A Poisson sprinkling has one length, the discreteness length . Do the means and variances of counts built on it fix a dimensionless constant? We consider the causal-set action, a sum over pairs of points weighted by the number of points between them, and the horizon count, a sum over pairs straddling a horizon weighted by the number of points in the future of the lower one. Both means are linear in the weights, so the Newton constant in units of appears only with a free coupling, and the entropy per horizon molecule only with the weight normalisation if the count carries no entropy unit of its own (Proposition 1). The squared mean per unit area divided by the variance per unit area has no upper bound over weightings that give the area law exactly (Proposition 3). In backgrounds conformal to a static metric with round spatial sections the mean horizon count has no term proportional to a volume (Theorem 1).
The free coupling is in the action of Benincasa and Dowker [BD], and the cutoff dependence of the area coefficient of an entropy, together with the renormalisation of Newton’s constant, is known from entanglement entropy [Susskind–Uglum]. Horizon molecules and their area law are due to [Barton et al.]. The kernels with extra Mellin zeros are those of Boundary and corner terms of the causal-set action and the exact weights those of Horizon counts: which weights give the area law. Ahmed, Dodelson, Greene and Sorkin [Ahmed et al.] obtain a fluctuating cosmological constant of the observed order from the fluctuation of the number of points in a sprinkling, and section 6 replaces that number by the horizon count. The result not contained in these references is the unbounded variance ratio, Proposition 3 with the finite variance constant of Lemma 1. Section 5 concerns a different source of numbers: a finite rigid configuration defined over the rationals has algebraic Gram entries, and the nonzero rapidities and angles between its unit vectors are then transcendental under the conditions of Proposition 4(b); volumes and other integrals of the configuration are not constrained by the proposition.
1. The counts
A Poisson sprinkling of density into a -dimensional spacetime is a Poisson point process with intensity times the volume, ordered by the causal relation . The multivariate Mecke equation [Last] says that the expectation of a sum over -tuples of distinct points equals the integral over the positions of the tuple of the expectation, with the process in the integrand replaced by the process plus the tuple. All statements concern means and variances of the counts below.
The layer action
In four dimensions let be the number of points and the number of pairs with exactly points between them. A layer action is
with real and . The kernel has the Mellin symbol , where and . The Benincasa–Dowker action has , and . These are the negatives of the weights on the point itself and on the layers of the Benincasa–Dowker d’Alembertian, which the pages on mixed-Poisson and signed substrates and on kernel-scale kinematics of a single realisation use, and which are those of the Remark on the causal-set d’Alembertian in horizon counts and the area law; the symbol printed there for those operator weights is in the notation of this page. For the Benincasa–Dowker weights their Mellin transform, which contains the factor , equals .
The mean of tends to a multiple of the Einstein–Hilbert action when and ; these are the conditions (K1)–(K3) of boundary and corner terms. That page also imposes to remove a divergence at timelike boundaries. The Benincasa–Dowker symbol has , and the extra zero is not used below.
Horizon molecules
Let be a spacelike slice, a geodesic ball, the top tip of the domain of dependence of , and . The horizon is the part of below , and is the region inside it. For a point let be the set of sprinkled points in the causal future of and the past of . A horizon molecule is a pair with and , and its type is , the number of points of other than , wherever they lie. A molecule of type is a link across the horizon. In two dimensions, with and the horizon along , a molecule is a related pair with the lower point under the null line and the upper point above it, both below , and its type counts the further points in the future of the lower point below . With weights the weighted count is , being the number of molecules of type , and the Mellin symbol of the weighting is .
The expected count depends on the weights only through . In a conformally static background (section 6) the term of the expansion in the curvature and in the size of the ball that carries the density is multiplied by . The leading flat-space term sits at the area exponent , and for a flat ball of radius
with constants and the volume of the unit -ball; and . In the formula is the count per horizon point and is replaced by . A weighting with and for is an exact weighting; it may have finite support or, like the thinned weights (5) below, infinite support with convergent sums. For it the flat count is exactly at every ball size, and the first-order response to uniform curvature has the form of the area response. The exact weighting of lowest finite degree is unique up to normalisation, with , and in four dimensions its weights are . These statements are Proposition 1 and Theorem 2 of horizon counts and the area law, proved there for ; the converse, that the zeros are necessary, is proved there for . In the same formulas are used below as definitions.
