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Measurable (α,β) -spaces with vanishing S-curvature
Differential Geometry and its Applications 30 (2012) 549–561
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Measurable (α , β)-spaces with vanishing S-curvature Yi-Bing Shen, Huangjia Tian ∗ CMS, Zhejiang University, Hangzhou 310027, China
a r t i c l e
i n f o
Article history: Received 28 February 2011 Received in revised form 9 June 2012 Available online 4 October 2012 Communicated by Z. Shen MSC: 53B40
a b s t r a c t In this paper, we give a necessary and sufficient condition that an (α , β)-space admits a measure μ with vanishing S-curvature everywhere. 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. © 2012 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 metricF : T M → [0, ∞) can be expressed in the form F = α φ(s), β s = α , where φ is a C ∞ positive function on (−b0 , b0 ), α = ai j y i y j is a Riemannian metric, and β = b i y i is a 1-form with βx α < b0 . The S-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 1.2. Let ( M n , F , μ) (n 3) be a measurable non-Riemannian (α , β)-space with a measure μ and a Finsler metric F = α φ(s), s = β/α . Let b := βα . Then F has vanishing S-curvature everywhere if and only if there is a constant c ∈ R such that
*
Corresponding author. E-mail addresses: [email protected] (Y.-B. Shen), [email protected] (H. Tian).
0926-2245/$ – see front matter © 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.difgeo.2012.07.005
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
¯= b = constant, β is a Killing form with respect to the Riemannian metric α
(ai j + cbi b j ) y i y j , and μ coincides 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 = α φ(s), s = β/α , b := βα . Assume that ( M , F ) is not of Randers type. Then F admits a measure μ with vanishing S-curvature everywhere 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 , where c (x) is a scalar function on M and η = ηi y i is a 1-form on M. β For an (α , β)-metric F = α φ(s), s = α , b := βα , let
Φ(s) := − Q − s Q (n + s Q + 1) − b2 − s2 (1 + s Q ) Q ,
where
Q = r i j :=
φ , φ − sφ 1 2
= 1 + s Q + b 2 − s2 Q ,
(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 {xi }ni=1 , the admitted measure μ on M and the Riemannian volume measure of be expressed respectively as
μ(dx) = σ (x) dx1 ∧ dx2 ∧ · · · ∧ dxn , dV α = σα (x) dx1 ∧ · · · ∧ dxn ,
α , i.e., dV α can
σα = det(ai j ).
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 = α φ(s), s = β/α , b := βα . 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 , if and only if b = constant, β is a Killing form with respect to α and η = d(ln σσ(αx) )( y ). In this case, c = 0. Corollary 1.5. Let ( M n , F , μ) (n 3) be a measurable non-Riemannian (α , β)-space with the Finsler metric F = α φ(s), s = β/α , μ being Busemann–Hausdorff volume measure or Holmes–Thompson volume measure and b := βα . 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 1.2 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 (−b0 , b0 ). So we can rewrite their result through Corollary 1.5. 2. Preliminaries Let M be an n-dimensional differential manifold and T M be the tangent bundle. A Finsler metric on M is the function F = F (x, y ) : T M → R satisfying the following conditions [1]: (a) F (x, y ) is a C ∞ function on T M \ {0}; (b) F (x, y ) 0 and F (x, y ) = 0 ⇔ y = 0; (c) F (x, λ y ) = λ F (x, y ), λ > 0; (d) the fundamental tensor g i j (x, y ) := 12 ( F 2 ) y i y j =
1 ∂2(F 2) 2 ∂ yi ∂ y j
is positively defined.
Let
C i jk :=
1 4
F2
y i y j yk
=
1 ∂ gi j 2 ∂ yk
.
Define symmetric trilinear form C := C i jk dxi ⊗ dx j ⊗ dxk on T M \ {0}. We call C the Cartan torsion. The mean Cantor torsion I := I i dxi is defined by
I i := g jk C i jk .
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
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Let F be a Finsler metric on an n-dimensional manifold M. The canonical geodesics of F are characterized by
d2 c i (t ) dt 2
+ 2G i c (t ), c˙ (t ) = 0,
where G i are the geodesic coefficients having the expression G i := 14 g il {[ F 2 ]xk yl yk − [ F 2 ]xl } with ( g i j ) = ( g i j )−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 = α φ(s),
s=
where
β
α
,
α = ai j y i y j is a Riemannian metric, β = b s y s is a 1-form with βx α < b0 . φ = φ(s) is a C ∞ positive function on
an open interval (−b0 , b0 ) satisfying [9]
φ(s) − sφ (s) + ρ 2 − s2 φ (s) > 0,
∀|s| ρ < b0 .
= s yields that φ(s) − sφ (s) > 0,
Letting ρ ∀|s| < b0 . Denote the covariant derivatives of b i with respect to
r i j :=
1 2
(b i | j + b j |i ),
r j := b i r i j ,
si j :=
1 2
(b i | j − b j |i ),
si j := aih shj ,
s j := b i si j = b i si j ,
s0 := si y i ,
si 0 := si j y j ,
si0 := si j y j .
r00 := r i j y i y j ,
i Let G α be the geodesic coefficients of
α by bi| j . Let
α . The geodesic coefficients of an (α , β)-metric F have the following expression [9]
i Gi = Gα + α Q si 0 + α −1 Θ(−2α Q s0 + r00 ) y i + Ψ (−2α Q s0 + r00 )b i , φ Q − sQ Q Q = , Θ= , Ψ= , φ − sφ 2 2
φ[φ(s)−sφ (s)+(b2 −s2 )φ (s)] > 0, (φ−sφ )2 is a Killing form if b i | j + b j |i = 0 holds on M.
where := 1 + s Q + (b2 − s2 ) Q =
(1)
∀|s| b.
We say that β Fixing a local coordinate {xi }ni=1 , the admitted measure on M can be expressed as
μ(dx) = σ (x) dx1 ∧ dx2 ∧ · · · ∧ dxn . Then the local expression of S-curvature is given by
S=
∂ Gm ∂ − ym m (ln σ ). ∂ ym ∂x
The Busemann–Hausdorff volume form dV BH = σBH (x) dx is given by
σBH (x) =
ωn Vol{( y i ) ∈ R n : F (x, y i ∂ i ) < 1} ∂x
and the Holmes–Thompson volume from dV HT = σHT (x) is given by
σHT (x) =
1
det( g i j ) dy .
ωn {( y i )∈ R n :
F (x, y i ∂ i ∂x
)<1}
Here Vol denotes the Euclidean volume and
ωn := Vol B n (1) =
1 n
Vol S n−1
denotes the Euclidean volume of the unit ball in R n . Definition 2.1. 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)c F + η, where c = c (x) is a function on M and
η = ηi y i is a 1-form on M.
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
(b) The S-curvature is isotropic if c = c (x) is a function on M and (c) The S-curvature is constant if c is constant and η = 0.
η = 0.
Proposition 2.2. (See [5].) Let F = α φ(s), s = β/α be an (α , β)-metric on an n-dimensional manifold M and b := βα . Let dV = dV BH or dV HT . Let
f (b) :=
⎧ π n −2 t dt 0 sin ⎪ ⎪ ⎨ π sinn−2 t dt 0
if dV = dV BH ,
φ(b cos t )n
π n −2 ⎪ ⎪ ⎩ 0 (sin π t )nT−(2b cos t ) dt sin
0
if dV = dV HT ,
dt
where T (s) := φ(φ − sφ )n−2 [(φ − sφ ) + (b2 − s2 )φ ]. Then the volume form
dV = f (b) dv α ,
where dv α = det(ai j ) dx denotes the Riemannian volume form of α . To compute the S-curvature, we need the following identity [4]:
∂ Gm ∂ Φ = ym m (ln σα ) + 2Ψ (r0 + s0 ) − α −1 (r00 − 2α Q s0 ), ∂ ym ∂x 2 2 where
Φ := − Q − s Q (n + 1 + s Q ) − b2 − s2 (1 + s Q ) Q , σα (x) = det(ai j ). Then the S-curvature is given by
S = 2Ψ (r0 + s0 ) − α −1 where
tm =
∂ ∂ xm
Φ 2 2
(r00 − 2α Q s0 ) + ym tm ,
(2)
σα . ln σ
3. Proof of Theorem 1.2
β
Firstly, we consider such (α , β)-metrics F = α φ( α ) where φ = k1 1 + k2 s2 + k3 s, i.e.,
F = k1
α 2 + k2 β 2 + k3 β,
where α is a Riemannian metric, β is a 1-form on M, k1 > 0, k2 and k3 = 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
¯ + β¯ = F =α
k21 (ai j + k2 b i b j ) y i y j + (k3 b i ) y i .
