text:
\begin{document}

\begin{center}
{\large\bf A further $q$-analogue of Van Hamme's (H.2)
supercongruence for $p\equiv1\pmod{4}$
 \footnote{This work is supported by the National Natural Science
Foundation of China (No. 11661032).}}
\end{center}

\renewcommand{\thefootnote}{$\dagger$}

\vskip 2mm \centerline{Chuanan Wei}
\begin{center}
{School of Biomedical Information and Engineering,\\ Hainan Medical University, Haikou 571199, China\\
{\tt weichuanan78@163.com  } }
\end{center}

\vskip 0.7cm \noindent{\bf Abstract.} Several years ago, Long and
Ramakrishna [Adv. Math. 290 (2016), 773--808] extended Van Hamme's
(H.2) supercongruence to the modulus $p^3$ case. Recently, Guo [Int.
J. Number Theory, to appear] found a $q$-analogue of the
Long--Ramakrishna formula for $p\equiv 3\pmod 4$. In this note, a
$q$-analogue of the Long--Ramakrishna formula for $p\equiv 1\pmod 4$
is derived through the $q$-Whipple formulas and the Chinese
remainder theorem for coprime polynomials.

\vskip 3mm \noindent {\it Keywords}: basic hypergeometric series;
$q$-Whipple formula; $q$-supercongruence

 \vskip 0.2cm \noindent{\it AMS
Subject Classifications:} 33D15; 11A07; 11B65

\section{Introduction}
For any complex variable $x$, define the shifted-factorial to be
\[(x)_{0}=1\quad \text{and}\quad (x)_{n}
=x(x+1)\cdots(x+n-1)\quad \text{when}\quad n\in\mathbb{N}.\]
 In 1997, Van Hamme
\cite[(H.2)]{Hamme} conjectured that
\begin{equation}\label{eq:hamme}
\sum_{k=0}^{(p-1)/2}\frac{(1/2)_k^3}{k!^3}\equiv
\begin{cases} \displaystyle -\Gamma_p(1/4)^4  \pmod{p^2}, &\text{if $p\equiv 1\pmod 4$,}\\[10pt]
 0\pmod{p^2}, &\text{if $p\equiv 3\pmod 4$.}
\end{cases}
\end{equation}
Here and throughout the paper, $p$ always denotes an odd prime and
$\Gamma_p(x)$ is the $p$-adic Gamma function. In 2016, Long and
Ramakrishna \cite[Theorem 3]{LR} gave the following extension of
\eqref{eq:hamme}:
\begin{equation}\label{eq:long}
\sum_{k=0}^{(p-1)/2}\frac{(1/2)_k^3}{k!^3}\equiv
\begin{cases} \displaystyle -\Gamma_p(1/4)^4  \pmod{p^3}, &\text{if $p\equiv 1\pmod 4$,}\\[10pt]
 \displaystyle -\frac{p^2}{16}\Gamma_p(1/4)^4\pmod{p^3}, &\text{if $p\equiv 3\pmod 4$.}
\end{cases}
\end{equation}

 For any complex numbers $x$ and $q$, define the $q$-shifted factorial
 as
 \begin{equation*}
(x;q)_{0}=1\quad\text{and}\quad
(x;q)_n=(1-x)(1-xq)\cdots(1-xq^{n-1})\quad \text{when}\quad
n\in\mathbb{N}.
 \end{equation*}
For simplicity, we also adopt the compact notation
\begin{equation*}
(x_1,x_2,\dots,x_m;q)_{n}=(x_1;q)_{n}(x_2;q)_{n}\cdots(x_m;q)_{n}.
 \end{equation*}
Following Gasper and Rahman \cite{Gasper}, define the basic
hypergeometric series $_{r+1}\phi_{r}$ by
$$
_{r+1}\phi_{r}\left[\begin{array}{c}
a_1,a_2,\ldots,a_{r+1}\\
b_1,b_2,\ldots,b_{r}
\end{array};q,\, z
\right] =\sum_{k=0}^{\infty}\frac{(a_1,a_2,\ldots, a_{r+1};q)_k}
{(q,b_1,b_2,\ldots,b_{r};q)_k}z^k.
$$
Then the $q$-Whipple formula due to Andrews \cite{Andrews} and the
$q$-Whipple formula due to Jain \cite{Jain} can be stated as
\begin{align}
& _{4}\phi_{3}\!\left[\begin{array}{cccccccc}
 q^{-n},  q^{1+n}, b,  -b \\
 -q, c,  b^2q/c
\end{array};q,\, q \right]
=q^{\binom{n+1}{2}}\frac{(b^2q^{1-n}/c, cq^{-n};q^2)_{n}}
 {(b^2q/c, c;q)_{n}}, \label{eq:q-whipple-a}
 \\[5pt]
& _{4}\phi_{3}\!\left[\begin{array}{cccccccc}
 a,  q/a, q^{-n},  -q^{-n} \\
 c, q^{1-2n}/c,  -q
 \end{array};q,\, q \right]
=\frac{(ac, cq/a;q^2)_{n}}
 {(c;q)_{2n}}. \label{eq:q-whipple-b}
\end{align}

Recently, Guo and Zudilin \cite[Theorem 2]{GuoZu2} displayed a
$q$-analogue of \eqref{eq:hamme}: for any positive odd integer $n$,
\begin{align}
&\sum_{k=0}^{(n-1)/2}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k}
\notag\\[5pt]
&\equiv
\begin{cases} \displaystyle\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}\pmod{\Phi_n(q)^2}, &\text{if $n\equiv 1\pmod 4$,}\\[10pt]
 \displaystyle 0\pmod{\Phi_n(q)^2}, &\text{if $n\equiv 3\pmod 4$.}
\end{cases}
\label{eq:guo-a}
\end{align}
 Here and throughout the
paper, $\Phi_n(q)$ stands for  the $n$-th cyclotomic polynomial in
$q$:
\begin{equation*}
\Phi_n(q)=\prod_{\substack{1\leqslant k\leqslant n\\
\gcd(k,n)=1}}(q-\zeta^k),
\end{equation*}
where $\zeta$ is an $n$-th primitive root of unity. Further, Guo
\cite[Theorem 1]{Guo-new} provided the following partial
$q$-analogue of \eqref{eq:long}: for any positive integer
$n\equiv3\pmod{4}$,
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(q,;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k}
\equiv[n]\frac{(q^3;q^4)_{(n-1)/2}}{(q^5;q^4)_{(n-1)/2}}\pmod{\Phi_n(q)^3}.
\label{eq:guo-b}
\end{align}
For more $q$-analogues of supercongruences, we refer the reader to
\cite{Guo-rima,Guo-jmaa,Guo-rama,Guo-a2,GS1,GuoZu,LP,NP,Tauraso,WY-a,Zu19}.

Motivated by the work just mentioned, we
shall establish the following result.

\begin{thm}\label{thm-a}
Let $n\equiv 1\pmod 4$ be a positive integer. Then, modulo
$\Phi_n(q)^3$,
\begin{align*}
\sum_{k=0}^{(n-1)/2}\frac{(q;q^2)_k^2(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k}
\equiv
q^{(n-1)/2}\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}\bigg\{1+2[n]^2\sum_{i=1}^{(n-1)/4}\frac{q^{4i-2}}{[4i-2]^2}\bigg\}.
\end{align*}
\end{thm}

Obviously, Theorem \ref{thm-a} is an extension of \eqref{eq:guo-a}
for $n\equiv1\pmod{4}$. Letting $n=p$ be a prime and taking $q\to 1$
in this theorem, we obtain the conclusion:
\begin{align}\label{eq:wei}
\sum_{k=0}^{(p-1)/2}\frac{(1/2)_k^3}{k!^3} \equiv
\frac{(1/2)_{(p-1)/4}^2}{\big((p-1)/4\big)!^2}\bigg\{1+\frac{p^{2}}{2}H_{(p-1)/2}^{(2)}-\frac{p^{2}}{8}H_{(p-1)/4}^{(2)}\bigg\}\pmod{p^3},
\end{align}
where the harmonic numbers of $2$-order are given by
\[H_{m}^{(2)}
  =\sum_{k=1}^m\frac{1}{k^{2}}.\]
Using the known formula (cf. \cite[Page 7]{Sun}):
\begin{align*}
&H_{(p-1)/2}^{(2)}\equiv0\pmod{p}\quad\text{with}\quad p>3,
\end{align*}
we deduce the following supercongruence from \eqref{eq:wei}.

\begin{cor}\label{cor-a}
Let $p\equiv 1\pmod 4$ be a prime. Then
\begin{align}\label{eq:wei-a}
\sum_{k=0}^{(p-1)/2}\frac{(1/2)_k^3}{k!^3} \equiv
\frac{(1/2)_{(p-1)/4}^2}{\big((p-1)/4\big)!^2}\bigg\{1-\frac{p^{2}}{8}H_{(p-1)/4}^{(2)}\bigg\}\pmod{p^3}.
\end{align}
\end{cor}

For the sake of explaining the equivalence of \eqref{eq:long} for
$p\equiv 1\pmod 4$ and \eqref{eq:wei-a}, we need to verify the
following relation.

\begin{prop}\label{prop-a}
Let $p\equiv 1\pmod 4$ be an odd prime. Then
\begin{align*}
\frac{(1/2)_{(p-1)/4}^2}{\big((p-1)/4\big)!^2}\bigg\{1-\frac{p^{2}}{8}H_{(p-1)/4}^{(2)}\bigg\}\equiv
-\Gamma_p(1/4)^4\pmod{p^3}.
\end{align*}
\end{prop}

The rest of the paper is arranged as follows. By means of the
Chinese remainder theorem for coprime polynomials, a
$q$-supercongruence modulo $(1-aq^n)(a-q^n)(b-q^n)$ will be derived
in Section 2. Then it is utilized to provide a proof of Theorem
\ref{thm-a} in the same section. Finally, the proof of Proposition
\ref{prop-a} will be given in Section 3.

