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Reading this paper of Masahito Yamazaki and Kazuya Yonekura, I am having trouble turning the physics jargon into mathematical statements. He is talking about Yang-Mills theory over $\mathbb{R}^4$.

Perturbative expansions are ill-defined if observables suffer from infrared divergences. For example, in the vacuum energy is computed as: $$ E(\theta) \sim - \int_0^\infty \frac{d\rho}{\rho^5}(\mu \rho)^{b_1}e^{-8\pi^2/g^2(\mu) } \cos \theta $$ The paper then says this integral is divergence in the infrared scale, which is $\rho \to \infty$. That range actually has good decay, so he could possibly have meant $\rho \to 0$. Here $\rho$ is the size of the modulus of the instanton. Introducting an IR cuttoff one obtains: $$ E(\theta) = \Lambda^4 \cos \theta $$ which is still wrong because another computation (by Witten, on a possibly-related QCD theory) shows the vaccum energy is given by: $$ E(\theta) = \min_{\mathbf{e} \in \mathbb{Z}} \Lambda^4 (\theta - 2\pi \mathbf{e})^2 $$ Having read all of this I still don't know what infrared divergence is (or strong coupling), or how resurgence played a role in any of this.

After some work I was able to locate an action for $SU(N)$ Yang-Mills theory over $\mathbb{R}^4$

$$ S = \int d^4 x \, \mathrm{Tr}\left( \frac{-1}{4g^2} F \wedge \ast F + \frac{i\theta}{8\pi^2} F \wedge F \right) $$

and they are asking about the $\theta$ dependence. The action is periodic in $\theta$ but the vacuum energies they find have severe problems (such as discontinuities at $\theta = 0, \pi$).

These authors take as a given the definitions of Yang-Mills theory, but also that their eigenvalue estimation procedures work. So that being un-familiar could have an advantages.

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  • $\begingroup$ In terms of the infrared divergence: $b_1$ is usually a pretty big integer. In 't Hooft's article which your article cites for the vacuum energy, the leading term has $b_1 = 3 N^f$ where $N^f$ is generally some integer at least $2$ (it is the number of doublets). (I don't know what any of this means, but the divergence of the integral is pretty clear as soon as you factor in $b_1$.) $\endgroup$
    – Willie Wong
    Commented Oct 9, 2017 at 16:56
  • $\begingroup$ Thanks @WillieWong if $b_1 \geq 4$ there's an issue. I guess the "cutoff prescription" is to approximate $\int_0^\infty$ with $\int_0^\Lambda$ and truncate the infrared range $\rho \gg \Lambda$. Strong coupling and Weak coupling usually have to do with the size of $g$. He says "Of course, Yang-Mills theory on $\mathbb{R}^4$ is strongly-coupled in the IR and hence the perturbative expansion breaks down." so I'd like to understand what is going wrong with perturbation theory here. $\endgroup$
    – john mangual
    Commented Oct 9, 2017 at 17:01
  • $\begingroup$ He is saying Taylor expand $e^{-8\pi^2/g^2}$ around $g=0$, and then interchange the integral and summation $\int_0^\infty \sum_0^\infty \approx\sum_0^\infty \int_0^\infty \approx \sum_0^\infty \int_0^\Lambda $ and finally approximate $ \Lambda \approx \infty$ which can only end in disaster. He's saying the coefficients are infinite so you can't do resurgence at all. And the cutoff prescription gives the wrong answer. $\endgroup$
    – john mangual
    Commented Oct 9, 2017 at 18:07

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