Differential Geometry and its Applications

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1 Differential Geometry and its Applications 30 (01) Contents lists available at SciVerse ScienceDirect Differential Geometry and its Applications Measurable (α,β)-spaces with vanishing S-curvature Yi-Bing Shen, Huangjia Tian CMS, Zhejiang University, Hangzhou 31007, China article info abstract Article history: Received 8 February 011 Received in revised form 9 June 01 Available online 4 October 01 Communicated by Z. Shen MSC: 53B40 In this paper, we give a necessary and sufficient condition that an (α,β)-space admits a measure μ with vanishing S-curvature every. It is shown that the measure of such an (α,β)-space must coincide with the Riemannian volume measure of α up to a constant multiplication. Moreover, the characterization of (α,β)-spaces admitting a measure μ with weak isotropic S-curvature is given. 01 Elsevier B.V. All rights reserved. Keywords: (α,β)-spaces S-curvature Spray coefficients 1. Introduction This paper is concerned with the characterization of (α,β)-spaces admitting a measure with special S-curvatures. An (α,β)-space is a special Finsler manifold whose Finsler metric F : TM [0, ) can be expressed in the form F = αφ(s), s = α β, φ is a C positive function on ( b 0, b 0 ), α = a ij y i y j is a Riemannian metric, and β = b i y i is a 1-form with β x α < b 0.TheS-curvature is an important non-riemannian quantity in Finsler geometry [8] introduced firstly by Z. Shen when he studied the volume comparison in Riemann Finsler geometry [7]. It is well known that the S-curvature is associated with the volume measure. There are two important volume measures in Finsler geometry. One is the Busemann Hausdorff volume measure and the other is the Holmes Thompson volume measure. By considering these two volume measures, the complete classification of (α,β)-spaces with isotropic S-curvature has been shown by Cheng and Shen [5]. An interesting question is when does an (α,β)-space admit a measure μ with S 0? For Randers spaces this question has been solved by S. Ohta [3] recently. Theorem 1.1. (See [3].) A Randers space (M, α + β) admits a measure μ with S 0 if and only if β is a Killing form of constant length with respect to α and the μ coincides with the Busemann Hausdorff volume measure up to a constant multiplication. The main purpose of this paper is to give a complete answer to the above question. We obtain the following. Theorem 1.. Let (M n, F, μ) (n 3) be a measurable non-riemannian (α,β)-space with a measure μ and a Finsler metric F = αφ(s), s= β/α. Letb:= β α. Then F has vanishing S-curvature every if and only if there is a constant c R such that * Corresponding author. addresses: yibingshen@zju.edu.cn (Y.-B. Shen), hjtian1984@163.com (H. Tian) /$ see front matter 01 Elsevier B.V. All rights reserved.

2 550 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) b = constant, β is a Killing form with respect to the Riemannian metric ᾱ = (a ij + cb i b j )y i y j,andμcoincides with the Riemannian volume measure of α up to a constant multiplication. Corollary 1.3. Let (M n, F, μ) (n 3) be a measurable non-riemannian (α,β)-space with the Finsler metric F = αφ(s), s= β/α, b := β α. Assume that (M, F ) is not of Randers type. Then F admits a measure μ with vanishing S-curvature every if and only if b = constant, β is a Killing form with respect to α and μ coincides with the Riemannian volume measure of α up to a constant multiplication. In general, we consider the case of (α,β)-spaces admitting a measure μ with weak isotropic S-curvature, i.e., S = (n + 1)c(x)F + η, c(x) is a scalar function on M and η = η i y i is a 1-form on M. For an (α,β)-metric F = αφ(s), s = β α, b := β α,let Φ(s) := ( Q sq ) (n + sq + 1) ( b s ) (1 + sq)q, Q = φ φ sφ, = 1 + sq + ( b s ) Q, r ij := 1 (b i j + b j i ). Here denotes the derivative with respect to s and denotes the covariant derivative with respect to α. Fixing a local coordinate {x i } n i=1, the admitted measure μ on M and the Riemannian volume measure of α, i.e., dv α can be expressed respectively as