$\left(\det\left(\mathbb{I}_{n}-QUQ^{*}\right)\right)^{-n-r}=1+O\left(\|Q\|^{2}\right).$

As for the second factor, the averaged product of characteristic polynomials
has actually been verified in Theorem 3.1 (i), except that some minor
correction will be needed because of $Q$ and the scaling factor $(N-n)/N$. So
we repeat almost the same procedure quickly, with emphasis on the differences.

Note that

$\sqrt{\mathbb{I}_{r}-PQ^{*}QP^{*}}=\mathbb{I}_{r}+O\left(\|Q\|^{2}\right),$

we have

$\widetilde{A}_{0}=A_{0}+O\left(N^{-1}+\|Q\|^{2}\right).$ (5.21)

Obviously,

$\sqrt{\frac{N}{N-n}}z_{i}=z_{0}+\frac{\hat{z}_{i}}{\sqrt{N}}+\frac{nz_{0}}{2N}+O\big{(}N^{-\frac{3}{2}}\big{)}.$
(5.22)

Using the duality relation (2.9) in Corollary 2.5, we have

$\mathbb{E}\bigg{(}\prod_{i=1}^{n}\Big{|}\det\Big{(}\sqrt{\frac{N}{N-n}}z_{i}-\widetilde{X}_{N-n}\Big{)}\Big{|}^{2}\bigg{)}=\int_{\mathbb{C}^{n^{2}}}Q_{2}\big{(}A;\widetilde{X}_{0},Y\big{)}\widehat{P}_{K,2}(1;Y,0){\rm
d}Y,$

where $Q_{2}\big{(}A;\widetilde{X}_{0},Y\big{)}$, $\widehat{P}_{K,2}(1;Y,0)$
and $A$ are defined in (2.6), (2.3), (2.10) respectively, but with $N\to N-n$,
$\widetilde{X}_{0}={\rm diag}(\widetilde{A}_{0},0_{N-n-r})$ and
$Z=\sqrt{N/(N-n)}{\rm diag}(z_{1},\cdots,z_{n})$.

Proceeding as in the proof of Proposition 3.5, we can prove that for
sufficiently small $\delta>0$ and for $|z_{0}|=1$

$F_{n}^{(2)}(Z,Z)=\frac{1}{\widehat{Z}_{K,2}}\Big{(}I_{\delta,N}^{(2)}+O\big{(}e^{-\frac{1}{2}N\Delta^{(2)}}\big{)}\Big{)},$

with

$I_{\delta,N}^{(2)}=\int\limits_{{\rm Tr}(YY^{*})\leq\delta}\
g_{2}(Y)\exp\left\\{-Nf_{2}(Y,Z_{0})+Nh_{2}(Y)\right\\}{\rm d}Y,$ (5.23)

where

$g_{2}(Y)=\det(B_{11}^{(2)})\left(\det\left(A_{2}\right)\right)^{-n-r},$
(5.24) $f_{2}(Y)=\Big{(}1-\frac{n}{N}\Big{)}{\rm
Tr}(YY^{*})-\log\det\left(\mathbb{I}_{n}+YY^{*}\right).$

Noting that we need to replace $\hat{Z}$ with
$\hat{Z}+\frac{1}{2}nz_{0}N^{-\frac{1}{2}}\mathbb{I}_{n}+O\left(N^{-1}\right)$
both in (3.34) and (3.38), we then get

$h_{2}(Y)=\log\det\big{(}\mathbb{I}_{2n}+N^{-\frac{1}{2}}H_{2}\big{)}.$

Moreover, we can obtain

$f_{2}(Y)=\frac{1}{2}{\rm
Tr}\left((YY^{*})^{2}\right)+O\left(\|Y\|^{6}+N^{-1}\|Y\|^{2}\right),$

and

$\displaystyle
Nh_{2}(Y)=h_{2,0,0}-\sqrt{N}h_{2,0}(Y)+O\left(\|Y\|^{2}+\sqrt{N}\|Y\|^{4}+N^{-\frac{1}{2}}\right)$
(5.25)

where

$h_{2,0,0}=\sqrt{N}{\rm Tr}(Z_{0}^{-1}\hat{\Lambda})+n^{2}-\frac{1}{2}{\rm
Tr}(Z_{0}^{-2}\hat{\Lambda}^{2}),\,h_{2,0}(Y)=z_{0}^{-1}{\rm
Tr}(Y^{*}Y\hat{Z})+\overline{z_{0}}^{-1}{\rm Tr}(YY^{*}\hat{Z}^{*}).$

For the factors of $g_{2}(Y)$ in (5.24), noting (5.22), we have

$A_{2}=Z_{0}+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}.$

from which

$\left(\det(A_{2})\right)^{-n-r}=1+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}.$
(5.26)

On the other hand, like Lemma 4.1, taking $\hat{\theta}_{i}=0$
($i=1,\cdots,m$), and with (5.21) in mind, we have

$\displaystyle\det\left(B_{11}^{(2)}\right)$
$\displaystyle=\prod_{i=m+1}^{\gamma}|z_{0}-\theta_{i}|^{2n\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}$
(5.27)
$\displaystyle\times\bigg{\\{}\det\begin{bmatrix}N^{-\frac{1}{4}}\Omega_{11}&\Omega_{12}\\\
\Omega_{21}&N^{-\frac{1}{4}}\Omega_{22}\end{bmatrix}+O\Big{(}\sum_{(\alpha_{1},\alpha_{2},\alpha_{3})}N^{-\alpha_{1}}\|Y\|^{\alpha_{2}}\|Q\|^{2\alpha_{3}}\Big{)}\bigg{\\}},$

where $\alpha_{1},\alpha_{2},\alpha_{3}$ satisfy the restriction

$4\alpha_{1}+\alpha_{2}+\alpha_{3}\geq
2n\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}\beta_{i,j}+\delta_{\alpha_{3},0},$

where $\Omega_{ij}$ are defined in (3.2) and (3.3). Also noting
$\Omega_{11}=\Omega_{22}=0$ and

$\det\begin{bmatrix}0&\Omega_{12}\\\
\Omega_{21}&0\end{bmatrix}=\Big{(}\det\big{(}U_{I_{1}}\big{)}\det\big{(}\big{(}U^{-1})_{I_{2}}\big{)}\Big{)}^{n}\big{(}\det(Y^{*}Y)\big{)}^{\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}\beta_{i,j}},$

where $I_{2}$ is the index set consisting of numbers corresponding to last
columns of all blocks $R_{p_{i,j}}\left(\theta_{i}\right)$ from $J_{2}$ in
(1.21) where $1\leq i\leq m,1\leq j\leq\alpha_{i}$.

Combine (5.24), (5.26) and (5.27), we get

$g_{2}(Y,\Lambda)=\left(D^{(2,c)}+O\big{(}N^{-\frac{1}{4}}+\|Y\|\big{)}\right)\Big{(}\big{(}\det(Y^{*}Y)\big{)}^{t}+O\Big{(}\sum_{(\alpha_{1},\alpha_{2},\alpha_{3})}N^{-\alpha_{1}}\|Y\|^{\alpha_{2}}\|Q\|^{2\alpha_{3}}\Big{)}\Big{)},$

where $t=\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}\beta_{i,j}$ and

$D^{(2,c)}=\Big{(}\det\big{(}U_{I_{1}}\big{)}\det\big{(}\big{(}U^{-1})_{I_{2}}\big{)}\Big{)}^{n}|z_{0}-\theta_{i}|^{2n\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}.$

By the change of variable $Y\to N^{-\frac{1}{4}}Y$ in (5.23), as in the proof
of Theorem 3.1 (i) we can get

$F^{(2)}_{n}(Z,Z)=\frac{e^{n^{2}}}{\hat{Z}_{K,2}}|z_{0}-\theta_{i}|^{2n\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}N^{-\frac{1}{2}n^{2}-\frac{1}{2}\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}n\beta_{i,j}}e^{\sqrt{N}{\rm
Tr}(z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}})}\\\ \times
e^{-\frac{1}{2}{\rm
Tr}\big{(}z_{0}^{-2}\hat{Z}^{2}+\overline{z_{0}^{-2}\hat{Z}^{2}}\big{)}}\Big{(}I^{(2)}(z_{0}^{-1}\hat{Z})+O\big{(}\sum_{(\alpha_{1},\alpha_{3})}N^{-\alpha_{1}}\|Q\|^{2\alpha_{3}}\big{)}\Big{)},$
(5.28)

where
$I^{(2)}(z_{0}^{-1}\hat{Z}):=I^{(2)}(0,\ldots,0;z_{0}^{-1}\hat{Z},z_{0}^{-1}\hat{Z})$
is defined in (3.4) and $\alpha_{1},\alpha_{3}$ satisfy the restriction
$4\alpha_{1}+\alpha_{3}\geq\delta_{\alpha_{3},0}$.

So this completes a modified estimate of $\widetilde{g}_{2}(Q)$.

Step 3: Summary and matrix integrals.

For the matrix integral (5.9), make the change of variables

$Q_{R}\to N^{-\frac{1}{2}}Q_{R},\quad Q_{L}\to
N^{-\frac{1}{4}}Q_{L}(\hat{U}^{(1,1)})^{-\frac{1}{2}},$

and then integrate out $Q_{R}$, through the calculation we obtain

$\frac{1}{N^{n}}R_{N}^{(n)}\left(A_{0};z_{1},\cdots,z_{n}\right)=(2\pi)^{-\frac{n}{2}}\pi^{-tn-
n-n^{2}}\rho^{n}\\\ \times\prod_{1\leq i<j\leq
n}\left|\hat{z}_{i}-\hat{z}_{j}\right|^{2}e^{-\frac{1}{2}{\rm
Tr}\left(z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\right)^{2}}\left(I^{(2)}(z_{0}^{-1}\hat{Z})\tilde{I}^{(2)}(z_{0}^{-1}\hat{Z})+O\big{(}N^{-\frac{1}{4}}\big{)}\right),$
(5.29)

where for an $n\times t$ matrix $Q$

$\tilde{I}^{(2)}(\hat{Z})=\int\exp\\!\Big{\\{}-\frac{1}{2}\sum_{i=1}^{n}\left(QQ^{*}\right)_{ii}^{2}+{\rm
Tr}\left(\big{(}\hat{Z}^{*}+\hat{Z}\big{)}QQ^{*}\right)\Big{\\}}{\rm d}Q,$
(5.30)

and

$\rho:=|z_{0}-\theta_{i}|^{2\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}\frac{1}{\det(\widetilde{U}^{(22)})}\det(U)\det\big{(}\big{(}U^{-1})_{I_{2}}\big{)}=1.$

Here the second identity in the latter follows from $\widetilde{U}^{(22)}$ in
(5.16), $(U^{-1})_{I_{2}}$ and the Schur complement technique.

