Complete Manifolds with Harmonic Curvature and Finite $L^p$-Norm Curvature
Received:January 25, 2016  Revised:February 27, 2017
Key Word: Harmonic curvature   trace-free curvature tensor   constant curvature space  
Fund ProjectL:Supported by the National Natural Science Foundations of China (Grant Nos.11261038; 11361041) and the Natural Science Foundation of Jiangxi Province (Grant No.20132BAB201005).
Author NameAffiliation
Haiping FU Department of Mathematics, Nanchang University, Jiangxi 330031, P. R. China 
Pingping DAN Department of Mathematics, Nanchang University, Jiangxi 330031, P. R. China 
Shulin SONG School of IOT Engineering, Jiangnan University, Jiangsu 214122, P. R. China 
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Abstract:
      Let $(M^n, g)~(n\geq3)$ be an $n$-dimensional complete Riemannian manifold with harmonic curvature and positive Yamabe constant. Denote by $R$ and $\mathring{Rm}$ the scalar curvature and the trace-free Riemannian curvature tensor of $M$, respectively. The main result of this paper states that $\mathring{Rm}$ goes to zero uniformly at infinity if for $p\geq n$, the $L^{p}$-norm of $\mathring{Rm}$ is finite. As applications, we prove that $(M^n, g)$ is compact if the $L^{p}$-norm of $\mathring{Rm}$ is finite and $R$ is positive, and $(M^n, g)$ is scalar flat if $(M^n, g)$ is a complete noncompact manifold with nonnegative scalar curvature and finite $L^{p}$-norm of $\mathring{Rm}$. We prove that $(M^n, g)$ is isometric to a spherical space form if for $p\geq \frac n2$, the $L^{p}$-norm of $\mathring{Rm}$ is sufficiently small and $R$ is positive. In particular, we prove that $(M^n, g)$ is isometric to a spherical space form if for $p\geq n$, $R$ is positive and the $L^{p}$-norm of $\mathring{Rm}$ is pinched in $[0,C)$, where $C$ is an explicit positive constant depending only on $n, p$, $R$ and the Yamabe constant.
Citation:
DOI:10.3770/j.issn:2095-2651.2017.03.011
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