# Infinitesimal Operators and the Distribution of Anticommutators and
Commutators

James A. Mingo  Department of Mathematics and Statistics, Queen’s University,
Jeffery Hall, Kingston, Ontario, Canada K7L 3N6<EMAIL_ADDRESS>and
Pei-Lun Tseng  Department of Mathematics
New York University Abu Dhabi
Saadiyat Marina District, Abu Dhabi, United Arab Emirates<EMAIL_ADDRESS>
###### Abstract.

In an infinitesimal probability space we consider operators which are
infinitesimally free and one of which is infinitesimal, in that all its
moments vanish. Many previously analysed random matrix models are captured by
this framework. We show that there is a simple way of finding non-commutative
distributions involving infinitesimal operators and apply this to the
commutator and anticommutator. We show the joint infinitesimal distribution of
an operator and an infinitesimal idempotent gives us the Boolean cumulants of
the given operator. We also show that Boolean cumulants can be expressed as
infinitesimal moments thus giving matrix models which exhibit asymptotic
Boolean independence and monotone independence.

††Research was supported by a Discovery Grant from the Natural Sciences and
Engineering Research Council of Canada and the Fields Institute for Research
in the Mathematical Sciences.

##  1\. Introduction

The largest eigenvalues of a random covariance matrix has been much studied in
recent years, motivated in many cases by problems in principle component
analysis. An important question in principal component analysis is to decide
how many eigenvalues can be considered significant. To analyse this problem a
spike or signal-plus-noise model is often used where we have a source of noise
represented by a random matrix $X_{1}$ and a signal represented by a non-
random matrix $X_{2}$. Moreover it is usually assumed that the rank of the
signal is much much smaller than the rank of the noise, which increases with
the sample size. Even though the rank of $X_{2}$ is small, it still affects
the largest eigenvalue of $X_{1}+X_{2}$. In this paper we idealize this by
saying that the signal, $X_{2}$, is infinitesimal.

The tools of free probability give explicit rules (see [41, 32, 34]) for
computing the distributions of the sum or product of freely independent random
variables. The method of linearization extends these rules to the case of
polynomials (see [9] and [32, Ch. 10]) or rational functions (see [29]) of
freely independent random variables, but explicit solutions are often
infeasible.

An example where explicit calculations are feasible is the case of a
commutator: $i(x_{1}x_{2}-x_{2}x_{1})$ and an anti-commutator:
$x_{1}x_{2}+x_{2}x_{1}$. In [34, Lect. 19] the distribution of the commutator
of freely independent operators is given; the case of the anticommutator is
more involved (see [34, Rem. 15.21]); however a graph theoretic description
was given by Perales in [35]. A related case is the recent work of Campbell
[12] on the finite free distribution of a commutator an anticommutator. Also
relevant are the papers of Ejsmont, Lehner [18, 19], and Hęćka [17] on the
tangent law.

In [13] Collins, Hasebe, and Sakuma investigated the spectrum of the
commutator $i(X_{1}X_{2}-X_{2}X_{1})$ and the anticommutator
$X_{1}X_{2}+X_{2}X_{1}$ when the rank $X_{1}$ tended to infinity and the rank
of $X_{2}$ was fixed. The regime of [13] can be modelled by the infinitesimal
free independence of Belinschi and Shlyakhtenko [9] and its application to
study asymptotic independence given by Shlyakhtenko [38]. Expanding on this
Cébron and Dahlqvist, and Gabriel [15], used infinitesimal freeness to give a
rigorous prove the BBP phase transition phenomena of outliers found by Baik,
Ben Arous, and Péché [6]. Finally in [40] it was shown that infinitesimal free
independence was just a special case of freeness with amalgamation over a
commutative algebra. This was the inspiration for our work. As we shall see in
this paper, many of the algebraic and combinatorial tools of [34] transfer
directly to the infinitesimal case. Moreover this transfer also extends to the
Boolean independence of Speicher and Wourodi [39], as was shown by Perales and
Tseng in [36]. An important point to notice here is that our methods give the
full distribution of the commutator and anticommutator, not just the support
of the distribution as was done in [13].

From the point of view of free probability, the key property of $X_{2}$ used
in [13] was that its limit distribution was $\delta_{0}$, the Dirac mass at
$0$. An operator with this property we call infinitesimal.

There has been much recent work connecting free and Boolean cumulants (see for
example the papers of Février, Mitja, Nica, and Szpojankowski [23] and [22]).
We show that if one further assumes that the infinitesimal operator is an
idempotent. Then there are new and simple formulas relating infinitesimal
moments to Boolean cumulants (see Prop. 5.3). This enables us to produce
matrix models exhibiting asymptotic Boolean independence and asymptotic
monotone independence, which until now has been difficult to do, see [27] and
[30]. Also in Remark 5.4 we show how infinitesimal operators give an
abstraction for the Markov-Krein transform of Kerov [25].

In §2 we collect the basic notions from [34], [24], and [32] that we shall
need for our results. In §3 we consider the distribution of the commutator and
the anticommutator when $x_{1}$ and $x_{2}$ are infinitesimally free and
$x_{2}$ is infinitesimal. In §4 we consider infinitesimal $R$-diagonal
operators. In §5 we show how to write a Boolean cumulant as an infinitesimal
moment. In §6 we revisit the problem in §3 with infinitesimal free
independence replaced by Boolean independence.

##  2\. Preliminaries

###  2.1. Free Independence

A non-commutative probability space $(\mathcal{A},\varphi)$ consists of a
unital algebra over $\mathbb{C}$ and a linear functional
$\varphi:\mathcal{A}\to\mathbb{C}$ with $\varphi(1)=1$. We let
$\mathbb{C}\langle X_{1},\dots,X_{k}\rangle$ be the algebra on polynomials in
the non-commuting variables $X_{1},\dots,X_{n}$.

For $a_{1},\dots,a_{k}$ be elements in $\mathcal{A}$, the mapping
$\mathbb{C}\langle X_{1},\dots,X_{k}\rangle\ni
p\mapsto\varphi(p(a_{1},\dots,a_{k}))\in\mathbb{C}$ is the joint distribution
of $\\{a_{1},\dots,a_{k}\\}$. We denote this linear functional by
$\mu_{a_{1},\dots,a_{k}}$.

For a given non-commutative probability space $(\mathcal{A},\varphi)$, unital
subalgebras $(\mathcal{A}_{i})_{i\in I}$ of $\mathcal{A}$ are freely
independent (or free for short) if for all $n\in\mathbb{N}$,
$a_{1},\dots,a_{n}\in\mathcal{A}$ are such that $a_{j}\in\mathcal{A}_{i_{j}}$
where $i_{1},\dots,i_{n}\in I$, $i_{1}\neq i_{2}\neq\cdots\neq i_{n}$ and
$\varphi(a_{j})=0$ for all $j$, then

$\varphi(a_{1}\cdots a_{n})=0.$

A set $\\{a_{i}\in\mathcal{A}\mid i\in I\\}$ is free if the unital subalgebras
$\mathrm{alg}(1,a_{i})$ generated by $a_{i}\ (i\in I)$ form a free family.

For $n\geq 1$, a partition of $[n]$ is a set $\pi=\\{V_{1},\dots,V_{r}\\}$ of
pairwise disjoint non-empty subsets of $[n]:=\\{1,2,\dots,n-1,n\\}$ such that
$V_{1}\cup\cdots\cup V_{r}=[n]$. The set of all partitions of $[n]$ is denoted
by $\mathcal{P}(n)$, and $NC(n)$ denotes all non-crossing partitions of $[n]$
in the sense that we cannot find distinct blocks $V_{r}$ and $V_{s}$ with
$a,c\in V_{r}$ and $b,d\in V_{s}$ such that $a<b<c<d$.

######  Notation 2.1.

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space and
$a\in\mathcal{A}$. Let $m_{n}(a)=\varphi(a^{n})$ be the moments of $a$.
Suppose $a_{1},\dots,a_{n}$ are elements in $\mathcal{A}$ and
$V=\\{i_{1}<\cdots<i_{s}\\}$ is a block of some partition of $[n]$, we set
$(a_{1},\dots,a_{n})|_{V}=(a_{i_{1}},\dots,a_{i_{s}}).$ Let
$\\{f_{n}\\}_{n\geq 1}$ is a sequence of multilinear functionals
$f_{n}:\mathcal{A}^{n}\to\mathbb{C}$, then for each partition $\pi$, we set

$f_{\pi}(a_{1},\dots,a_{n})=\prod_{V\in\pi}f_{|V|}((a_{1},\dots,a_{n})|_{V}).$

The free cumulants $\\{\kappa_{n}:\mathcal{A}^{n}\to\mathbb{C}\\}_{n\geq 1}$
are defined inductively in terms of moments via

$\varphi(a_{1}\cdots a_{n})=\sum_{\pi\in
NC(n)}\kappa_{\pi}(a_{1},\dots,a_{n}).$

Note that the free cumulants can also be defined via the following formula

$\kappa_{n}(a_{1},\dots,a_{n})=\sum_{\pi\in
NC(n)}\mu(\pi,1_{n})\varphi_{\pi}(a_{1},\dots,a_{n}),$

where $\mu$ is the Möbius function of $NC(n)$ (see [34, Lect. 11]).

Free cumulants are crucial to characterizing freeness. More precisely, unital
subalgebras $(\mathcal{A}_{i})_{i\in I}$ are freely independent if and only if
for all $n\geq 2$, and for all $a_{j}\in\mathcal{A}_{i_{j}}$ for all
$j=1,\dots,n$ with $i_{1},\dots,i_{n}\in I$ we have
$\kappa_{n}(a_{1},\dots,a_{n})=0$ whenever there exist $1\leq l,k\leq n$ with
$i_{l}\neq i_{k}$ (see [34, Thm. 11.16]). This characterization of freeness is
called the vanishing of mixed cumulants.

For a given $a\in\mathcal{A}$, the Cauchy transform of $a$ is defined by

$G_{a}(z)=\sum_{n=0}^{\infty}\frac{m_{n}(a)}{z^{n+1}},$

and the $R$-transform of $a$ is defined as

$R_{a}(z)=\sum_{n=0}^{\infty}\kappa_{n+1}(a)z^{n}\text{ where
}\kappa_{n}(a)=\kappa_{n}(a,\dots,a).$

Note that following the vanishing of mixed cumulants property, we have if
$a,b\in\mathcal{A}$ are free, then $R_{a+b}=R_{a}+R_{b}$. We shall always work
with formal power series, however if our distributions arise from Borel
measures on $\mathbb{R}$ with compact support, then all of our series converge
to an analytic function on an appropriate domain.

###  2.2. $\bm{R}$-diagonal Operators

A non-commutative probability space $(\mathcal{A},\varphi)$ is called a _$*$
-probability space_ if $\mathcal{A}$ is a $*$-algebra and $\varphi$ is
positive in the sense that $\varphi(a^{*}a)\geq 0$ for all $a\in\mathcal{A}$.
For $\varepsilon\in\\{-1,1\\}$ and $a\in\mathcal{A}$ we let
$a^{(\varepsilon)}=a$ if $\varepsilon=1$ and $a^{(\varepsilon)}=a^{*}$ if
$\varepsilon=-1$. Recall from [34, Lect. 15] that $a\in\mathcal{A}$ is
$R$-diagonal if for all $\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$
we have

$\kappa_{n}(a^{(\varepsilon_{1})},\dots,a^{(\varepsilon_{n})})=0$

unless $n$ is even and $\varepsilon_{i}+\varepsilon_{i+1}=0$ for $1\leq i<n$.
In addition, we recall that $a\in\mathcal{A}$ is even if $a=a^{*}$ and
$\varphi(a^{2n+1})=0$ for all $n\geq 0$.

Note that the $R$-diagonal element is uniquely determined by its determining
sequences (see [34, Cor. 15.7]). To be precise, if $a$ is an $R$-diagonal
element in a $*$-probability space $(\mathcal{A},\varphi)$, then the
$*$-distribution of $a$ can be uniquely determined by $(\alpha_{n})_{n}$ and
$(\beta_{n})_{n}$ where

$\alpha_{n}=\kappa_{2n}(a,a^{*},a,a^{*},\dots,a,a^{*})\text{ and
}\beta_{n}=\kappa_{2n}(a^{*},a,a^{*},a\cdots,a^{*},a).$

Moreover, the distribution of $aa^{*}$ and $a^{*}a$ can be described by its
determining sequences $\\{\alpha_{n}\\}_{n}$ and $\\{\beta_{n}\\}_{n}$ in the
following relation

$\displaystyle\kappa_{n}(aa^{*},\dots,aa^{*})$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\pi\in NC(n)\\\
\pi=\\{V_{1},\dots,V_{r}\\}\end{subarray}}\alpha_{|V_{1}|}\beta_{|V_{2}|}\cdots\beta_{|V_{r}|}$
$\displaystyle\kappa_{n}(a^{*}a,\dots,a^{*}a)$ $\displaystyle=$
$\displaystyle\sum_{\begin{subarray}{c}\pi\in NC(n)\\\
\pi=\\{V_{1},\dots,V_{r}\\}\end{subarray}}\beta_{|V_{1}|}\alpha_{|V_{2}|}\cdots\alpha_{|V_{r}|}.$

where $V_{1}$ is the block of $\pi$ which contains the first element $1$.

In addition, if $a$ and $b$ are $*$-free elements in a $*$-probability space
such that $a$ is $R$-diagonal, then $ab$ is also $R$-diagonal (see [34, Prop.
15.8]).

###  2.3. Infinitesimal Free Independence

######  Definition 2.2.

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space. Let
$\varphi^{\prime}:\mathcal{A}\rightarrow\mathbb{C}$ be a linear functional
such that $\varphi^{\prime}(1)=0$. Then we call the triple
$(\mathcal{A},\varphi,\varphi^{\prime})$ an infinitesimal probability space.

For $a_{1},\dots,a_{k}\in\mathcal{A}$, the pair of linear functionals
$(\mu,\mu^{\prime})$ which both map $\mathbb{C}\langle
X_{1},\dots,X_{k}\rangle$ into $\mathbb{C}$ are the joint infinitesimal
distribution of $\\{a_{1},\dots,a_{k}\\}$ if $\mu=\mu_{a_{1},\dots,a_{k}}$ and

$\mu^{\prime}(p)=\varphi^{\prime}(p(a_{1},\dots,a_{k})).$

Similarly, we denote the linear functional $\mu^{\prime}$ by
$\mu^{\prime}_{a_{1},\dots,a_{k}}.$

######  Definition 2.3.

Let $(\mathcal{A},\varphi,\varphi^{\prime})$ be an infinitesimal probability
space, unital subalgebras $(\mathcal{A}_{i})_{i\in I}$ of $\mathcal{A}$ are
infinitesimally freely independent (or infinitesimally free for short) with
respect to $(\varphi,\varphi^{\prime})$ if whenever
$a_{1},\dots,a_{n}\in\mathcal{A}$ are such that
$a_{k}\in\mathcal{A}_{i_{k}},i_{1},\dots,i_{n}\in I,i_{1}\neq\cdots\neq i_{n}$
and $\varphi(a_{j})=0$ for all $j$, then we have

$\displaystyle\varphi(a_{1}\cdots a_{n})$ $\displaystyle=$ $\displaystyle 0$
$\displaystyle\varphi^{\prime}(a_{1}\cdots a_{n})$ $\displaystyle=$
$\displaystyle\sum_{k=1}^{n}\varphi^{\prime}(a_{k})\varphi(a_{1}\cdots
a_{k-1}a_{k+1}\cdots a_{n}).$

######  Notation 2.4.

Let $(\mathcal{A},\varphi,\varphi^{\prime})$ be an infinitesimal probability
space and $a\in\mathcal{A}$. Let $m_{n}^{\prime}(a)=\varphi^{\prime}(a^{n})$
be the infinitesimal moments of $a$. Let
$\\{f_{n}:\mathcal{A}^{n}\to\mathbb{C}\\}_{n\geq 1}$ and
$\\{f_{n}^{\prime}:\mathcal{A}^{n}\to\mathbb{C}\\}_{n\geq 1}$ be two sequences
of multi-linear functionals. Given a partition $\pi\in NC(n)$, then for a
fixed $V\in\pi$, we define $\partial f_{\pi,V}$ be the map that is equal to
$f_{\pi}$ except for the block $V$, where we replace $f_{|V|}$ by
$f^{\prime}_{|V|}$. Then we define $\partial f_{\pi}$ by

$\partial f_{\pi}(a_{1},\dots,a_{n})=\sum_{V\in\pi}\partial
f_{\pi,V}(a_{1},\dots,a_{n}).$

######  Definition 2.5.

Suppose $(\mathcal{A},\varphi,\varphi^{\prime})$ is an infinitesimal
probability space, the infinitesimal free cumulants
$\\{\kappa_{n}^{\prime}:\mathcal{A}^{n}\to\mathbb{C}\\}$ are defined
inductively in terms of moments and infinitesimal moments via

$\varphi^{\prime}(a_{1},\dots,a_{n})=\sum_{\pi\in
NC(n)}\partial\kappa_{\pi}(a_{1},\dots,a_{n}).$

Note that the infinitesimal free cumulants can also be described via

$\kappa_{n}^{\prime}(a_{1},\dots,a_{n})=\sum_{\pi\in
NC(n)}\mu(\pi,1_{n})\partial\varphi_{\pi}(a_{1},\dots,a_{n})$

(see [24, 31]). The infinitesimal free cumulants can be used to characterize
infinitesimal freeness.

