# Exponencial of a connected, compact non-abelian Lie Group.

I am trying to solve the following true or false question:

(True or False) Let $$G$$ be a connected, compact non-abelian Lie Group $$\Longrightarrow$$ $$\mathrm{exp}:\mathfrak{g} \to G$$ is not a local diffeomorphism map.

Once $$G$$ is compact, then $$\mathfrak{g}= \mathfrak{z}(\mathfrak{g})\oplus\mathfrak{s}$$ where $$\mathfrak{z}(\mathfrak{g})$$ is the center of the Lie Algebra $$\mathfrak{g}$$ and $$\mathfrak{s}$$ is a semi-simple ideal. Moreover $$\mathrm{exp}$$ is a surjetive map.

How should I proceed?

• $\exp$ may not be a local homeomorphisms, see this question with $G=SU(2)$. – Dietrich Burde Jun 10 at 18:09
• But it always occours when $G$ is connected, compact and non-abelian? – Matheus Manzatto Jun 10 at 18:55
• Hint: Every noncommutative compact Lie algebra contains a copy of ${\mathfrak su}(2)$. – Moishe Kohan Jun 10 at 21:59

It is well known that $$\mathrm{exp}:\mathfrak{g}\to G,$$ satisfies $$\begin{eqnarray}\mathrm{d}\left(\mathrm{exp}\right)_X = \mathrm{d}L_{e^X}\circ T_X,\quad (*)\end{eqnarray}$$ where $$T_X = \frac{ e^{\mathrm{ad}(X)} -1}{\mathrm{ad}(X)} =\sum_{k\geq0}\frac{1}{(k+1)!}(\mathrm{ad}(X))^k.$$
Since $$G$$ is compact, all eigenvalues of $$\mathrm{ad}(X)$$ are purely imaginary numbers, moreover, once $$G$$ is non-abelian there exists an eigenvalue $$i \lambda\neq 0$$ of $$\mathrm{ad}(X)$$ for some $$X\in \mathfrak{g}$$ and $$\lambda \in \mathbb{R}$$. Let $$c = 2\pi/\lambda$$ then $$2\pi i$$ is an eigenvalue of $$\mathrm{ad}(cX)$$ $$\Rightarrow$$ $$0$$ is eigenvalue of $$T_X$$ $$\Rightarrow$$ using $$(*)$$ $$\mathrm{d}(\mathrm{exp})_X$$ is not an isomorphism $$\Rightarrow$$ $$\mathrm{exp}$$ is not a local diffeomorphism at the point $$cX \in \mathfrak{g}$$.