# Central Limit Theorem for tensor products of free variables

Cécilia Lancien and Patrick Oliveira Santos and Pierre Youssef  Cécilia
Lancien. CNRS & Institut Fourier UMR 5582, Université Grenoble Alpes,
Grenoble, France<EMAIL_ADDRESS>Patrick Oliveira
Santos. Laboratoire d’Analyse et de Mathématiques Appliquées UMR 8050,
Université Gustave Eiffel, Université Paris Est Créteil, Marne-la-Vallée,
France<EMAIL_ADDRESS>Pierre Youssef. Division of Science,
NYU Abu Dhabi, Abu Dhabi, UAE & Courant Institute of Mathematical Sciences,
New York University, New York, USA<EMAIL_ADDRESS>
###### Abstract.

We establish a central limit theorem for tensor product random variables
$c_{k}:=a_{k}\otimes a_{k}$, where $(a_{k})_{k\in\mathbb{N}}$ is a free family
of variables. We show that if the variables $a_{k}$ are centered, the limiting
law is the semi-circle. Otherwise, the limiting law depends on the mean and
variance of the variables $a_{k}$ and corresponds to a free interpolation
between the semi-circle law and the classical convolution of two semi-circle
laws.

##  1\. Introduction

The Free Central Limit Theorem serves as a foundational principle in free
probability [15], [11, Lecture 8]. It asserts that as the number of freely
independent operators summed together approaches infinity, the distribution of
the normalized sum tends towards an asymptotically semi-circular shape. This
mirrors the classical Central Limit Theorem but with independence conditions
replaced by free independence (also known as freeness) and the Gaussian limit
substituted with a semi-circular limit. More precisely, let
$(\mathcal{A},\tau)$ be a unital noncommutative probability space equipped
with a faithful tracial state $\tau$ [11, Lecture 1]. We say that subalgebras
$\mathcal{A}_{1},\ldots,\mathcal{A}_{d}\subset\mathcal{A}$ are free if

$\displaystyle\tau(a_{1}\ldots a_{p})=0,$

whenever $p\geq 1$, $a_{i}\in\mathcal{A}_{j_{i}}$, $\tau(a_{i})=0$ for all
$i\in[p]$ and $j_{1}\neq j_{2}\neq\cdots\neq j_{p}$. We say that random
variables $a_{1},\ldots,a_{d}\in\mathcal{A}$ are free if their generated
algebras are free. We say that a sequence of (self-adjoint) variables
$a_{n}\in(\mathcal{A}_{n},\tau_{n})$ converges in distribution to a variable
$a\in(\mathcal{A},\tau)$ if

$\displaystyle\tau_{n}(a_{n}^{p})\to\tau(a^{p}),$

for all integers $p\geq 0$ and we denote it $a_{n}\Rightarrow a$. We denote
$a-\lambda:=a-\lambda\mathbf{1}$, where $\mathbf{1}\in\mathcal{A}$ is the unit
in the algebra. As usual, $\tau(a)$ is the mean of $a$ and the variance is
given by

$\displaystyle\operatorname{var}(a)=\tau((a-\tau(a))^{2}).$

The Free Central Limit Theorem states that if
$a_{1},\ldots,a_{n}\in(\mathcal{A},\tau)$ are free self-adjoint i.i.d random
variables with mean $\lambda$ and variance $\sigma^{2}$, then

$\frac{1}{\sigma\sqrt{n}}\sum_{k\in[n]}(a_{k}-\lambda)\Rightarrow\mu_{sc},$

where $\mu_{sc}$ denotes the semi-circle distribution whose density is given
by

$\displaystyle f_{sc}(x)=\frac{1}{2\pi}\sqrt{4-x^{2}}\mathbf{1}_{|x|\leq 2}.$

The goal of this paper is to establish a central limit theorem for the tensor
product of free random variables. Concretely, given
$a_{1},\ldots,a_{n}\in(\mathcal{A},\tau)$ free self-adjoint i.i.d random
variables, we aim at studying the convergence of the normalized sequence

$\frac{1}{\sqrt{n}}\sum_{k\in[n]}\left(a_{k}\otimes
a_{k}-\tau\otimes\tau(a_{k}\otimes a_{k})\right),$ (1)

in the product space $(\mathcal{A}\otimes\mathcal{A},\tau\otimes\tau)$.

Just as free probability captures the limiting behavior of random matrices,
the above expression appears naturally as the limiting object corresponding to
several models of random quantum channels [7]. Indeed, given
$M_{1},\ldots,M_{n}\in\mathcal{M}_{d}(\mathbb{C})$ independent random self-
adjoint matrices, it was shown in [7] that the empirical spectral distribution
(ESD) of the quantum channel

$\displaystyle\Delta_{d,n}:=\frac{1}{\sqrt{n}}\sum_{k\in[n]}\left(M_{k}\otimes
M_{k}-\operatorname{\mathbb{E}}[M_{k}\otimes M_{k}]\right),$

having the $M_{k}$’s as random Kraus operators and with fixed Kraus rank $n$,
converges as $d\to\infty$ to the expression in (1) with the $a_{k}$’s being
the corresponding limits of the ESD of the $M_{k}$’s. Moreover, it was in
particular shown that if the random matrices $M_{k}$ are centered, then the
ESD of $\Delta_{d,n}$ converges as $n,d\to\infty$ to the semi-circle
distribution. These two statements combined suggest that, in the case where
the $a_{k}$’s are centered, an analogue of the free central limit theorem
should hold for the $a_{k}\otimes a_{k}$’s. Whereas these heuristics indicate
that the semi-circle distribution should appear as the limit of (1) when the
$a_{k}$’s are centered, the convergence and the explicit limit are not clear
in the general case. The goal of this paper is to address this by establishing
the convergence of the expression in (1) and identifying the limiting object.
The latter, as we show, depends on the mean and variance of the variables
$a_{k}$’s and represents a free interpolation between a semi-circle
distribution and the classical convolution of two semi-circle distributions.

Random matrix models of the form

$\displaystyle M=\sum_{k\in[n]}M_{k}\otimes M_{k},$ (2)

for $M_{1},\ldots,M_{n}\in\mathcal{M}_{d}(\mathbb{C})$ independent random
self-adjoint matrices, are in fact useful in other areas of Quantum
Information Theory. When the $M_{k}$’s are positive semidefinite matrices,
normalizing $M$ by its trace produces a model for a random separable quantum
state. Little is known about the typical asymptotic spectrum of separable
states, contrary to that of entangled ones [1]. Moreover, a random matrix $M$
of the form (2) appears naturally when performing a so-called realignment
operation on a quantum state. Understanding the spectrum of the realignment of
a state is important as it gives information on the entanglement of the state.
In [2], this was done in the particular case where the $M_{k}$’s are Gaussian
matrices (corresponding to the case where the state is a normalized Wishart
matrix). The results and techniques we develop here (combined with those in
[7]) could be useful in addressing the questions mentioned above.

Given a measure $\mu$, we denote

$\displaystyle(t\mu)(A):=\mu(t^{-1}A),$

its dilation by $t\neq 0$, where $A$ is any Borel set in $\mathbb{R}$.
Equivalently, if $a$ is a random variable with distribution $\mu$, then $ta$
has distribution $t\mu$.

The following is our main theorem.

######  Theorem 1.1.

Let $a\in(\mathcal{A},\tau)$ be a self-adjoint random variable with mean
$\tau(a)=\lambda$ and variance $\operatorname{var}(a)=\sigma^{2}\neq 0$.
Denote

$\delta^{2}:=\operatorname{var}(a\otimes
a)=\sigma^{2}(\sigma^{2}+2\lambda^{2}),$

and

$\displaystyle q:=\frac{2\lambda^{2}}{\sigma^{2}+2\lambda^{2}}\in[0,1].$

Given $(a_{k})_{k\in\mathbb{N}}$ a sequence of free copies of $a$, the
normalized sum

$\displaystyle S_{n}:=\frac{1}{\delta\sqrt{n}}\sum_{k\in[n]}(a_{k}\otimes
a_{k}-\lambda^{2})$

converges in distribution as $n\to\infty$ to

$\mu_{q}:=\sqrt{q}\left(\frac{1}{\sqrt{2}}\mu_{sc}+\frac{1}{\sqrt{2}}\mu_{sc}\right)\boxplus\sqrt{1-q}\,\mu_{sc},$
(3)

where $+$ denotes the classical convolution and $\boxplus$ denotes the free
convolution.

The difficulty in analyzing $S_{n}$ stems from the complicated dependence
structure exhibited by tensors, combining classical independence (between the
two legs of the tensor) and freeness (between the variables across tensors).
In the centered case, similar computations were made for semi-circle random
variables [10, 5]. It would be of interest to design a general notion of
independence corresponding to the tensor case, analyze its properties, derive
the corresponding limit theorems, and characterize the corresponding universal
objects. One particular generalization is by replacing the tensor product with
the product of $\varepsilon$-independent random variables [14, 9, 13]. A
direct consequence of Theorem 1.1 is that such a notion cannot, in general,
reduce to freeness.

######  Corollary 1.2.

Let $a_{1},\ldots,a_{n}\in(\mathcal{A},\tau)$ be self-adjoint free i.i.d
noncentered random variables. Then $\\{a_{k}\otimes a_{k}:k\in[n]\\}$ are not
free.

The above corollary trivially follows from Theorem 1.1, since if the
$a_{k}\otimes a_{k}$’s were free, the limit of their normalized sum would be
the semi-circle distribution contradicting the conclusion of Theorem 1.1 (when
$\lambda\neq 0$). This fact was originally proved in [4] where, more
generally, the freeness of tensors of free variables was characterized.

This paper is organized as follows. In Section 2, we recall some definitions
and notations. In Section 3, we provide some properties of the limiting
measure appearing in Theorem 1.1. Section 4 establishes the existence of the
limit, while Section 5 is dedicated to the proof of Theorem 1.1.

### Acknowledgments

The last named author would like to thank Guillaume Cébron and Roland Speicher
for helpful discussions. The second named author also thanks Philippe Biane
for helpful discussions. Part of this work was initiated during a stay of the
second named author at New York University in Abu Dhabi, partly funded by a
doctoral mobility grant delivered by Université Gustave Eiffel; he would like
to thank both institutions for their support and the excellent working
assumptions. The first named author was supported by the ANR projects
ESQuisses (grant number ANR-20-CE47-0014-01), STARS (grant number
ANR-20-CE40-0008), and QTraj (grant number ANR-20-CE40-0024-01).

##  2\. Preliminaries and notations

Given $p\in\mathbb{N}$, a partition $\pi=\\{V_{1},\ldots,V_{k}\\}$ of $[p]$ is
a collection of disjoint sets $V_{1},\ldots,V_{k}$ called blocks such that

$\displaystyle V_{1}\cup\cdots\cup V_{k}=[p].$

We denote by $P(p)$ the set of partitions of $[p]$. We say that a partition
$\pi\in P(p)$ is connected (also referred to as a linked diagram in [12]) if
no proper subinterval of $[p]$ can be written as the union of blocks of $\pi$.
A partition $\pi\in P(p)$ has a crossing $i<k<j<l$ if there exist two disjoint
blocks $V_{1},V_{2}\in\pi$ such that $\\{i,j\\}\subset V_{1}$ and
$\\{k,l\\}\subset V_{2}$. In this case, we say that $V_{1}$ crosses $V_{2}$. A
block $V\in\pi$ is crossing if there exists another $V^{\prime}\in\pi$ such
that $V^{\prime}$ crosses $V$, and noncrossing if it does not cross any other
block. We say that a partition $\pi\in P(p)$ is a noncrossing partition if all
its blocks are noncrossing. We denote by $P^{\operatorname{con}}(p)$ (resp.
$NC(p)$) the set of all connected (resp. noncrossing) partitions of $[p]$; see
Figure 1. We note that the cardinal $|NC(p)|$ is equal to $C_{p}$, the $p$-th
Catalan number.

(a) Connected partition

(b) General partition

(c) Noncrossing partition

Figure 1. Examples of partitions.

