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Let $(\mathcal{M},g)$ be a $C^{\infty}$-Riemannian manifold. A basic fact is that $g$ endows the manifold $\mathcal{M}$ with a metric space structure, that is, we can define a distance function $d:\mathcal{M}\times\mathcal{M}\longrightarrow\mathbb{R}$ (the distance between two points will be the infimum of the lengths of the curves which join the points) which is compatible with the topology of $\mathcal{M}$. Of course $d$ is continuous function, but what can we say about the differentiability of $d$?, is it smooth?. If not, Is there some criterion to know when it is?

Thanks in advance.

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    $\begingroup$ It's never smooth for a compact manifold. Consider a diameter -- the maximum of the function. It's not smooth there. For example, consider $S^n$. You get cone-type singularities. $\endgroup$ – Ryan Budney Apr 14 '10 at 4:02
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    $\begingroup$ Similarly, it's not smooth along the diagonal $\{(x,x) : x \in \mathcal{M}\}$ $\endgroup$ – Ryan Budney Apr 14 '10 at 4:04
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    $\begingroup$ One often looks at $d^2$ instead of $d$. This at least fixes the smoothness issues along the diagonal mentioned by Ryan. Still, on any comapct manifold, $d^2$ will necessarily fail to be smooth at some points. $\endgroup$ – Jason DeVito Apr 14 '10 at 5:04
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    $\begingroup$ I know we all know what you're talking about but presumeably you want $\mathbb{R}$ as your codomain rather than the manifold. $\endgroup$ – Spencer Apr 14 '10 at 10:34
  • $\begingroup$ Spencer. You are right. I have corrected the post. $\endgroup$ – Mauricio Apr 14 '10 at 16:45
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As others mentioned, you have to remove the diagonal of $M\times M$ or square the distance function. Then, for a complete $M$, the answer is the following.

The distance function is differentiable at $(p,q)\in M\times M$ if and only if there is a unique length-minimizing geodesic from $p$ to $q$. Furthermore, the distance function is $C^\infty$ in a neighborhood of $(p,q)$ if and only if $p$ and $q$ are not conjugate points along this minimizing geodesic.

Thus, the function is smooth everywhere if and only if $M$ is simply connected and the geodesics have no conjugate points. This property has numerous equivalent reformulations, including the following

  • for every pair of points, there is a unique minimizing geodesic between them;

  • for every pair of points, there is a unique geodesic between them;

  • every geodesic is minimizing;

  • the exponential map at every point $p\in M$ is a diffeomorphism from $T_pM$ to $M$.

In general, the distance function has one-sided directional derivatives everywhere. This derivative has a nice description in the case when you fix $p\in M$ and study the function $f=d(p,\cdot)$. Namely let $q\in M$, $q\ne p$, and denote by $\vec{qp}$ the set of initial velocity vectors (in $T_qM$) of unit-speed minimizing geodesics from $q$ to $p$. Then, for a vector $v\in T_qM$, the one-sided derivative $f'_v$ of $f$ in the direction of $v$ is $$ f'_v=\min\{-\langle v,\xi\rangle:\xi\in \vec{qp}\} . $$ This follows from the first variation formula and holds not only in Riemannian manifolds but also in Alexandrov spaces. It is not hard to derive the above differentiablity properties from this.

I don't have a textbook reference for this precise formulation in the Riemannian case, but any book that covers Berger's lemma about geodesics realizing the diameter probably has directional derivatives as a sublemma. For Alexandrov spaces, the standard reference is Burago-Gromov-Perelman's paper. An intro-level proof (not in a full generality) can be found in (a shameless advertisement follows) "A course in metric geometry" by Burago, Burago and myself, section 4.5.

