Does the matrix exponential take open sets into open sets?

Question

This is from Hall's Lie Groups, Lie Algebras, and Representations, in theorem $2.13$:

Let $B_\varepsilon$ be the open ball of radius $\varepsilon$ about zero in $M_n (\mathbb{C})$ [$= \mathbb{C}^{n\times n}$] (...). Assume that $\varepsilon < \log 2$. Then, we have shown that "exp" takes $B_\varepsilon$ injectively into $M_n (\mathbb{C})$, with continuous inverse that we denote "log". Now, let $U =\exp(B_{\varepsilon /2})$, which is an open set in $GL(n; \mathbb{C})$.

The last sentence is not obvious to me. A general continuous function doesn't take open sets into open sets, so it probably has to do with the exponential being injective or with the inequality $||e^X|| \le e^{||X||}$. What is the proof of this?

Answer 1

Here you go :)

It is also clear from what is written in the text alone: If you accept that 1) $ \exp $ is injective on the domain $ B_\epsilon $ and 2) the inverse $ \log $ is continuous, then you get $ \exp( B_{\epsilon /2} ) = \log^{-1}( B_{\epsilon /2} ) $ is open, as it is a preimage of an open set under a continuous function.

Answer 2

I will cite the theorem from his same book which Hall is using on its proof of theorem 2.13.

Theorem 2.8. The function $$ \log A =\sum_{m=1}^\infty (-1)^{m+1}\frac{(A-I)^m}{m}$$ is defined and continuous on the set of all $n\times n$ complex matrices $A$ with $||A-I||<1$.

For all $A$ with $||A-I||<1$, $$\tag{1}\label{explog} e^{\log A} = A.$$ For all $X$ with $||X||<\log 2$, $||e^X-I||<1$ and $$\tag{2}\label{logexp} \log e^X = X. $$

Observe that the domain of the $\log$ function is equal to $B(I,1)$, where $B(C,δ)$ denotes the open ball in $M_n(\mathbb{C})$ (the set of all $n\times n$ complex matrices) of center $C$ and radius $δ$, in the metric induced by the Frobenius norm.

Two things are immediately deduced from \eqref{explog}:

• Since the exponential of any matrix is always invertible, $B(I,1)\subset \operatorname{GL}(n;\mathbb{C})$. That is, $\log$ is really defined on a subset of the set of complex invertible matrices.
• The $\log$ function is injective.

Let's see that $U=\exp(B_{ε/2})$ is open in $\operatorname{GL}(n;\mathbb{C})$. Firstly, from \eqref{logexp} is easy to show that $\log (U)=B_{ε/2}$. Then, from this last equality it only deduces that $U\subset \log^{-1}(B_{ε/2})$. However, it is precisely thanks to the fact that $\log$ is injective that the $\subset$ must be an equality. Thus $U = \log^{-1}(B_{ε/2})$ is open in $B(I,1)\subset \operatorname{GL}(n;\mathbb{C})$ and therefore open in $\operatorname{GL}(n;\mathbb{C})$ (in fact, since also $\operatorname{GL}(n;\mathbb{C})$ is an open subset of $M_n(\mathbb{C})$ —think about preimages of the determinant function— the set $U$ is open in $M_n(\mathbb{C})$).