On m-th root metrics with special curvature properties

TL;DR

This paper proves that m-th root Finsler metrics with certain curvature properties simplify to more specific types, revealing their geometric structure and curvature behavior.

Contribution

It demonstrates that m-th root Finsler metrics with isotropic Landsberg curvature are actually Landsberg metrics and those with almost vanishing H-curvature have vanishing H-curvature.

Findings

m-th root Finsler metrics with isotropic Landsberg curvature reduce to Landsberg metrics

m-th root metrics with almost vanishing H-curvature have vanishing H-curvature

clarifies curvature properties of m-th root Finsler metrics

Abstract

In this Note, we prove that every m-th root Finsler metric with isotropic Landsberg curvature reduces to a Landsberg metric. Then, we show that every m-th root metric with almost vanishing H-curvature has vanishing H-curvature.

Equations

$$A:=a_{i_1\dots i_m}(x)y^{i_1}{y}^{i_2}\dots y^{i_m}$$

$$A_i=\frac{\partial A}{\partial y^i},A_{ij}=\frac{{\partial }^{2}A}{\partial y^i\partial y^j},A_{x^k}=\frac{\partial A}{\partial x^k},A_0=A_{x^k}{y}^k,A_{0j}=A_{x^k{y}^j}{y}^k.$$

$$\displaystyle g_{ij}=\frac{A^{\frac{2}{m}-2}}{m^{2}}[mAA_{ij}+(2-m)A_{i}A_{j}],\ \ \ g^{ij}=A^{-\frac{2}{m}}[mAA^{ij}+\frac{m-2}{m-1}y^{i}y^{j}],\ \$$

$$\displaystyle y^{i}A_{i}=mA,\ \ y^{i}A_{ij}=(m-1)A_{j},\ \ y_{i}=\frac{1}{m}A^{\frac{2}{m}-1}A_{i},\ \ A^{ij}A_{i}=\frac{1}{m-1}y^{j},\ \ A_{i}A_{j}A^{ij}=\frac{m}{m-1}A.\ \$$

$$R_k^i=2\frac{\partial G^i}{\partial x^k}-y^j\frac{{\partial }^2{G}^i}{\partial x^j\partial y^k}+2G^j\frac{{\partial }^2{G}^i}{\partial y^j\partial y^k}-\frac{\partial G^i}{\partial y^j}\frac{\partial G^j}{\partial y^k}.$$

$$G^i=\frac{1}{2}(A_{0j}-A_{x^j})A^{ij}.$$

$$C_{ijk}=\frac{1}{m}A^{\frac{2}{m}-3}\Big{[}A^{2}A_{ijk}+(\frac{2}{m}-1)(\frac{2}{m}-2)A_{i}A_{j}A_{k}+(\frac{2}{m}-1)A\{A_{i}A_{jk}+A_{j}A_{ki}+A_{k}A_{ij}\}\Big{]}.$$

$$A_{s}G^{s}_{\ y^{i}y^{j}y^{k}}=-2cA^{\frac{1}{m}-2}\Big{[}A^{2}A_{ijk}+(\frac{2}{m}-1)\Big{\{}(\frac{2}{m}-2)A_{i}A_{j}A_{k}+A\{A_{i}A_{jk}+A_{j}A_{ki}+A_{k}A_{ij}\}\Big{\}}\Big{]}.$$

$$A^2{A}_{ijk}+(\frac{2}{m}-1)(\frac{2}{m}-2)A_i{A}_j{A}_k+(\frac{2}{m}-1)A\{A_i{A}_{jk}+A_j{A}_{ki}+A_k{A}_{ij}\}=0$$

$$H_{ij}=\frac{n+1}{2F}\theta h_{ij},$$

$$h_{ij}=\frac{1}{m^2}[mA{A}_{ij}+(1-m)A_i{A}_j]A^{\frac{2}{m}-2}.$$

$$H_{ij}=\frac{(n+1)\theta }{2m^2}[mA{A}_{ij}+(1-m)A_i{A}_j]A^{\frac{1}{m}-2}.$$

$$mA{A}_{ij}+(1-m)A_i{A}_j=0.$$

Full text

1 Introduction

Let $M$ be an $n$-dimensional $C^{\infty}$ manifold. Denote by $TM=\cup_{x\in M}T_{x}M$ the tangent space of $M$. Let $TM_{0}=TM\setminus\{0\}$. Let $F=\sqrt[m]{A}$ be a Finsler metric on $M$, where $A$ is given by

[TABLE]

with $a_{i_{1}\dots i_{m}}$ symmetric in all its indices [2]. Then $F$ is called an $m$-th root Finsler metric. Let $F$ be an $m$-th root Finsler metric on an open subset $U\subset\mathbb{R}^{n}$. Put

[TABLE]

Suppose that $(A_{ij})$ is a positive definite tensor and $(A^{ij})$ denotes its inverse. Then the following hold

[TABLE]

Let $(M,F)$ be a Finsler manifold. The second derivatives of ${1\over 2}F_{x}^{2}$ at $y\in T_{x}M_{0}$ is an inner product ${\bf g}_{y}$ on $T_{x}M$. The third order derivatives of ${1\over 2}F_{x}^{2}$ at $y\in T_{x}M_{0}$ is a symmetric trilinear forms ${\bf C}_{y}$ on $T_{x}M$. We call ${\bf g}_{y}$ and ${\bf C}_{y}$ the fundamental form and the Cartan torsion, respectively. The rate of change of the Cartan torsion along geodesics is the Landsberg curvature ${\bf L}_{y}$ on $T_{x}M$ for any $y\in T_{x}M_{0}$. $F$ is said to be Landsbergian if ${\bf L}=0$. The quotient ${\bf L}/{\bf C}$ is regarded as the relative rate of change of Cartan torsion ${\bf C}$ along Finslerian geodesics. Then $F$ is said to be isotropic Landsberg metric if ${\bf L}=cF\bf C$, where $c=c(x)$ is a scalar function on $M$. In this paper, we prove the following.

