E,M_{k}(\mathcal{B}))$. Hence, we get for all
$\mathfrak{b}\in\mathbb{H}^{+}(M_{k}(\mathcal{B}))$,

$\|\mathcal{G}^{M_{k}(\mathcal{B})}_{\mathbf{1}_{k}\otimes
A_{0}}(\mathfrak{b})-\mathcal{G}^{M_{k}(\mathcal{B})}_{\mathbf{1}_{k}\otimes
A_{1}}(\mathfrak{b})\|\leq\|{\Im}m(\mathfrak{b})^{-1}\|^{3}\|\operatorname{id}_{k}\otimes\eta_{0}-\operatorname{id}_{k}\otimes\eta_{1}\|.$

Using the fact that
$\|\operatorname{id}_{k}\otimes\eta_{0}-\operatorname{id}_{k}\otimes\eta_{1}\|\leq
k\|\eta_{0}-\eta_{1}\|$, which follows from [26, Exercise 3.10], we arrive at
the desired bound for general $k$. Now to prove (4.6) and (4.7), we observe
that for $z\in\mathbb{C}^{+},$

$\displaystyle\big{|}\varphi\big{[}G_{\mathbf{z}^{0}_{i}}(z)\big{]}(\eta_{0}-\eta_{1})(z)\varphi\big{[}G_{\mathbf{z}^{0}_{i}}(z)\big{]}\big{|}$
$\displaystyle\leq$
$\displaystyle\frac{1}{{\Im}m(z)}\|\eta_{0}-\eta_{1}\|\big{|}\varphi\big{[}G_{\mathbf{u}_{i}}(\varphi[G_{\mathbf{v}_{i}}(z)]^{-1})\big{]}\big{|}^{2}$
$\displaystyle\leq$
$\displaystyle\frac{1}{{\Im}m(z)}\|\eta_{0}-\eta_{1}\|\|G_{\mathbf{z}^{0}_{i}}(z)\|_{L^{2}(\mathcal{A},\varphi)}^{2}.$

Note that the last inequality holds due to the Cauchy Schwarz inequality and
following (3.10):
$\|G_{\mathbf{u}_{i}}(\varphi[G_{\mathbf{v}_{i}}(z)]^{-1})\|_{L^{2}(\mathcal{A},\varphi)}\leq\|G_{\mathbf{u}_{i}+\mathbf{v}_{i}}(z)\|_{L^{2}(\mathcal{A},\varphi)}$.
Then, the remaining argument follows the analogous proof in [6, Theorem 3.5].
∎

###  4.3. Fourth moment theorem for monotone infinitely divisible measures

Fourth moment theorems refer to a simplification of the moment method to prove
the weak convergence of a sequence of random variables $y_{n}$ to a given
random variable $y$. Such theorems state that if the fourth moment of $y_{n}$,
$\varphi[y_{n}^{4}]$, approaches the fourth moment of a random variable $y$,
$\varphi[y^{4}]$, then $y_{n}\to y$ weakly. In certain cases, one can even
quantify such convergence in terms of the difference
$|\varphi[y_{n}^{4}]-\varphi[y^{4}]|$.

Here we are concerned with the class of infinitely divisible measures with
respect to the monotone convolution, which we denote by $ID(\rhd)$. This
completes the main theorems from [2] where the authors give a quantitative
version of the results in [1], in the cases of free and tensor independence. A
fourth moment theorem was given in [3] for the Boolean case keeping the
monotone case open. Our results complete the picture by including the monotone
case and proving quantitative bounds for the fourth moment theorem in this
setting.

The main observation is that a Berry-Esseen estimate may be used to prove such
a theorem. Indeed, using the fact that $\varphi[x^{4}]\geq\varphi[x^{2}]^{2}$,
we deduce by Theorem 4.3 that if the $x_{i}$’s are monotonically independent
and identically distributed variables with mean $0$, variance $1$ and finite
fourth moment $\varphi[x_{i}^{4}]$, then

$L(\mu_{X_{n}},\mu_{A_{n}})\leq
K\Big{(}\frac{\varphi[x_{i}^{4}]}{n}\Big{)}^{1/14},$ (4.9)

for some constant $K>0$. In order to connect the above bound to a fourth
moment theorem we will use the monotone cumulants. It was shown by Hasebe and
Saigo [15] that the sequence of monotone cumulants $\\{h_{n}\\}_{n\geq 1}$
satisfies the property that for any $m\in\mathbb{N}$:
$h_{n}(\sum_{i=1}^{m}x_{i},\dots,\sum_{i=1}^{m}x_{i})=mh_{n}(x_{1},\dots,x_{1}),$
whenever $\\{x_{i}\\}^{m}_{i=1}$ is a collection of identically distributed
monotone independent variables. The first four cumulants are written in terms
of the first four moments as follows:

$\begin{split}h_{1}(x)&=\varphi[x],\\\
h_{2}(x,x)&=\varphi[x^{2}]-\varphi[x]^{2},\\\
h_{3}(x,x,x)&=\varphi[x^{3}]-\frac{5}{2}\varphi[x^{2}]\varphi[x]+\frac{3}{2}\varphi[x]^{3},\\\
h_{4}(x,x,x,x)&=\varphi[x^{4}]-\frac{3}{2}\varphi[x^{2}]^{2}-3\varphi[x^{3}]\varphi[x]+\frac{37}{6}\varphi[x^{2}]\varphi[x]^{2}-\frac{8}{3}\varphi[x]^{4}.\end{split}$
(4.10)

Note that if we normalize so that $\varphi[x]=0$ and $\varphi[x^{2}]=1$, then
$h_{2}(x,x)=\varphi[x^{2}]=1$ and
$h_{4}(x,x,x,x)=\varphi[x^{4}]-\frac{3}{2}\varphi[x^{2}]^{2}=\varphi[x^{4}]-1.5$.
Having this in hand, we now state our fourth moment theorem for the scalar-
valued monotone case, which is completely new. The key point is that now we
have a bound in terms of the second and fourth moment, which was not the case
in previous works, see [4, Theorem 1.1].

######  Theorem 4.5.

Let $Y$ be a monotone infinitely divisible random variable with mean $0$ and
variance $1$ and $A$ be a standard arcsine element. Then
$L(\mu_{Y},\mu_{A})\leq K(\varphi[Y^{4}]-1.5)^{1/14}$, for some $K>0$.

###### Proof.

Since $Y$ is infinitely divisible, then for each $n\in\mathbb{N}$, we may
write $Y$ as a sum of $n$ identically monotone independent random variables

$Y=y_{1}+\dots+y_{n}=\frac{1}{\sqrt{n}}(\sqrt{n}y_{1}+\dots+\sqrt{n}y_{n}).$

Noting that
$\varphi[(\sqrt{n}y_{i})^{2}]=h_{2}(\sqrt{n}y_{i},\sqrt{n}y_{i})=nh_{2}(y_{i},y_{i})=h_{2}(Y,Y)=1$
and
$h_{4}(\sqrt{n}y_{i})=n^{2}h_{4}(y_{i},y_{i},y_{i},y_{i})=nh_{4}(Y,Y,Y,Y),$
then

$\varphi[(\sqrt{n}y_{i})^{4}]=h_{4}(\sqrt{n}y_{i},\sqrt{n}y_{i},\sqrt{n}y_{i},\sqrt{n}y_{i})+1.5=nh_{4}(Y,Y,Y,Y)+1.5\,.$

For a standard arcsine element $A$, we may assume, without loss of generality,
that $A=A_{n}=\frac{1}{\sqrt{n}}(x_{1}+\dots+x_{n})$ where the $x_{i}$’s are
also standard arcsine variables. With this observation, Theorem 4.3 holds and
hence the bound (4.9) yields that

$\displaystyle L(\mu_{Y},\mu_{A})$ $\displaystyle=$ $\displaystyle
L(\mu_{Y},\mu_{A_{n}})\leq
K\left(\frac{\varphi[(\sqrt{n}y_{i})^{4}]}{n}\right)^{1/14}$ $\displaystyle=$
$\displaystyle
K\left(\frac{h_{4}(\sqrt{n}y_{i},\sqrt{n}y_{i},\sqrt{n}y_{i},\sqrt{n}y_{i})+1.5}{n}\right)^{1/14}=K\left(h_{4}(Y,Y,Y,Y)+\frac{1.5}{n}\right)^{1/14}.$

Since $n$ is arbitrary, we let it go to infinity to obtain as desired

$L(\mu_{A},\mu_{Y})\leq
K(h_{4}(Y,Y,Y,Y))^{1/14}=K(\varphi[Y^{4}]-1.5)^{1/14}.$

∎

In the case of operator-valued monotone independent variables, we can give a
bound for the difference of the Cauchy transforms in terms of the fourth
$\mathcal{B}$-valued cumulants.

For this aim, we need analogue operator-valued formulas to those in (4.10). As
before, when we restrict to the centered case, i.e. $E[x]=0$, the formulas
simplify and we get that $h_{1}^{\mathcal{B}}(x)=E(x)=0$,
$h_{2}^{\mathcal{B}}(xb,x)=E[xbx]=\eta(b)$,
$h_{3}^{\mathcal{B}}(xb_{1},xb_{2},x)=E[xb_{1}xb_{2}x]$ and

$h_{4}^{\mathcal{B}}(xb_{1},xb_{2},xb_{3},x)=E[xb_{1}xb_{2}xb_{3}x]-E[xb_{1}x]b_{2}E[xb_{3}x]-\frac{1}{2}E\left[xb_{1}E[xb_{2}x]b_{3}x\right].$

######  Theorem 4.6.

Let $(\mathcal{A},E,\mathcal{B})$ be a $C^{\ast}$-probability space and $Y$ be
monotone $n$-divisible element over $\mathcal{B}$ with $E[Y]=0$ and
$E[YbY]=\eta(b)$. Then, for any any $k\in\mathbb{N}$ and
$\mathfrak{b}\in\mathbb{H}^{+}(M_{k}(\mathcal{B}))$,

$\big{\|}E[G^{M_{k}(\mathcal{B})}_{1_{k}\otimes
Y}(\mathfrak{b})]-E[G^{M_{k}(\mathcal{B})}_{1_{k}\otimes
A}(\mathfrak{b})]\big{\|}\leq
k\|{\Im}m(\mathfrak{b})^{-1}\|^{4}\sqrt{\|\eta(1)\|}\sqrt{\sup\|h_{4}^{\mathcal{B}}(Yb,Y,Yb^{\prime},Y)\|},$

where $A$ is the $\mathcal{B}$-valued arcsine element with $E[A]=0$ and
$E[AbA]=\eta(b)$ and the supremum is taken over all
$b,b^{\prime}\in\mathcal{B}$ with $\|b\|=\|b^{\prime}\|=1$.

