*(α, β)-metrics and Kropina metrics*

### Feng Mu and Xinyue Cheng

Abstract. In this paper, we get the necessary and sufficient conditions
*that the (α, β)-metrics in the form F = α + ²β + kα*^{2}*/β are projectively*
equivalent to a Kropina metric. More generally, we characterize the pro-
*jective equivalence between an (α, β)-metric and a Kropina metric.*

M.S.C. 2010: 53B40, 53C60.

*Key words: (α, β)-metric; Kropina metric; projective equivalence; Douglas metric.*

### 1 Introduction

In Finsler geometry, it is an important topic to study projectively equivalent Finsler
*metrics on a manifold. Two Finsler metrics F and ¯F on a manifold M are said to*
be projectively equivalent if they have the same geodesics as point sets. It is well-
*known that two Finsler metrics F and ¯F are projectively equivalent if and only if*
their geodesic coefficients have the following relation

*G** ^{i}*= ¯

*G*

^{i}*+ P (x, y)y*

^{i}*,*

*where P (x, y) is a scalar function on T M \ {0} with P (x, λy) = λP (x, y), λ > 0.*

*(α, β)-metrics form a special and very important class of Finsler metrics which can*
*be expressed in the form F = αφ(s), s = β/α, where α is a Riemannian metric and β is*
*a 1-form and φ is a C** ^{∞}*positive function on the definition domain. In particular, when

*φ = 1/s, the Finsler metric F = α*

^{2}

*/β is called Kropina metric. Kropina metrics were*first introduced by L. Berwald in connection with a two-dimensional Finsler space with rectilinear extremal and were investigated by V. K. Kropina [6]. They together with Randers metrics are C-reducible [9]. However, Randers metrics are regular Finsler metrics but Kropina metrics are non-regular Finsler metrics. Kropina metrics seem to be among the simplest nontrivial Finsler metrics with many interesting application in physics, electron optics with a magnetic field, dissipative mechanics and irreversible thermodynamics (see [5][12]). Also, they have interesting applications in relativistic field theory, control theory, evolution and developmental biology.

Based on Stavrinos’ work on Finslerian structure of anisotropic gravitational field [13], we know that the anisotropy is an intrinsic issue of the background radiation,

### D

ifferential Geometry - Dynamical Systems, Vol.14, 2012, pp. 105-116.*° Balkan Society of Geometers, Geometry Balkan Press 2012.*c

*for all possible (α, β)-metrics. Then the 1-form β represents the same direction of the*
*observed anisotropy of the microwave background radiation. That is, if two (α, β)-*
*metrics F = αφ(β/α) and ¯F = ¯αψ( ¯β/¯α) are the same anisotropy directions (or,*
*they have the same axis rotation to their indicatrices), then their 1-form β and ¯β are*
*collinear, i.e, there is a function µ ∈ C*^{∞}*(M ) such that β(x, y) = µ(x) ¯β(x, y).*

In [10] , the projectively related Randers metrics are studied. It has been shown
that two Randers metrics are projectively equivalent if and only if they have the
same Douglas tensors and the corresponding Riemannian metrics are projectively
related [10]. The purpose of this paper is to study the projective equivalence between
*an (α, β)-metric and Kropina metric ¯F = ¯α*^{2}*/ ¯β. We will first prove the following*
theorem.

*Theorem 1.1. Let F = α + ²β + k*^{α}_{β}^{2} *(² and k 6= 0 are constant) be an (α, β)-metric*
*and ¯F = ¯α*^{2}*/ ¯β be a Kropina metric on an n-dimensional manifold M (n ≥ 3), where*
*α and ¯α are two Riemannian metrics, β and ¯β are two nonzero collinear 1-forms.*

*Then F is projectively equivalent to ¯F if and only if they are Douglas metrics and the*
*geodesic coefficients of α and ¯α have the following relation:*

(1.1) *G*^{i}_{α}*+ 2kτ α*^{2}*b** ^{i}*= ¯

*G*

^{i}

_{α}_{¯}+ 1

*2¯b*

^{2}

³

¯

*α*^{2}¯*s** ^{i}*+ ¯

*r*00

*¯b*

*´*

^{i}*+ θy*

^{i}*,*

*where b*^{i}*:= a*^{ij}*b**j**, ¯b*^{i}*:= ¯a*^{ij}*¯b**j**, ¯b*^{2} *:= k ¯βk*^{2}_{α}_{¯} *and τ = τ (x) is a scalar function and*
*θ = θ**i**y*^{i}*is a 1-form on M.*

By [7] and [8], we obtain immediately from Theorem 1.1 that

*Proposition 1.2. Let F = α + ²β + k*^{α}_{β}^{2} *(² and k 6= 0 are constant) be an (α, β)-*
*metric and ¯F = ¯α*^{2}*/ ¯β be a Kropina metric on an n-dimensional manifold M (n ≥ 3),*
*where α and ¯α are two Riemannian metrics, β and ¯β are two nonzero collinear 1-*
*forms. Then F is projectively equivalent to ¯F if and only if the following equations*
*hold*

(1.2) *G*^{i}_{α}*+ 2kτ α*^{2}*b** ^{i}*= ¯

*G*

^{i}

_{α}_{¯}+ 1

*2¯b*^{2}(¯*α*^{2}¯*s** ^{i}*+ ¯

*r*00

*¯b*

^{i}*) + θy*

^{i}*,*

(1.3) *b**i|j**= 2τ*£

*(1 + 2kb*^{2}*)a**ij**− 3kb**i**b**j*

¤*,*

(1.4) *s*¯*ij* = 1

*¯b*^{2}*(¯b**i**s*¯*j**− ¯b**j*¯*s**i**),*

*where b*_{i|j}*denote the coefficients of the covariant derivatives of β with respect to α.*

*Further, for the projective equivalence between a general (α, β)-metric and a*
Kropina metric, we have the following theorem.

