Conformal Transformations on General (α,β)-Spaces

Abstract

In this paper, we study conformal transformations between two almost regular general $(\alpha ,\beta )$-metrics. By using the method of special coordinate system, the necessary and sufficient conditions for conformal transformations preserving the mean Landsberg curvature are obtained. Further, a rigidity theorem for regular general $(\alpha ,\beta )$-metrics is proved.

1. Introduction

In Finsler geometry, the Weyl theorem states that the projective and conformal properties of a Finsler space determine the metric properties uniquely [1]. Therefore, the conformal properties of Finsler metrics deserve extra attention. Let F and $\tilde{F}$ be two Finsler metrics on a manifold M. A conformal transformation between F and $\tilde{F}$ is defined by L: $F\to \tilde{F}$, $\tilde{F}=e^{\sigma \left(x\right)}F$, where the conformal factor $\sigma :=\sigma \left(x\right)$ is a scalar function on M. The metrics F and $\tilde{F}$ are conformally related. If $\sigma$ is a constant, then the conformal transformation is called a homothetic transformation.

A natural problem is knowing how to determine, given a Finsler metric with some properties on a manifold M, all conformally related Finsler metrics with the given properties. Bácsó-Cheng [2] characterized conformal transformations that preserve the Riemann curvature, the Ricci curvature, the (mean) Landsberg curvature, or the $\mathbf{S}$-curvature, respectively. Chen-Cheng-Zou [3] proved that if both conformally related $(\alpha ,\beta )$-metrics are of the Douglas type or of isotropic $\mathbf{S}$-curvature, then the conformal transformations between them are homothetic. Later, Chen-Liu [4] characterized conformal transformations between two almost regular $(\alpha ,\beta )$-metrics that preserve the mean Landsberg curvature. Furthermore, they proved that conformal transformations between non-Riemannian regular $(\alpha ,\beta )$-metrics, which preserve the mean Landsberg curvature, must be homothetic.

Li-Shen [5] studied $(\alpha ,\beta )$-metrics with the mean Landsberg curvature and obtained its characterizing equation. Cheng–Wang–Wang [6] characterized $(\alpha ,\beta )$-metrics with the relative isotropic mean Landsberg curvature. Zou-Cheng [7] explored an $(\alpha ,\beta )$-metric whose $\varphi \left(s\right)$ is a polynomial about s, and they proved that it has vanishing mean Landsberg curvature if and only if it is a Berwald metric. Under the condition that the 1-form $\beta$ is a conformal field of the Riemannian metric $\alpha$, Behzadi–Rafiei [8] proved that the general $(\alpha ,\beta )$-metric has vanishing mean Landsberg curvature if and only if it is of the Landsberg type. Najafi–Saberali [9] explored the special $(\alpha ,\beta )$-metric and obtained that it has an isotropic mean Landsberg curvature that is equivalent to that its isotropic Landsberg curvature.

The general $(\alpha ,\beta )$-metric was first introduced by Yu-Zhu [10] in the following form:

$$F=\alpha \varphi (b^2,s),s=\frac{\beta }{\alpha },$$

(1)

where $\varphi =\varphi (b^2,s)$ is a ${\mathbb{C}}^{\infty }$ positive function, $\alpha =\alpha (x,y)=\sqrt{a_{ij}\left(x\right)y^i{y}^j}$ is a Riemannian metric, $\beta =\beta (x,y)=b_i\left(x\right)y^i$ is a 1-form, and $b:={\parallel \beta \parallel }_{\alpha }$. It is known that $F=\alpha \varphi (b^2,\frac{\beta }{\alpha })$ is a Finsler metric with $b<b_0$ if and only if $\varphi (b^2,s)$ is a positive ${\mathbb{C}}^{\infty }$ function satisfying [10]

$$\varphi -s{\varphi }_s>0,\varphi -s{\varphi }_s+(b^2-s^2){\varphi }_{ss}>0$$

for $n\ge 3$. If we consider a 1-form $\beta$ with $b\le b_0$ where $b_0:={sup}_{x\in M}b$, then $F=\alpha \varphi (b^2,s)$ might be singular in the two extremal directions $y\in T_{x}M$ with $\beta (x,y)=\pm b_0\alpha (x,y)$. Such metrics are called almost regular general $(\alpha ,\beta )$-metrics. In particular, when $\varphi =\varphi \left(s\right)$ in $\left(1\right)$, the Finsler metric $F=\alpha \varphi \left(s\right)$ is called an $(\alpha ,\beta )$-metric.

