Timeline for Smoothness of the fourth power of the geodesic distance in a Finsler geometry
Current License: CC BY-SA 3.0
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| Oct 20, 2015 at 14:06 | comment | added | Julien Bernard | R. Bryant is right. As I explained in the question, I don't take an arbitrary Finsler metric. I take the fourth root of an homogeneous polynomial of degree 4. But the coefficients can smoothly vary from one point to another. The question is then still open. Thanks for your attempts. | |
| Oct 17, 2015 at 21:15 | comment | added | Vladimir S Matveev | Robert, I read his answer differently, I assumed that the author asks whether the 4th power of the distance function in an arbitrary finsler metric is smooth. If your version of his answer is true, my answer is indeed irrelevant. Let us wait the comment or the explanation of the author | |
| Oct 17, 2015 at 20:57 | comment | added | Robert Bryant | I'm sorry, but I don't understand this answer. The OP is not considering an arbitrary smooth norm, but one that is a 4th root of a smooth, strictly convex, quartic differential form. For example, when $$ds = \bigl(dx^4 + 6A dx^2dy^2+dy^4\bigr)^{1/4}$$ (in the standard $xy$-plane) with $A$ a constant satisfying $0<A<1$ (which ensures strict convexity), then this defines a Minkowski-Finsler metric on the plane for which the distance from the origin is $$\rho=\bigl (x^4 + 6A x^2y^2+y^4\bigr)^{1/4},$$ so $\rho^4$ is a smooth function. | |
| Oct 17, 2015 at 20:42 | history | answered | Vladimir S Matveev | CC BY-SA 3.0 |