On Ricci curvature of isotropic and Lagrangian submanifolds in complex space forms
Let \((M^{n},g)\) be a Riemannian manifold isometrically immersed in \((\tilde{M}^{m},g).\) Let us denote by \(S,H^{2}\) the Ricci tensor and the squared mean curvature, respectively. In a previous paper [\textit{B.-Y. Chen}, Glasg. Math. J. 41, 33--41 (1999; Zbl 0962.53015)], the author proved the following basic inequality for any submanifold \((M^{n},g)\) of a real space form \(R^{m}(c):\) \(S\leq \left( \left( n-1)c\right) +\frac{n^{2}}{4}H^{2}\right) ,\) with equality holding, if and only if, either \((M^{n},g)\) is totally geodesic or \(n=2\) and \((M^{2},g)\) is a totally umbilical surface. Continuing a series of important results concerning basic inequalities for Riemannian submanifolds, the author first proves in this paper, a similar result to the above one for totally real submanifolds \((M^{n},g)\) of complex space forms \((\tilde{M}^{m}(4c),g).\) Then, he obtains that a Lagrangian submanifold \((M^{n},g)\) satisfying the equality \(\overline{\text{Ric}}=\left(\left( n-1)c\right) +\frac{n^{2}}{4}H^{2}\right) \) ( where \(\overline{\text{Ric}}\) is the maximum Ricci curvature function) must be a minimal submanifold. As a consequence of the proof, he shows that these submanifolds must have a relative null space of dimension one at least. The final part is devoted to the geometric description of the Lagrangian submanifolds satisfying the above equality and such that the dimension of the relative null space is constant.
scientific article; zbMATH DE number 1437102
scientific article