Existence of a symmetric bipodal phase in the edge-triangle model

In the edge-triangle model with edge density close to 1/2 and triangle density below 1/8 we prove that the unique entropy-maximizing graphon is symmetric bipodal. We also prove that, for any edge density e less than e0=(3−3)/6≈0.2113 and triangle density slightly less than e 3, the entropy-maximizing graphon is not symmetric bipodal. We also discuss the implications for an old idea of Landau for using symmetry to give an intrinsic difference between solid and fluid phases of matter.


Introduction and Results
1.1.Results.We study emergent smoothness with respect to change of competing constraints in asymptotically large dense random graphs.More specifically, we determine and study smooth phases separated by sharp transitions.We derive a new phase in the model with sharp constraints on edge and triangle densities.The phase is "symmetric bipodal" and we show how to use its symmetry to distinguish the phase intrinsically from other phases.Unlike in previous work, graphs in this phase are not small perturbations of Erdős-Rényi graphs; this requires new techniques, which we develop.
Let g(x, y) be a graphon, a measurable symmetric function g : [0, 1] 2 → [0, 1].Let (1) ε(g) = g(x, y) dx dy, τ (g) = g(x, y)g(y, z)g(z, x) dx dy dz, and The integrals ε(g) and τ (g) represent the overall edge and triangle densities of graphs whose adjacency matrices are close to the graphon in the "cut metric" [17], while S(g) is proportional to the entropy of a random process for generating such graphs.We refer to S(g) as the "entropy of the graphon g," distinct but related to the Boltzmann entropy defined in the next paragraph.The quantities ε(g), τ (g) and S(g) are all invariant under measurepreserving transformations of [0, 1].All statements about uniqueness of graphons should be understood to mean "unique up to measure-preserving transformations of [0, 1]." We consider constrained systems, where the edge and triangle densities are constrained to a vanishingly small tolerance as follows.For each achievable ordered pair (e, t), we define the Boltzmann entropy in terms of the partition function Z n,δ e,t , which is the cardinality of the set of graphs on n nodes with edge density in the interval (e − δ, e + δ) and triangle density in the interval (t − δ, t + δ).From the partition function we define the Boltzmann entropy [27] as (3) B(e, t) = lim δ↘0 lim n→∞ 1 n 2 ln Z n,δ e,t .
Using [6] we established [27] the variational formula If the supremum of S(g) is attained by a unique graphon g e,t ∈ G(e, t), then all but exponentially few large graphs with the constrained edge and triangle densities are described by g e,t .In particular, the density of all possible subgraphs are given by integrals involving g e,t .
See Figures 1 and 2. The following is immediate by inspection.
Corollary 2. B(e, t) and the densities of all subgraphs are real analytic in (e, t) in O.Our second major result describes a region where the optimizing graphon is not symmetric bipodal.
Theorem 3. Let e 0 = (3 − √ 3)/6 ≈ 0.2113.For any fixed edge density e < e 0 and any sufficiently small positive σ, the symmetric bipodal graphon (6) does not maximize S among graphons with triangle density t = e 3 − σ 3 .e − (e 3 − t) 1/3  e + (e 3 − t) 1/3   e − (e 3 − t) 1/3 e + (e 3 − t) 1/3   Figure 2. The piecewise-constant optimal graphon of Theorem 1 1.2.Background.This work is concerned with emergent features [1] of large dense simple graphs, features that are only meaningful in the limit as the number n of vertices goes to infinity.We will concentrate on graphs with competing constraints, specifically the prescribed densities e and t, edges and triangles, respectively.The emergent behavior is smoothness as a function of (e, t), of the Boltzmann entropy B(e, t) and of all subgraph densities.
The study of graphs with competing constraints is an old topic in extremal combinatorics.For graphs, the range of achievable values of the pair (e, t), and the graphs that achieve them, was completed in 2012 by Purkurko and Razborov [23]: see Figure 3 for a distorted view of the "Razborov triangle".
A graphon formalism was developed starting around 2006 [2,3,14,15,16] to give a useful meaning to the asymptotic limits of dense graphs.A large deviation principle (LDP) was added by Chatterjee and Varadhan in 2010 [6].Using graphons we define a phase as a maximal connected open subset of the Razborov triangle on which B(e, t), and the density of every fixed subgraph (e.g., the density of squares, pentagons, tetrahedra, . ..) of a typical graph, is an analytic function of (e, t).The system is said to have a phase transition wherever such quantities are not analytic.Usually this occurs at the boundary of two or more phases, but sometimes there is an analytic path from one side of a phase transition to another, as discussed in Section 7. Note that if the optimal graphon in the variational formula (4) is not unique, the densities of some subgraphs are not even well-defined, much less analytic; by definition, such points of non-uniqueness can never lie within a phase.Even when uniqueness does hold, it can be difficult to prove.On the other hand, where there is a unique entropyoptimizing graphon g e,t , all but exponentially few large graphs are close to g e,t ; this facilitates the analysis of emergent features.