---
abstract: 'We give a necessary and sufficient condition on a Randers space for the existence of a measure for which Shen’s ${\ensuremath{\mathbf{S}}}$-curvature vanishes everywhere. Moreover, such a measure coincides with the Busemann-Hausdorff measure up to a constant multiplication.'
author:
- |
    Shin-ichi OHTA[^1]\
    [Department of Mathematics, Faculty of Science, Kyoto University,]{}\
    [Kyoto 606-8502, JAPAN (e-mail: [sohta@math.kyoto-u.ac.jp]{})]{}
title: 'Vanishing S-curvature of Randers spaces'
---

Introduction {#sc:intr}
============

This short article is concerned with a characterization of Randers spaces admitting measures with vanishing ${\ensuremath{\mathbf{S}}}$-curvature. A Randers space (due to Randers [@Ra]) is a special kind of Finsler mainfold $(M,F)$ whose Finsler structure $F:TM {\longrightarrow}[0,\infty)$ is written as $F(v)=\alpha(v)+\beta(v)$, where $\alpha$ is a norm induced from a Riemannian metric on $M$ and $\beta$ is a one-form on $M$. Randers spaces are important in applications and reasonable for concrete calculations. See [@AIM] and [@BCS Chapter 11] for more on Randers spaces.

We equip a Finsler manifold $(M,F)$ with an arbitrary smooth measure $m$. Then the ${\ensuremath{\mathbf{S}}}$-curvature ${\ensuremath{\mathbf{S}}}(v) \in {\ensuremath{\mathbb{R}}}$ of $v \in TM$ introduced by Shen (see [@Shlec §7.3]) measures the difference between $m$ and the volume measure of the Riemannian structure induced from the tangent vector field of the geodesic $\eta$ with $\dot{\eta}(0)=v$ (see §\[ssc:S\] for the precise definition). The author’s recent work [@Oint], [@OS] on the weighted Ricci curvature (in connection with optimal transport theory) shed new light on the importance of this quantity.

A natural and important question arising from the theory of weighted Ricci curvature is: when $(M,F)$ admits a measure $m$ with ${\ensuremath{\mathbf{S}}}\equiv 0$? This is because, if such a measure exists, then we can choose it as a good reference measure. Our main result provides a complete answer to this question for Randers spaces.

\[th:main\] A Randers space $(M,F)$ admits a measure $m$ with ${\ensuremath{\mathbf{S}}}\equiv 0$ if and only if $\beta$ is a Killing form of constant length. Moreover, then $m$ coincides with the Busemann-Hausdorff measure up to a constant multiplication.

It has been observed by Shen [@Shlec Example 7.3.1] that a Randers space with the Busemann-Hausdorff measure satisfies ${\ensuremath{\mathbf{S}}}\equiv 0$ if $\beta$ is a Killing form of constant length. Our theorem asserts that his condition on $\beta$ is also necessary for the existence of $m$ with ${\ensuremath{\mathbf{S}}}\equiv 0$, and then it immediately follows that $m$ must be a constant multiplication of the Busemann-Hausdorff measure.

We observe from Theorem \[th:main\] that many Finsler manifolds have no measure with ${\ensuremath{\mathbf{S}}}\equiv 0$. This means that there is no canonical (reference) measure (in respect of the weighted Ricci curvature) on such a Finsler manifold. Therefore, for a general Finsler manifold, it is natural to start with an arbitrary measure, as was discussed in [@Oint] and [@OS].

Preliminaries for Finsler geometry {#sc:pre}
==================================

We first review the basics of Finsler geometry. Standard references are [@BCS] and [@Shlec]. We will follow the notations in [@BCS] with a little change (e.g., we use $v^i$ instead of $y^i$).

Finsler structures
------------------

Let $M$ be a connected $n$-dimensional $C^{\infty}$-manifold with $n \ge 2$, and $\pi:TM {\longrightarrow}M$ be the natural projection. Given a local coordinate $(x^i)_{i=1}^n:U {\longrightarrow}{\ensuremath{\mathbb{R}}}^n$ on an open set $U \subset M$, we will always denote by $(x^i;v^i)_{i=1}^n$ the local coordinate on $\pi^{-1}(U)$ given by $v=\sum_i v^i({\partial}/{\partial}x^i)|_{\pi(v)}$.

