$\displaystyle\quad-\frac{Nb_{c}^{2}}{b_{c}+\overline{v}}\Big{(}\frac{-\gamma_{c}q}{2\overline{v}N^{1/3}}-\frac{\gamma_{c}^{2}(qr+q^{\prime})}{4\overline{v}^{2}N^{2/3}}+\frac{\gamma_{c}^{2}qr}{4u\overline{v}N^{2/3}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}\Big{)}+{\cal O}(N^{2/3})\bigg{)}$
$\displaystyle+{\rm
e}^{-Na(u+\overline{v})}\Big{(}\frac{\sqrt{c}}{\sqrt{c}+u}\frac{\sqrt{c}}{\sqrt{c}+\overline{v}}\Big{)}^{Nc}\Big{(}\frac{b_{c}+u}{b_{c}}\frac{b_{c}+\overline{v}}{b_{c}}\Big{)}^{Nb_{c}^{2}+N^{\frac{1}{3}}\frac{2b_{c}s}{\gamma_{c}}}\bigg{(}\frac{Nb_{c}\sqrt{c}}{\sqrt{c}+\overline{v}}\Big{(}\frac{\gamma_{c}^{2}q^{2}}{4u\overline{v}N^{2/3}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}\Big{)}$
$\displaystyle\quad-\frac{Nb_{c}^{2}}{b_{c}+\overline{v}}\Big{(}1-\frac{\gamma_{c}r}{N^{1/3}}\frac{u+\overline{v}}{2u\overline{v}}+\frac{\gamma_{c}^{2}r^{2}}{4u\overline{v}N^{2/3}}-\frac{\gamma_{c}^{2}(q^{2}-r^{2})}{N^{2/3}}\frac{(u^{2}+\overline{v}^{2})}{8u^{2}\overline{v}^{2}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}\Big{)}+{\cal
O}(N^{1/3+\epsilon})\bigg{)}.$

Here the first error bound ${\cal O}(N^{1/3+\epsilon})$ comes from the terms
${\cal C}_{1}P_{11}^{z}(P_{21}^{\zeta})^{*}$ and ${\cal
C}_{4}P_{21}^{z}(P_{11}^{\zeta})^{*}$ in (5.16), and the second error bound
${\cal O}(N^{1/3+\epsilon})$ comes from the terms ${\cal
C}_{1}P_{12}^{z}(P_{22}^{\zeta})^{*}$ and ${\cal
C}_{4}P_{22}^{z}(P_{12}^{\zeta})^{*}$ in (5.16), the first error bound ${\cal
O}(N^{2/3})$ comes from the term ${\cal C}_{1}P_{11}^{z}(P_{22}^{\zeta})^{*}$
in (5.16), and the second error bound ${\cal O}(N^{2/3})$ comes from the term
${\cal C}_{4}P_{22}^{z}(P_{11}^{\zeta})^{*}$ in (5.16). And the error bound
${\cal O}(N^{3\tau-1})$ stands for ${\cal
O}\big{(}\frac{1}{Nu^{3}},\frac{1}{N\overline{v}^{3}},\frac{1}{Nu^{2}\overline{v}},\frac{1}{Nu\overline{v}^{2}}\big{)}$.

Using the fact $1/u={\cal O}(N^{\tau})$, we have, for any non-zero constant
$k$,

$\Big{(}1+\frac{u}{k}\Big{)}^{Nk^{2}}=\exp\Big{(}Nk^{2}\log\Big{(}1+\frac{u}{k}\Big{)}\Big{)}=\exp\bigg{(}Nk^{2}\Big{(}\frac{u}{k}-\frac{u^{2}}{2k^{2}}+\frac{u^{3}}{3k^{3}}-\frac{u^{4}}{4k^{4}}+\frac{u^{5}}{5k^{5}}\Big{)}\bigg{)}\big{(}1+{\cal
O}\big{(}Nu^{6}\big{)}\big{)}$ (5.28)

and

$\Big{(}1+\frac{u}{b_{c}}\Big{)}^{N^{\frac{1}{3}}\frac{2b_{c}s}{\gamma_{c}}}=\exp\bigg{(}\frac{2suN^{1/3}}{\gamma_{c}}\bigg{)}\big{(}1+{\cal
O}\big{(}N^{1/3}u^{2}\big{)}\big{)}.$ (5.29)

It follows that

$\displaystyle{\rm
e}^{-Nau}\Big{(}\frac{\sqrt{c}}{\sqrt{c}+u}\Big{)}^{Nc}\Big{(}1+\frac{u}{b_{c}}\Big{)}^{Nb_{c}^{2}}$
$\displaystyle=\exp\bigg{(}N\Big{(}\frac{u^{3}}{3}\Big{(}\frac{1}{b_{c}}-\frac{1}{\sqrt{c}}\Big{)}-\frac{u^{4}}{4}\Big{(}\frac{1}{b_{c}^{2}}-\frac{1}{c}\Big{)}+\frac{u^{5}}{5}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}\Big{)}\bigg{)}\big{(}1+{\cal
O}\big{(}Nu^{6}\big{)}\big{)},$ (5.30) $\displaystyle{\rm
e}^{-Nk(u+\overline{v})}\Big{(}1+\frac{u}{k}\Big{)}^{Nk^{2}}\Big{(}1+\frac{\overline{v}}{k}\Big{)}^{Nk^{2}}$
$\displaystyle=\exp\bigg{(}-N\Big{(}\frac{u^{2}+\overline{v}^{2}}{2}-\frac{u^{3}+\overline{v}^{3}}{3k}+\frac{u^{4}+\overline{v}^{4}}{4k^{2}}-\frac{u^{5}+\overline{v}^{5}}{5k^{3}}\Big{)}\bigg{)}\big{(}1+{\cal
O}\big{(}Nu^{6},N\overline{v}^{2}\big{)}\big{)}$ (5.31)

and

$\Big{(}1+\frac{u}{b_{c}}\Big{)}^{N^{\frac{1}{3}}\frac{2b_{c}s}{\gamma_{c}}}\Big{(}1+\frac{\overline{v}}{b_{c}}\Big{)}^{N^{\frac{1}{3}}\frac{2b_{c}s}{\gamma_{c}}}=\exp\bigg{(}\frac{2s(u+\overline{v})N^{1/3}}{\gamma_{c}}\bigg{)}\big{(}1+{\cal
O}\big{(}N^{1/3}u^{2},N^{1/3}\overline{v}^{2}\big{)}\big{)}.$ (5.32)

Combining (5.13) and (5.27) with (5.30), (5.31) and (5.32), we have (5.2).
This ends the proof. ∎

###  5.2 Integrating the C-D identity: Proof of Theorem 1.2

Let us define

$\displaystyle z:$ $\displaystyle=b_{c}+u,\quad u:=\frac{{\rm
i}(2cb_{c}^{2})^{\frac{1}{4}}Y}{N^{\tau}}+\frac{\nu}{\sqrt{N}},$ (5.33)
$\displaystyle\zeta:$ $\displaystyle=b_{c}+v,\quad v:=\frac{{\rm
i}(2cb_{c}^{2})^{\frac{1}{4}}Y}{N^{\tau}}+\frac{\eta}{\sqrt{N}},$

where $(\nu,\eta)\in{\mathbb{C}}^{2}$ and $Y\in{\mathbb{R}}$.

The following lemma holds for all positive $\tau$.

######  Lemma 5.2.

Let $u$ and $v$ be given above. Assuming $\nu$ and $\eta$ are all ${\cal
O}(N^{\epsilon})$ with $\epsilon<1/2-\tau$ and $Y$ is bounded, we have the
following identities:

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{u^{3}+\overline{v}^{3}}{3k}-\frac{u^{4}+\overline{v}^{4}}{4k^{2}}+\frac{u^{5}+\overline{v}^{5}}{5k^{3}}\Big{)}\bigg{)}$
(5.34)
$\displaystyle=\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{\sqrt{c}b_{c}Y^{2}\sqrt{N}}{kN^{2\tau}}\Big{)}^{2}\bigg{)}\Big{(}1+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}}\Big{)}\Big{)},$
$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{1}{3}\Big{(}\frac{u^{3}}{b_{c}}+\frac{\overline{v}^{3}}{\sqrt{c}}\Big{)}-\frac{1}{4}\Big{(}\frac{u^{4}}{b_{c}^{2}}+\frac{\overline{v}^{4}}{c}\Big{)}+\frac{1}{5}\Big{(}\frac{u^{5}}{b_{c}^{3}}+\frac{\overline{v}^{5}}{c^{3/2}}\Big{)}\Big{)}+\frac{2sN^{\frac{1}{3}}u}{\gamma_{c}}\bigg{)}$
(5.35) $\displaystyle=\exp\bigg{(}\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}-\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}+\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\bigg{)}$
$\displaystyle\quad\times\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}b_{c}\overline{\eta}+\sqrt{c}\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}+b_{c}\sqrt{c}\Big{)}\bigg{)}\Big{(}1+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{\epsilon}}{N^{1/6}}\Big{)}\Big{)},$

and

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{1}{3}\Big{(}\frac{\overline{v}^{3}}{b_{c}}+\frac{u^{3}}{\sqrt{c}}\Big{)}-\frac{1}{4}\Big{(}\frac{\overline{v}^{4}}{b_{c}^{2}}+\frac{u^{4}}{c}\Big{)}+\frac{1}{5}\Big{(}\frac{\overline{v}^{5}}{b_{c}^{3}}+\frac{u^{5}}{c^{3/2}}\Big{)}\Big{)}+\frac{2sN^{1/3}\overline{v}}{\gamma_{c}}\bigg{)}$
(5.36) $\displaystyle=\exp\bigg{(}-\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}+\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}-\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\bigg{)}$
$\displaystyle\quad\times\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}\sqrt{c}\overline{\eta}+b_{c}\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}+b_{c}\sqrt{c}\Big{)}\bigg{)}\Big{(}1+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{\epsilon}}{N^{1/6}}\Big{)}\Big{)}.$

###### Proof.

The assumption that $\nu$ and $\eta$ are all ${\cal O}(N^{\epsilon})$ with
$\epsilon+\tau<1/2$ simply means that, in the expressions of $u$ and $v$
(5.33), the term containing $Y$ scales bigger than the term containing $\nu$
or $\eta$ in the limit of large $N$. The former scales $N^{-\tau}$ while the
latter scales $N^{\epsilon-1/2}$.