Now we need to evaluate the two matrix integrals $\tilde{I}^{(2)}$ and
$I^{(2)}$. For $\tilde{I}^{(2)}$ in (5.30), treated as $n$ integrals
separately according to its rows it is simplified to

$\displaystyle\tilde{I}^{(2)}\big{(}z_{0}^{-1}\hat{Z}\big{)}$
$\displaystyle=\prod_{i=1}^{n}\int\cdots\int\exp\Big{\\{}-\frac{1}{2}\big{(}\sum_{j=1}^{t}\left|q_{1j}\right|^{2}\big{)}^{2}+\left(z_{0}^{-1}\hat{z}_{i}+\overline{z_{0}}^{-1}\overline{\hat{z}_{i}}\right)\sum_{j=1}^{t}\left|q_{1j}\right|^{2}\Big{\\}}\prod_{j=1}^{t}{\rm
d}\Re q_{1j}{\rm d}\Im q_{1j}.$

Apply the spherical polar coordinates and we can get

$\tilde{I}^{(2)}\big{(}z_{0}^{-1}\hat{Z}\big{)}=\pi^{(t+\frac{1}{2})n}\prod_{i=1}^{n}\Big{(}e^{\frac{1}{2}(z_{0}^{-1}\hat{z}_{i}+\overline{z_{0}}^{-1}\overline{\hat{z}_{i}})^{2}}\mathrm{IE}_{t-1}\big{(}z_{0}^{-1}\hat{z}_{i}+\overline{z_{0}}^{-1}\overline{\hat{z}_{i}}\big{)}\Big{)},$
(5.31)

where $\mathrm{IE}_{n}(z)$ is defined in (1.24).

For
$I^{(2)}(z_{0}^{-1}\hat{Z}):=I^{(2)}(0,\ldots,0;z_{0}^{-1}\hat{Z},z_{0}^{-1}\hat{Z})$
defined in (3.4), by the singular value decomposition:

$Y=U\sqrt{R}V,\quad R={\rm diag}\left(r_{1},\cdots,r_{n}\right),\quad
r_{1}\geq\cdots\geq r_{n}\geq 0,$ (5.32)

the Jacobian reads

${\rm d}Y=\pi^{n^{2}}\Big{(}\prod_{i=1}^{n-1}i!\Big{)}^{-2}\prod_{1\leq
i<j\leq n}(r_{j}-r_{i})^{2}{\rm d}R{\rm d}U{\rm d}V,$ (5.33)

where $U$ and $V$ are chosen from the unitary group $\mathcal{U}(n)$ with the
Haar measure. Use Harish-Chandra-Itzykson-Zuber integration formula [29, 37]
(see also [43])

$\int_{\mathcal{U}(n)}\exp\\!\left\\{{\rm
Tr}\left(AUBU^{*}\right)\right\\}{\rm
d}U=\frac{\det\big{(}[e^{a_{i}b_{j}}]_{i,j=1}^{n}\big{)}}{\prod_{1\leq i<j\leq
n}(a_{j}-a_{i})(b_{j}-b_{i})}\prod_{i=1}^{n-1}i!,$ (5.34)

with $A={\rm diag}\left(a_{1},\cdots,a_{n}\right)$ and $B={\rm
diag}\left(b_{1},\cdots,b_{n}\right)$, and also Andréief’s integration formula
[4]

$\int\det\big{(}[f_{i}(x_{j})]_{i,j=1}^{n}\big{)}\det\big{(}[g_{i}(x_{j})]_{i,j=1}^{n}\big{)}\prod_{i=1}^{n}{\rm
d}\mu(x_{i})=n!\det\Big{(}\Big{[}\int f_{i}(x)g_{j}(x){\rm
d}\mu(x)\Big{]}_{i,j=1}^{n}\Big{)},$ (5.35)

we obtain

$\displaystyle(2\pi)^{-\frac{n}{2}}\pi^{-n^{2}}$ $\displaystyle\prod_{1\leq
i<j\leq n}|\hat{z}_{i}-\hat{z}_{j}|^{2}e^{-\frac{1}{2}{\rm
Tr}\big{(}z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\big{)}^{2}}I^{(2)}(z_{0}^{-1}\hat{Z})$
(5.36)
$\displaystyle=\det\left(\left[\Gamma(t+1)e^{-\frac{1}{2}\left(\left|\hat{z}_{i}\right|^{2}+\left|\hat{z}_{j}\right|^{2}\right)+\hat{z}_{i}\overline{\hat{z}_{j}}}\mathrm{IE}_{t}\left(z_{0}^{-1}\hat{z}_{i}+\overline{z_{0}}^{-1}\overline{\hat{z}_{j}}\right)\right]_{i,j=1}^{n}\right).$

Combining (5.29), (5.31) and (5.36), we thus give a complete proof of Theorem
1.5. ∎

###  5.3  Proof of Proposition 1.4

###### Proof of Proposition 1.4.

Step 1. To verify (1.13), we use the fact that any real quaternion matrix can
be triangularized by a unitary quaternion matrix; see e.g. [41]. That is, the
Schur decomposition gives rise to

$X=V^{*}\left(\begin{bmatrix}\Lambda&\\\
&\overline{\Lambda}\end{bmatrix}+T\right)V,$ (5.37)

where $\Lambda=\left(z_{1},\cdots,z_{N}\right)$ is diagonal and

$V=\begin{bmatrix}V^{(1)}&V^{(2)}\\\
-\overline{V^{(2)}}&\overline{V^{(1)}}\end{bmatrix},\quad
T=\begin{bmatrix}T_{1}&T_{2}\\\
-\overline{T}_{2}&\overline{T}_{1}\end{bmatrix},$

in which $V$ is a unitary matrix, both $T_{1}$ and $T_{2}$ are strictly upper
triangular. The Jacobian is given by

${\rm d}X=\prod_{i=1}^{N}\left|z_{i}-\overline{z}_{i}\right|^{2}\prod_{1\leq
i<j\leq
N}\left|z_{i}-z_{j}\right|^{2}\left|z_{i}-\overline{z}_{j}\right|^{2}{\rm
d}\Lambda{\rm d}T_{1}{\rm d}T_{2}{\rm d}V,$ (5.38)

see e.g. [2].

Put

$B=VX_{0}V^{*},\quad V^{(i)}=\begin{bmatrix}V_{i1}&V_{i2}\end{bmatrix},\quad
V_{1}=\begin{bmatrix}V_{11}&V_{21}\\\
-\overline{V_{21}}&\overline{V_{11}}\end{bmatrix},$

where for $i=1,2$, $V_{i1}$ and $V_{i2}$ are of sizes $N\times r$ and
$N\times(N-r)$ respectively, then

$B=V_{1}A_{0}V_{1}^{*}.$

By (5.37) we arrive at

$\displaystyle{\rm Tr}\left(X-X_{0}\right)$
$\displaystyle\left(X-X_{0}\right)^{*}=2{\rm
Tr}\left(\Lambda\Lambda^{*}\right)+{\rm Tr}\left((T-B)(T-B)^{*}\right)$
$\displaystyle-{\rm Tr}\left(\begin{bmatrix}\Lambda&\\\
&\Lambda^{*}\end{bmatrix}B^{*}+B\begin{bmatrix}\Lambda^{*}&\\\
&\Lambda\end{bmatrix}\right).$

Integrate out $V_{12}$ and $V_{22}$, we get

$\displaystyle
f_{N,4}\left(A_{0};z_{1},\cdots,z_{N}\right)=\frac{1}{\widetilde{Z}_{N,r;4}}e^{-2N\sum_{k=1}^{N}\left|z_{k}\right|^{2}}\prod_{i=1}^{N}\left|z_{i}-\overline{z_{i}}\right|^{2}\prod_{1\leq
i<j\leq N}\left|z_{i}-z_{j}\right|^{2}\left|z_{i}-\overline{z_{j}}\right|^{2}$
(5.39)
$\displaystyle\times\int\delta\left(V_{1}^{*}V_{1}-\mathbb{I}_{2r}\right)e^{N\Phi_{N}(\Lambda,B,T)}{\rm
d}V_{11}{\rm d}V_{21}{\rm d}T_{1}{\rm d}T_{2},$

where

$\Phi_{N}(\Lambda,B,T)=-{\rm Tr}\big{(}(T-B)(T-B)^{*}\big{)}+{\rm
Tr}\bigg{(}\begin{bmatrix}\Lambda&\\\
&\Lambda^{*}\end{bmatrix}B^{*}+B\begin{bmatrix}\Lambda^{*}&\\\
&\Lambda\end{bmatrix}\bigg{)}.$

Again, integrate out the strictly upper triangular matrices $T_{1}$ and
$T_{2}$, which are indeed Gaussian matrix integrals, we immediately get
(1.13).

To compute constants $Z_{N,r;4}$ and $\widetilde{Z}_{N,r;4}$, noting that they
are independent from $A_{0}$, choose $A_{0}=0$. For $Z_{N,r;4}$, the integral
in (1.13) reduces to the volume

$\int\delta\left(V_{1}^{*}V_{1}-\mathbb{I}_{2r}\right){\rm d}V_{11}{\rm
d}V_{21}=\frac{\pi^{2Nr+r-r^{2}}}{\prod_{j=0}^{r-1}\Gamma(2(N-j))}$

which follows from the formula [39, Proposition 3.3]. Combining the fact that
$f_{N}\left(0;z_{1},\cdots,z_{N}\right)$ is a density and the integration
formula (see e.g. [43])

$\displaystyle\int_{\Im z_{1},\ldots,\Im z_{N}\geq
0}e^{-2N\sum_{k=1}^{N}\left|z_{k}\right|^{2}}$
$\displaystyle\prod_{i=1}^{N}\left|z_{i}-\overline{z_{i}}\right|^{2}\prod_{1\leq
i<j\leq
N}\left|z_{i}-z_{j}\right|^{2}\left|z_{i}-\overline{z_{j}}\right|^{2}d^{N}z$
$\displaystyle=(2N)^{-N(N+1)}\pi^{N}N!\prod_{k=1}^{N}(2k-1)!,$

we thus arrive at (1.14). Similarly, we can obtain

$\widetilde{Z}_{N,r;4}=(2N)^{-2N^{2}}N!\pi^{N(N+2r)+r-r^{2}}\prod_{k=1}^{N-r}\Gamma(2k).$

Step 2.  The goal is to obtain integral representations for correlation
functions.

For $i=1,2$, set

$\displaystyle B=\begin{bmatrix}B_{1}&B_{2}\\\
-\overline{B_{2}}&\overline{B_{1}}\end{bmatrix},\quad
B_{i}=\left[B_{i;k,l}\right]_{k,l=1}^{2};V_{i1}=\begin{bmatrix}Q_{i}\\\
V_{D;i1}\end{bmatrix},$ $\displaystyle\quad
V_{D}=\begin{bmatrix}V_{D;11}&V_{D;21}\\\
-\overline{V_{D;21}}&\overline{V_{D;11}}\end{bmatrix};\quad\Lambda=\begin{bmatrix}\Lambda_{1}&\\\
&\Lambda_{2}\end{bmatrix};\quad T_{i}=\begin{bmatrix}T_{i;11}&T_{i;12}\\\
0_{(N-n)\times n}&T_{i;22}\end{bmatrix},$

where $B_{1}$ and $B_{i;11}$ are of size $N\times N$ and $n\times n$
respectively; $Q_{i}$ is of size $n\times r$; $T_{i;11}$ is of size $n\times
n$ and $\Lambda_{1}={\rm diag}\left(z_{1},\cdots,z_{n}\right)$.