A [*$C^{\infty}$-Finsler structure*]{} is a function $F:TM {\longrightarrow}[0,\infty)$ satisfying the following conditions:

(I) $F$ is $C^{\infty}$ on $TM \setminus \{ 0 \}$;

(II) $F(cv)=cF(v)$ for all $v \in TM$ and $c \ge 0$;

(III) The matrix $$g_{ij}(v):=\frac{1}{2} \frac{{\partial}(F^2)}{{\partial}v^i {\partial}v^j}(v)$$ is positive definite for all $v \in TM \setminus \{0\}$.

The positive definite matrix $(g_{ij}(v))$ defines a Riemannian structure $g_v$ of $T_xM$ through $$\label{eq:gv}
g_v\bigg( \sum_i a^i \frac{{\partial}}{{\partial}x^i},\sum_j b^j \frac{{\partial}}{{\partial}x^j} \bigg)
 :=\sum_{i,j} g_{ij}(v) a^i b^j.$$ Note that $g_v(v,v)=F(v)^2$. This inner product $g_v$ is regarded as the best approximation of $F|_{T_xM}$ in the direction $v$. Indeed, the unit sphere of $g_v$ is tangent to that of $F|_{T_xM}$ at $v/F(v)$ up to the second order. If $(M,F)$ is Riemannian, then $g_v$ always coincides with the original Riemannian metric. As usual, $(g^{ij})$ will stand for the inverse matrix of $(g_{ij})$.

Define the [*Cartan tensor*]{} $$A_{ijk}(v):=\frac{F(v)}{2} \frac{{\partial}g_{ij}}{{\partial}v^k}(v)$$ for $v \in TM \setminus \{0\}$, and recall that $A_{ijk} \equiv 0$ if and only if $(M,F)$ is Riemannian. We also define the [*formal Christoffel symbol*]{} $$\gamma^i{}_{jk}(v) :=\frac{1}{2}\sum_l g^{il}(v) \bigg\{ \frac{{\partial}g_{lj}}{{\partial}x^k}(v)
 +\frac{{\partial}g_{kl}}{{\partial}x^j}(v) -\frac{{\partial}g_{jk}}{{\partial}x^l}(v) \bigg\}$$ for $v \in TM \setminus \{0\}$. Then the geodesic equation is written as $\ddot{\eta}+G(\dot{\eta})=0$ with the [*geodesic spray coefficients*]{} $$G^i(v):=\sum_{j,k} \gamma^i{}_{jk}(v) v^j v^k$$ for $v \in TM$ ($G^i(0):=0$ by convention). Using these, we further define the [*nonlinear connection*]{} $$N^i{}_j(v):=\sum_k \bigg\{ \gamma^i{}_{jk}(v)v^k
 -\frac{1}{F(v)} A^i{}_{jk}(v)G^k(v) \bigg\}$$ for $v \in TM$ ($N^i{}_j(0):=0$ by convention), where $A^i{}_{jk}(v):=\sum_l g^{il}(v)A_{ljk}(v)$. We can rewrite this as (see [@BCS Exercise 2.3.3]) $$N^i{}_j(v)=\frac{1}{2} \frac{{\partial}G^i}{{\partial}v^j}(v).$$

${\ensuremath{\mathbf{S}}}$-curvature and weighted Ricci curvature {#ssc:S}
------------------------------------------------------------------

We choose an arbitrary positive $C^{\infty}$-measure $m$ on a Finsler manifold $(M,F)$. Fix a unit vector $v \in F^{-1}(1)$ and let $\eta:(-{\varepsilon},{\varepsilon}) {\longrightarrow}M$ be the geodesic with $\dot{\eta}(0)=v$. Along $\eta$, the tangent vector field $\dot{\eta}$ defines the Riemannian metric $g_{\dot{\eta}}$ as $(\ref{eq:gv})$. Denoting the volume form of $g_{\dot{\eta}}$ by $\vol_{\dot{\eta}}$, we decompose $m$ into $m(dx)=e^{-\Psi(\dot{\eta})} \vol_{\dot{\eta}}(dx)$ along $\eta$. Then we define the [*${\ensuremath{\mathbf{S}}}$-curvature*]{} of $v$ by $${\ensuremath{\mathbf{S}}}(v):=\frac{d(\Psi \circ \dot{\eta})}{dt}(0).$$ We extend this definition to all $w=cv$ with $c \ge 0$ by ${\ensuremath{\mathbf{S}}}(w):=c{\ensuremath{\mathbf{S}}}(v)$. Clearly ${\ensuremath{\mathbf{S}}}\equiv 0$ holds on Riemannian manifolds with the volume measure.