######  Theorem 2.6.

[24, Thm. 1.2] Suppose that $(\mathcal{A},\varphi,\varphi^{\prime})$ is an
infinitesimal probability space, and $\mathcal{A}_{i}$ is a unital subalgebra
of $\mathcal{A}$ for each $i\in I$. Then $(\mathcal{A}_{i})_{i\in I}$ are
infinitesimally free if and only if for each $s\geq 2$ and
$i_{1},\dots,i_{s}\in I$ which are not all equal, and for
$a_{1}\in\mathcal{A}_{i_{1}},\dots,a_{s}\in\mathcal{A}_{i_{s}}$, we have
$\kappa_{n}(a_{1},\dots,a_{s})=\kappa_{n}^{\prime}(a_{1},\dots,a_{s})=0.$

For a given $a\in\mathcal{A}$, we define the infinitesimal Cauchy transform of
$a$ by

$g_{a}(z)=\sum_{n=1}^{\infty}\frac{m_{n}^{\prime}(a)}{z^{n+1}},$

and the infinitesimal $r$-transform of $a$ is defined by

$r_{a}(z)=\sum_{n=0}^{\infty}\kappa^{\prime}_{n+1}(a)z^{n}\text{ where
}\kappa^{\prime}_{n}(a)=\kappa^{\prime}_{n}(a,\dots,a).$

Following [31], the infinitesimal Cauchy and $R$-transforms are related by the
equations

(2.1)
$r_{a}(z)=-g_{a}\left(G^{\langle-1\rangle}_{a}(z)\right)(G^{\langle-1\rangle}_{a})^{\prime}(z)\text{
and }g_{a}(z)=-r_{a}(G_{a}(z))G_{a}^{\prime}(z)$

where $G^{\langle-1\rangle}_{a}$ is the composition inverse of $G_{a}$.

###  2.4. Applications to Random Matrices

In [13, §4] Collins, Hasebe and Sakuma proved asymptotic infinitesimal
freeness for some random matrix ensembles, using the language of cyclic
monotone independence. In particular they considered two ensembles of $N\times
N$ matrices $\\{A_{1,N},\dots,A_{k,N}\\}$ and $\\{B_{1,N},\dots,B_{l,N}\\}$.
Let $\varphi_{\scriptscriptstyle N}=\mathrm{E}\circ\mathrm{tr}$ and
$\varphi^{\prime}_{\scriptscriptstyle N}=\mathrm{E}\circ\mathrm{Tr}$ where
$\mathrm{Tr}$ is the un-normalized trace and $\mathrm{tr}=N^{-1}\mathrm{Tr}$
is the normalized trace. They assumed that there was a limit joint
distribution for the $B_{i,N}$’s with respect to $\varphi_{\scriptscriptstyle
N}$ and a limit joint distribution of the $A_{j,N}$’s with respect to
$\varphi^{\prime}_{\scriptscriptstyle N}$. More precisely that for all
$i_{1},\dots,i_{m}\in[l]$ the limit $\lim_{N}\varphi_{\scriptscriptstyle
N}(B_{i_{1},N}\cdots B_{i_{m},N})$ exists, and for all
$j_{1},\dots,j_{n}\in[k]$ the limit
$\lim_{N}\varphi^{\prime}_{\scriptscriptstyle N}(A_{j_{1},N}\cdots
A_{j_{n},N})$ exists.

Then in [13, Thm. 4.1], they showed that when $\\{U_{N}\\}_{N}$ is a Haar
unitary independent from $\\{A_{1,N},\dots,\allowbreak A_{k,N},\allowbreak
B_{1,N},\dots,B_{l,N}\\}$ then the ensembles
$\\{U_{N}B_{1,N}U_{N}^{*},\dots,\allowbreak U_{N}B_{l,N}U_{N}^{*}\\}$ and
$\\{A_{1,N},\dots,\allowbreak A_{k,N}\\}$ are asymptotically infinitesimally
free with respect to $(\varphi,\varphi^{\prime})$. The special case when the
$A_{i,N}$’s are stationary was already presented by Shlyakhtenko in [38]. In
both these models the limit operators of $\\{A_{1,N},\dots,A_{k,N}\\}$ are
infinitesimal operators in the sense of Definition 3.1, so the limit
distributions of this model are covered by our theory. In addition, the
results of [13, Thm. 5.1] are spectral results, i.e. about the support of the
distribution, and do not give information about the distributions themselves.

###  2.5. Boolean Independence

Here we first review some basic concepts of Boolean independence, introduced
in [39].

For a given non-commutative probability space $(\mathcal{A},\varphi)$, (non-
unital) subalgebras $(\mathcal{A}_{i})_{i\in I}$ are _Boolean independent_ if
for all $n\geq 0$, $a_{1},\dots,a_{n}\in\mathcal{A}$ are such that
$a_{j}\in\mathcal{A}_{i_{j}}$ where $i_{1},\dots,i_{n}\in I$,
$i_{1}\neq\cdots\neq i_{n}$, then

(2.2) $\varphi(a_{1}\cdots a_{n})=\varphi(a_{1})\cdots\varphi(a_{n}).$

The elements $\\{a_{i}\in\mathcal{A}\mid i\in I\\}$ are Boolean independent if
the subalgebras $\mathrm{alg}(a_{i})$ generated by $a_{i}\ (i\in I)$ are
Boolean independent.

For $n\geq 1$, a partition $\pi$ is an _interval partition_ if each block of
$\pi$ is an interval of the form $\\{i,i+1,\dots,i+s\\}$ for some $1\leq i\leq
n$ and $0\leq s\leq n-i$. $I(n)$ denotes the set of all interval partitions.
Then the _Boolean cumulants_
$\\{\beta_{n}:\mathcal{A}^{n}\to\mathbb{C}\\}_{n\geq 1}$ are defined
inductively in terms of moments via

(2.3) $\varphi(a_{1}\cdots a_{n})=\sum_{\pi\in
I(n)}\beta_{\pi}(a_{1},\dots,a_{n}).$

As in the free case, Boolean cumulants characterize Boolean independence. To
be precise, $(\mathcal{A}_{i})_{i\in I}$ are Boolean independent if and only
if for all $n\geq 2$ and for all
$a_{1},\dots,a_{n}\in\cup_{i}\mathcal{A}_{i}$, we have
$\beta_{n}(a_{1},\dots,a_{n})=0$, whenever there are at least two elements
from different subalgebras; i.e. there exists $1\leq k,l\leq n$ with
$i_{k}\neq i_{l}$ and $a_{k}\in\mathcal{A}_{i_{k}}$ and
$a_{l}\in\mathcal{A}_{i_{l}}$.

The relation between moments and Boolean cumulants can also be understood via
the moment-generating function and $\eta$-transform. To be precise, for a
given $a\in\mathcal{A}$, the _moment generating function_ $\psi_{a}$ and the
_$\eta$ -transform_ $\eta_{a}$ are defined by

$\psi_{a}(z)=\sum_{n=1}^{\infty}m_{n}(a)z^{n}\text{ and
}\eta_{a}(z)=\sum_{n=1}^{\infty}\beta_{n}(a)z^{n}$

where $\beta_{n}(a)=\beta_{n}(a,\dots,a)$. Then we have

(2.4) $\eta_{a}(z)(1+\psi_{a}(z))=\psi_{a}(z).$

###  2.6. Infinitesimal Boolean Independence

The notion of infinitesimal Boolean independence was introduced in [36]. We
recall the definition of it as follows.

######  Definition 2.7.

Suppose that $(\mathcal{A},\varphi,\varphi^{\prime})$ is an infinitesimal
probability space. Non-unital subalgebras $(\mathcal{A}_{i})_{i\in I}$ are
infinitesimally Boolean independent if for all $n\geq
0,a_{1},\dots,a_{n}\in\mathcal{A}$ are such that $a_{j}\in\mathcal{A}_{i_{j}}$
where $i_{1},\dots,i_{n}\in I$, $i_{1}\neq\cdots\neq i_{n}$, then

$\displaystyle\varphi(a_{1}\cdots a_{n})$ $\displaystyle=$
$\displaystyle\varphi(a_{1})\cdots\varphi(a_{n});$
$\displaystyle\varphi^{\prime}(a_{1}\cdots a_{n})$ $\displaystyle=$
$\displaystyle\sum_{j=1}^{n}\varphi(a_{1})\cdots\varphi(a_{j-1})\varphi^{\prime}(a_{j})\varphi(a_{j+1})\cdots\varphi(a_{n}).$

Note that the infinitesimal Boolean independence can be understood as Boolean
independence with amalgamation. To be precise, for a given infinitesimal
probability space $(\mathcal{A},\varphi,\varphi^{\prime})$, we define

$\widetilde{\mathcal{A}}=\Big{\\{}\begin{bmatrix}a&a^{\prime}\\\
0&a\end{bmatrix}\Big{|}a,a^{\prime}\in\mathcal{A}\Big{\\}},\
\widetilde{\mathbb{C}}=\Big{\\{}\begin{bmatrix}c&c^{\prime}\\\
0&c\end{bmatrix}\Big{|}c,c^{\prime}\in\mathbb{C}\Big{\\}}$

and then
$\widetilde{\varphi}:\widetilde{\mathcal{A}}\to\widetilde{\mathbb{C}}$ is
defined by

$\widetilde{\varphi}\Big{(}\begin{bmatrix}a&a^{\prime}\\\
0&a\end{bmatrix}\Big{)}=\begin{bmatrix}\varphi(a)&\varphi^{\prime}(a)+\varphi(a^{\prime})\\\
0&\varphi(a)\end{bmatrix}.$

Note that
$(\widetilde{\mathcal{A}},\widetilde{\mathbb{C}},\widetilde{\varphi})$ forms
an operator-valued probability space (see [40]). Moreover, in [36], the
authors showed the following result.

######  Theorem 2.8.

Non-unital subalgebras $(\mathcal{A}_{i})_{i\in I}$ are infinitesimally
Boolean independent with respect to $(\varphi,\varphi^{\prime})$ if and only
if $(\widetilde{\mathcal{A}}_{i})_{i\in I}$ are Boolean independent with
respect to $\widetilde{\varphi}$.

###  2.7. Monotone Independence [33]

Suppose that $(\mathcal{A},\varphi)$ is a non-commutative probability space.
Let $\mathcal{B}$ be a non-unital subalgebra and $\mathcal{C}$ be a (non-
unital) subalgebra of $\mathcal{A}$. $(\mathcal{B},\mathcal{C})$ are _monotone
independent_ if whenever $a_{j}\in\mathcal{B}\cup\mathcal{C}$ for all $j$, and
$a_{k-1}\in\mathcal{B}$, $a_{k}\in\mathcal{C}$, and $a_{k+1}\in\mathcal{B}$
for some $k$, one has

(2.5) $\displaystyle\varphi(a_{1}\cdots a_{n})$ $\displaystyle=$
$\displaystyle\varphi(a_{1}\cdots a_{k-1}\varphi(a_{k})a_{k+1}\cdots a_{n})$

where one of the inequalities is eliminated if $k=1$ or $k=n.$ We denote it by
$\mathcal{B}\prec\mathcal{C}$.

##  3\. The Infinitesimal Distribution of Some Polynomials

######  Definition 3.1.

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space. If
$a\in\mathcal{A}$ is such that $\varphi(a^{n})=0$ for all $n\geq 1$, we say
that $a$ is an infinitesimal operator.

We observe that if $a_{1}$ and $a_{2}$ are free elements and $a_{1}$ is
infinitesimal, then

$\displaystyle\varphi(a_{1}a_{2}a_{1}a_{2})$ $\displaystyle=$
$\displaystyle\varphi(a_{1}^{2})\varphi(a_{2})^{2}+\varphi(a_{2}^{2})\varphi(a_{1})^{2}-\varphi(a_{1})^{2}\varphi(a_{2})^{2}=0$

since $\varphi(a_{1})=\varphi(a_{1})^{2}=0$. In fact, we can obtain general
result as the following Lemma.

######  Lemma 3.2.

Suppose $a_{1}$ and $a_{2}$ are free elements in a ncps
$(\mathcal{A},\varphi)$ that $a_{1}$ is infinitesimal. Then any polynomial in
$a_{1}$ and $a_{2}$ is infinitesimal, provided each term contains at least one
$a_{1}$. In particular, $a_{1}a_{2}+a_{2}a_{1}$ and $i(a_{1}a_{2}-a_{2}a_{1})$
will be infinitesimal.

###### Proof.

By linearity, it suffices to show that any word $a_{i_{1}}\cdots a_{i_{n}}$ is
also infinitesimal where there is at least one $j$ with $i_{j}=1$. Note that
$\varphi((a_{i_{1}}\cdots a_{i_{n}})^{m})$ may be expanded, by freeness, into
a sum of products of moments of $a_{1}$ and moments of $a_{2}$. Then every
term in this sum will be $0$ because $a_{1}$ is infinitesimal. ∎

######  Remark 3.3.

With this freeness assumption, infinitesimal elements form a weak kind of
ideal in $\mathcal{A}$. In [13], the infinitesimal elements were trace class
operators, which form an ideal in the usual sense.

Going a little further, if we assume that $a_{1}$ and $a_{2}$ are
infinitesimally free and both infinitesimal then all mixed moments
$\varphi^{\prime}(a_{i_{1}}\cdots a_{i_{n}})$ vanish whenever
$i_{1},\dots,i_{n}$ are not all equal. We get this by writing
$\varphi^{\prime}(a_{i_{1}}\cdots a_{i_{n}})$ as a sum of cumulants
$\partial\kappa_{\pi}(a_{i_{1}},\dots,a_{i_{n}})$ with $\pi\in NC(n)$, and
then observing that there can be only one $\kappa^{\prime}$ in
$\partial\kappa_{\pi}(a_{i_{1}},\dots,a_{i_{n}})$ and all other $\kappa$’s
vanish.

######  Proposition 3.4.

Let $a$ be an infinitesimal operator, then we have
$r_{a}(z)=z^{-2}g_{a}(z^{-1})$. Furthermore,
$m_{n}^{\prime}(a)=\kappa_{n}^{\prime}(a)$ for all $n\geq 1.$

###### Proof.

We observe that $\varphi(a^{n})=0$ implies that $G_{a}(z)=z^{-1}$ and then
$G_{a}^{\langle-1\rangle}(z)=z^{-1}.$ Then we have

$r_{a}(z)=-g_{a}\left(G^{\langle-1\rangle}_{a}(z)\right)(G^{\langle-1\rangle}_{a})^{\prime}(z)=\frac{1}{z^{2}}g_{a}(\frac{1}{z}),$

which implies that

$\sum_{n=1}^{\infty}\kappa^{\prime}_{n}(a)z^{n-1}=\frac{1}{z^{2}}\sum_{n=1}^{\infty}\frac{m_{n}^{\prime}(a)}{(z^{-1})^{n+1}}.$

Thus, we obtain $\kappa_{n}^{\prime}(a)=m_{n}^{\prime}(a)$ for all $n\geq 1$.
∎

###  3.1. The Anticommutator Case

Let $p=x_{1}x_{2}+x_{2}x_{1}$ where $x_{1}$ and $x_{2}$ are infinitesimally
free and $x_{2}$ is infinitesimal. We shall find the infinitesimal moments of
$p$ by tracking the corresponding $\varphi^{\prime}(x_{i_{1}}\cdots
x_{i_{2n}})$, and we summarize them in the following theorem. Note that this
shows that $p$ is the sum of two infinitesimally freely independent copies of
$x_{2}$, scaled by $\alpha$ and $\beta$ respectively.

Before we state the theorem, let us introduce the following notation:

######  Notation 3.5.

Given a string $(j_{1},\dots,j_{k})\in[n]^{k}$, this can be read as a map
$j:[k]\to[n]$ by $j(l)=j_{l}$. Then $ker(j)$ is the partition on $[k]$ such
that $r,s$ are in the same block of $ker(j)$ if and only if $j_{r}=j_{s}.$ For
instance, $(j_{1},j_{2},j_{3},j_{4})=(1,2,1,2)$, then
$ker(j)=\\{(1,3),(2,4)\\}.$

######  Theorem 3.6.

Suppose that $x_{1}$ and $x_{2}$ are infinitesimally free with $x_{2}$ an
infinitesimal operator. Let $p=x_{1}x_{2}+x_{2}x_{1}$ and
$\alpha=m_{1}(x_{1})+\sqrt{m_{2}(x_{1})}$ and
$\beta=m_{1}(x_{1})-\sqrt{m_{2}(x_{1})}$. Then $p$ is an infinitesimal
operator with infinitesimal cumulants and moments

$\kappa^{\prime}_{n}(p)=m_{n}^{\prime}(p)=\kappa_{n}^{\prime}(\alpha
x_{2})+\kappa_{n}^{\prime}(\beta x_{2}).$

Moreover, $g_{p}=g_{\alpha x_{2}}+g_{\beta x_{2}}$ and $r_{p}=r_{\alpha
x_{2}}+r_{\beta x_{2}}$.