(Such uniqueness is known to fail in similar models; see Section 5 in [11].)This paper is part of a project begun in 2013 [27], following [5], to study emergent analyticity of typical graphs with edge and triangle constraints in the Razborov triangle.Using [6] we derived [27] the variational formula (4) relating the Boltzmann entropy B(e, t) to the graphon entropy S(g).Let G be a fixed subgraph, such as a square or pentagon or tetrahedron.When S(g) achieves the value B(e, t) at a unique graphon, all but exponentially few graphs have the same density t G of G, so we can speak of t G being a function of (e, t) and ask whether that function is analytic.(Our method was based on large deviations of G(n, p) [6]; for an approach based on G(n, m) see [7].) In 2015 we established [11] the existence of two open subsets of the Razborov triangle, both with t > e 3 , in which B(e, t) and all subgraph densities were real-analytic in (e, t).To prove this we proved that the constrained S-optimizing graphons were unique, and determined that they have a 2-block ("bipodal") [11] structure whose parameters were analytic in (e, t).More recently, we proved [21] a complementary result in the more difficult case of undersaturated triangles (t < e 3 ) when e > 1  2 .This yielded the satisfying result of a pair of open sets, O 1 , O 2 , on which the Boltzmann entropy B(e, t) and all subgraph densities are analytic, separated by the bounding curve t = e 3 , on which the entropy isn't even differentiable [28].See Figure 4.
Put another way, in [21] we proved the existence of two phases with e > 1  2 , one just above the curve t = e 3 and one just below, with a phase transition on the curve itself.As noted earlier, within each phase there must be a unique S-optimizing graphon for each (e, t).This  optimal graphon is called the "state" of the system.As discussed above, models of dense graphs with sharp competing constraints are a natural extension of extremal graph theory.We note there have been parallel studies within other parts of extremal combinatorics, for instance permutations and sphere packings; see Section 8.
One may reasonably ask how we know that the two open sets O 1 and O 2 actually belong to different phases; how can we rule out the possibility that there is an analytic path between them, going around the phase transition?Neither the O 1 phase nor the O 2 phase seems to have any intrinsic property that clearly rules out an analytic continuation between them.Such an analytic continuation does not actually exist, since we previously showed [28] that B(e, t) cannot be differentiable at any point on the curve t = e 3 for any 0 < e < 1. However that isn't a very satisfying explanation.
The "symmetric bipodal" phase whose existence we prove in this paper is different.The symmetry provides an intrinsic difference between the new phase and the O 1 and O 2 phases.In Section 7 we discuss the connection between our symmetry argument and the use of "order parameters" in equilibrium statistical mechanics, and as a by-product we clarify a problematic argument of Landau from the 1950s.
Another key difference between our new phase and the previously proven phases is that proof of the symmetric bipodal phase is not limited to a small neighborhood of the curve t = e 3 .In [11], and again in [21], we studied small perturbations of the Erdős-Rényi graph G(n, p) and attempted to get the greatest possible change in triangle count for the smallest possible entropy cost.The results are closely related to moderate deviation estimates [20]; depending on the sizes of n and e 3 − t, a finite graph with triangle density slightly less than e 3 can be viewed either as a typical (e, t) graph or as a deviation of an Erdős-Rényi graph.When e < 1/2, moderate deviations estimates that apply when n −1 ≪ e 3 − t ≪ 1 agree to leading order with large deviations estimates.This is reflected in the different method of proof.In Theorem 3, σ is not a small parameter.We can still do a power series expansion in σ, but we have to estimate all terms, not just the first few.A graphon is said to be bipodal if it is equivalent to a graphon with the block structure shown in Figure 5.It is symmetric bipodal if c = 1/2 and a = b.To avoid questions about graphons being equivalent under measure-preserving transformations of [0, 1], we restate the definition using arbitrary measurable subsets I 1 and I 2 of [0, 1], rather than intervals [0, c] and (c, 1].A graphon is bipodal if there exist complementary measurable subsets I 1 and I 2 such that g(x, y) is constant on I 1 × I 1 , constant on I 2 × I 2 , and constant on