Finally, for a partition $\pi\in P(p)$, we denote by $G(\pi)$ its intersection
graph. It is the graph over the blocks of $\pi$ such that two blocks are
connected if they cross, under some arbitrary labeling. We say that a
partition $\pi$ is a bipartite partition if its intersection graph is
bipartite and denote it $\pi\in P^{\operatorname{bi}}(p)$; see Figure 2 for
the partitions in Figure 1.

(a) Connected partition

(b) General partition

(c) Noncrossing partition

Figure 2. Examples of intersection graphs.

Given $\pi\in P(p)$, we denote $|\pi|$ its number of blocks and
$\operatorname{\bf cr}(\pi)$ its number of crossing blocks. Therefore, the
number of noncrossing blocks of $\pi$ is

$\displaystyle\operatorname{\bf ncr}(\pi):=|\pi|-\operatorname{\bf cr}(\pi).$

Given $\pi\in P(p)$, we denote $\operatorname{\bf cc}(\pi)$ its number of
connected components. We denote $P_{2}(p),P_{2}^{\operatorname{bi}}(p)$,
$P_{2}^{\operatorname{con}}(p)$, $P_{2}^{\operatorname{bicon}}(p)$ and
$NC_{2}(p)$ the set of pair partitions, bipartite pair partitions, connected
pair partitions, bipartite connected pair partitions, and noncrossing pair
partitions, respectively, that is, those partitions such that all of their
blocks have cardinality two.

A pair partition $\pi\in P_{2}(p)$ can be decomposed into its crossing
connected components, namely, let $\hat{\pi}\in P(p)$ be the choice of
connected components and, for each block $T\in\hat{\pi}$, draw a connected
pair partition $\pi_{T}\in P_{2}^{\operatorname{con}}(T)$. By definition,
$\hat{\pi}\in NC(p)$ as otherwise two disjoint components would meet
($\hat{\pi}$ is called the noncrossing closure of $\pi$ in [8]). The mapping

$\displaystyle\Phi:\pi\mapsto(\hat{\pi},(\pi_{T})_{T\in\hat{\pi}})$ (4)

is a bijection that will be used throughout the proof of Theorem 1.1; see
Figure 3.

(a) $\pi$

(b) $\hat{\pi}$

(c) $\pi_{\\{1,2,3,4\\}}$

(d) $\pi_{\\{5,8\\}}\cong\pi_{\\{6,7\\}}$

Figure 3. A partition $\pi$ and its image $\Phi(\pi)$.

Note, for instance, that $|\hat{\pi}|=\operatorname{\bf cc}(\pi)$. We denote

$\displaystyle\operatorname{Proj}(\hat{\pi}):=\\{((\pi_{T})_{T\in\hat{\pi}}):\pi_{T}\in
P_{2}^{\operatorname{con}}(T),\ \forall T\in\hat{\pi}\\}.$

Given $a_{1},\ldots,a_{n}\in\mathcal{A}$, we denote by
$\kappa^{\operatorname{free}}_{n}(a_{1},\ldots,a_{n})$ their free cumulants,
namely, for any $i\in[n]^{k}$, we have

$\displaystyle\tau(a_{i_{1}}\cdots a_{i_{k}})=\sum_{\pi\in
NC(k)}\kappa^{\operatorname{free}}_{\pi}(a_{i_{1}},\ldots,a_{i_{k}}),$
$\displaystyle\kappa^{\operatorname{free}}_{\pi}(a_{i_{1}},\ldots,a_{i_{k}})=\prod_{V=\\{v_{1}<\cdots<v_{l}\\}\in\pi}\kappa^{\operatorname{free}}_{|V|}(a_{i_{v_{1}}},\ldots,a_{i_{v_{l}}}).$

This is known as the moment-cumulant formula [11, Notation 11.5]. Here and
throughout the paper, we denote $V=\\{v_{1}<\cdots<v_{l}\\}$ a block
$V=\\{v_{1},\ldots,v_{l}\\}$ such that $v_{1}<\cdots<v_{l}$; see [11, Lecture
11]. Note also that if the variables $a_{1},\ldots,a_{n}$ are free, the free
mixed cumulants vanish [11, Proposition 11.15]. We denote
$\kappa^{\operatorname{free}}_{n}(a)$ the free cumulants of a random variable
$a$. The moment-cumulant formula implies the following, which is going to be
used extensively in Subsection 5.3; see [11, Lecture 5, Equation 5.6].

######  Lemma 2.1.

Let $a,c_{1},c_{2}$ be variables such that $a$ is free from
$\\{c_{1},c_{2}\\}$. Then

$\displaystyle\tau(ac_{1}ac_{2})=\operatorname{var}(a)\tau(c_{1})\tau(c_{2})+\tau^{2}(a)\tau(c_{1}c_{2}).$

We will equivalently denote $\kappa^{\operatorname{free}}_{n}(\mu)$ the free
cumulants of a random variable $a$ with distribution $\mu$. Given two measures
$\mu_{a}$ and $\mu_{b}$, the free convolution $\mu_{a}\boxplus\mu_{b}$ denotes
the distribution of $a+b$, where $a$ and $b$ are free random variables with
distribution $\mu_{a}$ and $\mu_{b}$, respectively. The classical convolution
$\mu_{a}+\mu_{b}$ denotes the distribution of $a+b$, where now $a$ and $b$ are
classical independent random variables with distribution $\mu_{a}$ and
$\mu_{b}$, respectively.

##  3\. Properties of the limiting measure

In this section, we summarize some of the properties of the measure $\mu_{q}$
appearing in (3). We start by computing the free cumulants and the moments of
$\mu_{1}$.

######  Proposition 3.1.

Let

$\displaystyle\mu_{1}=\frac{1}{\sqrt{2}}\mu_{sc}+\frac{1}{\sqrt{2}}\mu_{sc}.$

Then the following hold.

  1. (1)

Its odd moments vanish and, for every $p\in\mathbb{N}$, its $2p$-th moment is
given by

$\displaystyle\int x^{2p}\,\text{d}\mu_{1}$
$\displaystyle=2^{-p}\sum_{l=0}^{p}\binom{2p}{2l}C_{l}C_{p-l}=2^{-p}\sum_{\pi\in
P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf cc}(\pi)},$

where we recall that $C_{l}$ denotes the $l$-th Catalan number.

  2. (2)

Its odd free cumulants vanish, $\kappa^{\operatorname{free}}_{2}(\mu_{1})=1$
and for any even integer $n\geq 4$, we have

$\displaystyle\kappa^{\operatorname{free}}_{n}(\mu_{1})=2\left(\frac{1}{2}\right)^{n/2}|P_{2}^{\operatorname{bicon}}(n)|.$

###### Proof.

Let $x_{1},x_{2}$ be two classical i.i.d semi-circle random variables. Then

$\displaystyle\int
x^{2p}\,\text{d}\mu_{1}=2^{-p}\tau\left((x_{1}+x_{2})^{2p}\right).$

Since the variables are classical independent, and in particular, they
commute, we get

$\displaystyle\tau\left((x_{1}+x_{2})^{2p}\right)=2^{-p}\sum_{l=0}^{2p}\binom{2p}{l}\tau(x_{1}^{l})\tau(x_{2}^{2p-l}).$

The odd moments of $x_{1}$ vanish and, for any $l\geq 1$,
$\tau(x_{1}^{2l})=C_{l}$, hence

$\displaystyle\int
x^{2p}\,\text{d}\mu_{1}=2^{-p}\sum_{l=0}^{p}\binom{2p}{2l}C_{l}C_{p-l}.$

Now consider the sequence $m_{p}$ given by

$\displaystyle m_{p}:=2^{-p}\sum_{\pi\in
P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf cc}(\pi)}.$

For every bipartite pair partition $\pi$, we can decompose it into its
bipartite sets. Namely, let $\mathcal{V}_{1}=\\{V_{1},\ldots,V_{m}\\}$ and
$\mathcal{V}_{2}=\\{V_{m+1},\ldots,V_{p}\\}$ be the bipartition of its blocks
in two disjoint families of vertices. Then, the blocks $V_{1},\ldots,V_{m}$
are noncrossing from one another, and so are $V_{m+1},\ldots,V_{p}$. Let

$\displaystyle I=\bigcup_{i=1}^{m}V_{i}\subseteq[2p].$

Then

$\displaystyle\pi_{1}:=\\{V_{1},\ldots,V_{m}\\}\in NC_{2}\left(I\right);$
$\displaystyle\pi_{2}:=\\{V_{m+1},\ldots,V_{p}\\}\in
NC_{2}\left(I^{c}\right).$

We denote $(\pi_{1},\pi_{2},I,I^{c})$ a left-right noncrossing representation
of $\pi$. We note that if $(\pi_{1},\pi_{2},I,I^{c})$ is a left-right
noncrossing representation of $\pi$, so is $(\pi_{2},\pi_{1},I^{c},I)$. Let
$R(\pi)$ be the set of all left-right noncrossing representations of $\pi$. We
note that $|R(\pi)|$ corresponds to the number of ways the vertices of
$G(\pi)$ can be split into two independent families of vertices. Therefore, it
is clear that

$\displaystyle|R(\pi)|=2^{\operatorname{\bf cc}(\pi)}.$

Hence

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)}=\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}|R(\pi)|.$ (5)

Since $R(\pi)$ is the set of all left-right noncrossing representations of
$\pi$ and we sum over all $\pi\in P_{2}^{\operatorname{bi}}(2p)$, we get

$\displaystyle\sum_{\pi\in
P_{2}^{\operatorname{bi}}(2p)}|R(\pi)|=|\\{(\pi_{1},\pi_{2},I,I^{c}):I\subseteq[2p],\pi_{1}\in
NC_{2}(I),\pi_{2}\in NC_{2}(I^{c})\\}|.$

Note that for fixed $I\subseteq[2p]$, each $(\pi_{1},\pi_{2})\in
NC_{2}(I)\times NC_{2}(I^{c})$ will correspond to a unique partition $\pi\in
P_{2}^{\operatorname{bi}}(2p)$ such that $(\pi_{1},\pi_{2},I,I^{c})$ is a
left-right noncrossing representation of $\pi$. Hence we can first sum over
$I\subseteq[2p]$ so that

$\displaystyle\sum_{\pi\in
P_{2}^{\operatorname{bi}}(2p)}|R(\pi)|=\sum_{I\subseteq[2p]}|\\{(\pi_{1},\pi_{2}):\pi_{1}\in
NC_{2}(I),\pi_{2}\in NC_{2}(I^{c})\\}|.$

Therefore, by (5), we have

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)}=\sum_{I\subseteq[2p]}|\\{(\pi_{1},\pi_{2}):\pi_{1}\in
NC_{2}(I),\pi_{2}\in NC_{2}(I^{c})\\}|.$

The latter cardinality can be written as a product over the cardinals and only
depends on the length of $I$. More precisely, we have

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)}=\sum_{l=0}^{p}\binom{2p}{2l}C_{l}C_{p-l},$

where the cardinal $|NC_{2}(2l)|$ is equal to $C_{l}$. Hence

$\displaystyle\int x^{2p}\,\text{d}\mu_{1}=m_{p},$

and the first statement follows. For the second, we use the bijection $\Phi$
in (4) to write $\Phi(\pi)=\left(\hat{\pi},(\pi_{T})_{T\in\hat{\pi}}\right)$.
We note that

$\displaystyle\sum_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}1=\operatorname{\bf cc}(\pi)-\operatorname{\bf ncr}(\pi),$
$\displaystyle\sum_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}\left(\frac{|T|}{2}\right)=\operatorname{\bf cr}(\pi).$

Hence

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)-p}$ $\displaystyle=\sum_{\pi\in
P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf cc}(\pi)-\operatorname{\bf
ncr}(\pi)}\left(\frac{1}{2}\right)^{\operatorname{\bf cr}(\pi)}$
$\displaystyle=\sum_{\hat{\pi}\in
NC(2p)}\sum_{\begin{subarray}{c}T\in\hat{\pi}\\\ \pi_{T}\in
P_{2}^{\operatorname{bicon}}(T)\end{subarray}}\prod_{\begin{subarray}{c}T\in\hat{\pi}\\\
|T|\geq 4\end{subarray}}2\prod_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}\left(\frac{1}{2}\right)^{|T|/2},$

where we use that $\operatorname{\bf ncr}(\pi)+\operatorname{\bf cr}(\pi)=p$
in the first equality. We thus have

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)-p}=\sum_{\hat{\pi}\in
NC(2p)}\prod_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}\left\\{2\left(\frac{1}{2}\right)^{|T|/2}|P_{2}^{\operatorname{bicon}}(|T|)|\right\\}.$

Since the free cumulants $\kappa^{\operatorname{free}}_{n}(\mu_{1})$ are
uniquely characterized by the moments, it follows that for any odd integer
$n$, we have $\kappa^{\operatorname{free}}_{n}(\mu_{1})=0$,
$\kappa^{\operatorname{free}}_{2}(\mu_{1})=1$ and for any even integer $n\geq
4$, we have

$\displaystyle\kappa^{\operatorname{free}}_{n}(\mu_{1})=2\left(\frac{1}{2}\right)^{n/2}|P_{2}^{\operatorname{bicon}}(n)|.$

∎

######  Remark 3.2.

For any $p\geq 1$, it was shown in [6] that

$\displaystyle\sum_{l=0}^{p}\binom{2p}{2l}C_{l}C_{p-l}=C_{p}C_{p+1}.$

Therefore, the $2p$-th moment of $\mu_{1}$ is also characterized by
$2^{-p}C_{p}C_{p+1}$. The sequence $(C_{p}C_{p+1})_{p\geq 1}$ is A005568 in
Sloane’s encyclopedia (https://oeis.org/A005568), where several combinatorial
objects counted by it are shown.

