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I'm studying equivariant cohomology on three references:

  1. Szabo's review about equivariant localization (S);
  2. Libine's note on equivariant cohomology (L);
  3. Berline, Getzler, Vigne's book "Heat Kernels and Dirac Operators", Springer (BGV).

I'm deriving all formulas by myself. I face two problems:

  1. In (S) when he derive (2.83) he uses a commutator while (L), in the longest formula below Definition 34 writes

$$ i_X\nabla + \nabla i_X. $$

I agree with (L) because it's just the square:

$$ (\nabla - i_X)^2, $$

so why (S) put the commutator? (BGV) seems to agree with (S).

  1. Then I compute

$$ \begin{align} (i_X\nabla + \nabla i_X)\alpha & = i_X(d\alpha + \omega\wedge\alpha) + (d + \omega)i_X\alpha\\ & = i_Xd\alpha + i_X\omega\wedge\alpha - \omega\wedge i_X\alpha + di_X\alpha + \omega\wedge i_X\alpha\\ & = (i_Xd\alpha + \omega(X)\alpha) +di_X\alpha\\ & = \nabla_X\alpha + di_X\alpha \end{align}. $$

So I get an extra $di_X\alpha$. This must be correct because also (BGV) write (page 211)

$$ [\nabla,i_X]=\nabla_X $$

which is, apart for the mysterious commutator instead anticommutator, the same formula.

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  • $\begingroup$ Would this do better on the math site? $\endgroup$ – DanielSank Nov 18 '15 at 9:46
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The main issue seems to be that BGV uses a supercommutator notation

$$\tag{1} [a,b]~:=~ab-(-1)^{|a||b|}ba,$$

where $|a|$ and $|b|$ denote the $\mathbb{Z}_2$-grading of operators $a$ and $b$, respectively. So if $a$ and $b$ have odd gradings, then $[a,b]$ is actually the anticommutator.

References:

[BGV] N. Berline, E. Getzler & M. Vergne, Heat Kernels and Dirac Operators, 1991; p.39.

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  • $\begingroup$ Thank you. but I still cannot derive $$ [\nabla, i_X] = \nabla_X $$ even using coordinates. $\endgroup$ – MaPo Nov 18 '15 at 14:40
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    $\begingroup$ With $\nabla=d+\omega$, we have $[\nabla, i_X] = {\cal L}_X+ \omega(X)=\nabla_X$, where we in the last equality used that $X\in\mathfrak{g}$ belongs to the Lie algebra, cf. L p. 14 or BGV p. 211. $\endgroup$ – Qmechanic Nov 19 '15 at 22:02
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    $\begingroup$ I still cannot understand the last equality, since $$ \mathcal{L}_X = i_Xd + di_X $$ how can you set $di_X=0$? $\endgroup$ – MaPo Nov 20 '15 at 12:55
  • $\begingroup$ Is it true that $$ \nabla_X = i_Xd + \omega(X) $$ ? $\endgroup$ – MaPo Nov 20 '15 at 13:10
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    $\begingroup$ Let me restate my doubt. In page 22, it is written that $\nabla_X = i_X \nabla$ which I find to be equal to $i_X d + \omega(X)$. On the other hand, following page 211, $\nabla_X = [\nabla,i_X] = \mathcal{L}_X + \omega(X)$, like you wrote. There seems to be an extra term, $di_X$. $\endgroup$ – MaPo Jan 14 '16 at 13:08

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