# Are distance functions necessarily nonsmooth on the cut locus?

Let $$(M,g)$$ be a complete Riemannian manifold and fix $$p\in M$$. Consider the distance function $$r(x):=d(p,x)$$. It is well-known that $$r$$ is smooth outside $$\operatorname{cut}(p)\cup\{p\}$$ where $$\operatorname{cut}(p)$$ is the cut locus of $$p$$. My question is:

Is $$r$$ necessarily nonsmooth on every point of $$\operatorname{cut}(p)$$?

It is well-known that $$x\in\operatorname{cut}(p)$$ if and only if either (a) there are two distinct unit-speed minimizing geodesics $$\gamma_1,\gamma_2:[0,\ell]\to M$$ joining $$p$$ and $$x$$, or (b) $$x$$ is a critical value of $$\exp_p$$. In Peter Petersen's Riemannian Geometry, the author gave a remark on this: In case (a), $$\nabla r$$ could be either $$\gamma_1'(\ell)$$ or $$\gamma_2'(\ell)$$ and hence does not exist; in case (b), $$\operatorname{Hess}r$$ is undefined since it must tend to $$-\infty$$ along certain fields.

I know that the part about (a) is intuitive, but is there any way to make the argument rigorous? O the other hand, I don't see why $$\operatorname{Hess}r$$ must blow up.

For a): Let $$U$$ be the set on which $$r$$ is differentiable. Since $$r$$ is 1-Lipschitz, we have $$\Vert \nabla r \Vert \le 1$$. Anyway, what I want to show is that for any shortest geodesic $$\gamma$$ with $$\gamma(0) = p$$, we have $$(\nabla r)_{\gamma(t)} = \gamma'(t).$$ For this, let $$v \in T_{\gamma(t)}M$$ be arbitrary and $$\tilde{\gamma}$$ the geodesic with $$\tilde{\gamma}(0) = \gamma(t)$$ and $$\tilde{\gamma}'(0) = v$$. Then we can compute $$\langle(\nabla r)_{\gamma(t)},v\rangle = (dr)_{\gamma(t)} \cdot v = \frac{d}{dt}_{\vert t=0} r(\tilde{\gamma}(t)) = \frac{d}{dt}_{\vert t=0} d(p,\tilde{\gamma}(t)) = \langle \gamma'(t), \tilde{\gamma}'(0)\rangle,$$ where the last equality follows from the first variation formula. By the uniqueness of the gradient we gain our claim.

Also, here's another way of computing $$(\nabla r)_{\gamma(t)}$$ without using the first variation formula: $$\langle \nabla r, \gamma' \rangle = \frac{d}{dt} r(\gamma(t)) = \frac{d}{dt} t = 1$$ but also by Cauchy Schwarz $$\langle \nabla r, \gamma' \rangle \le 1 \cdot 1 = 1$$ and hence we have $$\nabla r = \gamma'$$.

Notice that we computed $$\frac{d}{dt} r(\gamma(t))$$ with the limit from below, assuming it was differentiable. So, if you are in case a) and assumed $$r$$ was differentiable in $$x$$, then you would get $$\nabla r(x) = \gamma_1'(x)$$ but also $$\nabla r(x) = \gamma_2'(x)$$, which is a contradiction.

For b): If $$\gamma:[0,L] \to M^n$$ is a geodesic and $$x = \gamma(L)$$ its first conjugated point to $$p = \gamma(0)$$, then the Weingarten map $$A(t) = \nabla_\cdot N$$ (where $$N = \nabla r$$ is a normed normal field along the distance spheres $$S_t(p)$$) has a pole in $$t = L$$. This is because $$A(t) \cdot J(t) = J'(t)$$ for $$0 and $$J$$ any Jacobi field along $$\gamma$$ with $$J(0)=0$$ and $$J'(0) \neq 0$$. But since the two points are conjugated, there exists such a $$J$$ with also $$J(L)=0$$ and $$J'(L) \neq 0$$ (otherwise $$J\equiv 0$$). Thus $$\lim\limits_{t \to L} A(t) \cdot J(t) = \lim\limits_{t \to L} J'(t) = J'(L) \neq 0,$$ but $$\lim\limits_{t \to L} J(t) = 0,$$ so $$A(t)$$ must blow up for $$t \rightarrow L$$.

Since the Hessian of $$r$$ is (tangentially to the distance spheres) given by the Weingarten map, the claim follows, since if $$r$$ were smooth at $$x = \gamma(T)$$ then $$A(t)$$ was continuous which is impossible since $$\lim\limits_{t \to T} A(t)$$ blows up. Thus, $$r$$ can't be contiuously differentiable at $$x$$.