\section{Proof of Theorem \ref{thm-a}}

In order to prove Theorem \ref{thm-a}, we need the following
parameter extension of it.

\begin{thm}\label{thm-b}
Let $n\equiv 1\pmod 4$ be a positive integer. Then, modulo
$(1-aq^n)(a-q^n)(b-q^n)$,
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a,q/b,-q/b;q^2)_k}{(q^2,q^2,-q^2,q^2/b^2;q^2)_k}q^{2k}
\equiv \Omega_n(a,b),\label{eq:wei-aa}
\end{align}
where
\begin{align*}
\Omega_n(a,b)&=\frac{(b-q^n)(ab-1-a^2+aq^n)}{(a-b)(1-ab)}\frac{(b/q)^{(1-n)/2}(q^2,b^2q^2;q^4)_{(n-1)/4}}{(q^4,q^4/b^2;q^4)_{(n-1)/4}}
\\[5pt]
&+\frac{(1-aq^n)(a-q^n)}{(a-b)(1-ab)}\frac{(aq^3,q^3/a;q^4)_{(n-1)/2}}{(q^2;q^2)_{n-1}}.
\end{align*}
\end{thm}

\begin{proof}
When $a=q^{-n}$ or $a=q^n$, the left-hand side of  \eqref{eq:wei-aa}
is equal to
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(q^{1-n},q^{1+n},q/b,-q/b;q^2)_k}{(q^2,q^2,-q^2,q^2/b^2;q^2)_k}q^{2k}
= {_{4}\phi_{3}}\!\left[\begin{array}{cccccccc}
 q^{1-n},  q^{1+n}, q/b,  -q/b \\
 q^2, -q^2,  q^2/b^2
\end{array};q^2,\, q^2 \right].
 \label{eq:whipple-aa}
\end{align}
According to \eqref{eq:q-whipple-a}, the right-hand side of
\eqref{eq:whipple-aa} can be expressed as
\begin{align*}
(b/q)^{(1-n)/2}\frac{(q^2,b^2q^2;q^4)_{(n-1)/4}}{(q^4,q^4/b^2;q^4)_{(n-1)/4}}.
\end{align*}
Since $(1-aq^n)$ and $(a-q^n)$ are relatively prime polynomials, we
get the following result: Modulo $(1-aq^n)(a-q^n)$,
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a,q/b,-q/b;q^2)_k}{(q^2,q^2,-q^2,q^2/b^2;q^2)_k}q^{2k}\equiv
 (b/q)^{(1-n)/2}\frac{(q^2,b^2q^2;q^4)_{(n-1)/4}}{(q^4,q^4/b^2;q^4)_{(n-1)/4}}. \label{eq:wei-bb}
\end{align}

When $b=q^{n}$, the left-hand side of  \eqref{eq:wei-aa} is equal to
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a,q^{1-n},-q^{1-n};q^2)_k}{(q^2,q^2,-q^2,q^{2-2n};q^2)_k}q^{2k}
= {_{4}\phi_{3}}\!\left[\begin{array}{cccccccc}
 aq,  q/a, q^{1-n},  -q^{1-n} \\
 q^2, -q^2,  q^{2-2n}
\end{array};q^2,\, q^2 \right].
 \label{eq:whipple-bb}
\end{align}
In terms of \eqref{eq:q-whipple-b}, the right-hand side of
\eqref{eq:whipple-bb} can be written as
\begin{align*}
\frac{(aq^3,q^3/a;q^4)_{(n-1)/2}}{(q^2;q^2)_{n-1}}.
\end{align*}
Therefore, we are led to the following conclusion: Modulo $(b-q^n)$,
\begin{align}
\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a,q/b,-q/b;q^2)_k}{(q^2,q^2,-q^2,q^2/b^2;q^2)_k}q^{2k}
 \equiv\frac{(aq^3,q^3/a;q^4)_{(n-1)/2}}{(q^2;q^2)_{n-1}}.\label{eq:wei-cc}
\end{align}

It is clear that the polynomials $(1-aq^n)(a-q^n)$ and $(b-q^n)$ are
relatively prime. Noting the $q$-congruences
\begin{align*}
&\frac{(b-q^n)(ab-1-a^2+aq^n)}{(a-b)(1-ab)}\equiv1\pmod{(1-aq^n)(a-q^n)},
\\[5pt]
&\qquad\qquad\frac{(1-aq^n)(a-q^n)}{(a-b)(1-ab)}\equiv1\pmod{(b-q^n)}
\end{align*}
and employing the Chinese remainder theorem for coprime polynomials,
we deduce Theorem \ref{thm-b} from \eqref{eq:wei-bb} and
\eqref{eq:wei-cc}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm-a}]
It is not difficult to see that
\begin{align*}
(q^2;q^2)_{n-1}&=(q^2, q^{n+1}, q^4, q^{n+3};q^4)_{(n-1)/4}
\\[5pt]
&=q^{(n-1)(3n-1)/4}(q^2, q^4, q^{2-2n}, q^{4-2n};q^4)_{(n-1)/4}\\[5pt]
&\equiv
b^{n-1}q^{(1-n^2)/4}(q^{2},q^{4},q^2/b^2,q^4/b^2;q^4)_{(n-1)/4}\pmod{(b-q^n)},
\\[5pt]
(aq^3;q^4)_{(n-1)/2}&=(aq^3;q^4)_{(n-1)/4}(aq^{n+2};q^4)_{(n-1)/4}
\\[5pt]
&\equiv(abq^{3-n};q^4)_{(n-1)/4}(abq^{2};q^4)_{(n-1)/4}
\\[5pt]
&=(-ab)^{(n-1)/4}q^{-(n-1)^2/8}(abq^2,q^2/ab;q^4)_{(n-1)/4}\pmod{(b-q^n)},
\\[5pt]
(q^3/a;q^4)_{(n-1)/2}&=(-b/a)^{(n-1)/4}q^{-(n-1)^2/8}(bq^2/a,aq^2/b;q^4)_{(n-1)/4}\pmod{(b-q^n)}.
\end{align*}
Thus, the $q$-supercongruence \eqref{eq:wei-aa} may be rewritten as
follows: Modulo $(1-aq^n)(a-q^n)(b-q^n)$,
\begin{align*}
&\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a,q/b,-q/b;q^2)_k}{(q^2,q^2,-q^2,q^2/b^2;q^2)_k}q^{2k}  \\[5pt]
&\quad\equiv
\frac{(b-q^n)(ab-1-a^2+aq^n)}{(a-b)(1-ab)}\frac{(b/q)^{(1-n)/2}(q^2,b^2q^2;q^4)_{(n-1)/4}}{(q^4,q^4/b^2;q^4)_{(n-1)/4}}
\\[5pt]
&\quad\quad+\frac{(1-aq^n)(a-q^n)}{(a-b)(1-ab)}\frac{(b/q)^{(1-n)/2}(abq^2,bq^2/a,aq^2/b,q^2/ab;q^4)_{(n-1)/4}}{(q^2,q^4,q^2/b^2,q^4/b^2;q^4)_{(n-1)/4}}.
\end{align*}
Letting $b\to 1$, we arrive at the following formula: Modulo
$\Phi_n(q)(1-aq^n)(a-q^n)$,
\begin{align}
&\sum_{k=0}^{(n-1)/2}\frac{(aq,q/a;q^2)_k(q^2;q^4)_k}{(q^2;q^2)_k^2(q^4;q^4)_k}q^{2k}
\notag\\[5pt]
&\:\:\:\equiv
q^{(n-1)/2}\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}+q^{(n-1)/2}\frac{(1-aq^n)(a-q^n)}{(1-a)^2}
\notag\\[5pt]
&\quad\:\:\times\bigg\{\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}-\frac{(aq^2,q^2/a;q^4)_{(n-1)/4}^2}{(q^2,q^4;q^4)_{(n-1)/4}^2}\bigg\}.
\label{eq:wei-dd}
\end{align}
By the L'Hospital rule, we have
\begin{align*}
&\lim_{a\to1}\frac{(1-aq^n)(a-q^n)}{(1-a)^2}\bigg\{\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}
-\frac{(aq^2,q^2/a;q^4)_{(n-1)/4}^2}{(q^2,q^4;q^4)_{(n-1)/4}^2}\bigg\}\\[5pt]
&=2[n]^2\frac{(q^2;q^4)_{(n-1)/4}^2}{(q^4;q^4)_{(n-1)/4}^2}\sum_{i=1}^{(n-1)/4}\frac{q^{4i-2}}{[4i-2]^2}.
\end{align*}
Letting $a\to1$ in \eqref{eq:wei-dd} and utilizing the above limit,
we complete the proof of  Theorem~\ref{thm-a}.

