Nomenclature: Yang-Mills theory vs Gauge theory

If you're writing about a theory with Yang-Mills/Gauge fields for an arbitrary reductive gauge group coupled to arbitrary matter fields in some representation, is it best to call it a Yang-Mills theory or a Gauge theory?

I've heard that one is more likely to refer to a theory with no matter sector - but I can't remember which one! Or are the terms basically interchangeable in the context of quantum field theory?

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Yang-Mills is a gauge theory, with or without matter. So I'm not sure what the doubt is. – user346 Apr 15 '11 at 4:31
@Deepak: Sure. So in general is it best to call it Yang-Mills theory or Gauge theory or does it not make any difference? – Simon Apr 15 '11 at 4:54
If the action is of Yang-Mills form $F^2$ then its Yang-Mills. If the gauge group is $U(1)$ rather than $SU(2)$ or $SU(3)$ then it reduces to E&M and its simpler to call it that. Regardless these are all examples of "gauge" theories. Any theory which is made out of the curvature $F_{\mu\nu}$ of a connection $A_{\mu}^I$ (in E&M this is called the potential 4-vector with components $(\phi,\textbf{A}$) is a gauge theory. – user346 Apr 15 '11 at 4:57
Any Yang-Mills theory is a gauge theory, but not every gauge theory is a Yang-Mills theory. (Gauge theories may have abelian gauge groups, or may have additional matter fields, and then are not Yang-Mills.) – Arnold Neumaier May 9 '12 at 12:56

Very briefly, a classical theory is a gauge theory if its field variables $\varphi^i(\vec{x},t)$ have a non-trivial local gauge transformation that leaves the action $S[\varphi]$ gauge invariant. Usually, a gauge transformation is demanded to be a continuous transformation.

[Gauge theory is a huge subject, and I only have time to give some explanation here, and defer a more complete answer to, e.g., the book "Quantization of Gauge Systems" by M. Henneaux and C. Teitelboim. By the word local is meant that the gauge transformation in different space-time point are free to be transformed independently without affecting each others transformation (as opposed to a global transformation). By the word non-trivial is meant that the gauge transformation does not vanish identically on-shell. Note that an infinitesimal gauge transformation does not have to be on the form

$$\delta_{\varepsilon}A_{\mu}(\vec{x},t) = D_{\mu}\varepsilon(\vec{x},t),$$

nor does it have to involve a $A_{\mu}$ field. More generally, an infinitesimal gauge transformation is of the form

$$\delta_{\varepsilon}\varphi^i(x) = \int d^d y \ R^i{}_a (x,y)\varepsilon^a(y),$$

where $R^i{}_a (x,y)$ are Lagrangian gauge generators, which form a gauge algebra, which, in turn, may be open and reducible, and $\varepsilon^a$ are infinitesimal gauge parameters. Besides gauge transformations that are continuously connected to the identity transformation, there may be so-called large gauge transformations, which are not connected continuously to the identity transformation, and the action may not always be invariant under those. Ultimately, physicists want to quantize the classical gauge theories using, e.g., Batalin-Vilkovisky formalism, but let's leave quantization for a separate question. Various subtleties arise at the quantum level as, e.g., pointed out in the comments below. Moreover, some quantum theories do not have classical counterparts.]

Yang-Mills theory is just one example out of many of a gauge theory, although the most important one. To name a few other examples: Chern-Simons theory and BF theory are gauge theories. Gravity can be viewed as a gauge theory.

Yang-Mills theory without matter is called pure Yang-Mills theory.

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+1, although I would add that a gauge transformation does not just leave the action invariant but also the Hilbert space. This makes it distinct from an ordinary symmetry transformation, which also leaves the action invariant but maps states to other states in the Hilbert space (active sense). – Olaf Apr 15 '11 at 12:12
+1, but just a pedantic comment. The action need not be invariant under gauge transformations, but all observables must be! For example a $SU(2)$ Chern-Simons theory $S_{CS}[a] = \frac k{4\pi}\int_M\text{tr}\left(a\wedge da + \frac 23 a\wedge a\wedge a\right)$, transforms under a gauge transformations as $S_{CS}[a]\rightarrow S_{CS}[a] + 2\pi km$, where $m\in\mathbb N_0$ labels the homotopy class of the gauge transformation (if $M$ has a boundary, another term will be present). Thus if $k$ is an integer, the theory is gauge invariant on quantum level but not the classical action. – Heidar Apr 15 '11 at 13:11
@4tnemele that's hardly pedantic ;) – user346 Apr 15 '11 at 13:58

What I've usually heard in practice is that a Yang-Mills theory is a gauge theory with a non-Abelian gauge group. That would disqualify classical E&M, for example, where the gauge group is $U(1)$.

Wikipedia has a slightly more restrictive definition, saying that a Yang-Mills theory is a gauge theory based on $SU(N)$ specifically.

I'm not quite sure which definition to go with. It probably depends on who you're talking to. Regardless, the only distinction I'm familiar with is based on the nature of the gauge group and has nothing to do with whether there happen to be matter fields or not.

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This doesn't seem to make sense. If I have a $U(1)$ gauge symmetry, but a Chern-Simons lagrangian, that doesn't make it Yang-Mills at all! There are an infinitude of gauge invariant theories (gauge theories for short), of which Yang-Mills is a special case, picked out by its dynamics (action). – genneth Apr 15 '11 at 7:32
Actually if you really want to be pedantic. SU(2) with the usual lagrangian is the Yang Mills theory. Of course the nomenclature is far more general now. But yes, I'd say it typically involves any SO or SU gauge group with the appropriate Yang Mills term included in the action. Eg F wedge F – Columbia Apr 15 '11 at 8:31
@Columbia is correct. The original paper by Yang-Mills was an attempt to construct a gauge theory of nuclear interactions by locally gauging the symmetry - isospin - under which a proton and neutron exchange places. – user346 Apr 15 '11 at 12:00

"Yang-Mills theory" is pretty much equivalent to "gauge theory". You could say gauge theories are more general, since we usually think of Yang-Mills theories as being gauge theories of SO or SU groups, or sometimes one of the other classical lie groups. But you could make a distinction by saying something like "gravity is a gauge theory but not a Yang-Mills theory," since its gauge group would be local Poincare transformations.

The term you're thinking of without matter is probably "pure Yang-Mills theory" or "pure gauge theory."

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Actually in the connection formulation GR turns out to be a gauge theory with the exact same kinematical phase space as Yang-Mills. Pretty neat, eh? – user346 Apr 15 '11 at 5:57
I think saying that "Yang-Mills theory is pretty much equivalent to gauge theory" is simply wrong. There are plenty of gauge theories which are not of the Yang-Mills type (Topological Quantum Field Theories, lattice gauge theories, gravity, ..) – Olaf Apr 15 '11 at 12:12
Yes, that's why I said "we usually think of..." and that whole specific example I gave of gravity not being a Yang-Mills theory... But the point is there's not well-defined agreed upon definition of what a Yang-Mills theory is--it's just a gauge theory that we usually think of as being SO or SU, which is exactly what I said. And I have certainly heard lattice people call their theories Yang-Mills, btw. I think the terminology is probably more of a cultural one than a definitional one, anyway. – Mr X Apr 15 '11 at 20:31