Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with cohomology
Search options not deleted
user 40804
A branch of algebraic topology concerning the study of cocycles and coboundaries. It is in some sense a dual theory to homology theory. This tag can be further specialized by using it in conjunction with the tags group-cohomology, etale-cohomology, sheaf-cohomology, galois-cohomology, lie-algebra-cohomology, motivic-cohomology, equivariant-cohomology, ...
5
votes
Finding cocycles that square to zero
To manage the higher cases one has to also ask that certain higher cohomology classes are equal, where the higher cohomology classes are made out of iterated cup-1 products of lower ones. … In particular you can extend every odd cohomology class on $SU(n)$ to a twisting sequence. …
- 9,580
3
votes
What is this cochain complex about, whose $H^1 = \mathbb{R}$?
H^i(C) \to H^i(\overline C) \to \cdots$$
Now the group cohomology of the integers, $H^i(C)$, is $\Bbb Z$ in degrees $0$ and $1$ and is zero otherwise. … In general your complex $C$ interpolates the usual group cohomology and the bounded cohomology. …
- 9,580
5
votes
Dual surfaces of a first cohomology class of a 3-manifold
$L(4,1)$ is a counterexample to your conjecture, taking $\alpha$ to be the nontrivial element of $H^1(L;\Bbb Z/2)$. Notice that this element squares to zero (the square is the same as the Bockstein, a …
- 9,580