We are ready to compute the moments and free cumulants of $\mu_{q}$.

######  Proposition 3.3.

Given $q\in[0,1]$, let

$\mu_{q}:=\sqrt{q}\left(\frac{1}{\sqrt{2}}\mu_{sc}+\frac{1}{\sqrt{2}}\mu_{sc}\right)\boxplus\sqrt{1-q}\,\mu_{sc}.$

Then the following hold.

  1. (1)

The odd free cumulants of $\mu_{q}$ vanish,
$\kappa^{\operatorname{free}}_{2}(\mu_{q})=1$ and for any even integer $n\geq
4$, we have

$\displaystyle\kappa^{\operatorname{free}}_{n}(\mu_{q})=2\Big{(}\frac{q}{2}\Big{)}^{n/2}|P_{2}^{\operatorname{bicon}}(n)|.$

  2. (2)

The odd moments of $\mu_{q}$ vanish and, for every $p\in\mathbb{N}$, its
$2p$-th moment is given by

$\displaystyle\sum_{\pi\in P_{2}^{\operatorname{bi}}(2p)}2^{\operatorname{\bf
cc}(\pi)-p}q^{\operatorname{\bf cr}(\pi)}.$

###### Proof.

We will use the notion of $R$-transform; see [11, Lecture 16]. For a random
variable $a\in\mathcal{A}$ with distribution $\mu$, let

$\displaystyle R_{a}(z)=R_{\mu}(z)=\sum_{n\geq
1}\kappa^{\operatorname{free}}_{n}(a)z^{n},$

be its $R$-transform, defined as a formal series. The $R$-transform of a
standard semi-circle law is given by

$\displaystyle R_{\mu_{sc}}(z)=z^{2},$

whereas Proposition 3.1 shows that the $R$-transform of $\mu_{1}$ is given by

$\displaystyle R_{\mu_{1}}(z)=z^{2}+2\sum_{n\geq
2}\left(\frac{1}{2}\right)^{n}|P_{2}^{\operatorname{bicon}}(2n)|z^{2n}.$

Since the free cumulants linearize the free convolution, we deduce that

$\displaystyle R_{\mu_{q}}(z)$
$\displaystyle=R_{\sqrt{q}\mu_{1}}(z)+R_{\sqrt{1-q}\,\mu_{sc}}(z)$
$\displaystyle=R_{\mu_{1}}\big{(}\sqrt{q}\,z\big{)}+R_{\mu_{sc}}\big{(}\sqrt{1-q}\,z\big{)}$
$\displaystyle=z^{2}+2\sum_{n\geq
2}\Big{(}\frac{q}{2}\Big{)}^{n}|P_{2}^{\operatorname{bicon}}(2n)|z^{2n}.$

This proves the first part of the proposition. To prove the second part, we
use the moment-cumulant formula to deduce that the odd moments vanish, while
the $2p$-th moment can be expressed as

$\displaystyle\sum_{\hat{\pi}\in
NC(2p)}\prod_{T\in\hat{\pi}}\kappa^{\operatorname{free}}_{|T|}(\mu_{q})$
$\displaystyle=\sum_{\hat{\pi}\in NC(2p)}\prod_{\underset{|T|\geq
4}{T\in\hat{\pi}}}\left\\{2\Big{(}\frac{q}{2}\Big{)}^{|T|/2}|P_{2}^{\operatorname{bicon}}(T)|\right\\}$
$\displaystyle=\sum_{\hat{\pi}\in
NC(2p)}\sum_{(\pi_{T})_{T\in\hat{\pi}}}\prod_{\underset{|T|\geq
4}{T\in\hat{\pi}}}\left\\{2\Big{(}\frac{q}{2}\Big{)}^{|T|/2}\right\\},$

where the second summation is over
$(\pi_{T})_{T\in\hat{\pi}}\in\operatorname{Proj}(\hat{\pi})$ such that
$\pi_{T}\in P_{2}^{\operatorname{bicon}}(T)$ for $T\in\hat{\pi}$. Now noting
that

$\displaystyle|\\{T\in\hat{\pi}:|T|\geq 4\\}|=\operatorname{\bf
cc}(\pi)-\operatorname{\bf ncr}(\pi),$

and

$\sum_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}\frac{|T|}{2}=\operatorname{\bf cr}(\pi),$

we finish the proof after using the bijection $\Phi$ from (4) to rewrite the
above expression. ∎

##  4\. Existence of the limit

The goal of this section is to show that the expression in (1) admits a limit
that depends only on the first and second moments of the variables at hand.
Let us first prove the following centering lemma.

######  Lemma 4.1.

Let $(\mathcal{A},\tau)$ be a unital faithful tracial noncommutative
probability space. Let
$\\{a_{k}:k\in[n]\\},\\{d_{k}:k\in[n]\\}\subset\mathcal{A}$ be two collections
of free random variables and let

$\displaystyle c_{k}:=a_{k}\otimes d_{k}-\tau\otimes\tau(a_{k}\otimes d_{k}),$

for every $k\in[n]$. Then for any $m\geq 1$ and $i_{1},\ldots,i_{m}\in[n]$
such that there exists an index $l^{*}\in[m]$ satisfying $i_{l^{*}}\neq i_{j}$
for all $j\neq l^{*}$, we have

$\displaystyle\tau\otimes\tau(c_{i_{1}}\cdots c_{i_{m}})=0.$

In particular, we have

$\displaystyle\tau\otimes\tau\left(c_{i_{1}}\cdots
c_{i_{l^{*}-1}}a_{i_{l^{*}}}\otimes d_{i_{l^{*}}}c_{i_{l^{*}+1}}\cdots
c_{i_{m}}\right)=\tau(a_{i_{l^{*}}})\tau(d_{i_{l^{*}}})\tau\otimes\tau(c_{i_{1}}\cdots
c_{i_{l^{*}-1}}c_{i_{l^{*}+1}}\cdots c_{i_{m}}).$

###### Proof.

By cyclicity of the trace, we can assume that $l^{*}=1$. We write

$\displaystyle\tau\otimes\tau(c_{i_{1}}\cdots
c_{i_{m}})=\tau\otimes\tau(a_{i_{1}}\otimes d_{i_{1}}c_{i_{2}}\cdots
c_{i_{m}})-\tau\otimes\tau(a_{i_{1}}\otimes
d_{i_{1}})\tau\otimes\tau(c_{i_{2}}\cdots c_{i_{m}}).$ (6)

For the first term, we open the expressions for $c_{i_{j}}$ for $j\geq 2$ to
get that

$\displaystyle\tau\otimes\tau(a_{i_{1}}\otimes d_{i_{1}}c_{i_{2}}\cdots
c_{i_{m}})$
$\displaystyle=\sum_{I\subseteq\\{2,\ldots,m\\}}(-1)^{|I|}\prod_{j\in
I}\left(\tau\otimes\tau(a_{i_{j}}\otimes
d_{i_{j}})\right)\tau\otimes\tau\left(a_{i_{1}}\otimes d_{i_{1}}\prod_{j\in
J^{c}}^{\to}a_{i_{j}}\otimes d_{i_{j}}\right),$

where $\overset{\to}{\prod}$ denotes the product respecting the ordering.
Since $a_{i_{1}},d_{i_{1}}$ are free from $a_{i_{j}},d_{i_{j}}$ for all $j\geq
2$, freeness implies that

$\displaystyle\tau\otimes\tau(a_{i_{1}}\otimes d_{i_{1}}c_{i_{2}}\cdots
c_{i_{m}})$ $\displaystyle=\tau\otimes\tau(a_{i_{1}}\otimes
d_{i_{1}})\sum_{I\subseteq\\{2,\ldots,m\\}}(-1)^{|I|}\prod_{j\in
I}\left(\tau\otimes\tau(a_{i_{j}}\otimes
d_{i_{j}})\right)\tau\otimes\tau\left(\prod_{j\in J^{c}}^{\to}a_{i_{j}}\otimes
d_{i_{j}}\right).$

The summation can be easily written as $\tau\otimes\tau(c_{i_{2}}\cdots
c_{i_{m}})$, hence

$\displaystyle\tau\otimes\tau(a_{i_{1}}\otimes d_{i_{1}}c_{i_{2}}\cdots
c_{i_{m}})=\tau\otimes\tau(a_{i_{1}}\otimes
d_{i_{1}})\tau\otimes\tau(c_{i_{2}}\cdots c_{i_{m}}).$

By (6), we then have

$\displaystyle\tau\otimes\tau(c_{i_{1}}\cdots c_{i_{m}})=0.$

Finally, since $\tau\otimes\tau(a_{i_{1}}\otimes
d_{i_{1}})=\tau(a_{i_{1}})\tau(d_{i_{1}})$, we get that

$\displaystyle\tau\otimes\tau(a_{i_{1}}\otimes d_{i_{1}}c_{i_{2}}\cdots
c_{i_{m}})$ $\displaystyle=\tau\otimes\tau(a_{i_{1}}\otimes
d_{i_{1}})\tau\otimes\tau(c_{i_{2}}\cdots c_{i_{m}})$
$\displaystyle=\tau(a_{i_{1}})\tau(d_{i_{1}})\tau\otimes\tau(c_{i_{2}}\cdots
c_{i_{m}}).$

∎

We are now ready to prove the existence of the limit and that it only depends
on the first and second moments of the $a_{k}$’s.

######  Proposition 4.2 (Existence).

Let $(a_{n})_{n\in\mathbb{N}}\in(\mathcal{A},\tau)$ be self-adjoint free i.i.d
random variables with mean $\lambda$, variance $\sigma^{2}$, and denote
$\delta^{2}:=\operatorname{var}(a_{1}\otimes
a_{1})=\sigma^{2}(\sigma^{2}+2\lambda^{2})$. For every $k\in\mathbb{N}$,
denote

$\displaystyle b_{k}:=\frac{1}{\delta}\big{(}a_{k}\otimes
a_{k}-\tau\otimes\tau(a_{k}\otimes a_{k})\big{)},$

and for every $n\in\mathbb{N}$, denote

$\displaystyle S_{n}:=\frac{1}{\sqrt{n}}\sum_{k\in[n]}b_{k}.$

Then there exists a random variable
${\bf{S}}\in(\mathcal{A}^{\prime},\tau^{\prime})$ such that
$S_{n}\Rightarrow{\bf{S}}$. Moreover, the law of ${\bf{S}}$ depends only on
$\lambda$ and $\sigma$, its odd moments vanish and, for every
$p\in\mathbb{N}$,

$\displaystyle\tau^{\prime}({\bf{S}}^{2p})=\sum_{\pi\in
P_{2}(2p)}\tau\otimes\tau(b_{i_{1}}\cdots b_{i_{2p}}),$

where $i\in[p]^{2p}$ is any sequence such that $i_{j}=i_{k}$ if and only
$\\{j,k\\}\in\pi$.