The proof is done by a straightforward computation, but let us explain the
error bounds. The error comes from ignoring the terms containing
$Y\nu^{2},Y^{3}\nu$ and $Y^{4}\nu$. The other homogeneous terms are smaller
than one of these; for example, the term containing $Y^{2}\nu^{2}$ is smaller
than $Y^{3}\nu$ because one of $\nu$ is replaced by $Y$ to go from the former
to the latter. Furthermore, since it is also clear that the term containing
$Y^{3}\nu$ dominates over the term containing $Y^{4}\nu$, hence the final
error bounds in (5.34) come from the two terms $Y\nu^{2}$ and $Y^{3}\nu$,
which give $N{\cal O}(N^{-\tau}N^{2(\epsilon-1/2)})$ and $N{\cal
O}(N^{-3\tau}N^{\epsilon-1/2})$ respectively.

∎

####  5.2.1 Case: $1/6<\tau\leq 1/4$

If $1/6<\tau\leq 1/4$, let us redefine $\nu$ and $\eta$ such that

$\displaystyle\nu_{\text{old}}=\nu-\frac{AY^{2}\sqrt{N}}{\sqrt{2}N^{2\tau}},\quad\eta_{\text{old}}=\eta-\frac{AY^{2}\sqrt{N}}{\sqrt{2}N^{2\tau}},$
(5.37)

where $\nu_{\text{old}}$ and $\mu_{\text{old}}$ refer to the variables, $\nu$
and $\mu$, showing up in (5.33). Since $\nu_{\text{old}}$ and
$\mu_{\text{old}}$ are ${\cal O}(N^{\epsilon})$ in Lemma 5.2 this definition
requires that the redefined $\nu$ and $\eta$ are also ${\cal O}(N^{\epsilon})$
and $\epsilon\geq 1/2-2\tau.$ Here we assume $Y$ is bounded and finitely away
from $0$.

Using Lemma 5.1, Lemma 5.2, and the facts that

$\displaystyle\frac{1}{N^{1/3}}\Big{(}\frac{1}{u}+\frac{1}{\overline{v}}\Big{)}\sim{\cal
O}\big{(}\frac{N^{4\tau}}{N^{4/3}}\big{)},\quad\frac{1}{N^{1/3}}\Big{(}\frac{1}{u}-\frac{1}{\overline{v}}\Big{)}\sim{\cal
O}\big{(}\frac{N^{\tau}}{N^{1/3}}\big{)},$ (5.38)

we have

$\displaystyle\overline{\partial}_{\zeta}{\cal
K}_{n}(z,\zeta)=\frac{N^{3/2}}{\sqrt{2}\pi^{3/2}}\bigg{[}\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}-\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}\bigg{]}$
(5.39)
$\displaystyle\quad+\frac{N^{3/2}}{\sqrt{2}\pi^{3/2}}\bigg{[}\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(b_{c}-A)\overline{\eta}-(A-\sqrt{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}\bigg{)}$
$\displaystyle\quad\quad\times\exp\Big{(}\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}-\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}+\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\Big{)}{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle\quad+\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N}{N^{6\tau}},\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle\quad+\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(\sqrt{c}-A)\overline{\eta}-(A-b_{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}\bigg{)}$
$\displaystyle\quad\quad\times\exp\Big{(}-\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}+\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}-\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\Big{)}{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle\quad+\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{\frac{1}{3}}}{N^{2\tau}},\frac{N^{3\tau}}{N},\frac{N^{\epsilon}}{N^{1/6}}\Big{)}\bigg{]}.$

Let us define $F_{n}$ such that

$\displaystyle\frac{1}{n}{\cal K}_{n}(z,\zeta)$ $\displaystyle=\frac{1}{2\pi
t_{c}}\bigg{[}{\rm
erfc}\bigg{(}\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}-{\rm
erfc}\bigg{(}\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}+F_{n}(z,\zeta)\bigg{]}.$
(5.40)

Taking the derivative with respect to $\overline{\eta}$ on both sides, we have

$\displaystyle\frac{1}{n}\overline{\partial}_{\eta}{\cal K}_{n}(z,\zeta)$
$\displaystyle=\frac{1}{\sqrt{2}t_{c}\pi^{3/2}}\bigg{[}\exp\bigg{(}-\bigg{(}\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}^{2}\bigg{)}-\exp\bigg{(}-\bigg{(}\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}^{2}\bigg{)}$
(5.41) $\displaystyle\quad+\overline{\partial}_{\eta}F_{n}\bigg{]}.$

Moreover, using (5.39) and the following relation,

$\displaystyle\frac{1}{n}\overline{\partial}_{\eta}{\cal K}_{n}(z,\zeta)$
$\displaystyle=\frac{1}{\sqrt{N}}\frac{1}{n}\overline{\partial}_{\zeta}{\cal
K}_{n}(z,\zeta),$ (5.42)

we have

$\displaystyle\overline{\partial}_{\eta}F_{n}$
$\displaystyle=\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(b_{c}-A)\overline{\eta}-(A-\sqrt{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}\bigg{)}$
(5.43) $\displaystyle\quad\times\exp\Big{(}\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}-\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}+\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\Big{)}{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle+\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N}{N^{6\tau}},\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle+\exp\bigg{(}-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(\sqrt{c}-A)\overline{\eta}-(A-b_{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}\bigg{)}$
$\displaystyle\quad\times\exp\Big{(}-\frac{{\rm
i}N(2cb_{c}^{2})^{\frac{1}{4}}\sqrt{2}aY^{3}}{3N^{3\tau}}+\frac{{\rm
i}N}{5}\frac{(2cb_{c}^{2})^{\frac{5}{4}}Y^{5}}{N^{5\tau}}\Big{(}\frac{1}{b_{c}^{3}}-\frac{1}{c^{3/2}}\Big{)}-\frac{{\rm
i}2s(2cb_{c}^{2})^{\frac{1}{4}}Y}{\gamma_{c}N^{\tau-1/3}}\Big{)}{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}$
$\displaystyle+\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}.$

We define $\zeta_{0}=b_{c}+\frac{{\rm
i}(2cb_{c}^{2})^{\frac{1}{4}}Y}{N^{\tau}}-\frac{AY^{2}}{\sqrt{2}N^{2\tau}}+\frac{\eta_{0}+i\mathop{\rm
Im}(\eta)}{\sqrt{N}}$ for some positive $\eta_{0}$, such that $\mathop{\rm
Im}\zeta=\mathop{\rm Im}\zeta_{0}$ and $\zeta_{0}$ sits outside the droplet.
Note that the droplet boundary can be at most ${\cal O}(N^{-2\tau},N^{-2/3})$
from $b_{c}$ in the horizontal direction. The first error comes from the
second order approximation of the droplet boundary and the second error comes
from moving droplet boundary by ${\cal O}(t-t_{c})$. For $\zeta_{0}$ to be
outside the droplet, $|\eta_{0}/\sqrt{N}|$ needs to scale greater than ${\cal
O}(N^{-2\tau})$ for $\tau\leq 1/4$. This can be satisfied if $\eta_{0}\sim
N^{\delta}$ with

$\delta\geq 1/2-2\tau.$ (5.44)

Such $\delta$ exists within the range of our consideration because
$1/2-2\tau$, the right hand side, is smaller than $1/2-\tau$, the upper bound
of $\epsilon$ in Lemma 5.2. As a result $F_{n}(z,\zeta_{0})$ is exponentially
suppressed for large $N$ by (5.40) and Theorem 4.1.

