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Linear embeddings of random complexes

Andrew Newman Carnegie Mellon University
Abstract

For XX(n;1,nα1,nα2,)X\sim X(n;1,n^{-\alpha_{1}},n^{-\alpha_{2}},...) in the multiparameter random simplicial complex model we establish necessary and sufficient strict inequalities on the αi\alpha_{i}’s to linearly embed the complex into 2d\mathbb{R}^{2d}.

1 Introduction

Recall the multiparameter random simplicial complex model introduced by Costa and Farber [5]. Given nn\in\mathbb{N} and a sequence of probabilities p=(pi)i0\textbf{p}=(p_{i})_{i\geq 0} we sample XX(n;p)X\sim X(n;\textbf{p}) starting with the ground set [n][n]. We first include each element of the ground set as a vertex independently with probability p0p_{0}, from here we include each edge between exisiting vertices with probability p1p_{1}, and then each triangle whose boundary is included with probability p2p_{2}, and so on. This model interpolates between the clique complex model, introduced first by Kahle [11], which is the case where p1p_{1} can vary but all other pip_{i}’s are 1 and the Linial–Meshulam–Wallach dd-dimensional model [12, 15] where pdp_{d} can vary, pi=1p_{i}=1 for i<di<d and pi=0p_{i}=0 for i>di>d (note that the d=1d=1 case for Linial–Meshulam–Wallach is the Erdős–Rényi random graph).

Here we establish thresholds for linear embeddings of random simplicial complexes in Euclidean space. Recall that a linear map from a simplicial complex Δ\Delta into Euclidean space d\mathbb{R}^{d} is a map ϕ:Δd\phi:\Delta\rightarrow\mathbb{R}^{d} that is determined linearly by the image of vertices. That is, the image of any face σΔ\sigma\in\Delta under ϕ\phi is the convex hull in d\mathbb{R}^{d} of the image of the vertices of σ\sigma. We say that Δ\Delta embeds linearly in d\mathbb{R}^{d} if there is an injective linear map ϕ:Δd\phi:\Delta\hookrightarrow\mathbb{R}^{d}.

Perhaps the most well known situation concerning linear embeddings of simplicial complexes is graph planarity. A graph GG is planar if and only if it admits a linear embedding into 2\mathbb{R}^{2}. We could take this as a definition of planar graphs, but it turns out that when it comes to planar graphs it doesn’t matter if we characterize them by existence of linear embeddings or existence of topological embeddings. The result that a graph is planar if and only if it can be drawn in the plane with all edges as straight-line segments is known as Fáry’s theorem. The analogous statement for higher dimensions was first demonstrated to be false by Brehm [3] who gave a triangulation of the Möbius strip on nine vertices that does not linearly embed in 3\mathbb{R}^{3}. Results and discussion in [6] examine certain cases of equivalence or nonequivalence of topological and linear embeddings.

In between linear embeddings and topological embeddings are PL-embeddings (piecewise-linear embeddings). A simplicial complex is PL-embeddable in d\mathbb{R}^{d} if it admits a subdivision that is linearly embeddable in d\mathbb{R}^{d}. It is known that PL-emdeddability is strictly weaker than linear embeddability in general. Brehm and Sarkaria [4] disproved a conjecture of Grünbaum that PL-embeddability of a dd-complex in 2d2d-dimensional space implies linear embeddability. A result of Matoušek, Tancer, and Wagner [14] establishing complexity results on PL-embeddings of simplicial complexes gives another proof that PL-embeddability does not imply linear embeddability by demonstrating that the two decision problems are not in the same complexity class. Here as a corollary to the main result we give a proof of this same nonequivalence via the probabilistic method by establishing threshold results for linear embeddability of random complexes and contrasting them with known threshold results for PL-embeddability, see the concluding remarks for a precise statement on this.

We consider multiparameter random complexes in a sparse regime where each pi=nαip_{i}=n^{-\alpha_{i}} for some fixed αi[0,]\alpha_{i}\in[0,\infty] (with αi=\alpha_{i}=\infty corresponding to pi=0p_{i}=0). Furthermore, we assume that p0=1p_{0}=1. This is a convenience for the proofs, but by a straightforward but perhaps somewhat tedious application of large deviation inequalities on the number of vertices the natural extension of the main result also holds when p0p_{0} is allowed to depend on nn provided that p0np_{0}n\rightarrow\infty as nn\rightarrow\infty.

To state our main result we first introduce some notation. For α¯=(α1,α2,)\operatorname{\overline{\alpha}}=(\alpha_{1},\alpha_{2},...) we denote X(n;1,nα1,nα2)X(n;1,n^{-\alpha_{1}},n^{-\alpha_{2}}...) by X(n;nα¯)X(n;n^{-\operatorname{\overline{\alpha}}}). For each s1s\geq 1 we let vsv_{s} denote the vector of binomial coefficients vs=((si+1))i1{v}_{s}=\left(\binom{s}{i+1}\right)_{i\geq 1}. By linearity of expectation the expected number of (s1)(s-1)-simplices in X(n,nα¯)X(n,n^{-\operatorname{\overline{\alpha}}}) is

Θ(nsα¯vs).\Theta(n^{s-\operatorname{\overline{\alpha}}\cdot v_{s}}).

Our main result establishes a threshold in α¯\operatorname{\overline{\alpha}} for embeddability in even dimensions.

Theorem 1.

Fix d1d\geq 1 we have the following for random simplicial complexes in the multiparameter model XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}). If d+1α¯vd+1<1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}<1 then with high probability XX linearly embeds in 2d\mathbb{R}^{2d} while if d+1α¯vd+1>1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}>1 with high probability XX does not linearly embed in 2d\mathbb{R}^{2d}.

Stated another way, our main result says if the α¯\operatorname{\overline{\alpha}} is selected so that the expected number of dd-faces is at most n1εn^{1-\varepsilon} for some ε>0\varepsilon>0 then the random complex will embed in 2d\mathbb{R}^{2d} with high probability, while an average number of dd-faces of order at least n1+εn^{1+\varepsilon} implies that the random complex will not embed in 2d\mathbb{R}^{2d}. Somewhat surprisingly then the threshold to embed a random complex in 2d\mathbb{R}^{2d} turns out to only depends on the dd-skeleton of the random complex.

2 Preliminaries

Recall that a dd-complex Δ\Delta is alway embeddable in 2d+1\mathbb{R}^{2d+1}. A generic set of points in m\mathbb{R}^{m} is one where for any kk there is no affine subspace of m\mathbb{R}^{m} containing more than k+1k+1 of the points and for any two disjoint subsets AA and BB of the point set each of size at most mm, the affine subspace of dimension |A|1|A|-1 spanned by the points in AA and the affine subspace of dimension |B|1|B|-1 spanned by the points in BB intersect as an affine subspace of dimension |A|+|B|2m|A|+|B|-2-m, if this quantity is nonnegative and don’t intersect at all otherwise. If we place the points of a dd-complex Δ\Delta in 2d+1\mathbb{R}^{2d+1} generically then it will be an embedding as two dd-dimensional subspaces generically don’t intersect in 2d+1\mathbb{R}^{2d+1}.

