The answer to step 1 is in ── and the reason is that the supertranslation weight was "a half"

There is no central charge Up to now we have narrowed the question to one ── does the supertranslation sector of BMS₄ have an analogue of \(c_M\)?
Here is the answer. No. Not "not yet found," but structurally unable to enter.
Put the central term in as unknowns and solve the Jacobi identity as a linear system, and the solution vanishes.
And the reason came out too ── because the four-dimensional supertranslation weight is the awkward number \(1/2\).

A reader made a fair point about the previous article ── "the usual loop of swapping one problem for another?" Half right. Episode 12 merely arrived at Episode 11's wall from another direction; it was not progress.
There are only two ways out. Compute, or go and get the negative. This episode does both. We actually ran step 0 (the positive control) and step 1 (the branch point) from the procedure table placed at the end of last time.
The result first. Step 0 failed once and spat out a bug in my own formula. And the answer to step 1 is NO.
This is not a rephrasing. One door has closed.

01First, confirm the tool is not broken

Episode 11 wrote down the template for the proof ── asymptotic symmetry → central charge → Cardy. If we intend to use it in four dimensions, we must first see that the tool works correctly in the three-dimensional case where the answer is known. Step 0.

Doing it for real, it failed. The BMS-Cardy formula written in Episode 11 was mixing conventions ── the algebra was written with \(\frac{c}{12}m^3\) while the formula used the form from another convention that writes the central term bare. The entropy was overestimated by \(\sqrt{12}=2\sqrt3\approx3.46\).

Corrected and re-run, it agrees with the geometry side to a relative deviation of \(2.2\times10^{-16}\) across six orders of magnitude in \(GM\). The details and the corrected formula are in Episode 11, §03.

The tool passed. Having earned the right to proceed to step 1, we proceed.

02Make the central term an unknown

The question was this ── can a central term enter the bracket of BMS₄ that straddles superrotations and supertranslations?

"I looked and did not find it" is no answer. Write down every form it could take and solve the equations. The central term is a number (the coefficient of the central element \(Z\)), so there are infinitely many unknowns rather than finitely many, but cut off by mode it is a system of simultaneous linear equations.

First, write the algebra in a form covering both three and four dimensions. With superrotations \(L_m\) and supertranslations \(T_r\) ──

Here \(a\) is the weight the supertranslations carry under the superrotations. This single number distinguishes three dimensions from four.

The \(1/2\) in four dimensions comes from the supertranslations carrying weight \((-\tfrac12,-\tfrac12)\) on the celestial sphere ── an established fact, to the point that some write \(\mathfrak{bms}_4\) as \(W(-\tfrac12,-\tfrac12,-\tfrac12,-\tfrac12)\). That awkward number becomes the protagonist of this episode.

Now, \(C(m,r)\) is the coefficient of a central element, so it can only be non-zero when the modes sum to zero (otherwise \(Z\) would carry a mode). So the unknowns line up in a single row ──

The unknowns are \(\{c_m\}\). The stage is set.

03The Jacobi identity becomes the equations

You may not put in arbitrary \(c_m\). A Lie algebra must satisfy the Jacobi identity. Written with the three elements \(L_m,\,L_n,\,T_r\) ──

The central element commutes with everything, so we can pick out just the central part. Only \(r=-(m+n)\) survives ──

Completely linear in \(c_m\). Which is to say, this is just a system of simultaneous linear equations. All we have to do is count the dimension of the solution space.

But one subtraction is needed, because there are merely apparent solutions. Redefine a generator as \(T_0\to T_0+\beta Z\) and a central term appears although nothing was done ──

This is nothing but a change of coordinates, so it is not a physical central charge. Mathematics calls it a coboundary. What we want is

If this is \(0\), "there is no central charge"; if \(1\) or more, "there is." The answer is an integer, and binary.

04Positive control ── does \(m^3-m\) come out for bms₃?

The same discipline again. Try it first where the answer is known. Three dimensions really has \(c_M=3/G\), so if putting in \(a=1,\,b=0\) does not give \(\dim H^2=1\), this computation cannot be trusted.

Cutting the modes at \(|m|\le8\) gives 17 unknowns and 217 equations. Solved exactly over the rationals (with floating point, judging the rank becomes a question of thresholds).

