Timeline for a result of soul theorem,right?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 17, 2012 at 7:21 | comment | added | jiangsaiyin | @Renato:My algebraic topology is poor.Why H_1(X,X-Y) iso to Z?And the exact sequence is ...-H_1(X-Y)-H_1(X)-H_1(X,X-Y)-H_0(X-Y)-0,I just know X-Y is connected if and only if H_0(X-Y)=Z,how can you get your last statement? | |
| Sep 14, 2012 at 15:58 | comment | added | Renato G. Bettiol | @jiangsaiyin: Yes, take, e.g., the equatorial embedding of $Y=\mathbb RP^n$ inside $X=\mathbb RP^{n+1}$. Then $X\setminus Y$ only has one connected component. However, up to a double cover, the embedded hypersurface disconnects the manifold. The number of connected components of $X\setminus Y$ is the rank of $\tilde H_0(X\setminus Y)$ plus $1$. You can use the long exact sequence on the homology of the pair $(X,X\setminus Y)$ to prove that $X\setminus Y$ is connected if and only if the morphism $H_1(X)\to H_1(X,X\setminus Y)\cong\mathbb Z$ is nontrivial. | |
| Sep 14, 2012 at 7:21 | comment | added | jiangsaiyin | @Renato:If X is a closed manifold of positive sectional curvature and Y⊆X is a codimension one totally geodesic submanifold.Is it possible that Y doen not disconnect X? | |
| Sep 12, 2012 at 9:28 | comment | added | J. GE | @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). | |
| Sep 12, 2012 at 7:50 | comment | added | jiangsaiyin | @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? | |
| Sep 11, 2012 at 23:26 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
Answer GB's comment
|
| Sep 11, 2012 at 20:30 | comment | added | J. GE | @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? | |
| Sep 11, 2012 at 20:22 | comment | added | Renato G. Bettiol | @GB: Yes, you're right, thank you. I edited the post to mention this. | |
| Sep 11, 2012 at 20:21 | history | edited | Renato G. Bettiol | CC BY-SA 3.0 |
Incorporated that Cheeger-Gromoll's version is enough
|
| Sep 11, 2012 at 19:45 | comment | added | J. GE | @Renato, This argument is considered by Cheeger-Gromoll first. And here we don't need Alexandrov space theory to get the conclusion. | |
| Sep 11, 2012 at 18:31 | history | answered | Renato G. Bettiol | CC BY-SA 3.0 |