The [*weighted Ricci curvature*]{} (introduced in [@Oint]) is defined in a similar manner as follows:

(i) $\Ric_n(v):=\Ric(v)+(\Psi \circ \eta)''(0)$ if ${\ensuremath{\mathbf{S}}}(v)=0$, $\Ric_n(v):=-\infty$ otherwise;

(ii) $\Ric_N(v):=\Ric(v)+(\Psi \circ \eta)''(0)-{\ensuremath{\mathbf{S}}}(v)^2/(N-n)$ for $N \in (n,\infty)$;

(iii) $\Ric_{\infty}(v):=\Ric(v)+(\Psi \circ \eta)''(0)$.

Here $\Ric(v)$ is the usual (unweighted) Ricci curvature of $v$. The author [@Oint] shows that bounding $\Ric_N$ from below by $K \in {\ensuremath{\mathbb{R}}}$ is equivalent to the curvature-dimension condition ${\mathsf{CD}}(K,N)$, and then there are rich analytic and geometric applications. Observe that $\Ric_n \ge K >-\infty$ makes sense only when the ${\ensuremath{\mathbf{S}}}$-curvature vanishes everywhere. Therefore the class of such special triples $(M,F,m)$ deserves a particular interest. We remark that, if two measures $m_1, m_2$ on $(M,F)$ satisfy ${\ensuremath{\mathbf{S}}}\equiv 0$, then $m_1=c \cdot m_2$ for some positive constant $c$.

We rewrite ${\ensuremath{\mathbf{S}}}(v)$ according to [@Shlec §7.3] for ease of later calculation. Fix a local coordinate $(x^i)_{i=1}^n$ containing $\eta$ and represent $m$ along $\eta$ as $$m(dx)=\sigma(\eta)\, dx^1 dx^2 \cdots dx^n
 =\frac{\sigma(\eta)}{\sqrt{\det(g_{\dot{\eta}})}} \vol_{\dot{\eta}}(dx).$$ Then by definition we have $${\ensuremath{\mathbf{S}}}(v) =\frac{d}{dt}\Big|_{t=0}
 \log\bigg( \frac{\sqrt{\det(g_{\dot{\eta}(t)})}}{\sigma(\eta(t))} \bigg)
 =\frac{1}{2\det(g_v)} \frac{d}{dt}\Big|_{t=0} \big[ \det(g_{\dot{\eta}(t)}) \big]
 -\sum_i \frac{v^i}{\sigma(x)} \frac{{\partial}\sigma}{{\partial}x^i}(x).$$ Since $\eta$ solves the geodesic equation $\ddot{\eta}+G(\dot{\eta})=0$, the first term is equal to $$\begin{aligned}
&\frac{1}{2} \sum_{i,j,k} \bigg\{ g^{ij}(v) \frac{{\partial}g_{ij}}{{\partial}x^k}(v) v^k
 +g^{ij}(v) \frac{g_{ij}}{{\partial}v^k}(v) \ddot{\eta}^k(0) \bigg\} \\
&=\sum_{i,k} \bigg\{ \gamma^i{}_{ik}(v) v^k -\frac{1}{F(v)}A^i{}_{ik}(v)G^k(v) \bigg\}
 =\sum_i N^i{}_i(v).\end{aligned}$$ Thus we obtain $$\label{eq:SN}
{\ensuremath{\mathbf{S}}}(v)=\sum_i \bigg\{ N^i{}_i(v) -\frac{v^i}{\sigma(x)} \frac{{\partial}\sigma}{{\partial}x^i}(x) \bigg\}.$$ Note that ${\ensuremath{\mathbf{S}}}(cv)=c{\ensuremath{\mathbf{S}}}(v)$ indeed holds for $c \ge 0$ in this form.

