# What is relevant for Yang-Mills theory below $d = 4$?

There are two ways to write the Lagrangian for Yang-Mills, differing by the scaling of the Yang-Mills field. Fancy theorists tend to write $$S = \int d^dx \, \frac{1}{4e^2} \, \text{tr}(F^2)$$ while people who do practical calculations tend to write $$S = \int d^dx\, \frac{1}{4} \, \text{tr}(F^2).$$ This is a totally trivial difference; it's just absorbing a factor of $e$ into the field.

However, in less than four dimensions this totally changes the infrared behavior of the theory, because the mass dimension of $e$ is positive. In the 'practical' setup, the kinetic term is marginal, so it just stays the same under renormalization group flow, like every other theory. For $d < 4$ the coupling $e$ is relevant, getting stronger in the infrared just like it does in $d = 4$ quantum chromodynamics.

But in the 'theory' setup, the kinetic term is irrelevant for $d < 4$, since its coefficient has negative mass dimension, which means that in the infrared the theory has no propagating degrees of freedom! I'm told that the only term you get in $d = 3$ is the Chern-Simons term, and we end up with a topological field theory which looks completely unlike quantum chromodynamics.

How can a simple rescaling of the field lead to such different conclusions? Is one of these choices simply invalid? Which of these setups describes what would actually happen in $d < 4$?

It's helpful to make a stronger notational distinction. I'm going to work in the Abelian Maxwell-Chern-Simons theory, since the nonlinearities only obscure what's going on. Let's write the 'theory' action as:

$$S(A) = \frac{1}{g^2}\int dA \wedge *dA + \kappa \int A \wedge dA.$$

Here the mass dimension $[A]=1$ and $[g]=\frac{1}{2}$. If we make the substitution $A = g B$, we obtain the 'practical' version:

$$S'(B) = S(A) = \int dB \wedge * dB + g^2 \kappa \int B\wedge dB$$

where $[B] = \frac{1}{2}$. (We also alter the gauge transformations: $A \mapsto A + d \alpha$ becomes $B \mapsto B + \frac{1}{g} d\alpha$. And this transformation changes the form of the observables, sending the Wilson loop $exp(\oint A)$ to $exp(g\oint B)$.)

This is, as you observe, just a change of variable in the path integral. Nothing happened, no physics was changed.

The renormalization flow is defined (even for the first version of the action) by integrating out slices and rescaling to fix the normalization of the kinetic term. There're no interactions, so renormalization amounts to scaling. The coupling constant $g$ is small in the UV, and large in the IR. This means that the Chern-Simons term is relatively unimportant at short distances, but dominates the Maxwell term at long distances. So if you're studying the theory in the IR, it's a good idea to switch back to the 'theory' variables, where the Maxwell term drops out and the Wilson loop observables don't contain a coupling constant $g$ which is headed to infinity.

So, there's no contradiction. Both actions predict the same physics in the IR.

If I am not mistaken, a vector (or scalar) field generally has mass dimension $\frac {(D-2)}{2}$. Derivatives always have mass dimension 1. In three dimensions, that would mean $F^2$ has mass dimension $\frac {6}{2}=3$ and $e$ would be marginal again? Maybe you should explain your "practical" setup.

• That’s true in the “practical” convention. The point is that in the “theory” convention the vector does not have the usual dimension, instead it always has dimension 1. – knzhou Mar 2 '18 at 14:40
• Note that $e$ has the same dimensions in either convention. In the "practical" convention the covariant derivative is $D = \partial + i e A$ so $[e A] = 1$ and hence $[e]$ is nonzero for $d \neq 4$. – knzhou Mar 2 '18 at 16:43
• The nice thing about the 'theory' normalization is that it maintains the classical limit. In the classical theory, you can integrate $A$ along a line to make a Wilson loop. In the 'practical' normalization, you'd have to integrate $A^2$. – user1504 Mar 2 '18 at 19:32
• @user1504 Could you elaborate on that a bit more? – knzhou Mar 2 '18 at 21:22
• Sorry, a bit vague. The point is that operator with mass dimension [1/2] aren't naturally integrated along lines; they don't scale correctly. For Wilson lines, you can integrate $gA$ around a loop (the reasonable choice), or you can integrate $A^2$ (which has the right mass dimension, modulo quantum corrections). – user1504 Mar 2 '18 at 21:57