*Theorem 1.3. Let F = αφ(*^{β}_{α}*) be an (α, β)-metric on an n-dimensional manifold*
*M (n ≥ 3) satisfying that β is not parallel with respect to α, db 6= 0 everywhere or*
*b=constant and F is not of Randers type. Let ¯F = ¯α*^{2}*/ ¯β be a Kropina metric on the*

*manifold M , where ¯α = λ(x)α, ¯β = µ(x)β. Then F is projectively equivalent to ¯F if*
*and only if the following equations hold*

(1.5) h

*1 + (k*1*+ k*2*s*^{2}*)s*^{2}*+ k*3*s*^{2}i

*φ*^{00}*= (k*1*+ k*2*s*^{2}*)(φ − sφ*^{0}*),*

(1.6) *G*^{i}* _{α}*= ¯

*G*

^{i}

_{α}_{¯}

*+ θy*

^{i}*− σ(k*1

*α*

^{2}

*+ k*2

*β*

^{2}

*)b*

^{i}*,*

(1.7) *b**i|j**= 2σ*£

*(1 + k*1*b*^{2}*)a**ij**+ (k*2*b*^{2}*+ k*3*)b**i**b**j*

¤*,*

(1.8) *s*¯*ij* = 1

*¯b*^{2}

¡*¯b**i**s*¯*j**− ¯b**j*¯*s**i*

¢*,*

*where σ = σ(x) is a scalar function and θ is a 1-form, k*_{1}*, k*_{2}*, k*_{3} *are constants. In*
*this case, F = αφ(*^{β}_{α}*) and ¯F = ¯α*^{2}*/ ¯β are both Douglas metrics.*

### 2 Preliminaries

*For a given Finsler metric F = F (x, y), the geodesics of F are characterized locally*
by a system of 2nd ODEs as follows ([4]),

*d*^{2}*x*^{i}*dt*^{2} *+ 2G*^{i}

³
*x,dx*

*dt*

´

*= 0,*
*where G*^{i}*= G*^{i}*(x, y) are defined by*

*G** ^{i}*= 1
4

*g*

*©*

^{il}*[F*^{2}]_{x}*m**y*^{l}*y*^{m}*− [F*^{2}]_{x}*l*

ª*.*

*We call G*^{i}*the geodesic coefficients of F .*
Definition 2.1. Let

(2.1) *D*_{jkl}* ^{i}* :=

*∂*

^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
µ

*G*^{i}*−* 1
*n + 1*

*∂G*^{m}

*∂y*^{m}*y*^{i}

¶
*,*

*The tensor D := D*_{jkl}^{i}*∂**i**⊗ dx*^{j}*⊗ dx*^{k}*⊗ dx** ^{l}* is called the Douglas tensor. A Finsler
metric is called the Douglas metric if the Douglas tensor vanishes.

We can check easily that the Douglas tensor is a projectively invariant. A funda- mental fact is that all Berwald metrics must be Douglas metrics.

*Given an (α, β)-metric*

*F = αφ(s), s =* *β*
*α,*
*where α =*p

*a**ij**y*^{i}*y*^{j}*is a Riemannian metric and β = b**i**(x)y*^{i}*is a 1-form with kβk**α**<*

*b*0*. It is known that F = αφ(β/α) is a regular Finsler metric if and only if the function*
*φ(s) is a positive C*^{∞}*function on an open interval (−b*0*, b*0) satisfying ([1][4])

*φ(s) − sφ*^{0}*(s) + (b*^{2}*− s*^{2}*)φ*^{00}*(s) > 0, |s| ≤ b < b*0*.*

*For Kropina metric F = α*^{2}*/β, it is easy to see that it is not a regular (α, β)-metric*
*but the relation φ(s) − sφ*^{0}*(s) + (b*^{2}*− s*^{2}*)φ*^{00}*(s) > 0 is still true for |s| > 0.*

*Let G*^{i}*(x, y) and G*^{i}_{α}*(x, y) denote the geodesic coefficients of F and α, respectively.*

*Let ∇β = b**i|j**dx*^{i}*⊗ dx*^{j}*denote the covariant derivative of β with respect to α. Define*
*r**ij*:= 1

2*(b*_{i|j}*+ b*_{j|i}*), s**ij* :=1

2*(b*_{i|j}*− b*_{j|i}*), r**i**:= r**ij**b*^{j}

*and put r*00 *:= r**ij**y*^{i}*y*^{j}*, r*0 *:= r**j**y*^{j}*, s**l0* *:= s**li**y*^{i}*, s*0 *:= b*^{l}*s**l0*, etc.. We have the
*following formula for the geodesic coefficients G*^{i}*(x, y) of F ([11])*