By a direct computation, we have
¯ α¯ = k23 a¯ i j b i b j = β dV α¯ =
k23 k21
det(¯ai j ) dx =
k2 b i b j
ij
a −
1 + k2 b 2
bi b j =
k23
b2
k21 1 + k2 b2
,
det k21 (ai j + k2 b i b j ) dx = kn1 1 + k2 b2 det(ai j ) dx = kn1 1 + k2 b2 dV α ,
¯ = where (ai j ) and (¯ai j ) are inverse matrix of (ai j ) and (¯ai j ) respectively, b i := ai j b j . Since α
k21 (ai j + k2 b i b j ) y i y j has the
(ai j + cbi b j ) y i y j where c = k2 , then β¯ = k3 β being the Killing form with respect to α¯ ˆ . If b is constant and μ coincides with the Riemannian volume is equivalent to β being the Killing form with respect to α ¯ α¯ = constant and μ coincides with the Riemannian measure of α up to a constant multiplication, we can see that β ¯ up to a constant multiplication, and vise versa. Thus, according to Theorem 1.1, Theorem 1.2 is true volume measure of α ˆ= same Christoffel symbols with α
for metrics of Randers type. From now on, we should exclude the case of Randers type in the following.
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553
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) := βα = ai j b i b j is constant if and only if β satisfies
r i + s i = 0. Proof. Noticing the following identity:
2bb|k = b2 |k = 2ai j b|ik b j = 2b j |k b j = 2(r jk + s jk )b j = 2(rk + sk ),
2
we immediately obtain the conclusion.
With the help of Lemma 3.1, we see that if β satisfies b = constant and r i j = 0 (β is a Killing form with respect to
α = ai j y i y j , i.e., c = 0 in Theorem 1.2), then si = 0. Combining with σ = kσα for some constant k, it follows from (2) that S = 0. Thus, the sufficiency of Theorem 1.2 has been shown. Now we concentrate on the necessity of Theorem 1.2. First we take an orthonormal basis at x with respect to
n 2 α= yi ,
α so that
β = by 1 ,
i =1
and take the following coordinate transformation [4] in T x M,
s y1 = √ α¯ , yA = ξ A, 2 b − s2 n A 2 ¯= where α A =2 (ξ ) . We have b
α= √
b2
−
s2
α¯ ,
β=√
A = 2, 3, . . . , n ,
bs
b2
ϕ : (s, ξ A ) → ( y i )
− s2
α¯ ,
(3)
and the Jacobian of the transformation ϕ is non-degenerate. We now compute r0 , s0 and r00 in this new coordinate.
¯ ¯ sα sα r0 = y i ri = √ br11 + by A r1 A = √ br11 + br¯10 , 2 2 2 b −s b − s2
(4)
s0 = y i si = y 1 bs11 + y A bs1 A = by A s1 A = b s¯ 10 ,
(5)
r00 =
s
2
¯2
α
b 2 − s2
¯ sα r11 + 2 √ r¯10 + r¯00 , 2 b − s2
(6)
where
r¯10 :=
n
r1 A ξ A ,
s¯ 10 =
A =2
n A =2
s1 A ξ A ,
r¯00 =
n
rABξ Aξ B.
A , B =2
Substituting (4), (5), (6) into (2) and using (3), we see that (2) is equivalent to the following two equations.
2 Φ sΦ 2 2 ¯ + − r b − s sr − 2 Ψ b sbt α¯ 2 = 0, 00 11 1 2 2 2 2 Φ sΦ 2 − 2Ψ b (r1 A + s1 A ) − b2 Q + s 2 s1 A − bt A = 0. 2
(7) (8)
Lemma 3.2. (See [6].) Let F = α φ(β/α ) be an (α , β)-metric on a manifold M. The mean Carton torsion is given by
Ii = −
Φ 2F
φ − sφ h i ,
where h i = b i − α −1 sy i . By Deick’s theorem [1] and Lemma 3.2, 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 1.2, we can suppose that Φ = 0 in the following. d sΦ We divide our proof into two cases: (1) Λ = 0 everywhere and (2) Λ = 0, where Λ := ds [ 2 − 2Ψ b2 ].
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
The case (1): Λ = 0 everywhere.
Lemma 3.3. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M. Assume that φ = k1 1 + k2 s2 + k3 s for arbitrary constants k1 > 0, k2 and k3 = 0. If Λ = 0, then b = constant. sΦ 2 Proof. Since Λ = 0, i.e., is independent of s, we have 2 − 2Ψ b
sΦ
2
− 2Ψ b2 = −2Ψ (0)b2 = −
Q (0)b2 1 + Q (0)b2
.
If b = constant, we view b as a variable on M. Note that
2
sΦ
2
− 2Ψ b 2 +
Q (0)b2 1+
Q (0)b2
1 + Q (0)b2 = 0.
The left side of the above equation is a polynomial of b. Exactly, we have
Γ1 (s)b4 + Γ2 (s)b2 + Γ3 (s) = 0, where
Γ1 (s) = − Q 2 + Q 1 + s Q − s2 Q Q (0) + s −n Q Q − s Q − (1 + s Q ) Q Q (0), Γ2 (s) = 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q + s −n Q Q − s Q − Q (1 + s Q ) + s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q Q (0), Γ3 (s) = s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q . Since Γ1 (s) = Γ2 (s) = Γ3 (s) = 0, then, by a direct computation, we get that
0 = Q (0)Γ2 (s) − Γ1 (s) − Q (0)Γ3 (s) = Q − Q (0) 1 + s Q − s2 Q
2
,
i.e.,
Q − Q (0) 1 + s Q − s2 Q = 0. The general solution of the above ODE is
Q (s) = Q (0)s + C 1 1 + Q (0)s2 , where C 1 is a constant. From Γ3 (s) = 0, we have
− Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) 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 2 implies φ = C 2 1 + Q (0)s , where C 2 is a constant. This is contradictory to our assumption. Hence, b must be constant. 2
By Lemma 3.3, Eqs. (7) and (8) are reduced to
Φ 2 2
¯ 2, r¯00 b2 − s2 = sbt 1 α
Φ − s + b2 Q s1 A − bt A = 0. 2
(9) (10)
Letting s = b yields
t 1 = 0,
(11)
and
Φ 2 2
r¯00 b2 − s2 = 0.
Since Φ = 0, we obtain
r¯00 = 0.
(12)
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
555
Set
f (s) := s + b2 Q =
(b2 − s2 )φ + sφ . φ − sφ
We note
bφ(b)
f (b) =
φ(b) − bφ (b)
> 0,
f (−b) =
−bφ(−b) < 0, φ(−b) + bφ (−b)
since φ > 0, φ(b) − bφ (b) > 0, φ(−b) + bφ (−b) > 0. So there exists a point k0 in [−b, b] such that
k0 + b2 Q (k0 ) = 0. Letting s = k0 in (10) we get
t A = 0,
(13)
and
Φ − s + b2 Q s 1 A = 0. 2
√
It is easy to see that if φ satisfies s + b2 Q = 0, then φ = c b2 − s2 , which is impossible as we exclude this case. Consequently, we have
s 1 A = 0.
(14)
Finally, it follows from Lemma 3.3, (11), (12), (13) and (14) that r i j = 0 and t i = 0. The case (2): Λ = 0. Lemma 3.4. Let F = α φ(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 everywhere. Then
Φ −2s k − b2 Ψ + k − s2 − s γ = 0, 2 2
(15)
and in addition, if s0 = 0, then
−2Ψ −
sΦ 2 = δ, −λ − 2 Ψ b 2
QΦ
2
(16)
where λ, k, , γ , and δ are functions of x. Proof. By our assumption, Φ = 0 and (7), there is a function k = k(x) independent of s, such that
¯ 2, r¯00 = kα
s
Φ 2 2 − 2 Ψ b r11 + k b − s2 = sbt 1 . 2 2 2 2 sΦ
(17)
Put
r11 = k − b2 ,
t 1 = bγ ,
where = (x), γ = γ (x) are independent of s. Plugging them into (17) yields (15). Suppose that s0 = 0. Then there exists an integer A k ∈ [2, n] such that s Ak = bs1 Ak = 0. Differentiating (8) with respect to s yields
Λr1 A −
d
QΦ
2
ds
+ 2Ψ b 2 s 1 A = 0.
