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. 



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> **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 $\mathbb{C}P^2$. 
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**Reference :** [Convex sets in Alexandrov spaces](https://mathoverflow.net/questions/289356/convex-sets-in-alexandrov-spaces)



Totally geodesic submanifold of codimension 1
There is no three-dimensional totally geodesic submanifold in $\mathbb{CP}^2$. 


