Homology of Multi-Parameter Random Simplicial Complexes

Abstract

We consider a multi-parameter model for randomly constructing simplicial complexes that interpolates between random clique complexes and Linial–Meshulam random k-dimensional complexes. Unlike these models, multi-parameter complexes exhibit nontrivial homology in numerous dimensions simultaneously. We establish upper and lower thresholds for the appearance of nontrivial cohomology in each dimension and characterize the behavior at criticality.

Appendices

Appendix A: Boundaries of Simplices

Proof of Lemma 5.2

We consider the case

$$\begin{aligned} 1 \le \sum _{l=1}^k \left( {\begin{array}{c}k\\ l\end{array}}\right) \, \alpha _l \end{aligned}$$

where (from (12) in Sect. 5) we have ${\mathbb {E}}\, [M_{k-1} ] \rightarrow \infty$. By Chebyshev’s inequality,

$$\begin{aligned} {\mathbb {P}} \Big [ \big |M_{k-1} - {\mathbb {E}}\,[M_{k-1}] \big |\ge {\mathbb {E}}\,[M_{k-1}] \Big ] \le \frac{\text {Var}\,[M_{k-1}]}{{\mathbb {E}}\,[M_{k-1}]^2}. \end{aligned}$$

Thus if we can show $\text {Var}\, [M_{k-1}] = o( {\mathbb {E}}\, [M_{k-1}]^2 )$, then we may conclude

$$\begin{aligned} {\mathbb {P}}[ M_{k-1} > 0 ] \rightarrow 1. \end{aligned}$$

Considering $M_{k-1}$ as a sum of indicator random variables,

$$\begin{aligned} \text {Var}\,[M_{k-1}]&\le {\mathbb {E}}\,[M_{k-1}] + \sum _{i,j \in \left( {\begin{array}{c}[n]\\ k\end{array}}\right) } \text {Cov}[1_{A_i}, 1_{A_j}] \\&={\mathbb {E}}\,[M_{k-1}] + \sum _{i,j \in \left( {\begin{array}{c}[n]\\ k\end{array}}\right) } \left( {\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i]\,{\mathbb {P}}\,[A_j] \right) . \end{aligned}$$

Clearly ${\mathbb {E}}\,[M_{k-1}] = o ( {\mathbb {E}}\,[M_{k-1}]^2 )$, to handle the sum we consider pairs $i, j \in \left( {\begin{array}{c}[n]\\ k+1\end{array}}\right)$ and break them into three cases depending on $I = |i \cap j |$. To make the calculations more readable we introduce some useful notation, defining $\eta _k$ to be

$$\begin{aligned} \eta _k = (1-p_k) \prod _{l=1}^{k-1} p_l^{\left( {\begin{array}{c}k+1\\ l+1\end{array}}\right) }, \end{aligned}$$

the probability that our complex contains the unfilled boundary of a specific k-simplex. We define $\gamma _k$ as

$$\begin{aligned} \gamma _k = \prod _{l=1}^k p_l^{\left( {\begin{array}{c}k\\ l\end{array}}\right) }, \end{aligned}$$

the probability that a fixed $(k-1)$-face and vertex form a k-simplex.

1.1 A.1: $I = 0$

We begin by calculating ${\mathbb {P}}\,[A_i \cap A_j]$. The probability that both boundaries are in our complex but unfilled is $\eta _k^2$. By inclusion–exclusion principles the probability that neither $\sigma _i$ nor $\sigma _j$, the associated first $(k-1)$-faces of these subcomplexes, form a k-simplex with a vertex outside of $i \cup j$ is $1 - 2\gamma _k + \gamma _k^2$, and there are $n-2k-2$ such vertices. Finally, we must have that no k-face is formed between one subcomplex and a single vertex of the other. While this probability can be explicitly calculated, every term that is not 1 will contain a copy of $\gamma _k$, so this probability is $1 - O(\gamma _k)$. Thus

$$\begin{aligned} {\mathbb {P}}\,[A_i \cap A_j] = \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-2k-2} \left( 1- O(\gamma _k) \right) , \end{aligned}$$

and by (11) in Sect. 5 we know

$$\begin{aligned} {\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j]&= \left( \eta _k \left( 1-\gamma _k\right) ^{n-k-1} \right) ^2 \\&= \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2\right) ^{n-k-1} \\&= \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-2k-2} \left( 1-2\gamma _k +\gamma _k^2 \right) ^{k+1} \\&= \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-2k-2} \left( 1 - O(\gamma _k) \right) . \end{aligned}$$

