# Cartan-Hadamard's Theorem, Completeness

Theorem: Let $(M^n,g)$ complete, simply connected Riemannian manifold with non-positive sectional curvature. Then $M$ is diffeomorphic to $\mathbb{R}^n$.

Question: Can we change complete to "every two points can be joined by a minimizing geodesic"?

-
I do not have an answer, just few remarks. Answer is positive if $M$ is homogeneous (by using the developing map to the model homogeneous space). It is probably also positive for surfaces (using Gauss-Bonet formula, but I did not check the details). In general, I expect negative answer, but it is hard to construct an example. One can first try to construct a locally $CAT(0)$ example which is simply-connected and where every two points are connected by a minimizing geodesic,but even this is hard. – Misha Mar 22 '12 at 8:48
In the case $n=2$ I meant the question about injectivity of the exponential map, since the topological statement is clear. – Misha Mar 22 '12 at 10:32
Cartan-Hadamard does hold for some classes of incomplete nonpositively curved manifold, see Bowditch'es paper, projecteuclid.org/euclid.pjm/1102366182, Pacific J. Math. (1996) – Igor Belegradek Mar 22 '12 at 11:44
@Igor: nice reference! – alvarezpaiva Mar 22 '12 at 12:13

Take a point $x \in M$ and consider the star-shaped open set $U \subset T_xM$ where the exponential $$exp_x : U \rightarrow M$$ is defined. $U$ is diffeomorphic to ${\mathbb R}^n$ and the map ${\rm exp_x}$ is

(1) a local diffeomorphism (because of the absence of conjugate points = singularities of ${\rm exp}_x$);

(2) surjective (because I can join any other point of $M$ to $x$ by a geodesic).

I guess there is no problem and $M$ is diffeomorphic to $U$. Right??

Edit. As Claudio points out, the fact that ${\rm exp}_x$ is surjective and a local diffeomorphism does not imply that it is a covering map. Hence we cannot use the homotopy lifting property to conclude that it is injective and a diffeomorphism.

-
It is not clear to me that $\exp_x$ is injective (equivalently, a covering). – Claudio Gorodski Mar 21 '12 at 17:05
@Claudio: You're probably right. A surjective local homeomorphism is not necesarily a covering. I just can't see right now where this comes up in the standard proof of the Cartan-Hadamard theorem, but I've been grading all day ... – alvarezpaiva Mar 21 '12 at 18:47
@Claudio: You're right. That is the issue. – alvarezpaiva Mar 21 '12 at 19:08
If you look at Karcher's proof in his paper on Riemannian Comparison Constructions he observes that the exponential map is locally expanding when the curvature is non-positive and this should imply that the exponential map is a covering . – Mohan Ramachandran Mar 21 '12 at 19:29
Mohan: This type of argument need completeness. There are local isometries which are not covering maps. – Misha Mar 22 '12 at 8:25

Let $X$ be completion of $(M,g)$. Note that $X$ is simply connected. It follows since any loop in $X$ is a limit of a broken geodesic in $(M,g)$.

Therefore $X$ is CAT(0), in particular any two points are joined by unique geodesic. Therefore $M$ is diffeomorphic to $\mathbb R^n$ as any star-shaped domain (thanks to alvarezpaiva).

-
@Jason: to get $S^1$ as completion you have to take the subspace metric on $M$, but the metric is not geodesic: if $x,y$ are points near $p$ in $S^1$ such that $p$ is between $x$ and $y$, then any geodesic in $M$ that joints $x$, $y$ has length about $2\pi$. The correct metric on $M$ is the one induced by Riemannian metric on $S^1$ whose completion is $[0,2\pi]$. – Igor Belegradek Mar 22 '12 at 0:17
@epsilon-delta: The space $X$ (a priori) need not be locally $CAT(0)$. One cn easily construct (non-simply-connected) incomplete flat surfaces whose completion is not even locally contractible. – Misha Mar 22 '12 at 8:23
@Misha, I am sure this is true, but do not see a simple proof. I will delete my answer once the true answer is found. – ε-δ Mar 23 '12 at 1:00
@epsilon-delta: Take $X$ be the complement of the closed round ball in ${\mathbb R}^3$. Then $X$ is (locally) flat. Its metric completion is not a $CAT(0)$ space. Of course, $X$ is not convex... – Misha Mar 25 '12 at 13:30

You can't change it, but complete implies that there exists such a geodesic.

http://en.wikipedia.org/wiki/Hopf%E2%80%93Rinow_theorem

but the conditions are NOT equivalent. Regard an open ball in the euclidean space.

-
Yes, but an open ball is diffeomorphic to $\mathbb{R}^n$. Is there a counterexample to the theorem if the adjective "complete" is replaced by that assumption that "every two points can be joined by a minimizing geodesic"? – Deane Yang Mar 21 '12 at 9:43