1 Introduction
Let K be an oriented knot in the three-sphere
$S^3$
. For a positive integer N, we denote by
$J_N(K;q)$
the colored Jones polynomial associated with the irreducible N-dimensional representation of the Lie algebra
$\mathfrak {sl}(2;\mathbb {C})$
. Here, we normalize
$J_N(K;q)$
so that
$J_N(U;q)=1$
for the unknot U.
Let us consider an evaluation
$J_N\left (K;e^{2\pi \sqrt {-1}/N}\right )$
. It is well known that it coincides with Kashaev’s invariant
$\langle K\rangle _N$
[Reference Kashaev12, Reference Murakami and Murakami25]. Kashaev conjectured that his invariant grows exponentially as
$N\to \infty $
and that its growth rate gives the hyperbolic volume of the knot complement when K is a hyperbolic knot, that is,
$S^3\setminus {K}$
possesses a (unique) complete hyperbolic structure with finite volume [Reference Kashaev13]. In [Reference Murakami and Murakami25], Kashaev’s conjecture was generalized to any knot replacing the hyperbolic volume with simplicial volume (also known as Gromov’s norm [Reference Gromov8]).
Conjecture 1.1 (Volume conjecture)
Let
$K\subset S^3$
be any knot. Then we have
$$ \begin{align*} \lim_{N\to\infty} \frac{1}{N}\log\bigl|J_N(K;e^{2\pi\sqrt{-1}/N})\bigr| = \frac{1}{2\pi}\operatorname{Vol}(S^3\setminus{K}), \end{align*} $$
where
$\operatorname {Vol}(S^3\setminus {K})$
is the simplicial volume of
$S^3\setminus {K}$
.
So far, Kashaev’s conjecture is proved for the figure-eight knot by Ekholm, and for knots with up to seven crossings [Reference Ohtsuki31–Reference Ohtsuki and Yokota33]. The volume conjecture is proved for hyperbolic knots with up to seven crossings as above, for all the torus knots by Kashaev and Tirkkonen [Reference Kashaev and Tirkkonen14], for the Whitehead doubles of the torus knots by Zheng [Reference Zheng37], and the
$(2,2k+1)$
-cable of the figure-eight knot by T. Le and Tran [Reference Le and Tran18].
J. Murakami, Okamoto, Takata, Yokota, and the author complexified Kashaev’s conjecture as follows [Reference Murakami, Murakami, Okamoto, Takata and Yokota26, Conjecture 1.2].
Conjecture 1.2 For a hyperbolic knot K in
$S^3$
, we have
$$ \begin{align*} \log{J_N(K;e^{2\pi\sqrt{-1}/N})} \underset{N\to\infty}{\sim} \frac{N}{2\pi}\operatorname{CV}(K), \end{align*} $$
where
$\operatorname {CV}(K):=\operatorname {Vol}(S^3\setminus {K})+\sqrt {-1}\operatorname {CS}^{\mathrm {SO}(3)}(S^3\setminus {K})$
is the complex volume with
$\operatorname {CS}^{\mathrm {SO}(3)}$
the
$\mathrm {SO}(3)$
Chern–Simons invariant [Reference Meyerhoff21].
For a hyperbolic knot
$K\subset S^3$
, let
$\rho \colon \pi _1(S^3\setminus {K})\to \mathrm {SL}(2;\mathbb {C})$
be an irreducible representation that is a small deformation of the holonomy representation corresponding to the complete hyperbolic structure. Note that
$\rho $
corresponds to an incomplete hyperbolic structure [Reference Thurston35]. Up to conjugation, we may assume that
$\rho $
sends the meridian of K to
$\begin {pmatrix}e^{u/2}&\ast \\0&e^{-u/2}\end {pmatrix}$
and the preferred longitude to
$\begin {pmatrix}-e^{v(u)/2}&\ast \\0&-e^{-v(u)/2}\end {pmatrix}$
(see, for example, [Reference Neumann and Zagier30]). Associated with u, we can define the
$\mathrm {SL}(2;\mathbb {C})$
Chern–Simons invariant
$\operatorname {CS}_{u,v(u)}(\rho )$
and the cohomological adjoint Reidemeister torsion
$T_K(u)$
. See [Reference Murakami and Yokota29, Chapter 5] for example. Note that, in [Reference Murakami and Yokota29], we define the homological adjoint Reidemeister torsion (it is called the twisted Reidemeister torsion there). So we need to take its inverse to define the cohomological torsion. Note also that
$\operatorname {Vol}(S^3\setminus {K})+\sqrt {-1}\operatorname {CS}^{\mathrm {SO}(3)}(S^3\setminus {K})$
in Conjecture 1.2 coincides with
$\sqrt {-1}\operatorname {CS}_{0,0}(\rho _0)$
for a hyperbolic knot K with holonomy representation
$\rho _0$
.
In [Reference Murakami and Yokota28], Yokota and the author proved that for the figure-eight knot E, the limit
$\lim _{N\to \infty }\frac {1}{N}\log J_{N}\left (E;e^{(u+2\pi \sqrt {-1})/N}\right )$
exists if the complex number u is in a small neighborhood of
$0$
(and not a rational multiple of
$\pi \sqrt {-1}$
). Moreover, the limit determines the holomorphic function
$f(u)$
introduced in [Reference Neumann and Zagier30, Theorem 2]. In other words, the asymptotic behavior of
$J_{N}(E;e^{(u+2\pi \sqrt {-1})/N})$
determines the
$\mathrm {SL}(2;\mathbb {C})$
Chern–Simons invariant associated with u.
For a general hyperbolic knot K, the following conjecture was proposed in [Reference Murakami24] (see also [Reference Dimofte and Gukov2, Reference Gukov and Murakami9]).
Conjecture 1.3 Let
$K\subset S^3$
be a hyperbolic knot. Then there exists a neighborhood
$U\subset \mathbb {C}$
of
$0$
such that if
$u\in U\setminus \pi \sqrt {-1}\mathbb {Q}$
, then we have
$$ \begin{align*} &J_{N}\left(K;e^{(u+2\pi\sqrt{-1})/N}\right) \\ &\qquad \underset{N\to\infty}{\sim} \frac{\sqrt{-\pi}}{2\sinh(u/2)} T_K(u)^{1/2}\left(\frac{N}{u+2\pi\sqrt{-1}}\right)^{1/2} \exp\left(\frac{N\times S_K(u)}{u+2\pi\sqrt{-1}}\right), \end{align*} $$
where
$T_K(u)$
is the cohomological adjoint Reidemeister torsion, and
$\operatorname {CS}_{u,v(u)}(\rho )=S_K(u)-u\pi \sqrt {-1}-\frac {1}{4}uv(u)$
is the Chern–Simons invariant, both associated with u.
In [Reference Murakami24], we proved that the conjecture is true for the figure-eight knot and a positive real number
$u<\operatorname {arccosh}(3/2)$
.
In this paper, we study the colored Jones polynomial of the figure-eight knot evaluated at
$q=\exp \bigl ((u+2p\pi \sqrt {-1})/N\bigr )$
for a real number u with
$0<u<\operatorname {arccosh}(3/2)$
and a positive integer p. We will show the following theorem.
Theorem 1.4 Let E be the figure-eight knot and put
$\xi :=u+2p\pi \sqrt {-1}$
. Then we have
$$ \begin{align} &J_N\left(E;e^{\xi/N}\right) \nonumber \\ &\quad= \frac{\sqrt{-\pi}}{2\sinh(u/2)}T_E(u)^{1/2} J_{p}\left(E;e^{4N\pi^2/\xi}\right) \left(\frac{N}{\xi}\right)^{1/2} e^{\frac{N}{\xi}\times S_E(u)} \bigl(1+O(N^{-1})\bigr), \end{align} $$
as
$N\to \infty $
, where we put
$$ \begin{align*} S_E(u) &:= \operatorname{Li}_2\left(e^{-u-\varphi(u)}\right)-\operatorname{Li}_2\left(e^{-u+\varphi(u)}\right) + u\bigl(\varphi(u)+2\pi\sqrt{-1}\bigr), \\ T_E(u) &:= \frac{2}{\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)}}. \end{align*} $$
Here,
$\operatorname {Li}_2(z):=-\int _{0}^{z}\frac {\log (1-w)}{w}\,dw$
is the dilogarithm function and we put
$$ \begin{align*} \varphi(u) := \log \left( \cosh{u}-\frac{1}{2}-\frac{1}{2}\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)} \right). \end{align*} $$
Remark 1.5 The case where
$p=1$
was proved in [Reference Murakami24].
Remark 1.6 When
$p=0$
, the author proved that for the figure-eight knot E,
$J_N\left (E;e^{u/N}\right )$
converges to
$1/\Delta (E;e^{u})$
, where
$\Delta (K;t)$
is the Alexander polynomial of a knot K normalized so that
$\Delta (K;t)=\Delta (K;t^{-1})$
and
$\Delta (U;t)=1$
[Reference Murakami23]. Soon after, it was generalized by Garoufalidis and Le to any knot in
$S^3$
. See [Reference Garoufalidis and Le4–Reference Garoufalidis and Lê6].
As a corollary we have the following asymptotic equivalence.
Corollary 1.7 We have
$$ \begin{align} \frac{J_N\left(E;e^{\xi/N}\right)}{J_p\left(E;e^{4N\pi^2/\xi}\right)} \underset{N\to\infty}{\sim} \frac{\sqrt{-\pi}}{2\sinh(u/2)}T_E(u)^{1/2} \left(\frac{N}{\xi}\right)^{1/2} e^{\frac{N}{\xi}\times S_E(u)}. \end{align} $$
This indicates a quantum modularity for the colored Jones polynomial.
Conjecture (Conjecture 7.3)
Let K be a hyperbolic knot. For a small complex number u that is not a rational multiple of
$\pi \sqrt {-1}$
, and positive integers p and N, put
$\xi :=u+2p\pi \sqrt {-1}$
and
$X:=\frac {2N\pi \sqrt {-1}}{\xi }$
. Then, for any
$\eta =\begin {pmatrix}a&b\\c&d\end {pmatrix}\in \mathrm {SL}(2;\mathbb {Z})$
with
$c>0$
, the following asymptotic equivalence holds:
$$ \begin{align*} \frac{J_{cN+dp}\left(K;e^{2\pi\sqrt{-1}\eta(X)}\right)}{J_{p}\left(K;e^{2\pi\sqrt{-1} X}\right)} \underset{N\to\infty}{\sim} C_{K,\eta}(u)\frac{\sqrt{-\pi}}{2\sinh(u/2)} \left(\frac{T_K(u)}{\hbar_{\eta}(X)}\right)^{1/2} \exp\left(\frac{S_K(u)}{\hbar_{\eta}(X)}\right) \end{align*} $$
for
$C_{K,\eta }(u)\in \mathbb {C}$
that does not depend on p, where we put
$\eta (X):=\frac {aX+b}{cX+d}$
and
$\hbar _{\eta }(X):=\frac {2c\pi \sqrt {-1}}{cX+d}$
.
Compare it with Zagier’s quantum modularity conjecture for Kashaev’s invariant [Reference Zagier36].
Conjecture (Conjecture 7.1)
Let K,
$\eta $
, N, and p as above. If we put
$X_0:=\frac {N}{p}$
, the following holds:
$$ \begin{align*} \frac{J_{cN+dp}\left(K;e^{2\pi\sqrt{-1}\eta(X_0)}\right)}{J_{p}\left(K;e^{2\pi\sqrt{-1} X_0}\right)} \underset{N\to\infty}{\sim} C_{K,\eta} \left(\frac{2\pi}{\hbar_{\eta}(X_0)}\right)^{3/2} \exp\left(\frac{\sqrt{-1}\operatorname{CV}(K)}{\hbar_{\eta}(X_0)}\right), \end{align*} $$
where
$C_{K,\eta }$
is a complex number depending only on
$\eta $
and K.
The paper is organized as follows: In Section 2, we define the colored Jones polynomial and introduce a quantum dilogarithm. We express the colored Jones polynomial as a sum of the quantum dilogarithms assuming
$(p,N)=1$
in Section 3. In Section 4, we approximate it by using the dilogarithm function by using the fact that the quantum dilogarithm converges to the dilogarithm. We use the Poisson summation formula á la Ohtsuki [Reference Ohtsuki31] to replace the sum with an integral in Section 5. In Section 6, we prove the main theorem (Theorem 1.4). We discuss a quantum modularity of the colored Jones polynomial in Section 7. Section 8 is devoted to proofs of lemmas used in the other sections. In the Appendix, we calculate the colored Jones polynomial in the case where
$(p,N)\ne 1$
.
2 Preliminaries
Let
$J_N(K;q)$
be the N-dimensional colored Jones polynomial of
$K\subset S^3$
associated with the N-dimensional irreducible representation of the Lie algebra
$\mathfrak {sl}_2(\mathbb {C})$
, where N is a positive integer and q is a complex parameter [Reference Kirby and Melvin15, Reference Kirillov and Reshetikhin16, Reference Reshetikhin and Turaev34]. It is normalized so that
$J_N(U;q)=1$
for the unknot U. In particular,
$J_2(K;q)$
is (a version) of the original Jones polynomial [Reference Jones11]. More precisely,
$J_2(K;q)$
satisfies the following skein relation:
Habiro [Reference Habiro10, p. 36(1)] (see also [Reference Masbaum19, Theorem 5.1]) and Le [Reference Le17, Example 1.2.2, p. 129] gave a simple formula for the colored Jones polynomial of the figure-eight knot E as follows:
$$ \begin{align} J_{N}(E;q) &= \sum_{k=0}^{N-1}\prod_{l=1}^{k} \left(q^{(N+l)/2}-q^{-(N+l)/2}\right)\left(q^{(N-l)/2}-q^{-(N-l)/2}\right) \end{align} $$
$$ \begin{align} &= \sum_{k=0}^{N-1}q^{-kN} \prod_{l=1}^{k}\left(1-q^{N+l}\right)\left(1-q^{N-l}\right). \end{align} $$
For a real number u with
$0<u<\kappa :=\operatorname {arccosh}(3/2)=0.962424\dots $
, and a positive integer p, we put
$\xi :=u+2p\pi \sqrt {-1}$
. Then we have
$$ \begin{align} J_N\left(E;e^{\xi/N}\right) = \sum_{k=0}^{N-1}e^{-k\xi} \prod_{l=1}^{k}\left(1-e^{(N+l)\xi/N}\right)\left(1-e^{(N-l)\xi/N}\right). \end{align} $$
We want to replace the products in (2.3) with some values of a continuous function. To do that, we introduce a so-called quantum dilogarithm following [Reference Faddeev3].
Put
and orient it from left to right. We consider the integral
, where
$\gamma :=\frac {\xi }{2N\pi \sqrt {-1}}$
.
Lemma 2.1 The integral
converges if
$-p/(2N)<\operatorname {Re}{z}<1+p/(2N)$
.
A proof is given in Section 8. Note that the poles of the integrand is
$$ \begin{align*} \{x\in\mathbb{C}\mid x=k\pi\sqrt{-1}\;\;(k\in\mathbb{Z}) \} \cup \{x\in\mathbb{C}\mid x=l\pi\sqrt{-1}/\gamma\;\;(l\in\mathbb{Z})\} \end{align*} $$
and so
avoids the poles.