###### Proof.

We begin by writing

$\displaystyle\tau\otimes\tau(S_{n}^{p})=\frac{1}{n^{p/2}}\sum_{i\in[n]^{p}}\tau\otimes\tau(b_{i_{1}}\cdots
b_{i_{p}}).$

Since $b_{1},\ldots,b_{n}$ are identically distributed, the expression

$\displaystyle\tau\otimes\tau(b_{i_{1}}\cdots b_{i_{p}})$ (7)

depends only on the partition $\pi=\pi(i)\in P(p)$ given by $l\sim_{\pi}k$
(that is, $l,k$ belong to the same block of $\pi$) if and only if
$i_{l}=i_{k}$. Denote the common value of (7) by $\tau\otimes\tau(\pi)$. Then
we have

$\displaystyle\tau\otimes\tau(S_{n}^{p})=\frac{1}{n^{p/2}}\sum_{\pi\in
P(p)}\tau\otimes\tau(\pi)|\\{i\in[n]^{p}:\pi(i)=\pi\\}|.$

To count the cardinality, we choose an index for each block. Therefore, we
have

$\displaystyle|\\{i\in[n]^{p}:\pi(i)=\pi\\}|=n(n-1)\cdots(n-|\pi|+1)\sim
n^{|\pi|}.$

By Lemma 4.1, if $\pi$ has a block of size $1$, $\tau\otimes\tau(\pi)=0$.
Thus, we have

$\displaystyle\tau\otimes\tau(S_{n}^{p})=\sum_{\begin{subarray}{c}\pi\in
P(p)\\\ |V|\geq 2;\forall
V\in\pi\end{subarray}}\tau\otimes\tau(\pi)\frac{n(n-1)\cdots(n-|\pi|+1)}{n^{p/2}}.$

Since $|V|\geq 2$ for all blocks $V\in\pi$, we have $|\pi|\leq p/2$. If there
exists a block $V\in\pi$ such that $|V|\geq 3$, we immediately have
$|\pi|<p/2$, and its contribution is negligible. In particular, this implies
that the odd moments of $S_{n}$ are asymptotically vanishing. We deduce that

$\displaystyle\lim_{n\to\infty}\tau\otimes\tau(S_{n}^{p})=\lim_{n\to\infty}\sum_{\begin{subarray}{c}\pi\in
P(p)\\\ |V|=2;\forall
V\in\pi\end{subarray}}\tau\otimes\tau(\pi)\frac{n(n-1)\cdots(n-|\pi|+1)}{n^{p/2}}.$

In this case, $\pi$ is a pair partition and $|\pi|=p/2$, hence we deduce the
formula

$\displaystyle\lim_{n\to\infty}\tau\otimes\tau(S_{n}^{p})=\sum_{\pi\in
P_{2}(p)}\tau\otimes\tau(\pi),$

which shows that $S_{n}$ converges. To prove that the limit depends only on
$\lambda$ and $\sigma$, we write

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(b_{i_{1}}\cdots
b_{i_{p}})=\frac{1}{\delta^{p}}\sum_{I\subseteq[p]}(-1)^{|I|}\lambda^{2|I|}\tau^{2}\left(\prod_{l\in
I^{c}}^{\to}a_{i_{l}}\right),$

where $\pi(i)=\pi$. By the moment-cumulant formula, we have

$\displaystyle\tau\left(\prod_{l\in
I^{c}}^{\to}a_{i_{l}}\right)=\sum_{\sigma\in
NC(I^{c})}\kappa_{\sigma}((a_{i_{l}})_{l\in I^{c}}).$

Since the mixed cumulants of free variables vanish, the only partitions
$\sigma\in NC(I^{c})$ that contribute are those such that every block
$V\in\sigma$ has cardinality at most two. We then have

$\displaystyle\kappa_{\sigma}((a_{i_{l}})_{l\in I^{c}})$
$\displaystyle=\prod_{\begin{subarray}{c}V\in\sigma\\\
|V|=2\end{subarray}}\kappa_{|V|}(a,a)\prod_{\begin{subarray}{c}V\in\sigma\\\
|V|=1\end{subarray}}\kappa_{|V|}(a)=\sigma^{2\\#\\{V\in\sigma:|V|=2\\}}\lambda^{\\#\\{V\in\sigma:|V|=1\\}}.$

This concludes the proof. ∎

##  5\. Proof of Theorem 1.1

After proving the existence of the limit in the previous section, the goal
here is to identify this limit as stated in Theorem 1.1. In all this section,
$(a_{n})_{n\in\mathbb{N}}$ denote free copies of a random variable $a$ with
mean $\lambda$ and variance $\sigma^{2}$. Moreover, the common law of the
normalized tensors will be denoted by

$\displaystyle b=\frac{1}{\delta}\big{(}a\otimes a-\lambda^{2}\big{)},$ (8)

where $\delta^{2}=\operatorname{var}(a\otimes
a)=\sigma^{2}(\sigma^{2}+2\lambda^{2})$.

Throughout the proof, we will assume $p$ is an even integer. Following
Proposition 4.2, we denote ${\bf{S}}$ the limit of $S_{n}$ and note that

$\displaystyle\tau^{\prime}({\bf{S}}^{p})=\sum_{\pi\in
P_{2}(p)}\tau\otimes\tau(\pi),$

where $\tau\otimes\tau(\pi)=\tau\otimes\tau(b_{i_{1}}\cdots b_{i_{p}})$ with
$\pi(i)=\pi$.

###  5.1. Contribution of noncrossing blocks

We begin by removing interval blocks.

######  Lemma 5.1.

Let $\pi\in P_{2}(p)$ and suppose that there exists $l\in[p]$ such that
$\\{l,l+1\\}\in\pi$ (with the convention that $p+1:=1$). Then

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(\pi\setminus\\{l,l+1\\}).$

###### Proof.

By cyclicity, we can assume $l=p-1$. We then have

$\displaystyle\delta^{2}\tau\otimes\tau(\pi)$
$\displaystyle=\tau\otimes\tau\left(b_{i_{1}}\cdots
b_{i_{p-2}}\cdot\left(a_{i_{p}}^{2}\otimes a_{i_{p}}^{2}+\lambda^{4}1\otimes
1-2\lambda^{2}a_{i_{p}}\otimes a_{i_{p}}\right)\right)$
$\displaystyle=\tau\otimes\tau\left(b_{i_{1}}\cdots b_{i_{p-2}}\cdot
a_{i_{p}}^{2}\otimes
a_{i_{p}}^{2}\right)+\lambda^{4}\tau\otimes\tau\left(b_{i_{1}}\cdots
b_{i_{p-2}}\right)$
$\displaystyle\quad-2\lambda^{2}\tau\otimes\tau\left(b_{i_{1}}\cdots
b_{i_{p-2}}a_{i_{p}}\otimes a_{i_{p}}\right)$ $\displaystyle=:I+II+III.$

We immediately recognize

$\displaystyle II=\lambda^{4}\tau\otimes\tau(\pi\setminus\\{p-1,p\\}).$

By Lemma 4.1, we have

$\displaystyle I=\tau^{2}(a^{2})\tau\otimes\tau(\pi\setminus\\{p-1,p\\});$
$\displaystyle III=-2\lambda^{4}\tau\otimes\tau(\pi\setminus\\{p-1,p\\}).$

Hence

$\displaystyle\delta^{2}\tau\otimes\tau(\pi)=\left(\tau^{2}(a^{2})-\lambda^{4}\right)\tau\otimes\tau(\pi\setminus\\{p-1,p\\}).$

To conclude, we note that $\tau^{2}(a^{2})-\lambda^{4}=\delta^{2}$ and finish
the proof. ∎

Recall that a noncrossing pair partition $\pi\in NC_{2}(p)$ always has an
interval block $V=\\{l,l+1\\}\in\pi$ such that $\pi\setminus V$ is a
noncrossing pair partition. In particular, by induction, Lemma 5.1 implies the
following.

######  Corollary 5.2.

For any $\pi\in NC_{2}(p)$, we have $\tau\otimes\tau(\pi)=1$.

###  5.2. Decomposition of pair partitions

In order to capture the contribution of crossing partitions, we need to
decompose a partition $\pi\in P_{2}(p)\setminus NC_{2}(p)$ using smaller
partitions. We denote $\pi=\pi_{1}\oplus\cdots\oplus\pi_{k}$ if $[p]$ can be
decomposed into $k$ intervals $I_{1},\ldots,I_{k}$ such that $\pi_{k}\in
P_{2}(I_{k})$ and $V\in\pi$ if $V\in\pi_{l}$ for some $1\leq l\leq k$. For
$I\subseteq[p]$, let

$\displaystyle a_{I}:=\prod_{l\in I}^{\to}a_{i_{l}},$ (9)

and $a_{\varnothing}:=1$.

######  Lemma 5.3.

Let $\pi\in P_{2}(p)$. Then, the following hold.

  1. (5.3.i)

If $\pi=\pi_{1}\oplus\cdots\oplus\pi_{l}$, then

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(\pi_{1})\cdots\tau\otimes\tau(\pi_{l}).$

  2. (5.3.ii)

If $\pi=\\{1,p\\}\cup\pi_{1}$, where $\pi_{1}\in P_{2}(\\{2,\ldots,p-1\\})$,
then

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(\pi_{1}).$

Moreover, if there exists an interval $I\subseteq[p]$ such that $\pi|_{I}$ is
a pair partition, then

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(\pi|_{I})\tau\otimes\tau(\pi|_{I^{c}}).$

  3. (5.3.iii)

If $\pi$ has a block $V=\\{r,s\\}$ such that for any block $U=\\{l,k\\}\in\pi$
with $r<l<s$, we have $r<k<s$ (i.e., every point inside $V$ matches another
one inside $V$), we have

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(\pi|_{V_{-}})\tau\otimes\tau(\pi|_{V_{+}}),$

where $\pi|_{V_{-}}$ is the restriction of $\pi$ to inside of $V$ and
$\pi|_{V_{+}}$ is the restriction of $\pi$ to outside of $V$.

###### Proof.

(5.3.i) By induction, it suffices to prove the case
$\pi=\pi_{1}\oplus\pi_{2}$. Let $I_{1},I_{2}$ be the disjoint decomposition of
$[p]$ given by $\pi_{1}$ and $\pi_{2}$. Then, we can write

$\displaystyle\tau\otimes\tau(\pi)=\frac{1}{\delta^{p}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq
I_{2}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|}\lambda^{2(|J_{1}|+|J_{2}|)}\tau^{2}\left(\prod_{l\in
J_{1}^{c}}a_{i_{l}}\prod_{l\in J_{2}^{c}}a_{i_{l}}\right).$

Since $\pi$ is the direct sum of $\pi_{1},\pi_{2}$, the variables
$a_{J_{1}^{c}},a_{J_{2}^{c}}$ are free and we can write

$\displaystyle\tau\otimes\tau(\pi)=\frac{1}{\delta^{p}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq
I_{2}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|}\lambda^{2(|J_{1}|+|J_{2}|)}\tau^{2}\left(\prod_{l\in
J_{1}^{c}}a_{i_{l}}\right)\tau^{2}\left(\prod_{l\in
J_{2}^{c}}a_{i_{l}}\right).$

It is immediate to check that the right-hand-side is equal to
$\tau\otimes\tau(\pi_{1})\tau\otimes\tau(\pi_{2})$.

(5.3.ii) By cyclicity, we have

$\displaystyle\tau\otimes\tau(\pi)$
$\displaystyle=\tau\otimes\tau(b_{i_{2}}\cdots
b_{i_{p-1}}b_{i_{p}}b_{i_{1}}).$

Since $\\{1,p\\}\in\pi$, Lemma 5.1 implies that

$\displaystyle\tau\otimes\tau(\pi)=\tau\otimes\tau(b_{i_{2}}\cdots
b_{i_{p-1}})=\tau\otimes\tau(\pi_{1}).$

The second part follows again by cyclicity as we can assume
$I=\\{1,\ldots,k\\}$ for some $k\in[p]$, and the variables are free.