• The problem is that $d(p,\tilde{\gamma}(t))$ may not be realized by the curves in your variation, so you cannot apply the first variation formula. In other words, it may not be possible to construct a variation consisting all of shortest geodesics. Consider the case of spheres. Choose any shortest geodesic $\gamma$ connecting the north and south poles, then choose any vector $v$ at the north pole orthogonal to the chosen geodesic. As in your answer, we have $\tilde{\gamma}$. But the shortest geodesic from the south pole to $\tilde{\gamma}(t)$ changes drastically as $t$ goes away from $0$. Feb 6, 2020 at 1:59
• Yes, exactly. That's why I wrote "this helps you for (a)". You pointed out the problem. I assumed $U$ to be the set on which $r$ is differentiable. In that case, the curves $\tilde{\gamma}$ realize the distance. In other words: In case (a), the answer is yes: $r$ is nonsmooth in these points. Feb 6, 2020 at 7:27
• The "problem" was referring to your argument... The derivative of $d(p,\tilde{\gamma}(t))$ is not computable from the first variation formula, because you simply can't find a variation with all nearby curves shortest, as in the example I gave in the previous comment. So it does not constitute a proof for nonsmoothness. Feb 6, 2020 at 9:23
• I added some stuff. Feb 11, 2020 at 17:54
• For the case of a first conjugate point to a fixed point but with a unique minimal geodesic connecting these two points, the distance function may be differentiable at this conjugate point but not up to 2nd order differentiability. Please see post here: mathoverflow.net/questions/403600/…
– Chee
Aug 23, 2023 at 4:31

I'd like to offer another answer on why $$\mathrm{Hess}r$$ blows up. Actuarally the following arguement has already been contained on Petersen's book.

Suppose $$\gamma:[0,l]\to M$$ be a unique minimizing geodesic connecting $$\gamma(0)=p$$ and $$\gamma(l)$$ with $$\gamma(l)$$ being the first conjugate point to $$p$$. Suppose $$J$$ is a nontrivial Jacobi field along $$\gamma$$ such that $$J(0)=0=J(l)$$.

If $$r$$ is smooth at $$\gamma(0)$$, then as (a) implies $$\nabla r=\gamma'(l),$$ \begin{aligned} \mathrm{Hess}r(J,J)&=JJr-\nabla_JJr=J\langle\gamma'(l),J\rangle-\langle\gamma'(l),\nabla_JJ\rangle\\ &=\langle \nabla_J\gamma'(l),J\rangle=\langle \nabla_{\gamma'(l)}J,J\rangle\\ &=\frac{d}{dt}\Big{|}_{t=l}\langle J,J\rangle. \end{aligned}

But also note that $$|J(t)|^2\to 0$$ as $$t\to l$$ which implies $$\log|J|^2\to -\infty$$ as $$t\to l$$. By elementary calculus there exists a sequence $$\{t_k\}$$ such that $$\lim\limits_{k\to\infty}\frac{2\mathrm{Hess}r(J(t_k),J(t_k))}{|J(t_k)|^2}=\lim\limits_{k\to\infty}\frac{d}{dt}\Big{|}_{t=t_k}\log |J|^2= -\infty.$$

Define a smooth vector field along $$\gamma(t)$$ by $$X(t)=\frac{J(t)}{|J(t)|}.$$ The norm of $$X(t)$$ equals $$1$$.

In local coordinates around $$\gamma(l)$$, we can assume $$X(t)=\sum\limits_{i=1}^nX^i(t)\frac{\partial}{\partial x_i}\Big{|}_{\gamma(t)}$$. Since the norm of $$X$$ is bounded, we can subtract a subsequence $$\{t_{k_l}\}$$of $$\{t_k\}$$ such that $$X^i(t_{k_l})$$ converges to some number $$X^i$$. Now $$X(t_{k_l})$$ converges to some $$X=X^i\frac{\partial}{\partial x_i}\Big{|}_{\gamma(l)}$$

However, by what we previously argued $$\lim\limits_{l\to\infty}\mathrm{Hess}r(X(t_{k_l}),X(t_{k_l}))=-\infty,$$ thus $$\mathrm{Hess}r(X,X)$$ is not well defined.

• a comment to your answer: (a) $Hess(r) (J,J) = (1/2) \partial_r \langle J, J \rangle$
– Chee
Aug 23, 2023 at 5:05