2.1. The graph alignment problem as a Bayesian inference problem

The graph alignment problem is a Bayesian inference problem. As defined previously, it involves sampling p random graphs $(G^{(1)}, G^{(2)}, \ldots, G^{(p)})$ from some fixed joint distribution; an alignment $\boldsymbol{\pi}^* \in \mathcal{S}_n^{p-1}$ is then picked uniformly at random, and we try to recover $\boldsymbol{\pi}^*$ from the observation of the graphs $(\boldsymbol{\pi}^*)^{-1}(G) = ((\boldsymbol{\pi}^*_1)^{-1}(G^{(1)}), (\boldsymbol{\pi}^*_2)^{-1}(G^{(2)}), \ldots, (\boldsymbol{\pi}^*_p)^{-1}(G^{(p)}))$ . Another way to formulate this is that we have observed a p-tuple of random graphs $\tilde{G} =(\tilde{G}^{(1)}, \ldots, \tilde{G}^{(p)})$ (where $\tilde{G} = (\boldsymbol{\pi}^*)^{-1}(G)$ ), sampled from a distribution $\mathbb{P}_{\boldsymbol{\pi}^*}$ for some unknown $\boldsymbol{\pi}^* \in \mathcal{S}_n^{p-1}$ whose prior distribution is assumed to be the uniform distribution $\mathcal{U}(\mathcal{S}_n^{p-1})$ ; again, we wish to estimate $\boldsymbol{\pi}^*$ . This reformulation of the problem places it squarely within the zoo of Bayesian inference problems. Specifically, we are trying to construct an estimator $\hat{\boldsymbol{\pi}}$ , which is a $\tilde{G}$ -measurable random variable, such that the expected loss

(4) \begin{equation} L = \mathbb{E}_{\boldsymbol{\pi}^* \sim \mathcal{U}(\mathcal{S}_n^{p-1})}[\mathbb{E}_{(\tilde{G}^{(1)}, \ldots, \tilde{G}^{(p)}) \sim \mathbb{P}_{\boldsymbol{\pi}^*}}[l(\boldsymbol{\pi}^*, \hat{\boldsymbol{\pi}})]]\end{equation}

is as small as possible, where $l: (\mathcal{S}_n^{p-1})^2 \to \mathbb{R}_+$ is a predetermined loss function.

What would be an adequate choice for l? A natural default choice could be the all-or-nothing loss function $l(\boldsymbol{\pi}, \boldsymbol{\pi}') = \mathbf{1}_{\boldsymbol{\pi} \neq \boldsymbol{\pi}'}$ ; with this loss function, this inference problem is known as exact alignment. However, we have a notion of closeness for permutations, and we would like to acknowledge estimators which are ‘close enough’ to the true alignment.

In the case where $p=2$ , we may define

(5) \begin{equation} ov(\pi, \pi') = \frac{1}{n} \sum_{i=1}^n \mathbf{1}_{\pi(i) = \pi'(i)} \in [0, 1]\end{equation}

and $d = (1 - ov)$ is a distance over $\mathcal{S}_n$ . This means that we now have other natural loss functions: for instance, the functions $l_r(\pi, \pi') = \mathbf{1}_{d(\pi, \pi') \geq r}$ , for $0 \leq r < 1$ . This allows us to ask slightly different questions: namely, the so-called almost exact and partial alignment problems, which we describe in the next section. We may similarly extend the notion of overlap to $\mathcal{S}_n^{p-1}$ when $p \geq 3$ ; there are multiple ways to do this, including the definition (3) which we adopted to state our main results. We discuss this further in Section 2.3.

Before proceeding, let us describe what the metric space $\mathcal{S}_n$ actually looks like. One way to conceptualise it is as a hyperbolic space: for any $\pi \in \mathcal{S}_n$ , the size of the ball $B(\pi, r)$ grows roughly like $(n^n)^r$ , for $0 \leq r \leq 1$ .

Another, more explicit, description of this space is the following. Within the space $\mathcal{M}_n(\mathbb{R})$ of $n \times n$ matrices, the set $B_n$ of doubly stochastic matrices is a polytope (known as the Birkhoff polytope), whose vertices are the permutation matrices (which we may identify with $\mathcal{S}_n$ ). The canonical Euclidean structure on this space gives us the metric

(6) \begin{equation} \langle \pi, \pi' \rangle = ov(\pi, \pi') \text{ i.e. } d_B(\pi, \pi') = \sqrt{2(1 - ov(\pi, \pi'))}, \end{equation}

which largely serves the same purpose as our previous distance d in this context. This description also shows quite nicely why we often relax the alignment problem to solve a more general quadratic assignment problem over $B_n$ , as in [Reference Varma, Waldspurger and Massoulié13, Reference Fan, Mao, Wu and Xu34].

2.2. Metric estimation problems

What follows is fairly general; as motivation for our definitions, it may be helpful to think of the specific case of the graph alignment problem, as defined above.

Fix a sequence $(A_n, d = d_n, \mu = \mu_n)$ , which are metric spaces equipped with an a priori probability distribution. Fix a sequence $(B_n)$ of (measurable) observation spaces, equipped with probability distributions $(\nu_x)_{x \in A_n}$ such that $x \mapsto \nu_x$ is measurable. To fix ideas, we assume that $\textrm{diam} A_n = 1$ , so that $d(x, y) \simeq 1$ means that x, y are far apart.

We set $X_n \in A_n$ a random variable with law $\mu$ , and $Y_n \in B_n$ a random variable with law $\nu_{X_n}$ ; from the observation of $Y_n$ , we are trying to estimate the value of $X_n$ . We define $\mathbb{P} = \mathbb{P}_n$ to be the distribution of $(X_n, Y_n)$ , and, for any measurable $S_n \subseteq A_n$ ,

(7) \begin{equation} \mathbb{P}_{\rm post}(S_n) = \mathbb{P}(S_n | Y_n)\end{equation}

the a posteriori probability of $S_n$ .

Definition 1. A partial estimator $\hat{X}_n$ of $X_n$ is a $Y_n$ -measurable random variable such that, for some $0 \leq r < 1$ ,

(8) \begin{equation} \mathbb{E}[l_r(\hat{X}_n, X_n)] \underset{n \to +\infty}{\longrightarrow} 0, \quad \text{i.e.} \ \ \mathbb{P}(d(\hat{X}_n, X_n) > r) \underset{n \to +\infty}{\longrightarrow} 0. \end{equation}

We say that partial estimation is feasible if a partial estimator exists.

On the other hand, we say that partial estimation of $X_n$ is intractable if, for any $0 \leq r < 1$ , and for any sequence $(\hat{X}_n)$ of estimators,

(9) \begin{equation} \mathbb{E}[l_r(\hat{X}_n, X_n)] \underset{n \to +\infty}{\longrightarrow} 1. \end{equation}

An almost exact estimator $\hat{X}_n$ of $X_n$ is a $Y_n$ -measurable random variable such that, for any $\varepsilon > 0$ ,

(10) \begin{equation} \mathbb{E}[l_\varepsilon(\hat{X}_n, X_n)] \underset{n \to +\infty}{\longrightarrow} 0 \quad \text{i.e.} \ \ \mathbb{P}(d(\hat{X}_n, X_n) > \varepsilon) \underset{n \to +\infty}{\longrightarrow} 0. \end{equation}

We say that almost exact estimation is feasible if an almost exact estimator exists.

On the other hand, we say that almost exact estimation of $X_n$ is intractable if, for some $\varepsilon > 0$ , and for any sequence $(\hat{X}_n)$ of estimators,

(11) \begin{equation} \mathbb{E}[l_\varepsilon(\hat{X}_n, X_n)] \underset{n \to +\infty}{\longrightarrow} 1. \end{equation}

A classical result from Bayesian decision theory (see [Reference Berger11, Section 4] for instance) is the following.

Proposition 1. Fix $r \geq 0$ : we define

(12) \begin{equation} \hat{X}_n^{\rm opt} = \underset{x \in A_n}{\arg \min}\mathbb{E}[l_r(x,X_n) | Y_n] = \underset{x \in A_n}{\arg \max} \mathbb{P}_{\rm post}(B(x, r)), \end{equation}

where B(x, r) is the (closed) ball of radius r centred at x, for the distance d. Then, $\hat{X}_n^{\rm opt}$ minimises the expected loss $\mathbb{E}[l_r]$ .