As the above bound holds on the level of the fully matricial extensions of
Cauchy transforms, it is sufficient to capture convergence in distribution
over $\mathcal{B}$ which yields the following operator-valued monotone fourth
moment theorem:

######  Corollary 4.7.

Let $(\mathcal{A},E,\mathcal{B})$ be a $C^{\ast}$-probability space and
$(Y_{N})_{N\in\mathbb{N}}$ a family of monotone $n$-divisible elements over
$\mathcal{B}$ such that $E[Y_{N}]=0$, $E[Y_{N}bY_{N}]=\eta(b)$ and
$\sup_{N\in\mathbb{N}}\|Y_{N}\|<\infty$. Provided that
$\sup\|h_{4}^{\mathcal{B}}(Y_{N}b,Y_{N},Y_{N}b^{\prime},Y_{N})\|\rightarrow 0$
as $N\rightarrow\infty$ where the supremum is taken over all
$b,b^{\prime}\in\mathcal{B}$ with $\|b\|=\|b^{\prime}\|=1$ , then $Y_{N}$
converges in distribution over $\mathcal{B}$ to a $\mathcal{B}$-valued
centered arcsine element $A$ with variance $\eta$.

If, in addition, $(\mathcal{A},\varphi)$ is a $C^{\ast}$-probability space and
$\varphi=\varphi\circ E$, then the distribution of $Y_{N}$ with respect to
$\varphi$, converges to that of $A$.

###### Proof of Theorem 4.6.

Let $Y$ be monotone $n$-divisible over $\mathcal{B}$. We write $Y$ as a sum of
$n$ identically distributed monotone independent random variables

$Y=y_{1}+\dots+y_{n}=\frac{1}{\sqrt{n}}(\sqrt{n}y_{1}+\dots+\sqrt{n}y_{n}).$

For a $\mathcal{B}$-valued arcsine element $A$, we may assume that
$A=A_{n}=\frac{1}{\sqrt{n}}(x_{1}+\cdots+x_{n})$ where the $x_{i}^{\prime}s$
are themselves $\mathcal{B}$-valued arcsine variables with mean zero and
variance map $\eta$. With this observation, Theorem 4.3 holds and yields
following bound: for any $n\in\mathbb{N}$ and any
$\mathfrak{b}\in\mathbb{H}^{+}(\mathcal{B})$,

$\displaystyle\big{\|}E[G_{Y}(\mathfrak{b})]-E[G_{A}(\mathfrak{b})]\big{\|}\leq\frac{1}{\sqrt{n}}\|{\Im}m(\mathfrak{b})^{-1}\|^{4}B_{n}(y),$

where

$B_{n}(y)=\sqrt{\alpha_{2}(\sqrt{n}y_{1})}\left(\sqrt{\alpha_{4}(\sqrt{n}y_{1})}+\sqrt{\frac{3}{2}\alpha_{2}(\sqrt{n}y_{1})^{2}}\right).$

To control the above estimate, we note that

$\alpha_{2}(\sqrt{n}y_{1})=\|E[\sqrt{n}y_{1}1\sqrt{n}y_{1}]\|=\|h_{2}^{\mathcal{B}}(\sqrt{n}y_{1}1,\sqrt{n}y_{1})\|=\|h_{2}^{\mathcal{B}}(Y1,Y)\|=\|\eta(1)\|.$

Moreover, we observe that

$\displaystyle E[\sqrt{n}y_{1}b^{*}(\sqrt{n}y_{1})^{2}b\sqrt{n}y_{1}]$
$\displaystyle=h_{4}^{\mathcal{B}}(\sqrt{n}y_{1}b^{*},\sqrt{n}y_{1},\sqrt{n}y_{1}b,\sqrt{n}y_{1})+\eta(b^{*})\eta(b)+\frac{1}{2}\eta(b^{*}\eta(1)b)$
$\displaystyle=nh_{4}^{\mathcal{B}}(Yb^{*},Y,Yb,Y)+\eta(b^{*})\eta(b)+\frac{1}{2}\eta(b^{*}\eta(1)b),$

which yields that

$\displaystyle\alpha_{4}(\sqrt{n}y_{1})$ $\displaystyle=$
$\displaystyle\sup\limits_{b\in\mathcal{B},\
\|b\|=1}\|E[\sqrt{n}y_{1}b^{*}(\sqrt{n}y_{1})^{2}b\sqrt{n}y_{1}]\|$
$\displaystyle\leq$ $\displaystyle\sup\limits_{b\in\mathcal{B},\
\|b\|=1}n\|h_{4}^{\mathcal{B}}(Yb^{*},Y,Yb,Y)\|+\|\eta(1)\|^{2}+\frac{1}{2}\|\eta(1)\|.$

Hence, we for any $n\in\mathbb{N}$

$\displaystyle\frac{1}{\sqrt{n}}B_{n}(y)\\!\leq\\!\sqrt{\|\eta(1)\|}\\!\left(\\!\sqrt{\sup\limits_{b\in\mathcal{B},\
\|b\|=1}\\!\|h_{4}^{\mathcal{B}}(Yb^{*},Y,Yb,Y)\|+\frac{\|\eta(1)\|^{2}}{n}+\frac{\|\eta(1)\|}{2n}}+\\!\sqrt{\frac{3\|\eta(1)\|^{2}}{2n}}\right).$

Finally letting $n\to\infty$ and putting the above bounds together, we
conclude that

$\big{\|}E[G_{Y}(\mathfrak{b})]-E[G_{A}(\mathfrak{b})]\big{\|}\leq\|{\Im}m(\mathfrak{b})^{-1}\|^{4}\sqrt{\|\eta(1)\|}\sqrt{\sup\limits_{b\in\mathcal{B},\
\|b\|=1}\|h_{4}^{\mathcal{B}}(Yb^{*},Y,Yb,Y)\|}.$ (4.11)

This proves the assertion for $k=1$. To prove it for general $k\in\mathbb{N}$,
we note first that, by Proposition 2.6, monotone independence is preserved
under amplification with the identity matrix, we prove the assertion for
general $k\in\mathbb{N}$, by applying (4.11) for $1_{k}\otimes Y$ and
$1_{k}\otimes A$ in the framework of the operator-valued $C^{*}$-probability
space $(M_{k}(\mathcal{A}),id_{k}\otimes E,M_{k}(\mathcal{B}))$. we first note
that by [6, Lemma 5.4], we have that
$\|(id_{k}\otimes\eta)(1)\|\leq\|\eta(1)\|$. Moreover, for any $B\in
M_{k}(\mathcal{B})$, one can easily check that

$\displaystyle h_{4}^{M_{k}(\mathcal{B})}\big{(}(1_{k}\otimes
Y)B^{*},(1_{k}\otimes Y),(1_{k}\otimes Y)B,(1_{k}\otimes
Y)\big{)}=\sum_{i,j,\ell=1}^{k}E_{ij}\otimes
h_{4}^{\mathcal{B}}(YB_{i\ell}^{*},Y,YB_{\ell j},Y),$

and hence, by [26, Exercise 3.10 (i)], we get

$\displaystyle\Big{\|}$ $\displaystyle
h_{4}^{M_{k}(\mathcal{B})}\big{(}(1_{k}\otimes Y)B^{*},(1_{k}\otimes
Y),(1_{k}\otimes Y)B,(1_{k}\otimes Y)\big{)}\Big{\|}$
$\displaystyle\leq\Big{(}\sum_{i,j=1}^{k}\big{\|}\sum_{\ell=1}^{k}h_{4}^{\mathcal{B}}(YB_{i\ell}^{*},Y,YB_{\ell
j},Y)\big{\|}^{2}\Big{)}^{1/2}\leq
k^{2}\|B\|^{2}\sup\limits_{\begin{subarray}{c}b\in\mathcal{B},\ \|b\|=1\\\
b^{\prime}\in\mathcal{B},\
\|b^{\prime}\|=1\end{subarray}}\big{\|}h_{4}^{\mathcal{B}}(Yb,Y,Yb^{\prime},Y)\big{\|}.$

From this, we infer that

$\displaystyle\sup_{B\in M_{k}(\mathcal{B}),\ \|B\|=1}$
$\displaystyle\Big{\|}h_{4}^{M_{k}(\mathcal{B})}\big{(}(1_{k}\otimes
Y)B^{*},(1_{k}\otimes Y),(1_{k}\otimes Y)B,(1_{k}\otimes Y)\big{)}\Big{\|}$
$\displaystyle\leq\Big{(}\sum_{i,j=1}^{k}\big{\|}\sum_{\ell=1}^{k}h_{4}^{\mathcal{B}}(YB_{i\ell}^{*},Y,YB_{\ell
j},Y)\big{\|}^{2}\Big{)}^{1/2}\leq
k^{2}\sup\limits_{\begin{subarray}{c}b\in\mathcal{B},\ \|b\|=1\\\
b^{\prime}\in\mathcal{B},\
\|b^{\prime}\|=1\end{subarray}}\big{\|}h_{4}^{\mathcal{B}}(Yb,Y,Yb^{\prime},Y)\big{\|}.$

Putting the above terms together, we prove the assertion for any
$k\in\mathbb{N}$.

