On Landsberg (α, β)-Metrics Zhongmin Shen Dept. of Math., IUPUI 402 N. Blackford Street Indianapolis, IN46202-3216, USA
[email protected] July 24, 2004
1
Introduction
In Finsler geometry, there are three important classes of Finsler metrics: Riemannian metrics, Berwald metrics and Landsberg metrics. Riemannian metrtics are special Berwald metrics and Berwald metrics are special Landsberg metrics. The local structures of Berwald metrics have been completely determined by Z. I. Szabo [7]. However, it is a long existing open problem whether or not any Landsberg metric is a Berwald metric. Randers metrics are the most simple non-Riemannian Finsler metrics. They p are expressed in the form F = α + β, where α = aij (x)yi yj is a Riemannian metric and β = bi (x)yi is a 1-form on a manifold M . It is well-known that a Randers metric is a Landsberg metric if and only if it is a Berwald metric [3], [6]. Then it is natural to consider more general Finsler metrics on a manifold M . For a positive function φ = φ(s) on (−bo , bo ) with φ(0) = 1, define F = αφ(s), where α =
s=
β , α
p aij yi yj and β = biyi with kβkx < bo , ∀x ∈ M . If φ satisfies φ(s) − sφ0 (s) + (b2 − s2 )φ00(s) > 0,
(|s| ≤ b < bo ),
(1)
then F is a Finsler metric (Lemma 2.1). Finsler metrics in this form are called (α, β)-metrics. Note that Randers metrics are special (α, β)-metrics with φ = 1 + s. In this paper, we are going to show the following Theorem 1.1 Let φ = φ(s) be a C ∞ positive function on√an interval I = (−bo , bo ) with φ(0) = 1. Suppose that φ satisfies (1), but φ 6= 1 + ks2 on I for any constant k. For an (α, β)-metric F = αφ(β/α) with β 6= 0 on a manifold of dimension n ≥ 3, the following hold
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(a) F is a Landsberg metric; (b) F is a Berwald metric; (c) β is parallel with respect to α. It is known that an (α, β)-metric F = αφ(β/α) is a Berwald metric if and only if β is parallel with respect to α [2] [4]. Thus it suffices to prove the equivalence between (a) and (c). The same conclusion might hold in dimension two. However, in my approach, the ODEs are too complicated to be conclusive. Recently, the author has learned from H. Shimada that M.Matstumoto has already proved the above result in dimension two. √ Note that the (α, β)-metric F defined by φ = 1 + ks2 is a Riemannian √ 2 metric. Why do we assume that p φ 6= 1 + ks on I? The reason is as follows. For a Riemannian metric F = gij (x)yi yj , let β = bi yi be an arbitrary 1-form p and let aij := gij −kbi bj where k is a small constant. Then for α := aij (x)yi yj , √ F can be expressed as F = αφ(β/α) with φ = 1 + ks2 . Note that F is a Riemannian metric, hence a Landsberg metric. But β is not parallel with respect to α in general.
2
Landsberg Curvature
∂ i ∂ For a Finsler metric F = F (x, y) on a manifold M , the spray G = yi ∂x i −2G ∂y i is a vector field on T M , where Gi = Gi(x, y) are defined by
Gi =
o gil n 2 [F ]xk yl yk − [F 2]xl , 4
(2)
where gij := 12 [F 2]yiyj and (gij ) = (gij )−1 . By definition, F is called a Berwald ∂ metric if Gi = 12 Γijk (x)yj yk are quadratic in y = yi ∂x i |x ∈ Tx M for any x ∈ M . On a Berwald manifold (M, F ), all tangent spaces Tx M with the induced Minkowski norm Fx are linearly isometric. Moreover, F is affinely equivalent to a Riemannian metric g, namely, F and g have the same spray. Using this fact, one can determine the local structure of a Berwald metric [7]. The Landsberg tensor L = Lijk dxi ⊗ dxj ⊗ dxk is defined by h i 1 Ljkl := − F Fyi Gi . 2 yj yk yl
(3)
∂ Clearly, if Gi = 12 Γijk (x)yj yk are quadratic in y = yi ∂x i |x ∈ Tx M for any x ∈ M , then Ljkl = 0. A Finsler metric is called a Landsberg metric if Ljkl = 0. It is known that on a Landsberg manifold M , all tangent spaces Tx M with the induced Riemannian metric gˆx = gij (x, y)dyi ⊗ dyj are isometric. ∞ Now we consider (α, β)-metrics. Let φ = φ(s) be a pC function on (−bo , bo) with φ(0) = 1 and φ(s) > 0, ∀s ∈ (−bo , bo). Let α = aij yi yj be a Riemannian
2
metric and β = bi yi a 1-form on a manifod M . Assume that q kβkx := aij (x)bi (x)bj (x) < bo , ∀x ∈ M. Define F := αφ(s),
s=
β . α
(4)
By a direct computation, we obtain gij = ρaij + ρ0 bi bj − α−1ρ1 (bi yj + bj yi ) + sα−2 ρ1 yi yj , where yi := aij yj , and ρ: = ρ0 : = ρ1 : =
φ(φ − sφ0 ), φφ00 + φ0 φ0, s(φφ00 + φ0φ0 ) − φφ0.
