Why is matter-free gravity not ultraviolet finite?

[Jim Kata]

TL;DR Summary

Matter-free gravity should be locally conformally invariant. Conformal field theories should have a beta function equal to zero. Which should imply a finite theory.

In reading Weinberg volume 1 I learned gravity is not renormalizable by Dyson power counting. This means that it has an infinite number of free parameters, and such theories lose their predictive power at energies of the common mass scale. This being said, T Hooft and Veltman showed miraculously at the one-loop level pure gravity is renormalizable. However, about ten years later, Goroff and Sagnotti showed that gravity at the two-loop level is not renormalizable. That is to say, there is no way to introduce counterterms to cancel the ultraviolet divergences at the two-loop level. Is this the end of the story?

Einstein's field equations for pure gravity imply that the Ricci tensor is zero. The Ricci tensor is constructed out of the Riemann curvature tensor. The decomposition of the Riemann curvature tensor into irreducible representations of the Poincare group takes the form (2,0)+(0,2)+(1,1)+(0,0). The only parts of the Riemann curvature tensor itself not constructed from the Ricci tensor are the (2,0) and (0,2) parts which represent the Weyl curvature tensor. The Weyl curvature tensor is locally conformally invariant.
All of that is just a long-winded way of saying that the field equations for pure gravity are locally conformally invariant. The significance of this is quoting Wikipedia:

"Conformal symmetry is associated with the vanishing of the beta function. This can occur naturally if a coupling constant is attracted, by running, toward a fixed point at which β(g) = 0. "

or to quote physicist Jean Zinn-Justin

"The scale dependence of a quantum field theory (QFT) is characterized by the way its coupling parameters depend on the energy scale of a given physical process. This energy dependence is described by the renormalization group and is encoded in the beta functions of the theory.

For a QFT to be scale-invariant, its coupling parameters must be independent of the energy scale, and this is indicated by the vanishing of the beta-functions of the theory. Such theories are also known as fixed points of the corresponding renormalization group flow."Jean-Zinn-Justin

The vanishing of the beta function , β(g) = 0 , is synonymous with a theory being UV finite.

To quote Weinberg in volume 3 of his QFT books about N = 4 SYM.
" The beta function vanishes. This is therefore a finite theory with no renormalizations at all."

In summary, why have physicists not been able to prove asymptotic safety for pure gravity? They can prove it conformally invariant. They act as though conformal invariance guarantees β(g) = 0. So what am I missing? What is the problem?

 

[renormalize]

Matter-free gravity should be locally conformally invariant.

Can you cite a reputable reference that states this? Local conformal-transformations of the metric tensor are normally defined by $g_{\mu\nu}\rightarrow g_{\mu\nu}^{\prime}=\Omega^{2}\left(x\right)g_{\mu\nu}$. Even in the absence of matter, the pure Einstein-Hilbert action $\int d^{4}x\sqrt{-g}R$ is not invariant under this transformation (see, e.g., https://en.wikipedia.org/wiki/List_of_formulas_in_Riemannian_geometry).

Conformal field theories should have a beta function equal to zero. Which should imply a finite theory.

This is true, but simple conformal-invariance of the action for a field is not sufficient to make it a conformal field theory. For instance, the action of pure Yang-Mills theory (e.g., QCD without quarks) is conformally-invariant in 4D but nevertheless the theory has a non-vanishing beta function (see, e.g., https://en.wikipedia.org/wiki/Beta_function_(physics)).

 

[Jim Kata]

To answer your question about citation I am quoting Penrose and Rindler volume 1 page 240. Here is a paraphrase of the quote :

"The tensor C_abcd is called the Weyl conformal tensor ( and Psi_abcd is often called the Weyl conformal spinor). It comprises the conformally invariant part of the curvature tensor. ... In another section they prove this ... we shall show that its vanishing is actually necessary and sufficient for space-time to be piecewise conformally flat. In space-time restricted by the Einstein field equations (without a cosmological term) R_abcd reduces to C_abcd in vacuum."

I don't understand the nuance between conformal invariance and a conformal field theory, but in the example you cite isn't pure Yang-Mills a quadradic theory? I mean the technicality is an algebraic one in taking traces not a fundamental one in not knowing how to do calculations. Correct my misunderstandings.

