On weak Berwald (α,β) -metrics of scalar flag curvature

Journal of Geometry and Physics 86 (2014) 112–121

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On weak Berwald (α, β)-metrics of scalar flag curvature Guangzu Chen ∗ , Qun He, Shengliang Pan Department of Mathematics, Tongji University, Shanghai 200092, PR China

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Article history: Received 26 October 2013 Accepted 22 July 2014 Available online 30 July 2014

abstract In this paper, we study weak Berwald (α, β)-metrics of scalar flag curvature. We prove that this kind of (α, β)-metrics must be Berwald metric and their flag curvatures vanish. In this case, they are locally Minkowskian. © 2014 Elsevier B.V. All rights reserved.

MSC: 53B40 53C60 Keywords: (α, β)-metric Weak Berwald Finsler metric Flag curvature Berwald metric Minkowskian metric

1. Introduction It is well known that the spray coefficients Gi of a Riemannian metric are quadratic in y ∈ Tx M. It is a natural question whether or not there are non-Riemannian metrics whose spray coefficients Gi are quadratic in y. There are plenty of such Finsler metrics firstly investigated by L. Berwald. Thus we call Finsler metrics whose spray coefficients are quadratic in y Berwald metrics. Let F be a Finsler metric on an n-dimensional manifold M and Gi be the geodesic coefficients of F , consider the following quantity Bj ikl :=

∂ 3 Gi . ∂ ∂ yk ∂ yl yj

We obtain a well-defined tensor B := Bj ikl dxj ⊗ dxk ⊗ dxl ⊗ ∂i on TM /{0}. B is called Berwald curvature. It is clear that F is a Berwald metric if and only if B = 0. Define the mean Berwald curvature E := Eij dxi ⊗ dxj by Eij :=

1 2

Bmmkl .

F is called a weak Berwald metric if E = 0. It is obvious that Berwald metrics must be weak Berwald metrics but the reverse is not true. The flag curvature, a natural extension of the sectional curvature in Riemannian geometry, plays the central role in Finsler geometry. Generally, the flag curvature depends not only on the section but also on the flagpole. A Finsler metric is of scalar

∗

Corresponding author. Tel.: +86 13524345535. E-mail addresses: [email protected] (G. Chen), [email protected] (Q. He), [email protected] (S. Pan).

http://dx.doi.org/10.1016/j.geomphys.2014.07.031 0393-0440/© 2014 Elsevier B.V. All rights reserved.

G. Chen et al. / Journal of Geometry and Physics 86 (2014) 112–121

113

flag curvature if its flag curvature depends only on the flagpole. It is one of hot and difficult problems to characterize Finsler metrics with scalar flag curvature. In Finsler geometry, there is an important class of Finsler metrics expressed in the following form F = αφ(s),

s=

β , α

where α = aij yi yj is a Riemann metric and β = bi (x)yi is a 1-form with b := ∥βx ∥α = aij bi bj < b0 . It is proved that F = αφ(β/α) is a positive definite Finsler metric if and only if the function φ = φ(s) is a C ∞ positive function on an open interval (−b0 , b0 ) satisfying [1]





φ(s) − sφ ′ (s) + (b2 − s2 )φ ′′ (s) > 0, |s| ≤ b < b0 . Such a metric is called an (α, β)-metric. In  particular, when φ = 1 + s, Finsler metric F = α + β is Randers metric with 2 ∥β∥ α < 1. More generally, when φ = k1 1 + k2 s + k3 s, where k1 > 0, k2 and k3 ̸= 0 are constant, Finsler metrics F = 2 2 k1 α + k2 β + k3 β are called Finsler metrics of Randers type. Let rij :=

