3
$\begingroup$

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 $p$.

$\endgroup$
0

1 Answer 1

2
$\begingroup$
  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.

$\endgroup$
11
  • $\begingroup$ 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? $\endgroup$
    – Yuxiao Xie
    Feb 9, 2020 at 10:32
  • $\begingroup$ @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 $x$. $\endgroup$ Feb 10, 2020 at 0:04
  • $\begingroup$ What do you mean by ∡[xxnp]+∡[xnxp]? (sorry, don't even know the latex commands) $\endgroup$
    – Mathy
    Feb 12, 2020 at 7:23
  • $\begingroup$ @Mathy angles. $ $ $ $ $ $ $ $ $\endgroup$ Feb 13, 2020 at 6:30
  • $\begingroup$ Yeah, what's the angle? I don't get your notation, an angle is something between 2 things, and you just out one argument. $\endgroup$
    – Mathy
    Feb 13, 2020 at 6:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.