Enumeration of regular multipartite hypergraphs

Mikhail Isaev
School of Mathematics and Statistics
University of New South Wales
Sydney, NSW, Australia
m.isaev@unsw.edu.au Supported by the Australian Research Council grant DP250101611. Tamás Makai
Department of Mathematics
LMU Munich
Munich 80333, Germany
makai@math.lmu.de
Supported by the Australian Research Council grant DP190100977. Brendan D. McKay¹¹footnotemark: 1
School of Computing
Australian National University
Canberra, ACT 2601, Australia
brendan.mckay@anu.edu.au

Abstract

We determine the asymptotic number of regular multipartite hypergraphs, also known as multidimensional binary contingency tables, for all values of the parameters.

1 Introduction

Let $[n]:=\{1,\ldots,n\}$ be a vertex set partitioned into $r$ disjoint classes $V_{1},\ldots,V_{r}$ . We consider multipartite $r$ -uniform hypergraphs such that every edge has exactly one vertex in each class and there are no repeated edges. We call such hypergraphs $(r,r)$ -graphs. Note that $(2,2)$ -graphs are just bipartite graphs. These objects are also known as $r$ -dimensional binary contingency tables.

The degree of a vertex is the number of edges that contain it. If every vertex has degree $d$ then the hypergraph is called $d$ -regular, which implies (unless there are no edges) that all the classes have the same size. We are interested in the number of labelled $d$ -regular $(r,r)$ -graphs with $n=mr$ vertices where every partition class contains exactly $m$ vertices. We denote this number by $H_{r}(d,m)$ .

In order to motivate our answer, we start with a non-rigorous argument. Consider a $(r,r)$ -graph $G$ with classes of size $m$ and $md$ edges, created by choosing uniformly at random $md$ distinct edges out of the $m^{r}$ available. Let $\mathcal{R}$ be the event that $G$ is $d$ -regular. Then

H_{r}(d,m)=\binom{m^{r}}{md}\operatorname{\mathbb{P}}\nolimits(\mathcal{R}).

To estimate $\operatorname{\mathbb{P}}\nolimits(\mathcal{R})$ , we first look at one class $V_{i}$ , $i\in[r]$ . The probability of the event $\mathcal{R}_{i}$ that every vertex in $V_{i}$ has degree $d$ is

\operatorname{\mathbb{P}}\nolimits(\mathcal{R}_{i}):=\frac{\binom{m^{r-1}}{d}^{m}}{\binom{m^{r}}{md}},

since $\binom{m^{r-1}}{d}$ is the number of ways to choose $d$ edges of a $(r,r)$ -graph incident to one vertex in $V_{i}$ and these choices are independent. If the events $\mathcal{R}_{i}$ were also independent, we would have $\operatorname{\mathbb{P}}\nolimits(\mathcal{R})=\prod_{i=1}^{r}\operatorname{\mathbb{P}}\nolimits(\mathcal{R}_{i})$ , providing the estimate

\hat{H}_{r}(d,m):=\binom{m^{r}}{md}\prod_{i\in[r]}\operatorname{\mathbb{P}}\nolimits(\mathcal{R}_{i})=\frac{\binom{m^{r-1}}{d}^{rm}}{\binom{m^{r}}{md}^{r-1}}.

(1.1)

Of course, the events $\mathcal{R}_{i}$ are not independent, but comparing $\hat{H}_{r}(d,m)$ with the correct value $H_{r}(d,m)$ will be instructive.

In the case of $(2,2)$ -graphs, that is for $d$ -regular bipartite graphs, we have as $n=2m\to\infty$ that

H_{2}(d,m)=(1+o(1))\,e^{-1}\hat{H}_{2}(d,m)

except in the trivial cases $d=0$ and $d=m$ . This is the result of three previous investigations. The sparse range was solved by McKay [7] (see also [4]), the intermediate range of densities by Liebenau and Wormald [6], and the dense range by Canfield and McKay [3].

In this paper we show that, for $r\geqslant 3$ , the estimate $\hat{H}_{r}(d,m)$ is asymptotically correct.

Theorem 1.1.

Let $n=rm\to\infty$ , where $m=m(n)$ and $r=r(n)\geqslant 3$ . Then for any $0\leqslant d\leqslant m^{r-1}$ we have

H_{r}(d,m)=(1+o(1))\,\hat{H}_{r}(d,m).

In the sparse range we employ a combinatorial model introduced in 1972 by Békéssy, Békéssy and Komlós [1] and later developed under the name “configurations” by Bollobás and others [2, 9]. For $r\geqslant 4$ we show that this model produces a uniform random $d$ -regular $(r,r)$ -graph with probability $1-o(1)$ . When $r=3$ , this is insufficient to cover all of the sparse cases. However, the result can be extended to the whole sparse range by applying the method of switchings [7]. Furthermore, for the dense regime we use the complex-analytic approach, relying on the machinery developed by Isaev and McKay [5].

In Section 2 we introduce our key lemmas for both the sparse and the dense regimes and prove Theorem 1.1. We then consider the sparse regime in Section 3 and the dense regime in Section 4.

2 The four regimes covering all possibilities

We prove Theorem 1.1, considering the following four regimes separately.

(a)

$d=o(m)$ for $r=3$ ;
(b)

$rd^{2}=o(m^{r-2})$ for $r\geqslant 3$ ;
(c)

The complements of cases (a) and (b); i.e., replacing $d$ by $m^{r-1}-d$ .
(d)

$\min\{d,m^{r-1}-d\}=\Omega(r^{16}m)$ .

Each of the following lemmas covers one of the regimes (a), (b), and (d), while region (c) follows from (a) and (b) as $H_{r}(m^{r-1}{-}d,m)=H_{r}(d,m)$ and $\hat{H}_{r}(m^{r-1}{-}d,m)=\hat{H}_{r}(d,m)$ .

Lemma 2.1.

If $d=o(m)$ then $H_{3}(d,m)=(1+o(1))\hat{H}_{3}(d,m)$ .

Lemma 2.2.

If $r\geqslant 3$ and $rd^{2}=o(m^{r-2})$ then $H_{r}(d,m)=(1+o(1))\hat{H}_{r}(d,m)$ .

Lemma 2.3.

If $r\geqslant 3$ and $\min\{d,m^{r-1}-d\}=\Omega(r^{16}m)$ , then $H_{r}(d,m)=(1+o(1))\hat{H}_{r}(d,m)$ .

We prove Lemmas 2.1 and 2.2 in Section 3 and Lemma 2.3 in Section 4. To see that four regimes (a)–(d) together cover $0\leqslant d\leqslant m^{r-1}$ , we employ the following short lemma. It is stated in a slightly more general form as it will be useful in a few other places in the paper.

Lemma 2.4.

Let $a,b,c,r_{0}$ be constants with $c>0$ . Then, as $n\to\infty$ with $r\geqslant r_{0}$ and $m\geqslant 2$ ,

r^{a}m^{-cr+b}=O(n^{b-cr_{0}}).

Proof.

Recall that $n=mr$ . If $m=2$ then $r^{a}m^{-cr+b}=e^{-\Omega(n)}$ so the stated bound is immediate. For $m\geqslant 3$ , to maximise the function $f_{n}(r):=r^{a}(n/r)^{-cr+b}$ over $r$ , we observe that its derivative

f^{\prime}_{n}(r)=-r^{cr+a-b-1}n^{b-cr}\bigl{(}cr(\log n-\log r-1)+b-a\bigr{)}

is negative for $r>\frac{a-b}{c(\log 3-1)}$ since $\log n-\log r=\log m\geqslant\log 3$ by assumption. Thus, the maximum occurs for $r=O(1)$ , where the implicit constant in $O(1)$ depends only on $a,b,c$ . The result follows as for every $r\geqslant r_{0}$ satisfying $r=O(1)$ we have $r^{a}m^{-cr+b}=O(n^{b-cr_{0}})$ . ∎

Proof of Theorem 1.1.

First note that the statement holds trivially when $m=1$ . For the remainder of the proof assume that $m\geqslant 2$ .

For $r=3$ the result holds by Lemma 2.1 for the regime $d=o(m)$ , and when $m^{2}-d=o(m)$ , by taking the complement hypergraph. On the other hand Lemma 2.3 proves the result for $\min\{d,m^{2}-d\}=\Omega(m)$ , covering the remaining possible degrees.

For $r\geqslant 4$ , Lemma 2.2 applied to the hypergraph or its complement, proves the result for the regimes $d=o(m^{(r-2)/2}r^{-1/2})$ and $m^{r-1}{-}d=o(m^{(r-2)/2}r^{-1/2})$ . Furthermore Lemma 2.3 proves the result for $\min\{d,m^{r-1}-d\}=\Omega(r^{16}m)$ . Observe that all possible degrees are covered if $m^{(r-2)/2}r^{-1/2}=\Omega(r^{16}m)$ or equivalently

r^{33}m^{4-r}=O(1),

which follows from Lemma 2.4, by taking $a=33$ , $b=r_{0}=4$ , and $c=1$ . ∎

3 Sparse regimes

In this section we prove Lemma 2.1 and Lemma 2.2. First, we introduce the configuration model. Take $r$ classes $V_{1},\ldots,V_{r}$ of $m$ vertices each, and attach $d$ spines to each vertex. A spine set is a set of $r$ spines, one from each class. A configuration is an unordered partition of the $rmd$ spines into $md$ spine sets. Thus, there are $((md)!)^{r-1}$ possible configurations. Each configuration provides a $d$ -regular multi-hypergraph where each edge consists of the vertices to which the spines in a spine set of the configuration are attached.

If $G$ is a simple $d$ -regular $(r,r)$ -graph, the number of configurations which provide $G$ is $(d!)^{mr}$ . Therefore,

H_{r}(d,m)=\frac{((md)!)^{r-1}}{(d!)^{mr}}P_{r}(d,m),

(3.1)

where $P_{r}(d,m)$ is the probability that a uniform random configuration provides a simple hypergraph, that is, no two spine sets give the same hyperedge.

3.1 Proof of Lemma 2.2

Observe that

\hat{H}_{r}(d,m)=\frac{\binom{m^{r-1}}{d}^{rm}}{\binom{m^{r}}{md}^{r-1}}=\frac{\mathopen{}\mathclose{{}\left((m^{r-1})_{d}}\right)^{rm}}{\mathopen{}\mathclose{{}\left((m^{r})_{md}}\right)^{r-1}}\cdot\frac{((md)!)^{r-1}}{(d!)^{rm}}=e^{O(rd^{2}m^{2-r})}\cdot\frac{((md)!)^{r-1}}{(d!)^{rm}}.

For the last equality in the above, we used the bounds

(m^{r-1})_{d}=m^{d(r-1)}e^{O(d^{2}m^{1-r})}\quad\text{and}\quad(m^{r})_{md}=m^{rmd}e^{O(d^{2}m^{2-r})}.

Therefore, by assumptions and (3.1), it is sufficient to show that $P_{r}(d,m)=1-o(1)$ . This is obvious for $d=0,1$ so we can assume $d\geqslant 2$ .

Consider any two spines $s_{1}$ and $s_{2}$ attached to the same vertex from $V_{1}$ . The probability that two spine sets of a random configuration that contain $s_{1}$ and $s_{2}$ coincide is

\raise 0.21529pt\hbox{\small$\displaystyle\frac{(md)^{r-1}(d-1)^{r-1}((md-2)!)^{r-1}}{((md)!)^{r-1}}$}=\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{d-1}{md-1}$}}\right)^{r-1}\leqslant m^{1-r}.

Indeed, $(md)^{r-1}$ represents the number of choices for the spine set containing $s_{1}$ . Then, we have $(d-1)^{r-1}$ ways to form the spine set containing $s_{2}$ which corresponds to the same edge. Finally, $(md-2)!)^{r-1}$ is the number of ways to complete the configuration. By the union bound over all possible choices for $s_{1}$ and $s_{2}$ , we obtain

1-P_{r}(d,m)\leqslant m\binom{d}{2}m^{1-r}=o(1),

completing the proof.

3.2 The case $r=3$

For $r=3$ we need to cover the range $d=o(m)$ , in order to complement the range given by the complex-analytic approach detailed in the next section. To do this, we use a switching argument to estimate the probability that a uniform random configuration provides a simple $(r,r)$ -graph. Throughout this section, we can also assume that $d\geqslant 2$ since Lemma 2.2 already covers $d=O(1)$ and even much more.

Define

M:=\lfloor 8d^{2}m^{-1}+\log m\rfloor.

Lemma 3.1.

Suppose $r=3$ and $2\leqslant d=o(m)$ . Then, with probability $1-o(1)$ , a uniform random configuration:

(a)

provides no sets of 3 equal edges; and
(b)

provides at most $M$ sets of 2 equal edges.

Proof.

Similarly to Lemma 2.2, we consider any three spines $s_{1}$ , $s_{2}$ , $s_{3}$ attached to the same vertex from $V_{1}$ . The probability that the corresponding three edges of a random configuration coincide is

\raise 0.21529pt\hbox{\small$\displaystyle\frac{(md)^{2}(d-1)^{2}(d-2)^{2}((md-3)!)^{2}}{((md)!)^{2}}$}=\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{(d-1)(d-2)}{(md-1)(md-2)}$}}\right)^{2}\leqslant m^{-4}.

