1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ $\exp$ may not be a local homeomorphisms, see this question with $G=SU(2)$. $\endgroup$ – Dietrich Burde Jun 10 at 18:09
  • $\begingroup$ But it always occours when $G$ is connected, compact and non-abelian? $\endgroup$ – Matheus Manzatto Jun 10 at 18:55
  • 1
    $\begingroup$ Hint: Every noncommutative compact Lie algebra contains a copy of ${\mathfrak su}(2)$. $\endgroup$ – Moishe Kohan Jun 10 at 21:59
1
$\begingroup$

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 $cX \in \mathfrak{g}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.