Busemann-Hausdorff measure and Berwald spaces
---------------------------------------------

Different from the Riemannian case, there are several constructive measures on a Finsler manifold. Each of them is canonical in some sense, and coincides with the volume measure on Riemannian manifolds. Among them, here we treat only the Busemann-Hausdorff measure which is actually the Hausdorff measure associated with the suitable distance structure of $(M,F)$.

Using a basis $w_1,w_2,\ldots,w_n \in T_xM$ and its dual basis $\theta^1,\theta^2,\ldots,\theta^n \in T_x^*M$, the [*Busemann-Hausdorff measure*]{} $m_{BH}(dx)=\sigma_{BH}(x)\, \theta^1 \wedge \theta^2 \wedge \cdots \wedge \theta^n$ is defined as $$\frac{\omega_n}{\sigma_{BH}(x)} =
 \vol_n \bigg( \bigg\{ (c^i) \in {\ensuremath{\mathbb{R}}}^n \,\Big|\, F\bigg( \sum_i c^i w_i \bigg) <1 \bigg\} \bigg),$$ where $\vol_n$ is the Lebesgue measure and $\omega_n$ is the volume of the unit ball in ${\ensuremath{\mathbb{R}}}^n$.

Let $(M,F)$ be a [*Berwald space*]{} (see [@BCS Chapter 10] for the precise definition). Then it is well known that ${\ensuremath{\mathbf{S}}}\equiv 0$ for the Busemann-Hausdorff measure (see [@Shlec Proposition 7.3.1]). In fact, along any geodesic $\eta:[0,l] {\longrightarrow}M$, the parallel transport $T_{0,t}:T_{\eta(0)}M {\longrightarrow}T_{\eta(t)}M$ with respect to $g_{\dot{\eta}}$ preserves $F$. Therefore choosing parallel vector fields along $\eta$ as a basis yields that $\sigma_{BH}$ is constant on $\eta$. This implies ${\ensuremath{\mathbf{S}}}\equiv 0$.

Proof of Theorem \[th:main\]
============================

Let $(M,F)$ be a Randers space, i.e., $F(v)=\alpha(v)+\beta(v)$ such that $\alpha$ is a norm induced from a Riemannian metric and that $\beta$ is a one-form. In a local coordinate $(x^i)_{i=1}^n$, we can write $$\alpha(v) =\sqrt{\sum_{i,j} a_{ij}(x) v^i v^j}, \qquad
 \beta(v) =\sum_i b_i(x)v^i$$ for $v \in T_xM$. The [*length*]{} of $\beta$ at $x$ is defined by $\| \beta \|(x):=\sqrt{\sum_{i,j}a^{ij}(x)b_i(x)b_j(x)}$, which is necessarily less than $1$ in order to guarantee $F>0$ on $TM \setminus \{0\}$.

We denote the Christoffel symbol of $(a_{ij})$ by $\tilde{\gamma}^i{}_{jk}$. We also define $$b^i(x):=\sum_j a^{ij}(x)b_j(x), \qquad b_{i|j}(x):=
 \frac{{\partial}b_i}{{\partial}x^j}(x)-\sum_k b_k(x)\tilde{\gamma}^k{}_{ij}(x).$$ Note that $$\label{eq:beta}
\frac{{\partial}(\| \beta \|^2)}{{\partial}x^i}(x) =2\sum_j b_{j|i}(x) b^j(x).$$ We say that $\beta$ is a [*Killing form*]{} if $b_{i|j}+b_{j|i} \equiv 0$ on $M$. The geodesic spray coefficients of $F$ are given by (see [@BCS (11.3.11)]) $$\begin{aligned}
G^i(v) &= \sum_{j,k} \gamma^i{}_{jk}(v) v^j v^k \nonumber\\
&= \sum_{j,k} \bigg[ \tilde{\gamma}^i{}_{jk}(x) v^j v^k
 +b_{j|k}(x)\big( a^{ij}(x)v^k -a^{ik}(x)v^j \big) \alpha(v) \nonumber\\
&\quad +b_{j|k}(x) \frac{v^i}{F(v)}
 \big\{ v^j v^k +\big( b^k(x)v^j -b^j(x)v^k \big) \alpha(v) \big\} \bigg] \nonumber\\
&=: \sum_{j,k} \tilde{\gamma}^i{}_{jk}(x) v^j v^k +X^i(v)+Y^i(v). \label{eq:G}\end{aligned}$$