Thus

$$\begin{aligned} {\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i]\, {\mathbb {P}}\,[A_j] = \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-2k-2} O(\gamma _k) \end{aligned}$$

and there are $O(n^{2k+2} )$ such pairs i, j, so the overall contribution of these pairs to our sum is

$$\begin{aligned} S_0&= O\left( n^{2k+2} \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-2k-2} \gamma _k \right) \\&= O\left( n^{2k+2} \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-k-1} \gamma _k \right) . \end{aligned}$$

The second equality holds by restricting our consideration to $n > k$, then $\gamma _k \le n^{-1} < k^{-1}$ and there is some $C > 0$ such that

$$\begin{aligned} (1-2\gamma _k + \gamma _k^2)^{k+1}> (1-2\gamma _k)^{k+1}> (1-2k^{-1})^k > C, \end{aligned}$$

so removing this term does not affect our big-O calculations.

Since

$$\begin{aligned} {\mathbb {E}}\, [M_{k-1}]^2 = \left( {\begin{array}{c}n\\ k+1\end{array}}\right) ^2 \eta _k^2 (1-\gamma _k)^{2(n-k-1)} = O \left( n^{2k+2} \eta _k^2 (1-2\gamma _k + \gamma _k^2)^{n-k-1} \right) \end{aligned}$$

and $\gamma _k \rightarrow 0$ we conclude

$$\begin{aligned} \frac{S_0}{{\mathbb {E}}\,[M_{k-1}]^2} = O(\gamma _k) = o(1). \end{aligned}$$

Hence the contribution of these pairs to the variance is seen to be $o({\mathbb {E}}\,[M_{k-1}]^2 )$.

1.2 A.2: $I = 1$

The probability of both subcomplexes being in X is again $\eta _k^2$ since the two do not share a face of dimension greater than 0. We again use inclusion–exclusion to calculate the probability that $\sigma _i$ and $\sigma _j$ do not form k-simplices with another vertex. However, these faces may or may not both contain the shared vertex: if they do not then the calculations are identical to above, so we assume the alternative. In this case the two k-faces formed with some new vertex would share a common edge. So the probability is $1 - 2\gamma _k + \gamma _k^2 p_1^{-1}$ for each of the $n-2k-1$ remaining vertices. Similarly, the probability we do not have a k-face consisting of $\sigma _i$ or $\sigma _j$ and a vertex in $i \bigtriangleup j$ is $1-O(\gamma _k p_1^{-1})$. We then calculate ${\mathbb {P}}\, [A_i \cap A_j]$ to be

$$\begin{aligned} {\mathbb {P}}\,[A_i \cap A_j] = \eta _k^2 \left( 1- 2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-2k-1} \left( 1 - O(\gamma _k p_1^{-1})\right) . \end{aligned}$$

Before calculating ${\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j]$, we observe

$$\begin{aligned} 1-2\gamma _k +\gamma _k^2&= \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) \frac{1-2\gamma _k +\gamma _k^2}{1-2\gamma _k + \gamma _k^2 p_1^{-1}} \\&= \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) \left( 1 - \frac{\gamma _k^2(p_1^{-1} -1)}{1-2\gamma _k + \gamma _k^2 p_1^{-1}} \right) \\&= \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) \left( 1 - O(\gamma _k^2 p_1^{-1})\right) . \end{aligned}$$

The last equality holds by an identical argument to the one in the first case: we can bound $1-2\gamma _k+\gamma _k^2p_1^{-1}$, and consequently its inverse, from above and below by constants. We use this to calculate

$$\begin{aligned} {\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j]&= \eta _k^2 \left( 1-2\gamma _k +\gamma _k^2 \right) ^{n-k-1} \\&= \eta _k^2 \left[ \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) \left( 1 - O (\gamma _k^2 p_1^{-1} ) \right) \right] ^{n-k-1} \\&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-k-1} \left( 1 - O(\gamma _k^2 p_1^{-1}) \right) ^{n-k-1}. \end{aligned}$$