∎

###  4.4. Infinitesimal CLTs

This section is devoted to studying operator-valued infinitesimal central
limit theorems, that were introduced in [27], see Theorem 2.17. We provide in
this paper the first quantitative results in this setting that are on the
level of operator-valued infinitesimal Cauchy transforms.

Suppose that $(\mathcal{A},E,E^{\prime},\mathcal{B})$ is an OV
$C^{*}$-infinitesimal probability space. Let $x:=\\{x_{1},\dots,x_{n}\\}$ be a
family of self-adjoint elements in $\mathcal{A}$ such that
$E[x_{j}]=E^{\prime}[x_{j}]=0$ for all $j$ and set

$X_{n}:=\frac{1}{\sqrt{n}}\sum_{j=1}^{N}x_{j}.$

Given the setting we consider, we will denote by $S_{n}$ a
$\mathcal{B}$-valued infinitesimal semicircle, Bernoulli or generalized
arcsine element (see [37, Section 3.3]) with the infinitesimal variance
$(\eta_{n},\eta_{n}^{\prime})$ given by

$\eta_{n}(b)=\frac{1}{n}\sum_{j=1}^{n}E[x_{j}bx_{j}]\qquad\text{ and
}\qquad\eta^{\prime}_{n}(b)=\frac{1}{n}\sum_{j=1}^{n}E^{\prime}[x_{j}bx_{j}].$

We assume that our OVI probability space is rich enough to contain a family
$y:=\\{y_{1},\dots,y_{n}\\}$ of independent $\mathcal{B}$-valued infinitesimal
semicircular/Bernoulli/arcsine elements, that is itself independent of $x$ and
is such that $E[y_{i}by_{i}]=E[x_{i}bx_{i}]$ and
$E^{\prime}[y_{i}by_{i}]=E^{\prime}[x_{i}bx_{i}]$ for any $i=1,\dots,n$. In
the monotone case, the independence of $y$ from $x$ is assumed to hold in the
sense that
$\mathcal{A}_{x_{1}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\dots\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\mathcal{A}_{x_{n}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\mathcal{A}_{y_{1}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\dots\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\mathcal{A}_{y_{n}}$
over $\mathcal{B}$.

We state now the first quantitative estimates in the infinitesimal setting
that follow directly from Theorem 1.3 and that provide analogue estimates to
those in Section 4.

######  Theorem 4.8.

Let $x:=\\{x_{1},\dots,x_{n}\\}$ be a family of self-adjoint elements in
$\mathcal{A}$ that are infinitesimally free/Boolean/monotone independent with
amalgamation over $\mathcal{B}$ and that are such that
$E[x_{j}]=E^{\prime}[x_{j}]=0$. Then, for any
$\mathfrak{b}\in\mathbb{H}^{+}(\mathcal{B})$,

$\big{\|}E^{\prime}\big{[}G_{X_{n}}(\mathfrak{b})\big{]}-E^{\prime}\big{[}G_{S_{n}}(\mathfrak{b})\big{]}\big{\|}\leq\frac{2}{\sqrt{n}}\|{\Im}m(\mathfrak{b})^{-1}\|^{4}\left(\max_{1\leq
i\leq N}\|x_{i}\|^{3}+8\max_{1\leq i\leq
N}\|E[x_{i}^{2}]\|^{\frac{3}{2}}\right),$

where $S_{n}$ is a $\mathcal{B}$-valued infinitesimal
semicircular/Bernoulli/generalized arcsine element with the infinitesimal
variance $(\eta_{n},\eta_{n}^{\prime})$. The bounds can be slightly improved
in the Boolean case by removing the factor $8$ on the right-hand side.

###### Proof.

To obtain the above estimates, we choose the family
$y=\\{y_{1},\dots,y_{n}\\}$ as above and then apply Theorem 1.3. We give a
complete proof for the free case whereas we only indicate what choices of the
family $y$ we do in the Boolean and monotone cases.

For the free case, we choose $y=\\{y_{1},\dots,y_{n}\\}$ to be a family of
$\mathcal{B}$-valued infinitesimal semicircular elements that are
infinitesimally free and that are such that $E[y_{j}]=E^{\prime}[y_{j}]=0$,
$E[x_{j}bx_{j}]=E[y_{j}by_{j}]$, and
$E^{\prime}[x_{j}bx_{j}]=E^{\prime}[y_{j}by_{j}]$ for any $b\in\mathcal{B}$
and $j\in[n]$. Finally, we set $S_{n}=\frac{1}{\sqrt{n}}\sum_{j=1}^{n}y_{j}$,
and note that by Lemma 2.19, $S_{n}$ is a $\mathcal{B}$-valued infinitesimal
semicircular element with infinitesimal variance
$(\eta_{n},\eta^{\prime}_{n})$. Hence, Theorem 1.3 yields

$\big{\|}E^{\prime}\big{[}G_{X_{n}}(\mathfrak{b})\big{]}-E^{\prime}\big{[}G_{S_{n}}(\mathfrak{b})\big{]}\big{\|}\leq\frac{2}{\sqrt{n}}\|{\Im}m(\mathfrak{b})^{-1}\|^{4}\left(\max_{1\leq
i\leq N}\|x_{i}\|^{3}+\max_{1\leq i\leq N}\|y_{i}\|^{3}\right).$

Finally, we note that $\|y_{i}\|\leq
2\|\eta_{y_{i}}(1)\|^{1/2}=2\|E[x_{i}^{2}]\|^{1/2}$ for each $i=1,\dots,n$
(see [17, Proposition 6.2.1]) which completes the proof of the free case.

For the Boolean case, we choose $y=\\{y_{1},\dots,y_{n}\\}$ to be a family of
$\mathcal{B}$-valued infinitesimal Bernoulli elements that are infinitesimally
Boolean independent over $\mathcal{B}$ with matching moments as described
above. Finally, we note that for any $j\in[n]$,
$\|y_{j}\|=\|\eta_{y_{j}}(1)\|^{1/2}$ which is the reason why the factor $8$
can be improved to $1$, see [17, Proposition 6.2.4]).

Finally, for the monotone case, we choose $y=\\{y_{1},\dots,y_{n}\\}$ to be a
family of $\mathcal{B}$-valued infinitesimal arcsine elements that are
infinitesimally monotone independent in the sense that
$\mathcal{A}_{x_{1}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\cdots\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}A_{x_{n}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\mathcal{A}_{y_{1}}\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}\cdots\mathrel{\scalebox{0.9}[1.0]{$\prec$}\mkern-5.0mu\scalebox{0.4}[1.0]{$\prec$}\mkern-5.5mu\scalebox{0.4}[1.0]{$\prec$}}A_{y_{n}}$
over $\mathcal{B}$ with matching moments as described above. Finally, we note
that for any $j\in[n]$, $\|y_{j}\|\leq 2\|\eta_{y_{j}}(1)\|^{1/2}$, see [17,
Proposition 6.2.6]. ∎

Conjecture: _Let
$(\mathcal{A}_{1},E_{1},E_{1}^{\prime}),\ldots,(\mathcal{A}_{k},E_{k},E_{k}^{\prime})$
be OVI $C^{\ast}$-probability spaces over $\mathcal{B}$ and let
$(\mathcal{A},E)$ be their OV $C^{\ast}$-free/Boolean/monotone product. Then
there exists a linear $\mathcal{B}$-bimodule self-adjoint completely bounded
map $E^{\prime}:\mathcal{A}\rightarrow\mathcal{B}$ with $E^{\prime}(1)=0$ such
that $\left.E^{\prime}\right|_{\mathcal{A}_{j}}=E_{j}^{\prime}$ for $1\leq
j\leq k$ and such that $\mathcal{A}_{1},\dots,\mathcal{A}_{k}$ are
infinitesimally free/Boolean/monotone independent over $\mathcal{B}$. _

######  Remark 4.9.

We give an algebraic construction of the OVI free/Boolean/monotone products in
Appendix A. The mapping $E^{\prime}$ that we construct, satisfies all the
properties in the above conjecture except for complete boundedness which is
still open.

######  Theorem 4.10.

Let $(\mathcal{A},E,E^{\prime},\mathcal{B})$ be an OV $C^{*}$-infinitesimal
probability space, and let $c_{0}$ and $c_{1}$ be infinitesimal
semicircular/Bernoulli/arcsine elements with respective infinitesimal
variances $(\eta_{0},\eta_{0}^{\prime})$ and $(\eta_{1},\eta_{1}^{\prime})$.
Provided that the above conjecture holds, then for any
$\mathfrak{b}\in\mathbb{H}^{+}(\mathcal{B})$,

$\|E^{\prime}[G_{c_{0}}(\mathfrak{b})]-E^{\prime}[G_{c_{1}}(\mathfrak{b})]\|\leq
9\|{\Im}m(\mathfrak{b})^{-1}\|^{3}(\|\eta_{0}-\eta_{1}\|+\|\eta^{\prime}_{0}-\eta^{\prime}_{1}\|).$

###### Proof.

We shall prove only the free case and comment at the end on the Boolean and
monotone cases. Let $s_{0}$ and $s_{1}$ be infinitesimal semicircular elements
with respective infinitesimal variances $(\eta_{0},\eta_{0}^{\prime})$ and
$(\eta_{1},\eta_{1}^{\prime})$. Given $m\in\mathbb{N}$, we let
$x=\\{x_{1},\dots,x_{m}\\}$ and $y=\\{y_{1},\dots,y_{m}\\}$ be two
infinitesimally free families whose elements are themselves infinitesimally
free copies of $\frac{1}{\sqrt{m}}s_{0}$ and $\frac{1}{\sqrt{m}}s_{1}$
respectively. We consider the upper triangular probability space
$(\widetilde{A},\widetilde{E},\widetilde{B})$ that is induced by
$(\mathcal{A},E,E^{\prime},\mathcal{B})$, set

$S_{0}=\begin{bmatrix}s_{0}&0\\\ 0&s_{0}\end{bmatrix}\text{ and
}S_{1}=\begin{bmatrix}s_{1}&0\\\ 0&s_{1}\end{bmatrix},$

and finally define $\widetilde{\eta}_{0}(B):=\widetilde{E}[S_{0}BS_{0}]$ and
$\widetilde{\eta}_{1}(B):=\widetilde{E}[S_{1}BS_{1}]$ for any
$B\in\widetilde{B}$.