Definitions and Notation
A graphon is symmetric bipodal if there exist complementary subsets I 1 and I 2 , each of measure 1/2, and a positive number σ < e, such that the graphon is (7) g(x, y) = e − σ (x, y) The edge density, triangle density and entropy of a symmetric bipodal graphon are Of course, this only makes sense if σ ≤ min(e, 1−e).Another characterization of a symmetric bipodal graphon is that it is a rank-1 perturbation of a constant graphon, with g(x, y) = e − σv 1 (x)v 1 (y), where 1 0 v 1 (x)dx = 0 and v 1 (x) 2 = 1 everywhere.It was previously known that the unique optimal graphon on the open line interval e = 1/2, t ∈ (0, 1/8) was symmetric bipodal [28].Our main result, Theorem 1, extends this to an open set O containing the line interval.It is convenient for our proofs to reformulate Theorem 1 as Theorem 4. For fixed σ ∈ 0, 1  2 and for all sufficiently small δ (of either sign), the unique S-maximizing graphon with edge density e = 1  2 + δ and triangle density t = e 3 − σ 3 is symmetric bipodal.Furthermore, the size of the allowed interval of δ's varies continuously with σ.
Corollary 5.The Boltzmann entropy B(e, t) and the densities of all subgraphs are real analytic functions of (e, t) in the open set thus defined.
We expect that the region where the optimal graphon is symmetric bipodal is not limited to the small open set described in Theorems 1 and 4.There is considerable numerical evidence that this region, called the A(2,0) phase in [13], is much bigger than that.However, there are provable limits to its extent.Theorem 3 says that it does not extend to the curve t = e 3 when e < e 0 ≈ 0.2113.Theorem 1 from [21], which we restate here, says that it does not extend to the curve t = e 3 when e > 1/2.It is an open question whether the phase extends to the curve t = e 3 when e 0 < e < 1/2.