Note that ${(.)}_s$ denotes the partial derivative of the quantity $(.)$ with respect to s.

In this paper, we mainly study conformal transformations preserving the mean Landsberg curvature. It is known that a homothetic transformation must preserve the mean Landsberg curvature. Thus, we focus on non-homothetic conformal transformations.

Theorem 1.

Let F be an almost regular general $(\alpha ,\beta )$-metric on an $n(\ge 3)$-dimensional manifold M. Assume that F and $\tilde{F}$ are two conformally related metrics with the conformal factor $\sigma =\sigma \left(x\right)$. Then, F and $\tilde{F}$ have the same mean Landsberg curvature if and only if one of the following cases holds:

(1) 

ϕ must be

$$\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}},$$

where $k_1=k_1\left(b^2\right)$ and $k_2=k_2\left(b^2\right)>0$ are arbitrary differential functions. In this case, the conformal factor is arbitrary;

(2) 

The conformal factor σ satisfies ${\sigma }_i\left(x\right)=\frac{{\sigma }_b}{b^2}{b}_i\left(x\right)$, and ϕ satisfies

$$\mathsf{\Phi }=\frac{k_3{\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}},$$

(2)

where ${\sigma }_i:=\frac{\partial \sigma }{\partial x^i}$, ${\sigma }_b:={\sigma }_i{b}^i$, $b^2:=a^{ij}{b}_i{b}_j$, $Q:=\frac{{\varphi }_s}{\varphi -s{\varphi }_s}$, $\Delta :=1+sQ+(b^2-s^2)Q_s$, $\mathsf{\Phi }:=(1+sQ)(b^2-s^2)Q_{ss}+(n\Delta +1+sQ)(Q-s{Q}_s)$, and $k_3=k_3\left(b^2\right)$ is a differential positive function.

This theorem generalizes the results obtained by Chen-Liu about conformal transformations preserving the mean Landsberg curvature on $(\alpha ,\beta )$-spaces [4].

Based on Theorem 1, we obtain a rigidity theorem for regular general $(\alpha ,\beta )$-metrics as follows.

Theorem 2.

Let F be a regular general $(\alpha ,\beta )$-metric on an $n(\ge 3)$-dimensional manifold M. Assume that F and $\tilde{F}$ are two conformally related metrics. Then, F and $\tilde{F}$ have the same mean Landsberg curvature if and only if F is Riemannian.

2. Preliminaries

Let F be a non-Riemannian Finsler metric on a manifold M of dimension $n(\ge 3)$. Its spray coefficients $G^i$ are defined by

$$G^i:=\frac{1}{4}{g}^{ij}\{{\left[F^2\right]}_{x^k{y}^j}{y}^k-{\left[F^2\right]}_{x^j}\},$$

where $g_{ij}:=\frac{1}{2}{\left(F^2\right)}_{y^i{y}^j}$ and $\left(g^{ij}\right):={\left(g_{ij}\right)}^{-1}.$

The Cartan tensor is defined by $\mathbf{C}:=C_{ijk}d{x}^i\otimes d{x}^j\otimes d{x}^k$, where

$$C_{ijk}:=\frac{1}{4}\frac{{\partial }^3{F}^2}{\partial y^i\partial y^j\partial y^k}=\frac{1}{2}\frac{\partial g_{ij}}{\partial y^k}.$$

The mean Cartan torsion $\mathbf{I}:=I_{i}d{x}^i:T_{x}M\to \mathbf{R}$ is defined by

$$I_i:=g^{jk}{C}_{ijk}.$$

Deicke’s theorem shows that $\mathbf{I}=0$ if and only if F is Riemannian.