(5.3.iii) By cyclicity, we can assume that $V=\\{1,k\\}$ for some $k\in[p]$.
Note that $V$ creates a direct sum
$\pi=(V\cup\pi|_{V_{-}})\oplus\pi|_{V_{+}}$. The result follows by (5.3.i) and
(5.3.ii). ∎

Note that any block $V\in\pi$ that does not cross any other block of $\pi$ is
either an interval block or a block that satisfies (5.3.iii). In particular,
its removal does not affect the value of $\tau\otimes\tau(\pi)$. It is clear
then that $\tau\otimes\tau(\pi)$ is a multiplicative function [3] over the
connected components of $\pi$. Using the mapping $\Phi$ defined in (4), we can
write

$\displaystyle\tau\otimes\tau(\pi)=\prod_{T\in\hat{\pi}}\tau\otimes\tau(\pi_{T}),$
(10)

where $(\hat{\pi},(\pi_{T})_{T\in\hat{\pi}})=\Phi(\pi)$. In view of this, we
will now focus on the case where $\pi\in P_{2}^{\operatorname{con}}(p)$, for
$p\geq 4$, as the case $p=2$ corresponds to noncrossing blocks.

###  5.3. Contribution of connected partitions

Given $\pi\in P_{2}^{\operatorname{con}}(p)$, for an even integer $p\geq 4$,
we recall that $\pi\in P_{2}^{\operatorname{bicon}}(p)$ if its intersection
graph $G(\pi)$ is a bipartite connected graph.

The following is the main proposition of this subsection.

######  Proposition 5.4.

Let $p\geq 4$ be an even integer and $\pi\in P_{2}^{\operatorname{con}}(p)$.
Then the following hold.

  1. (1)

If $\pi\notin P_{2}^{\operatorname{bicon}}(p)$, then $\tau\otimes\tau(\pi)=0$
;

  2. (2)

If $\pi\in P_{2}^{\operatorname{bicon}}(p)$, then

$\displaystyle\tau\otimes\tau(\pi)=2\Big{(}\frac{q}{2}\Big{)}^{\frac{p}{2}},$

where $q=\frac{2\lambda^{2}}{\sigma^{2}+2\lambda^{2}}$.

To prove Proposition 5.4, our goal is to remove the blocks $V\in\pi$ one at a
time and track its influence on the rest of the partition $\pi\setminus V$. To
this end, we will define two main quantities associated with $V$. First, we
will define a coloring $w_{V}\in\\{0,1\\}$ of $V$ that encodes how the removal
of $V$ affects the other blocks. Secondly, we will define a binary vector
$\theta_{V}\in\\{0,1\\}^{4}$ satisfying

$\displaystyle\sum_{j\in[4]}\theta_{j}=1,$

that is, only one coordinate of $\theta_{V}$ is equal to $1$. It encodes how
we remove the block $V$ from $\pi$, that is, its weight to
$\tau\otimes\tau(\pi)$. Let us begin this description now.

Let $\pi\in P_{2}^{\operatorname{con}}(p)$ and $i\in[p]^{p}$ such that
$\pi(i)=\pi$. Given an integer $t\geq 0$, blocks
$V_{1},\ldots,V_{t},V_{t+1}\in\pi$, and a coloring $w\in\\{0,1\\}^{t+1}$, we
denote $V_{j}^{(w_{j})}$ the block $V_{j}$ under the color $w_{j}$. We then
define the iterated joint law of $b_{i_{j}}$

$\displaystyle
B_{k}:=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\left(b_{i_{j},k}\right)_{j\in[p]}$

for $k\leq t+1$ as follows. Let $(\tilde{a}_{j})_{j\in\mathbb{N}}$ be a family
of free independent copies of $a$, free from $(a_{j})_{j\in\mathbb{N}}$. For
$k=0$, we define $B_{0}=(b_{i_{j}})_{j\in[p]}$ and for any $1\leq k\leq t+1$,
we have

$\displaystyle B_{k}=\frac{1}{\delta}\left(a_{i_{j}}^{(l)}\otimes
a_{i_{j}}^{(r)}-\lambda^{2}\right)_{j\in[p]},$

where

$\displaystyle
a_{i_{j}}^{(l)},a_{i_{j}}^{(r)}\in\\{a_{i_{j}},\tilde{a}_{i_{j}}\\},$ (11)

for any $j\in[p]$. We then recursively define the choices in (11) as follows.
Given

$\displaystyle B_{k}=\frac{1}{\delta}\left(a_{i_{j}}^{(l)}\otimes
a_{i_{j}}^{(r)}-\lambda^{2}\right)_{j\in[p]},$

we consider the block $V_{k+1}=\\{s_{1}<s_{2}\\}$ and its color $w_{k+1}$, for
$1\leq k\leq t$. If $w_{k+1}=0$, for all blocks $V=\\{t_{1},t_{2}\\}\in\pi$
that cross $V_{k+1}$, we replace the left legs of
$(b_{i_{t_{1}},k},b_{i_{t_{2}},k})$ by a pair of free variables. Concretely,
we define

$\displaystyle
b_{i_{t_{1}},k+1}=\frac{1}{\delta}\left(\tilde{a}_{i_{t_{1}}}\otimes
a_{i_{t_{1}}}^{(r)}-\lambda^{2}\right);$ $\displaystyle
b_{i_{t_{2}},k+1}=\frac{1}{\delta}\left(a_{i_{t_{1}}}\otimes
a_{i_{t_{2}}}^{(r)}-\lambda^{2}\right).$

Otherwise, if $w_{k+1}=1$, we replace the right legs of
$(b_{i_{t_{1}},k},b_{i_{t_{2}},k})$ by a pair of free variables instead,

$\displaystyle
b_{i_{t_{1}},k+1}=\frac{1}{\delta}\left(a_{i_{t_{1}}}^{(l)}\otimes\tilde{a}_{i_{t_{1}}}-\lambda^{2}\right);$
$\displaystyle
b_{i_{t_{2}},k+1}=\frac{1}{\delta}\left(a_{i_{t_{2}}}^{(l)}\otimes
a_{i_{t_{1}}}-\lambda^{2}\right).$

All the other blocks remain unchanged. Recall that for a vector $B$ and a set
$I\subseteq[p]$, we set as in (9)

$\displaystyle B_{I}=\prod_{l\in I}^{\to}B_{i_{l}},$

and $B_{\varnothing}=1$. We then define

$\displaystyle\tau\otimes\tau\left(\pi,V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right):=\tau\otimes\tau\left(\left(B_{k}\right)_{[p]}\right)$
$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{U_{1},\ldots,U_{m}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right):=\tau\otimes\tau\left(\left(B_{k}\right)_{[p]\setminus\left(U_{1}\cup\cdots\cup
U_{m}\right)}\right),$

where $U_{1},\ldots,U_{m},V_{1},\ldots,V_{k}\in\pi$ and

$\displaystyle B_{k}=B_{k}\left(V_{1}^{(w_{1})},\dots,V_{k}^{(w_{k})}\right).$

Note first that at each step $k$, the distribution of
$(b_{i_{v_{1}},k},b_{i_{v_{2}},k})$ for a block $V=\\{v_{1}<v_{2}\\}\in\pi$
can be written as one of the following four cases.

  1. (1)

The blocks $V_{1},\ldots,V_{k}$ do not cross $V$, and we have

$\displaystyle b_{i_{v_{1}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right);$ $\displaystyle
b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right),$

where we recall that $i_{v_{1}}=i_{v_{2}}$.

  2. (2)

All blocks $V_{j}$ that cross $V$ have the same color $w_{j}=0$, and we have

$\displaystyle
b_{i_{v_{1}},k}=\frac{1}{\delta}\left(\tilde{a}_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right);$ $\displaystyle
b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right).$

  3. (3)

All blocks $V_{j}$ that cross $V$ have the same color $w_{j}=1$, and we have

$\displaystyle
b_{i_{v_{1}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes\tilde{a}_{i_{v_{1}}}-\lambda^{2}\right);$
$\displaystyle b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right).$

  4. (4)

Otherwise, there exist two blocks $V_{j_{1}},V_{j_{2}}$ that cross $V$ such
that $w_{j_{1}}\neq w_{j_{2}}$, that is, they have different colors, and then

$\displaystyle
b_{i_{v_{1}},k}=\frac{1}{\delta}\left(\tilde{a}_{i_{v_{1}}}\otimes\tilde{a}_{i_{v_{1}}}-\lambda^{2}\right);$
$\displaystyle b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right).$

We remark that even if the choices of replacements do not always happen for
$b_{i_{v_{1}}}$, the joint distribution of $(b_{i_{v_{1}},k},b_{i_{v_{2}},k})$
can always be written as one of those four cases since the variables are free
i.i.d. For instance, we have

$\displaystyle(a\otimes
a,a\otimes\tilde{a})\overset{d}{=}(a\otimes\tilde{a},a\otimes
a)\overset{d}{=}(\tilde{a}\otimes a,\tilde{a}\otimes\tilde{a}),$

and so on. Let

$\displaystyle\theta\in\\{0,1\\}^{4},$
$\displaystyle\sum_{j\in[4]}\theta_{j}=1,$

and denote

$\displaystyle
b_{i_{v_{1}}}(\theta)=\frac{1}{\delta}\left(\theta_{1}a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}+\theta_{2}\tilde{a}_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}+\theta_{3}a_{i_{v_{1}}}\otimes\tilde{a}_{i_{v_{1}}}+\theta_{4}\tilde{a}_{i_{v_{1}}}\otimes\tilde{a}_{i_{v_{1}}}-\lambda^{2}\right);$
$\displaystyle
b_{i_{v_{2}}}(\theta)=\frac{1}{\delta}\left(a_{i_{v_{1}}}\otimes
a_{i_{v_{1}}}-\lambda^{2}\right).$

The choice of
$\theta:=\theta\left(V,V_{1}^{(w_{1})}\ldots,V_{k}^{(w_{k})}\right)$
corresponds to the case in which we are, that is, we define $\theta$ as the
binary vector such that

$\displaystyle\left(b_{i_{v_{1}}}(\theta),b_{i_{v_{2}}}(\theta)\right)\overset{d}{=}\left(b_{i_{v_{1}},k},b_{i_{v_{2}},k}\right).$

Let $V=\\{v_{1}<v_{2}\\}\in\pi\setminus\\{V_{1},\ldots,V_{k}\\}$ and

$\displaystyle
I_{1}:=\\{1,\ldots,v_{1}-1\\}\setminus\left(\bigcup_{j\in[k]}V_{j}\right);$
$\displaystyle
I_{2}:=\\{v_{1}+1,\ldots,v_{2}-1\\}\setminus\left(\bigcup_{j\in[k]}V_{j}\right);$
$\displaystyle
I_{3}:=\\{v_{2}+1,\ldots,p\\}\setminus\left(\bigcup_{j\in[k]}V_{j}\right).$

Our first goal is to obtain the contribution and coloring of $V$ to

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\tau\otimes\tau\left(\left(B_{k}\right)_{I_{1}}b_{i_{v_{1}},k}\left(B_{k}\right)_{I_{2}}b_{i_{v_{2}},k}\left(B_{k}\right)_{I_{3}}\right),$

according to the value of
$\theta=\theta\left(V,V_{1}^{(w_{1})}\ldots,V_{k}^{(w_{k})}\right)\in\\{0,1\\}^{4}$,
where

$\displaystyle
B_{k}=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$

We recall that

$\displaystyle\tau\otimes\tau\left(\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{2}}\left(B_{k}\right)_{I_{3}}\right)$
$\displaystyle=\tau\otimes\tau\left(\left(B_{k}\right)_{[p]\setminus\left(V_{1}\cup\cdots\cup
V_{k}\cup V\right)}\right)$
$\displaystyle=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$

To simplify the notation, we denote

$\displaystyle h:=|I_{1}|+|I_{2}|+|I_{3}|=p-2(k+1).$ (12)

We divide the four cases into the next four lemmas.

