Topology of random $2$-dimensional cubical complexes

Proof. Fix an $n>0$. By definition of the relation $\parallel $, a necessary condition for two $4$-cycles to be related is to have their two $*$s in the same position. There are $\binom {n}{2}$ ways of choosing the positions of two $*$s in a vector of size n, which implies that there are at most $\binom {n}{2}$ components in $G\left [Q^n_2\right ]$. This gives us a partition of the set of vertices, which we represent by

$$ \begin{align*} V^n=\bigcup _{i=1}^{\binom{n}{2}} V_i^n. \\[-17pt] \end{align*} $$

Set $1\leq i\leq {\binom {n}{2}}$; in what follows we prove that the induced subgraph of $V^n_i$ in $G\left [Q^n_2\right ]$ is isomorphic to $Q^{n-2}_1$. Any element in $V^n_i$ has the two $*$s in the same position, and the rest of the $n-2$ entries have all possible binary entries. This implies that $\left \lvert V^n_{i}\right \rvert = \left \lvert Q_0^{n-2}\right \rvert $. Let $\phi : V^n_{i} \to Q_0^{n-2}$ be the natural bijection between these two sets of vertices. It is clear from Definition 2.3 that two $4$-cycles x and y in $V^n_{i}$ are connected in $G\!\left [Q^n_2\right ]$ if and only if the Hamming distance of $\phi (x)$ and $\phi (y)$ is $1$. This implies that $G\left [V_i^{n}\right ] \equiv Q^{n-2}_1$.

Proof. Set $Q \sim Q_2(n,p)$. From Lemma 2.5, we know that each $G_i[Q]$ is a random graph on $Q^{n-2}_1$. Let x and y be two vertices in $G_i[Q]$ that are connected in $G[Q^n_2]$, and represent this edge by $\overline {xy}$. This implies in particular that x and y are $4$-cycles that have, with the previously defined star notation, the $*$ in the same entries and Hamming distance equal to $1$. Let $c_{\overline {xy}} \in Q^n$ be the unique $3$-dimensional cube that contains x and y.

The probability of $\overline {xy}$ being an edge in $G_i[Q]$ is equal to the probability of the other $4$-cycles in $c_{\overline {xy}}$ being covered by $2$-faces in Q. This event happens with probability $p^4$ because in $Q_2(n,p)$ each $2$-face is added independently with probability p.

Moreover, observe that any of these $4$-cycles in $c_{\overline {xy}}$ are not vertices in $G_i[Q]$, because they do not have the two $*$s in the same location as x (or y). This implies the independence between the colouring of the vertices and the inclusion of the edges in $G_i[Q]$.

Let $\mathcal {C}$ be the set of all $3$-dimensional cubes $c_{\overline {xy}}$ with x and y varying among all unordered pairs of vertices in $G_i[Q]$ that are connected in $G\left [Q^n_2\right ]$. Then, by the uniqueness of the cube $c_{\overline {xy}} \in Q^n$, we have that $\lvert \mathcal {C} \rvert $ is equal to the number of edges in $G_i[Q]$. Finally, edges in $G_i[Q]$ are added independently with probability $p^4$ because each $4$-face in a cube in $\mathcal {C}$ appears in only one $3$-dimensional cube in $\mathcal {C}$.

Proof. Set $s \geq 1$; then for any $S \in \mathscr {Q}_s$ we have (from [Reference Hart16])

(6)$$ \begin{align} b(S) \geq s(n-\lfloor \log_2(s)\rfloor ). \end{align} $$

Also, using the fact that the degree of each vertex of $Q^n$ is at most n,

(7)$$ \begin{align} \lvert \mathscr{Q}_s \rvert \leq 2^n (n) (2n) (3n) \dotsm ((s-1)n) \leq 2^n(ns)^s. \end{align} $$

Hence,

(8)$$ \begin{align} g(s)\leq \sum_{S\in \mathscr{Q}_s}(1-p)^{ s\left(n-\left\lfloor \log_2(s)\right\rfloor \right)} \leq 2^n(ns)^s (1-p)^{ s\left(n-\left\lfloor \log_2(s)\right\rfloor\right)}. \\[-36pt]\nonumber \end{align} $$

Proof. Let $T_p$ be defined by

$$ \begin{align*} T_p = \inf_{T \in \mathbb{N}} \left\{ 2\cdot (1-p)^T \right\} < 1. \end{align*} $$

