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?


I believe that this is discussed in Zhongmin Shen's book "Lectures on Finsler Geometry". The result, if I remember correctly, is that the square of the geodesic distance function is generally not smooth (i.e., infinitely differentiable) at the origin in the Finsler case (as opposed to the Riemannian case), though it is smooth in a punctured neighborhood of the origin.

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 $d(0,x) = f(x)$.

  • $\begingroup$ 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). $\endgroup$ – Julien Bernard Aug 11 '14 at 14:33
  • $\begingroup$ Oh, sorry, when I wrote 'differentiable' there, I meant 'smooth', i.e., 'differentiable to all orders', which is the case in Riemannian geometry but not in Finsler geometry. I'll fix the answer. $\endgroup$ – Robert Bryant Aug 11 '14 at 16:12
  • $\begingroup$ 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 $F(X, dx)=\sqrt{g(X)_{\alpha\beta}dx^\alpha dx^beta}$. $\endgroup$ – Julien Bernard Aug 13 '14 at 6:46
  • $\begingroup$ 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 $C\infty$? $\endgroup$ – Julien Bernard Aug 13 '14 at 6:52
  • $\begingroup$ I have now answered this question, which you posed separately as a new question. See mathoverflow.net/a/224704/13972 $\endgroup$ – Robert Bryant Nov 28 '15 at 10:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.