1 2

(bi|j + bj|i ),

sij :=

1 2

(bi|j − bj|i ),

bi := aij bj ,

si := bj sji ,

si j := ail slj ,

s0 := si yi ,

si 0 := si j yj ,

r00 := rij yi yj

ri := bl rli ,

where ‘‘|’’ denotes the horizontal covariant derivative with respect to α . In [2], Z. Shen shows that any regular (α, β)-metrics are Berwald metric if and only if β is parallel with respect to α , that is, bi|j = 0. However, how to characterize the weak Berwald (α, β)-metrics is still open. In 2009, X. Cheng and C. Xiang study a class of (α, β)-metrics in the form F = (α + β)m+1 /α m , where m ̸= −1, 0, −1/n and prove that F is weak Berwald if and only if it satisfies that β is a killing 1-form with constant length with respect to α , that is, rij = si = 0 [3]. In [4], X. Cheng and C. Lu show that two kinds of weak Berwald (α, β)-metrics in the form F = α + ϵβ + k(β 2 /α) (ϵ and k ̸= 0 are constants) and F = α 2 /(α − β) are weak Berwald if and only if they satisfy that β is a killing 1-form with constant length with respect to α . Further, they prove that the two kinds of weak Berwald (α, β)-metrics with scalar flag curvature must be locally Minkowskian. In this paper, we have Theorem  1.1. Let F = αφ(s), s = β/α be a non-Riemannian (α, β)-metric on a manifold M of dimension n ≥ 3. Suppose φ ̸= k1 1 + k2 s2 + k3 s for any constant k1 > 0, k2 and k3 ̸= 0. Then F is a weak Berwald metric if and only if β is a killing 1-form with constant length with respect to α . Further, we obtain Theorem 1.2. Let F = αφ(s ), s = β/α be a non-Riemannian (α, β)-metric of scalar flag curvature on a manifold M of dimension n ≥ 3. Suppose φ ̸= k1 1 + k2 s2 + k2 s for any constant k1 > 0, k2 and k3 ̸= 0 on M. Then F is a weak Berwald metric if and only if F is a Berwald metric and the flag curvature K = 0. In this case, F must be locally Minkowskian. 2. Preliminaries For a given Finsler F = F (x, y), the geodesics of F are characterized locally by a system of 2nd ODEs as follows [1], d2 xi dt 2

  dx + 2Gi x, = 0, dt

where Gi =

1 4





g il [F 2 ]xm yl ym − [F 2 ]xl .

Gi are called the geodesic coefficients of F . There are many interesting non-Riemannian quantities in Finsler geometry. For a non-zero vector y ∈ Tp M, the Cartan torsion Cy = Cijk dxi ⊗ dxj ⊗ dxk : Tp M ⊗ TpM ⊗ TpM −→ R is defined by Cijk :=

1 4

[F 2 ]yi yj yk =

1 ∂ gij 2 ∂ yk

(x, y).

The mean Cartan torsion Iy = Ii (x, y)dxi : Tp M −→ R is defined by Ii := g jk Cijk , where (g ij ) := (gij )−1 and gij := 12 [F 2 ]yi yj . It is obvious that Cijk = 0 if and only if F is Riemannian. According to Deicke’s theorem [5], a Finsler metric is Riemannian if and only if the mean Cartan torsion vanishes.

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Express the volume form of F by dVF = σ (x)dx1 · · · dxn . For a non-zero vector y ∈ Tp M, the S-curvature S(y) is defined by S(y) :=

∂ Gi yi ∂σ (x, y) − (x). i ∂y σ (x) ∂ xi

From the definition, we have 1

Syi yj . 2 We say that F is of isotropic S-curvature if Eij =

(1)

S = (n + 1)cF , where c = c (x) is a scalar function on M. The S-curvature was first introduced by Z. Shen when he studied volume comparison. The S-curvature has important influence on the geometric structure of Finsler metrics. Lemma 2.1 ([6]). Let (M , F ) be an n-dimensional Finsler manifold of scalar flag curvature. Suppose that the S-curvature is isotropic, i.e., S = (n + 1)c (x)F . Then the flag curvature must be of the form 3cxm ym