The proof relies as before on the operator-valued Lindeberg method with the
only difference that it is now applied with respect to $\widetilde{E}$ instead
of $E$. With this aim we set, for each $i\in[m]$,

$\widetilde{x}_{i}=\begin{bmatrix}x_{i}&0\\\ 0&x_{i}\end{bmatrix}\text{ and
}\widetilde{y}_{i}=\begin{bmatrix}y_{i}&0\\\ 0&y_{i}\end{bmatrix},$

and note that the $\widetilde{x}_{i}$’s and $\widetilde{y}_{i}$’s are centered
with respect to $\widetilde{E}$ but do not have matching moments of second
order. Thus, we have for any $B=\begin{bmatrix}\mathfrak{b}&b^{\prime}\\\
0&\mathfrak{b}\end{bmatrix}\in\widetilde{\mathcal{B}}$ with
$\mathfrak{b}\in\mathbb{H}^{+}(\mathcal{B})$,

$\displaystyle\widetilde{E}[G_{S_{0}}(B)]-\widetilde{E}[G_{S_{1}}(B)]$
$\displaystyle=$
$\displaystyle\frac{1}{m}\sum\limits_{i=1}^{m}\left(\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]-\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]\right)$
$\displaystyle+$
$\displaystyle\frac{1}{m\sqrt{m}}\sum\limits_{i=1}^{m}\left(\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}]-\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i-1}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}]\right)$

where

$\widetilde{\mathbf{z}}_{i}=\sum\limits_{j=1}^{i}\widetilde{x}_{j}+\sum\limits_{j=i+1}^{m}\widetilde{y}_{j}\qquad\text{and}\qquad\widetilde{\mathbf{z}}^{0}_{i}=\sum\limits_{j=1}^{i-1}\widetilde{x}_{j}+\sum\limits_{j=i+1}^{m}\widetilde{y}_{j}.$

Hence,

$\|\widetilde{E}[G_{S_{0}}(B)]-\widetilde{E}[G_{S_{1}}(B)]\|\\\
\leq\frac{1}{m}\sum\limits_{i=1}^{m}\left\|\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]-\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]\right\|\\\
+\frac{1}{m\sqrt{m}}\sum\limits_{i=1}^{m}\left(\left\|\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i-1}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}]\right\|+\left\|\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i-1}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}]\right\|\right).$
(4.12)

To control the above terms, we observe first that

$\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|=\left\|\begin{bmatrix}(b-\mathbf{z}^{0}_{i})^{-1}&-(b-\mathbf{z}^{0}_{i})^{-1}b^{\prime}(b-\mathbf{z}^{0}_{i})^{-1}\\\
0&(b-\mathbf{z}^{0}_{i})^{-1}\end{bmatrix}\right\|\leq\|{\Im}m(b)^{-1}\|(1+\|b^{\prime}\|\|{\Im}m(b)^{-1}\|),$

and that if we let
$\widetilde{\eta}=\widetilde{\eta}_{0}-\widetilde{\eta}_{1}$,
$\eta=\eta_{0}-\eta_{1}$, and
$\eta^{\prime}=\eta^{\prime}_{0}-\eta^{\prime}_{1}$, then
$\|\widetilde{\eta}\|\leq\|\eta\|+\|\eta^{\prime}\|$. Indeed, we have

$\displaystyle\left\|\widetilde{\eta}(B)\right\|$ $\displaystyle\leq$
$\displaystyle\|\eta(b)\|+\|\eta(b^{\prime})\|+\|\eta^{\prime}(b)\|$
$\displaystyle\leq$
$\displaystyle\|\eta\|\|b\|+\|\eta\|\|b^{\prime}\|+\|\eta^{\prime}\|\|b\|$
$\displaystyle\leq$
$\displaystyle(\|\eta\|+\|\eta^{\prime}\|)(\|b\|+\|b^{\prime}\|)=(\|\eta\|+\|\eta^{\prime}\|)\|B\|,$

By Remark 3.7, the third order terms are controlled as follows: for each
$i\in[m]$,

$\left\|\widetilde{E}\big{[}G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}\big{]}\right\|\leq
3\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|^{4}\|\widetilde{x}_{i}\|^{3}\leq
3\|{\Im}m(b)^{-1}\|^{4}(1+\|b^{\prime}\|\|{\Im}m(b)^{-1}\|)^{4}\|x_{i}\|^{3},$

and

$\left\|\widetilde{E}\big{[}G_{\widetilde{\mathbf{z}}_{i-1}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{3}\big{]}\right\|\leq
3\|{\Im}m(b)^{-1}\|^{4}(1+\|b^{\prime}\|\|{\Im}m(b)^{-1}\|)^{4}\|y_{i}\|^{3}.$

Following the lines of proof [6, Theorem 3.1] for free case and Remark 3.7, we
then obtain for each $i\in[m]$,

$\displaystyle\left\|\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]-\widetilde{E}[G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}]\right\|$
$\displaystyle=\left\|\widetilde{E}\left[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\left(\widetilde{\eta}_{0}-\widetilde{\eta}_{1}\right)(\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)])G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\right]\right\|$
$\displaystyle\leq
3\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|^{2}\|\widetilde{\eta}_{0}-\widetilde{\eta}_{1}\|\|\widetilde{E}[G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)]\|$
$\displaystyle\leq
9\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|^{3}\|\widetilde{\eta}_{0}-\widetilde{\eta}_{1}\|.$

Now, choosing $B=\begin{bmatrix}\mathfrak{b}&0\\\ 0&\mathfrak{b}\end{bmatrix}$
(i.e, $b^{\prime}=0$), we get

$\displaystyle\|E^{\prime}[G_{s_{0}}(\mathfrak{b})]-E^{\prime}[G_{s_{1}}(\mathfrak{b})]\|$
$\displaystyle\leq$
$\displaystyle\left\|\begin{bmatrix}E[G_{s_{0}}(\mathfrak{b})]-E[G_{s_{0}}(\mathfrak{b})]&E^{\prime}[G_{s_{0}}(\mathfrak{b})]-E^{\prime}[G_{s_{1}}(\mathfrak{b})]\\\
0&E[G_{s_{0}}(\mathfrak{b})]-E[G_{s_{0}}(\mathfrak{b})]\end{bmatrix}\right\|$
$\displaystyle=$
$\displaystyle\|\widetilde{E}[G_{S_{0}}(B)]-\widetilde{E}[G_{S_{1}}(B)]\|.$

Therefore,

$\displaystyle\|E^{\prime}[G_{s_{0}}(\mathfrak{b})]-E^{\prime}[G_{s_{1}}(\mathfrak{b})]\|$
(4.13) $\displaystyle\leq$ $\displaystyle
9\|{\Im}m(\mathfrak{b})^{-1}\|^{3}\|\widetilde{\eta}_{0}-\widetilde{\eta}_{1}\|+\frac{3}{\sqrt{m}}\|{\Im}m(\mathfrak{b})^{-1}\|^{4}\big{(}\max_{1\leq
i\leq m}\|x_{i}\|^{3}+\max_{1\leq i\leq m}\|y_{i}\|^{3}\big{)}.$

Finally, note that (4.13) holds for any $m\in\mathbb{N}$, then we let
$m\to\infty$ to complete the proof.

For Boolean and monotone cases, we apply the proofs of (1.1 and 1.3
respectively,

$\displaystyle\left\|\widetilde{E}\big{[}G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{x}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}\big{]}-\widetilde{E}\big{[}G_{\widetilde{\mathbf{z}}_{i}}(B)(\widetilde{y}_{i}G_{\widetilde{\mathbf{z}}^{0}_{i}}(B))^{2}\big{]}\right\|\leq
9\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|^{2}\|B^{-1}\|\|\widetilde{\eta}_{0}-\widetilde{\eta}_{1}\|$

and note that

$\|B^{-1}\|\leq\|\mathfrak{b}^{-1}\|(1+\|b^{\prime}\|\|\mathfrak{b}^{-1}\|)\text{
and
}\|G_{\widetilde{\mathbf{z}}^{0}_{i}}(B)\|\leq\|{\Im}m(\mathfrak{b})^{-1}\|(1+\|b^{\prime}\|\|{\Im}m(\mathfrak{b})^{-1}\|).$

The remaining arguments follow the analogous ones of the free case. ∎

##  5\. Matrices with Boolean Entries

Let $(\mathcal{A},\varphi)$ be a $W^{\ast}$-probability space and denote by
$M_{n}(\mathcal{A})$ the algebra of $n\times n$ matrices whose entries are
elements of $\mathcal{A}$, i.e.
$M_{n}(\mathcal{A})=M_{n}(\mathbb{C})\otimes\mathcal{A}$. The aim of this
section is to study the distribution of self-adjoint matrices whose entries
are Boolean independent with a variance profile. In Section 5.1, we will first
illustrate some properties of $\mathcal{B}$-valued Bernoulli elements and then
study distributions of matrices whose entries are in particular Boolean
independent _$\eta$ -circular elements_. Finally, in Section 5.2, we extend
the study to the general Boolean Wigner case.

###  5.1. Matrix Valued Bernoulli

Let $(\mathcal{A},E,\mathcal{B})$ be an OV $C^{*}$-probability space, and
$(\mathcal{A},\varphi)$ be a $C^{*}$-probability space such that
$\varphi=\varphi\circ E$. Let $B$ be a $\mathcal{B}$-valued Bernoulli element
with variance map $\eta(b)=E[BbB]$. We illustrate in the following lemma what
the scalar-valued distribution of $B$ is. For a measure $\mu$, let us denote
by $\mu^{(2)}$ the push forward of the measure $\mu$ along the map $t\to
t^{2}.$

######  Lemma 5.1.

Let $B$ be a $\mathcal{B}$-valued Bernoulli element with $\eta(b)=E[BbB]$ as
variance map, then for any $n\in\mathbb{N}$

$\varphi[B^{2n}]=\varphi[\eta(1)^{n}]\quad\text{ and
}\quad\varphi[B^{2n+1}]=0.$ (5.1)

In particular, if $\mu_{B}$ and $\mu_{\eta(1)}$ are the distributions of $B$
and $\eta(1)$ with respect to $\varphi$, respectively, then $\mu_{B}$ is the
unique symmetric measure such that $\mu^{(2)}_{B}=\mu_{\eta(1)}$.

