Gauge invariance is just a redundancy. Why is massive abelian gauge field renormalizable but massive non-abelian gauge field nonrenormalizable?

For example, Kaku's QFT pp. 214-215:

Massive vector theory with non-Abelian group is non-renormalizable.

Massive vector Abelian theory is renormalizable.

I heard about the following arguments, but I don't find them satisfactory.

1. 
   
   

   Someone will say the propagator of massive vector field is like $$ \frac{g_{\mu \nu} - k_{\mu} k_{\nu}/m^2}{k^2-m^2}. $$ In large $k$, it will not decay like $1/k^2$, so the power counting law breaks down. Certainly I admit that power counting law is violated, but why does a violation of power-counting have a relation with renormalizablity? And we already know that a massive $U(1)$ gauge field is still renormalizable even though it violates the power-counting.

   
   


2. 
   

   Someone will say mass term $m^2 \operatorname{tr} A^\mu A_\mu$ will break gauge invariance. But why is gauge invariance important? Guage invariance is not a symmetry, and it's nothing but a redundancy to describe the true physical degrees of freedom. Any theory without gauge invariance can be rewritten by the Stückelberg trick as a gauge theory which describes the same physics.

   
   

   For example, the massive Maxwell field $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 A^\mu A_\mu \tag{1}$$ under the replacement, $$A_\mu\rightarrow A_\mu +\partial_\mu \phi \tag{2}$$ becomes $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 (A_\mu +\partial_\mu \phi)^2 \tag{3}$$

   
   

   Now $(3)$ has local gauge invariance, $$\delta A_\mu =\partial _\mu \Lambda,\quad \delta \phi = -\Lambda \tag{4}$$

   
   

   It's obvious that $(3)$ with local gauge invariance $(4)$ describes the same theory $(1)$.

   
   

   Rescaling $\phi\rightarrow \frac{1}{m}\phi$, $$\mathcal{L}= -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\frac{1}{2}m^2 A^\mu A_\mu + \frac{1}{2}\partial_\mu \phi \partial ^\mu \phi +m A_\mu \partial^\mu \phi\tag{5}$$ with local gauge invariance, $$\delta A_\mu = \partial_\mu \Lambda,\quad \delta \phi = -m \Lambda \tag{6}$$

   
   

   By the same way, any theory without gauge invariance, like massive non-abelian gauge field, can be rewritten as a gauge theory. So why is there a relation between gauge invariance and renormalizablity?

   
   

So it seems that the above two handwaving arguments are untenable. In general how to prove that massive abelian gauge theory is renormalizable but massive non-abelian gauge theory non-renormalizable?

Answer

Certainly I admit that power counting law is violated, but why does violation of power-counting have relation with renormalizablity?

As per the Dyson-Weinberg power-counting theorem (see Ref.1, chapter 12 and Ref.2, chapter 8-1), a diagram is convergent if and only if each of its subdiagrams has a negative superficial degree of divergence $\omega$. The latter is defined as \begin{equation} \begin{aligned} \omega&=\phantom{-}\left(\text{# of factors of momentum in the numerator}\right)\\ &\,\phantom{-}-\,\left(\text{# of factors of momentum in the denominator}\right)\\ &\phantom{-}\,+d\cdot \left(\text{# of independent momentum variables}\right) \end{aligned} \end{equation} where $d$ is the number of spacetime dimensions. Therefore, the number of powers of momentum in the numerator is an essential ingredient in the analysis of potentially divergent diagrams in a certain theory. This number contributes with a plus sign to $\omega$, so the higher the former the higher the latter, and there are more diagrams that are superficially divergent. In fact, it is possible to argue that the number of divergent diagrams is finite if and only if all the interactions are power-counting renormalisable. This is why power-counting is a key part of renormalisability.

And we already know that a massive $U(1)$ gauge field is still renormalizable even though it violates the power-counting.

This is, to some extent, a happy coincidence. Recall that each vertex carries a factor of the current ($J^\mu\sim e\bar\psi\gamma^\mu\psi$). In (massive or massless) QED the current is conserved, and therefore the factors of $k^\mu$ in the numerators don't contribute to scattering amplitudes. Formally speaking, the propagator behaves as $\mathcal O(k^{-2})$ for high momentum (instead of being $\mathcal O(1)$, as one may expect). Therefore, even if massive QED violates power-counting renormalisability, the theory is in fact renormalisable.

In non-abelian gauge theories the current is not conserved (but covariantly conserved instead). Therefore, the factors of $k^\mu$ do contribute and the general analysis of power-counting renormalsiability applies: the propagator is $\mathcal O(1)$ for high momentum, and the theory is non-renormalisable. Only in the massless case do the longitudinal modes decouple, and therefore the propagator effectively behaves as $\mathcal O(k^{-2})$. Renormalisability is thus recovered (and only because the ghosts cancel the longitudinal contribution; without them, the longitudinal part contributes as well).

So why is there relation between gauge invariance and renormalizablity?

This is a matter of semantics. Any divergence, in any local theory, can be eliminated by introducing counter-terms. When the counter-terms have the same form as the original Lagrangian, we say that the theory is renormalisable. Therefore, if the initial theory is gauge invariant, then the theory will be renormalisable only if the counter-terms are gauge-invariant, by definition. If you happen to need a non-gauge-invariant counter-term, that means you need a counter-term that was not initially in the Lagrangian, and the theory is non-renormalisable.

In the case of Yang-Mills, one can prove that the counter-terms are indeed gauge-invariant, and they have the same form as the terms originally present in the initial Lagrangian (see Ref.2, chapter 12-4 and Ref.3, chapter 23). Therefore, the theory is renormalisable. The proof is rather non-trivial and is best understood in the context of Batalin-Vilkovisky (based on the early work of Zinn-Justin). In the case of naïve Quantum Gravity, one can prove that the counter-terms are gauge invariant as well but they are not of the same form as the original Lagrangian (cf. this PSE post). Therefore, the theory is non-renormalisable. Finally, in a gauge-theory where anomalies are present, the counter-terms are not gauge-invariant and the theory is non-renormalisable.

In general how to prove that massive abelian gauge field is renormalizable but massive non-abelian gauge field nonrenormalizable?

You can find an explicit proof of the renormalisability of massive QED in Ref.2, chapter 8-4. You can find the discussion of the non-renormalisability of massive non-abelian gauge theories in Ref.2, chapter 12-5-2 (the essential points are summarised in this PSE post).

References.

1. 
   

   Weinberg's QFT, Vol.1.

   
   


2. 
   

   Itzykson & Zuber's QFT.

   
   


3. 
   
   

   DeWitt's The Global Approach to QFT, Vol.1.