# Gradient of distance function at cut points on Alexandrov spaces

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.

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

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.
• 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 $|\nabla_q d_p|=-\cos \alpha$. – mafan Jul 1 '15 at 3:00
• 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 $\pi$. – John Harvey Jul 1 '15 at 8:20