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The map $\phi\colon M_n(\mathbb C) \to\operatorname{GL}_n(\mathbb C)$ is given by $\phi(A) = e^A$ for fixed $n\in \mathbb N$.

Part $1$. Show that there exists an open set $U\subset M_n(\Bbb C)$ containing $\mathbf 0_n$, and an open set $V \subset\operatorname{GL}_n(\Bbb C)$ containing $I_n$, such that (i) for all $A\in U$, $\phi(A) \in V$ (ii) the restriction map $\phi: U\to V$ is bijective, and (iii) both $\phi, \phi^{-1}$ are $C^\infty$.

I have solved Part 1 above.

Part $2$. Let $H$ be $\langle V\rangle$, the subgroup of $\operatorname{GL}_n(\Bbb C)$ generated by $V$. Show that $H = \operatorname{GL}_n(\Bbb C)$. Finally, show that $\phi(M_n(\Bbb C)) \subset\operatorname{GL}_n(\Bbb C)$ generates the group $\operatorname{GL}_n(\Bbb C)$.

Question. We have $H = \langle V\rangle \subset\operatorname{GL}_n(\Bbb C)$. As $\operatorname{GL}_n(\Bbb C)$ is connected (in fact, it is path-connected), my intuition is to show that $H$ is both open and closed in $GL_n(\Bbb C)$ - which would ensure $H =\operatorname{GL}_n(\Bbb C)$, since $H \ne \varnothing$. How should I do this?

To show that $\phi(M_n(\Bbb C))$ generates (and actually equals, since the map is surjective) $\operatorname{GL}_n(\Bbb C)$, the linked question (by @orangeskid) contains a proof.

Any help and hints would be greatly appreciated. Thanks a lot!

Samuel Adrian Antz
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stoic-santiago
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1 Answers1

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An open subgroup $H$ of a topological group $G$ is also closed. Indeed, $G$ can be written as the disjoint union $G= \bigcup _{a\in G/H} a.H$. But all the terms in this disjoint union are open, and $H ^c= \bigcup _{a\in G/H, a\not = H} a.H$, thus $H^c$ is open.

Thomas
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  • Thanks a lot! The fact that $H = \langle V\rangle$ is open follows from $H = \bigcup_{g\in \langle V\rangle} gV$ right? Then using your argument, $H$ is also closed. Therefore, $H$ being non-empty, is equal to $GL_n(\Bbb C)$. – stoic-santiago Jan 25 '22 at 16:24