Properties of Berwald scalar curvature
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Properties of Berwald scalar curvature Ming LI1 , Lihong ZHANG2 1 Mathematical Science Research Center, Chongqing University of Technology, Chongqing 400054, China 2 School of Sciences, Chongqing University of Technology, Chongqing 400054, China
c Higher Education Press 2020
Abstract We prove that a Finsler manifold with vanishing Berwald scalar curvature has zero E-curvature. As a consequence, Landsberg manifolds with vanishing Berwald scalar curvature are Berwald manifolds. For (α, β)-metrics on manifold of dimension greater than 2, if the mean Landsberg curvature and the Berwald scalar curvature both vanish, then the Berwald curvature also vanishes. Keywords Landsberg curvature, Berwald curvature, E-curvature, S-curvature Berwald scalar curvature MSC2020 53B40, 53C60, 53A15 1
Introduction
Let (M, F ) be an n-dimensional Finsler manifold. Let SM = {F (x, y) = 1} be the unit sphere bundle (or indicatrix bundle) with the natural projection π : SM → M. Let ω = Fyi dxi be the Hilbert form, which defines a contact structure on SM. The dual of ω with respect to the Sasaki-type metric on SM is the Reeb vector filed ξ. It is verified that ξ is the restriction of the spray G on SM. Let D be the contact distribution {ω = 0}. The Berwald curvature B as a part of the curvature endomorphism of the Berwald connection is divided into four parts. Along the Reeb vector filed direction B(ξ) = 0, however, the transpose of B along ω gives twice the Landsberg curvature B t (ω) = 2L. The remain part of B on D is in general not a symmetric endomorphism. B is totally symmetric on D when L = 0. M is called a Berwald manifold if B = 0. Let E = F · tr B be the mean Berwald curvature or the E-curvature. There are examples with vanishing E-curvature which are not Berwald manifolds. Li [6] proved that L = 0 and E = 0 imply B = 0. It is natural to consider the scalar e := tr E, which will be called the Berwald scalar curvature. In general, e is a function on SM, and the symmetric tensor Received June 18, 2020; accepted October 8, 2020 Corresponding author: Ming LI, E-mail: [email protected]
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Ming LI, Lihong ZHANG
E is not isotropic. The main result of this paper is the following property about the Berwald scalar curvature. Theorem 1 Let (M, F ) be an n-dimensional Finsler manifold. If e = tr E is a function on M, then E is isotropic: E=
1 eh(·, J·), n−1
where h is the angular metric, J the almost complex structure on D. In this case, the Finsler manifold has weak isotropic S-curvature. A direct consequence is that vanishing Berwald scalar curvature implies vanishing E-curvature. The following theorem improves the above mentioned result in [6]. Theorem 2 Let (M, F ) be an n-dimensional Landsberg manifold. If the Berwald scalar curvature vanishes, then M is a Berwald manifold. For Finsler manifolds with (α, β)-metrics, we obtain the following result. Theorem 3 For n > 3, let M be an n-dimensional manifold equipped an (α, β)-metric F = αφ(s), where s = β/α, α is a Riemannian metric, β an one form on M, and φ a smooth function of real
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