Why doesn't the $\theta$ Angle Renormalize? The $\theta$ term for Yang-Mills takes the form $$L_{\theta}=\frac{\theta}{64\pi^2}\varepsilon^{\mu\nu\rho\sigma}F^a_{{\mu\nu}}F^a_{\rho\sigma}$$
A fact that I have heard is that $\theta$ does not run under renormalization. I understand that this term is topological. I have been told by peers that this explains why it does not run, however I would like a much better explanation for this fact. Is there a detailed proof of this statement? Perhaps a reference would help!
 A: The reason is that it is a total derivative. When people say that that's because it is topological they mean total derivative.${}^1$
A reference that explains is is, for example, Marcos Mariño's notes on large $N$. See around $(5.29)$.
The argument is roughly as follows:
$$
q(x) \equiv \frac{1}{64\pi^2}\epsilon^{\mu\nu\rho\sigma}F^a_{{\mu\nu}}F^a_{\rho\sigma} = \partial_\mu K^\mu\,,
$$
where $K^\mu$ is actually identical to the Chern-Simons term in $d=3$, times $\epsilon_{\mu\nu\rho\lambda}$. In Fourier transform this means
$$
\tilde{q}(p) = p^\mu \tilde{K}_\mu(p)\,\underset{p\to0}{\longrightarrow} 0\,.
$$
Now the $\theta$ dependent partition function would be
$$
Z[\theta] = \frac1{\mathcal{N}}\int \mathcal{D}[A,\psi,\ldots]\,\exp\left({\mathcal{L}_{\mathrm{QCD}} + \theta\int \mathrm{d}x\,q(x)}\right) \,.
$$
The perturbative expansion looks like
$$
Z[\theta] = Z[0] + \theta Z^{(1)} + \frac{\theta^2}{2!} Z^{(2)} + \cdots\,,
$$
and each term is of the form
$$
Z^{(n)} = \int \prod_{i=1}^{n-1} \frac{\mathrm{d}^4k_i}{(2\pi)^4}\langle \tilde{q}(k_1)\cdots  \tilde{q}(k_{n-1})\tilde{q}(k_1+\cdots + k_{n})\rangle\,.
$$
The last argument shows up when Fourier transforming $\langle q(x_1) \cdots q(x_n)\rangle$ as a consequence of translational invariance, because one could send $x_i \to x_i - x_n$. Due to momentum conservation the sum of the momenta vanishes and thus
$$
\tilde{q}(k_1+\cdots + k_n) = \tilde{q}(0) = 0\,.
$$
This means that $Z[\theta]$ does not depend on $\theta$ at any order in perturbation theory and thus one can ignore it in all perturbative computations. In particular, there cannot be perturbative corrections to $\theta$ itself.

$\qquad{}^1$ As a note on the comment under the original post: the Chern-Simons action is famously topological but not a total derivative. So the seeming inconsistency with the statement that topological terms do not give perturbative effects is due to a terminology issue.

Edit:
Even though it is very clearly written in the linked notes and it has been discussed in the comments, I still want to emphasize that the true partition function $Z[\theta]$ does depend on $\theta$ through non perturbative effects. For instance by summing infinitely many diagrams (large $N$ limit) or via lattice computations. A complete review about the subject is Ettore Vicari, Haralambos Panagopoulos.