Theorem 6. (Theorem 1 from [21])
There is an open subset O 1 in the planar set of achievable parameters (e, t), whose upper boundary is the curve t = e 3 , 1/2 < e < 1, such that at (e, t) in O 1 there is a unique entropy-optimizing graphon g e,t .This graphon is bipodal and for fixed (e, t), the values of a, b, c, d can be approximated to arbitrary accuracy via an explicit iterative scheme.These parameters can also be expressed via asymptotic power series in σ = (e 3 − t) 1/3 whose leading terms are: H ′ (e) − e − This corollary was proven in the last paragraph of the proof of Theorem 1 in [21], although not included in the statement of the theorem.
The bulk of this paper is devoted to proving Theorem 4, which is tantamount to proving Theorem 1.To explain the steps, we need some more notation.We diagonalize g(x, y) − e, viewed as an integral operator, and write (10) g where Our goal is to show that g 2 = 0 and that v 1 (x) = ±1, taking each value on a set of measure 1/2.We do this in stages: • In Section 3, we prove a priori entropy bounds on any graphon having the given values of (e, t).We show that the symmetric bipodal graphon comes within O(δ 2 ) of saturating those bounds.This implies that any entropy-maximizing graphon must be L 2 -close to a symmetric bipodal graphon.Specifically, g 2 must be L 2 -small and v 1 must be L 2 -close to the desired step function.• In Section 4 we show that g 2 is pointwise small and that v 1 (x) 2 is pointwise close to 1.More precisely, the L ∞ norms of g 2 and v 2 1 − 1 must go to zero as δ → 0. • In Section 5 we expand the entropy S(g) using a convergent Taylor series for H(u) around u = 1 2 .Using the fact that g 2 is pointwise small, we express the difference between S(g) and the entropy of a symmetric bipodal graphon as a quadratic function of the L 2 norm of g 2 , the L 2 -norm of v 1 (x) 2 − 1 and the integral 1 0 v 1 (x)dx, plus higher-order corrections.We show that the quadratic function is negative-definite, implying that g 2 must be zero, v 1 (x) 2 must be 1, and 1 0 v 1 (x)dx must be zero.In other words, our graphon must be symmetric bipodal.
• In Section 6 we turn our attention to Theorem 3. We construct a family of tripodal graphons and we express the entropy of both this tripodal graphon and the symmetric bipodal graphon as power series in σ.When e < e 0 , we can choose the parameters of the tripodal graphon such that the tripodal graphon has more entropy at order σ 2 than the symmetric bipodal graphon.This does not prove that the optimal graphon is tripodal!However, it does proves that, for σ sufficiently small, the symmetric bipodal graphon is not optimal.
We use big-O and little-o notation throughout.When we say that a certain quantity is O(δ n ), we mean that there exist positive numbers C and δ 0 (which may depend on σ) such that our quantity is bounded by C|δ| n whenever |δ| < δ 0 .When we say that a quantity is o(δ n ), we mean that there exists a constant δ 0 and function f (δ), going to zero as δ → 0, such that the quantity is bounded by f (δ)|δ n | when |δ| < δ 0 .

A priori estimates
We begin with an upper bound on entropy.Theorem 8.If g is a graphon with edge density e = 1 2 + δ and triangle density e 3 − σ 3 , with σ > 0, then Proof.Let g be our arbitrary graphon, which we expand as in equations ( 10) and (11).For i = 1, 2, let d i (x) = 1 0 g i (x, y)dy.A direct computation of the triangle density gives with equality if and only if Next we maximize the entropy for a fixed ∥g 1 + g 2 ∥ 2 2 .We use an absolutely convergent power series for ( 14) The terms with n odd are identically zero, while the terms with n nonzero and even are negative.As a result, ( 16) The second moment depends only on the size of g 1 and g 2 .Since g 1 (x, y)+g 2 (x, y) dx dy = 0, there are no cross terms between δ and g 1 + g 2 , leaving us with Maximizing S(g) is equivalent to minimizing all of the higher moments µ 2k with k > 1.This happens when (δ + g 1 (x, y) + g 2 (x, y)) 2 is constant, equaling µ 2 .In that case, g is everywhere equal to 1 2 ± √ µ 2 and ( 18) Since If g is an entropy-maximizing graphon, then Proof.The symmetric bipodal graphon has entropy By contrast, the upper bound ( 12) is Since the symmetric bipodal graphon comes within O(δ 2 ) of achieving the upper bound ( 12), the entropy-maximizing graphon must also come within O(δ 2 ) of that bound.In particular, ∥g 1 + g 2 ∥ 2 2 must be within O(δ 2 ) of σ 2 and the fourth moment µ 4 can be no more than O(δ 2 ) greater than ( .

This can only be within
and since Since this must be within O(δ 2 ) of σ 2 , we must have ∥g 2 ∥ 2 2 = O(δ 2 ).Finally, we consider the fourth moment µ 4 .The leading contribution is For this to be within O(δ 2 ) of σ 4 ,