###### Proof.

To compute the moments of a $\mathcal{B}$-valued Bernoulli variable $B$, we
apply the Boolean moment-cumulant formula and use the fact that the only non-
vanishing cumulant of $B$ is $\beta_{2}(Bb,B)=\eta(b)$. Since there is only
one pair interval partition on $[2n]$, namely
$\pi=\\{\\{1,2\\},\dots,\\{2n-1,2n\\}\\}$, we get

$E[B^{2n}]=\eta(1)^{n}\text{ and }E[B^{2n+1}]=0.$

Now applying $\varphi$ on both sides, we obtain (5.1) and end the proof. ∎

Having Lemma 5.1 in hand, we obtain immediately the following result. We give
a proof for the convenience of the reader.

######  Corollary 5.2.

Let $B$ and $\\{B_{n}\\}_{n}\geq 0$ be $\mathcal{B}$-valued Bernoulli elements
with variance maps $\eta$ and $\\{\eta_{n}\\}_{n}\geq 0$, respectively. Then,
$\mu_{B_{n}}\to\mu_{B}$ weakly, if and only if
$\mu_{\eta_{n}(1)}\to\mu_{\eta(1)}$, weakly.

###### Proof.

To relate $\mu_{\eta_{n}(1)}$ and $\mu_{\eta(1)}$, we first note that
$\mu_{B_{n}}([-t,t])=\mu_{\eta_{n}(1)}([0,t^{2}])$, and similarly,
$\mu_{B}([-t,t])=\mu_{\eta(1)}([0,t^{2}])$. If $\mu_{B_{n}}\to\mu_{B}$, then
$\mu_{B_{n}}([-t,t])\to\mu_{B}([-t,t])$ and then
$\mu_{\eta_{n}(1)}([0,t^{2}])$ $\to\mu_{\eta(1)}([0,t^{2}])$, which implies
that $\eta_{n}(1)\to\eta(1)$, since by definition $\eta_{n}(1)\to\eta(1)$ are
measure supported on $\mathbb{R^{+}}$. The converse follows the same lines by
the observations that symmetric measures are determined by the measure on the
intervals of the form $[-t,t]$.

∎

In order to study matrices with Boolean entries, we consider the quadruple
$(M_{n}(\mathcal{A}),\operatorname{tr}_{n}\otimes\varphi,$
$id_{n}\otimes\varphi,M_{n}(\mathbb{C}))$ where $(\mathcal{A},\varphi)$ is a
given $W^{*}$-probability space. For this aim, we call an element
$x\in\mathcal{A}$ a _$\eta$ -circular element_ if $x$ is $\eta$-diagonal (see
[9, Definition 2.6]) and its Boolean cumulants satisfy in addition the
property that
$\beta_{n}(x,x^{*},\cdots,x,x^{*})=\beta_{n}(x^{*},x,\cdots,x^{*},x)=0$ unless
$n=2.$

######  Remark 5.3 (Norm of $\eta$-circular operator).

We note that one can construct $\eta$-circular elements such that
$\beta_{2}(A,A^{*})=\sqrt{\alpha}$, $\beta_{2}(A^{*},A)=\tilde{\alpha}$ and
$||A||=max(\sqrt{\alpha},\sqrt{\tilde{\alpha})}$. Indeed, in [9, Remark 5.3 ],
a construction is given for $\eta$-diagonal elements. Their construction is of
the form $A=V(X\otimes Y)$, where $V$ is the flip operator, $X=T_{1}\oplus
T_{2}$ and $Y$ is a partial isometry. Since the norm of $\|V(X\otimes
Y)\|=\|X\otimes Y\|$ and $\|X\otimes Y\|=\|X\|~{}\|Y\|$, then $\|V(X\otimes
Y)\|=\|X\|$, because $Y$ is a partial isometry. In the specific case of the
$\eta$-circular elements with $\beta_{2}(A,A^{*})=\alpha$ and
$\beta_{2}(A^{*},A)=\tilde{\alpha}$, we choose $T_{1}$ and $T_{2}$ to be the
Bernoulli variables given by $T_{1}=\begin{bmatrix}0&\sqrt{\alpha}\\\
\sqrt{\alpha}&0\end{bmatrix}$ and
$T_{2}=\begin{bmatrix}0&\sqrt{\tilde{\alpha}}\\\
\sqrt{\tilde{\alpha}}&0\end{bmatrix}$. Hence, $\|X\|=\|T_{1}\otimes
T_{2}\|=\max\\{\|T_{1}\|,\|T_{2}\|\\}=max(\sqrt{\alpha},\sqrt{\tilde{\alpha}})$.

Let $b=\\{b_{ij}\mid 1\leq j\leq i\leq n\\}$ be a family of _Boolean
independent elements_ such that

  * •

for any $1\leq i\leq n$, $b_{ii}$ is a Bernoulli element with
$\varphi(b_{ii})=0$ and $\varphi(b_{ii}^{2}):=\sigma_{i}$.

  * •

for any $1\leq j<i\leq n$, $b_{ij}$ is a $\eta$-circular element with
$\varphi(b_{ij})=0$ and $\varphi(b_{ij}b_{ij}^{*}):=\alpha_{ij}$ and
$\varphi(b_{ij}^{\ast}b_{ij}):=\tilde{\alpha}_{ij}$.

Let us set $e_{ii}=\frac{1}{2\sqrt{n}}E_{ii}$ and
$e_{ij}=\frac{1}{\sqrt{n}}E_{ij}$ for $j<i$, with $(E_{ij})_{1\leq i,j\leq n}$
are the standard matrix units in $M_{n}(\mathbb{C})$. Our main interest in
this section is the matrix

$B_{n}=\sum_{1\leq j\leq i\leq n}\big{(}e_{ij}\otimes b_{ij}+e_{ij}^{*}\otimes
b_{ij}^{*}\big{)},$ (5.2)

which we will prove to be a $M_{n}(\mathbb{C})$-valued Bernoulli element in
the space of matrices over $\mathcal{A}$, see Proposition 5.5 for the precise
statement. To do this, we need the following lemma whose proof follows
similarly to the free case, see [20, Section 9.3].

######  Lemma 5.4.

Let $(\mathcal{A},\varphi)$ be a non-commutative probability space. We denote
by $\beta_{n}$ the scalar-valued Boolean cumulants with respect to $\varphi$
and by $\beta^{M_{n}(\mathbb{C})}_{n}$ the operator-valued cumulants with
respect to $id\otimes\varphi$. Consider the matrices
$A_{n}^{(k)}=(a^{(k)}_{ij})_{i,j=1}^{n}\in M_{n}(\mathcal{A})$. Then the
operator-valued cumulants of the family $\\{A^{(k)}\\}_{k}$ are given in terms
of the cumulants of their entries as follows:

$\big{[}\beta^{M_{n}(\mathbb{C})}_{m}\big{(}A_{n}^{(r_{1})},A_{n}^{(r_{2})},\dots,A_{n}^{(r_{m})}\big{)}\big{]}_{ij}=\sum^{n}_{i_{2},\ldots,i_{m}=1}\beta_{m}\big{(}a_{ii_{2}}^{(r_{1})},a_{i_{2}i_{3}}^{(r_{2})},\ldots,a_{i_{m}j}^{(r_{m})}\big{)}.$

With Lemma 5.4 in hand, we are ready to describe the OV-distribution of
$B_{n}$.

######  Proposition 5.5.

In the quadruple
$(M_{n}(\mathcal{A}),\operatorname{tr}_{n}\otimes\varphi,id_{n}\otimes\varphi,M_{n}(\mathbb{C}))$,
the matrix $B_{n}$ is an operator valued Bernoulli element over
$\mathcal{D}_{n}$, the subalgebra of diagonal matrices with variance map

$\eta_{n}:\mathcal{D}_{n}\rightarrow\mathcal{D}_{n},\qquad
D\mapsto\eta_{n}(D)$

where for any $D=(d_{ij})_{i,j=1}^{n}\in\mathcal{D}_{n}$,

$\big{(}\eta_{n}(D)\big{)}_{i,j}=\delta_{ij}\Big{(}\sum^{i-1}_{k=1}d_{kk}\alpha_{ik}+d_{ii}\sigma_{i}+\sum^{n}_{k=i+1}d_{kk}\tilde{\alpha}_{ik}\Big{)}.$

In particular, the distribution of $B_{n}$ with respect to
$tr_{n}\otimes\varphi$ is given by

$\mu_{B_{n}}=\frac{1}{2n}\sum^{n}_{i=1}(\delta_{\sqrt{\lambda_{i}}}+\delta_{-\sqrt{\lambda_{i}}}).$
(5.3)

###### Proof.

The fact that $B_{n}$ is an operator-valued Bernoulli element follows directly
from Lemma 5.4. Indeed, as the diagonal entries are Bernoulli elements and the
off-diagonal entries are $\eta$-circular elements that are all Boolean
independent up to symmetry, it follows that for any $m\in\mathbb{N}$ and
$C,C_{1},\dots,C_{m-1}\in M_{n}(\mathbb{C})$, we have that
$\beta^{M_{n}(\mathbb{C})}_{m}(B_{n}C_{1},\dots,B_{n-1}C_{n-1},B_{n})\big{]}=0$
whenever $m\neq 2$ and
$\beta^{M_{n}(\mathbb{C})}_{2}(B_{n}C,B_{n}):=\eta_{n}(C)$ is given for any
$i,j\in[n]$ by

$\displaystyle\big{[}\beta^{M_{n}(\mathbb{C})}_{2}(B_{n}C,B_{n})\big{]}_{ij}$
$\displaystyle=\sum^{n}_{k,\ell=1}\beta_{2}(b_{ik}c_{k\ell},b_{\ell
j})=\delta_{ij}\sum^{n}_{k=1}c_{kk}\beta_{2}(b_{ik},b_{ki})$
$\displaystyle=\delta_{i,j}\big{(}\sum^{i-1}_{k=1}c_{kk}\alpha_{ik}+c_{ii}\sigma_{i}+\sum^{n}_{k=i+1}c_{kk}\tilde{\alpha}_{ik}\big{)}.$

As any $M_{n}(\mathbb{C})$-valued joint Boolean cumulants of order larger than
$2$ is equal to $0$, we deduce that $B_{n}$ is an operator-valued Bernoulli
element over $M_{n}(\mathbb{C})$. It suffices to remark that the covariance
map $\eta_{n}$ sends the subalgebra of diagonal matrices
$\mathcal{D}_{n}\subset M_{n}(\mathbb{C})$ into itself, to see that $B_{n}$ is
in fact an operator-valued Bernoulli element over $\mathcal{D}_{n}$. We omit
the proof of this statement in the Boolean case, which can be proven in the
same way as its free analogue in [25, Theorem 3.1].

