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.

  • $\begingroup$ Matrix exponential is not an open map for $G$ noncompact. See [this question][1]. [1]: math.stackexchange.com/questions/189619 $\endgroup$ Commented Feb 12, 2013 at 19:53
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    $\begingroup$ Even for $G$ compact, $\exp$ is not a local homeomorphism: When $G=SU(2)$, exponential collapses the sphere $\left \{ \begin{pmatrix} a+bi & c+di \\ -c+di & a-bi \end{pmatrix} : a^2+b^2+c^2+d^2 = \pi^2 \right\}$. I don't know whether or not it is an open map. $\endgroup$ Commented Feb 12, 2013 at 19:59
  • $\begingroup$ @David: thank you for the first example, I'll think about it. As for the second, I don't follow. Where does the set you've given lie and how is it related to the property of local homeomorphism? $\endgroup$
    – Marek
    Commented Feb 12, 2013 at 20:09
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    $\begingroup$ The example should be the following. The Lie algebra of $SU(2)$ is the skew-Hermitian matrices. Inside the skew Hermitian matrices, we have the sphere $\left\{ \begin{pmatrix} bi & c+di \\ -c+di & -bi \end{pmatrix} : b^2+c^2+d^2=\pi^2 \right\}$. Every point on this sphere exponentiates to $- \mathrm{Id}$. This shows that the map is not locally injective, and hence not a local homeomorphism. $\endgroup$ Commented Feb 12, 2013 at 20:11
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    $\begingroup$ Right, that doesn't rule out $\exp$ being open. But it does rule out $\exp$ being locally injective, since any neighborhood of a point on the sphere will contain other points on the sphere. So the map is not locally injective, and hence not a local homeomorphism. It also turns out not to be open, see the answer I just wrote. $\endgroup$ Commented Feb 12, 2013 at 20:24

3 Answers 3


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.


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 $\left( \begin{smallmatrix} 1 \\ \pm i \end{smallmatrix} \right)$.

  • $\begingroup$ Is it clear that it is possible to prove surjectivity in this generality without using some structure theory, e.g. knowing that every element is contained in a maximal torus? I think what remains of the arguments in the question after your answer only shows that every element of a compact and connected Lie group $G$ is a product of finitely many elements which are in the image of the exponential map. $\endgroup$
    – Martin
    Commented Feb 12, 2013 at 20:34
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    $\begingroup$ As you say, you can use differential geometry to avoid the structure theory. In terrytao.wordpress.com/2011/06/25/… , Terry Tao asks whether there is a proof which avoids both of these. Although he precedes the statement with a lot of modest comments about how little he knows about Lie groups, I would take this as evidence that there isn't an easy way. $\endgroup$ Commented Feb 12, 2013 at 20:42
  • $\begingroup$ I seem to have dropped an "along these lines"... Thanks for the link! $\endgroup$
    – Martin
    Commented Feb 12, 2013 at 20:49

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 $\{(at,t)+\mathbb{Z}^2 : t\in\mathbb{R}\} \subset \mathbb{R}^2/\mathbb{Z}^2$.


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