2. The action and the entropy: only a product is fixed
The conclusions of this section assume (H1) that the count contains no entropy unit, so that the constant in the horizon entropy is not determined by the sprinkling, and (H2) that the normalisation of a weighting is a convention, so that and with describe the same functional in different units. An intrinsic unit for weighted molecules would remove the conclusion; we do not address whether a signed count can be an entropy.
Proposition 1 (homogeneity). Consider the layer action (1) with a kernel satisfying (K1)–(K3), and put , so that for Benincasa–Dowker.
(a) The conditions (K1)–(K3), and the ratio of any two coefficients of the large-density expansion of (for instance a corner value divided by the Einstein–Hilbert coefficient), are unchanged by and by for .
(b) The coefficient of in the limit of is and
which changes by under the first rescaling and by under the second. The conditions in (a) therefore do not determine .
(c) Take . The area law for gives , in which only the product appears. Requiring this to equal the one in (3) gives
a single relation between the free quantities and .
The proposition is bookkeeping. The mean of the action is linear in and each condition of (K1)–(K3) is either a zero of or a ratio (), so a common factor drops out of all of them while the overall size of the kernel remains free.
Proof. (a) By the Mecke equation and , so is linear in . Zeros of and the relation are unchanged by , and a ratio of coefficients is unchanged because numerator and denominator scale alike. (b) Aslanbeigi, Saravani and Sorkin [ASS] show that for kernels with (K1)–(K3) the coefficient of the Einstein–Hilbert term is proportional to , and Belenchia [Belenchia] extends the curvature limit to the whole family. For Benincasa–Dowker the mean tends to [BD], and , so the proportionality constant is fixed by there. With the coefficient of is , which is (3). (c) By (2), at leading order, linear in , which gives the first statement. Equating from (3) with gives (4).
The factor in (4) comes from the normalisations of the Einstein–Hilbert coefficient and of the area law, and it relates the two free quantities. The area law with the coefficient of holds for every once is chosen by (4).
As an example, suppose one postulates with (one bit per unit of weighted count). Then (4) gives and , that is . Any other postulated gives another value.
Two routes lie outside (H1) and (H2). The entropy of the short-distance fields could supply an absolute unit, as in induced-gravity pictures where the area coefficient and renormalise together [Susskind–Uglum]; the induced Newton constant discusses their scheme dependence. A dynamical selection of , for example a critical value in a path integral over causal sets, is a question about dynamics that the kernel family does not address.
3. Weightings fixed by interpolation conditions
The exact weights of section 1 are fixed by a normalisation and a list of zeros. We call a weighting interpolating if its symbol is fixed by conditions of this type: at one exponent and zeros at the other points of a set of exponents.
Proposition 2 (interpolating weightings are Lagrange polynomials). Let be a finite set of distinct real numbers and . The weightings of degree at most with and for form an affine space of dimension if and are empty if . For there is exactly one, with , the Lagrange basis polynomial of the node set at .
Proof. The map is a linear bijection from weightings of degree at most onto polynomials of degree at most , because has degree exactly . The conditions prescribe the values of at distinct nodes. For polynomial interpolation gives a unique solution. For the solutions form a coset of the space of polynomials of degree at most vanishing at all nodes, of dimension . For a nonzero polynomial of degree at most cannot have zeros, so the only candidate is , which fails the normalisation.
For the horizon count in , and give the integer weights of section 1. Adding the nodes and gives the next member of the tower (see horizon counts and the area law).
Corollary 1 (no relation between interpolating functionals from first moments). Let be finitely many first-moment functionals of interpolating weightings, each identified with a physical quantity with a unit . By (H2) the normalisations are conventions. The zero conditions that remain are invariant under with independent , and is linear in . The units are therefore independent free parameters, and no relation among the (for instance a value of ) follows from the conditions. An invariant combination involves a single functional and is a ratio of outputs of that functional, such as a corner value divided by the Einstein–Hilbert coefficient.
This extends the homogeneity of section 2 to any finite number of functionals. Under (H2) it excludes a procedure in which two or more interpolating count functionals fix each other’s units.
4. The variance ratio
The mean of is linear in and the variance is quadratic, so the ratio of the squared mean to the variance is invariant under and is a pure number attached to each exact weighting. Thinning shows that it has no upper bound.