Then for $s \leq 2^{\epsilon n}$, by Lemma 2.7,

$$ \begin{align*} g(s) \leq 2^n(ns)^s(1-p)^{s\left(n-\left\lfloor\log_2(s)\right\rfloor\right)} \leq 2^n(1-p)^{-2\epsilon n} (n 2^{\epsilon n}(1-p)^{n})^s \end{align*} $$

for all n sufficiently large. Then

$$ \begin{align*} \sum_{s=T_p}^{\lfloor 2^{\epsilon n}\rfloor} g(s) &\leq 2^n(1-p)^{-2\epsilon n} \sum_{s=T_p}^\infty (n 2^{\epsilon n}(1-p)^{n})^s \\ &\leq 2^n(1-p)^{-2\epsilon n} (n 2^{\epsilon n}(1-p)^{n})^{T_p}(1+o(1)), \end{align*} $$

provided $\epsilon $ is chosen so that $2^\epsilon (1-p) <1$ and n is taken large. Taking $\epsilon $ sufficiently small,

$$ \begin{align*} \alpha = 2^{1+\epsilon T_p} (1-p)^{T_p-2\epsilon} < 1. \end{align*} $$

Hence, in terms of $\alpha $,

$$ \begin{align*} \sum_{s=T_p}^{\left\lfloor 2^{\epsilon d}\right\rfloor} g(s) \leq \alpha^n n^{T_p} (1+o(1)) \leq 2^{-\delta n} \end{align*} $$

for some $\delta> 0$ sufficiently small and all n sufficiently large.

Proof of Theorem 1.2. We first establish that for $p> \tfrac 12$, $\mathcal {I} = 0$ w.h.p., and for $p \leq \tfrac 12$, $\mathcal {I} \geq 2$ w.h.p. For the first claim, the expectation (10) tends to $0$. For the second, again from equation (10), if $(1-p) \geq 1/2$ for a random $2$-cubical complex, $Q \sim Q_2(n,p)$, then

$$ \begin{align*} \mathbb{E}[\mathcal{I}]=2^{n-1}d(1-p)^{n-1} \geq n. \end{align*} $$

Thus $\mathbb {E}[\mathcal {I}] \to \infty $ as $n \to \infty $. Now, we use a second moment argument (see [Reference Alon and Spencer1, Corollary 4.3.5]) to prove that $\mathbb {P}[\mathcal {I} \geq 2]\to 1$ as $n\to \infty $.

Fix an edge $e_i$. Any other edge $e_j$ such that $I_j$ is not independent from $I_i$ – we represent this nonindependent relation between edges $e_i$ and $e_j$ by $j \sim i$ – will be an edge of one and only one of the $(n-1)\, 4$-cycles that contain $e_i$. There are $3(n-1)$ such edges and $\mathbb {P}\left [I_j \mid I_i\right ] = (1-p)^{n-2}$. If we define $\Delta ^*_i=\sum _{j\sim i} \mathbb {P}\left [I_j \mid I_i\right ]$, then

$$ \begin{align*} \Delta^*_i=\sum_{j \sim i} \mathbb{P}\left[I_j\mid I_i\right]=3(n-1)(1-p)^{n-2}. \end{align*} $$

Thus $\Delta ^*_i=o(\mathbb {E}[I_n])$, which implies that $\mathbb {P}[\mathcal {I}>n/2] \to 1$ as $n \to \infty $.

Hence from Theorem 1.3, we have that for $p> \tfrac 12$, $\pi _1(Q_2(n,p))=0$ with high probability, and for any $p<\tfrac 12$, $\pi _1(Q_2(n,p)) = G * \mathbb {Z} * \mathbb {Z}$ for some group G with high probability. The event that $Q_2(n,p)$ has such a free factorisation is a decreasing event, in that for any complex Q that satisfies $\pi _1(Q) = G * \mathbb {Z} * \mathbb {Z}$ for some group G, removing any $2$-face (i.e., removing relations from $\pi _1(Q)$) yields a complex $Q'$ so that $\pi _1(Q') = G' * \mathbb {Z} * \mathbb {Z}$ for some other group $G'$. It follows that for any $p \leq \tfrac 12$,

$$ \begin{align*} \mathbb{P}[\exists~G: \pi_1(Q_2(n,p)) = G * \mathbb{Z} * \mathbb{Z}] \geq \mathbb{P}\left[\exists~G: \pi_1\left(Q_2\left(n,\tfrac12\right)\right) = G * \mathbb{Z} * \mathbb{Z}\right] \to 1 \end{align*} $$

as $n \to \infty $, which completes the proof.