Moreover, since $\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\geq a^{2}/4$, we obtain
that

$\displaystyle-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(b_{c}-A)\overline{\eta}-(A-\sqrt{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}<-\frac{a^{2}}{8}\frac{Y^{4}N}{N^{4\tau}}$
(5.45)

and

$\displaystyle-\frac{(\nu+\overline{\eta})^{2}}{2}-\frac{\sqrt{2}Y^{2}\sqrt{N}}{N^{2\tau}}\big{(}(\sqrt{c}-A)\overline{\eta}-(A-b_{c})\nu\big{)}-\frac{Y^{4}N}{N^{4\tau}}\Big{(}\frac{a^{2}}{2}-(b_{c}-A)(A-\sqrt{c})\Big{)}<-\frac{a^{2}}{8}\frac{Y^{4}N}{N^{4\tau}}$
(5.46)

for a sufficiently large $N$ with $\tau<1/4$. This guarantees that the real
exponents in (5.43) are negative.

Assuming $\eta={\cal O}(N^{\delta})$, we have

$\displaystyle|F_{n}(z,\zeta)|$
$\displaystyle=\Big{|}\int_{\eta_{0}}^{\overline{\eta}}\overline{\partial}_{\eta}F+F_{n}(z,\zeta_{0})\Big{|}\leq{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}+|F_{n}(z,\zeta_{0})|={\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}.$

Here the error bound is uniform for $0<\epsilon<\min\\{\tau/2,3\tau-1/2\\}$.

Similarly, we exchange the roles of $\eta$ and $\nu$. Taking $\nu_{0}>0$ such
that $z_{0}=b_{c}+\frac{{\rm
i}(2cb_{c}^{2})^{\frac{1}{4}}Y}{N^{\tau}}-\frac{AY^{2}}{\sqrt{2}N^{2\tau}}+\frac{\nu_{0}}{\sqrt{N}}$
remains outside the droplet, we have

$\displaystyle|F_{n}(z,\zeta)|$ $\displaystyle={\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}+|F_{n}(z_{0},\zeta)|={\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}.$

It follows that

$\displaystyle\frac{1}{n}{\cal K}_{n}(z,\zeta)$ $\displaystyle=\frac{1}{2\pi
t_{c}}\bigg{[}{\rm
erfc}\bigg{(}\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}-{\rm
erfc}\bigg{(}\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}\bigg{]}$
(5.47) $\displaystyle\quad+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}.$

Moreover, by using (5.28) we obtain

$\displaystyle\frac{{\rm e}^{Nz\overline{\zeta}}}{{\rm
e}^{\frac{N}{2}|z|^{2}+\frac{N}{2}|\zeta|^{2}}}\frac{|z-a|^{Nc}|\zeta-a|^{Nc}}{(z-a)^{Nc}(\overline{\zeta}-a)^{Nc}}$
$\displaystyle=\frac{\exp(Nu\overline{v})}{\exp\big{(}\frac{N}{2}|u|^{2}+\frac{N}{2}|v|^{2}\big{)}}\frac{\exp\big{(}\frac{Nb_{c}}{2}(u-\overline{u})\big{)}}{\exp\big{(}\frac{Nb_{c}}{2}(v-\overline{v})\big{)}}\frac{(\sqrt{c}+\overline{u})^{\frac{Nc}{2}}}{(\sqrt{c}+u)^{\frac{Nc}{2}}}\frac{(\sqrt{c}+v)^{\frac{Nc}{2}}}{(\sqrt{c}+\overline{v})^{\frac{Nc}{2}}}$
(5.48) $\displaystyle=\frac{G(\nu,\eta)}{C_{N,\tau}(X,Y,\nu,\eta)},$

where $G(\nu,\eta):={\rm
e}^{\nu\overline{\eta}-\frac{|\nu|^{2}}{2}-\frac{|\eta|^{2}}{2}}$ is the
Ginibre kernel, and

$\displaystyle C_{N,\tau}(X,Y,\nu,\eta)$ $\displaystyle:={\rm
e}^{\frac{(\overline{\nu}-\nu+\eta-\overline{\eta})(X+\sqrt{2}\sqrt{N}b_{c})}{2\sqrt{2}}}{\rm
e}^{\frac{\nu+\overline{\nu}-\eta-\overline{\eta}}{2}\frac{{\rm
i}(2cb_{c}^{2})^{\frac{1}{4}}Y\sqrt{N}}{N^{2\tau}}}\frac{(\sqrt{c}+u(\nu))^{\frac{Nc}{2}}}{(\sqrt{c}+\overline{u(\nu)})^{\frac{Nc}{2}}}\frac{(\sqrt{c}+\overline{v(\eta)})^{\frac{Nc}{2}}}{(\sqrt{c}+v(\eta))^{\frac{Nc}{2}}}.$

Using the relation between the correlation kernel ${\bf K}_{n}$ and ${\cal
K}_{n}$ in (2.3) and (5.48), we get

$\displaystyle\frac{1}{n}{\bf K}_{n}(z,\zeta)$ $\displaystyle=\frac{{\rm
e}^{Nz\overline{\zeta}}}{{\rm
e}^{\frac{N}{2}|z|^{2}+\frac{N}{2}|\zeta|^{2}}}\frac{|z-a|^{Nc}|\zeta-a|^{Nc}}{(z-a)^{Nc}(\overline{\zeta}-a)^{Nc}}\frac{1}{n}{\cal
K}_{n}(z,\zeta)$ (5.49)
$\displaystyle=\frac{G(\nu,\eta)}{C_{N,\tau}(-AY^{2}\sqrt{N}/N^{2\tau},Y,\nu,\eta)}\frac{1}{2\pi
t_{c}}\bigg{[}{\rm
erfc}\bigg{(}\frac{(\sqrt{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}$
$\displaystyle\quad-{\rm
erfc}\bigg{(}\frac{(b_{c}-A)Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\bigg{)}\bigg{]}+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}},\frac{N^{1/3}}{N^{2\tau}},\frac{N^{2\tau}}{N^{2/3}}\Big{)}.$

Here the error is uniform for any $1/6<\tau\leq 1/4$ and for any
$0<\epsilon<\min\\{\tau/2,3\tau-1/2\\}$. Taking $A=-X/Y^{2}$ and
$n,N\to\infty$, we have (1.14).

In particular, when $\tau=1/4$ and $A=-X/Y^{2}$, we have

$\displaystyle\frac{1}{n}{\bf K}_{n}(z,\zeta)$
$\displaystyle=\frac{G(\nu,\eta)}{C_{N,1/4}(X,Y,\nu,\eta)}\frac{1}{2\pi
t_{c}}\bigg{[}{\rm
erfc}\Big{(}X+\sqrt{c}Y^{2}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\Big{)}-{\rm
erfc}\Big{(}X+b_{c}Y^{2}+\frac{\nu+\overline{\eta}}{\sqrt{2}}\Big{)}\bigg{]}+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{1/4}},\frac{1}{N^{1/6}}\Big{)}.$ (5.50)

Here the error is uniform over bounded regions
$\\{(X,Y)\in{\mathbb{R}}^{2}\\}$ and
$\\{(N^{-\epsilon}\nu,N^{-\epsilon}\eta)\in{\mathbb{C}}^{2}\\}$ that is
finitely away from $Y=0$ with $0<\epsilon<1/8$.

For the case of $0<\tau\leq\frac{1}{6}$, one can see from (5.49) that ${\cal
O}\big{(}N^{1/3-2\tau}\big{)}$-term also contributes. It follows that the
leading term and the error bar in (5.49) need to be further modified with
additional corrections. We will not pursue such case in the current paper. See
Remark 1.

####  5.2.2 Case: $1/4<\tau<3/10$

If $1/4<\tau<3/10$, we denote $\nu$ and $\eta$ in (5.33) as

$\displaystyle\nu:$ $\displaystyle=x+\frac{{\rm i}\sqrt{2}\pi
y}{aY^{2}}\frac{N^{2\tau}}{\sqrt{N}}-\frac{b_{c}+\sqrt{c}}{2\sqrt{2}}\frac{Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{s}{\sqrt{2}\gamma_{c}N^{1/6}},$
(5.51) $\displaystyle\eta:$ $\displaystyle=x^{\prime}+\frac{{\rm i}\sqrt{2}\pi
y^{\prime}}{aY^{2}}\frac{N^{2\tau}}{\sqrt{N}}-\frac{b_{c}+\sqrt{c}}{2\sqrt{2}}\frac{Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{s}{\sqrt{2}\gamma_{c}N^{1/6}}.$

Since $\nu$ and $\mu$ are ${\cal O}(N^{\epsilon})$ in Lemma 5.2 the above
definition implies that $\epsilon=2\tau-1/2$ and we are allowed to have
$x\sim{\cal O}(N^{2\tau-1/2})$ and $x^{\prime}\sim{\cal O}(N^{2\tau-1/2})$ to
be able to use the lemma.

In the discussion below, however, we only need $x$ and $x^{\prime}$ to be
${\cal O}(N^{\delta})$ with an arbitrarily small $\delta>0$. In fact we will
assume $\delta=0$ until we need to turn on nonzero $\delta$. We assume that
$(Y,y,y^{\prime})\in{\mathbb{R}}^{3}$ in a bounded region and away from $Y=0$.