The Landsberg curvature $\mathbf{L}=L_{ijk}d{x}^i\otimes d{x}^j\otimes d{x}^k$ is a horizontal tensor on $TM\setminus \left\{0\right\}$, defined by

$$L_{ijk}:=-\frac{1}{2}F{F}_{y^l}\frac{{\partial }^3{G}^l}{\partial y^i\partial y^j\partial y^k}.$$

The mean Landsberg curvature is defined by

$$\mathbf{J}:=J_{i}d{x}^i,J_i:=g^{jk}{L}_{ijk}.$$

Lemma 1

([10]). For a general $(\alpha ,\beta )$-metric $F=\alpha \varphi (b^2,\frac{\beta }{\alpha })$, the coefficients of the fundamental tensor are

$$g_{ij}=\rho a_{ij}+\overline{\rho }{b}_i{b}_j+\frac{1}{\alpha }\tilde{\rho }(b_i{y}_j+b_j{y}_i)-\frac{1}{{\alpha }^2}s\tilde{\rho }{y}_i{y}_j,$$

where

$$y_i:=a_{ij}{y}^j,\rho :=\varphi (\varphi -s{\varphi }_s),\overline{\rho }:=\varphi {\varphi }_{ss}+{\varphi }_s{\varphi }_s,\tilde{\rho }:=(\varphi -s{\varphi }_s){\varphi }_s-s\varphi {\varphi }_{ss}.$$

And $\left(g^{ij}\right):={\left(g_{ij}\right)}^{-1}$ are as follows:

$$g^{ij}={\rho }^{-1}[a^{ij}+\eta b^i{b}^j+\overline{\eta }{\alpha }^{-1}(b^i{y}^j+b^j{y}^i)+\tilde{\eta }{\alpha }^{-2}{y}^i{y}^j],$$

where $b^i:=a^{ij}{b}_j$,

$$\begin{array}{cc}\hfill \eta :=& -\frac{\varphi }{\varphi -s{\varphi }_s+(b^2-s^2){\varphi }_{ss}},\overline{\eta }:=-\frac{(\varphi -s{\varphi }_s){\varphi }_s-s\varphi {\varphi }_{ss}}{\varphi [\varphi -s{\varphi }_s+(b^2-s^2){\varphi }_{ss}]},\hfill \\ \hfill \tilde{\eta }:=& \frac{[s\varphi +(b^2-s^2){\varphi }_s][(\varphi -s{\varphi }_s){\varphi }_s-s\varphi {\varphi }_{ss}]}{{\varphi }^2[\varphi -s{\varphi }_s+(b^2-s^2){\varphi }_{ss}]}.\hfill \end{array}$$

Lemma 2

([11]). For a general $(\alpha ,\beta )$-metric $F=\alpha \varphi (b^2,\frac{\beta }{\alpha })$, the coefficients of the Cartan tensor and the mean Cartan torsion are as follows:

$$\begin{array}{cc}\hfill C_{ijk}=& \frac{1}{2}[{\alpha }^{-1}\tilde{\rho }(a_{ij}{b}_k+a_{ik}{b}_j+a_{jk}{b}_i)-{\alpha }^{-2}s\tilde{\rho }(a_{ij}{y}_k+a_{ik}{y}_j+a_{jk}{y}_i)\hfill \\ \hfill & +{\alpha }^{-4}(3s\tilde{\rho }-s^3{\overline{\rho }}_s)y_i{y}_j{y}_k+{\alpha }^{-3}(s^2{\overline{\rho }}_s-\tilde{\rho })(b_i{y}_j{y}_k+b_j{y}_i{y}_k+b_k{y}_i{y}_j)\hfill \\ \hfill & -{\alpha }^{-2}s{\overline{\rho }}_s(b_i{b}_j{y}_k+b_j{b}_k{y}_i+b_k{b}_i{y}_j)+{\alpha }^{-1}{\overline{\rho }}_s{b}_i{b}_j{b}_k],\hfill \\ \hfill I_i=& \frac{1}{2{\alpha }^2\rho }A(\alpha b_i-s{y}_i),\hfill \end{array}$$

where $A:=\frac{{\varphi }^2\mathsf{\Phi }}{\Delta {(1+sQ)}^2}$.

By Deicke’s theorem, a general $(\alpha ,\beta )$-metric is Riemannian if and only if $A=0$.

