Let $M$ be a complete, connected Riemannian manifold without boundary. Given a point $p\in M$ and a subset $K$ of $S\_p M$, the unit sphere in $T\_p M$, define the $K$-cone of directions $C(K)$ around by $C(K) := \{w \in T\_p M| \ w/|w| \in K\}$.


Given $p$ and $K$ as above, define $\text{inj}(p, K)$ to be the supremum of values $r \geq 0$ for which the exponential map at $p$ is injective on $C(K) \cap B\_r (0)$.


Suppose that


$\sup\_{p \in M} \sup\_{K \subset S\_p M, K \text{ non-null}} \text{inj}(p, K) < \infty$.


Does it follow that the manifold $M$ is compact?


*Note: By non-null, I mean not of null measure as a subset of the smooth manifold $S\_p M$.*



If every non null set of geodesics intersects itself in uniformly bounded finite time, is the manifold compact?
Such a manifold has bounded diameter, and is therefore compact. To see this, let $0 < T < \infty$ be the quantity you define. I claim that $\mathrm{diam} M \leq T$.


Additionally for each point $p \in M$ and unit tangent vector $v \in T\_p M$ let $\tau(v)$ be the first conjugate time along the geodesic $\gamma: t \mapsto \exp\_p(tv)$. Explicitly, $\tau(v) > 0$ is the first time $t > 0$ so that $\gamma(t)$ is a conjugate point of $p$ along the geodesic.


Then $\tau(v) \leq T$ for all unit vectors, which can be obtained as follows. Let $\rho\_j \downarrow 0$ and $K\_j = \{ w \in T\_p M \mid \lvert w \rvert = 1, \lvert w - v \rvert \leq \rho\_j \}$. By definition of $T$ there are vectors $v\_j \neq w\_j \to v$ so that the corresponding geodesics intersect before this time. Therefore $\gamma$ has a conjugate point before or at $\gamma(T)$.


It is well-known that a geodesic is not length-minimising past its first conjugate point. Therefore every length-minimising geodesic has length at most $T$, and the claim is proved.


**Edit.** To see why $\gamma$ must reach a conjugate point before time $T$, one can argue as follows. Let $q\_j \neq p$ be the first intersection point of the geodesics directed by $v\_j,w\_j$. Because $v\_j,w\_j \to v$ we may extract a subsequence so that $q\_j \to q$ as $j \to \infty$, where $q = \gamma(t)$ for some $t > 0$. This point $q$ is conjugate to $p$. If $D \exp\_p(t v)$ were non-singular then there would exist a neighbourhood $U \subset T\_p M$ of $tv$ so that $\exp\_p \mid U$ is a diffeomorphism onto its image. But eventually $t v\_j \neq t w\_j$ both lie in $U$, and $q\_j$ in its image. In particular $\exp\_p$ is not injective in $U$.