But since $\gamma _k < n^{-1}$ we have

$$\begin{aligned} \left( 1 - O(\gamma _k^2 p_1^{-1})\right) ^{n-k-1}&= 1 - O( n \gamma _k^2 p_1^{-1}) \\&= 1 - O(\gamma _k p_1^{-1}). \end{aligned}$$

We calculate

$$\begin{aligned} {\mathbb {P}}\,[A_i]\, {\mathbb {P}}\,[A_j]&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-k-1} \left( 1 - O(\gamma _k p_1^{-1}) \right) \\&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-2k-1} \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^k \left( 1 - O(\gamma _k p_1^{-1}) \right) \\&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-2k-1} \left( 1-O(\gamma _k) \right) \left( 1 - O(\gamma _k p_1^{-1}) \right) \\&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-2k-1} \left( 1 - O(\gamma _k p_1^{-1}) \right) . \end{aligned}$$

Therefore

$$\begin{aligned} {\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i]\, {\mathbb {P}}\,[A_j] = \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 p_1^{-1}\right) ^{n-2k-1}O(\gamma _k p_1^{-1} ) \end{aligned}$$

with $O(n^{2k+1})$ such pairs i, j, making the total contribution of these pairs to the variance

$$\begin{aligned} S_1&= O \left( n^{2k-1} \eta _k^2 (1-2\gamma _k + \gamma _k^2 p_1^{-1})^{n-2k+1}\gamma _k p_1^{-1}) \right) \\&= O \left( n^{2k-1} \eta _k^2 (1-2\gamma _k + \gamma _k^2 p_1^{-1})^{n-k-1}\gamma _k p_1^{-1}) \right) . \end{aligned}$$

As before, the second equality follows from bounding $(1-2\gamma _k + \gamma _k^2 p_1^{-1})^{k-1}$ by constants on either side.

Since

$$\begin{aligned} {\mathbb {E}}\, [M_{k-1}]^2 = O \left( n^{2k+2} \eta _k^2 (1-2\gamma _k + \gamma _k^2)^{n-k-1} \right) \end{aligned}$$

it follows that

$$\begin{aligned} \frac{S_1}{{\mathbb {E}}\, [M_{k-1}]^2}&= O \left( \frac{(1-2\gamma _k + \gamma _k^2 p_1^{-1})^{n-k-1}\gamma _k p_1^{-1}}{n (1-2\gamma _k + \gamma _k^2)^{n-k-1}} \right) \\&= O \left( \frac{\gamma _k p_1^{-1}}{n} \left( 1 + \frac{\gamma _k^2(p_1^{-1}-1)}{1-2\gamma _k +\gamma _k^2} \right) ^{n-k-1} \right) \\&= O \left( \frac{\gamma _k p_1^{-1}}{n} \left( 1 + \frac{\gamma _k^2p_1^{-1}}{1-2\gamma _k} \right) ^{n-k-1} \right) . \end{aligned}$$

We proceed by bounding the right term by a constant.

$$\begin{aligned} \left( 1 + \frac{\gamma _k^2p_1^{-1}}{1-2\gamma _k} \right) ^{n-k-1}&\le \exp \left( (n-k-1)\, \frac{\gamma _k^2p_1^{-1}}{1-2\gamma _k} \right) \\&\le \exp \left( \frac{ n \gamma _k^2 p_1^{-1}}{1-k} \right) \\&\le e^{1/(1-k)}. \end{aligned}$$

Then

$$\begin{aligned} \frac{S_1}{{\mathbb {E}}\, [ M_{k-1} ] ^2}&= O \left( \frac{\gamma _k p_1^{-1}}{n} \right) \\&= O\left( \frac{1}{n}\right) \\&= o(1). \end{aligned}$$

Thus the contribution of these pairs is also $o({\mathbb {E}}\, [ N_{k-1} ]^2 )$, as desired.

