Suppose $\mathbb{R}^n$ is equipped with the p-norm $\left\Vert x \right\Vert_p$. Let $x\in \mathbb{R}^n$ and let $y$ be in a neighborhood of $x$. The distance between $x$ and $y$ can be defined as $\left\Vert x-y \right\Vert_p$. Is this setting an example of a Finsler manifold? If so, it should be a very simple example of a Finsler manifold. Can you point out some references about the study of the geometry of this type of p-norm spaces, like the curvature and geodesics? (Apparently when $p=2$, it is the trivial case of a Euclidean space, which is a trivial case of Riemannian manifold, which is a special case of Finsler manifold. But what about other $p\neq 2$?)
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$\begingroup$ it should be true for $p>1$ : math.stackexchange.com/questions/144449/… $\endgroup$– Bipolar MindsSep 13, 2016 at 19:32
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It is a Finsler metric for $p>1$. I don't know a reference. But you can easily see that its indicatrix is smooth and strictly convex and symmetric. The triangle inequality says that the geodesics are the usual straight lines. The Finsler flag curvature vanishes by the Jacobi equation. I don't know the complete description of the curvature.
answered Sep 14, 2016 at 7:12
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2$\begingroup$ It depends on your definition of Finsler manifold, but most books actually require something a bit stronger: That the norm be strongly convex, i.e., that the Hessian of the squared norm on each tangent space be positive definite away from the origin (where it may not be defined). This does not hold, for example, for $p=4$, so, according to that definition, $\mathbb{R}^n$ with the $4$-norm is not a Finsler manifold. $\endgroup$ Sep 14, 2016 at 9:41
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$\begingroup$ @RobertBryant: doesn't your objection hold whenever $p > 2$? Also, $C^\infty$ smoothness of the indicatrix will require that $p$ is an even integer. If finite regularities are admitted, to have enough regularity to define the Hessian requires $p \geq 2$. The convexity requirement would in fact rule out all other $p$ besides $2$. $\endgroup$ Sep 14, 2016 at 14:27
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$\begingroup$ @WillieWong: Yes, indeed, according to this stricter definition, none of the $p$-norms for $p>2$ define Finsler metrics. However, let me point out that I didn't object, I merely pointed out that the answer depends on what you take to be the definition of a Finsler manifold, and there are different definitions in the literature. The stricter definition is needed if you want to define the usual apparatus of a (smooth) Finsler connection and its curvature. The weaker definition is OK if you only want to use the Finsler norm to define a metric (by shortest curves). $\endgroup$ Sep 14, 2016 at 15:02
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$\begingroup$ @RobertBryant: apologies for the poor word choice. Quick question: for the metric definition, would the smoothness be even necessary (e.g. what about the cases $p = 1$ or $\infty$?) $\endgroup$ Sep 14, 2016 at 15:55
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$\begingroup$ @WillieWong: Actually, I don't know a lot about what is usually assumed when you only care about the metric properties. Certainly smoothness is not required; you can get many of the main theorems about Finsler metrics as metric spaces without assuming any smoothness at all, just continuity and convexity. However, I don't normally think about those things, so I won't be of much help with the literature on such things. $\endgroup$ Sep 14, 2016 at 16:31