Proof. The proof will follow from two key steps. Let $A_s$ be the event that a quasicomponent of size s in $G\left (V^n_1\right )$ exists.

1. (Step i) For any $p \in (0,1)$, there exist an integer $T_p$ and $\epsilon ,\delta>0$ such that for all n sufficiently large,

   (11)$$ \begin{align} \mathbb{P}\left[ \cup_{s=T_p}^{2^{\epsilon n}}A_s \right]<2^{-\delta n}. \end{align} $$

2. (Step ii) For any $\epsilon> 0$, the probability that there exists a component of $G\left (V_1^n\right )$ bigger than $2^{\epsilon n}$ with no vertex in $V^n_1$ tends to zero with n.

We show first how to complete the proof from this point. We now suppose that there is no quasicomponent of $G\left (V_1^n\right )$ of size between $T_p$ and $2^{\epsilon n}$ for some integer $T_p$ and some $\epsilon>0$, and we further suppose that there is no component of $G\left (V_1^n\right )$ of size $2^{\left (\epsilon /2\right )n}$ with no vertex in $V_1^n$.

Let X be a minimal quasicomponent of $G\left (V_1^n\right )$ which is bigger than $T_p$ and which is disjoint from $V_1^n$, if it exists. Then the size of X must be at least $2^{\epsilon n}$. As it is so large (in particular larger than $2^{\left (\epsilon /2\right )n}$), it cannot be that X is connected. So we must be able to remove a connected component (say by building a spanning tree in $G(V)$ of X by connecting spanning trees of its components and removing a component which contains a leaf) to form a smaller quasicomponent $X'$. The component we remove must be smaller than $2^{\left (\epsilon /2\right )n}$, and so for all n sufficiently large, we conclude that $X'$ is a strictly smaller quasicomponent of $G\left (V_1^n\right )$ which is bigger than $T_p$ and which is disjoint from $V_1^n$ – a contradiction.

We turn to proving Step (i). For any s, we have $\mathbb {P}[A_s]\leq g(s)$ (recalling g from equation (4)), as $g(s)$ is the expected number of quasicomponents of size s in $Q_2(n,p)$, and hence from a union bound,

$$ \begin{align*} \mathbb{P}\left[ \cup_{s=T_p}^{2^{\epsilon n}}A_s \right] \leq \sum_{s} g(s), \end{align*} $$

with the sum over all $T_p \leq s \leq 2^{\epsilon n}$. The existence of $\epsilon ,\delta ,T_p$ such that formula (11) holds is then a direct consequence of Lemma 2.8.

We turn to showing Step (ii). Let W be the event that there exists a component in $G\left (V_1^n\right )$ of size greater than or equal than $2^{\epsilon n}$ that does not intersect $V_1^n$. We show in what follows that $\mathbb {P}[W]\to 0$ as $n \to \infty $. First, we observe that $G\left (V_1^n\right )=G[Q]$ and that the vertices in $V_1^n$ are precisely the coloured vertices in $G[Q]$, which by Lemma 2.6 are coloured independently with probability equal to p. For $1 \leq i \leq {\binom {n}{2}}$, define $W_i$ as the event that there exists in $G_i[Q]$ a component of size greater than or equal to $2^{\epsilon n}$ that has all its vertices uncoloured. Thus, by Lemma 2.6,

(12)$$ \begin{align} W = \bigcup_{i=1}^{\binom{n}{2}} W_i. \end{align} $$

Set $1 \leq i \leq {\binom {n}{2}}$. Conditioned on knowing $G_i[Q]$ – in particular on knowing that there are exactly l components with uncoloured vertices and with sizes $s_1, s_2, \dotsc , s_l$ greater than $2^{\epsilon n}$ in $G_i[Q]$ – we get

(13)$$ \begin{align} \mathbb{P}[W_i \mid G_i] \leq \sum_{k=1}^{l}(1-p)^{s_j}. \end{align} $$

Observing that $s_1 + s_2 + \dotsb + s_l \leq 2^{n-2}$, it has to be the case that $l\leq 2^{n-2}$, and because $(1-p)< 1$, we have $(1-p)^{s_{k}} \leq (1-p)^{2^{\epsilon n}}$ for all $1\leq k \leq l$. Thus from formula (13) we get

(14)$$ \begin{align} \mathbb{P}[W_i \mid G_i] \leq \sum_{k=1}^{l}(1-p)^{2^{\epsilon n}} \leq 2^{n-2}(1-p)^{2^{\epsilon n}}. \end{align} $$

