# Conditions on the covariance operator in Gaussian Path Integrals

In field theory, one typically encounters integrals of the form:

$$\mathcal{Z}[J] = \int \mathcal{D}[\phi] \exp \left( - \frac{1}{2} \int d^Dx d^Dx' \ \phi(x)A(x,x')\phi(x')+ \int d^Dx \phi(x) J(x)\right).$$

This can easily be done by discretizing space, and assuming that $$A(x,x')$$ is real and symmetric so that it is diagonalizable. Suppose however that there are multiple fields $$\phi_a(x)$$, and a path integral of the form

$$\mathcal{Z}[J] = \int \mathcal{D}[\phi] \exp \left( - \frac{1}{2}\int d^Dx d^Dx' \ \phi_i(x)A^{ij}(x,x')\phi_j(x')+ \int d^Dx \phi(x) J(x) \right).$$

What is the condition on $$A^{ij}(x,x')$$ such that I can write

$$\mathcal{Z}[J] \propto \exp \left( \frac{1}{2}\int d^Dx d^Dx' \ J_i(x)[A^{-1}]^{ij}(x,x')J_j(x') \right)?$$ Any help will be appreciated.

My ideas: Fixing two $$x$$ and $$x'$$, and requiring $$A^{ij}(x,x')=A^{ji}(x,x')$$, we could write $$A(x,x') = O^T D(x,x') O$$, diagonalizing the $$3 \times 3$$ matrix. Then, performing the transformation $$\tilde{\phi}_a = O_{ab} \phi_b$$, would break the integral into three independent components, one for each $$\tilde{\phi}_a$$. Relabelling $$\tilde{J} = O^T J$$, yields

$$\mathcal{Z}[J] \propto \exp \left( \frac{1}{2}\int d^Dx d^Dx' \ J(x)[A^{-1}]^{ij}(x,x')J_j(x') \right).$$

However, is it sufficient if $$A^{ij}(x,x')=A^{ji}(x',x)$$ instead of just $$A^{ij}(x,x')=A^{ji}(x,x')$$?

1. We may assume w.l.o.g. that the integral kernel $$A(x,x^{\prime})_{ik}=A(x^{\prime},x)_{ki}\tag{1}$$ is symmetric (because an antisymmetric part would not be visible under the integral/summation).
2. Assuming that the fields $$\phi^i\in\mathbb{R}$$ are real, in order for the Gaussian functional integral to formally be convergent, the real integral kernel $${\rm Re}A(x,x^{\prime})_{ik}\tag{2}$$ should correspond to a positive definite (and hence invertible) operator$$^1$$.
3. One may show that the condition (2) in principle implies that the complex integral kernel $$A(x,x^{\prime})_{ik}$$ corresponds to an invertible operator, so that the Gaussian formula is well-defined.
$$^1$$ There are subtleties with unbounded operators, domains, selfadjoint extensions, etc., that we ignore in this answer.