(1) If $N^k$ is a submanifold in a compact Riemannian manifold $M^{k+m},\ m\geq 1$ s.t. each $p\in N$ has the following property : There exists independent set $\{ X_i\}_{i=1}^k$ tangent to $T_pN$ s.t. $$\sum_{i=1}^k( \nabla_{X_i}n,{X_i})=0$$ for any $n$ which is any unit normal to $N$, then $N$ is not totally geodesic in general. Is this true ?

If so, can you give an example ?

In $M={\bf R}^3$, i.e., $M$ is noncompact, hyperbolid is not totally geodesic but it satisfies the above condition.

(2) We can give an example under more strong condition ? : $ (\nabla_{X_i}n,{X_i})=0$ for all $i$

Thank you in anticipaction

  • 1
    $\begingroup$ Is $N$ an hypersurface ? And what is the counter example in $\mathbb{R}^3$ ? $\endgroup$ Mar 17 '16 at 9:45
  • 2
    $\begingroup$ No just a submanifold. In ${\bf R}^3$, hyperbolid is minimal but not totally geodesic $\endgroup$ Mar 17 '16 at 9:47
  • 3
    $\begingroup$ You mean the catenoid right ? and minimal (say for hyper surfaces for simplicity) is by definition that the trace of the Weingarten map $X\mapsto \nabla_X n$ vanishes, so a minimal surface will not satisfy $\langle\nabla_Xn,X\rangle=0$ for every $X$. If I understand correctly your condition is the vanishing of the Weingarten map (or second fundamental form for that matter form), which is equivalent to being totally geodesic. $\endgroup$ Mar 17 '16 at 9:53
  • $\begingroup$ Just to be sure, you ask for your conditions to hold for every normal vector field $n$ ? $\endgroup$ Mar 17 '16 at 10:13
  • $\begingroup$ Yes any unit normal vector field on $N$ $\endgroup$ Mar 17 '16 at 10:15

Let us consider the case of hypersufaces for the sake of simplicity. The higher codimension case follows by taking products with $(S^1)^k$.

Observation : Minimal submanifolds satisfy both of your conditions. In fact in this case the vectors $X_i$ can be taken orthonormal. In fact if you ask for $X_i$ to be orthonormal your first condition asks for the second fundamental form $B(X,Y)=\langle \nabla_Xn,Y\rangle$ to have zero trace (so the mean curvature is minimal), while the second condition asks for the existence of an orthonormal basis of vectors which are isotropic for $B$ which is equivalent to the vanishing of the trace for a quadratic form.

So if we can find a compact manifold with a non totally geodesic minimal hypersurface, we can answer yes to both of your questions. Schwarz's $P$-surface in $\mathbb{R}^3$ is invariant by the action of $\mathbb{Z}^3$ by translation, and is not totally geodesic, so that it gives a non totally geodesic minimal surface in $\mathbb{T}^3$.

See here for Schwartz surface : wiki.

I suspect there are examples of non totally geodesic minimal submanifolds of any codimension between $1$ and $n-2$ in any compact manifold $M$ of dimension bigger than $3$ but I don't know a way to prove it on top of my head. However the construction I can think of may produce singularities, I am not sure how one can avoid that.


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