I am learning about Lie groups by myself, and I tried to prove that $GL_n(\mathbb{C})$ and $SL_n(\mathbb{C})$ are smooth manifolds.
I tried to use the definition of smooth manifold: Let $M$ be an $n$-dimension manifold, let $\{$$u_\alpha$$\}_{\alpha\in I}$ be an open cover of $M$,
and let $\varphi_\alpha$$:$$u_\alpha$$\rightarrow$$v_\alpha$$\subseteq$$\mathbb{R}^n$ ($v_\alpha$ is an open set) be a homeomorphism.
$M$ is a "smooth manifold" if: $\forall \alpha,\beta\in I$: $u_\alpha \cap u_\beta \neq \emptyset$ $ \Longrightarrow$ $\varphi_\beta \circ \varphi_\alpha^{-1}: \varphi_\alpha(u_\alpha \cap u_\beta)\rightarrow \varphi_\beta(u_\alpha \cap u_\beta)$ is smooth.
I tried to use the definition, but I even dont know were to start.
I would really appreciate if anyone could help.
Best Answer
Hint(s):
$\operatorname{GL}(n,\Bbb{C})$: As suggested above, note that $\operatorname{GL}(n,\Bbb{C})$ is an open subset of $M(n,\Bbb{C})\cong \Bbb{C}^{n^2}\cong \Bbb{R}^{2n^2},$ and is thus an open submanifold of $\Bbb{R}^{2n^2}$. Indeed, note that an open subset of a manifold has a natural structure as a submanifold.
$\operatorname{SL}(n,\Bbb{C})$: This is the submanifold of $\operatorname{GL}(n,\Bbb{C})$ cut out by the condition $\det(A)=1$. It is an exercise to check that $1$ is a regular value of $\det:\operatorname{GL}(n,\Bbb{C})\to \Bbb{C}$, and from this the result follows by appealing to the "Regular Value Theorem" which you can find in any standard manifolds text.