To use this, we define a random variable

(13) \begin{equation} C_n(r) \underset{{\rm def}}{=} \max_{x \in A_n} \mathbb{P}_{\rm post}(B(x, r)) = 1 - \mathbb{E}[l_r(\hat{X}_n^{\rm opt}, X_n) | Y_n],\end{equation}

which measures to what extent the random measure $\mathbb{P}_{\rm post}$ concentrates around a single point. Then, owing to the choices made in defining feasibility and intractability, the following result is an immediate consequence of the proposition.

Remark 2. Assume that almost exact estimation is feasible, and let $\varepsilon > 0$ . Then,

(14) \begin{equation} \mathbb{P}_{\rm post}(B(X_n, 2 \varepsilon)) \geq \mathbb{P}_{\rm post}(B(\hat{X}_n^{\rm opt}, \varepsilon) )\mathbf{1}_{d(X_n, \hat{X}_n^{\rm opt}) < \varepsilon} = 1 - o_{\mathbb{P}}(1).\end{equation}

Otherwise put, with high probability, the mass of $\mathbb{P}_{\rm post}$ will tend to concentrate around $X_n$ .

2.3. A brief aside on multi-graph alignment

As we previously suggested, there are a few natural extensions of ov to $\mathcal{S}_n^{p-1}$ . Two particularly relevant ones here are

(15) \begin{equation} ov(\boldsymbol{\pi}, \boldsymbol{\pi}') = \frac{1}{n} \sum_{i=1}^n \mathbf{1}_{\pi_2(i) = \pi'_2(i), \ldots, \pi_p(i) = \pi'_p(i)} \text{ (as defined earlier)},\end{equation}

(16) \begin{align} ov_w(\boldsymbol{\pi}, \boldsymbol{\pi}') = \frac{1}{p-1} \sum_{i=2}^p ov(\pi_i, \pi'_i).\\[6pt] \nonumber \end{align}

Essentially, ov measures how often the $(\pi_i, \pi_i')$ all agree at the same time; while $ov_w$ measures how often they will agree pairwise. Thus, informally, graph alignment for $d = 1 - ov$ involves determining $(\pi_2^*, \ldots, \pi_p^*)$ on some given set X, while graph alignment for $d_w = 1 - ov_w$ involves determining $\pi_2^*$ on some (large) set $X_2, \ldots$ , and $\pi_p^*$ on some (large) set $X_p$ . With this in mind, it seems that the distance d should be more natural as a choice, motivating the following definition.

The same definition applies to almost exact alignment. However, in this case, we do not need to make the distinction between normal and weak alignment.

2.4. Applying metric estimation

If we want to understand partial, or almost exact, alignment, our job is thus to control the random variable $C_n(r)$ defined earlier. In the Gaussian case, in order to show intractability of partial alignment, we can directly perform calculations to estimate $C_n(r)$ .

In the Erdös–Rényi case, however, our arguments are much more combinatorial in nature. The general idea is quite simple: if $B = B(\boldsymbol{\pi}^*, r)$ is a ball, we would like to take the alignments within B (which are close to the true alignment $\boldsymbol{\pi}^*$ ) and modify them to obtain a large number of other alignments, which are nowhere near $\boldsymbol{\pi}^*$ but which seem just as plausible to an outside observer. In the case where $p=2$ , this is done in [Reference Ganassali, Massoulié and Lelarge2]: they construct a large number of $(\sigma_i) \in S_n$ which are far apart and which exactly preserve $\mathbb{P}_{\rm post}$ , so that $\mathbb{P}_{\rm post}(\sigma_i(B)) = \mathbb{P}_{\rm post}(B)$ and the $\sigma_i(B)$ are all disjoint. This is fairly technical, and not quite necessary: at the cost of some abstraction, we can provide a significantly more robust argument, which holds for any metric estimation problem, and does not require that we preserve $\mathbb{P}_{\rm post}$ exactly.

Here, F takes on the same role as the map $\pi \mapsto (\sigma_i \circ \pi)_{i}$ , in the 2-graph argument from [Reference Ganassali, Massoulié and Lelarge2].

Proof. Let $B = B(x, r)$ be a ball with radius $r < 1$ . Then:

(18) \begin{align} \mathbb{P}_{\rm post}(B) & \leq \mathbb{P}_{\rm post}(\textrm{Neg}_n) + \sum_{y \in B \setminus \textrm{Neg}_n} \mathbb{P}_{\rm post}(y) \nonumber\\[3pt] & \leq o_{\mathbb{P}}(1) + \varepsilon_n(r)\sum_{y \in B \setminus \textrm{Neg}_n} \mathbb{P}_{\rm post}(F(y)) \nonumber\\[3pt] & \leq o_{\mathbb{P}}(1) + \varepsilon_n(r) \sum_{y' \in A_n} \sum_{y \text{ s.t. }y' \in F(y)} \mathbb{P}_{\rm post}(y') \nonumber\\[3pt] & \leq o_{\mathbb{P}}(1) + \varepsilon_n(r) \sum_{y' \in A_n} K_n(r) \mathbb{P}_{\rm post}(y') \nonumber\\[3pt] & = o_{\mathbb{P}}(1) + \varepsilon_n(r) K_n(r) \mathbb{P}_{\rm post}(A_n) = o_{\mathbb{P}}(1).\end{align}

Applying Corollary 1, the lemma is shown.

3.1. The Gibbs measure

The main appeal of the Gaussian model is that the a posteriori distribution has a nice expression. Specifically, it is as follows.

Proposition 3. For any $\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}$ , we may write

(20) \begin{equation} \mathbb{P}_{\rm post}(\boldsymbol{\sigma}) = \frac{1}{\mathcal{Z}} \mathrm{e}^{- \beta \mathcal{H}(\sigma)} \end{equation}

where the partition function $\mathcal{Z}$ is independent from $\boldsymbol{\sigma}$ . Here,

(21) \begin{equation} \mathcal{H}(\boldsymbol{\sigma}) = - \sum_{e \in E} \sum_{1 \leq i \neq j \leq p} G_e^{(i)} G_{\sigma_{ij}(e)}^{(j)}; \quad \beta = \dfrac{\rho}{2(1 - \rho)(1 + (p-1)\rho)} = \dfrac{\rho}{2}(1 + o(1)). \end{equation}

Proof. By definition, the probability density function for $(\boldsymbol{\sigma}, G^{(1)}, \ldots, G^{(p)})$ is proportional to

(22) \begin{equation} p(\boldsymbol{\sigma}, G^{(1)}, \ldots, G^{(p)}) = \frac{1}{(n!)^{p-1}} \prod_{e \in E} \exp\bigg( \!-\!\frac{1}{2}\sum_{1 \leq i,j \leq p} (M^{-1})_{ij} G^{(i)}_{e} G^{(j)}_{e}\bigg), \end{equation}

where $M = (1 - \rho)I_p + \rho J_p$ is our covariance matrix. Consequently, if we observe the graphs $H^{(i)} = (\pi_i^*)^{-1}(G^{(i)})$ , the distribution $\mathbb{P}_{\rm post}$ is given by

(23) \begin{align} \mathbb{P}_{\rm post}(\boldsymbol{\sigma}) & = \mathbb{P}(\boldsymbol{\pi}^* = \boldsymbol{\sigma} | H^{(1)}, \ldots, H^{(p)}) \nonumber\\[3pt] & = \frac{1}{\mathcal{Z}'} p(\boldsymbol{\sigma}, \sigma_1(H^{(1)}), \ldots, \sigma_p(H^{(p)})) \nonumber\\[3pt] & = \frac{1}{\mathcal{Z}'} \prod_{e \in E} \exp\bigg( \!-\!\frac{1}{2}\sum_{1 \leq i,j \leq p} (M^{-1})_{ij} G^{(i)}_{(\pi_i^*)^{-1} \sigma_i(e)} G^{(j)}_{(\pi_j^*)^{-1} \sigma_j(e)}\bigg) \nonumber\\[3pt] & = \frac{1}{\mathcal{Z}} \exp \bigg( \!- \!\frac{1}{2}\sum_{e \in E} \sum_{1 \leq i \neq j \leq p} (M^{-1})_{ij} G^{(i)}_{(\pi_i^*)^{-1} \sigma_i(e)} G^{(j)}_{(\pi_j^*)^{-1} \sigma_j(e)}\bigg), \end{align}

since the product over the diagonal terms ( $i=j$ ) is independent from $\boldsymbol{\sigma}$ . (Here, $\mathcal{Z}, \mathcal{Z'}$ are quantities which may depend upon the $(H^{(i)})$ but not upon $\boldsymbol{\sigma}$ .) However, for $i \neq j$ , $(M^{-1})_{ij} = - 2 \beta$ ; we thus obtain our desired expression, with

