# Non-local field redefinition and effects on path-integral measure

Consider the partition function

$$Z[0] = \int \left[\mathcal{D}A_\mu\right]\left[\mathcal{D}\pi\right] e^{-i \int d^4x \left(-\frac{1}{2}(\partial\pi)^2-\frac{1}{4}F_{\mu\nu}F^{\mu\nu}+ \frac{a}{M^2} \pi^2\left(F_{\mu\nu}F^{\mu\nu}\right)\right)}.$$ where $$a$$ is $$O(1)$$ coefficient, suppressed by some mass-scale $$M$$.

The action is gauge invariant. Is the path-integral measure invariant under the following field redefinition?

$$A_\mu \rightarrow A_\mu + \frac{\partial_\mu \pi}{\square}$$

Or it gives an anomalous contribution to the action, being a non-local transformation?

So, I would start by giving a precise definition of $$\frac{\partial_\mu \pi}{\square}$$, which is \begin{align} \frac{\partial_\mu \pi(x)}{\square}&=\square^{-1}_x\int d^4y\,\delta^{(4)}(x-y)\partial^y_\mu \pi(y)=\int d^4y\ \Delta(x-y)\,\partial^y_\mu \pi(y)\\ &=-\int d^4y\ \partial^y_\mu \Delta(x-y)\,\pi(y)=\int d^4y\ \partial^x_\mu \Delta(x-y)\,\pi(y) \end{align} where $$\Delta(x-y)$$ is defined as $$\square\, \Delta(x-y)=\delta^{(4)}(x-y)$$ the fields transformations are the following \begin{align} &A'_\mu = A_\mu+\frac{\partial_\mu \pi}{\square}=A_\mu+\int d^4y\ \partial^x_\mu \Delta(x-y)\,\pi(y)\\ &\pi'=\pi \end{align} the action is invariant under this transformation.
It is now possible to write the Jacobian matrix $$J^i_j(x,x') = \frac{\delta \phi'_i(x)}{\delta \phi_j(x')}=\left[ \begin{array}{cc} \delta^{(4)}(x-x') & \partial_\mu\Delta(x-x') \\ 0 & \delta^{(4)}(x-x')\end{array} \right]$$ remember that functional derivatives generate distributions and must be evaluated at different points. Furthermore the Jacobian can be considered a $$2\times 2$$ matrix where in each entry there is a matrix with continuos indices $$x,x'$$. Once the theory is regularized the determinant can be evaluated.
The formula for the determinant of a matrix of matrices is $$\det\left[ \begin{array}{cc} A & B \\ 0 & D\end{array} \right]=\det(A)\det(D)$$ since in our case we have that $$C = 0$$ and so $$\det J = \det(\delta^{(4)}(x-x'))^2 = \exp\left(2\,tr\log \delta^{(4)}(x-x')\right)$$ this determinant is clearly an highly singular object and the theory must be regularized in order to evaluate it, we can choose $$\delta^{(4)}(0)=\lim_{\Lambda\rightarrow\infty} \int \frac{d^4p}{(2\pi)^4}\, \exp\left({-\frac{D^\mu D_\mu}{\Lambda}}\right)$$ where $$D_\mu=\partial_\mu+i A_\mu$$ and $$D^\mu D_\mu$$ is the gauge invariant laplacian. You see that $$\det J=\exp\left( 2\,\int d^4x\,\delta^{4}(0)\, tr\, \log 1\right)=1$$ where $$1$$ is the trivial transformation $$\pi'=\pi$$ and $$A'_\mu =A_\mu$$. So you see that the $$tr\log 1=0$$ and the transformation is not anomalous since the regularizating function does not contribute to the trace. The chiral anomaly has a nontrivial trace of the log with the regularized delta $$\int d^4x\,tr\,\delta^{4}(0) \log M\neq0$$ and therefore you get the anomaly etc, in this case it is zero.
Another way of seeing that I think is that the $$\pi$$ field can be exactly integrated out $$Z=\int \mathcal{D} A\, \det\left(\square-\frac{2a}{M^2} F^{\mu\nu}F_{\mu\nu}\right)^{-\frac{1}{2}}\, e^{i\int d^4x\, \frac{1}{4}F^{\mu\nu}F_{\mu\nu}}$$
and so the field transformation becomes a simple shift in the field which leaves the path integral measure invariant. Furthermore a gauge anomaly would change the gauge invariance of the effective action $$\Gamma[A] = -\frac{i}{2}\log\det\left(\square-\frac{2a}{M^2} F^{\mu\nu}F_{\mu\nu}\right)$$ but it is gauge invariant since the field $$\pi$$ is left unchanged by the tranformation.
Fun fact: if you calculate the determinant and integrate away the the field modes between $$M$$ and $$\Lambda in the Wilsonian way you get from the expansion of the determinant $$\det\left(\square-\frac{2a}{M^2} F^{\mu\nu}F_{\mu\nu}\right) = \det(\square)\,\exp\left(\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{n}(\frac{2a}{M^2})^n\,tr (\square^{-1}F^{\mu\nu}F_{\mu\nu})^n\right)$$ which becomes up to third order of the effective action (numerical factors not guaranteed) \begin{align} i\Gamma[A]=&\frac{a}{16\pi^2M^2}(M^2-\Lambda^2)\int d^4x\, F^{\mu\nu}F_{\mu\nu}-\frac{a^2}{16\pi^2M^4}\log\frac{\Lambda}{M}\int d^4x\, (F^{\mu\nu}F_{\mu\nu})^2\\ &+\frac{a^2}{48\pi^2M^4}\left(\frac{1}{\Lambda^2}-\frac{1}{M^2}\right)\int d^4x\,F^{\mu\nu}F_{\mu\nu}\,\square F^{\mu\nu}F_{\mu\nu}\\ &+\frac{a^2}{160\pi^2M^4}\left(\frac{1}{\Lambda^4}-\frac{1}{M^4}\right)\int d^4x\, \square^2 ( F^{\mu\nu}F_{\mu\nu}\, F^{\mu\nu}F_{\mu\nu})\\ &+\frac{a^3}{12\pi^2M^6}\left(\frac{1}{\Lambda^2}-\frac{1}{M^2}\right)\int d^4x\, (F^{\mu\nu}F_{\mu\nu})^3+\ldots \end{align}