In view of (5.39), integrate out $T_{i;12}$ and we get an extra factor
$(2N/\pi)^{2n(n-N)}$. If we set

$\displaystyle B_{U}=\begin{bmatrix}B_{1;11}&B_{2;11}\\\
-\overline{B_{2;11}}&\overline{B_{1;11}}\end{bmatrix},\quad
B_{D}=\begin{bmatrix}B_{1;22}&B_{2;22}\\\
-\overline{B_{2;22}}&\overline{B_{1;22}}\end{bmatrix},\quad
B_{21}=\begin{bmatrix}B_{1;21}&B_{2;21}\\\
-\overline{B_{2;21}}&\overline{B_{1;21}}\end{bmatrix};$ $\displaystyle
T_{U}=\begin{bmatrix}T_{1;11}&T_{2;11}\\\
-\overline{T_{2;11}}&\overline{T_{1;11}}\end{bmatrix},\quad
T_{D}=\begin{bmatrix}T_{1;22}&T_{2;22}\\\
-\overline{T_{2;22}}&\overline{T_{1;22}}\end{bmatrix},$

then $\Phi_{N}(\Lambda,B,T)$ can be divided into three parts

$\Phi_{N}(\Lambda,B,T)=\Phi_{n}(\Lambda_{1},B_{U},T_{U})+\Phi_{N-n}(\Lambda_{1},B_{D},T_{D})-{\rm
Tr}\left(B_{21}B_{21}^{*}\right).$

It’s easy to see

$B_{D}=V_{D}A_{0}V_{D}^{*},$

and

${\rm Tr}\left(B_{21}B_{21}^{*}\right)={\rm
Tr}\left(V_{D}^{*}V_{D}A_{0}Q^{*}QA_{0}^{*}\right),\quad
V_{D}^{*}V_{D}+Q^{*}Q=\mathbb{I}_{2r}.$

After a change of variables
$V_{D}=\widetilde{V}_{D}\sqrt{\mathbb{I}_{2r}-Q^{*}Q}$ the integral in (5.39)
can be rewritten as

$\int\int{\rm d}V_{11}{\rm
d}V_{21}\delta\left(V_{1}^{*}V_{1}-\mathbb{I}_{2r}\right)\Big{(}\cdot\Big{)}=\\\
\int_{Q^{*}Q\leq\mathbb{I}_{2r}}{\rm d}Q_{1}{\rm
d}Q_{2}\left(\det\left(\mathbb{I}_{2r}-Q^{*}Q\right)\right)^{N-n-r+\frac{1}{2}}\int{\rm
d}\widetilde{V}_{D;11}{\rm
d}\widetilde{V}_{D;21}\delta\left(\widetilde{V}_{D}^{*}\widetilde{V}_{D}-\mathbb{I}_{2r}\right)\Big{(}\cdot\Big{)}.$
(5.40)

So if we set $\tilde{z}_{k}=\sqrt{N/(N-n)}z_{k+n},k=1,\cdots,N-n,$ and

$\displaystyle\widetilde{A}_{0}=\sqrt{N/(N-n)}\sqrt{\mathbb{I}_{2r}-Q^{*}Q}A_{0}\sqrt{\mathbb{I}_{2r}-Q^{*}Q},\quad\widetilde{T}_{D}=\sqrt{N/(N-n)}T_{D},$

and treat $\\{\tilde{z}_{k}\\}$ as eigenvalues of a $(N-n)\times(N-n)$
quaternion Ginibre matrix $\widetilde{X}$ perturbed by a $2r\times 2r$ matrix
$\widetilde{A}_{0}$, we immediately obtain the desired integral representation
from (5.39) and (5.40).

This thus completes the proof. ∎

###  5.4 Proof of Theorem 1.6

To verify Theorem 1.6, we will use the following lemma.

######  Lemma 5.2.

Given $z_{0}\in\mathbb{C}$ such that $|z_{0}|\leq 1$, let

$Z_{0,n}=\begin{bmatrix}z_{0}\mathbb{I}_{n}&\\\
&\overline{z_{0}}\mathbb{I}_{n}\end{bmatrix},\quad
A_{0}=\begin{bmatrix}A_{1}&A_{2}\\\
-\overline{A_{2}}&\overline{A_{1}}\end{bmatrix},\quad
Q=\begin{bmatrix}Q_{1}&Q_{2}\\\
-\overline{Q_{2}}&\overline{Q_{1}}\end{bmatrix},$ (5.41)

where $Q_{1}$ and $Q_{2}$ are $n\times r$ complex matrices, and both $A_{1}$
and $A_{2}$ are $r\times r$ complex matrices. Put

$\displaystyle J_{n}(Q,A_{0}):={\rm Tr}$
$\displaystyle\big{(}Z_{0,n}^{*}QA_{0}Q^{*}+Z_{0,n}QA_{0}^{*}Q^{*}\big{)}-{\rm
Tr}\left(QA_{0}^{*}A_{0}Q^{*}\right)$ (5.42)
$\displaystyle+2\sum_{i<j}^{n}\left(\big{|}(QA_{0}Q^{*})_{i,j}\big{|}^{2}+\big{|}(QA_{0}Q^{*})_{i,j+n}\big{|}^{2}\right)+\log\det\left(\mathbb{I}_{2n}-QQ^{*}\right),$

then $J_{n}(Q)\leq 0$ whenever $QQ^{*}<\mathbb{I}_{2n}$, with equality if and
only if $Q=0$.

###### Proof.

We proceed by induction on $n$. For $n=1$, note that both $Q_{1}$ and $Q_{2}$
are row vectors, so the summation term vanishes and we have

$\displaystyle J_{1}(Q;A_{0})$ $\displaystyle\leq{\rm
Tr}\big{(}Z_{0,n}^{*}QA_{0}Q^{*}+Z_{0,n}QA_{0}^{*}Q^{*}\big{)}-{\rm
Tr}\left(QA_{0}^{*}A_{0}Q^{*}\right)-{\rm
Tr}\left(QQ^{*}\right)-\frac{1}{2}{\rm Tr}\left(QQ^{*}\right)^{2}$
$\displaystyle=-{\rm
Tr}\big{(}\left(Z_{0,n}^{*}Q-QA^{*}\right)\left(Z_{0,n}^{*}Q-QA_{0}^{*}\right)^{*}\big{)}-(1-|z_{0}|^{2}){\rm
Tr}(QQ^{*})-\frac{1}{2}{\rm Tr}\left(QQ^{*}\right)^{2}$ $\displaystyle\leq 0.$

Moreover, $J_{1}(Q;A_{0})=0$ if and only if $Q=0$.

Suppose that the lemma is true for $1,\ldots,n-1$. Rewrite $Q_{1}$ and $Q_{2}$
as $2\times 1$ block matrices

$Q_{i}=\begin{bmatrix}Q_{i1}\\\ Q_{i2}\end{bmatrix},\quad
Q_{11}\,\&\,Q_{21}\in\mathbb{C}^{(n-1)\times r},\quad
Q_{12}\,\&\,Q_{22}\in\mathbb{C}^{1\times r},$

and set

$\hat{Q}_{i}=\begin{bmatrix}Q_{1i}&Q_{2i}\\\
-\overline{Q_{2i}}&\overline{Q_{1i}}\end{bmatrix},\quad i=1,2.$

One can directly verify that

$\displaystyle
2\sum_{i<j}^{n}\left(\big{|}(QA_{0}Q^{*})_{i,j}\big{|}^{2}+\big{|}(QA_{0}Q^{*})_{i,j+n}\big{|}^{2}\right)=$
$\displaystyle
2\sum_{i<j}^{n}\left(\big{|}(\hat{Q}_{1}A_{0}\hat{Q}_{1}^{*})_{i,j}\big{|}^{2}+\big{|}(\hat{Q}_{1}A_{0}\hat{Q}_{1}^{*})_{i,j+n}\big{|}^{2}\right)$
(5.43) $\displaystyle+{\rm
Tr}\left(\hat{Q}_{2}A_{0}^{*}\hat{Q}_{1}^{*}\hat{Q}_{1}A_{0}\hat{Q}_{2}^{*}\right),$

and

$\displaystyle{\rm Tr}$
$\displaystyle\left(Z_{0,n}^{*}QA_{0}Q^{*}+Z_{0,n}QA_{0}^{*}Q^{*}\right)-{\rm
Tr}\left(QA_{0}^{*}A_{0}Q^{*}\right)=$ (5.44) $\displaystyle{\rm
Tr}\left(Z_{0,n-1}^{*}\hat{Q}_{1}A_{0}\hat{Q}_{1}^{*}+Z_{0,n-1}\hat{Q}_{1}A_{0}^{*}\hat{Q}_{1}^{*}\right)-{\rm
Tr}\left(\hat{Q}_{1}A_{0}^{*}A_{0}\hat{Q}_{1}^{*}\right)$ $\displaystyle+{\rm
Tr}\left(Z_{0,1}^{*}\hat{Q}_{2}A_{0}\hat{Q}_{2}^{*}+Z_{0,1}\hat{Q}_{2}A_{0}^{*}\hat{Q}_{2}^{*}\right)-{\rm
Tr}\left(\hat{Q}_{2}A_{0}^{*}A_{0}\hat{Q}_{2}^{*}\right).$

As $QQ^{*}<\mathbb{I}_{2n}$, we know that
$\hat{Q}_{1}\hat{Q}_{1}^{*}<\mathbb{I}_{2n-2}$, from which use of Schur
complement technique shows

$\log\det\left(\mathbb{I}_{2n}-QQ^{*}\right)=\log\det\left(\mathbb{I}_{2n-2}-\hat{Q}_{1}\hat{Q}_{1}^{*}\right)+\log\det\left(\mathbb{I}_{2}-\hat{Q}_{2}F\hat{Q}_{2}^{*}\right),$
(5.45)

where

$F=\mathbb{I}_{2r}+\hat{Q}_{1}^{*}\left(\mathbb{I}_{2n-2}-\hat{Q}_{1}\hat{Q}_{1}^{*}\right)^{-1}\hat{Q}_{1}.$

Combine(5.43), (5.44) and (5.45), we arrive at

$J_{n}(Q;A_{0})=J_{n-1}(\hat{Q}_{1};A_{0})+\widetilde{J}(Q),$ (5.46)

where

$\displaystyle\widetilde{J}(Q)=\log\det\left(\mathbb{I}_{2}-\hat{Q}_{2}F\hat{Q}_{2}^{*}\right)$
$\displaystyle-{\rm
Tr}\hat{Q}_{2}A_{0}^{*}\left(\mathbb{I}_{2r}-\hat{Q}_{1}^{*}\hat{Q}_{1}\right)A_{0}\hat{Q}_{2}^{*}$
$\displaystyle+{\rm
Tr}\left(Z_{0,1}^{*}\hat{Q}_{2}A_{0}\hat{Q}_{2}^{*}+Z_{0,1}\hat{Q}_{2}A_{0}^{*}\hat{Q}_{2}^{*}\right).$

Noting that

$F^{\frac{1}{2}}\left(\mathbb{I}_{2r}-\hat{Q}_{1}^{*}\hat{Q}_{1}\right)F^{\frac{1}{2}}=\mathbb{I}_{2r},$

which can be verified through the identity

$\hat{Q}_{1}\left(\mathbb{I}_{2r}-\hat{Q}_{1}^{*}\hat{Q}_{1}\right)=\hat{Q}_{1}-\hat{Q}_{1}\hat{Q}_{1}^{*}\hat{Q}_{1}=\left(\mathbb{I}_{2n-2}-\hat{Q}_{1}\hat{Q}_{1}^{*}\right)\hat{Q}_{1},$

if introducing

$\widetilde{Q}_{2}=\hat{Q}_{2}F^{\frac{1}{2}},\quad\widetilde{A_{0}}=F^{-\frac{1}{2}}A_{0}F^{-\frac{1}{2}},$

then

$\widetilde{J}(Q)=J_{1}(\hat{Q}_{1};\widetilde{A_{0}}).$

Finally, we see immediately from (5.46) and the induction hypothesis that the
lemma holds true for $n$. ∎

Now we are ready to prove Theorem 1.6.