(24) \begin{align} \mathcal{H}(\sigma) & = -\sum_{e \in E} \sum_{1 \leq i \neq j \leq p} G^{(i)}_{(\pi_i^*)^{-1} \sigma_i(e)} G^{(j)}_{(\pi_j^*)^{-1} \sigma_j(e)} \nonumber\\[3pt] & = -\sum_{e \in E} \sum_{1 \leq i \neq j \leq p} G^{(i)}_{e} G^{(j)}_{\sigma_{ij}(e)}. \end{align}

This Hamiltonian is quite descriptive, and invites a natural visual interpretation of the graph alignment problem: we have p weighted graphs which we ‘stick’ together with our alignment $\boldsymbol{\sigma}$ , and each edge has a fixed ‘spin’ such that edges with similar spins try to stick to each other.

From a technical perspective however, this Hamiltonian is not quite what we want. Recall that our ultimate goal is to understand to what extent the measure $\mathbb{P}_{\rm post}$ concentrates around any given point; and, if it is to be concentrated around a point, we should expect that point to be the ground truth $\pi^*$ . As a result, it makes sense to renormalise our system, writing

(25) \begin{equation} \mathbb{P}_{\rm post}(\boldsymbol{\sigma}) = \frac{1}{Z} \mathrm{e}^{- \beta V(\boldsymbol{\sigma})},\end{equation}

where the potential $V(\boldsymbol{\sigma})$ is equal to

(26) \begin{align} V(\boldsymbol{\sigma}) & = \mathcal{H}(\boldsymbol{\sigma}) - \mathcal{H}(\pi^*) \nonumber\\[3pt] & = \sum_{e \in E, 1 \leq i \neq j \leq p} G_e^{(i)} (G_e^{(j)} - G_{\sigma_{ij}(e)}^{(j)}) \nonumber\\[3pt] & = \sum_{\underset{\sigma_{ij}(e) \neq e}{e \in E, 1 \leq i \neq j \leq p}}G_e^{(i)} G_e^{(j)} - \sum_{\underset{\sigma_{ij}(e) \neq e}{e \in E, 1 \leq i \neq j \leq p}}G_e^{(i)} G_{\sigma_{ij}(e)}^{(j)} \nonumber\\[3pt] & \underset{{\rm def}}{=} - V_{\mathrm{diag}}(\boldsymbol{\sigma}) + V_{\mathrm{off}}(\boldsymbol{\sigma}).\end{align}

We have split up the potential V into two terms for a couple of reasons. The first is that, viewing V as a quadratic form of $G \in \mathbb{R}^{E} \otimes \mathbb{R}^p$ , $V_{\mathrm{diag}}$ contains the block-diagonal terms while $V_{\mathrm{off}}$ contains the off-diagonal blocks. (The expression of V as a quadratic form is made explicit in Appendices C and B.) It is typical, when studying quadratic forms of Gaussian vectors, to separate the two; they tend to require different tools. The second reason is that $V_{\mathrm{diag}}(\boldsymbol{\sigma})$ concentrates around its mean quite uniformly in $\boldsymbol{\sigma}$ , in a sense which we make explicit in a moment.

Before moving on, let us briefly explain the structure of the rest of this section. Our ultimate goal is to prove that $\mathbb{P}_{\rm post}$ concentrates around the true alignment $\boldsymbol{\pi}^*$ (respectively, is spread out over the entire state space) if $\eta > 0$ (respectively, $\eta < 0$ ). Doing this will require a few steps:

• • first, we do a bit of combinatorics in order to understand the structure of $(\mathcal{S}_n^{p-1}, d)$ a bit better;

• • next, we compute the ‘annealed free energy’ $\log \mathbb{E}[Z]$ ;

• • finally, we use this to compute the quenched free energy $\mathbb{E}[\log Z]$ .

This last step is the hardest; as is fairly standard in systems derived from Gibbs measures, once we have $\mathbb{E}[\log Z]$ it is not too hard to obtain most relevant properties of the distribution $\mathbb{P}_{\rm post}$ .

3.2. Some combinatorics

For the rest of this section, we define $d_{ij}(\boldsymbol{\sigma}) =\# \{ u \in \left[ \! \left[ 1, n \right] \! \right]\!, \sigma_{ij}(u) = u \}$ : this combinatorial quantity appears a fair amount in our analysis, as it is closely linked to the number of terms in the sums defining $V_{\mathrm{diag}}$ and $V_{\mathrm{off}}$ . In this section, we control the behaviour of $(d_{ij}(\boldsymbol{\sigma}))$ as a function of $\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}$ .

Theorem 3. Let $(d_{ij})_{1 \leq i \neq j \leq p}$ be positive integers, with $d_{ij} = d_{\rm ji}$ for any i, j. We set

(27) \begin{equation} F((d_{ij})) = \# \{\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}, \ {for\, all }\, 1 \leq i \neq j \leq p,\ d_{ij}(\boldsymbol{\sigma}) \geq d_{ij} \}. \end{equation}

Then, for any permutation $\tau \in \mathcal{S}_p$ ,

(28) \begin{equation} F((d_{ij})) \leq \frac{(n!)^{p-1}}{(\!\displaystyle\max_{j < 2} d_{\tau(j)\tau(2)})!(\!\displaystyle\max_{j < 3} d_{\tau(j) \tau(3)})! \ldots (\!\displaystyle\max_{j < p} d_{\tau(j) \tau(p)})!}. \end{equation}

Proof. It is sufficient to handle the case where $\boldsymbol{\pi}^* = \textrm{Id}$ ; the general case follows by replacing $\boldsymbol{\sigma}$ by $\boldsymbol{\pi}^* \circ \boldsymbol{\sigma}$ in the following proof.

We begin by noting that F is a symmetric function of $(d_{ij})$ , in the sense that if $\tau \in \mathcal{S}_p$ , $F((d_{\tau(i)\tau(j)})) = F((d_{ij}))$ . As a result, we just need to show the theorem when $\tau = \textrm{Id}$ ; the general statement then follows.

We now note that, for any $(d_{ij})$ , setting $Y_{ij}(\boldsymbol{\sigma}) \underset{{\rm def}}{=} \{ u \in \left[ \! \left[ 1, n \right] \! \right]\!, \sigma_{ij}(u) = u \}$ ,

(29) \begin{equation} F((d_{ij})) = \frac{1}{n!}\# \{\boldsymbol{\sigma} \in \mathcal{S}_n^{p}, \mbox{ for all }1 \leq i \neq j \leq p,\ |Y_{ij}(\boldsymbol{\sigma})| \geq d_{ij} \} \end{equation}

(i.e. counting the p-tuples of permutations where we do not enforce that $\sigma_1 = \textrm{Id}$ ). To see this, note that the map

\begin{align*} \mathcal{S}_n^{p} & \longrightarrow \mathcal{S}_n^{p-1} \simeq \{ \textrm{Id} \} \times \mathcal{S}_n^{p-1} \\[2pt] (\sigma_1, \sigma_2, \ldots, \sigma_p) & \longmapsto (\textrm{Id}, \sigma_1^{-1} \sigma_2, \ldots, \sigma_1^{-1} \sigma_p) \end{align*}

preserves the values of $(d_{ij})_{1 \leq i \neq j \leq p}$ , and any given element of $\mathcal{S}_n^{p-1}$ has exactly $n!$ preimages. We therefore count the number of elements of the subset defined in (29).