Then F Fyi = gij yj = ρyi + φφ0αbi.
(5)
By further computation, one obtains h i n−2 det (gij ) = φn+1 (φ − sφ0 ) (φ − sφ0 ) + (kβk2x − s2 )φ00 det (aij ) . Using the continuity, one can easily show that Lemma 2.1 Let bo > 0 and φ = φ(s) be a C ∞ positive function on (−bo , bo) with φ(0) = 1. Let F = αφ(β/α). Then F is a Finsler metric on M for any pair {α, β} with kβkx < bo , ∀x ∈ M , if and only if φ = φ(s) satisfies (1). Proof: Suppose that φ = φ(s) satisfies (??) and (1). For any s with |s| < bo , taking b = |s| in (1), we obtain φ(s) − sφ0(s) > 0,
(|s| < bo ).
(6)
Let φλ(s) := (1 − λ) + λφ(s). φλ satisfies (??), (1) and (6) for any 0 ≤ λ ≤ 1. Let {α, β} be an arbitrary pair with kβkx < bo , ∀x ∈ M . Then the determinant λ of the fundamental matrix gij of F λ := αφλ(β/α) is given by λ det(gij ) =
h in+1 h in−2 (1 − λ) + λ(φ − sφ0 ) (1 − λ) + λφ × n h io × (1 − λ) + λ (φ − sφ0 ) + (kβk2x − s2 )φ00 det(aij ).
Thus λ det(gij ) 6= 0. 0 1 Since (gij ) = (aij ) is postive definite, by the continuity, we conclude that (gij )= (gij ) is positive definite.
3
Suppose that F = αφ(β/α) is a Finsler metric for any pair {α, β} with kβkx < bo , ∀x ∈ M . First, for an arbitrary number b with 0 ≤ b < bo , one can choose {α, β} such that (i) kβkxo = b at some point xo ∈ M and (ii) kβkx < bo ∀x ∈ M . For any s with |s| ≤ b, we can always find a vector y ∈ Tx M such that β/α = s. Since F = αφ(β/α) is a Finsler metric at xo , we have det(gij (xo , y)) > 0. Thus φ(s) − sφ0 (s) + (b2 − s2 )φ00 (s) > 0,
(|s| ≤ b).
This proves the lemma.
Q.E.D.