We define
We also consider three related integrals
(
$k=0,1,2$
), which converge if
$0<\operatorname {Re}{z}<1$
by similar reasons to Lemma 2.1.
Definition 2.2 We put
for z with
$0<\operatorname {Re}{z}<1$
.
Their derivatives are given as follows:
$$ \begin{align*} \frac{d\,\mathcal{L}_2(z)}{d\,z} &= -2\pi\sqrt{-1}\mathcal{L}_1(z), \\ \frac{d\,\mathcal{L}_1(z)}{d\,z} &= -\mathcal{L}_0(z). \end{align*} $$
We also have the following lemma.
Lemma 2.3 If
$0<\operatorname {Re}{z}<1$
, then we have
$$ \begin{align*} \mathcal{L}_0(z) &= \frac{-2\pi\sqrt{-1}}{1-e^{-2\pi\sqrt{-1} z}}, \\ \mathcal{L}_1(z) &= \log\left(1-e^{2\pi\sqrt{-1} z}\right), \\ \mathcal{L}_2(z) &= \operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right). \end{align*} $$
Here, we use the branch of
$\log {w}$
so that
$-\pi <\operatorname {Im}\log {w}\le \pi $
and
$\operatorname {Li}_2(w)$
has branch cut at
$(1,\infty )$
.
Proof As [Reference Murakami and Tran27, Lemma 2.5], we can prove the following equalities:
$$ \begin{align*} \mathcal{L}_0(z) &= \frac{-2\pi\sqrt{-1}}{1-e^{-2\pi\sqrt{-1} z}}, \\ \mathcal{L}_1(z) &= \begin{cases} \log\left(1-e^{2\pi\sqrt{-1} z}\right), &\text{if } \operatorname{Im}{z}\ge0, \\ \pi\sqrt{-1}(2z-1)+\log\left(1-e^{-2\pi\sqrt{-1} z}\right),&\text{if } \operatorname{Im}{z}<0, \end{cases} \\ \mathcal{L}_2(z) &= \begin{cases} \operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right), &\text{if }\operatorname{Im}{z}\ge0, \\ \frac{\pi^2}{3}(6z^2-6z+1)-\operatorname{Li}_2\left(e^{-2\pi\sqrt{-1} z}\right),&\text{if }\operatorname{Im}{z}<0. \end{cases} \end{align*} $$
So we need to prove the lemma for the case where
$\operatorname {Im}{z}<0$
.
There is nothing to prove for
$\mathcal {L}_0(z)$
.
If
$0<\operatorname {Re}{z}<1$
, then using the identity (see, for example, [Reference Maximon20])
$$ \begin{align} \operatorname{Li}_2(w^{-1}) = -\operatorname{Li}_2(w)-\frac{\pi^2}{6}-\frac{1}{2}\bigl(\log(-w)\bigr)^2, \end{align} $$
we have
$$ \begin{align*} \operatorname{Li}_2\left(e^{-2\pi\sqrt{-1} z}\right) &= -\operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right)-\frac{\pi^2}{6} -\frac{1}{2}\bigl(2\pi\sqrt{-1} z-\pi\sqrt{-1}\bigr)^2 \\ &= -\operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right) +2\pi^2z^2+\frac{\pi^2}{3}-2\pi^2 z, \notag \end{align*} $$
where we use the fact that
$0<\operatorname {Im}(2\pi \sqrt {-1} z)<2\pi $
. Therefore, we have
$$ \begin{align*} \mathcal{L}_2(z) = -\operatorname{Li}_2\left(e^{-2\pi\sqrt{-1} z}\right)+\frac{\pi^2}{3}\bigl(6z^2-6z+1\bigr) = \operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right), \end{align*} $$
as required.
As for
$\mathcal {L}_1(z)$
, since
$\log \left (e^{\pi \sqrt {-1}(2z-1)}\right )=2\pi \sqrt {-1} z-\pi \sqrt {-1}$
, we have
$$ \begin{align*} \log\left(1-e^{-2\pi\sqrt{-1} z}\right)+\pi\sqrt{-1}(2z-1) &= \log\left(1-e^{-2\pi\sqrt{-1} z}\right)+\log\left(e^{\pi\sqrt{-1}(2z-1)}\right) \\ &= \log\left(1-e^{2\pi\sqrt{-1} z}\right), \end{align*} $$
completing the proof.
We can prove that
$T_N(z)$
converges to
$\frac {N}{\xi }\operatorname {Li}_2\left (e^{2\pi \sqrt {-1} z}\right )$
. More precisely, we have the following.
Lemma 2.4 For any positive real number M and a sufficiently small positive real number
$\nu $
, we have
$$ \begin{align*} T_N(z) = \frac{N}{\xi}\operatorname{Li}_2\left(e^{2\pi\sqrt{-1} z}\right)+O(1/N), \end{align*} $$
as
$N\to \infty $
in the region
$$ \begin{align*} \{z\in\mathbb{C}\mid\nu\le\operatorname{Re}{z}\le1-\nu,|\operatorname{Im}{z}|\le M\}. \end{align*} $$
In particular,
$T_N(z)$
uniformly converges to
$\frac {N}{\xi }\operatorname {Li}_2\left (e^{2\pi \sqrt {-1} z}\right )$
in the region above.
A proof is also given in Section 8.
The following lemma is essential in the paper. Put
$E_N(z):=e^{T_N(z)}$
.
Lemma 2.5 If
$0<\operatorname {Re}{z}<1$
, then we have
$$ \begin{align*} \frac{E_N(z-\gamma/2)}{E_N(z+\gamma/2)} = 1-e^{2\pi\sqrt{-1} z}. \end{align*} $$
Proof Recalling that
$\gamma =\frac {\xi }{2N\pi \sqrt {-1}}$
, we have
Taking the exponentials of both sides, the lemma follows from Lemma 2.3.
As a corollary, we have the following.
Corollary 2.6 Let n be an integer. If
$nN/p<j<(n+1)N/p$
, we have
$$ \begin{align*} \frac{E_N\bigl((j-1/2)\gamma-n\bigr)}{E_N\bigl((j+1/2)\gamma-n\bigr)} &= 1-e^{2j\gamma\pi\sqrt{-1}} \end{align*} $$
and
$$ \begin{align*} \frac{E_N\bigl(n+1-(j+1/2)\gamma\bigr)}{E_N\bigl(n+1-(j-1/2)\gamma\bigr)} &= 1-e^{-2j\gamma\pi\sqrt{-1}}. \end{align*} $$
Proof Since
$\operatorname {Re}\gamma =p/N$
, we have
$0<\operatorname {Re}(j\gamma -n)<1$
. Therefore, putting
$z:=j\gamma -n$
in Lemma 2.5, we have the first equality. Similarly, putting
$z:=n+1-j\gamma $
, we have the second equality.
We prepare other two lemmas.
Lemma 2.7 For a complex number z with
$|\operatorname {Re}{z}|<\operatorname {Re}\gamma /2$
, we have
$$ \begin{align*} \frac{E_{N}(z)}{E_{N}(z+1)} = 1+e^{2\pi\sqrt{-1} z/\gamma}. \end{align*} $$
Lemma 2.8 For a complex number
$w\ne 0$
with
$|\operatorname {Re}{w}|<\operatorname {Re}\gamma $
, we have
$$ \begin{align*} \frac{E_N\bigl(w+\gamma/2\bigr)}{E_N\bigl(w-\gamma/2+1\bigr)} = \frac{1-e^{2\pi\sqrt{-1} w/\gamma}}{1-e^{2\pi\sqrt{-1} w}}. \end{align*} $$
For
$w=0$
, we have
$$ \begin{align*} \frac{E_N\bigl(\gamma/2\bigr)}{E_N\bigl(-\gamma/2+1\bigr)} = 1/\gamma. \end{align*} $$
Proof If
$\operatorname {Re}{w}>0$
, then from Lemmas 2.5 and 2.7, we have
$$ \begin{align*} \frac{E_{N}(w+\gamma/2)}{E_{N}(w-\gamma/2)} &= \frac{1}{1-e^{2\pi\sqrt{-1} w}}, \\ \frac{E_{N}(w-\gamma/2)}{E_{N}(w-\gamma/2+1)} &= 1-e^{2\pi\sqrt{-1} w/\gamma}, \end{align*} $$
and the lemma follows.
If
$\operatorname {Re}{w}<0$
, in a similar way, we have
$$ \begin{align*} \frac{E_{N}(w+1+\gamma/2)}{E_{N}(w+1-\gamma/2)} &= \frac{1}{1-e^{2\pi\sqrt{-1} w}}, \\ \frac{E_{N}(w+\gamma/2)}{E_{N}(w+\gamma/2+1)} &= 1-e^{2\pi\sqrt{-1} w/\gamma}, \end{align*} $$
and the lemma also follows.
If
$\operatorname {Re}{w}=0$
, then consider the limit
$$ \begin{align*} \lim_{\varepsilon\to0} \frac{E_{N}(w+\varepsilon+\gamma/2)}{E_{N}(w+\varepsilon-\gamma/2+1)}, \end{align*} $$
and the proof is complete.
3 Summation
In this section, we express
$J_N\left (E;e^{\xi /N}\right )$
in terms of the quantum dilogarithm
$T_N(z)$
.
We assume that p and N are coprime. See the Appendix for the case with
$(p,N)\ne 1$
.
If
$k<N/p$
, then from Corollary 2.6 with
$(j,n)=(N-l,p-1)$
and
$(j,n)=(N+l,p)$
, we have
$$ \begin{align*} &\prod_{l=1}^{k}(1-e^{(N-l)\xi/N})(1-e^{(N+l)\xi/N}) \\ &\quad = \prod_{l=1}^{k}(1-e^{2(N-l)\gamma\pi\sqrt{-1}})(1-e^{2(N+l)\gamma\pi\sqrt{-1}}) \\ &\quad = \prod_{l=1}^{k} \frac{E_N\bigl((N-l-1/2)\gamma-p+1\bigr)}{E_N\bigl((N-l+1/2)\gamma-p+1\bigr)} \\ &\qquad\times \prod_{l=1}^{k} \frac{E_N\bigl((N+l-1/2)\gamma-p\bigr)}{E_N\bigl((N+l+1/2)\gamma-p\bigr)} \\ &\quad = \frac{E_N\bigl((N-k-1/2)\gamma-p+1\bigr)}{E_N\bigl((N-1/2)\gamma-p+1\bigr)} \frac{E_N\bigl((N+1/2)\gamma-p\bigr)}{E_N\bigl((N+k+1/2)\gamma-p\bigr)} \\ &\quad= \frac{1-e^{4p\pi^2N/\xi}}{1-e^{\xi}} \times \frac{E_N\bigl((N-k-1/2)\gamma-p+1\bigr)}{E_N\bigl((N+k+1/2)\gamma-p\bigr)}, \end{align*} $$
where we use Lemma 2.8 with
$w=N\gamma -p$
in the last equality.
Similarly, if k satisfies
$mN/p<k<(m+1)N/p$
, then we have
$$ \begin{align} &\prod_{l=1}^{k}(1-e^{(N-l)\xi/N})(1+e^{(N+l)\xi/N}) \\ &\quad = \prod_{j=0}^{m-1} \left( \prod_{l=\lfloor jN/p\rfloor+1}^{\lfloor(j+1)N/p\rfloor} \frac{E_N\bigl((N-l-1/2)\gamma-p+j+1\bigr)}{E_N\bigl((N-l+1/2)\gamma-p+j+1\bigr)} \right. \notag \\ &\qquad \left. \times \prod_{l=\lfloor jN/p\rfloor+1}^{\lfloor(j+1)N/p\rfloor} \frac{E_N\bigl((N+l-1/2)\gamma-p-j\bigr)}{E_N\bigl((N+l+1/2)\gamma-p-j\bigr)} \right) \notag \\ &\qquad \times \prod_{l=\lfloor mN/p\rfloor+1}^{k} \frac{E_N\bigl((N-l-1/2)\gamma-p+m+1\bigr)}{E_N\bigl((N-l+1/2)\gamma-p+m+1\bigr)} \notag \\ &\qquad \times \prod_{l=\lfloor mN/p\rfloor+1}^{k} \frac{E_N\bigl((N+l-1/2)\gamma-p-m\bigr)}{E_N\bigl((N+l+1/2)\gamma-p-m\bigr)} \notag \\ &\quad = \prod_{j=0}^{m-1} \frac{E_N\bigl((N-\lfloor(j+1)N/p\rfloor-1/2)\gamma-p+j+1\bigr)} {E_N\bigl((N-\lfloor jN/p\rfloor-1/2)\gamma-p+j+1\bigr)} \notag \\ &\qquad\times \prod_{j=0}^{m-1} \frac{E_N\bigl((N+\lfloor jN/p\rfloor+1/2)\gamma-p-j\bigr)} {E_N\bigl((N+\lfloor(j+1)N/p\rfloor+1/2)\gamma-p-j\bigr)} \notag \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+m+1\bigr)}{E_N\bigl((N-\lfloor mN/p\rfloor-1/2)\gamma-p+m+1\bigr)} \notag \\ &\qquad\times \frac{E_N\bigl((N+\lfloor mN/p\rfloor+1/2)\gamma-p-m\bigr)}{E_N\bigl((N+k+1/2)\gamma-p-m\bigr)} \notag \\ &\quad = \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \left( \prod_{j=1}^{m} \left(1-e^{4(p-j)N\pi^2/\xi}\right)\left(1-e^{4(p+j)N\pi^2/\xi}\right) \right) \notag \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+m+1\bigr)}{E_N\bigl((N+k+1/2)\gamma-p-m\bigr)}, \notag \end{align} $$
where we use Lemma 2.8 with
$w=N\gamma -p$
, and Lemma 2.7 with
$z=(N-\lfloor lN/p\rfloor -1/2)\gamma -p+l$
and
$z=(N+\lfloor lN/p\rfloor +1/2)\gamma -p-l$
(
$l=1,2,\dots ,m$
).
Remark 3.1 Since
$\operatorname {Re}{\gamma }=p/N$
, we have
$\operatorname {Re}\bigl ((N-\lfloor lN/p\rfloor -1/2)\gamma -p+l\bigr )=-\frac {p}{N}\lfloor \frac {lN}{p}\rfloor -\frac {p}{2N}+l$
. Since
$lN/p$
is not an integer, we have
$lN/p-1<\lfloor lN/p\rfloor <lN/p$
(the equality
$\lfloor lN/p\rfloor =lN/p$
does not hold). So
$\left |\operatorname {Re}\bigl ((N-\lfloor lN/p\rfloor -1/2)\gamma -p+l\bigr )\right |<\operatorname {Re}{\gamma }/2$
, and the assumption of Lemma 2.7 holds.
