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

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?

• The only proofs I know of surjectivity of $exp$ for compact connected Lie groups either 1) use some Riemannian geometry (Hopf-Rinow, etc.) or 2) the maximal torus theorem. How does one prove it using Lefschetz? Jan 29, 2018 at 23:38

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 $SL(2,\mathbb{C})$.
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:
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.