Timeline for What is the top cohomology group of a non-compact, non-orientable manifold?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2021 at 21:35 | vote | accept | Georges Elencwajg | ||
| May 28, 2021 at 21:32 | comment | added | Georges Elencwajg | Thank you, @Z.M. (Although I'm afraid I don't know Borel-Moore homology) | |
| May 28, 2021 at 16:26 | comment | added | Z. M | @GeorgesElencwajg As the case for $H_n(M)$, this should be a consequence of Poincaré duality, which identifies $H^n(M)$ with the Borel-Moore homology $H_0^{\operatorname{BM}}(M)\cong\operatorname{Hom}_{\mathbb Z}(H_c^0(M),\mathbb Z)$. | |
| May 28, 2021 at 15:57 | comment | added | Z. M | Sorry for being stupid. This sequence is essentially $0\to\mathbb Z\to\mathbb Z[C_2]\to\mathbb Z\to0$ of $C_2$-groups, which is indeed non-split. | |
| May 28, 2021 at 7:23 | comment | added | Georges Elencwajg | Dear Tom, thank you very much again for quickly (and very satisfyingly!) answering the question in my comment. If I may be so bold as to abuse your patience once more: I "knew" that orientable non-compact manifolds satisfy $H^n(M,\mathbb Z)=0$, a fact indeed applied in your answer to $\tilde M$. Do you have an easy explanation or a reference for that statement, since I must confess that I don't remember why it is true! | |
| May 28, 2021 at 6:36 | comment | added | Z. M | It should be possible to find a short exact sequence of sheaves on $M$ which recovers the exact sequence above. The first should be the orientation sheaf and the third should be constant $\mathbb Z$. What is their extension? I am a bit confused. | |
| May 28, 2021 at 0:19 | comment | added | Tom Goodwillie | The local system that I am calling $\mathbb Z^t$ provides a group $\mathbb Z^t(x)$ at each point $x\in M$, namely the infinite cyclic group whose two generators are the two orientations of $M$ at $x$. An element of $C^k(M;\mathbb Z^t)$ is a function that assigns to each singular $k$-simplex $\sigma$ of $M$ an element of the group associated with a point of $\sigma$ (say, the number $0$ vertex). | |
| May 27, 2021 at 18:14 | comment | added | Georges Elencwajg | Thank you for this answer, Tom. What exactly are the elements of the group $C^k (M;\mathbb Z^t)$? | |
| May 27, 2021 at 0:55 | history | answered | Tom Goodwillie | CC BY-SA 4.0 |