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 equivariant-cohomology
Search options answers only
not deleted
user 102519
11
votes
Accepted
Equivariant bundles invisible in K-theory and Borel cohomology
The simplest example of what you are looking for occurs when $G = S^1$ and $X=S^1/C_6$, where $C_6$ is the group of 6th roots of unity. Then the map
$$ K_G(X) \rightarrow K(EG\times_G X)$$
identifies …
- 11.2k
2
votes
Equivariant cohomology of fixed points using the localisation theorem
Like Andy's answer, this is about alternate proofs of the inequality in Theorem 1, which is, as was mentioned, perhaps due to Ed Floyd in a 1952 paper.
This Floyd theorem obviously implies the 1941 re …
- 11.2k
2
votes
What is equivariant chains on a representation sphere?
Exercise 10 of section 1 of Chapter II of tom Dieck's book Transformation Groups gives you one answer to your question. It reads:
Let $S(V)$ be the representation sphere of a finite group $G$. Show di …
- 11.2k