μ(dx) = σ (x) dx 1 dx dx n, dv α = σ α (x) dx 1 dx n, σ α = det(a ij ). Theorem 1.4. Let (M n, F, σ (x) dx) (n 3) be a measurable non-riemannian (α,β)-space with the Finsler metric F = αφ(s), s= β/α,b:= β α. Assume that (M, F ) is not of Randers type. Then F is of weak isotropic S-curvature, i.e., S = (n + 1)c(x)F + η,ifand )(y). In this case, c = 0. only if b = constant, β is a Killing form with respect to α and η = d(ln σ α σ (x) Corollary 1.5. Let (M n, F, μ) (n 3) be a measurable non-riemannian (α,β)-space with the Finsler metric F = αφ(s), s= β/α, μ being Busemann Hausdorff volume measure or Holmes Thompson volume measure and b := β α. Assume that (M, F ) is not of Randers type. Then F is of isotropic S-curvature, i.e., S = (n + 1)c(x)F if and only if S = 0. Remark 1. If the referenced measure μ is either Busemann Hausdorff volume measure or Holmes Thompson volume measure with isotropic S-curvature, i.e., S = (n + 1)c(x)F, the complete classification of (α,β)-spaces has been shown by X. Cheng and Z. Shen in [5]. It is worth mentioning that the case (i) of Theorem 1. of [5] is just the Riemannian case and the case (ii) is exactly the case (iii) by a simple reasoning if we emphasize the regularity of φ on the whole interval ( b 0, b 0 ). So we can rewrite their result through Corollary Preliminaries Let M be an n-dimensional differential manifold and TM be the tangent bundle. A Finsler metric on M is the function F = F (x, y) : TM R satisfying the following conditions [1]: (a) F (x, y) is a C function on TM\{0}; (b) F (x, y) 0 and F (x, y) = 0 y = 0; (c) F (x,λy) = λf (x, y), λ>0; (d) the fundamental tensor g ij (x, y) := 1 (F ) y i y j = 1 (F ) y i y j is positively defined. Let C ijk := 1 4 [ F ] y i y j y k = 1 g ij y k. Define symmetric trilinear form C := C ijk dx i dx j dx k on TM\{0}. WecallC the Cartan torsion. Themean Cantor torsion I := I i dx i is defined by I i := g jk C ijk.

3 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) Let F be a Finsler metric on an n-dimensional manifold M. The canonical geodesics of F are characterized by d c i (t) dt + G i( c(t), ċ(t) ) = 0, G i are the geodesic coefficients having the expression G i := 1 4 gil {[F ] x k y l yk [F ] x l} with (g ij ) = (g ij ) 1. In Finsler geometry, there is a quite important class of metrics called (α,β)-metrics. An (α,β)-metric can be expressed as F = αφ(s), s = β α, α = a ij y i y j is a Riemannian metric, β = b s y s is a 1-form with β x α < b 0. φ = φ(s) is a C positive function on an open interval ( b 0, b 0 ) satisfying [9] φ(s) sφ (s) + ( ρ s ) φ (s)>0, s ρ < b 0. Letting ρ = s yields that φ(s) sφ (s)>0, s < b 0. Denote the covariant derivatives of b i with respect to α by b i j.let r ij := 1 (b i j + b j i ), s ij := 1 (b i j b j i ), s i j := a ih s hj, s j := b i s i j = b i s ij, r j := b i r ij, r 00 := r ij y i y j, s 0 := s i y i, s i 0 := s i j y j, s i0 := s ij y j. Let Gα i be the geodesic coefficients of α. The geodesic coefficients of an (α,β)-metric F have the following expression [9] G i = G i α + α Qsi 0 + α 1 Θ( α Qs 0 + r 00 )y i + Ψ( α Qs 0 + r 00 )b i, Q = φ φ sφ, Θ = Q sq, Ψ = Q, (1) := 1 + sq + (b s )Q = φ[φ(s) sφ (s)+(b s )φ (s)] > (φ sφ 0, s b. ) We say that β is a Killing form if b i j + b j i = 0 holds on M. Fixing a local coordinate {x i } n, the admitted measure on M can be expressed as i=1 μ(dx) = σ (x) dx 1 dx dx n. Then the local expression of S-curvature is given by S = Gm y m ym (ln σ ). xm The Busemann Hausdorff volume form dv BH = σ BH (x) dx is given by ω n σ BH (x) = Vol{(y i ) R n : F (x, y i )<1} x i and the Holmes Thompson volume from dv HT = σ HT (x) is given by σ HT (x) = 1 det(g ij ) dy. ω n {(y i ) R n : F (x,y i x i )<1} Here Vol denotes the Euclidean volume and ω n := Vol ( B n (1) ) = 1 n Vol( S n 1) denotes the Euclidean volume of the unit ball in R n. Definition.1. Let F be a Finsler metric on an n-dimensional manifold M. Let S denote the S-curvature of F with respect to the volume measure μ(dx). (a) The S-curvature is weak isotropic if S = (n + 1)cF + η, c = c(x) is a function on M and η = η i y i is a 1-form on M.

