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This question is inspired by question in reference.

Question : If $M$ is a simply connected closed Riemannian manifold of nonnegative sectional curvature, then there is a totally geodesic submanifold $S$ of codimension 1

Def : ${\rm conv}\ X$ is a smallest closed convex set containing $X$. And a subset $X$ is convex if for $x,\ y\in X$, any shortest geodesic between them is in $X$.

Construction of $S$ : Let $X_1=B(p,\delta)$ to be a closed ball, which is convex. If $p_1$ is not in $X_1$, then let $X_2={\rm conv}\ X_1\bigcup\{p_1\}$ s.t. $d_H(X_1,X_2) >0$ is small and ${\rm vol}\ X_2-{\rm vol}\ X_1 >0$ is small. Hence if we repeat this process, then $X_n$ goes to $ X_\infty$.

If $p_\infty$ is not in $X_\infty$, and $d_H(X_\infty,{\rm conv}\ X_\infty\bigcup\{p_\infty\})$ is large, then $\partial X_\infty$ is a desired one.

If not, we let $X_1=X_\infty$ and repeat these process. That is, ${\rm vol}\ X_\infty$ is supremum of volumes of convex sets containing $p$.

Example : If we starts this process at a point of negative sectional curvature in torus of revolution, then $X_\infty$ is closure of set of all points of negative sectional curvature.

Additional question : We have alvarezpaiva's answer. But I want to know in dimension $3$. Or I want to know why my argument fails in $\mathbb{C}P^2$.

Reference : Convex sets in Alexandrov spaces

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    $\begingroup$ As stated it is unclear (to me) what you are asking and what you are proving in the "proof". Also (to me) the proof does not make sense: you never explain why $X_2$ that satisfies your assumptions exists. What seems to be happening is that you propose some construction for a convex subset, and then ask whether the construction produces a totally geodesic submanifold of codimension $1$. $\endgroup$
    – Igor Belegradek
    Commented Sep 1, 2018 at 17:25
  • $\begingroup$ "you never explain why $ X_2$ that satisfies your assumptions exists" In $\mathbb{C}P^2$, by considering following answer, this may not hold. This is my question. $\endgroup$
    – Hee Kwon Lee
    Commented Sep 1, 2018 at 17:50
  • $\begingroup$ What I am saying is that the question does not logically make sense. What is ``small'' in the definition of $X_2$? What do you mean by the boundary of $X_\infty$? Why do you think the construction of $S$ yields a submanifold? If it is a submanifold, why do you expect it have codimension one? Why is it totally geodesic? $\endgroup$
    – Igor Belegradek
    Commented Sep 1, 2018 at 18:54
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    $\begingroup$ The main result in arxiv.org/pdf/1703.09240.pdf gives an answer to your question (the way I understand it). Also in the introduction they mention that Tsukada found left-invariant metrics on 3–dimensional Lie groups without totally geodesic surfaces. $\endgroup$
    – Igor Belegradek
    Commented Sep 1, 2018 at 19:32

2 Answers 2

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There is no three-dimensional totally geodesic submanifold in $\mathbb{CP}^2$.

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About you second question.

If $S$ is a totally geodesic submanifold, then for any geodesic in $S$ the tangent planes $\mathrm{T}S$ are invariant with respect to parallel translations and Jacobi fields with ends in $\mathrm{T}S$ have to be in $\mathrm{T}S$. If $\dim S>1$, then it gives a nontrivial equation on the curvature tensor along the geodesic.

So generic 3-dimensional manifold does not admit a geodesic sumbmanifold of dimension at least 2. In particular, any metric on a 3-manifold admits arbitrary small $C^\infty$ deformation such that the obtained Riemannian manifold has no totally geodesic surfaces, even locally.

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