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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.

AccidentalFourierTransform
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  • 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. – Ultima Jan 14 '20 at 06:21
  • 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? –  Jan 15 '20 at 00:13
  • 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? – knzhou Jan 15 '20 at 08:02
  • @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. – AccidentalFourierTransform Jan 16 '20 at 00:54
  • @knzhou Well, if the rescaling does have consequences, then one has to redo several chapters of many QFT textbooks. Rescalings are very common, e.g., when discussing the scale invariance of $\phi^4$ theory: if $\phi\to\lambda\phi$ affected the measure, this would not be a symmetry of the theory, contradicting virtually every textbook, right? Also, one could define the measure to be $\prod_x \frac{g_{\mu\nu}(x)}{\hbar^{2/(d-2)}}$, and now the argument in the OP should work, no? Finally, yes: I choose GR because it is easy to write down, but any homogeneous Lagrangian works too. – AccidentalFourierTransform Jan 16 '20 at 00:58
  • @AccidentalFourierTransform Inspired by your question, I was recently surprised to learn that for $d \geq 3$, every compact manifold admits a Riemannian metric with negative scalar curvature, including the sphere $S^d$. Using a conformal factor, we can then make the euclidean Einstein-Hilbert action arbitrarily negative. If I'm not mistaken, this breaks the argument that classical solutions dominate when $\hbar\to 0$. Would you agree? – Chiral Anomaly Aug 02 '20 at 16:26
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    @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? – AccidentalFourierTransform Aug 02 '20 at 19:32
  • @AccidentalFourierTransform I haven't heard of it, but it seems like it should work as long as the manifold $M$ is fixed (as you specified in a comment), and as long as $R$ has a finite lower bound $R_0(M)$. If we take the action to be $S=\int \sqrt{g} (R-R_0)$, then $S>0$, and rescaling the metric can't change that. So I guess my argument evaporates... assuming that $R_0(M)$ exists, which seems plausible. Apparently my plausibility-meter needs a tune-up, though, because I wouldn't have guessed that $R$ could be negative everywhere on a $d$-sphere. :P – Chiral Anomaly Aug 03 '20 at 00:15
  • @AccidentalFourierTransform On second thought, maybe my argument stands. Theorem 1.1 in Scalar curvature and conformal deformation of Riemannian structure seems to say that $R_0(M)$ doesn't exist. (Reviewed here.) Gibbons, Hawking, and Perry wrote about this, and the text below equation (3) in arXiv:1304.6946 says "Several schemes have been proposed for addressing this problem". – Chiral Anomaly Aug 03 '20 at 02:23

2 Answers2

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Here are a couple of words to be said about this.

  • 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.
  • As of yet there is no good way of defining the path integral as a sum over geometries. This is because gravity is somehow different from QFTs (and we don’t really know what the off-shell formulation should look like; for example 3d Einstein gravity can be written as a gauge theory which is easier to quantize, but the spaces over which the path integral needs to be done seems to include things like degenerate/complex metrics). Georgi was known to have an argument for ruling out QG as a worthwhile thing to study back in the days. Basically if somehow you managed to find a UV fixed point for Einstein gravity, coupling to matter will destroy this. So ‘pure’ QG is useless for predictions unless you know all the matter content in the universe. Even in the nicest example, asymptotically AdS3 quantum gravity, there seem to be issues making sense of the theory as an honest quantum system. There is also the issue of wormholes, which seem to be at odds with the ‘goo’ theories of quantum gravity that we do have.
  • There are arguments that say that any correlation function of curvature tensors vanishes. Basically You can use diffeomorphisms to move the positions of the insertions around on some neighborhood. This is a gauge symmetry, to the correlation functions should not be affected. This means that all these types of correlators are locally constant, and then by moving the open ball around you can get them to be constant. But if they are constant, you might as well set them ti zero. This is usual lore that gravity has no local degrees of freedom, and that everything lives on some type of ‘asymptotic boundary’. This is even true classically; things like energy are not well defined locally, and the best you can do is to build things like the ADM and Bondi masses.
Non-invertibleSym
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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.

Ivan Burbano
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    Thanks for the input! This seems a bit circular, right? The fact that counterterms are required stems from the fact that the path-integral is divergent. So you basically assumed the conclusion. I gave a (wrong) argument as to why the path-integral is in fact convergent, so that argument (wrongly) proves that no counterterm is required. Once we agree the path-integral is ill-defined, we can talk about counterterms. But the question is, precisely, why it is ill-defined to begin with. The counterterms are not the reason, but rather the consequence. Agree? Did I perhaps misunderstand your answer? – AccidentalFourierTransform Aug 05 '20 at 10:59
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    That is a very good point! However, there is a second reason why terms of higher order in the action are required. This is the gauge invariance of the path integral. The cleanest treatment of this (much of which can be understood with finite-dimensional toy models) is in the BV formalism. In that formalism the BV action has to satisfy something called the quantum master equation, whose solution will in general require the addition of extra terms. I would suggest taking a look at Henneaux and Teitelboim, Quantization of Gauge Systems 18.2.1 – Ivan Burbano Aug 05 '20 at 11:48