Observe that any set of points in m\mathbb{R}^{m} are moved to generic position by an arbitrarily small perturbation. The proof of the sparse side of Theorem 1 will give a generic embedding of our random complex in 2d\mathbb{R}^{2d}, and we will rule out the existence of a generic embedding on the dense side of the claimed threshold. If there is no generic embedding of our finite simplicial complex there is no embedding at all as a small perturbation of an embedding will still be an embedding.

The proof for the sparse side will follow a collapsibility argument. We will prove a linear embedding analogue of a result of Horvatić from [10] about PL-embeddability. A face of a simplicial complex is said to be free if it is properly contained in only one other face. The removal of a free face and the face that properly contains it is called an elementary collapse, and a complex is dd-collapsible if there is a sequence of elementary collapse on the complex that leave behind only faces of dimension at most d1d-1. Theorem 3.2 of [10] implies that a dd-complex which is dd-collapsible PL embeds in 2d\mathbb{R}^{2d}. We’ll prove on the sparse side that if a stronger collapsibility property holds for a dd-complex it will linearly embed in 2d\mathbb{R}^{2d} and then show that with high probability our random complex satisfies that strong collapsibility property.

For the dense side we make use of Radon’s Theorem. Recall that Radon’s Theorem states that for any set of d+2d+2 points in d\mathbb{R}^{d} there is a partition into two sets AA and BB so that the convex hull of AA intersects the convex hull of BB. For the dense side we start with a generic embedding of the points into 2d\mathbb{R}^{2d} and then generate the complex and count how many sets of 2d+22d+2 points, in expectation, have a Radon partition (A,B)(A,B) with the simplex on AA and the simplex on BB both included in the complex. We show that the expected number of such point sets is large and then prove a concentration of measure inequality that survives multiplication by the number of combinatorially distinct ways to place the nn points in 2d\mathbb{R}^{2d} (the number of order types).

Both for the proof of the sparse side and later for the proof of the concentration inequality on the dense side, we need the following lemma that gives basic facts about the ff-vector of a random complex in the multiparameter model.

Lemma 2.

For any s2s\geq 2 if sα¯vs<1s-\operatorname{\overline{\alpha}}\cdot v_{s}<1 then s+1α¯vs+1<0s+1-\operatorname{\overline{\alpha}}\cdot v_{s+1}<0, and for s3s\geq 3 if sα¯vs>1s-\operatorname{\overline{\alpha}}\cdot v_{s}>1 then s1α¯vs1>1s-1-\operatorname{\overline{\alpha}}\cdot v_{s-1}>1.

Proof.

Fix ε>0\varepsilon>0 and suppose that sα¯vs=1εs-\operatorname{\overline{\alpha}}\cdot v_{s}=1-\varepsilon. We maximize s+1α¯vs+1s+1-\operatorname{\overline{\alpha}}\cdot v_{s+1} subject to this constraint and the constraints that the αi\alpha_{i}’s are all nonnegative. Obviously if one of the αi\alpha_{i}’s is infinity so that α¯vs+1=\overline{\alpha}\cdot v_{s+1}=\infty there is nothing to prove, so we may assume all the relevant αi\alpha_{i}’s are finite. The region of the hyperplane determined by sα¯vs=1εs-\operatorname{\overline{\alpha}}\cdot v_{s}=1-\varepsilon in the first orthant is a simplex on s1s-1 vertices in s1\mathbb{R}^{s-1}. By the simplex method we only have to check s+1α¯vs+1s+1-\operatorname{\overline{\alpha}}\cdot v_{s+1} at the vertices of this simplex. Moreover we may assume that αs=0\alpha_{s}=0. At the vertices of the feasible region exactly one of the αi\alpha_{i}’s is nonzero. Let αj10\alpha_{j-1}\neq 0 for some j{2,,s}j\in\{2,...,s\} with the rest of the αi\alpha_{i}’s in this range being 0. So we have that sαj1(sj)=1εs-\alpha_{j-1}\binom{s}{j}=1-\varepsilon. Now

s+1αj1(s+1j)\displaystyle s+1-\alpha_{j-1}\binom{s+1}{j} =\displaystyle= s+1αj1((sj)+(sj1))\displaystyle s+1-\alpha_{j-1}\left(\binom{s}{j}+\binom{s}{j-1}\right)
=\displaystyle= sαj1(sj)+1αj1(sj1)\displaystyle s-\alpha_{j-1}\binom{s}{j}+1-\alpha_{j-1}\binom{s}{j-1}
=\displaystyle= 1ε+1αj1(sj1)\displaystyle 1-\varepsilon+1-\alpha_{j-1}\binom{s}{j-1}
=\displaystyle= 1ε+1αj1(sj)((sj1)(sj))\displaystyle 1-\varepsilon+1-\alpha_{j-1}\binom{s}{j}\left(\frac{\binom{s}{j-1}}{\binom{s}{j}}\right)
=\displaystyle= 2ε(s1+ε)jsj+1\displaystyle 2-\varepsilon-(s-1+\varepsilon)\frac{j}{s-j+1}

Now we have to verify that this is negative, as js+1j\frac{j}{s+1-j} is an increasing function and j2j\geq 2, we have

2ε(s1+ε)jsj+12ε(s1+ε)2s1\displaystyle 2-\varepsilon-(s-1+\varepsilon)\frac{j}{s-j+1}\leq 2-\varepsilon-(s-1+\varepsilon)\frac{2}{s-1}

The righthand side is negative provided that

2ε<2(s1+εs1)2-\varepsilon<2\left(\frac{s-1+\varepsilon}{s-1}\right)

And this holds clearly since the right hand side is bigger than 2.

For the other part, we minimize s1α¯vs1s-1-\operatorname{\overline{\alpha}}\cdot v_{s-1} subject to sα¯vs=1+εs-\operatorname{\overline{\alpha}}\cdot v_{s}=1+\varepsilon. As before it suffices to check in the cases that a single αj10\alpha_{j-1}\neq 0 for some j{2,,s}j\in\{2,...,s\} with the rest of the αi\alpha_{i}’s being zero and so sαj1(sj)=1+εs-\alpha_{j-1}\binom{s}{j}=1+\varepsilon. We have

s1αj1(s1j)\displaystyle s-1-\alpha_{j-1}\binom{s-1}{j} =\displaystyle= s1αj1((sj)(s1j1))\displaystyle s-1-\alpha_{j-1}\left(\binom{s}{j}-\binom{s-1}{j-1}\right)
=\displaystyle= s1αj1(sj)+αj1(s1j1)\displaystyle s-1-\alpha_{j-1}\binom{s}{j}+\alpha_{j-1}\binom{s-1}{j-1}
=\displaystyle= 1+(1+ε)+αj1(sj)(js)\displaystyle-1+(1+\varepsilon)+\alpha_{j-1}\binom{s}{j}\left(\frac{j}{s}\right)
=\displaystyle= 1+(1+ε)+(s1ε)(js)\displaystyle-1+(1+\varepsilon)+(s-1-\varepsilon)\left(\frac{j}{s}\right)

Again it suffices to check in the case that j=2j=2 as that’s what makes final expression above the smallest. So we check

