How badly does the geodesic exponential map fail to be $C^2$ on Finsler manifolds

Question

Consider a Finsler manifold $M$. Then for each $x \in M$, we can consider the partial map $\exp_x: T_x M \to M$, which is $C^\infty$ away from the origin, $C^1$ at the origin, but never $C^2$ at the origin for all $x \in M$ unless $M$ is a Berwald space, in which case $\exp_x$ will actually be $C^\infty$.

Are there any results that control how badly $\exp_x$ fails to be $C^2$ at the origin for a Finsler manifold that is not a Berwald space. For instance, in some smooth coordinate system defined in a neighborhood of $x$, do we have bounds on the partial derivatives $(\partial_i \partial_j \exp_x^k)(v)$ as $v \to 0$? Perhaps a bound of the form $$|(\partial_i \partial_j \exp_x^k)(v)| \leq C |v|^{-N} \quad\text{as $v \to 0$} $$ for some $N > 0$, and some constant $C > 0$ possibly depending on $x$?

Edit: For those unfamiliar with Finsler manifolds, the exponential map $\exp_x$ is the unique map with the property that for each $v \in T_x M$, the curve $c(t) = \exp_x(tv)$ is the unique constant speed geodesic on $M$ with $c(0) = x$ and $c'(0) = v$, i.e. an extremal for the length functional $L(c) = \int F(c,\dot{c})$, or equivalently, a (necessarily smooth) curve which in coordinates satisfies the geodesic equation $$ \ddot{c}^i + \dot{c}^j \dot{c}^k\ \gamma^i_{jk}(c, \dot{c}) = 0, $$ where $\dot{c} = \frac{dc}{dt}$, $\ddot{c} = \frac{d^2c}{dt^2}$, $$ \gamma^i_{jk} = \frac{g^{is}}{2} \left( \frac{\partial g_{sj}}{\partial x^k} - \frac{\partial g_{jk}}{\partial x^s} + \frac{\partial g_{ks}}{\partial x_j} \right), $$ the metric coefficients $g_{ij}$ and their inverses $g^{ij}$ are defined by $$ g_{ij} = \frac{1}{2} \frac{\partial^2 F^2}{\partial v^i \partial v^j} $$ and Einstein notation is used throughout.

Answer 1

Note: I am making the standard assumptions in Finsler geometry, i.e., that $F:TM\to\mathbb{R}$ is smooth away from the zero section and $F$ is 'strictly convex'.

Although $\partial_i\partial_j\exp^k_p(v)$ is usually not continuous at $v=0$, there is a bound $$ |\partial_i\partial_j\exp^k_p(v)|\le C(p), $$ where $C(p)$ depends on $p\in M$. Thus, one can take $N=0$ in the above desired bound.

To see this, for simplicity, take coordinates centered at $p$, and identify $M$ with a neighborhood of $0$ in $\mathbb{R}^n$. Let $\Sigma\subset T_0\mathbb{R}^n$ denote the Finsler unit sphere (i.e., $F(u)=1$ for $u\in \Sigma$) and use `polar coordinates', then there exists a smooth map $E:(-\epsilon,\epsilon)\times\Sigma\to\mathbb{R}^n$ so that the Finsler exponential map satisfies $$ \exp_0(ru) = ru + r^2 E(r,u) $$ for $u\in\Sigma$ and $r\ge0$. (In fact, for each $u\in\Sigma$, the curve $\gamma_u(t) = tu + t^2 E(t,u)$ is a smooth geodesic for $|t|<\epsilon$.) However, $E(r,u)$ for $0<r<\epsilon$, while it does define a bounded smooth function $E^*$ on a punctured neighborhood of $0\in\mathbb{R}^n$ (after shrinking $\epsilon$ if necessary), that function does not (usually) extend continuously to $0$, since $E(0,u):\Sigma\to\mathbb{R}^n$ is not usually constant.

Consider the 'polar coordinates' map $P:\mathbb{R}^+\times\Sigma\to\mathbb{R}^n$ given by $P(r,u)=ru$. This is a diffeomorphism onto $\mathbb{R}^n$ minus the origin. Using this to identify the two spaces, we can express the vector fields $\partial_i$ as vector fields on $\mathbb{R}^+\times \Sigma$: We have $$ \partial_i = p_i\,\partial_r + r^{-1} X_i $$ where the $p_i$ are the smooth functions on $\Sigma$ satisfying $p_i\,\mathrm{d}u^i = 0$ and $p_iu^i=1$ (note the use of the summation convention) and where the $X_i$ are the smooth vector fields on $\Sigma$ that satisfy $X_i(u^j)= \delta_i^j-u^jp_i$. Via this identification, it follows that $$ \partial_i\partial_j(r^2 E^*) = (p_i\,\partial_r + r^{-1} X_i)(p_j\,\partial_r + r^{-1} X_j) \bigl(r^2 E(r,u)\bigr). $$ The latter expression is then seen to be a smooth function on $(-\epsilon,\epsilon)\times \Sigma$ and hence is bounded (after one shrinks $\epsilon$ a little bit, if necessary).

However, there is no reason to believe that this function will be constant on the slice $r=0$ (and, indeed, in general it is not), so this function does not generally push forward under $P$ to extend to a continuous function on a neighborhood of $0\in\mathbb{R}^n$ (which, of course, is why the exponential map is not $C^2$ at the origin in general).