Taking the union bound over all choices of the spines $s_{1},s_{2},s_{3}$ , we get that the probability of having $3$ equal edges is at most $m\binom{d}{3}\cdot m^{-4}=o(1)$ , which proves part (a).

For $i=1,\ldots,M+1$ , let $\{s_{1}^{i},s_{2}^{i}\}$ be distinct pairs of spines, such that each spine in a pair is attached to the same vertex of $V_{1}$ . The probability that the edges corresponding to the spine sets containing $s_{1}^{i}$ and $s_{2}^{i}$ of a random configuration coincide for all $i$ is at most

\frac{(md)^{2(M+1)}d^{2(M+1)}((md-2(M+1))!)^{2}}{((md)!)^{2}}\leqslant m^{-2(M+1)}e^{O(M^{2}/md)}=(1+o(1))m^{-2(M+1)},

where we used the assumption that $d=o(m)$ and the definition of $M$ . The number of ways to choose all pairs $\{s_{1}^{i},s_{2}^{i}\}_{i\in[M+1]}$ is at most $\binom{dm}{M+1}d^{M+1}\leqslant(d^{2}m)^{M+1}/(M+1)!$ . Applying the union bound implies that the probability of having $M+1$ pairs of 2 equal edges is at most

\frac{\bigl{(}(1+o(1))m^{-1}d^{2}\bigr{)}^{M+1}}{(M+1)!}\leqslant\biggl{(}\frac{e(1+o(1))d^{2}m^{-1}}{M+1}\biggr{)}^{\!M+1}\leqslant((1+o(1))e/8)^{\log m}=o(m^{-1}),

proving part (b). ∎

From now on, we only consider configurations satisfying Lemma 3.1 (a) and (b). Our task is to estimate the probability that the configuration provides no double edges.

For a given configuration, with a slight abuse of notation, we will call a spine set in the configuration an edge. It is a simple edge if no other edge uses the same vertices, and half of a double edge if it is an edge and there is a different edge with the same vertices. Two spine sets are parallel if they use the same vertices.

Refer to caption — Figure 1: Switching operations for $r=3$ .

We define forward and backward switchings involving 21 distinct spines as follows

\{a,b,c\}\cup\{a_{j},b_{j},c_{j}\mathrel{:}1\leqslant j\leqslant 6\}.

In all cases, the letter in a spine name (and the colour) indicates which class $V_{i}$ it belongs to. All of these spines must be attached to distinct vertices, except that there may be coincidences within the sets

\{v(a),v(a_{1}),v(a_{2})\},\quad\{v(b),v(b_{1}),v(b_{2})\},\quad and\quad\{v(c),v(c_{1}),v(c_{2})\},

(3.2)

where $v(x)$ is the vertex to which spine $x$ is attached. To describe our switching operations, we define two families of labelled spine sets (see Figure 1).

	$\displaystyle\mathcal{F}_{1}$	$\displaystyle=\bigl{\{}\{a,b,c\},\{a_{1},b_{5},c_{3}\},\{a_{2},b_{4},c_{6}\},\{a_{3},b_{1},c_{5}\},\{a_{4},b_{6},c_{2}\},\{a_{5},b_{3},c_{1}\},\{a_{6},b_{2},c_{4}\}\bigr{\}}$
	$\displaystyle\mathcal{F}_{2}$	$\displaystyle=\bigl{\{}\{a,b_{2},c_{1}\},\{a_{1},b,c_{2}\},\{a_{2},b_{1},c\}\bigr{\}}\cup\bigl{\{}\{a_{j},b_{j},c_{j}\}\mathrel{:}3\leqslant j\leqslant 6\bigr{\}}.$

Informally, a forward switching takes a configuration $C_{1}$ whose edges include $\mathcal{F}_{1}$ , with $\{a,b,c\}$ being half of a double edge, and replaces the edges in $\mathcal{F}_{1}$ by the edges in $\mathcal{F}_{2}$ , thereby creating a new configuration $C_{2}$ without that double edge. A reverse switching is the inverse operation that changes $C_{2}$ into $C_{1}$ . We will restrict $C_{1}$ and $C_{2}$ so that neither the forward nor reverse switchings create or destroy any double edge other than $\{a,b,c\}$ and its parallel spine set.

In detail, the requirements for $C_{1}$ are as follows.

(F1)

$\{a,b,c\}$ is half of a double edge, while all the other spine sets in $\mathcal{F}_{1}$ are simple edges;
(F2)

none of the spine sets of $\mathcal{F}_{2}$ are parallel to edges of $C_{1}$ ;
(F3)

no two of the spine sets $\{a,b_{2},c_{1}\},\{a_{1},b,c_{2}\},\{a_{2},b_{1},c\}$ are parallel.

The requirements for $C_{2}$ are as follows.

(R1)

$\{a,b,c\}$ is parallel to a simple edge of $C2$ , while all the other spine sets in $\mathcal{F}_{2}$ are simple edges;
(R2)

none of the spine sets in $\mathcal{F}_{1}$ are parallel to edges of $C_{2}$ .

We will need the following summation result from [4, Cor. 4.5].

Lemma 3.2 (Greenhill, McKay, Wang [4]).

Let $M\geqslant 2$ be an integer and, for $1\leqslant i\leqslant M$ , let real numbers $A(i)$ , $B(i)$ be given such that $A(i)\geqslant 0$ and $1-(i-1)B(i)\geqslant 0$ . Define

A_{1}:=\min_{i\in[M]}A(i),\quad A_{2}:=\max_{i\in[M]}A(i),\quad C_{1}:=\min_{i\in[M]}A(i)B(i),\quad C_{2}:=\max_{i\in[M]}A(i)B(i).

Suppose that there exists $\hat{c}$ with $0<\hat{c}<\frac{1}{3}$ such that $\max\{A/M,|C|\}\leqslant\hat{c}$ for all $A\in[A_{1},A_{2}]$ , $C\in[C_{1},C_{2}]$ . Define $n_{0},\ldots,n_{M}$ recursively by $n_{0}=1$ and

n_{i}:=\frac{n_{i-1}}{i}A(i)(1-(i-1)B(i)),\qquad\text{for $i\in[M]$.}

Then, the following bounds hold:

\exp\bigl{(}A_{1}-\lower 0.6458pt\hbox{\large$\frac{1}{2}$}A_{1}C_{2}\bigr{)}-(2e\hat{c}\bigr{)}^{M}\leqslant\sum_{i=0}^{M}n_{i}\leqslant\exp\bigl{(}A_{2}-\lower 0.6458pt\hbox{\large$\frac{1}{2}$}A_{2}C_{1}+\lower 0.6458pt\hbox{\large$\frac{1}{2}$}A_{2}C_{1}^{2}\bigr{)}+(2e\hat{c})^{M}.

Now we use Lemma 3.2 to estimate the probability $P_{3}(d,m)$ that the configuration model gives a simple hypergraph.

Lemma 3.3.

If $2\leqslant d=o(m)$ , then $P_{3}(d,m)=(1+o(1))\exp\Bigl{(}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}\Bigr{)}.$

Proof.

For $0\leqslant\ell\leqslant M$ , let $\mathcal{T}(\ell)$ be the set of configurations that have no edges of multiplicity greater than 2 and exactly $\ell$ double edges. By Lemma 3.1, a random configuration belongs to $\bigcup_{\ell=0}^{M}\mathcal{T}(\ell)$ with probability $1-o(1)$ . Note that our assumption $d=o(m)$ implies that

\varepsilon_{\ell}:=\frac{\ell+d}{md}\leqslant\frac{M+d}{md}=o(1).

(3.3)

Consider a configuration $C_{1}$ in $\mathcal{T}(\ell)$ for $1\leqslant\ell\leqslant M$ . We claim that

the number of available forward switchings is

(1+O(\varepsilon_{\ell}))\,2\ell m^{6}d^{6}

(3.4)

Indeed, half of a double edge, $\{a,b,c\}$ , can be chosen in $2\ell$ ways. Then we can choose 6 labelled simple edges, vertex-disjoint apart from the allowed coincidences described in (3.2), in $(md-2\ell-O(d))^{6}$ ways.

The possibility forbidden by (F2) that an edge $e$ in $C_{1}$ is parallel to $\{a_{j},b_{j},c_{j}\}$ for some $3\leqslant j\leqslant 6$ eliminates $O(m^{4}d^{7})$ cases, since we have $O(md)$ choices for the edge $e$ , then there are $O(d^{3})$ ways to select the three spines $a_{j},b_{j}$ and $c_{j}$ (and thus also the edges in $\mathcal{F}_{1}$ containing these spines), such that $\{a_{j},b_{j},c_{j}\}$ is parallel to $e$ , while the choice of the remaining edges in $\mathcal{F}_{1}$ is at most $(md)^{3}$ . Similarly, the possibility that one of $\{a,b_{2},c_{1}\}$ , $\{a_{1},b,c_{2}\}$ , or $\{a_{2},b_{1},c\}$ violates (F2) because it is parallel to an edge eliminates $O(m^{4}d^{7})$ cases: there are $O(d)$ ways to pick an edge $e$ in $C_{1}$ sharing a vertex with $a,b$ or $c$ , at most $d^{2}$ ways to choose two edges of $\mathcal{F}_{1}$ sharing a vertex with $e$ , and at most $(md)^{4}$ ways to pick the remaining edges for $\mathcal{F}_{1}$ . Finally, the number of ways to violate (F3) is bounded by $O(m^{3}d^{6})$ : for example, for $\{a,b_{2},c_{1}\}$ and $\{a_{1},b,c_{2}\}$ to be parallel we can pick the three edges of $\mathcal{F}_{1}$ containing $c_{2}$ , $a_{2}$ and $b_{1}$ in at most $m^{3}d^{3}$ ways, and then the number of ways to pick the remaining edges of $\mathcal{F}_{1}$ is at most $d^{3}$ (as each of these edges has to share a vertex with $a,b$ and $c_{2}$ respectively). Using (3.3), we find that the number of available forward switchings is

2\ell(md-2\ell-O(d))^{6}-O(m^{4}d^{7})-O(m^{3}d^{6})=(1+O(\varepsilon_{\ell}))\,2\ell m^{6}d^{6},

which proves claim (3.4).

Next, consider a configuration $C_{2}$ in $\mathcal{T}(\ell-1)$ for $1\leqslant\ell\leqslant M$ . We claim that

the number of available reverse switchings is

(1+O(\varepsilon_{\ell}))\,m^{5}d^{5}(d-1)^{3}

(3.5)

Indeed, we can choose a simple edge $\{a^{\prime},b^{\prime},c^{\prime}\}$ in $md-2\ell+2$ ways, and then choose one additional spine $a,b,c$ on $v(a^{\prime}),v(b^{\prime}),v(c^{\prime})$ respectively in $(d-1)^{3}$ ways. This leads to three edges, all of which are allowed, even if they coincide in a vertex, unless they are halves of double edges. The latter happens in at most $O(\ell(d-1)^{3})$ cases. So we have $(md-O(\ell))(d-1)^{3}$ choices for $\{a,b_{2},c_{1}\},\{a_{1},b,c_{2}\},\{a_{2},b_{1},c\}$ . For each such choice, we have $(md-2\ell+2-O(d))^{4}$ ways to choose the 4 simple edges $\{a_{j},b_{j},c_{j}\}$ for $3\leqslant j\leqslant 6$ . Of those choices, $O(m^{2}d^{5})$ violate condition (R2): for example, for $\{a_{1},b_{5},c_{3}\}$ to be parallel to an edge $e$ , the number of choices of $e$ and the edges containing $c_{3}$ and $b_{5}$ in $\mathcal{F}_{2}$ is at most $d^{3}$ while the number of ways to pick the remaining two edges is at most $(md)^{2}$ . Using (3.3), we find that the number of reverse switchings is

(md-O(\ell))(d-1)^{3}\bigl{(}(md-2\ell+2-O(d))^{4}-O(m^{2}d^{5})\bigr{)}=(1+O(\varepsilon_{\ell}))\,m^{5}d^{5}(d-1)^{3},

which proves claim (3.5).

Consequently, by a simple double counting argument for the number of forward/reverse switchings between $\mathcal{T}(\ell)$ and $\mathcal{T}(\ell-1)$ , we deduce that

\frac{\mathopen{|}\mathcal{T}(\ell)\mathclose{|}}{\mathopen{|}\mathcal{T}(\ell-1)\mathclose{|}}=(1+O(\varepsilon_{\ell}))\raise 0.21529pt\hbox{\small$\displaystyle\frac{(d-1)^{3}}{2\ell dm}$}=\bigl{(}1+O(m^{-1}+(\ell-1)(md)^{-1})\bigr{)}\raise 0.21529pt\hbox{\small$\displaystyle\frac{(d-1)^{3}}{2\ell dm}$}.