The planar horizon
For the second moment we use the flat geometry that a large ball presents near its boundary: the horizon in Minkowski space of dimension , with , , transverse coordinates , and (for lengths are measured in ). For a box of transverse area let be the weighted count of the molecules whose lower point has . In there is no transverse direction, , and is the count for the single horizon point. The mean is for any weighting, being the planar limit of the leading term in (2). We write (in , ) and
which is unchanged by . If one postulates that a count with variance equal to its mean is the horizon entropy, the modular area law discussed in [Verlinde–Zurek], then , the mean of is , and . This is a postulate, and a classical sprinkling variance and a modular fluctuation are different objects. The question here is only whether is a universal number.
Thinning
Let be the independent -thinning of a sprinkling , in which each point is kept with probability . A molecule survives with probability , and each of the further points of its type is kept independently, so that with . The weights
(Proposition 5 of horizon counts and the area law) have infinite support, the same zeros as and the same value at the area exponent, so they have the same mean as in the flat count and in the first-order response to uniform curvature, where every other exponent is a zero. By construction . They differ from at exponents where , which occur at higher order in the curvature.
The variance of is controlled by the add-one cost through the Poincaré inequality for Poisson functionals, [Last] (Theorem 8.3). The lemma shows that the right side is of order .
Lemma 1 (finite variance constant). Let be a weighting with for . In the planar model with ,
the supremum being over boxes of any transverse area (in there is one count and ). Consequently and .
The difficulty is that the sum defining runs over an unbounded region, so the add-one cost has to be shown to be localised near the horizon. The localisation comes from two facts: a molecule requires that the cone of the lower point reach outside before the slice, and a weighting of degree ignores lower points with more than points in their future, which become exponentially rare with depth.
Proof. Put and, for a point at depth , , the volume of its future below . The number of sprinkled points in that future is Poisson with mean , so the probability that there are at most of them is at most , using . Let be the rim. A point at depth has , and its future below leaves exactly when , so only points of can be lower points of molecules. A molecule of type has weight unless .
Adding with changes nothing. For there are two effects. First, may be the lower point of new molecules; this contributes at most and requires and . Second, may lie in the future of an existing lower point , raising by one and the number of points of outside by one if . The change in the contribution of is at most , and it vanishes unless and , which forces (for the whole future of below lies in ). Let be the number of with , and . Then and
We integrate the three terms over . The first gives , since the rim has width at depth . By the Mecke equation , and integrating over (the volume of above and below is ) gives . For the second factorial moment the Mecke equation gives a double integral over with of the probability that both have at most further points. The extra points only increase the counts, and the probability of the intersection is at most . Integrating over multiplies by the volume of the common future of below , at most , and it is zero unless , because a common future point has and so lies at spatial distance less than from . The integral over is thus at most with the volume of the unit ball of (equal to for ), and over at most . The integral gives . The result is
Every integrand is a polynomial in times or , so all three integrals are finite and independent of .
Proposition 3 (the variance ratio is unbounded). For every weighting of finite degree with (in particular every exact weighting of finite degree) and every ,
If is exact, the weightings are exact weightings of infinite support, outside the finite-degree class of Proposition 2 and Lemma 1.
Averaging a count over thinnings keeps the mean and removes variance, because a point of the sprinkling influences the thinned average only when it is retained, which has probability , and the rescaled problem is again a sprinkling with a larger discreteness length.
Proof. Write . Adding a point to changes the thinned configuration only if is retained, so , and by the conditional Jensen inequality , where is a Poisson process of intensity . The Poincaré inequality gives . The slice and the planar horizon are invariant under the dilation and depends only on the causal order, so a process of intensity and the box become a process of intensity and a box of area , with volume element scaled by . By Lemma 1, which holds for every box, . So . Finally and . The mean of equals that of by (5), which gives the bound on .
Computed variances
In we have , , and the exact weighting is with . We computed and the variance of the thinned weights (5), truncated at , by Monte Carlo over sprinklings of the strip , with the horizon ; the strip is deep enough that the mean of every row is to within . The table gives the variance and the ratio .
| factor over | |||
|---|---|---|---|
Proposition 3 guarantees growth of order in relative to a constant , and the computed growth is faster, so the bound is not saturated. In a Monte Carlo evaluation on a transverse torus of area and depth cut , with sprinklings, gives for , whose mean per unit area is (here ), hence .
So has no universal value on exact weightings: it is a different pure number for each weighting and has no upper bound over weightings of infinite support. Under (H1) and (H2), the first and second moments of Poisson kinematics therefore give no pure number without a rule that selects one weighting.