3. The Proof of Main Theorems

Before proving Theorem 1, we need following Lemmas.

Lemma 3

([2]). Let F be a Finsler metric on a manifold M. Assume that F and $\tilde{F}$ are two conformally related metrics with the conformal factor $\sigma =\sigma \left(x\right)$. Then, their mean Landsberg curvature must satisfy

$${\tilde{J}}_i=J_i+{\sigma }_0{I}_i+F^2{\sigma }^j{I}_{i.j}+\left({\sigma }^j{I}_j\right)y_i^F-F^2{\sigma }^j{I}_k{C}_{ij}^k,$$

where ${\sigma }_0:={\sigma }_k{y}^k$, ${\sigma }^i:=g^{ik}{\sigma }_k$, $I_{i.j}:=\frac{\partial I_i}{\partial y^j}$, $y_i^F:=g_{ij}{y}^j$, $C_{ij}^k:=g^{kl}{C}_{ijl}$.

Based on Lemma 3, F and $\tilde{F}=e^{\sigma \left(x\right)}F$ have the same mean Landsberg curvature if and only if the following holds:

$${\sigma }_0{I}_i+F^2{\sigma }^j{I}_{i.j}+\left({\sigma }^j{I}_j\right)y_i^F-F^2{\sigma }^j{I}_k{C}_{ij}^k=0.$$

Assume that F is a general $(\alpha ,\beta )$-metric. By direct computations, the above equation is equivalent to

$$(T_1{\sigma }_0+\alpha T_2{\sigma }_b)y_i+\alpha (T_3{\sigma }_0+\alpha T_4{\sigma }_b)b_i+{\alpha }^2{T}_5{\sigma }_i=0,$$

(3)

where

$$\begin{array}{cc}\hfill T_1=& \left[\frac{(s+b^{2}Q)(1+sQ+2s^2{Q}_s)}{(1+sQ)\Delta }+\frac{3s(1+b^2{Q}_s)}{\Delta }-\frac{2s}{1+sQ}-\frac{s(s+b^{2}Q){\Delta }_s}{{\Delta }^2}\right]{\varphi }^{2}A\hfill \\ \hfill & +\frac{2s(b^{2}Q+s)}{\Delta }{\varphi }^2{A}_s,\hfill \\ \hfill T_2=& \frac{[(1+sQ)s{\Delta }_s-\Delta (sQ+5s^2{Q}_s)]}{{\Delta }^2}{\varphi }^{2}A-\frac{2s(1+sQ)}{\Delta }{\varphi }^2{A}_s,\hfill \\ \hfill T_3=& \left[-\frac{2Q_s(2s^2+b^{2}sQ-b^2)}{(1+sQ)\Delta }-\frac{2b^2{Q}_s+s^2{Q}_s-sQ}{\Delta }+\frac{(s+b^{2}Q){\Delta }_s}{{\Delta }^2}\right]{\varphi }^{2}A\hfill \\ \hfill & -\frac{2(b^{2}Q+s)}{\Delta }{\varphi }^2{A}_s,\hfill \\ \hfill T_4=& -\frac{[(1+sQ){\Delta }_s-\Delta (Q+5s{Q}_s)]}{{\Delta }^2}{\varphi }^{2}A+\frac{2(1+sQ)}{\Delta }{\varphi }^2{A}_s,\hfill \\ \hfill T_5=& -\frac{s\Delta +(s+b^{2}Q)}{\Delta }{\varphi }^{2}A.\hfill \end{array}$$

Note that $T_1+s{T}_3+T_5=0$ and $T_2+s{T}_4=0$ hold.

Lemma 4.

Let a positive $C^{\infty }$ function $\varphi =\varphi (b^2,s)$ satisfy $s\Delta +s+b^{2}Q=0$. Then,

$$\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}},$$

where $k_1=k_1\left(b^2\right)$ and $k_2=k_2\left(b^2\right)>0$ are arbitrary differential functions.

Proof. 