Therefore, from (2.3), we have
$$ \begin{align} &J_N\left(E;e^{\xi/N}\right) \\ &\quad= \sum_{m=0}^{p-1} \sum_{mN/p<k<(m+1)N/p} e^{-k\xi} \prod_{l=1}^{k}\left(1-e^{(N+l)\xi/N}\right)\left(1-e^{(N-l)\xi/N}\right) \notag \\ &\quad= \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \sum_{m=0}^{p-1} \left( \prod_{j=1}^{m} \left(1-e^{4(p-j)N\pi^2/\xi}\right)\left(1-e^{4(p+j)N\pi^2/\xi}\right) \right. \notag \\ &\qquad\times \left.\vphantom{\sum_{m=0}^{p-1}} \sum_{mN/p<k<(m+1)N/p} e^{-k\xi} \frac{E_N\bigl((N-k-1/2)\gamma-p+m+1\bigr)}{E_N\bigl((N+k+1/2)\gamma-p-m\bigr)} \right) \notag \\ &\quad = \frac{1-e^{-4pN\pi^2/\xi}}{2\sinh(u/2)} \notag \\ &\qquad\times \sum_{m=0}^{p-1} \left( \beta_{p,m} \sum_{mN/p<k<(m+1)N/p} \exp \left( N\times f_{N}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) \right) \right), \notag \end{align} $$
where we put
$$ \begin{align} \beta_{p,m} &:= e^{-4mpN\pi^2/\xi} \prod_{j=1}^{m} \left(1-e^{4(p-j)N\pi^2/\xi}\right)\left(1-e^{4(p+j)N\pi^2/\xi}\right), \notag \\ f_{N}(z) &:= \frac{1}{N}T_N\left(\frac{\xi(1-z)}{2\pi\sqrt{-1}}-p+1\right) - \frac{1}{N}T_N\left(\frac{\xi(1+z)}{2\pi\sqrt{-1}}-p\right) \\ &\quad -uz+\frac{4p\pi^2}{\xi}. \notag \end{align} $$
Remark 3.2 Since we have
$$ \begin{align*} \operatorname{Re}\left(\frac{\xi(1\pm z)}{2\pi\sqrt{-1}}\right) = p(1\pm\operatorname{Re}{z})\pm\frac{u}{2\pi}\operatorname{Im}{z}, \end{align*} $$
the function
$f_{N}(z)$
is defined in the region
$$ \begin{align*} \left\{ z\in\mathbb{C}\Bigm|-\frac{1}{2N}<\frac{u}{2p\pi}\operatorname{Im}{z}+\operatorname{Re}{z}<\frac{1}{p}+\frac{1}{2N} \right\} \end{align*} $$
from Lemma 2.1.
4 Approximation
In the previous section, we express
$J_N\left (E;e^{\xi /N}\right )$
as a sum of the function
$f_N(z)$
. In this section, we approximate it by using a function that does not depend on N.
Since
$T_N(z)/N$
uniformly converges to
$\operatorname {Li}_2\left (e^{2\pi \sqrt {-1} z}\right )/\xi $
(Lemma 2.4),
$f_{N}(z)$
uniformly converges to
$$ \begin{align*} F(z) := \frac{1}{\xi}\operatorname{Li}_2\left(e^{\xi(1-z)}\right) - \frac{1}{\xi}\operatorname{Li}_2\left(e^{\xi(1+z)}\right) -uz+\frac{4p\pi^2}{\xi} \end{align*} $$
in the region
$$ \begin{align} \left\{ z\in\mathbb{C} \Bigm| \frac{\nu}{p}\le\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}\le\frac{1}{p}-\frac{\nu}{p}, \left|\operatorname{Re}{z}-\frac{2p\pi}{u}\operatorname{Im}{z}\right|\le\frac{2M\pi}{u}+1 \right\}. \end{align} $$
By using the identity (2.4), if z is in the region
$$ \begin{align*} U_0 := \left\{ z\in\mathbb{C} \Bigm| 0<\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}<\frac{1}{p} \right\}, \end{align*} $$
we have
$$ \begin{align*} \operatorname{Li}_2\left(e^{\xi(1-z)}\right) &= -\operatorname{Li}_2\left(e^{-\xi(1-z)}\right)-\frac{\pi^2}{6} -\frac{1}{2}\left(\log\left(-e^{-\xi(1-z)}\right)\right)^2 \\ &= -\operatorname{Li}_2\left(e^{-\xi(1-z)}\right)-\frac{\pi^2}{6} -\frac{1}{2}(-\xi(1-z)+(2p-1)\pi\sqrt{-1})^2 \end{align*} $$
since
$\operatorname {Im}\xi (1-z)=2p\pi -(uy+2p\pi x)$
. Similarly, we have
$$ \begin{align*} \operatorname{Li}_2\left(e^{\xi(1+z)}\right) &= -\operatorname{Li}_2\left(e^{-\xi(1+z)}\right)-\frac{\pi^2}{6} -\frac{1}{2}\left(\log\left(-e^{-\xi(1+z)}\right)\right)^2 \\ &= -\operatorname{Li}_2\left(e^{-\xi(1+z)}\right)-\frac{\pi^2}{6} -\frac{1}{2}(-\xi(1+z)+(2p+1)\pi\sqrt{-1})^2 \end{align*} $$
since
$\operatorname {Im}\xi (1+z)=2p\pi +(uy+2p\pi x)$
. Therefore,
$F(z)$
can also be written as
$$ \begin{align*} F(z) = \frac{1}{\xi}\operatorname{Li}_2\left(e^{-\xi(1+z)}\right) - \frac{1}{\xi}\operatorname{Li}_2\left(e^{-\xi(1-z)}\right) +uz-2\pi\sqrt{-1} \end{align*} $$
in
$U_0$
.
The first derivative of
$F(z)$
is
$$ \begin{align} \frac{d}{d\,z}F(z) = \log\left(1-e^{-u-\xi z}\right) + \log\left(1-e^{-u+\xi z}\right) +u = \log\left(e^u+e^{-u}-e^{\xi z}-e^{-\xi z}\right) \end{align} $$
because
$-\pi <\arg \left (1-e^{-u-\xi z)}\right )+\arg \left (1-e^{-u+\xi z)}\right )<\pi $
when u is real from the lemma below. Here, we choose the branch of
$\arg $
so that
$-\pi <\arg \zeta \le \pi $
for any
$\zeta \in \mathbb {C}$
. Note that
$e^{\pm \xi z}\in \mathbb {R}$
if and only if
$\operatorname {Im}(\xi z)=u\operatorname {Im}{z}+2p\pi \operatorname {Re}{z}=2k\pi $
for some
$k\in \mathbb {Z}$
, which implies that if
$z\in U_0$
then
$e^{\pm \xi z}\not \in \mathbb {R}$
.
Lemma 4.1 Let a be a positive real number, and let w be a complex number with
$w\not \in \mathbb {R}$
. Then we have
$-\pi <\arg (1-aw)+\arg (1-aw^{-1})<\pi $
.
Proof We may assume that
$\operatorname {Im}{w}>0$
without loss of generality. Then we can easily see that
$-\pi <\arg (1-aw)<0$
and that
$0<\arg (1-aw^{-1})<\pi $
, which implies the result.
The second derivative of
$F(z)$
equals
$$ \begin{align*} \frac{d^2}{d\,z^2}F(z) = \frac{\xi\left(e^{-\xi z}-e^{\xi z}\right)}{e^{u}+e^{-u}-e^{\xi z}-e^{-\xi z}}. \end{align*} $$
Now, define
$$ \begin{align} \varphi(u) := \log \left( \cosh{u}-\frac{1}{2}-\frac{1}{2}\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)} \right), \end{align} $$
where we take the square root as a positive multiple of
$\sqrt {-1}$
, recalling that
$\cosh {u}<3/2$
. Note that
$\varphi (u)$
satisfies the equality
$$ \begin{align*} e^{u}+e^{-u}-e^{\varphi(u)}-e^{-\varphi(u)}=1. \end{align*} $$
Lemma 4.2 If
$0<u<\kappa =\operatorname {arccosh}(3/2)$
, then
$\varphi (u)$
is purely imaginary with
$-\pi /3<\operatorname {Im}\varphi (u)<0$
.
Proof First, note that
$e^{\varphi (u)}$
is a solution to the following quadratic equation:
$$ \begin{align*} x^2-(2\cosh{u}-1)x+1=0. \end{align*} $$
Therefore,
$\left |e^{\varphi (u)}\right |=1$
, and we conclude that
$\varphi (u)$
is purely imaginary. Put
$\theta :=\operatorname {Im}\varphi (u)$
.
Since
$0<u<\kappa $
, we see that
$1<2\cosh {u}-1<2$
. Then, since
$e^{-\theta \sqrt {-1}}$
is the other solution to the quadratic equation above, we have
$2\cos \theta =2\cosh {u}-1$
. Therefore, we see that
$-\pi /3<\theta <0$
because the argument of
$\log $
in (4.3) is in the fourth quadrant.
As in the proof above, we put
$\theta :=\operatorname {Im}\varphi (u)$
. We also put
$\sigma _0:=\frac {(\theta +2\pi )\sqrt {-1}}{\xi }$
. Since we have
$$ \begin{align*} \operatorname{Re}\sigma_0+\frac{u}{2p\pi}\operatorname{Im}\sigma_0 = \frac{\theta+2\pi}{2p\pi} \end{align*} $$
and
$0>\theta >-\pi /3$
, we see that
$\sigma _0\in U_0$
.
We have
$$ \begin{align*} \frac{d}{d\,z}F(\sigma_0) = \log\left(e^u+e^{-u}-e^{\varphi(u)}-e^{-\varphi(u)}\right) =0. \end{align*} $$
We also have
$$ \begin{align*} \frac{d^2}{d\,z^2}F(\sigma_0) = \xi\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)}. \end{align*} $$
Therefore, we conclude that
$F(z)$
is of the form
$$ \begin{align} F(z) = F(\sigma_0) +a_2(z-\sigma_0)^2 +a_3(z-\sigma_0)^3 +a_4(z-\sigma_0)^4+\cdots \end{align} $$
with
$a_2:=\frac {1}{2}\xi \sqrt {(2\cosh {u}+1)(2\cosh {u}-3)}$
.
Now, the sum
$$ \begin{align} \sum_{m/p<k/N<(m+1)/p} \exp\left(N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \end{align} $$
can be approximated by the sum
$$ \begin{align*} \sum_{m/p<k/N<(m+1)/p} \exp\left(N\times\Phi_m\left(\frac{2k+1}{2N}\right)\right), \end{align*} $$
where we put
$$ \begin{align*} \Phi_m(z) := F\left(z-\frac{2m\pi\sqrt{-1}}{\xi}\right). \end{align*} $$
Moreover, in the next section, we approximate the sum (4.5) by the integral
$N\int _{m/p}^{(m+1)/p}e^{N\Phi _m(z)}\,dz$
.
Note that the function
$\Phi _m(z)$
is defined in the region
$$ \begin{align*} U_m := \left\{ z\in\mathbb{C} \Bigm| \frac{m}{p}<\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}<\frac{m+1}{p} \right\}. \end{align*} $$
Put
$\sigma _m:=\sigma _0+\frac {2m\pi \sqrt {-1}}{\xi }$
. Then we see that
$$ \begin{align*} \operatorname{Re}{\sigma_m}+\frac{u}{2p\pi}\operatorname{Im}{\sigma_m} = \operatorname{Re}{\sigma_0}+\frac{u}{2p\pi}\operatorname{Im}{\sigma_0}+\frac{m}{p} = \frac{\theta+2(m+1)\pi}{2p\pi}, \end{align*} $$
and so we have
$\sigma _m\in U_m$
. From (4.4), we conclude that
$\Phi _m(z)$
is of the form
$$ \begin{align} \Phi_m(z) = F(\sigma_0) +a_2(z-\sigma_m)^2 +a_3(z-\sigma_m)^3 +a_4(z-\sigma_m)^4+\cdots. \end{align} $$
5 The Poisson summation formula
First of all, note that the function
$f_N\left (z-\frac {2m\pi \sqrt {-1}}{\xi }\right )$
uniformly converges to
$\Phi _m(z)$
in the region
$$ \begin{align} \left\{ z\in\mathbb{C} \Bigm| \frac{m}{p}+\frac{\nu}{p}\le\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}\le\frac{m+1}{p}-\frac{\nu}{p}, \left|\operatorname{Re}{z}-\frac{2p\pi}{u}\operatorname{Im}{z}\right|\le\frac{2M\pi}{u}+1 \right\} \end{align} $$
from (4.1). So we expect that the sum (4.5) is approximated by the integral
$N\int _{m/p}^{(m+1)p}e^{N\Phi _m(z)}\,dz$
by using the Poisson summation formula [Reference Ohtsuki31, Proposition 4.2]. To do that, we will show the following proposition, which confirms the assumption of [Reference Ohtsuki31, Proposition 4.2].
Proposition 5.1 Let m be an integer with
$0\le m\le p-1$
. Put
$b_m^{-}:=m/p+\nu /p$
and
$b_m^{+}:=(m+1)/p-\nu /p$
.
Define
$$ \begin{align*} B_m &:= \left\{\frac{k}{N}\in\mathbb{R}\Bigm|k\in\mathbb{Z},b_m^{-}\le\frac{k}{N}\le b_m^{+}\right\}, \\ C_m &:= \{t\in\mathbb{R}\mid b_m^{-}\le t\le b_m^{+}\}, \\ D_m &:= \{z\in\mathbb{C}\mid\operatorname{Re}{\Phi_m(z)}<\operatorname{Re}{\Phi_m(\sigma_m)}\}, \\ E_m &:= \{z\in\mathbb{C}\mid b_m^{-}\le\operatorname{Re}{z}\le b_m^{+},|\operatorname{Im}{z}|\le2\operatorname{Im}\sigma_m\}\cap U_m. \end{align*} $$
Then the following hold.
-
1. The region
$E_m$
contains
$\sigma _m$
and
$\Phi _m(z)$
is a holomorphic function in
$E_m$
of the form
$$ \begin{align*} F(\sigma_0)+a_2(z-\sigma_m)^2+a_3(z-\sigma_m)^3+a_4(z-\sigma_m)^4+\cdots \end{align*} $$
with
$\operatorname {Re}{a_2}<0$
. -
2.
$D_m\cap E_m$
has two connected components. -
3.
$b_m^{+}$
and
$b_m^{-}$
are in different components of
$D_m\cap E_m$
and, moreover,
$\operatorname {Re}{\Phi _m(b_m^{\pm })}<\operatorname {Re}{\Phi _m(\sigma _m)}-\varepsilon _{m}$
for some
$\varepsilon _{m}>0$
. -
4. Both
$b_m^{+}$
and
$b_m^{-}$
are in a connected component of
$$ \begin{align*} &\overline{R}_m := \{x+y\sqrt{-1}\in\mathbb{C}\mid b_m^{-}\le x\le b_m^{+}, \\ &\qquad\qquad y\in[0,2\operatorname{Im}\sigma_m],\operatorname{Re}{\Phi_m(x+y\sqrt{-1})}<\operatorname{Re}{\Phi_m(\sigma_m)}+2\pi y\} \cap U_m. \end{align*} $$
-
5. Both
$b_m^{+}$
and
$b_m^{-}$
are in a connected component of
$$ \begin{align*} &\underline{R}_m := \{x-y\sqrt{-1}\in\mathbb{C}\mid b_m^{-}\le x\le b_m^{+}, \\ &\qquad\qquad y\in[0,2\operatorname{Im}\sigma_m],\operatorname{Re}{\Phi_m(x-y\sqrt{-1})}<\operatorname{Re}{\Phi_m(\sigma_m)}+2\pi y\} \cap U_m. \end{align*} $$
See Figure 1 for a contour plot of
$\operatorname {Re}{\Phi _m(z)}$
with
$p=3$
,
$m=2$
, and
$u=0.5$
.