\end{proof}

\section{Proof of Proposition \ref{prop-a}}

Via the congruence due to Wang and Pan \cite[Page 6]{Wang}:
\begin{align*}
H_{(p-1)/4}^{(2)}\equiv\frac{\Gamma_p^{''}(1/4)}{\Gamma_p(1/4)}-\bigg\{\frac{\Gamma_p^{'}(1/4)}{\Gamma_p(1/4)}\bigg\}^2\pmod{p},
\end{align*}
where $\Gamma_p^{'}(x)$ and $\Gamma_p^{''}(x)$ are respectively the
first derivative and second derivative of $\Gamma_p(x)$, we obtain
\begin{align}\label{eq:wei-b}
1-\frac{p^{2}}{8}H_{(p-1)/4}^{(2)}\equiv1-\frac{p^{2}}{8}\frac{\Gamma_p^{''}(1/4)}{\Gamma_p(1/4)}+
\frac{p^{2}}{8}\bigg\{\frac{\Gamma_p^{'}(1/4)}{\Gamma_p(1/4)}\bigg\}^2\pmod{p^3}.
\end{align}
In terms of the properties of the $p$-adic Gamma function, we get
\begin{align}
\frac{(1/2)_{(p-1)/4}^2}{\big((p-1)/4\big)!^2}&=\bigg\{\frac{\Gamma_p((1+p)/4)\Gamma_p(1)}{\Gamma_p(1/2)\Gamma_p((3+p)/4)}\bigg\}^2
\notag\\[5pt]
&=\bigg\{\frac{\Gamma_p((1+p)/4)\Gamma_p((1-p)/4)}{\Gamma_p(1/2)}\bigg\}^2
\notag\\[5pt]
&\equiv-\bigg\{\Gamma_p(1/4)+\Gamma_p^{'}(1/4)\frac{p}{4}+\Gamma_p^{''}(1/4)\frac{p^2}{2\times4^2}\bigg\}^2
\notag\\[5pt]
&\quad\times\bigg\{\Gamma_p(1/4)-\Gamma_p^{'}(1/4)\frac{p}{4}+\Gamma_p^{''}(1/4)\frac{p^2}{2\times4^2}\bigg\}^2\pmod{p^3}.
\label{eq:wei-c}
\end{align}
The combination of  \eqref{eq:wei-b} and  \eqref{eq:wei-c} produces
\begin{align*}
&\frac{(1/2)_{(p-1)/4}^2}{\big((p-1)/4\big)!^2}\bigg\{1-\frac{p^{2}}{8}H_{(p-1)/4}^{(2)}\bigg\}
 \\[5pt]
 &\quad\equiv-\bigg\{\Gamma_p(1/4)+\Gamma_p^{'}(1/4)\frac{p}{4}+\Gamma_p^{''}(1/4)\frac{p^2}{2\times4^2}\bigg\}^2
\notag\\[5pt]
&\qquad\times\bigg\{\Gamma_p(1/4)-\Gamma_p^{'}(1/4)\frac{p}{4}+\Gamma_p^{''}(1/4)\frac{p^2}{2\times4^2}\bigg\}^2
\\[5pt]
&\qquad\times\bigg\{1-\frac{p^{2}}{8}\frac{\Gamma_p^{''}(1/4)}{\Gamma_p(1/4)}+
\frac{p^{2}}{8}\bigg\{\frac{\Gamma_p^{'}(1/4)}{\Gamma_p(1/4)}\bigg\}^2\bigg\}
\\[5pt]
&\quad\equiv -\Gamma_p(1/4)^4 \pmod{p^3}.
\end{align*}


TITLE: Why is elastic modulus greater than shear modulus?
QUESTION [2 upvotes]: I was looking at data for elastic modulus $E$ and shear modulus $G$, and found that $G$ is always lower than $E$. So I'm wondering what are the underlying principles that may be the cause of this.
$$G = \dfrac{T\cdot L}{J \cdot \phi}$$
where $T= \text{torque}, \quad  J = \text{polar moment of inertia}, \quad \phi = \text{angle of twist},  \quad L = \text{lever arm}$.
$$E = \dfrac{F \cdot L^3}{4bd^3 \cdot v}$$
Where $F = \text{force}, \quad L = \text{length of beam}, \quad v =\text{deflection}, \quad b = \text{width}, \quad d =\text{depth}$

REPLY [4 votes]: Strictly speaking, Young's modulus is not always greater than the shear modulus, but it does tend to work out that way. You can see the reason why if you look at the relation between the two quantities (and Poisson's ratio).
$$
G = \frac{E}{2(1+\nu)}
$$
Combined with the knowledge that $\nu$ can be anywhere in the range $(-1, \frac{1}{2})$, one can see that G can be greater than E for $\nu < -1/2$. That being said, materials with such a negative Poisson's ratio are extremely uncommon, and it is safe to assume that the shear modulus is less than half of Young's modulus.
For an explanation of why the Poisson's ratio must fall within the above range, I invite you to check out some of my previous answers.
High-level: Range of poissons ratio
Detailed explanation: Limits of Poisson's ratio in isotropic solid


TITLE: When Is Refinement of Open Covers a Partial Ordering on the Set of Open Covers?
QUESTION [2 upvotes]: Rick Miranda's Algebraic Curves and Riemann Surfaces has a problem in the $\check{C}ech$ $Cohomology$ section (Problem IX.3.C) which says:
Show that refinement gives a partial ordering on the set of all open coverings of a space.
I proved the transitivity and reflexivity properties, but I don't know what to do for a proof of antisymmetry. Letting $\mathcal{U},\mathcal{V}$ be two covers of a topological space $X$ such that $\mathcal{U}\prec\mathcal{V}$ and $\mathcal{V}\prec\mathcal{U}$ (i.e. $\mathcal{U}$ is a refinement of $\mathcal{V}$ and $\mathcal{V}$ is a refinement of $\mathcal{U}$), I suppose that I have a set/element $U\in\mathcal{U}$ and would like to show that $U\in\mathcal{V}$ as well to show set containment $\mathcal{U}\subset\mathcal{V}$ (and then to show the other direction and thus the set equality of the open covers by a similar argument). But... I'm stuck. By the refinement conditions I have that there exists $V\in\mathcal{V}$ such that $U\subset V$ and there exists similarly a $U'\in\mathcal{U}$ such that $V\subset U'$. But a) I don't know how to prove that $V = U$ or anything like that from here, and b) the Wiki for "Open Cover" explicitly states that refinement is a preorder (rather than a partial ordering) so that antisymmetry is probably the condition which fails.
I'm thinking there is just a slight typo in the text but wanted to ask the community. Do note that I am working in a Riemann surfaces book, so perhaps some of the conditions on Riemann surfaces like Hausdorff, second countable, etc. could come into play here to make refinement into a partial ordering. (However, the problem doesn't say we have a Riemann surface $X$ but rather just a space $X$, so I'm thinking he meant just a topological space in general considering we're in the chapter on sheaves which just require a topological space as their domain.)

REPLY [2 votes]: It's just a mistake. The family of open covers is pre-ordered, but except in trivial cases not partially ordered by the refinement relation.
Any two open covers containing the whole space are refinements of each other.
For a less trivial example, consider the two covers of $\mathbb{R}$ by the families of bounded open intervals with rational endpoints, where in the first family we require the denominator of both endpoints (in the reduced form) to be odd, and in the second family the denominator of both end points shall be even. These are mutual refinements, but they don't have a set in common.


TITLE: Modulo fix-point by induction
QUESTION [0 upvotes]: Suppose we have a function that iterates an another function from some starting point. The function is defined as follows: $$iterate\ f\ x\ (n + 1) = f\ (iterate\ f\ x\ n)$$
Of course, with no iterations left we're constant. $$iterate\ f\ x\ 0 = x$$
Now suppose we find a fixed point after $m$ iterations. $$iterate\ f\ x\ m = x$$
And we would like to state that doing $n$ iterations with a fixpoint after $m$ iterations is the same as doing $n\ mod\ m$ iterations. $$iterate\ f\ x\ m = x \rightarrow \forall n, iterate\ f\ x\ n = iterate\ f\ x\ (n\ mod\ m)$$
I was trying to prove this by induction on $n$, with the inductive hypothesis: $$iterate\ f\ x\ k = iterate\ f\ x\ (k\ mod\ m)$$
However, I am stuck at transforming $$iterate\ f\ x\ (k + 1) = iterate\ f\ x\  ((k + 1)\ mod\  m)$$ to anything more usable. I have tried playing with the lhs, first using the definition of $iterate$, getting to lhs: $$f\ (itereate\ f\ x\ k)$$where I can use the inductive hypothesis to get: $$f\ (iterate\ f\ x\ (k\ mod\ m)) = iterate\ f\ x\  ((k + 1)\ mod\ m)$$
However, I still find the right hand side incredibly clumsy. I also tried proceeding by trichotomy $n < m \vee n = m \vee n > m$ where it's easy to show this holds for $n = m$ and $n < m$ from the definition of $iterate$, but I am stuck on the last case where we need to roll over the modulo.
Is induction not a good way to proceed here? If so, why? What am I missing here?

REPLY [0 votes]: Here's what I'd do.  I'll use the notation $f^{\circ n}(x)$ for your $iterate \; f \; x\; n$.
So you want to prove $f^{\circ n}(x) = x \Rightarrow f^{\circ m}(x) = f^{\circ m\,\mathrm{mod}(n)}(x)$, for $m \geq 0$.
If $0 \leq m < n$, this is obvious since then $m\,\mathrm{mod}n = m$ and obviously $f^{\circ m}(x) = f^{\circ m}(x)$.  Also, the case where $m = n$ is the hypothesis.  Hence you have $n+1$ base cases for $m = 0, 1, \dots, n$.
For the inductive case, let's prove $f^{\circ m+1}(x) = f^{\circ (m+1)\,\mathrm{mod}(n)}(x)$.  Let's separate when $m+1\equiv 0 \mod n$ or not.
If $m+1\not\equiv0\mod n$, then $m+1\,\mathrm{mod}(n) = (m\,\mathrm{mod}(n)) + 1$.  In that case
$$f^{\circ m+1}(x) = f\left(f^{\circ m}(x)\right) = f\left(f^{\circ m\,\mathrm{mod}(n)}\right)(x) = f^{\circ (m\,\mathrm{mod}(n))+1}(x) = f^{\circ (m+1)\,\mathrm{mod}(n)}(x).$$
If $m + 1 \equiv 0 \mod n$, then $m = k \cdot n$ for some $k \geq 1$.  Therefore
$$f^{\circ m+1}(x) = f^{\circ kn}(x) = f^{\circ n}\left(f^{\circ (k-1)n}(x)\right) = f^{\circ n}(x) = x =f^{\circ m\,\mathrm{mod}(n)}(x).$$
PS: Note that $f^{\circ0}$ is not constant, but the identity.


TITLE: Why we can't take vertical cylinder as Gaussian surface to calculate the electric field of an uniformly charged infinite sheet?
QUESTION [0 upvotes]: To calculate the electric field due to an uniformly charged infinite sheet, we take a horizontal cylinder as Gaussian surface. But why we can't take a vertical cylinder to do the same?

REPLY [4 votes]: When constructing a Gaussian surface, our main goal is to take advantage of the inherent symmetry of the charge distribution.
For an infinite sheet aligned in the $x-z$ plane, there is symmetry along the $x$ and $z$ directions. In other words, we cannot expect any physical characteristics of the system to depend on $x$ or $z$. This includes the $\mathbf{E}$ field generated by the sheet. This symmetry, along with the fact that $\mathbf{E}$ is conservative, allows us to deduce that $\mathbf{E}$ can only have a $y$ component.