1.3 A.3: $2 \le I \le k$

In this final case the probability of the two subcomplexes being contained is $\eta _k^2 \eta _I^{-1}$ where $\eta _I := \prod _{l=1}^{I-1} p_l^{\left( {\begin{array}{c}I\\ l+1\end{array}}\right) }$. The $\eta _I^{-1}$ accounts for all faces common to i and j, which would otherwise be counted twice. We note $\sigma _i$ and $\sigma _j$ share between $I-2$ and I vertices, and assuming maximal overlap provides an upper bound on ${\mathbb {P}}\,[A_i \cap A_j]$. Hence the probability that neither will form a k-simplex with some other vertex is at most $\left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k-2+I}$ with $\gamma _I := \prod _{l=1}^{I} p_l^{\left( {\begin{array}{c}I\\ l\end{array}}\right) }$. The probability of one not forming a k-simplex with one vertex of the other is $1-O(\gamma _k \gamma _I^{-1} )$. We see

$$\begin{aligned} {\mathbb {P}}\,[A_i \cap A_j] = \eta _k^2 \eta _I^{-1} \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1} \right) ^{n-2k-2+I} \left( 1-O(\gamma _k \gamma _I^{-1}) \right) . \end{aligned}$$

Just as in the previous case,

$$\begin{aligned} 1 - 2\gamma _k + \gamma _k^2&= \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) \left( 1 - O(\gamma _k^2 \gamma _I^{-1}) \right) . \end{aligned}$$

We now calculate

$$\begin{aligned} {\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j]&= \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2 \right) ^{n-k-1} \\&= \eta _k^2 \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-k-1} \left( 1 - O( \gamma _k^2 \gamma _I^{-1}) \right) ^{n-k-1} \\&= \eta _k^2 \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k-2+I} \left( 1 - O(\gamma _k \gamma _I^{-1}) \right) . \end{aligned}$$

It then follows that

$$\begin{aligned} \frac{{\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j]}{{\mathbb {P}}\,[A_i \cap A_j]}&= \frac{ \eta _k^2 \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k-2+I} \left( 1 - O(\gamma _k \gamma _I^{-1})\right) }{ \eta _k^2 \eta _I^{-1} \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k-2+I} \left( 1-O(\gamma _k \gamma _I^{-1}) \right) } \\&= O(\eta _I). \end{aligned}$$

Thus if $\eta _I \ne 1$ then ${\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i] \,{\mathbb {P}}\,[A_j] = (1 - o(1)) \,{\mathbb {P}}\,[A_i \cap A_j]$, and otherwise ${\mathbb {P}}\,[A_i \cap A_j] - {\mathbb {P}}\,[A_i]\, {\mathbb {P}}\,[A_j] = \eta _k^2 \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k+I} O(\gamma _k \gamma _I^{-1})$. There are $O(n^{2k+2-I})$ such pairs, so their total contribution to the variance is either

$$\begin{aligned} S_I&= O \left( n^{2k+2-I} \eta _k^2 \eta _I^{-1} \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-2k-2+I} \right) \\&= O \left( n^{2k+2-I} \eta _k^2 \eta _I^{-1} \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-k-1} \right) , \end{aligned}$$

or

$$\begin{aligned} S_I = O \left( n^{2k+2-I} \eta _k^2 \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-k-1} \gamma _k \gamma _I^{-1} \right) . \end{aligned}$$

In the first case we have

$$\begin{aligned} \frac{S_I}{{\mathbb {E}}\, [ M_{k-1}]^2}&= O\left( \frac{n^{2k+2-I} \eta _k^2 \eta _I^{-1} \left( 1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}\right) ^{n-k-1} }{n^{2k+2} \eta _k^2 \left( 1-2\gamma _k + \gamma _k^2\right) ^{n-k-1}} \right) \\&= O\left( \frac{\eta _I^{-1}}{n^I} \left( \frac{1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}}{1 - 2\gamma _k + \gamma _k^2} \right) ^{n-k-1} \right) . \end{aligned}$$