###### Proof of Theorem 1.6.

In Proposition 1.4, let $A_{0}=PJ_{q}P^{-1}$ where
$J_{q}=J_{2}\bigoplus\overline{J}_{2}$ and set $D=P^{*}P$. First, make a
change of variables $Q\to QP^{*}$, and then by Lemma 5.2 we can proceed as in
Proposition 3.5 to show that for sufficiently small $\delta>0$ there exists
$\widetilde{\Delta}^{(4)}>0$ such that

$R_{N,4}^{(n)}\left(A_{0};z_{1},\cdots,z_{n}\right)=\widetilde{C}_{N}^{(4)}(z_{1},\ldots,z_{n})\left(\widetilde{I}_{\delta,N}^{(4)}+O\big{(}e^{-N\widetilde{\Delta}^{(4)}}\big{)}\right)$

where

$z_{k}:=z_{0}+\hat{z}_{k}/\sqrt{N},\quad k=1,2,\ldots,n,$ (5.47)
$\widetilde{C}_{N}^{(4)}(z_{1},\ldots,z_{n})=\frac{1}{C_{N,4}}e^{-2N\sum_{k=1}^{N}\left|z_{k}\right|^{2}}\prod_{i=1}^{N}\left|z_{i}-\overline{z_{i}}\right|^{2}\prod_{1\leq
i<j\leq
N}\left|z_{i}-z_{j}\right|^{2}\left|z_{i}-\overline{z_{j}}\right|^{2}\left(\det
D\right)^{n},$

and

$\widetilde{I}_{\delta,N}^{(4)}=\int_{{\rm
Tr}Q^{*}Q\leq\delta}\widetilde{g}_{4}(Q)\exp\\!\big{\\{}-N\widetilde{f}_{4}(Q)+\sqrt{N}\widetilde{h}_{4}(Q)\big{\\}}{\rm
d}Q.$ (5.48)

Here

$\widetilde{g}_{4}(Q)=\big{(}\det\left(\mathbb{I}_{2n}-QDQ^{*}\big{)}\right)^{-n-r+\frac{1}{2}}{{\mathbb{E}}_{\mathrm{GinSE}_{N-n}}(\widetilde{A}_{0})}\bigg{[}\prod_{i=1}^{n}\left|\det\left(\sqrt{\frac{N}{N-n}}z_{i}-\widetilde{X}_{N-n}\right)\right|^{2}\bigg{]},$
(5.49)

in which

$\widetilde{A}_{0}=\sqrt{N/(N-n)}\sqrt{\mathbb{I}_{2r}-PQ^{*}QP^{*}}A_{0}\sqrt{\mathbb{I}_{2r}-PQ^{*}QP^{*}},$

$\displaystyle\widetilde{f}_{4}(Q)=$ $\displaystyle-{\rm
Tr}\left(Z_{0}QJ_{q}^{*}DQ^{*}\right)-{\rm
Tr}\left(Z_{0}^{*}QDJ_{q}Q^{*}\right)+{\rm
Tr}\left(QJ_{q}^{*}DJ_{q}Q^{*}\right)$
$\displaystyle-\log\det\left(\mathbb{I}_{2n}-QDQ^{*}\right)-2\sum_{1\leq
i<j\leq
n}\left(\big{|}(QDJ_{q}Q^{*})_{i,j}\big{|}^{2}+\big{|}(QDJ_{q}Q^{*})_{i,j+n}\big{|}^{2}\right)$

and

$\widetilde{h}_{4}(Q)={\rm
Tr}\big{(}\hat{\Lambda}^{*}QDJ_{q}Q^{*}+\hat{\Lambda}QJ_{q}^{*}DQ^{*}\big{)}.$
(5.50)

Moreover, by the Taylor expansion for the log determinant, we have

$\displaystyle\widetilde{f}_{4}(Q)=$ $\displaystyle{\rm
Tr}\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)D\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)^{*}+\frac{1}{2}{\rm
Tr}(QDQ^{*})^{2}$ (5.51) $\displaystyle-2\sum_{1\leq i<j\leq
n}\left(\big{|}(QDJ_{q}Q^{*})_{i,j}\big{|}^{2}+\big{|}(QDJ_{q}Q^{*})_{i,j+n}\big{|}^{2}\right)+O\left(\|Q\|^{6}\right).$

By the Stirling formula we easily see from (1.16) that the following estimates
hold: for $z_{0}=\pm 1$,

$\widetilde{C}_{N}^{(4)}(z_{1},\ldots,z_{n})=\Big{(}1+O\Big{(}N^{-1}\Big{)}\Big{)}N^{n(2r+1-n)}e^{-2\sqrt{N}{\rm
Tr}\big{(}z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\big{)}}e^{-2n^{2}}(\pi
N)^{-\frac{n}{2}}\\\
\times\pi^{-n-2nr}2^{2nr-n}\left(\det(D)\right)^{n}e^{-2\sum_{i=1}^{n}|\hat{z}_{i}|^{2}}\prod_{1\leq
i<j\leq
n}|\hat{z}_{i}-\hat{z}_{j}|^{2}|\hat{z}_{i}-\overline{\hat{z}}_{j}|^{2}\prod_{i=1}^{n}|\hat{z}_{i}-\overline{\hat{z}}_{i}|^{2},$
(5.52)

and for $|z_{0}|=1$ and $\Im z_{0}>0$,

$\widetilde{C}_{N}^{(4)}(z_{1},\ldots,z_{n})=N^{n\left(2r+1-\frac{n-1}{2}\right)}e^{-2\sqrt{N}{\rm
Tr}\big{(}z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\big{)}}e^{-2n^{2}}(\pi
N)^{-\frac{n}{2}}\pi^{-n-2nr}\\\ \times
2^{2nr-n}\left(\det(D)\right)^{n}e^{-2\sum_{i=1}^{n}|\hat{z}_{i}|^{2}}\Big{(}|z_{0}-\overline{z}_{0}|^{n(n+1)}\prod_{1\leq
i<j\leq
n}|\hat{z}_{i}-\hat{z}_{j}|^{2}+O\Big{(}N^{-\frac{1}{2}}\Big{)}\Big{)},$
(5.53)

uniformly for all $\hat{z}_{k}$ in a compact subset of $\mathbb{C}$.

The remaining task is to analyze $\widetilde{I}_{\delta,N}^{(4)}$. We mainly
focus on the real case of $z_{0}=\pm 1$. For this, we proceed in three steps
as in the proof of Theorem 1.5.

Step 1: Taylor expansions of $\widetilde{f}_{4}(Q)$ and
$\widetilde{h}_{4}(Q)$.

Noting $J_{2}$ in (1.21) and the assumption (4.2) where $\hat{\theta}_{i}=0$,
$i=1,\cdots,m$. Combine (4.6)-(4.9) and use the notation therein, we get
$J_{2}=J_{2}^{(0)}$. Next, we are devoted to establishing leading terms in
(5.50) and (5.51).

For $\widetilde{f}_{4}(Q)$ in (5.51), noting

$Z_{0}^{*}Q-QJ_{q}^{*}=Q\left(z_{0}\mathbb{I}_{2r}-J_{q}\right)^{*},$

for convenience let’s introduce

$D:=\begin{bmatrix}D_{1}&D_{2}\\\
-\overline{D}_{2}&\overline{D}_{1}\end{bmatrix},\quad\widetilde{D}=\big{(}z_{0}\mathbb{I}_{2r}-J_{q}\big{)}^{*}D\big{(}z_{0}\mathbb{I}_{2r}-J_{q}\big{)}:=\begin{bmatrix}\widetilde{D}_{1}&\widetilde{D}_{2}\\\
-\overline{\widetilde{D}}_{2}&\overline{\widetilde{D}}_{1}\end{bmatrix},$

then we have

$\widetilde{D}_{1}=\big{(}z_{0}\mathbb{I}_{r}-J_{2}^{(0)}\big{)}^{*}D_{1}\big{(}z_{0}\mathbb{I}_{r}-J_{2}^{(0)}\big{)},\quad\widetilde{D}_{2}=\big{(}z_{0}\mathbb{I}_{r}-J_{2}^{(0)}\big{)}^{*}D_{2}\big{(}z_{0}\mathbb{I}_{r}-\overline{J_{2}^{(0)}}\big{)}.$

According to the Jordan block structure of $J_{2}^{(0)}$, for $\alpha=1,2$,
both $D_{\alpha}$ and $\widetilde{D}_{\alpha}$ admit a three-layer structure

$D_{(\alpha)}=[D_{k;l}^{(\alpha)}]_{k,l=1}^{m+1},\quad\widetilde{D}_{(\alpha)}=[\widetilde{D}_{k;l}^{(\alpha)}]_{k,l=1}^{m+1},$

where for the former $D_{m+1;m+1}^{(\alpha)}$ is of size $r_{0}\times r_{0}$
with $r_{0}=r-\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}\beta_{i,j}p_{i,j}$, and
for $k,l=1,\cdots,m$

$\displaystyle
D_{k;l}^{(\alpha)}=\left[D_{k,a;l,b}^{(\alpha)}\right]_{\alpha_{k}\times\alpha_{l}},\quad
D_{k,a;l,b}^{(\alpha)}=\left[D_{k,a,c;l,b,d}^{(\alpha)}\right]_{\beta_{k,a}\times\beta_{l,b}},$
$\displaystyle
D_{m+1;l}^{(\alpha)}=\left[D_{m+1;l,b}^{(\alpha)}\right]_{1\times\alpha_{l}},\quad
D_{m+1;l,b}^{(\alpha)}=\left[D_{m+1;l,b,d}^{(\alpha)}\right]_{1\times\beta_{l,b}},$
$\displaystyle
D_{k;m+1}^{(\alpha)}=\left[D_{k,a;m+1}^{(\alpha)}\right]_{\alpha_{k}\times
1},\quad
D_{k,a;m+1}^{(\alpha)}=\left[D_{k,a,c;m+1}^{(\alpha)}\right]_{\beta_{k,a}\times
1},$

with $D_{k,a,c;l,b,d}^{(\alpha)}$, $D_{m+1;l,b,d}^{(\alpha)}$ and
$D_{k,a,c;m+1}^{(\alpha)}$ being of sizes $p_{k,a}\times p_{l,b}$,
$r_{0}\times p_{l,b}$ and $p_{k,a}\times r_{0}$ respectively. For the latter,
the notations are similar. Moreover, it’s easy to establish the following
identities for $\alpha=1,2$,