Figure 1: A contour plot of
$\operatorname {Re}{\Phi _m(z)} $
in
$E_m$
by Mathematica for
$p=3$
,
$m=2$
, and
$u=0.5$
. The region
$\overline {R}_m$
(resp.
${\underline {R}}_m$
) is indicated by yellow (resp. green). The region
$D_m$
is indicated by red, which overwrites a part of
${\overline {R}}_m\cup {\underline {R}}_m$
.
Before we give a proof, let us define several lines as indicated in Figure 2.
Figure 2: The region
$U_m$
is between
$L_E$
and
$L_W$
.
$$ \begin{align*} L_{\sigma}&:\operatorname{Re}{z}-\frac{2p\pi}{u}\operatorname{Im}{z}=0, \\ L_E&:\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}=\frac{m+1}{p}, \\ L_M&:\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}=\frac{2m+1}{2p}, \\ L_W&:\operatorname{Re}{z}+\frac{u}{2p\pi}\operatorname{Im}{z}=\frac{m}{p}, \\ \overline{H}&:\operatorname{Im}{z}=2\operatorname{Im}\sigma_m, \\ \underline{H}&:\operatorname{Im}{z}=-2\operatorname{Im}\sigma_m, \\ V_E&:\operatorname{Re}{z}=\frac{m+1}{p}, \\ V_W&:\operatorname{Re}{z}=\frac{m}{p}. \end{align*} $$
Note that
$E_m$
is the hexagonal region surrounded by
$\overline {H}$
,
$L_E$
,
$V_E$
,
$\underline {H}$
,
$L_W$
, and
$V_W$
. Strictly speaking, we need to push
$L_E$
and
$L_W$
slightly inside. We name the vertices of its boundary as indicated in Figure 3. Their coordinates are given as follows:
$$ \begin{align*} P_{0}&: \frac{m}{p}, \\ P_{1}&: \frac{m}{p}+\frac{\overline{\xi}}{p\pi}\operatorname{Im}\sigma_m, \\ P_{2}&: \frac{m+1}{p}-2\operatorname{Im}\sigma_m\sqrt{-1}, \\ P_{3}&: \frac{m+1}{p}, \\ P_{4}&: \frac{m+1}{p}-\frac{\overline{\xi}}{p\pi}\operatorname{Im}\sigma_m, \\ P_{5}&: \frac{m}{p}+2\operatorname{Im}\sigma_m\sqrt{-1}, \end{align*} $$
where
$\overline {\xi }$
is the complex conjugate of
$\xi $
.
Figure 3: The region
$E_m$
.
We also put
$P_{12}:=L_M\cap \underline {H}$
,
$P_{34}:=L_E\cap L_{\sigma }$
,
$P_{45}:=L_M\cap \overline {H}$
, and
$P_{50}:=L_W\cap L_{\sigma }$
. Their coordinates are given as follows:
$$ \begin{align*} P_{12}&: \frac{2m+1}{2p}+\frac{\overline{\xi}}{p\pi}\operatorname{Im}\sigma_m, \\ P_{34}&: \frac{2(m+1)\pi\sqrt{-1}}{\xi}, \\ P_{45}&: \frac{2m+1}{2p}-\frac{\overline{\xi}}{p\pi}\operatorname{Im}\sigma_m, \\ P_{50}&: \frac{m\overline{\xi}\sqrt{-1}}{2p^2\pi}. \end{align*} $$
We use the following lemmas in the proof of Proposition 5.1.
Lemma 5.2 We have the inequalities
$0<\operatorname {Re}{F(0)}<\operatorname {Re}{F(\sigma _0)}$
.
Lemma 5.3 We have the inequality
$\operatorname {Re}{\Phi _m\left (P_{12}\right )}<\operatorname {Re}{\Phi _m(\sigma _m)}$
.
Proofs of the lemmas are given in Section 8.
Proof of Proposition 5.1
In the following proof, we assume that
$\nu $
is sufficiently small. We may need to modify the argument below slightly if necessary.
(1) We know that
$\Phi _m(z)$
is of the form (4.6). Since
$a_2=\frac {1}{2}\xi \sqrt {-1}\sqrt {(2\cosh {u}+1)(3-2\cosh {u})}$
and
$0<u<\operatorname {arccosh}(3/2)$
, we see that
$\operatorname {Re}{a_2}=-p\pi \sqrt {(2\cosh {u}+1)(3-2\cosh {u})}<0$
. So we conclude that
$\Phi _m(z)$
is of this form.
(2) Writing
$z=x+y\sqrt {-1}$
, we have
$$ \begin{align*} \frac{\partial}{\partial\,y}\operatorname{Re}{\Phi_m(x+y\sqrt{-1})} = -\arg\tau(x,y) \end{align*} $$
from (4.2), where we put
$\tau (x,y):=2\cosh (u)-2\cosh \bigl (\xi (x+y\sqrt {-1})\bigr )$
. Since we have
$$ \begin{align*} \operatorname{Im}\tau(x,y) = -2\sinh(ux-2p\pi y)\sin(uy+2p\pi x), \end{align*} $$
we see that
$\operatorname {Im}\tau (x,y)>0$
(resp.
$\operatorname {Im}\tau (x,y)<0$
) if and only if
$ux<2p\pi y$
and
$2k\pi <uy+2p\pi x<(2k+1)\pi $
for some integer k, or
$ux>2p\pi y$
and
$(2l-1)\pi <uy+2p\pi x<2l\pi $
for some integer l (resp.
$ux>2p\pi y$
and
$2k\pi <uy+2p\pi x<(2k+1)\pi $
for some integer k, or
$ux<2p\pi y$
and
$(2l-1)\pi <uy+2p\pi x<2l\pi $
for some integer l). Since
$z\in U_m$
, we have
$2m\pi <uy+2p\pi x<2(m+1)\pi $
. So we have
$$ \begin{align*} &\frac{\partial}{\partial\,y}\operatorname{Re}{\Phi_m(x+y\sqrt{-1})}>0 \quad\text{if and only if} \\ &\quad\phantom{\text{or }}ux>2p\pi y\text{ and }2m\pi<uy+2p\pi x<(2m+1)\pi \\ &\quad\text{or }ux<2p\pi y\text{ and }(2m+1)\pi<uy+2p\pi x<2(m+1)\pi, \end{align*} $$
and
$$ \begin{align*} &\frac{\partial}{\partial\,y}\operatorname{Re}{\Phi_m(x+y\sqrt{-1})}<0 \quad\text{if and only if} \\ &\quad\phantom{\text{or }}ux<2p\pi y\text{ and }2m\pi<uy+2p\pi x<(2m+1)\pi \\ &\quad\text{or }ux>2p\pi y\text{ and }(2m+1)\pi<uy+2p\pi x<2(m+1)\pi. \end{align*} $$
Therefore, fixing x,
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
is monotonically increasing (resp. decreasing) with respect to y in the red region (resp. yellow region) in Figure 3.
Next, we will show that (i) the segment
$\overline {P_{50}P_{34}}\subset L_{\sigma }$
except
$\sigma _m$
, (ii) the segment
$\overline {P_3P_{34}}\subset L_E$
, and (iii) the segment
$\overline {P_{12}P_{45}}\subset L_M$
are in
$D_m$
. See Figure 4.
Figure 4: The red segments are in
$D_m$
.
(i): Consider the segment of
$L_{\sigma }$
between
$L_W$
and
$L_E$
that is parametrized as
$\ell _{\sigma }(t):=t\sigma _m$
(
$\frac {2m\pi }{2(m+1)\pi +\theta }\le t\le \frac {2(m+1)\pi }{2(m+1)\pi +\theta }$
). Then we have
$$ \begin{align*} \frac{d}{d\,t}\operatorname{Re}\Phi_m\left(\ell_{\sigma}(t)\right) &= \operatorname{Re}\left(\sigma_m\log\bigl(2\cosh(u)-2\cosh(t\sigma_m\xi\bigr)\right) \\ &= (\operatorname{Re}\sigma_m) \log\left(2\cosh(u)-2\cos\left(\bigl(\theta+2(m+1)\pi\bigr)t\right)\right). \end{align*} $$
Since
$2m\pi \le \bigl (2(m+1)\pi +\theta \bigr )t\le 2(m+1)\pi $
and
$\cosh {u}-1/2=\cosh \varphi (u)=\cos \theta $
, we see that
$\frac {d}{d\,t}\operatorname {Re}\Phi _m\left (\ell _{\sigma }(t)\right )>0$
if and only if
$\frac {2m\pi -\theta }{2(m+1)\pi +\theta }<t<1$
, and that
$\frac {d}{d\,t}\operatorname {Re}\Phi _m\left (\ell _{\sigma }(t)\right )<0$
if and only if
$\frac {2m\pi }{2(m+1)\pi +\theta }<t<\frac {2m\pi -\theta }{2(m+1)\pi +\theta }$
or
$1<t<\frac {2(m+1)\pi }{2(m+1)\pi +\theta }$
.
Let
$P_W$
be the point
$L_{\sigma }\cap L_W$
with coordinate
$\frac {2m\pi \sqrt {-1}}{\xi }$
. Since
$\Phi _m(P_W)=F(0)$
and
$\Phi _m(\sigma _m)=F(\sigma _0)$
, Lemma 5.2 implies that
$\operatorname {Re}\Phi _m\left (\ell _{\sigma }(t)\right )$
takes its maximum
$\operatorname {Re}\Phi _m(\sigma _m)$
at
$t=1$
. This shows that
$L_{\sigma }\cap E_m$
is in
$D_m$
except for
$\sigma _m$
.
(ii): Consider the segment
$\overline {P_3P_4}$
that is parametrized as
$\ell _E(t):=\frac {m+1}{p}-\frac {u}{2p\pi }t+t\sqrt {-1}=\frac {m+1}{p}-\frac {\overline {\xi }}{2p\pi }t$
(
$0\le t\le 2\operatorname {Im}\sigma _m$
). We have
$$ \begin{align*} &\frac{d}{d\,t}\operatorname{Re}{\Phi_m\left(\ell_E(t)\right)} \\ =& -\operatorname{Re} \left( \frac{\overline{\xi}}{2p\pi} \log\left(2\cosh{u}-2\cosh\left(\xi\ell_E(t)\right)\right) \right) \\ =& - \frac{u}{2p\pi} \log\left(2\cosh{u}-2\cosh\left(\frac{(m+1)u}{p}-\frac{|\xi|^2t}{2p\pi}\right)\right)>0, \end{align*} $$
because
$$ \begin{align*} &\left| \frac{(m+1)u}{p}-\frac{|\xi|^2t}{2p\pi} \right| \\ \le& \max \left\{ \frac{(m+1)u}{p},\left|\frac{(m+1)u}{p}-\frac{u\bigl(\theta+2(m+1)\pi\bigr)}{p\pi}\right| \right\} \\ =& \max \left\{ \frac{(m+1)u}{p},\frac{(m+1)u}{p}+\frac{u\theta}{p\pi} \right\} = \frac{(m+1)u}{p} \le u. \end{align*} $$
Since the point
$P_{34}$
is in
$D_m$
, we conclude that
$\overline {P_3P_{34}}\subset D_m$
.
(iii): The line
$L_M$
between
$\underline {H}$
and
$\overline {H}$
is parametrized as
$\ell _M(t):=\frac {2m+1}{2p}-\frac {u}{2p\pi }t+t\sqrt {-1}=\frac {2m+1}{2p}-\frac {\overline {\xi }}{2p\pi }t$
(
$-2\operatorname {Im}\sigma _m\le t\le 2\operatorname {Im}\sigma _m$
). Now, we have
$$ \begin{align*} &\frac{d}{d\,t}\operatorname{Re}{\Phi_m\left(\ell_M(t)\right)} \\ =& -\operatorname{Re} \left( \frac{\overline{\xi}}{2p\pi} \log \left( 2\cosh(u)-2\cosh\left(\frac{(2m+1)}{2p}\xi-\frac{|\xi|^2}{2p\pi}t\right) \right) \right) \\ =& - \frac{u}{2p\pi} \log \left( 2\cosh(u)+2\cosh\left(\frac{(2m+1)u}{2p}-\frac{|\xi|^2}{2p\pi}t\right) \right) <0. \end{align*} $$
Since
$\ell _M(-2\operatorname {Im}\sigma _m)=P_{12}$
, from Lemma 5.3, we see that
$\operatorname {Re}{\Phi _m(P_{12})}<\operatorname {Re}{\Phi _m(\sigma _m)}$
. Therefore, every point z on
$\overline {P_{12}P_{45}}$
satisfies
$\operatorname {Re}{\Phi _m(z)}<\operatorname {Re}{\Phi _m(\sigma _m)}$
.
Now, we split
$E_m$
into five pieces:
$$ \begin{align*} E_{m,1} &:= \{z\in E_m\mid b_m^{-}\le\operatorname{Re}{z}\le\operatorname{Re}{P_{45}}\}, \\ E_{m,2} &:= \{z\in E_m\mid\operatorname{Re}{P_{45}\le\operatorname{Re}{z}}\le\operatorname{Re}{Q}\}, \\ E_{m,3} &:= \{z\in E_m\mid\operatorname{Re}{Q}\le\operatorname{Re}{z}\le\operatorname{Re}{P_{12}}\}, \\ E_{m,4} &:= \{z\in E_m\mid\operatorname{Re}{P_{12}}\le\operatorname{Re}{z}\le\operatorname{Re}{\sigma_m}\}, \\ E_{m,5} &:= \{z\in E_m\mid\operatorname{Re}{\sigma_m}\le\operatorname{Re}{z}\le\operatorname{Re}{P_{34}}\}, \\ E_{m,6} &:= \{z\in E_m\mid\operatorname{Re}{P_{34}}\le\operatorname{Re}{z}\le b_m^{+}\}, \end{align*} $$
where Q is the intersection of
$L_M$
and
$L_{\sigma }$
. See Figure 5.
Figure 5: The red region is
$D_m$
.