5. Rigid Lorentz-invariant configurations
A random substrate is not the only possible source of a number. A finite configuration of vectors that is rigid, fixed up to a Lorentz transformation by rational conditions, has Lorentz invariants that are determined with no sprinkling involved. The question is which numbers these can be. Whether a constant of a sprinkling is of this kind is not addressed here.
Proposition 4 (algebraic Gram entries). Let be a nonempty set defined by finitely many polynomial equations and inequalities with rational coefficients, and suppose that the Gram matrix takes a single value on (for instance because is a single orbit of the Lorentz group). The form is .
(a) Every entry of is a real algebraic number, and so is every real number that is algebraic over the field generated by the entries.
(b) If are timelike unit vectors () in the same cone, the rapidity with is transcendental unless . If are spacelike unit vectors () with , the angle with is transcendental.
Proof. (a) The set is the image of a semialgebraic set under a polynomial map with rational coefficients, hence is semialgebraic and defined over (Tarski–Seidenberg). By hypothesis it is the single point . The statement “this set is nonempty” holds over , hence over the real closed field of real algebraic numbers by Tarski transfer, so the set contains a point with algebraic coordinates; that point is . (b) For timelike vectors is algebraic by (a). If were algebraic, would be transcendental by the Hermite–Lindemann theorem. For spacelike vectors, is algebraic, and the same theorem excludes an algebraic ; is excluded by .
Two timelike unit vectors with rational coordinates, and , have , so , and . The pair is a single Lorentz orbit defined over (its Gram matrix is the only datum), and its rapidity is transcendental by (b), as is every nonzero rapidity or angle covered by (b).
A richer family comes from circle packings. A disc on the celestial sphere is represented by a unit spacelike vector , the normal of the hyperbolic plane bounded by its circle, oriented towards the disc. Two discs are tangent when and disjoint when with the inversive distance , the hyperbolic cosine of the distance between the corresponding hyperbolic planes. By the Koebe–Andreev–Thurston theorem a triangulation of the sphere has a circle packing with tangency graph , unique up to Möbius maps, that is, up to the Lorentz group. The set of packings with prescribed tangencies is defined by rational equations and inequalities, so Proposition 4(a) applies and the inversive distances of non-adjacent pairs are algebraic. The parameter count matches: discs have parameters, a triangulation of the sphere has edges and so imposes contacts, and .
For a stacked triangulation (a vertex repeatedly inserted into a face) the inversive distances are integers. Start from four mutually tangent discs, whose Gram matrix (diagonal , off-diagonal ) has eigenvalues . The disc inserted in the face , on the side away from the fourth disc , is , for which , for and . All vectors are integer combinations of the initial basis, so all Gram entries are integers. A computer enumeration of all insertion sequences up to vertices, with the rule above applied in every face, finds no inversive distance below ; this is an observation and is not proved for more vertices. For the bipyramid on an -gon the ring discs have angular radius and the two polar discs , and the inversive distance between the polar discs is
This is a Möbius function of with rational coefficients, so ; for the values are , , , , , .
The proposition makes the Gram data of a rigid configuration over algebraic and, by (b), the nonzero rapidities and angles between its unit vectors transcendental. Volumes and other integrals of the configuration are not constrained by it. For a configuration that is rigid over a number field other than the invariants lie in that field. A single configuration is one orbit of the Lorentz group, on which an average is not normalisable, and the proposition says nothing about the Poisson sprinkling, which is not a rigid configuration. The relation to exact discrete Lorentz symmetry is in lattices with an arithmetic Lorentz group and Lorentz-invariant point sets: what they can and cannot carry.
6. The cosmological constant and the volume degree of the horizon count
A cosmological constant would appear as a term of the count proportional to a volume. Let be the size of the ball, so that is the volume of , and let a term of the expected count carry a curvature factor of dimension (a power of inverse length). Its area degree is . The ball-volume degree is , and the spacetime-volume degree is . At they are the terms proportional to the area of , to the volume of , and to the volume of its domain of dependence, and the last is the form of a cosmological-constant term.
Theorem 1 (no volume degree). Let and . Take the conformally static metric , a time-dependent conformal factor times a static metric, with the round metric of radius on the -sphere, , , and a geodesic ball in of fixed volume . Expand the expected weighted horizon count as a formal power series in and the ; a term of order has curvature dimension . For every finite weighting , every term of curvature dimension is a multiple of with . In particular there is no term of ball-volume or spacetime-volume degree, at any order.