Now, the Gaussian surface must take advantage of this symmetry we've identified. The choice of a horizontal cylinder does this as follows.

The cylinder is constructed such that the rounded sides are parallel to $\mathbf{E}$, while the caps are perpendicular to $\mathbf{E}$. This heavily simplifies the flux through the whole cylindrical surface, which can just be written as the magnitude of $\mathbf{E}$ multiplied by the areas of the caps.
$$\oint\mathbf{E}\cdot\mathbf{ds}=2EA$$
In fact, the surface does not need to be a cylinder. It could simply be a surface that has parallel sides and perpendicular caps.

A vertical cylinder, on the other hand, would not take advantage of this symmetry.

This is because $\mathbf{E}\cdot\mathbf{ds}$ would be different along the surface, and you would need to compute the surface integral without any simplifications.
P.S. I have not indicated the $\mathbf{E}$ fields on the left side of the sheets out of convenience. They still exist.


TITLE: Proof that $\frac{d}{dx}\sin(ax+b) = a\cos(ax+b)$ by first principles
QUESTION [2 upvotes]: I have been trying to prove that the deriative of  $\sin(ax+b)$ is equal to $\:a\cos(ax+b)$. I tried to use the first principle to prove for $\sin(2x)$ and it has worked using the sum to product identity.
I tried to use the sum to product identity to prove a general term of $\frac{\mathrm dy}{\mathrm dx}$ by first principle with the sum to product rule but it hasn't worked.
How can this be proven by first principle? Thanks in advance :)

REPLY [1 votes]: Let $f(x) = \sin(ax+b)$.
$$\begin{align*} f'(x) &= \lim_{h\to 0}\frac{\sin(a(x+h)+b) - \sin(ax+b)}{h}\\
&= \lim_{h\to 0}\frac{\sin(ax+b + ah) - \sin(ax+b)}{h}\\
&= \lim_{h\to 0}\frac{2\cos\left(ax+b + \frac{ah}{2}\right)\sin\left(\frac{ah}{2}\right)}{h}\\
&= \lim_{h\to 0}\cos\left(ax+b + \frac{ah}{2}\right)\cdot \lim_{h\to 0}\frac{2a\sin\left(\frac{ah}{2}\right)}{ah}\\
&= \cos(ax+b)\cdot a\cdot \lim_{ah/2\to 0}\frac{\sin\left(\frac{ah}{2}\right)}{ah/2}\\
&= a\cos(ax+b)
\end{align*}$$


TITLE: Find $F'(x)$ of $F(x)=\int_{\pi e^x}^{\cos(x)} \cos (t) dt$
QUESTION [0 upvotes]: Let $F: \mathbb{R} \to \mathbb{R}$ be a function, $F(x)=\int_{\pi e^x}^{\cos(x)} \cos (t) dt$.
Find $F'(x)$.
I can't get to the correct answer which is $\pi (-e^x) \cos(\pi e^x) - \sin(x) \cos(\cos(x))$.
I used the chain rule, first with the upper bound:
$\cos(\cos(x))(-\sin(x))$.
And then with the lower bound:
$\cos(\pi e^x)(e^x)$.
So $F'(x)=\cos(\cos(x))(-\sin(x))-\cos(\pi e^x)(e^x)$?
Am I missing something?

REPLY [1 votes]: for some function:
$$F(x)=\int_{a(x)}^{b(x)}f(x,t)dt$$
$$F'(x)=f(x,b(x)).\frac{d}{dx}[b(x)]-f(x,a(x)).\frac{d}{dx}[a(x)]+\int_{a(x)}^{b(x)}\frac{\partial}{\partial_x}f(x,t)dt$$
so for our function:
$$F(x)=\int_{\pi e^x}^{\cos(x)}\cos(t)dt$$
$$F'(x)=-\cos(\cos(x))\sin(x)-\pi\cos(\pi e^x)e^x+\int_{\pi e^x}^{\cos(x)}0.dt$$
so in your case it all appears correct apart from a missing $\pi$ in the second term, due to the fact that $D[\pi e^x]=\pi e^x$


TITLE: Evaluating Specific Heat at low and high temperature limit
QUESTION [1 upvotes]: Solving a problem in statistical mechanics, I obtained the following expression:
$$C_v=36Nk_B\frac{T^3}{\theta_D^3}\int_0^{\theta_D/T}\frac{x^3}{e^{x}-1}dx-\frac{9Nk_B\theta_D}{T(e^{\theta_D/T}-1)}$$
where $k_B$ is the Boltzmann constant, $\theta_D$ is the debye temperature and N the number of particles.
I wish to evaluate at High and low temperature limits, but I am running intro trouble.
For $T>>\theta_D$, the upper limit of the integral becomes 0, and thus the first term is 0. Then, the second term is divided by T, but T -> infinity, so it is 0.
How can this be? What am I doing wrong here?
In the second case, for $T<<\theta_D$, the upper limit of the first integral goes to infinity, and thus the contribution of the first term is of the order $T^3$. However, the second therm is divided by T, which is almost 0 and an exponential that is infinite. What can I do?
I'm sure I am doing something very wrong, but I cannot figure out what

REPLY [0 votes]: If upper integration limit is small you can expand the integrand around x=0 and keep only the first order terms (x does not go too far from zero). You will end up with an easy to do integral and the result will be some power of $ (\theta_D/T) $ . For the second term, you can use the same approximation so the denominator will be just $ \theta_D $
For very high temperatures you could neglect the second term, I believe. It goes like $ x e^{-x} $ where $x =\theta_D/T $ when x goes to zero. Or you can expand it around x=0 and keep the linear term only.


text:
\begin{document}
\title[Duality relations for hypergeometric series]{Duality relations for hypergeometric series}

\author[Frits Beukers]{Frits Beukers}
\address{Utrecht University, Department of Mathematics, P.O. Box 80.010, 3508 TA Utrecht, Netherlands}
\email{f.beukers@uu.nl}
\urladdr{http://www.staff.science.uu.nl/{\textasciitilde}beuke106/}

\author[Fr\'ed\'eric Jouhet]{Fr\'ed\'eric Jouhet}
\address{Institut Camille Jordan, Universit\'e Claude Bernard Lyon 1,
69622 Villeurbanne Cedex, France}
\email{jouhet@math.univ-lyon1.fr}
\urladdr{http://math.univ-lyon1.fr/{\textasciitilde}jouhet/}

\thanks{}

\date{}

\subjclass[2010]{}

\keywords{hypergeometric series, basic hypergeometric series,
difference equations, difference modules, differential equations, D-modules.}

\begin{abstract}
We  explicitly give the relations between the hypergeometric solutions of the general
hypergeometric equation and their duals, as well as similar relations for q-hypergeometric
equations. They form a family of very general identities for hypergeometric series. Although
they were foreseen already by N. M. Bailey in the 1930's on analytic grounds, we give a purely
algebraic treatment based on general principles in general differential and difference modules.
\end{abstract}

\maketitle

\section{Introduction and notations}

Hypergeometric series in one variable were introduced by Euler and extensively studied
by Gauss~\cite[pp. 123, 207]{Ga1866} as solutions of the so-called hypergeometric differential
equation.
Later Thomae~\cite{Th1870} generalized the Gauss hypergeometric functions to higher order versions.
In this paper a one variable hypergeometric function of order $r$ depends on $2r$ complex parameters
$a_1,\ldots,a_r,b_1,\ldots,b_r$ where we take $b_r=1$. The
hypergeometric function is given by a power series in $z$ of the form
$$\sum_{n=0}^{\infty}{(a_1)_n\cdots(a_r)_n\over (b_1)_n\cdots(b_{r-1})_nn!}\ z^n.$$
Here $(x)_n$ stands for the Pochhammer symbol (or rising factorial) defined by $(x)_0:=1$ and
$$(x)_n:=x(x+1)\cdots(x+n-1),\quad n\ge1.$$
In order for the coefficients to exist we must assume that $b_i\not\in \bbbz_{\le0}$ for
all $i$. In that case the radius of convergence of the series is $1$, unless $a_i\in\bbbz_{\le0}$
for some $i$, in which case we have a polynomial. The corresponding function is denoted by
$${}_rF_{r-1}\!\left(\begin{matrix}a_1,\ldots,a_r\\
b_1,\ldots,b_{r-1}\end{matrix};z\right).$$
It satisfies the $r$-th order linear differential equation
\begin{equation}\label{eq:hypergeq}
(\theta+b_1-1)\cdots(\theta+b_r-1) f=z(\theta+a_1)\cdots(\theta+a_r)f,
\end{equation}
where $\theta=z{d\over dz}$. The singularities of this equation are given by $z=0,1,\infty$.
Around $z=0$ one can easily give a basis of solutions if one assumes that the $b_i$ (including $b_r=1$) are
distinct modulo $\bbbz$. They read
$$f_i(z):=z^{1-b_i}\,{}_rF_{r-1}\!\left(\begin{matrix}a_1+1-b_i,\ldots,a_r+1-b_i\\
b_1+1-b_i,\ldots,\vee,
\ldots,b_r+1-b_i\end{matrix};z\right),\qquad 1\leq i\leq r,$$
where $\vee$ denotes deletion of the term with index $i$.
Note that in this notation $f_r(z)$ is the hypergeometric function we started with.
For details concerning irreducibility, rigidity and monodromy of the equation we refer to the paper
\cite{BH89}.

Basic hypergeometric series, or q-hypergeometric series, have appeared in the eighteenth
century when Euler wrote the generating function for integer partitions as an infinite product.
However, the  systematic study of basic hypergeometric series and equations appeared quite later in Heine's paper~\cite{He}. We refer the reader to the classical books of  Gasper and Rahman~\cite{GR} and
Slater~\cite{Sl} for additional historical information.  The generalized basic hypergeometric equations
are quantizations (or $q$-analogues) of the previous generalized hypergeometric equations.

