Abstract Proof that Exponential Map is Surjective onto $\mathrm{GL}_n(\mathbb{C})$

Question

It is well-known that the exponential map associated to any compact connected Lie group is surjective (the proof is a simple application of the Lefschetz fixed point theorem). As it happens, the exponential map associated to $\mathrm{GL}_n(\mathbb{C})$ is also surjective, although $\mathrm{GL}_n(\mathbb{C})$ fails to be compact. The only way I know how to prove this latter claim is by expressing each element of $\mathrm{GL}_n(\mathbb{C})$ in its Jordan canonical form and then showing that any Jordan block is the exponential of some matrix.

My question is this: is there a wider class of Lie groups (more general than just compact and connected, and perhaps including examples like $\mathrm{GL}_n(\mathbb{C})$) for which we can say that the exponential map is surjective?

Answer 1

I doubt that there is a nice answer like the one you want, since the exponential map fails to be surjective in the case of $SL(2,\mathbb{C})$.

Answer 2

There are several articles about the surjectivity of the exponential function for reductive Lie groups (including $GL_n(\mathbb{C})$), and other classes, like solvable and nilpotent Lie groups:

Historical Remarks on the Surjectivity of the Exponential Function of Lie Groups

The surjectivity question for the exponential function of real Lie groups: A status report

Answer 3

Not an answer, but another proof for surjectivity of $\exp$ for $\operatorname{GL}_n(\mathbb{C})$ is by the Cauchy integral formula for Banach space valued holomorphic functions:

If $x\in A$, a complex Banach algebra, and the spectrum $\sigma(x)$ of x does not separate $0$ from $\infty$, then $x$ has a logarithm (and also nth roots) in $A$. That is, there is a $y\in A$ with $x=\sum\frac{y^n}{n!}$. I seen this on page 264 of 'functional analysis' by Rudin.