If ${\ensuremath{\mathbf{S}}}\equiv 0$ on $T_xM$, then we deduce from $(\ref{eq:SN})$ that $\sum_i N^i{}_i(v)$ is linear in $v \in T_xM$. We shall see that only this infinitesimal constraint is enough to imply the condition on $\beta$ stated in Theorem \[th:main\]. To see this, we calculate $2N^i{}_i={\partial}G^i/{\partial}v^i$ using $(\ref{eq:G})$. As the first term $\sum_{j,k}\tilde{\gamma}^i{}_{jk}(x) v^j v^k$ comes from a Riemannian structure, it suffices to consider only the linearly of $\sum_i\{ {\partial}X^i/{\partial}v^i(v) +{\partial}Y^i/{\partial}v^i(v) \}$. For the sake of simplicity, we will omit evaluations at $x$ and $v$ in the following calculations.

We first observe $$\sum_i \frac{{\partial}X^i}{{\partial}v^i} =\sum_{i,j} (b_{j|i}-b_{i|j})a^{ij} \alpha
 +\sum_{i,j,k,l}b_{j|k}(a^{ij}v^k -a^{ik}v^j) \frac{a_{il}v^l}{\alpha}
 =0.$$ As Euler’s theorem [@BCS Theorem 1.2.1] ensures $$\sum_i \frac{{\partial}}{{\partial}v^i} \frac{v^i}{F}
 =\frac{1}{F^2} \sum_i \bigg( F- v^i\frac{{\partial}F}{{\partial}v^i} \bigg)=\frac{n-1}{F},$$ we next obtain $$\begin{aligned}
\sum_i \frac{{\partial}Y^i}{{\partial}v^i} &=\sum_{i,j} \frac{v^i}{F} \bigg\{
(b_{i|j}+b_{j|i})v^j +(b_{i|j}-b_{j|i})b^j \alpha
 +\sum_{k,l} b_{j|k}(b^k v^j -b^j v^k) \frac{a_{il}v^l}{\alpha} \bigg\} \\
&\qquad +\frac{n-1}{F}\sum_{j,k}b_{j|k}
 \big\{ v^j v^k+(b^k v^j-b^j v^k)\alpha \big\} \\
&=\frac{n+1}{2} \sum_{i,j} (b_{i|j}+b_{j|i}) \frac{v^i v^j}{F}
 +(n+1)\sum_{i,j} (b_{i|j}-b_{j|i})b^j \frac{\alpha v^i}{F}.\end{aligned}$$ By comparing the evaluations at $v$ and $-v$, the coefficients $b_{i|j}+b_{j|i}$ in the first term must vanish for all $i,j$, namely $\beta$ is a Killing form. For the second term, we find that $(\alpha/F) \sum_j(b_{i|j}-b_{j|i})b^j$ must be constant on each $T_xM$. If $\alpha/F$ is not constant on some $T_xM$ (i.e., $\| \beta \|(x) \neq 0$), then it holds that $\sum_j (b_{i|j}-b_{j|i})b^j=0$. Since $\beta$ is a Killing form, we deduce from $(\ref{eq:beta})$ that $$0=\sum_j(b_{i|j}-b_{j|i})b^j =-2\sum_j b_{j|i}b^j
 =-\frac{{\partial}(\| \beta \|^2)}{{\partial}x^i}.$$ Therefore $\beta$ has a constant length as required, for $\|\beta\| \neq 0$ is an open condition. If $\alpha/F$ is constant on some $T_xM$, then the above argument yields that $\beta \equiv 0$ on $M$. This completes the proof of the “only if” part of Theorem \[th:main\].