$s\Delta +s+b^{2}Q=0$ is equivalent to

$$(b^2-s^2)(Q+s{Q}_s)+2s(1+sQ)=0.$$

It can be rewritten as

$${\left(\frac{b^2-s^2}{1+sQ}\right)}_s=0.$$

Integrating this equation with respect to s yields

$$Q=\frac{k_1(b^2-s^2)-1}{s},$$

(4)

where $k_1=k_1\left(b^2\right)$ is an arbitrary differential function. Since $Q:=\frac{{\varphi }_s}{\varphi -s{\varphi }_s}$, the above equation leads to

$$\frac{{\varphi }_s}{\varphi }=\frac{Q}{1+sQ}.$$

It is equivalent to

$${\left(\ln \varphi \right)}_s=\frac{Q}{1+sQ}.$$

Integrating the above equation with respect to s yields

$$\varphi =k_{2}exp\left(\int \frac{Q}{1+sQ}ds\right),$$

where $k_2=k_2\left(b^2\right)>0$ is a $C^{\infty }$ function.

Substituting $\left(4\right)$ into the above equation yields

$$\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}}.$$

This completes the proof of Lemma 4. □

Lemma 5.

Let a positive $C^{\infty }$ function $\varphi =\varphi (b^2,s)$ satisfy $s{T}_1+b^2{T}_2=0$. Then,

$$\mathsf{\Phi }=\frac{k_3{\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}},$$

where $k_3=k_3\left(b^2\right)$ is an arbitrary positive function.

Proof. 

The direct computation yields

$$s{T}_1+b^2{T}_2=\frac{2s^2\Delta [1+sQ-2(b^2-s^2)Q_s]+s(1+sQ)(b^2-s^2){\Delta }_s}{(1+sQ){\Delta }^2}{\varphi }^{2}A-\frac{2s(b^2-s^2)}{\Delta }{\varphi }^2{A}_s.$$

Thus, $s{T}_1+b^2{T}_2=0$ is equivalent to

$$\{2s\Delta [1+sQ-2(b^2-s^2)Q_s]+(1+sQ)(b^2-s^2){\Delta }_s\}A-2(b^2-s^2)(1+sQ)\Delta A_s=0.$$

It can be rewritten as

$$\frac{A_s}{A}=\frac{Q-s{Q}_s}{1+sQ}-\frac{Q}{1+sQ}-\frac{s{Q}_s}{1+sQ}+\frac{2s\Delta +(b^2-s^2){\Delta }_s}{2(b^2-s^2)\Delta },$$

i.e.,

$$\frac{A_s}{A}=\frac{{\rho }_s}{\rho }-\frac{{\varphi }_s}{\varphi }-\frac{s{\varphi }_{ss}}{\varphi -s{\varphi }_s}+\frac{2s\Delta +(b^2-s^2){\Delta }_s}{2(b^2-s^2)\Delta }.$$

That means

$${\left(\ln A\right)}_s={\left(\ln \rho \right)}_s-{\left(\ln \varphi \right)}_s+{\left[\ln (\varphi -s{\varphi }_s)\right]}_s+\frac{1}{2}{\left(\ln \frac{\Delta }{b^2-s^2}\right)}_s.$$

Integrating it with respect to s yields

$$A=k_3\frac{{\varphi }^2}{{(1+sQ)}^2}\sqrt{\frac{\Delta }{(b^2-s^2)}},$$

where $k_3=k_3\left(b^2\right)$ is an arbitrary positive function. This equation is equivalent to

$$\mathsf{\Phi }=\frac{k_3{\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}}.$$

This completes the proof of Lemma 5. □

Using above Lemmas, we can prove Theorem 1.

Proof of Theorem 1. 

“Sufficiency”. Assume that $\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}}$ and the conformal factor $\sigma$ is arbitrary. Note that $0=T_1=T_2=T_3=T_4=T_5$ and $\left(3\right)$ hold. Thus, the conclusion is obvious. Assume that ${\sigma }_i=\frac{{\sigma }_b}{b^2}{b}_i$ and $\varphi$ satisfies $\left(2\right)$. Then, $\left(3\right)$ holds. Hence, F and $\tilde{F}$ have the same mean Landsberg curvature.

