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A subset $S\in M$ is called strongly convex, or geodesically convex, if for any $p,q\in S$ there is a unique normal minimal geodesic $\gamma$ joining $p$ to $q$, and $\gamma$ is contained in $S$.

For example, on $(\mathbb{S}^{2},g_{\mathbb{S}^{2}})$, any geodesic ball of radius $r<\frac{\pi}{2}$ is strongly convex, (since, as I see, I take the unit sphere $\mathbb{S}^{1}$ on $\mathbb{R}^{2}$ his parametrization is $(\cos(\theta),\sin(\theta))$ then for any $r<\frac{\pi}{2}$ I will be still in the sphere, and as the sphere in $\mathbb{R}^{3}$ is the "same" the sphere $\mathbb{S}^{1}$ we're done).

And the book's claim, each strongly convex set is contractible, and the intersection of a family of strongly convex sets in $M$ is again strongly convex if it is not empty, I don't think is so trivial see this. And finally, can you give me an example of geodesically convex that does not belong to space, Thanks!!

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  • $\begingroup$ Out of curiosity, what book is this from? $\endgroup$ – littleO May 3 '16 at 22:16
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    $\begingroup$ The last paragraph is from the book Riemannian Geometry By Takashi Sakai. $\endgroup$ – User166523 May 3 '16 at 22:48
  • $\begingroup$ I think he (or you) is missing further assumption. Without this assumption it is unclear why the unique geodesic varies continuously. For instance, if you assume that $S$ is compact this will be the case. Or you need no conjugate points assumption. $\endgroup$ – Moishe Kohan May 5 '16 at 15:23
  • $\begingroup$ Related: math.stackexchange.com/questions/479022/…. From this answer you can conclude that your $S$ is contractible if, say, $(M,g)$ is complete and $S$ is closed. $\endgroup$ – Moishe Kohan May 5 '16 at 15:30
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The intersection rule is trivial. For any two points $p$ and $q$ in the intersection, they are both in each set, and therefore the unique geodesic $\gamma$ between them must have been in each set. But if the geodesic is in each set it is in the intersection. Therefore the definition is satisfied.

Contractibility seems harder. Let $p$ be a point in $M$. For $q$ in $M$ and $t$ in $[0, 1]$ define $f(t, q)$ to be the point a portion $t$ of the way along the unique geodesic $\gamma$ from $q$ to $p$. This is well-defined. Also $f(0, q) = q$ and $f(1, q) = p$. That has to be the contraction. The question is whether this map is continuous.

The challenge is this. If you have a triangle between two "nearby" points and a distant one, need the edges remain close? Well, in a general manifold, no. The "unique geodesic" condition has to come in somehow. But I can't see how to do it.

As for the last part of your question, the hyperbolic plane does not fit in space and is strongly convex.

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