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X is an n-dim positively curved manifold and Y is a totally geodesic submanifold of codimension 1.Then cutting along Y we get n-dim positively curved manifolds without boundary,by soul theorem these manifolds should be homeomorphic to R^n.Am I right?If not,please give counterexamples.

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yes, it is actually diffeomorphic to $\mathbb R^n$. –  J. GE Sep 11 '12 at 12:21
    
Not sure what GB meant, but surely removing a geodesic from a paraboloid one can make it disconnected, so it is not homeomorphic to $\mathbb R^2$. On a more serious note the soul theorem needs completeness (or total convexity), and even if $X$ were assumed complete, $X-Y$ won't be, so the soul theorem is irrelevant. –  Igor Belegradek Sep 11 '12 at 12:31
    
@Igor, I think he meant closed manifold and the distance function to the boundary, which is $Y$ is strictly concave, hence the interior of the component is diffeomorphic to $\mathbb R^n$. Although this is not called Soul Theorem, but follows the same idea. –  J. GE Sep 11 '12 at 12:36
    
@GB: removing an equator from the round sphere makes it disconnected. Do you imply that components of $X-Y$ are always totally convex? If so, why? –  Igor Belegradek Sep 11 '12 at 12:50
    
I mean X-Y may be more than one connected component,but each of them is manifold homeo to R^n,right? –  jiangsaiyin Sep 11 '12 at 12:58
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1 Answer

If I understood your question correctly, the answer is yes. More precisely, the following statement should hold:

If $X$ is a closed manifold of positive sectional curvature and $Y\subset X$ is a codimension one totally geodesic submanifold that disconnects $X$, then $X$ is homeomorphic to a sphere.

This follows, as the OP suggests, from the Soul Argument of Cheeger-Gromoll, extended to Alexandrov spaces by Perelman (see, e.g., Section 6 of Perelman's notes). As mentioned in the comments, Cheeger-Gromoll's version of the argument actually suffices to get the conclusion.

A few details: denote by $C_1$ and $C_2$ the closure of the two connected components of $X\setminus Y$. These are positively curved compact Alexandrov spaces with boundary $Y$. On each of them, since the curvature is positive, the distance function to the boundary is concave. Therefore, the set of points at maximal distance (the soul) consists of a unique point. This implies that each $C_i$ is homeomorphic to a disk, hence $X=C_1\cup_{Y} C_2$ is a twisted sphere.


edit (to answer GB's comment): As discussed above, if $C$ a compact Alexandrov space with curvatures $\geq k>0$, then the soul $S=\{p\}$ is a point. Moreover, according to Perelman, the pairs $(C,\partial C)$ and $(\overline K(\Sigma_S),\Sigma_S)$ are homeomorphic (see 6.2 for proof), where $\Sigma_S$ is the space of directions at the soul and $\overline K(\Sigma_S)$ is the closure of the topological cone over $\Sigma_S$, i.e., the join of $\Sigma_S$ and a point. If $C$ is a manifold, the space of directions are spheres, so $(\overline K(\Sigma_S),\Sigma_S)$ is simply a pair $(D,\partial D)$, where $D$ is a disk.

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@Renato, This argument is considered by Cheeger-Gromoll first. And here we don't need Alexandrov space theory to get the conclusion. –  J. GE Sep 11 '12 at 19:45
    
@GB: Yes, you're right, thank you. I edited the post to mention this. –  Renato G Bettiol Sep 11 '12 at 20:22
    
@Renato, it seems you need more argument to conclude that Ci is homeomorphic to a closed disk. Since for Alexandrov space this is not true. It is not trivial, right? –  J. GE Sep 11 '12 at 20:30
    
@Renato,your answer is very good!But you assume X is closed,what is the result when X is not closed?And the Alexandrov space case? –  jiangsaiyin Sep 12 '12 at 7:50
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@jiangsaiyin. When $X$ is not closed, $X$ is diffeomorphic to $\mathbb R^n$ by Gromoll-Meyer. For ALexandrov space, Renato already pointed out it is homeomorphic to the closed cone over the space of directions at the soul point. (interior is homeomorphic to the open cone). –  J. GE Sep 12 '12 at 9:28
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