To compute the Landsberg curvature of an (α, β)-metric F = αφ(β/α), we first find the spray coefficients of F . Let 1 1 rij := bi|j + bj|i , sij := bi|j − bj|i . 2 2 rj := bi rij ,
sj := bisij ,
where bi|j denote the coefficients of the covariant derivatives of β = bi yi with respect to α. Let ri0 := rij yj , si0 := sij yj , r0 := rj yj and s0 := sj yj . ¯i = G ¯ i (x, y) denote the coefficients of F and α Let Gi = Gi(x, y) and G respectively in the same coordinate system. By (2), we obtain the following identity. ¯ i + P yi + Qi , Gi = G (7) where P
=
Qi
=
o n α−1Θ − 2Qαs0 + r00 o n αQsi 0 + Ψ − 2Qαs0 + r00 bi
where Q = Θ
=
Ψ
=
φ0 φ − sφ0 φφ0 − s(φφ00 + φ0 φ0) 2φ (φ − sφ0 ) + (b2 − s2 )φ00 1 φ00 . 2 (φ − sφ0 ) + (b2 − s2 )φ00
The formula (7) in a different form has been given in [1], [5]. Clearly, if β is parallel with respect to α (rij = 0 and sij = 0), then P = 0 ¯ i are quadratic in y, and F is a Berwald metric. and Qi = 0. In this case, Gi = G In fact, the converse is true too. This is part of Theorem 1.1. By (3), (5) and (7), we obtain the following formula for Ljkl . o 1 n Ljkl = − 5 hj hk Cl + hj hl Ck + hk hl Cj + 3Ej hkl + 3Ek hjl + 3El hkl , (8) 6α 4
where hj : = hjk : = Cj : =
αbj − syj α2ajk − yj yk , (A1 r00 + A2 αs0 )hj + 3ΛJj
Ej : = Jj : =
(B1 r00 + B2 αs0 )hj + 3µJj α2(sj0 + Γrj0 + Παsj ) − (Γr00 + Παs0 )yj (φ − sφ0 )2 φ[φ − sφ0 + (b2 − s2 )φ00] φ0(φ − sφ0 ) − , φ[φ − sφ0 + (b2 − s2 )φ00] φ[φφ0 − s(φφ00 + φ02)] − , 3(φ − sφ0 ) (sφφ0 − φ2 )φφ000 + (sφ02 − φφ0 − 2sφφ00)φφ00 (φ − sφ0 )2 o n ρ (1 + Qs)Θsss + 3QΘss + (s + Qb2)Ψsss o n ρ (1 + Qs)Ξsss + 3QΞss + (s + Qb2 )Φsss + QQsss n o −ρ Q(Θ + sΘs ) + (1 + Qs)[2Θs + Θss] + (s + Qb2)[sΨss + Ψs ] o n −ρ (1 + Qs)(Ξs + sΞss) + s(s + Qb2)Φss + sQΞs + sQQss ,
Γ: = Π: = µ: = Λ: = A1
=
A2
=
B1
=
B2
=
ρ: = Ξ = Φ =
3
φ(φ − sφ0 ), −2QΘ, −2QΨ.
Landsberg Metrics
By definition, a Finsler metric is called a Landsberg metric if Ljkl = 0. In this section, we are going to derive three sufficient and necessary conditions for a Finsler metric be Landsbergian. First note that hj bj = α(b2 − s2 ), hjk bk = αhj ,
hjkyk = 0, Cj yj = 0,
hj yj = 0, hjk bj bk = α2(b2 − s2 ).
Ej yj = 0.
Let J: =
n o bj Jj = α α(s0 + Γr0) − (Γr00 + Παs0 )s .
C: = E: =
bj Cj = (A1r00 + A2 αs0 )α(b2 − s2 ) + 3ΛJ, bj Ej = (B1 r00 + B2 αs0 )α(b2 − s2 ) + 3µJ. 5
It follows from the definitions of Cj and Ej that o n α(b2 − s2 )Cj − Chj = 3Λ α(b2 − s2 )Jj − Jhj , o n α(b2 − s2 )Ej − Ehj = 3µ α(b2 − s2 )Jj − Jhj .
(9) (10)
Lemma 3.1 The following three conditions are equivalent. (i) Ljkl = 0, (ii) Cj = 0 and Ej = 0, (iii) C = 0, E = 0 and µ[α(b2 − s2 )Jj − Jhj ] = 0. Proof: (i) ⇒ (ii) Assume that Ljkl = 0. Then by (8), hj hk Cl + hj hl Ck + hk hl Cj + 3Ej hkl + 3Ek hjl + 3El hkl = 0.
(11)
Contracting (11) with bj , bk and bl , we obtain (b2 − s2 )C + 3E = 0. Contracting (11) with bk and bl yields (b2 − s2 )Cj + 3Ej = 0.
(12)
Wjk := (b2 − s2 )hjk − hj hk .
(13)
Let By (12), (11) is equivalent to the following equation: Wjk Cl + Wjl Ck + Wkl Cj = 0.
(14)
Contracting (14) with bl yields Wjk C = 0. Since Wjk 6= 0, we obtain that C = 0. From (13) and (14), we get akl Wkl = (n − 2)α2 (b2 − s2 ),
akl Wjk Cl = αChj − α2 (b2 − s2 )Cj .
