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Let $(M, g)$ be a Riemannian manifold. For $p \in M$, the convexity radius at $p$, denoted by $\mathrm{conv}(p)$, is defined as $$\mathrm{conv}(p) := \sup \{r > 0 \ \mid \ B_g(p, r) \text{ is a geodesically convex geodesic ball} \} \in (0, \infty]. $$ It is known that $\mathrm{conv}(p) > 0$ for all $p \in M$.

I would like to show that the map $$\mathrm{conv} : M \to (0, \infty], \quad p \mapsto \mathrm{conv}(p) $$ is continuous.

I tried showing that, around a point $p \in M$ where $\mathrm{conv}(p) < \infty$ is finite, the $\mathrm{conv}$ map is $1$-Lipschitz. For $q \in M$, due to symmetry, is suffices to show that $$\mathrm{conv}(q) \geq \mathrm{conv}(p) - d_g(p,q). $$ If $d_g(p, q) \geq \mathrm{conv}(p)$, we cleary have $$\mathrm{conv}(q) \geq 0 \geq \mathrm{conv}(p) - d_g(p,q). $$ If, however, $d_g(p, q) < \mathrm{conv}(p)$, I am not sure how to prove that $$\mathrm{conv}(q) \geq \mathrm{conv}(p) - d_g(p,q), $$ because I am not sure that the geodesic ball of radius $\mathrm{conv}(p) - d_g(p,q)$ around $q$ is geodesically convex.

Also, I do not how to prove continuity in the case that $\mathrm{conv}(p) = \infty$.

user1003515
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  • The proof of continuity has two parts: Lower semicontinuity and upper semicontinuity. Can you prove either one of these? – Moishe Kohan Dec 11 '21 at 15:51
  • @MoisheKohan I do not think so. – user1003515 Dec 13 '21 at 08:39
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    A supplement on reference: this is problem 6-6 in Introduction to Riemannian Manifolds by John M.Lee. Maybe we should @Jack Lee :) – Juggler Dec 16 '21 at 14:42
  • I think this is a related discussion: https://math.stackexchange.com/questions/745813/is-there-any-relationship-between-the-convexity-radii-of-two-near-points-in-a – Yunmath Dec 17 '21 at 16:39
  • @Yunmath I looked into the proof of this result in Klingenberg's book, but it is not useful...It states the it is "trivial from the definition" that the $\mathrm{conv}$ map is $1$-Lipschitz. Moreover, the book does not treat the case $\mathrm{conv}(p) = \infty$ from my understanding. – user1003515 Dec 18 '21 at 21:28
  • @user1003515 I think the definition of convexity radius in Klingenberg's book is different from yours, he use a notion called strongly geodesically convex. By the way, I think in the link above, it has an example showing that the convexity radius is not continuous if we use your definition. – Yunmath Dec 19 '21 at 05:50
  • Some thoughts here. By applying a limit process, we can show the value at each point is no less than the limsup of values around the point since we are dealing with open geodesic ball. But I cannot address the other side. – Juggler Dec 21 '21 at 04:49
  • @juggler could you please explain more detail? I think the issue is that, in the definition above, we do not know if we have a geodesically convex ball B, whether other metric balls contained in B (may not have the same center) is also convex. I think this is the reason why Klingerberg introduce the notion of strongly geodesically convex in his book. – Yunmath Dec 21 '21 at 06:22
  • @Yunmath To clarify, I follow Lee’s definition that being geodesically convex means any two points in it have a unique minimizing geodesic segment that lies in the ball. – Juggler Dec 21 '21 at 06:42
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    If $x_i$ converge to $x$ with convex geodesic balls $B_{r_i}(x_i)=:B_i$ and $r_i$ converge to $r$, then any two points in $B_r(x)$ lie in $B_i$ both for sufficiently large $i$, so their minimizing geodesic segment lies in the ball at $x$. – Juggler Dec 21 '21 at 06:49
  • Thank you very much! I just think about a small issue in your argument, how can we show that the minimizing geodesic does not touch the boundary of $B_r(x)$? – Yunmath Dec 21 '21 at 07:32
  • @Yunmath You’re right. I don’t know how to adjust it. Maybe the issue is not as small. – Juggler Dec 21 '21 at 10:45

1 Answers1

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The key part of the proof is the fact that if $\gamma$ is a minimizing geodesic segment joining two points of a geodesically convex geodesic ball $B_r(p)$, then $d_g(p,\gamma(t))$ attains a maximum at one of its endpoints. This is from Problem 6-5 in Introduction to Riemannian Manifolds by John M.Lee or Lemma 4.1 from Riemannian Geometry by Do Carmo.

To prove $$\text{conv}(p)\geq \text{conv}(q)-d_g(p,q)$$ when the RHS is positive, let $\text{conv}(q) = R$. For any $r<R$, the ball $B_r(q)$ is geodesically convex and $B_{r-d_g(q,p)}(p) \subseteq B_r(q)$. For any two points $x_1,x_2\in B_{r-d_g(q,p)}(p) \subseteq B_r(q)$ there is a minimizing curve $\gamma$ segment in $B_r(q)$ joining these two points. Now $$ d_g(p,\gamma(t)) \leq d_g(q,p) + d_g(q,\gamma(t)) $$ Since $d_g(q,\gamma(t))$ attains its maximum at $x_1$ or $x_2$, so does $d_g(p,\gamma(t))$ i.e., the curve $\gamma(t)\in B_{r-d_g(q,p)}(p)$. This argument is wrong, see comment by @Amd below. Therefore $$ conv(p) \geq r-d_g(p,q) $$ Since this is true for all $r<R$, the desired result follows.

Note that this proof also works for the case $conv(q)=\infty$.

Shubham Yadav
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  • Are you sure about your argument "since $(,())$ attains its maximum ..., so does $(,())$ "? Basically, you have two continuous functions $_1$ and $_2$ on $[0,1]$ such that $_1\leq +_2$ ( a constant), and you claim that since $_2$ attains its max at a boundary, so does $_1$. – Amd May 14 '24 at 07:41
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    @Amd you are correct. The argument is wrong and I don't know if it can be rectified. – Shubham Yadav May 30 '24 at 04:46