(2.2) *G*^{i}*= G*^{i}_{α}*+ αQs*^{i}_{0}+ Θn

*− 2αQs*_{0}*+ r*_{00}*o y*^{i}*α* + Ψ©

*− 2αQs*_{0}*+ r*_{00}ª
*b*^{i}*,*
*where s*^{i}_{j}*:= a*^{ik}*s**kj*and

*Q =* *φ*^{0}*φ − sφ*^{0}*,*

Θ = *φφ*^{0}*− s(φφ*^{00}*+ φ*^{0}*φ** ^{0}*)

*2φ((φ − sφ*

^{0}*) + (b*

^{2}

*− s*

^{2}

*)φ*

*)*

^{00}*,*

Ψ = *φ*^{00}

2£

*(φ − sφ*^{0}*) + (b*^{2}*− s*^{2}*)φ*^{00}*¤ .*

*In [7], the authors characterized the (α, β)-metrics of Douglas type.*

*Lemma 2.1. ([7]) Let F = αφ(*^{β}_{α}*) be a regular (α, β)-metric on an n-dimensional*
*manifold M (n ≥ 3). Assume that β is not parallel with respect to α and db 6= 0*
*everywhere or b=constant, and F is not of Randers type. Then F is a Douglas metric*
*if and only if the function φ = φ(s) with φ(0) = 1 satisfies following ODE:*

(2.3) £

*1 + (k*1*+ k*2*s*^{2}*)s*^{2}*+ k*3*s*^{2}¤

*φ*^{00}*= (k*1*+ k*2*s*^{2}*)(φ − sφ** ^{0}*)

*and β satisfies*

(2.4) *b**i|j**= 2σ[(1 + k*1*b*^{2}*)a**ij**+ (k*2*b*^{2}*+ k*3*)b**i**b**j**],*

*where b*^{2}*:= kβk*^{2}_{α}*and σ = σ(x) is a scalar function and k*1*, k*2*, k*3 *are constants with*
*(k*2*, k*3*) 6= (0, 0).*

For Kropina metrics, we have the following

*Lemma 2.2. ([8]) Let F = α*^{2}*/β be a Kropina metric on an n-dimensional manifold*
*M . Then*

*(1) (n ≥ 3) Kropina metric F with b*^{2}*6= 0 is a Douglas metric if and only if*

(2.5) *s**ik*= 1

*b*^{2}

¡*b**i**s**k**− b**k**s**i*

¢*.*

*(2) (n = 2) Kropina metric F is a Douglas metric.*

### 3 The proof of Theorem 1.1

*We will first compute the Douglas tensor of (α, β)-metrics. Let*
*G*˜^{i}*= G*^{i}_{α}*+ αQs*^{i}_{0}*+ Ψ{−2αQs*0*+ r*00*}b*^{i}*.*
Then (2.2) becomes

*G** ^{i}*= ˜

*G*

^{i}*+ Θ{−2Qαs*0

*+ r*00

*}α*

^{−1}*y*

^{i}*.*

*Clearly, the sprays G** ^{i}* and ˜

*G*

*are projectively equivalent. Then they have the same Douglas tensor. Denote*

^{i}(3.1) *T*^{i}*:= αQs*^{i}_{0}*+ Ψ{−2Qαs*0*+ r*00*}b*^{i}*.*
Then ˜*G*^{i}*= G*^{i}_{α}*+ T** ^{i}*. By (2.1), we have

*D*^{i}* _{jkl}* =

*D*˜

_{jkl}

^{i}= *∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y*^{l}

³

*G*^{i}_{α}*−* 1
*n + 1*

*∂G*^{m}_{α}

*∂y*^{m}*y*^{i}*+ T*^{i}*−* 1
*n + 1*

*∂T*^{m}

*∂y*^{m}*y*^{i}

´

= *∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
µ

*T*^{i}*−* 1
*n + 1*

*∂T*^{m}

*∂y*^{m}*y*^{i}

¶
*.*
(3.2)

By (3.1), we have

*∂T*^{m}

*∂y*^{m}*= Q*^{0}*s*0*+ 2Ψ[r*0*− Qss*0*− Q*^{0}*(b*^{2}*− s*^{2}*)s*0]

*−Ψ*^{0}*α*^{−1}*(b*^{2}*− s*^{2}*)[2Qαs*0*− r*00*].*

(3.3)

*Thus, if F and ¯F have the same Douglas tensor, i.e. D*^{i}* _{jkl}*= ¯

*D*

^{i}*, by (3.2), we have*

_{jkl}*∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
h

*T*^{i}*− ¯T*^{i}*−* 1
*n + 1*

¡*T*_{y}^{m}*m**− ¯T*_{y}^{m}*m*

¢*y** ^{i}*
i

*= 0.*

*Then there exists a class of scalar functions H*_{jk}^{i}*:= H*_{jk}^{i}*(x) such that*

(3.4) *T*^{i}*− ¯T*^{i}*−* 1

*n + 1*

¡*T*_{y}^{m}*m**− ¯T*_{y}^{m}*m*

¢*y*^{i}*= H*_{00}^{i}*,*

*where H*_{00}^{i}*:= H*_{jk}^{i}*y*^{j}*y** ^{k}*.

