CONFORMAL TRANSFORMATIONS THAT CLICKEPISODE 9 ／ The thing that was supposed to be bookkeeping bares its teeth

The "wrong sign" found in Episode 4 and collected in Episode 5 — here comes the real bill

The conformal factor
was a ghost Classically it was harmless, because \(\phi\) was pure gauge.
But a quantum path integral will not allow "it is gauge, so do not look".

What you need: watching a sign, and imagining an exponential explode The action becomes unbounded below

When we rewrote gravity in Episode 4, the dilaton \(\phi\) had a kinetic term with the wrong sign. In Episode 5 we saw that the same sign makes every potential a "hill". Both times we added "harmless classically" and moved on. This episode is the bill. In 1978 Gibbons, Hawking and Perry pointed out that because of this one sign, a quantum theory of gravity cannot even be defined. The conformal factor problem — the oldest sore spot in quantum gravity — head on.

01Recap: where was the sign wrong?

The form obtained in Episode 4, once more.

$$S=\frac{c^4}{16\pi G}\int\!\sqrt{-\tilde g}\,\Big[\phi^2\tilde R+6(\tilde\partial\phi)^2\Big]d^4x$$

For a healthy scalar field the kinetic term would begin with a minus, \(-\tfrac12(\partial\phi)^2\). Here it is \(+6\). The sign points the wrong way.

　A healthy scalar fieldThe conformal factor \(\phi\)
Kinetic term\(-\tfrac12(\partial\phi)^2\)\(+6(\partial\phi)^2\)
Kinetic energypositivenegative
Move it violently andthe energy goes up (it is suppressed)the energy goes down (it is encouraged)
Namean ordinary fielda ghost

What is bad about a ghost fits in one line — the more violently it oscillates, the better off it is. An ordinary field consumes energy as it moves faster, so it settles down; this field does the opposite, losing energy the more it thrashes. There is no reason for it to stop.

02Why was this harmless classically?

As section 05 of Episode 4 showed, \(\phi\) was pure gauge — a single conformal transformation fixes it to any value you like. Set \(\phi=1\) and you have the standard picture; set \(\tilde g=\) Minkowski and you have the growing-mass picture; either way the physics is the same.

Something with no freedom to move has no way to run away. So classically this sign was "unpleasant but harmless". When the ball rolled down the hill in Episode 5, how it rolled was fixed uniquely by the Friedmann equation; it did not accelerate of its own accord.

The counting was doing the work Episode 4's "one field, minus one symmetry, net zero" — the ghost \(\phi\) is eaten precisely by the symmetry and does not survive as a physical degree of freedom. It was safe despite being a ghost because it was gauge. Once that collapses, the story changes.

03Quantum mechanically, "do not look" is not allowed

The most straightforward way to build quantum gravity is the path integral — sum over every conceivable shape of spacetime, weighting each by \(e^{-S_E}\).

$$Z=\int\!\mathcal{D}g\;e^{-S_E[g]}$$

Here \(S_E\) is the Euclidean action (time rotated onto the imaginary axis — the Wick rotation of Episode 9 of the previous series). For this integral to mean anything, \(S_E\) must be bounded below. With no lower bound, \(e^{-S_E}\) can grow without limit and the sum diverges.

But the Euclidean gravitational action contains that reversed sign, unchanged.

Calculation — confirming it can be made arbitrarily negative

① The conformal-factor part of the Euclidean action

$$S_E\;\supset\;-\frac{6}{16\pi G}\int\!(\partial\phi)^2\,d^4x\qquad(\text{in Euclidean signature }(\partial\phi)^2\ge0)$$

② Let the conformal factor oscillate finely

$$\phi(x)=1+\varepsilon\sin(2\pi k x)\qquad\Longrightarrow\qquad \int_0^1\!(\partial_x\phi)^2dx=2\pi^2k^2\varepsilon^2$$

③ Therefore

$$S_E\;\propto\;-2\pi^2k^2\varepsilon^2\qquad\xrightarrow{\;k\to\infty\;}\;-\infty$$

Simply making the oscillation finer drives the action arbitrarily negative. The weight \(e^{-S_E}\) grows without limit and the path integral diverges. And \(\varepsilon\) may stay small — a tiny ripple wins, so long as you make it fine enough.

Figure: above, a ripple in the conformal factor \(\phi=1+0.15\sin(2\pi kx)\). Below, the Euclidean action it produces. Making it finer alone drops the action and blows up the path-integral weight
k = 1　S_E = −0.44　→　path-integral weight e^(−S_E) = 10^0.2
Conformal factor φ(x) Euclidean action S_E (falling)

Push the knob right and the ripple above grows only a little, while the action below plunges. At \(k=20\) the weight is \(10^{77}\). Finer still, and there is no limit. With this, "sum over all spacetimes" cannot be defined.

04Gibbons, Hawking and Perry's prescription — and its limits

In 1978 Gibbons, Hawking and Perry stated the problem clearly and proposed a first-aid measure: take the contour of integration for the conformal factor along the imaginary axis rather than the real one (rotating in the direction \(\phi\to i\phi\)). The sign flips and the integral converges, at least in the one-loop approximation.

But this is a prescription, not a solution.