Using (5.34) in Lemma 5.2 with $k=\sqrt{c}$ and $b_{c}$, respectively, we have

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{u^{3}+\overline{v}^{3}}{3\sqrt{c}}-\frac{u^{4}+\overline{v}^{4}}{4c}+\frac{u^{5}+\overline{v}^{5}}{5c^{3/2}}\Big{)}\bigg{)}$
(5.52)
$\displaystyle=\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{b_{c}Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}\bigg{)}\Big{(}1+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}}\Big{)}\Big{)}$
$\displaystyle={\rm e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1-\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{1}{N^{\tau}},\frac{N}{N^{4\tau}}\Big{)}\Big{)}$
$\displaystyle={\rm e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1-\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}$

and

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{u^{3}+\overline{v}^{3}}{3b_{c}}-\frac{u^{4}+\overline{v}^{4}}{4b_{c}^{2}}+\frac{u^{5}+\overline{v}^{5}}{5b_{c}^{3}}\Big{)}+\frac{2sN^{1/3}(u+\overline{v})}{\gamma_{c}}\bigg{)}$
(5.53)
$\displaystyle=\exp\bigg{(}-\Big{(}\frac{\nu+\overline{\eta}}{\sqrt{2}}+\frac{\sqrt{c}Y^{2}\sqrt{N}}{N^{2\tau}}\Big{)}^{2}+\frac{2s(\nu+\overline{\eta})}{\gamma_{c}N^{1/6}}\bigg{)}\Big{(}1+{\cal
O}\Big{(}\frac{N^{2\epsilon}}{N^{\tau}},\frac{\sqrt{N}N^{\epsilon}}{N^{3\tau}}\Big{)}\Big{)}$
$\displaystyle={\rm e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}{\rm e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1+\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}.$

Here the error ${\cal O}(N^{3\tau-1},N^{-\tau})$ comes from the error ${\cal
O}(N^{2\epsilon-\tau},N^{\epsilon+1/2-3\tau})$ with $\epsilon=2\tau-1/2$, the
error ${\cal O}(N^{1-4\tau})$ comes from the term containing $Y^{4}$. The
final error bound is obtain by the fact that $1/4<\tau<3/10$.

Similarly, using (5.35) in Lemma 5.2 we have

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{1}{3}\Big{(}\frac{u^{3}}{b_{c}}+\frac{\overline{v}^{3}}{\sqrt{c}}\Big{)}-\frac{1}{4}\Big{(}\frac{u^{4}}{b_{c}^{2}}+\frac{\overline{v}^{4}}{c}\Big{)}+\frac{1}{5}\Big{(}\frac{u^{5}}{b_{c}^{3}}+\frac{\overline{v}^{5}}{c^{3/2}}\Big{)}\Big{)}+\frac{2sN^{1/3}u}{\gamma_{c}}\bigg{)}$
(5.54) $\displaystyle={\rm e}^{\frac{{\rm
i}Y(2cb_{c}^{2})^{1/4}}{N^{\tau-1/3}}\big{(}\frac{\sqrt{2}aY^{2}}{3N^{2\tau-2/3}}+\frac{2s}{\gamma_{c}}\big{)}}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}{\rm e}^{{\rm
i}\pi(y+y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1+\frac{aY^{2}\sqrt{N}(x-x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N}{N^{4\tau}},\frac{N^{3\tau}}{N}\Big{)}\Big{)}$

and using (5.36) in Lemma 5.2, we have

$\displaystyle\quad\exp\Big{(}-Nu\overline{v}-\frac{N}{2}(u^{2}+\overline{v}^{2})\Big{)}\exp\bigg{(}N\Big{(}\frac{1}{3}\Big{(}\frac{\overline{v}^{3}}{b_{c}}+\frac{u^{3}}{\sqrt{c}}\Big{)}-\frac{1}{4}\Big{(}\frac{\overline{v}^{4}}{b_{c}^{2}}+\frac{u^{4}}{c}\Big{)}+\frac{1}{5}\Big{(}\frac{\overline{v}^{5}}{b_{c}^{3}}+\frac{u^{5}}{c^{3/2}}\Big{)}\Big{)}+\frac{2sN^{1/3}\overline{v}}{\gamma_{c}}\bigg{)}$
(5.55) $\displaystyle={\rm e}^{-\frac{{\rm
i}Y(2cb_{c}^{2})^{1/4}}{N^{\tau-1/3}}\big{(}\frac{\sqrt{2}aY^{2}}{3N^{2\tau-2/3}}+\frac{2s}{\gamma_{c}}\big{)}}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}{\rm e}^{-{\rm
i}\pi(y+y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1-\frac{aY^{2}\sqrt{N}(x-x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N}{N^{4\tau}},\frac{N^{3\tau}}{N}\Big{)}\Big{)}.$

Using Lemma 5.1, the facts in (5.38) and above equations, we have

$\displaystyle\overline{\partial}_{\zeta}{\cal K}_{n}(z,\zeta)$
$\displaystyle=\frac{N^{3/2}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{\sqrt{2}\pi^{3/2}}\bigg{[}{\rm
e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1-\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}$ (5.56)
$\displaystyle\quad-{\rm e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1+\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}\bigg{]}.$

Moreover, using the following relation, we have

$\displaystyle\frac{1}{n}\partial_{x^{\prime}}{\cal K}_{n}(z,\zeta)$
$\displaystyle=\frac{1}{n}\overline{\partial}_{\eta}{\cal
K}_{n}(z,\zeta)=\frac{1}{n}\frac{1}{\sqrt{N}}\overline{\partial}_{\zeta}{\cal
K}_{n}(z,\zeta)$ (5.57) $\displaystyle=\frac{{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{\sqrt{2}t_{c}\pi^{3/2}}\bigg{[}{\rm
e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1-\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}$
$\displaystyle\quad-{\rm e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\Big{(}1+\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\Big{)}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N}\Big{)}\bigg{]}.$

Let us define $F_{n}$ such that

$\frac{1}{n}{\cal K}_{n}(z,\zeta)=\frac{{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{2\pi
t_{c}}\frac{{\rm i}aY^{2}\sqrt{N}}{\pi^{3/2}N^{2\tau}}\bigg{(}\frac{{\rm
e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}}{y-y^{\prime}}+F_{n}(z,\zeta)\bigg{)}.$

Taking the derivative with respect to $x^{\prime}$ on both sides, we have

$\displaystyle\frac{1}{n}\partial_{x^{\prime}}{\cal K}_{n}(z,\zeta)$
$\displaystyle=\frac{{\rm e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{\sqrt{2}t_{c}\pi^{3/2}}\bigg{(}{\rm
e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}$
$\displaystyle\quad-\frac{1}{\sqrt{2}\pi}\frac{{\rm
i}aY^{2}\sqrt{N}\big{(}{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\big{)}(x+x^{\prime})}{(y-y^{\prime})N^{2\tau}}\bigg{)}$
$\displaystyle\quad+\partial_{x^{\prime}}\Big{(}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}\frac{1}{2\pi
t_{c}}\frac{{\rm i}aY^{2}\sqrt{N}}{\pi^{3/2}N^{2\tau}}F_{n}(z,\zeta)\Big{)}$

It follows that

$\displaystyle\quad\partial_{x^{\prime}}\Big{(}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}\frac{1}{2\pi
t_{c}}\frac{{\rm i}aY^{2}\sqrt{N}}{\pi^{3/2}N^{2\tau}}F_{n}(z,\zeta)\Big{)}$
(5.58) $\displaystyle=\frac{{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{\sqrt{2}t_{c}\pi^{3/2}}\bigg{[}\frac{aY^{2}\sqrt{N}(x+x^{\prime})}{\sqrt{2}N^{2\tau}}\bigg{(}\frac{{\rm
i}\big{(}{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\big{)}}{\pi(y-y^{\prime})}$
$\displaystyle\quad-{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}\bigg{)}+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\bigg{]}$
$\displaystyle=\frac{{\rm e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{\sqrt{2}t_{c}\pi^{3/2}}{\cal
O}\Big{(}\frac{\sqrt{N}(y-y^{\prime})}{N^{2\tau}},\frac{N^{3\tau}}{N}\Big{)}.$

Setting
$\widehat{X}:=\frac{\sqrt{2}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}$, then
(5.58) reads

$\frac{\sqrt{N}}{N^{2\tau}}\partial_{x^{\prime}}\Big{(}{\rm
e}^{-\frac{1}{2}(x+x^{\prime}+{\rm i}\widehat{X})^{2}}F_{n}\Big{)}={\rm
e}^{-\frac{1}{2}(x+x^{\prime}+{\rm
i}\widehat{X})^{2}}\bigg{[}\frac{\sqrt{N}}{N^{2\tau}}(x+x^{\prime})(y-y^{\prime})\hat{g}(y-y^{\prime})+{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\bigg{]},$ (5.59)

where $\hat{g}:{\mathbb{C}}\to{\mathbb{C}}$ is a bounded analytic function. So
far we have assumed $x$ and $x^{\prime}$ in (5.58) are bounded. Now we will
let them be ${\cal O}(N^{\delta})$ for an arbitrarily small nonzero
$\delta>0$, to be able to integrate both sides of the equation (5.58) in the
$x^{\prime}$ variable from $x^{\prime}$ to $N^{\delta}$.

