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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?

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  • $\begingroup$ Presumably $A=S$? $\endgroup$
    – Didier
    Commented Nov 25, 2022 at 12:55
  • $\begingroup$ Yes, sorry :). I edit it $\endgroup$ Commented Nov 25, 2022 at 12:57
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    $\begingroup$ 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? $\endgroup$
    – Thomas
    Commented Nov 25, 2022 at 13:10
  • $\begingroup$ 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 $t>0$. $\endgroup$
    – Didier
    Commented Nov 25, 2022 at 13:10
  • $\begingroup$ 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. $\endgroup$ Commented Nov 25, 2022 at 13:14

1 Answer 1

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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.

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