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Hi to all!

Let $M$ be a compact smooth manifold without boundary and let $$d:M\times M\rightarrow [0,+\infty)$$ a distance on $M$ compatible with its topology. Suppose there exist $\varepsilon\in (0,+\infty)$ s.t. on the open set $\Delta_{\varepsilon}\subset M\times M$ $$\Delta_{\varepsilon}:=d^{-1}\left([0,\varepsilon)\right)$$ the function $$d^{2}:\Delta_{\varepsilon}\rightarrow [0,+\infty)$$ is smooth. Does a distance like this one come from a riemannian metric on $M$?

Thank you in advance.

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up vote 7 down vote accepted

This is not even true on the circle. In that case, given any metric that comes from a Riemannian metric there is a constant $c>0$ such that there is a diffeomorphism of the circle with itself that carries the metric to $c$ times the standard metric: $$ d\bigl((x_1,y_1),(x_2,y_2)\bigr) = \cos^{-1}(x_1x_2+y_1y_2). $$ If you now take $d' = h\circ d$ where $h$ is any continuous, strictly increasing subadditive function on $[0,\infty)$ that satisfies $h(0)=0$, then you'll get a new metric $d'$. If $h$ is smooth and satisfies $h(x)^2 = x^2g(x^2)$ where $g$ is a smooth function with $g(0)\not=0$, then the square of $d'$ will be smooth (and nondegenerate) near the diagonal in $S^1\times S^1$, but, in general, $d'$ will not come from a Riemannian metric, for the same reasons that Vladimir gave in his answer. (The main advantage of this counterexample is that it produces examples that do not come from a Riemannian metric in any neighborhood of the diagonal, unlike Vladimir's example. For example, just take $h^{-1}(x) = x+x^3$.)

A second source of examples: Another way to construct examples is to embed $M$ into another Riemannian manifold so that the image is not totally geodesic (for example, any smooth embedding into $\mathbb{E}^m$ will do since you assumed that $M$ is compact). Then the restriction of the ambient Riemannian distance function to $M\times M$ will have the properties that you want, but it won't make $M$ into a length space (since the shortest curves joining to points in $M$ that lies entirely in $M$ won't be a geodesic in the ambient space), so such a distance doesn't come from Riemannian metrics on $M$.

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The Riemannian distance has the following property: it is the so called length space. I recall (one of) the definitions of the length space. Having a distance function $d$, we can define the length of any curve $c(t)$, $t\in [0,1]$ , as the supremum of the sum $\sum_{i}d(c(t_i), c(t_{i+1}))$ over all $0=t_0\le t_1\le ... \le t_k=1$. Having this definition of the length of the curve, you can construct a new distance $\hat d$ by putting $\hat d(x,y)= $infimum of the lengths of all curves connecting $x$ and $y$. Then, the length spaces are characterized by the property $d=\hat d$.

Riemannian distance is evidently a length space. One could easily construct a distance such that it satisfies your condition but is not a length space and therefore not a Riemannian metric: put $d(x,y)= min(d_{riem}(x,y), 1)$. It still satisfies your property since we changed nothing in a small neightborhood, and is not a length space and hence not a Riemannian metric anymore.

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As pointed out in other answers, only length metrics can be Riemannian and there are plenty of non-length metrics satisfying your condition.

If your metric is a length metric, then yes, it is Riemannian. Indeed, for every $x\in M$ the function $f_x=d(x,\cdot)^2$ is smooth near $x$. Since it attains its minimum at $x$, its derivative at $x$ is zero. Therefore its second derivative at $x$ is well-defined as a quadratic form on $T_xM$. Define the metric tensor at $x$ as one half of this second derivative. Now from the Taylor expansion it is easy to see that for every $\varepsilon>0$ there is a neighborhood of $x$ where the original distance and the new Riemannian distance are Lipschitz equivalent with Lipschitz constant $1+\varepsilon$. Since both metrics are length metrics, this property implies that they are $(1+\varepsilon)$-Lipschitz equivalent globally. Since $\varepsilon$ is arbitrary, this means that they are equal.

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