We estimate the integrals over the first and second summand in (5.59)
separately. For the second summand we use its size and the Gaussian decay in
$x^{\prime}$ to see that

$\int_{x^{\prime}}^{N^{\delta}}{\rm e}^{-\frac{1}{2}(x+t^{\prime})^{2}-{\rm
i}(x+t^{\prime})\widehat{X}}{\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}{\rm d}t^{\prime}={\cal
O}\Big{(}\frac{N^{3\tau}}{N},\frac{N}{N^{4\tau}}\Big{)}\,.$

The second summand does not contribute when $y=y^{\prime}$. In case $y\neq
y^{\prime}$ we use a standard stationary phase approximation and that
$\widehat{X}\gg 1$ to see that

$\int_{x^{\prime}}^{N^{\delta}}{\rm e}^{-\frac{1}{2}(x+t^{\prime})^{2}-{\rm
i}(x+t^{\prime})\widehat{X}}(x+t^{\prime}){\rm d}t^{\prime}={\cal
O}\Big{(}\frac{N^{\delta}}{\widehat{X}}\Big{)}\,.$

Together we have

$F_{n}={\cal
O}\bigg{(}\frac{|y-y^{\prime}|N^{\delta}}{\widehat{X}},\frac{N^{5\tau}}{N^{3/2}},\frac{\sqrt{N}}{N^{2\tau}}\bigg{)}={\cal
O}\Big{(}\frac{N^{\delta}\sqrt{N}}{N^{2\tau}},\frac{N^{5\tau}}{N^{3/2}}\Big{)}$

where $\delta>0$ is arbitrarily small.

Consequently, we have

$\frac{1}{n}{\cal K}_{n}(z,\zeta)=\frac{{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}}{2\pi
t_{c}}\frac{{\rm i}aY^{2}\sqrt{N}}{\pi^{3/2}N^{2\tau}}\bigg{(}\frac{{\rm
e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}}{y-y^{\prime}}+N^{\delta}{\cal
O}\Big{(}\frac{\sqrt{N}}{N^{2\tau}},\frac{N^{5\tau}}{N^{3/2}}\Big{)}\bigg{)}.$

Using the relation between the correlation kernel ${\bf K}_{n}$ and ${\cal
K}_{n}$ in (2.3) and (5.48), we get

$\displaystyle\frac{1}{n}{\bf K}_{n}(z,\zeta)$ $\displaystyle=\frac{{\rm
e}^{Nz\overline{\zeta}}}{{\rm
e}^{\frac{N}{2}|z|^{2}+\frac{N}{2}|\zeta|^{2}}}\frac{|z-a|^{Nc}|\zeta-a|^{Nc}}{(z-a)^{Nc}(\overline{\zeta}-a)^{Nc}}\frac{1}{n}{\cal
K}_{n,N}(z,\zeta)$ (5.60)
$\displaystyle=\frac{1}{\widehat{C}_{N}(Y,x,y,x^{\prime},y^{\prime})}{\rm
e}^{-\frac{(x-x^{\prime})^{2}}{2}+\frac{{\rm
i}\sqrt{2}\pi(x+x^{\prime})(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}-\frac{\pi^{2}(y-y^{\prime})^{2}N^{4\tau}}{a^{2}Y^{4}N}}{\rm
e}^{-\big{(}\frac{x+x^{\prime}}{\sqrt{2}}+\frac{{\rm
i}\pi(y-y^{\prime})N^{2\tau}}{aY^{2}\sqrt{N}}\big{)}^{2}}$
$\displaystyle\times\frac{1}{2\pi t_{c}}\frac{{\rm
i}aY^{2}\sqrt{N}}{\pi^{3/2}N^{2\tau}}\bigg{(}\frac{{\rm e}^{-{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}-{\rm
e}^{{\rm
i}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}}}{y-y^{\prime}}+N^{\delta}{\cal
O}\Big{(}\frac{\sqrt{N}}{N^{2\tau}},\frac{N^{5\tau}}{N^{3/2}}\Big{)}\bigg{)}$
$\displaystyle=\frac{1}{\widehat{C}_{N}(Y,x,y,x^{\prime},y^{\prime})}\frac{aY^{2}\sqrt{N}{\rm
e}^{-x^{2}-(x^{\prime})^{2}}}{\pi^{3/2}t_{c}N^{2\tau}}\bigg{(}\frac{\sin\big{(}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}\big{)}}{\pi(y-y^{\prime})}+N^{\delta}{\cal
O}\Big{(}\frac{\sqrt{N}}{N^{2\tau}},\frac{N^{5\tau}}{N^{3/2}}\Big{)}\bigg{)},$

where

$\widehat{C}_{N}(Y,x,y,x^{\prime},y^{\prime}):=C_{N,\tau}\Big{(}-\frac{b_{c}+\sqrt{c}}{2}\frac{Y^{2}\sqrt{N}}{N^{2\tau}}+\frac{s}{\sqrt{2}\gamma_{c}N^{1/6}},Y,x+\frac{{\rm
i}\sqrt{2}\pi y}{aY^{2}}\frac{N^{2\tau}}{\sqrt{N}},x^{\prime}+\frac{{\rm
i}\sqrt{2}\pi y^{\prime}}{aY^{2}}\frac{N^{2\tau}}{\sqrt{N}}\Big{)},$

and the last error bar comes from the fact that $1/4<\tau<3/10$. It follows
that

$\displaystyle\frac{\widehat{C}_{N}(Y,x,y,x^{\prime},y^{\prime})N^{2\tau}}{n\sqrt{N}}{\bf
K}_{n}(z,\zeta)$ $\displaystyle=\frac{aY^{2}{\rm
e}^{-x^{2}-(x^{\prime})^{2}}}{\pi^{3/2}t_{c}}\bigg{(}\frac{\sin\big{(}\pi(y-y^{\prime})\big{(}1+\frac{\sqrt{2}sN^{2\tau}}{aY^{2}\gamma_{c}N^{2/3}}\big{)}\big{)}}{\pi(y-y^{\prime})}+N^{\delta}{\cal
O}\Big{(}\frac{\sqrt{N}}{N^{2\tau}},\frac{N^{5\tau}}{N^{3/2}}\Big{)}\bigg{)}.$
(5.61)

The error is uniform over a bounded region $\\{Y\in{\mathbb{R}}\\}$ that is
finitely away from $Y=0$ for any $1/4<\tau<3/10$ and an arbitrarily small
$\delta>0$.

This ends the proof of Theorem 1.2.

For $3/10\leq\tau<1/3$, one can see that ${\cal O}(N^{5\tau+\delta-3/2})$-term
in the error becomes larger as $N$ grows. To handle such term we need
additional corrections. However we still believe that the limiting kernel in
(1.16) holds. See Remark 1.

##  Appendix A Relations in the Painlevé II Riemann-Hilbert problem

Let $\widetilde{\Psi}:=\widetilde{\Psi}(\xi;\hat{s})$. Since the Riemann-
Hilbert problem (1.8) has constant jump conditions, we write the following lax
pair,

$\frac{{\rm d}}{{\rm d}\xi}\widetilde{\Psi}=A\widetilde{\Psi},\quad\frac{{\rm
d}}{{\rm d}\hat{s}}\widetilde{\Psi}=B\widetilde{\Psi},$ (A.1)

where

$A={A}(\xi,\hat{s}):=\begin{bmatrix}-4{\rm i}\xi^{2}-{\rm
i}(\hat{s}+2q^{2})&4\xi q+2{\rm i}q^{\prime}\\\ 4\xi q-2{\rm
i}q^{\prime}&4{\rm i}\xi^{2}+{\rm i}(\hat{s}+2q^{2})\end{bmatrix}$ (A.2)

and

${B}={B}(\xi,\hat{s}):=\begin{bmatrix}-{\rm i}\xi&q\\\ q&{\rm
i}\xi\end{bmatrix}.$ (A.3)

The compatibility condition of the linear system (A.1) gives

$A_{\hat{s}}-B_{\xi}+[A,B]=0.$

It follows that $q(\hat{s})$ is the Hastings-McLeod solution of the Painlevé
II equation $q^{\prime\prime}=\hat{s}q+2q^{3}$.