Pointwise estimates
The upshot of Section 3 is that g must be L 2 -close to a symmetric bipodal graphon, with λ 1 being close to −σ, with the sum of the other λ 2 j being small, and with v 1 (x) being close to 1 on a set of measure approximately 1/2 and close to −1 on a set of measure approximately 1/2.In this section we upgrade those L 2 estimates into pointwise estimates: We prove Propositions 10 and 11 with a series of lemmas.We begin by showing that g 1 and g 2 are pointwise bounded.
Lemma 12. Let g be an entropy-maximizing graphon.For all x, y ∈ [0, 1], the following bounds apply: Proof.We use the fact that 0 ≤ e + g 1 (x, y) + g 2 (x, y) ≤ 1 for all (x, y).The only way for g 2 (x, y) be be big and positive (resp.negative) is for g 1 (x, y) to be big and negative (resp.positive).This can only occur if |v 1 (x)| is large for some x.
Suppose that there is a point We already know that the set of points with v 1 (x) close to ±1 each have measure close to 1/2, since Since λ 1 is negative, g 1 (x, y) is less than or equal to -0.9 for all y ∈ I + and is greater than or equal to 0.9 for all y ∈ I − .Since e is close to 1/2, e + g 1 (x 0 , y) is close to or less than -1.4 when y ∈ I + (and in particular is less than -1.3) and is close to or greater than 1.4 (and in particular is greater than 1.3) when y ∈ I − .As a result, g 1 (x 0 , y) has magnitude at least 0.3, and sign opposite to that of g 2 (x 0 , y), for all y ∈ I + ∪ II − .This means that g 1 (x 0 , y)g 2 (x 0 , y) < −0.27 for all y ∈ I + ∪ I − .When y ̸ ∈ I − ∪ I + , |g 2 (x 0 , y)| ≤ 0.9 < 1, so |g 2 (x 0 , y)| < 2, so g 1 (x 0 , y)g 2 (x 0 , y) < 2. Since by the orthogonality of the functions v j (y) in L 2 ([0, 1]).This is a contradiction, so x 0 does not exist.
The same argument, with signs reversed, rules out the possibility that v 1 (x) is ever less than Since e and 1 − e are both less than 1, this implies that □ Lemma 12 is stated in terms of λ 1 , which of course depends on the graphon g.However, , so for small δ we can replace our bounds involving λ 1 with uniform bounds in terms of σ, at the cost of replacing the constant 1 with a slightly smaller number.For instance, whenever δ is sufficiently small.In practice, we do not need the specific bounds of Lemma 12.All we really need is for v 1 (x), g 1 (x, y) and g 2 (x, y) to be bounded.
We next turn to showing that g 2 (x, y) is not only bounded but small.Since ∥g 2 ∥ 2 2 is O(δ 2 ), the set of points where |g 2 (x, y)| is not small (say, smaller than a fixed ϵ) has measure O(δ 2 ).We now establish a similar result for vertical strips.
Proof.Let G(x, y) = 1 0 g(x, z)g(z, y) dz.As as operator, this is the square of g.Expanding that square using g(x, y) = e + g 1 (x, y) + g 2 (x, y), we G 1 be the portion of G that comes from e + g 1 , and let G 2 be the additional contributions that involve g 2 .We next turn to G 2 .Since 1 0 v 1 (z)g 2 (y, z) dz = 0, there is no contribution from the product of g 1 and g 2 .We only have eg 2 and g 2 2 terms, specifically The function d 2 (y) has small L 2 -norm, and so must be o(1) except on a set of small measure.(Since x is fixed, we cannot similarly argue that d 2 (x) is small.)Finally, since g 2 is bounded and has small L 2 norm, the integral 1 0 g 2 (x, z)g 2 (y, z)dz for fixed x is small except for a set of y's that has small measure.The result is an estimate that is true for y in the complement of a set of measure o(1), where that small set may depend on x.
Combining this with our estimate of G 1 , we have for all but a small set of y's.
Since g is assumed to maximize entropy subject to constraints on ε(g) and τ (g), the functional derivative of S(g) must be a linear combination of the functional derivatives of ε(g) and τ (g).This yields the pointwise equations ( 21) where Λ e and Λ t are Lagrange multipliers.
For all but a small set of y's, equation ( 21) takes the form , For most values of (x, y), e + g 1 (x, y) is close to e ± σ and g 2 (x, y) and d 2 (x) are small.This fixes Λ e and Λ t , and therefore µ, ν, and ρ, to within a small error.Since e = 1  2 + δ and , we obtain with solution where all of the approximations are "+o(1)" as δ → 0.
From the explicit form of H ′ (g) = ln(1 − g) − ln(g), we see that there are only three roots to the equation H ′ (g) = µ + νg, which are located near g = e ± σ and g = e.Our immediate goal is to show that v 1 (x) only takes values close to 0 and ±1, which implies that g 1 (x, y) is only close to e and e ± σ.
Since v 1 (x)dx and (v 1 (x) 2 − 1) 2 dx are small, the function v 1 must be close to 1 on an interval (call it I 1 ) of measure close to 1/2, must be close to −1 on an interval I 2 of measure close to 1/2, may be close to 0 on a third interval I 3 of small measure, and may take on other values on a fourth interval I 4 of small measure.
Let x be an arbitrary point in [0, 1], and let y 1 and y 2 be generic points in I 1 and I 2 .Equation ( 22) then determines a = g 2 (x, y 1 ) and b = g 2 (x, y 2 ) in terms of d 2 (x).What's more, g 2 takes values close to a on all of {x} × I 1 (excepting those values of y where equation ( 22) doesn't apply), and takes values close to b on all of {x} × I 2 .We then compute However, this integral must be zero, since v 1 (y) is orthogonal to all of the functions v i (y) that make up g 2 (x, y).We conclude that a ≈ b.
If b ≈ a, then d 2 (x) = 1 0 g 2 (x, y)dy also equals a and our equations at (x, y 2 ) and (x, y 1 ) become Adding these equations, and using the fact that H By the mean value theorem, the left hand side of equation ( 26) is 2aH ′′ (u) for some u between 1  2 + σv 1 (x) − a and 1 2 + σv 1 (x) + a. Regardless of the value of u, this is a negative multiple of a.However, the right hand side is a positive multiple of a, since H ′ ( 1 2 + σ) < 0. Since a negative multiple of a equals a positive multiple, a must be (approximately) zero.
In particular, g 2 (x, y) ≈ 0 for all y such that equation ( 22) applies.That is, for fixed x, g 2 (x, y) is close to zero except on a set of y's of small measure.□ Proof of Proposition 10.By Lemma 12, g 2 (x, y) is bounded.By Lemma 13, for each x, g 2 (x, y) is small for all but a small set of y's.Combining these results, we see that the degree function d 2 (x) is everywhere small, as is the integral y) is pointwise small.But then equation ( 22) applies everywhere, so g 2 (x, y) is small everywhere.□ Proof of Proposition 11.In the notation of the proof of Lemma 13, we must show that the intervals I 3 and I 4 are empty, implying that v 1 (x) is everywhere close to ±1.
Since G 2 (x, y) is small for all (x, y), we must have for all y ∈ I 2 .However, the only solutions to this equation are (approximately) σv 1 (x) = ±σ or 0, implying that v 1 (x) ≈ ±1 or 0. In other words, x ∈ I 1 ∪ I 2 ∪ I 3 and I 4 is empty.
Showing that I 3 is empty requires a completely different argument, since equation ( 21) is indeed satisfied when x ∈ I 3 .However, equation ( 21) only defines stationary points with respect to pointwise small changes in g(x, y).We also have to consider infinitesimal changes in the boundary between I 1 , I 2 and I 3 .
So suppose that we increase the size of I 1 by an amount ϵ at the expense of I 3 .That is, we change the value of v 1 (x) from near 0 to near 1 on a set of small measure ϵ.To first order in ϵ, the change in entropy is 2ϵ(H )), since we are changing g(x, y) from near zero to near 1  2 ± σ on a set of measure 2ϵ + O(ϵ 2 ), and since H( 1 2 − σ) = H( 1 2 + σ).The change in edge density is −2ϵλ 1 1 0 v 1 (y)dy = O(δϵ).To leading order, the change in the triangle density is 3λ 3  1 ϵ, since the λ 3 1 contribution to τ (g) is actually λ 3 1 ( ).The variational equations dS = Λ e dε + Λ t dτ /3 then become We expand both sides of equation ( 27) as power series in σ.The left-hand side is The right-hand side is The coefficients agree when k = 1, but are strictly greater for the right-hand side when k > 1.Since all terms are strictly negative (insofar as all even derivatives of H are negative-definite), the right-hand side strictly smaller than the left-hand side.
Since varying the size of I 3 does not satisfy the variational equation ( 27), we cannot be in the interior of our parameter space.Rather, the measure of I 3 must be zero.□