Since $d=o(m)$ , to apply Lemma 3.2 for $n_{\ell}:=\lower 0.6458pt\hbox{\large$\frac{\mathopen{|}\mathcal{T}(\ell)\mathclose{|}}{\mathopen{|}\mathcal{T}(\ell-1)\mathclose{|}}$}$ , we can take

A(\ell)=(1+O(m^{-1}))\raise 0.21529pt\hbox{\small$\displaystyle\frac{(d-1)^{3}}{2dm}$}\leqslant(1+o(1))\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}\quad\text{and}\quad B(\ell)=O((md)^{-1}).

Recalling that $M=\lfloor 8d^{2}m^{-1}+\log m\rfloor\geqslant 8d^{2}m^{-1}$ , we obtain

A(\ell)/M\leqslant\lower 0.6458pt\hbox{\large$\frac{1}{8}$}\qquad\text{and}\qquad C_{1},C_{2}=O\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{d}{m^{2}}$}}\right)=o(1)\leqslant\lower 0.6458pt\hbox{\large$\frac{1}{8}$}.

Therefore, we can take $\hat{c}=\lower 0.6458pt\hbox{\large$\frac{1}{8}$}$ . Applying Lemma 3.2, we obtain

	$\displaystyle\raise 0.21529pt\hbox{\small$\displaystyle\frac{\mathopen{\|}\mathcal{T}(0)\mathclose{\|}}{\sum_{\ell=0}^{M}\,\mathopen{\|}\mathcal{T}(\ell)\mathclose{\|}}$}=\biggl{(}\sum_{\ell=0}^{M}n_{\ell}\biggr{)}^{\!-1}$	$\displaystyle=\exp\Bigl{(}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{(d-1)^{3}}{2dm}$}+o(1)\Bigr{)}+O((e/4)^{M})$
		$\displaystyle=\exp\Bigl{(}-\frac{(d-1)^{3}}{2dm}+o(1)\Bigr{)},$

where we used

M\log(4/e)\geqslant 3d^{2}m^{-1}+\Theta(\log m)\rightarrow\infty,

to derive the last equality. ∎

Now we can compare $H_{3}(d,m)$ with the estimate $\hat{H}_{3}(d,m)$ given in (1.1).

Proof of Lemma 2.1.

Combining (3.1) and Lemma 3.3, we have

H_{3}(d,m)=\raise 0.21529pt\hbox{\small$\displaystyle\frac{((md)!)^{2}}{(d!)^{3m}}$}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}+o(1)}\right).

To estimate $\hat{H}_{3}(d,m)$ we argue similarly to Lemma 2.2. First, using $d=o(m)$ , observe that

	$\displaystyle(m^{2})_{d}=m^{2d}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d(d-1)}{2m^{2}}$}+O\mathopen{}\mathclose{{}\left(d^{3}m^{-4}}\right)}\right)=m^{2d}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m^{2}}$}+o(m^{-1})}\right)$
	$\displaystyle(m^{3})_{md}=m^{3md}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}+O(dm^{-2}+d^{3}m^{-3})}\right)=m^{3md}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}+o(1)}\right).$

We obtain

\hat{H}_{3}(d,m)=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\binom{m^{2}}{d}^{3m}}{\binom{m^{3}}{md}^{2}}$}=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\mathopen{}\mathclose{{}\left((m^{2})_{d}}\right)^{3m}}{\mathopen{}\mathclose{{}\left((m^{3})_{md}}\right)^{2}}$}\cdot\raise 0.21529pt\hbox{\small$\displaystyle\frac{((md)!)^{2}}{(d!)^{3m}}$}=\frac{((md)!)^{2}}{(d!)^{3m}}\exp\mathopen{}\mathclose{{}\left(-\raise 0.21529pt\hbox{\small$\displaystyle\frac{d^{2}}{2m}$}+o(1)}\right),

as required. ∎

4 Dense range

Throughout this section we always assume that the partition classes of the vertex set $[n]$ are

V_{t}:=\{(t-1)m+1,\ldots,tm\},\qquad\text{for $t\in[r]$.}

Let ${\mathcal{S}_{r}(m)}$ denote the set of all possible edges of $(r,r)$ -graphs. Clearly, we have

n=mr\qquad\text{and}\qquad|{\mathcal{S}_{r}(m)}|=m^{r}.

Let

\lambda=\raise 0.21529pt\hbox{\small$\displaystyle\frac{d}{m^{r-1}}$}\qquad\text{ and }\qquad\varLambda=\lambda(1-\lambda).

In this section we consider the dense range defined by $\varLambda=\Omega(r^{16}m^{2-r})$ , which is equivalent to $\min\{d,m^{r-1}{-}d\}=\Omega(r^{16}m)$ , using the complex-analytic approach. We establish a generating function for $(r,r)$ -graphs by degrees, then extract the required coefficient via Fourier inversion and perform asymptotic analysis on the resulting multidimensional integrals.

We will use $\mathopen{\|}\cdot\mathclose{\|}_{p}$ for the vector $p$ -norm and its induced matrix norm for $p=1,2,\infty$ .

4.1 An integral for the number of $(r,r)$ -graphs

The generating function for $(r,r)$ -graphs by degree sequence is

\prod_{e\in{\mathcal{S}_{r}(m)}}\Bigl{(}1+\prod_{j\in e}x_{j}\Bigr{)}.

Using Cauchy’s coefficient formula, the number of $d$ -regular $(r,r)$ -graphs is

	$\displaystyle H_{r}(d,m)$	$\displaystyle=[x_{1}^{d}\cdots x_{n}^{d}]\prod_{e\in{\mathcal{S}_{r}(m)}}\Bigl{(}1+\prod_{j\in e}x_{j}\Bigr{)}$
		$\displaystyle=\frac{1}{(2\pi i)^{rm}}\,\oint\cdots\oint\,\raise 0.21529pt\hbox{\small$\displaystyle\frac{\prod_{e\in{\mathcal{S}_{r}(m)}}\bigl{(}1+\prod_{j\in e}x_{j}\bigr{)}}{\prod_{j\in[rm]}x_{j}^{d+1}}$}\,d{\boldsymbol{x}}.$

Considering the contours $x_{j}=\mathopen{}\mathclose{{}\left(\lower 0.6458pt\hbox{\large$\frac{\lambda}{1-\lambda}$}}\right)^{1/r}e^{i\theta_{j}}$ for $j\in[n]$ , we obtain

	$\displaystyle H_{r}(d,m)$	$\displaystyle=(2\pi)^{-n}\,\mathopen{}\mathclose{{}\left(\lower 0.6458pt\hbox{\large$\frac{\lambda}{1-\lambda}$}}\right)^{-dm}\ \int_{-\pi}^{\pi}\!\cdots\!\int_{-\pi}^{\pi}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\prod_{e\in{\mathcal{S}_{r}(m)}}\bigl{(}1+\frac{\lambda}{1-\lambda}\prod_{j\in e}e^{i\theta_{j}}\bigr{)}}{\exp\bigl{(}i\sum_{j\in[rm]}d\theta_{j}\bigr{)}}$}\,d\boldsymbol{\theta}$
		$\displaystyle=\raise 0.21529pt\hbox{\small$\displaystyle\frac{(2\pi)^{-n}}{\mathopen{}\mathclose{{}\left(\lambda^{\lambda}(1-\lambda)^{1-\lambda}}\right)^{m^{r}}}$}\int_{U_{n}(\pi)}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\prod_{e\in{\mathcal{S}_{r}(m)}}\bigl{(}1+\lambda\bigl{(}e^{i\sum_{j\in e}\theta_{j}}-1\bigr{)}\bigr{)}}{\exp\bigl{(}id\sum_{j\in[rm]}\theta_{j}\bigr{)}}$}\,d\boldsymbol{\theta},$		(4.1)

where

U_{n}(\rho)=\{{\boldsymbol{x}}\in{\mathbb{R}}^{n}\mathrel{:}\mathopen{\|}{\boldsymbol{x}}\mathclose{\|}_{\infty}\leqslant\rho\}.

Denote the integrand of (4.1) by $F(\boldsymbol{\theta})$ , that is,

F(\boldsymbol{\theta}):=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\prod_{e\in{\mathcal{S}_{r}(m)}}\bigl{(}1+\lambda\bigl{(}e^{i\sum_{j\in e}\theta_{j}}-1\bigr{)}\bigr{)}}{\exp\bigl{(}id\sum_{j\in[n]}\theta_{j}\bigr{)}}$}\,d\boldsymbol{\theta}.

The absolute value of $F(\boldsymbol{\theta})$ is

\prod_{e\in{\mathcal{S}_{r}(m)}}\sqrt{1+2\varLambda\Bigl{(}\cos\Bigl{(}\textstyle\sum_{j\in e}\theta_{j}\Bigr{)}-1\Bigr{)}}\;.

From this we can see that $\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\leqslant 1$ , with equality if and only if $\sum_{j\in e}\theta_{j}=0\pmod{2\pi}$ for each $e\in{\mathcal{S}_{r}(m)}$ . This is equivalent to the existence of some constants $c_{1},\ldots,c_{r}$ whose sum is $0$ modulo $2\pi$ such that $\theta_{j}=c_{t}$ for all $t\in[r]$ and $j\in V_{r}$ . Consider the transformation $\Phi_{{\boldsymbol{c}}}(\boldsymbol{\theta})=(\varphi_{1},\ldots,\varphi_{n})^{\mathrm{t}}\in(-\pi,\pi]^{n}$ , where ${\boldsymbol{c}}=(c_{1},\ldots,c_{r})^{\mathrm{t}}\in{\mathbb{R}}^{r}$ defined by

\varphi_{j}=\theta_{j}+c_{t}\pmod{2\pi}\qquad\text{ for all }t\in[r],\ j\in V_{t}.

Note that if $c_{1}+\cdots+c_{r}=0$ modulo $2\pi$ then

F(\Phi_{{\boldsymbol{c}}}(\boldsymbol{\theta}))=F(\boldsymbol{\theta}).

(4.2)

Using this symmetry we can reduce the integral in (4.1) to an integral over a subset $\mathcal{B}$ of dimension $n-r+1$ , where

\mathcal{B}=\{\boldsymbol{\theta}\in U_{n}(\pi)\mathrel{:}\theta_{2m}=\theta_{3m}=\cdots=\theta_{rm}=0\}.

Lemma 4.1.

We have

H_{r}(d,m)=\frac{(2\pi)^{r-n-1}}{\mathopen{}\mathclose{{}\left(\lambda^{\lambda}(1-\lambda)^{1-\lambda}}\right)^{m^{r}}}\int_{\mathcal{B}}F(\boldsymbol{\theta})\,d\boldsymbol{\theta}.

Proof.

Define ${\boldsymbol{c}}={\boldsymbol{c}}(\theta_{2m},\ldots,\theta_{rm})$ by

c_{1}=\sum_{j=2}^{r}\theta_{jm}\quad\text{ and }\quad c_{j}=-\theta_{jm}\quad\text{for all $j\geqslant 2$}.

Then $\Phi_{{\boldsymbol{c}}}(\boldsymbol{\theta})\in\mathcal{B}$ . Using (4.2), we get $F(\Phi_{{\boldsymbol{c}}}(\boldsymbol{\theta}))=F(\boldsymbol{\theta})$ . Integrating over $\theta_{2m},\ldots,\theta_{rm}$ and then over the remaining coordinates separately, we get

\int_{U_{n}(\pi)}F(\boldsymbol{\theta})\,d\boldsymbol{\theta}=(2\pi)^{r-1}\int_{\mathcal{B}}F(\boldsymbol{\theta})\,d\boldsymbol{\theta}.

The result follows from (4.1). ∎

The advantage of considering the integral over a set $\mathcal{B}$ of smaller dimension is that there is only one point for which $|F(\boldsymbol{\theta})|=1$ , namely the nullvector. Let

\rho_{0}:=\rho_{0}(\lambda)=\frac{r^{5/2}\log{n}}{(\varLambda m^{r-1})^{1/2}}.

(4.3)

In Section 4.2, we show that the contribution of the region $\mathcal{B}\setminus U_{n}(\rho_{0})$ to the integral $\int_{\cal B}F(\boldsymbol{\theta})\,d\boldsymbol{\theta}$ is asymptotically negligible.

A theoretical framework for asymptotically estimating integrals over truncated multivariate Gaussian distributions, including cases where the covariance matrix is not of full rank, was developed by Isaev and McKay in [5, Section 4]. Using the tools from [5], we estimate the integral of $F(\boldsymbol{\theta})$ over $\mathcal{B}\cap U_{n}(\rho_{0})$ in Section 4.3.

4.2 Outside the main region

In this section, we estimate the contribution of $|F(\boldsymbol{\theta})|$ over $\mathcal{B}\setminus U_{n}(\rho_{0})$ , where $\rho_{0}$ is defined in (4.3). Recall also that $\varLambda=\lambda(1-\lambda)$ . Let

|x|_{2\pi}=\min_{k\in{\mathbb{Z}}}|x-2\pi k|.

Note that $|x-y|_{2\pi}$ is the circular distance between $x,y\in[-\pi,\pi]$ . In particular, it never exceeds $\pi$ and satisfies the triangle inequality. Our arguments rely on the next two lemmas.

Lemma 4.2.