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    $\begingroup$ Your advertisement would only have been shameless if there was another book which covered anywhere near the same material. $\endgroup$ – Paul Siegel Apr 14 '10 at 11:58
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It is a cute thoerem of Franz-Erick Wolter that a complete $n$-dimensional Riemannian manifold $M$ is necessarily diffeomorphic to $\mathbb R^n$ if there is a point $p\in M$ such that the squared-distance function $d(p,\mathord-)^2:M\to\mathbb R$ has directional derivatives at all points and in all directions. This provides examples.

See [Wolter, Franz-Erich. Distance function and cut loci on a complete Riemannian manifold. Arch. Math. (Basel) 32 (1979), no. 1, 92--96. MR0532854]

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The purpose of this answer is to provide a proof for the following result which Sergei mentioned in the accepted answer:

Proposition. Let $M$ be a complete Riemannian manifold and $x,y \in M\times M$, $x\neq y$. Then the following are equivalent:

  • The Riemannian distance $d:M\times M\rightarrow[0,\infty)$ is smooth in a neighbourhood of $(x,y)$.

  • There is only one length minimising geodesic connecting the points $x$ and $y$ and they are not conjugate along that geodesic.


Proof of Proposition. The Proposition follows from the three Lemmas below which freely use some properties of the so called segment domains $\Sigma_x=\{w\in T_xM: d(x,\exp_x(w))=\vert w \vert\}$:

  1. $\exp_x: \mathrm{int} \Sigma_x\rightarrow M$ is a diffeomorphism onto its image [Gallot-Lafontaine, Corollary 3.77 or Petersen, Lemma 5.7.8 and Proposition 5.7.10]
  2. $M= \exp_x(\mathrm{int}\Sigma_x)\cup\exp_x(\partial \Sigma_x)$ and the union is disjoint [Gallot-Lafontaine, Proposition 2.113].
  3. Denote $\partial^1\Sigma_x = \{w\in \partial \Sigma_x: \exp_x(w)=\exp_x(w')$ for some $w'\in \partial \Sigma_x\backslash\{w\}\}$. Then:
    • If $w\in \partial \Sigma_x\backslash \partial^1\Sigma_x$, then $D\exp_x\vert_w$ is singular. [Gallot-Lafontaine, Scholium 3.78 or Petersen, Lemma 5.78 ]
    • $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is dense [Sakai, Remark 4.9].


Warning. The denseness of $\exp_x(\partial^1\Sigma_x) \subset \exp_x(\partial \Sigma_x)$ is crucial for the proof of Lemma 1, however Sakai does not prove this result in Remark 4.9, but rather states that It is known that.... To me the statement is not trivial, so I would appreciate if someone could explain this result or show how one can work without the density argument.

Lemma 1. $d^2(x,\cdot):M\rightarrow [0,\infty)$ is smooth in a neighbourhood of $y$ if and only if $y\in \exp_x(\mathrm{int} \Sigma_x).$

Proof. On (the open set) $\exp_x(\mathrm{int}\Sigma_x)$ we have $d(x,y) = \vert \exp_x^{-1}(y)\vert^2$ which is clearly a smooth function in $y$. For the converse assume that $d(x,\cdot)^2$ is smooth on some open set $U\subset M$ and note that it suffices to prove $$U \cap \exp_x(\partial \Sigma_x) = \emptyset \tag{1}.$$ Without loss of generality we may assume that $x\notin U$, then also $d(x,\cdot)$ is smooth in $U$ and has a gradient $G\in C^\infty(U;TM)$. Let $\gamma:[0,l]\rightarrow M$ be a length minimising unit-speed geodesic with $\gamma(0)=x$ and $y=\gamma(1)\in U$. Then $d(x,\gamma(t))=t$ and differentiation yields $\langle G_y, \dot \gamma(l)\rangle = 1$. In particular $d(x,\cdot)$ has non-vanising gradient at $\gamma(l)$ and thus it is a submersion in a neighbourhood of $\gamma(l)$. This implies that the orthogonal complement of $\dot\gamma(l)\in T_yM$ is spanned by vectors $\dot c(0)$, where $c:(-\epsilon,\epsilon)\rightarrow M$ are curves with $c(0)=y$ and $d(x,c(s))=\mathrm{const}$. Since $\langle G_y , \dot c(0)\rangle = 0$ for such curves, we have $ G_y = \dot \gamma(l). $ We conclude: $$\text{Length minimising geodesics which start in $x$ don't intersect in $U$.} \tag{2}$$ Now we can prove (1). Assume to the contrary that $U \cap \exp_x(\partial \Sigma_x) \neq \emptyset$. Then by densitity of $\exp_x(\partial^1 \Sigma_x)$ we also have $U \cap \exp_x(\partial^1 \Sigma_x) \neq \emptyset$ and there are $w,w'\in \partial \Sigma_x$ with $w\neq w'$ and $\exp_x(w) = \exp_x(w')\in U$, which is in contradiction to (2).

