Episode 1 wrote that "we need something to count" ── a candidate for that something appears
Episode 8 split the list of unsolved items into three kinds: technical, a matter of choice, and essential. Only two were left in the last box ── "does the state count of the horizon's Carrollian theory give \(e^{A/4G}\)?" and "what is holography for \(\Lambda>0\)?"
This episode asks whether those two are different problems. To give the conclusion first: there is a clear reason they look like the same question, and if they are the same, the answer fits in one sentence.
But there are also three ways they might not be the same. And the field itself is split into two camps on this point.
Recall what Episode 7 established.
null hypersurface \(\Longrightarrow\) degenerate metric \(\Longrightarrow\) Carroll manifold
That argument uses only nullness. And the cosmological horizon of de Sitter space is also a null hypersurface. The edge of the static patch, the surface beyond which an observer can in principle receive no information.
Furthermore, as Gibbons–Hawking (1977) showed, this surface has a temperature and an entropy.
Exactly the same formulae as for a black hole.
So although Episode 7 said "the universe has two Carroll surfaces" ── there were three.
| null surface | area | if a theory sits there |
|---|---|---|
| \(\mathscr{I}\) (null infinity) | infinite | holography for flat spacetime |
| black-hole horizon | finite | black-hole entropy |
| de Sitter horizon | finite | holography for the whole universe |
And the two blanks left by Episode 8 are the bottom two rows of this table. The same kind of question about the same kind of surface.
A Carrollian theory living on a null surface of area \(A\)
has \(e^{A/4G}\) states.
That one sentence settles all three. And it is self-consistent.
| null surface | area | consequence |
|---|---|---|
| \(\mathscr{I}\) | infinite | an infinite-dimensional Hilbert space = ordinary field theory. no contradiction |
| black-hole horizon | finite | Bekenstein–Hawking. reproduces the known answer |
| de Sitter horizon | finite | the Hilbert space of the whole universe is finite-dimensional |
The first row is doing work. In flat spacetime the area of \(\mathscr{I}\) is infinite, so the state count is infinite ── which is the same as ordinary field theory, and nothing is odd about it. The same formula drops to the known conclusion in flat spacetime.
The third row is an entirely different claim. It says the universe is a finite-state machine.
Let us put the numbers in. The de Sitter horizon radius is \(r_H=c/H_0\).
\(r_H=c/H_0=1.367\times10^{26}\,\mathrm{m}\) (14.45 Gly). Area \(A=4\pi r_H^2=2.35\times10^{53}\,\mathrm{m^2}\).
divide by the Planck area \(\ell_P^2=\hbar G/c^3=2.61\times10^{-70}\,\mathrm{m^2}\)
$$\frac{S}{k_B}=\frac{A}{4\ell_P^2}=\mathbf{2.25\times10^{122}}$$hence
$$\dim\mathcal{H}=e^{S/k_B}=10^{\,9.76\times10^{121}}$$For comparison, look at the ladder of entropies. All are computed with the same formula \(S=A/4\ell_P^2\).
Even M87*'s supermassive black hole (\(6.5\times10^9\,M_\odot\)) is 25.7 orders smaller than the de Sitter horizon.
Now back to Episode 1. But not to the result of the brute force ── to the order that episode placed at the end.
Section 08 of Episode 1 was about the wall that dimensional analysis cannot produce \(O(1)\) coefficients. It said this.
For black-hole entropy: dimensions get you as far as \(S\propto A\), and only the \(1/4\) refuses to come. Filling that in requires the thing being counted itself ── states, a mechanism, a symmetry.
What §02 of this episode did was precisely to name the thing being counted.
"the states of a Carrollian theory living on a null surface of area \(A\)"
Episode 1 stopped at "something must be counted." Episode 9 offers a candidate: "perhaps this is what we count." We have not counted it yet, but it does take the shape of a reply to that order.
The example Episode 1 §08 used was the agreement between the cosmological constant and the critical density ── \(\Lambda c^2/(G\rho_c)=17.3\) against \(8\pi=25.1\). The \(\Lambda\) that appeared there as an instance of "an \(O(1)\) dimensional analysis cannot produce" is, in this episode, what fixes the state count of the universe itself.
And let us write down the relation to the number \(\rho_\Lambda/\rho_{\rm Pl}\sim10^{-122}\) so often quoted for the cosmological-constant problem. Since \(S_{\rm dS}=3\pi/(\Lambda\ell_P^2)\) and \(\rho_\Lambda/\rho_{\rm Pl}\propto\Lambda\ell_P^2\) ──
Numerically too, \(1/(2.25\times10^{122})=4.4\times10^{-123}\) ── agreeing with \(10^{-122}\) up to the factor of \(3\pi\).
The question of why the cosmological constant is small was the same question as why the universe's bit count is \(10^{122}\).
Episode 2 wrote this ── what computability requires is a finite amount of information, not a spacing of space.
That "finite amount of information" now has a value.
The universe is a finite-state machine with \(e^{2.25\times10^{122}}\) states.
Not as a metaphor, but as a number computable from \(\Lambda\).