Just as before, the right-most term can be bounded above by a constant. We note

$$\begin{aligned} \sum _{l=1}^{I-1} \alpha _l \left( {\begin{array}{c}I\\ l+1\end{array}}\right)< \frac{I}{k}\, \sum _{l=1}^{k-1} \alpha _l \left( {\begin{array}{c}k\\ l+1\end{array}}\right) < \frac{I}{k}\, k = I \end{aligned}$$

and conclude

$$\begin{aligned} \frac{S_I}{{\mathbb {E}}\, [ M_{k-1}]^2}&= O \left( \frac{\eta _I^{-1}}{n^I} \right) \\&= O \left( n^{-I + \sum _{l=1}^{I-1} \alpha _l \left( {\begin{array}{c}I\\ l+1\end{array}}\right) } \right) \\&= o(1). \end{aligned}$$

In the second case we have

$$\begin{aligned} \frac{S_I}{{\mathbb {E}}\, [ M_{k-1}]^2}&= O\left( \frac{\gamma _k \gamma _I^{-1}}{n^I} \left( \frac{1 - 2\gamma _k + \gamma _k^2 \gamma _I^{-1}}{1 - 2\gamma _k + \gamma _k^2} \right) ^{n-k-1} \right) \\&= O\left( \frac{\gamma _k \gamma _I^{-1}}{n^I}\right) \\&= O\left( n^{-I}\right) \\&= o(1). \end{aligned}$$

Thus $S_I = o({\mathbb {E}} \,[ M_{k-1}]^2)$ for $2 \le I \le k$. We therefore have that ${\mathbb {E}}\,[M_{k-1}^2] = o ( {\mathbb {E}}\,[M_{k-1}]^2)$, and our result that $M_{k-1} \sim {\mathbb {E}}\,[M_{k-1}]$ follows by Chebyshev’s Inequality. $\square$

Appendix B: Factorial Moments of Maximal Faces

Proof of Lemma 7.1

Similarly to previous second moment calculations:

$$\begin{aligned} {\mathbb {E}}\,[N_{k-1}^2]&= \left( {\begin{array}{c}n\\ k\end{array}}\right) \sum _{m=0}^k \Bigg [ \left( {\begin{array}{c}k\\ m\end{array}}\right) \left( {\begin{array}{c}n-k\\ k-m\end{array}}\right) \left( \prod _{i=1}^{k-1} p_i^{2\left( {\begin{array}{c}k\\ i+1\end{array}}\right) - \left( {\begin{array}{c}m\\ i+1\end{array}}\right) }\right) \\&\quad \times \left( 1- 2\prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } + \prod _{i=1}^k p_i^{2\left( {\begin{array}{c}k\\ i\end{array}}\right) - \left( {\begin{array}{c}m\\ i\end{array}}\right) } \right) ^{n-2k+m} \left( 1- o(1)\right) \Bigg ]. \end{aligned}$$

We can simplify this slightly to

$$\begin{aligned} {\mathbb {E}}\,[N_{k-1}^2]&= \left( 1+o(1) \right) \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \prod _{i=1}^{k-1} p_i^{2\left( {\begin{array}{c}k\\ i+1\end{array}}\right) } \right) \sum _{m=0}^k \Bigg [ \left( {\begin{array}{c}k\\ m\end{array}}\right) \left( {\begin{array}{c}n-k\\ k-m\end{array}}\right) \left( \prod _{i=1}^{m-1}p_i^{- \left( {\begin{array}{c}m\\ i+1\end{array}}\right) } \right) \\&\quad \times \left( 1- 2\prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } + \prod _{i=1}^k p_i^{2\left( {\begin{array}{c}k\\ i\end{array}}\right) - \left( {\begin{array}{c}m\\ i\end{array}}\right) } \right) ^{n-2k+m} \Bigg ]. \end{aligned}$$

Pulling out the $m=0$ summand, asymptotically

$$\begin{aligned}&\left( {\begin{array}{c}n\\ k\end{array}}\right) \left( {\begin{array}{c}n-k\\ k\end{array}}\right) \left( \prod _{i=1}^{k-1} p_i^{2\left( {\begin{array}{c}k\\ i+1\end{array}}\right) } \right) \left( 1- \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) }\right) ^{2(n-2k)} \\&\quad = \left( 1 + o(1)\right) \left( \left( {\begin{array}{c}n\\ k\end{array}}\right) \left( \prod _{i=1}^{k-1} p_i^{\left( {\begin{array}{c}k\\ i+1\end{array}}\right) } \right) \left( 1- \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) }\right) ^{n-k} \right) ^2 \\&\quad = ( 1 + o(1) ) \,{\mathbb {E}}\,[ N_{k-1}]^2 . \end{aligned}$$