###### Proof.

We have

$\varphi^{\prime}(p^{n})=\sum_{j_{1},\dots,j_{2n}=1}^{2}\varphi^{\prime}(x_{j_{1}}x_{j_{2}}\cdots
x_{j_{2n-1}}x_{j_{2n}})=\sum_{\pi\in
NC(2n)}\sum_{j_{1},\dots,j_{2n}=1}^{2}\partial\kappa_{\pi}(x_{j_{1}},x_{j_{2}},\dots,x_{j_{2n-1}},x_{j_{2n}})$

where the sum over $j_{1},\dots,j_{2n}$ only includes those tuples such that
for each $1\leq l\leq n$ we have $(j_{2l-1},j_{2l})$ is either $(1,2)$ or
$(2,1)$. Then by our assumption on $x_{1}$ and $x_{2}$,
$\partial\kappa_{\pi}(x_{j_{1}},\cdots,x_{j_{2n}})=0$ unless

  * $\scriptscriptstyle\bullet$

$\pi\leq\ker(j)$; (by infinitesimal freeness of $x_{1}$ and $x_{2}$)

  * $\scriptscriptstyle\bullet$

$j^{-1}(2)$ is exactly one block of $\pi$ (by our assumption that $x_{2}$ is
infinitesimal).

For instance, when $n=5$ and $(j_{1},\dots,j_{1})=(1,2,2,1,1,2,2,1,1,2)$

$\pi=\mbox{}$$x_{1}$$x_{2}$$x_{2}$$x_{1}$$x_{1}$$x_{2}$$x_{2}$$x_{1}$$x_{1}$$x_{2}$

$\partial\kappa_{\pi}(x_{1},x_{2},x_{2},x_{1},x_{1},x_{2},x_{2},x_{1},x_{1},x_{2})=\kappa_{1}(x_{1})\kappa_{2}(x_{1},x_{1})^{2}\kappa_{5}^{\prime}(x_{2},x_{2},x_{2},x_{2},x_{2}).$

Then, letting $m_{1}=m_{1}(x_{1})$ and $m_{2}=m_{2}(x_{1})$, we have for such
a tuple that

(3.1)
$\sum_{\begin{subarray}{c}\pi\leq\ker(j)\end{subarray}}\partial\kappa_{\pi}(x_{j_{1}},\cdots,x_{j_{2n}})=\kappa^{\prime}_{n}(x_{2})m_{1}^{n-2v}(\sqrt{m_{2}})^{2v}$

where $0\leq v\leq[n/2]$ counts the number of times in
$j=(j_{1},j_{2},\dots,j_{2n})$ we have two $``1"$s cyclically adjacent. Now
for a given $0\leq v\leq[n/2]$ the number of such tuples is counted as
follows. We parameterize such tuples by a string $l_{1}\cdots l_{n}$ where
$l_{k}=f$ ($f$ for first) when $(j_{2k-1},j_{2k})=(1,2)$ and $l_{k}=s$ ($s$
for second) when $(j_{2k-1},j_{2k})=(2,1)$. Then the possible strings are
$f^{k_{1}}s^{k_{2}}\cdots f^{k_{r}}$ (when $r$ is odd) or
$f^{k_{1}}s^{k_{2}}\cdots f^{k_{r-1}}s^{k_{r}}$ (when $r$ is even), where
$k_{1},\dots,k_{r}\geq 1$ and $k_{1}+\cdots+k_{r}=n$. Here we have only to
count strings that start with $f$ as, by cyclic invariance, the contribution
of the strings starting with $s$ will be the same; as long as we multiply by 2
at the end. With this parametrization $v$ will be either $r/2$, when $r$ is
even and $(r-1)/2$ when $r$ is odd, i.e. $v=[r/2]$. The number of $r$-tuples
$(k_{1},\dots,k_{r})$ is $\binom{n-1}{r-1}$. Thus

$\displaystyle\varphi^{\prime}(p^{n})\mbox{}=\mbox{}$ $\displaystyle
2\kappa^{\prime}_{n}(x_{2})\sum_{r=1}^{n}\binom{n-1}{r-1}m_{1}^{n-2[r/2]}(\sqrt{m_{2}})^{2[r/2]}=\mbox{}\kappa^{\prime}_{n}(x_{2})\
2\sum_{j=0}^{[n/2]}\binom{n}{2j}m_{1}^{n-2j}(\sqrt{m_{2}})^{2j}$
$\displaystyle\mbox{}=\mbox{}$
$\displaystyle\kappa^{\prime}_{n}(x_{2})[(m_{1}+\sqrt{m_{2}})^{n}+(m_{1}-\sqrt{m_{2}})^{n}]=\kappa_{n}^{\prime}(\alpha
x_{2})+\kappa_{n}^{\prime}(\beta x_{2}).$

∎

In order to state a corollary, let us recall some notation from [9]. Suppose
we have an infinitesimal probability space
$(\mathcal{A},\varphi,\varphi^{\prime})$ and $a_{1},a_{2}\in\mathcal{A}$ with
$a_{1}\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\mu_{1},\mu_{1}^{\prime})$
and
$a_{2}\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\mu_{2},\mu_{2}^{\prime})$.
If $a_{1}$ and $a_{2}$ are infinitesimally free then we denote the
infinitesimal distribution of $a_{1}+a_{2}$ by
$(\mu_{1},\mu_{1}^{\prime})\boxplus_{\scriptscriptstyle
B}(\mu_{2},\mu_{2}^{\prime})=(\mu,\mu^{\prime})$. Then
$\mu=\mu_{1}\boxplus\mu_{2}$ is the usual free additive convolution of
$\mu_{1}$ and $\mu_{2}$ , and $\mu^{\prime}$ is the infinitesimal free
additive convolution of $(\mu_{1},\mu_{1}^{\prime})$ and
$(\mu_{2},\mu_{2}^{\prime})$.

Suppose that $x_{1}$ and $x_{2}$ are self-adjoint elements in a C∗-probability
space that are infinitesimally free. Assume
$x_{2}\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\delta_{0},\nu)$, and we
let $\gamma^{\prime}$ be the infinitesimal law of $p=x_{1}x_{2}+x_{2}x_{1}$.

######  Corollary 3.7.

$\gamma^{\prime}(E)=\nu(\alpha E)+\nu(\beta E)$ for each Borel set $E$ where
$\alpha=m_{1}(x_{1})+\sqrt{m_{2}(x_{1})}$ and
$\beta=m_{1}(x_{1})-\sqrt{m_{2}(x_{1})}$. Equivalently, $g_{p}=g_{\alpha
x_{2}}+g_{\beta x_{2}}$ or $r_{p}=r_{\alpha x_{2}}+r_{\beta x_{2}}$.

###### Proof.

Recall that $x_{2}\in\mathcal{A}$ in an infinitesimal operator implies that
the infinitesimal $r$-transform and the infinitesimal Cauchy transform of
$x_{2}$ are related by the simple equation
$r_{x_{2}}(z)=z^{-2}g_{x_{2}}(z^{-1})$ (see Equation (2.1)). Then the
remaining proof is immediate from Proposition 3.4 and Theorem 3.6.

∎

###  3.2. The Commutator Case

Next we consider the commutator $q=i(x_{1}x_{2}-x_{2}x_{1})$ and compute its
infinitesimal law, where we assume that $x_{1}$ and $x_{2}$ are
infinitesimally free in an infinitesimal probability space
$(\mathcal{A},\varphi,\varphi^{\prime})$ and $x_{2}$ is an infinitesimal
operator.

Our theorem asserts that $q=i(x_{1}x_{2}-x_{2}x_{1})$ has the distribution of
the sum of infinitesimally free copies of $\omega x_{2}$ and $-\omega x_{2}$
with $\omega=\sqrt{\kappa_{2}(x_{1})}$. To deal with the minus signs in the
commutator we need a technical lemma; a proof of which is presented for the
reader’s convenience.

######  Lemma 3.8.

For $1\leq r\leq n$ with $n$ is even, we have

(3.2) $\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=(-1)^{[r/2]}\binom{l-1}{[\frac{r-1}{2}]}$

where $l=n/2$ we interpret the left hand side as $1$ when $r=1$.

###### Proof.

To find the sum (3.2) we let $u=k_{2}+k_{4}+\cdots+k_{2[r/2]}$ and condition
on $u$. We first consider the case $r=2s$ is even.

$\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=\sum_{u=s}^{n-s}(-1)^{u}\binom{u-1}{s-1}\binom{n-u-1}{s-1}$

because $\displaystyle\binom{n-u-1}{s-1}\binom{u-1}{s-1}$ is the cardinality
of

$\\{(k_{1},\dots,k_{r})\mid k_{1},\dots,k_{r}\geq
1,k_{1}+\cdots+k_{r-1}=n-u,k_{2}+\cdots+k_{r}=u\\}.$

Observe that

$(1-x)^{-s}(1+x)^{-s}=\Big{[}\sum\limits_{k=0}^{\infty}\tbinom{s+k-1}{s-1}x^{k}\Big{]}\cdot\Big{[}\sum\limits_{m=0}^{\infty}\tbinom{s+m-1}{s-1}(-x)^{m}\Big{]}$

If we let $k=u-s$ and $m=n-u-s$, then $s+k-1=u-1$ and $s+m-1=n-u-1$. Then the
coefficient of $x^{n-2s}$ in $(1-x)^{-s}(1+x)^{-s}$ is

$\mathop{\sum_{k,m\geq
0}}_{k+m=2n-s}(-1)^{m}\tbinom{s+k-1}{s-1}\tbinom{s+m-1}{s-1}=\sum_{u=s}^{n-s}(-1)^{u+s}\binom{u-1}{s-1}\binom{n-u-1}{s-1}$

On the other hand,
$\displaystyle(1-x)^{-s}(1+x)^{-s}=(1-x^{2})^{-s}=\sum\limits_{j=0}^{\infty}\tbinom{s+j-1}{s-1}x^{2j}.$
Note that $2j=n-2s$ implies that $j=\frac{n}{2}-s$ (recall that $n$ is assumed
to be even). Then the coefficient of $x^{n-2s}$ of $(1-x)^{-s}(1+x)^{-s}$ is
$\tbinom{s+(n/2-s)-1}{s-1}=\tbinom{n/2-1}{s-1}.$ Thus, we conclude that (still
assuming $r=2s$)

$\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=(-1)^{s}\binom{l-1}{s-1}=(-1)^{[r/2]}\binom{l-1}{[\frac{r-1}{2}]}$

Finally assume that $r=2s+1$ is odd.

$\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=\sum_{u=s}^{n-(s+1)}(-1)^{u}\binom{u-1}{s-1}\binom{n-u-1}{s}$

because $\displaystyle\binom{n-u-1}{(s+1)-1}\binom{u-1}{s-1}$ is the
cardinality of

$\\{(k_{1},\dots,k_{r})\mid k_{1},\dots,k_{r}\geq
1,k_{1}+\cdots+k_{r}=n-u,k_{2}+\cdots+k_{r-1}=u\\}.$

Observe that

$(1-x)^{-s}(1+x)^{-(s+1)}=\Big{[}\sum\limits_{k=0}^{\infty}\tbinom{s+k-1}{s-1}x^{k}\Big{]}\cdot\Big{[}\sum\limits_{m=0}^{\infty}\tbinom{s+m}{s}(-x)^{l}\Big{]}$

If we let $k=u-s$ and $m=n-u-s-1$, then $s+k-1=u-1$ and $s+m=n-u-1$. Then the
coefficient of $x^{n-2s-1}$ in $(1-x)^{-s}(1=s)^{-(s+1)}$ is

$\mathop{\sum_{k,l\geq
0}}_{k+m=n-2s-1}(-1)^{l}\tbinom{s+k-1}{s-1}\tbinom{s+m}{s}=\sum_{u=s}^{n-(s+1)}(-1)^{u+s+1}\binom{u-1}{s-1}\binom{n-u-1}{s}.$

On the other hand,
$\displaystyle(1-x)^{-s}(1+x)^{-(s+1)}=(1-x)(1-x^{2})^{-(s+1)}=\sum\limits_{j=0}^{\infty}\tbinom{s+j}{s}x^{2j}-\sum\limits_{j=0}^{\infty}\tbinom{s+j}{s}x^{2j+1}.$
Since $n-2s-1$ is odd, so the coefficient of $x^{n-2s-1}$ in
$(1-x)^{-s}(1+x)^{-(s+1)}$ is $-\tbinom{s+j}{s}$ when $n-2s-1=2j+1$, i.e.
$j=n/2-(s+1)$. Thus

$\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=(-1)^{s}\binom{s+n/2-(s+1)}{s}=(-1)^{s}\binom{l-1}{s}=(-1)^{[r/2]}\binom{l-1}{[\frac{r-1}{2}]}.$

∎

######  Theorem 3.9.

Suppose $x_{1},x_{2}\in(\mathcal{A},\varphi,\varphi^{\prime})$ are
infinitesimally free with $x_{2}$ an infinitesimal operator. Let
$q=i(x_{1}x_{2}-x_{2}x_{1})$ and $\omega=\sqrt{\kappa_{2}(x_{1})}$. Then the
infinitesimal moments of $q$ are given by

(3.3)
$\kappa_{n}^{\prime}(q)=m_{n}^{\prime}(q)=\left.\begin{cases}2\kappa_{n}^{\prime}(\omega
x_{2})&\text{ if }n\text{ is even}\\\ 0&\text{ if }n\text{ is
odd}\end{cases}\right\\}=\kappa_{n}^{\prime}(\omega
x_{2})+\kappa_{n}^{\prime}(-\omega x_{2}).$

Moreover, $g_{q}=g_{\omega x_{2}}+g_{-\omega x_{2}}$ and $r_{q}=r_{\omega
x_{2}}+r_{-\omega x_{2}}.$

###### Proof.

Writing $-iq=x_{1}x_{2}-x_{2}x_{1}$ we have

$\displaystyle\varphi^{\prime}((-iq)^{n})$
$\displaystyle=\kern-10.00002pt\sum_{j_{1},\dots,j_{2n}=1}^{2}\kern-5.0pt(-1)^{\psi(j)}\varphi^{\prime}(x_{j_{1}}x_{j_{2}}\cdots
x_{j_{2n-1}}x_{j_{2n}})$ (3.4) $\displaystyle=\mbox{}$
$\displaystyle\kern-5.0pt\sum_{\pi\in
NC(2n)}\sum_{j_{1},\dots,j_{2n}=1}^{2}(-1)^{\psi(j)}\partial\kappa_{\pi}(x_{j_{1}},x_{j_{2}},\dots,x_{j_{2n-1}},x_{j_{2n}})$

where the sum over $j_{1},\dots,j_{2n}$ only includes those tuples such that
for each $1\leq l\leq n$ we have $(j_{2l-1},j_{2l})$ is either $(1,2)$; or
$(2,1)$ and $\psi(j)$ is the number of times that $(j_{2l-1},j_{2l})=(2,1)$.
Then by our assumption on $x_{1}$ and $x_{2}$,
$\partial\kappa_{\pi}(x_{j_{1}},\cdots,x_{j_{2n}})=0$ unless

  * $\scriptscriptstyle\bullet$

$\pi\leq\ker(j)$; (by infinitesimal freeness of $x_{1}$ and $x_{2}$)

  * $\scriptscriptstyle\bullet$

$j^{-1}(2)$ is exactly one block of $\pi$ (by our assumption that $x_{2}$ is
infinitesimal).

Then, letting $m_{1}=m_{1}(x_{1})$ and $m_{2}=m_{2}(x_{1})$, we have for such
a tuple that

$\sum_{\begin{subarray}{c}\pi\leq\ker(j)\end{subarray}}\partial\kappa_{\pi}(x_{j_{1}},\cdots,x_{j_{2n}})=\kappa^{\prime}_{n}(x_{2})m_{1}^{n-2v}(\sqrt{m_{2}})^{2v}$

where $0\leq v\leq[n/2]$ counts the number of times in
$j=(j_{1},j_{2},\dots,j_{2n})$ we have two $``1"$s cyclically adjacent. Now
for a given $0\leq v\leq[n/2]$ the set of such tuples is parameterized as
follows. We parameterize such tuples by a string $l_{1}\cdots l_{n}$ where
$l_{k}=f$ ($f$ for first) when $(j_{2k-1},j_{2k})=(1,2)$ and $l_{k}=s$ ($s$
for second) when $(j_{2k-1},j_{2k})=(2,1)$. Then the possible strings are
$f^{k_{1}}s^{k_{2}}\cdots f^{k_{r}}$ and $s^{k_{1}}f^{k_{2}}\cdots s^{k_{r}}$
(when $r$ is odd) or $f^{k_{1}}s^{k_{2}}\cdots f^{k_{r-1}}s^{k_{r}}$ and
$s^{k_{1}}f^{k_{2}}\cdots s^{k_{r-1}}f^{k_{r}}$ (when $r$ is even), where
$k_{1},\dots,k_{r}\geq 1$ and $k_{1}+\cdots+k_{r}=n$. The number of $r$-tuples
$(k_{1},\dots,k_{r})$ is $\tbinom{n-1}{r-1}$. In each of these cases
(depending on the parity of $r$), $v$ is the same; indeed with this
parametrization $v$ will be either $r/2$, when $r$ is even and $(r-1)/2$ when
$r$ is odd, i.e. $v=[r/2]$. Thus replacing $f$ by $s$ and $s$ by $f$ does not
change the contribution, but this switch is equivalent to replacing $q$ by
$-q$. Hence $\varphi^{\prime}(q^{n})=-\varphi^{\prime}(q^{n})$ for $n$ odd.
Thus the odd moments are $0$ and from now on we shall assume that $n$ is even.