Cost-benefit analysis
In Sections 3 and 4 we showed that g 2 (x, y) is pointwise small, as is v 1 (x) 2 − 1.In this section we show that they are zero, completing the proof of Theorem 1.The key measures of how far they are from being zero are ( 28) By Corollary 9, α, β and γ are all O(δ).The symmetric bipodal graphon is characterized by α = β = γ = 0.
We use the expansion (16) and compare the moments µ 2k to those of the symmetric bipodal graphon.We will estimate costs (terms that increase µ 2k ) and benefits (terms that decrease µ 2k ).We will show that having α or β or γ nonzero comes with costs that go as α 2 , β 2 , and γ 2 , while the benefits are o(α 2 + β 2 + γ 2 ).When δ is sufficiently small, the costs exceed the benefits, so the symmetric bipodal graphon has more entropy than any graphon that isn't symmetric bipodal.
We first establish the costs.Our triangle density is This implies that ).That is, there are α 2 and γ 2 costs associated with µ 2 .
We next look at µ 4 .This contains a term That is, there is a cost proportional to β 2 .Having established costs proportional to α 2 , β 2 , and γ 2 , we just have to show that the benefits of having α, β, γ nonzero are smaller.
We are looking at even moments We expand out the power, getting terms proportional to a power of δ times a power of g 1 times a power of g 2 .We make repeated use of the following trick: This means that and We divide the terms obtained by expanding (δ + g 1 + g 2 ) 2k into several classes.
(1) Terms with three or more powers of g 2 .Since g 2 is pointwise small and g 1 is bounded, these are bounded by small multiples of g 2 2 = α 2 .In other words, they are o(α 2 ).(2) Terms with two powers of g 2 , an even number of powers of δ and an even number of powers of g 1 .These are manifestly positive and represent costs, not benefits.Aside from the δ 0 g 0 1 g 2 2 = α 2 contribution to µ 2 that we already considered, we do not keep track of these.
(3) Terms with two powers of g 2 , an odd number of powers of δ and an odd number of powers of g 1 .The integrand is a positive power of δ times a bounded quantity times g 2 (x, y) 2 , making the integral O(δα 2 ).
(4) Terms with one power of g 2 , an odd power of δ and an even power of g 1 .We expand these using equation ( 29) and consider each piece separately.First, we compute Next, the L 2 norm of v 1 (x) 2 − 1 is β and the L 2 norm of g 2 is α, so The (v 1 (y) 2 − 1) piece is similar, while the (v 1 (x (5) Terms with one power of g 2 , an even power of δ and an odd power of g 1 .We expand these using equation (30), noting that the first line in ( 30) has a single factor of v 1 (y) and the second line has a single factor of v 1 (x).However, where x is arbitrary in the first integral and y is arbitrary in the second.Thus the first two lines contribute nothing and g 1 (x, y) 2m+1 g 2 (x, y) dx dy is equal to The factor v 1 (x) 2 − 1 v 1 (y) 2 − 1 has L 2 norm β 2 , the factor g 2 (x, y) has L 2 norm α, and the factor ). ( 6) Terms with no powers of g 2 , an even number of powers of δ and an even number of powers of g 1 .These are all positive and are at least as big as the corresponding terms for the symmetric bipodal graphon.We have already taken into account the costs associated with µ 2 and µ 4 .There are additional costs associated with higher moments, but they are not needed for this proof.(7) Finally, there are terms with no powers of g 2 , an odd number of powers of δ and an odd number of powers of g 1 .Note that where and multiplying by an odd power of δ gives O(δγ Putting everything together, we have identified costs proportional to α 2 , β 2 and γ 2 .Other costs only add to that total, so the total cost is at least a constant times α 2 + β 2 + γ 2 .All of the potential benefits are smaller, either involving three or more powers of α, β and γ, or δ times a quadratic function of α, β and γ, or the sup norm of g 2 times α 2 .When δ is sufficiently small, the costs outweigh the benefits, so the optimal graphon is symmetric bipodal.