$\displaystyle\widetilde{D}_{k,a,c;l,b,d}^{(\alpha)}=R_{p_{k,a}}(0)^{*}D_{k,a,c;l,b,d}^{(\alpha)}R_{p_{l,b}}(0),$
$\displaystyle\widetilde{D}_{m+1;l,b,d}^{(\alpha)}=-\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)}^{*}D_{m+1;l,b,d}^{(\alpha)}R_{p_{l,b}}(0);$
$\displaystyle\widetilde{D}_{k,a,c;m+1}^{(1)}=-R_{p_{k,a}}(0)^{*}D_{k,a,c;m+1}^{(1)}\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)};$
$\displaystyle\widetilde{D}_{k,a,c;m+1}^{(2)}=-R_{p_{k,a}}(0)^{*}D_{k,a,c;m+1}^{(2)}\big{(}z_{0}\mathbb{I}_{r_{0}}-\overline{J_{2}^{(2)}}\big{)};$
$\displaystyle\widetilde{D}_{m+1;m+1}^{(1)}=\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)}^{*}D_{m+1;m+1}^{(1)}\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)};$
$\displaystyle\widetilde{D}_{m+1;m+1}^{(2)}=\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)}^{*}D_{m+1;m+1}^{(2)}\big{(}z_{0}\mathbb{I}_{r_{0}}-\overline{J_{2}^{(2)}}\big{)};$

Again, by the structure of $R_{p_{k,a}}(0)$, for $k,l=1,\cdots,m$ set

$\displaystyle
D_{k,a,c;l,b,d}^{(\alpha)}=\begin{bmatrix}D_{kac;lbd}^{(\alpha;11)}&D_{kac;lbd}^{(\alpha;12)}\\\
D_{kac;lbd}^{(\alpha;21)}&D_{kac;lbd}^{(\alpha;22)}\end{bmatrix},\quad
D_{k,a,c;m+1}^{(\alpha)}=\begin{bmatrix}D_{kac;m+1}^{(\alpha;U)}\\\
D_{kac;m+1}^{(\alpha;D)}\end{bmatrix},$ (5.54) $\displaystyle
D_{m+1;l,b,d}=\begin{bmatrix}D_{m+1;lbd}^{(\alpha;L)}&D_{m+1;lbd}^{(\alpha;R)}\end{bmatrix},\quad$

where $D_{kac;lbd}^{(\alpha;11)}$, $D_{m+1;lbd}^{(\alpha;L)}$ and
$D_{kac;m+1}^{(\alpha;U)}$ are of sizes $(p_{k,a}-1)\times(p_{l,b}-1)$,
${r_{0}}\times(p_{l,b}-1)$ and $(p_{k,a}-1)\times{r_{0}}$, respectively. Then
simple calculations show that

$\displaystyle\widetilde{D}_{k,a,c;l,b,d}^{(\alpha)}=\begin{bmatrix}0&0\\\
0&D_{kac;lbd}^{(\alpha;11)}\end{bmatrix},\quad\widetilde{D}_{k,a,c;m+1}^{(1)}=\begin{bmatrix}0\\\
-D_{kac;m+1}^{(1;U)}\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)}\end{bmatrix},$
(5.55) $\displaystyle\widetilde{D}_{k,a,c;m+1}^{(2)}=\begin{bmatrix}0\\\
-D_{kac;m+1}^{(2;U)}\big{(}z_{0}\mathbb{I}_{r_{0}}-\overline{J_{2}^{(2)}}\big{)}\end{bmatrix},\widetilde{D}_{m+1;l,b,d}^{(\alpha)}=\begin{bmatrix}0&-\big{(}z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\big{)}^{*}D_{m+1;lbd}^{(\alpha;L)}\end{bmatrix}.$

For

$Q=\begin{bmatrix}Q_{1}&Q_{2}\\\
-\overline{Q}_{2}&\overline{Q}_{1}\end{bmatrix},$

put for $\alpha=1,2$,

$Q_{\alpha}=\begin{bmatrix}Q_{1}^{(\alpha)}&\cdots&Q_{m}^{(\alpha)}&Q_{m+1}^{(\alpha)}\end{bmatrix},$

where $Q_{m+1}^{(\alpha)}$ is of size $n\times r_{0}$ and for $l=1,\cdots,m$

$Q_{l}^{(\alpha)}=[Q_{l,b}^{(\alpha)}]_{1\times\alpha_{l}},\quad
Q_{l,b}^{(\alpha)}=\big{[}Q_{l,b,d}^{(\alpha)}\big{]}_{1\times\beta_{l,b}},\quad
Q_{l,b,d}^{(\alpha)}=\begin{bmatrix}Q_{lbd}^{(\alpha;L)}&Q_{lbd}^{(\alpha;R)}\end{bmatrix},$

while $Q_{l,b,d}^{(\alpha)}$ and $Q_{lbd}^{(\alpha;L)}$ are of size $n\times
p_{l,b}$ and $n\times 1$, respectively. Now we can draw some blocks from
$Q_{\alpha}$ and restructure it into two new matrices

$\displaystyle
Q_{R}=\begin{bmatrix}Q_{1}^{(1;R)}&\cdots&Q_{m}^{(1;R)}&Q_{m+1}^{(1)}&Q_{1}^{(2;R)}&\cdots&Q_{m}^{(2;R)}&Q_{m+1}^{(2)}\end{bmatrix},$
(5.56) $\displaystyle Q_{L}=\begin{bmatrix}Q_{1;L}&Q_{2;L}\\\
-\overline{Q_{2;L}}&\overline{Q_{1;L}}\end{bmatrix},\quad
Q_{\alpha;L}=\begin{bmatrix}Q_{1}^{(\alpha;L)}&\cdots&Q_{m}^{(\alpha;L)}\end{bmatrix},$

where

$Q_{l}^{(\alpha;L)}=[Q_{l,b}^{(\alpha;L)}],\
Q_{l,b}^{(\alpha;L)}=[Q_{lbd}^{(\alpha;L)}],\quad
Q_{l}^{(\alpha;R)}=[Q_{l,b}^{(\alpha;R)}],\
Q_{l,b}^{(\alpha;R)}=[Q_{lbd}^{(\alpha;R)}].$

Similarly, with (5.55) in mind, define

$\widetilde{D}^{(11)}=\begin{bmatrix}\widetilde{D}_{1;11}&\widetilde{D}_{2;11}\\\
-\overline{\widetilde{D}_{2;11}}&\overline{\widetilde{D}_{1;11}}\end{bmatrix},\quad\widetilde{D}_{\alpha;11}=\big{[}\widetilde{D}_{k,l}^{(\alpha;11)}\big{]}_{k,l=1}^{m+1},$
(5.57)

where for $k,l=1,\cdots,m$,

$\displaystyle\widetilde{D}_{k,l}^{(\alpha;11)}=\begin{bmatrix}D_{kac;lbd}^{(\alpha;11)}\end{bmatrix},\quad\widetilde{D}_{m+1,l}^{(\alpha;11)}=\begin{bmatrix}-\left(z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)}\right)^{*}D_{m+1;lbd}^{(\alpha;L)}\end{bmatrix},\quad\widetilde{D}_{m+1,m+1}^{(\alpha;11)}=\widetilde{D}_{m+1,m+1}^{(\alpha)}$
$\displaystyle\widetilde{D}_{k,m+1}^{(1;11)}=\begin{bmatrix}-D_{kac;m+1}^{(1;U)}\left(z_{0}\mathbb{I}_{s}-J_{2}^{(2)}\right)\end{bmatrix},\quad\widetilde{D}_{k,m+1}^{(2;11)}=\begin{bmatrix}-D_{kac;m+1}^{(2;U)}\left(z_{0}\mathbb{I}_{s}-\overline{J_{2}^{(2)}}\right)\end{bmatrix}.$

Also, in (5.54), choose the (1,1)-entry of $D_{kac;lbd}^{(\alpha;11)}$ and
construct a matrix $\hat{D}^{(1,1)}$ as follows

$\widehat{D}_{11}=\begin{bmatrix}\widehat{D}_{1;11}&\widehat{D}_{2;11}\\\
-\overline{\widehat{D}_{2;11}}&\overline{\widehat{D}_{1;11}}\end{bmatrix},\quad\widehat{D}_{\alpha;11}=[\hat{D}_{k,l}^{(\alpha)}],\quad\hat{D}_{k,l}^{(\alpha)}=\left[\big{(}D_{kac;lbd}^{(\alpha;11)}\big{)}_{1,1}\right]_{\alpha_{k}\times\alpha_{l}}.$
(5.58)

If we set

$D^{-1}:=\begin{bmatrix}D_{1}^{(-1)}&D_{2}^{(-1)}\\\
-\overline{D^{(-1)}_{2}}&\overline{D^{(-1)}_{1}}\end{bmatrix},$

then $U$ given in (3.7) and $D=P^{*}P$ satisfy the following relations

$\displaystyle\mathbb{J}_{r}U=-\mathrm{i}D,\quad\overline{U}_{2}=\mathrm{i}D_{1},\
\ \overline{U}_{1}=-\mathrm{i}D_{2};$ (5.59) $\displaystyle
V\mathbb{J}_{r}=-\mathrm{i}D^{-1},\quad V_{2}=\mathrm{i}D_{1}^{(-1)},\ \
V_{1}=-\mathrm{i}D_{2}^{(-1)}.$

Therefore, we have

$-{\rm
i}\widehat{D}_{11}=\mathbb{J}_{t}\begin{bmatrix}(U_{1})_{I_{1}}&(U_{2})_{I_{1}}\\\
-\overline{(U_{2})_{I_{1}}}&\overline{(U_{1})_{I_{1}}}\end{bmatrix};$ (5.60)

see Section 2.1 for the meaning of the relevant symbols.

We are ready to do some calculation. First,

${\rm
Tr}\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)D\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)^{*}={\rm
Tr}(Q\widetilde{D}Q^{*})=2{\rm
Tr}\big{(}Q_{R}\widetilde{D}^{(11)}Q_{R}^{*}\big{)}.$ (5.61)

Secondly, as in the $\beta=2$ case we can obtain

$QDJ_{q}Q^{*}=z_{0}Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}+O\left(\|Q_{R}\|\|Q\|\right).$
(5.62)

Finally, noting that combination of (5.56) and (5.58) gives

$QDQ^{*}=Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}+O\left(\|Q_{R}\|\|Q\|\right),$

we immediately see from $|z_{0}|=1$ and
$(\hat{D}^{(1,1)})^{*}=\hat{D}^{(1,1)}$ that

$\displaystyle\frac{1}{2}{\rm
Tr}\left(QDQ^{*}\right)^{2}-2\sum_{i<j}^{n}\left(\big{|}(QDJ_{q}Q^{*})_{i,j}\big{|}^{2}+\big{|}(QDJ_{q}Q^{*})_{i,j+n}\big{|}^{2}\right)=$
(5.63)
$\displaystyle\frac{1}{2}\sum_{i=1}^{2n}\left(Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}\right)_{i,i}^{2}+O\left(\|Q_{R}\|\|Q\|^{3}\right),$

since $\left(Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}\right)_{i,i+n}=0$, $i=1,\cdots,n$.