We recall some standard notation for basic (or q-) hypergeometric series
(see for instance~\cite{GR} for a
comprehensive study of their theory).
Let $q$ be a fixed complex parameter (the ``base'') with $0<|q|<1$.
The $q$-shifted factorial is defined for any complex
parameter $a$ and any non-negative integer $n$ by $(a;q)_0:=1$ and
$$
(a;q)_n:=(1-a)(1-aq)\cdots(1-aq^{n-1}),\qquad n\geq1.
$$

Further,  for a positive integer $r$ and complex  parameters $a_1,\dots,a_r$,  $b_1,\dots,b_{r}$ with $b_r=q$, recall the definition of the corresponding basic hypergeometric series,
$$
{}_r\phi_{r-1}\!\left[\begin{matrix}a_1,\dots,a_r\\
b_1,\dots,b_{r}\end{matrix};q,z\right]:=
\sum_{n=0}^\infty\frac{(a_1;q)_n\cdots(a_r;q)_n}{(b_1;q)_n\cdots(b_r;q)_n}z^n.
$$

Again, in order for the coefficients to exist we must assume that $b_i\not\in q^{\bbbz_{\le0}}$ for
all $i$. In that case the radius of convergence of the series is $1$, unless $a_i\in q^{\bbbz_{\le0}}$
for some $i$, in which case we have a polynomial.
The $q$-hypergeometric series is a solution of the following $r$th order difference equation
\begin{equation}\label{eq:qhyperg}
(1-b_1\sigma_q/q)\cdots(1-b_{r}\sigma_q/q)f=z(1-a_1\sigma_q)\cdots(1-a_r\sigma_q)f,
\end{equation}
where $\sigma_q$ is the automorphism on the field of formal Laurent series $\bbbc((z))$ given by
$\sigma_q(f)(z):=f(qz)$. Note that replacing $a_i$ by $q^{a_i}$ and $b_i$ by $q^{b_i}$ for all $i$ and letting $q$ tends to $1$, the basic hypergeometric series tends to the previous hypergeometric series. Moreover, performing these replacements and dividing both sides of~\eqref{eq:qhyperg} by $(1-q^r)$ before letting $q$ tends to $1$, one gets back the hypergeometric differential equation~\eqref{eq:hypergeq}.

Suppose that no quotient of two different $b_i$'s (including $b_r=q$) is an integral power of $q$. Define for $i=1,\dots,r$ the
formal solution $z^{1-\beta_i}$ of $\sigma_q(f)=(q/b_i)f$. This means that $\beta_i$ can
be thought of as $\log(b_i)/\log(q)$.
It is then easy to show~\cite{Ja} that a basis of solutions of~\eqref{eq:qhyperg} reads
$$f_i(q;z):=z^{1-\beta_i}\,{}_r\phi_{r-1}\!\left[
\begin{matrix}qa_1/b_i,\ldots,qa_r/b_i\\qb_1/b_i,\ldots,\vee,
\ldots,qb_r/b_i\end{matrix};q,z\right],\qquad 1\leq i\leq r.$$

For the general theory of irreducibility and (strong) rigidity of the $q$-difference
equation~\eqref{eq:qhyperg}, we refer the reader to~\cite{Ro11, Ro14}.

In this paper we shall also consider the duals of the (basic) hypergeometric equations and
their solutions. For the differential case these solutions read
$$g_i(z):=z^{b_i-1}\,{}_rF_{r-1}\!\left(\begin{matrix}b_i-a_1,\ldots,b_i-a_r\\
b_i+1-b_1,\ldots,\vee,
\ldots,b_1+1-b_r\end{matrix};z\right),\qquad 1\leq i\leq r.$$

For the basic hypergeometric case, we introduce
$$g_i(q;z):=z^{\beta_i-1}\,{}_r\phi_{r-1}\!\left[
\begin{matrix}b_i/a_1,\dots,b_i/a_r\\qb_i/b_1,\dots,\lor,\dots,qb_i/b_r\end{matrix};q;\frac{a_1\dots a_rzq^{r-2}}{b_1\dots b_{r-1}}\right],\qquad 1\leq i\leq r.$$

The main goal of the present paper is to exhibit an explicit form of the duality relations
between the $f_i(z),g_i(z)$ and of the $q$-versions $f_i(q;z),g_i(q;z)$. Here is our first theorem.

\begin{theorem}\label{duality}
With the previous notations, set
$$
c_i:=\prod_{j=1\atop j\neq i}^r\frac{1}{b_j-b_i},
$$
for $i=1,\dots,r$, where $r\ge2$. Then we have for any $k,l=0,1,\ldots,r-1$,
$$\sum_{i=1}^rc_i\theta^k(f_i)(z)\,\theta^l(g_i)(z)=:M_{kl}\in H(z),$$
where $H$ is the field generated over $\bbbq$ by the $a_i,b_j$.

In particular we have
$M_{kl}=0$ if $k+l\le r-2$ and $M_{kl}={(-1)^{k}\over 1-z}$ if $k+l=r-1$.
\end{theorem}

As an example, the case $k=l=0,r=2$ yields
\begin{multline*}
{}_2F_1\left(\begin{matrix}a_1,a_2\\
b_1\end{matrix};z\right)
{}_2F_1\left(\begin{matrix}1-a_1,1-a_2\\
2-b_1\end{matrix};z\right)\\
=
{}_2F_1\left(\begin{matrix}a_1+1-b_1,a_2+1-b_1\\
2-b_1\end{matrix};z\right)
{}_2F_1\left(\begin{matrix}b_1-a_1,b_1-a_2\\
b_1\end{matrix};z\right).
\end{multline*}
This follows directly from Euler's standard identity
$${}_2F_1\left(\begin{matrix}a,b\\
c\end{matrix};z\right)=
(1-z)^{c-a-b}
{}_2F_1\left(\begin{matrix}c-a,c-b\\
c\end{matrix};z\right).$$
The case $k=l=0,r=3$ can be found in the paper \cite{Da32} by Darling,
published in 1932.
Very soon after, Bailey~\cite{Ba33} in 1933 found another method to prove this identity
and its generalizations. It is based on explicit calculation of the coefficients
of the sum $\sum_ic_if_ig_i$, which are then shown to vanish by a contour integration argument.
The approach we take in this paper is entirely different.
It is purely algebraic and based on the general relations that
arise from differential hypergeometric modules and their dual modules.

To give an idea of a complete set of relations we reproduce the matrix $M:=(M_{kl})_{0\leq k,l\leq r-1}$
for the case $r=2$ here,
$$M=\begin{pmatrix} 0 & {1\over1-z}\\
-{1\over 1-z} & {-2+b_1+b_2+(1-a_1-a_2)z\over (1-z)^2}
\end{pmatrix}
$$
and for the case $r=3$,
$$M=\begin{pmatrix} 0 & 0 & {1\over1-z}\\ 0 & -{1\over1-z} &
{3-e_1(\boldb)+(-2+e_1(\bolda))z\over (1-z)^2}\\
{1\over1-z} &  {-3+e_1(\boldb)+(1-e_1(\bolda))z\over (1-z)^2}& {A+Bz+Cz^2\over(1-z)^3}\\
\end{pmatrix},$$
where
\begin{eqnarray*}
A&=&(e_1(\boldb)-1)^2-e_2(\boldb)+2,\\
B&=&e_2(\boldb)-2(e_1(\bolda)-1/2)(e_1(\boldb)-5/2)+e_2(\bolda)-5/2,\\
C&=&(e_1(\bolda)-1)^2-e_2(\bolda),
\end{eqnarray*}
and
$$e_1(\v x)=x_1+x_2+x_3,\quad e_2(\v x)=x_1x_2+x_1x_3+x_2x_3$$
are elementary symmetric functions.

Such matrices can be computed with the formula $M=\Psi^{-1}$
coming from equation~\eqref{matrixduality}.
There is an analogous result in the q-hypergeometric case.

\begin{theorem}\label{qduality}
With the previous notations, set
$$
c_i(q):=qb_i^{r-2}\prod_{j=1\atop j\neq i}^r\frac{1}{b_i-b_j},
$$
for $i=1,\dots,r$ where $r\ge2$. Then we have for any $k,l=0,1,\ldots,r-1$,
$$\sum_{i=1}^rc_i(q)\,f_i(q;zq^k)\,g_i(q;zq^l)=:M_{kl}(q)\in H_q(z),$$
where $H_q$ is the field generated by $q,a_i,b_j$ over $\bbbq$.

In particular we have $M_{kl}(q)=0$ if $l\le k\le r-1$, except when $(k,l)=(r-1,0)$. In the latter
case,
$$M_{r-1,0}(q)=\frac{(-1)^{r+1}q^r}{b_1\dots b_{r}-a_1\dots a_rq^{r-1}z}.$$
In addition we have
$$M_{k,k+1}(q)={1\over 1-q^kz},\quad k=0,1,\ldots,r-2.$$
\end{theorem}

For $k=l=0,r=2$ we obtain the relation
\begin{multline*}
{}_2\phi_1\!\left[
\begin{matrix}a_1,a_2\\b_1\end{matrix};q,z\right]
{}_2\phi_1\!\left[
\begin{matrix}q/a_1,q/a_2\\q^2/b_1\end{matrix};q;\frac{a_1a_2z}{b_1}\right]\\
={}_2\phi_1\!\left[
\begin{matrix}qa_1/b_1,qa_2/b_1\\q^2/b_1\end{matrix};q,z\right]
{}_2\phi_1\!\left[
\begin{matrix}b_1/a_1,b_1/a_2\\b_1\end{matrix};q;\frac{a_1a_2z}{b_1}\right],
\end{multline*}

where we explicitly set $b_2=q$.
Note that this identity can be proved by applying Heine's transformation formula for a ${}_2\phi_1$ series~\cite[Appendix III, (III.3)]{GR}. As communicated to us by S.~O.~Warnaar,
extracting coefficients of $z^n$ in this identity yields a special case of Sears' transformation
formula for a terminating balanced ${}_4\phi_3$ series~\cite[Appendix III, (III.16)]{GR}.
In~\cite{Ba33}, Bailey explained how to prove Theorem~\ref{qduality}
for $k=l=0$ and $r=3$ by using contour integration techniques. Moreover, the case $k=r-2, l=0$ and general $r$ is both proved in Sears' paper~\cite{Se51} by recursion and  classical techniques of $q$-series, and in Shukla's article~\cite{Sh57} by use of previous work due to Slater on bilateral basic hypergeometric series.

