# Totally geodesic submanifold of codimension 1 in noncompact Riemannian manifold

Assume that $M$ is a noncompact complete simply connected manifold of nonnegative sectional curvature. Then by Soul theorem, it has a soul $S$.

Question 1 : Fix a point $p\in S$. Then there is a totally geodesic submanifold of codimension 1 passing through $p$

example : In $\mathbb{E}^3$, $z=x^2+y^2$ has a non-selfintersecting infinite geodesic.

Question 2 : If $M$ is a noncompact complete Riemannian manifold and it contains a line $l$, then for any $p\in l$, $M$ has a totally geodesic submanifold $N$ of codimension 1 s.t. $N$ intersects the line transversally at $p$.

• I don't understand the "proof" part. What are you trying to prove ? What is $p_i$ ? – M. Dus Aug 15 '18 at 9:45
• Sorry. I editted. – Hee Kwon Lee Aug 15 '18 at 9:53
• I still don't understand the proof part. What does "Yim" mean? What is the verb in the sentence comprising the proof? – Ben McKay Aug 15 '18 at 10:54
• After arbitrarily small perturbation (in your favourite topology) of any Riemannian metric on any manifold of dimension 3 or more, I believe that there are no codimension 1 totally geodesic submanifolds, although I don't have a proof at hand. – Ben McKay Aug 15 '18 at 10:56