Note the following:
-
•
$\operatorname {Re}{P_1}<\operatorname {Re}{P_{45}}$
: This is because
$\operatorname {Re}{P_1}-\operatorname {Re}{P_{45}}=-\frac {1}{2p}+2\frac {u}{p\pi }\operatorname {Im}\sigma _m$
, which can be proved to be negative. -
•
$\operatorname {Re}{P_{12}}<\operatorname {Re}{P_4}$
: This is because
$\operatorname {Re}{P_{12}}-\operatorname {Re}{P_4}=-\frac {1}{2p}+2\frac {u}{p\pi }\operatorname {Im}\sigma _m<0$
as above. -
•
$\operatorname {Re}{P_{12}}<\operatorname {Re}{\sigma _m}$
: This is because
$\operatorname {Re}{P_{12}}-\operatorname {Re}\sigma _m=\frac {2m+1}{2p}+\frac {u}{p\pi }\operatorname {Im}\sigma _m-\operatorname {Re}\sigma _m<0$
. -
•
$\operatorname {Re}{\sigma _m}$
can be greater than, less than, or equal to
$\operatorname {Re}{P_4}$
.
In the following, we will show that any point in
$\left (E_{m,1}\cup E_{m,2}\cup E_{m,3}\cup E_{m,4}\right )\cap D_m$
can be connected to a point on
$L_{\sigma }$
by a segment contained in
$D_m$
and that any point in
$\left (E_{m,5}\cup E_{m,6}\right )\cap D_m$
can also be connected to a point on
$L_{\sigma }$
by a segment contained in
$D_m$
. We will also show that the vertical line through
$\sigma _m$
does not intersect with
$D_m$
. Then, we conclude that
$D_m\cap E_m$
has two connected components
$\left (E_{m,1}\cup E_{m,2}\cup E_{m,3}\cup E_{m,4}\right )\cap D_m$
and
$\left (E_{m,5}\cup E_{m,6}\right )\cap D_m$
because
$L_{\sigma }\setminus \{\sigma _m\}$
has two connected components.
Figure 6: The vertical axis is
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
, and the horizontal axis is y with fixed x. The red part is included in
$D_m$
. Note that the local maximum is less than
$\operatorname {Re}{\Phi _m(\sigma _m)}$
.
-
•
$E_{m,1}$
: Since
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}<\operatorname {Re}{\Phi _m(\sigma _m)}$
when
$x+y\sqrt {-1}$
is on
$L_{\sigma }$
and
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
decreases whether y increases or decreases fixing
$x\in \left [b_m^{-},\operatorname {Re}{P_{45}}\right ]$
, we conclude that
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}<\operatorname {Re}{\Phi _m(\sigma )}$
for any
$x+y\sqrt {-1}\in E_{m,1}$
. So we can connect any point in
$E_{m,1}$
to a point on
$L_{\sigma }$
. -
•
$E_{m,2}$
: Figure 6 indicates a graph of
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
for
$x+y\sqrt {-1}\in E_{m,2}$
with fixed x. This figure shows that any point in
$E_{m,2}\cap D_m$
can be connected to a point on
$L_{\sigma }$
by a vertical segment in
$D_m$
.Figure 7: The vertical axis is
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
, and the horizontal axis is y for fixed x. The red part is included in
$D_m$
. -
•
$E_{m,3}$
: A graph of
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
for
$x+y\sqrt {-1}\in E_{m,3}$
with fixed x looks like Figure 7 because
$\overline {P_{12}P_{45}}\subset D_m$
. Therefore, the argument as before shows that any point in
$E_{m,3}\cap D_m$
can be connected to a point on
$L_{\sigma }$
by a vertical segment in
$D_m$
. -
•
$E_{m,4}$
: Starting at a point on
$L_{\sigma }$
, whether y increases or decreases,
$\operatorname {Re}\Phi _m(x+y\sqrt {-1})$
increases. Therefore, any point in
$E_{m,4}\cap D_m$
can be connected to a point on
$L_{\sigma }$
by a vertical segment in
$D_m$
. -
•
$E_{m,5}$
: This follows by the same reason as
$E_{m,4}\cap D_m$
. -
•
$E_{m,6}$
: By the same argument as
$E_{m,4}$
, we can connect any point z in
$E_{m,6}\cap D_m$
to a point
$z'$
in
$\overline {P_3P_{34}}$
by a vertical segment in
$D_m$
, and then connect
$z'$
to a point in
$L_{\sigma }$
by a segment in
$\overline {P_3P_{34}}$
. (Precisely speaking, we need to push these segments in
$E_{m,6}$
.)
The fact that the vertical segment through
$\sigma _m$
does not intersect with
$D_m$
easily follows because
$\sigma _m\not \in D_m$
, and
$\frac {\partial }{\partial \,y}\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}$
is increasing (resp. decreasing) if
$x+y\sqrt {-1}$
is above
$\sigma _m$
(resp. below
$\sigma _m$
).
See Figure 5.
(3) From the definition, we know that
$b_m^{-}\in E_{m,1}$
and
$b_m^{+}\in E_{m,6}$
. Therefore, we can choose
$\varepsilon _{m}$
such that
$\operatorname {Re}{\Phi _m(b_m^{\pm })}<\operatorname {Re}{\Phi _m(\sigma _m)}-\varepsilon _{m}$
.
(4) Since any point z (
$z\ne \sigma _m$
) on the polygonal chain
$\overline {P_0P_{50}P_{34}}P_3$
satisfies
$\operatorname {Re}{\Phi _m(z)}<\operatorname {Re}{\Phi _m(\sigma _m)}$
, and
$\operatorname {Im}\sigma _m>0$
, we conclude that this is in
$\overline {R}_m$
. Therefore, we can connect
$b_m^{-}$
and
$b_m^{+}$
in
$\overline {R}_m$
.
(5) We know that if z is on the polygonal chain
$\overline {P_0P_1P_{12}}$
, then
$\operatorname {Re}{\Phi _m(z)}<\operatorname {Re}{\Phi _m(\sigma _m)}$
, which shows that
$\overline {P_0P_1P_{12}}$
is in
$\underline {R}_m$
.
We will show that the segment
$\overline {P_{12}P_2}$
is also in
$\underline {R}_m$
. From the proof of (2), we have
$0>\frac {\partial }{\partial \,y}\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}>-\pi $
if
$x+y\sqrt {-1}\in \overline {P_{12}P_2}$
. We know that if
$x+y\sqrt {-1}$
is on the polygonal chain
$\overline {QP_{34}P_3}$
, then
$\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}\le \operatorname {Re}{\Phi _m(\sigma _m)}$
. Since the difference of the imaginary part of
$x-2\operatorname {Im}\sigma _m\sqrt {-1}$
and
$x+y\sqrt {-1}$
is less than
$4\operatorname {Im}\sigma _m$
if
$x+y\sqrt {-1}$
is on the polygonal chain
$\overline {QP_{34}P_3}$
, we have
$\operatorname {Re}{\Phi _m(x-2\operatorname {Im}\sigma _m\sqrt {-1})}-\operatorname {Re}{\Phi _m(x+y\sqrt {-1})}<4\pi \operatorname {Im}\sigma _m$
. Therefore,
$\operatorname {Re}{\Phi _m(x-2\operatorname {Im}\sigma _m\sqrt {-1})}-\operatorname {Re}{\Phi _m(\sigma _m)}<2\pi \times 2\operatorname {Im}\sigma _m$
, proving that
$z\in \underline {R}_m$
if z is on
$\overline {P_{12}P_2}$
.
The segment
$\overline {P_2P_3}$
is also in
$\underline {R}_m$
. This is because
$\frac {\partial }{\partial \,y}\bigl (\operatorname {Re}{\Phi _m\bigl ((m+1)/p+y\sqrt {-1}\bigr )}+2\pi y\bigr )=\frac {\partial }{\partial \,y}\operatorname {Re}{\Phi _m\bigl ((m+1)/p+y\sqrt {-1}\bigr )}+2\pi>0$
and
$P_3\in \underline {R}_m$
.
Now, we can connect
$b_m^{-}$
and
$b_m^{+}$
by the polygonal chain
$\overline {P_0P_1P_2P_3}$
.
The proof is complete.
6 Proof of Theorem 1.4
Now, we can prove Theorem 1.4
Proof of Theorem 1.4
Since
$f_N(z)$
uniformly converges to
$F(z)$
in the region (4.1),
$f_N\left (z-\frac {2m\pi \sqrt {-1}}{\xi }\right )$
uniformly converges to
$\Phi _m(z)$
in (5.1). So we can use [Reference Ohtsuki31, Proposition 4.2] (see also Remark 4.4 there) to conclude that
$$ \begin{align} &\frac{1}{N}\sum_{m/p+\nu/p\le k/N\le(m+1)/p-\nu/p} \exp\left(N\times f_{N}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \\ &\quad = \int_{m/p+\nu/p}^{(m+1)/p-\nu/p} e^{N\Phi_m(z)} \,dz + O(e^{-N\varepsilon^{\prime}_{m}}) \notag \end{align} $$
for some
$\varepsilon ^{\prime }_{m}>0$
from Proposition 5.1.
Since
$\Phi _m(z)$
is of the form (4.6) in
$E_m$
, we can apply the saddle point method (see [Reference Ohtsuki31, Proposition 3.2 and Remark 3.3]) to obtain
$$ \begin{align} \int_{m/p+\nu/p}^{(m+1)/p-\nu/p}e^{N\Phi_m(z)}\,dz = \frac{\sqrt{\pi}e^{N\times F(\sigma_0)}} {\sqrt{-\frac{1}{2}\xi\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)}}\sqrt{N}} \bigl(1+O(N^{-1})\bigr), \end{align} $$
where we choose the sign of the outer square root so that its real part is positive (recall that we choose the sign of the inner square root so that it is a positive multiple of
$\sqrt {-1}$
). From (6.1) and (6.2), we have
$$ \begin{align} &\sum_{m/p+\nu/p\le k/N\le(m+1)/p-\nu/p} \exp\left(N\times f_{N}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \\ =& \frac{\sqrt{2\pi}e^{\pi\sqrt{-1}/4}}{\bigl((1+2\cosh{u})(3-2\cosh{u})\bigr)^{1/4}} e^{N\times F(\sigma_0)} \times \sqrt{\frac{N}{\xi}} \bigl(1+O(N^{-1})\bigr), \notag \end{align} $$
since
$\operatorname {Re}{F(\sigma _0)}>0$
from Lemma 5.2.
Now, we use the following lemma, a proof of which is given in Section 8.
Lemma 6.1 There exists
$\varepsilon>0$
such that
$\operatorname {Re}{\Phi _m\left (\frac {m+1}{p}\right )}<\operatorname {Re}{\Phi _m(\sigma _m)}-2\varepsilon $
for
$m=0,1,2,\dots ,p-1$
. Moreover, there exists
$\tilde {\delta }_m>0$
such that if
$\frac {m}{p}\le \frac {k}{N}<\frac {m}{p}+\tilde {\delta }_m$
or
$\frac {m+1}{p}-\tilde {\delta }_m<\frac {k}{N}<\frac {m+1}{p}$
, then we have
$$ \begin{align} \operatorname{Re}{f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)} < \operatorname{Re}{F(\sigma_0)}-\varepsilon \end{align} $$
for sufficiently large N.
If we choose
$\nu $
so that
$\nu /p\le \tilde {\delta }_m$
, the sums
$$ \begin{align*} \sum_{m/p\le k/N<m/p+\nu/p} \exp\left(N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \end{align*} $$
and
$$ \begin{align*} \sum_{(m+1)/p-\nu/p<k/N\le(m+1)/p} \exp\left(N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \end{align*} $$
are both of order
$O\left (Ne^{N(\operatorname {Re}{F(\sigma _0)}-\varepsilon )}\right )$
from Lemma 6.1. Therefore, we have
$$ \begin{align*} &\sum_{m/p\le k/N\le(m+1)/p} \exp \left( N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) \right) \\ &\quad = \sum_{m/p+\nu/p\le k/N\le(m+1)/p-\nu/p} \exp \left( N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) \right) \\ &\qquad + O\left(Ne^{N(\operatorname{Re}{F(\sigma_0)}-\varepsilon)}\right) \\ &\quad= \frac{\sqrt{2\pi}e^{\pi\sqrt{-1}/4}}{\bigl((1+2\cosh{u})(3-2\cosh{u})\bigr)^{1/4}} e^{N\times F(\sigma_0)} \times \sqrt{\frac{N}{\xi}} \bigl(1+O(N^{-1})\bigr), \end{align*} $$
where the second equality follows from (6.3).
It follows that
$$ \begin{align*} J_N\left(E;e^{\xi/N}\right) &= \frac{1}{2\sinh(u/2)} \left(\sum_{m=0}^{p-1}\beta_{p,m}\right) \times \frac{\sqrt{2\pi}e^{\pi\sqrt{-1}/4}}{\bigl((1+2\cosh{u})(3-2\cosh{u})\bigr)^{1/4}} \\ &\quad\times \sqrt{\frac{N}{\xi}} e^{N\times F(\sigma_0)} \bigl(1+O(N^{-1})\bigr) \end{align*} $$
from (3.2). Using (2.1) with
$N=p$
and
$q=e^{4N\pi ^2/\xi }$
, we have
$$ \begin{align*} \sum_{m=0}^{p-1} \beta_{m,p} = J_{p}\left(E;e^{4N\pi^2/\xi}\right). \end{align*} $$
Therefore, we have
$$ \begin{align*} J_N\left(E;e^{\xi/N}\right) =& \frac{1}{2\sinh(u/2)} J_{p}\left(E;e^{4N\pi^2/\xi}\right) \times \frac{\sqrt{2\pi}e^{\pi\sqrt{-1}/4}}{\bigl((1+2\cosh{u})(3-2\cosh{u})\bigr)^{1/4}} \\ &\quad\times \sqrt{\frac{N}{\xi}} e^{N\times F(\sigma_0)} \bigl(1+O(N^{-1})\bigr). \end{align*} $$
Putting
$$ \begin{align*} S_E(u) &:= \xi\bigl(F(\sigma_0)+2\pi\sqrt{-1}\bigr) \\ &= \operatorname{Li}_2\left(e^{-u-\varphi(u)}\right)-\operatorname{Li}_2\left(e^{-u+\varphi(u)}\right) + u\bigl(\varphi(u)+2\pi\sqrt{-1}\bigr), \\ T_E(u) &:= \frac{2}{\sqrt{(2\cosh{u}+1)(2\cosh{u}-3)}}, \end{align*} $$
we finally have
$$ \begin{align*} &J_N\left(E;e^{\xi/N}\right) \\ &\quad = \frac{\sqrt{-\pi}}{2\sinh(u/2)}T_E(u)^{1/2} J_{p}\left(E;e^{4N\pi^2/\xi}\right) \left(\frac{N}{\xi}\right)^{1/2} e^{\frac{N}{\xi}\times S_E(u)} \bigl(1+O(N^{-1})\bigr), \end{align*} $$
which proves Theorem 1.4.
We can see that the cohomological adjoint Reidemeister torsion
$T_E(u)$
equals
$\pm T_E(u)$
and the Chern–Simons invariant
$\operatorname {CS}_{u,v(u)}(\rho )$
is given by
$S_E(u)-u\pi \sqrt {-1}-\frac {1}{4}uv(u)\pmod {\pi ^2\mathbb {Z}}$
. See, for example, [Reference Murakami and Yokota29, Chapter 5] for the calculation of the adjoint Reidemeister torsion and the Chern–Simons invariant.
