Skip to main content
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
Search options answers only not deleted user 391
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 …
Allen Knutson's user avatar
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 …
Allen Knutson's user avatar
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. …
Allen Knutson's user avatar