4 55 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) (b) The S-curvature is isotropic if c = c(x) is a function on M and η = 0. (c) The S-curvature is constant if c is constant and η = 0. Proposition.. (See [5].) Let F = αφ(s), s= β/α be an (α,β)-metric on an n-dimensional manifold M and b := β α.letdv = dv BH or dv HT.Let π 0 sinn tdt π if dv = dv sin n t BH, 0 φ(b cost) f (b) := n dt π 0 (sinn t)t (b cos t) dt π if dv = dv HT, 0 sinn dt T (s) := φ(φ sφ ) n [(φ sφ ) + (b s )φ ]. Then the volume form dv = f (b) dv α, dv α = det(a ij ) dx denotes the Riemannian volume form of α. To compute the S-curvature, we need the following identity [4]: G m y = m ym x m (ln σ α) + Ψ(r 0 + s 0 ) α 1 Φ (r 00 α Qs 0 ), Φ := ( Q sq ) (n sq) ( b s ) (1 + sq)q, σ α (x) = det(a ij ). Then the S-curvature is given by S = Ψ(r 0 + s 0 ) α 1 Φ (r 00 α Qs 0 ) + y m t m, () t m = ( x m ln σ α σ 3. Proof of Theorem 1. ). Firstly, we consider such (α,β)-metrics F = αφ( β α ) φ = k 1 F = k 1 α + k β + k 3 β, 1 + k s + k 3 s, i.e., α is a Riemannian metric, β is a 1-form on M, k 1 > 0, k and k 3 0 are constants. A Finsler metric in this form is said to be of Randers type. Obviously, these Finsler metrics are essentially Randers metrics. In fact, it can be rewritten as F = ᾱ + β = k 1 (a ij + k b i b j )y i y j + (k 3 b i )y i. By a direct computation, we have β ᾱ = k 3āij b i b j = k 3 (a ij k b i b j ) k 1 + k 1 b dvᾱ = det(ā ij ) dx = b i b j = k 3 k 1 det ( k 1 (a ij + k b i b j ) ) dx = k n 1 b 1 + k b, 1 + k b det(a ij ) dx = k n k b dv α, (a ij ) and (ā ij ) are inverse matrix of (a ij ) and (ā ij ) respectively, b i := a ij b j.sinceᾱ = k 1 (a ij + k b i b j )y i y j has the same Christoffel symbols with ˆα = (a ij + cb i b j )y i y j c = k,then β = k 3 β being the Killing form with respect to ᾱ is equivalent to β being the Killing form with respect to ˆα. Ifb is constant and μ coincides with the Riemannian volume measure of α up to a constant multiplication, we can see that β ᾱ = constant and μ coincides with the Riemannian volume measure of ᾱ up to a constant multiplication, and vise versa. Thus, according to Theorem 1.1, Theorem 1. is true for metrics of Randers type. From now on, we should exclude the case of Randers type in the following.