Combining (5.51), (5.61) and (5.63), we obtain

$\widetilde{f}_{4}(Q)=2{\rm
Tr}\big{(}Q_{R}\widetilde{D}^{(11)}Q_{R}^{*}\big{)}+\frac{1}{2}\sum_{i=1}^{2n}\left(Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}\right)_{i,i}^{2}+O\left(\|Q_{R}\|\|Q\|^{3}+\|Q\|^{6}\right).$

Meanwhile, by (5.62) we get

$\widetilde{h}_{4}(Q)=z_{0}{\rm
Tr}\Big{(}\big{(}\hat{\Lambda}+\hat{\Lambda}^{*}\big{)}Q_{L}\hat{D}^{(1,1)}Q_{L}^{*}\Big{)}+O\left(\|Q_{R}\|\|Q\|\right).$

Step 2: Modified estimate of $\widetilde{g}_{4}(Q)$. For the first factor in
(5.49), it’s easy to see

$\left(\det\left(\mathbb{I}_{2n}-QDQ^{*}\right)\right)^{-n-r+\frac{1}{2}}=1+O\left(\|Q\|^{2}\right).$

As for the second factor, the averaged product of characteristic polynomials
has actually been verified in Theorem 3.2 (ii), except that some minor
correction will be needed because of $Q$ and the scaling factor $(N-n)/N$. So
we repeat almost the same procedure quickly, with emphasis on the differences.
As in the $\beta=2$ case and Proposition 3.5, denote $Z=\sqrt{N/(N-n)}{\rm
diag}\left(z_{1},\cdots,z_{n}\right)$ with $K_{1}=n$, we can prove that for
sufficiently small $\delta>0$ and for $|z_{0}|=1$,

$F_{n}^{(4)}(Z,Z)=\frac{1}{\widehat{Z}_{2K_{1},1}}\Big{(}I_{\delta,N}^{(4)}+O\big{(}e^{-\frac{1}{2}N\Delta^{(4)}}\big{)}\Big{)},$

with

$I_{\delta,N}^{(4)}=\int\limits_{{\rm Tr}(YY^{*})\leq\delta}\
g_{4}(Y)\exp\left\\{-Nf_{4}(Y,Z_{0})+Nh_{4}(Y)\right\\}{\rm d}Y,$ (5.64)

where

$g_{4}(Y)={\rm Pf}\begin{bmatrix}R&S\\\
-S^{t}&T\end{bmatrix}\left(\det\left(A_{4}\right)\right)^{-n-r},$ (5.65)
$f_{4}(Y)=\Big{(}1-\frac{n}{N}\Big{)}{\rm
Tr}(YY^{*})-\log\det\left(\mathbb{I}_{2n}+YY^{*}\right).$

Noting that we need to replace $\hat{Z}$ with
$\hat{Z}+\frac{1}{2}nz_{0}N^{-\frac{1}{2}}\mathbb{I}_{n}+O\left(N^{-1}\right)$
both in (3.34) and (3.38), we then get

$h_{4}(Y)=\log\det\big{(}\mathbb{I}_{4n}+N^{-\frac{1}{2}}H_{4}\big{)}.$

Moreover, we can obtain

$f_{4}(Y)=\frac{1}{2}{\rm
Tr}\left((YY^{*})^{2}\right)+O\left(\|Y\|^{6}+N^{-1}\|Y\|^{2}\right),$

and

$\displaystyle
Nh_{4}(Y)=h_{4,0,0}-\sqrt{N}h_{4,0}(Y)+O\left(\|Y\|^{2}+\sqrt{N}\|Y\|^{4}+N^{-\frac{1}{2}}\right)$
(5.66)

where

$h_{4,0,0}=2\sqrt{N}{\rm Tr}(Z_{0}^{-1}\hat{\Lambda})+2n^{2}-{\rm
Tr}(Z_{0}^{-2}\hat{\Lambda}^{2}),\,h_{4,0}(Y)=2z_{0}^{-1}{\rm
Tr}(YY^{*}\hat{\Lambda}).$

For the factors of $g_{4}(Y)$ in (5.65), we have

$\left(\det(A_{2})\right)^{-n-r}=1+O\big{(}N^{-\frac{1}{2}}+\|Y\|\big{)}.$
(5.67)

On the other hand, with Lemma 4.3 taking $\Xi=N^{-1}+\|Q\|^{2}$ and
$\hat{\theta}_{i}=0$ ($i=1,\cdots,m$), we have

$\displaystyle{\rm Pf}\begin{bmatrix}R&S\\\
-S^{t}&T\end{bmatrix}=\prod_{i=m+1}^{\gamma}\prod_{j=1}^{\alpha_{i}}\left|z_{0}-\theta_{i}\right|^{4np_{i,j}\beta_{i,j}}$
$\displaystyle{\rm
Pf}\begin{bmatrix}R\left(Y;(V_{1})_{I_{2}},(U_{1})_{I_{1}}\right)&N^{-\frac{1}{4}}S\left(N^{\frac{1}{4}}Y\right)\\\
-N^{-\frac{1}{4}}S\left(N^{\frac{1}{4}}Y\right)^{t}&R\left(Y^{*};(U_{1})_{I_{1}},(V_{1})_{I_{2}}\right)\end{bmatrix}$
$\displaystyle+O\left(\sum
N^{-\alpha_{1}}\|Y\|^{\alpha_{2}}\Xi^{\alpha_{3}}\right),$

where $R\left(Y;A,B\right)$ and $S\left(Y\right)$ are defined in (3.8) and
(3.9), and $(\alpha_{1},\alpha_{2},\alpha_{3})$ satisfies

$4\alpha_{1}+\alpha_{2}+\alpha_{3}\geq 4nt+\delta_{\alpha_{3},0}.$

Noting $\Omega_{11}^{{\rm(re)}}=0$, using the Pfaffian identity in Section 2.1
we can get

$\displaystyle{\rm Pf}$
$\displaystyle\begin{bmatrix}R\left(Y;(V_{1})_{I_{2}},(U_{1})_{I_{1}}\right)&N^{-\frac{1}{4}}S\left(N^{\frac{1}{4}}Y\right)\\\
-N^{-\frac{1}{4}}S\left(N^{\frac{1}{4}}Y\right)^{t}&R\left(Y^{*};(U_{1})_{I_{1}},(V_{1})_{I_{2}}\right)\end{bmatrix}=\left(\det(YY^{*})\right)^{t}$
(5.68)
$\displaystyle\times\left(\det\begin{bmatrix}(V_{1})_{I_{2}}&(V_{2})_{I_{2}}\\\
-\overline{(V_{2})}_{I_{2}}&\overline{(V_{1})}_{I_{2}}\end{bmatrix}\det\begin{bmatrix}(U_{1})_{I_{1}}&(U_{2})_{I_{1}}\\\
-\overline{(U_{2})}_{I_{1}}&\overline{(U_{1})}_{I_{1}}\end{bmatrix}\right)^{n},$

where $t=\sum_{i=1}^{m}\sum_{j=1}^{\alpha_{i}}\beta_{i,j}$.

Combine (5.65), (5.67) and (5.68), we get

$g_{4}(Y,\Lambda)=\left(D^{(4,{\rm
re},c)}+O\big{(}N^{-\frac{1}{4}}+\|Y\|\big{)}\right)\Big{(}\big{(}\det(Y^{*}Y)\big{)}^{t}+O\Big{(}\sum_{(\alpha_{1},\alpha_{2},\alpha_{3})}N^{-\alpha_{1}}\|Y\|^{\alpha_{2}}\Xi^{\alpha_{3}}\Big{)}\Big{)},$

where

$D^{(4,{\rm
re},c)}=\left(\det\begin{bmatrix}(V_{1})_{I_{2}}&(V_{2})_{I_{2}}\\\
-\overline{(V_{2})}_{I_{2}}&\overline{(V_{1})}_{I_{2}}\end{bmatrix}\det\begin{bmatrix}(U_{1})_{I_{1}}&(U_{2})_{I_{1}}\\\
-\overline{(U_{2})}_{I_{1}}&\overline{(U_{1})}_{I_{1}}\end{bmatrix}\right)^{n}\prod_{i=m+1}^{\gamma}|z_{0}-\theta_{i}|^{4n\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}.$

By the change of variable $Y\to N^{-\frac{1}{4}}Y$ in (5.64), as in the proof
of Theorem 3.2 (i) we get

$F^{(4)}_{n}(Z,Z)=\frac{e^{2n^{2}}}{\hat{Z}_{2K_{1},1}}D^{(4,{\rm
re},c)}N^{-n^{2}-\frac{n}{2}-nt}e^{2z_{0}^{-1}\sqrt{N}{\rm
Tr}(\hat{Z}+\overline{\hat{Z}})}\\\ \times e^{-{\rm
Tr}\big{(}\hat{Z}^{2}+\overline{\hat{Z}^{2}}\big{)}}\Big{(}I^{(4,{\rm
re},c)}\left(2z_{0}^{-1}\hat{\Lambda}\right)+O\big{(}\sum_{(\alpha_{1},\alpha_{3})}N^{-\alpha_{1}}\Xi^{\alpha_{3}}\big{)}\Big{)},$
(5.69)

where $(\alpha_{1},\alpha_{3})$ satisfies the restriction

$4\alpha_{1}+\alpha_{3}\geq\delta_{\alpha_{3},0}$

and

$I^{(4,{\rm
re},c)}(\hat{\Lambda})=\int\left(\det(YY^{*})\right)^{t}\exp\Big{\\{}-\frac{1}{2}{\rm
Tr}(YY^{*})^{2}-{\rm Tr}\big{(}\hat{\Lambda}YY^{*}\big{)}\Big{\\}}{\rm d}Y.$
(5.70)

So this completes a modified estimate of $\widetilde{g}_{4}(Q)$.

Step 3: Summary and matrix integrals.