Since $n$ is even we have
$(-1)^{k_{1}+k_{3}+\cdots+k_{r-1}}=(-1)^{k_{2}+k_{4}+\cdots+k_{r}}$ (when $r$
is even) and $(-1)^{k_{1}+\cdots+k_{r}}=(-1)^{k_{2}+\cdots+k_{r-1}}$ (when $r$
is odd). Hence $(-1)^{\psi(j)}$ is unchanged when we replace $f$ by $s$. Thus
we have only to count strings that start with $f$; as long as we multiply by 2
at the end.

Now assuming $n=2l$ is even, we rewrite the sum (3.2), conditioning on $r$:

$\varphi^{\prime}((-iq)^{n})=2\kappa_{n}^{\prime}(x_{2})\sum_{r=1}^{n}m_{1}^{n-2[r/2]}(\sqrt{m_{2}})^{2[r/2]}\kern-10.00002pt\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}\kern-10.00002pt(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}},$

where we set $\kern-10.00002pt\displaystyle\mathop{\sum_{k_{1},\dots,k_{r}\geq
1}}_{k_{1}+\cdots+k_{r}=n}\kern-10.00002pt(-1)^{k_{2}+k_{4}+\cdots+k_{2[r/2]}}=1$
when $r=1$. By Lemma 3.8 we have that

$\varphi^{\prime}((-iq)^{n})=2\kappa_{n}^{\prime}(x_{2})\sum_{s=0}^{l}\binom{l}{s}(m_{1}^{2})^{l-s}(-m_{2})^{s}=2\kappa_{n}^{\prime}(x_{2})(m_{1}^{2}-m_{2})^{l}=2\kappa_{n}^{\prime}(\omega
x_{2})$

because $\binom{l-1}{[\frac{r-1}{2}]}+\binom{l-1}{[\frac{r}{2}]}=\binom{l}{s}$
when $r=2s$. ∎

######  Remark 3.10.

In the anticommutator case with
$x_{2}\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\delta_{0},\nu)$ for $\nu$
is a signed measure on $\mathbb{R}$, we have, by Stieltjes inversion,
$p\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\delta_{0},\gamma_{p})$ where
$\gamma_{p}(E)=\nu(\alpha E)+\nu(\beta E)$ for each Borel set
$E\subseteq\mathbb{R}$ and $\alpha=m_{1}(x_{1})+\sqrt{m_{2}(x_{1})}$ and
$\beta=m_{1}(x_{1})-\sqrt{m_{2}(x_{1})}$. In the commutator case we have
$q\stackrel{{\scriptstyle\mathcal{D}}}{{\sim}}(\delta_{0},\gamma_{q})$ where
$\gamma_{q}(E)=\nu(\omega E)+\nu(-\omega E)$ for each Borel set
$E\subseteq\mathbb{R}$ and $\omega=\sqrt{\kappa_{2}(x_{1})}$. Moreover the
distribution of $q$ depends only on $\nu$ and the second moment of $x_{1}$ as
centering $x_{1}$ does not change $q$. In addition we see that the
infinitesimal distribution of both $p$ and $q$ is the distribution of a sum of
infinitesimally freely independent copies of $x_{2}$, suitably rescaled.

######  Remark 3.11.

Suppose $x=b_{1}a_{1}b_{1}^{*}+\cdots+b_{k}a_{k}b_{k}^{*}$ with
$a_{1},\dots,a_{k}$ infinitesimal operators in
$(\mathcal{A},\varphi,\varphi^{\prime})$ infinitesimally free from
$\\{b_{1},\dots,b_{k}\\}$. Let $A=\textrm{diag}(a_{1},\dots,a_{k})$,
$B=(\beta_{ij})_{i,j=1}^{k}$ with $\beta_{ij}=\varphi(b_{i}b_{j}^{*})$, and
$X=\sqrt{B}A\sqrt{B}\in M_{k}(\mathcal{A})$. Then

$\varphi^{\prime}(x^{n})=\mathrm{Tr}(K^{\prime}_{n}(AB,\dots,AB))=\mathrm{Tr}(K^{\prime}_{n}(\sqrt{B}A\sqrt{B},\dots,\sqrt{B}A\sqrt{B}))=\mathrm{Tr}\otimes\varphi^{\prime}(X^{n}),$

where $K^{\prime}_{n}(AB,\dots,AB)$ is the $n^{th}$ $M_{k}(\mathbb{C})$-valued
cumulant of $AB$ and the last equality follows from Proposition 3.4. Hence $x$
and $X$ have the same distribution.

######  Remark 3.12.

Note that the infinitesimal law of anti-commutators and commutators can also
be computed via the linearization trick. Recall that for a given a polynomial
$p\in\mathbb{C}\langle x_{1},\dots,x_{n}\rangle$ in non-commuting variables
$x_{1},\dots,x_{n}$, a matrix $L_{p}\in M_{m}(\mathbb{C}\langle
x_{1},\dots,x_{n}\rangle)$, is called a _linearization of $p$_ if there are
$b_{0},\dots,b_{n}\in M_{m}(\mathbb{C})$, $u\in M_{m-1\times
1}(\mathbb{C}\langle x_{1},\dots,x_{n}\rangle)$, $v\in M_{1\times
m-1}(\mathbb{C}\langle x_{1},\dots,x_{n}\rangle)$ and $Q\in
M_{m-1}(\mathbb{C}\langle x_{1},\allowbreak\dots,x_{n}\rangle)$ such that

$L_{p}=\begin{bmatrix}0&u\\\ v&Q\end{bmatrix}=b_{0}\otimes 1+b_{1}\otimes
x_{1}+\cdots+b_{n}\otimes x_{n},\mbox{ with }Q\mbox{ invertible, and
}p=-uQ^{-1}v.$

A linearization can always be found, see [32, Ch. 9] and [8]. Given a
linearization, let

$X=\begin{bmatrix}0&-uQ^{-1}\\\ v&0\end{bmatrix};\mbox{ then
}X^{2}=\begin{bmatrix}-uQ^{-1}v&0\\\ 0-vuQ^{-1}\end{bmatrix}.$

This leads to the following result:

(3.5) $p^{n}=[X^{2n}]_{1,1}\textrm{, thus
}\varphi^{\prime}(p^{n})=\left[E^{\prime}(X^{2n})\right]_{1,1}\text{ for all
}n\in\mathbb{N}.$

where $[A]_{1,1}$ denotes the $(1,1)$-entry of the matrix $A$ and
$E^{\prime}=\varphi^{\prime}\otimes Id$. In the case of the commutator and
anticommutator the matrices $b_{0}$, $b_{1}$, and $b_{2}$ are very simple; so
one could obtain a proof of Theorems 3.6 and 3.9 in this way. However we found
the proof presented here simpler in these special cases.

###  3.3. Random Matrix Examples

Note that there are now many results which show that finite rank matrices are
asymptotically infinitesimally free from standard random matrix models, see Au
[4, 5], Dallaporta and Février [16], Collins, Hasebe, and Sakuma [13],
Collins, Leid, and Sakuma [14], and Shlyakhtenko [38]. This gives the results
of this section a wide application.

In [13, Thm. 5.1] Collins, Hasebe and Sakuma considered the limit spectrum of
the commutator and anticommutator:
$i(A_{N}U_{N}B_{N}U_{N}^{*}-U_{N}B_{N}U_{N}^{*}A_{N})$ and
$A_{N}U_{N}B_{N}U_{N}^{*}+U_{N}B_{N}U_{N}^{*}A_{N}$ respectively. Now we show
that their results are a consequence of our Theorem 3.6 and Theorem 3.9. If
$a$ is infinitesimal and in addition infinitesimally free from $b$ then as
observed in Remark 3.10, when $\alpha=m_{1}(b)+\sqrt{m_{2}(b)}$,
$\beta=m_{1}(b)-\sqrt{m_{2}(b)}$ and $\omega=\sqrt{\kappa_{2}(b)}$, we have

$g_{ab+ba}(z)=g_{\alpha a}(z)+g_{\beta a}(z)\mathrm{\ and\ }g_{i(ab-
ba)}(z)=g_{\omega a}(z)+g_{-\omega a}(z),\mbox{\ and\ }$ (3.6)
$r_{ab+ba}(z)=r_{\alpha a}(z)+r_{\beta a}(z)\mathrm{\ and\ }r_{i(ab-
ba)}(z)=r_{\omega a}(z)+r_{-\omega a}(z).$

The first set of equations gives us the spectral results of [13] and the
second equation gives us that the distribution is the free additive
convolution of two free copies of $a$, after rescaling.

##  4\. Infinitesimal $R$-diagonal Operators

In this Section, we extend $R$-diagonal properties (see Subsection 2.2) to the
infinitesimal setting in the obvious way.

Recall that $(\mathcal{A},\varphi,\varphi^{\prime})$ is a infinitesimal
$*$-probability space if $(\mathcal{A},\varphi)$ is a $*$-probability space
with $\varphi^{\prime}$ is self-adjoint in the sense that
$\varphi^{\prime}(a^{*})=\overline{\varphi^{\prime}(a)}$ for all
$a\in\mathcal{A}.$

######  Definition 4.1.

Let $(\mathcal{A},\varphi,\varphi^{\prime})$ be a infinitesimal
$*$-probability space. We say $a\in\mathcal{A}$ is _infinitesimal
$R$-diagonal_ if it is $R$-diagonal and in addition for all
$\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$, we have that

$\kappa_{n}^{\prime}(a^{(\varepsilon_{1})},\dots,a^{(\varepsilon_{n})})=0$

unless $n$ is even and $\varepsilon_{i}+\varepsilon_{i+1}=0$ for $1\leq i<n$.

We say that $a\in\mathcal{A}$ is infinitesimally even if $a$ is even and
$\varphi^{\prime}(a^{2n+1})=0$ for all $n\in\mathbb{N}$.

######  Proposition 4.2.

Suppose $(\mathcal{A},\varphi,\varphi^{\prime})$ is a $*$-infinitesimal
probability space, and $x_{1}=x_{1}^{*}$ and $x_{2}=x_{2}^{*}$ are
infinitesimally free elements in $\mathcal{A}$ such that $x_{2}$ is an
infinitesimal operator. We further assume that $m_{1}(x_{1})=0$, i.e. $x_{1}$
is centered. Then $c=x_{1}x_{2}$ is infinitesimally $R$-diagonal, and

$\kappa_{n}^{\prime}(c,c^{*},c,c^{*},\dots,c,c^{*})=\kappa_{n}^{\prime}(c^{*},c,c^{*},c,\dots,c^{*},c)=\kappa_{n}^{\prime}(x_{2})\kappa_{2}(x_{1})^{n/2}.$

###### Proof.

We first observe that the $*$-distribution of $c$ is 0, by this we mean that
all $*$-moments $\varphi(c^{(\varepsilon_{1})}\cdots c^{(\varepsilon_{n})})$
equal $0$. Indeed, all cumulants of $x_{2}$ are 0, and every $*$-cumulant of
$c$ can be written as a linear combination of products of cumulants of $x_{1}$
and cumulants of $x_{2}$ using the formula for cumulants with products as
entries, c.f. [34, Thm. 11.12], and the vanishing of mixed cumulants, c.f.
[34, Thm. 11.16]. Thus all $*$-cumulants of $c$ are 0, and in particular $c$
is $R$-diagonal.

Now suppose that $\varepsilon_{1},\dots,\varepsilon_{k}\in\\{-1,1\\}$; we must
now show that
$\kappa_{n}^{\prime}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{k})})=0$
unless $k$ is even and $\varepsilon_{i}+\varepsilon_{i+1}=0$ for $1\leq i<k$.
For the purpose of this part of the proof let us set $(j_{2l-1},j_{2l})=(1,2)$
if $\varepsilon_{l}=1$ and $(j_{2l-1},j_{2l})=(2,1)$ if $\varepsilon_{l}=-1$.
Let $\delta=\\{(1,2),\dots,(2k-1,2k)\\}$. Again, by the formula for cumulants
with products as entries (this time in the infinitesimal case, see ([24, Prop.
3.15]), we have

$\kappa_{n}^{\prime}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})=\mathop{\sum_{\pi\in
NC(2k)}}_{\pi\vee\delta=1_{2k}}\partial\kappa_{\pi}(x_{j_{1}},x_{j_{2}},\dots,x_{j_{2k-1}},x_{j_{2k}})$

After the discussion following Equation (3.1), we observe that
$\partial\kappa_{\pi}(x_{j_{1}},x_{j_{2}},\dots,x_{j_{2k-1}},x_{j_{2k}})=0$
unless $j^{-1}(2)$ is a block of $\pi$ of size $k$ and all remaining blocks of
$\pi$ are either a pair connecting one $x_{1}$ to another $x_{1}$, or a
singleton with a single $x_{1}$. Since we have assumed that $m_{1}(x_{1})=0$,
all the $x_{1}$-blocks of $\pi$ must have size $2$, or else the contribution
vanishes. This can only happen if each $c$ is followed cyclically by a
$c^{*}$, and hence $k$ must be even and the $\varepsilon$’s must alternate.
This shows that $c$ is infinitesimally $R$-diagonal. Moreover, there is only
one such $\pi$ and for this $\pi$ we have
$\partial\kappa_{\pi}(x_{j_{1}},x_{j_{2}},\dots,x_{j_{2k-1}},x_{j_{2k}})=\kappa_{k}^{\prime}(x_{2})\kappa_{2}(x_{1})^{k/2}$.
Hence

$\kappa_{n}^{\prime}(c,c^{*},c,c^{*},\dots,c,c^{*})=\kappa_{n}^{\prime}(c^{*},c,c^{*},c,\dots,c^{*},c)=\kappa_{n}^{\prime}(x_{2})\kappa_{2}(x_{1})^{n/2}$

as claimed. ∎

Recall that if $a$ is an $R$-diagonal element in a $*$-probability space
$(\mathcal{A},\varphi)$, then the $*$-distribution of $a^{*}a$ and $aa^{*}$
can be uniquely determined by $(\alpha_{n})_{n}$ and $(\beta_{n})_{n}$ where

$\alpha_{n}=\kappa_{2n}(a,a^{*},a,a^{*},\dots,a,a^{*})\text{ and
}\beta_{n}=\kappa_{2n}(a^{*},a,a^{*},a\cdots,a^{*},a).$

Here we will generalize this idea to the infinitesimal free framework. For a
given element $a\in\mathcal{A}$, we denote

$\alpha^{\prime}_{n}=\kappa_{2n}^{\prime}(a,a^{*},a,a^{*},\cdots,a,a^{*})\text{
and
}\beta^{\prime}_{2n}=\kappa^{\prime}_{2n}(a^{*},a,a^{*},a,\cdots,a^{*},a).$

Let us first observe that we have the following Lemma immediately from [24,
Prop. 3.15].

######  Lemma 4.3.

Fix $n\in\mathbb{N}$, let $\sigma=\\{(1,2),(3,4),\dots,(2n-1,2n)\\}$. Then we
have

$\kappa_{n}^{\prime}(aa^{*},\dots,aa^{*})=\sum_{\begin{subarray}{c}\pi\in
NC(2n)\\\
\pi\vee\sigma=1_{2n}\end{subarray}}\partial\kappa_{\pi}(a,a^{*},\dots,a,a^{*})$

and

$\kappa_{n}^{\prime}(a^{*}a,\dots,a^{*}a)=\sum_{\begin{subarray}{c}\pi\in
NC(2n)\\\
\pi\vee\sigma=1_{2n}\end{subarray}}\partial\kappa_{\pi}(a^{*},a,\dots,a^{*},a).$

######  Proposition 4.4.

Let $a$ be an infinitesimal $R$-diagonal element. Then we have

(4.1)
$\kappa^{\prime}_{n}(aa^{*},\dots,aa^{*})=\sum_{\begin{subarray}{c}\pi\in
NC(n)\\\
\pi=\\{V_{1},\dots,V_{r}\\}\end{subarray}}\left(\alpha^{\prime}_{|V_{1}|}\prod_{j=2}^{r}\beta_{|V_{j}|}+\alpha_{|V_{1}|}\sum_{j=2}^{r}\beta_{|V_{j}|}^{\prime}\prod_{i=2,i\neq
j}^{r}\beta_{|V_{i}|}\right)$

and

(4.2)
$\kappa_{n}^{\prime}(a^{*}a,\dots,a^{*}a)=\sum_{\begin{subarray}{c}\pi\in
NC(n)\\\
\pi=\\{V_{1},\dots,V_{r}\\}\end{subarray}}\left(\beta^{\prime}_{|V_{1}|}\prod_{j=2}^{r}\alpha_{|V_{j}|}+\beta_{|V_{1}|}\sum_{j=2}^{r}\alpha_{|V_{j}|}^{\prime}\prod_{i=2,i\neq
j}^{r}\alpha_{|V_{i}|}\right).$

where $V_{1}$ is the block of $\pi$ which contains the first element $1$.