ε+(s1ε)(2s)>1\varepsilon+(s-1-\varepsilon)\left(\frac{2}{s}\right)>1

However our assumption that sα¯vs=1+εs-\operatorname{\overline{\alpha}}\cdot v_{s}=1+\varepsilon gives that ε\varepsilon is at most s1s-1 and the above expression attains a minimum at ε=0\varepsilon=0 where we just have to check that

2(s1s)>12\left(\frac{s-1}{s}\right)>1

and this holds for s3s\geq 3 as required. ∎

3 Sparse side

For the sparse side we will prove a certain collapsibility condition on our random complexes that implies the complex is embeddable. For XX a dd-complex, we associate a (d+1)(d+1)-uniform hypergraph \mathcal{H} whose hyperedges are the dd-faces of XX. If \mathcal{H} has no 2-core then XX will embed in 2d\mathbb{R}^{2d}. By 2-core we mean a subhypergraph of \mathcal{H} in which every vertex is contained in at least two hyperedges. If \mathcal{H} has no 2-core then there is a sequence of vertex deletions obtained by removing a vertex of degree at most 1, along with the hyperedge that contains it, that removes all the vertices. We will say the pure part of XX has no 2-core to mean that the associated hypergraph has no 2-core. Recall that the pure part of a dd-complex is the subcomplex generated by the downward closure of the dd-faces. The next theorem shows that the absences of a 2-core in dimension dd implies embeddability in dimension 2d2d.

Theorem 3.

For d1d\geq 1, if XX is a finite dd-complex and the pure part of XX has no 2-core then XX linearly embeds into 2d\mathbb{R}^{2d}.

Proof.

If we can embed the pure part of XX into 2d\mathbb{R}^{2d} generically then this embedding can be extended to an embedding of the entire complex. Indeed if the vertices of XX are mapped into 2d\mathbb{R}^{2d} generically then the only faces of XX that might intersect one another in their interior are pairs of dd-faces, each pair of which could possibly intersect at a single point. Therefore once we have placed the points into 2d\mathbb{R}^{2d} generically so that the pure part of XX embeds, the lower dimensional faces can be added without obstructing the embedding.

Let f1,,fkf_{1},...,f_{k} be an ordering of the dd-faces of XX obtained by removing vertices of degree one (called free vertices) with fkf_{k} as the first dd-face removed and f1f_{1} as the last face removed. We show that if we have embedded f1ff_{1}\cup\cdots\cup f_{\ell} then we may extend the embedding to include f+1f_{\ell+1}. Obviously f1f_{1} as a dd-simplex embeds into 2d\mathbb{R}^{2d}. Now suppose we have embedded f1ff_{1}\cup\cdots\cup f_{\ell}. When f+1f_{\ell+1} is added to XX it comes with a new vertex that belongs to f+1f_{\ell+1} but not to any other fif_{i}’s we’ve embedded so far by how the ordering is defined. Thus

(f1f)f+1(f_{1}\cup\cdots\cup f_{\ell})\cap f_{\ell+1}

contains at most dd vertices. If the intersection contains less than dd vertices then we place all but one of the new vertices of f+1f_{\ell+1} in d\mathbb{R}^{d} generically with the rest of the embedding. Now we fill in the convex hull of these dd vertices of f+1f_{\ell+1}. By general position this will be compatible with the embedding as everything we are adding is of dimension at most d1d-1. Now we want to add the last vertex and the simplex f+1f_{\ell+1} attaching it along the (d1)(d-1)-simplex on the first dd-vertices. Call this (d1)(d-1)-simplex σ\sigma. We take an arbitrary affine dd-space through σ\sigma in 2d\mathbb{R}^{2d}, call this space KK. Now KK will intersect the complex we have embedded so far only at σ\sigma and at some finite number of points. Therefore there is some thickening of σ\sigma within KK that avoids all the finitely many points of intersection with the rest of the complex. This can be obtained by thickening σ\sigma by half the distance to the nearest of the finitely many points. Now put the new vertex vv inside this this thickening and take the convex of vv and σ\sigma and we have extended the embedding to include vσ=f+1v\cup\sigma=f_{\ell+1}. Therefore inductively the pure part of XX can be embedded in 2d\mathbb{R}^{2d} generically and this enough to conclude that all of XX can be embedded in 2d\mathbb{R}^{2d}. ∎

By Theorem 3 we want to prove a collapsibility result about XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}). This is a different type of collapsibility than is studied in [13] and [16]. Here we are collapsing (vertex, dd-simplex)-pairs and these other papers deal with collapsing ((d1)(d-1)-simplex, dd-simplex)-pairs. The overall strategy though is the very similar: we prove that there are no large obstructions to collapsibility, and that local obstructions to collapsibility are too dense to appear in this sparse regime. We start with a definition.

Definition 4.

For XX a dd-complex, associate a graph G(X)G(X) whose vertices are the dd-simplices of XX with σ\sigma adjacent to τ\tau exactly when στ\sigma\cap\tau\neq\emptyset. We say that a subcomplex YY of XX is weakly connected if for any σ\sigma and τ\tau in YY there is a path within G(X)G(X) between σ\sigma and τ\tau. The weakly connected components in XX are the maximal weakly connected subcomplexes.

We could also refer to weak connectivity as path connectivity but we use weak connectivity to contrast with strong connectivity from the other papers in the literature on collapsibility. Clearly a minimal 2-core of XX is weakly connected. We first rule out large weakly connected components as that will rule out large minimal 2-cores.

Lemma 5.

Fix d1d\geq 1 and ε>0\varepsilon>0, for vector α¯\operatorname{\overline{\alpha}} so that d+1α¯vd+1=1εd+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}=1-\varepsilon there is some constant L=L(d,ε)L=L(d,\varepsilon) so that with high probability XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) contains no weakly connected subcomplex on more than LL vertices.

Proof.

If ε>1\varepsilon>1 a simple first moment argument shows that there are no dd-faces at all with high probability, so we assume ε1\varepsilon\leq 1. The idea is to show that for LL large, but constant independent of nn, if YY is any weakly connected subcomplex on LL vertices there is some δ>0\delta>0 so that the dot product of the ff-vector of YY denoted (L,fY)(L,f_{Y}), with fYf_{Y} as the ff-vector omitting the number of vertices, with the vector (1,α¯):=(1,α1,,αd)(1,\operatorname{\overline{\alpha}}):=(1,\alpha_{1},...,\alpha_{d}) satisfies

(L,fY)(1,α¯)=LfYα¯<δ.(L,f_{Y})\cdot(1,\operatorname{\overline{\alpha}})=L-f_{Y}\cdot\operatorname{\overline{\alpha}}<-\delta.

If we have this then the expected number of weakly connected subcomplexes with LL vertices in XX is at most

(nL)22Lnα¯fY22LnLα¯fY22Lnδ=o(1).\binom{n}{L}2^{2^{L}}n^{-\operatorname{\overline{\alpha}}\cdot f_{Y}}\leq 2^{2^{L}}n^{L-\operatorname{\overline{\alpha}}\cdot f_{Y}}\leq 2^{2^{L}}n^{-\delta}=o(1).