Episode 2 concluded "no lattice is needed." None is needed here either ── the source of the finiteness is not a spacing of space but the area of a horizon. It is defined covariantly and creates no preferred frame.
But being finite-dimensional carries a harsh price too. A finite-dimensional Hilbert space has no observable with an exactly continuous spectrum. An exact S-matrix cannot be defined either (since scattering's "infinite future" is cut off by the horizon). In exchange for "computable," we lose "exactly measurable."
Here comes the honest line. §01 gave the reason to expect the two problems to be the same, but there are three ways they might not be.
| asymmetry | black hole | de Sitter |
|---|---|---|
| ① inside vs outside | the observer is outside the horizon | the observer is inside the horizon (the static patch) |
| ② observer dependence | the horizon is essentially unique | a different horizon per observer. the "whose horizon?" problem |
| ③ the quality of the finiteness | the total Hilbert space is infinite-dimensional | the whole thing is finite-dimensional ── qualitatively different |
③ is especially heavy. Even with a black hole present, the Hilbert space of an asymptotically flat spacetime as a whole remains infinite-dimensional. In de Sitter the universe itself becomes finite-dimensional. That is not "a different size" but "a different kind."
There is one more thing to state honestly. De Sitter holography has two programmes that have not been unified.
| programme | where the dual lives | character |
|---|---|---|
| dS/CFT (Strominger 2001) | \(\mathscr{I}^+\) (a spacelike boundary) | Euclidean, non-unitary, imaginary central charge |
| static patch / horizon | the horizon (a null surface) | finite-dimensional. tied to an observer |
Only the latter connects to this episode's argument. The former's \(\mathscr{I}^+\) is a spacelike surface, hence not null, hence not a Carroll manifold either. The tools built in Episode 7 do not apply.
So the reading that "the two blanks were one" holds only after choosing one of the two de Sitter-holography camps. Whether that choice is right is not settled.
There were three Carroll surfaces. Episode 7 said two (\(\mathscr{I}\) and the horizon), but the de Sitter cosmological horizon is a null hypersurface too, hence a Carroll manifold. And by Gibbons–Hawking it has an entropy given by the same formula \(S=A/4G\) as a black hole.
So Episode 8's two blanks look like the same question. The tentative answer is one sentence ── "a Carrollian theory living on a null surface of area \(A\) has \(e^{A/4G}\) states." That sentence settles all three, and in flat spacetime it drops to the known conclusion (infinite-dimensional = ordinary field theory).
The number for our universe. \(S_{\rm dS}=2.25\times10^{122}\), \(\dim\mathcal{H}=10^{9.76\times10^{121}}\). Even M87*'s supermassive black hole is 25.7 orders below. The mass giving \(S=S_{\rm dS}\) is 0.61 times the mass of the observable universe ── consistent.
And Episode 1's order has a reply in shape. Episode 1 §08 stopped at "filling in the \(1/4\) requires something to count." §02's sentence names that something ── the states of a Carrollian theory living on a null surface of area \(A\). Together with \(S_{\rm dS}\propto1/(\rho_\Lambda/\rho_{\rm Pl})\), the cosmological-constant problem's \(10^{-122}\) is the reciprocal of the de Sitter entropy, and the cosmological-constant problem is rephrased as a problem about an amount of information.
But they have not been shown to be the same. Inside vs outside, observer dependence, the quality of the finiteness ── three asymmetries remain. And de Sitter holography itself is split into two camps, with only the horizon-side camp connecting to this episode's argument. §02's sentence is a candidate, not a solution.
This document is Episode 9 of the "Lattice We Build" series, a reading piece for high-school and university students who love physics. Where the sister series "That Clicks" explains known physics, this series shows the work itself.
Established material: that the de Sitter cosmological horizon is a null hypersurface; the Gibbons–Hawking temperature and entropy \(S=A/4G\) (1977); the algebraic identity \(S_{\rm dS}=3\pi/(\Lambda\ell_P^2)\); the numbers in the text (computed from \(H_0=2.1927\times10^{-18}\,\mathrm{s^{-1}}\)); and the coexistence of the two programmes dS/CFT and static-patch holography.
On the other hand, §02's "a Carrollian theory living on a null surface of area \(A\) has \(e^{A/4G}\) states" is a conjecture and has not been shown. Whether the state count of the horizon's Carrollian theory reproduces the Bekenstein–Hawking entropy is precisely the open problem listed in Episode 8, and this episode does not solve it. And the reading that the two blanks are the same problem holds only after choosing the horizon side of the two de Sitter-holography camps; whether that choice is right is also unsettled. The numbers move by tens of percent depending on the choice of current cosmological parameters.
Main series: Episode 1｜Episode 2｜Episode 3｜Episode 4｜Episode 5｜Episode 6｜Episode 7｜Episode 8 ｜ bonus: ①／②／③ ｜ sister series: Black Holes That Click／The Universe Is a Computer ── to print, use your browser's "Print" → "Save as PDF."
Print / PDF: ⌘+P (Ctrl+P on Windows). In the figure you can confirm that black holes never reach de Sitter.