7 Quantum modularity
For
$\eta :=\begin {pmatrix}a&b\\c&d\end {pmatrix}\in \mathrm {SL}(2;\mathbb {Z})$
and a complex number z, define
$\eta (z):=\dfrac {az+b}{cz+d}$
as usual. We also define
$\hbar _{\eta }(z):=\frac {2\pi \sqrt {-1}}{z-\eta ^{-1}(\infty )}=\frac {2c\pi \sqrt {-1}}{cz+d}$
.
In [Reference Zagier36], Zagier conjectured the following.
Conjecture 7.1 (Quantum modularity conjecture)
Let K be a hyperbolic knot in
$S^3$
and
$\eta :=\begin {pmatrix}a&b\\c&d\end {pmatrix}\in \mathrm {SL}(2;\mathbb {Z})$
with
$c>0$
. Putting
$X_0:=N/p$
for positive integers N and p, the following asymptotic equivalence holds:
$$ \begin{align} \frac{J_{cN+dp}\left(K;e^{2\pi\sqrt{-1}\eta(X_0)}\right)}{J_{p}\left(K;e^{2\pi\sqrt{-1} X_0}\right)} \underset{N\to\infty}{\sim} C_{K,\eta} \left(\frac{2\pi}{\hbar_{\eta}(X_0)}\right)^{3/2} \exp\left(\frac{\sqrt{-1}\operatorname{CV}(K)}{\hbar_{\eta}(X_0)}\right), \end{align} $$
where
$C_{K,\eta }$
is a complex number depending only on
$\eta $
and K.
Note that Conjecture 7.1 is just a part of Zagier’s original quantum modularity conjecture. See [Reference Bettin and Drappeau1, Reference Garoufalidis and Zagier7, Reference Zagier36] for more details.
Remark 7.2 The modularity conjecture was proved by Garoufalidis and Zagier [Reference Garoufalidis and Zagier7] in the case of the figure-eight knot, and by Bettin and Drappeau [Reference Bettin and Drappeau1] for hyperbolic knots with at most seven crossings except for
$7_2$
.
Bettin and Drappeau also proved that for the figure-eight knot E,
$C_{E,\eta }$
is given as follows:
$$ \begin{align*} C_{E,\eta} = \frac{ce^{3\pi\sqrt{-1}/4}}{3^{1/4}} \prod_{g=1}^{c}|\omega_g|^{2g/c} \left(\sum_{r=1}^{c}\prod_{g=1}^{r}|\omega_g|^2\right), \end{align*} $$
where
$\omega _g:=1-\exp \left (2\pi \sqrt {-1}(\frac {ag}{c}-\frac {5}{6c})\right )$
.
Since
$$ \begin{align*} S_E(0) = \operatorname{Li}_2\left(e^{\pi\sqrt{-1}/3}\right)-\operatorname{Li}_2\left(e^{-\pi\sqrt{-1}/3}\right) = \operatorname{Vol}\left(S^3\setminus{E}\right)\sqrt{-1} \end{align*} $$
(see, for example, [Reference Milnor22, Appendix]), if K is the figure-eight knot E and
$\eta =\begin {pmatrix}0&-1\\1&0\end {pmatrix}$
, (7.1) turns out to be
$$ \begin{align} \frac{J_{N}\left(E;e^{2p\pi\sqrt{-1}/N}\right)}{J_{p}\left(E;e^{2N\pi\sqrt{-1}/p}\right)} \underset{N\to\infty}{\sim} -2\pi^{3/2}T_E(0)^{1/2}\left(\frac{N}{2p\pi\sqrt{-1}}\right)^{3/2} \exp\left(\frac{NS_E(0)}{2p\pi\sqrt{-1}}\right). \end{align} $$
Here, we use the fact that E is amphicheiral, that is, E is equivalent to its mirror image, to conclude
$J_{N}(E;q)=J_{N}(E;q^{-1})$
. Compare (7.2) with (1.2), noting that
$\xi =2p\pi \sqrt {-1}$
when
$u=0$
.
We can regard (1.2) as a kind of quantum modularity with
$\eta =\begin {pmatrix}0&-1\\1&0\end {pmatrix}$
as follows.
Put
$X:=\frac {2N\pi \sqrt {-1}}{\xi }$
. Note that
$\operatorname {Re}{X}\to \infty $
as
$N\to \infty $
. We have
$\eta (X)=\frac {-\xi }{2N\pi \sqrt {-1}}$
,
$\exp (2\pi \sqrt {-1} X)=e^{-4N\pi ^2/\xi }$
,
$\exp \bigl (2\pi \sqrt {-1}\eta (X)\bigr )=e^{-\xi /N}$
, and
$\hbar _{\eta }(X)=\xi /N$
. Since the figure-eight knot is amphicheiral, (1.2) can be written as
$$ \begin{align*} \frac{J_N\left(E;e^{2\pi\sqrt{-1}\eta(X)}\right)}{J_p\left(E;e^{2\pi\sqrt{-1} X}\right)} \sim \frac{\sqrt{-\pi}}{2\sinh(u/2)} \left(\frac{T_E(u)}{\hbar_{\eta}(X)}\right)^{1/2} \exp\left(\frac{S_E(u)}{\hbar_{\eta}(X)}\right). \end{align*} $$
We would like to generalize this to other elements of
$\mathrm {SL}(2;\mathbb {Z})$
and other hyperbolic knots in
$S^3$
. Some computer experiments indicate the following conjecture stated in the Introduction.
Conjecture 7.3 (Quantum modularity conjecture for the colored Jones polynomial)
Let
$K\subset S^3$
be a hyperbolic knot, and let u be a small complex number that is not a rational multiple of
$\pi \sqrt {-1}$
. For positive integers p and N, put
$\xi :=u+2p\pi \sqrt {-1}$
and
$X:=\frac {2N\pi \sqrt {-1}}{\xi }$
. Then, for any
$\eta =\begin {pmatrix}a&b\\c&d\end {pmatrix}\in \mathrm {SL}(2;\mathbb {Z})$
with
$c>0$
, the following asymptotic equivalence holds:
$$ \begin{align} \frac{J_{cN+dp}\left(K;e^{2\pi\sqrt{-1}\eta(X)}\right)}{J_{p}\left(K;e^{2\pi\sqrt{-1} X}\right)} \underset{N\to\infty}{\sim} C_{K,\eta}(u)\frac{\sqrt{-\pi}}{2\sinh(u/2)} \left(\frac{T_K(u)}{\hbar_{\eta}(X)}\right)^{1/2} \exp\left(\frac{S_K(u)}{\hbar_{\eta}(X)}\right), \end{align} $$
where
$C_{K,\eta }(u)\in \mathbb {C}$
does not depend on p.
Note that
$cN+dp$
comes from the denominator of
$\eta (N/p)=\eta \left (X\bigm |_{u=0}\right )$
.
Remark 7.4 Compare the exponent
$1/2$
of
$1/\hbar _{\eta }(X)=\frac {cX+d}{2c\pi \sqrt {-1}}$
in (7.3) with
$3/2$
in (7.1). Our modularity would have weight
$1/2$
rather than
$3/2$
.
Remark 7.5 Since
$(-\eta )(X)=\eta (X)$
, we may assume that
$c\ge 0$
.
If
$c=0$
, then
$\eta =\pm \begin {pmatrix}1&k\\0&1\end {pmatrix}$
for some integer k. Since
$\eta (X)=X+k$
, we have
$\exp \bigl (2\pi \sqrt {-1}\eta (X)\bigr )=\exp (2\pi \sqrt {-1} X)$
and so
$J_{p}\left (E;e^{2\pi \sqrt {-1}\eta (X)}\right )=J_{p}\left (E;e^{2\pi \sqrt {-1} X}\right )$
.
Remark 7.6 When
$p=1$
and
$\eta =\begin {pmatrix}0&-1\\1&0\end {pmatrix}$
, (7.3) becomes
$$ \begin{align*} &J_{N}\left(K;e^{-(u+2\pi\sqrt{-1})/N}\right) \\ &\quad\underset{N\to\infty}{\sim} C_{K,\eta}(u)\frac{\sqrt{-\pi}}{2\sinh(u/2)} \left(T_K(u)\right)^{1/2}\left(\frac{N}{u+2\pi\sqrt{-1}}\right)^{1/2} \exp\left(\frac{N\times S_K(u)}{u+2\pi\sqrt{-1}}\right), \end{align*} $$
which coincides with [Reference Murakami24, Conjecture 1.6] with
$C_{K,\eta }(u)=1$
. See also [Reference Dimofte and Gukov2, Reference Gukov and Murakami9]. Strictly speaking, we need to take the mirror image
$\overline {K}$
of K because
$J_N\left (\overline {K};e^{-(u+2\pi \sqrt {-1})/N}\right )=J_N\left (K;e^{(u+2\pi \sqrt {-1})/N}\right )$
.
8 Lemmas
In this section, we prove lemmas that we use.
Proof of Lemma 2.1
Recall that
$\xi =u+2p\pi \sqrt {-1}$
and
$\gamma =\frac {\xi }{2N\pi \sqrt {-1}}$
.
Since
$\operatorname {Re}{\gamma }=p/N>0$
,
$\sinh (\gamma x)\underset {x\to \infty }{\sim }\frac {e^{\gamma x}}{2}$
and
$\sinh (\gamma x)\underset {x\to -\infty }{\sim }\frac {-e^{-\gamma x}}{2}$
. So we have
$$ \begin{align*} \frac{e^{(2z-1)x}}{x\sinh(x)\sinh(\gamma x)} &\underset{x\to\infty}{\sim} \frac{4}{x}e^{(2z-\gamma-2)x} \end{align*} $$
and
$$ \begin{align*} \frac{e^{(2z-1)x}}{x\sinh(x)\sinh(\gamma x)} &\underset{x\to-\infty}{\sim} \frac{4}{x}e^{(2z+\gamma)x}. \end{align*} $$
Therefore, if
$-\operatorname {Re}\gamma /2<\operatorname {Re}{z}<1+\operatorname {Re}\gamma /2$
, then the integral converges, completing the lemma.
The following proof is almost the same as [Reference Murakami and Tran27, Proposition 2.8]. See also [Reference Ohtsuki31, Proposition A.1].
Proof of Lemma 2.4
We will show that
$T_N(z)=\frac {N}{\xi }\mathcal {L}_2(z)+O(1/N)$
.
Recalling that
$\xi =2N\pi \gamma \sqrt {-1}$
, we have
Since the Taylor expansion of
$\frac {\sinh (y)}{y}$
around
$y=0$
is
$1+\frac {y^2}{6}+\cdots $
, we have
$\frac {y}{\sinh (y)}=1-\frac {y^2}{6}+o(y^2)$
as
$y\to 0$
. Therefore, we have
$\left |\frac {\gamma x}{\sinh (\gamma x)}-1\right |\le \frac {c|x|^2}{N^2}$
for some constant
$c>0$
and so
where we put
$c':=\frac {c\pi }{2|\xi |}$
.
We put
$$ \begin{align*} I_{+} &:= \int_{1}^{\infty}\left|\frac{e^{(2z-1)x}}{\sinh(x)}\right|\,dx, \\ I_{-} &:= \int_{-\infty}^{-1}\left|\frac{e^{(2z-1)x}}{\sinh(x)}\right|\,dx, \\ I_{0} &:= \int_{|x|=1,\operatorname{Im}{x}\ge0}\left|\frac{e^{(2z-1)x}}{\sinh(x)}\right|\,dx. \end{align*} $$
We have
$$ \begin{align*} I_{+} &\le \int_{1}^{\infty}\frac{2e^{2x\operatorname{Re}{z}-x}}{e^{x}-e^{-x}}\,dx = \int_{1}^{\infty}\frac{2e^{2x(\operatorname{Re}{z}-1)}}{1-e^{-2x}}\,dx \le \frac{2}{1-e^{-2}}\int_{1}^{\infty}e^{-2\nu x}\,dx \\ &= \frac{e^{-2\nu}}{\nu(1-e^{-2})}, \end{align*} $$
where we use the assumption
$\operatorname {Re}{z}\le 1-\nu $
.
Similarly, we have
$$ \begin{align*} I_{-} &\le \int_{-\infty}^{-1}\frac{2e^{2x\operatorname{Re}{z}-x}}{e^{-x}-e^{x}}\,dx = \int_{-\infty}^{-1}\frac{2e^{2x\operatorname{Re}{z}}}{1-e^{2x}}\,dx \le \frac{2}{1-e^{-2}}\int_{-\infty}^{-1}e^{2\nu x}\,dx \\ &= \frac{e^{-2\nu}}{\nu(1-e^{-2})}, \end{align*} $$
where we use the assumption
$\operatorname {Re}{z}\ge \nu $
.
Putting
$x=e^{t\sqrt {-1}}$
(
$0\le t\le \pi $
) and
$L:=\min _{|x|=1,\operatorname {Im}{x}\ge 0}|\sinh (x)|$
, we have
$$ \begin{align*} I_{0} = \int_{0}^{\pi} \left|\frac{e^{(2z-1)e^{t\sqrt{-1}}}}{\sinh\left(e^{t\sqrt{-1}}\right)}\right| \times\left|\sqrt{-1} e^{t\sqrt{-1}}\right|\,dt \le \frac{1}{L} \int_{0}^{\pi} e^{(2\operatorname{Re}{z}-1)\cos{t}-2\operatorname{Im}{z}\sin{t}}\,dt, \end{align*} $$
which is bounded from the above because both
$\operatorname {Re}{z}$
and
$\operatorname {Im}{z}$
are bounded.
Therefore, we see that
$I_{+}+I_{-}+I_{0}$
is bounded from above, which implies that
$\left |T_N(z)-\frac {N}{\xi }\mathcal {L}_2(z)\right |=O(1/N)$
.
Proof of Lemma 5.2
Since
$\operatorname {Re}{F(0)}$
coincides with
$\operatorname {Re}{\Phi (w_0)}$
in [Reference Murakami24] (see [Reference Murakami and Tran27, Remark 1.6]), we have
$\operatorname {Re}{F(0)}>0$
from [Reference Murakami24, Lemma 3.5].
Next, we will show that
$\xi \bigl (F(\sigma _0)-F(0)\bigr )$
is purely imaginary with positive imaginary part. Then we conclude that
$\operatorname {Re}\bigl (F(\sigma _0)-F(0)\bigr )>0$
, since
$\xi $
is in the first quadrant.
Since
$\varphi (u)$
is purely imaginary, we have
$\overline {\operatorname {Li}_2\left (e^{-u-\varphi (u)}\right )}=\operatorname {Li}_2\left (e^{-u+\varphi (u)}\right )$
. So we see that
$\xi \bigl (F(\sigma _0)-F(0)\bigr )=\operatorname {Li}_2\left (e^{-u-\varphi (u)}\right )-\operatorname {Li}_2\left (e^{-u+\varphi (u)}\right )+u(\theta +2\pi )\sqrt {-1}$
is purely imaginary with imaginary part
$2\operatorname {Im}\operatorname {Li}_2\left (e^{-u-\varphi (u)}\right )+u(\theta +2\pi )$
, which coincides with
$\operatorname {Im}\bigl (\xi \Phi (w_0)\bigr )+2u\pi>0$
in [Reference Murakami24, p. 214].
