# Ball and geodesics in a riemannian manifold

Let $$(M,\langle\cdot,\cdot\rangle)$$ be a riemannian manifold. Let $$S \subset M$$ be a compact submanifold with corners and $$p \in M$$ a point belonging to $$\partial A$$. Can we ensure that exists $$\varepsilon > 0$$ and $$\delta > 0$$ (depending on $$p$$) such that for every $$q \in B(p,\varepsilon)\cap S$$ and $$0 \leq t \leq \delta$$ it is satisfied $$\exp_p(t\exp_{p}^{-1}(q)) \in S \cap B(p,\varepsilon)?$$

I know that if $$p \in \operatorname{int}(S)$$ the result is obvious. If $$S$$ is locally convex the result is obvious too. Although, I was drawing some polyhedras in $$\mathbb{R}^3$$ and I thought that it is true in a this context too. Anyone can help me?

PS. I've just realized that the condition that I'm asking is: are compact submanifold with corners locally star-shaped?

• Presumably $A=S$? Commented Nov 25, 2022 at 12:55
• Yes, sorry :). I edit it Commented Nov 25, 2022 at 12:57
• there are some other things you may want to fix. I assume $x= p$? Are you assuming that $q\in S$? And is $\exp$ the exponentional map or $M$, or of $S$? If I look at the northern half of the $n-1$ dimensional sphere in $\mathbb{R}^n$ (which does not even have corners) I wonder why such a result should be true, anyway? Commented Nov 25, 2022 at 13:10
• What about $S=[-1,1]^2$ in $M=\Bbb R^2$, $p = (0,1)$, $q = (0,1+\varepsilon/2)$? In that case, $\exp_p(tv) = p + tv$, to that $\exp_p(t\exp^{-1}(q)) = p+t(0,\varepsilon/2)$, which is never in $S$ for $t>0$. Commented Nov 25, 2022 at 13:10
• The example it not valid because I said $q \in B(p,\varepsilon) \cap S$. Your example, in fact, has the property that I want to prove. It is a very interesting question and I can't find anything about it. Commented Nov 25, 2022 at 13:14

Consider $$M=\Bbb R^2$$ endowed with the euclidean metric, and $$S = \left\{(x,y) \in M \mid x\in [-1,1] \quad \text{and} \quad 0\leqslant y \leqslant 1 + x^2\right\}.$$
$$S$$ is a compact submanifold with corners (the corners are $$(\pm 1,0)$$ and $$(\pm 1,2)$$). Let $$p= (0,1)\in S$$, and $$q = (x,1+x^2)\in S$$ for some $$x\in [-1,1]$$. By strict convexity of the function $$f(x) = 1+x^2$$, we have $$S \cap [p,q] = \{p,q\}.$$ From this, it should be clear that the statement you wish to be true is false.