“Necessity”. In general, it is impossible to solve $\left(3\right)$ if $\varphi =\varphi \left(s\right)$ is an unknown function. To overcome this difficulty, we choose a special coordinate system at a point x as in [12]. First, we assume that

$${\alpha }_x=\sqrt{\sum _{i=1}^n{\left(y^i\right)}^2},{\beta }_x=b{y}^1.$$

Then, we take another special coordinates: $(s,y^a)\to \left(y^i\right)$ given by

$$y^1=\frac{s}{\sqrt{b^2-s^2}}\overline{\alpha },y^a=y^a,$$

where

$$\overline{\alpha }=\sqrt{\sum _{a=2}^n{\left(y^a\right)}^2}.$$

We make the following agreement

$$1\le i,j,k,\cdots \le n,2\le a,b,c,\cdots \le n.$$

We have

$$\alpha =\frac{b}{\sqrt{b^2-s^2}}\overline{\alpha },\beta =\frac{bs}{\sqrt{b^2-s^2}}\overline{\alpha }.$$

When we take $i=1$ in (3), by the rational and irrational terms of y, $\left(3\right)$ is equal to

$$(s{T}_1+b^2{T}_3)\overline{{\sigma }_0}=0$$

(5)

and

$$(s{T}_1+b^2{T}_2){\sigma }_1=0.$$

(6)

Similarly, when we take $i=a$ in (3), $\left(3\right)$ leads to

$$(s{T}_1+b^2{T}_2){\sigma }_1=0$$

and

$${\overline{\sigma }}_0{y}_a{T}_1+\frac{b^2}{b^2-s^2}{\overline{\alpha }}^2{\sigma }_a{T}_5=0.$$

(7)

We divide the problem into two cases:

Case 1: $s\Delta +s+b^{2}Q=0$. By Lemma 4, we have

$$\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}},$$

where $k_1=k_1\left(b^2\right)$ and $k_2=k_2\left(b^2\right)\ge 0$ are any differentiable functions. By direct calculations, we have $T_1,T_2,T_3,T_4,T_5$ all equal to zero. In this case, $\left(3\right)$ holds for any conformal factor $\sigma$.

Case 2: $s\Delta +s+b^{2}Q\ne 0$. It implies that $T_5\ne 0$.

Case 2-1: ${\overline{\sigma }}_0\ne 0$. Differentiating $\left(7\right)$ with respect to $y^b$ and $y^c$ yields

$$({\sigma }_b{\delta }_{ac}+{\sigma }_c{\delta }_{ab})T_1+2\frac{b^2}{b^2-s^2}{\sigma }_a{\delta }_{bc}{T}_5=0.$$

(8)

Contracting it with ${\delta }^{bc}$ yields

$$[T_1+(n-1)\frac{b^2}{b^2-s^2}{T}_5]{\sigma }_a=0.$$

(9)

On the other hand, contracting $\left(8\right)$ with ${\delta }^{ac}$ yields

$$\left(n{T}_1+\frac{b^2}{b^2-s^2}{T}_5\right){\sigma }_b=0.$$

(10)

For ${\overline{\sigma }}_0\ne 0$, by $\left(9\right)$ and $\left(10\right)$, we obtain $T_1=0$ and $T_5=0$. This contradicts $T_5\ne 0$. Thus, it is discarded.

Case 2-2: ${\overline{\sigma }}_0=0$, ${\sigma }_1\ne 0$. In this case, $\left(5\right)$ and $\left(7\right)$ hold constantly. By $\left(6\right)$, we can obtain $s{T}_1+b^2{T}_2=0$. Then, by Lemma 5, $\varphi$ satisfies

$$\mathsf{\Phi }=\frac{k_3{\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}},$$

(11)

where $k_3=k_3\left(b^2\right)$ is a differentiable positive function.