This proves the lemma.
Proof of Lemma 5.3
We have
$$ \begin{align*} &\xi\left(\Phi_m(P_{12})-\Phi_m(\sigma_m)\right) \\ =& \operatorname{Li}_2\left(-e^{-u-\frac{u((6m+5)\pi+2\theta)}{2p\pi}}\right) - \operatorname{Li}_2\left(-e^{-u+\frac{u((6m+5)\pi+2\theta)}{2p\pi}}\right) \\ &- \operatorname{Li}_2\left(e^{-u-\varphi(u)}\right) + \operatorname{Li}_2\left(e^{-u+\varphi(u)}\right) \\ &+ \frac{(2m+1)u\xi}{2p}+\frac{u^2(2(m+1)\pi+\theta)}{p\pi} - u\bigl(2(m+1)\pi+\theta\bigr)\sqrt{-1}. \end{align*} $$
Its real part is
$$ \begin{align*} \operatorname{Li}_2\left(-e^{-u-q_m(u)}\right) - \operatorname{Li}_2\left(-e^{-u+q_m(u)}\right) + uq_m(u), \end{align*} $$
where we put
$q_m(u):=\frac {u((6m+5)\pi +2\theta )}{2p\pi }$
, and its imaginary part is
$$ \begin{align*} -2\operatorname{Im}\operatorname{Li}_2\left(e^{-u-\varphi(u)}\right) - u(\pi+\theta). \end{align*} $$
Then we have
$$ \begin{align*} &\frac{|\xi|^2}{u}\operatorname{Re}\left(F(P_{12})-F(\sigma_m)\right) \\ =& \operatorname{Re}\left(\xi\bigl(F(P_{12})-F(\sigma_m)\bigr)\right) + \frac{2p\pi}{u}\operatorname{Im}\left(\xi\bigl(F(P_{12})-F(\sigma_m)\bigr)\right) \\ =& \operatorname{Li}_2\left(-e^{-u-q_m(u)}\right) - \operatorname{Li}_2\left(-e^{-u+q_m(u)}\right) +uq_m(u) \\ &- \frac{2p\pi}{u} \left(2\operatorname{Im}\operatorname{Li}_2\left(e^{-u-\varphi(u)}\right)+u(\pi+\theta)\right). \end{align*} $$
By using the inequality
$2\operatorname {Im}\operatorname {Li}_2\left (e^{-u-\varphi (u)}\right )+u\theta>0$
in [Reference Murakami24, Section 7], this is less than
$c_{p,m}(u)$
, where we put
$$ \begin{align*} c_{p,m}(u) := \operatorname{Li}_2\left(-e^{-u-q_m(u)}\right) - \operatorname{Li}_2\left(-e^{-u+q_m(u)}\right) +uq_m(u)-2p\pi^2. \end{align*} $$
Now, we have
$$ \begin{align*} \frac{d}{d\,u}c_{p,m}(u) = q^{\prime}_m(u)\log\bigl(2\cosh{u}+2\cosh{q_m(u)}\bigr) + \log\left(\frac{e^{q_m(u)}+e^{-u}}{1+e^{-u+q_m(u)}}\right), \end{align*} $$
which can be easily seen to be positive. Since
$u<\kappa $
, it suffices to prove
$c_{p,m}(\kappa )<0$
. Since
$\varphi (\kappa )=0$
, we have
$$ \begin{align*} c_{p,m}(\kappa) = \operatorname{Li}_2\left(-e^{-\kappa(1+\frac{6m+5}{2p})}\right) - \operatorname{Li}_2\left(-e^{-\kappa(1-\frac{6m+5}{2p})}\right) +\frac{(6m+5)\kappa^2}{2p}-2p\pi^2, \end{align*} $$
which is increasing with respect to m, fixing p. We will prove that
$c_{p,p-1}(\kappa )<0$
.
We calculate
$$ \begin{align*} c_{p,p-1}(\kappa) = \operatorname{Li}_2\left(-e^{\kappa(\frac{1}{2p}-4)}\right) - \operatorname{Li}_2\left(-e^{\kappa(-\frac{1}{2p}+2)}\right) +\left(3-\frac{1}{2p}\right)\kappa^2-2p\pi^2. \end{align*} $$
The derivative of
$c_{p,p-1}(\kappa )$
with respect to p equals
$$ \begin{align*} \frac{\kappa}{2p^2}\log\left(3+2\cosh\left(\kappa\left(3-\frac{1}{2p}\right)\right)\right) -2\pi^2, \end{align*} $$
which is less than
$-2\pi ^2+\log (6+2\cosh (3\kappa )=-18.274\ldots <0$
. It follows that
$c_{p,p-1}(\kappa )<c_{1,0}(\kappa )=-14.9942\ldots <0$
.
This shows that
$\operatorname {Re}\bigl (F(P_{12})-F(\sigma _m)\bigr )<0$
, proving the lemma.
Before proving Lemma 6.1, we prepare the following lemma.
Lemma 8.1 Put
$g(x):=4\sinh \left (\frac {\xi }{2}(1+x)\right )\sinh \left (\frac {\xi }{2}(1-x)\right )$
. For an integer
$0\le m\le p$
, there exists
$\delta _m>0$
such that
$|g(l/N)|<1$
if
$\frac {m}{p}-\delta _m<\frac {l}{N}<\frac {m}{p}+\delta _m$
.
Proof For an integer
$0\le m\le p$
, we can easily see that
$$ \begin{align*} g(m/p) = 2\bigl(\cosh{u}-\cosh(mu/p)\bigr). \end{align*} $$
So we conclude that
$g(m/p)$
is monotonically decreasing with respect to m. Therefore, we have
$0=g(1)\le g(m/p)\le g(0)=2\bigl (\cosh (u)-1\bigr )<2\cosh (\kappa )-2=1$
. So we have
$0\le g(m/p)<1$
.
Therefore, there exists
$\delta _m>0$
such that
$|g(x)|<1$
if
$|x-m/p|<\delta _m$
, completing the proof.
Proof of Lemma 6.1
From (2)(ii) of the proof of Proposition 5.1, we know that
$\frac {m+1}{p}\in D_{m}$
, that is,
$\operatorname {Re}{\Phi _m\left (\frac {m+1}{p}\right )}<\operatorname {Re}{\Phi _m(\sigma _{m})}$
. Therefore, there exists
$\varepsilon>0$
such that
$\operatorname {Re}{\Phi _m\left (\frac {m+1}{p}\right )}<\operatorname {Re}{\Phi _m(\sigma _m)}-2\varepsilon $
for
$m=0,1,2,\dots ,p-1$
.
Next, we show that there exists
$\tilde {\delta }_m>0$
such that if
$\frac {m+1}{p}-\tilde {\delta }_m<\frac {k}{N}<\frac {m+1}{p}$
, then (6.4) holds.
We can choose
$\delta ^{\prime }_m>0$
so that
$\operatorname {Re}{\Phi _m\left (\frac {k}{N}\right )}<\operatorname {Re}{\Phi _m\left (\frac {m+1}{p}\right )}+\varepsilon $
if
$(m+1)/p-\delta ^{\prime }_m<k/N<(m+1)/p$
. So we have
$\operatorname {Re}{\Phi _m\left (\frac {k}{N}\right )}<\operatorname {Re}{\Phi _m(\sigma _m)}-\varepsilon $
. Now, recall that
$f_N(z)$
converges to
$F(z)$
in the region (4.1). Since we have
$$ \begin{align*} &\operatorname{Re}\left(\frac{2k+1}{2N}-\frac{2(m-1)\pi\sqrt{-1}}{\xi}\right) + \frac{u}{2p\pi} \operatorname{Im}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) \\ &\quad = \frac{2k+1}{2N}-\frac{m}{p}, \\ &\operatorname{Re}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) - \frac{2p\pi}{u} \operatorname{Im}\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) \\ &\quad= \frac{2k+1}{2N}, \end{align*} $$
if
$\nu /p+m/p-1/(2N)\le k/N\le (m+1)/p-\nu /p-1/(2N)$
and
$k/N\le 2M\pi /u+1-1/(2N)$
, then
$f_N\left (\frac {2k+1}{2N}-\frac {2m\pi \sqrt {-1}}{\xi }\right )$
converges to
$$ \begin{align*} F\left(\frac{k}{N}-\frac{2m\pi\sqrt{-1}}{\xi}\right) = \Phi_m\left(\frac{k}{N}\right), \end{align*} $$
as
$N\to \infty $
. Therefore, we see
$$ \begin{align*} \operatorname{Re}{f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)} &< \operatorname{Re}\Phi_m(\sigma_m)-\varepsilon \\ &= \operatorname{Re}{F(\sigma_0)}-\varepsilon \end{align*} $$
if we choose
$\nu $
small enough so that
$\delta ^{\prime }_m>\frac {\nu }{p}+\frac {1}{2N}$
(and N is large enough). Note that so far k should satisfy the inequalities
$$ \begin{align} \frac{m+1}{p}-\delta^{\prime}_m<\frac{k}{N}\le\frac{m+1}{p}-\frac{\nu}{p}-\frac{1}{2N}. \end{align} $$
On the other hand, putting
$h_N(k):=\prod _{l=1}^{k}g\left (\frac {l}{N}\right )$
, we have
$$ \begin{align} \left|h_N(k)\right|>\left|h_N(k')\right| \end{align} $$
if
$\frac {m}{p}-\delta _m<\frac {k}{N}<\frac {k'}{N}<\frac {m}{p}+\delta _m$
from Lemma 8.1. Note that if
$\frac {m}{p}\le \frac {k}{N}<\frac {m+1}{p}$
, we have
$$ \begin{align} h_N(k) = \frac{1-e^{-4pN\pi^2/\xi}}{2\sinh(u/2)}\beta_{p,m} \exp\left(N\times f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)\right) \end{align} $$
from (3.2). From (8.2) and (8.3), if
$\frac {m+1}{p}-\delta _{m+1}<\frac {k}{N}<\frac {k'}{N}<\frac {m+1}{p}$
, then we have
$$ \begin{align*} \operatorname{Re}{f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)} &= \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\beta_{p,m}^{-1}h_N(k)\right| \\ &> \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\beta_{p,m}^{-1}h_N(k')\right| \\ &= \operatorname{Re}{f_N\left(\frac{2k'+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)}, \end{align*} $$
which means that
$\operatorname {Re}{f_N\left (\frac {2k+1}{2N}-\frac {2m\pi \sqrt {-1}}{\xi }\right )}$
is monotonically decreasing with respect to k if
$\frac {m+1}{p}-\delta _{m+1}<\frac {k}{N}<\frac {m+1}{p}$
. Combined with (8.1), we conclude that (6.4) holds if
$\frac {m+1}{p}-\delta ^{\prime }_m<\frac {k}{N}<\frac {m+1}{p}$
, choosing
$\delta ^{\prime }_m$
less than
$\delta _{m+1}$
if necessary.
Now, we show that for
$m=1,2,\dots ,p-1$
, (6.4) holds if
$\frac {m}{p}\le \frac {k}{N}<\frac {m}{p}+\delta _m$
.
From (8.2) and (8.3), if
$\frac {m}{p}-\delta ^{\prime }_m<\frac {k'}{N}<\frac {m}{p}\le \frac {k}{N}<\frac {m}{p}+\delta _m$
, we have
$$ \begin{align*} \operatorname{Re}{f_N\left(\frac{2k+1}{2N}-\frac{2m\pi\sqrt{-1}}{\xi}\right)} &= \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\beta_{p,m}^{-1}h_N(k)\right| \\ &< \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\beta_{p,m}^{-1}h_N(k')\right| \\ &< \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\beta_{p,m-1}^{-1}h_N(k')\right| \\ &= \operatorname{Re}{f_N\left(\frac{2k'+1}{2N}-\frac{2(m-1)\pi\sqrt{-1}}{\xi}\right)}, \end{align*} $$
which is less than
$\operatorname {Re}{F(\sigma _0)}-\varepsilon $
from the argument above. Here, the second inequality follows since
$$ \begin{align*} \left|\frac{\beta_{p,m}}{\beta_{p,m-1}}\right| &= 2\Bigl|\cosh(4pN\pi^2/\xi)-\cosh(4mN\pi^2/\xi)\bigr| \\ &\underset{N\to\infty}{\sim} \frac{1}{2}\exp\left(\frac{4pu\pi^2\times N}{|\xi|^2}\right). \end{align*} $$
So (6.4) holds.
Finally, we consider the case where
$m=0$
. Since
$h_N(0)=\beta _{p,0}=1$
, we have
$$ \begin{align*} \operatorname{Re}{f_N\left(\frac{1}{2N}\right)} &= \frac{1}{N}\log\left|\frac{2\sinh(\xi/2)}{1-e^{-4pN\pi^2/\xi}}\right| \le \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1+\left|e^{-4pN\pi^2/\xi}\right|}\right| \\ &= \frac{1}{N} \log\left|\frac{2\sinh(\xi/2)}{1+e^{-4puN\pi^2/|\xi|^2}}\right| \to0 \quad(N\to\infty). \end{align*} $$
Since
$\operatorname {Re}{F(\sigma _0)}>0$
from Lemma 5.2, (6.4) holds if
$k/N<\delta _0$
and N is sufficiently large.
As a result, if we put
$\tilde {\delta }_m:=\min \{\delta ^{\prime }_m,\delta _m\}$
, (6.4) holds.
Appendix. The case where
$(p,N)\ne 1$
In this appendix, we will calculate
$\prod _{l=1}^{k}\left (1-e^{(N-l)\xi /N}\right )\left (1-e^{(N+l)\xi /N}\right )$
assuming
$(p,N)=c>1$
. Put
$N':=N/c\in \mathbb {N}$
and
$p':=p/c\in \mathbb {N}$
.
Note that
$jN/p$
(
$1\le j\le N-1$
,
$j\in \mathbb {N}$
) is an integer if and only j is a multiple of
$p'$
.
If
$k<N'$
, then we can choose an integer
$m<p'$
so that
$mN/p<k<(m+1)N/p$
because
$N/p,2N/p,\dots ,(p'-1)N/p$
are not integers. Therefore, from (3.1), we have
$$ \begin{align*} &\prod_{l=1}^{k}(1-e^{(N-l)\xi/N})(1+e^{(N+l)\xi/N}) \\ &\quad = \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \left( \prod_{j=1}^{m} \left(1-e^{4(p-j)N\pi^2/\xi}\right)\left(1-e^{4(p+j)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+m+1\bigr)}{E_N\bigl((N+k+1/2)\gamma-p-m\bigr)}. \end{align*} $$
If
$k=N'$
, we have
$$ \begin{align*} &\prod_{l=1}^{N'}\left(1-e^{(N-l)\xi/N}\right)\left(1-e^{(N+l)\xi/N}\right) \\ &\quad= \left(\prod_{l=1}^{N'-1}\left(1-e^{(N-l)\xi/N}\right)\left(1-e^{(N+l)\xi/N}\right)\right) \left(1-e^{(N-N')\xi/N}\right)\left(1-e^{(N+N')\xi/N}\right) \\ &\quad = \left(1-e^{(c-1)\xi/c}\right)\left(1-e^{(c+1)\xi/c}\right) \\ &\qquad \times \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \left( \prod_{j=1}^{p'-1} \left(1-e^{4(p-j)N\pi^2/\xi}\right) \left(1-e^{4(p+j)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N-p'+1/2)\gamma-p+p'\bigr)}{E_N\bigl((N+p'-1/2)\gamma-p-p'+1\bigr)}, \end{align*} $$
since
$(p'-1)N/p<N'-1<p'N/p$
.