*Now, we consider (α, β)-metrics in the form F = α + ²β + k*^{α}_{β}^{2}*. Let b*0*= b*0*(k) > 0*
*be the largest number such that 1 + 2kb*^{2}*− 3ks*^{2} *> 0, |s| ≤ b < b*0*. Then F =*
*α + ²β + k*^{α}_{β}^{2} *is a Finsler metric if and only if β satisfies b := kβk**α**< b*0. By (2.2),
*the geodesic coefficients of F are determined by*

*Q =* *² + 2ks*
*1 − ks*^{2}*,*

Θ = *² − 3k²s*^{2}*− 4k*^{2}*s*^{3}

*2(1 + 2kb*^{2}*− 3ks*^{2}*)(1 + ²s + ks*^{2})*,*

Ψ = *k*

*1 + 2kb*^{2}*− 3ks*^{2}*.*
(3.5)

For Kropina metric ¯*F = ¯α*^{2}*/ ¯β, the geodesic coefficients are given by (2.2) with*

(3.6) *Q = −*¯ 1

*2s, ¯Θ = −s*

*¯b*^{2}*, ¯*Ψ = 1
*2¯b*^{2}*.*

*Lemma 3.1. Let F = α + ²β + k*^{α}_{β}^{2} *(² and k 6= 0 are constant) be an (α, β)-metric*
*and ¯F = ¯α*^{2}*/ ¯β be a Kropina metric on an n-dimensional manifold M (n ≥ 2), where*
*α and ¯α are Riemannian metrics, β and ¯β are two nonzero collinear 1-forms. Then*
*F and ¯F have the same Douglas tensor if and only if they are all Douglas metrics.*

*Proof. The sufficiency is obvious. Suppose that F and ¯F have the same Douglas*
*tensor on an n-dimensional manifold M when n ≥ 3. Then (3.4) holds. Plugging*
(3.5) and (3.6) into (3.4), we obtain

(3.7) *A*^{i}*α*^{7}*+ B*^{i}*α*^{6}*+ C*^{i}*α*^{5}*+ D*^{i}*α*^{4}*+ E*^{i}*α*^{3}*+ F*^{i}*α*^{2}*+ H*^{i}

*Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L* +*A*¯^{i}*α*¯^{2}+ ¯*B*^{i}

*2¯b*^{2}*β*¯ *= H*_{00}^{i}*,*
where

*A*^{i}*= ²(2kb*^{2}*+ 1)(2kb*^{2}*s*^{i}_{0}*− 2ks*0*b*^{i}*+ s*^{i}_{0}*),*

*B*^{i}*= k(2kb*^{2}*+ 1)(4kb*^{2}*βs*^{i}_{0}*− 4kβs*0*b*^{i}*+ r*00*b*^{i}*− 2λs*0*y*^{i}*− 2λr*0*y*^{i}*+ 2βs*^{i}_{0}*),*
*C*^{i}*= 6k²β(−2kb*^{2}*βs*^{i}_{0}*+ kβs*0*b*^{i}*+ 2kb*^{2}*µs*0*y*^{i}*− βs*^{i}_{0}*),*

*D*^{i}*= 2k*^{2}*β(6kβ*^{2}*s*0*b*^{i}*− 12kb*^{2}*β*^{2}*s*^{i}_{0}*− kβb*^{2}*r*00*b*^{i}*+ 2kβb*^{2}*λr*0*y*^{i}

*+14kβλb*^{2}*s*0*y*^{i}*− 6β*^{2}*s*^{i}_{0}*+ 4βλr*0*y*^{i}*− 2βr*00*b*^{i}*+ 4λβs*0*y*^{i}*− 3λb*^{2}*r*00*y*^{i}*),*
*E*^{i}*= 3k*^{2}*²β*^{3}*(3βs*^{i}_{0}*− 4λs*0*y*^{i}*),*

*F*^{i}*= 3k*^{2}*β*^{3}*(6kβ*^{2}*s*^{i}_{0}*− 2kλβr*_{0}*y*^{i}*+ kβr*_{00}*b*^{i}*− 10kλβs*_{0}*y*^{i}*+ 2kb*^{2}*λr*_{00}*y*^{i}*+ 2λr*_{00}*y*^{i}*),*
*H*^{i}*= −6k*^{3}*λβ*^{5}*r*00*y*^{i}

and

*I* *= (2kb*^{2}+ 1)^{2}*,*

*J* *= −kβ*^{2}*(2kb*^{2}*+ 7)(2kb*^{2}*+ 1),*
*K* *= 3k*^{2}*β*^{4}*(4kb*^{2}*+ 5),*

*L = −9k*^{3}*β*^{6}*,*
and

*A*¯^{i}*= ¯b*^{2}*s*¯^{i}_{0}*− ¯b*^{i}*s*¯0*,*

*B*¯* ^{i}* =

*β[2λy*¯

*(¯*

^{i}*r*0+ ¯

*s*0

*) − ¯b*

^{i}*r*¯00

*].*

*Here λ :=*_{n+1}^{1} . Further, (3.7) is equivalent to

*(A*^{i}*α*^{7}*+ B*^{i}*α*^{6}*+ C*^{i}*α*^{5}*+ D*^{i}*α*^{4}*+ E*^{i}*α*^{3}*+ F*^{i}*α*^{2}*+ H*^{i}*)(2¯b*^{2}*β)*¯
+( ¯*A*^{i}*α*¯^{2}+ ¯*B*^{i}*)(Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L)*

*= H*_{00}^{i}*(2¯b*^{2}*β)(Iα*¯ ^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L).*

