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 herehere 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)
