On some 3-dimensional Riemannian manifolds
On some 3-dimensional Riemannian manifolds
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关于一些 3 维黎曼流形
DOI：
10.14492/hokmj/1381758986
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发表时间：
1973
期刊：
影响因子：
--
通讯作者：
K. Sekigawa
中科院分区：
文献类型：
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作者：
K. Sekigawa
其中R(X，Y)在R上运算，作为M的每个点的张量代数的导数。相反，曲率张量场R上的这个代数条件是否意味着R=0？K.Nomizu猜想，当(M，g)是完全不可约且维Mgeqq3时，答案是肯定的。但最近，H.Takagi[9]给出了满足(^{*})的三维完备不可约实解析黎曼流形(M，g)的一个例子，并使R\neq0成为4维欧氏空间E^{4}中的一个超曲面。此外，本文证明了在(m+1)维欧氏空间中
where R(X, Y) operates on R as a derivation of the tensor algebra at each point of M. Conversely, does this algebraic condition on the curvature tensor field R imply that \nabla R=0 ? K. Nomizu conjectured that the answer is positive in the case where (M, g) is complete irreducible and dim M\geqq 3 . But, recently, H. Takagi [9] gave an example of 3-dimensional complete, irreducible real analytic Riemannian manifold (M, g) satisfying (^{*}) and \nabla R\neq 0 as a hypersurface in a 4-dimensional Euclidean space E^{4} . Furthermore, the present author proved that, in an (m+1)-dimensional Euclidean space