5 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) Lemma 3.1. Let β = b i y i be a 1-form on a Riemannian manifold (M, α). Then the length of β with respect to α, b(x) := β α = a ij b i b j is constant if and only if β satisfies r i + s i = 0. Proof. Noticing the following identity: bb k = ( b ) k = a ijb i k b j = b j k b j = (r jk + s jk )b j = (r k + s k ), we immediately obtain the conclusion. With the help of Lemma 3.1, we see that if β satisfies b = constant and r ij = 0 (β is a Killing form with respect to α = a ij y i y j, i.e., c = 0inTheorem 1.), then s i = 0. Combining with σ = kσ α for some constant k, it follows from () that S = 0. Thus, the sufficiency of Theorem 1. has been shown. Now we concentrate on the necessity of Theorem 1.. First we take an orthonormal basis at x with respect to α so that α = n ( ) y i, β = by 1, i=1 and take the following coordinate transformation [4] in T x M, ϕ : (s,ξ A ) (y i ) y 1 = s b s ᾱ, y A = ξ A, A =, 3,...,n, n ᾱ = A= (ξ A ).Wehave α = b b ᾱ, β = bs s b ᾱ, s (3) and the Jacobian of the transformation ϕ is non-degenerate. We now compute r 0, s 0 and r 00 in this new coordinate. r 0 = y i r i = sᾱ b s br sᾱ 11 + by A r 1A = b s br 11 + b r 10, (4) s 0 = y i s i = y 1 bs 11 + y A bs 1A = by A s 1A = b s 10, (5) r 00 = s ᾱ b s r sᾱ 11 + b r s 10 + r 00, (6) r 10 := n r 1A ξ A, s 10 = A= n s 1A ξ A, r 00 = A= n A,B= r AB ξ A ξ B. Substituting (4), (5), (6) into () and using (3), we see that () is equivalent to the following two equations. Φ r ( 00 b s ) [ ( ) ] + sr 11 Ψ b sbt 1 ᾱ = 0, (7) ( ) Ψ b (r 1A + s 1A ) ( b Q + ) s Φ s 1A bt A = 0. (8) Lemma 3.. (See [6].) Let F = αφ(β/α) be an (α,β)-metric on a manifold M. The mean Carton torsion is given by I i = Φ ( φ sφ ) h i, F h i = b i α 1 sy i. By Deick s theorem [1] and Lemma 3., it is easy to see that an (α,β)-metric is Riemannian metric if and only if Φ = 0. Since we exclude the case that F is a Riemannian metric in Theorem 1., we can suppose that Φ 0 in the following. We divide our proof into two cases: (1) Λ = 0 every and () Λ 0, Λ := d ds [ Ψ b ].

6 554 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) The case (1): Λ = 0 every. Lemma 3.3. Let F = αφ(β/α) be a non-riemannian (α,β)-metric on a manifold M. Assume that φ k k s +k 3 s for arbitrary constants k 1 > 0,k and k 3 0.IfΛ = 0,thenb= constant. Proof. Since Λ = 0, i.e., Ψ b is independent of s, wehave Ψ b = Ψ(0)b = Q (0)b 1 + Q (0)b. If b constant, we view b as a variable on M. Note that ( Ψ b + Q (0)b ) (1 + Q 1 + Q (0)b (0)b ) = 0. The left side of the above equation is a polynomial of b. Exactly,wehave Γ 1 (s)b 4 + Γ (s)b + Γ 3 (s) = 0, Γ 1 (s) = Q + Q ( 1 + sq s Q ) Q (0) + s [ nq ( Q sq ) (1 + sq)q ] Q (0), Γ (s) = ( 1 + sq s Q )[ Q + Q (0) ( 1 + sq s Q )] + s [ nq ( Q sq ) Q (1 + sq) ] + s { ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q } Q (0), Γ 3 (s) = s { ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q }. Since Γ 1 (s) = Γ (s) = Γ 3 (s) = 0, then, by a direct computation, we get that i.e., 0 = Q (0)Γ (s) Γ 1 (s) Q (0)Γ 3 (s) = [ Q Q (0) ( 1 + sq s Q )], Q Q (0) ( 1 + sq s Q ) = 0. The general solution of the above ODE is Q (s) = Q (0)s + C Q (0)s, C 1 is a constant. From Γ 3 (s) = 0, we have ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q = 0. Letting s = 0 in the above equation yields Q (0) = 0. So we must have C 1 = 0. We conclude that Q (s) = Q (0)s, which implies φ = C 1 + Q (0)s, C is a constant. This is contradictory to our assumption. Hence, b must be constant. By Lemma 3.3, Eqs. (7) and (8) are reduced to Φ r ( 00 b s ) = sbt 1 ᾱ, ( s + b Q ) Φ s 1A bt A = 0. (9) (10) Letting s = b yields t 1 = 0, (11) and Φ r ( 00 b s ) = 0. Since Φ 0, we obtain r 00 = 0. (1)

