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I had a couple of related questions on the cut locus, conjugate points and smoothness of distance function. Let $(M,g)$ be a smooth complete Riemannian manifold and $r(x) = d(p,x)$ the distance function to a fixed point $p\in M$. Recall that the cut locus $Cut(p)$ consists of two kinds of points - points that are conjugate to $p$ along some geodesic, or points that have multiple minimal geodesics connecting them to $p$.

1) I know that $r$ is smooth on $M\backslash (Cut(p)\cup \{p\}$. Is it true that $r$ is in fact smooth at $x$ as long as there is a unique minimal geodesic connecting it to $p$? The only proof of smoothness involves the exponential map, but the exponential map is not invertible at conjugate points, so that approach wont work.

2) Is there a simple example of $(M,g,p)$ with points conjugate to $p$ but with unique minimal geodesics to $p$.

Poincare-Lelong
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1 Answers1

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  1. The distance function has to be differentiable at $x$ if there is a unique minimizing geodesic $[p,x]$. It means that the differential $d_xr$ is well defined linear function on the tangent space $\mathrm{T}_x$. The proof is straightforward. But, it does not mean that $r$ is smooth in a neighborhood of $x$ --- evidently it does not hold if $x$ is conjugate to $p$ along the geodesic.

  2. Take a surface of revolution for an even function $f$; say $f(0)=1$, $f''<0$. You can make it so that the curvature on the equator $(f(0)\cdot \sin t,f(0)\cdot\cos t,0)$ is positive and it takes maximal value on the surface. Then the maximal minimizing arc on the equator is unique minimal geodesics between its ends.

Anton Petrunin
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  • I'm sorry to disturb six years later, but I really don't see why 1) is true. That there is a unique minimizing geodesic from $p$ to $x$ does not mean the same is true in a neighborhood of $x$, so we can't apply the first variation formula. So how do we prove the distance function is differentiable? – Yuxiao Xie Feb 09 '20 at 10:32
  • @Colescu If the function is not differentiable at $x$, then there is a sequence $x_n\to x$ such that $\measuredangle[x_p^{x_n}]+\measuredangle[{x_n}_p^{x}]>\pi+\varepsilon$ for some fixed $\varepsilon>0$. Observe that the geodesics $[p, x_n]$ do not converge to $[p,x]$, whence its limit is another geodesic from $p$ to $x$. – Anton Petrunin Feb 10 '20 at 00:04
  • What do you mean by ∡[xxnp]+∡[xnxp]? (sorry, don't even know the latex commands) – Mathy Feb 12 '20 at 07:23
  • @Mathy angles. $ $ $ $ $ $ $ $ – Anton Petrunin Feb 13 '20 at 06:30
  • Yeah, what's the angle? I don't get your notation, an angle is something between 2 things, and you just out one argument. – Mathy Feb 13 '20 at 06:37
  • @Mathy these are angles of triangle $[xx_np]$ at vertexes $x$ and $x_n$. – Anton Petrunin Feb 13 '20 at 06:40
  • I still don't get the way you are arguing. And anyway, are you showing that there exists a second geodesic from x to p, or is it also shortest? – Mathy Feb 13 '20 at 07:51
  • @AntonPetrunin: I do not think that "there is a unique minimal geodesic connecting $p$ and $x$ for $p \ne x$" is sufficient for the "differentiability of $r(x)$ at $x$", because Klingenberg's famous book on Riemannian geometry gives an example of a conjugate point $a$ to a fixed point $b \ne a$ for which there exists exactly one minimal geodesic that connects $a$ and $b$. For such $b$, $r(b)$ will never be differentiable at $b$; see also this post: https://mathoverflow.net/questions/403600/riemannian-geometry-differentiability-of-distance-function-at-first-conjugate – Chee Feb 23 '23 at 04:40
  • @Chee, I guess your definition of differetiability is different from mine. I just say that differential exists at the point. – Anton Petrunin Feb 23 '23 at 16:30
  • @Chee I added a clarification --- hope it helps. – Anton Petrunin Feb 23 '23 at 17:22
  • @AntonPetrunin: thank you for your clarification – Chee Feb 23 '23 at 19:28