This implies that $ \mathbb {E}[\mathbb {P}[W_i \mid G_i]] \leq 2^{n-2}(1-p)^{2^{\epsilon n}}$, and thus

(15)$$ \begin{align} \mathbb{P}[W_i]\leq 2^{n-2}(1-p)^{2^{\epsilon n}} \end{align} $$

for all $1 \leq i \leq {\binom {n}{2}}$. Finally, by a union bound argument on equation (12) and inequality (15), we have that

(16)$$ \begin{align} \mathbb{P}[W] \leq {\binom{n}{2}} 2^{n-2}(1-p)^{2^{\epsilon n}}, \end{align} $$

with

(17)$$ \begin{align} \lim_{n \to \infty} {\binom{n}{2}} 2^{n-2}(1-p)^{2^{\epsilon n}}= 0. \\[-36pt]\nonumber\end{align} $$

Proof. Suppose $\mathcal {F}$ holds, and let v be any $4$-cycle. Suppose that v has at least $T_p$ neighbors in $G\left (V_1^n\right )$. Then the connected component of v in $G(V_1^n)$ has at least $T_p$ neighbors, and therefore this connected component intersects $V_1^n$. It follows by the definition of $V_2^n$ that $v \in V_2^n$.

Suppose now that v has at least $T_p$ neighbors in $G\left (V_2^n\right )$. We would like to show that at least one of these neighbors is also in $V_2^n$, for then in the next step of the parallel transport algorithm, we would have $v \in V_3^n$. We may also suppose that v is not in a component of $G\left (V_1^n\right )$ that intersects $V_1^n$, for if it is, then $v \in V_2^n$ and we are done.

Suppose by way of a contradiction that none of the neighbors of v is in $V_2^n$, which implies that each is in a component of $G\left (V_1^n\right )$ disjoint from $V_1^n$. Hence the union of these components and the component of $G\left (V_1^n\right )$ containing v is a quasicomponent of $G\left (V_1^n\right )$ that is disjoint from $V_1^n$. Moreover, it is a quasicomponent which is larger than $T_p$, which is disjoint from $V_1^n$. This does not exist in $\mathcal {F}$, and therefore v has a neighbor in $V_2^n$. Hence $v \in V_3^n$.

Proof. The possible neighbors of $\left ( \{1,2\}^*,\emptyset ^1\right )$ in $G(V^n)$ all have the form $\left ( \{1,2\}^*,\{j\}^1\right )$ for some $3 \leq j \leq n$. To have an edge between these $4$-cycles in $G\left (V^n_2\right )$, we must have

$$ \begin{align*} &\left( \{1,j\}^*, \emptyset^1\right) \in V^n_2, \qquad \left( \{1,j\}^*, \{2\}^1\right) \in V^n_2, \\ &\left( \{2,j\}^*, \emptyset^1\right) \in V^n_2, \qquad \left( \{2,j\}^*, \{1\}^1\right) \in V^n_2. \end{align*} $$

In the event $\mathcal {F}$, we must only lower-bound the degree of these $4$-cycles in $G\left (V^n_1\right )$ to ensure that they are in $V^n_2$. Hence, define

(19)$$ \begin{align} \begin{aligned} Y_{1j} & = \mathbf{1}\left\{ \deg\left( \left(\{1,j\}^*, \emptyset^1\right) \right) \geq T_p \right\}, \qquad Y_{2j} = \mathbf{1}\left\{\deg\left( \left(\{1,j\}^*, \{2\}^1\right) \right) \geq T_p \right\}, \\ Y_{3j} & = \mathbf{1}\left\{\deg\left( \left(\{2,j\}^*, \emptyset^1\right) \right) \geq T_p\right\}, \qquad Y_{4j} = \mathbf{1}\left\{\deg\left( \left(\{2,j\}^*, \{1\}^1\right) \right) \geq T_p\right\}. \end{aligned} \end{align} $$

Here, ‘$\deg $’ refers to the degree of the $4$-cycle in $G\left (V^n_1\right )$. We would like to show that there are at least $T_p$ choices j for which all $Y_{\ell j}$, $\ell \in \{1,2,3,4\}$, are $1$.