K=

F

+ σ,

where σ = σ (x) is a scalar function on M. The Landsberg curvature L := Lijk dxi ⊗ dxj ⊗ dxk and the mean Landsberg curvature J := Ji dxi are defined respectively by 1 Lijk := − FFym Bi mjk , Ji := g jk Lijk . 2 Finsler metrics with (J = 0)L = 0 are called (weak)Landsberg metrics. Consider the (α, β)-metrics F = αφ(s), s = β/α on a manifold. Let Gi and Giα denote the spray coefficients of α and F , respectively, then we have [1] Gi = Giα + α Qsi 0 + {−2Q α s0 + r00 }{Ψ bi + Θ α −1 yi },

(2)

where Q :=

φ′ , φ − sφ ′

Θ :=

Q − sQ ′ 2∆

,

Ψ :=

Q′ 2∆

(3)

and ∆ := 1 + sQ + (b2 − s2 )Q ′ . Put yi := aij yj ,

hi := α bi − syi ,

Φ := −(Q − sQ ′ ){n∆ + 1 + sQ } − (b2 − s2 )(1 + sQ )Q ′′ , √ ′  2 − ss Φ b Φ 1 Ψ1 := b2 − ss ∆ 2 , Ψ2 := 2(n + 1)(Q − sQ ′ ) + 3 . 3 ∆ ∆2 By a direct computation, we obtain the following formula about the mean Cartan torsion of (α, β)-metrics [7] Ii := −

Φ (φ − sφ ′ ) hi . 2∆φα 2

(4)

By Deicke’s theorem, an (α, β)-metric is Riemannian if and only if Φ ≡ 0. In [8], X. Cheng and Z. Shen have obtained the formula of S-curvature for an (α, β)-metric



S = 2Ψ −

f ′ (b) bf (b)



(r0 + s0 ) −

Φ 2 α ∆2

(r00 − 2α Qs0 ),

where

π f (b) :=

0

sinn−2 tT (b cos t )dt

π 0

sinn−2 tdt

and T (s) := φ(φ − sφ ′ )n−2 [(φ − sφ ′ ) + (b2 − s2 )φ ′′ ].

(5)

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115

Further, the mean Landsberg curvature of an (α, β)-metric is given by [9]



2α 2

 Φ ′ Ji = − + (n + 1)(Q − sQ ) (r0 + s0 )hi 2∆α 4 b2 − s2 ∆   α Φ + 2 Ψ + s (r00 − 2α Qs0 )hi + α[−α Q ′ s0 hi + α Q (α 2 si − yi s0 ) 1 b − s2 ∆  Φ 2 2 + α ∆si0 + α (ri0 − 2α Qsi ) − (r00 − 2α Qs0 )yi ] . ∆ 1



(6)

Contracting Ji with bi , we obtain 1 J¯ := Ji bi = − {Ψ1 (r00 − 2α Qs0 ) + α Ψ2 (r0 + s0 )}. 2∆α 2 If Finsler metric F is of scalar flag curvature, we have the following Bianchi identity [1] 1 F 2 K·i , Ji;m ym + KF 2 Ii = − n+1 where ‘‘;’’ denotes the horizontal covariant derivative with respect to F and K·i = Kyi . For an (α, β)-metric F = αφ(s), s = β/α , it is easy to get [10,4]

∂(Gl − Glα ) i ∂ J¯ n+1 2 2 b − 2 l (Gl − Glα ) + Kα 2 φ 2 Ii bi = − α φ K·i bi , ∂ yi ∂y 3 where J¯|0 := J¯|m ym , rl0 := rlk yk and sl0 := slk yk . J¯|0 − Ji ail (rl0 + sl0 ) − Jl

(7)

3. Weak Berwald (α, β)-metrics Let F = αφ(s), s = β/α be an (α, β)-metric on a manifold M, where α = aij (x)yi yj is a Riemannian metric and β = bi (x)yi is a 1-form on M. Assuming that F is a weak Berwald metric, by (1), we have that its S-curvature satisfies S = ξ + κ , where ξ := ξi (x)yi is a 1-form and κ := κ(x) is a scalar function on manifold M. The definition of S-curvature implies that S(y) is positively homogeneous of degree one for any non-zero vector y ∈ Tx M, then S = ξi (x)yi . It follows from (5) that



f ′ (b)



2Ψ −



( r 0 + s0 ) −

Φ (r00 − 2α Qs0 ) = ξi (x)yi . 2α ∆2

bf (b) To simplify the computations, we take an orthonormal basis at x with respect to α such that

  n  α =  (yi )2 ,

β = by1 ,

i=1

and take the following coordinate transformation [8] in Tx M , ψ : (s, uA ) → (yi ): s α, ¯ y A = uA , y1 = √ 2 b − s2 where α¯ =

 n

i=2

(uA )2 . Here, our index conventions are

1 ≤ i, j, k, · · · ≤ n,

2 ≤ A, B, C , · · · ≤ n.

