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This fact basically tells us that all matrix Lie groups which are subgroups of $GL(n,\mathbb{C})$ have Lie algebras associated to them spanned by special sets of matrices where the Lie bracket is just the commutator.

It is clear that the matrix Lie groups are closed subgroups of $GL(n,\mathbb{C})$ and thus submanifolds. The differential of the inclusion $$(d \iota)_e : T_e (G) \to T_e (GL(n,\mathbb{C}))$$ is an inclusion of the tangent spaces in this case. Thus, $T_e (G) $ is definitely a vector subspace of $ T_e (GL(n,\mathbb{C}))$.

However, I am not sure how to tackle bracket. I understand that it could in principle be that $\mathfrak{gl}(n,\mathbb{C})$ is closed under the commutator as the bracket while smaller subspaces of it aren't. Please help me in showing the vector subspaces are indeed all closed under the bracket (the commutator). :)

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If $G$ and $H$ are Lie groups, if their Lie algebras are $\mathfrak g$ and $\mathfrak h$ respectively, and if $f\colon G\longrightarrow H$ is a Lie group homomorphism, them $f_e'\colon\mathfrak{g}\longrightarrow\mathfrak h$ is a Lie algebra homomorphism. Now, apply this to $G$ and to $GL(n,\mathbb{C})$. In your case, $f$ is the inclusion, and therefore $f_e'$ is the inclusion map too.

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    $\begingroup$ I would say $f$ is an inclusion rather than the identity. $\endgroup$
    – Pedro ♦
    Dec 18, 2018 at 14:15
  • $\begingroup$ Of course! I've edited my answer. Thank you. $\endgroup$
    – José Carlos Santos
    Dec 18, 2018 at 14:17
  • $\begingroup$ I dont know if you have even read my full question. I have pointed out almost everything you have said, so as for now your answer doesnt help me a lot. What I dont see is: Even if your $f'_e$ or my $(d \iota)_e$ is a Lie algebra homomorphism, then I know the bracket is preserved under this map, but how does this tell me that my subspace is gonna be closed??? $\endgroup$
    – Marsl
    Dec 18, 2018 at 14:49
  • $\begingroup$ We are talking about the inclusion of $\mathfrak g$ in $\mathfrak{gl}(n,\mathbb{C})$ here. So, if $X,Y\in\mathfrak{gl}(n,\mathbb{C})$, then $[X,Y]$ means the same thing either when you compute it with the Lie bracket of $\mathfrak g$ or with the Lie bracket of $\mathfrak{gl}(n,\mathbb{C})$. But then, and since $\mathfrak g$ is a Lie algebra, this means that if you compute the Lie bracket in $\mathfrak{gl}(n,\mathbb{C})$, then you will get again an element of $\mathfrak g$. In other words, $\mathfrak g$ is closed for the Lie bracket operation in $\mathfrak{gl}(n,\mathbb{C})$. $\endgroup$
    – José Carlos Santos
    Dec 18, 2018 at 14:55
  • $\begingroup$ Ok, got it finally, was missing that by proposition $\mathfrak{g}$ is already a Lie algebra and thus is closed under its own bracket. Now, for me it remains to prove your first statement: " their Lie algebras are $g$ and $h$ respectively, and if $f:G⟶H$ is a Lie group homomorphism, them $f′_e:g⟶h$ is a Lie algebra homomorphism. $\endgroup$
    – Marsl
    Dec 18, 2018 at 15:29

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