# Smoothness of the fourth power of the geodesic distance in a Finsler geometry

The simplest form of Finsler metric is: $ds (X, dx)=\sqrt{g_{\alpha, \beta, \gamma, \delta}(X).dx^{\alpha}dx^{\beta}dx^{\gamma}dx^{\delta}}$, 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 $ds (X, dx)=\sqrt{g_{\alpha, \beta}(X).dx^{\alpha}dx^{\beta}}$, 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.

• ^{4}\sqrt $\mapsto$ \sqrt $\:$ ? $\;\;\;\;\;$ – user5810 Oct 14 '15 at 13:52
• Can you explain a bit more? The Riemannian case, the usual metric on the circle. Why is the distance squared smooth at the antipode of the chosen origin? – Gerald Edgar Oct 14 '15 at 14:39
• I have added the precision of locality (neighboordhood of $O$) to answer your remark. Thanks. – Julien Bernard Oct 15 '15 at 7:15

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 $C^6$.

Looking on the Minkowski Finlser metric (when the Finsler function does not depend on the point of the manifold) we see that the 4th power of the distance function can not be very smooth. Indeed, for the Minkowski metric generated by the norm $F:R^n \to R^n$ the function $s$ for $O=\vec 0$ is simply $s(y)=F(y)$. So in this special case your question is how smooth is the 4th degree of an arbitrary smooth norm. Observing that
it vanishes up to the order $4$ we conclude that it is $C^3$-smooth but not $C^4$-smooth.

This answer survives in the general case assuming the Finsler metric is not very wild (one needs its smoothness and strict convexity).

• I'm sorry, but I don't understand this answer. The OP is not considering an arbitrary smooth norm, but one that is a 4th root of a smooth, strictly convex, quartic differential form. For example, when $$ds = \bigl(dx^4 + 6A dx^2dy^2+dy^4\bigr)^{1/4}$$ (in the standard $xy$-plane) with $A$ a constant satisfying $0<A<1$ (which ensures strict convexity), then this defines a Minkowski-Finsler metric on the plane for which the distance from the origin is $$\rho=\bigl (x^4 + 6A x^2y^2+y^4\bigr)^{1/4},$$ so $\rho^4$ is a smooth function. – Robert Bryant Oct 17 '15 at 20:57
• Robert, I read his answer differently, I assumed that the author asks whether the 4th power of the distance function in an arbitrary finsler metric is smooth. If your version of his answer is true, my answer is indeed irrelevant. Let us wait the comment or the explanation of the author – Vladimir S Matveev Oct 17 '15 at 21:15
• R. Bryant is right. As I explained in the question, I don't take an arbitrary Finsler metric. I take the fourth root of an homogeneous polynomial of degree 4. But the coefficients can smoothly vary from one point to another. The question is then still open. Thanks for your attempts. – Julien Bernard Oct 20 '15 at 14:06

The paper by Li and Nirenberg seems relevant, though I don't see whether it gives your conjecture result immediately.

• Thank you for your anwer. I don't think that the paper answers my conjecture, because it deals with the properties of the geodesic distance at distance, from a point inside a domain and the frontier of the domain. My question is rather linked with the properties of the geodesic distance for infinitesimally close points. – Julien Bernard Oct 15 '15 at 6:10