(18)
Set
λ := −
r1 Ak b 2 s1 A k
.
Substituting it into (18) yields
−λΛ −
d ds
QΦ
2
+ 2Ψ
= 0.
(19)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
It follows from (19) that
δ := −
QΦ
2
sΦ 2 − 2Ψ − λ − 2 Ψ b 2
is independent of s.
2
Lemma 3.5. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M n , n 3. Assume that φ = k1 1 + k2 s2 + k3 s for arbitrary constants k1 > 0, k2 and k3 = 0. If Λ = 0, then
r i + s i = 0. Proof. We use reductio ad absurdum. Suppose that r i + si = 0. Then b := βx α = constant in a neighborhood U ∈ M. By Lemma 3.4, φ = φ(s) satisfies (15). Putting s = 0 in (15) yields
k
Φ(0) = 0. 2 (0) β
We claim Φ(0) = 0. If this not true, i.e., Φ(0) = 0, then Φ( α ) = 0 holds for any y i orthogonal to b i with respect to ai j in T x U . On one hand, from Lemma 3.2 we get
Φ(0) I i (x, y ) = − b i = 0, 2 ai j y i y j (0) for any y i orthogonal to b i with respect to ai j 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
sΦ 2 2
− 2Ψ b
2
= −γ .
(20)
Φ Φ 2 2 We claim that 2s must depend on s. If this is not true, i.e., [ 2s 2 − 2Ψ b 2 − 2Ψ b ] = 0, then
sΦ 2 2
− 2Ψ b 2 = −
Q (0)b2 1 + Q (0)b2
.
Note that
2
sΦ 2 2
− 2Ψ b 2 +
Q (0)b2 1+
Q (0)b2
1 + Q (0)b2 = 0.
The left side of the above equation is a polynomial of b since b = constant. Exactly, we have
D 1 (s)b4 + D 2 (s)b2 + D 3 (s) = 0, where
D 1 (s) = 2Q 1 + s Q − s2 Q Q (0) + s −n Q Q − s Q − (1 + s Q ) Q Q (0),
+ s −n Q Q − s Q − Q (1 + s Q ) + s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q Q (0), D 3 (s) = s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q .
D 2 (s) = 2 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
557
Since D 1 (s) = D 2 (s) = D 3 (s) = 0, then, by a direct computation, we obtain
0 = Q (0) D 2 (s) − D 1 (s) − Q (0) D 3 (s) = 2 1 + s Q − s2 Q −2Q + Q (0) 1 + s Q − s2 Q
Q (0),
i.e.,
1 + s Q − s2 Q = 0
or
Q (0) = 0
or −2Q + Q (0) 1 + s Q − s2 Q = 0.
Setting s = 0 in 1 + s Q − s2 Q = 0, we get a contradiction. If Q (0) = 0, then
D 2 (s) = 2 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q
+ s −n Q Q − s Q − Q (1 + s Q ) .
Solving this ODE, we get Q (s) = C 1 , where C 1 is a constant. It implies that φ = C 2 s + contradictory to our assumption. If
C2 C1
, where C 2 is a constant. This is
−2Q + Q (0) 1 + s Q − s2 Q = 0, the general solution of this ODE is
1
Q (s) =
2
Q (0)s + C 3 1 +
1 2
Q (0)s2 ,
where C 3 is a constant. From D 3 (s) = 0, we have
− Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q = 0.
Putting s = 0 yields that Q (0) = 0. So we must have C 3 = 0. We obtain that Q (s) =
1 2
Q (0)s, which implies φ =
Φ 2 2 Q (0)s2 , where C 4 is a constant. This case is excluded in the assumption of the lemma. Therefore, 2s 2 b − 2Ψ b must depend on s. Differentiating (20) with respect to s yields
C4 1 +
1 2
sΦ
− 2Ψ b
2 2
2
= 0,
= 0 since [ 2sΦ2 − 2Ψ b2 ] = 0. As a result, we have proved that
which implies
k==γ =0 by using Eq. (15). Now we claim that si = 0. In fact, if si = 0, combining with k = 0 and
r11 = k − b2 = 0,
sΦ
2
= 0, we obtain (21)
− 2Ψ b2 r1 A = bt A .
(22)
Since Λ = 0, from (22) we have r1 A = 0. It follows that si + r i = 0, which is contradictory to that b = constant. Consequently, we have proved that si = 0. By Lemma 3.4, φ = φ(s) satisfies (16). Let the Taylor series of Q (s) in a neighborhood of s = 0 be
Q (s) = q 0 + q 1 s + q 2 s2 + q 3 s3 + q 4 s4 + q 5 s5 + q 6 s6 + o s6 .
(23)
Differentiating (16) with respect to s yields
λΛ = −2Ψ −
QΦ
2
.
We get that
λ=
[−2Ψ − Λ
QΦ
2
]
.
Putting s = 0 yields that
λ=
E1 E0
,
(24)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
where
E 1 = q0 q1 2q0 q2 − q21 n + 8q0 q22 − 6q1 q0 q3 − 2q21 q2 − 2q1 q20 q2 b4 + q0 q1 q20 − 2q21 + 2q0 q2 n
− 2q21 q0 − 6q0 q3 + 6q20 q2 − q1 q30 b2 + q0 q20 − q1 n − 2q1 q0 + q30 + 2q2 , E 0 = 1 + b2 q1 (−q1 q0 + 4q2 + nq0 q1 )b2 + (n + 1)q0 .
Since Λ = 0 and b = constant, we conclude that E 0 = 0, i.e., q20 + q22 = 0. Plugging (24) into (16), we find that
δ=
E2 E0
(25)
,
where
E 2 = q0 q1 2q0 q2 − q21 n − 6q1 q0 q3 + 8q0 q22 − 2q21 q2 − 2q1 q20 q2 b4 + q30 q1n2 + q0 6q0 q2 − 2q21 n
− q1 q30 − q21 q0 + 6q20 q2 − 2q1 q2 b2 + q0 n2 q20 + 2q20 − q1 n + q20 − q1 .
Let
G := 2 2Ψ +
QΦ
2
+λ
sΦ
2
− 2Ψ b 2 + δ .
(26)
Substituting (23), (24) and (25) into (26), we have that
E 0 G = f 2 s2 + f 3 s3 + f 4 s4 + · · · , where f i (i = 2, 3, 4, . . .) are polynomials of b. It follows from (16) that G = 0. Thus
f 2 = 0,
f 3 = 0,
f 4 = 0.
We can express f 2 in the form
f 2 = f 20 + f 21 b2 + f 22 b4 + f 23 b6 . Noting that b = constant, we know
f 20 = f 21 = f 22 = f 23 = 0. Solving the above system of linear equations yields that
q2 =
1 2
q 3 = 0,
q0 q1 ,
1 q4 = − q0 q21 , 8
q0 = 0
(27)
or
q2 =
1 2
q30 (n + 1) +
1 2
q30 ,
1 q3 = − (n + 1)q40 , 3
1 q4 = − q0 (n + 1)(n − 3)q40 + 2(n + 1)q1 q20 + q21 , 8
q0 = 0.
