Are there any smooth manifolds $M$ with the following property:

There exist a realizing metric $d$ (i.e $d$ induces the topology on $M$), and $d$ is smooth on all of $M \times M$?

If not, is it possibe to guarantee smoothness of the function $x \mapsto d(x,y)$ (for a fixed $y$ ), or smoothness of $d^2$ even on a compact manifold?

(I am trying to see if we can achieve "improved smoothness" if we do not force the metric to be Riemannian.)

Of course, such a metric cannot be induced by a Riemannian metric. (see here and here).

Update and further questions:

(1) Joonas Ilmavirta showed $d$ cannot be smooth at a neighbourhood of points on the diagonal. Actually, the proof shows $d$ cannot even be twice continuously differentiable. (This is the regularity needed to bound from above the Taylor remainder*).

Now a natural quesion is whether this regularity can be achieved by some metric? (I suspect not, in fact I think the distance should not even be differentiable once at the diagonal, the intuition is based on the example of absolute value on $\mathbb{R}$).

(2) Is it also necessary for a singularity to exist at the diameter of the metric (for compact manifolds)?

*In fact the proof works even if we only assume $x \mapsto d(x,y)$ is continuously twice differentiable, and continuity of the partial derivatives (as functions of two variables).

$ \frac{d}{dt}f(t,t) = \lim_{\Delta \to 0} \frac {f(t+\Delta,t+\Delta)-f(t,t)}{\Delta} = \lim_{\Delta \to 0} \left( \frac {f(t+\Delta,t+\Delta)-f(t+\Delta,t)}{\Delta} + \frac {f(t+\Delta,t)-f(t,t)}{\Delta} \right) = \lim_{\Delta \to 0} \left( \frac {f(t+\Delta,t+\Delta)-f(t+\Delta,t)}{\Delta} + \frac {f(t+\Delta,t)-f(t,t)}{\Delta} \right) = \lim_{\Delta \to 0} \frac{\partial f}{\partial s}(t+\Delta,t+\alpha(\Delta) \cdot \Delta) + \lim_{\Delta \to 0} \frac{\partial f}{\partial t}(t+\beta(\Delta) \cdot \Delta,t) = \frac{\partial f}{\partial s} (t,t) + \frac{\partial f}{\partial t} (t,t)$

($0 \le \alpha(\Delta), \beta(\Delta) \le 1$ , Lagrange mean value theorem)

  • $\begingroup$ You probably know this, but, just in case: It is more reasonable to ask that the square of the metric be smooth (or, more generally, that $F(d):M\times M\to\mathbb{R}$ be smooth, where $F:[0,\infty)\to[0,\infty)$ is a strictly increasing function satisfying certain other properties). This turns out to be just as good for most purposes (for example, in information geometry) and, for many purposes, is much better than asking that $d$ itself be smooth. $\endgroup$ Aug 24, 2015 at 11:14
  • $\begingroup$ I updated my answer to discuss the existence of a metric which is as smooth as possible. $\endgroup$ Nov 8, 2016 at 18:29

1 Answer 1


In short: Non-smoothness at the diagonal is inevitable but that is the only obstruction. (I added the second part much later.)

Diagonal singularity

It is not possible to have a smooth metric. Non-smoothness at the diagonal is inevitable. In fact, any smooth semimetric is zero whenever the two points are in the same connected component.

Suppose you had such a metric on a manifold $M$. Take a smooth curve $\gamma:(-1,1)\to M$ with $\gamma'\neq0$. Consider the function $f(t,s)=d(\gamma(t),\gamma(s))$, which is now smooth.

Since $$ \begin{split} 0 &= \frac{d}{dt}0 \\&= \frac{d}{dt}f(t,t) \\&= \partial_1f(t,t)+\partial_2f(t,t), \end{split} $$ we have $\partial_1f=-\partial_2f$ on the diagonal. But since $f(t,s)=f(s,t)$, we also have $\partial_1f=\partial_2f$ on the diagonal. Thus $\partial_1f(t,t)=\partial_2f(t,t)=0$.

This implies that $f(t,s)\leq C(t-s)^2$ for some constant $C$, for all $t,s\in[-1/2,1/2]$. Pick any number $a\in[0,\frac12]$ and a large integer $N$ and observe that by triangle inequality $$ \begin{split} f(0,a) &\leq f(0,\frac{a}{N})+f(\frac{a}{N},\frac{2a}{N})+\dots+f(\frac{(N-1)a}{N},a) \\&\leq NC\left(\frac aN\right)^2 \\&= \frac{Ca^2}{N}. \end{split} $$ This holds for any $N$, so in fact $f(0,a)=0$. This implies that as long as $x,y\in M$ are in the same connected component, they must satisfy $d(x,y)=0$.

In fact, this argument works for any $C^{1,\alpha}$ semimetric for $\alpha>0$ (a similar calculation leads to an estimate $f(0,a)\lesssim N^{-\alpha}$), and I guess the claim is true for $C^1$, too.

Off-diagonal regularity

For any differentiable manifold $M$ there is a metric $d\colon M\times M\to\mathbb R$ which is smooth everywhere outside the diagonal and gives the usual topology of $M$. This is based on two observations:

  • There is a Riemannian metric $g$ on $M$ with injectivity radius at least one. (If I am mistaken, this is at least possible on compact manifolds. On non-compact manifolds one should be able to start with any metric and multiply it with a slowly varying conformal factor to force injectivity radius above one everywhere.)
  • There is a smooth concave function $\phi\colon[0,\infty)\to[0,\infty]$ so that $\phi(0)=0$ and $\phi(x)=1$ for $x\geq1$.

Now if $d_g$ is the distance function of $g$, the metric $d(x,y)=\phi(d_g(x,y))$ is smooth outside the diagonal. The distance from $x$ to nearby points is smooth (except at $x$) within the injectivity radius. Outside that radius $d_g$ may be singular, but $\phi$ is constant at that scale, removing all singularities from $d$. The composition of a metric and a concave function is always a metric. It follows from the assumptions that $\phi$ is a strictly increasing bi-Lipschitz diffeomorphism in some neighborhood of zero, and thus $d$ and $d_g$ give the same topologies. Any Riemannian metric gives the original manifold topology.

The metric $d$ is non-smooth at the diagonal, but its square $d^2$ is smooth everywhere.


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