Fix $(d_{ij})_{i\neq j}$ . We are going to bound the number of possible choices for $\sigma_1$ , then bound the number of possible choices for $\sigma_{2}$ given a choice of $\sigma_{1}$ and so on; finally, bounding the number of choices for $\sigma_{p}$ given a choice of $\sigma_{1}, \ldots, \sigma_{p-1}$ .

There are $n!$ choices for $\sigma_{1}$ . In order to fix $\sigma_{2}$ given $\sigma_{1}$ , it is sufficient to determine the set $X_2 = \left[ \! \left[ 1, n \right] \! \right] \setminus Y_{1 2}(\sigma)$ of points where $\sigma_{2} \neq \sigma_1$ , as well as $\sigma_{2}|_{X_2}^{\sigma_1(X_2)}$ , giving at most

(30) \begin{equation} \binom{n}{n - d_{12}}(n - d_{12})! = \frac{n!}{d_{12}!} \end{equation}

possible choices.

Recursively, given $\sigma_{1}, \sigma_{2}, \ldots, \sigma_{l-1}$ , in order to fix $\sigma_{l}$ , it is sufficient to choose $j < l$ , and determine the set $X_{l} = \left[ \! \left[ 1, n \right] \! \right] \setminus Y_{jl}(\sigma)$ of points where $\sigma_{l} \neq \sigma_{j}$ as well as the restriction $\sigma_{l}|_{X_{l}}^{\sigma_{j}(X_{l})}$ , giving at most ${n!}/{d_{jl}!}$ possibilities. Since j can be chosen arbitrarily, we can pick j such that $d_{jl}$ is maximal. The total number of possible choices is thus at most

(31) \begin{equation} \frac{1}{n!} \cdot n! \cdot \frac{n!}{(\!\displaystyle\max_{j < 2} d_{j2})!} \cdot \cdots \cdot \frac{n!}{(\!\displaystyle\max_{j < p} d_{jp})!} \end{equation}

giving us the desired inequality and concluding the proof of Theorem 3.

Theorem 3 can be used by itself, but the introduction of the quantities $\max_{j < k} d_{\tau(j)\tau(k)}$ for $1 < k \leq p$ makes it somewhat unwieldy to use. Instead, we will mostly use the following corollary.

Corollary 2. Let $(d_{ij})_{1 \leq i \neq j \leq p}$ be positive integers, with $d_{ij} = d_{\rm ji}$ for any i,j. Then,

(32) \begin{equation} F((d_{ij})) \leq \frac{(n!)^{p-1}}{\bigg( \displaystyle\prod_{1 \leq i \neq j \leq p} (d_{ij}!) \bigg)^{{1}/{p}}} \end{equation}

Proof. For $\tau \in \mathcal{S}_p$ , set $D_{i}(\tau) = \max_{j < i} d_{\tau(j)\tau(i)}$ . In order to deduce Corollary 2 from Theorem 3, we need to show that, for some choice of $\tau$ ,

(33) \begin{equation} \prod_{i=2}^p D_i(\tau)! \geq \bigg( \prod_{1 \leq i \neq j \leq p} (d_{ij}!) \bigg)^{{1}/{p}}.\end{equation}

In order to show this, we use the probabilistic method, showing that if $\tau \in \mathcal{S}_p$ is chosen uniformly at random, then

(34) \begin{equation} \frac{1}{p} \sum_{1 \leq i \neq j \leq p} \log (d_{ij}!) \leq \mathbb{E}_\tau\bigg[ \sum_{i=2}^p \log (D_i(\tau)!) \bigg].\end{equation}

Indeed, more generally, if $\varphi$ is any increasing function,

(35) \begin{align} \mathbb{E}_\tau\bigg[ \sum_{i=2}^p \varphi(D_i(\tau)) \bigg] & = \mathbb{E}_\tau\bigg[ \sum_{i=2}^p \max_{1 \leq j \leq i-1} \varphi(d_{\tau(i) \tau(j)}) \bigg]\nonumber\\[3pt] & \geq \mathbb{E}_\tau\bigg[ \sum_{i=2}^p \frac{1}{i-1} \sum_{j=1}^{i-1} \varphi(d_{\tau(i) \tau(j)}) \bigg] \nonumber\\[3pt] & = \sum_{i=2}^p \frac{1}{i-1} \sum_{j=1}^{i-1} \mathbb{E}_\tau[ \varphi(d_{\tau(i) \tau(j)})].\end{align}

However, for any $i,j \in \left[ \! \left[ 1, p \right] \! \right]$ ,

\begin{align*}\mathbb{E}_\tau[\varphi(d_{\tau(i) \tau(j)})] = \dfrac{1}{p(p-1)}\displaystyle \sum_{1 \leq i \neq j \leq p} \varphi(d_{ij}).\end{align*}

Plugging this into (35), we obtain

(36) \begin{equation} \frac{1}{p} \sum_{1 \leq i \neq j \leq p} \varphi(d_{ij}) \leq \mathbb{E}_\tau\bigg[ \sum_{i=2}^p \varphi(D_i(\tau)) \bigg]\end{equation}

and this holds in particular for the function $\varphi(d) = \log(d!)$ , proving the corollary.

3.3. The annealed free energy

The propositions in this subsection are consequences of some sharp controls on the behaviour of their Gaussian components; their proofs are all deferred to the appendix.

In order to compute $\log \mathbb{E}[Z]$ , we begin by computing $V_{\mathrm{diag}}$ and $V_{\mathrm{off}}$ . The behaviour of $V_{\mathrm{diag}}$ is the simplest to bound: it is essentially deterministic.

Proposition 4. For some $c > 0$ , the following property $(\mathcal{C})$ holds with probability at least $1 - \exp(-c n \sqrt{\log n})$ :

(37) \begin{equation} {for\, all\, }\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}, \left|V_{\mathrm{diag}}(\boldsymbol{\sigma}) - \mathbb{E}[V_{\mathrm{diag}}(\boldsymbol{\sigma})]\right| \leq (\!\log n)^{{1}/{4}} n^{{3}/{2}} \end{equation}

Consequently, for any $\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}$ ,

(38) \begin{equation} V(\boldsymbol{\sigma})\mathbf{1}_\mathcal{C} = \rho \sum_{1 \leq i \neq j \leq p} \bigg( \binom{n}{2} - \binom{d_{ij}(\boldsymbol{\sigma})}{2} \bigg)\mathbf{1}_\mathcal{C} + V_{\mathrm{off}}(\boldsymbol{\sigma}) \mathbf{1}_{\mathcal{C}} + O(n^{{3}/{2}} (\!\log n)^{{1}/{4}}), \end{equation}

where the error term is bounded uniformly in $\boldsymbol{\sigma}$ .

The fluctuations of $V_{\mathrm{off}}$ are somewhat less nice but can still be controlled.

Proposition 5. Let $\boldsymbol{\sigma} \in \mathcal{S}_n^{p-1}$ . Then,

(39) \begin{equation} \log \mathbb{E}[e^{-\beta V_{\mathrm{off}}(\boldsymbol{\sigma})}] = \beta^2\displaystyle\sum_{1 \leq i \neq j \leq p} \bigg(\displaystyle\binom{n}{2} - \displaystyle \binom{d_{ij}(\boldsymbol{\sigma})}{2} \bigg) + O(n), \end{equation}

where, again, the error term is bounded uniformly in $\boldsymbol{\sigma}$ .

Putting these together allows us to obtain the behaviour of the annealed free energy. To this end, for any $\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}$ and $0 < r \leq 1$ , we define

(40) \begin{equation} Z_r(\boldsymbol{\pi}) = \sum_{\boldsymbol{\sigma} \in B(\boldsymbol{\pi}, r)} e^{-\beta V(\boldsymbol{\sigma})}. \end{equation}

Proposition 6. For any $0 < r \leq 1$ ,

(41) \begin{equation} \log \mathbb{E}[Z_r(\boldsymbol{\pi}^*) \mathbf{1}_\mathcal{C}] \leq (p-1) n \log n \max_{0 \leq \rho \leq r} \varphi(\rho) + O(n(\!\log n)^{{3}/{4}}) \end{equation}

and

(42) \begin{align} \log \mathbb{E}[(Z - Z_r(\boldsymbol{\pi}^*))\mathbf{1}_{\mathcal{C}} ] &\leq (p-1) n \log n \bigg( \frac{p-1}{p} \max_{r' \in [0,1]} \varphi(r') + \frac{1}{p} \max_{r' \in [{r}/({p-1}), 1] } \varphi(r') \bigg) \end{align}

(43) \begin{align} &\quad + O(n (\!\log n)^{3/4}), \end{align}

where $\varphi(r') = (1 + \eta)(r')^2 - (1 + 2 \eta)r'$ for $r' \in [0,1]$ .