The expansion is that of horizon counts and the area law, from which the proof takes its input: by its Lemma 1,
where is the kernel of the weighting. The factor is the conformal volume factor at depth below , and is the volume of the future of a point at that depth, so is the weight of a lower point after averaging over the further points of its type. The function is the volume of pairs, the lower point in the slice of at depth and the upper point in its future outside : is the -volume of a geodesic ball of radius , and is the radius of . The size enters only through inside the bracket.
Proof. The cap volume is with , and fixing the volume of gives with . The bracket expands as , with no term , and the part of order of is a constant times . All other factors are independent of and carry non-negative powers of , so in a term of order of the product the cap factor has , and its exponent satisfies . A term of order has , so .
The proof uses only the form of the reduction formula, so the conclusion holds for every finite weighting. Outside the conformally static class the degree bound is a conjecture; the rim localisation of Lemma 1 (a molecule exists only if its lower point lies within its own depth of the horizon) is the supporting observation.
The mean of the horizon count therefore contains no term that could play the role of a cosmological constant, for every finite weighting. A cosmological constant enters the count only through the curvature of the background, inside the area-degree response (the response to and to the area deficit of horizon counts and the area law), and that response matters for balls of the size of the curvature radius.
The fluctuation of the count is a different matter. For a region let be the number of points in .
Proposition 5 (covariance with the number of points). For any integrable functional of the sprinkling and any region ,
If outside , then . For an exact weighting in dimensions, , and for the region of the planar model of section 4, where is the box with a collar of width , up to an error of at most times a polynomial in times (lengths in units of ; see the proof). The regression coefficient of on is then . In particular and are uncorrelated in up to that error.
Proof. By the Mecke equation, , and . For the second statement, a Poisson process of intensity is the sum of independent processes of intensities and ; adding the second to the first adds a point at with probability to first order, so . When vanishes outside this equals . For the horizon count the dependence region is not bounded, and we bound the first moment outside by the pointwise bound of Lemma 1, . A point with lies in if its depth is at most . A point above a lower point of depth lies in too, since is at depth less than and at spatial distance less than from . The integral of the bound over is therefore at most the tails and of the first two integrals in the proof of Lemma 1, which are a polynomial in times .
By Lemma 1, (with the -scaling of the proof of Proposition 3). The Cauchy–Schwarz inequality gives for the region , whose volume is with the transverse area of , of order for in units of , which is a lower bound of order in those units. The standard deviation of the count is therefore of order the square root of the area.
Comparison with the observed vacuum energy density
The comparison needs a rule, which we state as (H3): a fluctuating count plays the role that the number of points plays in the fluctuating cosmological constant of [Ahmed et al.], with , where is the standard deviation of and the four-volume. For this gives in units of (setting ; the estimates of this section drop factors of order one, among them the in ), of the order of the observed value when the Hubble four-volume contains points.
For we keep , for km s Mpc [DESI], , and normalise by (the convention of (H2), with the unit of the count taken as one in these units), so that its mean is of the order of the horizon area in units of . A count with Poisson-like fluctuations, , would give , a fraction of . For the four-dimensional exact weights the variance per unit area is (section 4), so the standard deviation exceeds by the factor , and with it is . The fraction becomes of order (the leading digit depends on order-one conventions, here ).
The component of along the number of points is smaller still. Take for the Hubble four-volume, so that . By Proposition 5 the covariance is of order (it equals in when contains the region on which depends). The part of that predicts linearly then has standard deviation of order in counts, smaller by the factor than the Poisson-like value , and through it is a fraction of order of . In each case the ratio to the observed value is far below one, because the horizon fluctuation grows with the area and not with the volume. The estimate rests on (H3) and on ; what is proved is Theorem 1 and the area bounds on the variance.
7. Finite-degree variance and balls of curvature size
The infimum of over exact weightings of degree at most , at fixed normalisation , is non-increasing in since the families are nested, and Proposition 3 shows that the infimum over exact weightings of infinite support is zero. Whether the infimum over finite degree tends to zero or to a positive limit as is not known, and a lower bound on that is uniform in the degree would settle it. A positive limit would still be a number attached to a family of weightings, available as a constant of the sprinkling only once a rule selects the weighting.
Theorem 1 concerns the formal expansion in the curvature. For balls comparable to the curvature radius, evaluating the reduction formula of section 6 without expanding in would show whether the degree bound persists.
References
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