In the event $\mathcal {E}$, there are four disjoint sets $R_\ell \subset \{3,4, \dotsc , d\}$ for $\ell \in \{1,2,3,4\}$ of size $\left \lfloor M_p/4 \right \rfloor $ such that

$$ \begin{align*} &\left( \{1,k\}^*, \emptyset^1\right) \in V^n_1 \text{ for } j \in R_1, \qquad \left( \{1,k\}^*, \{2\}^1\right) \in V^n_1 \text{ for } j \in R_2, \\ &\left( \{2,k\}^*, \emptyset^1\right) \in V^n_1 \text{ for } j \in R_3, \qquad \left( \{2,k\}^*, \{1\}^1\right) \in V^n_1 \text{ for } j \in R_4. \end{align*} $$

Observe that the possible neighbors of $\left (\{1,j\}^*, \emptyset ^1\right )$, $j \in \{3,4, \dotsc , n\}$, are given by $\left (\{1,j\}^*, \{k\}^1\right )$ for $k \not \in \{1,j\}$. For simplicity, we will also discard the case $k=2$. To have this edge in $G\left (V^n_1\right )$, we would need

$$ \begin{align*} &\left( \{1,k\}^*, \emptyset^1\right) \in V^n_1, \qquad \left( \{1,k\}^*, \{j\}^1\right) \in V^n_1, \\ &\left( \{j,k\}^*, \emptyset^1\right) \in V^n_1, \qquad \left( \{j,k\}^*, \{1\}^1\right) \in V^n_1. \end{align*} $$

In particular, for $k \in R_1$ the first of these requirements is guaranteed. Hence we can define

$$ \begin{align*} Z_{1jk} = \mathbf{1}\left\{ \left( \{1,k\}^*, \{j\}^1\right) \in V^n_1, \left( \{j,k\}^*, \emptyset^1\right) \in V^n_1, \left( \{j,k\}^*, \{1\}^1\right) \in V^n_1 \right\}, \end{align*} $$

and define

$$ \begin{align*} Z_{1j} = \sum_{k \in R_1} Z_{1jk}. \end{align*} $$

Then $Z_{1j}$ is a lower bound for $\deg \left ( \left (\{1,j\}^*, \emptyset ^1\right ) \right )$, and so if $Z_{1j}$ is at least $T_p$, then $Y_{1j}=1$.

We do a similar construction for $\ell \in \{2,3,4\}$, making appropriate modifications. We list them for clarity:

$$ \begin{align*} Z_{2jk} & = \mathbf{1}\left\{ \left( \{1,k\}^*, \{2,j\}^1\right) \in V^n_1, \left( \{j,k\}^*, \{2\}^1\right) \in V^n_1, \left( \{j,k\}^*, \{1,2\}^1\right) \in V^n_1 \right\}, \\ Z_{3jk} & = \mathbf{1}\left\{ \left( \{2,k\}^*, \{j\}^1\right) \in V^n_1, \left( \{j,k\}^*, \emptyset^1\right) \in V^n_1, \left( \{j,k\}^*, \{1\}^1\right) \in V^n_1 \right\}, \\ Z_{4jk} & = \mathbf{1}\left\{ \left( \{2,k\}^*, \{1,j\}^1\right) \in V^n_1, \left( \{j,k\}^*, \{1\}^1\right) \in V^n_1, \left( \{j,k\}^*, \{1,2\}^1\right) \in V^n_1 \right\}. \end{align*} $$

In terms of these, we set $Z_{\ell j} = \sum _{k \in R_\ell } Z_{\ell jk}$.

Let $J = \{3,4, \dotsc , d\} \setminus \left (\cup _1^4 R_\ell \right )$. Then the family

$$ \begin{align*} \left\{ Z_{\ell jk} : \ell \in \{1,2,3,4\}, \ j \in J, \ k \in R_\ell \right\} \end{align*} $$

is independent random variables. Moreover, for any $\ell \in \{1,2,3,4\}$ and $j \in J$, from equation (18) we have

$$ \begin{align*} \mathbb{P}\left(Z_{\ell j} < T_p\right) < \mathbb{P} \left( \text{Binomial}\left( \left\lfloor M_p/4 \right\rfloor, p^3\right) < T_p\right) \leq \left(\tfrac{1}{2}\right)^{1/4}. \end{align*} $$

It follows that with

$$ \begin{align*} Z = \sum_{j \in J} \prod_{\ell = 1}^4 \mathbf{1}\left\{ Z_{\ell j} \geq T_p\right\} \end{align*} $$

and with $q = \mathbb {P}\left (Z_{\ell j} \geq T_p\right )^4> \tfrac 12$, we have

$$ \begin{align*} \mathbb{P}\left(Z < T_p\right) \leq \mathbb{P}\left( \text{Binomial}\left( n - 3 - M_p, q\right) < T_p\right) =n^{O(1)}(1-q)^{n}, \end{align*} $$

which completes the proof.