###### Proof.

Following Lemma 4.3, we have

$\kappa_{n}^{\prime}(aa^{*},\dots,aa^{*})=\sum_{\begin{subarray}{c}\pi\in
NC(2n)\\\
\pi\vee\sigma=1_{2n}\end{subarray}}\partial\kappa_{\pi}(a,a^{*},\dots,a,a^{*}).$

where $\sigma=\\{(1,2),\dots,(2n-1,2n)\\}$. We note that if $\pi\in NC(2n)$
such that $\pi\vee\sigma=1_{2n}$, then $\pi$ is canonical bijection with
$NC(n)$ (see [34, Prop. 11.25]). Thus (4.1) can be deduced immediately from
the fact that

$\partial\kappa_{\pi}(a,a^{*},\dots,a,a^{*})=\alpha^{\prime}_{|V_{1}|}\prod_{j=2}^{r}\beta_{|V_{j}|}+\alpha_{|V_{1}|}\sum_{j=2}^{r}\beta_{|V_{j}|}^{\prime}\prod_{i=2,i\neq
j}^{r}\beta_{|V_{i}|}.$

The proof of (4.2) is the same as above. ∎

By applying Lemma 4.3 and combining the vanishing of mixed infinitesimal free
cumulants property with the analogous argument in the proof of [34, Prop.
15.8], we have the following result.

######  Proposition 4.5.

Let $a$ and $b$ are elements in a infinitesimal $*$-probability space that $a$
and $b$ are $*$-infinitesimally free. Suppose that $a$ is infinitesimal
$R$-diagonal, then $ab$ is also infinitesimal $R$-diagonal.

##  5\. Infinitesimal Idempotents and Boolean Cumulants

In this section we consider the special case when our infinitesimal operator
is an idempotent. We get the surprising result that Boolean cumulants can be
expressed as infinitesimal moments. Theorem 5.6 means that any random matrix
model that produces asymptotic infinitesimal freeness (see [38] and [16]) will
automatically produce asymptotic Boolean independence. Theorem 5.7 will
accomplish the same for monotone independence. In particular, this allows us
to recover the asymptotic Boolean independence results of Male [30, Cor. 5.12]
and Lenczewski [27, Thm. 7.1]. Note that there are now many results which show
that finite rank matrices are asymptotically infinitesimally free from
standard random matrix models, see Au [4], Dallaporta and Février [16],
Collins, Hasebe, and Sakuma [13], Collins, Leid, and Sakuma [14], and
Shlyakhtenko [38]. This gives the results of this section a wide application.
An example is provided in [7, §5].

######  Notation 5.1.

Suppose that we have an infinitesimal probability space
$(\mathcal{A},\varphi,\varphi^{\prime})$ and $j\in\mathcal{A}$, an
infinitesimal idempotent with $\varphi^{\prime}(j)\not=0$. Suppose further we
have a unital subalgebra $\mathcal{B}\subseteq\mathcal{A}$ such that $\\{j\\}$
and $\mathcal{B}$ are infinitesimally free with respect to
$(\varphi,\varphi^{\prime})$. We let
$\psi(a)=\varphi^{\prime}(aj)/\varphi^{\prime}(j)$. Then $\psi(1)=1$ and we
may consider the new non-commutative probability space $(\mathcal{A},\psi)$.
Note that for $a\in\mathcal{B}$ we have

$\psi(a)=\varphi^{\prime}(aj)/\varphi^{\prime}(j)=\varphi^{\prime}(j)\varphi(a)/\varphi^{\prime}(j)=\varphi(a).$

Let $j^{(-1)}=j$ and $j^{(1)}=j^{\perp}:=1-j$. We define
$\mathcal{J}(\mathcal{B})$ to be the algebra generated by

$\\{j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}\cdots
j^{(\varepsilon_{k-1})}a_{k-1}j^{(\varepsilon_{k})}\mid k\geq 0,\
\varepsilon_{1},\dots,\varepsilon_{k}\in\\{-1,1\\},\text{ and
}a_{1},a_{2},\dots,a_{k}\in\mathcal{B}\\},$

and $\mathcal{J}_{a}(\mathcal{B})$ be the subalgebra of
$\mathcal{J}(\mathcal{B})$ where we require
$\varepsilon_{1}\neq\varepsilon_{2}\neq\cdots\neq\varepsilon_{k-1}\neq\varepsilon_{k}$.

######  Notation 5.2.

Given a string
$\varepsilon_{1},\varepsilon_{2},\dots,\varepsilon_{n}\in\\{-1,1\\}$ with
$\varepsilon_{1}=-1$, we let $\sigma_{\varepsilon}\in I(n)$ be the interval
partition where each $l$ such that $\varepsilon_{l}=-1$ is the start of a new
block. For example, suppose $n=9$ and
$\varepsilon_{1}=\varepsilon_{2}=-1,\varepsilon_{3}=\varepsilon_{4}=1,\varepsilon_{5}=\varepsilon_{6}=-1,\varepsilon_{7}=\varepsilon_{8}=\varepsilon_{9}=1$;
then $\sigma_{\varepsilon}=\\{(1),(2,3,4),(5),(6,7,8,9)\\}.$ Given two strings
$(\varepsilon_{1},\dots,\varepsilon_{n})$ and $(\eta_{1},\dots,\eta_{n})$ in
$\\{-1,1\\}^{n}$, we say that $\varepsilon\leq\eta$ if
$\varepsilon_{k}\leq\eta_{k}$ for $k=1,\dots,n$. Then $\varepsilon\leq\eta$
$\Leftrightarrow$ $\sigma_{\varepsilon}\leq\sigma_{\eta}$. The strings
$(\varepsilon_{1},\dots,\varepsilon_{n})$ parametrize $I(n)$.

Next suppose we have a string
$\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$ with $\varepsilon_{1}=1$;
we then create a non-crossing partition
$\sigma_{\varepsilon}\in\mathit{NC}(n)$ in which the blocks are cyclic
intervals. By a cyclic interval we mean one where we consider $n$ and $1$ to
be adjacent (see [3, Figure 5].) To obtain $\sigma_{\varepsilon}$, we start a
new interval at $l$ whenever $\varepsilon_{l}=-1$. If
$|\\{l\mid\varepsilon_{l}=-1\\}|\leq 1$ then $\sigma_{\varepsilon}=1_{n}$,
otherwise we will have $\sigma_{\varepsilon}\not\in I(n)$. So we let
$\mathit{CI}(n)$ (as in [3]) denote the set of the cyclic interval partitions,
i.e. non-crossing partitions which when viewed on the circle are interval
partitions. If we have $a_{1},\dots,a_{8}\in\mathcal{A}$ and
$(\varepsilon_{1},\varepsilon_{2},\varepsilon_{3},\varepsilon_{4},\varepsilon_{5},\varepsilon_{6},\varepsilon_{7},\varepsilon_{8})=(1,-1,-1,-1,1,1,-1,-1)$,
then $\sigma_{\varepsilon}=\big{\\{}(1,8),(2),(3),(4,5,6),(7)\big{\\}}$ with

$\beta_{\sigma_{\varepsilon}}(a_{1},a_{2},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8})=\beta_{2}(a_{1},a_{8})\beta_{1}(a_{2})\beta_{1}(a_{3})\beta_{3}(a_{4},a_{5},a_{6})\beta_{1}(a_{7}).$

This extends the notation of Boolean cumulants mentioned in equation (2.3) in
§2.5. We define the following Boolean-like cumulant using cyclic interval
partitions.

$\tilde{\beta}_{n}(a_{1},\dots,a_{n})\mbox{}:=\mbox{}(n-1)\,\varphi(a_{1}\cdots
a_{n})\,-\kern-5.0pt\sum_{\sigma\in\mathit{CI}(n)}(-1)^{\\#(\sigma)}\varphi_{\sigma}(a_{1},\dots,a_{n}).$

This is very similar to the definition of Boolean cumulants except we allow
cyclic intervals and we attach weight $n$ to the partition $1_{n}$ instead of
$1$, which is what the formula for the other partitions would produce.

######  Proposition 5.3.

Suppose $(\mathcal{A},\varphi,\varphi^{\prime})$ is an infinitesimal
probability space, $j\in\mathcal{A}$ is an infinitesimal idempotent with
$\varphi^{\prime}(j)\not=0$, and $\mathcal{B}\subseteq\mathcal{A}$ is a unital
subalgebra infinitesimally free from $j$. Then for all
$a_{1},\dots,a_{n}\in\mathcal{B}$ we have

  1. ($i$)

$\displaystyle\varphi^{\prime}(ja_{1}ja_{2}j\cdots
ja_{n})=\varphi^{\prime}(j)\varphi(a_{1})\varphi(a_{2})\cdots\varphi(a_{n})=\varphi^{\prime}(ja_{1}ja_{2}j\cdots
ja_{n}j)$

and $\displaystyle\varphi^{\prime}(a_{1}ja_{2}j\cdots
ja_{n})\allowbreak=\varphi^{\prime}(j)\varphi(a_{1}a_{n})\varphi(a_{2})\cdots\varphi(a_{n-1})$.

  2. ($ii$)

for any $\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$ with
$|\\{l\mid\varepsilon_{l}=-1\\}|\geq 1$ let $\sigma_{\varepsilon}$ be the
partition in Notation 5.2, then

$\displaystyle\varphi^{\prime}(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}\cdots
j^{(\varepsilon_{n})}a_{n})=\varphi^{\prime}(j)\beta_{\sigma_{\varepsilon}}(a_{1},\dots,a_{n})$.

  3. ($iii$)

if $\varepsilon_{1}=-1$ then $\sigma_{\varepsilon}\in I(n)$ and
$\displaystyle\varphi^{\prime}(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}\cdots
j^{(\varepsilon_{n})}a_{n}j)=\varphi^{\prime}(j)\beta_{\sigma_{\varepsilon}}(a_{1},\dots,a_{n})$.

  4. ($iv$)

$\displaystyle\varphi^{\prime}(j^{\perp}a_{1}j^{\perp}a_{2}j^{\perp}a_{3}j^{\perp}\cdots
j^{\perp}a_{n})=\varphi^{\prime}(a_{1}\cdots
a_{n})-\varphi^{\prime}(j)\tilde{\beta}_{n}(a_{1},\dots,a_{n}).$

###### Proof.

By the infinitesimal moment cumulant formula we have

$\varphi^{\prime}(ja_{1}ja_{2}j\cdots ja_{n})=\sum_{\pi\in
NC(2n)}\partial\kappa_{\pi}(j,a_{1},j,a_{2},j,\dots,j,a_{n}).$

We have $\partial\kappa_{\pi}(j,a_{1},j,a_{2},j,\dots,j,a_{n})=0$ unless
$\pi=\\{(1,3,\dots,2n-1),(2),(4),\dots,(2n)\\}$, because we cannot have an
$a_{i}$ and a $j$ in the same block (by infinitesimal freeness) and we can
have only one block for the $j$’s (by the infinitesimality of $j$). Thus

$\displaystyle\varphi^{\prime}$ $\displaystyle(ja_{1}ja_{2}j\cdots
ja_{n})=\kappa^{\prime}_{n}(j,\dots,j)\varphi(a_{1})\cdots\varphi(a_{n})=\varphi^{\prime}(j)\varphi(a_{1})\cdots\varphi(a_{n}).$

This proves the first equality in the first part of $($$i$$)$. For the second
equality we must have $\pi=\\{(1,3,\dots,2n+1),(2),(4),\dots,(2n)\\}$.

For the second part of $($$i$$)$ we must have
$\pi=\\{(2,4,\dots,2n-2),(1),(3),\dots,(2n-1)\\}$ in order to get a non-zero
contribution, in which case we get

$\displaystyle\varphi^{\prime}$ $\displaystyle(a_{1}ja_{2}j\cdots
ja_{n})=\kappa^{\prime}_{n-1}(j,\dots,j)\varphi(a_{1}a_{n})\cdots\varphi(a_{n-1})=\varphi^{\prime}(j)\varphi(a_{1}a_{n})\cdots\varphi(a_{n-1}).$

For the proof of $($$ii$$)$ we use the following notation. Let
$\lambda_{-1}=j$ and $\lambda_{1}=1$, then
$j^{\perp}=1-j=\lambda_{1}-\lambda_{-1}$. Then

$\displaystyle\varphi^{\prime}(j^{\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}j^{(\varepsilon_{3})}a_{3}j^{(\varepsilon_{4})}\cdots
j^{(\varepsilon_{n})}a_{n})=\sum_{i_{1},\dots,i_{n}\in\\{-1,1\\}}(-1)^{k(i)}\varphi^{\prime}(\lambda_{i_{1}}a_{1}\lambda_{i_{2}}a_{2}\lambda_{i_{3}}a_{3}\lambda_{i_{4}}\cdots\lambda_{i_{n}}a_{n})$

where in the sum we require that $i_{l}=-1$ whenever $\varepsilon_{l}=-1$ and
$k(i)$ is the number of times $i_{l}=-1$ minus the number of times
$\varepsilon_{l}=-1$.

Given a tuple $(i_{1},i_{2},\dots,i_{n})\in\\{-1,1\\}^{n}$ we create a non-
crossing partition $\sigma_{i}$ by starting a new block at $l$ if $i_{l}=-1$,
as in Notation 5.2. We next separate the proof of $($$ii$$)$ into two parts:
the first is when $\varepsilon_{1}=-1$ and then $\sigma_{\varepsilon}\in
I(n)$; the second is when $\varepsilon_{1}=1$ and then
$\sigma_{\varepsilon}\in\mathit{CI}(n)$.

To start, let us assume that $\varepsilon_{1}=-1$. With this assumption, the
tuples $(i_{1},\dots,i_{n})$ parametrize $I(n)$. The condition that $i_{l}=-1$
whenever $\varepsilon_{l}=-1$ is equivalent to the condition that
$\sigma_{i}\leq\sigma_{\varepsilon}$. So summing over $i_{1},\dots,i_{n}$ is
the same as summing over $\sigma\in I(n)$ with
$\sigma\leq\sigma_{\varepsilon}$. With this notation we have
$(-1)^{k(i)}=(-1)^{\\#(\sigma_{i})-\\#(\sigma_{\varepsilon})}\allowbreak=\mu_{B}(\sigma_{i},\sigma_{\varepsilon})$
where $\mu_{B}$ is the Möbius function of $I(n)$.

Let $\sigma\in I(n)$ be such that $\sigma\leq\sigma_{\varepsilon}$. For each
block $V=\\{l_{1},\dots,l_{m}\\}$ of $\sigma$ let $a_{V}=a_{l_{1}}\cdots
a_{l_{m}}$. Then we have by $(i)$

$\varphi^{\prime}(\lambda_{i_{1}}a_{1}\lambda_{i_{2}}a_{2}\lambda_{i_{3}}a_{3}\lambda_{i_{4}}\cdots\lambda_{i_{n}}a_{n})=\varphi^{\prime}(ja_{V_{1}}ja_{V_{2}}j\cdots
ja_{V_{k}})=\varphi^{\prime}(j)\varphi_{\sigma}(a_{1},\dots,a_{n}).$

Hence

$\varphi^{\prime}(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}j^{(\varepsilon_{3})}a_{3}j^{(\varepsilon_{4})}\cdots
j^{(\varepsilon_{n})}a_{n})=\varphi^{\prime}(j)\mathop{\sum_{\sigma\in
I(n)}}_{\sigma\leq\sigma_{\varepsilon}}\mu_{B}(\sigma,\sigma_{\varepsilon})\varphi_{\sigma}(a_{1},\dots,a_{n})=\varphi^{\prime}(j)\beta_{\sigma_{\varepsilon}}(a_{1},\dots,a_{n}),$

where the last equality follows from [34, Prop. 10.6].

