The simplest form of Finsler metric is:
$ds (X, dx)=\sqrt[4]{g\_{\alpha, \beta, \gamma, \delta}(X).dx^{\alpha}dx^{\beta}dx^{\gamma}dx^{\delta}}$, where $g\_{\alpha, \beta, \gamma, \delta}$ is a fourth degree polynomial that smoothly depends on $X$.


My question is: is it always true that $s^4(X)$, the fourth power of the geodesic distance from an origin point $O$ to $X$, constructed from the Finsler metric $ds$, is everywhere smooth on a neighboorhood of $O$?


My question arises from the fact that, when we take a Riemmanian metric $ds (X, dx)=\sqrt[2]{g\_{\alpha, \beta}(X).dx^{\alpha}dx^{\beta}}$, then we know that the squarred geodesic distance $s^2(X)$ is everywhere smooth on a neighboord of $O$. Smoothness is easy to prove outside the origin $O$. At the origin, we need to use this argument: the exponential map is smooth and radially isometric.


Unfortunately, this argument seems not to be decisive in the Finslerian case, where the exponential map is not anymore smooth at the origin. I don't find any answer to my question in classical books on Finsler geometry.


Thanks.



Smoothness of the fourth power of the geodesic distance in a Finsler geometry
It turns out that the answer is 'No, the fourth power of the geodesic distance from a point $p$ in a Finsler space $M$ whose norm, raised to the fourth power, is a smooth convex quartic form is *not* generally a smooth function on a neighborhood of $p$'. 


While the desired smoothness does hold in the Minkowski case (i.e., when, in some local coordinate system, the functions $g\_{\alpha\beta\gamma\delta}(X)$ are constant) and (of course) in the case that the norm is the square root of a positive definite quadratic form, it does not hold in general. (In fact, at least in the $2$-dimensional case, it appears that the above two cases are the *only* cases in which, for all points $p$, the fourth power of the geodesic distance from $p$ is $C^6$ near $p$. — RLB)


For example, let $M$ be the upper half-plane $y>0$ in the $xy$-plane, and let
$$
ds^4 = \frac{\mathrm{d}x^4 + 2A\,\mathrm{d}x^2\mathrm{d}y^2 + \mathrm{d}y^4}{y^4},
$$
where $A$ is a constant satisfying $0<A<1$. It is easy to show that this does define a strictly convex Finsler metric on $M$ that is, in fact, complete. Moreover, it can also be shown that the geodesic distance $s\_p:M\to [0,\infty)$ from any given point $p = (x\_0,y\_0)\in M$ has the property that $({s\_p})^4$ is $C^5$ at $p$ but it is not $C^6$ there. (It is smooth everywhere else, of course.)


It does not seem to be so easy to explain why, however. Even in this example, which is homogeneous under translation in $x$ and simultaneous scaling in $x$ and $y$, simply computing the geodesic flow and using this explicitly to compute $s\_p$ for some specific $p$ turns out to be surprisingly tricky. Ultimately, I had to use a different method.


Here is a sketch of the method I used; details can be supplied upon request: 


In a Finsler space $(M,\mathrm{d}s)$ that is smooth (i.e., $\mathrm{d}s:TM\to[0,\infty)$ is strictly convex fiber wise and smooth away from the zero section in $TM$), the geodesic distance function from either a point or a submanifold satisfies the so-called *eikonal equation* wherever it is $C^1$. Here, the eikonal equation is a first order PDE for functions $f:M\to\mathbb{R}$ that is characterized by the condition that, for each $x\in M$, the $1$-form $\mathrm{d}f\_x:T\_xM\to \mathbb{R}$ lies in the *Legendre transform* $\Sigma^\*\subset T^\*M$ of the tangent indicatrix $\Sigma\subset TM$ consisting of 'unit vectors' in $TM$, i.e., $\Sigma = \{ u\in TM\ |\ \mathrm{d}s(u)=1 \}$. (For $u\in \Sigma\_x = \Sigma\cap T\_xM$, its Legendre transform is the element $u^\*\in T^\*\_xM$ that satisfies $u^\*(u)=1$ and $u^\*(v)\le 1$ for all $v\in \Sigma\_x$.)


When $\mathrm{d}s^4$ is a smooth function on $TM$ that is a homogeneous quartic on each fiber $T\_pM$, one can write down the eikonal equation for $f:M\to\mathbb{R}$ as an explicit polynomial PDE on $F=f^4$. In the above example, the eikonal equation takes the form $P\bigl(F,\,\tfrac14yF\_x,\,\tfrac14yF\_y\bigr)=0$, where $P$ is an irreducible polynomial in its three arguments. In fact, one finds, after some calculation,
$$
P(r\_0,r\_1,r\_2) = \left(1-{A}^{2}\right)^{2}{r\_{{0}}}^{9}+ \left(1-{A}^{2}\right) 
\left( \left( {A}^{2}-3 \right) {r\_{{1}}}^{4}+12\,A{r\_{{1}}
}^{2}{r\_{{2}}}^{2}+ \left( {A}^{2}-3 \right) {r\_{{2}}}^{4} \right){r\_
{{0}}}^{6} - \left( \left( -2\,{A}^{2}+3 \right) {r\_{{1}}}^{8}+
 \left( 10\,{A}^{3}-6\,A \right) {r\_{{1}}}^{6}{r\_{{2}}}^{2}+ \left( {A
}^{4}+26\,{A}^{2}-21 \right) {r\_{{1}}}^{4}{r\_{{2}}}^{4}+ \left( 10\,{A
}^{3}-6\,A \right) {r\_{{1}}}^{2}{r\_{{2}}}^{6}+ \left( -2\,{A}^{2}+3
 \right) {r\_{{2}}}^{8} \right) {r\_{{0}}}^{3}- \left({r\_{{1}}}^{4}+ 2\,A{r\_{{1}}}^{2}{r\_{{2}}}^{2}+{r\_{{2}}}^{4} \right)^{3}
$$ 


Now, if $f = s\_{(0,1)}$ had the property that $F = f^4$ were smooth, then it is not hard to show that
$$
F = x^4 + 2Ax^2(y{-}1)^2 + (y{-}1)^4 + R\_5(x,y{-}1),
$$
where $R\_5(u,v)$ vanishes to order $5$ at $(u,v)=(0,0)$. Using the explicit form of the eikonal equation above, one can then show that
$$
F = x^4 + 2A\,x^2(y{-}1)^2 + (y{-}1)^4 - 2\,x^4(y{-}1) - 2A\,x^2(y{-}1)^3 + R\_6(x,y{-}1) 
$$
where $R\_6(u,v)$ vanishes to order $6$ at $(u,v)=(0,0)$. However, now going back and substituting this into the eikonal equation and examining the lowest order terms (which have degree $38$), one finds that there is no solution to these equations for any term of order $6$. Hence $F=f^4$ cannot be $C^6$.