Contracting (14) with akl yields that Cj = 0. By (12), we get Ej = 0. (ii) ⇒ (i): Obvious by (8). (ii) ⇒ (iii): Assume that Cj = 0 and Ej = 0. Then C = 0 and E = 0. Then it follows from (10) that h i µ α2(b2 − s2 )Jj − Jhj = 0. (iii) ⇒ (ii): Assume that C = 0, E = 0 and µ[α(b2 − s2 )Jj − Jhj ] = 0. Then it follows from (9) and (10) that Cj = 0 and Ej = 0. This proves (ii). Q.E.D. 6
4
Proof of Theorem 1.1
We are going to show that Cj = 0 and Ej = 0 if and only if rij = sij = 0. In fact, we just need to study the equation Ej = 0. the difficulty is how to deal with the terms involving φ(β/α) in the equation Ej = 0. To overcome this difficulty, we change the y-coordinates (yi ) at a point to “polar” coordinates (s, yα ), where i = 1, · · · , n and α = 2, · · ·, n. Fix an arbitrary point x ∈ M . Take an orthonormal basis {ei } at x such that v u n uX α = t (yi )2, β = by1 . i=1
Fix an arbitrary number s with |s| ≤ b. It follows from β = sα that s y1 = √ α ¯, b2 − s2 where
v u n uX α ¯ := t (yα )2. α=2
Then
b α= √ α ¯, 2 b − s2
Let r¯10 :=
n X
r1αyα ,
r¯00 :=
α=2
β= √
n X
bs α ¯. − s2
b2
rαβ yα yβ ,
α,β=2
s¯0 :=
n X
sα yα .
α=2
We have s¯0 = b¯ s10 ,
s1 = bs11 = 0.
By a direct compution we obtain h i (b2 − s2 )3/2 α ¯ −1E1 = s2 (b2 − s2 )B1 + 3s(b2 − 2s2 )µΓ r11α ¯2 h i +(b2 − s2 ) (b2 − s2 )B1 − 6sµΓ r¯00 nh i p − b2 − s2 2s2 (6µΓ + sB1 ) − b2(3µΓ + 2sB1 ) r¯10 h i o ¯ +b2 s(6µΠ + sB2 ) − (3µ + b2 B2 ) s¯10 α
(b2 − s2 )2 Ea
=
n p b2 − s2 3b2µ((b2 Π − s)s1a + sΓra1 )¯ α2 i o h ¯ −(b2 − s2 ) b2 (6µΠ + sB2 )¯ s10 + 2s(6µΓ + sB1 )¯ r10 ya α nh i −(b2 − s2 ) s2 (6µΓ + sB1 )r11ya − 3b2µ(¯ sa0 + Γ¯ ra0 ) α ¯2 o +(b2 − s2 ) 6µΓ + sB1 r¯00ya . 7
Before we derive some equations on rij and sij , let us state the following trivial fact. pPn a 2 Lemma 4.1 If α ¯= a=2 (y ) satisfies the following equation ϕ + ψα ¯ = 0, where ϕ = ϕ(¯ y ) and ψ = ψ(¯ y ) are homogeneous polynomials in y¯ = (ya ), then ϕ = 0 and ψ = 0. Proof: First, we assume that deg ϕ = even. Then deg ψ = odd. We have ϕ(¯ y ) ± ψ(¯ y )¯ α(¯ y ) = 0. Thus ϕ(¯ y ) = 0 and ψ(¯ y ) = 0. If deg ϕ = odd, by a similar argument, we still get the same conclusion. Q.E.D. E1 = 0 implies that h i (b2 − s2 ) (b2 − s2 )B1 − 6sµΓ r¯00 i h +s 3(b2 − 2s2 )µΓ + s(b2 − s2 )B1 r11α ¯ 2 = 0, n o 2s2(6µΓ + sB1 ) − b2(3µΓ + 2sB1 ) r1a n o +b2 s(6µΠ + sB2 ) − (3µ + b2B2 ) s1a = 0. Ea = 0 implies that i h 3b2 µ sΓra1 + (b2Π − s)s1a α ¯2 i h −(b2 − s2 ) b2 (6µΠ + sB2 )¯ s10 + 2s(6µΓ + sB1 )¯ r10 ya = 0, h i s2 (6µΓ + sB1 )r11ya − 3b2 µ(¯ sa0 + Γ¯ ra0) α ¯2 +(b2 − s2 ) 6µΓ + sB1 r¯00ya = 0.