Next, let us assume that $\varepsilon_{1}=1$, then
$\sigma_{\varepsilon}\in\mathit{CI}(n)$. The tuples
$i_{1},\dots,i_{n}\in\\{-1,1\\}$ parametrize $\mathit{CI}(n)$. The condition
that $i_{l}=-1$ whenever $\varepsilon_{l}=-1$ is equivalent to the condition
that $\sigma_{i}\leq\sigma_{\varepsilon}$. So summing over $i_{1},\dots,i_{n}$
is the same as summing over $\sigma\in\mathit{CI}(n)$ with
$\sigma\leq\sigma_{\varepsilon}$. Moreover, as above,
$(-1)^{k(i)}=(-1)^{\\#(\sigma_{i})-\\#(\sigma_{\varepsilon})}$. Let
$\sigma\in\mathit{CI}(n)$ be such that $\sigma\leq\sigma_{\varepsilon}$. For
each block $V=\\{l_{1},\dots,l_{m}\\}$ of $\sigma$ let $a_{V}=a_{l_{1}}\cdots
a_{l_{m}}$. Let us write $V_{1}=V_{1,+}\cup V_{1,-}$ where $V_{1}$ is the
block of $\sigma$ containing $1$ and $n$, $V_{1,+}$ contains 1 and the points
of $V_{1}$ moving clockwise, and $V_{1,-}$ contains $n$ and the points of
$V_{1}$ moving counterclockwise. Then we have by $($$i$$)$

$\varphi^{\prime}(\lambda_{i_{1}}a_{1}\lambda_{i_{2}}a_{2}\lambda_{i_{3}}a_{3}\lambda_{i_{4}}\cdots\lambda_{i_{n}}a_{n})=\varphi^{\prime}(a_{V_{1},+}ja_{V_{2}}j\cdots
ja_{V_{1,-}})=\varphi^{\prime}(j)\varphi_{\sigma}(a_{1},\dots,a_{n}).$

Hence

$\displaystyle\varphi^{\prime}(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}j^{(\varepsilon_{3})}a_{3}j^{(\varepsilon_{4})}$
$\displaystyle\cdots
j^{(\varepsilon_{n})}a_{n})=\varphi^{\prime}(j)\mathop{\sum_{\sigma\in\mathit{CI}(n)}}_{\sigma\leq\sigma_{\varepsilon}}(-1)^{\\#(\sigma_{i})-\\#(\sigma_{\varepsilon})}\varphi_{\sigma}(a_{1},\dots,a_{n})$
$\displaystyle\mbox{}\stackrel{{\scriptstyle(*)}}{{=}}\varphi^{\prime}(j)\beta_{\sigma_{\varepsilon}}(a_{1},\dots,a_{n}).$

To justify the last equality $(*)$, note that each block of
$\sigma_{\varepsilon}$ is an interval (or a cyclic interval), so we can apply
[34, Prop. 10.6] to each interval and use that
$\sigma\mapsto(-1)^{\\#(\sigma)-\\#(\sigma_{\varepsilon})}$ is multiplicative
across intervals in the sense of [34, Def. 10.16].

To prove $($$iii$$)$, note that the first part of the proof of $($$ii$$)$
shows that $\sigma_{\varepsilon}\in I(n)$ and here we have by the same
argument that

$\varphi^{\prime}(\lambda_{i_{1}}a_{1}\lambda_{i_{2}}a_{2}\lambda_{i_{3}}a_{3}\lambda_{i_{4}}\cdots\lambda_{i_{n}}a_{n}j)=\varphi^{\prime}(j)\varphi_{\sigma}(a_{1},\dots,a_{n}).$

This proves $($$iii$$)$.

Proof of $($$iv$$)$. As above, we let $k(i)$ be the number of times
$i_{l}=-1$. When $k(i)=0$ we have
$\varphi^{\prime}(\lambda_{i_{1}}a_{1}\cdots\lambda_{i_{n}}a_{n})=\varphi^{\prime}(a_{1}\cdots
a_{n})$ and when $k(i)\geq 1$ we have
$\varphi^{\prime}(\lambda_{i_{1}}a_{1}\cdots\lambda_{i_{n}}a_{n})=\varphi_{\sigma_{i}}(a_{1}\cdots
a_{n})$. Thus when $k(i)=1$ there are $n$ tuples $(i_{1},\dots,i_{n})$ that
all produce $\varphi(a_{1}\cdots a_{n})$, because in these cases
$\sigma_{i}=1_{n}$. Then

$\displaystyle\varphi^{\prime}(j^{\perp}a_{1}\cdots
j^{\perp}a_{n})=\sum_{i_{1},\dots,i_{n}\in\\{-1,1\\}}(-1)^{k(i)}\varphi^{\prime}(\lambda_{i_{1}}a_{1}\cdots\lambda_{i_{n}}a_{n})$
$\displaystyle\mbox{}=\varphi^{\prime}(a_{1}\cdots
a_{n})+\mathop{\sum_{i_{1},\dots,i_{n}\in\\{-1,1\\}}}_{k(i)\geq
1}(-1)^{k(i)}\varphi^{\prime}(\lambda_{i_{1}}a_{1}\cdots\lambda_{i_{n}}a_{n})$
$\displaystyle\mbox{}=\varphi^{\prime}(a_{1}\cdots
a_{n})-n\varphi^{\prime}(j)\varphi(a_{1}\cdots
a_{n})+\mathop{\sum_{i_{1},\dots,i_{n}\in\\{-1,1\\}}}_{k(i)\geq
2}(-1)^{k(i)}\varphi^{\prime}(\lambda_{i_{1}}a_{1}\cdots\lambda_{i_{n}}a_{n})$
$\displaystyle\mbox{}=\varphi^{\prime}(a_{1}\cdots
a_{n})-n\varphi^{\prime}(j)\varphi(a_{1}\cdots
a_{n})+\varphi^{\prime}(j)\mathop{\sum_{\sigma\in\mathit{CI}(n)}}_{\\#(\sigma)\geq
2}(-1)^{\\#(\sigma)}\varphi_{\sigma}(a_{1},\dots,a_{n})$
$\displaystyle\mbox{}=\varphi^{\prime}(a_{1}\cdots
a_{n})-(n-1)\varphi^{\prime}(j)\varphi(a_{1}\cdots
a_{n})+\varphi^{\prime}(j)\mathop{\sum_{\sigma\in\mathit{CI}(n)}}(-1)^{\\#(\sigma)}\varphi_{\sigma}(a_{1},\dots,a_{n})$
$\displaystyle\mbox{}=\varphi^{\prime}(a_{1}\cdots
a_{n})-\varphi^{\prime}(j)\tilde{\beta}_{n}(a_{1},\dots,a_{n}).$

∎

######  Remark 5.4.

We mention here a surprising relation between an spectral shift functions in
operator theory [28], and [26, see p. 138] and Boolean cumulants [2, Prop.
3.2.1], [21, §6.2], and the Markov transform [25, §6], [20], and [1] .

We can form an abstraction of these examples using infinitesimal operators;
indeed let $(\mathcal{A},\varphi,\varphi^{\prime})$ be an infinitesimal
probability space and $a,p\in\mathcal{A}$ infinitesimally free with $p=p^{2}$
an idempotent with $\varphi(p)=1$ and $\varphi^{\prime}(p)=-1$ (i.e.
$p^{\perp}$ is _infinitesimal_). Let
$G_{\tau}=\sum_{n=0}^{\infty}\tilde{m}_{n}z^{-(n+1)}$ be the formal power
series with
$\tilde{m}_{n}=\varphi^{\prime}(a^{n})-\varphi^{\prime}((pap)^{n})$ and
$G_{\mu}(z)=\sum_{n=0}^{\infty}m_{n}z^{-(n+1)}$ be the formal power series
with $m_{n}=\varphi(a^{n})$, then

$G_{\tau}=-\frac{G_{\mu}^{\prime}}{G_{\mu}}.$

Krein [26] found an analytic representation of this relation in the form of
spectral shift functions. Let $A$ be a self-adjoint bounded operator and $J$
be a rank 1 projection on a Hilbert space $\mathcal{H}$. Let
$\xi\in\mathcal{H}$ be such that $\|\xi\|=1$ and $J\xi=\xi$. Let $\mu$ be the
spectral measure of $A$ with respect to the vector state
$\omega_{\xi}(T)=\langle T\xi,\xi\rangle$: for all continuous functions $f$ on
$\mathrm{Sp}(A)$ we have $\int f(t)\,d\mu(t)=\omega_{\xi}(f(A))$. The theorem
of Krein, [26, see p. 138], states that there is a finite signed Borel measure
$\tau$ such that $G_{\tau}=-G_{\mu}^{\prime}/G_{\mu}$ and

$\mathrm{Tr}\big{(}\,f(A)-f(J^{\perp}AJ^{\perp})\,\big{)}=\int
f(t)\,d\tau(t).$

In Kerov’s notation, $\tau$ is a Raleigh measure and $\mu$ is the Markov
transform of $\tau$. This same result is discussed in Kerov [25, §6], where
another proof is given. Another example of a Markov transform is given by
Bufetov in [11, Thm. 2] where $\mu$ is the Marchenko-Pastur law and $\tau$ is
the limiting eigenvalue distribution of a principal minor of a Wishart matrix.

Let us recall that $\psi(a)=\varphi^{\prime}(aj)/\varphi^{\prime}(j)$. Then,
by applying Proposition 5.3, we obtain the following result.

######  Corollary 5.5.

Suppose $(\mathcal{A},\varphi,\varphi^{\prime})$ is an infinitesimal
probability space, $j\in\mathcal{A}$ is an infinitesimal idempotent with
$\varphi^{\prime}(j)\not=0$, and $\mathcal{B}\subseteq\mathcal{A}$ is a unital
subalgebra infinitesimally free from $j$. Then for all
$a_{1},\dots,a_{n}\in\mathcal{B}$, and
$\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$, we have

(5.1) $\psi(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}\cdots
j^{(\varepsilon_{n})}a_{n})=\psi(j^{(\varepsilon_{1})}a_{1}j^{(\varepsilon_{2})}a_{2}\cdots
j^{(\varepsilon_{n})}a_{n}j)=\beta_{\sigma_{\varepsilon}}(a_{1},\dots,a_{n}).$

######  Theorem 5.6.

Suppose $(\mathcal{A}_{i})_{i\in I}$ are unital algebras of $\mathcal{A}$ that
are infinitesimally free from $j$. Then
$(j\mathcal{J}(\mathcal{A}_{i})j)_{i\in I}$ are Boolean independent with
respect to $\psi.$ If we further assume that $(\mathcal{A}_{i})_{i\in I}$ are
free, then we have $(\mathcal{J}_{a}(\mathcal{A}_{i}))_{i\in I}$ are Boolean
independent with respect to $\psi$.

###### Proof.

Following Corollary 5.5, we immediately obtain that
$(j\mathcal{J}(\mathcal{A}_{i})j)_{i\in I}$ are Boolean independent Indeed, if
we let

$x_{k}=ja_{1}^{(k)}j^{(\varepsilon_{2}^{(k)})}\cdots
j^{(\varepsilon_{m_{k}}^{(k)})}a_{m_{k}}^{(k)}j\quad\text{ where
}\quad\varepsilon_{2}^{(k)},\dots,\varepsilon_{m_{k}}^{(k)}=\pm 1,\text{ and
}a_{1}^{(k)},\dots,a_{m_{k}}^{(k)}\in\mathcal{A},$

then by applying Equation 5.1, we can easily to see that

$\psi(x_{1}\cdots x_{n})=\psi(x_{1})\cdots\psi(x_{n})\text{ for each }1\leq
i\leq n.$

Now we assume that $(\mathcal{A}_{i})_{i\in I}$ are free. Let
$x_{1},\dots,x_{n}$ be elements in $\mathcal{J}_{a}(\mathcal{A})$ where
$x_{k}\in\mathcal{J}({A}_{i_{k}})$ with $i_{1}\neq i_{2}\neq\cdots\neq i_{n}$.
To prove Boolean independence it is sufficient, by linearity, to show that

(5.2) $\psi(x_{1}\cdots x_{n})=\psi(x_{1})\cdots\psi(x_{n}).$

Following the definition of $\mathcal{J}_{a}(\mathcal{B})$, we note that for
each $k$,

$x_{k}=j^{(\varepsilon^{(k)})}a_{1}^{(k)}j^{(\varepsilon_{1}^{(k)})}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})}\text{ for some
}a_{1}^{(k)},\dots,a_{m_{k}}^{(k)}\in\mathcal{A}_{i_{k}}\text{ and
}\varepsilon^{(k)}\neq\varepsilon_{1}^{(k)}\neq\cdots\neq\varepsilon_{m_{k}}^{(k)}.$

Firstly, we may assume $\varepsilon^{(k+1)}=\varepsilon_{m_{k}}^{(k)}$ for all
$k$; otherwise both sides of Equation (2.2) are trivially zero. In addition,
if all $\varepsilon^{(k)}=\varepsilon_{m_{k}}^{(k)}=-1$ for all $k$, then
$(\mathcal{J}_{a}(\mathcal{A})_{i_{k}})_{k}$ are Boolean independent by
applying Corollary 5.5 directly. Indeed,

$\psi(x_{1}\cdots
x_{n})=\prod_{k=1}^{n}\prod_{s=1}^{m_{k}/2}\beta_{2}(a_{2s-1}^{(k)},a_{2s}^{(k)})=\psi(x_{1})\cdots\psi(x_{n}).$

Now, we will show that both sides of Equation (5.2) are zero whenever there is
$k$ such that $\varepsilon^{(k)}=1$ or $\varepsilon_{m_{k}}^{(k)}=1$. Without
loss of generality, we assume $\varepsilon^{(k)}=1$. Then we note that

$\displaystyle\psi(x_{1}\cdots x_{n})$ $\displaystyle=$
$\displaystyle\psi(Z_{1}\cdot ja_{m_{k-1}}^{(k-1)}j^{\perp}a_{1}^{(k)}j\cdot
Z_{2})$ $\displaystyle=$
$\displaystyle\beta_{\sigma_{1}}(a_{1}^{(1)},\dots,a_{m_{k-1}-1}^{(k-1)})\cdot\beta_{2}(a_{m_{k-1}}^{(k-1)},a_{1}^{(k)})\cdot\beta_{\sigma_{2}}(a_{2}^{(k)},\dots,a_{m_{n}}^{(n)})$

where $Z_{1}=x_{1}\cdots
x_{k-2}j^{(\varepsilon^{(k-1)})}a_{1}^{(k-1)}j^{(\varepsilon_{1}^{(k-1)})}\cdots
a_{m_{k-1}-1}^{(k-1)}j$ and $Z_{2}=ja_{2}^{(k)}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})}x_{k+1}\cdots x_{n},$ and
$\sigma_{i}$ is the interval partition respect to $Z_{i}$ for $i=1,2$. Since
$a_{m_{k-1}}^{(k-1)}$ and $a_{1}^{(k)}$ are free, we have
$\beta_{2}(a_{m_{k-1}}^{(k-1)},a_{1}^{(k)})=0$, which implies that
$\psi(x_{1}\cdots x_{n})=0$.

On the other hand,

$\displaystyle\psi(x_{k})$ $\displaystyle=$
$\displaystyle\psi(j^{\perp}a_{1}^{(k)}ja_{2}^{(k)}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})})$ $\displaystyle=$
$\displaystyle\psi(a_{1}^{(k)}ja_{2}^{(k)}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})})-\psi(ja_{1}^{(k)}ja_{2}^{(k)}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})})=0$

due to the fact that $\psi(x)=\psi(jx)$ for all $x\in\mathcal{J}(\mathcal{B})$
whenever $\mathcal{B}$ is any algebra infinitesimally free from $j$. Hence, we
conclude that both sides of Equation (5.2) are zero, which complete the proof.
∎

######  Theorem 5.7.

Suppose that $\mathcal{B}$ and $\mathcal{C}$ are unital subalgebras of
$\mathcal{A}$ and that $\\{\mathcal{B},\mathcal{C}\\}$ are infinitesimally
free from $j$. Then $(j\mathcal{J}(\mathcal{B})j,\mathcal{C})$ are monotone
independent. Furthermore, if $\mathcal{B}$ and $\mathcal{C}$ are free, then
$(\mathcal{J}_{a}(\mathcal{B}),\mathcal{C})$ are monotone independent.

###### Proof.

That $(j\mathcal{J}(\mathcal{B})j)\prec\mathcal{C}$ follows directly from
Corollary 5.5 and the following argument. Indeed, if we let

$x_{k}=ja_{1}^{(k)}j^{(\varepsilon_{2}^{(k)})}\cdots
j^{(\varepsilon_{m_{k}}^{(k)})}a_{m_{k}}^{(k)}j\quad\text{ where
}\quad\varepsilon_{2}^{(k)},\dots,\varepsilon_{m_{k}}^{(k)}=\pm 1,\text{ and
}a_{1}^{(k)},\dots,a_{m_{k}}^{(k)}\in\mathcal{A},$

then by applying Equation 5.1, we can easily to see that

$\psi(x_{1}\cdots x_{n})=\psi(x_{1}\cdots x_{i-1}\psi(x_{i})x_{i+1}\cdots
x_{n})\text{ for each }1\leq i\leq n.$

Now assume $\mathcal{B}$ and $\mathcal{C}$ are free, we are going to show
$(\mathcal{J}_{a}(\mathcal{B}),\mathcal{C}$) are monotone independent.