The extent of the symmetric bipodal phase
We have proven that the entropy-maximizing graphon is unique and symmetric bipodal on a region containing the interval e = 1/2, 0 < t < 1/8.It is natural to ask how far this symmetric bipodal phase extends.There is considerable evidence that this phase contains much of the region e ≤ 1/2, t < e 3 .
• When e < 1/2 and t < e 3 , the symmetric bipodal graphon has entropy strictly higher than any other graphon of the form g(x, y) = e + λ 1 v 1 (x)v 1 (y).This is Proposition 14, proven below.• When e < 1/2 and t sufficiently close to but below e 3 , the symmetric bipodal graphon has entropy strictly higher than any other bipodal graphon.This is Proposition 16, proven below.• Numerical investigations of the region e < 1/2, t < 1/8 [24] did not turn up any regions where the symmetric bipodal graphon was not optimal.That is not a proof that such regions don't exist, of course, but it does suggest that such regions are likely to be small.
Despite this evidence, Theorem 3 says that there is a (possibly very small) open subset of the region e < 1/2, t < 1/8 on which the symmetric bipodal graphon is not optimal.In this section we state and prove Propositions 14 and 16 and then prove Theorem 3. Proposition 14. Suppose that e < 1/2 and that g is a graphon of the form with edge density e and triangle density t = e 3 − σ 3 < e 3 .Then S(g) is bounded above by Proof.First note that 1 0 v 1 (x) dx = 0, since the overall edge density is exactly e.The triangle density is then e 3 + λ 3  1 , so λ 1 = −σ.Since 0 ≤ g(x, y), v 1 (x) is bounded in magnitude by e/σ and λ 1 v 1 (x)v 2 (x) is bounded in magnitude by e.This implies that the power series converges absolutely.Integrating over x and y then gives Since e < 1/2, all the odd derivatives of H are positive at e, while all the even derivatives are negative.Multiplying by (−σ) j , all of the terms with j > 0 are negative.We maximize the entropy by minimizing 1 0 v 1 (x) j dx for j even and by having 1 0 v 1 (x) j dx = 0 for j odd.The second moment 1 0 v 1 (x) 2 dx is always equal to 1.The fourth and higher even moments are minimized when (and only when!) v 1 (x) 2 is constant and equal to 1. Since 1 0 v 1 (x) dx = 0, this means that v 1 (x) is +1 on a set of measure 1/2 and −1 on a set of measure 1/2, which makes all of the odd integrals zero, as desired.In other words, the symmetric bipodal graphon is the unique entropy maximizer among graphons of this form.□ Before turning to what happens just below the line t = e 3 , we establish constraints on the form of any entropy-maximizing bipodal graphon with t < e 3 .We use the parameters (a, b, c, d) of Figure 5 to describe bipodal graphons.Without loss of generality we can assume that a ≤ b, since otherwise we could just swap c and 1 − c while swapping a and b.
Proposition 15.Suppose that t = e 3 − σ 3 < e 3 and that a graphon g maximizes entropy among all bipodal graphons with edge density e and triangle density t. (We do not assume that g maximizes entropy among all graphons, just that it is the best bipodal graphon.)Then either a = b and c = 1/2 (a symmetric bipodal graphon) or a < b < d and c < 1/2.
Proof.In this setting, the variational equations ( 21) become Subtracting the second equation from the first gives If a = b, then the left hand side is zero and the right hand side is a nonzero multiple of (1 − 2c)(a 2 − d 2 ), implying that either c = 1/2 or a = d.But if a = b = d, then the triangle density is exactly e 3 , which is a contradiction.Thus a = b implies that the graphon must by symmetric bipodal.
We now turn to the possibility that a < b.The left hand side is then positive, since Let λ 1 and λ 2 be the two nonzero eigenvalues of g.The trace of g is From this we can compute If d were less than a and b, this would be positive, meaning that both eigenvalues would be positive.Moreover, one of the two eigenvalues is at least e, so the triangle density, which is λ 3 1 + λ 3 2 , would be greater than e 3 .This rules out the possibility that d < a, and we conclude that a < b < d and c < 1/2.
□ When e > 1/2 and t is slightly less than e 3 , the optimal graphon has been proven to take this form, with a ≈ 1 − e, b slightly less than e, and d slightly greater than e, and with c small.The situation is different when e <1 2 .Proposition 16.Suppose that e < 1/2 and that g is a bipodal graphon with edge density e and triangle density t = e 3 − σ 3 < e 3 .Then, for σ sufficiently small, S(g) is bounded above by 1  2 (H(e + σ) + H(e − σ)), with equality if and only if g is symmetric bipodal.
The leading term is α 1 , while η measures the extent to which the degree function fails to be constant.We can then express all of our quantities in terms of α, η and c.
In terms of these parameters, the triangle density works out to be (34) Since the triangle density is less than e 3 , η 2 must be O(α 3 ).That is, η is much smaller than α.
We now compute the entropy where Since e < 1/2, the odd derivatives of H at e are positive, while the even derivatives are negative, so we want to minimize the even moments and maximize the odd moments.The symmetric bipodal graphon (uniquely) minimizes the even moments and has all the odd moments equal to zero.For an asymmetric bipodal graphon to do as well, it must have some positive odd moments.
The moment µ k is a k-th order homogeneous polynomial in α and η with coefficients that depend on c.Since η = O(α 3/2 ), The coefficient of α k is zero when k = 1, is 1 when k = 2, is negative when k is an odd number greater than 1, and is greater than 1 when k is an even number greater than 2. In other words, all moments with k > 2 are worse, to leading order, than the moments of the symmetric bipodal graphon.
There is one more point we must account for.For k odd, 2c(1 goes to zero as (1−2c) 2 as c → 1/2.We must rule out the possibility that other contributions to µ k might become greater than the α k term as c approaches 1/2.
This requires estimates on η.From the formula for the triangle density, we have that and hence that That is, there is a cost proportional to η 2 /σ that does not vanish as c → 1/2.Meanwhile, all contributions to moments involving odd powers of η are proportional to (1 − 2c).This is because the graphon is invariant under the transformation The leading such contribution comes from µ 3 and goes as (1 − 2c)α 2 η ≈ (1 − 2c)σ 2 η times a polynomial in c that does not vanish at c = 1/2.Setting the derivative of the entropy with respect to η equal to zero tells us that η = O(σ 3 (1 − 2c)).All contributions from odd powers of η are thus O(σ 5 (1 − 2c) 2 ), and so are dominated by the α 3 contribution to µ 3 , while all contributions from even powers of η are dominated by the µ 2 /σ contribution to µ 2 .