The 22L2^{2^{L}} is simply a trivial upper bound on the number of simplicial complexes on LL vertices.

Now we verify the claimed bound on LfYα¯L-f_{Y}\cdot\operatorname{\overline{\alpha}}. Suppose that YY is a weakly connected dd-complex, then by taking a spanning tree on G(Y)G(Y) we can find an ordering σ1,,σK\sigma_{1},...,\sigma_{K} on the dd-faces of YY so that for each 2kK2\leq k\leq K, σk(σ1σk1)\sigma_{k}\cap(\sigma_{1}\cup\cdots\cup\sigma_{k-1}) is nonempty. Let β=d+1α¯vd+1\beta=d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}, this is the value of the dot product of the ff-vector of σ1\sigma_{1} with (1,α¯)(1,\operatorname{\overline{\alpha}}). Throughout the process of adding the σi\sigma_{i}’s β\beta will denote the value of f0f1α1fdαdf_{0}-f_{1}\alpha_{1}-\cdots-f_{d}\alpha_{d} as the ff-vector changes. We keep track of how β\beta changes as we add the σi\sigma_{i}’s in their given order. Specifically we want to argue that if LL is large enough, that β\beta eventually becomes negative. We immediately see that if we add a σi\sigma_{i} that doesn’t contain any vertices that aren’t already in σ1σi1\sigma_{1}\cup\cdots\cup\sigma_{i-1} that β\beta cannot increase, so only have to consider how it changes when we add a face that includes new vertices. Suppose adding σi\sigma_{i} adds tt new vertices to YY, then the effect adding σi\sigma_{i} on β\beta is adding to it at most the quantity

γ(t):=tα1((t2)+g(t,d,1))α2((t3)+g(t,d,2))αd1((td)+g(t,d,d1))\gamma(t):=t-\alpha_{1}\left(\binom{t}{2}+g(t,d,1)\right)-\alpha_{2}\left(\binom{t}{3}+g(t,d,2)\right)-\cdots-\alpha_{d-1}\left(\binom{t}{d}+g(t,d,d-1)\right)

where g(t,d,k)g(t,d,k) is the number of kk-dimensional faces in the dd-simplex containing at least one vertex from a fixed subset of size tt and at least one vertex outside that set, so

g(t,d,k)=(d+1k+1)(tk+1)((d+1)tk+1)g(t,d,k)=\binom{d+1}{k+1}-\binom{t}{k+1}-\binom{(d+1)-t}{k+1}

So we maximize (over α¯\operatorname{\overline{\alpha}}, not over tt) γ(t)\gamma(t) subject to α¯0\operatorname{\overline{\alpha}}\geq 0 and dα¯vd+1=εd-\operatorname{\overline{\alpha}}\cdot v_{d+1}=-\varepsilon. Similar to the proof of Lemma 2 it suffices to check when exactly one entry of (α1,,αd)(\alpha_{1},...,\alpha_{d}) is nonzero. Set jj so that αj0\alpha_{j}\neq 0 and αi=0\alpha_{i}=0 for i{1,,d}i\in\{1,...,d\}, with iji\neq j and dαj(d+1j+1)=εd-\alpha_{j}\binom{d+1}{j+1}=-\varepsilon. In this case then for any t{1,,d}t\in\{1,...,d\},

γ(t)\displaystyle\gamma(t) =\displaystyle= tαj((tj+1)+(d+1j+1)(tj+1)((d+1)tj+1))\displaystyle t-\alpha_{j}\left(\binom{t}{j+1}+\binom{d+1}{j+1}-\binom{t}{j+1}-\binom{(d+1)-t}{j+1}\right)
=\displaystyle= tαj(d+1j+1)+αj((d+1)tj+1)\displaystyle t-\alpha_{j}\binom{d+1}{j+1}+\alpha_{j}\binom{(d+1)-t}{j+1}

We want to verify that this is always negative and bounded away from zero. Since we’ve assumed that

αj=d+ε(d+1j+1),\alpha_{j}=\frac{d+\varepsilon}{\binom{d+1}{j+1}},

it suffices to prove that for all t{1,,d}t\in\{1,...,d\} and j{1,,d}j\in\{1,...,d\}

t(d+1j+1)((d+1)tj+1)d(d+1j+1).\frac{t}{\binom{d+1}{j+1}-\binom{(d+1)-t}{j+1}}\leq\frac{d}{\binom{d+1}{j+1}}.

This is equivalent to showing that

td(d+1j+1)((d+1)tj+1)(d+1j+1).\frac{t}{d}\leq\frac{\binom{d+1}{j+1}-\binom{(d+1)-t}{j+1}}{\binom{d+1}{j+1}}.

We prove this as the following claim:

Claim 6.

For any nn, t{1,,n1}t\in\{1,...,n-1\}, and k{2,,n}k\in\{2,...,n\}

tn1(nk)(ntk)(nk)\frac{t}{n-1}\leq\frac{\binom{n}{k}-\binom{n-t}{k}}{\binom{n}{k}}
Proof.

The righthand side of the inequality is the probability that a uniform random kk-element subset of {1,,n}\{1,...,n\} contains at least one element of a fixed subset of {1,,n}\{1,...,n\} of size tt. Obviously that probability gets larger as kk increases. So it suffices to verify the inequality when k=2k=2.

(n2)(nt2)(n2)tn1\displaystyle\frac{\binom{n}{2}-\binom{n-t}{2}}{\binom{n}{2}}\geq\frac{t}{n-1}
<=>\displaystyle<=> n(n1)(nt)(nt1)n(n1)tn1\displaystyle\frac{n(n-1)-(n-t)(n-t-1)}{n(n-1)}\geq\frac{t}{n-1}
<=>\displaystyle<=> n2n(n2ntnnt+t2+t)tn\displaystyle\frac{n^{2}-n-(n^{2}-nt-n-nt+t^{2}+t)}{t}\geq n
<=>\displaystyle<=> 2ntt2ttn\displaystyle\frac{2nt-t^{2}-t}{t}\geq n
<=>\displaystyle<=> 2nt1n\displaystyle 2n-t-1\geq n
<=>\displaystyle<=> n1t\displaystyle n-1\geq t

and we have assumed that tn1t\leq n-1 so we have the claim ∎

It follows that for every σi\sigma_{i} that includes a new vertex β\beta decreases by at least some fixed positive constant depending on dd and ε\varepsilon, and β\beta never increases. Thus β\beta can be made arbitrarily negative by setting LL arbitrarily large. This is sufficient for the first moment argument covered at the beginning of the proof to hold. ∎

Next we show that small subcomplexes on at most LL vertices are too sparse to be 2-cores. To that end we prove the following lemma.

Lemma 7.

If α¯\operatorname{\overline{\alpha}} satisfies d+1α¯vd+1<1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}<1 and YY is a dd-complex so that for every subcomplex YYY^{\prime}\subseteq Y,

f0(Y),f1(Y),,fd(Y)1,α1,,αd>0\langle f_{0}(Y^{\prime}),f_{1}(Y^{\prime}),...,f_{d}(Y^{\prime})\rangle\cdot\langle 1,-\alpha_{1},...,-\alpha_{d}\rangle>0

then YY has a vertex in at most one dd-face, in particular YY does not contain a 2-core.