7 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) Set We note f (s) := s + b Q = (b s )φ + sφ φ sφ. f (b) = bφ(b) φ(b) bφ (b) > 0, f ( b) = bφ( b) φ( b) + bφ ( b) < 0, since φ>0, φ(b) bφ (b)>0, φ( b) + bφ ( b)>0. So there exists a point k 0 in [ b, b] such that k 0 + b Q (k 0 ) = 0. Letting s = k 0 in (10) we get and t A = 0, ( s + b Q ) Φ s 1A = 0. (13) It is easy to see that if φ satisfies s + b Q = 0, then φ = c b s, which is impossible as we exclude this case. Consequently, we have s 1A = 0. Finally, it follows from Lemma 3.3, (11), (1), (13) and (14) that r ij = 0 and t i = 0. The case (): Λ 0. (14) Lemma 3.4. Let F = αφ(β/α) be an (α,β)-metric on a manifold M. Suppose that φ = φ(s) satisfies Φ 0, Λ 0 and S-curvature vanishes every. Then ( s k ɛb ) Ψ + ( k ɛs ) Φ sγ = 0, (15) and in addition, if s 0 0, then Ψ Q Φ ( ) λ Ψ b = δ, λ,k,ɛ, γ,andδ are functions of x. Proof. By our assumption, Φ 0 and (7), there is a function k = k(x) independent of s, such that (16) Put r 00 = kᾱ, ( ) s Ψ b r 11 + k Φ ( b s ) = sbt 1. (17) r 11 = k ɛb, t 1 = bγ, ɛ = ɛ(x), γ = γ (x) are independent of s. Plugging them into (17) yields (15). Suppose that s 0 0. Then there exists an integer A k [,n] such that s Ak = bs 1Ak 0. Differentiating (8) with respect to s yields Λr 1A d ( ) Q Φ ds + Ψ b s 1A = 0. (18) Set λ := r 1A k b s 1Ak. Substituting it into (18) yields λλ d ( ) Q Φ ds + Ψ = 0. (19)

8 556 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) It follows from (19) that ( Ψ λ δ := Q Φ is independent of s. Ψ b ) Lemma 3.5. Let F = αφ(β/α) be a non-riemannian (α,β)-metric on a manifold M n,n 3. Assume that φ k k s + k 3 sfor arbitrary constants k 1 > 0,k and k 3 0.IfΛ 0,then r i + s i = 0. Proof. We use reductio ad absurdum. Suppose that r i + s i 0. Then b := β x α constant in a neighborhood U M. By Lemma 3.4, φ = φ(s) satisfies (15). Putting s = 0in(15) yields k Φ(0) (0) = 0. We claim Φ(0) 0. If this not true, i.e., Φ(0) = 0, then Φ( β α ) = 0holdsforanyyi orthogonal to b i with respect to a ij in T x U. On one hand, from Lemma 3. we get Φ(0) I i (x, y) = b i = 0, a ij y i y j (0) for any y i orthogonal to b i with respect to a ij in T x U. On the other hand, if we chose y i = b i, h i = b i α 1 sy i = b i b i = 0, from which we have I i (x, b) = 0. Consequently, we have I i (x, y) = 0, (x, y) T U. By Deick s theorem, we conclude that F is a Riemannian metric. It is impossible as we exclude this case. Therefore, Φ(0) 0. Since Φ(0) 0, we have k = 0. Then (15) is reduced to ( ) ɛ Ψ b = γ. We claim that Ψ b must depend on s. If this is not true, i.e., [ Ψ b ] = 0, then Ψ b = Q (0)b 1 + Q (0)b. Note that ( Ψ b + Q (0)b ) (1 + Q 1 + Q (0)b (0)b ) = 0. The left side of the above equation is a polynomial of b since b constant. Exactly, we have (0) D 1 (s)b 4 + D (s)b + D 3 (s) = 0, D 1 (s) = Q ( 1 + sq s Q ) Q (0) + s [ nq ( Q sq ) (1 + sq)q ] Q (0), D (s) = ( 1 + sq s Q )[ Q + Q (0) ( 1 + sq s Q )] + s [ nq ( Q sq ) Q (1 + sq) ] + s { ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q } Q (0), D 3 (s) = s { ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q }.