We shall show that for any $k$ we have

(5.3) $\psi(c_{0}x_{1}c_{1}\cdots x_{n}c_{n})=\psi(c_{0}x_{1}c_{1}\cdots
x_{k}\psi(c_{k})x_{k+1}\cdots c_{n-1}x_{n}c_{n}).$

where $c_{0},\dots,c_{n}\in\mathcal{C}$ and
$x_{1},\dots,x_{n}\in\mathcal{J}_{a}(\mathcal{B})$. Again following the
definition of $\mathcal{J}_{a}(\mathcal{B})$, we have

$x_{k}=j^{(\varepsilon^{(k)})}a_{1}^{(k)}j^{(\varepsilon_{1}^{(k)})}\cdots
a_{m_{k}}^{(k)}j^{(\varepsilon_{m_{k}}^{(k)})}\text{ for some
}a_{1}^{(k)},\dots,a_{m_{k}}^{(k)}\in\mathcal{B}\text{ and
}\varepsilon^{(k)}\neq\varepsilon_{1}^{(k)}\neq\cdots\neq\varepsilon_{m_{k}}^{(k)}.$

By linearity and induction on $n$ this will prove monotone independence.

Without lost of generality, we assume
$\varepsilon^{(k+1)}=\varepsilon_{m_{k}}^{(k)}$; otherwise, both sides of
Equation (5.3) are trivially zero. For the case
$\varepsilon^{(k+1)}=\varepsilon_{m_{k}}^{(k)}=-1$,

$\displaystyle\psi(c_{0}x_{1}c_{1}\cdots x_{k}c_{k}x_{k+1}\cdots x_{n}c_{n})$
$\displaystyle=$
$\displaystyle\beta_{\sigma_{1}}(c_{0},a_{1}^{(1)},\dots,a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})})\beta_{1}(c_{k})\beta_{\sigma_{2}}(a_{1}^{(k+1)},\dots,c_{n})$
$\displaystyle=$
$\displaystyle\beta_{\sigma_{1}}(c_{0},a_{1}^{(1)},\dots,a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})})\psi(c_{k})\beta_{\sigma_{2}}(a_{1}^{(k+1)},\dots,c_{n})$
$\displaystyle=$ $\displaystyle\psi(c_{k})\psi(c_{0}x_{1}c_{1}\cdots
x_{k}x_{k+1}\cdots x_{n}c_{n})$

where $\sigma_{1}$ is the interval partition with respect to
$c_{0}x_{1}c_{1}\cdots x_{k}$ and $\sigma_{2}$ is the interval partition with
respect to $x_{k+1}c_{k+1}\cdots x_{n}c_{n}$. Now, let us consider the case
$\varepsilon^{(k+1)}=\varepsilon_{m_{k}}^{(k)}=1$, we have by Corollary 5.5

$\displaystyle\psi(c_{0}x_{1}c_{1}\cdots x_{k}c_{k}x_{k+1}\cdots x_{n}c_{n})$
$\displaystyle=$ $\displaystyle\psi(c_{0}x_{1}c_{1}\cdots
a_{m_{k}-1}^{(\varepsilon_{m_{k}-1}^{(k)})}ja_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})}j^{\perp}c_{k}j^{\perp}a_{1}^{(k+1)}ja_{2}^{(k+1)}\cdots
x_{n}c_{n})$ $\displaystyle=$
$\displaystyle\beta_{\sigma_{1}}(c_{0},x_{1},\dots,a_{m_{k}-1}^{(\varepsilon_{m_{k}-1}^{(k)})})\beta_{3}(a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})},c_{k},a_{1}^{(k+1)})\beta_{\sigma_{2}}(a_{2}^{(k+1)},\dots,c_{n})$

where $\sigma_{1}$ is the interval partition with respect to
$c_{0}x_{1}c_{1}\cdots a_{m_{k}-1}^{(\varepsilon_{m_{k}-1}^{(k)})}j$ and
$\sigma_{2}$ is the interval partition with respect to $ja_{2}^{(k+1)}\cdots
x_{n}c_{n}$. The freeness of $\mathcal{B}$ and $\mathcal{C}$ implies that

$\beta_{3}(a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})},c_{k},a_{1}^{(k+1)})=\beta_{2}(a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})},a_{1}^{(k+1)})\beta_{1}(c_{k})=\beta_{2}(a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})},a_{1}^{(k+1)})\psi(c_{k}).$

Thus,

$\displaystyle\psi(c_{0}x_{1}c_{1}\cdots x_{k}c_{k}x_{k+1}\cdots x_{n}c_{n})$
$\displaystyle=$
$\displaystyle\psi(c_{k})\beta_{\sigma_{1}}(c_{0},x_{1},\dots,a_{m_{k}-1}^{(\varepsilon_{m_{k}-1}^{(k)})})\beta_{2}(a_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})},a_{1}^{(k+1)})\beta_{\sigma_{2}}(a_{2}^{(k+1)},\dots,c_{n})$
$\displaystyle=$ $\displaystyle\psi(c_{0}x_{1}c_{1}\cdots
a_{m_{k}-1}^{(\varepsilon_{m_{k}-1}^{(k)})}ja_{m_{k}}^{(\varepsilon_{m_{k}}^{(k)})}j^{\perp}1j^{\perp}a_{1}^{(k+1)}ja_{2}^{(k+1)}\cdots
x_{n}c_{n})$ $\displaystyle=$
$\displaystyle\psi(c_{k})\psi(c_{0}x_{1}c_{1}\cdots x_{k}x_{k+1}\cdots
x_{n}c_{n}).$

Hence, we show Equation (5.3) holds. ∎

##  6\. The Distributions of Some Polynomials of Boolean Independent Operators

Suppose that $(\mathcal{A},\varphi)$ is a non-commutative probability space
and $x_{1}$ and $x_{2}$ are elements in $\mathcal{A}$ that are Boolean
independent. In this final section, we show how to find the distribution of
$ax_{1}x_{2}+bx_{2}x_{1}$ where $a$ and $b$ are complex numbers. In
particular, we will obtain the distributions of the commutator and
anticommutator of Boolean independent random variables.

First note that by applying [34, Prop. 10.11] and the analogous argument of
[34, Thm. 11.12], we have the products as arguments formula in the Boolean, we
state it as follows:

######  Theorem 6.1.

Suppose $(\mathcal{A},\varphi)$ is a non-commutative probability space. Let
$m,n\in\mathbb{N}$ and $1\leq i_{1}<i_{2}<\cdots<i_{m}=n$ be given, and we set

$\hat{0}_{m}=\\{(1,\dots,i_{1}),\dots,(i_{m-1}+1,\dots,i_{m})\\}\in I(m).$

Then we have

$\beta_{m}(a_{1}\cdots a_{i_{1}},\dots,a_{i_{m-1}+1}\cdots
a_{i_{m}})=\sum_{\begin{subarray}{c}\pi\in I(n)\\\
\pi\vee\hat{0}_{m}=1_{n}\end{subarray}}\beta_{\pi}(a_{1},\dots,a_{n}).$

###  6.1. The linear span Case

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space. Assume
$x_{1}$ and $x_{2}$ are elements in $\mathcal{A}$ that are Boolean independent
and $a,b\in\mathbb{C}$. We let $g=ax_{1}x_{2}+bx_{2}x_{1}$. In this
Subsection, we will provide the distribution of $g$ by computing its Boolean
cumulants and also moments.

Let us define $c=x_{1}x_{2}$, and we set $c^{(1)}=x_{1}x_{2}$ and
$c^{(-1)}=x_{2}x_{1}.$ We first show the following result:

######  Lemma 6.2.

For all $\varepsilon_{1},\dots,\varepsilon_{n}\in\\{-1,1\\}$, we have

$\displaystyle\beta_{n}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})=0$

unless $\varepsilon_{i}+\varepsilon_{i+1}=0$ for $1\leq i<n$. Moreover,

$\displaystyle\beta_{n}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})=\begin{cases}\beta_{2}(x_{2})\beta_{1}(x_{1})^{2}\Big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{(n-2)/2}&\text{
if }(\varepsilon_{1},\dots,\varepsilon_{n})=(1,-1,\dots,1,-1)\\\
\beta_{2}(x_{1})\beta_{1}(x_{2})^{2}\Big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{(n-2)/2}&\text{
if }(\varepsilon_{1},\dots,\varepsilon_{n})=(-1,1,\dots,-1,1)\\\ \end{cases}$

for $n$ is even, and

$\beta_{n}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})=\beta_{1}(x_{1})\beta_{1}(x_{2})\Big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{(n-1)/2}$

for $n$ is odd with
$(\varepsilon_{1},\dots,\varepsilon_{n})=(1,-1,\dots,1,-1,1)$ or
$(-1,1,\dots,-1,1,-1)$.

###### Proof.

Let $\rho$ be the partition $\\{(1,2),(3,4),\dots,(2n-1,2n)\\}$:

$\rho=\mbox{}$$1$$2$$3$$4$$5$$6$$7$$8$$\cdots$$2n-1$$2n$

Next, let $x_{-1}=x_{2}$. Then
$c^{(\varepsilon)}=x_{\varepsilon}x_{-\varepsilon}$. So if $\varepsilon=1$ we
have $c^{(\varepsilon)}=x_{1}x_{2}$ and if $\varepsilon=-1$ we have
$c^{(\varepsilon)}=x_{2}x_{1}$. Then with this notation, Theorem 6.1 becomes

(6.1)
$\beta_{n}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})=\mathop{\sum_{\pi\in
I(2n)}}_{\pi\vee\rho=1_{2n}}\beta_{\pi}(x_{\varepsilon_{1}},x_{-\varepsilon_{1}},x_{\varepsilon_{2}},x_{-\varepsilon_{2}},\dots,x_{\varepsilon_{n}},x_{-\varepsilon_{n}}).$

In order to have

$\beta_{\pi}(x_{\varepsilon_{1}},x_{-\varepsilon_{1}},x_{\varepsilon_{2}},x_{-\varepsilon_{2}},\dots,x_{\varepsilon_{n}},x_{-\varepsilon_{n}})\neq
0$

we must have, by the vanishing of mixed cumulants, that $|V\cap W|\leq 1$, for
each block $V$ of $\pi$ and each block $W$ of $\rho$. On the other hand, in
order have $\pi\vee\rho=1_{2n}$, we must have that $2k+1\sim_{\pi}2k+2$, for
each $k$, since $\pi$ is an interval partition. These two conditions together
imply that for each $k$, $x_{-\varepsilon_{k}}$ and $x_{\varepsilon_{k+1}}$
are the same operator, which in turn means that
$-\varepsilon_{k}=\varepsilon_{k+1}$, as claimed.

Now suppose that $\varepsilon_{i}+\varepsilon_{i+1}=0$ for $1\leq i<n$, let us
compute $\beta_{n}(c^{(\varepsilon_{1})},\dots,c^{(\varepsilon_{n})})$. From
the discussion above, equation (6.1) means that there is only one $\pi$ which
contributes to (6.1); $\pi=\\{(1),(2,3),\dots,(2n-2,2n-1),(2n)\\}$:

$\pi=\mbox{}$$1$$2$$3$$4$$5$$6$$7$$\cdots$$2n-2$$2n-1$$2n$

When $n$ is odd, we must have $n-1$ blocks of size 2, half with $x_{1}$ and
half with $x_{2}$. This contributes
$(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-1)/2}$. Of the two singletons, one is
a $x_{1}$ and the other must be a $x_{2}$. Thus the total contribution is
$\beta_{1}(x_{1})\beta_{1}(x_{2})(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-1)/2}$.

When $n$ is even, we also have $n-1$ blocks of size $2$. If
$\varepsilon_{1}=1$, then we have $n/2$ blocks with $x_{2}$ and the remaining
$(n-2)/2$ are with $x_{1}$. The two singletons are both with $x_{1}$. Thus the
total contribution is
$\beta_{1}(x_{1})^{2}\beta_{2}(x_{1})^{(n-2)/2}\beta_{2}(x_{2})^{n/2}$.

When $n$ is even and $\varepsilon_{1}=-1$, it is the other way around. There
are $n/2$ blocks of size $2$ with $x_{1}$ and $(n-2)/2$ blocks of size $2$
with $x_{1}$, contributing $\beta_{2}(x_{1})^{n/2}\beta_{2}(x_{2})^{(n-2)/2}$.
Finally, there are two blocks of size $1$, one both with $x_{2}$. This proves
the claim. ∎

######  Remark 6.3.

If $(\mathcal{A},\varphi)$ is a $*$-probability space and if we further assume
$x_{1}$ and $x_{2}$ are self-adjoint, then by Lemma 6.2, we have $c$ is
$\eta$-diagonal (for more details on $\eta$-diagonal, see [10, 37]).

######  Theorem 6.4.

Suppose $x_{1}$ and $x_{2}$ are Boolean independent, and
$g=ax_{1}x_{2}+bx_{2}x_{1}$ for $a,b\in\mathbb{C}$, then

$\beta_{n}(g)=\begin{cases}ab(\beta_{2}(x_{1})\beta_{1}(x_{2})^{2}+\beta_{2}(x_{2})\beta_{1}(x_{1})^{2})\Big{(}ab\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{(n-2)/2}&\text{
if }n\text{ is even}\\\
(a+b)\beta_{1}(x_{1})\beta_{1}(x_{2})\Big{(}ab\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{(n-1)/2}&\text{
if }n\text{ is odd}.\end{cases}$

###### Proof.

Suppose $n$ is even, we note that

$\displaystyle\beta_{n}(g)$ $\displaystyle=$
$\displaystyle\beta_{n}(ac^{(1)}+bc^{(-1)})$ $\displaystyle=$
$\displaystyle\beta_{n}(ac^{(1)},bc^{(-1)},\cdots,ac^{(1)},bc^{(-1)})+\beta_{n}(bc^{(-1)},ac^{(1)},\dots,bc^{(-1)},ac^{(1)})+\text{
other terms}$ $\displaystyle=$
$\displaystyle(ab)^{\frac{n}{2}}[\beta_{n}(c^{(1)},c^{(-1)},\dots,c^{(1)},c^{(-1)})+\beta_{n}(c^{(-1)},c^{(1)},\dots,c^{(-1)},c^{(1)})]+\text{other
terms}.$

By Lemma 6.2, we further obtain

$\displaystyle\beta_{n}(g)$ $\displaystyle=$
$\displaystyle(ab)^{\frac{n}{2}}[\beta_{2}(x_{2})\beta_{1}(x_{1})^{2}\Big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{\frac{n-2}{2}}+\beta_{2}(x_{1})\beta_{1}(x_{2})^{2}\Big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{\frac{n-2}{2}}]+0$
$\displaystyle=$ $\displaystyle
ab(\beta_{2}(x_{1})\beta_{1}(x_{2})^{2}+\beta_{2}(x_{2})\beta_{1}(x_{1})^{2})\Big{(}ab\beta_{2}(x_{1})\beta_{2}(x_{2})\Big{)}^{\frac{n-2}{2}}.$

The proof for the case of odd $n$ is similar. ∎

Note that for the anti-commutator $p=x_{1}x_{2}+x_{2}x_{1}$ and the commutator
$q=i(x_{1}x_{2}-x_{2}x_{1})$, we can easily to see that

$\beta_{1}(p)=2\beta_{1}(x_{1})\beta_{1}(x_{2}),\text{ and
}\beta_{2}(p)=\beta_{2}(q)=\beta_{1}(x_{1})^{2}\beta_{2}(x_{2})+\beta_{1}(x_{2})^{2}\beta_{2}(x_{1}).$

Then following Theorem 6.4, we have the distributions of $p$ and $q$
immediately by letting $(a,b)=(1,1)$ and $(a,b)=(i,-i)$ respectively. We state
it as below.

######  Corollary 6.5.

With $p$ and $q$ as above

$\beta_{n}(p)=\begin{cases}\beta_{2}(p)(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-2)/2}&\text{if
}n\text{ is even}\\\
\beta_{1}(p)(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-1)/2}&\text{if }n\text{ is
odd}\end{cases}$

and

$\beta_{n}(q)=\begin{cases}\beta_{2}(q)(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-2)/2}&\text{if
}n\text{ is even}\\\ 0&\text{if }n\text{ is odd}\end{cases}.$

We will provide the formula for the moments of $g=ax_{1}x_{2}+bx_{2}x_{1}$ in
terms of cumulants via Theorem 6.4. Observe that following the moment-cumulant
formula, $m_{n}(g)=\sum_{\pi\in I(n)}\beta_{\pi}(g)$ where

$\beta_{2k}(g)=d^{k-1}\alpha_{2},\text{ and }\beta_{2k+1}(g)=d^{k}\alpha_{1}$

with $d=ab\beta_{2}(x_{1})\beta_{2}(x_{2})$,
$\alpha_{1}=\frac{1}{2}(a+b)\beta_{1}(p)$ and $\alpha_{2}=ab\beta_{2}(p)$. To
simplify our notation, now we will write $m_{n}$ for $m_{n}(g)$ and
$\beta_{n}$ for $\beta_{n}(g)$.

######  Notation 6.6.

Let $\pi\in I(n)$, where the first block of $\pi$ is denoted as $V_{1}$. Then
we define $I_{e}(n)$ (and $I_{o}(n)$ respectively) as the set of interval
partitions such that $|V_{1}|$ is even (odd respectively). We further define:

$\gamma_{o,n}=\sum_{\pi\in I_{o}(n)}\beta_{\pi}\quad\text{ and
}\quad\gamma_{e,n}=\sum_{\pi\in I_{e}(n)}\beta_{\pi}.$

We first show that $m_{k},\gamma_{e,k},$ and $\gamma_{o,k}$ satisfy the
following relation.