(3.8)

*Replacing y*^{i}*by −y** ^{i}* in (3.8) yields

*(−A*^{i}*α*^{7}*+ B*^{i}*α*^{6}*− C*^{i}*α*^{5}*+ D*^{i}*α*^{4}*− E*^{i}*α*^{3}*+ F*^{i}*α*^{2}*+ H*^{i}*)(−2¯b*^{2}*β)*¯

*−( ¯A*^{i}*α*¯^{2}+ ¯*B*^{i}*)(Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L)*

*= H*_{00}^{i}*(−2¯b*^{2}*β)(Iα*¯ ^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L).*

(3.9)

(3.8)+(3.9) yields

(3.10) *(A*^{i}*α*^{7}*+ C*^{i}*α*^{5}*+ E*^{i}*α*^{3}*)(¯b*^{2}*β) = 0.*¯
From (3.10), we have

*A*^{i}*α*^{7}*+ C*^{i}*α*^{5}*+ E*^{i}*α*^{3}*= 0.*

Therefore we conclude that (3.7) is equivalent to

(3.11) *B*^{i}*α*^{6}*+ D*^{i}*α*^{4}*+ F*^{i}*α*^{2}*+ H*^{i}

*Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L* +*A*¯^{i}*α*¯^{2}+ ¯*B*^{i}

*2¯b*^{2}*β*¯ *= H*_{00}^{i}*.*
(3.11) is equivalent to

*(B*^{i}*α*^{6}*+ D*^{i}*α*^{4}*+ F*^{i}*α*^{2}*+ H*^{i}*)(2¯b*^{2}*β) + ( ¯*¯ *A*^{i}*α*¯^{2}+ ¯*B*^{i}*)(Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L)*

*= H*_{00}^{i}*(2¯b*^{2}*β)(Iα*¯ ^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L).*

From this, we can see that ¯*A*^{i}*α*¯^{2}*(Iα*^{6}*+ Jα*^{4}*+ Kα*^{2}*+ L) can be devided by ¯β. Since*
*β = µ ¯β, then ¯A*^{i}*α*¯^{2}*Iα*^{6}can be devided by ¯*β. Because ¯β is prime with respect to α and*

¯

*α, therefore ¯A*^{i}*= ¯b*^{2}¯*s*^{i}_{0}*− ¯b*^{i}*s*¯0 can be devided by ¯*β. Hence, there is a scaler function*
*ϕ*^{i}*(x) such that*

(3.12) *¯b*^{2}*s*¯^{i}_{0}*− ¯b** ^{i}*¯

*s*0= ¯

*βϕ*

^{i}*.*

Contracting (3.12) by ¯*y**i* *:= ¯a**ij**y*^{j}*, we get that ϕ*^{i}*(x) = −¯s** ^{i}*. Then we have

(3.13) *s*¯*ij* = 1

*¯b*^{2}*(¯b**i**s*¯*j**− ¯b**j*¯*s**i**).*

Thus, by Lemma 2.2, ¯*F = ¯α*^{2}*/ ¯β is a Douglas metric. Since F and ¯F have the same*
Douglas tensor, both of them are Douglas metrics.

*When n = 2, ¯F = ¯α*^{2}*/ ¯β is a Douglas metric by Lemma 2.2. Thus F and ¯F having*
the same Douglas tensor means that thay are all Douglas metrics. This completes the

proof of Lemma 3.1. ¤

Now, we are in the position to prove Theorem 1.1.

*Proof of Theorem 1.1. First we proof the necessity. If F is projectively equivalent*
to ¯*F , they have the same Douglas tensor. By Lemma 3.1, we know that F and ¯F*
*are both Douglas metrics. By [7], we know that (α, β)-metric F = α + εβ + k*^{β}_{α}^{2} is a
Douglas metric if and only if

(3.14) *b*_{i|j}*= 2τ [(1 + 2kb*^{2}*)a**ij**− 3kb**i**b**j**],*

*where τ = τ (x) is a scalar function on M . In this case, β is closed. Plugging (3.14)*
and (3.5) into (2.2) yields

(3.15) *G*^{i}*= G*^{i}* _{α}*+

*εα*

^{3}

*− 3kεβ*

^{2}

*α − 4k*

^{2}

*β*

^{3}

*α*^{2}*+ εαβ + kβ*^{2} *τ y*^{i}*+ 2kτ α*^{2}*b*^{i}*.*
On the other hand, plugging (3.13) and (3.6) into (2.2) yields

(3.16) *G*¯* ^{i}*= ¯

*G*

^{i}

_{α}_{¯}

*−*1

*2¯b*

^{2}

·

*−¯α*^{2}¯*s** ^{i}*+ (2¯

*s*0

*y*

^{i}*− ¯r*00

*¯b*

*) + 2*

^{i}*r*¯00

*βy*¯

^{i}¯
*α*^{2}

¸
*.*

*By the projective equivalence of F and ¯F again, there is a scalar function P = P (x, y)*
*on T M \{0} such that*