Meanwhile, the $m= k$ term is seen to be ${\mathbb {E}}\,[N_{k-1}]$. We claim the $k-1$ other summands do not contribute in the limit. For a fixed $m = 1, \ldots , k-1$ let $d_m < 1$ be some constant value such that

$$\begin{aligned} d_m > \text {max} \left\{ 1 - \frac{m(m-1)}{k(k-1)} , 1 - \frac{m - \sum _{i=1}^{m-1} \alpha _i \left( {\begin{array}{c}m\\ i+1\end{array}}\right) }{k - \sum _{i=1}^{k-1} \alpha _i \left( {\begin{array}{c}k\\ i+1\end{array}}\right) } \right\} . \end{aligned}$$

Both fraction terms are between 0 and 1, so such a $d_m$ exists. For sufficiently large n we have u

$$\begin{aligned} 1- 2\prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } + \prod _{i=1}^k p_i^{2\left( {\begin{array}{c}k\\ i\end{array}}\right) - \left( {\begin{array}{c}m\\ i\end{array}}\right) }&= 1 - \left( 2 - \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) - \left( {\begin{array}{c}m\\ i\end{array}}\right) } \right) \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } \\&\le 1 -(1+ d_m) \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) }. \end{aligned}$$

Thus there exists a constant D such that

$$\begin{aligned} \left( 1- 2\prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } + \prod _{i=1}^k p_i^{2\left( {\begin{array}{c}k\\ i\end{array}}\right) - \left( {\begin{array}{c}m\\ i\end{array}}\right) } \right) ^{n-2k+m}&\le \left( 1 -(1+ d_m) \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } \right) ^{n-2k+m} \\&\le D e^{-n(1+d_m) \left( \prod _{i=1}^k p_i^{\left( {\begin{array}{c}k\\ i\end{array}}\right) } \right) } \\&= D e^{-(1+d_m) (\rho _1 \log n +\frac{k-1}{2} \log \log n + c)} \\&= D n^{-(1+d_m) \rho _1} (\log n)^{-(1+d_m) \frac{k-1}{2}} e^{-(1+d_m)c}. \end{aligned}$$

Then by our construction of $d_m$,

$$\begin{aligned} n^{-(1+d_m) \rho _1} = o \left( n^{-2k + m + \sum _{i=1}^{k-1} \alpha _i \left( {\begin{array}{c}k\\ i\end{array}}\right) - \sum _{i=1}^{m-1} \alpha _i \left( {\begin{array}{c}m\\ i\end{array}}\right) } \right) \end{aligned}$$

and

$$\begin{aligned} \left( \log n \right) ^{-(1+d_m) \frac{k-1}{2}} = o \left( (\log n ) ^{ -(k-1) + \frac{m(m-1)}{2k}} \right) . \end{aligned}$$

It follows that the corresponding summand is bounded by

$$\begin{aligned} D n^{2k - m} \prod _{i=1}^{k-1} p_i^{2 \left( {\begin{array}{c}k\\ i+1\end{array}}\right) - \left( {\begin{array}{c}m\\ i+1\end{array}}\right) } n^{-(1+d_m) \rho _1} (\log n)^{-(1+d_m) \frac{k-1}{2}} = o(1), \end{aligned}$$

thereby contributing nothing as $n \rightarrow \infty$. Therefore,

$$\begin{aligned} {\mathbb {E}}\,[( N_{k-1} )_2 ] = {\mathbb {E}}\,[ N_{k-1}^2] - {\mathbb {E}}\,[N_{k-1}] = {\mathbb {E}}\,[N_{k-1}]^2 (1-o(1)) \rightarrow {\mathbb {E}}\,[N_{k-1}]^2 \end{aligned}$$

as $n \rightarrow \infty$. We will now establish a similar result for each factorial moment.

