### Maybe this time …

For many years, I tried keeping up with the LQG literature. Though it provided occasional fodder for blogging, it mostly was an exercise in frustration. Years ago, I gave up the effort. Still, occasionally, an LQG paper crosses my radar screen with claims interesting enough to cause me to suspend my better judgement.

One such paper, by Gomes et al, purports to be a significant breakthrough in the understanding of AdS/CFT. They claim to reproduce the conformal anomaly of a boundary CFT from some Loopy formulation (“Shape Dynamics”) of the bulk theory, thereby shedding light on the 1998 computation of Henningson and Skenderis who first reproduced the conformal anomaly from AdS/CFT (a more careful and thorough derivation can be found in a followup paper).

How could I resist?

The central object of study is the trace of the 1-point function of the stress tensor on a curved background^{1}

Naïvely, this vanishes in a CFT. In fact, when the spacetime dimension, $d$, is odd, it does vanish. But for $d$ even, there is an anomaly

where^{2}

The trace anomaly coefficients ($c$ in $d=2$, $(a,c)$ in $d=4$) are constants which characterize, in part, the conformal field theory.

Note two obvious features

- $T(x)d\text{Vol}_g$ is invariant under
*constant*rescalings of $g$. As a consequence, the dependence of the effective action, $W[g]$, on the overall volume of $M$ is logarithmic, where the coefficient of the logarithm is proportional to the trace anomaly coefficient(s). - The group of Weyl transformations has a subgroup which preserves the overall volume of $M$ (the authors call these VPCTs). These act nontrivially, and integrating up that action yields a nontrivial dependence of $W[g]$ on the conformal factor (in $d=2$, this is the famous Liouville action).

Unfortunately, the authors got this *backwards*.

- They demanded
*invariance*of $W[g]$ under VPCTs. This lead them to postulate**Axiom 5’**of their paper, which states (in the notation of (1)) that $T(x)$ is a position-independent*constant*, for any choice of $g$. While that’s true for $d$ odd (where $T(x)=0$), it’s manifestly*untrue*for $d$ even (where it’s given by (2)). - Conversely, they were labouring under the mistaken impression that $W[g]$ has some complicated dependence on the overall volume of $M$, which can — for large volume — be expanded in an asymptotic expansion. And it is this latter dependence that they proceed to compute from their loopy considerations.

In other words, their ‘significant breakthrough’ was to get the entire story *precisely backwards.*

*Aughh!*

After picking myself up off the field, and brushing off the dirt, I emailed the authors. After a lengthier-than-necessary back-and-forth, they finally agreed that there was an issue, but *assured me* that they knew how to resolve it. Weeks passed. Eventually, they sent me a revised version. In it, they withdrew the central claim, namely that they could reproduce the correct expression for (1) from some loopy bulk computation. Instead, they admitted that their loopy prescription computes a quantity entirely *unrelated* to the stress tensor of the boundary theory but claimed (*incorrectly*) that, in the large volume limit, both quantities vanish — so they’re morally the same.

An even lengthier discussion ensued, which slowly devolved into a tutorial on the most *elementary* aspects of AdS/CFT. They had no clue, for instance, how the metric $g$, of the boundary theory, is supposed to be related to the metric $G$, of the bulk theory (the most *basic* part of the AdS/CFT correspondence). Rather than picking up any of the well-written introductions to AdS/CFT, they apparently had simply decided to *guess* how the correspondence is supposed to work.

It seems to me that, if you

*Guess*what the AdS/CFT dictionary is supposed to be.- Use your guess to compute the first obvious observable (the trace anomaly of the boundary theory).
- Obtain the wrong answer for that observable.

you might wish to re-examine your guess, instead of writing a paper claiming it to be a great conceptual breakthrough.

But that’s the kind of naïveté that keeps getting me (and Charlie Brown) in trouble.

^{1} For definiteness, we work in Euclidean signature, with a closed Riemannian $d$-manifold, $M$.

^{2} The Euler densities are normalized so that $\chi(M) = \int_M {(\text{Euler})}_{d} d\text{Vol}_g$

## Re: Maybe this time …

Very good analysis.