Hi, I'm trying to show that if ## M^n ## is orientable and connected, with boundary (say with just one boundary component), then its top homology is zero. Sorry, I have not done much differential topology/geometry in a while.(adsbygoogle = window.adsbygoogle || []).push({});

I'm trying to avoid using Mayer-Vietoris, by using this argument; but I don't think this is too rigorous; please critique: let B be the boundary. Now, B is an (n-1)-submanifold of M, and so it must bound some n-dimensional manifold. Since M is orientable and n-dimensional, then B must bound M itself. But then M, as the top cycle is a cycle that bounds, so the top cycle is a trivial cycle, and so the homology is zero.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Top Homology of Connected , Orientable Manifolds with Boundary

Loading...

Similar Threads for Homology Connected Orientable |
---|

A Connection 1-forms to Christoffel symbols |

I Connections on principal bundles |

A Is the Berry connection a Levi-Civita connection? |

A Can you give an example of a non-Levi Civita connection? |

I Arbitrariness of connection and arrow on sphere |

**Physics Forums | Science Articles, Homework Help, Discussion**