For the matrix integral (5.48), using the change of variables

$Q_{R}\to N^{-\frac{1}{2}}Q_{R},\quad Q_{L}\to
N^{-\frac{1}{4}}Q_{L}(\hat{D}^{(1,1)})^{-\frac{1}{2}},$

integrating out $Q_{R}$, we obtain

$\frac{1}{N^{n}}R_{N}^{(n)}\left(A_{0};z_{1},\cdots,z_{n}\right)=(\pi)^{-\frac{n}{2}}\pi^{-2(tn+n+n^{2})}2^{2(tn-n+n^{2})}\rho^{n}e^{-{\rm
Tr}(\hat{Z}+\overline{\hat{Z}})^{2}}\prod_{i=1}^{n}|\hat{z}_{i}-\overline{\hat{z}}_{i}|^{2}\\\
\times\prod_{1\leq i<j\leq
n}|\hat{z}_{i}-\hat{z}_{j}|^{2}|\hat{z}_{i}-\overline{\hat{z}}_{j}|^{2}\left(I^{(4,{\rm
re},c)}(2z_{0}^{-1}\hat{\Lambda})\tilde{I}^{(4)}(z_{0}^{-1}\hat{\Lambda})+O\big{(}N^{-\frac{1}{4}}\big{)}\right),$
(5.71)

where for

$Q_{L}=\begin{bmatrix}Q_{1;L}&Q_{2;L}\\\
-\overline{Q_{2;L}}&\overline{Q_{1;L}}&\end{bmatrix},\quad
Q_{1;L},Q_{2;L}:n\times t,$

,

$\tilde{I}^{(4)}(\hat{\Lambda})=\int\exp\\!\Big{\\{}-\frac{1}{2}\sum_{i=1}^{2n}\left(Q_{L}Q_{L}^{*}\right)_{ii}^{2}+{\rm
Tr}\left(\big{(}\hat{\Lambda}^{*}+\hat{\Lambda}\big{)}Q_{L}Q_{L}^{*}\right)\Big{\\}}{\rm
d}Q_{1;L}{\rm d}Q_{2;L},$ (5.72)

and

$\displaystyle\rho:=\prod_{i=m+1}^{\gamma}$
$\displaystyle|z_{0}-\theta_{i}|^{2\sum_{j=1}^{\alpha_{i}}p_{i,j}\beta_{i,j}}\frac{1}{\det(\widetilde{D}^{(11)})\det(\hat{D}^{(1,1)})}\det(D)$
$\displaystyle\times\det\begin{bmatrix}(V_{1})_{I_{2}}&(V_{2})_{I_{2}}\\\
-\overline{(V_{2})}_{I_{2}}&\overline{(V_{1})}_{I_{2}}\end{bmatrix}\det\begin{bmatrix}(U_{1})_{I_{1}}&(U_{2})_{I_{1}}\\\
-\overline{(U_{2})}_{I_{1}}&\overline{(U_{1})}_{I_{1}}\end{bmatrix}=1.$

Here the second identity in the latter follows from $\widetilde{D}^{(11)}$ in
(5.57), (5.60), (5.59) and the Schur complement technique.

Now we need to evaluate the two matrix integrals $\tilde{I}^{(4)}$ and
$I^{(4,{\rm re},c)}$. For $\tilde{I}^{(4)}$ in (5.72), treated as $n$
integrals separately according to its rows it is simplified to

$\displaystyle\tilde{I}^{(4)}\big{(}z_{0}^{-1}\hat{\Lambda}\big{)}$
$\displaystyle=\prod_{i=1}^{n}\int\cdots\int\exp\Big{\\{}-\big{(}\sum_{j=1}^{2t}\left|\widehat{q}_{1,j}\right|^{2}\big{)}^{2}+2z_{0}^{-1}\left(\hat{z}_{i}+\overline{\hat{z}_{i}}\right)\sum_{j=1}^{2t}\left|\widehat{q}_{1,j}\right|^{2}\Big{\\}}\prod_{j=1}^{2t}{\rm
d}\Re\widehat{q}_{1,j}{\rm d}\Im\widehat{q}_{1,j},$

where

$\widehat{Q}_{L}:=\left[Q_{1;L},Q_{2;L}\right]:=\left[\widehat{q}_{1,j}\right],\quad
l=1,\cdots,n,\ j=1,\cdots,2t.$

Apply the spherical polar coordinates and we can get

$\tilde{I}^{(4)}\big{(}z_{0}^{-1}\hat{\Lambda}\big{)}=\pi^{(2t+\frac{1}{2})n}2^{(-t+\frac{1}{2})n}\prod_{i=1}^{n}\Big{(}e^{(\hat{z}_{i}+\overline{\hat{z}_{i}})^{2}}\mathrm{IE}_{2t-1}\big{(}z_{0}^{-1}\sqrt{2}\hat{z}_{i}+\overline{z_{0}}^{-1}\sqrt{2}\overline{\hat{z}_{i}}\big{)}\Big{)},$
(5.73)

where $\mathrm{IE}_{n}(z)$ is defined in (1.24).

For $I^{(4,{\rm re},c)}\left(2z_{0}^{-1}\hat{\Lambda}\right)$ defined in
(5.70), by the singular value decomposition:

$Y=U\sqrt{R}U^{t},\quad R={\rm diag}\left(r_{1},\cdots,r_{2n}\right),\quad
r_{1}\geq\cdots\geq r_{2n}\geq 0,$ (5.74)

the Jacobian reads

${\rm d}Y=\frac{\pi^{n(n+1)}}{\prod_{i=1}^{2n-1}i!}\left|\Delta(r)\right|{\rm
d}R{\rm d}U,$ (5.75)

where $U$ is chosen from the unitary group $\mathcal{U}(2n)$ with the Haar
measure, and $\Delta(r)=\prod_{1\leq i<j\leq 2n}(r_{j}-r_{i})$. Use Harish-
Chandra-Itzykson-Zuber integration formula (5.34) and also the Pfaffian
integration formula [22]

$\int\det\big{(}[f_{i}(x_{j})]_{i,j=1}^{2n}\big{)}{\rm
sgn}\left(\Delta(x)\right)\prod_{i=1}^{2n}{\rm d}\mu(x_{i})=(2n)!{\rm
Pf}\Big{(}\Big{[}a_{i,j}\Big{]}_{i,j=1}^{n}\Big{)},$ (5.76)

where

$a_{i,j}=\int_{x\leq y}\left(f_{i}(x)f_{j}(y)-f_{j}(x)f_{i}(y)\right){\rm
d}\mu(x){\rm d}\mu(y),$

we obtain

$\displaystyle I^{(4,{\rm re},c)}\left(2z_{0}^{-1}\hat{\Lambda}\right)=$
$\displaystyle\frac{\pi^{n(2n+1)}(-z_{0})^{n}}{2^{n(2n-1)}\prod_{1\leq i<j\leq
n}|\hat{z}_{i}-\hat{z}_{j}|^{2}|\hat{z}_{i}-\overline{\hat{z}}_{j}|^{2}\prod_{i=1}^{n}(\overline{\hat{z}}_{i}-\hat{z}_{i})}$
(5.77) $\displaystyle\times{\rm
Pf}\begin{bmatrix}\widetilde{f}(\hat{z}_{i},\hat{z}_{j})&\widetilde{f}(\hat{z}_{i},\overline{\hat{z}}_{j})\\\
\widetilde{f}(\overline{\hat{z}}_{i},\hat{z}_{j})&\widetilde{f}(\overline{\hat{z}}_{i},\overline{\hat{z}}_{j})\end{bmatrix}_{i,j=1}^{n},$

where

$\widetilde{f}(z,w)=\int\int_{0\leq x\leq
y<\infty}(xy)^{t}\exp\\{-\frac{1}{2}(x^{2}+y^{2})\\}\left(e^{-2z_{0}(zx+wy)}-e^{-2z_{0}(wx+zy)}\right){\rm
d}x{\rm d}y.$

After a change of variables $x\to(u-v)/\sqrt{2}$, $y\to(u+v)/\sqrt{2}$,
recalling (1.27) we have

$\widetilde{f}(z,w)=2^{2-t}\pi
f_{t}(\sqrt{2}z_{0}^{-1}z,\sqrt{2}z_{0}^{-1}w).$

Combining (5.71), (5.73) and (5.77), changing
$\hat{z}_{i}\to\hat{z}_{i}/\sqrt{2},\quad i=1,\cdots,n,$ in (5.47), we can
thus give a complete proof of Theorem 1.6 at the real edge.

At the complex edge, although $Z_{0}$ is no longer a scalar matrix, we still
have

$\displaystyle{\rm
Tr}\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)D\left(Z_{0}^{*}Q-QJ_{q}^{*}\right)^{*}$
$\displaystyle=2{\rm
Tr}\left[Q_{1}\left(\overline{z}_{0}\mathbb{I}_{r}-J_{2}^{*}\right),Q_{2}\left(\overline{z}_{0}\mathbb{I}_{r}-J_{2}^{t}\right)\right]D\left[Q_{1}\left(\overline{z}_{0}\mathbb{I}_{r}-J_{2}^{*}\right),Q_{2}\left(\overline{z}_{0}\mathbb{I}_{r}-J_{2}^{t}\right)\right]^{*}$
$\displaystyle=2{\rm
Tr}\left[Q_{1},Q_{2}\right]\widetilde{D}\left[Q_{1},Q_{2}\right]^{*},$

where

$\widetilde{D}=\begin{bmatrix}z_{0}\mathbb{I}_{r}-J_{2}&\\\
&z_{0}\mathbb{I}_{r}-\overline{J_{2}}\end{bmatrix}^{*}D\begin{bmatrix}z_{0}\mathbb{I}_{r}-J_{2}&\\\
&z_{0}\mathbb{I}_{r}-\overline{J_{2}}\end{bmatrix}.$

Therefore, we can still admit a three-layer structure according to the Jordan
block structure of $J_{2}^{(0)}$. Based on this observation, following almost
the same procedure as in the proof of Theorem 1.5, we can complete the proof
of Theorem 1.6 at the complex edge. Here we omit the detailed derivation. ∎

##  6 Interpolating point processes

In this section we further investigate the nature of the transition at, when
some eigenvalues of the perturbed matrix $A_{0}$ go to the transition point
$z_{0}$ at a certain rate, there are interpolating point processes. For this,
we need more definitions.

Set

$J_{2}^{(1)}={\rm
diag}\Big{(}\hat{\theta}_{1}\mathbb{I}_{\beta_{1,1}},\cdots,\hat{\theta}_{m}\mathbb{I}_{\beta_{m,1}}\Big{)},$
(6.1)

and $t=\sum_{i=1}^{m}\beta_{i,1}$. Define two more matrix integrals over the
space of $n\times t$ and $n\times 2t$ complex matrices respectively

$\widetilde{I}^{(2)}(\hat{Z})=\int\exp\Big{\\{}-{\rm
Tr}\Big{(}Y\big{(}(P^{*}P)_{I_{1}}\big{)}^{-\frac{1}{2}}\big{(}J_{2}^{(1)}\big{)}^{*}\big{(}\big{(}(P^{*}P)^{-1}\big{)}_{I_{1}}\big{)}^{-1}J_{2}^{(1)}\big{(}(P^{*}P)_{I_{1}}\big{)}^{-\frac{1}{2}}Y^{*}\Big{)}\Big{\\}}\\\
\times\exp\Big{\\{}-\frac{1}{2}\sum_{i=1}^{n}\big{(}YY^{*}\big{)}_{i,i}^{2}+\mathrm{Tr}\big{(}\big{(}\hat{Z}+\hat{Z}^{*}\big{)}YY^{*}\big{)}\Big{\\}}{\rm
d}Y,$ (6.2)

with $I_{1}=\left\\{1,\cdots,t\right\\}$ and

$\widetilde{I}^{(4,\mathrm{re})}\big{(}\hat{Z}\big{)}=\int\exp\Big{\\{}-\sum_{i=1}^{n}\big{(}YY^{*}\big{)}_{i,i}^{2}+\mathrm{Tr}\Big{(}\big{(}\hat{Z}+\hat{Z}^{*}\big{)}YY^{*}\Big{)}\Big{\\}}\times\\\
\exp\bigg{\\{}-2{\rm
Tr}\bigg{(}Y\big{(}(P^{*}P)_{\widetilde{I}_{1}}\big{)}^{-\frac{1}{2}}\begin{bmatrix}\overline{J_{2}^{(1)}}&\\\
&J_{2}^{(1)}\end{bmatrix}\Big{(}\big{(}(P^{*}P)^{-1}\big{)}_{\widetilde{I}_{1}}\Big{)}^{-1}\begin{bmatrix}J_{2}^{(1)}&\\\
&\overline{J_{2}^{(1)}}\end{bmatrix}\big{(}(P^{*}P)_{\widetilde{I}_{1}}\big{)}^{-\frac{1}{2}}Y^{*}\bigg{)}\bigg{\\}}{\rm
d}Y,$ (6.3)

with
$\widetilde{I}_{1}=\left\\{1,2,\cdots,t\right\\}\bigcup\left\\{n+1,n+2,\cdots,n+t\right\\}.$
Here it should be noted that $P$ in (6.2) can be defined as in (1.21) or
(1.22), depending on the complex or quaternion ensembles, but $P$ in (6.3)
only as in (1.22).

