### Asymptotic Safety and the Gribov Ambiguity

Recently, an old post of mine about the Asymptotic Safety program for quantizing gravity received a flurry of new comments. Inadvertently, one of the pseudonymous commenters pointed out *yet another* problem with the program, which deserves a post all its own.

Before launching in, I should say that

- Everything I am about to say was known to Iz Singer in 1978. Though, as with the corresponding result for nonabelian gauge theory, the import seems to be largely unappreciated by physicists working on the subject.
- I would like to thank Valentin Zakharevich, a very bright young grad student in our Math Department for a discussion on this subject, which clarified things greatly for me.

### 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 integrating^{1} 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 case^{2} $G=SU(2)$, 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

- $\pi_k(\mathcal{A}_0)=0,\,\forall k\gt 0$. But
- $\pi_k(\mathcal{G})\neq 0$ 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 isometries^{3}. $\mathcal{Diff}$ acts freely on the fibers of $\mathcal{Met}_0\to \mathcal{B}_0$. Back in his 1978 paper, Singer already noted that

- $\pi_k(\mathcal{Met}_0)=0,\,\forall k\gt 0$, but
- $\mathcal{Diff}$ 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.

^{1} To get the right measure on $\mathcal{Q}$, we need to use the Fadeev-Popov trick. But, as long as $\mathcal{Q}$ is transverse to the gauge orbits, that’s all fine, and the prescription can be found in any textbook.

^{2} For more general choice of $M$, we would also have to require $H^2(M,\mathbb{Z})=0$.

^{3} When $dim(M)\gt 1$, $\mathcal{Met}_0(M)$ is dense in $\mathcal{Met}(M)$. But for $dim(M)=1$, $\mathcal{Met}_0=\emptyset$. In that case, we actually *can* choose a global section of $\mathcal{Met}(S^1) \to \mathcal{Met}(S^1)/\mathcal{Diff}(S^1)$.

## Re: Asymptotic Safety and the Gribov Ambiguity

It’s actually not clear whether there’s a Gribov problem for quantum gravity. This has ben investigated most carefully for the canonical path integral, where the relevant integration variables are the three-metric mod diffeos and the corresponding momentum. Here, Fischer and Moncrief showed back in 1996 that for many spatial topologies, the bundle is, in fact, trivial, and there is no Gribov problem – see Gen. Rel. Grav. 28 (1996) 221.

I know much less about the full four-dimensional case. But it’s a plausible conjecture that for a manifold with the topology Rx(one of the manifolds considered by Fischer and Moncrief), there is no four-dimensional Gribov problem.

(One minor subtlety: this all breaks if you include diffeos that are not in the identity component as “gauge” symmetries. But there are reasonable arguments that you shouldn’t.