3.4. The quenched free energy

Recall that we have written the parameter $\rho$ of our system as

\begin{align*}\rho = \sqrt{\frac{8(1 + \eta) \log n}{p n}};\end{align*}

we wish to estimate the quenched free energy $\mathbb{E}[\log Z]$ as a function of $\eta$ . This quantity captures the typical behaviour of Z better than the annealed free energy; controlling it is the difficult part in bounding $\mathbb{P}_{\rm post}$ . In these systems, we cannot usually expect to have $\mathbb{E}[\log Z] \sim \log \mathbb{E}[Z]$ in general; however, when $\eta > 0$ the system fluctuates sufficiently little for this to hold.

Lemma 2. In general, we have

(44) \begin{equation} 0 \leq \mathbb{E}[(\!\log Z)\mathbf{1}_{\mathcal{C}}] \leq \log \mathbb{E}[Z\mathbf{1}_{\mathcal{C}}](1 + o(1)). \end{equation}

In particular, if $\eta > 0$ ,

(45) \begin{equation} \mathbb{E}[(\!\log Z)\mathbf{1}_{\mathcal{C}}]= o(n \log n). \end{equation}

Proof. The first inequality simply follows from the fact that

(46) \begin{equation} Z \geq \mathrm{e}^{-\beta V(\boldsymbol{\pi}^*)} = 1 \end{equation}

since $V(\boldsymbol{\pi}^*) = \mathcal{H}(\boldsymbol{\pi}^*) - \mathcal{H}(\boldsymbol{\pi}^*) = 0$ .

The second inequality follows from Jensen’s inequality:

(47) \begin{equation} \mathbb{E}[(\!\log Z)\mathbf{1}_\mathcal{C}] \leq \mathbb{P}(\mathcal{C}) \log \mathbb{E}[Z | \mathcal{C}] = \log \mathbb{E}[Z \mathbf{1}_{\mathcal{C}}](1 + o(1)). \end{equation}

The final inequality follows from the previous proposition.

The difficult part comes when $\eta < 0$ : typically, we believe that $\mathbb{E}[\log Z]$ will be significantly smaller than $\log \mathbb{E}[Z]$ . To get around this, we take a small value $\eta = (\!\log n)^{-{1}/{5}}$ in the hope that the annealed free energy still provides some information in this regime. Indeed, we have the following.

Proposition 7. If $\eta = -(\!\log n)^{-{1}/{5}}$ ,

(48) \begin{equation} \mathbb{E}[\log Z] \sim \log \mathbb{E}[Z \mathbf{1}_{\mathcal{C}}] \sim (p-1)|\eta| n \log n. \end{equation}

Lemma 3. For some constant $c_0 > 0$ , if $0 \leq K \leq n^{{3}/{2}}$ ,

(49) \begin{equation} \mathbb{P}(|\log Z - \mathbb{E}[\log Z]| > K) \leq 2\exp\bigg(\!-\! c_0 \frac{K^2}{n \log n}\bigg). \end{equation}

Lemma 4. With probability at least $\exp(\!-\!O(n))$ ,

(50) \begin{equation} (\!\log Z) \mathbf{1}_{\mathcal{C}} \geq (p-1)|\eta| n \log n - (p-1) n (\!\log n)^{{3}/{5}}. \end{equation}

The proof of these lemmas is deferred to the appendix.

With these results in hand, we may note that

(51) \begin{align} & \mathbb{P}((\!\log Z)\mathbf{1}_{\mathcal{C}} > \mathbb{E}[\log Z] + n (\!\log n)^{{3}/{5}} ) \nonumber\\ &\quad \leq 2\exp(\! -\! C' n (\!\log n)^{{1}/{5}}) + \mathbb{P}(\mathcal{\overline{C}}) \end{align}

(52) \begin{align} & \quad < \exp(\!-\! O(n)) \end{align}

(53) \begin{align} & \quad \leq \mathbb{P}((\!\log Z)\mathbf{1}_{\mathcal{C}} \geq (p-1)|\eta| n \log n - (p-1) n (\!\log n)^{{3}/{5}}), \end{align}

so that

(54) \begin{align} \mathbb{E}[\log Z] + n (\!\log n)^{3/5}& \geq (p-1)|\eta| n \log n - (p-1) n (\!\log n)^{{3}/{5}} \\[-30pt] \nonumber \end{align}

(55) \begin{align} &\qquad \qquad \qquad \geq \log \mathbb{E}[Z\mathbf{1}_\mathcal{C}] + O(n (\!\log n)^{{3}/{4}}).\end{align}

This proves the difficult part of Proposition 7. The full proposition follows if we prove that $\mathbb{E}[\log Z] \sim \mathbb{E}[(\!\log Z)\mathbf{1}_{\mathcal{C}}]$ since, in this case, we may apply Lemma 2 and

(56) \begin{equation} \mathbb{E}[\log Z] \sim \mathbb{E}[(\!\log Z)\mathbf{1}_\mathcal{C}] \leq \log \mathbb{E}[Z \mathbf{1}_\mathcal{C}](1 + o(1)).\end{equation}

However, if we define $t_0 > 0$ by the condition

\begin{align*}\displaystyle\int_{t_0}^{+\infty} \exp \bigg(\!-\! c_0 \frac{u^2}{n \log n} \bigg) \,du = \mathbb{P}(\overline{\mathcal{C}}),\end{align*}

then by Lemma 3:

(57) \begin{align} \mathbb{E}[\log Z] - \mathbb{E}[(\!\log Z)\mathbf{1}_\mathcal{C}] & = \mathbb{E}[(\!\log Z)\mathbf{1}_{\overline{\mathcal{C}}}]\nonumber\\[3pt] & \leq 2\int_{t_0}^{+\infty} (u + \mathbb{E}[\log Z]) \exp \bigg(\!-\! c_0 \frac{u^2}{n \log n} \bigg) \,du \nonumber\\[3pt] & \ll \mathbb{E}[\log Z].\end{align}

This concludes the proof.

With this result, we have most of the required elements to prove Theorem 1.

Proof of Theorem 1. Assume that $\eta > 0$ : we begin by showing the existence of an almost exact estimator. In order to apply Corollary 1, we simply need to prove that

(58) \begin{equation} \mathbb{P}_{\rm post}(B(\boldsymbol{\pi}^*, \varepsilon)) \underset{\mathbb{P}}{\longrightarrow} 1,\quad \text{i.e.}\ Z_\varepsilon(\boldsymbol{\pi}^*) \underset{\mathbb{P}}{\sim} Z. \end{equation}

This is true since, by Proposition 6,

(59) \begin{equation} \log \mathbb{E}[(Z - Z_{\varepsilon}(\boldsymbol{\pi}^*)) \mathbf{1}_{\mathcal{C}}] \leq -\frac{\varepsilon}{p} n \log n(1 + o(1)) \end{equation}

and so $Z - Z_{\varepsilon}(\boldsymbol{\pi}^*) \ll_{\mathbb{P}} 1 \leq Z$ .

Conversely, when $\eta = -(\!\log n)^{-{1}/{5}}$ , our job is to prove that for any $r < 1$ , $\sup_{\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}} Z_r(\boldsymbol{\pi}) \ll_{\mathbb{P}} Z$ : by Corollary 1, this will imply that partial alignment is intractable. Here:

(60) \begin{equation} \log \mathbb{E}[Z_r(\boldsymbol{\pi}^*)\mathbf{1}_{\mathcal{C}}] \leq c_r (p-1)n (\!\log n)^{{4}/{5}}(1 + o(1))\end{equation}

for some $0 \leq c_r < 1$ , by Proposition 6. However, by Proposition 7 and Lemma 3,

(61) \begin{equation} \log Z \geq (p-1)|\eta|n\log n(1 - o_{\mathbb{P}}(1)) = (p-1)n (\!\log n)^{{4}/{5}}(1 -o_{\mathbb{P}}(1)) \end{equation}

so that $Z_r(\boldsymbol{\pi}^*) \ll_{\mathbb{P}} Z$ .