For the “if” part, it is sufficient to show that the Busemann-Hausdorff measure satisfies ${\ensuremath{\mathbf{S}}}\equiv 0$, that can be found in [@Shlec Example 7.3.1]. We briefly repeat his discussion for completeness. We first observe from [@Shlec (2.10)] that $$m_{BH}(dx)=\big( 1-\| \beta \|(x)^2 \big)^{(n+1)/2} \sqrt{\det(a_{ij}(x))}\, dx^1 \cdots dx^n
 =: \sigma_{BH}(x)\, dx^1 \cdots dx^n.$$ Since $\beta$ has a constant length, we have $$\sum_k \frac{v^k}{\sigma_{BH}(x)} \frac{{\partial}\sigma_{BH}}{{\partial}x^k}(x)
 =\frac{1}{2} \sum_{i,j,k} v^k a^{ij}(x) \frac{{\partial}a_{ij}}{{\partial}x^k}(x)
 =\sum_{i,j} \tilde{\gamma}^i{}_{ij}(x) v^j.$$ Therefore we conclude $${\ensuremath{\mathbf{S}}}(v)=\frac{1}{2} \sum_{i,j,k} \frac{{\partial}}{{\partial}v^i}
 \Big[ \tilde{\gamma}^i{}_{jk}(x) v^j v^k \Big]
 -\sum_k \frac{v^k}{\sigma_{BH}(x)} \frac{{\partial}\sigma_{BH}}{{\partial}x^k}(x) =0.$$ ${\hfill \Box}$

We finally remark related known results and several consequences of Theorem \[th:main\].

\[rm:last\] (a) A Randers space is a Berwald space if and only if $\beta$ is [*parallel*]{} in the sense that $b_{i|j} \equiv 0$ for all $i,j$ (see [@BCS Theorem 11.5.1]). Thanks to [@Shlec Example 7.3.2], we know that a Killing form of constant length is not necessarily parallel.

\(b) In [@De], Deng gives a characterization of vanishing ${\ensuremath{\mathbf{S}}}$-curvature for homogeneous Randers spaces with the Busemann-Hausdorff measure.

\(c) It is easy to construct a Randers space whose $\beta$ does not have a constant length. Hence many Finsler manifolds do not admit a measure with ${\ensuremath{\mathbf{S}}}\equiv 0$ (in other words, with $\Ric_n \ge K>-\infty$).

\(d) Another consequence of Theorem \[th:main\] is that only (constant multiplications of) the Busemann-Hausdorff measures can satisfy ${\ensuremath{\mathbf{S}}}\equiv 0$ on Randers spaces. Then a natural question is the following:

Is there a Finsler manifold $(M,F)$ on which some measure $m$ other than $($a constant multiplication of$)$ the Busemann-Hausdorff measure satisfies ${\ensuremath{\mathbf{S}}}\equiv 0$? If yes, what kind of measure is $m$?

If such a measure exists, then it is more natural than the Busemann-Hausdorff measure in respect of the weighted Ricci curvature.

[AIM]{} P. L. Antonelli, R. S. Ingarden and M. Matsumoto, The theory of sprays and Finsler spaces with applications in physics and biology, Kluwer Academic Publishers Group, Dordrecht, 1993.

D. Bao, S.-S. Chern and Z. Shen, An introduction to Riemann-Finsler geometry, Springer-Verlag, New York, 2000.

S. Deng, [*The S-curvature of homogeneous Randers spaces*]{}, Differential Geom. Appl. [**27**]{} (2009), 75–84.

S. Ohta, [*Finsler interpolation inequalities*]{}, to appear in Calc. Var. Partial Differential Equations.

S. Ohta and K.-T. Sturm, [*Heat flow on Finsler manifolds*]{}, Comm. Pure Appl. Math. [**62**]{} (2009), 1386–1433.

G. Randers, [*On an asymmetrical metric in the fourspace of general relativity*]{}, Phys. Rev. (2) [**59**]{} (1941), 195–199.

Z. Shen, Lectures on Finsler geometry, World Scientific Publishing Co., Singapore, 2001.

[^1]: Supported in part by the Grant-in-Aid for Young Scientists (B) 20740036.