(28)
If (27) holds, we plug q2 = 12 q0 q1 and q3 = 0 into (24) and (25). It follows that
λ = q20 − q1 ,
δ=
(1 + q1 b2 )(q20 − q1 ) + nq20 . 1 + q1 b 2
Substituting them into (26), we get
1 + q1 b 2 G = g 0 + g 1 b 2 + g 2 b 4 ,
where
g0 =
q20 − q1 Q Q + nq1 Q 2 + q20 − q1 Q 2 s4 −
Q 2 + q20 − q1 Q − n Q Q 2 + q1 − 2nq1 − q20 Q Q s3
+ Q Q − Q 2 + (2n + 1) Q 2 − (n − 1)q1 − (n + 1)q20 Q + nq1 Q 2 s2 + (n + 2) Q Q + (n + 1)q20 + (n − 1)q1 Q − (n + 1) Q 3 s + Q − (n + 1) Q 2 + (n + 1)q20 − q1 , g 1 = q1 q20 − q1 Q Q − (n − 1) Q 2 s4 + q1 Q 2 + q20 − q1 Q − n Q Q 2 + (2n − 1) q20 − q1 Q Q s3 + 2q1 − q20 Q Q − (n + 1)q20 + nq1 Q 2 + (2n + 1)q1 Q 2 Q + (n − 1)q1 q20 − q1 Q
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
559
− nq1 q20 − q1 Q 2 s2 + − Q 2 + q20 − q1 Q + n Q Q 2 + (2n + 1)q1 + (n + 1)q20 Q Q − (n + 1)q1 Q 3 − (n − 1)q1 q20 − q1 Q s − Q Q + Q 2 − n Q 2 Q + (2n + 1)q20 Q − (n + 1)q1 Q 2 + q1 q20 − q1 , g 2 = q1 q20 − q1 − Q Q + (n − 1) Q 2 s2 − q1 Q 2 + q20 − q1 Q − n Q Q 2 + (n − 1) q20 − q1 Q Q s − q1 Q Q − nq1 Q 2 Q + nq20 + q1 Q 2 + q1 q20 − q1 Q , and
Q =
φ (s) . φ(s) − sφ (s)
Since G = 0 and b = constant, solving the system of ODEs g 0 = g 1 = g 2 = 0 with φ(0) = 1 by Maple 12 [2] yields
φ(s) = q0 s +
1 + q 1 s2 .
It is impossible as we exclude this case. If (28) holds, we express f 3 in the form
f 3 = f 30 + f 31 b2 + f 32 b4 + f 33 b6 . Plugging (28) into f 30 = 0 yields
1
− (n + 1)2 (n − 2)q60 = 0. 3
Since n 3 and q0 = 0, we get a contradiction. Therefore, r i + si = 0. 2 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 i j = 0, t i = 0. So far, we complete the proof of the necessity of Theorem 1.2. Corollary 1.3 is the direct and brief result of Theorem 1.2 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)c F + η , where
(n + 1)c F + ηi y i = 2Ψ (r0 + s0 ) − α −1
Φ 2 2
η = ηi y i is a 1-form on M, we have from (2) that
(r00 − 2α Q s0 ) + t i y i ,
(29)
where tm = ∂ x∂m ln σσα . On putting
ρi = t i − ηi , we find that (29) is equivalent to the following two equations in a special coordinate system (s, ya ):
2 Φ sΦ 2 2 2 ¯ ¯ 2 = 0, + + ( r b − s sr − 2 Ψ b n + 1 ) cb φ − sb ρ 00 11 1 α 2 2 2 2 Φ sΦ 2 (r1 A + s1 A ) − b2 Q + s 2 s1 A − bρ A = 0. − 2 Ψ b 2
(30) (31)
Lemma 4.1. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M. If F has weak isotropic S-curvature, then β satisfies
r i j = kai j − τ b i b j +
1 b2
(r i b j + r j b i ),
(32)
where k = k(x) and τ = τ (x) are functions on M. Proof. By Lemma 3.2 and (30), we have that
r A B = kδ A B , where k = k(x) is a function on M.
(33)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
Set
r11 = − k − τ b2 , where
(34)
τ = τ (x) is a function on M. (33) and (34) give (32). 2
By Lemma 4.1, we see that (30) and (31) are equivalent to the following
Φ 2 sΦ 2 2 + + (n + 1)cb2 φ = sbρ1 , b − s sr − 2 Ψ b 11 2 2 2 2 Φ sΦ 2 (r1 A + s1 A ) − b2 Q + s 2 s1 A − bρ A = 0. − 2 Ψ b 2
k
(35)
We can also divide the argument into two cases like the case S = 0: (1) Λ = 0 everywhere and (2) Λ = 0, where d sΦ [ − 2Ψ b2 ]. ds 2 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
Λ :=
Φ 2 b − s2 + (n + 1)cb2 φ = sbρ1 , 2 2 Φ − b 2 Q + s 2 s 1 A − b ρ A = 0.
k
(36) (37)
Let
f (s) := s + b2 Q . We have known that
f (b) > 0,
f (−b) < 0.
So there exists a point k0 in [−b, b] such that
k0 + b2 Q (k0 ) = 0. Putting s = k0 in (37) with =
φ[φ(s)−sφ (s)+(b2 −s2 )φ (s)] (φ−sφ )2
> 0 we get
ρ A = 0,
(38)
and
Φ − s + b2 Q s 1 A = 0. 2
√
It is easy to see that if φ satisfies s + b2 Q = 0, then φ = c b2 − s2 , which is impossible as we exclude this case. Consequently, we have
s 1 A = 0,
ρ A = 0.
(39)
We conclude that
s i = 0. Since b = constant, then
r i = − s i = 0. By Lemma 4.1, we get
r i j = kai j − τ b i b j . Contracting (40) with b i gives
r j = k − τ b 2 b j = 0. Since β = 0, we obtain
k = τ b2
(40)
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
561
and (30) becomes
r i j = τ b 2 ai j − b i b j .
(41)
Setting s = b in (36) yields that
ρ1 = (n + 1)c φ(b), φ(b)
since (b) = 1 + b Q (b) = φ(b)−bφ (b) > 0 and the regularity of φ on interval (−b0 , b0 ). Plugging it into (36) yields
τ
Φ 2 φ(b) 2 = ( b − s n + 1 ) c s − φ( s ) . b 2 2
(42)
Letting s = −b in (42) yields that
−(n + 1)c φ(b) + φ(−b) = 0,
(43)
φ(−b) φ(−b)+bφ (−b)
since (−b) = > 0. We get that c = τ = 0 from (42) and (43) and (41) becomes r i j = 0. Due to ρi = 0, we obtain that
ηi = t i =
∂ ∂ xm
ln
σα . 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. If b is constant and r i j = 0, we get from (2) that
S = ym tm = ym
∂ σα ln . ∂ xm σ
This completes the proof of Theorem 1.4. Take the Busemann–Hausdorff volume form or the Holmes–Thompson volume form, and η = 0 in Theorem 1.4. We get b is constant. By Proposition 2.2, one can see that σσa is constant. The equation η = d(ln σσ(αx) )( y ) = 0 is satisfied automatically. We conclude c = 0 from in Theorem 1.4 directly. So we get Corollary 1.5. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
D. Bao, S.S. Chern, Z. Shen, An Introduction to Riemannian–Finsler Geometry, Springer-Verlag, 2000. Frank Garvan, The Maple Book, Chapman & Hall/CRC, 2001. S. Ohta, Vanishing S-curvature of Randers spaces, Differential Geometry and its Applications 29 (2) (2011) 174–178. S. Bacso, X. Cheng, Z. Shen, Curvature properties of (α , β)-metrics, in: S. Sabau, H. Shimada (Eds.), Finsler Geometry, Sapporo 2005—In Memory of Makoto Matsumoto, in: Advanced Studies in Pure Mathematics, vol. 48, Mathematical Society of Japan, 2007, pp. 73–110. X. Cheng, Z. Shen, A class of Finsler metrics with isotropic S-curvature, Israel Journal of Mathematics 169 (2009) 317–340. X. Cheng, H. Wang, M. Wang, (α , β)-metrics with relatively isotropic mean Landsberg curvature, Publicationes Mathematicae Debrecen 72 (2008) 475– 485. Z. Shen, Volume comparison and its applications in Riemannian–Finsler geometry, Advances in Mathematics 128 (1997) 306–328. Z. Shen, Lectures on Finsler Geometry, World Scientific Co., Singapore, 2001. 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, 2004.