For any $x\in{\mathbb{R}}$ and $\lambda\in[0,1]$ , we have

\mathopen{|}1+\lambda(e^{ix}-1)\mathclose{|}=\sqrt{1-2\varLambda(1-\cos x)}\leqslant\exp\mathopen{}\mathclose{{}\left(-\lower 0.6458pt\hbox{\large$\frac{\varLambda}{2}$}\mathopen{}\mathclose{{}\left(1-\lower 0.6458pt\hbox{\large$\frac{|x|^{2}_{2\pi}}{12}$}}\right)|x|^{2}_{2\pi}}\right).

Lemma 4.3.

For $\varLambda=\Omega(r^{16}m^{2-r})$ , we have

\int_{\mathcal{B}\cap U_{n}(\rho_{0})}|F(\boldsymbol{\theta})|d\boldsymbol{\theta}\leqslant m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2\pi}{\varLambda m^{r-1}}$}}\right)^{(n-r+1)/2}\exp\mathopen{}\mathclose{{}\left(O\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{r^{2}}{\varLambda m^{r-2}}$}}\right)+O(n^{-2/5})}\right).

Lemma 4.2 follows from Taylor’s theorem with remainder. We will prove Lemma 4.3 in Section 4.3 along with estimating $\int_{\mathcal{B}\cap U_{n}(\rho)}F(\boldsymbol{\theta})d\boldsymbol{\theta}$ .

First, we consider the region $\mathcal{B}_{1}$ , where the components $\theta_{j}$ ’s are widely spread apart within at least one class $V_{t}$ . Define

\mathcal{B}_{1}:=\Bigl{\{}\boldsymbol{\theta}\in\mathcal{B}\mathrel{:}\text{there is $t\in[r]$ that }\Bigl{|}\{\theta_{j}\}_{j\in V_{t}}\setminus\bigl{[}\theta_{k}\pm\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}\bigr{]}_{2\pi}\Bigr{|}>m/r\text{ for all $k\in V_{t}$}\Bigr{\}},

where $[x\pm\rho]_{2\pi}$ is the set of points at circular distance at most $\rho$ from $x$ on $[-\pi,\pi]$ :

[x\pm\rho]_{2\pi}=\{y\in[-\pi,\pi]\mathrel{:}|x-y|_{2\pi}\leqslant\rho\}.

Lemma 4.4.

If $\varLambda=\Omega(r^{16}m^{2-r})$ , we have

\int_{\mathcal{B}_{1}}|F(\boldsymbol{\theta})|\,d\boldsymbol{\theta}\leqslant e^{-\omega(rn\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}.

Proof.

Using Lemma 4.2 for each factor in $F(\boldsymbol{\theta})$ , we get

|F(\boldsymbol{\theta})|\leqslant\exp\biggl{(}-\Omega\biggl{(}\varLambda\sum_{e\in{\mathcal{S}_{r}(m)}}\Bigl{|}\sum_{j\in e}\theta_{j}\Bigr{|}_{2\pi}^{2}\biggr{)}\biggr{)}.

(4.4)

Let $V_{t}$ be a class such that $\Bigl{|}\{\theta_{j}\}_{j\in V_{t}}\setminus\bigl{[}\theta_{k}\pm\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}\bigr{]}_{2\pi}\Bigr{|}>m/r$ for all $k\in V_{t}$ . Let $a,b\in V_{t}$ . For any $e_{1},e_{2}\in{\mathcal{S}_{r}(m)}$ such that $e_{1}\mathbin{\triangle}e_{2}=\{a,b\}$ , we have

\biggl{|}\,\sum_{j\in e_{1}}\theta_{j}-\sum_{j\in e_{2}}\theta_{j}\biggr{|}_{2\pi}\!\!=|\theta_{a}-\theta_{b}|_{2\pi}.

This implies that

\biggl{|}\sum_{j\in e_{1}}\theta_{j}\biggr{|}_{2\pi}^{2}+\biggl{|}\sum_{j\in e_{2}}\theta_{j}\biggr{|}_{2\pi}^{2}\geqslant\frac{1}{2}\mathopen{}\mathclose{{}\left(\biggl{|}\sum_{j\in e_{1}}\theta_{j}\biggr{|}_{2\pi}-\biggl{|}\sum_{j\in e_{2}}\theta_{j}\biggr{|}_{2\pi}}\right)^{2}\geqslant\frac{|\theta_{a}-\theta_{b}|_{2\pi}^{2}}{2}.

(4.5)

Summing (4.5) over all choices of such pairs of edges $e_{1},e_{2}$ and $a,b\in V_{t}$ , we get

m\sum_{e\in{\mathcal{S}_{r}(m)}}\Bigl{|}\sum_{j\in e}\theta_{j}\Bigr{|}_{2\pi}^{2}\geqslant\frac{m^{r-1}}{2}\sum_{a,b\in V_{t}}|\theta_{a}-\theta_{b}|_{2\pi}^{2}.

By the choice of $V_{t}$ , for any $a\in V_{t}$ , there are at least $m/r$ components $b\in V_{t}$ such that $|\theta_{a}-\theta_{b}|_{2\pi}\geqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}$ , implying

\sum_{a,b\in V_{t}}|\theta_{a}-\theta_{b}|_{2\pi}^{2}\geqslant m\cdot\min_{a\in V_{t}}\sum_{b\in V_{t}}|\theta_{a}-\theta_{b}|_{2\pi}^{2}\geqslant\raise 0.21529pt\hbox{\small$\displaystyle\frac{m^{2}}{r}$}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}}{4r}$}}\right)^{2}.

Using the above bounds in (4.4), we find that

|F(\boldsymbol{\theta})|\leqslant\exp\mathopen{}\mathclose{{}\left(-\Omega\mathopen{}\mathclose{{}\left(\varLambda m^{r}\rho_{0}^{2}/r^{3}}\right)}\right).

Note that the volume of $\mathcal{B}_{1}$ can not exceed the volume of $\mathcal{B}$ , which is $(2\pi)^{n-r-1}$ . Then, recalling from (4.3) the definition of $\rho_{0}$ and that $n=mr$ , we get

\int_{\mathcal{B}_{1}}\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\,d\boldsymbol{\theta}\leqslant(2\pi)^{n-r-1}\exp\mathopen{}\mathclose{{}\left(-\Omega\mathopen{}\mathclose{{}\left(\varLambda m^{r}\rho_{0}^{2}/r^{3}}\right)}\right)\leqslant e^{-\omega(rn\log n)}.

Observing also that

m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}=e^{O(rn\log n)},

the claimed bound follows. ∎

The next region $\mathcal{B}_{2}$ consists of all $\boldsymbol{\theta}\in\mathcal{B}$ such that there is $(k_{1},\ldots,k_{r})\in V_{1}\times\cdots\times V_{r}$ satisfying the following two conditions:

|\theta_{k_{1}}+\cdots+\theta_{k_{r}}|_{2\pi}\geqslant\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}}{3}$}\quad\text{ and }\quad\Bigl{|}\{\theta_{j}\}_{j\in V_{t}}\cap\bigl{[}\theta_{k_{t}}\pm\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}\bigr{]}_{2\pi}\Bigr{|}\geqslant m-m/r\text{ for all $t\in[r]$.}

Lemma 4.5.

If $\varLambda=\Omega(r^{16}m^{2-r})$ , we have

\int_{\mathcal{B}_{2}}\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\,d\boldsymbol{\theta}\leqslant e^{-\omega(rn\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2\pi}{\varLambda m^{r-1}}$}}\right)^{(n-r+1)/2}.

Proof.

Let $\boldsymbol{\theta}\in\mathcal{B}_{2}$ . Consider any edge $e\in{\mathcal{S}_{r}(m)}$ such that, for all $t\in[r]$ , and $j\in e\cap V_{t}$ we have $|\theta_{j}-\theta_{k_{t}}|_{2\pi}\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}$ . Then, by the defintion of $\mathcal{B}_{2}$ , we have

\biggl{|}\sum_{j\in e}\theta_{j}\biggr{|}_{2\pi}\geqslant\biggl{|}\sum_{t\in[r]}\theta_{k_{t}}\biggr{|}_{2\pi}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}}{4}$}\geqslant\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}}{12}$}.

The number of such edges is at least $(m-m/r)^{r}\geqslant m^{r}/4$ . Using Lemma 4.2 for each corresponding factor, we obtain

|F(\boldsymbol{\theta})|\leqslant\exp\mathopen{}\mathclose{{}\left(-\Omega\mathopen{}\mathclose{{}\left(\varLambda m^{r}\rho_{0}^{2}}\right)}\right).

We complete the proof by repeating the arguments of Lemma 4.4. ∎

Next, for ${\boldsymbol{k}}=(k_{1},\ldots,k_{r})\in V_{1}\times\cdots\times V_{r}$ and $S\subseteq[n]\setminus\{k_{1},\ldots,k_{r}\}$ with $|S\cap V_{1}|=\ldots=|S\cap V_{r}|$ consider the regions $\mathcal{B}_{{\boldsymbol{k}},S}$ of $\boldsymbol{\theta}\in\mathcal{B}$ such that the following hold:

•

$|\theta_{k_{1}}+\cdots+\theta_{k_{r}}|_{2\pi}\leqslant\rho_{0}/3$ ;
•

$|\{\theta_{j}\}_{j\in V_{t}}\cap[\theta_{k_{t}}\pm\rho_{0}/(4r)]_{2\pi}|\geqslant m-m/r$ for all $t\in[r]$ ;
•

$S$ contains $\bigcup_{t\in[r]}\{j\in V_{t}:|\theta_{j}-\theta_{k_{t}}|_{2\pi}>\rho_{0}/(2r)\}$ ;
•

there is $t\in[r]$ such that $\{j\in V_{t}:|\theta_{j}-\theta_{k_{t}}|_{2\pi}>\rho_{0}/(2r)\}=S\cap V_{t}$ .

The second and the last property imply that if $\mathcal{B}_{{\boldsymbol{k}},S}\neq\emptyset$ then $|S\cap V_{t}|\leqslant m/r$ for some $t\in[r]$ , and therefore $|S|\leqslant m$ .

Lemma 4.6.

If $\varLambda=\Omega(r^{16}m^{2-r})$ , we have uniformly for every ${\boldsymbol{k}}\in V_{1}\times\cdots\times V_{r}$ and $S\subseteq[n]\setminus\{k_{1},\ldots,k_{r}\}$ with $|S|\leqslant m$ that

\int_{\mathcal{B}_{{\boldsymbol{k}},S}}\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\,d\boldsymbol{\theta}\leqslant e^{-\omega(|S|\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}.

Proof.

Due to the symmetry (4.2) and relabeling vertices, we can assume that $k_{1}=m$ , $k_{2}=2m$ , $\ldots$ , $k_{r}=rm$ .

The bounds indicated in this proof by the asymptotic notations $O(\,)$ , $\Omega(\,)$ and $\omega(\,)$ can be chosen independently of $S$ .

By definition, for $\boldsymbol{\theta}\in\mathcal{B}_{{\boldsymbol{k}},S}\subset\mathcal{B}$ , we have

\theta_{k_{2}}=\cdots=\theta_{k_{r}}=0\quad\text{ and }\quad|\theta_{k_{1}}|_{2\pi}\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{3}$}.

(4.6)

By definition there exists a $t\in[r]$ such that $|V_{t}\cap S|=|S|/r$ .

Consider any $e_{1},e_{2}\in{\mathcal{S}_{r}(m)}$ such that $e_{1}\mathbin{\triangle}e_{2}=\{a,b\}$ , where $a\in S\cap V_{t}$ and $b\in V_{t}$ satisfies $|\theta_{b}-\theta_{k_{t}}|_{2\pi}\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}$ . Let

\Sigma_{1}:=\sum_{\begin{subarray}{c}e\in{\mathcal{S}_{r}(m)}\\ e\cap S=\emptyset\end{subarray}}\biggl{|}\sum_{j\in e}\theta_{j}\biggr{|}_{2\pi}^{2}\qquad\text{and}\qquad\Sigma_{2}:=\sum_{\begin{subarray}{c}e\in{\mathcal{S}_{r}(m)}\\ e\cap S\neq\emptyset\end{subarray}}\biggl{|}\sum_{j\in e}\theta_{j}\biggr{|}_{2\pi}^{2}.

Arguing similarly to Lemma 4.4 and using (4.5), we get that

\displaystyle|S\cap V_{t}|\cdot\Sigma_{1}+m\cdot\Sigma_{2}\geqslant m^{r-1}\sum_{a\in S\cap V_{t}}\sum_{b\in V_{t}}\raise 0.21529pt\hbox{\small$\displaystyle\frac{|\theta_{a}-\theta_{b}|_{2\pi}^{2}}{2}$}\geqslant m^{r-1}|S\cap V_{t}|(m-m/r)\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}^{2}}{32r^{2}}$}.

(4.7)

If $|S\cap V_{t}|\cdot\Sigma_{1}\geqslant m\cdot\Sigma_{2}$ then from (4.7) we get $\Sigma_{1}=\Omega\mathopen{}\mathclose{{}\left(m^{r}\rho_{0}^{2}/r^{2}}\right)$ . Using (4.4), we find that

|F(\boldsymbol{\theta})|\leqslant\exp\mathopen{}\mathclose{{}\left(-\Omega\mathopen{}\mathclose{{}\left(\varLambda m^{r}\rho_{0}^{2}/r^{2}}\right)}\right).

