MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Has anyone read Atiyah and Bott's famous paper "The moment map and equivariant cohomology"?

I have some trouble with the equaations appearing between equations (4.18) and (4.19) in page 13 of the original paper. The paper claims $$D\lambda a = D(a-i(X)a\theta) = da-i(X)da \theta + i(X)au,$$ whereas I think that it should be $$D\lambda a=D(a-i(X)a\theta)=da+i(X)da\theta + i(X)au,$$ where the minus sign is due to $\mathcal{L}(X)a=i(X)da+di(X)a = 0$.

Can anyone tell me what it should be?

Thank you!

share|cite|improve this question
Sorry for the comment,i only persue the truth but not mend to comment someone needless to say Atiyah-Bott, so if you have the answer or you 're sure of it ,why not show me the correct.By the way , i also thinks Atiyah-Bott are of the most important celebrates in the 20'!! – HKSHLZW Dec 11 '10 at 1:05
I've taken the liberty of editing the question to make it more readable. – José Figueroa-O'Farrill Dec 11 '10 at 1:34
up vote 4 down vote accepted

I think that the map $\lambda$ should be defined as follows: $$\lambda a = a - \theta i(X)a.$$

In this way, $$D\lambda a = D( a - \theta i(X)a) = da + u i(X)a + \theta di(X)a$$ and using that $di(X) a + i(X) da = 0$ and that $\lambda(i(X) a) = i(X) a$, one finds $$D\lambda a = \lambda( da + i(X)a u ),$$ which is (4.19) in Atiyah-Bott.

By the way, the generalisation of $\lambda$ to non-abelian $\mathfrak{g}$ is the following. If $\theta^\alpha$, $e_\alpha$ and $u^\alpha$ are as in the paper, then $$\lambda b = \prod_\alpha (1 - \theta^\alpha i(e_\alpha) )b.$$ This is known as minimal coupling in the Physics literature.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.