This is close to what we wanted, but not quite there: we still need to go from $Z_r(\boldsymbol{\pi}^*) \ll_{\mathbb{P}} Z$ to the uniform bound $\sup_{\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}} Z_r(\boldsymbol{\pi}) \ll_{\mathbb{P}} Z$ . This requires another technical argument which we defer to Section H. With this extra argument, the case $\eta < 0$ is complete.

Finally, as for when $\eta > 0$ , we still wish to show that there exists an exact estimator. For this, we require the following extra lemma, whose proof is given in the appendix.

Since, for any $\boldsymbol{\sigma} \notin B(\boldsymbol{\pi}^*, \varepsilon)$ , $\mathrm{e}^{-\beta V(\boldsymbol{\sigma})} \leq Z - Z_{\varepsilon} \ll_{\mathbb{P}} 1$ , this means that with high probability,

(62) \begin{equation} \boldsymbol{\pi}^* = \underset{\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}}{\arg \max} \mathbb{P}_{\rm post}(\boldsymbol{\pi}),\end{equation}

concluding the proof of our main theorem.

4.1. A description of the system

As in the Gaussian case, we begin by evaluating the Hamiltonian of our system, and formulate a proof strategy based upon the behaviour of said Hamiltonian.

Proposition 8. Let $\boldsymbol{\pi} = (\pi_1 = \textrm{Id} , \pi_2, \ldots, \pi_p) \in \mathcal{S}_n^{p-1}$ . For any subset $X \subseteq \left[ \! \left[ 1, p \right] \! \right]$ , we define $e_X = e_X(\boldsymbol{\pi})$ to be the number of edges $e \in E$ ‘of type X’; i.e. such that for any $1 \leq i \leq p$ , $e \in \pi_i(\Gamma^{(i)})$ if and only if $i \in X$ . (In this sense, $\sum_X e_X = \binom{n}{2}$ .)

With this setup, we may write

(63) \begin{equation} \mathbb{P}_{\rm post}(\boldsymbol{\pi}^* = \boldsymbol{\pi}) = \frac{1}{Z} \mathrm{e}^{- \beta' \mathcal{H}(\boldsymbol{\pi})(1 + o(1))}\quad \textit{with}\ \beta' = \log n; { \mathcal{H}(\boldsymbol{\pi}) = -\sum_{X \neq \emptyset} e_X(\boldsymbol{\pi})}.\end{equation}

Here, Z is a random variable which does not depend on $\boldsymbol{\pi}$ , and the error term o(1) is bounded uniformly in $\boldsymbol{\pi}$ .

In other words, if we ‘glue’ together all of our graphs using the correspondence $\boldsymbol{\pi}$ , the associated energy for this configuration is equal to the total number of edges in the combined graph (plus a smaller, negligible term).

Also note that, in contrast to the Gaussian case, we are now in a low-temperature configuration: we care very much about correctly matching our edges since we have so much fewer of them.

Proof. Let $\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}$ , and let $(A^{(1)}, \ldots, A^{(p)})$ be graphs over $\left[ \! \left[ 1, n \right] \! \right]$ . Then:

(64) \begin{align} \mathbb{P}(\boldsymbol{\pi}^* = \boldsymbol{\pi}, G^{(1)} = A^{(1)}, \ldots, G^{(p)} = A^{(p)}) & = \frac{1}{(n!)^{p-1}} \mathbb{P}( G^{(1)} = A^{(1)}, \ldots, G^{(p)} = A^{(p)} | \boldsymbol{\pi}^* = \boldsymbol{\pi}) \nonumber\\[3pt] & = \frac{1}{(n!)^{p-1}} \prod_{e \in E} \mathbb{P}(\mbox{for all }i \in \left[ \! \left[ 1, p \right] \! \right]\!, e \in G^{(i)} \Leftrightarrow e \in A^{(i)})\end{align}

since the variables $((\mathbf{1}_{e \in G^{(1)}}, \ldots, \mathbf{1}_{e \in G^{(p)}}))_{e \in E}$ are all independent, and independent from $\boldsymbol{\pi}^*$ . In fact, we know the laws of these variables explicitly, and so we may write

(65) \begin{equation} \mathbb{P}(\mbox{for all }i \in \left[ \! \left[ 1, p \right] \! \right]\!, e \in G^{(i)} \Leftrightarrow e \in A^{(i)}) = \begin{cases} 1 - \frac{\lambda (1 - (1 - s)^{p})}{n} & \text{ if for all }1 \leq i \leq p,\ e \notin A^{(i)}, \\[3pt] \frac{\lambda}{n}s^k (1-s)^{p-k} & \text{ if } \# \{ 1 \leq i \leq p,\ e \in A^{(i)} \} = k > 0. \end{cases}\end{equation}

Consequently, if $B^{(1)}, \ldots, B^{(p)}$ are graphs over $\left[ \! \left[ 1, n \right] \! \right]$ :

(66) \begin{align} \mathbb{P} & (\boldsymbol{\pi}^* = \boldsymbol{\pi} | \Gamma^{(1)} = B^{(1)}, \ldots, \Gamma^{(p)} = B^{(p)}) \nonumber\\[3pt] & = \frac{1}{Z'} \mathbb{P}(\boldsymbol{\pi}^* = \boldsymbol{\pi}, G^{(1)} = \pi_1(B^{(1)}), \ldots, G^{(p)} = \pi_p(B^{(p)}))\end{align}

meaning that

(67) \begin{align} \mathbb{P}_{\rm post}(\boldsymbol{\pi}^* = \boldsymbol{\pi})& = \mathbb{P}(\boldsymbol{\pi}^* = \boldsymbol{\pi} | \Gamma) \nonumber\\[3pt] & = \frac{1}{Z'} \bigg( 1 - \frac{\lambda (1 - (1 - s)^{p})}{n} \bigg)^{e_{\emptyset}(\boldsymbol{\pi})} \prod_{X \subseteq \left[ \! \left[ 1, p \right] \! \right] \setminus \emptyset} \bigg(\frac{\lambda s^{|X|} (1- s)^{p - |X|}}{n} \bigg)^{e_X(\boldsymbol{\pi})}.\end{align}

Since $\sum_{X \subseteq \left[ \! \left[ 1, p \right] \! \right] }e_X(\boldsymbol{\pi}) = \binom{n}{2}$ is independent from $\boldsymbol{\pi}$ , we may therefore rewrite

(68) \begin{align} \mathbb{P}_{\rm post} (\boldsymbol{\pi}^* = \boldsymbol{\pi}) & = \frac{1}{Z'} \prod_{X \subseteq \left[ \! \left[ 1, p \right] \! \right] \setminus \emptyset} \bigg(\frac{\lambda s^{|X|} (1- s)^{p - |X|}}{n} \bigg)^{e_X(\boldsymbol{\pi})} \bigg( 1 - \frac{\lambda (1 - (1 - s)^{p})}{n} \bigg)^{-e_{X}(\boldsymbol{\pi})} \nonumber\\[3pt] & = \frac{1}{Z} \exp \bigg( \!-\! \log n\sum_{X \neq \emptyset} e_X(\boldsymbol{\pi}) \bigg(1 - \frac{1}{\log n} \log \bigg( \frac{1 - \frac{\lambda (1 - (1 - s)^{p})}{n} }{\lambda s^{|X|} (1- s)^{p - |X|}}\bigg)\bigg) \bigg)\nonumber\\[3pt] & = \frac{1}{Z} \exp ( \!-\! \mathcal{H}(\boldsymbol{\pi}) \log n (1 + o(1)) ),\end{align}

which is what we wanted to show.

