Let $M$ be a Riemannian manifold, and let $S$ be a hypersurface (codimension $1$). If $S$ is completely geodesic, does that imply that it is minimal? If not, what are the conditions? If yes, is there a stronger result (e.g. the normal curvature is zero)?
The converse is clearly false (think of any nontrivial minimal surface in $\Bbb{R}^3$).
What happens for general submanifolds?
A reference would also be welcome.
Thanks.
Yes, if $S$ is any totally geodesic submanifold of a Riemannian manifold, then $S$ is a minimal submanifold. This holds true in all codimensions, not just for hypersurfaces. (Note that the terms minimal submanifold, minimal hypersurface, and minimal surface all mean submanifolds that are critical points for the area functional, not necessarily ones that are area-minimizing.)
A totally geodesic submanifold is one for which the (vector-valued) second fundamental form is identically zero, while a minimal submanifold is one for which the mean curvature vector is everywhere zero. Since the mean curvature vector is the trace of the vector-valued second fundamental form, the result follows.