We also need more assumptions.

Assumption-D: For the Jordan form (1.21), all $\alpha_{i}=p_{i,1}=1$ for
$i=1,\ldots,\gamma$, and for a given integer $0\leq m\leq\gamma$,
$\theta_{j}\neq z_{0}$ is fixed for $j=m+1,\ldots,\gamma$ but

$\displaystyle\theta_{i}=z_{0}+N^{-\frac{1}{4}}\hat{\theta}_{i},\quad
i=1,2,\cdots,m.$ (6.4)

######  Theorem 6.1.

For the ${\mathrm{GinUE}}_{N}(A_{0})$ ensemble satisfying Assumption-D, let
$|z_{0}|=1$, as $N\to\infty$ scaled eigenvalue correlations hold uniformly for
all $\hat{z}_{k}$ in a compact subset of $\mathbb{C}$

$\frac{1}{N^{n}}R_{N}^{(n)}\Big{(}A_{0};z_{0}+\frac{\hat{z}_{1}}{\sqrt{N}},\cdots,z_{0}+\frac{\hat{z}_{n}}{\sqrt{N}}\Big{)}=\left(\det\big{(}(P^{*}P)_{I_{1}}\big{)}\det\big{(}\big{(}(P^{*}P)^{-1}\big{)}_{I_{1}}\big{)}\right)^{-n}\\\
\times(2\pi)^{-\frac{n}{2}}\pi^{-n(n+1+t)}e^{-\frac{1}{2}{\rm
Tr}\left(z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\right)^{2}}\prod_{1\leq
i<j\leq n}\left|\hat{z_{i}}-\hat{z}_{j}\right|^{2}\\\
\times\widetilde{I}^{(2)}\left(z_{0}^{-1}\hat{Z}\right)I^{(2)}\left(-\hat{\theta}_{1},\cdots,-\hat{\theta}_{m};z_{0}^{-1}\hat{Z},z_{0}^{-1}\hat{Z}\right)+O\big{(}N^{-\frac{1}{4}}\big{)},$
(6.5)

where $I^{(2)}$ is defined in (3.4).

###### A sketched proof.

Denote $U:=P^{*}P$, similar to (4.6) we rewrite $J_{2}$ as a sum

$J_{2}=J_{2}^{(0)}+N^{-\frac{1}{4}}J_{2}^{(1)}\bigoplus 0_{r_{0}},$

where

$J_{2}^{(0)}=z_{0}\mathbb{I}_{t}\bigoplus J_{2}^{(2)}$

with

$J_{2}^{(2)}={\rm
diag}\left(\theta_{m+1}\mathbb{I}_{\beta_{m+1,1}},\cdots,\theta_{\gamma}\mathbb{I}_{\beta_{\gamma,1}}\right).$

Not as in the proof of Theorem 1.5, now we have

$\displaystyle(z_{0}\mathbb{I}_{r}-J_{2})^{*}U(z_{0}\mathbb{I}_{r}-J_{2})$
$\displaystyle=(z_{0}\mathbb{I}_{r}-J_{2}^{(0)})^{*}U(z_{0}\mathbb{I}_{r}-J_{2}^{(0)})-N^{-\frac{1}{4}}\Big{(}\big{(}J_{2}^{(1)}\big{)}^{*}\bigoplus
0_{r_{0}}\Big{)}U(z_{0}\mathbb{I}_{r}-J_{2})$
$\displaystyle-N^{-\frac{1}{4}}(z_{0}\mathbb{I}_{r}-J_{2}^{(0)})^{*}U\Big{(}J_{2}^{(1)}\bigoplus
0_{r_{0}}\Big{)}+N^{-\frac{1}{2}}\Big{(}\big{(}J_{2}^{(1)}\big{)}^{*}\bigoplus
0_{r_{0}}\Big{)}U\Big{(}J_{2}^{(1)}\bigoplus 0_{r_{0}}\Big{)}.$

Denote

$U=\begin{bmatrix}U_{11}&U_{12}\\\ U_{21}&U_{22}\end{bmatrix},\quad
Q=\left[Q_{1},Q_{2}\right],$

where $U_{11}$, $U_{22}$, Q and $Q_{1}$ are of size $t\times t$, $r_{0}\times
r_{0}$, $n\times r$ and $n\times t$ respectively. In view of (5.8), we have

$\displaystyle\widetilde{f}_{2}(Q)={\rm
Tr}Q_{2}(z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)})^{*}U_{22}(z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)})Q_{2}^{*}$
$\displaystyle-N^{-\frac{1}{4}}{\rm
Tr}Q_{1}\big{(}J_{2}^{(1)}\big{)}^{*}U_{12}(z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)})Q_{2}^{*}-N^{-\frac{1}{4}}{\rm
Tr}Q_{2}(z_{0}\mathbb{I}_{r_{0}}-J_{2}^{(2)})^{*}U_{21}J_{2}^{(1)}Q_{1}^{*}$
$\displaystyle+N^{-\frac{1}{2}}{\rm
Tr}Q_{1}\big{(}J_{2}^{(1)}\big{)}^{*}U_{11}J_{2}^{(1)}Q_{1}^{*}+\frac{1}{2}\sum_{i=1}^{n}\big{(}Q_{1}U_{11}Q_{1}^{*}\big{)}_{i,i}^{2}+O\big{(}\|Q_{2}\|\|Q\|^{3}+\|Q\|^{6}\big{)},$

and

$\widetilde{h}_{2}(Q)={\rm
Tr}\left(\big{(}\overline{z_{0}^{-1}\hat{Z}}+z_{0}^{-1}\hat{Z}\big{)}Q_{1}U_{11}Q_{1}^{*}\right)+N^{-\frac{1}{2}}O\big{(}\|Q_{2}\|\|Q\|+N^{-\frac{1}{4}}\|Q\|^{2}\big{)}.$

Repeat almost the same procedure as in the proof of Theorem 1.5, we can get
the conclusion. ∎

Similarly, following almost the same procedure in the proof of Theorem 1.6, we
can verify

######  Theorem 6.2.

For the ${\mathrm{GinSE}}_{N}(A_{0})$ ensemble satisfying Assumption-D, as
$N\to\infty$ scaled eigenvalue correlations hold uniformly for all
$\hat{z}_{k}$ in a compact subset of $\mathbb{C}$.

(i)If $z_{0}=\pm 1$, then

$\frac{1}{(2N)^{n}}R_{N}^{(n)}\Big{(}A_{0};z_{0}+\frac{\hat{z}_{1}}{\sqrt{2N}},\cdots,z_{0}+\frac{\hat{z}_{n}}{\sqrt{2N}}\Big{)}=\left(\det\big{(}(P^{*}P)_{\widetilde{I}_{1}}\big{)}\det\big{(}(P^{*}P)^{-1}\big{)}_{\widetilde{I}_{1}}\right)^{-n}\\\
\times\pi^{-\frac{n}{2}}\pi^{-2n(n+1+t)}2^{n(n-3+2t)}e^{-\frac{1}{2}{\rm
Tr}\left(\hat{Z}+\overline{\hat{Z}}\right)^{2}}\prod_{1\leq i<j\leq
n}\left|\hat{z_{i}}-\hat{z}_{j}\right|^{2}\left|\hat{z_{i}}-\overline{\hat{z}}_{j}\right|^{2}\prod_{i=1}^{n}\left|\hat{z_{i}}-\overline{\hat{z}}_{i}\right|^{2}\\\
\times\widetilde{I}^{(4,\mathrm{re})}\left(\sqrt{2}z_{0}^{-1}\hat{Z}\right)I^{(4,\mathrm{re})}\left(-\hat{\theta}_{1},\cdots,-\hat{\theta}_{m};\sqrt{2}z_{0}^{-1}\hat{\Lambda}\right)+O\big{(}N^{-\frac{1}{4}}\big{)},$
(6.6)

where $I^{(4,\mathrm{re})}$ is defined in (3.11).

(ii)If $\Im z_{0}>0$ and $|z_{0}|=1$, then

$\frac{1}{(2N)^{n}}R_{N}^{(n)}\Big{(}A_{0};z_{0}+\frac{\hat{z}_{1}}{\sqrt{2N}},\cdots,z_{0}+\frac{\hat{z}_{n}}{\sqrt{2N}}\Big{)}=\left(\det\big{(}(P^{*}P)_{I_{1}}\big{)}\det\big{(}(P^{*}P)^{-1}\big{)}_{I_{1}}\right)^{-n}\\\
\times\pi^{-\frac{n}{2}}\pi^{-2n(n+1)-nt}2^{\frac{n(n-1-t)}{2}-n}(-1)^{nt}e^{-\frac{1}{2}{\rm
Tr}\left(z_{0}^{-1}\hat{Z}+\overline{z_{0}^{-1}\hat{Z}}\right)^{2}}\prod_{1\leq
i<j\leq n}\left|\hat{z_{i}}-\hat{z}_{j}\right|^{2}\\\
\times\widetilde{I}^{(2)}\left(z_{0}^{-1}\hat{Z}\right)I^{(4)}\left(-\hat{\theta}_{1},\cdots,-\hat{\theta}_{m};\sqrt{2}z_{0}^{-1}\hat{Z},\sqrt{2}z_{0}^{-1}\hat{Z}\right)+O\big{(}N^{-\frac{1}{4}}\big{)},$
(6.7)

where $I^{(4)}$ is defined in (3.10), $\widetilde{I}^{(2)}$ is defined in
(6.2), while we need to replace $J_{2}^{(1)}$ with
$2^{\frac{1}{4}}J_{2}^{(1)}$.

Acknowledgements We would like to thank Elton P. Hsu for his encouragement and
support, and also to thank Y.V. Fyodorov, J. Grela and Y. Wang for useful
comments on the first arXiv version. This work was supported by the National
Natural Science Foundation of China #12090012.

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