $\mathfrak h$ be their Lie algebras. Suppose that $Φ : G → H$ is a Lie group isomorphism. Prove that the induced map $φ : \mathfrak g → \mathfrak h$ is a Lie algebra isomorphism.

I am a beginner in Lie algebra and often become confused in Lie algebra homomorphism please help me in showing or giving steps as a hint.

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    $\begingroup$ Which definition of Lie algebra of a Lie group are you using? $\endgroup$ – José Carlos Santos Feb 13 '18 at 15:37
  • $\begingroup$ $\mathfrak g=\{X \in M(n, \Bbb C)| e^X \in G\}$ $\endgroup$ – user531052 Feb 13 '18 at 16:09
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    $\begingroup$ There are two problems with that definition. First one: it assumes that $G$ is a subgroup of some $GL(n,\mathbb{C})$. Second one: It doesn't work. With that definition, $\mathfrak g$ doesn't even have to be a vector space. $\endgroup$ – José Carlos Santos Feb 13 '18 at 16:12
  • $\begingroup$ try tweaking the proof of the Baker Campbell Hausdorff Theorem. $\endgroup$ – Vidit D Feb 15 '18 at 8:59
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One way to prove it is to show that the map $\phi$ is lie algebra homomorphism and it is surjective and bijective. Now it is lie algebra homomorphism that part I am leaving upto you. You can check by yourself or you can see Brian C Hall's book. It is given there.

Showing bijectiveness is bit tricky. Now in this context we use a definition

Let $A :\Bbb R \to \operatorname{GL}(n, \Bbb C)$. We call $A$ as a one-parameter subgroup of $\operatorname{GL}(n, \Bbb C)$ if $A$ is continuous lie group homomorphism i.e $(A(0)=1, A(t+s)=A(t)A(s)$

& one theorem:

Let $A$ be a one-parameter subgroup of $\operatorname{GL}(n, \Bbb C)$. Then $\exists ! X \in \operatorname{M}(n, \Bbb C)$ s.t $A(t)=e^{tx}; \forall t \in \Bbb R$

And this powerfull theorem gave birth to this operator $\phi$ because we take $A :\Bbb R \to \operatorname{GL}(n, \Bbb C)$ defined by $A(t)=\Phi(e^{tX})$. Then using that theorem we get that $\phi: \mathfrak g \to \mathfrak h$ such that $A(t)=\Phi(e^{tX})=e^{t\phi(X)}$

Now it is given that $\Phi$ is a group isomorphism then considering $\Phi^{-1}: H \to G$ then $B(t): \Bbb R \to \operatorname{GL}(n, \Bbb C)$ s.t $B(t)=\Phi^{-1}(e^{tY})$ we have that $\exists \psi: \mathfrak h \to \mathfrak g$ s.t $B(t)=\Phi^{-1}(e^{tY})=e^{t\psi(Y)}$.

Now claim that : $\psi=\phi^{-1}$

$\Phi(\Phi^{-1}(e^{tY}))=\Phi(e^{t\psi(Y)})=e^{t\phi(\psi(Y))}$

$\Rightarrow e^{tY}=e^{t\phi(\psi(Y))}$; Now differentiating w.r.t $t$ and putting $t=0$ we get $Y=\phi(\psi(Y))$.

Similarly, taking $\Phi^{-1}(\Phi(e^{tX}))=\Phi^{-1}(e^{t\phi(X)})=e^{t\psi(\phi(X))}$ we get $Y=\psi(\phi(X))$.

So, $\phi$ is a lie algebra automorphism. (Proved).

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