Positively curved manifold with a codimension 1 totally geodesic submanifold. Fact : Consider the inclusion $V^{n-1} \rightarrow M^n$ where $M$ is a closed orientable simply 
 connected positively curved manifold.
Then connectivity lemma implies that the inclusion is $(n-1)$-connected so that 
  $M$ is homeomorphic to a sphere.
Situation : As you know ther exists a $S^3$-action on $M={\bf CP}^2$ which is cohomogeneity one, i.e., $M/S^2=[0,1]$. 
Hence $M$ is the union of two disk bundles over two singular orbits $S^3\cdot x_1$, $S^3\cdot x_2$ : $S^3\cdot x_1$ is diffeomorphic to $S^2$ and $ S^3\cdot x_2 = \{ x_2\}$. 
Here think about two conditions : 
(1) A geodesic sphere of suitable radius around $x_2$ is totally geodesic. 
(2) $M$ is positively curved. 
By the above fact (1) and (2) cannot be compatible. 
Here I have a question : Is it true that 
  cannonical $S^5(1)/S^1={\bf CP}^2$ does not have a codimensional 
  1 totally geodesic submanifold. 
Thank you in advance. 
 A: $\mathbb{CP}^n$ (with $n>1$) does indeed not have any codimension $1$ totally geodesic manifold; neither does $\mathbb{CH}^n$. You can probably find a proof in Goldman's book on complex hyperbolic geometry.
(Added later: this is true even locally: there are no open codimension $1$ totally geodesic manifold in $\mathbb{CP}^n$ nor in $\mathbb{CH}^n$.)
Note that this is an important geometrical fact, as (as far as I know) all proofs of the isoperimetric inequality that work in the real hyperbolic space use reflexions with respect to a totally geodesic codimension one manifold. This explains why we still don't know if balls are optimal for the isoperimetric problem in $\mathbb{CH}^n$ and $\mathbb{CP}^n$ (small balls in the latter case, as for large volumes balls are known not to be optimal).
Also note that it is a source of great difficulty in the study of subgroups of isometries of $\mathbb{CH}^n$: for many groups $\Gamma$ acting isometrically on $\mathbb{CH}^n$, we do not know whether they are discrete; one cannot construct a fundamental domain with geodesic faces that could be used to prove discreteness, as it is done in real hyperbolic geometry. We are therefore mainly left with arithmetic methods, and to find non-arithmetic lattices of $\mathrm{SU}(1,n)$ is an important problem, see notably the work of Martin Deraux.
A: Compact positively curved manifold with totally geodesic hypersurface has to be sphere, or its double cover has to be a sphere.
(Search for "geodesic hypersurface" in my puzzles.)
To prove this cut along the hypersurface and apply the soul theorem to each part.
