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I recently read a footnote in Chern's article stating that a non-Riemmanian Finsler manifold does not possess normal coordinates.

As I'm still new to non-Riemmanian Finsler geometry I don't see why this is the case. Can someone provide me with a simple example in conjunction to an intuitive explanation why this fails in general?

Many thanks.

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    $\begingroup$ Could you provide what the definition of "normal coordinates" would be for a Finsler manifold? $\endgroup$ – Deane Yang Feb 5 '16 at 17:41
  • $\begingroup$ Well that's part of my question... ie: can we define an Exp map and if so why not define the normal coordinate of $Exp_p$ to be the interior of that function's Domain? $\endgroup$ – AIM_BLB Feb 5 '16 at 17:46
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I think it's important to keep two things separate here:

First, if $(M,F)$ is a smooth Finsler manifold (which means that $F^2:TM\to [0,\infty)$ is smooth and strongly convex away from the zero section of $TM$), then the unit sphere bundle $\Sigma\subset TM$ (aka the tangent indicatrix) is a smooth hypersurface in $TM$ and there is an open neighborhood $U$ of $\Sigma\times\{0\}$ in $\Sigma\times\mathbb{R}$ on which there is defined a smooth mapping $\exp:U\to M$ such that, for each fixed $u\in \Sigma$, the curve $\gamma_u(t) = \exp(u,t)$ (defined for all $t\in\mathbb{R}$ such that $(u,t)$ lies in $U$) is the maximally extended unit speed $F$-geodesic with initial velocity $u\in\Sigma$.

Second, for any given $p\in M$, there is a $\delta>0$ such that, if $B_\delta(p)\subset T_pM$ is the set of vectors $v\in T_pM$ satisfying $F(v)<\delta$, then there is a well-defined mapping $\exp_p:B_\delta(p)\to M$ such that $\exp_p(0_p) = p$ and $\exp_p(tu) = \exp(u,t)$ for all $u\in \Sigma_p$ and all $t\in(0,\delta)$. By taking $\delta$ sufficiently small, one can ensure that $\exp_p:B_\delta(p)\to M$ is a homeomorphism onto its (open) image, one that is a smooth diffeomorphism away from $0_p$.

Now, unless $F$ is reversible (i.e., $F(-v) = F(v)$), the map $\exp_p:B_\delta(p)\to M$ need not be $C^2$ at $0_p$, because it can easily happen that $\exp(-u,-t)\not=\exp(u,t)$, so the $\exp_p$-image of the line segment $\{tu\ |\ |t|<\delta\}$ need not be a $C^2$ curve in $M$ (it will be $C^1$, though). Even when $F$ is reversible (or, more generally, geodesically reversible), so that the exponential map $\exp_p$ is smooth on lines through $0_p\in B_\delta(p)$, it may still not be smooth at $0_p$.

Unfortunately, writing down a simple, explicit example for which smoothness clearly fails is not easy because, usually, it is not possible to integrate the geodesic equations and compute the map $\exp:U\to M$ explicitly. In one case where it is possible, namely the case of a Minkowski Finsler metric (i.e., $M=\mathbb{R}^n$ and $F$ is invariant under translations in space), the exponential map $$\exp:\Sigma\times \mathbb{R} \bigl(= (\mathbb{R}^n\times\Sigma_0)\times\mathbb{R}\bigr)\to\mathbb{R}^n$$ is $\exp((p,u),t) = p+tu$ for $p\in \mathbb{R}^n$ and $u\in \Sigma_0\subset T_0\mathbb{R}^n = \mathbb{R}^n$, so the map $\exp_p:T_p\mathbb{R}^n\to\mathbb{R}^n$ is a smooth diffeomorphism after all, regardless of how non-Riemannian $F$ is. (This shows that it's not just a matter of looking at the shapes of the unit spheres of $F$. The failure of smoothness is more subtle than that.)

However, in an answer to this question, I gave an example of a (homogeneous!) Finsler metric $F$ on a surface $M$ that is reversible and has the property that $F^4$ is a smooth function on $TM$, but the fourth power of the Finsler distance function from any given point of $M$ is not smooth. Thus, this is an example in which the map $\exp_p$ cannot be smooth at the origin $0_p$ (for any $p\in M$), because, if $s_p:M\to[0,\infty)$ is the function that gives the $F$-distance from $p\in M$, then, for $\delta>0$ sufficiently small, we have $F(v) = s_p\bigl(\exp_p(v)\bigr)$ for all $v\in B_\delta(p)\subset T_pM$.

Remark (28 Feb 2016): I was curious as to what are the necessary and sufficient conditions for a Finsler surface to have $\exp_p$ be at least $C^2$ at every point. It turns out that this is very restrictive: Either the Finsler structure must be Minkowskian (i.e., invariant under a $2$-dimensional abelian group of translations as discussed above) or else the Cartan scalar (usually denoted $I$) must be constant. The latter case includes the Riemannian case ($I$=0) as a special case; the cases where $I$ is a non-zero constant are actually 'generalized' Finsler structures (at each $p\in M$, the unit tangent vectors in $T_pM$ form a logarithmic spiral rather than a closed convex curve). In all these cases, the map $\exp_p$ is actually $C^\infty$.

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  • $\begingroup$ Really detailed and great answer! Thank you so much Robert! $\endgroup$ – AIM_BLB Feb 8 '16 at 11:43
  • $\begingroup$ So just to clarify if $F(-x)=F(x)$ then the exponential coordinates are well defined (as I expected)? $\endgroup$ – AIM_BLB Feb 27 '17 at 13:57
  • $\begingroup$ @CSA: Actually, the point of my remarks is that exponential normal coordinates are well-defined in general (you don't need $F(-x)=F(x)$ for this). It's just that they are (usually) not more than $C^1$ at the origin. $\endgroup$ – Robert Bryant Feb 27 '17 at 14:03
  • $\begingroup$ Hi Robert, I do have one critically important question on the above derivation. Do you take the Finsler function definition to require the Hessian convexity or do you only assume it is a norm on tangent spaces? If it is in the second case, do we still have (At least continuous..or maybe Bi-Lipschutz) exponential coordinates (I have come across the Bi-Lischutz case for low-regularity Riemannian manifolds. $\endgroup$ – AIM_BLB Apr 20 '17 at 17:01
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The exponential map is just $C^1$.

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  • $\begingroup$ Oh so it fails to be $C^{\infty}$, can you explain why? Or provide a simple example? $\endgroup$ – AIM_BLB Feb 5 '16 at 18:16
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    $\begingroup$ So there are $C^1$ normal coordinates? $\endgroup$ – Deane Yang Feb 5 '16 at 22:33
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    $\begingroup$ @Deane: Yes, the exponential map provides a $C^1$ set of 'normal coordinates'. To see this, assume that $M=\mathbb{R}^n$. Then using the notation in my answer, there exists a smooth map $E_2:U\to\mathbb{R^n}$ so that $\gamma_u(t) = tu + t^2E_2(u,t)$ for all $(u,t)\in U\subset\Sigma\times\mathbb{R}$. From this and the compactness of $\Sigma_p$, one sees that, for all $\delta>0$ sufficiently small, there is a constant $K>0$ such that $$|\,\exp_0(v)-v\,|\le K\,|v|^2$$ for all $v$ with $|v|<\delta$. Of course, this implies that $\exp_0$ is $C^1$ at $0\in\mathbb{R}^n$. $\endgroup$ – Robert Bryant Feb 8 '16 at 14:26

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