Theorem 1**.**

Let $(M,F)$ be an $n$-dimensional $m$-th root Finsler manifold. Suppose that $F$ is a non-Riemannian isotropic Landsberg metric. Then $F$ reduces to a Landsberg metric.

Let $F$ be a Finsler metric on a manifold $M$. The geodesics of $F$ are characterized locally by the equations ${d^{2}x^{i}\over dt^{2}}+2G^{i}(x,{dx\over dt})=0$, where G^{i}={1\over 4}g^{ik}\Big{\{}2{{\partial}g_{pk}\over{\partial}x^{q}}-{{\partial}g_{pq}\over{\partial}x^{k}}\Big{\}}y^{p}y^{q} are coefficients of the spray associated with $F$. A Finsler metric $F$ is called a Berwald metric if $G^{i}={1\over 2}\Gamma^{i}_{jk}(x)y^{j}y^{k}$ are quadratic in $y\in T_{x}M$ for any $x\in M$. Taking a trace of Berwald curvature gives rise the mean Berwald curvature ${\bf E}$. In [1], Akbar-Zadeh introduces the non-Riemannian quantity $\bf{H}$ which is obtained from the mean Berwald curvature by the covariant horizontal differentiation along geodesics. He proves that for a Finsler manifold of scalar flag curvature ${\bf K}$ with dimension $n\geq 3$, ${\bf K}=constant$  if and only if ${\bf{H}}=0$. It is remarkable that, the Riemann curvature $R_{y}=R^{i}_{\ k}dx^{k}\otimes{{\partial}\over{\partial}x^{i}}|_{x}:T_{x}M\to T_{x}M$ is a family of linear maps on tangent spaces, defined by

[TABLE]

A Finsler metric $F$ is said to be of scalar curvature if there is a scalar function ${\bf K}={\bf K}(x,y)$ such that $R^{i}_{\ k}={\bf K}(x,y)F^{2}h^{i}_{k}$. If ${\bf K}=constant$, then $F$ is called of constant flag curvature.

A Finsler metric is called of isotropic ${\bf H}$-curvature if $H_{ij}=\frac{n+1}{2F}\theta h_{ij}$, for some 1-form $\theta$ on $M$, where $h_{ij}$ is the angular metric. It is remarkable that in [5], Z. Shen with the authors prove that every Finsler metric of scalar flag curvature ${\bf K}$ and of isotropic ${\bf H}$-curvature has almost isotropic flag curvature, i.e., the flag curvature is in the form ${\bf K}=\frac{3\theta}{F}+\sigma$, for some scalar function $\sigma$ on $M$.

Theorem 2**.**

Let $(M,F)$ be an $n$-dimensional $m$-th root manifold with $n\geq 2$. Suppose that $F$ has isotropic ${\bf H}$-curvature. Then ${\bf H}=0$.

2 Proof of the Main Theorems

Lemma 1**.**

([6])* Let $F$ be an $m$-th root Finsler metric on an open subset $U\subset\mathbb{R}^{n}$. Then the spray coefficients of $F$ are given by*

[TABLE]

Proof of Theorem 1: Let $F=\sqrt[m]{A}$ be an $m$-th root isotropic Landsberg metric, i.e., $L_{ijk}=cFC_{ijk}$, where $c=c(x)$ is a scalar function on $M$. The Cartan tensor of $F$ is given by the following

[TABLE]

Since $L_{ijk}=-\frac{1}{2}y_{s}G^{s}_{\ y^{i}y^{j}y^{k}}$, then we have $L_{ijk}=-\frac{1}{2m}A^{\frac{2}{m}-1}A_{s}G^{s}_{\ y^{i}y^{j}y^{k}}$. Therefore, we have

[TABLE]

By Lemma 1, the left hand side of (2 .6) is a rational function in $y$, while its right hand side is an irrational function in $y$. Thus, $c=0$ or $A$ satisfies the following PDEs

[TABLE]

Plugging (2 .7) into (2 .5) implies that $C_{ijk}=0$. Hence, by Deicke’s theorem, $F$ is Riemannain metric, which contradicts our assumption. Therefore, $c=0$. This completes the proof. ∎

Proof of Theorem 2: Let $F=\sqrt[m]{A}$ be of isotropic ${\bf H}$-curvature, i.e.,

[TABLE]

where $\theta$ is a 1-form on $M$. The angular metric $h_{ij}=g_{ij}-F^{2}y_{i}y_{j}$ is given by the following

[TABLE]

Plugging (2 .9) into (2 .8), we get

[TABLE]

By (2 .4), one can see that $H_{ij}$ is rational with respect to $y$. Thus, (2 .10) implies that $\theta=0$ or

[TABLE]

By (2 .9) and (2 .11), we conclude that $h_{ij}=0$, which is impossible. Hence $\theta=0$ and then $H_{ij}=0$. ∎

By Schur Lemma, Theorem 2 and Theorem 1.1 of [5], we have the following.

Corollary 1**.**

Let $(M,F^{n})$ be an n-dimensional m-th root Finsler manifold of scalar flag curvature ${\bf K}$ with $n\geq 3$. Suppose that the flag curvature is given by ${\bf K}=\frac{3\theta}{F}+\sigma$, where $\theta$ is a 1-form and $\sigma=\sigma(x)$ is a scalar function on M. Then ${\bf K}=0$.