Substituting $\left(11\right)$ into $\left(3\right)$ yields

$$[(\alpha s{\sigma }_b-b^2{\sigma }_0)(s{y}_i-\alpha b_i)+\alpha (b^2-s^2)(\alpha s{\sigma }_i-{\sigma }_0{b}_i)]\frac{s\Delta +s+b^{2}Q}{\Delta (b^2-s^2)}=0.$$

Since $s\Delta +s+b^{2}Q\ne 0$,

$$(\alpha s{\sigma }_b-b^2{\sigma }_0)(s{y}_i-\alpha b_i)+\alpha (b^2-s^2)(\alpha s{\sigma }_i-{\sigma }_0{b}_i)=0.$$

Differentiating the above formula with respect to $y^j$ yields

$$2y_j(b^2{\sigma }_i-{\sigma }_b{b}_i)-2\alpha s{\sigma }_i{b}_j+b_j({\sigma }_0{b}_i+{\sigma }_b{y}_i)+\alpha s({\sigma }_j{b}_i+{\sigma }_b{a}_{ij})-b^2({\sigma }_j{y}_i+{\sigma }_0{a}_{ij})=0.$$

Contracting it with $a^{ij}$ yields

$$(n-2)(b^2{\sigma }_0-{\sigma }_b\beta )=0.$$

Because $n\ge 3$, it means that ${\sigma }_i\left(x\right)$ is proportional to $b_i\left(x\right)$, i.e., ${\sigma }_i=\frac{{\sigma }_b}{b^2}{b}_i$.

Case 2-3: ${\overline{\sigma }}_0=0$, ${\sigma }_1=0$. Then, conformal transformations between F and $\overline{F}$ are homothetic. □

Remark 1.

Note that $\varphi =1+s$ or $\varphi =1+s^2$ does not satisfy $\left(2\right)$. Thus, by the definition of general $(\alpha ,\beta )$-metrics and Theorem 1, conformal transformations that preserve the mean Landsberg curvature of Randers metrics $F=\alpha +\beta$ or square metrics $F=\frac{{(\alpha +\beta )}^2}{\alpha }$ are homothetic.

Remark 2.

Let $Q={\sum }_{i=0}^k{f}_i\left(b^2\right)s^i+g_1\left(b^2\right){(b^2-s^2)}^{\frac{1}{2}}+g_2\left(b^2\right){(b^2-s^2)}^{\frac{2m+1}{2}}$, where $k(\ge 0)$ and $m(\ge 1)$ are integers. If Q satisfies $\left(2\right)$, then $Q=f_1\left(b^2\right)s+g_1\left(b^2\right){(b^2-s^2)}^{\frac{1}{2}}$. If ${\sigma }_i=\frac{{\sigma }_b}{b^2}{b}_i$, then F and $\tilde{F}=e^{\sigma \left(x\right)}F$ have the same mean Landsberg curvature by Theorem 1.

Before proving Theorem 2, we need the following Lemma.

Lemma 6

([4]). Let the $(\alpha ,\beta )$-metric $F=\alpha \varphi \left(s\right)$ be a regular Finsler metric on an $n(\ge 3)$-dimensional manifold M. If ϕ satisfies

$$\mathsf{\Phi }=\frac{\lambda {\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}},$$

where λ is a constant, then F is Riemannian.

Remark 3.

When $\lambda =\lambda \left(b^2\right)$, the conclusion is still right.

Based on Lemma 6, we now give the proof of Theorem 2.

The Proof of Theorem 2. 

By Theorem 1, we divide the problem into two cases. If $\varphi =k_2{s}^{1-\frac{1}{b^2{k}_1}}{(b^2-s^2)}^{\frac{1}{2b^2{k}_1}}$, the general $(\alpha ,\beta )$-metric $F=\alpha \varphi (b^2,s)$ constructed by $\varphi$, is non-regular. We do not consider this case. If $\varphi$ satisfies $\mathsf{\Phi }=\frac{k_3\left(b^2\right){\Delta }^{\frac{3}{2}}}{\sqrt{b^2-s^2}}$, then $F=\alpha \varphi (b^2,s)$ is Riemannian by Lemma 6. □

4. Conclusions

In this paper, we study conformal transformations of general $(\alpha ,\beta )$-metrics preserving the mean Landsberg curvature. We obtain the necessity and sufficiency conditions for the mean Landsberg curvature and a rigidity theorem for the regular general $(\alpha ,\beta )$-metric case. The characterization equations for the general $(\alpha ,\beta )$-metrics with the mean Landsberg curvature are not yet completely solved, and only formal solutions are obtained.