We have

α= √

b

b2

−

s2

α, ¯

β= √

bs b2

− s2

α. ¯

Further s1 = 0,

sA = bs1A ,

r1 = br11 ,

rA = br1A ,

s0 = si yi = sA yA = bs¯10 , sα¯ r0 = √ br11 + br¯10 , 2 b − s2 r00 =

s2 α¯ 2 r11 b2

−

s2

2sα¯ ¯ r10

+√

b2 − s2

+ r¯00 ,

where s¯10 :=

n  A=2

s1A yA ,

r¯10 :=

n  A=2

r1A yA ,

r¯00 :=

n  A,B=2

rAB yA yB .

(8)

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Then (8) is equivalent to the following two equations in the special coordinate system (s, ya )

   Φ sΦ 2 2 2 ¯ 2, ( b − s )¯ r = − s − 2 Ψ b r − bt 00 11 1 α 2∆2 2∆2   sΦ Φ 2 − 2Ψ b (r1A + s1A ) − (b2 Q + s) 2 s1A − btA = 0 ∆2 ∆ f ′ (b)

(9)

(10)

f ′ (b)

where t1 := − bf (b) r1 − ξ1 and tA := − bf (b) (rA + sA ) − ξA . Lemma 3.1. Let F = αφ(s), s = β/α be a non-Riemannian (α, β)-metric on a manifold M. Suppose that F is weak Berwald metric, then

Φ 2∆2

(k − ϵ s2 ) = 2Ψ s(k − ϵ b2 ) + sυ,

(11)

and in addition, if there exists A0 such that sA0 ̸= 0, then

−2Ψ −

QΦ

∆2

−λ



sΦ

− 2Ψ b2

∆2



= δ,

(12)

where λ, k, ϵ, υ and δ are scalar functions of x. Proof. Since Φ ̸= 0 and notice that r¯00 and α¯ are independent of s, then (9) implies that there is a function k = k(x) independent of s such that rAB = kδAB ,

 s

sΦ

(13)

− 2Ψ b2

2∆2



Φ r11 + k (b2 − s2 ) = bst1 . 2∆2

(14)

Let r11 = k − ϵ b2 ,

t1 = bυ

where ϵ = ϵ(x), υ = υ(x) are independent of s. Then we have (11) from (14). Differentiating (12) with respect to s yields d ds



sΦ

∆2

− 2Ψ b

2

 r1A −

d



ds

QΦ

∆2



+ 2Ψ b2 s1A = 0.

(15)

Let

λ := −

r1A0 b2 s1A0

.

Then it follows from (15) that

δ := −2Ψ − is independent of s.

QΦ

∆2

−λ



sΦ

∆2

− 2Ψ b2





Noting that b := ∥βx ∥α =



aij bi bj , we can obtain

∂b b bi|m r i + si = = . i ∂x b b i

The following lemma is obvious. Lemma 3.2. Let β be a 1-form on a Riemannian manifold (M , α). Then b := ∥βx ∥α = satisfies the following equation



aij bi bj = constant if and only if β

r i + si = 0 . Next we are going to show that b must be constant under the assumption in Theorem 1.1. Firstly, we prove Proposition 3.3. Let F = αφ(s), s = β/α be a non-Riemannian (α, β)-metric on a manifold M of dimension n ≥ 3. Suppose b ̸= constant and φ satisfies (12). Then F must be Finsler metric of Randers type. Proof. By assumption b ̸= constant, i.e, db ̸= 0, then b ̸= constant in a neighborhood. We view b as a variable in (12). Let

    sΦ QΦ 2 eq := ∆ λ − 2Ψ b + 2Ψ + 2 + δ . ∆2 ∆ 2

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117

Denote the Taylor series of Q (s) in a neighborhood of s = 0 Q (s) = a0 + a1 s + a2 s2 + a3 s3 + a4 s4 + a5 s5 + a6 s6 + a7 s7 + a8 s8 + o(s8 ). We claim a0 ̸= 0. Suppose that a0 = 0, if ai = 0, i ≥ 2, we have Q (s) = a1 s, that is