Let us denote

$\Pi(\xi;\hat{s}):=I+\frac{\Pi_{1}(\hat{s})}{2{\rm
i}\xi}+\frac{\Pi_{2}(\hat{s})}{\xi^{2}}+\frac{\Pi_{3}(\hat{s})}{({\rm
i}\xi)^{3}}+{\cal O}\left(\frac{1}{\xi^{4}}\right),$ (A.4)

where

$\Pi_{1}(\hat{s})=\begin{pmatrix}r(\hat{s})&q(\hat{s})\\\
-q(\hat{s})&-r(\hat{s})\end{pmatrix},\quad\Pi_{2}(\hat{s})=\begin{pmatrix}p_{11}(\hat{s})&p_{12}(\hat{s})\\\
p_{21}(\hat{s})&p_{22}(\hat{s})\end{pmatrix},\quad\Pi_{3}(\hat{s})=\begin{pmatrix}q_{11}(\hat{s})&q_{12}(\hat{s})\\\
q_{21}(\hat{s})&q_{22}(\hat{s})\end{pmatrix}.$ (A.5)

Since the Painlevé II Riemann-Hilbert problem for $\Psi$ in (1.8) only has
constant jumps, we have

$\frac{{\rm d}\Psi}{{\rm d}\xi}=A(\xi;\hat{s})\Psi,$ (A.6)

where the matrix function $A(\xi;\hat{s})$ is meromorphic and can be
determined by identifying the singularities. Using $\Pi(\xi;\hat{s})$ in (A.4)
with (A.5), and (A.6), we have

$\displaystyle A(\xi;\hat{s})$ $\displaystyle=\begin{bmatrix}-4{\rm
i}\xi^{2}-{\rm i}(\hat{s}+2q^{2})&4\xi q-2{\rm i}qr+8{\rm i}p_{12}\\\ 4\xi
q+2{\rm i}qr-8{\rm i}p_{21}&4{\rm i}\xi^{2}+{\rm
i}(\hat{s}+2q^{2})\end{bmatrix}+\frac{A_{1}(\hat{s})}{\xi}+\frac{A_{2}(\hat{s})}{\xi^{2}}+\frac{A_{3}(\hat{s})}{\xi^{3}},$
(A.7)

where

$\displaystyle
A_{1}(\hat{s})=\begin{bmatrix}4q(p_{12}-p_{21})&q^{3}-8q_{12}+4p_{12}r+q(\hat{s}-4p_{22}-r^{2})\\\
q^{3}+8q_{21}+4p_{21}r+q(\hat{s}-4p_{11}-r^{2})&-4q(p_{12}-p_{21})\end{bmatrix},$
(A.8)

and the corresponding entries of $A_{2}$ and $A_{3}$ are given by

11 $\displaystyle=\frac{{\rm
i}}{2}(-16p_{21}^{2}+q^{4}+r+8p_{21}qr+q^{2}(\hat{s}-r^{2})),$ (A.9)
$\displaystyle[A_{2}(\hat{s})]_{22}$ $\displaystyle=-\frac{{\rm
i}}{2}(-16p_{21}^{2}+q^{4}+r+8p_{21}qr+q^{2}(\hat{s}-r^{2})),$
$\displaystyle[A_{2}(\hat{s})]_{12}$ $\displaystyle=\frac{{\rm
i}}{2q}(q^{4}r-16p_{21}(p_{21}r-2q_{12})+8qr(p_{21}r-q_{12})+q^{2}(1-8q_{22}+\hat{s}r-r^{3})),$
$\displaystyle[A_{2}(\hat{s})]_{21}$ $\displaystyle=\frac{-{\rm
i}}{2q}(q^{4}r-16p_{21}(p_{21}r+2q_{21})+8qr(p_{21}r+q_{21})+q^{2}(1+8q_{11}+\hat{s}r-r^{3})).$
11
$\displaystyle=\hat{s}p_{12}q-4p_{12}p_{22}q+\frac{q^{2}}{4}+2p_{12}q^{3}-2q^{2}\hat{q}_{11}+8p_{12}q_{21}-2q^{2}q_{22}+p_{22}q^{2}r$
(A.10)
$\displaystyle-\frac{(1+4p_{12}q)r^{2}}{4}-p_{11}(2+q(4p_{12}-4p_{21}+qr))+p_{21}(-2q^{3}+8q_{12}+q(-\hat{s}+4p_{22}+r^{2})),$
$\displaystyle[A_{3}(\hat{s})]_{22}$
$\displaystyle=-(\hat{s}-4p_{11})(p_{12}-p_{21})q+2(p_{21}-p_{12})q^{3}-8p_{21}q_{12}-8p_{12}q_{21}-\frac{r^{2}}{4}$
$\displaystyle+(p_{12}-p_{21})qr^{2}+q^{2}(\frac{1}{4}+2q_{11}+2q_{22}+p_{11}r)+p_{22}(-2-q(-4p_{12}+4p_{21}+qr)).$

By the fact that $A_{1}(\hat{s})=0$, we have

$\displaystyle p_{12}$ $\displaystyle=p_{21},$ (A.11) $\displaystyle p_{11}$
$\displaystyle=\frac{q^{3}+8q_{21}+4p_{21}r+q(\hat{s}-r^{2})}{4q},$
$\displaystyle p_{22}$
$\displaystyle=\frac{q^{3}-8q_{12}+4p_{21}r+q(\hat{s}-r^{2})}{4q}.$

Furthermore, by the fact that $A_{2}(\hat{s})=0$, we have

$\displaystyle r$ $\displaystyle=-\hat{s}q^{2}-q^{4}+(q^{\prime})^{2},$ (A.12)
$\displaystyle(q_{12}+q_{21})q^{\prime}$ $\displaystyle=q^{2}(q_{11}+q_{22}).$

Defining

$B(\xi;\hat{s}):=\frac{{\rm d}\Psi}{{\rm d}\hat{s}}\cdot\Psi^{-1},$ (A.13)

by the similar procedure as above, we have

$\displaystyle B(\xi;\hat{s})$ $\displaystyle=\begin{bmatrix}-{\rm i}\xi&q\\\
q&{\rm i}\xi\end{bmatrix}+\frac{B_{1}(\hat{s})}{\xi},$ (A.14)

where

$\displaystyle B_{1}(\hat{s})=\begin{bmatrix}-\frac{{\rm
i}(q^{2}+r^{\prime})}{2}&\frac{{\rm i}(4p_{12}-qr-q^{\prime})}{2}\\\
\frac{{\rm i}(4p_{21}-qr-q^{\prime})}{2}&\frac{{\rm
i}(q^{2}+r^{\prime})}{2}\end{bmatrix}.$ (A.15)

By the fact that $B_{1}(\hat{s})=0$, we have

$\displaystyle p_{12}=\frac{qr+q^{\prime}}{4},\quad r^{\prime}=-q^{2}.$ (A.16)

Using the fact that $A_{3}(\hat{s})=0$ and the relation
$(q_{12}+q_{21})q^{\prime}=q^{2}(q_{11}+q_{22})$ from (A.12), we have

$\displaystyle
q_{12}=-q_{21}=\frac{1}{16}(2\hat{s}q+q^{3}+\hat{s}^{2}q^{5}+2\hat{s}q^{7}+q^{9}-2q^{2}(\hat{s}+q^{2})q^{\prime}-2q^{3}(\hat{s}+q^{2})(q^{\prime})^{2}+2(q^{\prime})^{3}+q(q^{\prime})^{4}).$
(A.17)

Therefore,

$\displaystyle p_{12}(\hat{s})=p_{21}(\hat{s})$
$\displaystyle=\frac{q(\hat{s})r(\hat{s})+q^{\prime}(\hat{s})}{4},$ (A.18)
$\displaystyle p_{11}(\hat{s})=p_{22}(\hat{s})$
$\displaystyle=\frac{2q(\hat{s})q^{\prime}(\hat{s})(\hat{s}+q(\hat{s})^{2})+q(\hat{s})^{2}-r(\hat{s})^{2}}{8}+\frac{2q^{\prime}(\hat{s})(r(\hat{s})-q^{\prime}(\hat{s})^{2})}{8q(\hat{s})}$
(A.19) $\displaystyle=\frac{q(\hat{s})^{2}-r(\hat{s})^{2}}{8}.$ (A.20)

##  Appendix B Proof of Theorem 2.1

Let $p_{j}$ be the monic orthogonal polynomials satisfy the following
orthogonality conditions

$\displaystyle\int_{{\mathbb{C}}}p_{j}(z)\overline{p_{k}(z)}|z-a|^{2Nc}{\rm
e}^{-N|z|^{2}}\,{\rm d}A(z)=h_{j}\,\delta_{jk},$ (B.1)

where $h_{j}$ is the norming constant. Let us denote

$\psi_{j}(z):=(z-a)^{Nc}p_{j}(z),\qquad\phi_{j}(z):=(z-a)^{Nc}\frac{p_{j}(z)}{h_{j}}.$
(B.2)

We define the inner product

$\langle U|V\rangle:=\int_{\mathbb{C}}\overline{U(z)}\,V(z)\,{\rm
e}^{-N|z|^{2}}\,{\rm d}A(z)=\overline{\langle V|U\rangle}$ (B.3)

and denote $\Psi:=[\psi_{0},\psi_{1},\cdots]^{\rm T}$,
$\Phi:=[\phi_{0},\phi_{1},\cdots]^{\rm T}$, where ${\rm T}$ is the transpose
of a matrix. Let

$\Pi_{n}:=\textup{diag}(\underbrace{1,\cdots,1}_{n},0,\cdots)$ (B.4)

be the projection operator, we write

${\cal K}_{n}(z,\zeta)={\rm e}^{-Nz\bar{\zeta}}\Phi^{*}(\zeta)\Pi_{n}\Psi(z),$
(B.5)

where the superscript $*$ means the complex conjugation. Then we have

$\overline{\partial}_{\zeta}{\cal K}_{n}(z,\zeta)=-N{\rm
e}^{-Nz\bar{\zeta}}\Big{(}z-\frac{1}{N}\overline{\partial}_{\zeta}\Big{)}\Phi^{*}(\zeta)\Pi_{n}\Psi(z).$
(B.6)