To prove the last part of the proposition, we note that by Lemma 5.1, we have

$\operatorname{tr}_{n}\otimes\varphi[B_{n}^{2k}]=\operatorname{tr}_{n}\otimes\varphi[\eta(1)^{k}]=\frac{1}{n}\sum\limits_{i=1}^{n}\lambda_{i}^{k}\quad\text{and}\quad\operatorname{tr}_{n}\otimes\varphi[B_{n}^{2k+1}]=0.$

On the other hand, denoting by $\delta_{\lambda}$ the Dirac measure at
$\lambda$, we can easily check that the $m$-th moment of the probability
measure
$\mu=\frac{1}{2n}\sum\limits_{i=1}^{n}(\delta_{\sqrt{\lambda_{i}}}+\delta_{-\sqrt{\lambda_{i}}})$
is equal to $\operatorname{tr}_{n}\otimes\varphi[B_{n}^{m}]$, which proves
(5.3).

∎

######  Example 5.6.

We show in the following examples how we can explicitly compute the spectral
distribution of $B_{n}$ using Proposition 5.5 and find their limiting
distribution, as $n$ tends to infinity. We consider the cases when the entries
are identically distributed and when they have a variance profile.

  1. (i)

Identically distributed entries: For any $1\leq j\leq i\leq n$, let
$\sigma_{i}=\sigma$, $\alpha_{ij}=\alpha/n$ and
$\tilde{\alpha}_{ij}=\tilde{\alpha}/n$, and assume without loss of generality
that $0<\tilde{\alpha}\leq\alpha$. In this case,

$\lambda_{i}=\sigma+\frac{i-1}{n}\alpha+\frac{n-i}{n}\tilde{\alpha}=\sigma+\tilde{\alpha}-\frac{\alpha}{n}+\frac{i}{n}(\alpha-\tilde{\alpha}).$

Now as $\lambda_{i+1}-\lambda_{i}=(\alpha-\tilde{\alpha})/{n}$, we notice that
the $\lambda_{i}$’s are equally spaced with
$\lambda_{1}=\sigma+\frac{n-1}{n}\tilde{\alpha}$ and
$\lambda_{n}=\sigma+\frac{n-1}{n}\alpha$. Thus, in the limit as
$n\rightarrow\infty$,

$\mu_{X_{n}}:=\frac{1}{n}\sum_{i}\delta_{\lambda_{i}}\to\text{Uniform}\big{(}\sigma+\tilde{\alpha},\
\sigma+\alpha\big{)}.$

One can also see this by computing the cumulative distribution function of
$X_{n}$ for any $\omega\in\mathbb{R}$,

$P(X_{n}\leq
w)=\frac{w-(\sigma+\tilde{\alpha})+\alpha/n}{\alpha-\tilde{\alpha}}\longrightarrow\frac{w-(\sigma+\tilde{\alpha})}{\alpha-\tilde{\alpha}}\text{
as }n\to\infty.$

From this, one can easily show that
$\mu_{B_{n}}:=\frac{1}{2n}\sum^{n}_{i=1}(\delta_{\sqrt{\lambda_{i}}}+\delta_{-\sqrt{\lambda_{i}}})\xrightarrow[n\rightarrow\infty]{}\mu$
with density

$\mu(dt)=\frac{|t|}{(\alpha-\tilde{\alpha})}dt,\quad|t|\in(\sqrt{\sigma+\tilde{\alpha}},\sqrt{\sigma+\alpha}).$

Moreover, in the case where $\tilde{\alpha}=\alpha$, $B_{n}$ is a scalar
Bernoulli element with distribution

$\mu_{B_{n}}=\frac{1}{2}\delta_{\sqrt{\sigma+{\frac{n-1}{n}}\alpha}}+\frac{1}{2}\delta_{-\sqrt{\sigma+{\frac{n-1}{n}}\alpha}},$

which converges to
$\frac{1}{2}\delta_{\sqrt{\sigma+\alpha}}+\frac{1}{2}\delta_{-\sqrt{\sigma+\alpha}}.$

  2. (ii)

Variance profile: Assume for any $1\leq j\leq i\leq n$,
$\alpha_{ij}=\tilde{\alpha}_{ij}=|i-j|/n^{2}$ and $\sigma_{i}=0$. Then in this
case, it is easy to see that

$\lambda_{i}=\frac{1}{n^{2}}\sum^{n}_{k=1}|k-i|=\frac{(n-i+1)(n-i)}{2n^{2}}+\frac{i(i-1)}{2n^{2}}.$

This means that for $t\in[0,1]$, $\lambda_{[tn]}\to\frac{1}{2}-t(1-t)$. Thus
if $\rho$ is the limiting distribution of
$\frac{1}{n}\sum_{i}\delta_{\lambda_{i}}$, it satisfies that
$\rho((1/4,(t^{2}+1)/4))=t$, for $t\in(0,1)$. From where the the density of
$\rho$ is given by $\rho(dt)=\frac{4}{\sqrt{4t-1}}dt$ for $t\in(1/4,1/2)$ and
thus

$\mu(dt)=\frac{2|t|}{\sqrt{4t^{2}-1}}\quad|t|\in(1/2,\sqrt{1/2}).$

###  5.2. Wigner Matrices with Boolean entries and variance profile

Consider the $n\times n$ operator-valued matrix $A_{n}$ defined by

$A_{n}=\sum_{1\leq j\leq i\leq n}\big{(}e_{ij}\otimes a_{ij}+e_{ij}^{*}\otimes
a_{ij}^{*}\big{)}$ (5.4)

where we recall that $e_{ii}=\frac{1}{2\sqrt{n}}E_{ii}$ and
$e_{ij}=\frac{1}{\sqrt{n}}E_{ij}$ for $j<i$ with $(E_{ij})_{1\leq i,j\leq n}$
denoting the standard matrix units in $M_{n}(\mathbb{C})$.

Let $a=\\{a_{ij}\mid 1\leq j\leq i\leq n\\}$ be a family of elements of
$\mathcal{A}$ that are Boolean independent, centered with respect to $\varphi$
and are such that $a_{ii}=a_{ii}^{*}$, $\varphi[a_{ii}^{2}]:=\sigma_{i}$ for
all $i$ and $\varphi[a_{ij}^{2}]=0$, $\varphi[a_{ij}a_{ij}^{*}]:=\alpha_{ij}$
and $\varphi[a_{ij}^{\ast}a_{ij}]:=\tilde{\alpha}_{ij}$ for all $j<i$. Note
that the entries of $A_{n}$ are assumed to be, up to symmetry, Boolean
independent but do not need to be identically distributed. We will give
quantitative estimates on the distribution of such matrices in terms of Cauchy
transforms. Our results generalize the ones in Popa and Hao [30] in the case
of C*-algebras, since we do not impose a condition on the convergence of
moments of order larger than $4$ for the entries and more important we allow
having a variance profile in the matrix.

With this aim, we apply Theorem 1.1 in the quadruple $(M_{n}(\mathcal{A}),$
$\operatorname{tr}_{n}\otimes\varphi,id_{n}\otimes\varphi,M_{n}(\mathbb{C}))$
associated with $(\mathcal{A},\varphi)$.

######  Theorem 5.7.

Let $A_{n}$ be an operator-valued Wigner matrix described above and $B_{n}$ an
operator-valued Bernoulli $B_{n}$ as in (5.2). Then for any
$z\in\mathbb{C}^{+}$,

$\big{|}(\operatorname{tr}_{n}\otimes\varphi)[G_{A_{n}}(z)]-(\operatorname{tr}_{n}\otimes\varphi)[G_{B_{n}}(z)]\big{|}\leq\frac{16}{{\Im}m(z)^{4}}\Big{(}\max_{j\leq
i}\|a_{ij}\|^{3}\Big{)}\frac{1}{\sqrt{n}}.$ (5.5)

###### Proof.

Let $b=\\{b_{ij}\mid 1\leq j\leq i\leq n\\}$ be a family of _Boolean
independent elements_ such that

  * •

for any $1\leq i\leq n$, $b_{ii}$ is a Bernoulli element with
$\varphi[b_{ii}]=0$ and $\varphi[b_{ii}^{2}]=\varphi[a_{ii}^{2}]$,

  * •

for any $1\leq j<i\leq N$, $b_{ij}$ is a $\eta$-diagonal circular element with
$\varphi[b_{ij}]=0$ and
$\varphi[b_{ij}b_{ij}^{*}]=\varphi[a_{ij}a_{ij}^{*}]:=\alpha_{ij}$ and
$\varphi[b_{ij}^{\ast}b_{ij}]=\varphi[a_{ij}^{\ast}a_{ij}]:=\tilde{\alpha}_{ij}$.

