I was wondering if the following is true (and common knowledge):
Let $(M,F)$ be a Finsler manifold. Let d be the induced distance by the norm in the usual sense. That is, $d(x,y)=\inf${lenghts of all piece-wise smooth curves…}. We consider then $(M,d)$ as a metric space. My questions, which are probably quite easy, are these:
1) Suppose that $d$ is a righteous metric, in the sense that is also symmetric. Does that imply that $(M,F)$ was actually a Riemannian manifold for starters? If not could you please show me a counter example?
2) If the answer to 1) is negative then suppose $(M,d)$ happens to also be a $CAT(\kappa)$ space. does it follow that M is necessarily a Riemannian manifold? If not, what about $\kappa =0$?
Thanks a lot in advance for any help clearing this out. Of curse, any reference is gratefully welcomed.
The answer is no. The plane with taxicab distance $\|x\|=|x_1|+|x_2|$ is a counterexample.
A $CAT(0)$ Finsler space is Riemannian. Indeed, CAT(0) condition is scale-invariant, therefore passes to tangent spaces. And the only CAT(0) normed spaces are those with an inner product norm. One reference is: Ptolemaic spaces and CAT(0) by Buckley, Falk, and Wraith; Glasgow Math. J. 51 (2009) 301–314.
A $CAT(\kappa)$ Finsler space is also Riemannian, but I couldn’t locate a proof. The claim appears on MathOverflow and at the beginning of this paper where it’s said to be well-known.