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.



A question about a formula of Pfaffian
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](http://books.google.com/books?id=2yJIwWbh1isC&pg=PA204&lpg=PA204&dq=Pfaffian+%252B+%2522Berezin+integral%2522&source=bl&ots=ex-Shnj--R&sig=jRCLVTeXucsrtVVYi-rZDc1PCMI&hl=en&ei=ssBzTMqXMM_uOZuy2OUI&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBIQ6AEwAA#v=onepage&q=Pfaffian%2520%252B%2520%2522Berezin%2520integral%2522&f=false) by: José Gracia Bondía, Joseph C. Várilly, Héctor Figueroa.