Without loss of generality, $b$ can be assumed to be Boolean independent from
$a$. Now set $S_{N}$ to be the $N\times N$ operator-valued Wigner matrix whose
entries are given by the family $b$, i.e.

$B_{n}=\sum_{1\leq j\leq i\leq n}\big{(}e_{ij}\otimes b_{ij}+e_{ij}^{*}\otimes
b_{ij}^{*}\big{)}.$

Finally, set $x_{ij}=e_{ij}\otimes a_{ij}+e_{ij}^{*}\otimes a_{ij}^{*}$ and
$y_{ij}=e_{ij}\otimes b_{ij}+e_{ij}^{*}\otimes b_{ij}^{*}$ for all $1\leq
j\leq i\leq n$. By an application of Lemma 5.4, one sees that, when lifted to
matrices, Boolean independence is preserved, in the sense that independence of
the entries $\\{a_{ij},b_{ij}\mid 1\leq j\leq i\leq n\\}$ with respect to
$\varphi$ implies independence with amalgamation over $M_{n}(\mathbb{C})$ of
$\\{x_{ij},y_{ij}\mid 1\leq j\leq i\leq n\\}$ with respect to
$\operatorname{id}_{n}\otimes\varphi$. Moreover, the variables are centered
and the moments of second order match: for any $b\in M_{n}(\mathbb{C})$,

$(\operatorname{id}_{n}\otimes\varphi)[x_{ij}]=(\operatorname{id}_{n}\otimes\varphi)[y_{ij}]=0\quad\text{and}\quad(\operatorname{id}_{n}\otimes\varphi)\big{[}x_{ij}bx_{ij}\big{]}=(\operatorname{id}_{n}\otimes\varphi)\big{[}y_{ij}\,b\,y_{ij}\big{]}.$

Having this in hand and setting $N=n(n+1)/2$, we follow the same steps as
Theorem 1.1. By adopting its notation, we get for any $z\in\mathbb{C}^{+}$,

$\big{|}(\operatorname{tr}_{n}\otimes\varphi)[G_{A_{n}}(z)]-(\operatorname{tr}_{n}\otimes\varphi)[G_{B_{n}}(z)]\big{|}\leq\sum_{1\leq
j\leq i\leq
n}\Big{(}\big{|}(tr_{n}\otimes\varphi)[W_{ij}(z)]\big{|}+\big{|}(tr_{n}\otimes\varphi)[\widetilde{W}_{ij}(z)]\big{|}\Big{)},$

where

$\displaystyle W_{ij}(z)$
$\displaystyle=G_{\mathbf{z}_{ij}}(z)x_{ij}G_{\mathbf{z}^{0}_{ij}}(z)x_{ij}G_{\mathbf{z}^{0}_{ij}}(z)x_{ij}G_{\mathbf{z}^{0}_{ij}}(z),$
$\displaystyle\widetilde{W}_{ij}(z)$
$\displaystyle=G_{\mathbf{z}_{i-1,j}(z)}y_{ij}G_{\mathbf{z}^{0}_{ij}}(z)y_{ij}G_{\mathbf{z}^{0}_{ij}}(z)y_{ij}G_{\mathbf{z}^{0}_{ij}}(z).$

We start by controlling the term:

$\displaystyle\big{|}(tr_{n}\otimes$
$\displaystyle\varphi)\big{[}G_{\mathbf{z}_{ij}}(z)(e_{ij}\otimes
a_{ij})G_{\mathbf{z}^{0}_{ij}}(z)(e_{ij}\otimes
a_{ij})G_{\mathbf{z}^{0}_{ij}}(z)(e_{ij}\otimes
a_{ij})G_{\mathbf{z}^{0}_{ij}}(z)\big{]}\big{|}$
$\displaystyle\leq\|G_{\mathbf{z}_{ij}}(z)\|\|G_{\mathbf{z}^{0}_{ij}}(z)\|(tr_{n}\otimes\varphi)\big{[}\big{|}(e_{ij}\otimes
a_{ij})G_{\mathbf{z}^{0}_{ij}}(z)(e_{ij}\otimes
a_{ij})G_{\mathbf{z}^{0}_{ij}}(z)(e_{ij}\otimes a_{ij})\big{|}\big{]}$
$\displaystyle\leq\frac{1}{n\sqrt{n}}\|G_{\mathbf{z}_{ij}}(z)\|\|G_{\mathbf{z}^{0}_{ij}}(z)\|(tr_{n}\otimes\varphi)\big{[}\big{|}E_{ij}\otimes
a_{ij}[G_{\mathbf{z}_{ij}^{0}}(z)]_{ji}a_{ij}[G_{\mathbf{z}_{ij}^{0}}(z)]_{ji}a_{ij}\big{|}\big{]}$
$\displaystyle\leq\frac{1}{n^{2}\sqrt{n}}\|G_{\mathbf{z}_{ij}}(z)\|\|G_{\mathbf{z}^{0}_{ij}}(z)\|^{3}\|a_{ij}\|^{3}$
$\displaystyle\leq\frac{1}{n^{2}\sqrt{n}}\frac{1}{{\Im}m(z)^{4}}\|a_{ij}\|^{3}.$

As the other terms are treated in the same way, we get that

$\displaystyle\big{|}(tr_{n}\otimes\varphi)[W_{ij}(z)]\big{|}$
$\displaystyle\leq
8\frac{1}{n^{2}\sqrt{n}}\frac{1}{{\Im}m(z)^{4}}\|a_{ij}\|^{3},$
$\displaystyle\big{|}(tr_{n}\otimes\varphi)[\widetilde{W}_{ij}(z)]\big{|}$
$\displaystyle\leq
8\frac{1}{n^{2}\sqrt{n}}\frac{1}{{\Im}m(z)^{4}}\|b_{ij}\|^{3}.$

Summing over $1\leq j\leq i\leq n$, we finally get for any
$z\in\mathbb{C}^{+}$,

$\big{|}(\operatorname{tr}_{n}\otimes\varphi)[G_{A_{n}}(z)]-(\operatorname{tr}_{n}\otimes\varphi)[G_{B_{n}}(z)]\big{|}\leq\frac{8}{{\Im}m(z)^{4}}\max_{j\leq
i}\Big{(}\|a_{ij}\|^{3}+\|b_{ij}\|^{3}\Big{)}\frac{1}{\sqrt{n}}.$

As the $b_{ij}$’s are $\eta$-diagonal elements with the same variance as the
$a_{ij}$’s, then by Remark 5.3 we have that
$\|b_{ij}\|=\max\\{\sqrt{\alpha_{ij}},\sqrt{\widetilde{\alpha}_{ij}}\\}$. To
obtain 5.5 it remains to notice that
$||a_{ij}||^{2}=||a_{ij}a_{ij}^{*}||\geq\varphi[a_{ij}a_{ij}^{*}]=\alpha_{ij}$
and
$||a_{ij}||^{2}=||a_{ij}^{*}a_{ij}||\geq\varphi[a_{ij}^{*}a_{ij}]=\widetilde{\alpha}_{ij}$.

∎

Combining Theorem 5.7 with Theorem 4.2 we can get limit theorem for Wigner
Matrices with Boolean entries.

######  Proposition 5.8.

Let $A_{n}$ be as in (5.7) and suppose that $||a_{ij}||\leq C$, for some
constant $C$ independent of $N$. Furthermore let
$\eta_{n}(b)=(id_{n}\otimes\varphi)[A_{n}bA_{n}]$ and suppose that
$\eta_{n}(1)$ has distribution $\rho_{n}$ with respect $tr\otimes\varphi$. If,
$\rho_{n}$ to converges to $\rho$ in distribution, then, the distribution of
$A_{n}$ with respect to $tr\otimes\varphi$ converges to the unique symmetric
measure $\mu$ such that $\mu^{(2)}=\rho$.

###### Proof.

Recall that convergence in distribution is equivalent to uniform convergence
of the Cauchy transform on compact sets. Let $\mu_{A_{n}}$ be the distribution
of $A_{n}$ with respect to $tr\otimes\varphi$. We will prove that for any
$\varepsilon>0$, and any $K$ a compact sets of $\mathbb{C}^{+}$ there is an
$N$ such that, for any $n>N$,

$\big{|}\mathcal{G}_{\mu_{A_{n}}}(z)-\mathcal{G}_{\mu}(z)\big{|}\leq\varepsilon,\quad\text{for
all }z\in K.$

To do this, let $B_{n}$ be an operator-valued Bernoulli as in Theorem 5.7 with
distribution $\rho_{n}$ and let $B$ be an operator-valued Bernoulli random
variable with variance map $\eta$ such that $\eta(1)$ has distribution $\rho$.
Since $\rho_{n}$ converges weakly to $\rho$, then by Corollary 5.2,
$\mu_{B_{n}}$ converges to $\mu_{B}$, weakly. Hence, there exists an integer
$N_{1}>1$ such that, for all $n>N_{2}$,

$\big{|}\mathcal{G}_{\mu_{B_{n}}}(z)-\mathcal{G}_{\mu_{B}}(z)\big{|}\leq\varepsilon/2,\quad\text{for
all }z\in K.$

On the other hand, equation (5.5) yields that there exists $N_{2}>1$ such that
for all $n>N_{2}$,

$|\mathcal{G}_{\mu_{A_{n}}}(z)-\mathcal{G}_{\mu_{B_{n}}}(z)\big{|}\leq\varepsilon/2,\quad\text{for
all }z\in K.$ (5.6)

The result follows by taking $N=max(N_{1},N_{2})$. ∎

Let us finish by considering the case when $\\{a_{ij}\\}_{n\geq i>j\geq 0}$
are identically distributed, which corresponds to Theorem 5.1 of Popa and
Hao[30].

