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 391
1
vote
Accepted
Elementary question: Intuition for equivariant cohomology
I'm guessing that, unstated, $M,G$ are finite-dimensional and $G$ is connected Lie.
Then $H^*(M/G)$ vanishes for $* \gg 0$, but $H^*_G$ is positively graded, so $H^*(M/G)$ must be a torsion module. …
- 27.9k
5
votes
Accepted
Why is the equivariant Euler class a character ?
Well, there's going to be some map $T^* \to H^2_T(pt)$ taking a weight $\lambda$ to the equivariant Euler class of the corresponding line bundle over a point. If you add weights, that tensors the line …
- 27.9k
2
votes
Example equivariant Mayer-Vietoris for $H^{*}_{S^{1}}(S^{2})$
This is fine for $\mathbb Z$-coefficients; you don't need to go to $\mathbb R$.
The part of the sequence you need
is $$0\to H_{S^1}^0(S^2) \to H_{S^1}^0(U) \oplus H_{S^1}^0(V) \to H_{S^1}^0(U\cap V …
- 27.9k