4.2. Symmetries of $\mathcal{H}$

Recall that our aim is to apply Lemma 1 in order to show Theorem 2. To this end, we fix a ball $B = B(\boldsymbol{\pi}_0, r) \subseteq \mathcal{S}_n^{p-1}$ : we wish to construct a subset $\text{Neg} \subseteq \mathcal{S}_n^{p-1}$ and a function $F: B \setminus \text{Neg} \to \mathcal{P}(\mathcal{S}_{n}^{p-1})$ such that:

1. (i) for some $c > 0$ , for any $\boldsymbol{\pi} \in B \setminus \text{Neg}$ ,

   \begin{align*} \mathbb{P}_{\rm post}(\boldsymbol{\pi}) \leq \frac{1}{(cn)!}\mathbb{P}_{\rm post}(F(\boldsymbol{\pi})); \end{align*}

2. (ii) if $\boldsymbol{\pi}' \in \mathcal{S}_n^{p-1}$ , the set $F^{-1}(\boldsymbol{\pi}') \underset{{\rm def}}{=} \{ \boldsymbol{\pi} \in B \setminus \text{Neg}\,:\, \boldsymbol{\pi}' \in F(\boldsymbol{\pi}) \}$ has size at most $\exp O_\mathbb{P}(n)$ ;

3. (iii) $\mathbb{P}_{\rm post}(\text{Neg}) = o_\mathbb{P}(1)$ .

This guarantees that the prerequisites of Lemma 1 are verified, and thus prove Theorem 2.

In this subsection, we describe the symmetries which we intend to make use of, in order to construct our function F. In the following subsection, we define our set Neg verifying (iii), in such a way that, if $\boldsymbol{\pi} \notin \text{Neg}$ , there are many of these symmetries which we may exploit in order to build F. Then, in our final subsection, we explicitly construct F, and prove that conditions (i) and (ii) are verified, thus showing the theorem.

Thus, let us begin by describing the symmetries which we are interested in.

Proposition 9. Let $\boldsymbol{\pi} \in \mathcal{S}_n^{p-1}$ ; set $\mathcal{U}^{(1)} = \mathcal{U}^{(1)}(\boldsymbol{\pi}) = \displaystyle \bigcup_{i=2}^p \pi_i(\Gamma^{(i)})$ .

Then, if $\sigma \in \mathcal{S}_n$ is an automorphism of $\Gamma^{(1)} \cap \mathcal{U}^{(1)}$ , setting $\boldsymbol{\sigma} = (\textrm{Id}, \sigma, \ldots, \sigma) \in \{ \textrm{Id} \} \times \mathcal{S}_n^{p-1}$ ,

(69) \begin{equation} \mathcal{H}(\boldsymbol{\sigma} \boldsymbol{\pi}) \underset{{\rm def}}{=} \mathcal{H}(( \pi_1, \sigma \circ \pi_2, \ldots, \sigma \circ \pi_p)) \leq \mathcal{H}(\boldsymbol{\pi}). \end{equation}

The reason why such symmetries interest us is that $\mathcal{H}(\boldsymbol{\sigma} \boldsymbol{\pi}) \leq \mathcal{H}( \boldsymbol{\pi})$ corresponds (roughly) to having $\mathbb{P}_{\rm post}(\boldsymbol{\sigma} \boldsymbol{\pi}) \geq \mathbb{P}_{\rm post}(\boldsymbol{\pi})$ ; thus, if our function F is of the form

(70) \begin{equation} \boldsymbol{\pi} \mapsto \{ \boldsymbol{\sigma}_1 \boldsymbol{\pi}, \ldots, \boldsymbol{\sigma}_N \boldsymbol{\pi} \}\end{equation}

for some large N, then condition (i) from above will automatically be verified.

Proof. Let $\sigma$ be an automorphism of $\Gamma^{(1)} \cap \mathcal{U}^{(1)}(\boldsymbol{\pi})$ , and denote $\boldsymbol{\sigma} = (\textrm{Id}, \sigma, \ldots, \sigma) \in \{ \textrm{Id}\} \times \mathcal{S}_n^{p-1}$ . We note a few identities relating $e_X(\boldsymbol{\sigma} \boldsymbol{\pi})$ to $e_X(\boldsymbol{\pi})$ .

First of all,

(71) \begin{equation} \sum_{X \neq \emptyset, \{ 1 \} } e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) = e(\sigma(\mathcal{U}^{(1)})) = e(\mathcal{U}^{(1)}) = \sum_{X \neq \emptyset, \{ 1 \} } e_X(\boldsymbol{\pi})\end{equation}

and

(72) \begin{equation} \sum_{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) = e(\Gamma^{(1)}) = \sum_{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]} e_X(\boldsymbol{\pi}).\end{equation}

(These identities are true for any $\sigma \in \mathcal{S}_n$ .) Furthermore, since $\sigma$ is an automorphism of $\Gamma^{(1)} \cap \mathcal{U}^{(1)}$ ,

(73) \begin{align} \sum_{\underset{X \neq \{ 1 \}}{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]}}e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) & = e(\Gamma^{(1)} \cap \sigma(\mathcal{U}^{(1)})) \nonumber\\[3pt] & \geq e((\Gamma^{(1)} \cap \mathcal{U}^{(1)}) \cap \sigma(\Gamma^{(1)} \cap \mathcal{U}^{(1)})) \nonumber\\[3pt] & = e(\Gamma^{(1)} \cap \mathcal{U}^{(1)}) \nonumber\\[3pt] & = \sum_{\underset{X \neq \{ 1 \}}{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]}}e_X(\boldsymbol{\pi}). \end{align}

Putting these three identities together, we obtain:

(74) \begin{align*} \sum_{X \neq \emptyset} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) & = \bigg( \sum_{X \neq \emptyset, \{ 1\}} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) \bigg) + e_{\{ 1 \}}(\boldsymbol{\sigma} \boldsymbol{\pi})\nonumber\\[3pt] & = \sum_{X \neq \emptyset, \{ 1\}} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) + \sum_{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) - \sum_{\underset{X \neq \{ 1 \}}{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]}} e_X(\boldsymbol{\sigma} \boldsymbol{\pi}) \\[-30pt] & \leq \sum_{X \neq \emptyset, \{ 1\}} e_X(\boldsymbol{\pi}) + \sum_{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]} e_X(\boldsymbol{\pi}) - \sum_{\underset{X \neq \{ 1 \}}{1 \in X \subseteq \left[ \! \left[ 1, p \right] \! \right]}} e_X(\boldsymbol{\pi}) \nonumber\\ & = \sum_{X \neq \emptyset} e_X(\boldsymbol{\pi}) \end{align*}

concluding the proof.

4.3. Building many automorphisms

We are now tasked with finding such automorphisms $(\boldsymbol{\sigma}_k)_k$ of $A(\boldsymbol{\pi}) \underset{{\rm def}}{=} \Gamma^{(1)} \cap \mathcal{U}^{(1)}(\boldsymbol{\pi})$ . For the true alignment $\boldsymbol{\pi}^*$ , given that the intersection graph $\Gamma^{(1)} \cap \mathcal{U}^{(1)}(\boldsymbol{\pi}^*)$ is an Erdös–Rényi graph, this is possible when that graph is subcritical. Specifically, we have the following.

Essentially, a graph verifying the property $(P_{c, \delta})$ is one such that any reasonably large subgraph has a lot of symmetries: this is naturally the case for a subcritical Erdös–Rényi graph, which is locally a forest whose trees can be swapped around.

Proof. Let A be as in the proposition. Fix an ordering $(\mathcal{T}_1, \mathcal{T}_2, \ldots)$ of all isomorphism classes of trees. For any $i \geq 1$ , we set $n_{i}$ to be the number of connected components of A which are isomorphic to $\mathcal{T}_i$ . Then, for any i, there exists a constant $c_i$ with

(75) \begin{equation} \frac{n_i}{n} \underset{\mathbb{P}}{\longrightarrow} c_{i}\end{equation}

and $\sum_{i \geq 1} |\mathcal{T}_i|c_i = 1 $ . (This is shown in [Reference Bollobás12].)

Thus, let $\delta > 0$ . If $K > 0$ is large enough, with high probability,

(76) \begin{equation} \sum_{\underset{|\mathcal{T}_i| \leq K}{i \geq 1}} |\mathcal{T}_i| \frac{n_i}{n} \geq 1 - \frac{\delta^2}{2}.\end{equation}

As a result, any $X \subseteq \left[ \! \left[ 1, n \right] \! \right]$ with size at least $\delta n$ and with no outgoing edges, contains a subset $X' \subseteq X$ of size at least $(\delta - {\delta^2}/{2}) n$ which is a disjoint union of connected components of size at most K.