(3.17) *G** ^{i}*= ¯

*G*

^{i}*+ P y*

^{i}*.*

From (3.15), (3.16) and (3.17), we have

·

*P −²α*^{3}*− 3k²β*^{2}*α − 4k*^{2}*β*^{3}
*α*^{2}*+ ²αβ + kβ*^{2} *τ −* 1

*¯b*^{2}

¡¯*s*0+¯*r*00*β*¯

¯
*α*^{2}

¢¸
*y*^{i}

*= G*^{i}_{α}*− G*^{i}_{α}_{¯}*+ 2kτ α*^{2}*b*^{i}*−* 1
*2¯b*^{2}

¡*α*¯^{2}¯*s** ^{i}*+ ¯

*r*00

*b*

*¢*

^{i}*.*(3.18)

*Note that the right side of (3.18) is a quadratic in y. Then there exists a 1-form*
*θ = θ**i**(x)y*^{i}*on M such that*

*P −²α*^{3}*− 3k²β*^{2}*α − 4k*^{2}*β*^{3}
*α*^{2}*+ ²αβ + kβ*^{2} *τ −* 1

*¯b*^{2}

³

¯

*s*0+*r*¯00*β*¯

¯
*α*^{2}

´

*= θ.*

Thus we have

(3.19) *G*^{i}_{α}*+ 2kτ α*^{2}*b** ^{i}*= ¯

*G*

^{i}

_{α}_{¯}+ 1

*2¯b*

^{2}

¡*α*¯^{2}¯*s** ^{i}*+ ¯

*r*00

*¯b*

*¢*

^{i}*+ θy*

^{i}*.*This completes the proof of the necessity.

Conversely, from (3.15), (3.16) and (1.1) we have

(3.20) *G** ^{i}*= ¯

*G*

*+*

^{i}·

*θ +εα*^{3}*− 3kεβ*^{2}*α − 4k*^{2}*β*^{3}
*α*^{2}*+ ²αβ + kβ*^{2} *τ +* 1

*¯b*^{2}(¯*s*0+*r*¯00*β*¯

¯
*α*^{2} )

¸
*y*^{i}*.*

*Thus F is projectively equivalent to ¯F . This completes the proof of Theorem 1.1. ¤*

### 4 The proof of Theorem 1.3

*It is known that a Finsler metric F on a manifold M is a Douglas metric if and only*
*if the geodesic coefficients of F satisfy the following equations([2][3])*

*G*^{i}*y*^{j}*− G*^{j}*y** ^{i}*=1

2(Γ^{i}_{kl}*y*^{j}*− Γ*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*,*

where Γ^{i}* _{kl}* = Γ

^{i}

_{kl}*(x) are scalar functions on M . From (2.2), it is easy to see that an*

*(α, β)-metric is a Douglas metric if and only if the following equations hold,*

*αQ(s*^{i}_{0}*y*^{j}*− s*^{j}_{0}*y*^{i}*) + Ψ(−2Qαs*0*+ r*00*)(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)

=1

2*(G*^{i}_{kl}*y*^{j}*− G*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*,*
(4.1)

*where G*^{i}* _{kl}* := Γ

^{i}

_{kl}*− γ*

_{kl}

^{i}*and γ*

_{kl}*:=*

^{i}

_{∂y}

^{∂}^{2}

*k*

^{G}*∂y*

^{i}

^{α}*. Using (4.1), we can get the equivalent*

^{l}*conditions that two (α, β)-metrics have the same Douglas tensor.*

*Lemma 4.1. Two (α, β)-metrics F = αφ(α/β) and ¯F = ¯α ¯φ( ¯β/¯α) have the same*
*Douglas tensor if and only if*

*αQ(s*^{i}_{0}*y*^{j}*− s*^{j}_{0}*y*^{i}*) + Ψ(−2Qαs*0*+ r*00*)(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)

*−*©

¯

*α ¯Q(¯s*^{i}_{0}*y*^{j}*− ¯s*^{j}_{0}*y** ^{i}*) + ¯

*Ψ(−2 ¯Q¯α¯s*0+ ¯

*r*00

*)(¯b*

^{i}*y*

^{j}*− ¯b*

^{j}*y*

*)ª*

^{i}=1

2*(G*^{i}_{kl}*y*^{j}*− G*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*.*
(4.2)

*where G*^{i}_{kl}*are scalar function on M .*

*Proof. Let H*^{i}*:= G*^{i}*− ¯G*^{i}*. The H*^{i}*define a spray. If F and ¯F have the same*
*Douglas tensor, then H** ^{i}* satisfy the following

*∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
µ

*H*^{i}*−* 1
*n + 1*

*∂H*^{m}

*∂y*^{m}*y*^{i}

¶

= *∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
µ

*G*^{i}*−* 1
*n + 1*

*∂G*^{m}

*∂y*^{m}*y*^{i}

¶

*−* *∂*^{3}

*∂y*^{j}*∂y*^{k}*∂y** ^{l}*
µ

*G*¯^{i}*−* 1
*n + 1*

*∂ ¯G*^{m}

*∂y*^{m}*y*^{i}

¶

*= 0.*

*This shows that H** ^{i}* satisfy the following
(4.3)

*H*

^{i}*y*

^{j}*− H*

^{j}*y*

*=1*

^{i}2(Γ^{i}_{kl}*y*^{j}*− Γ*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*.*
From (2.2), it is easy to see that (4.3) is equivalent to the following

*αQ(s*^{i}_{0}*y*^{j}*− s*^{j}_{0}*y*^{i}*) + Ψ(−2Qαs*0*+ r*00*)(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)