Contents lists available at SciVerse ScienceDirect
Differential Geometry and its Applications www.elsevier.com/locate/difgeo
Measurable (α , β)-spaces with vanishing S-curvature Yi-Bing Shen, Huangjia Tian ∗ CMS, Zhejiang University, Hangzhou 310027, China
a r t i c l e
i n f o
Article history: Received 28 February 2011 Received in revised form 9 June 2012 Available online 4 October 2012 Communicated by Z. Shen MSC: 53B40
a b s t r a c t In this paper, we give a necessary and sufficient condition that an (α , β)-space admits a measure μ with vanishing S-curvature everywhere. 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. © 2012 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 metricF : T M → [0, ∞) can be expressed in the form F = α φ(s), β s = α , where φ is a C ∞ positive function on (−b0 , b0 ), α = ai j y i y j is a Riemannian metric, and β = b i y i is a 1-form with βx α < b0 . The S-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 1.2. Let ( M n , F , μ) (n 3) be a measurable non-Riemannian (α , β)-space with a measure μ and a Finsler metric F = α φ(s), s = β/α . Let b := βα . Then F has vanishing S-curvature everywhere if and only if there is a constant c ∈ R such that
*
Corresponding author. E-mail addresses: [email protected] (Y.-B. Shen), [email protected] (H. Tian).
0926-2245/$ – see front matter © 2012 Elsevier B.V. All rights reserved. http://dx.doi.org/10.1016/j.difgeo.2012.07.005
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
¯= b = constant, β is a Killing form with respect to the Riemannian metric α
(ai j + cbi b j ) y i y j , and μ coincides 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 = α φ(s), s = β/α , b := βα . Assume that ( M , F ) is not of Randers type. Then F admits a measure μ with vanishing S-curvature everywhere 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 , where c (x) is a scalar function on M and η = ηi y i is a 1-form on M. β For an (α , β)-metric F = α φ(s), s = α , b := βα , let
Φ(s) := − Q − s Q (n + s Q + 1) − b2 − s2 (1 + s Q ) Q ,
where
Q = r i j :=
φ , φ − sφ 1 2
= 1 + s Q + b 2 − s2 Q ,
(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 {xi }ni=1 , the admitted measure μ on M and the Riemannian volume measure of be expressed respectively as
μ(dx) = σ (x) dx1 ∧ dx2 ∧ · · · ∧ dxn , dV α = σα (x) dx1 ∧ · · · ∧ dxn ,
α , i.e., dV α can
σα = det(ai j ).
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 = α φ(s), s = β/α , b := βα . 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 , if and only if b = constant, β is a Killing form with respect to α and η = d(ln σσ(αx) )( y ). In this case, c = 0. Corollary 1.5. Let ( M n , F , μ) (n 3) be a measurable non-Riemannian (α , β)-space with the Finsler metric F = α φ(s), s = β/α , μ being Busemann–Hausdorff volume measure or Holmes–Thompson volume measure and b := βα . 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 1.2 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 (−b0 , b0 ). So we can rewrite their result through Corollary 1.5. 2. Preliminaries Let M be an n-dimensional differential manifold and T M be the tangent bundle. A Finsler metric on M is the function F = F (x, y ) : T M → R satisfying the following conditions [1]: (a) F (x, y ) is a C ∞ function on T M \ {0}; (b) F (x, y ) 0 and F (x, y ) = 0 ⇔ y = 0; (c) F (x, λ y ) = λ F (x, y ), λ > 0; (d) the fundamental tensor g i j (x, y ) := 12 ( F 2 ) y i y j =
1 ∂2(F 2) 2 ∂ yi ∂ y j
is positively defined.
Let
C i jk :=
1 4
F2
y i y j yk
=
1 ∂ gi j 2 ∂ yk
.
Define symmetric trilinear form C := C i jk dxi ⊗ dx j ⊗ dxk on T M \ {0}. We call C the Cartan torsion. The mean Cantor torsion I := I i dxi is defined by
I i := g jk C i jk .
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
551
Let F be a Finsler metric on an n-dimensional manifold M. The canonical geodesics of F are characterized by
d2 c i (t ) dt 2
+ 2G i c (t ), c˙ (t ) = 0,
where G i are the geodesic coefficients having the expression G i := 14 g il {[ F 2 ]xk yl yk − [ F 2 ]xl } with ( g i j ) = ( g i j )−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 = α φ(s),
s=
where
β
α
,
α = ai j y i y j is a Riemannian metric, β = b s y s is a 1-form with βx α < b0 . φ = φ(s) is a C ∞ positive function on
an open interval (−b0 , b0 ) satisfying [9]
φ(s) − sφ (s) + ρ 2 − s2 φ (s) > 0,
∀|s| ρ < b0 .
= s yields that φ(s) − sφ (s) > 0,
Letting ρ ∀|s| < b0 . Denote the covariant derivatives of b i with respect to
r i j :=
1 2
(b i | j + b j |i ),
r j := b i r i j ,
si j :=
1 2
(b i | j − b j |i ),
si j := aih shj ,
s j := b i si j = b i si j ,
s0 := si y i ,
si 0 := si j y j ,
si0 := si j y j .
r00 := r i j y i y j ,
i Let G α be the geodesic coefficients of
α by bi| j . Let
α . The geodesic coefficients of an (α , β)-metric F have the following expression [9]
i Gi = Gα + α Q si 0 + α −1 Θ(−2α Q s0 + r00 ) y i + Ψ (−2α Q s0 + r00 )b i , φ Q − sQ Q Q = , Θ= , Ψ= , φ − sφ 2 2
φ[φ(s)−sφ (s)+(b2 −s2 )φ (s)] > 0, (φ−sφ )2 is a Killing form if b i | j + b j |i = 0 holds on M.
where := 1 + s Q + (b2 − s2 ) Q =
(1)
∀|s| b.
We say that β Fixing a local coordinate {xi }ni=1 , the admitted measure on M can be expressed as
μ(dx) = σ (x) dx1 ∧ dx2 ∧ · · · ∧ dxn . Then the local expression of S-curvature is given by
S=
∂ Gm ∂ − ym m (ln σ ). ∂ ym ∂x
The Busemann–Hausdorff volume form dV BH = σBH (x) dx is given by
σBH (x) =
ωn Vol{( y i ) ∈ R n : F (x, y i ∂ i ) < 1} ∂x
and the Holmes–Thompson volume from dV HT = σHT (x) is given by
σHT (x) =
1
det( g i j ) dy .
ωn {( y i )∈ R n :
F (x, y i ∂ i ∂x
)<1}
Here Vol denotes the Euclidean volume and
ωn := Vol B n (1) =
1 n
Vol S n−1
denotes the Euclidean volume of the unit ball in R n . Definition 2.1. 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)c F + η, where c = c (x) is a function on M and
η = ηi y i is a 1-form on M.
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
(b) The S-curvature is isotropic if c = c (x) is a function on M and (c) The S-curvature is constant if c is constant and η = 0.
η = 0.
Proposition 2.2. (See [5].) Let F = α φ(s), s = β/α be an (α , β)-metric on an n-dimensional manifold M and b := βα . Let dV = dV BH or dV HT . Let
f (b) :=
⎧ π n −2 t dt 0 sin ⎪ ⎪ ⎨ π sinn−2 t dt 0
if dV = dV BH ,
φ(b cos t )n
π n −2 ⎪ ⎪ ⎩ 0 (sin π t )nT−(2b cos t ) dt sin
0
if dV = dV HT ,
dt
where T (s) := φ(φ − sφ )n−2 [(φ − sφ ) + (b2 − s2 )φ ]. Then the volume form
dV = f (b) dv α ,
where dv α = det(ai j ) dx denotes the Riemannian volume form of α . To compute the S-curvature, we need the following identity [4]:
∂ Gm ∂ Φ = ym m (ln σα ) + 2Ψ (r0 + s0 ) − α −1 (r00 − 2α Q s0 ), ∂ ym ∂x 2 2 where
Φ := − Q − s Q (n + 1 + s Q ) − b2 − s2 (1 + s Q ) Q , σα (x) = det(ai j ). Then the S-curvature is given by
S = 2Ψ (r0 + s0 ) − α −1 where
tm =
∂ ∂ xm
Φ 2 2
(r00 − 2α Q s0 ) + ym tm ,
(2)
σα . ln σ
3. Proof of Theorem 1.2
β
Firstly, we consider such (α , β)-metrics F = α φ( α ) where φ = k1 1 + k2 s2 + k3 s, i.e.,
F = k1
α 2 + k2 β 2 + k3 β,
where α is a Riemannian metric, β is a 1-form on M, k1 > 0, k2 and k3 = 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
¯ + β¯ = F =α
k21 (ai j + k2 b i b j ) y i y j + (k3 b i ) y i .