Call $m_i$ the number of connected components of X’ isomorphic to $\mathcal{T}_i$ , for any i: then, we may assume that $m_i \neq 1$ for any $i \geq 1$ , even if that means decreasing the size of X’ to be at least $(\delta - \delta^2) n$ . Consequently, we may build automorphisms of X’ by permuting its connected components which are isomorphic. The number of automorphisms with no fixed point which we can build in this way is at least equal to

(77) \begin{equation} \prod_{\underset{|\mathcal{T}_i| \leq K}{i \geq 1}} \frac{m_i!}{3}\end{equation}

since, for any $k \neq 1$ , there are at least ${k!}/{3}$ elements of $\mathcal{S}_k$ with no fixed points. Since K is independent from n and

\begin{align*}\sum_i m_i \geq \frac{(\delta-\delta^2) n}{K},\end{align*}

this provides us with at least

(78) \begin{equation} \exp\bigg({\frac{(\delta-\delta^2)}{K} n \log n + O(n)} \bigg) \geq (cn)!\end{equation}

automorphisms of X’, for large enough n (and some $c > 0$ independent from n).

Finally, we may extend these automorphisms to be $\textrm{Id}$ on the complement of X’, proving the proposition.

The previous proposition concerns the structure of $A(\boldsymbol{\pi}^*)$ . However, the measure $\mathbb{P}_{\rm post}$ puts a large weight on permutations which are ‘hard to distinguish’ from $\boldsymbol{\pi}^*$ ; such permutations $\boldsymbol{\pi}$ give us a graph $A(\boldsymbol{\pi})$ with a similar structure. This is captured in the following corollary.

Corollary 3. Let $\delta > 0$ . If $\lambda s (1 - (1 - s)^{p-1}) < 1$ , the graph $A(\boldsymbol{\pi}^*) = \Gamma^{(1)} \cap \mathcal{U}^{(1)}(\boldsymbol{\pi}^*)$ is subcritical: there exists $c > 0$ such that A verifies the property $(P_{c, \delta})$ defined above with high probability. Otherwise put, the set

(79) \begin{equation} \textrm{Neg} = \{ \boldsymbol{\pi} \in \mathcal{S}_n^{p-1} | \Gamma^{(1)} \cap \mathcal{U}^{(1)}(\boldsymbol{\pi}) \textit{ does not verify }(P_{c, \delta}) \} \end{equation}

verifies $\mathbb{P}(\pi^* \in \textrm{Neg}) = o(1)$ . Consequently, $\mathbb{P}_{\rm post}(\pi^* \in \textrm{Neg}) = o_{\mathbb{P}}(1)$ .

4.4. Applying the lemma

We now have most of the ingredients necessary to apply Lemma 1 and thus prove Theorem 2.

Proof of Theorem 2. Let $\boldsymbol{\pi}_0 \in \mathcal{S}_n^{p-1}$ and $\varepsilon > 0$ ; we build $F\,:\, B(\boldsymbol{\pi}_0, 1- \varepsilon) \setminus \textrm{Neg} \to \mathcal{P}(\mathcal{S}_n^{p-1})$ verifying the conditions of Lemma 1.

Thus, let $\boldsymbol{\pi} \in B(\boldsymbol{\pi}_0, 1-\varepsilon) \setminus \textrm{Neg}$ . We consider the set $X(\boldsymbol{\pi}) = \{ u \in \left[ \! \left[ 1, n \right] \! \right]\!, \mbox{ for all }i \in \left[ \! \left[ 1, n \right] \! \right]\!, \pi_i(u) = (\boldsymbol{\pi}_0)_i(u) \}$ ; by construction, $|X| \geq \varepsilon n$ . We then consider the set $\overline{X}(\boldsymbol{\pi})$ , which is the minimal subset of $\left[ \! \left[ 1, n \right] \! \right]$ containing X such that no edges of $A(\boldsymbol{\pi})$ link $\overline{X}$ and $\overline{X}^c$ . Since $\boldsymbol{\pi} \notin \textrm{Neg}$ , by Proposition 10, we may construct $(c n)!$ automorphisms $(\sigma_1, \ldots, \sigma_N)$ of the graph A which leave $\overline{X}^c$ fixed, and which have at most $\varepsilon^2 n$ fixed points within $\overline{X}$ , for some $c > 0$ independent of n. The function F is defined as mapping $\boldsymbol{\pi}$ to the set $\{ \boldsymbol{\sigma}_i \boldsymbol{\pi}, 1 \leq i \leq N \}$ (where, as defined earlier, $\boldsymbol{\sigma}_i = (\textrm{Id}, \sigma_i, \ldots, \sigma_i)$ ).

Does F satisfy the conditions of Lemma 1? Condition (i) is verified by construction: for any $\boldsymbol{\pi} \in B(\boldsymbol{\pi}_0, 1-\varepsilon)$ ,

(80) \begin{equation} \mathbb{P}_{\rm post}(\boldsymbol{\pi}) \leq \frac{e^{o(n \log n)}}{(cn)!} \mathbb{P}_{\rm post}(F(\boldsymbol{\pi})) \end{equation}

since $\mathcal{H}(\boldsymbol{\sigma}_i \boldsymbol{\pi}) \leq \mathcal{H}(\boldsymbol{\pi})$ for any $1 \leq i \leq N$ . Condition (ii), however, is less obvious. We prove that it holds with $K_n(r) = \exp O_{\mathbb{P}}(n)$ .

Let $\boldsymbol{\pi}' \in \mathcal{S}_n^{p-1}$ . We may partition $F^{-1}(\boldsymbol{\pi}') = \{ \boldsymbol{\pi}: \boldsymbol{\pi}' \in F(\boldsymbol{\pi}) \}$ as

(81) \begin{align} F^{-1}(\boldsymbol{\pi}') & = \bigsqcup_{\underset{S \text{ subgraph of } (\Gamma^{(1)})|_Z}{Y \subseteq Z \subseteq \left[ \! \left[ 1, n \right] \! \right]}} \{ \boldsymbol{\pi} \in F^{-1}(\boldsymbol{\pi}')\,:\, X(\boldsymbol{\pi}) = Y, \overline{X}(\boldsymbol{\pi}) = Z, (\Gamma^{(1)}\cap \mathcal{U}^{(1)}(\boldsymbol{\pi}))|_Z = S \} \end{align}

(82) \begin{align} & \underset{{\rm def}}{=} \bigsqcup_{\underset{S \text{ subgraph of } (\Gamma^{(1)})|_Z}{Y \subseteq Z \subseteq \left[ \! \left[ 1, n \right] \! \right]}} W(Y, Z, S).\end{align}

Then, there are only $\exp O_\mathbb{P}(n)$ possible values for (Y, Z, S), as $|\Gamma^{(1)}| = O_{\mathbb{P}}(n)$ . We therefore just need to show that $|W(Y, Z, S)| \leq \exp O_\mathbb{P}(n)$ for any triplet (Y, Z, S). To this end, let $\boldsymbol{\pi}_1, \boldsymbol{\pi}_2 \in W(Y, Z, S)$ : by construction, $\boldsymbol{\pi}_2 \boldsymbol{\pi}_1^{-1}$ is of the form $(\textrm{Id}, \sigma, \ldots, \sigma)$ for some $\sigma \in \mathcal{S}_n$ . Furthermore, given such a $\sigma$ , we know that $\sigma|_Y = \textrm{Id}$ , and $\sigma|_Z$ must be an automorphism of S. This allows us to use the following lemma, whose proof is deferred to the appendix.

Indeed, by construction of $\overline{X}(\boldsymbol{\pi}_1) = Z$ , any connected component of S intersects Y, and so $\sigma$ must send it to itself. This gives us at most $e^{{n}/{e}}\prod_{i=1}^n (d_i!) $ possible values for $\sigma$ , showing that (ii) is verified as

(83) \begin{equation} \log \prod_{i=1}^n (d_i!) \leq \sum_{i=1}^n d_i \log d_i = O_{\mathbb{P}}(n \mathbb{E}[d_i \log d_i ]) = O_{\mathbb{P}}(n), \end{equation}

given that $\mathbb{E}[d_i \log d_i ] = O(1)$ . Applying Lemma 1 thus proves Theorem 2.