By reasoning as in Lemma 4.4, we show that the contribution of such $\boldsymbol{\theta}$ is negligible.

Next, we consider the case when $|S\cap V_{t}|\cdot\Sigma_{1}<m\cdot\Sigma_{2}$ . Using (4.7) and Lemma 4.2 for each factor in $F(\boldsymbol{\theta})$ corresponding $\Sigma_{2}$ , we get that

|F(\boldsymbol{\theta})|\leqslant\exp\Bigl{(}-\Omega(\varLambda|S\cap V_{t}|m^{r-1}\rho_{0}^{2}/r^{2})\Bigr{)}|\hat{F}(\hat{\boldsymbol{\theta}})|,

where $\hat{\boldsymbol{\theta}}$ consists of the components of $\theta_{j}$ with $j\in[n]\setminus S$ and

|\hat{F}(\hat{\boldsymbol{\theta}})|=\prod_{\begin{subarray}{c}e\in{\mathcal{S}_{r}(m)}\\ e\cap S=\emptyset\end{subarray}}\bigl{|}1+\lambda(e^{i\sum_{j\in e}\theta_{j}}-1)\bigr{|}.

Recalling from (4.3) the definition of $\rho_{0}$ and that $|S\cap V_{t}|=|S|/r$ , we find that

\varLambda\,\mathopen{|}S\cap V_{t}\mathclose{|}\,m^{r-1}\rho_{0}^{2}/r^{2}\geqslant\mathopen{|}S\mathclose{|}r^{2}\log^{2}n.

Note also that the definition of $\mathcal{B}_{{\boldsymbol{k}},S}$ and (4.6) implies $\mathopen{\|}\hat{\boldsymbol{\theta}}\mathclose{\|}_{\infty}\leqslant\rho_{0}$ . Integrating over the components corresponding to $S$ , together with $(2\pi)^{|S|}=e^{o(|S|r^{2}\log^{2}n)}$ leads to

\int_{\mathcal{B}_{{\boldsymbol{k}},S}}\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\,d\boldsymbol{\theta}\leqslant e^{-\Omega(|S|r^{2}\log^{2}n)}\int_{\hat{\mathcal{B}}\cap U_{r\hat{m}}(\rho_{0})}\mathopen{|}\hat{F}(\hat{\boldsymbol{\theta}})\mathclose{|}\,d\hat{\boldsymbol{\theta}},

(4.8)

where $\hat{\mathcal{B}}$ is defined as $\mathcal{B}$ for reduced parameter

\hat{m}:=m-|S|/r\geqslant 2m/3.

(4.9)

Finally, using Lemma 4.3 to estimate the integral in the RHS of (4.8), we obtain

	$\displaystyle\int_{\mathcal{B}_{{\boldsymbol{k}}},S}\|F(\boldsymbol{\theta})\|d\boldsymbol{\theta}$	$\displaystyle\leqslant e^{-\Omega(\|S\|r^{2}\log^{2}n)}{\hat{m}}^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda\hat{m}^{r-1}}}\right)^{(n-\|S\|-r+1)/2}\exp\mathopen{}\mathclose{{}\left(O\mathopen{}\mathclose{{}\left(\frac{r^{2}}{\varLambda\hat{m}^{r-2}}+(r\hat{m})^{-2/5}}\right)}\right)$
		$\displaystyle\leqslant e^{-\Omega(\|S\|r^{2}\log^{2}n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}.$

To derive that last equality, we observe that

\hat{m}^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda\hat{m}^{r-1}}}\right)^{(n-|S|-r+1)/2}=\exp\Bigl{(}O\mathopen{}\mathclose{{}\left(|S|\log(\varLambda\hat{m}^{r-1})+|S|r}\right)\Bigr{)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}

and

r^{2}\log^{2}n\gg\frac{r^{2}}{\varLambda\hat{m}^{r-2}}+(r\hat{m})^{-2/5}+\log(\varLambda\hat{m}^{r-1})+r,

which is straightforward by (4.9) and the assumptions. ∎

Now we are ready to show that the region $\mathcal{B}\setminus U_{n}(\rho_{0})$ has a negligible contribution.

Lemma 4.7.

If $\varLambda=\Omega(r^{16}m^{2-r})$ , we have

\int_{\mathcal{B}\setminus U_{n}(\rho_{0})}|F(\boldsymbol{\theta})|d\boldsymbol{\theta}\leqslant e^{-\omega(r\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}.

Proof.

Combining Lemma 4.4 and Lemma 4.5, we get that

\int_{\mathcal{B}_{1}\cup\mathcal{B}_{2}}|F(\boldsymbol{\theta})|d\boldsymbol{\theta}\leqslant m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2}e^{-\omega(rn\log n)}.

By the definitions of $\mathcal{B}_{1}$ and $\mathcal{B}_{2}$ , we find that that if $\boldsymbol{\theta}\in\mathcal{B}\setminus(\mathcal{B}_{1}\cup\mathcal{B}_{2})$ then there is ${\boldsymbol{k}}=(k_{1},\ldots,k_{r})\in V_{1}\times\cdots\times V_{r}$ such that

|\theta_{k_{1}}+\cdots+\theta_{k_{r}}|_{2\pi}\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{3}$}\quad\text{ and }\quad\Bigl{|}\{\theta_{j}\}_{j\in V_{t}}\cap\bigl{[}\theta_{k_{t}}\pm\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{4r}$}\bigr{]}_{2\pi}\Bigr{|}\geqslant m-m/r\text{ for all $t\in[r]$.}

Next, if $|\theta_{j}-\theta_{k_{t}}|_{2\pi}\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{2r}$}$ for all $t\in[r]$ and $j\in V_{t}$ then, recalling $\theta_{2m}=\cdots\theta_{rm}=0$ , we have

\displaystyle|\theta_{k_{t}}|

\displaystyle=|\theta_{k_{t}}-\theta_{tm}|\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{2r}$},\;\text{ for $t=2,\ldots,r$,}\quad\text{and}\quad|\theta_{k_{1}}|\leqslant\lower 0.6458pt\hbox{\large$\frac{\rho_{0}}{3}$}+\sum_{t=2}^{r}|\theta_{k_{t}}|\leqslant\lower 0.6458pt\hbox{\large$\frac{(5r-3)\rho_{0}}{6r}$}.

Together with the fact that $\rho_{0}/(2r)+\rho_{0}/(2r)\leqslant\rho_{0}$ and $(5r-3)\rho_{0}/(6r)+\rho_{0}/(2r)\leqslant\rho_{0}$ this implies that all such $\boldsymbol{\theta}$ are in $U_{n}(\rho_{0})$ . Therefore, $\mathcal{B}\setminus(\mathcal{B}_{1}\cup\mathcal{B}_{2}\cup U_{n}(\rho_{0}))$ is covered by the sets $\mathcal{B}_{{\boldsymbol{k}},S}$ considered in Lemma 4.6, where $S\neq\emptyset$ and $|S|\leqslant m$ . Summing over $j=|S|/r$ and allowing $m^{r}$ for the choices of ${\boldsymbol{k}}$ and $m^{rj}$ for the choices of $S$ , we get

\int_{\mathcal{B}\setminus(\mathcal{B}_{1}\cup\mathcal{B}_{2}\cup U_{n}(\rho_{0}))}|F(\boldsymbol{\theta})|d\boldsymbol{\theta}\leqslant\sum_{j=1}^{m/r}m^{(j+1)r}e^{-\omega(jr\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2},

where the $\omega(jr\log n)$ in the exponent are uniform in $j$ . Thus

\int_{\mathcal{B}\setminus(\mathcal{B}_{1}\cup\mathcal{B}_{2}\cup U_{n}(\rho_{0}))}|F(\boldsymbol{\theta})|d\boldsymbol{\theta}\leqslant e^{-\omega(r\log n)}m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\frac{2\pi}{\varLambda m^{r-1}}}\right)^{(n-r+1)/2},

as claimed. ∎

4.3 Inside the main region

Turning to the integral inside $U_{n}(\rho_{0})$ , we see from the next lemma that the main term in the expansion of $\log F(\boldsymbol{\theta})$ around the origin is $-\boldsymbol{\theta}^{\mathrm{t}}\!A\boldsymbol{\theta}$ , where $A$ is $n\times n$ matrix defined by

A:=\lower 0.6458pt\hbox{\large$\frac{1}{2}$}\varLambda m^{r-1}(I-B/m+J_{n}/m).

(4.10)

Here $J_{k}$ is the $k\times k$ all one matrix, and $B$ is the block diagonal matrix consisting of $r$ blocks of $J_{m}$ . Let $a_{i}(\lambda)$ denote the coefficients of Taylor’s expansion of $\log(1+\lambda(e^{ix}-1))$ around the origin. In particular, we have

	$\displaystyle a_{1}(\lambda)$	$\displaystyle=i\lambda,\qquad a_{2}(\lambda)=-\lower 0.6458pt\hbox{\large$\frac{1}{2}$}\varLambda,\qquad a_{3}(\lambda)=-\lower 0.6458pt\hbox{\large$\frac{i}{6}$}\varLambda(1-2\lambda),$
	$\displaystyle a_{4}(\lambda)$	$\displaystyle=\lower 0.6458pt\hbox{\large$\frac{1}{24}$}\varLambda(1-6\varLambda),\qquad a_{5}(\lambda)=O(\varLambda).$

Lemma 4.8.

If $r\geqslant 3$ and $\varLambda=\Omega(r^{16}m^{2-r})$ , then for any $\boldsymbol{\theta}\in U_{n}(O(\rho_{0}))$ , we have

\log F(\boldsymbol{\theta})=-\boldsymbol{\theta}^{\mathrm{t}}\!A\boldsymbol{\theta}+\sum_{p=3}^{4}\,\sum_{e\in{\mathcal{S}_{r}(m)}}\!\!a_{p}(\lambda)\,\biggl{(}\,\sum_{j\in e}\theta_{j}\biggr{)}^{\!p}+O(n^{-1/2}\log^{5}n).

Proof.

Observing that for all $e\in{\mathcal{S}_{r}(m)}$ we have $|\sum_{j\in e}\theta_{j}|=O(r\rho_{0})$ for $\boldsymbol{\theta}\in U_{n}(O(\rho_{0}))$ and using Taylor’s theorem, we get that

\log\bigl{(}1+\lambda(e^{i\sum_{j\in e}\theta_{j}}-1)\bigr{)}=\sum_{p=1}^{4}\,a_{p}(\lambda)\biggl{(}\,\sum_{j\in e}\theta_{j}\biggr{)}^{\!p}+O\mathopen{}\mathclose{{}\left(\varLambda(r\rho_{0})^{5}}\right).

Summing over all $e\in{\mathcal{S}_{r}(m)}$ , we get that the linear term of $\log F(\boldsymbol{\theta})$ (which includes terms from the denominator of $F(\boldsymbol{\theta})$ ), is

i\,\sum_{j\in[n]}\theta_{j}\,(\,m^{r-1}\lambda-d),

which is zero as $d=\lambda m^{r-1}$ . Furthermore in $-\boldsymbol{\theta}^{\mathrm{t}}\!A\boldsymbol{\theta}$ the coefficient of $\theta_{j}^{2}$ is $-\varLambda m^{r-1}/2$ for any $j$ , while the coefficient of $\theta_{j},\theta_{k}$ is $-\varLambda m^{r-2}$ if $j,k$ belong to different partition classes, and 0 if $j,k$ are within the same partition class, matching the quadratic term of $\log F(\boldsymbol{\theta})$ . It remains to observe that the combined error term

O(\varLambda m^{r}(r\rho_{0})^{5})=O\mathopen{}\mathclose{{}\left(\frac{m^{-3r/2+5/2}r^{35/2}\log^{5}n}{\varLambda^{3/2}}}\right)=O(n^{-1/2}\log^{5}n),

using $\varLambda=\Omega(r^{16}m^{2-r})$ and Lemma 2.4. ∎

Note that $A$ is singular of nullity $r-1$ . We evaluate $\int_{\mathcal{B}\cap U(\rho_{0})}F(\boldsymbol{\theta})\,d\boldsymbol{\theta}$ using the methods from [5] to first raise it to an integral over a domain of full dimension, and then to evaluate the resulting integral.

The following matrices will play a relevant role in our proof.

W:=\biggl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\varLambda m^{r-3}}{2r}$}\biggr{)}^{\!1/2}(rB-J_{n})\quad\text{ and }\quad T:=\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2}{\varLambda m^{r-1}}$}}\right)^{\!1/2}\mathopen{}\mathclose{{}\left(I-\raise 0.21529pt\hbox{\small$\displaystyle\frac{\sqrt{r}-1}{\sqrt{r}m}$}B}\right).

(4.11)

Lemma 4.9.

The following identities and bounds hold.