Let us denote

$L_{j,j-1}=-\frac{\langle z\psi_{j}|\phi_{0}\rangle}{\langle
z\psi_{j-1}|\phi_{0}\rangle}\quad\mbox{for each $j$}.$ (B.7)

Notice that the denominator does not vanish due to the fact that
$p_{n+1}(0)\neq 0$ and $p_{n}(z)\neq zp_{n-1}(z)$ for all $n$, which is
obtained by the following relation

$h_{n}=-\frac{\Gamma(Nc+n+1)}{2{\rm
i}N^{Nc+n+1}}\frac{\widetilde{h}_{n}}{p_{n+1}(0)}$

with $\widetilde{h}_{n}\neq 0$ and $h_{n}\neq 0$ for all $n$ from [8,
Proposition 3.3, Proposition 7.1]. Note also that

$z\big{(}\psi_{j}(z)+L_{j,j-1}\psi_{j-1}(z)\big{)}\,\bot\,\phi_{0}$

with respect to the inner product (B.3). The numbers $L_{j,j-1}$ are building
blocks to define the lower diagonal matrix

$L:=\begin{bmatrix}0&0&0&0&\dots\\\ L_{2,1}&0&0&0&\dots\\\
0&L_{3,2}&0&0&\dots\\\ 0&0&L_{4,3}&0&\dots\\\
\vdots&\vdots&\vdots&\ddots&\vdots\end{bmatrix}.$

We write

$\widetilde{\psi}_{j}:=\psi_{j}+L_{j,j-1}\psi_{j-1}.$ (B.8)

Then if

$\phi(z)=(\text{polynomials of deg}\leq j-2)\cdot(z-a)\cdot(z-a)^{Nc},$

we have

$\langle\phi\,|\,z\widetilde{\psi}_{j}\rangle=\langle\partial\phi\,|\,\widetilde{\psi}_{j}\rangle=0.$
(B.9)

It follows that

$\mathrm{span}\\{\phi_{0},\phi_{1},\cdots,\phi_{j-1}\\}\,\bot\,z\widetilde{\psi}_{j},$

which leads to

$z\widetilde{\psi}_{j}(z)=\psi_{j+1}(z)+B_{j,j}\,\psi_{j}(z)$ (B.10)

for some $B_{j,j}$. Thus we obtain

$z(I+L)\Psi=B\,\Psi,\qquad B:=\begin{bmatrix}B_{1,1}&1&0&0&\dots\\\
0&B_{2,2}&1&0&\dots\\\ 0&0&B_{3,3}&1&\dots\\\ 0&0&0&B_{4,4}&\ddots\\\
\vdots&\vdots&\vdots&\ddots&\ddots\end{bmatrix}.$ (B.11)

Let us also write

$U_{j,j+1}=-\frac{h_{j+1}}{h_{j}}\frac{p_{j}(a)}{p_{j+1}(a)},$ (B.12)

which is well-defined due to the fact that $h_{j}\neq 0$ for all $j$ from [8,
Proposition 3.3, Proposition 7.1], and define the upper diagonal matrix

$U:=\begin{bmatrix}0&U_{1,2}&0&0&\dots\\\ 0&0&U_{2,3}&0&\dots\\\
0&0&0&U_{3,4}&\dots\\\ 0&0&0&0&\ddots\\\
\vdots&\vdots&\vdots&\vdots&\ddots\end{bmatrix}.$

Then the function

$\displaystyle\begin{split}\widehat{\phi}_{j}(z)&:=\phi_{j}(z)+U_{j,j+1}\phi_{j+1}(z)=(\text{polynomials
of deg}\leq j)\cdot(z-a)\cdot(z-a)^{Nc}\end{split}$ (B.13)

satisfies

$\langle\partial\widehat{\phi}_{j}\,|\,\psi_{k}\rangle=\langle\widehat{\phi}_{j}\,|\,z\,\psi_{k}\rangle=0\quad\text{if
}k\leq j-2.$ (B.14)

Thus we have

$\partial\widehat{\phi}_{j}=A_{j,j}\phi_{j}+A_{j,j-1}\phi_{j-1}$ (B.15)

for some $A_{j,k}$, equivalently,

$\partial(I+U)\Phi=A\,\Phi,\qquad A:=\begin{bmatrix}A_{1,1}&0&0&0&\dots\\\
A_{2,1}&A_{2,2}&0&0&\dots\\\ 0&A_{3,2}&A_{3,3}&0&\dots\\\
0&0&A_{4,3}&A_{4,4}&\dots\\\ \vdots&\vdots&\vdots&\ddots&\ddots\end{bmatrix}.$
(B.16)

We now determine $A_{j,j-1}$ and $A_{j,j}$. Note that integration by parts
gives

$\displaystyle\begin{split}\overline{B}(I+U)^{t}&=\overline{B}\langle\Psi\,|\,\Phi^{t}\rangle(I+U)^{t}=\langle
B\Psi\,|\,\Phi^{t}(I+U)^{t}\rangle\\\ &=\langle
z(I+L)\Psi\,|\,\Phi^{t}(I+U)^{t}\rangle=\frac{1}{N}\langle(I+L)\Psi\,|\,\partial\Phi^{t}(I+U)^{t}\rangle\\\
&=\frac{1}{N}\langle(I+L)\Psi\,|\,\Phi^{t}A^{t}\rangle=\frac{1}{N}(I+\overline{L})A^{t}.\end{split}$
(B.17)

Thus we obtain the relation

$\tfrac{1}{N}A(I+L^{*})=(I+U)B^{*},\qquad
B=\tfrac{1}{N}(I+L)A^{*}(I+U^{*})^{-1}.$ (B.18)

Comparing the terms involving $A_{j,j-1}$, one can observe that

$A_{j,j-1}=N.$ (B.19)

To determine $A_{j,j}$, note that

$\displaystyle\partial\widehat{\phi}_{j}(z)$
$\displaystyle=\partial\Big{(}\phi_{j}+U_{j,j+1}\phi_{j+1}\Big{)}=\frac{1}{h_{j}}\partial\Big{(}(z-a)^{Nc}p_{j}\Big{)}+\frac{U_{j,j+1}}{h_{j+1}}\partial\Big{(}(z-a)^{Nc}p_{j+1}\Big{)}$
$\displaystyle=(z-a)^{Nc-1}\frac{1}{h_{j}}\bigg{(}\Big{(}Ncp_{j}+(z-a)p_{j}^{\prime}\Big{)}-\frac{p_{j}(a)}{p_{j+1}(a)}\Big{(}Ncp_{j+1}+(z-a)p_{j+1}^{\prime}\Big{)}\bigg{)}$
$\displaystyle=(z-a)^{Nc}\frac{1}{h_{j}}\bigg{(}\frac{Nc}{z-a}\Big{(}p_{j}-\frac{p_{j}(a)}{p_{j+1}(a)}p_{j+1}\Big{)}+p_{j}^{\prime}-\frac{p_{j}(a)}{p_{j+1}(a)}p_{j+1}^{\prime}\bigg{)}.$

This gives

$A_{j,j}p_{j}+N\frac{h_{j}}{h_{j-1}}p_{j-1}=\frac{Nc}{z-a}\Big{(}p_{j}-\frac{p_{j}(a)}{p_{j+1}(a)}p_{j+1}\Big{)}+p_{j}^{\prime}-\frac{p_{j}(a)}{p_{j+1}(a)}p_{j+1}^{\prime}.$

Comparing the coefficient of $z^{j}$ term of this identity, we obtain

$A_{j,j}=-\frac{p_{j}(a)}{p_{j+1}(a)}(Nc+j+1).$ (B.20)

Notice in particular that $A_{j,k}$’s are real.

Now let us consider the decomposition

$A=N\,T_{-}+A_{0},$ (B.21)

where

$T_{-}:=\begin{bmatrix}0&0&0&0&\dots\\\ 1&0&0&0&\dots\\\ 0&1&0&0&\dots\\\
0&0&1&0&\dots\\\ \vdots&\vdots&\vdots&\ddots&\vdots\end{bmatrix},\qquad
A_{0}:=\textup{diag}(A_{1,1},A_{2,2},\dots)$

are the translation and the diagonal part respectively. Write

$A^{*}=T_{+}+A_{0}^{*},\qquad T_{+}:=T_{-}^{*}.$ (B.22)

Note also that we have

$\displaystyle\quad(T_{+}-\zeta)\Psi=\big{(}T_{+}-(I+L)^{-1}B\big{)}\Psi=\Big{(}T_{+}-(I+L)^{-1}\frac{1}{N}(I+L)A^{*}(I+U^{*})^{-1}\Big{)}\Psi$
$\displaystyle=\Big{(}T_{+}-\frac{1}{N}A^{*}(I+U^{*})^{-1}\Big{)}\Psi=(T_{+}(I+U^{*})-\frac{1}{N}A^{*})(I+U^{*})^{-1}\Psi=\Big{(}T_{+}U^{*}-\tfrac{1}{N}A_{0}^{*}\Big{)}(I+U^{*})^{-1}\Psi,$

where the second and the fourth identity follow from (B.18) and (B.21)
respectively.