We did not find in the literature identities from Theorem~\ref{qduality} in full generality.
As in the differential case, our proof is purely algebraic and based on relations that
exist between q-hypergeometric difference modules and their duals.

To give an idea of a complete set of relations we reproduce the matrix $M_q:=(M_{kl}(q))_{0\leq k,l\leq r-1}$
for the case $r=2$ here,
$$M_q=\begin{pmatrix}0 & {1\over 1-z}\\ {-q^2\over b_1b_2-a_1a_2qz} & 0
\end{pmatrix},$$
and the case $r=3$,
$$M_q=\begin{pmatrix}0 & {1\over 1-z} & {e_1(\v b)-e_1(\v a)qz\over (1-z)(1-qz)}\\
0 & 0 & {1\over 1-qz}\\
{q^3\over b_1b_2b_3-a_1a_2a_3q^2z} & 0 & 0
\end{pmatrix}
$$

Such matrices can be computed with the formula $M_q=\Psi_q^{-1}$
coming from equation~\eqref{matrixqduality}.

This paper is organized as follows. In Section~\ref{Dmodules},
we will recall the notion of D-modules associated to general linear differential equations and their corresponding duals.
The proof of Theorem \ref{duality} will then be given in Section~\ref{hypergeometric},
as a consequence of these general considerations. In a similar vein we recall the notion of difference modules associated to
general linear difference equations and their duals in Section~\ref{deltamodules}.
Theorem \ref{qduality} will then be proven
in Section~\ref{qhypergeometric}.

\section{D-modules and duality}\label{Dmodules}

Let $K$ be a differential field with a derivation
$D$. Let $K_0$ be the field of constants.

\begin{definition}
A D-module $M$ over $K$ is a $K$-vector space with a map
$\nabla:M\to M$ such that
\begin{enumerate}
\item $\nabla(m_1+m_2)=\nabla(m_1)+\nabla(m_2)$ for all $m_1,m_2\in M$,
\item $\nabla(fm)=D(f)m+f\nabla(m)$ for all $f\in K$ and $m\in M$.
\end{enumerate}
\end{definition}

In what follows we shall abbreviate $\nabla$ by $D$ again.
The major example is the D-module associated to a linear differential operator
$L\in K[D]$. Consider the left ideal $(L)$ generated by $L$.
The action of $D$ from the left on the quotient ring $K[D]/(L)$ turns it
into a D-module.

\begin{definition}\label{}
Let $M,M'$ be $D$-modules over $K$.

A $K$-linear map $\phi:M\to M'$ is called a $D$-homomorphism if
$D\circ \phi=\phi\circ D$. If $\phi$ is a $K$-vector space isomorphism
we call $\phi$ a $D$-isomorphism.
\smallskip

A $D$-module $M$ is called irreducible if $\{\v 0\}$ and $M$ are the
only D-submodules.
\smallskip

The tensor product $M\otimes M'$ can be given a $D$-module structure
via
$$D(m\otimes m')=D(m)\otimes m'+m\otimes D(m'),$$
for all $m\in M,m'\in M'$.

The dual $M^*$ of the vectorspace $M$ can be given a $D$-module structure
via
$$\langle D(m^*),m\rangle=D(\langle m^*,m\rangle)-\langle m^*,D(m)\rangle$$
for every $m^*\in M^*$ and $m\in M$. Here $\langle m^*,m\rangle$ denotes the evaluation
of $m^*$ at the point $m$.
\end{definition}

\begin{proposition}\label{explicitdual}
Let $M$ be a finite dimensional D-module over $K$. Let $m_1,\ldots,m_r$ be a
$K$-basis and let $A_{ij}\in K$ be such that
$$D(m_i)=\sum_{j=1}^rA_{ij}m_j$$
for all $i$. Define the dual basis $m^*_i$ by $\langle m^*_i,m_j\rangle=\delta_{ij}$ for
$i,j=1,\ldots,r$, where $\delta_{ij}$ is the Kronecker delta. Then,
\begin{enumerate}
\item $D(m^*_i)=-\sum_{j=1}^{r}A_{ji}m^*_j$ for $i=1,\ldots,r$,
\item $\omega=\sum_{i=1}^{r}m^*_i\otimes m_i\in M^*\otimes M$ does not depend
on the choice of basis $m_i$ and $D(\omega)=0$.
\end{enumerate}
\end{proposition}

\begin{proof}
To see the first assertion, evaluate the form $m_k^*$ on the equality
$D(m_i)=\sum_{j=1}^rA_{ij}m_j$. We get
$$\langle m_k^*,D(m_i)\rangle=\sum_{j=1}^r A_{ij}\langle m_k^*,m_j\rangle.$$
By definition of $D(m_k^*)$ and $\langle m_k^*,m_j\rangle=\delta_{jk}$ we get
$$-\langle D(m_k^*),m_i\rangle=A_{ik}.$$
From this it follows that $D(m_k^*)=-\sum_{i=1}^rA_{ik}m_i^*$.

The second assertion now follows by straightforward computation.
\end{proof}

In the following proposition we consider elements $\Omega\in N\otimes M$
where $N,M$ are D-modules of the same rank $r$. We say that $\Omega$ is
 non-degenerate if it cannot be written in the form $\sum_{i=1}^sn_i\otimes m_i$
with $m_i\in M,n_i\in N$ and $s<r$.

\begin{proposition}\label{correspondence}
Let $M,N$ be finite dimensional D-modules over $K$ of rank $r$.
Let $m_1,\ldots,m_r$ be a basis of $M$ and $m_1^*,\ldots,m_r^*$ the corresponding
dual basis of $M^*$.

Then the D-homomorphisms $\phi:M^*\to N$ are in one-to-one correspondence
with tensors $\Omega\in N\otimes M$ such that $D(\Omega)=0$.

The correspondence is given by $\phi$ goes to $\Omega=\sum_{i=1}^r
\phi(m_i^*)\otimes m_i$.

Conversely, the tensor $\Omega=\sum_{i=1}^rn_i\otimes m_i$ with
$D(\Omega)=0$ corresponds to the $K$-linear map generated by
$m_i^*\to n_i$ $(i=1,\ldots,r)$, which is a D-homomorphism.

Finally, $\phi$ is an isomorphism if and only if $\Omega$ is
non-degenerate.
\end{proposition}

\begin{proof}
Suppose we are given a $K$-linear map $\phi:M^*\to N$.
It is determined by its values $n_i:=\phi(m_i^*)$
for $i=1,\ldots,r$. We write the D-homomorphism condition in two ways.

First of all, by definition one should have $D(n_i)=D(\phi(m_i^*))=
\phi(D(m_i^*))$ for all $i$. Using Proposition \ref{explicitdual} this is the
same as
$$D(n_i)=\phi\left(-\sum_{j=1}^rA_{ji}m_j^*\right)=-\sum_{j=1}^rA_{ji}n_j, \quad i=1,\ldots,r.$$
Now define $\Omega=\sum_{i=1}^r n_i\otimes m_i$ and subsitute it into $D(\Omega)=0$.
We get
$$D(\Omega)=\sum_{i=1}^r D(n_i)\otimes m_i+n_i\otimes D(m_i)=0.$$
Using $D(m_i)=\sum_{j=1}^rA_{ij}m_j$ rewrite this as
$$\sum_{j=1}^r \left(D(n_j)+\sum_{i=1}^rA_{ij}n_i\right)\otimes m_j=0.$$
This is equivalent to $D(n_j)+\sum_{i=1}^rA_{ij}n_i=0$ for $j=1,\ldots,r$.
Here we recognize our formulation of the D-homomorphism condition.

Note that the tensor $\sum_{i=1}^rn_i\otimes m_i$ is non-degenerate if and
only if $n_1,\ldots,n_r$ are linearly independent. But this is equivalent to
$\phi$ being an isomorphism.
\end{proof}

The next result identifies (up to D-isomorphism) the dual differential operator
associated with any fixed differential operator $L\in K[D]$.

\begin{theorem}\label{theorem:dual}
Consider the $r$-th order differential operator
$$ L=A_rD^r+A_{r-1}D^{r-1}+\cdots+A_1D+A_0,$$
with $A_i\in K$ for all $i$. Then the dual of $K[D]/(L)$ is D-isomorphic
to $K[D]/(L^*)$ where
$$L^*=(-D)^r\circ A_r+(-D)^{r-1}\circ A_{r-1}+\cdots+(-D)\circ A_1+A_0.$$
\end{theorem}

\begin{proof}
According to Proposition \ref{correspondence} we need to find an
element $\Omega\in (K[D]/(L^*))\otimes
(K[D]/(L))$ such that $D(\Omega)=0$. We propose to take
$$\Omega=\sum_{k=1}^n[A_k,1]_{k-1}$$
where we define
$$[u,v]_m=u\otimes(D^m\circ v)-(D\circ u)\otimes(D^{m-1}\circ v) +
(D^2\circ u)\otimes (D^{m-2}\circ v)+\cdots+(-1)^m (D^m\circ u)\otimes v.$$
Note that
$$D([u,v]_m)=u\otimes(D^{m+1}\circ v)+(-1)^m (D^{m+1}\circ u)\otimes v.$$
Using this property we can show that $D(\Omega)=0$. The fact that $\Omega$
is non-degenerate follows from the invertibility of the matrix $\Psi$ defined
below.
\end{proof}

We get the following immediate consequence.
\begin{corollary}\label{corollary:omegafixed}
Let ${\mathcal K}$ be a differential extension of $K$ and suppose that
$f,g\in {\mathcal K}$ satisfy the equations $L(f)=0$ and $L^*(g)=0$, where $L$ and $L^*$ are defined in Theorem~\ref{theorem:dual}. Then
the element
\begin{eqnarray*}
\Omega(g,f)&:=&A_0gf\\
&&+(A_1g)D(f)-D(A_1g)f\\
&&+(A_2g)D^2(f)-D(A_2g)D(f)+D^2(A_2g)f\\
&&\vdots\\
&&+(A_ng)D^{n-1}(f)-D(A_ng)D^{n-2}(f)+\cdots+(-1)^{n-1}D^{n-1}(A_ng)f\\
\end{eqnarray*}
belongs to $K_0$, the subfield of constant elements of ${\mathcal K}$ under $D$.
\end{corollary}

The following lemma can be proved by direct computation.

\begin{lemma}
Let notations be as above. Then $\Omega(g,f)=\sum_{i,j=0}^{r-1}\tilde{A}_{ij}D^i(g)D^j(f)$
where
$$\tilde{A}_{ij}=(-1)^i\sum_{l=0}^{r-1-i-j}(-1)^j{l+i\choose i}D^l(A_{i+j+l+1}).$$
In particular we have that $\tilde{A}_{ij}=0$ if $i+j\ge r$ and
$\tilde{A}_{ij}=(-1)^iA_r$ if $i+j=r-1$.
\end{lemma}

Let $\Psi$ be the $r\times r$-matrix with the entries $\tilde{A}_{ij}$. Then $\Psi$ is
invertible, since its determinant is $A_r^r$. As a consequence $\Omega$ is non-degenerate.