Proof.

Suppose YY is a dd-complex satisfying the inequality for α¯\operatorname{\overline{\alpha}} so that d+1α¯vd+1<1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}<1. For a vertex ww let degi(w)\deg_{i}(w) denote the number of ii-dimensional faces containing ww, then

wY(112deg1(w)α113deg2(w)α21d+1degd(w)αd)=f0(Y)α1f1(Y)αdfd(Y)>0.\sum_{w\in Y}\left(1-\frac{1}{2}\deg_{1}(w)\alpha_{1}-\frac{1}{3}\deg_{2}(w)\alpha_{2}-\cdots-\frac{1}{d+1}\deg_{d}(w)\alpha_{d}\right)=f_{0}(Y)-\alpha_{1}f_{1}(Y)-\cdots-\alpha_{d}f_{d}(Y)>0.

Therefore there is some ww so that

i=1d1i+1degi(w)αi<1.\sum_{i=1}^{d}\frac{1}{i+1}\deg_{i}(w)\alpha_{i}<1.

We want to show that ww does not belong to at least two dd-simplices. Suppose that ww is contained in at least two dd-simplices, then the link of ww contains two (d1)(d-1)-simplices σ\sigma and τ\tau. If |στ|=md1|\sigma\cap\tau|=m\leq d-1 then

i=1d1i+1degi(w)αii=1d1i+1(2(di)(mi))αi\sum_{i=1}^{d}\frac{1}{i+1}\deg_{i}(w)\alpha_{i}\geq\sum_{i=1}^{d}\frac{1}{i+1}\left(2\binom{d}{i}-\binom{m}{i}\right)\alpha_{i}

The assumption that d+1α¯vd+1<1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}<1 gives us that

i=1d(d+1i+1)αi>d.\sum_{i=1}^{d}\binom{d+1}{i+1}\alpha_{i}>d.

Combining the previous three inequalities then there must be some 0md0\leq m\leq d so that

i=1d(d+1i+1)αi>i=1ddi+1(2(di)(mi))αi\sum_{i=1}^{d}\binom{d+1}{i+1}\alpha_{i}>\sum_{i=1}^{d}\frac{d}{i+1}\left(2\binom{d}{i}-\binom{m}{i}\right)\alpha_{i}

We next prove the following claim. Doing so contradicts our assumption that ww belongs to at least two dd-simplices, so ww can be taken to be our vertex of degree at most 1 and we then apply the same argument to Y{w}Y\setminus\{w\} and repeat until all the simplices have been deleted by removing free or isolated vertices verifying that YY does not contain a 2-core.

Claim 8.

For any d1d\geq 1, i{1,,d}i\in\{1,...,d\} and m{0,,d1}m\in\{0,...,d-1\},

(d+1i+1)di+1(2(di)(mi))\binom{d+1}{i+1}\leq\frac{d}{i+1}\left(2\binom{d}{i}-\binom{m}{i}\right)
Proof.

Clearly we just have to check m=d1m=d-1.

di+1(2(di)(d1i))(d+1i+1)0\displaystyle\frac{d}{i+1}\left(2\binom{d}{i}-\binom{d-1}{i}\right)-\left(\frac{d+1}{i+1}\right)\geq 0
<=>\displaystyle<=> d(2d!(di)!(d1)!(d1i)!)(d+1)!(di)!0\displaystyle d\left(2\frac{d!}{(d-i)!}-\frac{(d-1)!}{(d-1-i)!}\right)-\frac{(d+1)!}{(d-i)!}\geq 0
<=>\displaystyle<=> d(2d!(d1)!(di))(d+1)!0\displaystyle d\left(2d!-(d-1)!(d-i)\right)-(d+1)!\geq 0
<=>\displaystyle<=> d!(2d(di))(d+1)!\displaystyle d!\left(2d-(d-i)\right)\geq(d+1)!
<=>\displaystyle<=> d+id+1\displaystyle d+i\geq d+1
<=>\displaystyle<=> i1.\displaystyle i\geq 1.

The claim is verified. ∎

With the previous lemmas we are ready to prove the sparse side of Theorem 1

Proof of sparse side of Theorem 1.

Suppose that d+1αvd+1<1d+1-\alpha\cdot v_{d+1}<1. We show that with high probability XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) is at most dd-dimensional and the pure dd-part of XX has no 2-core. If this holds then by Theorem 3, XX will be embeddable in 2d\mathbb{R}^{2d}. From Lemma 2, d+1αvd+1<1d+1-\alpha\cdot v_{d+1}<1 implies that d+2αvd+2<0d+2-\alpha\cdot v_{d+2}<0 and the expected number of (d+1)(d+1)-faces is O(nd+2αvd+2)=o(1)O(n^{d+2-\alpha\cdot v_{d+2}})=o(1), so XX is dd-dimensional with high probability. Additionally Lemma 5 implies that with high probability there is some LL so that all minimal 2-cores in XX have at most LL vertices.

However, the possibility of small cores is handled by Lemma 7 and a first moment argument. By Lemma 7 for any YY that is a dd-dimensional 2-core there is a subcomplex YYY^{\prime}\subseteq Y so that

f0(Y),f1(Y),,fd(Y)1,α1,,αd0\langle f_{0}(Y^{\prime}),f_{1}(Y^{\prime}),...,f_{d}(Y^{\prime})\rangle\cdot\langle 1,-\alpha_{1},...,-\alpha_{d}\rangle\leq 0

However if equality holds, then by some small perturbation of α¯\operatorname{\overline{\alpha}} we could get a contradiction to Lemma 7 and so we actually have a strict inequality. Now as we have ruled out minimal 2-cores on more than LL vertices if we can show that XX does not contain any subcomplexes YY on at most LL vertices so that f0(Y),f1(Y),,fd(Y)1,α1,,αd<0\langle f_{0}(Y),f_{1}(Y),...,f_{d}(Y)\rangle\cdot\langle 1,-\alpha_{1},...,-\alpha_{d}\rangle<0 we will be done. Let \mathcal{F} be the family of such subcomplexes. As \mathcal{F} is finite there is some δ>0\delta>0 so that for all YY\in\mathcal{F} f0(Y),f1(Y),,fd(Y)1,α1,,αd<δ\langle f_{0}(Y),f_{1}(Y),...,f_{d}(Y)\rangle\cdot\langle 1,-\alpha_{1},...,-\alpha_{d}\rangle<-\delta. By linearity of expectation the expected number of complexes in \mathcal{F} that are subcomplexes of XX is at most

||L!nδ|\mathcal{F}|L!n^{-\delta}

because the number of copies of YY in XX for YY\in\mathcal{F} in expectation is at most

(nf0(Y))nf1(Y)α1f2(Y)α2fd(Y)αd<nδ.\binom{n}{f_{0}(Y)}n^{-f_{1}(Y)\alpha_{1}-f_{2}(Y)\alpha_{2}-\cdots-f_{d}(Y)\alpha_{d}}<n^{-\delta}.