(a)

$A+W^{\mathrm{t}}W$ has $r$ eigenvalues equal to $\frac{1}{2}\varLambda rm^{r-1}$ and $n-r$ eigenvalues equal to $\frac{1}{2}\varLambda m^{r-1}$ .
(b)

$\mathopen{|}A+W^{\mathrm{t}}W\mathclose{|}=\mathopen{}\mathclose{{}\left(\lower 0.6458pt\hbox{\large$\frac{\varLambda m^{r-1}}{2}$}}\right)^{n}r^{r}$ .
(c)

$T^{\mathrm{t}}(A+W^{\mathrm{t}}W)T=I$ .
(d)

$(A+W^{\mathrm{t}}W)^{-1}=\lower 0.6458pt\hbox{\large$\frac{2}{\varLambda m^{r-1}}$}\mathopen{}\mathclose{{}\left(I-\lower 0.6458pt\hbox{\large$\frac{r-1}{n}$}B}\right)$ ,
(e)

$\mathopen{\|}T\mathclose{\|}_{\infty}=\mathopen{\|}T\mathclose{\|}_{1}\leqslant\lower 0.6458pt\hbox{\large$\frac{3}{\mathopen{}\mathclose{{}\left(\varLambda m^{r-1}}\right)^{1/2}}$}$ .
(f)

$\mathopen{\|}T^{-1}\mathclose{\|}_{\infty}=2^{-1/2}\bigl{(}\varLambda rm^{r-1}\bigr{)}^{1/2}$ .

Proof.

Using $B^{2}=mB$ , $J_{n}B=BJ_{n}=mJ_{n}$ and $J_{n}^{2}=rmJ_{n}$ , it is routine to verify that

	$\displaystyle A+W^{\mathrm{t}}W$	$\displaystyle=\lower 0.6458pt\hbox{\large$\frac{1}{2}$}\varLambda m^{r-1}\Bigl{(}I+\raise 0.21529pt\hbox{\small$\displaystyle\frac{r-1}{m}$}B\Bigr{)},\quad\text{and}$
	$\displaystyle T^{-1}$	$\displaystyle=2^{-1/2}\varLambda^{1/2}m^{(r-1)/2}\Bigl{(}I+\raise 0.21529pt\hbox{\small$\displaystyle\frac{\sqrt{r}-1}{m}$}B\Bigr{)}.$

The eigenspaces of $B$ are those that are constant on each class (dimension $r$ ) and those that sum to 0 on each class (dimension $n-r)$ . This proves (a), and (b) immediately follows. Parts (c) and (d) are proved by direct multiplication, while (e) and (f) follow from the explicit forms of $T$ and $T^{-1}$ . ∎

The kernel of matrix $A$ from (4.10) consists of the set of all vectors which sum to 0 and are constant within each class. Note that $\ker A$ is the subspace spanned by the vectors ${\boldsymbol{v}}_{2},\ldots,{\boldsymbol{v}}_{r}$ defined by

{\boldsymbol{v}}_{j}=\sum_{\ell\in V_{j}}{\boldsymbol{e}}_{\ell}-\sum_{\ell\in V_{1}}{\boldsymbol{e}}_{\ell},\qquad\text{for $2\leqslant j\leqslant r$,}

(4.12)

where ${\boldsymbol{e}}_{\ell}\in{\mathbb{R}}^{n}$ is the standard basis vector where the $\ell$ -th element is 1 and all others 0. Let $Q$ denote a projection operator into the space $\{{\boldsymbol{x}}\in{\mathbb{R}}^{n}\mathrel{:}x_{2m}=\cdots=x_{rm}=0\}$ defined by

Q:=I-\sum_{j=2}^{r}{\boldsymbol{v}}_{j}{\boldsymbol{e}}_{jm}^{\mathrm{t}}.

Note that $U_{n}(\rho_{0})\cap\mathcal{B}$ = $U_{n}(\rho_{0})\cap Q({\mathbb{R}}^{n})$ .

We will need Lemma 4.6 from [5].

Lemma 4.10 (Isaev and McKay [5]).

Let $Q,W:\mathbb{R}^{n}\rightarrow\mathbb{R}^{n}$ be linear operators such that $\ker Q\cap\ker W=\{\boldsymbol{0}\}$ and $\operatorname{span}(\ker Q,\ker W)={\mathbb{R}}^{n}$ . Let $n_{\scriptscriptstyle\perp}$ denote the dimension of $\ker Q$ . Suppose $\varOmega\subseteq{\mathbb{R}}^{n}$ and $G:\varOmega\cap Q({\mathbb{R}}^{n})\to{\mathbb{C}}$ . For any $\tau>0$ , define

\bar{\varOmega}:=\bigl{\{}{\boldsymbol{x}}\in{\mathbb{R}}^{n}\mathrel{:}Q{\boldsymbol{x}}\in\varOmega\text{~{}and~{}}W{\boldsymbol{x}}\in U_{n}(\tau)\bigr{\}}.

Then, if the integrals exist,

\int_{\varOmega\cap Q({\mathbb{R}}^{n})}G({\boldsymbol{y}})\,d{\boldsymbol{y}}=(1-K)^{-1}\,\pi^{-n_{\scriptscriptstyle\perp}/2}\,\bigl{|}Q^{\mathrm{t}}Q+W^{\mathrm{t}}W\bigr{|}^{1/2}\int_{\bar{\varOmega}}G(Q{\boldsymbol{x}})\,e^{-{\boldsymbol{x}}^{\mathrm{t}}W^{\mathrm{t}}W{\boldsymbol{x}}}\,d{\boldsymbol{x}},

where

0\leqslant K<\min(1,ne^{-\tau^{2}}).

Moreover, if $U_{n}(\rho_{1})\subseteq\varOmega\subseteq U_{n}(\rho_{2})$ for some $\rho_{2}\geqslant\rho_{1}>0$ then

U_{n}\biggl{(}\min\biggl{(}\frac{\rho_{1}}{\mathopen{\|}Q\mathclose{\|}_{\infty}},\frac{\tau}{\mathopen{\|}W\mathclose{\|}_{\infty}}\biggr{)}\biggr{)}\subseteq\bar{\varOmega}\subseteq U_{n}\bigl{(}\mathopen{\|}P\mathclose{\|}_{\infty}\,\rho_{2}+\mathopen{\|}R\mathclose{\|}_{\infty}\,\tau\bigr{)}

for any linear operators $P,R:{\mathbb{R}}^{n}\rightarrow{\mathbb{R}}^{n}$ such that $PQ+RW$ is equal to the identity operator on ${\mathbb{R}}^{n}$ .

Applying this lemma for $F(\boldsymbol{\theta})$ and $|F(\boldsymbol{\theta})|$ gives the following results.

Lemma 4.11.

There is some region $\bar{\varOmega}$ with $U_{n}\bigl{(}\lower 0.6458pt\hbox{\large$\frac{2\rho_{0}}{3r}$}\bigr{)}\subseteq\bar{\varOmega}\subseteq U_{n}(4\rho_{0})$ such that for $G({\boldsymbol{x}})=F({\boldsymbol{x}})$ and $G({\boldsymbol{x}})=|F({\boldsymbol{x}})|$ we have

\int_{U_{n}(\rho_{0})\cap Q({\mathbb{R}}^{n})}\!\!G({\boldsymbol{x}})\,d{\boldsymbol{x}}=(1+e^{-\omega(\log n)})\pi^{-(r-1)/2}\bigl{|}Q^{\mathrm{t}}Q+W^{\mathrm{t}}W\bigr{|}^{1/2}\int_{\bar{\varOmega}}G({\boldsymbol{x}})e^{-{\boldsymbol{x}}^{\mathrm{t}}W^{\mathrm{t}}W{\boldsymbol{x}}}\,d{\boldsymbol{x}}.

Proof.

We can apply Lemma 4.10 with $\varOmega=U_{n}(\rho_{0})$ , $\rho_{1}=\rho_{2}=\rho_{0}$ , $\tau=r^{2}\log n$ and the matrices $Q,W$ we have defined. In addition, define the matrices

P:=I-\raise 0.21529pt\hbox{\small$\displaystyle\frac{1}{r}$}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2r}{\varLambda m^{r-1}}$}}\right)^{\!1/2}W\quad\text{ and }\quad R:=\raise 0.21529pt\hbox{\small$\displaystyle\frac{1}{r}$}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2r}{\varLambda m^{r-1}}$}}\right)^{\!1/2}I,

Note that for $2\leqslant j\leqslant r$ and the vectors $v_{j}$ defined in (4.12) we have

Wv_{j}=\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{\varLambda m^{r-1}}{2r}$}}\right)^{1/2}rv_{j},

implying $W(I-Q)=\mathopen{}\mathclose{{}\left(\lower 0.6458pt\hbox{\large$\frac{\varLambda m^{r-1}}{2r}$}}\right)^{1/2}r(I-Q)$ . Therefore

PQ+RW=Q+\frac{1}{r}\mathopen{}\mathclose{{}\left(\frac{2r}{\varLambda m^{r-1}}}\right)^{1/2}W(I-Q)=I.

We have $n_{\scriptscriptstyle\perp}=r-1$ , and (4.2) implies $G(Q{\boldsymbol{x}})=G({\boldsymbol{x}})$ for all ${\boldsymbol{x}}$ . From their explicit forms we have $\mathopen{\|}Q\mathclose{\|}_{\infty}=r$ , $\mathopen{\|}R\mathclose{\|}_{\infty}=2^{1/2}\varLambda^{-1/2}r^{-1/2}m^{-(r-1)/2}$ , $\mathopen{\|}W\mathclose{\|}_{\infty}=(r-1)r^{-1/2}(2\varLambda m^{r-1})^{1/2}$ and $\mathopen{\|}P\mathclose{\|}_{\infty}\leqslant 3$ follows from $\mathopen{\|}W\mathclose{\|}_{\infty}$ . Therefore

\min\biggl{(}\frac{\rho_{1}}{\mathopen{\|}Q\mathclose{\|}_{\infty}},\frac{\tau}{\mathopen{\|}W\mathclose{\|}_{\infty}}\biggr{)}=\min\biggl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}}{r}$},\raise 0.21529pt\hbox{\small$\displaystyle\frac{\rho_{0}\sqrt{2}}{2r-2}$}\biggr{)}\geqslant\raise 0.21529pt\hbox{\small$\displaystyle\frac{2\rho_{0}}{3r}$},

and

\mathopen{\|}P\mathclose{\|}_{\infty}\,\rho_{2}+\mathopen{\|}R\mathclose{\|}_{\infty}\,\tau\leqslant 3\rho_{0}+\sqrt{2}\,r^{-1}\rho_{0}\leqslant 4\rho_{0}.\qed

Recall that for a complex random variable $Z$ , the variance is defined by

\operatorname{Var}Z=\operatorname{\mathbb{E}}|Z-\operatorname{\mathbb{E}}Z|^{2}=\operatorname{Var}\Re Z+\operatorname{Var}\Im Z,

while the pseudovariance is

\operatorname{\mathbb{V\!}}Z=\operatorname{\mathbb{E}}(Z-\operatorname{\mathbb{E}}Z)^{2}=\operatorname{Var}\Re Z-\operatorname{Var}\Im Z+2i\,\operatorname{Cov}(\Re Z,\Im Z).

In addition, for a domain $\varOmega\subset{\mathbb{R}}^{n}$ and a twice continuously differentiable function $g:\varOmega\to{\mathbb{C}}$ , let

H(g,\varOmega):=(h_{jk})\quad\mbox{where}\quad h_{jk}=\sup_{{\boldsymbol{x}}\in\varOmega}\,\mathopen{}\mathclose{{}\left|\raise 0.21529pt\hbox{\small$\displaystyle\frac{\partial^{2}g}{\partial x_{j}\,\partial x_{k}}$}({\boldsymbol{x}})}\right|.

The following theorem is a simplified version of Theorem 4.4 in [5].

Theorem 4.12 (Isaev and McKay [5]).

Let $c_{1},c_{2},\rho_{1},\rho_{2},\phi_{1},\phi_{2}$ be nonnegative real constants. Let $\accentset{\sim}{A}$ be an $n\times n$ positive definite symmetric real matrix and let $T$ be a real matrix such that $T^{\mathrm{t}}\!\accentset{\sim}{A}T=I$ . Let $\varOmega$ be a measurable set such that $T(U_{n}(\rho_{1}))\subseteq\varOmega\subseteq T(U_{n}(\rho_{2}))$ , and let $f:{\mathbb{R}}^{n}\to{\mathbb{C}}$ and $h:\varOmega\to{\mathbb{C}}$ be measurable functions. Assume the following conditions:

(a)

$\log n\leqslant\rho_{1}\leqslant\rho_{2}$ .