Suppose we have a basis of solutions $f_1,\ldots,f_r$ of $L(f)=0$ and a basis
of solutions $g_1,\ldots,g_r$ of $L^*(g)=0$. Let us write $C_{ij}=\Omega(g_i,f_j)$
for all $i,j$ in $\{1,2,\ldots,r\}$ and denote the $r\times r$-matrix with these
entries by $C$. Let
$$\bbbf:=W(f_1,\ldots,f_r)=
\begin{pmatrix}
f_1 & f_2 & \ldots & f_r\\
D(f_1) & D(f_2) & \ldots & D(f_r)\\
\vdots & \vdots & &\vdots\\
D^{r-1}(f_1) & D^{r-1}(f_2) & \ldots & D^{r-1}(f_r)
\end{pmatrix}$$ be the $r\times r$ Wronskian matrix and similarly $\bbbg:=W(g_1,\ldots,g_r)$. Then the previous matrix relation can be rewritten as
\begin{equation}\label{prematrixduality}
\bbbg^t\Psi\bbbf=C.
\end{equation}
Since the matrices $\bbbf$ and $\bbbg$ are Wronskian matrices of sets of independent functions,
they are invertible and we get
\begin{equation}\label{matrixduality}
\bbbf C^{-1}\bbbg^t=\Psi^{-1}.
\end{equation}
Notice that $\Psi^{-1}$ has zeros above the anti-diagonal which goes from the left lower corner
to the right upper corner. Moreover, by appropriately choosing the basis $g_1,\ldots,g_r$, we can assume that $C$ is the identity matrix, therefore yielding the following immediate consequence of~\eqref{matrixduality}.

\begin{corollary}\label{diagonalidentity}
Let $f_1,\ldots,f_r$ be a basis of solutions of $L(f)=0$, where $L$ is defined in Theorem~\ref{theorem:dual}. Then there
exists a solution basis $g_1,\ldots,g_r$ of $L^*(g)=0$ such that
$$\sum_{i=1}^rD^k(g_i) D^l(f_i) \in K$$
for all $k,l\ge0$. In particular we have $\sum_{i=1}^rD^k(g_i)D^l(f_i)=0$
for all $k,l$ with $k+l<r-1$.
\end{corollary}

\section{Hypergeometric equation of order $r$}\label{hypergeometric}

Recall the hypergeometric equation~\eqref{eq:hypergeq} of order $r$,
$$z(\theta+a_1)\cdots(\theta+a_r)f=(\theta+b_1-1)\cdots(\theta+b_r-1) f$$
where $\theta=z{d\over dz}$ and we have the default parameter $b_r=1$. By Theorem~\ref{theorem:dual}, the dual equation reads
$$(\theta-a_1)\cdots(\theta-a_r)(zg)=(\theta-b_1+1)\cdots(\theta-b_r+1) g.$$
Hence
$$z(\theta-a_1+1)\cdots(\theta-a_r+1)g=(\theta-b_1+1)\cdots(\theta-b_r+1) g.$$
So the dual equation is again hypergeometric with parameters $a'_i=1-a_i$ for
$i=1,\ldots,r$ and $b'_j=2-b_j$ for $j=1,\ldots,r$.

Suppose that the $b_j$ are all distinct modulo $\bbbz$.
Then recall from the introduction that the hypergeometric equation has a standard basis of solutions around $z=0$ of
the form
$$f_i(z)=z^{1-b_i}\,{}_rF_{r-1}\!\left(\begin{matrix}a_1+1-b_i,\ldots,a_r+1-b_i\\
b_1+1-b_i,\ldots,\vee,
\ldots,b_r+1-b_i\end{matrix};z\right),\qquad 1\leq i\leq r.$$
Similarly we have the following basis of solutions for the dual equation.
$$g_i(z)=z^{b_i-1}\,{}_rF_{r-1}\!\left(\begin{matrix}b_i-a_1,\ldots,b_i-a_r\\
b_i+1-b_1,\ldots,\vee,
\ldots,b_1+1-b_r\end{matrix};z\right),\qquad 1\leq i\leq r.$$

To use our D-module language, the ground field we employ here is the field of rational functions
$K=H(z)$ where $H$ is the field $\bbbq$ extended with the parameters $a_i,b_j$. The field
extension ${\mathcal K}$ in which our solutions lie is the field of Laurent series $H((z))$ extended
with the functions $z^{b_j}$. The latter are defined as non-trivial solution of $\theta(f)=b_j f$.

\begin{proposition}\label{dualpairinggauss}
Let $\Psi$ and $C$ be the $r\times r$-matrices defined in Section~\ref{Dmodules},
but for the case of the hypergeometric equation of order $r$ and $D=\theta$.
For the solution bases of the equation
and its dual we take the bases just defined. Then $C$ is a diagonal matrix. The $i$-th diagonal
entry reads
$$C_{ii}=\prod_{j=1\atop j\neq i}^r(b_j-b_i),$$
where $b_r=1$.
\end{proposition}

\begin{proof}
The elements of $C$ belong to $H$, the constant field. In particular, $C_{ij}\in H$ for all $i\ne j$.
However, $C_{ij}$ is also of the form $z^{-b_i+b_j}$ times a power series in $z$.
Since $b_i\ne b_j$ if $i\ne j$ we can only conclude that $C_{ij}=0$ for all
distinct $i,j$. For the computation of $C_{ii}$ it suffices to look only at
constant terms after cancellation of the powers $z^{1-b_i}$ and $z^{b_i-1}$
in formula~\eqref{prematrixduality}.
This means $C_{ii}$ equals
$$\left(1,b_i-1,\ldots,(b_i-1)^{r-1}\right)
\left(\begin{matrix}
B_1 & B_2 & \cdots     & B_r\\
-B_1 & -B_2 &  \cdots    & 0\\
B_2 & B_3 &    \cdots      & 0\\
\vdots &\vdots & &\vdots\\
(-1)^{r-1}B_{r-1} & (-1)^{r-1}B_r &     \cdots                  & 0\\
(-1)^rB_r & 0 & \cdots               & 0\end{matrix}\right)
\left(\begin{matrix}1\\ 1-b_i\\ \vdots\\ (1-b_i)^{r-1}\end{matrix}\right),
$$
where the $B_j$'s are the constants in the coefficients of the hypergeometric
operator. That is,
$$B_0+B_1\theta+\cdots+B_r\theta^r=(\theta-1+b_1)\cdots(\theta-1+b_r).$$
Direct calculation of $C_{ii}$ yields
$$C_{ii}=B_1+2B_2(1-b_i)+3B_3(1-b_i)^2+\cdots+rB_r(1-b_i)^{r-1}.$$
Note that $C_{ii}$ is simply the derivative of $(x-1+b_1)\cdots(x-1+b_r)$
evaluated at $x=1-b_i$. Hence $C_{ii}=\prod_{j\ne i}(b_j-b_i)$,
as asserted.
\end{proof}

As an immediate consequence of Corollary~\ref{diagonalidentity} and Proposition~\ref{dualpairinggauss}
we get the following result, which implies Theorem~\ref{duality} through explicitation of the desired terms in the matrix $\Psi^{-1}$ from equation~\eqref{matrixduality}.
\begin{corollary}\label{finalgauss}
Let $(f_i)_i$ and $(g_j)_j$ be the basis of solutions of the hypergeometric equation and the
dual equation, as given above. Let $H$ be the field generated over $\bbbq$ by  the $a_i,b_j$ and $q$. Then, with
$C_{ii}$ as in Proposition~\ref{dualpairinggauss}, we get
$$\sum_{i=1}^r{1\over C_{ii}}\theta^k(f_i)(z)\,\theta^l(g_i)(z)\in H(z),$$
for all integers $k,l\ge0$. In particular we have $\sum_{i=1}^r\theta^k(f_i)(z)\,\theta^l(g_i)(z)=0$
for all $k,l$ with $k+l<r-1$.
\end{corollary}

\section{$\Delta$-modules and duality}\label{deltamodules}

Let $K$ be a field of characteristic zero and $\Delta:K\to K$ a fixed isomorphism. We denote by $K_0:=\{a\in K|\Delta(a)=a\}$ the subfield of constants.
\begin{definition}\label{definition:deltamodules}
A $K$-vector space $M$ is called a $\Delta$-module (over $K$) if there
is a bijective map $\nabla:M\to M$
such that
\begin{enumerate}
\item $\nabla(m_1+m_2)=\nabla(m_1)+\nabla(m_2)$ for all $m_1,m_2\in M$,
\item $\nabla(fm)=\Delta(f)\nabla(m)$ for all $f\in K$ and $m\in M$.
\end{enumerate}
\end{definition}

We denote $\nabla$ by $\Delta$ again.
Consider the skew ring $K[\Delta,\Delta^{-1}]$ and an operator $L\in K[\Delta,\Delta^{-1}]$. Let
$(L)=\{\mu L|\mu\in K[\Delta,\Delta^{-1}]\}$ be the left ideal generated by $L$.
Then $K[\Delta,\Delta^{-1}]/(L)$
is again a $\Delta$-module, the module associated to the operator $L$. The action of $\Delta$
is given by left multiplication with $\Delta$. In order for
$\Delta$ to be bijective on this module we need that $L(0)\ne0$, which we will assume from now on.

\begin{definition}\label{definition:deltatensor}
Let $M,M'$ be $\Delta$-modules over $K$.