 

[renormalize]

To answer your question about citation I am quoting Penrose and Rindler volume 1 page 240. Here is a paraphrase of the quote :

"The tensor C_abcd is called the Weyl conformal tensor ( and Psi_abcd is often called the Weyl conformal spinor). It comprises the conformally invariant part of the curvature tensor. ... In another section they prove this ... we shall show that its vanishing is actually necessary and sufficient for space-time to be piecewise conformally flat. In space-time restricted by the Einstein field equations (without a cosmological term) R_abcd reduces to C_abcd in vacuum."

I don't understand the nuance between conformal invariance and a conformal field theory, but in the example you cite isn't pure Yang-Mills a quadradic theory? I mean the technicality is an algebraic one in taking traces not a fundamental one in not knowing how to do calculations. Correct my misunderstandings.

Well, your Penrose & Rindler "quote" neither states nor implies that "Matter-free gravity should be locally conformally invariant." So I ask again, from where do you get that idea?
And yes, the pure Yang-Mills (YM) action is, to leading order, quadratic in the gauge-potential field, and is conformally-invariant. But the Einstein-Hilbert (EH) action is also quadratic in the metric-tensor field to leading order, but is not conformally-invariant.

 

[Jim Kata]

Since you're so fond of quoting Wikipedia, here's a quote directly from Wikipedia on the Weyl Curvature tensor page.

"In general relativity, the Weyl curvature is the only part of the curvature that exists in free space—a solution of the vacuum Einstein equation—and it governs the propagation of gravitational waves through regions of space devoid of matter. More generally, the Weyl curvature is the only component of curvature for Ricci-flat manifolds and always governs the characteristics of the field equations of an Einstein manifold."

 

[Jim Kata]

Here's my sketch of a solution. Quoting Weinberg volume 1 again :

"Although non-renormalizable theories involve an infinite number of free parameters, they retain considerable predictive power. They allow us to calculate the non-analytic parts of Feynman amplitudes, like the In q and
q In q terms in the one-dimensional examples at the beginning of the previous section."

He explicates how do this in one of his papers: S. Weinberg. Physics 96.A. 327 (1979) .

He then goes on to say:

"Although non-renormalizable theories can provide useful expansions in powers of energy. They inevitably lose all predictive power at energies of the order of the common mass scale M that characterizes the various couplings."

But not what Zinn-Justin said:

"For a QFT to be scale-invariant, its coupling parameters must be independent of the energy scale, and this is indicated by the vanishing of the beta-functions of the theory "

So does the predictability of gravity calculations depend on the energy scale?

 

[renormalize]

Since you're so fond of quoting Wikipedia, here's a quote directly from Wikipedia on the Weyl Curvature tensor page.

"In general relativity, the Weyl curvature is the only part of the curvature that exists in free space—a solution of the vacuum Einstein equation—and it governs the propagation of gravitational waves through regions of space devoid of matter. More generally, the Weyl curvature is the only component of curvature for Ricci-flat manifolds and always governs the characteristics of the field equations of an Einstein manifold."

What does the existence of the Weyl curvature tensor have to do with the conformal-covariance, or not, of a gravitational theory on a Riemannian spacetime? Please be specific in your answer, preferably with math.

 

[renormalize]

So does the predictability of gravity calculations depend on the energy scale?

Yes, a gravity theory depends on the energy scale if the calculated beta-function of the theory is non-zero. This is the case for pure Einstein gravity at two-loops and even conformally-invariant Weyl/conformal gravity at one-loop.

 

[Jim Kata]

What does the existence of the Weyl curvature tensor have to do with the conformal-covariance, or not, of a gravitational theory on a Riemannian spacetime? Please be specific in your answer, preferably with math.

