2
$\begingroup$

Let $M$ be an $n$-dim Alexandrov space with curvature bounded below $sec \geqslant k$, possibly non-compact. We assume that $M$ has no boundary for simplicity. For a compact subset $K \subset M$, the cut points $C_K$ are defined as the points $x$ such that any geodesics connecting $K$ and $x$ are not extendable.

Let $d_K$ be the distance function to $K$. We know that $d_K$ is semiconcave on $$(M \setminus K)\cap\{d_K < \frac{\pi}{2\sqrt k}\}.$$

For a semiconcave function $f$, the gradient $\nabla_x f=d_xf(\xi_{max})\cdot \xi_{max}$ exists, possibly $0$. Then at $$M \setminus (K\cap C_K)\cap\{d_K < \frac{\pi}{2\sqrt k}\}, |\nabla_x d_K|=1.$$ But at cut point $x$, $\nabla_x d_K$ may be $0$ (for example, the local maximal point for $d_K$). Otherwise not $0$, but $|\nabla_x d_K|<1?$ So can one give an example for this last case?

Beginning at a "good" point, the gradient flow is a unit speed geodesic for a short time, and stops or changes direction when meeting a cut point.

$\endgroup$
  • $\begingroup$ (I tried to fix the LaTeX, and the English. Rollback if I have changed the intended meaning.) $\endgroup$ – Joseph O'Rourke Jun 27 '15 at 15:00
3
$\begingroup$

The distance function is semi-concave on all of $M \setminus K$. The additional restriction is unnecessary.

The easiest example I can think of for a gradient strictly between zero and one is the cone. Consider a 2-dimensional cone, and let $K = \lbrace p \rbrace$ be a single point which is not the vertex.

Let $q$ be the point antipodal to $p$, by which I mean the point at the same distance from the origin as $p$, but as far away from $p$ as possible.

Now there are two geodesics from $p$ to $q$, so extension is not possible. The vector $\xi_{\textrm{max}}$ is the vector at $q$ which points away from the origin, and $0 < \left| \nabla_q d_K \right| < 1$.

In fact, varying $x$ along the ray from the origin through $q$, one can obtain every possible value $0 \leq \left| \nabla_x d_K \right| \leq 1$. You can see this by cutting the cone along the ray through $p$, and considering the two geodesics from $p$ to any point of the ray.

$\endgroup$
  • $\begingroup$ Maybe I misuderstand you, I don't know how to get every possible value between $0$ and $1$ for a fixed cone. Cut the cone along the ray through p, we get a sector with angle $2\theta<\pi$. Then the angel between $p$ and $q$ is $\theta$. So the angle formed by $qp$ and the ray from the vertex through $q$ is $\alpha=(\pi+\theta)/2$. So $|\nabla_q d_p|=-\cos \alpha$. $\endgroup$ – mafan Jul 1 '15 at 3:00
  • $\begingroup$ First off, the sector angle only has to be $2 \theta < 2 \pi$. Your calculation of the gradient at $q$ is correct. I have edited my answer a little to clarify -- if you allow $x$ to vary along the ray through $q$, then the gradient at $x$ varies through all possible values. Move $x$ so that it is $\left| p \right| \cos \theta$ from the vertex. Then the angle is $\pi/2$. Move it to infinity and the angle becomes $\pi$. $\endgroup$ – John Harvey Jul 1 '15 at 8:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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