As |||\mathcal{F}| and LL are constants independent of nn the expected number of subcomplexes of \mathcal{F} in XX is o(1)o(1), so with high probability XX contains no element of \mathcal{F} and hence no 2-core. Thus with high probability XX embeds in 2d\mathbb{R}^{2d} as it satisfies the assumptions of Theorem 3. ∎

4 Dense side

The proof for the dense side is based on Radon’s theorem. Recall that Radon’s theorem, originally proved in [17], states that any set of d+2d+2 points in d\mathbb{R}^{d} can be partitioned into two sets whose convex hulls intersect and if the point set is generic the Radon partition is unique. Thus if we have a linear embedding of XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) into 2d\mathbb{R}^{2d} then for any set of 2d+22d+2 vertices we get from Radon’s theorem a pair of simplices on those vertices that must be excluded from XX. Given an embedding of nn points in 2d\mathbb{R}^{2d} and a simplicial complex XX on [n][n] we say that S([n]2d+2)S\in\binom{[n]}{2d+2} is a Radon match if in the Radon partition on SS into two subsets S1S_{1} and S2S_{2}, the simplex on S1S_{1} and the simplex on S2S_{2} are both present in XX. If XX has a Radon match for every embedding of the vertices of XX into 2d\mathbb{R}^{2d} then XX is not linearly embeddable in 2d\mathbb{R}^{2d}. Our strategy is to show that with high probability when XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) on the dense side of the claimed phase transition every embedding of the vertices has at least one Radon match. We also will assume throughout that all embeddings have vertices placed generically in 2d\mathbb{R}^{2d} but it is clear that if there is a linear embedding of XX into 2d\mathbb{R}^{2d} then by a sufficiently small perturbation we have a generic embedding.

Turning the strategy into a proof requires three ingredients. First we need some control on which type of splits of our (2d+2)(2d+2)-sets we see. If, for example, they were all split as 2d+12d+1 vertices in convex position surrounding a single vertex in the interior, we would not expect to see any Radon partitions just beyond the phase transition. Second, for our purposes any two embeddings of the nn vertices in d\mathbb{R}^{d} can be regarded as equivalent if they induce the same Radon partition on all 2(d+2)2(d+2)-subsets. Therefore we need an upper bound on how many nonequivalent embeddings there are to consider. Third we need an upper bound on the probability that a particular embedding has no Radon match and this upper bound has to go to zero faster than the number of nonequivalent embeddings is going to infinity so that we can take a union bound.

For the first point we use the following stronger version of the classical van Kampen–Flores Theorem.

Theorem 9 (Theorem 6.6 of [2]).

Let d1d\geq 1. Then for every embedding of d+3d+3 points in d\mathbb{R}^{d} there exists two disjoint sets S1S_{1} and S2S_{2} with |S1|=(d+2)/2|S_{1}|=\lfloor(d+2)/2\rfloor and |S2|=(d+2)/2|S_{2}|=\lceil(d+2)/2\rceil so that the convex hull of S1S_{1} and S2S_{2} intersect.

As a corollary to this we have the following regarding an embedding of nn points in d\mathbb{R}^{d}

Corollary 10.

For any embedding of nn points in d\mathbb{R}^{d} there are at least

1d+3(nd+2)\frac{1}{d+3}\binom{n}{d+2}

(d+2)(d+2)-subsets of [n][n] that split into two sets S1S_{1} and S2S_{2} with |S1|=(d+2)/2|S_{1}|=\lfloor(d+2)/2\rfloor and |S2|=(d+2)/2|S_{2}|=\lceil(d+2)/2\rceil so that the convex hull of S1S_{1} and S2S_{2} intersect.

Proof.

By Theorem 9 for any embedding of nn points in d\mathbb{R}^{d} every subset of size d+3d+3 contributes at least one pair (S1,S2)(S_{1},S_{2}) satisfying the required condition. Such a pair has |S1S2|=d+2|S_{1}\cup S_{2}|=d+2, and each belongs to at most n(d+2)n-(d+2) (d+3)(d+3)-element subsets. Thus the number of such pairs (S1,S2)(S_{1},S_{2}) is at least

1n(d+2)(nd+3)=1d+3(nd+2).\frac{1}{n-(d+2)}\binom{n}{d+3}=\frac{1}{d+3}\binom{n}{d+2}.

We now turn our attention to bounding the number of embeddings of the vertices we have to consider. We consider two general position embeddings f,g:[n]df,g:[n]\rightarrow\mathbb{R}^{d} to be equivalent if for every S([n]d)S\in\binom{[n]}{d} the Radon partition of SS under the image of ff is the same as the Radon partition of SS under the image of gg.

For this we are really counting labeled, simple order types of a configuration of nn points in d\mathbb{R}^{d}. Recall that the order type of a labeled set of points p1,,pnp_{1},...,p_{n} in d\mathbb{R}^{d} is the (nd+1)\binom{n}{d+1}-tuple of signs of determinants of (d+1)×(d+1)(d+1)\times(d+1) submatrices of

(p1pn11).\begin{pmatrix}p_{1}&\cdots&p_{n}\\ 1&\cdots&1\end{pmatrix}.

That is the order type is

(sgndet(pi1pid+111))1i1<<id+1n\left(\text{sgn}\det\begin{pmatrix}p_{i_{1}}&\cdots&p_{i_{d+1}}\\ 1&\cdots&1\end{pmatrix}\right)_{1\leq i_{1}<\cdots<i_{d+1}\leq n}

By construction then for any set of d+2d+2 points in general position the order type of those d+2d+2 points determines the Radon partition. Indeed for p1p_{1}, …, pd+1p_{d+1}, pd+2p_{d+2} we have p1p_{1}, …, pd+1p_{d+1} are vertices of a simplex and for any ii, the order type (either 1-1 or 11) of p1,..,pi1,pi+1,..,pd+1,pd+2p_{1},..,p_{i-1},p_{i+1},..,p_{d+1},p_{d+2} determines on which side of the hyperplane through p1,..,pi1,pi+1,..,pd+1p_{1},..,p_{i-1},p_{i+1},..,p_{d+1}, pd+2p_{d+2} sits. Thus the order type of p1,,pd+1,pd+2p_{1},...,p_{d+1},p_{d+2} determines which region of the hyperplane arrangement on p1,,pd+1p_{1},...,p_{d+1} contains pd+2p_{d+2}. This region uniquely determines the Radon partition on p1,..,pd+2p_{1},..,p_{d+2}. For more background on order types see the survey of Goodman and Pollack [9]. By simple order types we mean order types where none of the determinant signs are zero, these are all the order types that have to be considered for generic point sets.

The following enumeration result of Goodman and Pollack on order types will be sufficient for our proof. There is also a stronger result for all labeled order types, not just simple ones, due to [1]

Theorem 11.

[8, Theorem 1] The number of simple order types for nn labeled points in general position in d\mathbb{R}^{d} is at most

nd(d+1)n.n^{d(d+1)n}.