######  Example 5.9.

Let $\\{a_{ij}\\}_{n\geq i>j\geq 0}$ be a family of Boolean independent
identically distributed random variables in some $C^{*}$-probability space
$(A,\varphi)$ and let $A_{n}$ be the $n\times n$ matrix defined by

$A_{n}=\sum_{1\leq j\leq i\leq n}\big{(}e_{ij}\otimes a_{ij}+e_{ij}^{*}\otimes
a_{ij}^{*}\big{)}.$ (5.7)

If we denote by $\varphi(a_{ij}a_{ij}^{\ast})=\alpha$ and
$\varphi(a_{ij}^{*}a_{ij})=\tilde{\alpha}$, then
$\rho_{n}:=\eta_{n}(b)=(id_{n}\otimes\varphi)[A_{n}bA_{n}]=(id_{n}\otimes\varphi)[B_{n}bB_{n}]$
where $B_{n}$ is the matrix-valued Bernoulli as in part (i) from Example 5.6
with $\sigma=0$. Thus $\rho_{n}$ converges to the uniform distribution
$\text{Uniform}\big{(}\tilde{\alpha},\alpha\big{)}$ and hence we conclude by
Theorem 5.7 the limiting distribution of $A_{n}$ with respect to
$\operatorname{tr}_{n}\otimes\varphi$ is

$\mu(dt)=\frac{|t|}{(\alpha-\tilde{\alpha})}dt,\quad|t|\in(\sqrt{\tilde{\alpha}},\sqrt{\alpha}).$

In the case $\alpha=\tilde{\alpha}$, $\rho_{n}$ converges to $\delta_{\alpha}$
and
$\mu_{A_{n}}\to\frac{1}{2}\delta_{\sqrt{\alpha}}+\frac{1}{2}\delta_{\sqrt{\alpha}}$.

## Acknowlegments

OA was supported by CONACYT Grant CB-2017-2018-A1-S-9764 and by the SFB-TRR
195 “Symbolic Tools in Mathematics and their Application” of the German
Research Foundation (DFG). This project started when OA and MB were visiting
Saarland University. They would like to thank Roland Speicher for his
hospitality.

##  Appendix A Operator-valued Infinitesimal Products

This appendix is dedicated to the study of operator-valued infinitesimal
products of operator-valued infinitesimal probability spaces. Note that the OV
free product was introduced in [34], while the OV Boolean and monotone cases
have been studied in [31] and [28], respectively. We also refer the reader to
[17, Chapter 7] for a nice summary on the subject. However, this is less
studied in the infinitesimal setting where only the (scalar-valued)
infinitesimal free product was constructed in [12]. In this appendix, we give
an algebraic construction of the operator-valued infinitesimal product
algebraically in the free, Boolean, and monotone settings.

First of all, let us recall the following result (see [17, Theorem 4.3.1]):
suppose that $H_{1},\dots,H_{n}$ are Hilbert-$\mathcal{B}$-bimodules and
$\xi_{j}$ is a unit vector in $H_{j}$ for each $j=1,\dots,n$. In addition, for
each $j$, we define the map $E_{j}[b]=\langle\xi_{j},b\xi_{j}\rangle$ for all
$b\in\mathcal{B}(H_{j})$. Then there exists a Hilbert $\mathcal{B}$-bimodule
$H$ and a unit vector $\xi$ in $H$, and an injective $*$-homomorphism
$\rho_{j}:\mathcal{B}(H_{j})\to\mathcal{B}(H)$ such that

$E[\rho_{j}(b)]=E_{j}[b]\text{ for each }b\in\mathcal{B}(H_{j}).$

Moreover,
$\rho_{1}(\mathcal{B}(H_{1})),\rho_{2}(\mathcal{B}(H_{2})),\dots,\rho_{n}(\mathcal{B}(H_{n}))$
are freely/Boolean/monotone independent with respect to $E$.

Suppose that for each $j$, there exists a self-adjoint $\mathcal{B}$-bimodule
linear map $E^{\prime}_{j}:\mathcal{B}(H_{j})\to\mathcal{B}$ that is
completely bounded with $E_{j}^{\prime}[1]=0$. Here we let $\mathcal{A}$ be be
the unital $C^{*}$-algebra generated by
$\rho_{1}(\mathcal{B}(H_{1})),\dots,\rho_{n}(\mathcal{B}(H_{n}))$. For each
notion of independence, our aim is to construct a self-adjoint
$\mathcal{B}$-bimodule linear map $E^{\prime}$ on $\mathcal{A}$ that vanishes
on $\mathcal{B}$ and for which
$\rho_{1}(\mathcal{B}(H_{1})),\rho_{2}(\mathcal{B}(H_{2})),$
$\dots,\rho_{n}(\mathcal{B}(H_{n}))$ are infinitesimally independent in the
OVI probability space $(\mathcal{A},E,E^{\prime},\mathcal{B})$. Note that for
each $j$, we shall define $E^{\prime}$ on $\rho_{j}(\mathcal{B}(H_{j}))$ as
follows:

$E^{\prime}[\rho_{j}(b)]:=E^{\prime}_{j}[b]\text{ for all
}b\in\mathcal{B}(H_{j}).$

To describe the construction of $E^{\prime}$ on $\mathcal{A}$, it is
sufficient to show how $E^{\prime}$ is defined on terms of the following form

$\rho_{j_{1}}(\mathcal{B}(H_{j_{1}}))\rho_{j_{2}}(\mathcal{B}(H_{j_{2}}))\dots\rho_{j_{k}}(\mathcal{B}(H_{j_{k}}))$

where $j_{1},\dots,j_{k}\in[n]$ and $j_{1}\neq j_{2}\neq\dots\neq j_{k}$.
The Infinitesimally Free Case:

$E^{\prime}[\rho_{j_{1}}(b_{1})\dots\rho_{j_{k}}(b_{k})]:=\sum\limits_{s=1}^{k}E[\rho_{j_{1}}(b_{1})\dots\rho_{j_{s-1}}(b_{s-1})E^{\prime}[\rho_{j_{s}}(b_{s})]\rho_{j_{s+1}}(b_{s+1})\dots\rho_{j_{k}}(b_{k})]$

where $b_{1}\in H_{j_{1}},\dots,b_{k}\in H_{j_{k}}$ for some $j_{1}\neq
j_{2}\neq\dots\neq j_{k}$ that $E_{j_{1}}(b_{1})=\dots=E_{j_{k}}(b_{k})=0.$
The Infinitesimally Boolean Case:

$E^{\prime}[\rho_{j_{1}}(b_{1})\dots\rho_{j_{k}}(b_{k})]:=\sum\limits_{s=1}^{k}E[\rho_{j_{1}}(b_{1})]\dots
E[\rho_{j_{s-1}}(b_{s-1})]E^{\prime}[\rho_{j_{s}}(b_{s})]E[\rho_{j_{s+1}}(b_{s+1})]\dots
E[\rho_{j_{k}}(b_{k})]$

where $b_{1}\in H_{j_{1}},\dots,b_{k}\in H_{j_{k}}$ for some $j_{1}\neq
j_{2}\neq\dots\neq j_{k}$.
The Infinitesimally Monotone Case:

$\displaystyle E^{\prime}[\rho_{j_{1}}(b_{1})\dots\rho_{j_{k}}(b_{k})]$
$\displaystyle:=$ $\displaystyle
E^{\prime}[\rho_{j_{1}}(b_{1})\dots\rho_{j_{s-1}}(b_{s-1})E[\rho_{j_{s}}(b_{s})]\rho_{j_{s+1}}(b_{s+1})\dots\rho_{j_{k}}(b_{k})]$
$\displaystyle+E[\rho_{j_{1}}(b_{1})\dots\rho_{j_{s-1}}(b_{s-1})E^{\prime}[\rho_{j_{s}}(b_{s})]\rho_{j_{s+1}}(b_{s+1})\dots\rho_{j_{k}}(b_{k})]$

whenever $b_{1}\in H_{j_{1}},\dots,b_{k}\in H_{j_{k}}$ and
$j_{s-1}<j_{s}>j_{s+1}$.

Note that the fact that $E^{\prime}$ is well-defined on $\mathcal{A}$ in the
Boolean case is obvious. For the free case, we observe that $\mathcal{A}$ can
be written as follows:

$\mathcal{A}=\mathcal{B}1\oplus\bigoplus\limits_{n\geq
0}\bigoplus\limits_{i_{1}\neq\cdots\neq
i_{n}}\rho_{i_{1}}(\mathcal{B}(H_{i_{1}}))^{\circ}\rho_{i_{2}}(\mathcal{B}(H_{i_{2}}))^{\circ}\dots\rho_{i_{n}}(\mathcal{B}(H_{i_{n}}))^{\circ}$

where $\rho_{i_{j}}(\mathcal{B}(H_{i_{j}}))^{\circ}$ is the set of centered
elements of $\rho_{i_{j}}(\mathcal{B}(H_{i_{j}}))^{\circ}$. Due to the direct
sum decomposition, we only need to define it on $\mathcal{B}1$ and each of
$\rho_{i_{1}}(\mathcal{B}(H_{i_{1}}))^{\circ}\rho_{i_{2}}(\mathcal{B}(H_{i_{2}}))^{\circ}\dots\rho_{i_{n}}(\mathcal{B}(H_{i_{n}}))^{\circ}$.
Therefore, for the free case, $E^{\prime}$ is well-defined on $\mathcal{A}$.
In addition, $E^{\prime}$ is also well-defined on $\mathcal{A}$ for the
monotone case by recursively computing the mixed moments.

Since $E_{j}$ and $E_{j}^{\prime}$ are both self-adjoint for each $j$, we can
easily see that $E^{\prime}$ is self-adjoint for each case. Moreover,
following the definition of $E^{\prime}$, it is clear that
$\rho_{1}(\mathcal{B}(H_{1})),\rho_{2}(\mathcal{B}(H_{2})),\dots,$
$\rho_{n}(\mathcal{B}(H_{n}))$ are infinitesimally freely/ Boolean/monotone
independent with respect to $(E,E^{\prime})$. However, it is not clear whether
we have any continuity property for $E^{\prime}$. Hence, we only have the
operator-valued infinitesimal products in the algebraic sense.

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