IJPAM: Volume 72, No. 3 (2011)

ON SUBMANIFOLDS WITH CONSTANT MEAN CURVATURE
IN A REAL SPACE FORM

Shingo Kikuchi$^1$, Yoshio Matsuyama$^2$
$^{1,2}$Department of Mathematics
Chuo University
1-13-27, Kasuga, Bunkyo-ku, Tokyo, 112-8551, JAPAN


Abstract. The purpose of this paper is to classify submanifolds with constant mean curvature in a real space form. We put $S$ the squared norm of the second fundamental form and $\vert\phi\vert^{2} = S - n H^{2}$. Denote by $A_{H}$ and $B_{H}$ the squares of the positive roots of the equations

\begin{displaymath}
x^{2} + \frac{n(n-2)}{\sqrt{n(n-1)}}Hx - n(H^{2} + c) = 0\ ...
...3}{2}x^{2} + \frac{n(n-2)}{\sqrt{n(n-1)}}Hx - n(H^{2} + c) = 0,\end{displaymath}

respectively. We prove the following: First, le $M^{n}$ be a complete, connected and orientable submanifold with nonzero constant mean curvature $H$ in $S^{n+p}(c) (p \geq 3)$. If $\vert\phi\vert$ satisfies $\vert\phi^{2} \leq B_{H}$ for all $x \in M^{n}$, then $M^{n}$ lies in a totally geodesic submanifold $S^{n+1}(c)$ of $S^{n+p}(c)$, and $\vert\phi\vert^{2} \equiv 0$ and $M^{n}$ is totally umbilic. Next, let $M^{n}$ be a complete, connected and orientable hypersurface with constant mean curvature $H > 1$ in $H^{n+1}(-1)$. Assume that $\vert\phi\vert^{2} \leq A_{H}$ for all $x \in M^{n}$. Then (i) either $\vert\phi\vert^{2} \equiv 0$ and $M^{n}$ is totally umbilic or $\vert\phi\vert^{2} \equiv A_{H}$. (ii) $\vert\phi\vert^{2} \equiv A_{H}$ if and only if $M^{n}$ is isometric to $S^{n-r} \times H^{1}(- \frac{1}{r^{2} + 1})$ for some $r > 0$. Moreover, we prove a gereralization of this result of the hypersurface in a hyperbolic space.

Received: September 1, 2011

AMS Subject Classification: 53C40, 53B25

Key Words and Phrases: reduction of the codimension, parallel second fundamental form, totally umbilic

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Source: International Journal of Pure and Applied Mathematics
ISSN printed version: 1311-8080
ISSN on-line version: 1314-3395
Year: 2011
Volume: 72
Issue: 3