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.


$\mathbb{CP}^n$ (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.

  • $\begingroup$ @Benoit, do you know any references or survey articles regarding the isoperemetric problem in $CP^n$ or $CH^n$? $\endgroup$ – Ralph May 1 '13 at 21:02
  • $\begingroup$ @Ralph: no. As far as I know, very little has been written on this and the best lower bounds we have are the optimal linear bound (good for large domains) in complex dimension $2$ the Euclidean inequality (both for $\mathbb{CH}^n$) and optimal asymptotic bounds for small domains in all dimension (in a non-explicit sense). All are consequences of more general result (Yau's linear isoperimetric inequality, Croke's inequality for $4$-manifold with non-positive curvature, and Druet's inequality for manifolds with scalar curvature bounded above). $\endgroup$ – Benoît Kloeckner May 2 '13 at 13:39
  • $\begingroup$ (...) I just remember that the problem is posed in a survey by Choe, Ritoré or both of them, but I could not get my hand on it. $\endgroup$ – Benoît Kloeckner May 2 '13 at 13:41

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.

  • $\begingroup$ Thank you first. The above fact is about open problem : Compact positively curved manifold with totally geodesic hypersurface is diffeomorphic to be sphere, or its double cover has to be a sphere. As far as I know soul theorem is that non-compact complete nonnegatively curved manifold is diffeomorphic to normal bundle over a soul. But if we cut closed manifold along hypersurface then it is not complete. Hence how can we apply the soul theorem ? $\endgroup$ – Hee Kwon Lee May 1 '13 at 2:38
  • 2
    $\begingroup$ @Hee Kwon Lee, there is a version of soul theorem for manifolds with boundary (the proof is exactly the same). Alternatively, you may take the boundary$\times\mathbb{R}_\ge$ and glue it in; this way you get complete manifold without boundary. $\endgroup$ – Anton Petrunin May 1 '13 at 4:06

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