By a direct computation, we have
¯ α¯ = k23 a¯ i j b i b j = β dV α¯ =
k23 k21
det(¯ai j ) dx =
k2 b i b j
ij
a −
1 + k2 b 2
bi b j =
k23
b2
k21 1 + k2 b2
,
det k21 (ai j + k2 b i b j ) dx = kn1 1 + k2 b2 det(ai j ) dx = kn1 1 + k2 b2 dV α ,
¯ = where (ai j ) and (¯ai j ) are inverse matrix of (ai j ) and (¯ai j ) respectively, b i := ai j b j . Since α
k21 (ai j + k2 b i b j ) y i y j has the
(ai j + cbi b j ) y i y j where c = k2 , then β¯ = k3 β being the Killing form with respect to α¯ ˆ . If b is constant and μ coincides with the Riemannian volume is equivalent to β being the Killing form with respect to α ¯ α¯ = constant and μ coincides with the Riemannian measure of α up to a constant multiplication, we can see that β ¯ up to a constant multiplication, and vise versa. Thus, according to Theorem 1.1, Theorem 1.2 is true volume measure of α ˆ= same Christoffel symbols with α
for metrics of Randers type. From now on, we should exclude the case of Randers type in the following.
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553
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) := βα = ai j b i b j is constant if and only if β satisfies
r i + s i = 0. Proof. Noticing the following identity:
2bb|k = b2 |k = 2ai j b|ik b j = 2b j |k b j = 2(r jk + s jk )b j = 2(rk + sk ),
2
we immediately obtain the conclusion.
With the help of Lemma 3.1, we see that if β satisfies b = constant and r i j = 0 (β is a Killing form with respect to
α = ai j y i y j , i.e., c = 0 in Theorem 1.2), then si = 0. Combining with σ = kσα for some constant k, it follows from (2) that S = 0. Thus, the sufficiency of Theorem 1.2 has been shown. Now we concentrate on the necessity of Theorem 1.2. First we take an orthonormal basis at x with respect to
n 2 α= yi ,
α so that
β = by 1 ,
i =1
and take the following coordinate transformation [4] in T x M,
s y1 = √ α¯ , yA = ξ A, 2 b − s2 n A 2 ¯= where α A =2 (ξ ) . We have b
α= √
b2
−
s2
α¯ ,
β=√
A = 2, 3, . . . , n ,
bs
b2
ϕ : (s, ξ A ) → ( y i )
− s2
α¯ ,
(3)
and the Jacobian of the transformation ϕ is non-degenerate. We now compute r0 , s0 and r00 in this new coordinate.
¯ ¯ sα sα r0 = y i ri = √ br11 + by A r1 A = √ br11 + br¯10 , 2 2 2 b −s b − s2
(4)
s0 = y i si = y 1 bs11 + y A bs1 A = by A s1 A = b s¯ 10 ,
(5)
r00 =
s
2
¯2
α
b 2 − s2
¯ sα r11 + 2 √ r¯10 + r¯00 , 2 b − s2
(6)
where
r¯10 :=
n
r1 A ξ A ,
s¯ 10 =
A =2
n A =2
s1 A ξ A ,
r¯00 =
n
rABξ Aξ B.
A , B =2
Substituting (4), (5), (6) into (2) and using (3), we see that (2) is equivalent to the following two equations.
2 Φ sΦ 2 2 ¯ + − r b − s sr − 2 Ψ b sbt α¯ 2 = 0, 00 11 1 2 2 2 2 Φ sΦ 2 − 2Ψ b (r1 A + s1 A ) − b2 Q + s 2 s1 A − bt A = 0. 2
(7) (8)
Lemma 3.2. (See [6].) Let F = α φ(β/α ) be an (α , β)-metric on a manifold M. The mean Carton torsion is given by
Ii = −
Φ 2F
φ − sφ h i ,
where h i = b i − α −1 sy i . By Deick’s theorem [1] and Lemma 3.2, 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 1.2, we can suppose that Φ = 0 in the following. d sΦ We divide our proof into two cases: (1) Λ = 0 everywhere and (2) Λ = 0, where Λ := ds [ 2 − 2Ψ b2 ].
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
The case (1): Λ = 0 everywhere.
Lemma 3.3. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M. Assume that φ = k1 1 + k2 s2 + k3 s for arbitrary constants k1 > 0, k2 and k3 = 0. If Λ = 0, then b = constant. sΦ 2 Proof. Since Λ = 0, i.e., is independent of s, we have 2 − 2Ψ b
sΦ
2
− 2Ψ b2 = −2Ψ (0)b2 = −
Q (0)b2 1 + Q (0)b2
.
If b = constant, we view b as a variable on M. Note that
2
sΦ
2
− 2Ψ b 2 +
Q (0)b2 1+
Q (0)b2
1 + Q (0)b2 = 0.
The left side of the above equation is a polynomial of b. Exactly, we have
Γ1 (s)b4 + Γ2 (s)b2 + Γ3 (s) = 0, where
Γ1 (s) = − Q 2 + Q 1 + s Q − s2 Q Q (0) + s −n Q Q − s Q − (1 + s Q ) Q Q (0), Γ2 (s) = 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q + s −n Q Q − s Q − Q (1 + s Q ) + s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q Q (0), Γ3 (s) = s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q . Since Γ1 (s) = Γ2 (s) = Γ3 (s) = 0, then, by a direct computation, we get that
0 = Q (0)Γ2 (s) − Γ1 (s) − Q (0)Γ3 (s) = Q − Q (0) 1 + s Q − s2 Q
2
,
i.e.,
Q − Q (0) 1 + s Q − s2 Q = 0. The general solution of the above ODE is
Q (s) = Q (0)s + C 1 1 + Q (0)s2 , where C 1 is a constant. From Γ3 (s) = 0, we have
− Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) 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 2 implies φ = C 2 1 + Q (0)s , where C 2 is a constant. This is contradictory to our assumption. Hence, b must be constant. 2
By Lemma 3.3, Eqs. (7) and (8) are reduced to
Φ 2 2
¯ 2, r¯00 b2 − s2 = sbt 1 α
Φ − s + b2 Q s1 A − bt A = 0. 2
(9) (10)
Letting s = b yields
t 1 = 0,
(11)
and
Φ 2 2
r¯00 b2 − s2 = 0.
Since Φ = 0, we obtain
r¯00 = 0.
(12)
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
555
Set
f (s) := s + b2 Q =
(b2 − s2 )φ + sφ . φ − sφ
We note
bφ(b)
f (b) =
φ(b) − bφ (b)
> 0,
f (−b) =
−bφ(−b) < 0, φ(−b) + bφ (−b)
since φ > 0, φ(b) − bφ (b) > 0, φ(−b) + bφ (−b) > 0. So there exists a point k0 in [−b, b] such that
k0 + b2 Q (k0 ) = 0. Letting s = k0 in (10) we get
t A = 0,
(13)
and
Φ − s + b2 Q s 1 A = 0. 2
√
It is easy to see that if φ satisfies s + b2 Q = 0, then φ = c b2 − s2 , which is impossible as we exclude this case. Consequently, we have
s 1 A = 0.
(14)
Finally, it follows from Lemma 3.3, (11), (12), (13) and (14) that r i j = 0 and t i = 0. The case (2): Λ = 0. Lemma 3.4. Let F = α φ(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 everywhere. Then
Φ −2s k − b2 Ψ + k − s2 − s γ = 0, 2 2
(15)
and in addition, if s0 = 0, then
−2Ψ −
sΦ 2 = δ, −λ − 2 Ψ b 2
QΦ
2
(16)
where λ, k, , γ , and δ are functions of x. Proof. By our assumption, Φ = 0 and (7), there is a function k = k(x) independent of s, such that
¯ 2, r¯00 = kα
s
Φ 2 2 − 2 Ψ b r11 + k b − s2 = sbt 1 . 2 2 2 2 sΦ
(17)
Put
r11 = k − b2 ,
t 1 = bγ ,
where = (x), γ = γ (x) are independent of s. Plugging them into (17) yields (15). Suppose that s0 = 0. Then there exists an integer A k ∈ [2, n] such that s Ak = bs1 Ak = 0. Differentiating (8) with respect to s yields
Λr1 A −
d
QΦ
2
ds
+ 2Ψ b 2 s 1 A = 0.