①
No established non-linear generalisationIt works as a prescription for small fluctuations, but how to define it rigorously and non-perturbatively is unsettled. Part of it rests on an unproven assumption known as the positive action conjecture.
②
It still shows up in real calculationsTry to add quantum corrections to black hole entropy and you meet this reversed sign every time. Forty years on, people are still working around it.
③
Several other routes are being triedTaking the renormalisation group of the conformal sector seriously; handling it within the asymptotic safety programme; removing the conformal factor from the dynamics at the outset, as in unimodular gravity. None has been decisive.
◇ ◇ ◇

05What this means for the series

Pause here. The thing running wild is the very field we have been moving throughout this series.

In Episode 3 we moved \(\phi=a\) to pin down what "light slowing down" was. In Episode 4 we showed it is gauge. In Episode 5 we rolled it down a potential. In Episode 6 we built a dimensionless ratio out of its gradient. That \(\phi\) was the most awkward degree of freedom in quantum gravity.

This episode in one line

"Expansion is only bookkeeping" — true, but only classically.
In quantum gravity, it is the bookkeeping that breaks first.

The finale of the previous series, "Can gravity be put into this picture?", ended where gravity fails to be renormalisable. Today we have looked at that same wall from another angle — before renormalisability is even at issue, the sum you are supposed to take cannot be defined. And the cause is precisely the degree of freedom the series kept calling "free to choose".

Being straight with you

The conformal factor problem is a problem that appears when you try to quantise gravity by a path integral. It is not a proof that no quantum theory of gravity exists. Indeed several frameworks that route around this difficulty have been proposed (string theory, loop quantum gravity, asymptotic safety) — they are simply undecided.

Also, the reversed sign is not a disease created by our rewriting. Even without Episode 4's manipulation, the conformal factor carries this sign inside the Einstein–Hilbert action from the start. The rewriting did not create the illness; it made it visible. That is perhaps the most honest result this series has to offer.

Exercises
  1. Explain from the sign of the kinetic term why a ghost "profits by moving violently".
    Show answer
    Because the sign of its kinetic energy is negative. An ordinary field's energy rises the faster it moves, so it is suppressed; a ghost's energy falls the faster it moves, so there is no reason for it to stop.
  2. With \(\phi=1+0.15\sin(2\pi kx)\), by what factor does \(\int(\partial_x\phi)^2dx\) grow when \(k\) goes from 1 to 10?
    Show answer
    \(2\pi^2k^2\varepsilon^2\) is proportional to \(k^2\), so by 100. Even with the ripple's height (\(\varepsilon\)) unchanged, making it finer alone drives the action 100 times more negative.
  3. Why was \(\phi\)'s ghost harmless classically?
    Show answer
    Because \(\phi\) was pure gauge and could be fixed to any value by a single conformal transformation (Episode 4, section 05). Not surviving as a physical degree of freedom, it had no room to run away. A quantum path integral sums over all values, so that escape route is closed.
  4. (Harder) Could the problem be avoided by not performing Episode 4's rewriting?
    Show answer
    No. The conformal factor is contained in the original Einstein–Hilbert action from the start, rewriting or not. Episode 4's manipulation did not create the problem; it only put it in plain view. So this difficulty is not a flaw of the tool called the conformal transformation but a property of gravity itself.

SUMMARYIt is the bookkeeping that breaks first

The kinetic term of the conformal factor \(\phi\) is \(+6(\partial\phi)^2\) — the opposite sign to a healthy field's \(-\tfrac12(\partial\phi)^2\) — which makes it a ghost. Classically this was harmless because \(\phi\) was entirely gauge. But a quantum path integral sums over every value of every field, so that escape route is closed. With \(-\!\int(\partial\phi)^2\) sitting in the Euclidean action, making the conformal factor oscillate finely drives the action arbitrarily negative (\(S_E\propto-k^2\)), the weight \(e^{-S_E}\) explodes, and the integral diverges — the conformal factor problem (Gibbons, Hawking and Perry, 1978).

Rotating the contour onto the imaginary axis works at one loop, but the non-linear generalisation is unsettled and the issue still surfaces in, for instance, quantum corrections to black hole entropy. And what matters is that the thing running wild is the very field this series has been moving all along. "Expansion is only bookkeeping" held only classically — in quantum gravity that bookkeeping breaks first. But the rewriting did not create the disease. It only made a sign that was there all along visible.

This document is Episode 9 of the series "Conformal Transformations That Click", written for physics-minded high-school and university students. That the kinetic term of the conformal factor in the Einstein–Hilbert action has the wrong sign, and that the Euclidean gravitational action is consequently unbounded below so that the path integral diverges (the conformal factor problem), have been standard since Gibbons, Hawking & Perry (1978, Nucl. Phys. B138, 141). That taking the contour for the conformal factor along the imaginary direction works at least at one loop, and that this depends on the positive action conjecture with no established non-linear generalisation, has likewise been understood since that paper. The identity \(\int_0^1(\partial_x\phi)^2dx=2\pi^2k^2\varepsilon^2\) for \(\phi=1+\varepsilon\sin2\pi kx\) is elementary, and the figures' numbers follow from it. The figure is a schematic calculation of the conformal-factor sector alone, not an evaluation of the full gravitational action. The conformal factor problem is a difficulty specific to quantisation by path integral and is not a proof that quantum gravity does not exist. — To print, use your browser's Print → Save as PDF (in the print version the slider is frozen and answers are hidden).

Print / PDF: ⌘+P (Ctrl+P on Windows). On screen the slider makes the ripple finer and shows the action falling. "Show answer" opens the solutions.