φ′ = a1 s . φ − sφ ′ Solving the above differential equation yields

 φ(s) = k1 1 + a1 s2 . Then F is Riemannian, it is a contraction. Thus there must be some minimal integer m(m ≥ 2) such that am ̸= 0. Expressing Q (s) as Q (s) = a1 s + am sm + am+1 sm+1 + am+2 sm+2 + o(sm+2 ).

(16)

Plugging (16) into eq1 and let qi be the coefficients of s in eq1. By q0 = (1 + a1 b )[−λa1 b + a1 + (1 + a1 b )δ] = 0, we have i

δ=

a1 (λb2 − 1) 1 + a1 b 2

2

2

2

.

Plugging it into qm−1 = 0 and qm+1 = 0, we can obtain P¯ λ + P¯ 1 = 0, T¯ λ + T¯1 = 0,

(17)

where P¯ , P¯ 1 , T¯ , T¯1 are polynomials in b. By maple program, we have P¯ T¯1 − T¯ P¯ 1 = (. . .)b8 + (. . .)b6 + (. . .)b4 + (. . .)b2 + a2m m(m − 1)(m + n + 1) where the omitted terms in the brackets of the above are all constants. Thus (17) implies λ = 0, then

δ=−

a1 1 + a1 b2

.

Plugging it and λ = 0 into qm−1 , we have qm−1 = −m[a1 b2 (m − 1) − 1]am . Noting that am ̸= 0, we get a contradiction. Thus a0 ̸= 0. By a direction computation, we have q0 = (1 + a1 b2 )2 δ − a1 b2 (1 + a1 b2 )λ + a1 − na20 + a21 b2 − 2a0 a2 b2 − na20 a1 b2 − a20 , q1 = 2(1 + a1 b2 )(a0 + 2a2 b2 )δ − (1 + a1 b2 )[(n + 1)a0 + 4a2 b2 ]λ

+ 2a2 − na0 a1 − 2a20 a2 b2 − (n + 1)a20 − na0 a21 b2 + 2a1 a2 b2 − 6a0 a3 b2 − 2na20 a2 b2 . By q0 = 0 and q1 = 0, we have

λ=

E1 E0

,

δ=

E2 E0

,

where E0 = (1 + a1 b2 )[(4a2 + (n − 1)a0 a1 )b2 + (n + 1)a0 ], E1 = [a0 a1 (2a0 a2 − a21 )n + 8a0 a22 − 2a20 a1 a2 − 6a0 a1 a3 − 2a21 a2 ]b4

+ a0 [(a20 a1 + 2a0 a2 − 2a21 )n − a20 a1 + 6a0 a2 − 2a21 − 6a3 ]b2 + a0 (a20 − a1 )n + a30 − 2a0 a1 + 2a2 ,

and E2 = [a0 a1 (2a0 a2 − a21 )n + 8a0 a22 − 2a20 a1 a2 − 6a0 a1 a3 − 2a21 a2 ]b4

+ [a30 a1 n2 + a0 (6a0 a2 − 2a21 )n − a30 a1 − a0 a21 + 6a20 a2 − 2a1 a2 ]b2 + a0 [a20 n2 + (220 − a1 )n + a20 − a1 ]. Substituting the Taylor series of Q (s) and (18) into eq, we have E0 (eq) = f2 s2 + f3 s3 + f4 s4 + · · · , where fi (i = 2, 3, 4, . . .) are polynomials of b. Express f2 in the following f2 = g0 + g2 b2 + g4 b4 + g6 b6 .