9 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) Since D 1 (s) = D (s) = D 3 (s) = 0, then, by a direct computation, we obtain i.e., 0 = Q (0)D (s) D 1 (s) Q (0)D 3 (s) = ( 1 + sq s Q )[ Q + Q (0) ( 1 + sq s Q )] Q (0), 1 + sq s Q = 0 or Q (0) = 0 or Q + Q (0) ( 1 + sq s Q ) = 0. Setting s = 0in1+ sq s Q = 0, we get a contradiction. If Q (0) = 0, then D (s) = ( 1 + sq s Q )[ Q + Q (0) ( 1 + sq s Q )] + s [ nq ( Q sq ) Q (1 + sq) ]. Solving this ODE, we get Q (s) = C 1, C 1 is a constant. It implies that φ = C s + C C 1, C is a constant. This is contradictory to our assumption. If Q + Q (0) ( 1 + sq s Q ) = 0, the general solution of this ODE is Q (s) = 1 Q (0)s + C Q (0)s, C 3 is a constant. From D 3 (s) = 0, we have ( Q sq )[ n ( 1 + sq s Q ) sq ] + s (1 + sq)q = 0. Putting s = 0 yields that Q (0) = 0. So we must have C 3 = 0. We obtain that Q (s) = 1 Q (0)s, which implies φ = C Q (0)s, C 4 is a constant. This case is excluded in the assumption of the lemma. Therefore, must depend on s. Differentiating (0) with respect to s yields [ ] ɛ Ψ b = 0, which implies ɛ = 0since[ Ψ b ] 0. As a result, we have proved that k = ɛ = γ = 0 by using Eq. (15). Now we claim that s i 0. In fact, if s i = 0, combining with k = 0 and ɛ = 0, we obtain r 11 = k ɛb = 0, ( ) Ψ b r 1A = bt A. b Ψ b Since Λ 0, from () we have r 1A = 0. It follows that s i + r i = 0, which is contradictory to that b constant. Consequently, we have proved that s i 0. By Lemma 3.4, φ = φ(s) satisfies (16). LettheTaylorseriesofQ (s) in a neighborhood of s = 0be Q (s) = q 0 + q 1 s + q s + q 3 s 3 + q 4 s 4 + q 5 s 5 + q 6 s 6 + o ( s 6). (3) Differentiating (16) with respect to s yields [ λλ = Ψ Q Φ ]. We get that λ = [ Ψ Q Φ Λ ] Putting s = 0 yields that. (1) () λ = E 1 E 0, (4)

10 558 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) E 1 = [ ( q 0 q 1 q0 q q1) n + 8q0 q 6q 1 q 0 q 3 q 1 q q 1 q 0 ] q b 4 + [ ( q 0 q1 q 0 q 1 + ) q 0q n ] ( q 1 q 0 6q 0 q 3 + 6q 0 q q 1 q 3 0 b + q 0 q 0 1) q n q1 q 0 + q q, E 0 = ( )[ ] 1 + b q 1 ( q1 q 0 + 4q + nq 0 q 1 )b + (n + 1)q 0. Since Λ 0 and b constant, we conclude that E 0 0, i.e., q 0 + q 0. Plugging (4) into (16), we find that δ = E E 0, (5) Let E = [ ( q 0 q 1 q0 q q1) n 6q1 q 0 q 3 + 8q 0 q q 1 q q 1 q 0 ] q b 4 + [ ( q 3 0 q 1n + q 0 6q0 q q1) n q 1 q 3 0 ] q 1 q 0 + 6q 0 q q 1 q b + [ q0 n q 0 + ( q 0 ) q 1 n + q 0 1] q. { G := Ψ + Q Φ ( ) } + λ Ψ b + δ. (6) Substituting (3), (4) and (5) into (6), we have that E 0 G = f s + f 3 s 3 + f 4 s 4 +, f i (i =, 3, 4,...) are polynomials of b. It follows from (16) that G = 0. Thus f = 0, f 3 = 0, f 4 = 0. We can express f in the form f = f 0 + f 1 b + f b 4 + f 3 b 6. Noting that b constant, we know f 0 = f 1 = f = f 3 = 0. Solving the above