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Let $M$is a closed oriented 2n-dimensional smooth manifold, $E$ is a 2n-dimensional oriented real vector bundle on $M$, with inner product on each fibers. Let $\tau=(\sqrt{-1})^{n}c(e_{1})c(e_{2})...c(e_{2n})$, use it we can define a $\mathbb{Z}_2$-grading on $\wedge(E^{*})\otimes\mathbb{C}=\wedge_{+}(E^{*})\otimes\mathbb{C}\oplus\wedge_{-}(E^{*})\otimes\mathbb{C}$, the Clifford action is defined by $c(v)\alpha=\varepsilon(v)\alpha-\iota(v)\alpha$.

I read a formula about Pfaffian like this:

$$Str[exp(-R^{\wedge(E^{*})\otimes\mathbb{C}})]=2^{n}(\sqrt{-1})^{-n}{\rm Pf}(-R^{E})$$

here $R^{E}$ and $R^{\wedge(E^{*})\otimes\mathbb{C}}$ are curvatures,

$E^{*}$ is the dual to $E$.

How to proof this formula? or any matrial about this? If anyone can tell me something I will be very thanks.

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Am I right in interpreting $\tau$ as the "volume element" in the Clifford bundle $Cl(E)$? –  José Figueroa-O'Farrill Aug 24 '10 at 2:20
    
Yes,it is complex volume element. –  Chen Aug 24 '10 at 3:41
    
Two comments: (1) this ought to be a linear-algebraic identity, so it should be true for skewsymmetric endomorphisms, not necessarily curvatures; and (2) did you try looking at the paper of Matthai and Quillen Superconnections, Thom classes and differential forms? They have similar formula in terms of "fermion Gaussian integrals". –  José Figueroa-O'Farrill Aug 24 '10 at 11:44
    
Thanks! Professor. I try to looking at that paper, but I didn't understand how the curvature $R^{\wedge(E^{*})\otimes\mathbb{C}}$ changed to be the curvature $R^{E}$, what is the relation? –  Chen Aug 24 '10 at 14:36
    
There is an embedding of $\mathfrak{so}(E)$ into the Clifford algebra $Cl(E)$ and hence an action of $\mathfrak{so}(E)$ on $Cl(E)$ via the Clifford commutator. The exterior algebra is the associated graded algebra to $Cl(E)$ relative to the filtration coming from writing $Cl(E)$ as a quotient of the tensor algebra. The action on $Cl(E)$ induces an action on the exterior algebra. Now read David Bar Moshe's answer below. –  José Figueroa-O'Farrill Aug 24 '10 at 14:52

1 Answer 1

up vote 2 down vote accepted

The supertrace can be evaluated either by Berezin Gaussian integration, or equivalently by summation over a Clifford algebra. Here is a description of the second method.

Let $\omega$ be a skew-symmetric 2n by 2n matrix. Let $\{{\ e_1, e_2, . . . . e_{2n} \}}$ be a real 2n-dimensional Clifford algebra. Then:

$exp(\Sigma_{k,l=1}^{2n} \omega_{kl} e_k e_l) = \Sigma_{|K| even} Pf(\omega_K) \hat{e}_K$

where: $K$ is a subset of $\{{ 1, 2, . . . . 2n \}}$ and $\hat{e}_K$ is the corresponding wedge product of the Clifford generators and $\omega_K$ is the skew-symmeterized submatrix containg only the rows and columns in $K$.

Now, the supertrace selects the coefficient of the top form, giving you the required formula.

A proof of this result can be found for example in the following book by: José Gracia Bondía, Joseph C. Várilly, Héctor Figueroa.

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