(18)

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G. Chen et al. / Journal of Geometry and Physics 86 (2014) 112–121

By f2 = 0, we have g0 = g2 = g4 = g6 = 0. Solving the linear system yields a2 =

1 2

a0 a1 ,

a3 = 0,

1 a4 = − a0 a21 8

(19)

or a2 =

1 2

a0 [(n + 1)a20 + a1 ],

1 a3 = − (n + 1)a40 , 3

1 a4 = − a0 [(n + 1)(n − 3)a40 + 2(n + 1)a20 a1 + a21 ]. 8

(20)

Case 1: (19) holds. Plugging (19) into (18), we have

λ = a20 − a1 ,

δ=

na20 + (a20 − a1 )(1 + a1 b2 ) 1 + a1 b2

.

Plugging them into eq, we get

(1 + a1 b2 )(eq) = Ξ0 + Ξ2 b2 + Ξ4 b4 , where Ξ0 , Ξ2 , Ξ4 are ODES about Q (s). By maple program, solving Ξ0 = 0, Ξ2 = 0 and Ξ4 = 0, we have Q ( s ) = a1 s ± a0



1 + a1 s 2 .

Noting that Q (s) =

φ′ , φ − sφ

we get the differential equation about φ(s)

 φ′ = a1 s ± a0 1 + a1 s 2 . φ − sφ Solving the above differential equation yields

φ(s) =



1 + a1 s2 ± a0 s.

Then F is Finsler metric of Randers type. Case 2: (20) holds. Plugging (20) into f3 , we have f3 = f30 + f32 b2 , where f30 = − 13 (n + 1)(n − 2)a50 . Since a0 ̸= 0 and n ≥ 3, it contradicts that f3 = 0.



Lemma 3.4. Let F = αφ(s), s = β/α be a non-Riemannian (α, β)-metric on a manifold M of dimension n ≥ 3. Suppose φ ̸=  k1 1 + k2 s2 + k2 s for any constant k1 > 0, k2 and k3 ̸= 0. If F is weak Berwald metric, then ri + si = 0. Proof. Suppose that r0 + s0 ̸= 0, by Lemma 3.2, then b ̸= constant in a neighborhood. We view b as a variable in (11). Put eq1 := 2∆

2



 Φ 2 2 (k − ϵ s ) − 2Ψ s(k − ϵ b ) − sυ . 2∆2

Noting that k, ϵ, υ may dependent on b, Plugging the Taylor series of Q (s) in a neighborhood of s = 0 Q (s) = a0 + a1 s + a2 s2 + a3 s3 + a4 s4 + a5 s5 + a6 s6 + a7 s7 + a8 s8 + o(s8 ) into eq1 and let pi be the coefficients of si in eq1, by a direction computation, we have p0 = −k[(na0 a1 + 2a2 )b2 + (n + 1)a0 ]. In the following we prove k = 0, ϵ = 0, υ = 0. Case 1: a0 ̸= 0. By p0 = 0, we have k = 0, then

  −a1 ϵ b2 + υ a1 b2 + υ = 0.

p1 = −2 a1 b2 + 1



The above equation implies

υ=

a1 b 2 ϵ 1 + a1 b 2

.

Then we get p2 = {[(n − 2)a0 a1 + 6a2 ]b2 + (n + 1)a0 }ϵ . By p2 = 0 and a0 ̸= 0, we have ϵ = 0. Thus k = ϵ = υ = 0.

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119

Case 2: a0 = 0. Since Q (s) ̸= a1 s, there must be some minimal integer m(m ≥ 2) such that am ̸= 0. When m = 2, we get p0 = −2a2 b2 k. By p0 = 0, we have k = 0. Noting that the expression of p1 do not contain a0 , we still have a1 b2 ϵ

υ=

1 + a1 b 2

.

