On surjectivity of exponential map for Lie groups

Question

A recent question made me realize I didn't know any proof that exponential of a Lie algebra $\mathfrak g$ of a compact connected Lie group $G$ is surjective. After a bit of thinking I've come up with two proofs. First is based on relation between exponential curves and geodesics. This is rather technical but also gives us other useful information. This is not what I want to talk about here though.

The second proof (which I find slicker) is based only on topology and goes like this: Since $\exp$ is a local homeomorphism, it is both open and closed. Therefore $\exp(\mathfrak g)$ is clopen and so equal to $G$.

The trouble with this "proof" is that it also proves the statement for $G$ non-compact (which is false). So I wonder (and this is my question) what precisely went wrong.

Can the above mentioned "proof" be made into a real proof?

My thoughts on this are that $\exp$ is closed and open only when $G$ is compact because then we can pick a bounded open subset $C \subset \mathfrak g$ such that $\exp(C) = G$ and we can use the relation $\exp(A+\epsilon B) \approx \exp(A)\exp(\epsilon B)$ to conclude that $\exp$ is a local homeomorphism everywhere in $C$ (not just around $0$). This implies that $\exp$ is open (since it is locally open) in $C$. Also, since any closed subset of $C$ is compact, it's image is also compact and therefore closed in $G$.

Where exactly does this argument break down when $G$ is not compact.

Answer 1

Here's a non-compact example of the non-surjectivity of $\exp$. Take $\mathfrak g=\mathfrak{sl}_2(\Bbb C)$ : the matrix $$T=\left(\begin{array}{rr} -1&1\\0&-1\end{array}\right)$$ is not in $\exp(\mathfrak g)$. For if $x\in\mathfrak g$, we can find a basis in which it is triangular, say $$uxu^{-1}=\left(\begin{array}{rr} a&b\\0&-a\end{array}\right)$$ There are now two cases. If $a=0$, then $$u\exp(x)u^{-1}=\left(\begin{array}{rr}1&b\\0&1\end{array}\right)$$ which has a different spectrum than $T$. If $a\neq0$, then $x$ is diagonalisable, and so is its exponential. Since $T$ isn't diagonalisable, this finishes the proof that $T$ does not lie in the image of the exponential map.

---

In any case, I think the problem with your proof (at least with your geodesic argument), is that the exponential map, while it is a local diffeomorphisme on some neighborhood of $0_{\mathfrak g}$, isn't necessarily a local diffeomorphism near all points of $\mathfrak g$. This is easily seen in the case of $\mathfrak{su}(2)$ : all points at distance $\pi$ from $0$ are sent to the south pole of $\Bbb S^3\simeq SU(2)$, so the exponential fails to be locally injective near any of those points.

Answer 2

Even in the compact case, $\exp$ is not open. Look at the Lie group $SU(2)$, whose Lie Algebra is the skew-Hermitian $2 \times 2$ matrices. Look at the point $x_0 = \left( \begin{smallmatrix} \pi i & 0 \\ 0 & - \pi i \end{smallmatrix} \right)$ in the Lie algebra. We can find an open neighborhood $U$ of $x_0$ where the eigenvalues are distinct, with one in the upper halfplane and one in the lower halfplane. Let the eigenvector with eigenvalue in the upper half plane be $\left( \begin{smallmatrix} 1 \\ z \end{smallmatrix} \right)$; then $z$ is a continuous function from $U$ to $\mathbb{C}$. By shrinking $U$, we can assume that $|z| < 0.1$. Similarly, let the eigenvector for the eigenvalue in the lower halfplane be $\left( \begin{smallmatrix} w \\ 1 \end{smallmatrix} \right)$. Again, shrink $U$ so that $|w| < 0.1$.

For $x \in U$, then, $\exp(x)$ will have eigenvectors of the form $\left( \begin{smallmatrix} 1 \\ z \end{smallmatrix} \right)$ and $\left( \begin{smallmatrix} w \\ 1 \end{smallmatrix} \right)$ with $|w|$ and $|z|<0.1$.

However, $\exp(x_0) = - \mathrm{Id}$. So an arbitrarily small neighborhood of $\exp(x_0)$ will contain matrices of the form $\left( \begin{smallmatrix} - \cos \theta & \sin \theta \\ -\sin \theta & -\cos \theta \end{smallmatrix} \right)$, whose eigenvectors are $\left( \begin{smallmatrix} 1 \\ \pm i \end{smallmatrix} \right)$.

Answer 3

I think a local diffeomorphism need not be a closed map?

For example, fix some irrational $a \in \mathbb{R}\setminus \mathbb{Q}$, and consider an irrational winding of the plane around the torus: $$ f : \mathbb{R}^2 \rightarrow \mathbb{R}^2/\mathbb{Z}^2 : (x,y) \mapsto (ay+x,y)+\mathbb{Z}^2 . $$

Then $f$ is a local diffeomorphism, but $f$ maps the closed subset $\{0\} \times \mathbb{R} \subset \mathbb{R}^2$ to the non-closed $\{(at,t)+\mathbb{Z}^2 : t\in\mathbb{R}\} \subset \mathbb{R}^2/\mathbb{Z}^2$.