(b)

For ${\boldsymbol{x}}\in T(U_{n}(\rho_{1}))$ and $1\leqslant j\leqslant n$ ,

	$\displaystyle 2\rho_{1}\,\mathopen{\\|}T\mathclose{\\|}_{1}\,\mathopen{}\mathclose{{}\left\|\lower 0.6458pt\hbox{\large$\frac{\partial f}{\partial x_{j}}$}({\boldsymbol{x}})}\right\|\leqslant\phi_{1}n^{-1/3}\leqslant\lower 0.6458pt\hbox{\large$\frac{2}{3}$},\qquad\text{and}$
	$\displaystyle 4\rho_{1}^{2}\,\mathopen{\\|}T\mathclose{\\|}_{1}\,\mathopen{\\|}T\mathclose{\\|}_{\infty}\,\mathopen{\\|}H(f,T(U_{n}(\rho_{1})))\mathclose{\\|}_{\infty}\leqslant\phi_{1}n^{-1/3}.$

(c)

For ${\boldsymbol{x}}\in T(U_{n}(\rho_{2}))$ and $1\leqslant j\leqslant n$ ,

2\rho_{2}\,\mathopen{\|}T\mathclose{\|}_{1}\,\mathopen{}\mathclose{{}\left|\lower 0.6458pt\hbox{\large$\frac{\partial\,\Re f}{\partial x_{j}}$}({\boldsymbol{x}})}\right|\leqslant(2\phi_{2})^{3/2}n^{-1/2}.

(d)

$\mathopen{|}f({\boldsymbol{x}})\mathclose{|}\leqslant n^{c_{1}}e^{c_{2}{\boldsymbol{x}}^{\mathrm{t}}\!\tilde{A}{\boldsymbol{x}}/n}$ for ${\boldsymbol{x}}\in{\mathbb{R}}^{n}$ .

Let $\boldsymbol{X}$ be a random variable with the normal density $\pi^{-n/2}\mathopen{|}\accentset{\sim}{A}\mathclose{|}^{1/2}e^{-{\boldsymbol{x}}^{\mathrm{t}}\!\accentset{\sim}{A}{\boldsymbol{x}}}$ . Then, provided $\operatorname{\mathbb{V\!}}f(\boldsymbol{X})$ and $\operatorname{\mathbb{V\!}}\,\Re f(\boldsymbol{X})$ are finite and $h$ is bounded in $\varOmega$ ,

\int_{\varOmega}e^{-{\boldsymbol{x}}^{\mathrm{t}}\!\tilde{A}{\boldsymbol{x}}+f({\boldsymbol{x}})+h({\boldsymbol{x}})}\,d{\boldsymbol{x}}=(1+K)\,\pi^{n/2}\mathopen{|}\accentset{\sim}{A}\mathclose{|}^{-1/2}e^{\operatorname{\mathbb{E}}f(\boldsymbol{X})+\frac{1}{2}\operatorname{\mathbb{V\!}}f(\boldsymbol{X})},

where, for large enough $n$ , $K=K(n)$ satisfies

\mathopen{|}K\mathclose{|}\leqslant e^{\frac{1}{2}\operatorname{Var}\Im f(\boldsymbol{X})}\,\bigl{(}e^{\phi_{1}^{3}+e^{-\rho_{1}^{2}/2}}+2e^{\phi_{2}^{3}+e^{-\rho_{1}^{2}/2}}-3+\sup_{{\boldsymbol{x}}\in\varOmega}\,\mathopen{|}e^{h({\boldsymbol{x}})}-1\mathclose{|}\bigr{)}.

Lemma 4.13.

If $\varLambda=\Omega(r^{16}m^{2-r})$ and $\bar{\varOmega}$ is the domain provided by Lemma 4.11,

\int_{\bar{\varOmega}}F(\boldsymbol{\theta})e^{-\boldsymbol{\theta}W^{\mathrm{t}}W\boldsymbol{\theta}}\,d\boldsymbol{\theta}=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\pi^{n/2}}{|A+W^{\mathrm{t}}W|^{1/2}}$}\,\exp\Bigl{(}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{r}{12\varLambda m^{r-2}}$}+O(n^{-2/5})\Bigr{)}.

Proof.

We will apply Theorem 4.12 with $\accentset{\sim}{A}=A+W^{\mathrm{t}}W$ , $\rho_{1}=\log n$ , $\rho_{2}=3r^{3}\log n$ , $\varOmega=\bar{\varOmega}$ and $T$ as in (4.11).

We first verify the condition $T(U_{n}(\rho_{1}))\subseteq\bar{\varOmega}\subseteq T(U_{n}(\rho_{2}))$ . Using the bounds for $\mathopen{\|}T\mathclose{\|}_{\infty}$ and $\mathopen{\|}T^{-1}\mathclose{\|}_{\infty}$ established in Lemma 4.9 gives

	$\displaystyle T(U_{n}(\rho_{1}))$	$\displaystyle\subseteq U_{n}(\mathopen{\\|}T\mathclose{\\|}_{\infty}\log n)\subseteq U_{n}(3\varLambda^{-1/2}m^{-(r-1)/2}\log n)\subseteq U_{n}\Bigl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{2\rho_{0}}{3r}$}\Bigr{)},$		(4.13)
	$\displaystyle U_{n}(4\rho_{0})$	$\displaystyle\subseteq T(U_{n}(4\mathopen{\\|}T^{-1}\mathclose{\\|}_{\infty}\rho_{0}))=T(U_{n}(2^{3/2}r^{3}\log n))\subseteq T(U_{n}(\rho_{2})).$

By Lemma 4.8, we can take $f({\boldsymbol{x}})=if_{3}({\boldsymbol{x}})+f_{4}({\boldsymbol{x}})$ , where

f_{3}({\boldsymbol{x}})=-\lower 0.6458pt\hbox{\large$\frac{1}{6}$}\,\varLambda(1-2\lambda)\sum_{e\in{\mathcal{S}_{r}(m)}}\biggl{(}\sum_{j\in e}x_{j}\biggr{)}^{\!3},\quad f_{4}({\boldsymbol{x}})=\lower 0.6458pt\hbox{\large$\frac{1}{24}$}\,\varLambda(1-6\varLambda)\sum_{e\in{\mathcal{S}_{r}(m)}}\biggl{(}\sum_{j\in e}x_{j}\biggr{)}^{\!4},

and $h({\boldsymbol{x}})=O(n^{-1/2}\log^{5}n)$ .

By (4.13), ${\boldsymbol{x}}\in T(U_{n}(\rho_{1}))$ implies $\mathopen{\|}{\boldsymbol{x}}\mathclose{\|}_{\infty}\leqslant 3\varLambda^{-1/2}m^{-(r-1)/2}\log n$ . Consequently, for any $j$ , since $m^{r-1}$ elements of ${\mathcal{S}_{r}(m)}$ are incident with vertex $j$

	$\displaystyle 2\rho_{1}\mathopen{\\|}T\mathclose{\\|}_{1}\Bigl{\|}\lower 0.6458pt\hbox{\large$\frac{\partial f}{\partial x_{j}}$}({\boldsymbol{x}})\Bigr{\|}$	$\displaystyle=O(1)\log n\,\mathopen{\\|}T\mathclose{\\|}_{1}\varLambda m^{r-1}\bigl{(}(r\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty})^{2}+(r\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty})^{3}\bigr{)}$
		$\displaystyle=O(\varLambda^{-1/2}r^{2}m^{-(r-1)/2}\log^{3}n)=O(n^{-1/2}\log^{3}n),$

where in the final step we used $\varLambda=\Omega(r^{16}m^{2-r})$ and Lemma 2.4. Similarly,
$\mathopen{\|}H(f,T(U_{n}(\rho_{1}))\mathclose{\|}_{\infty}=O(\varLambda^{1/2}r^{2}m^{(r-1)/2}\log n)$ , which implies

4\rho_{1}^{2}\mathopen{\|}T\mathclose{\|}_{1}\mathopen{\|}T\mathclose{\|}_{\infty}\mathopen{\|}H(f,T(U_{n}(\rho_{1}))\mathclose{\|}_{\infty}=O(\varLambda^{-1/2}r^{2}m^{-(r-1)/2}\log^{3}n)=O(n^{-1/2}\log^{3}n),

using $\varLambda=\Omega(r^{16}m^{2-r})$ . Consequently, part (b) of Theorem 4.12 is satisfied with $\phi_{1}=n^{-1/7}$ if $n$ is sufficiently large.

If ${\boldsymbol{x}}\in T(U_{n}(\rho_{2}))$ then $\mathopen{\|}{\boldsymbol{x}}\mathclose{\|}_{\infty}\leqslant\mathopen{\|}T\mathclose{\|}_{\infty}\rho_{2}\leqslant 9\varLambda^{-1/2}r^{3}m^{-(r-1)/2}\log n$ . Noting that $\Re f({\boldsymbol{x}})=f_{4}({\boldsymbol{x}})$ , we calculate

	$\displaystyle 2\rho_{2}\mathopen{\\|}T\mathclose{\\|}_{1}\Bigl{\|}\lower 0.6458pt\hbox{\large$\frac{\partial f_{4}}{\partial x_{j}}$}({\boldsymbol{x}})\Bigr{\|}$	$\displaystyle=O(1)r^{3}\log n\mathopen{\\|}T\mathclose{\\|}_{1}\varLambda m^{r-1}(r\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty})^{3}$
		$\displaystyle=O(\varLambda^{-1}r^{15}m^{1-r}\log^{4}n)=O(n^{-1}\log^{4}n),$

using $\varLambda=\Omega(r^{16}m^{2-r})$ , so part (c) of Theorem 4.12 is satisfied with $\phi_{2}=n^{-1/4}$ if $n$ is sufficiently large.

We next check condition (d) of Theorem 4.12. Define $\ell({\boldsymbol{x}}):=\lower 0.6458pt\hbox{\large$\frac{1}{n}$}\,{\boldsymbol{x}}^{\mathrm{t}}(A+W^{\mathrm{t}}W){\boldsymbol{x}}$ . Since the least eigenvalue of $A+W^{\mathrm{t}}W$ is $\frac{1}{2}\varLambda m^{r-1}$ , we have

\mathopen{\|}{\boldsymbol{x}}\mathclose{\|}_{\infty}^{2}\leqslant\mathopen{\|}{\boldsymbol{x}}\mathclose{\|}_{2}^{2}=O(\varLambda^{-1}rm^{2-r}\ell({\boldsymbol{x}})).

Consequently,

	$\displaystyle f({\boldsymbol{x}})$	$\displaystyle=O(\varLambda r^{3}m^{r}\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty}^{3}+\varLambda r^{4}m^{r}\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty}^{4})$
		$\displaystyle=O(\varLambda r^{2}m^{r}\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty}^{2}+\varLambda r^{4}m^{r}\mathopen{\\|}{\boldsymbol{x}}\mathclose{\\|}_{\infty}^{4})$
		$\displaystyle=O(r^{3}m^{2})\bigl{(}\ell({\boldsymbol{x}})+\varLambda^{-1}r^{3}m^{2-r}\ell({\boldsymbol{x}})^{2}\bigr{)},$

where the second line is because $x^{3}\leqslant x^{2}+x^{4}$ for all $x$ . By $\varLambda=\Omega(r^{16}m^{2-r})$ , $\varLambda^{-1}r^{3}m^{2-r}=O(1)$ . Therefore,

f({\boldsymbol{x}})=O(r^{3}m^{2})\bigl{(}\ell({\boldsymbol{x}})+\ell({\boldsymbol{x}})^{2}\bigr{)}=O(n^{5})e^{\ell({\boldsymbol{x}})},

as required.

Now let $\boldsymbol{X}=(X_{1},\ldots,X_{n})$ be a Gaussian random variable with density proportional to $e^{-{\boldsymbol{x}}^{\mathrm{t}}(A+W^{\mathrm{t}}W){\boldsymbol{x}}}$ . To complete our calculation, we need the expectation and pseudovariance of $f(\boldsymbol{X})$ and the variance of $f_{3}(\boldsymbol{X})$ . The covariance matrix of $\boldsymbol{X}$ is $\frac{1}{2}(A+W^{\mathrm{t}}W)^{-1}$ , and so by Lemma 4.9,

\operatorname{Cov}\mathopen{}\mathclose{{}\left[X_{j},X_{k}}\right]=\begin{cases}\lower 0.6458pt\hbox{\large$\frac{1}{\varLambda m^{r-1}}$}\Bigl{(}1-\lower 0.6458pt\hbox{\large$\frac{r-1}{rm}$}\Bigr{)},&\mbox{if }j=k;\\[1.29167pt] -\lower 0.6458pt\hbox{\large$\frac{r-1}{\varLambda rm^{r}}$},&\mbox{if $j\neq k$ in the same class;}\\[1.29167pt] 0,&\mbox{otherwise.}\end{cases}

For $e\in{\mathcal{S}_{r}(m)}$ define $X_{e}=\sum_{j\in e}X_{j}$ and, for $e,e^{\prime}\in{\mathcal{S}_{r}(m)}$ define

\sigma(e,e^{\prime}):=\operatorname{Cov}[X_{e},X_{e^{\prime}}].

Since covariance is additive and $e$ and $e^{\prime}$ contain one vertex from each class, we have

\sigma(e,e^{\prime})=\varsigma(\mathopen{|}e\cap e^{\prime}\mathclose{|}),\text{~{}~{}where~{}~{}}\varsigma(k):=\frac{1}{\varLambda m^{r-1}}\Bigl{(}k-\raise 0.21529pt\hbox{\small$\displaystyle\frac{r-1}{m}$}\Bigr{)}.