We note that the number of order types is a technically a refinement of the number of embeddings which induce the same Radon partitions on every (d+2)(d+2)-subset of nn points. If we have only d+2d+2 points in general position we have a distinct Radon partition, (A,B)(A,B), but we can change the order type by permuting the points relabeling the points in AA and the points in BB. However the direction that we need is that the order type determines the Radon partition discussed above. So we have the following lemma as an immediate corollary to Theorem 11.

Lemma 12.

For d1d\geq 1 the number of nonequivalent, general position embeddings of nn vertices in d\mathbb{R}^{d} is at most

nd(d+1)n.n^{d(d+1)n}.

For the last piece of the proof for the dense side of the phase transition we bound the probability that for any fixed embedding of the vertices of XX we have no Radon matches.

Lemma 13.

Fix d1d\geq 1 and let π:[n]2d\pi:[n]\rightarrow\mathbb{R}^{2d} be a general position embedding for nn vertices in 2d\mathbb{R}^{2d}. For d+1α¯vd+1>1d+1-\operatorname{\overline{\alpha}}\cdot v_{d+1}>1 the probability that XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) has no Radon matches under the mapping into 2d\mathbb{R}^{2d} induced by π\pi is at most

exp(n1+ε)\exp(-n^{1+\varepsilon})

for some ε>0\varepsilon>0 that depends only on α¯\operatorname{\overline{\alpha}} and dd.

For the proof we will make use of the following form of Janson’s equality from Theorem 23.13 of [7]:

Theorem 14 (Janson’s Inequality).

Let RR be a random subset of [N][N] such that for each s[N]s\in[N], qs(0,1)q_{s}\in(0,1) denotes the probability that sNs\in N is included in RR. Let D1,,DnD_{1},...,D_{n} be a family of nn subsets of RR. Suppose that SnS_{n} is a sum of indicator random variables Sn=I1++InS_{n}=I_{1}+\cdots+I_{n} where IiI_{i} is the indicator for the event DiD_{i}. Write iji\sim j is DiDjD_{i}\cap D_{j}\neq\emptyset let

Δ¯={i,j}:ij𝔼(IiIj)\overline{\Delta}=\sum_{\{i,j\}:i\sim j}\mathbb{E}(I_{i}I_{j})

then

Pr(Sn=0)exp((𝔼(Sn))22Δ¯).\Pr(S_{n}=0)\leq\exp\left(\frac{-(\mathbb{E}(S_{n}))^{2}}{2\overline{\Delta}}\right).

In our case NN is all the faces of the simplex on nn vertices other than the vertices and RR is the random complex in X(n;nα¯)X(n;n^{-\operatorname{\overline{\alpha}}}). We might hesitate for a moment regarding the independence assumption as the faces are not included independently in XX. However we can instead associate to σ\sigma in the simplex on nn vertices qσ=nα|σ|1q_{\sigma}=n^{-\alpha_{|\sigma|-1}}, in this way each face is set to be either on or off independently and then the complex XX consists of those faces that are switched on and have all their subfaces switched on as well. Each DiD_{i} will be a Radon match which corresponds to some collection of faces all switched on independently.

Proof of Lemma 13.

Let YY be the random variable counting the number of evenly split Radon matches in XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) under π\pi, i.e. Radon matches for Radon partitions with d+1d+1 vertices in each part. By linearity of expectation and Corollary 10 we have

𝔼(Y)1(2d+3)(2d+2)2d+2n2d+2n2α¯vd+1=Θ(n2((d+1)α¯vd+1)).\mathbb{E}(Y)\geq\frac{1}{(2d+3)(2d+2)^{2d+2}}n^{2d+2}n^{-2\operatorname{\overline{\alpha}}\cdot v_{d+1}}=\Theta(n^{2((d+1)-\operatorname{\overline{\alpha}}\cdot v_{d+1})}).

Under our assumptions then 𝔼(Y)\mathbb{E}(Y)\rightarrow\infty. Now we use Janson’s inequality to bound the probability that Y=0Y=0. Observe that YY is a sum of indicator random variables 1(A,B)\textbf{1}_{(A,B)} where AA, BB is an evenly split Radon partition of ABA\cup B coming from π\pi, with AA the lexicographically smaller of the two sets with respect to some ordering, and 1(A,B)\textbf{1}_{(A,B)} is the indicator random variable for the event that the simplex on AA and the simplex on BB are both included in XX. If (A,B)(A,B) and (A,B)(A^{\prime},B^{\prime}) don’t share any edges, i.e. if |AB||A\cap B^{\prime}|, |AA||A\cap A^{\prime}|, |BA||B\cap A^{\prime}|, and |BB||B\cap B^{\prime}| are all of size at most 1, then 1(A,B)\textbf{1}_{(A,B)} and 1(A,B)\textbf{1}_{(A^{\prime},B^{\prime})} are independent. To apply Janson’s inequality then we let

Δ¯={(A,B),(A,B)(AB),(AB) share an edge}𝔼(1(A,B)1(A,B)),\overline{\Delta}=\sum_{\{(A,B),(A^{\prime},B^{\prime})\mid(A\cup B),(A^{\prime}\cup B^{\prime})\text{ share an edge}\}}\mathbb{E}(\textbf{1}_{(A,B)}\textbf{1}_{(A^{\prime},B^{\prime})}),

and we compute an upper bound on Δ¯\overline{\Delta}. If (A,B)(A,B) and (A,B)(A^{\prime},B^{\prime}), with ABA\cap B and ABA^{\prime}\cap B^{\prime} both empty, share at least an edge then the intersection of the two pairs is a disjoint union of up to four nonempty sets ABA\cap B^{\prime}, AAA\cap A^{\prime}, BAB\cap A^{\prime}, and BBB\cap B^{\prime}. Let m1,m2,m3,m4m_{1},m_{2},m_{3},m_{4} denote the sizes of these sets respectively, note that at least one of the mim_{i}’s is at least 2 and all of them are at most d+1d+1. Given m1m_{1}, m2m_{2}, m3m_{3}, and m4m_{4} the probability that A,B,A,BA,B,A^{\prime},B^{\prime} are all included in XX is

𝔼(1(A,B)1(A,B))=n4α¯vd+1(i=14α¯vmi)\mathbb{E}(\textbf{1}_{(A,B)}\textbf{1}_{(A^{\prime},B^{\prime})})=n^{-4\operatorname{\overline{\alpha}}\cdot v_{d+1}-(\sum_{i=1}^{4}-\operatorname{\overline{\alpha}}\cdot v_{m_{i}})}

as the simplices in the intersection are counted exactly twice.

Now for each choice of m1,m2,m3,m4m_{1},m_{2},m_{3},m_{4} the number of way to pick for sets of d+1d+1 vertices AA, BB, AA^{\prime}, and BB^{\prime} so that |AB|=|AB|=0|A\cap B|=|A^{\prime}\cap B^{\prime}|=0 so that |AB|=m1|A\cap B^{\prime}|=m_{1}, |AA|=m2|A\cap A^{\prime}|=m_{2}, |BA|=m3|B\cap A^{\prime}|=m_{3}, and |BB|=m4|B\cap B^{\prime}|=m_{4} is at most

O(n4(d+1)m1m2m3m4).O(n^{4(d+1)-m_{1}-m_{2}-m_{3}-m_{4}}).