Let us claim that $(T_{+}U^{*}-\tfrac{1}{N}A_{0}^{*})$ is invertible. Suppose
that this is not the case. Then there exists some $k$ such that
$U^{*}_{k-1,k}-\frac{1}{N}A_{k-1,k-1}=0$. Consequently, we have
$\psi_{k}(z)=z\psi_{k-1}(z)$. This contradicts the assumption that $a\neq 0$.
Therefore, $(T_{+}U^{*}-\tfrac{1}{N}A_{0}^{*})$ is invertible, which is also
equivalent to

$h_{j+1}\not=\frac{Nc+j+1}{N}h_{j}$ (B.23)

by (B.12) and (B.20).

By letting

$\widehat{\Psi}:=\Big{(}T_{+}U^{*}-\frac{1}{N}A_{0}^{*}\Big{)}^{-1}(T_{+}-z)\Psi,$
(B.24)

we have

$(I+U^{*})\widehat{\Psi}=\Psi.$ (B.25)

Note that

$\displaystyle\begin{split}\widehat{\Psi}&:=[\widehat{\psi}_{0},\widehat{\psi}_{1},\cdots]^{t}=\Big{(}T_{+}U^{*}-\frac{1}{N}A_{0}^{*}\Big{)}^{-1}(T_{+}-z)[\psi_{0},\psi_{1},\cdots]^{t}\\\
&=\textup{diag}\Big{(}U^{*}_{1,2}-\frac{1}{N}A_{1,1},U^{*}_{2,3}-\frac{1}{N}A_{2,2},\cdots\Big{)}^{-1}[\psi_{1}-z\psi_{0},\psi_{2}-z\psi_{1},\cdots]^{t}.\end{split}$

Thus we have

$\widehat{\psi}_{j}=\frac{\psi_{j+1}-z\psi_{j}}{U_{j,j+1}^{*}-\frac{1}{N}A_{j,j}}.$
(B.26)

Here the denominator again does not vanish due to (B.12), (B.20) and (B.23).
Then by (B.11) and (B.18), we have

$\displaystyle(I+L)z(I+U^{*})\widehat{\Psi}$
$\displaystyle=(I+L)z\Psi=B\Psi=B(I+U^{*})\widehat{\Psi}=\frac{1}{N}(I+L)A^{*}\widehat{\Psi},$

which leads to

$z(I+U^{*})\widehat{\Psi}=\tfrac{1}{N}A^{*}\widehat{\Psi}.$ (B.27)

Combining (B.27), (B.16) and (B.18), we obtain

$\displaystyle\Big{(}z-\frac{1}{N}\partial_{\zeta}\Big{)}\Phi^{*}(\zeta)\Pi_{n}\Psi(z)$
$\displaystyle=\Big{(}z-\frac{1}{N}\partial_{\zeta}\Big{)}\Phi^{*}(\zeta)\Pi_{n}(I+U^{*})\widehat{\Psi}(z)$
$\displaystyle=\frac{1}{N}\Phi^{*}(\zeta)[\Pi_{n},A^{*}]\widehat{\Psi}(z)-\frac{1}{N}\partial_{\zeta}\Phi^{*}(\zeta)[\Pi_{n},I+U^{*}]\widehat{\Psi}(z).$

Moreover by (B.19) and (B.26), we have

$\displaystyle\begin{split}\Phi^{*}(\zeta)[\Pi_{n},A^{*}]\widehat{\Psi}(z)&=\overline{\phi_{n-1}(\zeta)}A_{n,n-1}\widehat{\psi}_{n}(z)-\overline{\phi_{n}(\zeta)}A_{n-1,n}\widehat{\psi}_{n-1}(z)\\\
&=\frac{N}{U_{n,n+1}^{*}-\frac{1}{N}A_{n,n}}\overline{\phi_{n-1}(\zeta)}(\psi_{n+1}(z)-z\psi_{n}(z)).\end{split}$

Similarly, we obtain

$\displaystyle\begin{split}\partial_{\zeta}\Phi^{*}(\zeta)[\Pi_{n},I+U^{*}]\widehat{\Psi}(z)&=\partial_{\zeta}\overline{\phi_{n-1}(\zeta)}U_{n,n-1}^{*}\widehat{\psi}_{n}(z)-\partial_{\zeta}\overline{\phi_{n}(\zeta)}U_{n-1,n}^{*}\widehat{\psi}_{n-1}(z)\\\
&=-\frac{U_{n-1,n}^{*}}{U_{n-1,n}^{*}-\frac{1}{N}A_{n-1,n-1}}\partial_{\zeta}\overline{\phi_{n}(\zeta)}(\psi_{n}(z)-z\psi_{n-1}(z)).\end{split}$

Combining all of the above identities with (B.6), the proof is complete.

##  Appendix C Proof of Theorem 4.1

Since $\Delta Q(z)=(\partial_{x}^{2}+\partial_{y}^{2})Q(z)=4$ away from the
singularity, the function, $2\log|u(z)|+4N|z-z_{0}|^{2}-NQ(z)$, is a
subharmonic function (and harmonic away from the singularities) for any
polynomial $u(z)$. Exponentiating the subharmonic function we get that
$|u(z)|^{2}{\rm e}^{4N|z-z_{0}|^{2}-NQ(z)}$ is also a subharmonic function,
using Jensen’s inequality. Integrating around a small disk of radius
$1/\sqrt{N}$ centered at $z_{0}$, that we denote by $D(z_{0};1/\sqrt{N})$
below, we get

$\displaystyle|u(z_{0})|^{2}{\rm e}^{-NQ(z_{0})}$
$\displaystyle\leq\frac{N}{\pi}\int_{D(z_{0};1/\sqrt{N})}|u(z)|^{2}{\rm
e}^{4N|z-z_{0}|^{2}-NQ(z)}{\rm d}A(z)$ (C.1) $\displaystyle<\frac{N{\rm
e}^{4}}{\pi}\int_{D(z_{0};1/\sqrt{N})}|u(z)|^{2}{\rm e}^{-NQ(z)}{\rm d}A(z)$
(C.2) $\displaystyle<\frac{N{\rm e}^{4}}{\pi}\int_{{\mathbb{C}}}|u(z)|^{2}{\rm
e}^{-NQ(z)}{\rm d}A(z).$ (C.3)

The above inequality holds for arbitrary $z_{0}\in{\mathbb{C}}$. Taking the
log of the above inequality, we get

$2\log|u(z)|-NQ(z)\leq\log\frac{N{\rm e}^{4}}{\pi}\|u\|_{L^{2}(NQ)},\qquad
z\in{\mathbb{C}},$

where we define

$\|u\|_{L^{2}(NQ)}=\int_{{\mathbb{C}}}|u(z)|^{2}{\rm e}^{-NQ(z)}{\rm d}A(z).$

Let us define $\widehat{Q}$, adopting the notation of [4], by

$\widehat{Q}(z)=\frac{2t}{\text{Area}({\cal S})}\int_{\cal S}\log|z-w|{\rm
d}A(w)-\ell_{\text{2D}},$

where $\ell_{\text{2D}}$ is chosen such that $Q=\widehat{Q}$ on the boundary
of ${\cal S}$. Since $\widehat{Q}$ is harmonic in ${\cal S}^{c}$, and
$\widehat{Q}$ satisfies the growth condition $\widehat{Q}(z)\sim 2t\log|z|$ as
$|z|\to\infty$, we have

$2\log|u(z)|-N\widehat{Q}(z)\leq\log\frac{N{\rm e}^{4}}{\pi}\|u\|_{L^{2}(NQ)}$

on the boundary of ${\cal S}$, and the left hand side goes to $-\infty$ if
$u(z)$ is a polynomial of degree less than $tN$. Since the left hand side is
harmonic on ${\cal S}^{c}$ the above inequality holds for all $z\in{\cal
S}^{c}\cup\partial{\cal S}$.

Let us take $n=tN$ and $u(z)={\rm e}^{\frac{N}{2}Q(z)}{\bf
K}_{n}(z,w)/\sqrt{{\bf K}_{n}(w,w)}$ which is a polynomial of degree $n-1$. We
have $\|u\|_{L^{2}(NQ)}=1$ and, by exponentiating the above inequality, we get

${\rm e}^{N(Q(z)-\widehat{Q}(z))}\frac{|{\bf K}_{n}(z,w)|^{2}}{{\bf
K}_{n}(w,w)}\leq\frac{N{\rm e}^{4}}{\pi},\quad\text{for}\quad z\in{\cal
S}^{c}\cup\partial{\cal S}\quad\text{and}\quad w\in{\mathbb{C}}.$

Taking $w=z$ we have

${\bf K}_{n}(z,z)\leq\frac{N{\rm e}^{4}}{\pi}{\rm
e}^{-N(Q(z)-\widehat{Q}(z))},\quad\text{for}\quad z\in{\cal
S}^{c}\cup\partial{\cal S}.$

Setting ${\cal U}_{\text{2D}}(z)=Q(z)-\widehat{Q}(z)$, we finish the proof
Theorem 4.1.

Acknowledgements: We thank Yacin Ameur, Sung-Soo Byun, Tom Claeys, and Arno
Kuijlaars for helpful discussions.

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