*−*©

¯

*α ¯Q(¯s*^{i}_{0}*y*^{j}*− ¯s*^{j}_{0}*y** ^{i}*) + ¯

*Ψ(−2 ¯Q¯α¯s*

_{0}+ ¯

*r*

_{00}

*)(¯b*

^{i}*y*

^{j}*− ¯b*

^{j}*y*

*)ª*

^{i}= 1 2

£(Γ^{i}_{kl}*− γ*_{kl}* ^{i}* + ¯

*γ*

_{kl}

^{i}*)y*

^{j}*− (Γ*

^{j}

_{kl}*− γ*

^{j}*+ ¯*

_{kl}*γ*

_{kl}

^{j}*)y*

*¤*

^{i}*y*

^{k}*y*

^{l}= 1

2*(G*^{i}_{kl}*y*^{j}*− G*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*.*

Conversely, if (4.2) hold, then, by (4.3), we know that the Douglas tensor defined
*by H** ^{i}* vanishes, that is,

_{∂y}*j*

*∂y*

^{∂}^{3}

^{k}*∂y*

^{l}³

*H*^{i}*−*_{n+1}^{1} ^{∂H}_{∂y}*m*^{m}*y*^{i}

´

*= 0. Thus F and ¯F have the*
same Douglas tensor. This completes the proof of Lemma 4.1. ¤
*Lemma 4.2. Let F = αφ(β/α) be an (α, β)-metric on an n-dimensional manifold*
*M (n ≥ 2). Let ¯F = ¯α*^{2}*/ ¯β be a Kropina metric on M , where ¯α = λ(x)α, ¯β = µ(x)β.*

*Then F and ¯F have the same Douglas tensor if and only if they are Douglas metrics.*

*Proof. The sufficiency is obvious. Suppose that F and ¯F have the same Douglas*
tensor. From (4.2), we get

*αQ(s*^{i}_{0}*y*^{j}*− s*^{j}_{0}*y*^{i}*) + Ψ(−2Qαs*0*+ r*00*)(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)

*−*©

*−* *α*¯

2¯*s*(¯*s*^{i}_{0}*y*^{j}*− ¯s*^{j}_{0}*y** ^{i}*) + 1

*2¯b*

^{2}

¡ 1

¯

*sα¯*¯*s*0+ ¯*r*00

¢*(¯b*^{i}*y*^{j}*− ¯b*^{j}*y** ^{i}*)ª

= 1

2*(G*^{i}_{kl}*y*^{j}*− G*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*.*
Because ¯*α = λ(x)α, ¯β = µ(x)β, we have*

*αQ(s*^{i}_{0}*y*^{j}*− s*^{j}_{0}*y*^{i}*) + Ψ(−2Qαs*0*+ r*00*)(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)

*−*©

*−λ(x)α*

2¯*s* (¯*s*^{i}_{0}*y*^{j}*− ¯s*^{j}_{0}*y** ^{i}*) +

*µ(x)*

*2¯b*

^{2}

¡ 1

¯

*sλ(x)α¯s*0+ ¯*r*00

¢*(b*^{i}*y*^{j}*− b*^{j}*y** ^{i}*)ª

=1

2*(G*^{i}_{kl}*y*^{j}*− G*^{j}_{kl}*y*^{i}*)y*^{k}*y*^{l}*.*
(4.4)

*In order to simplify (4.4), firstly, we take an orthonormal basis on T*_{x}*M with*
*respect to α so that*

*α =*qX

*(y** ^{i}*)

^{2}

*,*

*β = by*

^{1}

*,*

*where b = kβ**x**k**α**. Further we take the following coordinate transformation, ψ :*
*(s, u*^{A}*) → (y** ^{i}*):

*y*^{1}= *s*

*√b*^{2}*− s*^{2}*α,*˜ *y*^{A}*= u*^{A}*,*
where ˜*α =*pP_{n}

*A=2**(u** ^{A}*)

^{2}. Then

*α =* *b*

*√b*^{2}*− s*^{2}*α,*˜ *β =* *bs*

*√b*^{2}*− s*^{2}*α.*˜

Thus

*F = αφ(β/α) =* *bφ(s)*

*√b*^{2}*− s*^{2}*α*˜
In the following, our index conventions are always as follows:

*1 ≤ i, j, k, · · · ≤ n,* *2 ≤ A, B, C, · · · ≤ n.*

*Now, we are going to simplify (4.4). Let i = A and j = B, (4.4) is equivalent to*

©*Q(s*^{A}_{1}*y*^{B}*− s*^{B}_{1}*y*^{A}*) + λ*1

2¯*s*(¯*s*^{A}_{1}*y*^{B}*− ¯s*^{B}_{1}*y** ^{A}*)ª

*αy*

^{1}+©

*Q(˜s*^{A}_{0}*y*^{B}*− ˜s*^{B}_{0}*y*^{A}*) + λ*1

2¯*s*(˜¯*s*^{A}_{0}*y*^{B}*− ˜¯s*^{B}_{0}*y** ^{A}*)ª

*α*

=1 2

©( ˜*G*^{A}_{00}*y*^{B}*− ˜G*^{B}_{00}*y*^{A}*) + s*^{2}*α*˜^{2}*(G*^{A}_{11}*y*^{B}*− G*^{B}_{11}*y*^{A}*)y*^{1}*y*^{1}ª

