Musings

Yang-Mills Theory

Let’s start by reviewing Singer’s explication of the Gribov ambiguity.

Say we want to do the path integral for Yang-Mills Theory, with compact semi-simple gauge group $G$. For definiteness, we’ll talk about the Euclidean path integral, and take $M= S^4$. Fix a principal $G$-bundle, $P\to M$. We would like to integrate over all connections, $A$, on $P$, modulo gauge transformations, with a weight given by $e^{-S_{\text{YM}}(A)}$. Let $\mathcal{A}$ be the space of all connections on $P$, $\mathcal{G}$ the (infinite dimensional) group of gauge transformations (automorphisms of $P$ which project to the identity on $M$), and $\mathcal{B}=\mathcal{A}/\mathcal{G}$, the gauge equivalence classes of connections.

“Really,” what we would like to do is integrate over $\mathcal{B}$. In practice, what we actually do is fix a gauge and integrate over actual connections (rather than equivalence classes thereof). We could, for instance, choose background field gauge. Pick a fiducial connection, $\overline{A}$, on $P$, and parametrize any other connection $$A= \overline{A}+Q$$ with $Q$ a $\mathfrak{g}$-valued 1-form on $M$. Background field gauge is

which picks out a linear subspace $\mathcal{Q}\subset\mathcal{A}$. The hope is that this subspace is transverse to the orbits of $\mathcal{G}$, and intersects each orbit precisely once. If so, then we can do the path integral by integrating¹ over $\mathcal{Q}$. That is, $\mathcal{Q}$ is the image of a global section of the principal $\mathcal{G}$-bundle, $\mathcal{A}\to \mathcal{B}$ and integrating over $\mathcal{B}$ is equivalent to integrating over its image, $\mathcal{Q}$.

What Gribov found (in a Coulomb-type gauge) is that $\mathcal{Q}$ intersects a given gauge orbit more than once. Singer explained that this is not some accident of Coulomb gauge. The bundle $\mathcal{A}\to \mathcal{B}$ is nontrivial and no global gauge choice (section) exists.

A small technical point: $\mathcal{G}$ doesn’t act freely on $\mathcal{A}$. Except for the case² $G=SU(2)$, there are reducible connections, which are fixed by a subgroup of $\mathcal{G}$. Because of the presence of reducible connections, we should interpret $\mathcal{B}$ as a stack. However, to prove the nontriviality, we don’t need to venture into the stacky world; it suffices to consider the irreducible connections, $\mathcal{A}_0\subset \mathcal{A}$, on which $\mathcal{G}$ acts freely. We then have $\mathcal{A}_0\to \mathcal{B}_0$ of which $\mathcal{G}$ acts freely on the fibers. If we were able to find a global section of $\mathcal{A}_0\to \mathcal{B}_0$, then we would have established $$\mathcal{A}_0\cong \mathcal{B}_0\times \mathcal{G}$$ But Singer proves that

1. $\pi_k(\mathcal{A}_0)=0,\,\forall k\gt 0$. But
2. $\pi_k(\mathcal{G})\neq 0$ for some $k\gt 0$.

Hence $$\mathcal{A}_0\ncong \mathcal{B}_0\times \mathcal{G}$$ and no global gauge choice is possible.

What does this mean for Yang-Mills Theory?

• If we’re working on the lattice, then $\mathcal{G}= G^N$, where $N$ is the number of lattice sites. We can choose not to fix a gauge and instead divide our answers by $Vol(G)^N$, which is finite. That is what is conventionally done.
• In perturbation theory, of course, you never see any of this, because you are just working locally on $\mathcal{B}$.
• If we’re working in the continuum, and we’re trying to do something non-perturbative, then we just have to work harder. Locally on $\mathcal{B}$, we can always choose a gauge (any principal $\mathcal{G}$-bundle is locally-trivial). On different patches of $\mathcal{B}$, we’ll have to choose different gauges, do the path integral on each patch, and then piece together our answers on patch overlaps using partitions of unity. This sounds like a pain, but it’s really no different from what anyone has to do when doing integration on manifolds.

Gravity

The Asymptotic Freedom people want to do the path-integral over metrics and search for a UV fixed point. As above, they work in Euclidean signature, with $M=S^4$. Let $\mathcal{Met}$ be the space of all metrics on $M$, $\mathcal{Diff}$ the group of diffeomorphism, and $\mathcal{B}= \mathcal{Met}/\mathcal{Diff}$ the space of metrics on $M$ modulo diffeomorphisms.

Pick a (fixed, but arbitrary) fiducial metric, $\overline{g}$, on $S^4$. Any metric, $g$, can be written as $$g_{\mu\nu} = \overline{g}_{\mu\nu}+ h_{\mu\nu}$$ They use background field gauge,

where $\overline{\nabla}$ is the Levi-Cevita connection for $\overline{g}$, and indices are raised and lowered using $\overline{g}$. As before, (2) defines a subspace $\mathcal{Q}\subset \mathcal{Met}$. If it happens to be true that $\mathcal{Q}$ is everywhere transverse to the orbits of $\mathcal{Diff}$ and meets every $\mathcal{Diff}$ orbit precisely once, then we can imagine doing the path integral over $\mathcal{Q}$ instead of over $\mathcal{B}$.

In addition to the other problems with the asymptotic safety program (the most grievous of which is that the infrared regulator used to define $\Gamma_k(\overline{g})$ is not BRST-invariant, which means that their prescription doesn’t even give the right path-integral measure locally on $\mathcal{Q}$), the program is saddled with the same Gribov problem that we just discussed for gauge theory, namely that there is no global section of $\mathcal{Met}\to\mathcal{B}$, and hence no global choice of gauge, along the lines of (2).

As in the gauge theory case, let $\mathcal{Met}_0$ be the metrics with no isometries³. $\mathcal{Diff}$ acts freely on the fibers of $\mathcal{Met}_0\to \mathcal{B}_0$. Back in his 1978 paper, Singer already noted that

1. $\pi_k(\mathcal{Met}_0)=0,\,\forall k\gt 0$, but
2. $\mathcal{Diff}$ has quite complicated homotopy-type.

Of course, none of this matters perturbatively. When $h$ is small, i.e. for $g$ close to $\overline{g}$, (2) is a perfectly good gauge choice. But the claim of the Asymptotic Safety people is that they are doing a non-perturbative computation of the $\beta$-functional, and that $h$ is not assumed to be small. Just as in gauge theory, there is no global gauge choice (whether (2) or otherwise). And that should matter to their analysis.

Note: Since someone will surely ask, let me explain the situation in the Polyakov string. There, the gauge group isn’t $\mathcal{Diff}$, but rather the larger group, $\mathcal{G}= \mathcal{Diff}\ltimes \text{Weyl}$. And we only do a partial gauge-fixing: we don’t demand a metric, but rather only a Weyl equivalence-class of metrics. That is, we demand a section of $\mathcal{Met}/\text{Weyl} \to \mathcal{Met}/\mathcal{G}$. And that can be done: in $d=2$, every metric is diffeomorphic to a Weyl-rescaling of a constant-curvature metric.