摘要
证明了如下结果:设x:M^2→S^n(n≥3)是不含脐点的曲面,若x的法丛平坦,则x的Moebius形式φ平行(φ=0)当且仅当φ4=0。
In this paper we prove the result that if x :M2→Sn(n≥3) is an umbilic-free surface with flat Moebius normal bundle then its Moebius form is parallel (▽φ=0) if and only if the Moebius form vanishes (φ=0).
出处
《数学研究》
CSCD
2003年第3期235-242,共8页
Journal of Mathematical Study