Lemma 2. $d^2:M\times M\rightarrow [0,\infty)$ is smooth in a neighbourhood of $(x,y)$ if and only if $y \in \exp_x(\mathrm{int} \Sigma_x)$.

Proof. If $d^2$ is smooth near $(x,y)$, then $d^2(x,\cdot)$ is smooth near $y$ and the previous Lemma implies that $y\in \exp_x(\mathrm{int}\Sigma_x)$. For the converse define $\Sigma = \bigcup_x \Sigma_x \subset TM$ and note that $$ \Sigma \text{ is closed }\quad \text{ and } \quad \mathrm{int} \Sigma \cap T_xM = \mathrm{int} \Sigma_x \tag{3}. $$ Define $$ F: \mathrm{int} \Sigma \rightarrow M\times M, (x,w) \mapsto (x,\exp_x(w)) $$ and note that $$ DF\vert_{(x,w)} = \begin{bmatrix} \mathrm{id} & 0\\ \ast& D \exp_x\vert_w \end{bmatrix} $$ is invertible for all $(x,w)\in \mathrm{int} \Sigma$. Further $F$ is easily seen to be injective and thus it has a smooth inverse $F^{-1}:F(\mathrm{int} \Sigma) \rightarrow \mathrm{int} \Sigma$. Hence $d^2(x,y)= \vert F^{-1}(x,y)\vert ^2$ is smooth in a neighbourhood of every $(x,y) \in F(\mathrm{int} \Sigma)$, which concludes the proof.

Lemma 3. $y\in \exp_x(\mathrm{int} \Sigma_x)$ if and only if there exists a unique distance minimising geodesic between $x$ and $y$ and along this geodesic they are not conjugate.

Proof. Let $y=\exp_x(w)$ with $w \in \mathrm{int}\Sigma_x$. Then $t\mapsto \exp_x(tw)$, $0\le t\le 1$ is length minimising (because $w \in \Sigma_x$) and $x$ and $y$ are not conjugate along this geodesic ($D\exp_x\vert_w$ is invertible because $\exp_x$ is a diffeomorphism on $\mathrm{int}\Sigma_x$). If there was another length minimising geodesic from $x$ to $y$, then $y=\exp_x(w')$ for some $w'\in \Sigma_x \backslash \{w\}$. Since $\exp_x(\mathrm{int}\Sigma_x)\cap\exp_x(\partial \Sigma_x)=\emptyset$ we must have $w'\in \mathrm{int} \Sigma_x$, but this is false (since $\exp_x$ is injective on $\mathrm{int} \Sigma_x$).

Conversely assume that there is a unique distance minimising geodesic from $x$ to $y$ and that they are not conjugate along that geodesic. Then $y=\exp_x(w)$ for some $w\in \Sigma_x$. If we had $w\in \partial \Sigma_x$, then either there would be two length minimising geodesic between $x$ and $y$ (corresponding to $w\in \partial^1 \Sigma_x$) or $x$ and $y$ would be conjugate (corresponding to $D \exp_x\vert_w$ being singular).

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