######  Lemma 5.5 (Case $\theta_{1}=1$).

Let $p\geq 4$ be an even integer, $\pi\in P_{2}^{\operatorname{con}}(p)$. Let
$V_{1},\ldots,V_{k}\in\pi$ be blocks, $w\in\\{0,1\\}^{k}$ be a coloring of
$V_{1},\ldots,V_{k}$ and

$\displaystyle
B_{k}=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\left(b_{i_{j},k}\right)_{j\in[p]}.$

Let $V=\\{v_{1}<v_{2}\\}\in\pi\setminus\\{V_{1},\ldots,V_{k}\\}$ and the joint
law

$\displaystyle\left(b_{i_{v_{1}},k},b_{i_{v_{2}},k}\right)\overset{d}{=}\left(b_{i_{v_{1}}}(\theta),b_{i_{v_{2}}}(\theta)\right).$

If $\theta_{1}=1$, we have

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
$\displaystyle=\frac{q}{2}\sum_{w=0,1}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(w)}\right).$

###### Proof.

To simplify the notation, let

$\displaystyle
T:=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$

Then, we must prove that

$\displaystyle
T=\frac{q}{2}\sum_{w=0,1}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(w)}\right).$

As $\theta_{1}=1$, we can write

$\displaystyle b_{i_{v_{1}},k}=\frac{1}{\delta}\left(a\otimes
a-\lambda^{2}\right);$ $\displaystyle
b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a\otimes a-\lambda^{2}\right),$

where $a:=a_{i_{v_{1}}}$. We get by Lemma 4.1 that

$\displaystyle T$
$\displaystyle=\frac{1}{\delta^{2}}\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}a\otimes
a\left(B_{k}\right)_{I_{2}}a\otimes a\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle\quad-\frac{\lambda^{4}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$
(13)

Let us write

$\displaystyle B_{k}=\frac{1}{\delta}\left(a_{i_{j}}^{(l)}\otimes
a_{i_{j}}^{(r)}-\lambda^{2}\right)_{j\in[p]}.$

Then

$\displaystyle\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}a\otimes
a\left(B_{k}\right)_{I_{2}}a\otimes a\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle=\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}\tau\left(a_{J_{1}^{c}}^{(l)}\,a\,a_{J_{2}^{c}}^{(l)}\,a\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a\,a_{J_{2}^{c}}^{(r)}\,a\,a_{J_{3}^{c}}^{(r)}\right),$

where $h$ is defined as in (12). We use Lemma 2.1 and compute

$\displaystyle R(J_{1},J_{2},J_{3})$
$\displaystyle:=\tau\left(a_{J_{1}^{c}}^{(l)}\,a\,a_{J_{2}^{c}}^{(l)}\,a\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a\,a_{J_{2}^{c}}^{(r)}\,a\,a_{J_{3}^{c}}^{(r)}\right)$
$\displaystyle=\left(\sigma^{2}\tau(a_{J_{1}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)})\tau(a_{J_{2}^{c}}^{(l)})+\lambda^{2}\tau(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)})\right)\left(\sigma^{2}\tau(a_{J_{1}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)})\tau(a_{J_{2}^{c}}^{(r)})+\lambda^{2}\tau(a_{J_{1}^{c}}^{(r)}\,a_{J_{2}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)})\right).$

After the expansion, we have

$\displaystyle
R(J_{1},J_{2},J_{3})=:\sigma^{4}R_{1}(J_{1},J_{2},J_{3})+\sigma^{2}\lambda^{2}\left(R_{2}(J_{1},J_{2},J_{3})+R_{3}(J_{1},J_{2},J_{3})\right)+\lambda^{4}R_{4}(J_{1},J_{2},J_{3}),$

where we choose $R_{2}$ to be the term with factor
$\tau(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)})$. For
$R_{1}$, we have

$\displaystyle\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}R_{1}(J_{1},J_{2},J_{3})$
$\displaystyle=\tau\otimes\tau\left(\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{3}}\right)\tau\otimes\tau\left(\left(B_{k}\right)_{I_{2}}\right).$

Since $\pi$ is connected, $V$ crosses at least one block
$U=\\{j,j^{\prime}\\}\in\pi$. Therefore, we assume $j\in I_{2}$ whose matching
symbol $j^{\prime}\in I_{1}\cup I_{3}$. By Lemma 4.1, we have

$\displaystyle\tau\otimes\tau\left(\left(B_{k}\right)_{I_{2}}\right)=0,$

and the contribution of $R_{1}$ is zero. For $R_{4}$, we have

$\displaystyle\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}R_{4}(J_{1},J_{2},J_{3})$
$\displaystyle=\tau\otimes\tau\left(\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{2}}\left(B_{k}\right)_{I_{3}}\right).$

Since

$\displaystyle\tau\otimes\tau\left(\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{2}}\left(B_{k}\right)_{I_{3}}\right)=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right),$

its contribution cancels out with
$\frac{\lambda^{4}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
in (5.3). For $R_{2}$, let us first denote

$\displaystyle
B_{k+1}=B_{k+1}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right)=:\frac{1}{\delta}\left(\overline{a}_{i_{j}}^{(l)}\otimes\overline{a}_{i_{j}}^{(r)}-\lambda^{2}\right)_{j\in[p]}.$

We then have

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right)$
$\displaystyle=\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}\tau\left(\overline{a}_{J_{1}^{c}}^{(l)}\,\overline{a}_{J_{2}^{c}}^{(l)}\,\overline{a}_{J_{3}^{c}}^{(l)}\right)\tau\left(\overline{a}_{J_{1}^{c}}^{(r)}\,\overline{a}_{J_{2}^{c}}^{(r)}\,\overline{a}_{J_{3}^{c}}^{(r)}\right).$

By definition, the color $w_{k+1}=1$ does not affect the left legs, hence

$\displaystyle\overline{a}_{i_{j}}^{(l)}=a_{i_{j}}^{(l)}.$

For the right legs and a block $U=\\{t_{1}<t_{2}\\}$, let us divide it into
two cases.

  1. (1)

If $U$ does not cross $V$, we have

$\displaystyle\overline{a}_{i_{t_{1}}}^{(r)}=a_{i_{t_{1}}}^{(r)};$
$\displaystyle\overline{a}_{i_{t_{2}}}^{(r)}=a_{i_{t_{2}}}^{(r)}.$

  2. (2)

Otherwise, either $t_{1}\in I_{2}$ and $t_{2}\in I_{1}\cup I_{3}$ or $t_{1}\in
I_{1}\cup I_{3}$ and $t_{2}\in I_{2}$. By the definition of the color
$w_{k+1}=1$, we have

$\displaystyle\overline{a}_{i_{t_{1}}}^{(r)}=\tilde{a}_{i_{t_{1}}};$
$\displaystyle\overline{a}_{i_{t_{2}}}^{(r)}=a_{i_{t_{1}}}.$

Since $(a_{k})_{k\in\mathbb{N}}$ and $(\tilde{a}_{k})_{k\in\mathbb{N}}$ are
free i.i.d copies of $a$, the following holds. First, let $t\in I_{2}$.
Consider $t^{\prime}$ its matching symbol, that is, $\\{t,t^{\prime}\\}\in\pi$
and $i_{t}=i_{t^{\prime}}\neq i_{j}$, for all $j\notin\\{t,t^{\prime}\\}$.
Recall that, by the definition in (11), we have

$\displaystyle\overline{a}_{i_{t}}^{(r)},\overline{a}_{i_{t^{\prime}}}^{(r)}\in\\{a_{i_{t}},\tilde{a}_{i_{t}}\\}.$

If $t^{\prime}\in I_{2}$, Case (1) implies that the law of
$(\overline{a}_{i_{t}}^{(r)},\overline{a}_{i_{t^{\prime}}}^{(r)})$ is equal to
the law of $({a}_{i_{t}}^{(r)},{a}_{i_{t^{\prime}}}^{(r)})$. If $t^{\prime}\in
I_{1}\cup I_{3}$, as $a_{i_{t}},\tilde{a}_{i_{t}}$ are free i.i.d and free
from $(a_{k})_{k\neq i_{t}},(\tilde{a}_{k})_{k\neq i_{t}}$, we have

$\displaystyle\left((\overline{a}_{i_{j}}^{(r)})_{j\in
I_{2}\setminus\\{t\\}},\overline{a}_{i_{t}}^{(r)}\right)\overset{d}{=}\left((\overline{a}^{(r)}_{i_{j}})_{j\in
I_{2}\setminus\\{t\\}},a_{i_{t}}\right)\overset{d}{=}\left((\overline{a}^{(r)}_{i_{j}})_{j\in
I_{2}\setminus\\{t\\}},a_{i_{t}}^{(r)}\right).$ (14)

Applying the above equalities in distribution for all $t\in I_{2}$, namely,
Case (1) for $t^{\prime}\in I_{2}$ and (14) for $t^{\prime}\notin I_{2}$, we
get

$\displaystyle\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in
I_{2}}\overset{d}{=}\left(a_{i_{j}}^{(r)}\right)_{j\in I_{2}}.$

By symmetry, we also have

$\displaystyle\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup
I_{3}}\overset{d}{=}\left(a_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup I_{3}}.$

Additionally, $\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{2}}$ is now
free from $\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup I_{3}}$.
Indeed, let us show that each variable $\overline{a}_{i_{t}}^{(r)}$ is free
from $\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup I_{3}}$, for all
$t\in I_{2}$. Consider $t^{\prime}$ its matching symbol. First, if
$t^{\prime}\in I_{2}$, the fact that $(a_{k})_{k\in\mathbb{N}}$ and
$(\tilde{a}_{k})_{k\in\mathbb{N}}$ are free i.i.d collections implies that
$\overline{a}_{i_{t}}^{(r)}$ (and $\overline{a}_{i_{t^{\prime}}}^{(r)}$) is
free from $\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup I_{3}}$. If
$t^{\prime}\in I_{1}\cup I_{3}$, Case (2) implies that
$\overline{a}_{i_{t}}^{(r)}$ is free from
$\overline{a}_{i_{t^{\prime}}}^{(r)}$ and then again
$\overline{a}_{i_{t}}^{(r)}$ is free from
$\left(\overline{a}_{i_{j}}^{(r)}\right)_{j\in I_{1}\cup I_{3}}$.

In summary, by freeness, we have

$\displaystyle\tau\left(\overline{a}_{J_{1}^{c}}^{(l)}\,\overline{a}_{J_{2}^{c}}^{(l)}\,\overline{a}_{J_{3}^{c}}^{(l)}\right)\tau\left(\overline{a}_{J_{1}^{c}}^{(r)}\,\overline{a}_{J_{2}^{c}}^{(r)}\,\overline{a}_{J_{3}^{c}}^{(r)}\right)$
$\displaystyle=\tau\left(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)}\right)\tau\left(a_{J_{2}^{c}}^{(r)}\right)$
$\displaystyle=R_{2}(J_{1},J_{2},J_{3}).$

Therefore

$\displaystyle\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}R_{2}(J_{1},J_{2},J_{3})$
$\displaystyle=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right),$

Following the same reasoning, it follows that

$\displaystyle\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}R_{3}(J_{1},J_{2},J_{3})$
$\displaystyle=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(0)}\right).$

Combining the above relations in $T$, we get

$\displaystyle
T=\frac{\sigma^{2}\lambda^{2}}{\delta^{2}}\sum_{w=0,1}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(w)}\right).$

It remains to note that

$\displaystyle\frac{\sigma^{2}\lambda^{2}}{\delta^{2}}=\frac{q}{2},$

to finish the proof. ∎

######  Lemma 5.6 (Case $\theta_{2}=1$).

Let $p\geq 4$ be an even integer, $\pi\in P_{2}^{\operatorname{con}}(p)$. Let
$V_{1},\ldots,V_{k}\in\pi$ be blocks, $w\in\\{0,1\\}^{k}$ be a coloring of
$V_{1},\ldots,V_{k}$ and

$\displaystyle
B_{k}=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\left(b_{i_{j},k}\right)_{j\in[p]}.$

Let $V=\\{v_{1}<v_{2}\\}\in\pi\setminus\\{V_{1},\ldots,V_{k}\\}$ and the joint
law

$\displaystyle\left(b_{i_{v_{1}},k},b_{i_{v_{2}},k}\right)\overset{d}{=}\left(b_{i_{v_{1}}}(\theta),b_{i_{v_{2}}}(\theta)\right).$