(15)
(16)
(17)
(18)
Contracting (17) and (18) with ya yields i h 3b2µ (b2Π − s)¯ s10 + sΓ¯ r10 i h −(b2 − s2 ) b2(6µΠ + sB2 )¯ (19) s10 + 2s(6µΓ + sB1 )¯ r10 = 0, s2 (6µΓ + sB1 )r11α ¯ 2 − 3b2 µΓ¯ r00 + (b2 − s2 ) 6µΓ + sB1 r¯00 = 0. (20)
Claim I: r11 = 0, rab = 0 and sab = 0.
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Proof: Assume that (r11, rab) 6= 0. Then the determinant of the linear system (15) and (20) in r11α ¯ 2 and r¯00 is −9sb2 µ2 Γ ≡ 0. Note that Γ 6= 0 for any s with |s| ≤ b. By assumption, b = b(x) 6= 0. Thus we have µ ≡ 0. We get p φ = 1 + ks2 . (21) This contradicts our assumption. Thus r11 = 0 and rab = 0. Now (18) is reduced to −3b2µ¯ sa0 α ¯ 2 = 0. If (sab ) 6= 0, then µ ≡ 0. Then φ is given by (21). This is a contradiction. Thus sab = 0. Claim II: r1a = 0 and s1a = 0. Proof: Since r11 = 0 and rab = 0, we can only use (16), (17) and (19). By (17) and (19), we get h i i h sΓr1a + (b2Π − s)s1a α ¯ 2 = sΓr10 + (b2 Π − s)s10 ya , (22) h i 2s(6µΓ + sB1 )r1a + b2(6µΠ + sB2 )s1a α ¯2 i h = 2s(6µΓ + sB1 )¯ r10 + b2 (6µΠ + sB2 )¯ s10 ya . (23) Now we assume that n ≥ 3. It follows from (22) and (23) that sΓr1a + (b2 Π − s)s1a = 0,
(24)
2s(6µΓ + sB1 )r1a + b2(6µΠ + sB2 )s1a = 0.
(25)
By (25), the equation (16) can be reduced to (3µΓ + 2sB1 )r1a + (3µ + b2 B2 )s1a = 0.
(26)
It follows from (25) and (26) that 3sµΓr1a + µ(2b2 Π − s)s1a = 0. The determinant of linear equations (24) and (27) in r1a and s1a is sµΓ(2s − b2Π). If (r1a , s1a) 6= (0, 0), then µ(2s − b2Π) ≡ 0. Let
n o I := s ∈ (−bo , bo), µ(s) 6= 0. . 9
(27)
We assert that I = ∅. Grant √ this for a moment, we see that µ ≡ 0. Solving this equation, we get φ = 1 + ks2 , ∀s ∈ (−bo , bo ). This is impossible by assumption in Theorem 1.1. Now we prove the assertation that I = ∅ under the assumption that (r1a, s1a ) 6= (0, 0). Suppose that I 6= ∅. Then 2s − b2Π ≡ 0,
∀s ∈ I.
Then (24) is reduced to Γr1a + s1a = 0,
∀s ∈ I.
(28)
From the definition of Γ and Π, we have Π+Γ
φ0 = 0. φ − sφ0
(29)
Let us first prove that r10 = 0. If r1a 6= 0 for some 2 ≤ a ≤ n, then by (28), Γ is a constant on I. On the other hand, by (29), 2s φ0 + Γ = 0. b2 φ − sφ0 We get φ = C1
p 2s2 − Γb2,
s ∈ I.
By a direct computation, we can easily verify that µ(s) = 0 for all s ∈ I. This is imposible by the definition of I. Thus we conclude that r10 = 0. By (28), we further get s10 = 0. That is, (r1a , s1a) = 0. This contradicts the above assumption that (r1a, s1a ) 6= 0. Therefore I = ∅. Q.E.D.
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[6] C. Shibata, H. Shimada, M. Azuma and H. Yosuda, On Finsler spaces with Randers’ metric, Tensor, N. S. 31(1977), 219-226. [7] Z.I. Szab´ o, Positive definite Berwald spaces (Structure theorems on Berwald spaces), Tensor, N. S. 35(1981), 25-39.
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