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Suppose you have an action $S(\epsilon) = S_1 + S_2 + \epsilon\, S_\mathrm{int}$. Assume that $S_1$ is gauge invariant under the action of the group $G$ and $S_2$ is gauge invariant under the action of the group $H$, such that the action $S_1$ + $S_2$ is gauge invariant under the action of $G\times H$. Suppose that $S_\mathrm{int}$ breaks the gauge group down to $F \in G\times H$, that is, the action $S(\epsilon)$ is gauge invariant under the action of $F$ only.

This implies that $$ S(0)=S_1+S_2= \lim\limits_{\epsilon\, \rightarrow\, 0}\,S(\epsilon) $$ has a wider gauge group than $S(\epsilon)$, that is, the gauge symmetry of the action $S(\epsilon)$ is enhanced when sending $\epsilon$ to $0$. For clarity, by "gauge invariant" I mean that the theory has a redundancy of description.

Does this imply that the parameter $\epsilon$ is technically natural?

To clarify, I mean "natural" in the sense of 't Hooft, e.g. as discussed in

The question is motivated by the fact that I could only find the concept of technical naturalness associated with global symmetries in the literature. On the other hand, I did not find any statement saying that it does not hold in the case of gauge symmetries.

EDIT: I can provide a simpler example to clarify ever more what I mean. Consider the Proca lagrangian density for a real massive spin-1 field,

$$ \mathcal{L}=-\dfrac{1}{2}F^{\mu\nu}F_{\mu\nu}+m^2A_\mu A^\mu, \qquad F_{\mu\nu}= \partial_\mu A_\nu - \partial_\nu A_\mu. $$

The corresponding Proca action is not invariant under the gauge group $U(1)$, but taking the limit $m\rightarrow 0$ gives us the action for a free photon, which is gauge invariant under $U(1)$. Hence, sending $m\rightarrow 0$ enhances the gauge symmetry of the action.

In this particular case, my question becomes: is the Proca mass $m$ natural in the sense of 't Hooft? In other words, is a small Proca mass $m$ protected against large quantum corrections, the latter being proportional to the small mass itself?

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  • $\begingroup$ By a “gauge symmetry” do you mean a redundancy of description or a symmetry whose parameters can be spacetime functions? $\endgroup$
    – knzhou
    Jul 1, 2019 at 10:47
  • $\begingroup$ I mean a redundancy of description in the theory. $\endgroup$
    – Frank
    Jul 1, 2019 at 11:32

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You would expect this to be the case, and in massive electrodynamics it works.

In the limit $m\rightarrow 0$, the theory has a $U(1)$ gauge invariance. Therefore, any loop corrections which break $U(1)$ gauge invariance, must be proportional to $m$. This means if $m$ is small, then corrections which violate $U(1)$ invariance are parametrically small, which is the 't Hooft naturalness idea. For example, the current is conserved in electrodynamics, $\partial_\mu J^\mu =0 $, but in Proca theory in general $\partial_\mu J^\mu = m C$, for some correction term $C$. Therefore, it is self-consistent to assume $m$ is small -- loop corrections to the mass won't enter at a scale parametrically larger than $m$.

A crucial physics point is that a massless spin-1 field has 2 degrees of freedom, while a massive spin-1 field has 3. Therefore, you aren't guaranteed to have a smooth, continuous limit, because the number of degrees of freedom changes by a finite amount between a case when $m$ is small but non-zero, and when $m$ is zero. However, what you might expect (and what happens in massive electrodynamics) is that the extra degree of freedom decouples from the rest of the fields in the $m\rightarrow 0$ limit.

In other, more complicated theories, such as massive Yang-Mills (without a Higgs mechanism) and massive gravity, this decoupling does not work in a straightforward way (if at all). In the case of massive gravity, the failure of corrections to the massless theory to vanish perturbatively in the limit $m \rightarrow 0$ is known as the vDVZ discontinuity. It is thought that non-perturbative effects may lead to the decoupling of the extra degrees of freedom, this is the Vainshtein mechanism.

This review by de Rham is about massive gravity, but also explains the massive Proca theory in detail: https://arxiv.org/abs/1401.4173

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Gauge symmetries are redundancies of the theory, i.e. they remove degrees of freedom -- consider QED, where the four components of the gauge field give rise to only two degrees of freedom. Hence, gauge symmetries cannot be approximate: two states related by a gauge symmetry are strictly equivalent (as BobKnighton points out, at least locally; globally, there might be subtleties, hwich do not change to argument here), and when the gauge group would change, the number of degrees of freedom would change as well.

(Note that that is still true for "spontaneously broken" theories such as electroweak theory in the Higs phase: The (original) action is always invariant under gauge transformations, Higgs vev or not.)

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  • $\begingroup$ In the considered system (which I now describe in more detail in the question) the action changes when changing the gauge group. Hence, as you say, the degrees of freedom also change. The thing is that, if all of these symmetries were global symmetries and not gauge symmetries, then, if I am not mistaken, the parameter $\epsilon$ would be considered to be natural in the sense of 't Hooft. Is that the case also when these symmetries are gauge symmetries? $\endgroup$
    – Frank
    Jul 1, 2019 at 12:15
  • $\begingroup$ "...two states related by a gauge symmetry are strictly equivalent..." This statement isn't strictly true. For instance, two nonabelian gauge fields which differ by a gauge transformation with nontrivial falloff at infinity can correspond to two different physical states (see, for instance, the classification of instantons). $\endgroup$ Jul 1, 2019 at 12:32
  • $\begingroup$ @BobKnighton You're right, I've added a comment $\endgroup$
    – Toffomat
    Jul 1, 2019 at 13:26

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