What I mean by conformal invariance in equations is the following:
Given the Einstein Hilbert action:

$$S = \int d^{4}x \sqrt{-g}R$$

looking at its equation of motion you get:

$$\delta S = \int d^{4}x G^{\alpha \beta} \delta g_{\alpha \beta} = 0$$

where

$$G^{\alpha \beta} = R^{\alpha \beta} - \frac{1}{2}g^{\alpha \beta}R = 0$$

from which it follows that

$$G^{\alpha \beta} = R^{\alpha \beta} = 0$$

the Ricci tensor in terms of the Riemann curvature is

$$R_{\alpha \beta} = R^{\gamma}_{\alpha \gamma \beta} = 0$$

The Riemann curvature tensor can be written in terms of the Weyl tensor as

$$R^{\kappa}_{\alpha \gamma \beta} = C^{\kappa}_{\alpha \gamma \beta} + \cdots$$

plus other terms that vanish for a Ricci flat manifold. All this is just to say that for $R^{\alpha \beta} = 0$

you have

$$R^{\kappa}_{\alpha \gamma \beta} = C^{\kappa}_{\alpha \gamma \beta}$$

The Weyl tensor and by extension the Riemann curvature in this case are invariant under conformal transformations. That is for, $g_{\alpha \beta} \rightarrow \overline{g}_{\alpha \beta} = \Omega^{2}(x)g_{\alpha \beta}$ both $C^{\kappa}_{\alpha \gamma \beta}$ & $R^{\kappa}_{\alpha \gamma \beta}$ are mapped to themselves $C^{\kappa}_{\alpha \gamma \beta} \rightarrow \overline{C}^{\kappa}_{\alpha \gamma \beta} = C^{\kappa}_{\alpha \gamma \beta}$ & $R^{\kappa}_{\alpha \gamma \beta} \rightarrow \overline{R}^{\kappa}_{\alpha \gamma \beta} = R^{\kappa}_{\alpha \gamma \beta}$. So the Einstein field equations in a vacuum are classically invariant under conformal transformations.
$$G_{\alpha \beta} = R_{\alpha \beta} = R^{\gamma}_{\alpha \gamma \beta} = C^{\gamma}_{\alpha \gamma \beta} = 0$$

The Einstein-Hilbert action itself is not conformally invariant, but the equations it yields are. Let's show this:

$$\overline{g}_{\alpha \beta} = \Omega^{2}(x)g_{\alpha \beta}$$

so

$$\sqrt{-\overline{g}} = \Omega^{4}\sqrt{-g}$$

by

$$\overline{g}_{\alpha \beta} = g_{\mu \tau} \frac{\partial x^{\mu}}{\partial \overline{x}^{\alpha}}\frac{\partial x^{\tau}}{\partial \overline{x}^{\beta}}$$

we get

$$\sqrt{-\overline{g}} = \frac{1}{J}\sqrt{-g}$$

where J is the Jacobian.

So we can see from $\sqrt{-\overline{g}} = \Omega^{4}\sqrt{-g} =\frac{1}{J}\sqrt{-g}$ that $J = \frac{1}{\Omega^{4}}$ .

Which means that $\sqrt{-\overline{g}} d^{4}\overline{x} = \Omega^{4}\sqrt{-g} J d^{4}x = \Omega^{4}\sqrt{-g} \frac{1}{\Omega^{4}}d^{4}x= \sqrt{-g}d^{4}x$ .

The part of the action then that is obviously not invariant is $R = g^{\alpha \beta} R_{\alpha \beta}$ even in the Ricci flat case where $R_{\alpha \beta}$ is the metric transforms as $g^{\alpha \beta} \rightarrow \overline{g}^{\alpha \beta} = \frac{1}{\Omega^{2}}g^{\alpha \beta}$

 

[Mordred]

Have you looked at the one loop effective action ?

It's been a while since I last looked at the issue with renormalizing gravity but I do recall you need to identify the divergent parts and use some scheme such as dimensional regularization, numerous schemes exist. Same for renormalization to account for the divergence. It might be helpful to have the Hoof't article handy



https://cds.cern.ch/record/261104/files/CM-P00049196

The format is difficult to read but you did mention this paper above so thought it may help
Also if I recall the one loop effective action is 1PI (One particle irreducable ) but it's been too long to recall with surety

 

[Jim Kata]

The problem is this:
Paraphrasing Polchinski from String Theory volume 1

"A number of familiar field theories in four dimensions are classically invariant under rescaling, pure Yang-Mills, $\phi^{4}$ and so on. However, we know that due to divergences in the quantum theory, there is a nonzero renormalization group beta function in each theory, implying that the effective coupling constant is in fact a function of the length scale."

In the quantum theory, you have an anomaly that breaks scale invariance.