######  Lemma 6.7.

For each $k$, we have

$\displaystyle
m_{k}=\gamma_{o,k}+\gamma_{e,k},\quad\gamma_{e,k+1}=\frac{\alpha_{2}}{\alpha_{1}}\gamma_{o,k}\quad\text{
and
}\quad\gamma_{o,k+1}=\alpha_{1}m_{k}+d\frac{\alpha_{1}}{\alpha_{2}}\gamma_{e,k}.$

###### Proof.

It is clear that

(6.2) $m_{k}=\sum_{\pi\in I(k)}\beta_{\pi}=\sum_{\pi\in
I_{o}(k)}\beta_{\pi}+\sum_{\pi\in
I_{e}(k)}\beta_{\pi}=\gamma_{o,k}+\gamma_{e,k}.$

In addition, we note that each $\pi\in I_{e}(k+1)$ is generated from a
$\sigma\in I_{o}(k)$ by adding one element into the first block of $\sigma$.
To be precise, if we write

$\sigma=\\{(1,\dots,i_{1}),(i_{1}+1,\dots,i_{2}),\dots,(i_{k-1}+1,\dots,i_{k})\\}\text{
with }i_{1}=2s+1\text{ for some $s$, and }i_{k}=k.$

Then
$\pi=\\{(1,\dots,i_{1}+1),(i_{1}+2,\dots,i_{2}+1),\dots,(i_{k-1}+2,\dots,i_{k}+1)\\}.$
Moreover, we observe

$\beta_{i_{1}+1}=\beta_{2s+2}=d^{s}\alpha_{2}=\frac{\alpha_{2}}{\alpha_{1}}d^{s}\alpha_{1}\beta_{2s+1}=\frac{\alpha_{2}}{\alpha_{1}}\beta_{i_{1}}.$

Hence, we have

$\beta_{\pi}=\beta_{i_{1}+1}\beta_{i_{2}-i_{1}}\cdots\beta_{i_{k}-i_{k-1}}=\frac{\alpha_{2}}{\alpha_{1}}\beta_{i_{1}}\beta_{i_{2}-i_{1}}\cdots\beta_{i_{k}-i_{k-1}}=\frac{\alpha_{2}}{\alpha_{1}}\beta_{\sigma}.$

Thus, we conclude $\gamma_{e,k+1}=\frac{\alpha_{2}}{\alpha_{1}}\gamma_{o,k}$.

On the other hand, note that each $\pi\in I_{o}(k+1)$ is generated either from
a $\sigma_{1}\in I_{e}(k)$ by adding one element into the first block of
$\sigma_{1}$ or adding singleton as the first block for an interval partition
$\sigma_{2}\in I(k)$. Following the analogous argument, one gets

$\gamma_{o,k+1}=\alpha_{1}m_{k}+d\frac{\alpha_{1}}{\alpha_{2}}\gamma_{e,k}.$

∎

######  Theorem 6.8.

With $g=ax_{1}x_{2}+bx_{2}x_{1}$, we have

$\displaystyle
m_{k}(g)=\frac{1}{\omega}\big{(}\frac{\alpha_{1}+\omega}{2}\big{)}^{k-1}\big{(}\alpha_{1}\frac{\alpha_{1}+\omega}{2}+\alpha_{2}\big{)}-\frac{1}{\omega}\big{(}\frac{\alpha_{1}-\omega}{2}\big{)}^{k-1}\big{(}\alpha_{1}\frac{\alpha_{1}-\omega}{2}+\alpha_{2}\big{)}$

where $\omega=\sqrt{\alpha_{1}^{2}+4(d+\alpha_{2})}$ with
$d=ab\beta_{2}(x_{1})\beta_{2}(x_{2})$,
$\alpha_{1}=\frac{1}{2}(a+b)\beta_{1}(p)$ and $\alpha_{2}=ab\beta_{2}(p)$. In
other words,

$g\sim\frac{\alpha_{1}\theta_{1}+\alpha_{2}}{\omega\theta_{1}}\delta_{\theta_{1}}-\frac{\alpha_{1}\theta_{2}+\alpha_{2}}{\omega\theta_{2}}\delta_{\theta_{2}}\text{
where }\theta_{1}=\frac{\alpha_{1}+\omega}{2},\text{ and
}\theta_{2}=\frac{\alpha_{1}-\omega}{2}.$

###### Proof.

Firstly, it is clear that $\gamma_{o,1}=\alpha_{1}$ and
$\gamma_{o,2}=\alpha_{1}^{2}$. Then by Lemma 6.7, we can deduce that

$\frac{\gamma_{o,k+2}}{\alpha_{1}}-\frac{d}{\alpha_{1}}\gamma_{o,k}=\gamma_{o,k+1}+\frac{\alpha_{2}}{\alpha_{1}}\gamma_{o,k},$

which implies

$\gamma_{o,k+2}-\alpha_{1}\gamma_{o,k+1}-(d+\alpha_{2})\gamma_{o,k}=0.$

By solving the characteristic polynomial $x^{2}-\alpha_{1}x-(d+\alpha_{2})=0$,
we obtain the general solution

$\gamma_{o,k}=C_{1}\big{(}\frac{\alpha_{1}+\omega}{2}\big{)}^{k}+C_{2}\big{(}\frac{\alpha_{1}-\omega}{2}\big{)}^{k}$

for some constants $C_{1}$ and $C_{2}$, where
$\omega=\sqrt{\alpha_{1}^{2}+4(d+\alpha_{2})}.$ Then by plugging in the
initial conditions $\gamma_{o,1}=\alpha_{1}$ and
$\gamma_{o,2}=\alpha_{1}^{2}$, one gets

$\gamma_{o,k}=\frac{\alpha_{1}}{\omega}\big{(}\frac{\alpha_{1}+\omega}{2}\big{)}^{k}-\frac{\alpha_{1}}{\omega}\big{(}\frac{\alpha_{1}-\omega}{2}\big{)}^{k}.$

Therefore,

$\gamma_{e,k}=\frac{\alpha_{2}}{\alpha_{1}}\gamma_{o,k-1}=\frac{\alpha_{2}}{\omega}\big{(}\frac{\alpha_{1}+\omega}{2}\big{)}^{k-1}-\frac{\alpha_{2}}{\omega}\big{(}\frac{\alpha_{1}-\omega}{2}\big{)}^{k-1};$

hence, we finally get

$m_{k}=\gamma_{o,k}+\gamma_{e,k}=\frac{1}{\omega}\big{(}\frac{\alpha_{1}+\omega}{2}\big{)}^{k-1}\big{(}\alpha_{1}\frac{\alpha_{1}+\omega}{2}+\alpha_{2}\big{)}-\frac{1}{\omega}\big{(}\frac{\alpha_{1}-\omega}{2}\big{)}^{k-1}\big{(}\alpha_{1}\frac{\alpha_{1}-\omega}{2}+\alpha_{2}\big{)}.$

∎

###  6.2. The Infinitesimal cases

In this subsection, we will give the infinitesimal distribution of
$g=ax_{1}x_{2}+bx_{2}x_{1}$ whenever $x_{1}$ and $x_{2}$ are infinitesimally
Boolean independent in an infinitesimal probability space
$(\mathcal{A},\varphi,\varphi^{\prime})$.

Let $(\mathcal{A},\varphi,\varphi^{\prime})$ be an infinitesimal probability
space, and suppose elements $x_{1}$ and $x_{2}$ in $\mathcal{A}$ are
infinitesimally Boolean independent. By Theorem 2.8,
$X_{1}=\begin{bmatrix}x_{1}&0\\\ 0&x_{1}\end{bmatrix}$ and
$X_{2}=\begin{bmatrix}x_{2}&0\\\ 0&x_{2}\end{bmatrix}$ are Boolean independent
in $(\widetilde{\mathcal{A}},\widetilde{\mathbb{C}},\widetilde{\varphi})$. Let
$G=aX_{1}X_{2}+bX_{2}X_{1}$. We note that $\widetilde{\mathbb{C}}$ is a
commutative algebra; as a result, we have that
$\beta_{n}^{\widetilde{\varphi}}(G)$ follows the same computation as we proved
for $\beta_{n}(g)$. To be precise, $\beta_{n}^{\widetilde{\varphi}}(G)$ is now
$2\times 2$ upper triangular matrices, and we have

(6.3)
$\beta_{n}^{\widetilde{\varphi}}(G)=\begin{cases}ab(\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{1}^{\widetilde{\varphi}}(X_{2})^{2}+\beta_{2}^{\widetilde{\varphi}}(X_{2})\beta_{1}^{\widetilde{\varphi}}(X_{1})^{2})\Big{(}ab\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{2}^{\widetilde{\varphi}}(X_{2})\Big{)}^{(n-2)/2}&\text{if
}n\text{ is even}\\\
(a+b)\beta_{1}^{\widetilde{\varphi}}(X_{1})\beta_{1}^{\widetilde{\varphi}}(X_{2})\Big{(}ab\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{2}^{\widetilde{\varphi}}(X_{2})\Big{)}^{(n-1)/2}&\text{if
}n\text{ is odd}\end{cases}$

Suppose $p=x_{1}x_{2}+x_{2}x_{1}$, then we recall that
$\beta_{1}(p)=2\beta_{1}(x_{1})\beta_{1}(x_{2})$ and
$\beta_{2}(p)=\beta_{1}(x_{1})^{2}\beta_{2}(x_{2})+\beta_{1}(x_{2})^{2}\beta_{2}(x_{1})$.
It can easily to deduce that

$\displaystyle\beta_{1}^{\prime}(p)$ $\displaystyle=$ $\displaystyle
2\beta_{1}^{\prime}(x_{1})\beta_{1}(x_{2})+2\beta_{1}(x_{1})\beta_{1}^{\prime}(x_{2}),$
$\displaystyle\beta_{2}^{\prime}(p)$ $\displaystyle=$ $\displaystyle
2\beta_{1}(x_{1})\beta_{1}^{\prime}(x_{1})\beta_{2}(x_{2})+\beta_{1}(x_{1})^{2}\beta_{2}^{\prime}(x_{2})+2\beta_{1}(x_{2})\beta_{1}^{\prime}(x_{2})\beta_{2}(x_{1})+\beta_{1}(x_{2})^{2}\beta_{2}^{\prime}(x_{1}).$

######  Theorem 6.9.

Suppose that $x_{1}$ and $x_{2}$ are infinitesimally Boolean independent, and
let $g=ax_{1}x_{2}+bx_{2}x_{1}$, we have

$\beta_{n}^{\prime}(g)=\begin{cases}(ab)^{\frac{n}{2}}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\Big{[}\frac{n-2}{2}\beta_{2}(p)\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\Big{]}&\\\
\qquad\qquad\qquad\qquad\mbox{}+(ab)^{\frac{n}{2}}\beta_{2}^{\prime}(p)\,\big{[}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{]}^{\frac{n-2}{2}}&\text{if
}n\text{ is even}\\\
(a+b)(ab)^{\frac{n-1}{2}}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-3}{2}}\Big{[}\frac{n-1}{2}\beta_{1}(p)\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\Big{]}&\\\
\qquad\qquad\qquad\qquad\mbox{}+(a+b)(ab)^{\frac{n-1}{2}}\beta^{\prime}_{1}(p)\,\big{[}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{]}^{\frac{n-1}{2}}&\text{if
}n\text{ is odd}\end{cases}.$

###### Proof.

Observe that the $(1,2)$-entry of $\beta_{n}^{\widetilde{\varphi}}(G)$ is
$\beta_{n}^{\prime}(g)$. On the other hand,

$\displaystyle
ab(\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{1}^{\widetilde{\varphi}}(X_{2})^{2}+\beta_{2}^{\widetilde{\varphi}}(X_{2})\beta_{1}^{\widetilde{\varphi}}(X_{1})^{2})\Big{(}ab\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{2}^{\widetilde{\varphi}}(X_{2})\Big{)}^{(n-2)/2}$
$\displaystyle=$ $\displaystyle
ab\begin{bmatrix}\beta_{2}(p)&\beta_{2}^{\prime}(p)\\\
0&\beta_{2}(p)\end{bmatrix}\Big{(}ab\begin{bmatrix}\beta_{2}(x_{1})&\beta_{2}^{\prime}(x_{1})\\\
0&\beta_{2}(x_{1})\end{bmatrix}\cdot\begin{bmatrix}\beta_{2}(x_{2})&\beta_{2}^{\prime}(x_{2})\\\
0&\beta_{2}(x_{2})\end{bmatrix}\Big{)}^{(n-2)/2}$ $\displaystyle=$
$\displaystyle ab\begin{bmatrix}\beta_{2}(p)&\beta_{2}^{\prime}(p)\\\
0&\beta_{2}(p)\end{bmatrix}\Big{(}ab\begin{bmatrix}\beta_{2}(x_{1})\beta_{2}(x_{2})&\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\\\
0&\beta_{2}(x_{1})\beta_{2}(x_{2})\end{bmatrix}\Big{)}^{(n-2)/2}$
$\displaystyle=$
$\displaystyle(ab)^{\frac{n}{2}}\begin{bmatrix}\beta_{2}(p)&\beta_{2}^{\prime}(p)\\\
0&\beta_{2}(p)\end{bmatrix}$
$\displaystyle\mbox{}\times\begin{bmatrix}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-2)/2}&\frac{n-2}{2}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-4)/2}\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\\\
0&(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{(n-2)/2}\end{bmatrix}$

Thus, the $(1,2)$-entry of
$ab(\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{1}^{\widetilde{\varphi}}(X_{2})^{2}+\beta_{2}^{\widetilde{\varphi}}(X_{2})\beta_{1}^{\widetilde{\varphi}}(X_{1})^{2})\Big{(}ab\beta_{2}^{\widetilde{\varphi}}(X_{1})\beta_{2}^{\widetilde{\varphi}}(X_{2})\Big{)}^{(n-2)/2}$
is

$(ab)^{\frac{n}{2}}\frac{n-2}{2}\beta_{2}(p)(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\Big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\Big{)}+(ab)^{\frac{n}{2}}\beta_{2}^{\prime}(p)(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-2}{2}}$
$\mbox{}=(ab)^{\frac{n}{2}}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\bigg{(}\frac{n-2}{2}\beta_{2}(p)\Big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\Big{)}\bigg{)}$
$\mbox{}+(ab)^{\frac{n}{2}}\beta_{2}^{\prime}(p)\big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{)}^{\frac{n-2}{2}}.$

Applying Equation (6.3), we obtain

$\displaystyle\beta_{n}^{\prime}(g)$
$\displaystyle=(ab)^{\frac{n}{2}}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\bigg{(}\frac{n-2}{2}\beta_{2}(p)\Big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\Big{)}\bigg{)}$
$\displaystyle\quad+(ab)^{\frac{n}{2}}\beta_{2}^{\prime}(p)\big{(}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{)}^{\frac{n-2}{2}}.$

whenever $n$ is even. By applying Equations (6.3), the remaining cases can be
done via an analogous argument. ∎

Following Theorem 6.9, we have the infinitesimal distributions of the anti-
commutator $p=x_{1}x_{2}+x_{2}x_{1}$ and the commutator
$q=i(x_{1}x_{2}-x_{2}x_{1})$ as follows.

######  Corollary 6.10.

$\beta_{n}^{\prime}(p)=\begin{cases}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\Big{[}\frac{n-2}{2}\beta_{2}(p)\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\Big{]}&\\\
\qquad\qquad\qquad\qquad\qquad\qquad\mbox{}+\beta_{2}^{\prime}(p)\,\big{[}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{]}^{\frac{n-2}{2}}&\text{if
}n\text{ is even}\\\
(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-3}{2}}\Big{[}\frac{n-1}{2}\beta_{1}(p)\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\Big{]}&\\\
\qquad\qquad\qquad\qquad\qquad\qquad\mbox{}+\beta^{\prime}_{1}(p)\,\big{[}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{]}^{\frac{n-1}{2}}&\text{if
}n\text{ is odd}\end{cases}$

and

$\beta_{n}^{\prime}(q)=\begin{cases}(\beta_{2}(x_{1})\beta_{2}(x_{2}))^{\frac{n-4}{2}}\Big{[}\frac{n-2}{2}\beta_{2}(q)\big{(}\beta_{2}(x_{1})\beta_{2}^{\prime}(x_{2})+\beta_{2}^{\prime}(x_{1})\beta_{2}(x_{2})\big{)}\Big{]}&\\\
\qquad\qquad\qquad\qquad\qquad\qquad\mbox{}+\beta_{2}^{\prime}(q)\,\big{[}\beta_{2}(x_{1})\beta_{2}(x_{2})\big{]}^{\frac{n-2}{2}}&\text{if
}n\text{ is even}\\\ 0&\text{if }n\text{ is odd}\end{cases}.$

## Acknowledgement

The authors would like to express our sincere gratitude to the anonymous
reviewers for their insightful comments and constructive suggestions, which
greatly contributed to the improvement of this manuscript. The revision of
this paper has also benefited from the comments and suggestions of Yen Jen
Cheng, Nicolas Gilliers, and Takahiro Hasebe.

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M ††footnotetext: xxviii.viii.mmxxiv