We direct our attention to the lth factorial moment of $N_{k-1}$, assuming that ${\mathbb {E}}\,[(N_{k-1})_{j}] \rightarrow {\mathbb {E}}\,[N_{k-1}]^j$ for all $j<l$. Using the notation from Sect. 3 we have

$$\begin{aligned} {\mathbb {E}}\left[ N_{k-1}^l \right]&= {\mathbb {E}}\left[ \left( \sum _{\sigma \in \left( {\begin{array}{c}[n]\\ k\end{array}}\right) } 1_{C_\sigma } \right) ^l \right] \\&= \sum _{\sigma _1, \ldots , \sigma _l \in \left( {\begin{array}{c}[n]\\ k\end{array}}\right) } {\mathbb {P}}\left[ C_{\sigma _1} \cap \dots \cap C_{\sigma _l} \right] . \end{aligned}$$

We break up this sum into two parts: where no two $\sigma _i$’s are identical and where such two $\sigma _i$ are the same. Considering the first case, an identical argument for $l=2$ tells us the only summand contributing in the limit corresponds to when no two faces share any vertices, and this term converges to ${\mathbb {E}}\,[N_{k-1}]^l$.

Moving on to the second case, we let s(l, j) and S(l, j) denote Stirling numbers of the first and second kind, respectively. There are S(l, j) ways to break our $\sigma _i$ up into j groups where all faces in a group are the same. Moreover, for any such configuration into j groups, the corresponding contribution to ${\mathbb {E}}\,[N_{k-1}^l]$ would be ${\mathbb {E}}\,[N_{k-1}^j]$. We begin by pulling out $S(l,l-1)=-s(l,l-1)$ copies of ${\mathbb {E}}\,[N_{k-1}^{l-1}]$. However, the number of partitions of $\sigma _i$ into $k-2$ groups has now been overcounted. There should only be $S(l,l-2)$ such configurations, but we have just counted $-s(l,l-1) S(l-1,l-2)$ of them, so we add $S(l,l-2) + s(l,l-1)\,S(l-1,l-2) = -s(l,l-2)$ copies of ${\mathbb {E}}\,[N_{k-1}^{l-2}]$.

Fixing some $j< l-1$, we now assume attaching a coefficient of $-s(l,m)$ to ${\mathbb {E}}\,[N_{k-1}^{m}]$ for all $m > j$ ensures every partition of the $\sigma _i$ into $j+1, \ldots l-1$ sets is properly counted. Then for each $m>j$, the $-s(l,m)$ copies of ${\mathbb {E}}\,[N_{k-1}^{m}]$ count $-s(l,m) \,S(m,j)$ partitions into just j groups. Meanwhile we know there are actually only S(l, j) distinct partitions, so we must add:

$$\begin{aligned} S(l,j) + \sum _{m=j+1}^{l-1} s(l,m) \,S(m,j)&= \sum _{m=j+1}^{l} s(l,m)\, S(m,j) \\&= \sum _{m=j}^{l} s(l,m)\, S(m,j) - s(l,j)\,S(j,j) \\&= \delta _{l,j} -s(l,j) = -s(l,j). \end{aligned}$$

The last line follows from a well-known Stirling number identity. We use induction to conclude that as $n \rightarrow \infty$,

$$\begin{aligned} {\mathbb {E}}\,[N_{k-1}^l] \rightarrow {\mathbb {E}}\,[N_{k-1}]^l - \sum _{j=1}^{l-1} s(l,j)\, {\mathbb {E}}\,[N_{k-1}^j], \end{aligned}$$

thus

$$\begin{aligned} {\mathbb {E}}\,[(N_{k-1})_j] = \sum _{j=1}^l s(l,j)\, {\mathbb {E}}\,[ N_{k-1}^j ] \rightarrow {\mathbb {E}}\,[N_{k-1}]^l \end{aligned}$$

for any fixed l. It follows that $N_{k-1}$ converges in distribution to ${{\,\mathrm{Poi}\,}}( \mu )$ with $\mu = {\mathbb {E}}\,[N_{k-1}]$, completing our proof. $\square$