If $\theta_{2}=1$, we have

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
$\displaystyle=\frac{q}{2}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right).$

###### Proof.

We must prove that

$\displaystyle T$
$\displaystyle:=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
$\displaystyle=\frac{q}{2}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right).$

As $\theta_{2}=1$, we can write

$\displaystyle b_{i_{v_{1}},k}=\frac{1}{\delta}\left(\tilde{a}\otimes
a-\lambda^{2}\right);$ $\displaystyle
b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a\otimes a-\lambda^{2}\right),$

where $a:=a_{i_{v_{1}}}$. We have

$\displaystyle T$
$\displaystyle=\frac{1}{\delta^{2}}\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\tilde{a}\otimes
a\left(B_{k}\right)_{I_{2}}a\otimes a\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle\quad-\frac{\lambda^{4}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$
(15)

We write again

$\displaystyle B_{k}=\frac{1}{\delta}\left(a_{i_{j}}^{(l)}\otimes
a_{i_{j}}^{(r)}-\lambda^{2}\right)_{j\in[p]}.$

Then

$\displaystyle\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\tilde{a}\otimes
a\left(B_{k}\right)_{I_{2}}a\otimes a\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle=\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}\tau\left(a_{J_{1}^{c}}^{(l)}\,\tilde{a}\,a_{J_{2}^{c}}^{(l)}\,a\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a\,a_{J_{2}^{c}}^{(r)}\,a\,a_{J_{3}^{c}}^{(r)}\right),$

where $h$ is defined in (12). We use Lemmas 2.1 and 4.1 to compute

$\displaystyle R(J_{1},J_{2},J_{3})$
$\displaystyle:=\tau\left(a_{J_{1}^{c}}^{(l)}\,\tilde{a}\,a_{J_{2}^{c}}^{(l)}\,a\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a\,a_{J_{2}^{c}}^{(r)}\,a\,a_{J_{3}^{c}}^{(r)}\right)$
$\displaystyle=\lambda^{2}\tau\left(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)}\right)\left(\sigma^{2}\tau(a_{J_{1}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)})\tau(a_{J_{2}^{c}}^{(r)})+\lambda^{2}\tau(a_{J_{1}^{c}}^{(r)}\,a_{J_{2}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)})\right)$

Hence

$\displaystyle R(J_{1},J_{2},J_{3})$
$\displaystyle=\lambda^{2}\sigma^{2}\tau\left(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)}\right)\tau\left(a_{J_{2}^{c}}^{(r)}\right)$
$\displaystyle\quad+\lambda^{4}\tau\left(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a_{J_{2}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)}\right).$

Both terms already appeared in the proof of Lemma 5.5. The second term cancels
out with
$\frac{\lambda^{4}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
in (5.3), and the contribution of the first is equal to

$\displaystyle\frac{1}{\delta^{h}}\sum_{\begin{subarray}{c}J_{1}\subseteq
I_{1}\\\ J_{2}\subseteq I_{2}\\\ J_{3}\subseteq
I_{3}\end{subarray}}(-1)^{|J_{1}|+|J_{2}|+|J_{3}|}\lambda^{2(|J_{1}|+|J_{2}|+|J_{3}|)}\tau\left(a_{J_{1}^{c}}^{(l)}\,a_{J_{2}^{c}}^{(l)}\,a_{J_{3}^{c}}^{(l)}\right)\tau\left(a_{J_{1}^{c}}^{(r)}\,a_{J_{3}^{c}}^{(r)}\right)\tau\left(a_{J_{2}^{c}}^{(r)}\right)$
$\displaystyle=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right).$

We then have

$\displaystyle
T=\frac{\sigma^{2}\lambda^{2}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(1)}\right),$

and the result follows by the definition of $q$. ∎

The case $\theta_{3}=1$ is symmetric to the case $\theta_{2}=1$ in Lemma 5.6,
and we omit the proof.

######  Lemma 5.7 (Case $\theta_{3}=1$).

Let $p\geq 4$ be an even integer, $\pi\in P_{2}^{\operatorname{con}}(p)$. Let
$V_{1},\ldots,V_{k}\in\pi$ be blocks, $w\in\\{0,1\\}^{k}$ be a coloring of
$V_{1},\ldots,V_{k}$ and

$\displaystyle
B_{k}=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\left(b_{i_{j},k}\right)_{j\in[p]}.$

Let $V=\\{v_{1}<v_{2}\\}\in\pi\setminus\\{V_{1},\ldots,V_{k}\\}$ and the joint
law

$\displaystyle\left(b_{i_{v_{1}},k},b_{i_{v_{2}},k}\right)\overset{d}{=}\left(b_{i_{v_{1}}}(\theta),b_{i_{v_{2}}}(\theta)\right).$

If $\theta_{3}=1$, we have

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
$\displaystyle=\frac{q}{2}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})},V^{(0)}\right).$

Finally, the case $\theta_{4}=1$ has a null contribution.

######  Lemma 5.8 (Case $\theta_{4}=1$).

Let $p\geq 4$ be an even integer, $\pi\in P_{2}^{\operatorname{con}}(p)$. Let
$V_{1},\ldots,V_{k}\in\pi$ be blocks, $w\in\\{0,1\\}^{k}$ be a coloring of
$V_{1},\ldots,V_{k}$ and

$\displaystyle
B_{k}=B_{k}\left(V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=\left(b_{i_{j},k}\right)_{j\in[p]}.$

Let $V=\\{v_{1}<v_{2}\\}\in\pi\setminus\\{V_{1},\ldots,V_{k}\\}$ and the joint
law

$\displaystyle\left(b_{i_{v_{1}},k},b_{i_{v_{2}},k}\right)\overset{d}{=}\left(b_{i_{v_{1}}}(\theta),b_{i_{v_{2}}}(\theta)\right).$

If $\theta_{4}=1$, we have

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=0.$

###### Proof.

Similarly to the previous proofs, we must show that

$\displaystyle
T:=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k}\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)=0.$

As $\theta_{4}=1$, we can write

$\displaystyle
b_{i_{v_{1}},k}=\frac{1}{\delta}\left(\tilde{a}\otimes\tilde{a}-\lambda^{2}\right);$
$\displaystyle b_{i_{v_{2}},k}=\frac{1}{\delta}\left(a\otimes
a-\lambda^{2}\right),$

where $a:=a_{i_{v_{1}}}$. First note that

$\displaystyle T$
$\displaystyle=\frac{1}{\delta^{2}}\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\tilde{a}\otimes\tilde{a}\left(B_{k}\right)_{I_{2}}a\otimes
a\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle\quad-\frac{\lambda^{4}}{\delta^{2}}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$

Applying Lemma 4.1 to both tensors, we get that

$\displaystyle\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\tilde{a}\otimes\tilde{a}\left(B_{k}\right)_{I_{2}}a\otimes
a\left(B_{k}\right)_{I_{3}}\right\\}=\lambda^{4}\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{2}}\left(B_{k}\right)_{I_{3}}\right\\}.$

The result follows by recalling that

$\displaystyle\tau\otimes\tau\left\\{\left(B_{k}\right)_{I_{1}}\left(B_{k}\right)_{I_{2}}\left(B_{k}\right)_{I_{3}}\right\\}$
$\displaystyle=\tau\otimes\tau\left(\left(B_{k}\right)_{[p]\setminus\left(V_{1}\cup\cdots\cup
V_{k}\cup V\right)}\right)$
$\displaystyle=\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{k},V\\},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right).$

∎

###### Proof of Proposition 5.4.

Let $\pi\notin P_{2}^{\operatorname{bicon}}(p)$, and let $2l^{*}+1$ be the
length of the smallest odd cycle in the intersection graph of $\pi$. We denote
$V_{1},\ldots,V_{2l^{*}+1}\in\pi$ (any) ordered sequence of blocks in this
smallest odd cycle, where $V_{j}$ crosses $V_{j+1}$ for each
$j=1,\ldots,2l^{*}+1$ and $V_{2l^{*}+2}:=V_{1}$. We first remove the first
block $V_{1}$, and by Lemma 5.5 we get that

$\displaystyle\tau\otimes\tau(\pi)=\frac{q}{2}\sum_{w=0,1}\tau\otimes\tau\left(\pi\setminus
V_{1},V_{1}^{(w)}\right).$

However, as it is the first block we remove, the distribution of
$B_{0}=(b_{i_{j}})_{j\in[p]}$ is symmetric in both legs. In particular, we
have

$\displaystyle B_{1}(V_{1}^{(0)})\overset{d}{=}B_{1}(V_{1}^{(1)}),$

and

$\displaystyle\tau\otimes\tau\left(\pi\setminus
V_{1},V_{1}^{(0)}\right)=\tau\otimes\tau\left(\pi\setminus
V_{1},V_{1}^{(1)}\right).$

Hence

$\displaystyle\tau\otimes\tau(\pi)=q\tau\otimes\tau\left(\pi\setminus
V_{1},V_{1}^{(0)}\right).$

We then fix the color of $V_{1}$ to be $w_{1}=0$. We aim to show by induction
that the colors $w_{1},w_{2},\ldots,w_{2l^{*}}$ of $V_{1},\ldots,V_{2l^{*}}$
are deterministic and alternating,

$\displaystyle w=(w_{1},\ldots,w_{2l^{*}})=(0,1,0,\ldots,1),$

and so is
$\theta^{(k+1)}:=\theta\left(V_{k},V_{1}^{(w_{1})},\ldots,V_{k}^{(w_{k})}\right)$
for $1\leq k\leq 2l^{*}$ in the sense that

$\displaystyle\left(\theta^{(2)}_{2},\theta^{(3)}_{3},\theta^{(4)}_{2},\ldots,\theta^{(2l^{*})}_{2}\right)=(1,1,1,\ldots,1).$

First, as $V_{2}$ crosses $V_{1}$ and $w_{1}=0$, we have $\theta^{(2)}_{2}=1$.
By Lemma 5.6, we have $w_{2}=1$. Suppose then that the result holds for some
$1<k<2l^{*}$. Let us prove that it also holds for $k+1$. To simplify the
reading, let us divide into two cases whether $k$ is even or odd.

Case $k$ is even. The induction hypothesis implies that

$\displaystyle(w_{2},w_{3},\ldots,w_{k})=(1,0,\ldots,1);$
$\displaystyle\left(\theta^{(2)}_{2},\theta^{(3)}_{3},\theta^{(4)}_{2},\ldots,\theta^{(k)}_{2}\right)=(1,1,1,\ldots,1).$

Since $V_{k+1}$ crosses $V_{k}$, we automatically have that
$\theta^{(k+1)}_{1}=0$. Moreover, as $w_{k}=1$, the definition of $\theta$
implies that

$\displaystyle\theta^{(k+1)}_{2}=0.$

Hence either $\theta^{(k+1)}_{4}=1$ or $\theta^{(k+1)}_{3}=1$. Now consider
all blocks $V_{j_{t}}$ that might potentially cross $V_{k+1}$, for $j_{t}\leq
k$. Since $2l^{*}+1$ is the length of the smallest odd cycle, the cycles
$(V_{j_{t}},\ldots,V_{k},V_{k+1})$ are of even length. By a parity check,
$j_{t}$ must be even, and hence

$\displaystyle w_{j_{t}}=1.$

This implies that all crossing blocks $V_{j_{t}}$ of $V_{k+1}$ for $j_{t}\leq
k$ have the same color $w_{j_{t}}=1$. In particular, we have

$\displaystyle\theta^{(k+1)}_{3}=1.$

We then apply Lemma 5.7 to get that $w_{k+1}=0$.

Case $k$ is odd. This follows similarly to the even case. We first have that
$\theta^{(k+1)}_{1}=0$ and as $w_{k}=0$, we have

$\displaystyle\theta^{(k+1)}_{3}=0.$

By the parity check on the even cycles $V_{k+1}$ might belong to, all crossing
blocks $V_{j_{t}}$ of $V_{k+1}$ for $j_{t}\leq k$ have the same color
$w_{j_{t}}=0$. Hence

$\displaystyle\theta^{(k+1)}_{2}=1,$

and Lemma 5.6 implies that $w_{k+1}=1$. This finishes the induction.