 

[renormalize]

So the Einstein field equations in a vacuum are classically invariant under conformal transformations.

This statement is simply wrong. Under a local conformal-transformation in which the metric tensor $g$ transforms according to $g_{\mu\nu}\rightarrow g_{\mu\nu}^{\prime}=e^{-2\sigma\left(x\right)}g_{\mu\nu}$, the Einstein tensor $G$ transforms as:$$G_{\mu\nu}\rightarrow G_{\mu\nu}^{\prime}=G_{\mu\nu}-2\sigma_{;\mu}\sigma_{;\nu}-g_{\mu\nu}\sigma^{;\lambda}\sigma_{;\lambda}+2\sigma_{;\mu;\nu}-2g_{\mu\nu}\square\sigma\tag{1}$$If we start in the unprimed case from a vacuum spacetime $G_{_{\mu\nu}}=0$, we transform to a non-vacuum primed spacetime satisfying $$G_{\mu\nu}^{\prime}=-2\sigma_{;\mu}\sigma_{;\nu}-g_{\mu\nu}\sigma^{;\lambda}\sigma_{;\lambda}+2\sigma_{;\mu;\nu}-2g_{\mu\nu}\square\sigma\tag{2}$$ in which the Einstein tensor is sourced by the peculiar "energy-momentum" tensor on the right side of (2). Einstein's field equations are most definitely not "classically invariant under conformal transformations".

 

[Jim Kata]

Globally it is not invariant, as you have shown, but I believe locally it is invariant under conformal transformations. This is probably nothing special per se since the EH action is, modulo a total derivative, invariant under all gauge transformations. For all transformations of the form $\overline{x}^{\alpha} = x^{\alpha} + \xi^{\alpha}$ where $\xi^{\alpha}$ is an infinitesimal you get:

$$\int d^4x \sqrt{det \overline{g}}\overline{R} = \int d^4x \left[ \sqrt{g} R +\partial_{\alpha}\left( \xi^{\alpha}\sqrt{g} R \right) \right]$$

Conformal transformations being the case where $\xi_{\alpha ; \beta} + \xi_{\beta ; \alpha} = \frac{1}{2}g_{\alpha \beta}\xi^{\gamma}_{;\gamma}$

A proof of this identity can be found in appendix A of t' Hooft and Veltman. Mordred linked the article above.

 

[renormalize]

Globally it is not invariant, as you have shown, but I believe locally it is invariant under conformal transformations.

This makes no sense to me. The conformal-transformation of the metric tensor in post #12 that takes the Einstein tensor from a vacuum to a non-vacuum spacetime (thereby demonstrating the non-invariance of vacuum solutions) is entirely local: it holds at every point $x$ in the manifold.

For all transformations of the form $\overline{x}^{\alpha} = x^{\alpha} + \xi^{\alpha}$ where $\xi^{\alpha}$ is an infinitesimal you get:$$\int d^4x \sqrt{det \overline{g}}\overline{R} = \int d^4x \left[ \sqrt{g} R +\partial_{\alpha}\left( \xi^{\alpha}\sqrt{g} R \right) \right]$$Conformal transformations being the case where $\xi_{\alpha ; \beta} + \xi_{\beta ; \alpha} = \frac{1}{2}g_{\alpha \beta}\xi^{\gamma}_{;\gamma}$

What you describe here is an infinitesimal coordinate transformation $\overline{x}^{\alpha} = x^{\alpha} + \xi^{\alpha}$ that satisfies a particular requirement. Of course the Einstein action is invariant since it has been defined to be so under any general coordinate transformation. But this has nothing to do with conformal transformations in a curved spacetime, which are implemented, not by coordinate transforms, but rather by arbitrary position-dependent rescalings of the metric tensor. Also note that your transformation requirement $\xi_{\alpha ; \beta} + \xi_{\beta ; \alpha} = \frac{1}{2}g_{\alpha \beta}\xi^{\gamma}_{;\gamma}$ defines $\xi$ to be a conformal Killing vector. But a generic spacetime admits no Killing vectors, conformal or otherwise.

A proof of this identity can be found in appendix A of t' Hooft and Veltman. Mordred linked the article above.

I can't find any such proof in that article. Can you please cite the page and equation number(s)? Thanks.