# On the finiteness of quantum gravity$.$

Consider naïve quantum gravity, defined by $$Z=\int e^{-\frac{1}{\hbar}\int R}\mathrm dg$$ where $$R$$ denotes the Ricci scalar, and $$\mathrm dg$$ a path integral over all metrics. I have set $$G_N=1$$ to simplify the notation. I will further assume $$d>2$$ so that $$\int R$$ is non-topological, and so gravitons propagate.

Consider the following (wrong) argument.

• On the one hand, if we perform the change of variables $$g_{\mu\nu}\to \hbar^{2/(d-2)}g_{\mu\nu}$$, the partition function becomes $$Z=\int e^{-\int R}\mathrm dg$$ and so we see that $$Z$$ is actually $$\hbar$$-independent.

• On the other hand, if we let $$\hbar\to0$$, the path-integral localises on the space of solutions to the classical equations of motion, the Einstein equations: $$Z=\sum_{G_{\mu\nu}\,\equiv\,0} e^{-\int R}$$ up to the standard one-loop determinant.

Putting these two observations together, it would appear that naïve quantum gravity is perfectly well-defined, finite, and in fact identical to classical gravity (who'da thunk it?).

Needless to say, the conclusion is false, so there must be some mistake somewhere in the argument. Where is it? Here are my thoughts on the possible source of the error:

• The change of variables does not leave the measure invariant. This doesn't seem likely, because it is just a rescaling, and we perform these all the time, e.g. when discussing scaling dimensions and beta functions, etc. Even in basic QFT, where we introduce the wave-function renormalization $$\phi\to Z^{1/2}\phi$$ to set the residue of the two-point function to $$1$$ (the normalisation required for the LSZ).

• The localisation procedure is subtle. I assume the key is here. For example, the localisation includes a one-loop determinant. But this determinant is in fact known to be finite (QG is one-loop finite, cf. this PSE post), so the argument above seems to suggest that $$Z$$ is finite too, to all orders, which is false. Furthermore, SUSY localisation really does follow the steps above, so I don't know why GR should be different.

• The sum over classical solutions is ill-defined. This again seems unlikely, because we could consider e.g. compact spacetimes, where $$\int R$$ is finite, and/or consider boundary conditions restrictive enough to have good control over the classical solutions.

• In general we are not only interested in $$Z$$, but also in correlation functions. So we should also consider insertions. A natural family of correlation functions is $$\langle R_{\mu\nu}R_{\mu'\nu'}\cdots\rangle$$. But $$R_{\mu\nu}$$ is scale-invariant, and so the argument above also "proves" that these correlation functions are finite, just like $$Z$$ itself.

• My guess would be that the next-to-leading-order correction to a given saddle doesn't vanish in the $\hbar \to 0$ limit, as opposed to the case with supersymmetric localization. Jan 14 '20 at 6:21
• When you talk about "a path integral over all metrics" what measure are you using on metrics? Do you sum over metrics that have different topologies, and if so, how do you do that? If two metrics are the same under a diffeomorphism, do you somehow make sure they don't get double-counted?
– user4552
Jan 15 '20 at 0:13
• Can you really do this rescaling with no consequences? The measure is invariant up to an overall constant which depends on the rescaling factor -- which we usually don't care about at all, but which matters for this argument. Furthermore, couldn't you apply that rescaling to a whole bunch of field theories, giving definitely wrong results? Jan 15 '20 at 8:02
• @BenCrowell typically the topology is fixed, although I do recall people discussing topology-changing amplitudes. I do not consider that here. OTOH, the sum is understood modulo diffeos, so as to not overcount. This can be done by fixing the gauge (e.g., by imposing some condition like $\partial g=0$, which can be done by inserting a Dirac delta $\delta(\partial g)$; this doesn't affect the argument), or by formally dividing $Z$ by $\mathrm{vol}(Diff)$, the "volume" of the diffeo group. Jan 16 '20 at 0:54
• @ChiralAnomaly That definitely sounds reasonable. But I also recall that QG people usually introduce a "compensating" term to the EH action to make it finite, something like subtracting the action evaluated on a flat metric, or something like that. I don't remember the details. There are also boundary terms, although these don't matter for compact manifolds. Had you heard about these compensating factors before? Or am I misremembering? Do they interfere with your argument? Aug 2 '20 at 19:32

• You have to be careful about units when rescaling by $$\hbar$$. Usually you want the metric components to be dimensionless, and you can arrange for $$\hbar$$ to be unitless, but the actual expansion parameter is actually the Plank length because the scalar curvature has units of 1/Area. So you can re scale $$\hbar$$ or $$G$$ but not both. This means that re-scaling the metric to absorb $$\hbar$$ also changes the scale of the theory (that is the strength of the quantum corrections).
• Because changing $$\hbar$$ is not quite allowed, this only gives the classical limit of the Path integral. Also not all localizations are created equal! SUSY localization relies on being able to deform the action by a term that is supersymmetric. This usually reduces the path integral to a BPS sector. So this doesn’t capture everything about the theory. Another example is the quantization of coadjoint orbits. In a phase space path integral you can always trade the path integral ‘measure’ coming from the pfaffian of the symplectic form for a path integral over some auxiliary ghost fermions. It turns out every system has this hidden "supersymmetry". However, the localization argument doesn’t work to your benefit, unless the phase space has some nice properties, like being Kahler, that lets you add deformation terms that make non-trivial localization possible.
There is definitely a mistake being made with you addressed in your first proposal. In general, a partition function of the form $$\int\mathcal{D}\phi\,e^{-\frac{1}{\hbar}S}$$ for some actions $$S$$ of a classical theory does not define a proper QFT. Renormalization (and gauge invariance) demand that the action be supplemented with some counterterms which are of higher order in $$\hbar$$. These terms make the partition function $$\hbar$$ dependent.