Then by p2 = {[(n − 2)a0 a1 + 6a2 ]b2 + (n + 1)a0 }ϵ = 6a2 b2 ϵ = 0, we obtain k = ϵ = υ = 0. When m = 3, by a direction computation, we get p1 = 2a1 b2 (1 + a1 b2 )ϵ − 2[(a21 + 3a3 )b2 + a1 ]k − 2(1 + a1 b2 )2 υ , then p1 = 0 implies a1 b2 (1 + a1 b2 )ϵ − [(a21 + 3a3 )b2 + a1 ]k

υ=

. (1 + a1 b2 )2 Plugging (21) into p3 = 0 and p5 = 0, we have the following linear system about k and ϵ

(21)

Pk + P1 ϵ = 0, Tk + T1 ϵ = 0,

(22)

where P , P1 , T , T1 are polynomials in b. By maple program, we obtain

 P  T



P1  = Mb4 + Nb2 − 4(n + 4)(n + 1)a23 , T1 

where M = −(108n − 324)a23 − 600a25 + (52n − 140)a1 a3 a5 − 4(n − 3)(n − 5)a21 a23 + 504a3 a7 , and N = 52(n + 1)a3 a5 − 4(n + 1)(2n − 5)a1 a23 . It follows from (22) that k = ϵ = 0. Plugging it into (21), we get υ = 0. When m > 3, by (21) and a3 = 0, we have

υ=

a1 (−k + ϵ b2 ) 1 + a1 b 2

,

(23)

To prove k = ϵ = 0, we need to compute pm−2 and pm . Expressing Q (s) as Q (s) = a1 s + am sm + am+1 sm+1 + am+2 sm+2 + o(sm+2 ),

(24)

we have pm−2 = −m(m − 1)b am k. By pm−2 = 0, one gets k = 0. Plugging k = 0 into pm , we have pm = m(m + 1)b am ϵ = 0, i.e, ϵ = 0. Then (23) implies υ = 0. Now we claim that there exists A0 s.t sA0 ̸= 0. Noting that r11 = k − ϵ b2 and k = ϵ = 0, we have r1 + s1 = b(r11 + s11 ) = br11 = 0. By assumption that r0 + s0 ̸= 0, there exists A0 such that rA0 + sA0 ̸= 0 which implies rA0 ̸= 0 or sA0 ̸= 0. If sA0 = 0, we have r1A0 = (1/b)rA0 ̸= 0. By (10), the following holds 2



sΦ

∆2

− 2Ψ b2



= γ,

2

(25)

where γ := btA0 /r1A0 . Plugging the Taylor series of Q (s) Q (s) = a0 + a1 s + a2 s2 + a3 s3 + a4 s4 + a5 s5 + a6 s6 + a7 s7 + a8 s8 + o(s8 ) into (25), we have p0 = −(1 + a1 b2 )[γ (1 + a1 b2 ) + a1 b2 ] = 0. It implies

γ =−

a1 b 2 1 + a1 b 2

.

Then we have p1 = −[(n − 1)a0 a1 + 4a2 ]b2 − (n + 1)a0 = 0. It implies a0 = 0. Since Q (s) ̸= a1 s, there is some minimal integer m(m ≥ 2) such that am ̸= 0. We find that the coefficient of sm−1 , pm−1 = −m2 b2 am ̸= 0. It is a contradiction. By Lemma  3.1, (12) holds. Proposition 3.3 implies that F is Finsler metric of Randers type. It contradicts to the assumption that φ ̸= k1 1 + k2 s2 + k2 s for any constant k1 > 0, k2 and k3 ̸= 0. Hence b = constant, that is, ri + si = 0.  Proof of Theorem 1.1. By Lemma 3.4, (9) and (10) are reduced to

Φ r¯00 (b2 − s2 ) = sbt1 α¯ 2 , 2∆2 Φ −(s + b2 Q ) 2 s1A = btA . ∆

(26) (27)

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G. Chen et al. / Journal of Geometry and Physics 86 (2014) 112–121

Taking s = b in (26), we have t1 = 0. Then (26) becomes

Φ 2∆2

r¯00 (b2 − s2 ) = 0.

