# Derivability properties of the distance function in a Finsler Manifold

We know that, in a Riemannian manifold, the geodesic distance between a point O and a point P, when we fix O, is a function of P that is $C^\infty$ everywhere on a local neighborhood, except in P=O. If we consider the squarred function, we have $C^\infty$ everywhere.

(There is a topic on that subject)

Now, in Finsler's work in 1918, and in Riemann's habilitation thesis in 1854, the authors consider a Finsler's manifold of a particular kind. They consider that the metric is given by the $(2n)^{th}$ root of an homogeneous form of degree $2n$:

$ds (X, dx)=^{2n}\sqrt{g_{\mu_1, \cdots, \mu_{2n}}(X).dx^{\mu_1}\cdots dx^{\mu_{2n}}}$.

My question is:

What can we say, in this framework, about the differentiability properties of the geodesic distance, and of the $2n^{th}$ power of the geodesic distance?

You can see this easily in the case of a Finsler metric on $\mathbb{R}^n$ that is translation invariant, i.e., the Finsler function is $F(x,\dot x) = f(\dot x)$ where $f:\mathbb{R}^n\to\mathbb{R}$ has the properties that $f$ is homogeneous of degree $1$ and smooth away from $0\in\mathbb{R}^n$ and that $f^2$ is strictly convex away from the origin but not smooth there. Then the geodesic distance function from the origin in standard coordinates is just $d(0,x) = f(x)$.
• Thank you for your reference to Shen's book. Concerning your answer, I am not sure to understand. It seems to me that the square of the distance function is always differentiable (of derivative=0) because the local squarred Finsler metric is an homogeneous function of order 2, and because the exponential is $C^1$. Therefore, I thought that the question was pertinent only for second order derivatives (and higer orders). – Julien Bernard Aug 11 '14 at 14:33
• Thanks to you, now I see what is happening and I can precise my question. The reason why we take the squarred distance to get the smooth property, in the Riemannian case, is that the metric is of the type $F(X, dx)=\sqrt{g(X)_{\alpha\beta}dx^\alpha dx^beta}$. – Julien Bernard Aug 13 '14 at 6:46
• Now, if we take another classical Finsler's metric:$F(X, dx)=^4\sqrt{g(X)_{\alpha\beta\gamma\delta}dx^\alpha dx^\beta dx^\gamma dx^\delta}$, is it true that the fourth power of the geodesic distance is $C\infty$? – Julien Bernard Aug 13 '14 at 6:52