What is a "free" non-Abelian Yang-Mill's theory?

Question

I hope this question will not be closed down as something completely trivial!

I did not think about this question till in recent past I came across papers which seemed to write down pretty much simple looking solutions to "free" Yang-Mill's theory on $AdS_{d+1}$. The solutions looks pretty much like the electromagnetic fields!

• Aren't classical exact solutions to Yang-Mill's theory very hard to find? Don't they have something to do with what are called Hitchin equations? (..I would be grateful if someone can point me to some expository literature about that..)

I would have thought that non-Abelian Yang-Mill's theory has no genuine free limit since it always has the three and the four point gauge vertices at any non-zero value of the coupling however arbitrarily small. This seemed consistent with what is called "background gauge field quantization" where one looks at fluctuations about a classical space-time independent gauge fields which can't be gauged to zero at will since they come into the gauge invariant quantities which have non-trivial factors of structure constants in them which are fixed by the choice of the gauge group and hence nothing can remove them by any weak coupling limit.

But there is another way of fixing the scale in which things might make sense - if one is working the conventions where the Yang-Mill's Lagrangian looks like $-\frac{1}{g^2}F^2$ then the structure constants are proportional to $g$ and hence a weak coupling limit will send all the gauge commutators to zero!

• So in the second way of thinking the "free" limit of a non-Abelian $SU(N_c)$ Yang-Mill's theory is looking like an Abelian gauge theory with the gauge group $U(1)^{N_c^2 -1}$. So is this what is meant when people talk of "free" non-Abelian Yang-Mills theory? (..which is now actually Abelian!..)

I would grateful if someone can help reconcile these apparently conflicting points of view.

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Answer 1

Free nonabelian gauge theory is the limit of zero coupling. So it is true, in a sense, that free $SU(N_c)$ gauge theory is a theory of $N_c^2 -1$ free "photons." There's a crucial subtlety, though: it's a gauge theory, so we only consider gauge-invariant states to be physical. Thus, already in the free theory, there is a "Gauss law" constraint that renders calculations (e.g. of the partition function) nontrivial and the physics of nonabelian free theories different from that of abelian ones (at least in finite volume). See, for instance, this beautiful work of Aharony, Marsano, Minwalla, Papadodimas, and van Raamsdonk, which shows that many thermodynamic aspects of confinement appear already in free theories at finite volume.

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Comments

@Squark Can you kindly explain as to why you think the $g=0$ gauge group is not a power of $U(1)$? I don't know what you mean by an Lie algebra with an additive group operation. Seems to make no sense to me beyond the triviality that any Lie algebra is also an Abelian group just like a vector space. But once we have deformed (here set to zero) the Lie brackets its a different Lie algebra with the same Abelian group structure down there and hence it makes sense to ask as to of which Lie group does this new Lie algebra correspond to. (genereically there will be no unique answer)

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@Squark I can see your argument about why Gauss's law means that on a compact spatial manifold there is no net charge but I can't see the argument in your last line where from this you derive that physical observables in a quantum gauge theory are in 1 dimensional irreducible representations of the gauge group. It would be helpful if you explain in details.

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Indeed I'm referring to this triviality, namely the Lie algebra with vector addition considered as group product. This is the group that arises because the g -> 0 limit is essentially "zooming in" to the vicinity of the identity element on the original group. To see this note that g can be regarded as a multiplicative constant in the definition of the Killing form, i.e. the metric on G. As g -> 0, the metric -> infinity and thus it's the same as the R -> infinity limit above. In fact SU(2) is a 3-sphere so the analogy is exact in this case.

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Note that the original group G survives as a symmetry in this limit, acting on the Lie algebra in the adjoint representation. It is precisely invariance under this symmetry that comes from the Gauss law constraint. Now, a power of U(1) would result if you took the quotient of the Abelian group by a lattice, but this would violate the G-symmetry

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The quantum mechanical operators Qi corresponding the charges generate the G-symmetry. Because of Gauss law we need to impose the constraints Qi Psi = 0. This means Psi has to be G-invariant. Another way to understand this is remember G is a _gauge_ symmetry even though it becomes a global symmetry in this limit

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- unlike in any other theory here it seems that one is doing perturbative expansion about a point ($g=0$) where infact the symmetry group of the theory is apparently different than that of the full theory! In hindsight that looks a bit freaky - its almost like saying that $SU(N_c)$ non-Abelian processes are only perturbative effects on a $U(1)^{N_c^2 -1}$ Abelian theory - though as you point out the $g=0$ theory still has many of the non-Abelian effects like confinement! (..though it has an Abelian gauge group!..)

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Firsly, the g = 0 gauge group is not a power of U(1). Instead, it is the Lie algebra of the original gauge group with the group operation +. Secondly, it seems strange that the gauge group at g = 0 is different, but it actually makes sense. Think that you're looking at a fixed spot on a sphere. As the sphere radius R goes to infinity, it begins to look like a plane. So the topology at R = infinity is different from any finite R. The same happens here: the g = 0 group is the tangent space at identity to the full group.

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