system of linear equations yields that or q = 1 q 0q 1, q 3 = 0, q 4 = 1 8 q 0q 1, q 0 0 (7) q = 1 q3 0 (n + 1) + 1 q3 0, q 3 = 1 3 (n + 1)q4 0, q 4 = 1 [ 8 q 0 (n + 1)(n 3)q (n + 1)q 1q 0 + 1] q, q0 0. (8) If (7) holds, we plug q = 1 q 0q 1 and q 3 = 0into(4) and (5). It follows that λ = q 0 q 1, δ = (1 + q 1b )(q 0 q 1) + nq q 1 b. Substituting them into (6), we get ( 1 + q1 b ) G = g 0 + g 1 b + g b 4, g 0 = [( q 0 1) q QQ + nq 1 Q + ( q 0 1) q Q ] s 4 [( Q + q 0 1) q Q nq Q + ( ) q 1 nq 1 q 0 QQ ] s 3 + { QQ Q + [ ] (n + 1)Q (n 1)q 1 (n + 1)q 0 Q + nq 1 Q } s + { (n + )Q Q + [ (n + 1)q 0 + (n 1] } 1)q Q (n + 1)Q 3 s + Q (n + 1)Q + (n + 1)q 0 q 1, ( g 1 = q 1 q 0 1)[ q QQ (n 1)Q ] [( s 4 + q 1 Q + q 0 q Q nq Q + (n 1) ( q 0 ) q 1 QQ ] s 3 + {( q 1 q0) QQ [ (n + 1)q 0 + ] nq 1 Q + (n + 1)q 1 Q Q ( + (n 1)q 1 q 0 ) q 1 Q

11 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) and ( nq 1 q 0 1) } q Q s + { ( Q + q 0 1) q Q + nq Q + [ ] (n + 1)q 1 + (n + 1)q 0 QQ (n + 1)q 1 Q 3 ( (n 1)q 1 q 0 ) } q 1 Q s QQ + Q nq Q + (n + 1)q 0 Q ( (n + 1)q 1 Q + q 1 q 0 1) q, ( g = q 1 q 0 1)[ q QQ + (n 1)Q ] [( s q 1 Q + q 0 1) q Q nq Q + (n 1) ( q 0 1) q QQ ] s q 1 QQ nq 1 Q Q + ( nq 0 + ) ( q 1 Q + q 1 q 0 1) q Q, Q = φ (s) φ(s) sφ (s). Since G = 0 and b constant, solving the system of ODEs g 0 = g 1 = g = 0withφ(0) = 1 by Maple 1 [] yields φ(s) = q 0 s q 1 s. It is impossible as we exclude this case. If (8) holds, we express f 3 in the form f 3 = f 30 + f 31 b + f 3 b 4 + f 33 b 6. Plugging (8) into f 30 = 0yields 1 3 (n + 1) (n )q 6 0 = 0. Since n 3 and q 0 0, we get a contradiction. Therefore, r i + s i = 0. From Lemma 3.5 we obtain that Eqs. (7) and (8) under the condition of Lemma 3.4 are also reduced to (9) and (10). By the same argument, we conclude that r ij = 0, t i = 0. So far, we complete the proof of the necessity of Theorem 1.. Corollary 1.3 is the direct and brief result of Theorem 1. when we exclude metrics of Randers type. 4. Proof of Theorem 1.4 If F has weak isotropic S-curvature, i.e., S = (n + 1)cF + η, η = η i y i is a 1-form on M, wehavefrom() that (n + 1)cF + η i y i = Ψ(r 0 + s 0 ) α 1 Φ (r 00 α Qs 0 ) + t i y i, (9) t m = x m ln σ α σ. On putting ρ i = t i η i, we find that (9) is equivalent to the following two equations in a special coordinate system (s, y a ): Φ r ( 00 b s ) [ ( ) ] + sr 11 Ψ b + (n + 1)cb φ sbρ 1 ᾱ = 0, (30) ( ) Ψ b (r 1A + s 1A ) ( b Q + ) s Φ s 1A bρ A = 0. (31) Lemma 4.1. Let F = αφ(β/α) be a non-riemannian (α,β)-metric on a manifold M. If F has weak isotropic S-curvature, then β satisfies r ij = ka ij τ b i b j + 1 b (r ib j + r j b i ), (3) k = k(x) and τ = τ (x) are functions on M. Proof. By Lemma 3. and (30), we have that r AB = kδ AB, (33) k = k(x) is a function on M.

