# Show that the convex neighborhood in a Riemannian Manifold are subset contractibles

Show that the convex neighborhood in a Riemannian Manifold are subset contractibles (to any of their points).

A subset $A$ of the differential manifold $M$ is contractible to the point $a\in A$ when, the aplications $id_{A}$ (identity in $A$) and $\gamma_{a}:x\in A\to a\in A$ are homotopic (with a point basis $a$).
I know that "geodesic convexity" is a natural generalization of convexity for sets and functions to Riemannian manifold, so in a set the convexity implies the homotopy. My idea was, take a riemannian metric and take exponential map in convex neighborhood, but I don't see these set are contractible to a point. Any hint!

• To be clear, a convex subset of a Riemannian manifold is a subset where any pair of points are connected by unique minimizing geodesics which stay in the set? – Jason DeVito May 5 '16 at 17:30
• Yes, anyway a definition convex neighborhoods: for each $p\in M$ exist a number $\alpha>0$ such that the "geodesic ball" $B_{\alpha}(p)$ is strongly convex – User166523 May 5 '16 at 22:38
• Once you know that the exponential map $e=\exp_a : U \to A$ is a diffeomorphism, this is very simple - since $U$ is star-shaped, the map $\sigma(x,t) = tx$ is a contraction of $U$, and thus the composition $e(\sigma(e^{-1}(x),t))$ is a contraction of $A$. – Anthony Carapetis May 6 '16 at 12:41
• @AnthonyCarapetis your answer has sense to me ( anyway I can't prove), you can explain more your answer. Thanks! – User166523 May 7 '16 at 20:52
• @AnthonyCarapetis is just giving you an explicit contraction from $A$ to the point $a\in A$. He is just saying that the exponential map gives you a diffeo from $A$to a start-shaped set (there is a preferred point $z$ such that for any other point in $u\in U$ the line from $u$ to $z$ is contained in the set). Now this set is obviously contractible (move along this line), the map $\sigma$ is such a deformation retraction if you want. Then he is just conjugating it via $\exp$ to get a contraction for $A$ – Spotty May 12 '16 at 9:40

Let $U$ be a geodesically convex subset of $(M,g)$. Choose a point $p\in U$. For each $q\in U$ there exists a unique minimizing geodesic starting at $p$ and ending at $q$, i.e. there is a unique tangent vector $\hat{q}\in T_pM$ such that $$q = \exp_p\hat{q}.$$ It follows that there is a diffeomorphism $$\hat{U}:=\{\hat{q}\in T_pM\mid q\in U\}\cong U$$ via the exponential map (exercise: check it is indeed smooth with smooth inverse). The contraction is then induced by the map $I\times\hat{U}\to\hat{U}$ given by $$(t,\hat{q})\longmapsto t\hat{q}.$$ Exercise: Show that the set $\hat{U}$ is star-shaped with $0$ as a center. (Hint: a shorter piece of a minimizing geodesic is again a minimizing geodesic.)