To conclude the proof, the block $V_{2l^{*}+1}$ crosses both $V_{1}$ of color
$w_{1}=0$ and $V_{2l^{*}}$ of color $w_{2l^{*}}=1$. Hence

$\displaystyle\theta^{(2l^{*}+1)}_{4}=1.$

Lemma 5.8 applied to $V=V_{2l^{*}+1}$ implies that

$\displaystyle\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{2l^{*}}\\},V_{1}^{(0)},\ldots,V_{2l^{*}}^{(1)}\right)=0.$

Applying Lemma 5.6 for blocks $V=V_{k}$ when $k$ is even and Lemma 5.7 for
$V=V_{k}$ when $k$ is odd, we get that

$\displaystyle\tau\otimes\tau(\pi)$
$\displaystyle=q\tau\otimes\tau\left(\pi\setminus V_{1},V_{1}^{(0)}\right)$
$\displaystyle=q\cdot\frac{q}{2}\tau\otimes\tau\left(\pi\setminus\\{V_{1},V_{2}\\},V_{1}^{(0)},V_{2}^{(1)}\right)$
$\displaystyle=\cdots$
$\displaystyle=q\left(\frac{q}{2}\right)^{2l^{*}-1}\tau\otimes\tau\left(\pi\setminus\\{V_{1},\ldots,V_{2l^{*}}\\},V_{1}^{(0)},\ldots,V_{2l^{*}}^{(1)}\right)$
$\displaystyle=0.$

Therefore, for all $\pi\notin P_{2}^{\operatorname{bicon}}(p)$, we have
$\tau\otimes\tau(\pi)=0$. This proves the first part of the proposition.

Assume now that $\pi\in P_{2}^{\operatorname{bicon}}(p)$. Let
$V_{1},\ldots,V_{p/2}$ be an ordering of the blocks of $\pi$ such that $V_{k}$
crosses at least one $V_{j}$ for $j<k$ and all $k$. This can always be done as
$\pi$ is connected. Let $\mathcal{V}_{1},\mathcal{V}_{2}$ be the bipartite
sets of vertices of the intersection graph of $\pi$,

$\displaystyle\mathcal{V}_{1}=\\{V_{1},V_{j_{1}},\ldots,V_{j_{m}}\\};$
$\displaystyle\mathcal{V}_{2}=\\{V_{2},V_{h_{1}},\ldots,V_{h_{n}}\\}.$

Since $\pi$ is bipartite, if $V_{k}\in\mathcal{V}_{1}$, all crossing blocks
$V_{j}$ of $V_{k}$ belong to $\mathcal{V}_{2}$, and similarly if
$V_{k}\in\mathcal{V}_{2}$. We assume again that the color of $V_{1}$ is
$w_{1}=0$, since

$\displaystyle\tau\otimes\tau(\pi)=q\tau\otimes\tau\left(\pi\setminus
V_{1},V_{1}^{(0)}\right).$

We will prove that both the color $w_{k}$ and the binary vector $\theta^{(k)}$
of $V_{k}$ depend only on which bipartite set $V_{k}$ belongs to, namely, for
all $k\geq 2$, the following hold.

  1. (1)

If $V_{k}\in\mathcal{V}_{1}$, then $w_{k}=0$ and $\theta^{(k)}_{3}=1$;

  2. (2)

Otherwise, $V_{k}\in\mathcal{V}_{2}$, $w_{k}=1$ and $\theta^{(k)}_{2}=1$.

Indeed, note first that $V_{2}\in\mathcal{V}_{2}$, $\theta^{(2)}_{2}=1$ as it
crosses $V_{1}$ and $w_{1}=0$. Then, Lemma 5.6 implies that $w_{2}=1$. Assume
then that the result holds for some $l$ and for all $1\leq k\leq l$. Let us
prove that it holds for $V_{l+1}$ as well. Indeed, assume without loss of
generality that $V_{l+1}\in\mathcal{V}_{1}$. Then $V_{l+1}$ only crosses
blocks $V_{j_{t}}$ such that $V_{j_{t}}\in\mathcal{V}_{2}$, for $j_{t}\leq l$.
Since all colors $w_{j_{t}}=1$, we deduce that

$\displaystyle\theta^{(l+1)}_{3}=1.$

Then, Lemma 5.7 implies that $w_{l+1}=0$ and induction is proved. We then
apply Lemma 5.5 for $V_{1}$, Lemma 5.6 for all blocks
$V_{k}\in\mathcal{V}_{2}$ and Lemma 5.7 for all blocks
$V_{k}\in\mathcal{V}_{1}$ for $k\geq 2$, so that

$\displaystyle\tau\otimes\tau(\pi)$
$\displaystyle=q\tau\otimes\tau\left(\pi\setminus V_{1},V_{1}^{(0)}\right)$
$\displaystyle=q\cdot\frac{q}{2}\tau\otimes\tau\left(\pi\setminus\\{V_{1},V_{2}\\},V_{1}^{(0)},V_{2}^{(1)}\right)$
$\displaystyle=\cdots$ $\displaystyle=q\left(\frac{q}{2}\right)^{p/2-1}.$

It follows then that

$\displaystyle\tau\otimes\tau(\pi)=q\left(\frac{q}{2}\right)^{p/2-1}=2\left(\frac{q}{2}\right)^{p/2},$

and the result is proved. ∎

###  5.4. Proof of Theorem 1.1

We are now ready to prove the main theorem. Recall that

$\displaystyle\tau^{\prime}({\bf{S}}^{p})=\sum_{\pi\in
P_{2}(p)}\tau\otimes\tau(\pi)=\sum_{\hat{\pi},(\pi_{T})_{T\in\hat{\pi}}}\prod_{T\in\hat{\pi}}\tau\otimes\tau(\pi_{T}),$

where the second summation runs over
$(\hat{\pi},(\pi_{T})_{T\in\hat{\pi}})\in\Phi(P_{2}(p))$, using the bijection
$\Phi$ defined in (4). Note that if $|T|=2$, then by Corollary 5.2 we have
$\tau\otimes\tau(\pi_{T})=1$. Therefore, we deduce that

$\displaystyle\tau^{\prime}({\bf{S}}^{p})=\sum_{\hat{\pi},(\pi_{T})_{T\in\hat{\pi}}}\prod_{\underset{|T|\geq
4}{T\in\hat{\pi}}}\tau\otimes\tau(\pi_{T}).$

Using Proposition 5.4, we get

$\displaystyle\tau^{\prime}({\bf{S}}^{p})=\sum_{\hat{\pi}\in
NC(p)}\sum_{(\pi_{T})_{T\in\hat{\pi}}}\prod_{\underset{|T|\geq
4}{T\in\hat{\pi}}}2\Big{(}\frac{q}{2}\Big{)}^{\frac{|T|}{2}},$

where the second summation runs over bipartite connected pair partitions
$\pi_{T}$, for $T\in\hat{\pi}$. Finally, note that the number of size-two
blocks is precisely $\operatorname{\bf ncr}(\pi)$ and thus

$\displaystyle|\\{T\in\hat{\pi}:|T|\geq 4\\}|=\operatorname{\bf
cc}(\pi)-\operatorname{\bf ncr}(\pi).$

Since

$\sum_{\begin{subarray}{c}T\in\hat{\pi}\\\ |T|\geq
4\end{subarray}}\frac{|T|}{2}=\operatorname{\bf cr}(\pi),$

using again the bijection $\Phi$, we deduce that

$\displaystyle\tau^{\prime}({\bf{S}}^{p})=\sum_{\pi\in
P_{2}^{\operatorname{bi}}(p)}2^{\operatorname{\bf cc}(\pi)-\operatorname{\bf
ncr}(\pi)}\Big{(}\frac{q}{2}\Big{)}^{\operatorname{\bf cr}(\pi)}.$

This finishes the proof in view of Proposition 3.3 and of the fact that

$\displaystyle\operatorname{\bf ncr}(\pi)+\operatorname{\bf
cr}(\pi)=|\pi|=\frac{p}{2}.$

## References

  * Ambainis et al. [2012] A. Ambainis, A. W. Harrow, and M. B. Hastings.  Random tensor theory: Extending random matrix theory to mixtures of random product states.  _Communications in Mathematical Physics_ , 310(1):25–74, 2012.  doi: 10.1007/s00220-011-1411-x.
  * Aubrun and Nechita [2012] G. Aubrun and I. Nechita.  Realigning random states.  _J. Math. Phys._ , 53(10):102210, 2012.
  * Bożejko and Speicher [1996] M. Bożejko and R. Speicher.  Interpolations between bosonic and fermionic relations given by generalized Brownian motions.  _Math. Z._ , 222(1):135–159, 1996.  ISSN 0025-5874,1432-1823.  doi: 10.1007/PL00004258.  URL https://doi.org/10.1007/PL00004258.
  * Collins and Lamarre [2017] B. Collins and P. Y. G. Lamarre.  *-freeness in finite tensor products.  _Advances in Applied Mathematics_ , 83:47–80, 2017.
  * Dehornoy and Biane [2014] P. Dehornoy and P. Biane.  Dual garside structure of braids and free cumulants of products.  _Séminaire Lotharingien de Combinatoire_ , 72:15, 2014\.
  * Gouyou-Beauchamps [1986] D. Gouyou-Beauchamps.  _Chemins sous-diagonaux et tableaux de Young_ , page 112–125.  Springer Berlin Heidelberg, 1986.  ISBN 9783540474029.  doi: 10.1007/bfb0072513.  URL http://dx.doi.org/10.1007/BFb0072513.
  * Lancien et al. [2023] C. Lancien, P. O. Santos, and P. Youssef.  Limiting spectral distribution of random self-adjoint quantum channels, 2023.
  * Lehner [2002] F. Lehner.  Free cumulants and enumeration of connected partitions.  _European J. Combin._ , 23(8):1025–1031, 2002\.  ISSN 0195-6698,1095-9971.  doi: 10.1006/eujc.2002.0619.  URL https://doi.org/10.1006/eujc.2002.0619.
  * Młotkowski. [2004] W. Młotkowski.  Λ-free probability.  _Infinite Dimensional Analysis, Quantum Probability and Related Topics_ , 07(01):27–41, Mar. 2004.  ISSN 1793-6306.  doi: 10.1142/s0219025704001517.  URL http://dx.doi.org/10.1142/S0219025704001517.
  * Nica [2016] A. Nica.  Free probability aspect of irreducible meandric systems, and some related observations about meanders.  _Infinite Dimensional Analysis, Quantum Probability and Related Topics_ , 19(02):1650011, June 2016.  ISSN 1793-6306.  doi: 10.1142/s0219025716500119.  URL http://dx.doi.org/10.1142/S0219025716500119.
  * Nica and Speicher [2006] A. Nica and R. Speicher.  _Lectures on the combinatorics of free probability_ , volume 13.  Cambridge University Press, 2006.
  * Nijenhuis and Wilf [1979] A. Nijenhuis and H. S. Wilf.  The enumeration of connected graphs and linked diagrams.  _J. Combin. Theory Ser. A_ , 27(3):356–359, 1979\.  ISSN 0097-3165,1096-0899.  doi: 10.1016/0097-3165(79)90023-2.  URL https://doi.org/10.1016/0097-3165(79)90023-2.
  * Speicher and Weber [2019] R. Speicher and M. Weber.  Quantum groups with partial commutation relations.  _Indiana University Mathematics Journal_ , 68(6):1849–1883, 2019.
  * Speicher and Wysoczański [2016] R. Speicher and J. Wysoczański.  Mixtures of classical and free independence.  _Archiv der Mathematik_ , 107(4):445–453, Sept. 2016.  ISSN 1420-8938.  doi: 10.1007/s00013-016-0955-6.  URL http://dx.doi.org/10.1007/s00013-016-0955-6.
  * Voiculescu [1985] D. Voiculescu.  _Symmetries of some reduced free product C*-algebras_ , page 556–588.  Springer Berlin Heidelberg, 1985.  ISBN 9783540395140.  doi: 10.1007/bfb0074909.  URL http://dx.doi.org/10.1007/BFb0074909.