12 560 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) Set r 11 = ( k τ b ), τ = τ (x) is a function on M. (33) and (34) give (3). (34) By Lemma 4.1, we see that (30) and (31) are equivalent to the following k Φ ( b s ) ( ) + sr 11 Ψ b + (n + 1)cb φ = sbρ 1, ( ) Ψ b (r 1A + s 1A ) ( b Q + ) s Φ s 1A bρ A = 0. (35) We can also divide the argument into two cases like the case S = 0: (1) Λ = 0 every and () Λ 0, Λ := d ds [ Ψ b ]. We can work by similar process as in Lemmas 3.3, 3.4 and 3.5 to prove b = constant no matter what case we consider. That is to say, (31) and (35) are equivalent to Let k Φ ( b s ) + (n + 1)cb φ = sbρ 1, ( b Q + s ) Φ s 1A bρ A = 0. f (s) := s + b Q. We have known that (36) (37) f (b)>0, f ( b)<0. So there exists a point k 0 in [ b, b] such that k 0 + b Q (k 0 ) = 0. Putting s = k 0 in (37) with = φ[φ(s) sφ (s)+(b s )φ (s)] (φ sφ ) and ρ A = 0, > 0weget (38) ( s + b Q ) Φ s 1A = 0. It is easy to see that if φ satisfies s + b Q = 0, then φ = c b s, which is impossible as we exclude this case. Consequently, we have s 1A = 0, ρ A = 0. (39) We conclude that s i = 0. Since b = constant, then r i = s i = 0. By Lemma 4.1, weget r ij = ka ij τ b i b j. (40) Contracting (40) with b i gives r j = ( k τ b ) b j = 0. Since β 0, we obtain k = τ b

13 Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (01) and (30) becomes r ij = τ ( b a ij b i b j ). (41) Setting s = b in (36) yields that ρ 1 = (n + 1)cφ(b), φ(b) since (b) = 1 + bq(b) = φ(b) bφ (b) > 0 and the regularity of φ on interval ( b 0, b 0 ). Plugging it into (36) yields τ Φ ( b s ) ( = (n + 1)c s φ(b) ) φ(s). (4) b Letting s = b in (4) yields that (n + 1)c ( φ(b) + φ( b) ) = 0, (43) φ( b) since ( b) = φ( b)+bφ ( b) > 0. We get that c = τ = 0from(4) and (43) and (41) becomes r ij = 0. Due to ρ i = 0, we obtain that η i = t i = x m ln σ α σ. From what has been discussed above, we have showed the necessity of Theorem 1.4. Now let us complete the sufficiency of Theorem 1.4. Ifb is constant and r ij = 0, we get from () that S = y m t m = y m x m ln σ α σ. This completes the proof of Theorem 1.4. Take the Busemann Hausdorff volume form or the Holmes Thompson volume form, and η = 0inTheorem 1.4. Wegetb is constant. By Proposition., one can see that σ a σ is constant. The equation η = d(ln σ α σ (x) )(y) = 0 is satisfied automatically. We conclude c = 0frominTheorem 1.4 directly. So we get Corollary 1.5. References [1] D. Bao, S.S. Chern, Z. Shen, An Introduction to Riemannian Finsler Geometry, Springer-Verlag, 000. [] Frank Garvan, The Maple Book, Chapman & Hall/CRC, 001. [3] S. Ohta, Vanishing S-curvature of Randers spaces, Differential Geometry and its Applications 9 () (011) [4] S. Bacso, X. Cheng, Z. Shen, Curvature properties of (α,β)-metrics, in: S. Sabau, H. Shimada (Eds.), Finsler Geometry, Sapporo 005 In Memory of Makoto Matsumoto, in: Advanced Studies in Pure Mathematics, vol. 48, Mathematical Society of Japan, 007, pp [5] X. Cheng, Z. Shen, A class of Finsler metrics with isotropic S-curvature, Israel Journal of Mathematics 169 (009) [6] X. Cheng, H. Wang, M. Wang, (α,β)-metrics with relatively isotropic mean Landsberg curvature, Publicationes Mathematicae Debrecen 7 (008) [7] Z. Shen, Volume comparison and its applications in Riemannian Finsler geometry, Advances in Mathematics 18 (1997) [8] Z. Shen, Lectures on Finsler Geometry, World Scientific Co., Singapore, 001. [9] Z. Shen, Landsberg curvature, S-curvature and Riemann curvature, in: A Sampler of Finsler Geometry, in: Mathematical Sciences Research Institute Publications, vol. 50, Cambridge Univ. Press, 004.