It follows that

Δ¯=O(maxm1,m2,m3,m4n4((d+1)α¯vd+1)i=14(miα¯vmi))\overline{\Delta}=O\left(\max_{m_{1},m_{2},m_{3},m_{4}}n^{4((d+1)-\operatorname{\overline{\alpha}}\cdot v_{d+1})-\sum_{i=1}^{4}(m_{i}-\operatorname{\overline{\alpha}}\cdot v_{m_{i}})}\right)

where the maximum is taken over all choices of m1,m2,m3,m4m_{1},m_{2},m_{3},m_{4} so that each is at most d+1d+1 and at least one of them is more than 1. By Janson’s inequality then,

Pr(Y=0)exp(Θ(nminm1,m2,m3,m4{i=14(miα¯vmi)})).\Pr(Y=0)\leq\exp\left(-\Theta\left(n^{\min_{m_{1},m_{2},m_{3},m_{4}}\{\sum_{i=1}^{4}(m_{i}-\operatorname{\overline{\alpha}}\cdot v_{m_{i}})\}}\right)\right).

We now need to verify that

minm1,m2,m3,m4{i=14(miα¯vmi)}>1\min_{m_{1},m_{2},m_{3},m_{4}}\left\{\sum_{i=1}^{4}(m_{i}-\operatorname{\overline{\alpha}}\cdot v_{m_{i}})\right\}>1

This follows though since at least one of the mim_{i}’s is always at least 2 and they are all always at most d+1d+1. So by the second part of Lemma 2 for all allowable choices of m1,m2,m3,m4m_{1},m_{2},m_{3},m_{4}, i=14(miα¯vmi)>1\sum_{i=1}^{4}(m_{i}-\operatorname{\overline{\alpha}}\cdot v_{m_{i}})>1. As there is a constant bound on the possible values for m1,,m4m_{1},...,m_{4} for dd fixed, we have that there is some ε\varepsilon depending on dd so that

Pr(Y=0)exp(n1+ε).\Pr(Y=0)\leq\exp(n^{1+\varepsilon}).

Proof of dense side of Theorem 1.

Set d1d\geq 1. By Lemma 13 there is ε>0\varepsilon>0 so that for any embedding π\pi of the vertices of XX(n;nα¯)X\sim X(n;n^{-\operatorname{\overline{\alpha}}}) into 2d\mathbb{R}^{2d}, the probability that XX has no Radon matches with π\pi is at most exp(n1+ε)\exp(-n^{1+\varepsilon}). Taking a union bound over all nonequivalent embeddings of the vertices from Lemma 12 we have that the probability that there is an embedding of the vertices with no Radon match is at most

n2d(2d+1)nen1+ε=o(1).n^{2d(2d+1)n}e^{-n^{1+\varepsilon}}=o(1).

So with high probability every embedding of XX into 2d\mathbb{R}^{2d} has a Radon match, therefore XX is not linearly embeddable in 2d\mathbb{R}^{2d}. ∎

5 Concluding remarks

Theorem 1 establishes a threshold result for linear embeddings in even dimensions. Two important special cases are for the Linial–Meshulam–Wallach model where we see that (up to the right exponent) p=ndp=n^{-d} is the threshold for YYd(n,p)Y\sim Y_{d}(n,p) to be embeddable in 2d\mathbb{R}^{2d} and for the clique complex model where we see that p=n2/(d+1)p=n^{-2/(d+1)} is the threshold for embeddability of XX(n,p)X\sim X(n,p) in 2d\mathbb{R}^{2d}. Up to the exponent, this matches the threshold for XX to be at least (d+1)(d+1)-dimensional even though we found obstructions to embeddability already in the dd-skeleton.

PL-embeddings of random complexes have also been studied in the past, and together with a result of Wagner, Theorem 1 gives a new, probabilistic proof of the known result that PL-embeddablility is a strictly weaker notion than linearly-embeddability for complexes of dimension larger than 1. Let Yd(n,p)Y_{d}(n,p) denote the dd-dimensional Linial–Meshulam–Wallach model. Wagner’s result on PL-embeddability is the following

Theorem 15 (Theorem 2 of [18]).

The threshold for PL-embeddability of YYd(n,p)Y\sim Y_{d}(n,p) into 2d\mathbb{R}^{2d} is at p=Θ(1/n)p=\Theta(1/n).

From this and Theorem 1 the following is immediate.

Theorem 16.

For 1<α<d1<\alpha<d with high probability YYd(n,nα)Y\sim Y_{d}(n,n^{-\alpha}) is PL-embeddable in 2d\mathbb{R}^{2d}, but not linearly embeddable in 2d\mathbb{R}^{2d}.

Perhaps the most natural open question is to establish linear embedding thresholds for odd dimensions. An earlier version of this paper claimed such a threshold, however upon writing this revision an error was found for the proof in the odd dimensional case. For showing nonembeddability in 2d1\mathbb{R}^{2d-1} for sparser complexes than those that fail to embed in 2d\mathbb{R}^{2d}, some new ideas would be required. To explain why we’ll examine the case of embedding the random clique complex in 3\mathbb{R}^{3}.

For 2/3<α<12/3<\alpha<1 we know that X(n,nα)X(n,n^{-\alpha}) embeds in 4\mathbb{R}^{4} but not in 2\mathbb{R}^{2}. If we take α=1δ\alpha=1-\delta for some small δ\delta and try to use the same proof based on Radon matches to rule out embeddability into 3\mathbb{R}^{3}, we could still show that on average a fixed placement of the vertices in 3\mathbb{R}^{3} will have Θ(n54α)\Theta(n^{5-4\alpha}) Radon matches, so that part of the proof would still be consistent with the even dimensional case. The problem, however, is that we cannot get an upper bound on the probability of no Radon matches of exp(n1+ε)\exp(-n^{1+\varepsilon}) in this case, and this was used in the even dimensional case to take a union bound over the approximately nnn^{n} order types. The reason we can’t prove such a bound is that if our random complex happens to have no triangles at all for α=1δ\alpha=1-\delta then the complex is 1-dimensional and so it will embed in 3\mathbb{R}^{3} generically. The average number of triangles is n33α=n3δn^{3-3\alpha}=n^{-3\delta}, so the probability there are no triangles should be roughly exp(n3δ)\exp(-n^{3\delta}) (there would be some things to check here since the triangles aren’t included independently, but this is just a sketch of the idea). So there is a lower bound that exceed exp(n1+ε)\exp(-n^{1+\varepsilon}) until δ>1/3\delta>1/3, but at that point we are at the nonembeddability threshold for 4\mathbb{R}^{4} anyway.

Other remaining open problems include establishing sharp threshold results for linear embeddability in even dimensions; the main result here establishes thresholds for linear embeddability only up to the right exponents. Additionally, now that we know that the Linial–Meshulam–Wallach model has different thresholds for PL-embeddability and for linear embeddability, it would be interesting then to determine PL-embeddability thresholds for entire the multiparameter model.

Acknowledgments

The author thanks Florian Frick for discussions leading to the central question answered in this paper and for helpful comments on an early draft and Boris Bukh for pointing out an error in a previous version.

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