□
Thanks to Propositions 14 and 16, any graphon that does better than symmetric bipodal in the region just below t = e 3 with e < 1/2 must be at least tripodal (or perhaps not even multipodal at all) and the difference between that graphon and a constant graphon must have rank at least two.That is exactly what we construct in the proof of Theorem 3.
What remains is showing that we can get F (A, B) > H ′′ (e) when e < e 0 .Let A be a small positive number and let Since H ′′′ (e) > 0 > H ′′ (e), B is positive.Since ).We compute the numerator of F (A, B) to order A 4 by doing a 4-th order Taylor series expansion of H(e − A + B) and H(e + A + B) around e and keeping terms proportional to A 2 , A 4 , B, B 2 , and A 2 B. (The expression is even in A, so we only get even powers of A.) The result is Since e < e 0 , the coefficient of A 2 is positive, so F (A, B) > H ′′ (e) when A is small.□

Symmetry as an order parameter
Question: Considering Figure 3 and the two open subsets O and O 1 there defined by Figure 1 and Figure 4, could O and O 1 be part of the same phase?
Although there is no barrier between O and O 1 like the curve t = e 3 between O 2 and O 1 , the answer must be "no", thanks to the following symmetry argument.On O, the entropymaximizing graphon has constant degree function d(x) = 1 0 g(x, y) dy = e.The density T 2 of 2-stars is given by the integral identically zero on O.If O and O 1 were part of the same phase and we had any analytic curve c(s) running between them, then Q would have to be zero on the first part of the curve and then by analyticity it would have to be zero on the entire curve.However, it is easy to check that Q is not zero on all of O 1 , insofar as the degree function for the graphon of Theorem 6 is not constant.Instead, Q is a nonzero multiple of σ 5 plus O(σ 6 ).This contradiction proves our assertion.□ This argument has a very similar flavor to an argument that is common in statistical physics.There, if you can find an "order parameter" [1,29], a physical quantity which is identically zero on an open subset of the parameter space and nonzero in another, then it cannot be analytic along any path from the first region to the second, so the open subsets cannot be parts of the same phase.Finding such an order parameter can be very difficult, but once found it can be very useful as we now show.
To appreciate the subtleties associated with some phase transitions in real materials, consider water in various common states.First consider gaseous water (steam) at temperature T 1 just above 100 • Celsius and atmospheric pressure P , and liquid water at temperature T 2 just below 100 • Celsius and again pressure P .The mass density is much lower in that gaseous state than in that liquid state, so in any reasonable sense there is a sharp transition in state corresponding to the (arbitrarily) small change in temperature.Now consider the pair of states: liquid water at temperature T 3 just above 0 • Celsius and atmospheric pressure P and solid water (ice) at temperature T 4 just below 0 • Celsius and again pressure P .Again the mass density is different between these two states (ice floats on water) so again in any reasonable sense there is a sharp transition in state corresponding to the (arbitrarily) small change in temperature.
There is a big difference between these two transitions.Consider a different way of changing the state from that gaseous state to its 'neighboring' liquid state.It is expermentally possible, by slowly accessing high temperatures and high pressures to use a different path between these two states without making any sharp change in state!(One says there is a 'critical' point in the gas/liquid transition.)But again experimentally, there is NO critical point in the liquid/solid transition; however you vary temperature and pressure to move slowly between a state of liquid water and a state of solid ice you must go through a sharp transition.Percy Bridgman received the Nobel prize in 1946 for his extensive experiments on high pressure, one result of which was to demonstrate that there is no critical point on the transition between fluid and solid in any known material.It is an old problem to try to understand why this should be the case.Consider the following quote in [31, page 11] The most outstanding unsolved problem of equilibrium statistical mechanics is the problem of the phase transitions.Why do all substances occur in at least three phases, the solid, liquid, and vapor phase which can coexist in the triple point?Why is there, again for all substances, a critical point for the vapor-liquid equilibrium, while apparently there is no critical point for the fluid-solid transition.Note that since these are general phenomena, they must have a general explanation; the precise details of the molecular structure and of the intermolecular forces should not matter.
As described in [1, page 19], Lev Landau tried to use symmetry as an order parameter to solve this problem: It was Landau (1958) who, long ago, first pointed out the vital importance of symmetry in phase transitions.This, the First Theorem of solid-state physics, can be stated very simply: it is impossible to change symmetry gradually.A given symmetry element is either there or it is not; there is no way for it to grow imperceptibly.This means, for instance, that there can be no critical point for the melting curve as there is for the boiling point: it will never be possible to go continuously through some high-pressure phase from liquid to solid.
While appealing, Landau's argument is not universally accepted; see [22, page 122].Part of the problem is that there is no known model in equilibrium statistical mechanics which can be proven to exhibit both fluid and solid phases [4,31].Even if such phases were proven to exist, it isn't at all clear how to define an appropriate order parameter.
Yet that is exactly what we have done in the context of random graphs at the beginning of this section: we contrasted the subset O, part of a "symmetric" phase in which the order parameter Q vanishes identically, with part of a phase in which Q is not zero.
Of course this is not a solution of the classic problem of proving a solid-fluid phase transition in a reasonable statistical mechanics model; we are working with random graphs, not with configurations of atoms.But that's actually the point!Graph models are a wonderful laboratory for developing, with full mathematical rigor, techniques that are simply too hard in statistical physics, a laboratory where important structural questions can be successfully solved.

Summary
This paper is part of a series [27,28,24,25,26,11,12,13,10,13,26,19,21] studying combinatorial systems (graphs, permutations, sphere packing) under competing constraints, as an extension of extremal combinatorics but concentrating on nonextreme states of the systems, that is, states under nonextreme constraints.We study asymptotically large systems and for graphs and permutations we use a large deviation principle (LDP) to analyze "typical" (i.e.exponentially most) constrained states.(We do not know an LDP for sphere packing but analyze such systems using the hard sphere model [18] in equilibrium statistical mechanics.)In this paper we sharpened our notion of phase by the use of analyticity; see Section 7.
Our goal in studying typical nonextremely-constrained states in these combinatorial systems is to analyze emergent smoothness response to infinitesimal change in the constraints, and of the combinatorial systems we have considered we have found dense graphs the most amenable to development.
By design our graph modelling has many features in common with that of the equilibrium statistical mechanics of particles with short range forces, but has a significant difference: there is no "distance" between edges, so each edge has the same influence on any other edge.In statistical mechanics, models with this feature of the influence of particles on one another are called "mean-field", and although not part of the mathematical formalism [29], mean-field models such as Curie-Weiss and van der Waals [30] are used to study phase transitions where more physical models prove too difficult.The random graph model we have been discussing has, in this sense, more in common with mean-field models of statistical mechanics, and, as seen by our success in proving the existence of phases and phase transitions, may be able to provide a mathematical formalism for studying the asymptotics of graphs and other combinatorial objects which will be as fruitful mathematically as statistical mechanics has been.
We conclude with the following open problems in this edge/triangle model.
(1) What is the actual structure of the optimal graphon when e < e 0 and t is slightly below e 3 ?Does it resemble the example given in Section 6, or is its structure still wilder?How does the behavior of this graphon as σ → 0 compare to the moderate deviations results for G(n, m) in [20]?(2) Is there a succession of phases as e → 0 and t remains close to e 3 , with tripodal graphons giving way to 4-podal, 5-podal, and so on?(3) When e 0 < e < 1/2 and t is slightly less than e 3 , is the optimal graphon symmetric bipodal, or is it something else? (4) When e < 1/2 and t = 0, the optimal graphon is symmetric bipodal.What if e < 1/2 and t is slightly positive?(5) Proposition 16 is stated for t close to e 3 .However, the only step in the proof that uses t ≈ e 3 is the estimate that η is much smaller than α.Can the result be extended to the entire region e < 1/2, t < e 3 ?(6) In [12] it is proven that there are two open sets with supersatured triangles, t > e 3 , which extend to phases.One of these is bounded below by the curve t = e 3 , e < 1/2, while the other is bounded by t = e 3 , e > 1/2.Are these actually parts of the same phase, or are they distinct?

Figure 1 .
Figure 1.The open set O of Theorem 1

2 t = e 3 t
edge density e

Figure 4 .
Figure 4.The phase transition between O 1 and O 2 was proven in [21].

Figure 5 .
Figure 5.The parameters of a bipodal graphon