(18)
Set
λ := −
r1 Ak b 2 s1 A k
.
Substituting it into (18) yields
−λΛ −
d ds
QΦ
2
+ 2Ψ
= 0.
(19)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
It follows from (19) that
δ := −
QΦ
2
sΦ 2 − 2Ψ − λ − 2 Ψ b 2
is independent of s.
2
Lemma 3.5. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M n , n 3. Assume that φ = k1 1 + k2 s2 + k3 s for arbitrary constants k1 > 0, k2 and k3 = 0. If Λ = 0, then
r i + s i = 0. Proof. We use reductio ad absurdum. Suppose that r i + si = 0. Then b := βx α = constant in a neighborhood U ∈ M. By Lemma 3.4, φ = φ(s) satisfies (15). Putting s = 0 in (15) yields
k
Φ(0) = 0. 2 (0) β
We claim Φ(0) = 0. If this not true, i.e., Φ(0) = 0, then Φ( α ) = 0 holds for any y i orthogonal to b i with respect to ai j in T x U . On one hand, from Lemma 3.2 we get
Φ(0) I i (x, y ) = − b i = 0, 2 ai j y i y j (0) for any y i orthogonal to b i with respect to ai j 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
sΦ 2 2
− 2Ψ b
2
= −γ .
(20)
Φ Φ 2 2 We claim that 2s must depend on s. If this is not true, i.e., [ 2s 2 − 2Ψ b 2 − 2Ψ b ] = 0, then
sΦ 2 2
− 2Ψ b 2 = −
Q (0)b2 1 + Q (0)b2
.
Note that
2
sΦ 2 2
− 2Ψ b 2 +
Q (0)b2 1+
Q (0)b2
1 + Q (0)b2 = 0.
The left side of the above equation is a polynomial of b since b = constant. Exactly, we have
D 1 (s)b4 + D 2 (s)b2 + D 3 (s) = 0, where
D 1 (s) = 2Q 1 + s Q − s2 Q Q (0) + s −n Q Q − s Q − (1 + s Q ) Q Q (0),
+ s −n Q Q − s Q − Q (1 + s Q ) + s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q Q (0), D 3 (s) = s − Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q .
D 2 (s) = 2 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
557
Since D 1 (s) = D 2 (s) = D 3 (s) = 0, then, by a direct computation, we obtain
0 = Q (0) D 2 (s) − D 1 (s) − Q (0) D 3 (s) = 2 1 + s Q − s2 Q −2Q + Q (0) 1 + s Q − s2 Q
Q (0),
i.e.,
1 + s Q − s2 Q = 0
or
Q (0) = 0
or −2Q + Q (0) 1 + s Q − s2 Q = 0.
Setting s = 0 in 1 + s Q − s2 Q = 0, we get a contradiction. If Q (0) = 0, then
D 2 (s) = 2 1 + s Q − s2 Q − Q + Q (0) 1 + s Q − s2 Q
+ s −n Q Q − s Q − Q (1 + s Q ) .
Solving this ODE, we get Q (s) = C 1 , where C 1 is a constant. It implies that φ = C 2 s + contradictory to our assumption. If
C2 C1
, where C 2 is a constant. This is
−2Q + Q (0) 1 + s Q − s2 Q = 0, the general solution of this ODE is
1
Q (s) =
2
Q (0)s + C 3 1 +
1 2
Q (0)s2 ,
where C 3 is a constant. From D 3 (s) = 0, we have
− Q − s Q n 1 + s Q − s2 Q + 1 + s Q + s2 (1 + s Q ) Q = 0.
Putting s = 0 yields that Q (0) = 0. So we must have C 3 = 0. We obtain that Q (s) =
1 2
Q (0)s, which implies φ =
Φ 2 2 Q (0)s2 , where C 4 is a constant. This case is excluded in the assumption of the lemma. Therefore, 2s 2 b − 2Ψ b must depend on s. Differentiating (20) with respect to s yields
C4 1 +
1 2
sΦ
− 2Ψ b
2 2
2
= 0,
= 0 since [ 2sΦ2 − 2Ψ b2 ] = 0. As a result, we have proved that
which implies
k==γ =0 by using Eq. (15). Now we claim that si = 0. In fact, if si = 0, combining with k = 0 and
r11 = k − b2 = 0,
sΦ
2
= 0, we obtain (21)
− 2Ψ b2 r1 A = bt A .
(22)
Since Λ = 0, from (22) we have r1 A = 0. It follows that si + r i = 0, which is contradictory to that b = constant. Consequently, we have proved that si = 0. By Lemma 3.4, φ = φ(s) satisfies (16). Let the Taylor series of Q (s) in a neighborhood of s = 0 be
Q (s) = q 0 + q 1 s + q 2 s2 + q 3 s3 + q 4 s4 + q 5 s5 + q 6 s6 + o s6 .
(23)
Differentiating (16) with respect to s yields
λΛ = −2Ψ −
QΦ
2
.
We get that
λ=
[−2Ψ − Λ
QΦ
2
]
.
Putting s = 0 yields that
λ=
E1 E0
,
(24)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
where
E 1 = q0 q1 2q0 q2 − q21 n + 8q0 q22 − 6q1 q0 q3 − 2q21 q2 − 2q1 q20 q2 b4 + q0 q1 q20 − 2q21 + 2q0 q2 n
− 2q21 q0 − 6q0 q3 + 6q20 q2 − q1 q30 b2 + q0 q20 − q1 n − 2q1 q0 + q30 + 2q2 , E 0 = 1 + b2 q1 (−q1 q0 + 4q2 + nq0 q1 )b2 + (n + 1)q0 .
Since Λ = 0 and b = constant, we conclude that E 0 = 0, i.e., q20 + q22 = 0. Plugging (24) into (16), we find that
δ=
E2 E0
(25)
,
where
E 2 = q0 q1 2q0 q2 − q21 n − 6q1 q0 q3 + 8q0 q22 − 2q21 q2 − 2q1 q20 q2 b4 + q30 q1n2 + q0 6q0 q2 − 2q21 n
− q1 q30 − q21 q0 + 6q20 q2 − 2q1 q2 b2 + q0 n2 q20 + 2q20 − q1 n + q20 − q1 .
Let
G := 2 2Ψ +
QΦ
2
+λ
sΦ
2
− 2Ψ b 2 + δ .
(26)
Substituting (23), (24) and (25) into (26), we have that
E 0 G = f 2 s2 + f 3 s3 + f 4 s4 + · · · , where f i (i = 2, 3, 4, . . .) are polynomials of b. It follows from (16) that G = 0. Thus
f 2 = 0,
f 3 = 0,
f 4 = 0.
We can express f 2 in the form
f 2 = f 20 + f 21 b2 + f 22 b4 + f 23 b6 . Noting that b = constant, we know
f 20 = f 21 = f 22 = f 23 = 0. Solving the above system of linear equations yields that
q2 =
1 2
q 3 = 0,
q0 q1 ,
1 q4 = − q0 q21 , 8
q0 = 0
(27)
or
q2 =
1 2
q30 (n + 1) +
1 2
q30 ,
1 q3 = − (n + 1)q40 , 3
1 q4 = − q0 (n + 1)(n − 3)q40 + 2(n + 1)q1 q20 + q21 , 8
q0 = 0.
