Represent a square-root of determinant by Grassmann numbers

Question

I am thinking about the representation of $ \sqrt{\det A}$ for a Matrix $A$ . Since it is known that for the Grassmann numbers $ \eta, \eta ' $ the following relation holds: $$ \int \int d \eta d \eta' \exp\left[\eta'^T A \eta\right] = \det A $$ However, if a square root is applied on this determinant, there must be used another grassmann integral representation. I am thinking about the following integral:

$$ \int \int \int d\eta d\eta' dp \exp\left[\eta'^T A \eta + p^T A p\right] = \det A (\det A)^{-1/2} $$ Here, $p$ is a vector of ordinary commutative (complex) numbers and Fubini's Theorem is used. Am I right with my calculations (even if $A$ is a functional matrix)?

Are there other Grassmann (or more general hypercomplex number) representations of $ \sqrt{\det A}$?

Answer 1

I) We will assume that the real $n\times n$ matrix $A$ is not necessary symmetric. Therefore the Bosonic Gaussian integral in OP's last formula should read

$$ \int_{\mathbb{R}^n} \! d^nx~ \exp \left[-\frac{1}{2}x^i A_{ij} ~x^j\right] ~=~\int_{\mathbb{R}^n} \! d^nx~ \exp \left[-\frac{1}{4}x^i (A+A^T)_{ij} ~x^j\right] ~=~\sqrt{\frac{(2\pi)^n}{\det(\frac{A+A^T}{2})}},\tag{1} $$ where we assume that the symmetric part of the matrix $A$ is positive definite. Hence OP's last formula is basically an integral representation for

$$\frac{\det(A)}{\sqrt{\det(\frac{A+A^T}{2})}}.\tag{2}$$

II) OP may also be interested in the Grassmann integral representation of the Pfaffian

$$\tag{3} {\rm pf}(A) ~\propto ~\int \! d^n\theta~ \exp \left[\frac{1}{2}\theta^i A_{ij} ~\theta^j\right]~=~\int \! d^n\theta ~\exp \left[\frac{1}{4}\theta^i (A-A^T)_{ij} ~\theta^j\right], $$

where we leave it as an exercise to the reader to fix the correct normalization factor in eq. (3). In the second equality we have used that the Grassmann-numbers anticommute $\theta^i\theta^j~=~-\theta^j\theta^i.$

If we use the integral representation (3) as a definition, then the Pfaffian ${\rm pf}(A)$ only depends on the antisymmetric part of the $n\times n$ matrix $A$. Let us therefore assume that the matrix $A$ is antisymmetric from now on. One may then prove that the square of the Pfaffian is the determinant:

$$\tag{4} {\rm pf}(A)^2~=~\det(A).$$