# [L_m, T_r] = (m - r) T_{m+r} + c_m delta_{m+r,0}
# 17 unknowns (|m| <= 8), 217 equations from Jacobi, exactly over the rationals

  cocycle space         dim = 2
  coboundary            dim = 1     c_m = 2m
  ------------------------------------------
  dim H^2 = 1

  identifying the extra solution as a polynomial:
    c_m = m^3        satisfies Jacobi : YES
    c_m = m^3 - m    satisfies Jacobi : YES
    c_m = m          satisfies Jacobi : YES  (← coboundary. trivial)

  → there it is. this is what c_M = 3/G really is.
     matches the form of the literature's central term (c_M/12)(m^3 - m).

There it is. \(m^3-m\) satisfies Jacobi, and it is not a coboundary. This is the \(c_M\) that carried the entire entropy in Episode 11. The three-dimensional success has been reproduced inside my own computation.

The tool works. On to the real run.

05bms₄ ── nothing comes out

Put \(a=1/2,\,b=-1/2\) into the same code.

# [L_m, T_r] = ((m-1)/2 - r) T_{m+r} + c_m delta_{m+r,0}

  cocycle space         dim = 1
  coboundary            dim = 1     c_m = (3m - 1)/2
  ------------------------------------------
  dim H^2 = 0

  looking at the unique cocycle (|m| <= 12, m = -3..3):
    -2/7  -1/5  -4/35  -1/35  2/35  1/7  8/35
     = (3m - 1) / 35     ← the coboundary itself

  trying the form that worked in three dimensions:
    c_m = m^3 - m    satisfies Jacobi : NO (violated at 388 places)

  unchanged when the mode cutoff is varied:
    |m| <= 4   dim H^2 = 0
    |m| <= 6   dim H^2 = 0
    |m| <= 8   dim H^2 = 0
    |m| <= 12  dim H^2 = 0

  the sign convention is written both (m+1)/2 and (m-1)/2 in the literature, so both:
    a=1/2, b=+1/2   dim H^2 = 0
    a=1/2, b=-1/2   dim H^2 = 0

All that survived in the solution space is one coboundary. And that is merely apparent, produced by shifting a generator's definition. Subtract it and you get zero.

06So what kind of algebra can have one?

Stop here and it ends at "four dimensions failed." Sweep \(a\) with the same code and the condition itself should become visible. Moving the weight continuously, I counted \(\dim H^2\).

Move the slider and the bar stays low almost everywhere. And it jumps only at the integers. Tabulating the sweep ──

This is a shape the series has stepped on many times. Anomaly cancellation in Episode 2, the closing of the algebra in Episode 6, the twistorial lift in Episode 8, the central charge in Episode 11 ── a binary target that cannot be tuned continuously. This time it wore the face of integrality.

And four dimensions missed. Not narrowly, not by a whisker ── by exactly \(1/2\).

07Where does the four-dimensional "half" come from?

Here I write with inference included. The numerical fact (\(H^2=0\) at \(a=1/2\)) is as computed above, but the reading of why only four dimensions is awkward is mine.

A supertranslation was "the freedom to shift the time." How the shift function \(T\) transforms under coordinate changes on the boundary ── that weight is \(a\).

The total is \(-1\) in both cases. What differs is that four dimensions has two chiralities, so the two of them share that \(-1\). Shared, it becomes \(1/2\), and falls off the integer target.

Put differently ── the weight split because the four-dimensional celestial sphere is two-dimensional. In three dimensions the boundary is one-dimensional, so there was nobody to split with and it stayed an integer.

08Checking against the literature

Saying "there is none" from my own computation alone would be risky, so I checked. Three independent lines give the same answer.

Zero from the algebra side and zero from the charge side. This \(1/2\) is not the kind of wall that merely happens to be beyond present technique.

09The door that closed, and the door that remains

Let us refill the procedure table placed at the end of Episode 11.

Episode 11 wrote that "the question is well posed, including the possibility that the answer is 'no'." We have now drawn that "no."

It looks like a loss, but not quite. A door that has closed is a door you never have to knock on again. We treat it the same way Episode 1 treated \(p=0.67\) as a result ── a negative is information.

What remains is 2b. Route B from Episode 11, §05 ── the road that uses no modular invariance and builds a Cardy-type formula by dimensional reduction in the limit of a small thermal circle (the line Di Pietro–Komargodski followed in \(d=4,6\)). That route does not require \(c_M\) in the first place. The absence of a central charge is no blow to it.