(28)
If (27) holds, we plug q2 = 12 q0 q1 and q3 = 0 into (24) and (25). It follows that
λ = q20 − q1 ,
δ=
(1 + q1 b2 )(q20 − q1 ) + nq20 . 1 + q1 b 2
Substituting them into (26), we get
1 + q1 b 2 G = g 0 + g 1 b 2 + g 2 b 4 ,
where
g0 =
q20 − q1 Q Q + nq1 Q 2 + q20 − q1 Q 2 s4 −
Q 2 + q20 − q1 Q − n Q Q 2 + q1 − 2nq1 − q20 Q Q s3
+ Q Q − Q 2 + (2n + 1) Q 2 − (n − 1)q1 − (n + 1)q20 Q + nq1 Q 2 s2 + (n + 2) Q Q + (n + 1)q20 + (n − 1)q1 Q − (n + 1) Q 3 s + Q − (n + 1) Q 2 + (n + 1)q20 − q1 , g 1 = q1 q20 − q1 Q Q − (n − 1) Q 2 s4 + q1 Q 2 + q20 − q1 Q − n Q Q 2 + (2n − 1) q20 − q1 Q Q s3 + 2q1 − q20 Q Q − (n + 1)q20 + nq1 Q 2 + (2n + 1)q1 Q 2 Q + (n − 1)q1 q20 − q1 Q
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
559
− nq1 q20 − q1 Q 2 s2 + − Q 2 + q20 − q1 Q + n Q Q 2 + (2n + 1)q1 + (n + 1)q20 Q Q − (n + 1)q1 Q 3 − (n − 1)q1 q20 − q1 Q s − Q Q + Q 2 − n Q 2 Q + (2n + 1)q20 Q − (n + 1)q1 Q 2 + q1 q20 − q1 , g 2 = q1 q20 − q1 − Q Q + (n − 1) Q 2 s2 − q1 Q 2 + q20 − q1 Q − n Q Q 2 + (n − 1) q20 − q1 Q Q s − q1 Q Q − nq1 Q 2 Q + nq20 + q1 Q 2 + q1 q20 − q1 Q , and
Q =
φ (s) . φ(s) − sφ (s)
Since G = 0 and b = constant, solving the system of ODEs g 0 = g 1 = g 2 = 0 with φ(0) = 1 by Maple 12 [2] yields
φ(s) = q0 s +
1 + q 1 s2 .
It is impossible as we exclude this case. If (28) holds, we express f 3 in the form
f 3 = f 30 + f 31 b2 + f 32 b4 + f 33 b6 . Plugging (28) into f 30 = 0 yields
1
− (n + 1)2 (n − 2)q60 = 0. 3
Since n 3 and q0 = 0, we get a contradiction. Therefore, r i + si = 0. 2 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 i j = 0, t i = 0. So far, we complete the proof of the necessity of Theorem 1.2. Corollary 1.3 is the direct and brief result of Theorem 1.2 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)c F + η , where
(n + 1)c F + ηi y i = 2Ψ (r0 + s0 ) − α −1
Φ 2 2
η = ηi y i is a 1-form on M, we have from (2) that
(r00 − 2α Q s0 ) + t i y i ,
(29)
where tm = ∂ x∂m ln σσα . On putting
ρi = t i − ηi , we find that (29) is equivalent to the following two equations in a special coordinate system (s, ya ):
2 Φ sΦ 2 2 2 ¯ ¯ 2 = 0, + + ( r b − s sr − 2 Ψ b n + 1 ) cb φ − sb ρ 00 11 1 α 2 2 2 2 Φ sΦ 2 (r1 A + s1 A ) − b2 Q + s 2 s1 A − bρ A = 0. − 2 Ψ b 2
(30) (31)
Lemma 4.1. Let F = α φ(β/α ) be a non-Riemannian (α , β)-metric on a manifold M. If F has weak isotropic S-curvature, then β satisfies
r i j = kai j − τ b i b j +
1 b2
(r i b j + r j b i ),
(32)
where k = k(x) and τ = τ (x) are functions on M. Proof. By Lemma 3.2 and (30), we have that
r A B = kδ A B , where k = k(x) is a function on M.
(33)
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Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
Set
r11 = − k − τ b2 , where
(34)
τ = τ (x) is a function on M. (33) and (34) give (32). 2
By Lemma 4.1, we see that (30) and (31) are equivalent to the following
Φ 2 sΦ 2 2 + + (n + 1)cb2 φ = sbρ1 , b − s sr − 2 Ψ b 11 2 2 2 2 Φ sΦ 2 (r1 A + s1 A ) − b2 Q + s 2 s1 A − bρ A = 0. − 2 Ψ b 2
k
(35)
We can also divide the argument into two cases like the case S = 0: (1) Λ = 0 everywhere and (2) Λ = 0, where d sΦ [ − 2Ψ b2 ]. ds 2 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
Λ :=
Φ 2 b − s2 + (n + 1)cb2 φ = sbρ1 , 2 2 Φ − b 2 Q + s 2 s 1 A − b ρ A = 0.
k
(36) (37)
Let
f (s) := s + b2 Q . We have known that
f (b) > 0,
f (−b) < 0.
So there exists a point k0 in [−b, b] such that
k0 + b2 Q (k0 ) = 0. Putting s = k0 in (37) with =
φ[φ(s)−sφ (s)+(b2 −s2 )φ (s)] (φ−sφ )2
> 0 we get
ρ A = 0,
(38)
and
Φ − s + b2 Q s 1 A = 0. 2
√
It is easy to see that if φ satisfies s + b2 Q = 0, then φ = c b2 − s2 , which is impossible as we exclude this case. Consequently, we have
s 1 A = 0,
ρ A = 0.
(39)
We conclude that
s i = 0. Since b = constant, then
r i = − s i = 0. By Lemma 4.1, we get
r i j = kai j − τ b i b j . Contracting (40) with b i gives
r j = k − τ b 2 b j = 0. Since β = 0, we obtain
k = τ b2
(40)
Y.-B. Shen, H. Tian / Differential Geometry and its Applications 30 (2012) 549–561
561
and (30) becomes
r i j = τ b 2 ai j − b i b j .
(41)
Setting s = b in (36) yields that
ρ1 = (n + 1)c φ(b), φ(b)
since (b) = 1 + b Q (b) = φ(b)−bφ (b) > 0 and the regularity of φ on interval (−b0 , b0 ). Plugging it into (36) yields
τ
Φ 2 φ(b) 2 = ( b − s n + 1 ) c s − φ( s ) . b 2 2
(42)
Letting s = −b in (42) yields that
−(n + 1)c φ(b) + φ(−b) = 0,
(43)
φ(−b) φ(−b)+bφ (−b)
since (−b) = > 0. We get that c = τ = 0 from (42) and (43) and (41) becomes r i j = 0. Due to ρi = 0, we obtain that
ηi = t i =
∂ ∂ xm
ln
σα . 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. If b is constant and r i j = 0, we get from (2) that
S = ym tm = ym
∂ σα ln . ∂ xm σ
This completes the proof of Theorem 1.4. Take the Busemann–Hausdorff volume form or the Holmes–Thompson volume form, and η = 0 in Theorem 1.4. We get b is constant. By Proposition 2.2, one can see that σσa is constant. The equation η = d(ln σσ(αx) )( y ) = 0 is satisfied automatically. We conclude c = 0 from in Theorem 1.4 directly. So we get Corollary 1.5. References [1] [2] [3] [4] [5] [6] [7] [8] [9]
D. Bao, S.S. Chern, Z. Shen, An Introduction to Riemannian–Finsler Geometry, Springer-Verlag, 2000. Frank Garvan, The Maple Book, Chapman & Hall/CRC, 2001. S. Ohta, Vanishing S-curvature of Randers spaces, Differential Geometry and its Applications 29 (2) (2011) 174–178. S. Bacso, X. Cheng, Z. Shen, Curvature properties of (α , β)-metrics, in: S. Sabau, H. Shimada (Eds.), Finsler Geometry, Sapporo 2005—In Memory of Makoto Matsumoto, in: Advanced Studies in Pure Mathematics, vol. 48, Mathematical Society of Japan, 2007, pp. 73–110. X. Cheng, Z. Shen, A class of Finsler metrics with isotropic S-curvature, Israel Journal of Mathematics 169 (2009) 317–340. X. Cheng, H. Wang, M. Wang, (α , β)-metrics with relatively isotropic mean Landsberg curvature, Publicationes Mathematicae Debrecen 72 (2008) 475– 485. Z. Shen, Volume comparison and its applications in Riemannian–Finsler geometry, Advances in Mathematics 128 (1997) 306–328. Z. Shen, Lectures on Finsler Geometry, World Scientific Co., Singapore, 2001. 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, 2004.