If k is an integer with
$nN'\le k<(n+1)N'$
, writing l (
$0\le l\le k$
) as
$l=aN'+b$
with
$0\le a\le n$
and
$0\le b\le N'-1$
, we have
$$ \begin{align*} &\prod_{l=1}^{k}\left(1-e^{(N-l)\xi/N}\right)\left(1-e^{(N+l)\xi/N}\right) \\ &\quad = \prod_{b=1}^{N'-1} \left(1-e^{(N-b)\xi/N}\right)\left(1-e^{(N+b)\xi/N}\right) \\ &\qquad\times \prod_{a=1}^{n-1}\prod_{b=0}^{N'-1} \left(1-e^{(N-aN'-b)\xi/N}\right)\left(1-e^{(N+aN'+b)\xi/N}\right) \\ &\qquad\times \prod_{b=0}^{k-nN'} \left(1-e^{(N-nN'-b)\xi/N}\right)\left(1-e^{(N+nN'+b)\xi/N}\right) \\ &\quad= \prod_{a=1}^{n-1} \left(1-e^{(c-a)\xi/c}\right)\left(1-e^{(c+a)\xi/c}\right) \times \prod_{a=0}^{n-1}\prod_{b=1}^{N'-1}P_{a,b} \\ &\qquad\times \left(1-e^{(c-n)\xi/c}\right)\left(1-e^{(c+n)\xi/c}\right) \prod_{b=1}^{k-nN'}Q_b, \end{align*} $$
where we put
$$ \begin{align*} P_{a,b} &:= \left(1-e^{(N-aN'-b)\xi/N}\right)\left(1-e^{(N+aN'+b)\xi/N}\right) \\ &= \left(1-e^{2(N-aN'-b)\pi\sqrt{-1}\gamma}\right)\left(1-e^{2(N+aN'+b)\pi\sqrt{-1}\gamma}\right), \\ Q_b&:= \left(1-e^{(N-nN'-b)\xi/N}\right)\left(1-e^{(N+nN'+b)\xi/N}\right) \\ &= \left(1-e^{2(N-nN'-b)\pi\sqrt{-1}\gamma}\right)\left(1-e^{2(N+nN'+b)\pi\sqrt{-1}\gamma}\right). \end{align*} $$
If we choose i (
$0\le i\le p'-1$
) with
$iN'/p'<b<(i+1)N'/p'$
, then we have
$(p'-ap'-i-1)N/p<N-aN'-b<(p'-ap'-i)N/p$
and
$(p'+ap'+i)N/p<N+aN'+b<(p'+ap'+i+1)N/p$
. So, from Corollary 2.6, we have
$$ \begin{align*} &\prod_{b=1}^{N'-1}P_{a,b} = \prod_{i=0}^{p'-1}\left(\prod_{iN'/p'<b<(i+1)N'/p'}P_{a,b}\right) \\ &\quad= \prod_{i=0}^{p'-1} \left( \prod_{iN'/p'<b<(i+1)N'/p'} \frac{E_N\bigl((N-aN'-b-1/2)\gamma-p+ap'+i+1\bigr)} {E_N\bigl((N-aN'-b+1/2)\gamma-p+ap'+i+1\bigr)} \right. \\ &\qquad\times \left. \prod_{iN'/p'<b<(i+1)N'/p'} \frac{E_N\bigl((N+aN'+b-1/2)\gamma-p-ap'-i\bigr)} {E_N\bigl((N+aN'+b+1/2)\gamma-p-ap'-i\bigr)} \right) \\ &\quad= \prod_{i=0}^{p'-2} \left( \frac{E_N\bigl((N-aN'-\lfloor(i+1)N'/p'\rfloor-1/2)\gamma-p+ap'+i+1\bigr)} {E_N\bigl((N-aN'-\lfloor iN'/p'\rfloor-1/2)\gamma-p+ap'+i+1\bigr)} \right. \\ &\qquad\times \left. \frac{E_N\bigl((N+aN'+\lfloor iN'/p'\rfloor+1/2)\gamma-p-ap'-i\bigr)} {E_N\bigl((N+aN'+\lfloor(i+1)N'/p'\rfloor+1/2)\gamma-p-ap'-i\bigr)} \right) \\ &\qquad\times \frac{E_N\bigl((N-(a+1)N'+1/2)\gamma-p+(a+1)p'\bigr)} {E_N\bigl((N-aN'-\lfloor(p'-1)N'/p'\rfloor-1/2)\gamma-p+(a+1)p'\bigr)} \\ &\qquad\times \frac{E_N\bigl((N+aN'+\lfloor(p'-1)N'/p'\rfloor+1/2)\gamma-p-(a+1)p'+1\bigr)} {E_N\bigl((N+(a+1)N'-1/2)\gamma-p-(a+1)p'+1\bigr)}. \end{align*} $$
Note that the case where
$i=p'-1$
is exceptional.
Using Lemma 2.7 with
$z=(N-aN'-\lfloor iN'/p'\rfloor -1/2)\gamma -p+ap'+i$
(
$i=1,\dots ,p'-2$
) and
$z=(N+aN'+\lfloor (i+1)N'/p'\rfloor -1/2)\gamma -p-ap'-i-1$
(
$i=1,\dots ,p'-2$
), this becomes
$$ \begin{align*} & \prod_{i=1}^{p'-1} \left( \left(1-e^{4(p-ap'-i)N\pi^2/\xi}\right) \left(1-e^{4(p+ap'+i)N\pi^2/\xi}\right) \right) \\ &\quad\times \frac{E_N\bigl((N+aN'+1/2)\gamma-p-ap'\bigr)} {E_N\bigl((N-aN'-1/2)\gamma-p+ap'+1\bigr)} \\ &\quad\times \frac{E_N\bigl((N-(a+1)N'+1/2)\gamma-p+(a+1)p'\bigr)} {E_N\bigl((N+(a+1)N'-1/2)\gamma-p-(a+1)p'+1\bigr)}. \end{align*} $$
Therefore, we have
$$ \begin{align*} &\prod_{a=0}^{n-1}\prod_{b=1}^{N'-1}P_{a,b} \\ &\quad= \prod_{a=0}^{n-1} \prod_{i=1}^{p'-1} \left( \left(1-e^{4(p-ap'-i)N\pi^2/\xi}\right) \left(1-e^{4(p+ap'+i)N\pi^2/\xi}\right) \right) \\ &\qquad\times \prod_{a=0}^{n-1} \left( \frac{E_N\bigl((N+aN'+1/2)\gamma-p-ap'\bigr)} {E_N\bigl((N-aN'-1/2)\gamma-p+ap'+1\bigr)} \right. \\ &\qquad\times \left. \frac{E_N\bigl((N-(a+1)N'+1/2)\gamma-p+(a+1)p'\bigr)} {E_N\bigl((N+(a+1)N'-1/2)\gamma-p-(a+1)p'+1\bigr)} \right) \\ &\quad= \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \prod_{a=0}^{n-1} \prod_{i=1}^{p'-1} \left( \left(1-e^{4(p+ap'+i)N\pi^2/\xi}\right) \left(1-e^{4(p-ap'-i)N\pi^2/\xi}\right) \right) \\ &\qquad\times \prod_{a=1}^{n-1} \left( \frac{1-e^{4(p+ap')N\pi^2/\xi}}{1-e^{(c+a)\xi/c}} \times \frac{1-e^{4(p-ap')N\pi^2/\xi}}{1-e^{(c-a)\xi/c}} \right) \\ &\qquad\times \frac{E_N\bigl((N-nN'+1/2)\gamma-p+np'\bigr)} {E_N\bigl((N+nN'-1/2)\gamma-p-np'+1\bigr)} \\ &\quad= \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \times \frac{\prod_{l=1}^{np'-1}\left(1-e^{4(p+l)N\pi^2/\xi}\right) \left(1-e^{4(p-l)N\pi^2/\xi}\right)} {\prod_{a=1}^{n-1}\left(1-e^{(c+a)\xi/c}\right)\left(1-e^{(c-a)\xi/c}\right)} \\ &\qquad\times \frac{E_N\bigl((N-nN'+1/2)\gamma-p+np'\bigr)} {E_N\bigl((N+nN'-1/2)\gamma-p-np'+1\bigr)},\\[-18pt] \end{align*} $$
where we use Lemma 2.8 for
$w=(N+aN')\gamma -p-ap'$
(
$a=0,1,\dots ,n-1$
) and
$w=(N-aN')\gamma -p+ap'$
(
$a=1,2,\dots ,n-1$
) at the second equality.
Similarly, letting h (
$0\le h\le p'-1$
) be an integer with
$hN'/p'<k-nN'<(h+1)N'/p'$
, from Corollary 2.6, we have
$$ \begin{align*} &\prod_{b=1}^{k-nN'}Q_b = \prod_{i=0}^{h-1}\left(\prod_{iN'/p'<b<(i+1)N'/p'}Q_b\right) \times \prod_{hN'/p'<b\le k-nN'}Q_b \\ &\quad= \prod_{i=0}^{h-1} \left( \prod_{iN'/p'<b<(i+1)N'/p'} \frac{E_N\bigl((N-nN'-b-1/2)\gamma-p+np'+i+1\bigr)} {E_N\bigl((N-nN'-b+1/2)\gamma-p+np'+i+1\bigr)} \right. \\ &\qquad\times \left. \prod_{iN'/p'<b<(i+1)N'/p'} \frac{E_N\bigl((N+nN'+b-1/2)\gamma-p-np'-i\bigr)} {E_N\bigl((N+nN'+b+1/2)\gamma-p-np'-i\bigr)} \right) \\ &\qquad\times \prod_{hN'/p'<b\le k-nN'} \frac{E_N\bigl((N-nN'-b-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N-nN'-b+1/2)\gamma-p+np'+h+1\bigr)} \\ &\qquad\times \prod_{hN'/p'<b\le k-nN'} \frac{E_N\bigl((N+nN'+b-1/2)\gamma-p-np'-h\bigr)} {E_N\bigl((N+nN'+b+1/2)\gamma-p-np'-h\bigr)} \\ &\quad= \prod_{i=0}^{h-1} \left( \frac{E_N\bigl((N-nN'-\lfloor(i+1)N'/p'\rfloor-1/2)\gamma-p+np'+i+1\bigr)} {E_N\bigl((N-nN'-\lfloor iN'/p'\rfloor-1/2)\gamma-p+np'+i+1\bigr)} \right. \\ &\qquad\times \left. \frac{E_N\bigl((N+nN'+\lfloor iN'/p'\rfloor+1/2)\gamma-p-np'-i\bigr)} {E_N\bigl((N+nN'+\lfloor(i+1)N'/p'\rfloor+1/2)\gamma-p-np'-i\bigr)} \right) \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N-nN'-\lfloor hN'/p'\rfloor-1/2)\gamma-p+np'+h+1\bigr)} \\ &\qquad\times \frac{E_N\bigl((N+nN'+\lfloor hN'/p'\rfloor+1/2)\gamma-p-np'-h\bigr)} {E_N\bigl((N+k+1/2)\gamma-p-np'-h\bigr)}.\\[-18pt] \end{align*} $$
Using Lemma 2.7 with
$z=(N-nN'-\lfloor iN'/p'\rfloor -1/2)\gamma -p+np'+i$
and
$z=(N+nN'+\lfloor iN'/p'\rfloor +1/2)\gamma -p-np'-i$
(
$i=1,2,\dots ,h$
), we have
$$ \begin{align*} \prod_{b=1}^{k-nN'}Q_b &= \prod_{i=1}^{h} \left( \left(1-e^{4(p-np'-i)N\pi^2/\xi}\right) \left(1-e^{4(p+np'+i)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N+nN'+1/2)\gamma-p-np'\bigr)} {E_N\bigl((N-nN'-1/2)\gamma-p+np'+1\bigr)} \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N+k+1/2)\gamma-p-np'-h\bigr)}. \end{align*} $$
Therefore, we finally have
$$ \begin{align*} &\prod_{l=1}^{k}\left(1-e^{(N-l)\xi/N}\right)\left(1-e^{(N+l)\xi/N}\right) \\ &\quad= \left(1-e^{(c-n)\xi/c}\right)\left(1-e^{(c+n)\xi/c}\right) \\ &\qquad\times \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \times \prod_{l=1}^{np'-1}\left(1-e^{4(p-l)N\pi^2/\xi}\right)\left(1-e^{4(p+l)N\pi^2/\xi}\right) \\ &\qquad\times \frac{E_N\bigl((N-nN'+1/2)\gamma-p+np'\bigr)} {E_N\bigl((N+nN'-1/2)\gamma-p-np'+1\bigr)} \\ &\qquad\times \prod_{i=1}^{h} \left( \left(1-e^{4(p-np'-i)N\pi^2/\xi}\right) \left(1-e^{4(p+np'+i)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N+nN'+1/2)\gamma-p-np'\bigr)} {E_N\bigl((N-nN'-1/2)\gamma-p+np'+1\bigr)} \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N+k+1/2)\gamma-p-np'-h\bigr)} \\ &\quad= \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \times \prod_{l=1}^{np'}\left(1-e^{4(p-l)N\pi^2/\xi}\right)\left(1-e^{4(p+l)N\pi^2/\xi}\right) \\ &\qquad\times \prod_{i=1}^{h} \left( \left(1-e^{4(p-np'-i)N\pi^2/\xi}\right) \left(1-e^{4(p+np'+i)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N+k+1/2)\gamma-p-np'-h\bigr)} \\ &\quad= \frac{1-e^{4pN\pi^2/\xi}}{1-e^{\xi}} \times \prod_{l=1}^{np'+h} \left( \left(1-e^{4(p-l)N\pi^2/\xi}\right)\left(1-e^{4(p+l)N\pi^2/\xi}\right) \right) \\ &\qquad\times \frac{E_N\bigl((N-k-1/2)\gamma-p+np'+h+1\bigr)} {E_N\bigl((N+k+1/2)\gamma-p-np'-h\bigr)}, \end{align*} $$
where we use Lemma 2.8 for
$w=(N-nN')\gamma -p+np'$
and
$w=(N+nN')\gamma -p-np'$
at the second equality. Recalling that we choose n and h so that
$nN'\le k<(n+1)N'$
and
$hN'/p'<k-nN'<(h+1)N'/p'$
, we see that
$np'+h$
satisfies
$(np'+h)N/p<k<(np'+h+1)N/p$
. So, putting
$m:=np'+h$
, we see that if
$mN/p<k<(m+1)N/p$
, then the formula above coincides with (3.1) where
$(p,N)=1$
.
