I recently read a footnote in [Chern's article](http://www.ams.org/notices/199609/chern.pdf) 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. 



No normal coordinates on general Finsler manifolds
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](https://mathoverflow.net/questions/220887/smoothness-of-the-fourth-power-of-the-geodesic-distance-in-a-finsler-geometry/224704#224704), 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 $C^\infty$.


