Is it possible for a metric on a smooth manifold to be smooth?

Question

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)?

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*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)

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.