Since $f_{3}$ is an odd function, we have $\operatorname{\mathbb{E}}f_{3}(\boldsymbol{X})=0$ . Isserlis’ theorem (also known as Wick’s formula), see for example [8, Theorem 1.1] implies $\operatorname{\mathbb{E}}X_{e}^{4}=3\sigma(e,e)^{2}$ , so

\operatorname{\mathbb{E}}f(\boldsymbol{X})=\raise 0.21529pt\hbox{\small$\displaystyle\frac{(1-6\varLambda)(mr-r+1)^{2}}{8\varLambda m^{r}}$}=\raise 0.21529pt\hbox{\small$\displaystyle\frac{r^{2}}{8\varLambda m^{r-2}}$}+O(n^{-1}),

using $\varLambda=\Omega(r^{16}m^{2-r})$ and Lemma 2.4.

By Isserlis’ theorem, for any pair $e,e^{\prime}\in{\mathcal{S}_{r}(m)}$ ,

\operatorname{\mathbb{E}}[X_{e}^{3}X_{e^{\prime}}^{3}]=9\,\sigma(e,e)\,\sigma(e^{\prime},e^{\prime})\,\sigma(e,e^{\prime})+6\,\sigma(e,e^{\prime})^{3}.

The number of pairs with $\mathopen{|}e\cap e^{\prime}\mathclose{|}=k$ is $m^{r}\binom{r}{k}(m-1)^{r-k}$ . Summing over $e,e^{\prime}\in{\mathcal{S}_{r}(m)}$ , and applying $\varLambda=\Omega(r^{16}m^{2-r})$ , we obtain

$\displaystyle\operatorname{Var}f_{3}(\boldsymbol{X})$	$\displaystyle=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\varLambda^{2}(1-2\lambda)^{2}m^{r}(m-1)^{r}}{36}$}\sum_{k=0}^{r}\binom{r}{k}(m-1)^{-k}\bigl{(}9\varsigma(r)^{2}\varsigma(k)+6\varsigma(k)^{3}\bigr{)}$
	$\displaystyle=\raise 0.21529pt\hbox{\small$\displaystyle\frac{(1-4\varLambda)r(3r+2)m}{12\varLambda m^{r-1}}$}\Bigl{(}1-\raise 0.21529pt\hbox{\small$\displaystyle\frac{6(r-1)}{(3r+2)m}$}+\raise 0.21529pt\hbox{\small$\displaystyle\frac{(r-1)(3r-5)}{r(3r+2)m^{2}}$}\Bigr{)}$
	$\displaystyle=\raise 0.21529pt\hbox{\small$\displaystyle\frac{r(3r+2)}{12\varLambda m^{r-2}}$}+O(n^{-1})$	(4.14)
	$\displaystyle=O(1).$	(4.15)

Similarly,

	$\displaystyle\operatorname{Var}[X_{e}^{4},X_{e^{\prime}}^{4}]$	$\displaystyle=\operatorname{\mathbb{E}}(X_{e}^{4}X_{e^{\prime}}^{4})-(\operatorname{\mathbb{E}}X_{e}^{4})(\operatorname{\mathbb{E}}X_{e^{\prime}}^{4})$
		$\displaystyle=72\sigma(e,e)\sigma(e^{\prime},e^{\prime})\sigma(e,e^{\prime})^{2}+24\sigma(e,e^{\prime})^{4}.$

Summing over $e,e^{\prime}\in{\mathcal{S}_{r}(m)}$ with $\varLambda=\Omega(r^{16}m^{2-r})$ gives

\operatorname{Var}(f_{4}(\boldsymbol{X}))=O(n^{-1}).

(4.16)

Finally, since $\operatorname{Cov}(f_{3}(\boldsymbol{X}),f_{4}(\boldsymbol{X}))=0$ by Isserlis’ theorem, (4.14) and (4.16) together imply that

\operatorname{\mathbb{V\!}}f(\boldsymbol{X})=-\frac{r(3r+2)}{12\varLambda m^{r-2}}+O(n^{-1})

and so

\operatorname{\mathbb{E}}f(\boldsymbol{X})+\lower 0.6458pt\hbox{\large$\frac{1}{2}$}\operatorname{\mathbb{V\!}}f(\boldsymbol{X})=-\raise 0.21529pt\hbox{\small$\displaystyle\frac{r}{12\varLambda m^{r-2}}$}+O(n^{-1}).

(4.17)

The lemma now follows from Theorem 4.12 and equations (4.15) and (4.17). ∎

The above precise result relies on $d=\lambda m^{r-1}$ since otherwise the linear term doesn’t vanish. However, if we are concerned with the integral of $\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}$ we can ignore both the linear and cubic terms. So, with identical proof, the following is true whenever $\varLambda=\Omega(r^{16}m^{2-r})$ .

Corollary 4.14.

For $\bar{\varOmega}$ as in Lemma 4.11 we have

\int_{\bar{\varOmega}}e^{-{\boldsymbol{x}}^{\mathrm{t}}W^{\mathrm{t}}W{\boldsymbol{x}}}\,\mathopen{|}F(\boldsymbol{\theta})\mathclose{|}\,d\boldsymbol{\theta}=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\pi^{n/2}}{|A+W^{\mathrm{t}}W|^{1/2}}$}\,\exp\Bigl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{r^{2}}{8\varLambda m^{r-2}}$}+O(n^{-2/5})\Bigr{)}.

Our final task is to evaluate the determinant $\mathopen{|}Q^{\mathrm{t}}Q+W^{\mathrm{t}}W\mathclose{|}$ . Let ${\boldsymbol{j}}$ be the column vector of all 1s, and recall that ${\boldsymbol{e}}_{j}$ is the $j$ -th elementary column vector.

Lemma 4.15.

$\mathopen{|}Q^{\mathrm{t}}Q+W^{\mathrm{t}}W\mathclose{|}=r^{r}\Bigl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\varLambda m^{r}}{2}$}\Bigr{)}^{r-1}$ .

Proof.

We proceed in three stages.

Claim 1. Consider the block matrix

M=\begin{pmatrix}M_{1,1}&\!M_{1,2}\\ M_{2,1}&\!M_{2,2}\end{pmatrix},

where each block has size $m\times m$ and have the following form

	$\displaystyle M_{1,1}$	$\displaystyle=I+(r-1)yJ_{m}$	$\displaystyle M_{1,2}$	$\displaystyle=(r-1){\boldsymbol{j}}{\boldsymbol{e}}_{m}^{\mathrm{t}}-(r-1)yJ_{m}$
	$\displaystyle M_{2,1}$	$\displaystyle={\boldsymbol{e}}_{m}{\boldsymbol{j}}^{\mathrm{t}}-yJ_{m}$	$\displaystyle M_{2,2}$	$\displaystyle=I-{\boldsymbol{j}}{\boldsymbol{e}}_{m}^{\mathrm{t}}-{\boldsymbol{e}}_{m}{\boldsymbol{j}}^{\mathrm{t}}+mr{\boldsymbol{e}}_{m}{\boldsymbol{e}}_{m}^{\mathrm{t}}+yJ_{m}.$

Then, we have $|M|=r^{2}m^{2}y$ .

Proof of Claim 1. Subtract from the last row the first $m$ rows and add to it the consequent $m-1$ rows. Then the last row has the form

(-rmy,\ldots,-rmy,rmy,\ldots rmy).

After dividing this row by $rmy$ it can be used to eliminate the terms of the form $yJ_{m}$ .

Finally add to the last row the first $m$ rows and subtract from it the consequent $m-1$ rows from it to create an upper triangular matrix and the result follows.

Claim 2. Let the matrix $M$ have size $m\times m$ with the form

\displaystyle M=I-{\boldsymbol{j}}{\boldsymbol{e}}_{m}^{\mathrm{t}}-{\boldsymbol{e}}_{m}{\boldsymbol{j}}^{\mathrm{t}}+m{\boldsymbol{e}}_{m}{\boldsymbol{e}}_{m}^{\mathrm{t}}+yJ_{m}.

Then, we have $|M|=m^{2}y$ .

Proof of Claim 2. Add the first $m-1$ rows to the last row, which will subsequently have the form

(my,\ldots,my).

After dividing this row by $my$ it can be used to eliminate the terms of the form $yJ_{m}$ . Subsequently subtract the first $m-1$ rows from the last row to create an upper triangular matrix.

Proof of the lemma. Break down the matrix into blocks of size $m\times m$ and we will consider “block” operations. Starting with the last row and finishing with the third row subtract from each row the one above it. Then starting from the penultimate column and finishing with the second column add the column on the right to it. These operations lead to an upper triangular block matrix, where the diagonal consists of $r-1$ blocks. The first block has size $2m\times 2m$ and form as in Claim 1 with $y=\varLambda m^{r-1}/(2m)$ . The remaining $r-2$ blocks have size $m\times m$ and has the form as in Claim 2 when $y=r\varLambda m^{r-1}/(2m)$ . ∎

Now we can prove Lemmas 2.3 and 4.3.

Proof of Lemma 4.3.

Lemma 4.11 and Corollary 4.14 implies

\int_{\mathcal{B}\cap U_{n}(\rho_{0})}\!\!|F(\boldsymbol{\theta})|\,d\boldsymbol{\theta}=\bigl{|}Q^{\mathrm{t}}Q+W^{\mathrm{t}}W\bigr{|}^{1/2}\raise 0.21529pt\hbox{\small$\displaystyle\frac{\pi^{(n-r+1)/2}}{|A+W^{\mathrm{t}}W|^{1/2}}$}\,\exp\Bigl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{r^{2}}{8\varLambda m^{r-2}}$}+O(n^{-2/5})\Bigr{)}.

Using Lemmas 4.9 (b) and 4.15 for the values of the determinants gives the required outcome

\int_{\mathcal{B}\cap U_{n}(\rho_{0})}\!\!|F(\boldsymbol{\theta})|\,d\boldsymbol{\theta}=m^{(r-1)/2}\mathopen{}\mathclose{{}\left(\raise 0.21529pt\hbox{\small$\displaystyle\frac{2\pi}{\varLambda m^{r-1}}$}}\right)^{(n-r+1)/2}\exp\Bigl{(}\raise 0.21529pt\hbox{\small$\displaystyle\frac{r^{2}}{8\varLambda m^{r-2}}$}+O(n^{-2/5})\Bigr{)}.\qed

Proof of Lemma 2.3.

Lemmas 4.1, 4.7, 4.9(b), 4.11, 4.13 and 4.15 imply

H_{r}(d,m)=\mathopen{}\mathclose{{}\left(\lambda^{\lambda}(1-\lambda)^{1-\lambda}}\right)^{-m^{r}}\!(2\pi\varLambda)^{(r-n-1)/2}m^{-r(r-1)(m-1)/2}\exp\Bigl{(}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{r}{12\varLambda m^{r-2}}$}+O(n^{-2/5})\Bigr{)}.

So our remaining task is to verify that $H_{r}(d,m)$ matches $\hat{H}_{r}(d,m)$ from (1.1). For integer $N$ and $\lambda\in(0,1)$ , Stirling’s expansion gives

\binom{N}{\lambda N}=\raise 0.21529pt\hbox{\small$\displaystyle\frac{\bigl{(}\lambda^{\lambda}(1-\lambda)^{1-\lambda}\bigr{)}^{-N}}{\sqrt{2\pi N\varLambda}}$}\exp\Bigl{(}-\raise 0.21529pt\hbox{\small$\displaystyle\frac{1-\varLambda}{12\varLambda N}$}+O(\varLambda^{-3}N^{-3})\Bigr{)}.

Applying this expansion for $N=m^{r}$ and $N=m^{r-1}$ in (1.1), with the assumption $\varLambda=\Omega(r^{16}m^{2-r})$ and Lemma 2.4 gives the same expression as we have shown for $H_{r}(d,m)$ with error term $O(n^{-1})$ . This completes the proof. ∎

References

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Enumeration of regular multipartite hypergraphs

Abstract

1 Introduction

Theorem 1.1.

2 The four regimes covering all possibilities

Lemma 2.1.

Lemma 2.2.

Lemma 2.3.

Lemma 2.4.

Proof.

Proof of Theorem 1.1.

3 Sparse regimes

3.1 Proof of Lemma 2.2

3.2 The case r=3r=3

Lemma 3.1.

Proof.

Lemma 3.2 (Greenhill, McKay, Wang [4]).

Lemma 3.3.

Proof.

Proof of Lemma 2.1.

4 Dense range

4.1 An integral for the number of (r,r)(r,r)-graphs

Lemma 4.1.

Proof.

4.2 Outside the main region

Lemma 4.2.

Lemma 4.3.

Lemma 4.4.

Proof.

Lemma 4.5.

Proof.

Lemma 4.6.

Proof.

Lemma 4.7.

Proof.

4.3 Inside the main region

Lemma 4.8.

Proof.

Lemma 4.9.

Proof.

Lemma 4.10 (Isaev and McKay [5]).

Lemma 4.11.

Proof.

Theorem 4.12 (Isaev and McKay [5]).

Lemma 4.13.

Proof.

Corollary 4.14.

Lemma 4.15.

Proof.

Proof of Lemma 4.3.

Proof of Lemma 2.3.

References

3.2 The case $r=3$

4.1 An integral for the number of $(r,r)$ -graphs