Noting that Φ ̸= 0, we have rAB = 0. Put h(s) := s + b2 Q , we have h(s) =

(b2 − s2 )φ ′ + sφ . φ − sφ ′

Since φ(s) > 0 and φ(s) − sφ ′ (s) > 0, we have h(b) =

bφ(b) > 0, φ(b) − bφ ′ (b)

h(−b) =

−bφ(−b) < 0. φ(−b) + bφ ′ (−b)

Thus there exists c1 ∈ (−b, b) such that c1 + b2 Q (c1 ) = 0. Taking s = c1 in (25), we get tA = 0. Then (27) becomes

Φ (s + b2 Q )s1A = 0. Assume that s1A ̸= 0. Since Φ (s) ̸= 0, there exists c2 ∈ (−b, b) such that Φ (c2 ) ̸= 0, i.e, Φ (s) ̸= 0 in some neighborhood of c2 ∈ (−b, b). Then we have s + b2 Q = 0 in this neighborhood. It implies that F is Riemannian. We get a contraction. Thus s1A = 0. Then sA = bs1A = 0 and s1 = bs11 = 0 imply si = 0. By si = 0 and ri + si = 0, we have r1A = (1/b)rA = −(1/b)sA = 0. Noting that rAB = 0, we have rij = 0.  4. (α, β)-metrics with scalar flag curvature In this section, we prove Theorem 1.2. By Theorem 1.1 and (5), we have S = 0. Then Lemma 2.1 implies that the flag curvature K = σ (x). In particular, by Schur theorem [5], we have the following Proposition 4.1. Let F = αφ(s),s = β/α be a non-Riemannian (α, β)-metric of scalar flag curvature on a manifold M of dimension n ≥ 3. Suppose φ ̸= k1 1 + k2 s2 + k2 s for any constant k1 > 0, k2 and k3 ̸= 0. If F is weak Berwald metric, then F is of constant flag curvature. In order to prove Theorem 1.2, we need to use (7). By Theorem 1.1, (2), (6) and (7) are simplified respectively as follows: Gi − Giα = α Qsi0 , Φ si0 , Ji = − 2∆α J¯ = 0. Proof of Theorem 1.2. By a direction computation, it is easy to obtain the following

Φ si0 si 0 , 2∆α Φ ∂(Gl − Glα ) i Jl b =− [Qs + Q ′ (b2 − s2 )]si0 si0 , ∂ yi 2∆α

Ji ail (rl0 + sl0 ) = −

∂ J¯ i (G − Giα ) = 0. ∂ yi

J¯|0 = 0,

(28)

It follows from (4) that Ii b i = −

Φ (φ − sφ ′ )(b2 − s2 ). 2∆αφ

(29)

Substituting (28) and (29) into (7), by Proposition 4.1, we have si0 si0 −

φ(φ − sφ ′ ) 2 α σ (b 2 − s 2 ) = 0 . ∆

Rewrite the above equation as follows: si0 si0 − σ D(s)α 2 = 0, where D(s) :=

φ(φ − sφ ′ ) 2 (b − s2 ). ∆

(30)

G. Chen et al. / Journal of Geometry and Physics 86 (2014) 112–121

121

Differentiating (30) with respect to yj , we obtain sij si0 + si0 sij − σ D′ (s)(α bj − syj ) − 2σ D(s)yj = 0,

(31)

where yj := aij yi . Contracting (31) with bj and by si = 0, we have

σ [D′ (s)(b2 − s2 ) + 2D(s)s] = 0. If D′ (s)(b2 − s2 ) + 2D(s)s = 0, one gets D(s) = µ(b2 − s2 ), where µ = constant. Thus the following equation holds.

µ=

φ(φ − sφ ′ ) (φ − sφ ′ )3 = . ∆ φ − sφ ′ + (b2 − s2 )φ ′′

Solving the above differential equation about φ(s) yields

 φ(s) = k1 1 + k2 s2 + k3 s. We get a contraction, Thus σ = 0. Then (30) implies sij = 0. In this case we have rij = sij = 0. By (2), F is Berwald metric. Nothing that K = σ = 0, we conclude that F is Minkowskian.  Acknowledgment The third author was supported by the National Natural Science Foundation of China (11171254). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10]

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