+*s*
2

©( ˜*G*^{B}_{10}+ ˜*G*^{B}_{01}*)y*^{A}*− ( ˜G*^{A}_{10}+ ˜*G*^{A}_{01}*)y** ^{B}*)ª

*y*

^{1}

*.*

*Plugging y*^{1}=^{√}_{b}_{2}^{s}_{−s}_{2}*α and α =*˜ ^{√}_{b}_{2}^{b}_{−s}_{2}*α into above equation, we get*˜

*bs˜α*^{2}*Q(s*^{A}_{1}*y*^{B}*− s*^{B}_{1}*y*^{A}*) + λbs˜α*^{2} 1

2¯*s*(¯*s*^{A}_{1}*y*^{B}*− ¯s*^{B}_{1}*y** ^{A}*)

= 1 2

©( ˜*G*^{A}_{00}*y*^{B}*− ˜G*^{B}_{00}*y*^{A}*)(b*^{2}*− s*^{2}*) + s*^{2}*α*˜^{2}*(G*^{A}_{11}*y*^{B}*− G*^{B}_{11}*y** ^{A}*)ª

*,*(4.5)

*bQ(˜s*^{A}_{0}*y*^{B}*− ˜s*^{B}_{0}*y*^{A}*) + λb*1

2¯*s*(˜¯*s*^{A}_{0}*y*^{B}*− ˜¯s*^{B}_{0}*y** ^{A}*)

= *s*
2

©( ˜*G*^{B}_{10}+ ˜*G*^{B}_{01}*)y*^{A}*− ( ˜G*^{A}_{10}+ ˜*G*^{A}_{01}*)y** ^{B}*)ª

*,*(4.6)

where ˜*G*^{A}_{10}*:= G*^{A}_{1B}*y*^{B}*, ˜G*^{A}_{00}:= ˜*G*^{A}_{BC}*y*^{B}*y*^{C}*, ˜¯s*^{A}_{0}:= ¯*s*^{A}_{B}*y** ^{B}*, etc.

Since ¯*β = µ(x)β, we have ¯b**A*= 0, and then, ¯*s**A**= ¯b*^{1}¯*s**1A*. Thus we have

(4.7) *s*¯*1A*= 1

*¯b*^{2}*(¯b*1¯*s**A**− ¯b**A*¯*s*1*).*

*On the other hand, plugging s = β/α and ¯s = ¯β/¯α into (4.6) and by use of*

¯

*α = λ(x)α, ¯β = µ(x)β, we get*

*2µαβbQ(˜s*^{A}_{0}*y*^{B}*− ˜s*^{B}_{0}*y*^{A}*) + λ*^{2}*α*^{2}*b(˜¯s*^{A}_{0}*y*^{B}*− ˜¯s*^{B}_{0}*y** ^{A}*)

*= µβ*^{2}©

( ˜*G*^{B}_{10}+ ˜*G*^{B}_{01}*)y*^{A}*− ( ˜G*^{A}_{10}+ ˜*G*^{A}_{01}*)y** ^{B}*)ª

*.*(4.8)

From (4.8), we first obtain ˜*s*^{A}_{0}*y*^{B}*− ˜s*^{B}_{0}*y*^{A}*= 0. Further, because α*^{2} *and β*^{2} are
*relatively prime polynominals of (y** ^{i}*), we can get

(4.9) *˜¯s*^{A}_{0}*y*^{B}*− ˜¯s*^{B}_{0}*y*^{A}*= 0.*

*Differenting (4.9) with y*^{C}*, and then letting B = C, we can get n¯s*^{A}_{B}*y** ^{B}* = 0, which
implies that ¯

*s*

*AB*= 0.

Therefore, together with (4.7), we have proved

¯
*s**ij* = 1

*¯b*^{2}*(¯b**i**s*¯*j**− ¯b**j*¯*s**i**).*

Thus, by Lemma 2.2, ¯*F = ¯α*^{2}*/ ¯β is a Douglas metric. Since F and ¯F have the same*

Douglas tensor, both of them are Douglas metrics. ¤

Now, we are in the position to prove Theorem 1.3.

*Proof of Theorem 1.3. Suppose that F and ¯F are projectively equivalent. Then*
they have the same Douglas tensor. From Lemma 4.2, we know that they are both
Douglas metrics. It is known that Kropia metric is a Douglas metric if and only if

¯

*s**ij*=_{¯}_{b}^{1}2*(¯b**i*¯*s**j**−¯b**j**s*¯*i**). Thus (1.8) holds. By Lemma 2.1, we know that the (α, β)-metric*
is a Douglas metric if and only if (1.5) and (1.7) hold. Plugging (1.5) and (1.7) into
(2.2) gives (1.6).

Conversely, if (1.5)-(1.8) hold, plugging them into (2.2), we can easily get the
*projectively equivalence of F and ¯F . This completes the proof.* ¤
Acknowledgements. This work was supported by National Natural Science
Fundation of China (10971239). The second author is the Corresponding Author of
this paper.

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*Authors’ addresses:*

Mu Feng

School of Mathematics and Statistics, Chongqing University of Technology, Chongqing 400054, P. R. China.

E-mail: mufeng612@163.com Xinyue Cheng

School of Mathematics and Statistics, Chongqing University of Technology, Chongqing 400054, P. R. China.

E-mail: chengxy@cqut.edu.cn