# A question on convexity and conjugate points

Let $$(M,g)$$ be a compact smooth simply connected Riemannian manifold with a smooth boundary. Assume also that $$(M,g)$$ does not have any conjugate pairs of points. Let $$\Gamma \subset \partial M$$ be a smooth connected subset of the boundary that is strictly concave in the sense of the second fundamental form. Prove that there can be no inextendible geodesic in $$M$$ with both end points on $$\Gamma$$.

• Maybe I am misunderstanding some of the terminology but what about a flat torus delete a small ball? Aug 24, 2022 at 4:00

## 1 Answer

Here is a proof that works, I think—although it's probably a bit clumsier than necessary. The short version is that the geodesics in $$M \setminus \partial M$$ are minimizing. This means we can't connect two points $$p,q \in \Gamma$$ by a geodesic in $$M$$, because there is a strictly shorter curve inside $$\partial M$$.

Now for the long version: we start with the following claim.

Claim. Let $$x \in M$$. Then $$\operatorname{exp}_x: \operatorname{exp}_x^{-1}(M \setminus \partial M) \subset T_x M \to M \setminus \partial M$$ is a diffeomorphism.

This is proved by essentially following the proof of the Cartan–Hadamard theorem; one notable difference is that the domain of definition is restricted because we are on a manifold with boundary. The argument essentially goes as follows: as $$M$$ has no conjugate points, the exponential map is a local diffeomorphism. In fact it is a covering map, and $$M$$ being simply connected, it must be a diffeomorphism.

We rely on the claim for the following corollary. It implies that any two points in $$M \setminus \partial M$$ are connected by a unique geodesic, and therefore all geodesics in $$M \setminus \partial M$$ are minimizing.

All this preamble being completed, let's get to the proof.

Proof. We argue by contradiction. Let $$p,q \in \Gamma \subset \partial M$$ be two points in the boundary, connected by some geodesic $$\gamma: [0,L] \to M$$ in the interior of $$M$$, with only its endpoints on the boundary. Let $$\eta: [0,d] \to \partial M$$ be the curve realizing the distance $$d < L$$ between the two; as $$\Gamma$$ is strictly concave, this must lie inside the boundary.

By the (corollary to) the claim, the geodesic $$\gamma$$ must be minimizing, at least for all times $$t \in (0,L)$$. To obtain a contradiction, take $$\epsilon > 0$$ small enough that $$d + 5 \epsilon < L$$. Consider the curve connecting the two points $$\gamma(\epsilon)$$ and $$\gamma(L-\epsilon)$$ by:

• first going from $$\gamma(\epsilon)$$ to $$\gamma(0) = p$$ via $$\gamma^{-1}$$;
• then from $$p$$ to $$q$$ via $$\eta$$;
• finally from $$q = \gamma(L)$$ to $$\gamma(L-\epsilon)$$ via $$\gamma^{-1}$$ again.

This has length $$d + 2 \epsilon$$, whereas going along $$\gamma$$ would give length $$L - 2 \epsilon$$.

We chose $$\epsilon$$ so small that $$d + 3 \epsilon < L - 2\epsilon$$. We use the extra wiggle room to move $$\eta$$ into the interior, to a curve $$\eta'$$ with length at most $$d + \epsilon$$. The curve resulting by pasting $$\eta'$$ into the $$\eta$$-portion has length at most $$d + 3 \epsilon < L - 2\epsilon$$, which is absurd because $$\gamma$$ is minimizing.

• I was looking at this proof today and do not understand why $d$ must be less than $L$. Could you justify this rigorously? Definition of convexity is local (based on second fundamental form) but here the d<L is a more global statement
– Ali
May 21, 2023 at 11:49