# Example of the Manifold with a Non-Surjective Exponential Map at a Point

I need to find an example of a connected riemannian manifold $$(M,g)$$ and a point $$p \in M$$ such that the exponential map $$\exp_p : T_pM \to M$$ is well-defined, but is not surjective.

Taking $$\mathbb{R}^n$$ and removing a point won't work because the resulting space won't be geodesicly complete and $$\exp_p$$ won't be defined over whole tangent space $$T_p M$$.

Every compact riemannian manifold will have surjective $$\exp_p$$. So, I can focus only on non-compact manifolds

Here the first idea is to construct a surface similar to $$\mathbb{R}^2$$ but with a hole such that a geodesic which will normally connect two points from the different sides of the hole, will run down the hole (possibly to another side of the surface). The problem is that there are many such surfaces, for example, one can b e defined parametricly as $$r(x,y) = \Big(x,y, \ln\sqrt{x^2 + y^2}\Big)$$ (this one doesn't have "running to the other side" effect, by the way). But it still very hard for me to write explicitly equations for geodesics and analyse there behavior near the hole.

So, I need an example of a manifold or a surface where the rigorous proof for $$\exp_p$$ not being surjective can be carried out with a relative ease.

• It looks like the surfaces you're going to think of or write down in $\Bbb R^3$, will be complete, hence geodesically complete. Do you know an example of a matrix that cannot be written in the form $e^A$ for any matrix $A$? Commented May 10, 2019 at 20:57
• @Ted I thought about $SL(\mathbb{R},2)$, .As I heard operator exponent of Lie algebra is the same as riemannian exponent iff the metric on the lie group is bi-invariant. And unlike $O(\mathbb{R})$ I don't know how to construct any bi-invariant metric on $SL(\mathbb{R})$.Of course, I can just assume to have one, but would it be an honest solution? Commented May 10, 2019 at 21:12
• Aha. No, in fact it has none, as follows from the fact that the exponential isn't surjective. Commented May 10, 2019 at 21:19
• @TedShifrin Moreover, I think $SL(\mathbb{R},n)$ is also complete as the preimage of the closed set $\det^{-1}\{1\}$ in the complete space $\mathbb{R}^{n\times n}$. As I understand being geodesically complete does not imply surjictivity of the map $\exp_p$, just that $\exp_p$ is defined over whole $T_p M$. Thanks for help. Commented May 10, 2019 at 21:19
• Hopf-Rinow will tell you there's a distance-minimizing geodesics joining any two points :) ... Commented May 10, 2019 at 21:21

There is no such example. The typical proof of the Hopf-Rinow theorem shows that if $$\exp_p$$ is defined on all of $$T_pM$$ for some $$p\in M$$, then $$M$$ is metrically complete, and in that case, every $$q\in M$$ can be joined to $$p$$ by a distance-minimizing geodesic $$\gamma$$. If we set $$v = \gamma'(0)\in T_pM$$, we then have $$q = \exp_p(tv)$$ for some $$t\in \mathbb R$$.

See Corollaries 6.20 and 6.21 in my Introduction to Riemannian Manifolds (2nd ed.).

• Yes this is indeed true. Thank you for the answer. The only thing I want to add for the sake of completeness is that strong Hopf-Rinow theorem seems not to work in the infinite dimension. But it would be total overkill to embark on constructing a Hilbert manifold with this property. Commented May 10, 2019 at 23:02
• @NikPronko: Indeed, see academic.oup.com/blms/article-abstract/7/3/261/… Commented May 11, 2019 at 0:10
• Link points to: C.J. Atkin, The Hopf-Rinow Theorem is false in infinite Dimensions, Bulletin of the London Mathematical Society, Volume 7, Issue 3, November 1975, Pages 261–266, doi.org/10.1112/blms/7.3.261 Commented Sep 24, 2023 at 20:49