A $K$-linear map $\varphi:M\to M'$ is called a $\Delta$-homomorphism if
$\Delta\circ \varphi=\varphi\circ\Delta$. If $\varphi$ is a $K$-vector space isomorphism
we call $\varphi$ a $\Delta$-isomorphism.
\smallskip

A $\Delta$-module $M$ is called irreducible if $\{\v 0\}$ and $M$ are the
only $\Delta$-submodules.
\smallskip

The tensor product $M\otimes M'$ can be given a $\Delta$-module structure
via
$$\Delta(m\otimes m')=\Delta(m)\otimes\Delta(m'),$$
for all $m\in M,m'\in M'$.
\smallskip

The dual vector space $M^*$ can be given a $\Delta$-module structure via
$$\langle \Delta(m^*),m\rangle = \Delta(\langle m^*,\Delta^{-1}(m)\rangle).$$
Here $\langle m^*,m\rangle$ stands for the evaluation of $m^*$ in $m$.
\end{definition}

\begin{proposition}\label{explicitqdual}
Let $M$ be a finite dimensional $\Delta$-module over $K$. Let $m_1,\ldots,m_r$ be a
$K$-basis and let $A_{ij}\in K$ be such that
$$\Delta(m_i)=\sum_{j=1}^rA_{ij}m_j$$
for all $i$. Define the dual basis $m^*_i$ by $\langle m^*_i,m_j\rangle=\delta_{ij}$ for
$i,j=1,\ldots,r$, where $\delta_{ij}$ is the Kronecker delta. Let $(B_{ij})_{1\leq i,j\leq r}$
be the transposed inverse of $(A_{ij})_{1\leq i,j\leq r}$. Then,
\begin{enumerate}
\item $\Delta(m^*_i)=\sum_{j=1}^rB_{ij}m^*_j$ for $i=1,\ldots,r$,
\item $\omega=\sum_{i=1}^rm^*_i\otimes m_i\in M^*\otimes M$ does not depend
on the choice of basis $m_i$ and $\Delta(\omega)=\omega$.
\end{enumerate}
\end{proposition}

\begin{proof}
To prove the first assertion, let us define the matrix with entries $E_{ij}$ by
the relation $\Delta(m_i^*)=\sum_{i=1}^rE_{ij}m_j^*$. Combine this with
$m_k=\sum_{l=1}^rA_{kl}m_l$ to get
$$\delta_{ik}=\langle\Delta(m_i^*),\Delta(m_k)\rangle
=\sum_{j,l=1}^rE_{ij}A_{kl}\langle m_j^*,m_l\rangle$$
for any $i,k=1,\ldots,r$. Working out the right hand side yields
$\delta_{ik}=\sum_{j=1}^rE_{ij}A_{kj}$, which shows that $(E_{ij})_{1\leq i,j\leq r}$ is
the transposed inverse of $(A_{ij})_{1\leq i,j\leq r}$.

The second assertion then follows by straightforward computation.
\end{proof}

In the following proposition we consider elements $\Omega\in N\otimes M$
where $N,M$ are $\Delta$-modules of the same rank $r$. We say that $\Omega$ is
 non-degenerate if it cannot be written in the form $\sum_{i=1}^sn_i\otimes m_i$
with $m_i\in M,n_i\in N$ and $s<r$.

\begin{proposition}\label{proposition:deltaisomorphism}
Let $M,N$ be $\Delta$-modules of finite rank $r$. Let $m_1,\ldots,m_r$ be a
basis of $M$ and $m_1^*,\ldots,m_r^*$ the corresponding dual basis of $M^*$.

Then the
$\Delta$-homomorphisms $M^*\to N$ are in one-to-one correspondence with the tensors
$\Omega\in N\otimes M$ such that $\Delta(\Omega)=\Omega$.

The correspondence sends a $\Delta$-morphism
$\varphi:M^*\to N$ to $\Omega=\sum_{i=1}^r\varphi(m^*_i)\otimes m_i$.

Conversely, to a tensor $\Omega=\sum_{i=1}^r n_i\otimes m_i$ with $\Delta(\Omega)=\Omega$,
the $K$-linear map generated by $m^*_i\mapsto n_i$ is a $\Delta$-morphism from
$M^*$ to $N$.

Moreover, $\varphi$ is a $\Delta$-isomorphism if and only if $\Omega$ is non-degenerate.
\end{proposition}

\begin{proof}
Suppose we are given a $K$-linear map $\varphi:M^*\to N$.
It is determined by its values $n_i:=\varphi(m_i^*)$
for $i=1,\ldots,r$. We write the $\Delta$-homomorphism condition in two ways.

First of all, by definition one should have $\Delta(n_i)=\Delta(\varphi(m_i^*))=
\varphi(\Delta(m_i^*))$ for all $i$. Using Proposition \ref{explicitqdual} this is the
same as
$$\Delta(n_i)=\varphi\left(\sum_{j=1}^rB_{ij}m_j^*\right)=\sum_{j=1}^rB_{ij}n_j, \quad i=1,\dots,r.$$
Now define $\Omega=\sum_{i=1}^r n_i\otimes m_i$ and subsitute it into $\Delta(\Omega)=\Omega$.
We get
$$\Delta(\Omega)=\sum_{i=1}^r \Delta(n_i)\otimes \Delta(m_i)=\Omega.$$
Using $\Delta(m_i)=\sum_{j=1}^rA_{ij}m_j$ rewrite this as
$$\sum_{j=1}^r \left(\sum_{i=1}^r\Delta(n_j)A_{ij}\right)\otimes m_j=\Omega.$$
This is equivalent to $\sum_{i=1}^r\Delta(n_i)A_{ij}=n_j$ for $j=1,\ldots,r$.
Here we recognize our formulation of the $\Delta$-homomorphism condition.

Note that the tensor $\sum_{i=1}^rn_i\otimes m_i$ is non-degenerate if and
only if $n_1,\ldots,n_r$ are linearly independent. But this is equivalent to
$\varphi$ being an isomorphism.
\end{proof}

Let  ${\mathcal K}$ be a field extension of $K$
and suppose $\Delta$ extends to an isomorphism $\Delta:{\mathcal K}\to{\mathcal K}$.
Suppose also that the field of fixed elements under $\Delta$ is still $K_0$.

Let $h_1,\ldots,h_r\in{\mathcal K}$. Define the Wronskian matrix by
$$W(h_1,\ldots,h_r):=\left(\begin{matrix}
h_1 & h_2 & \ldots & h_r\\
\Delta(h_1) & \Delta(h_2) & \ldots & \Delta(h_r)\\
\vdots & & & \vdots\\
\Delta^{r-1}(h_1) & \Delta^{r-1}(h_2) & \ldots & \Delta^{r-1}(h_r)\end{matrix}\right).$$

\begin{lemma}\label{lemma:wronskian}We have that $\det(W(h_1,\ldots,h_r))=0$ if and only if
$h_1,\ldots,h_r$ are linearly dependent over $K_0$.
\end{lemma}

\begin{proof}
When $h_1,\ldots,h_r$ are $K_0$-linear dependent the vanishing of the determinant  is
straightforward.

To prove the converse statement we proceed by induction on $r$. Suppose that
$\det(W(h_1,\ldots,h_r))=0$.
When $r=1$ we trivially get $h_1=0$.
Suppose $r>1$ and our statement holds for $h_1,\ldots,h_{r-1}$.

If $\rank_{\mathcal K}(W)<r-1$, then $W(h_1,\ldots,h_{r-1})$ has rank $<r-1$ and
the induction hypothesis implies that $h_1,\ldots,h_{r-1}$ are $K_0$-linear dependent.

Suppose $\rank_{\mathcal K}(W)=r-1$. Then there exist $\alpha_1,\ldots,\alpha_r\in{\mathcal K}$,
unique up to a common factor, such that $\sum_{i=1}^r\alpha_i\Delta^l(h_i)=0$ for
$l=0,1,\ldots,r-1$ (linear dependence of columns of $W$).

Let $s$ be the smallest index such that the $s$-th row is ${\mathcal K}$-linear dependent
of the previous rows. Then, by repeated applications of $\Delta$ we see that the
next rows are also dependent of the first $s-1$ rows. Since the rank of $W$ is $r-1$
we conclude that $s=r$ and there exist $\beta_0,\ldots,\beta_{r-2}\in{\mathcal K}$ such that
$\Delta^{r-1}(h_j)=\sum_{l=0}^{r-2}\beta_l\Delta^l(h_j)$ for $j=1,\ldots,r$. Apply $\Delta$ to obtain
$\Delta^r(h_j)=\sum_{l=0}^{r-2}\Delta(\beta_l)\Delta^{l+1}(h_j)$. It follows from this
that the above column relations also hold in case we take $l=r$.
Hence, after application of $\Delta^{-1}$,
$$\sum_{i=1}^r\Delta^{-1}(\alpha_i)\Delta^l(h_i)=0,$$
for $l=0,1,\ldots,r-1$. Since the coefficients of the column relations of $W$ are
unique up to a common factor, we find that there exists $\lambda\in{\mathcal K}$ such
that $\Delta^{-1}(\alpha_i)=\lambda\alpha_i$ for $i=1,2,\ldots,r$. Suppose that
$\alpha_j\ne0$. Then we find that $\alpha_i/\alpha_j\in K_0$ for all $i$.
This gives us the desired $K_0$-linear relation
between the $h_i$'s.
\end{proof}

The next result identifies (up to $\Delta$-isomorphism) the dual difference operator associated with any fixed difference operator  $L\in K[\Delta,\Delta^{-1}]$ satisfying $A_0:=L(0)\neq0$.

\begin{theorem}\label{theorem:qdual}
Consider the $r$-th order difference operator
$$ L=A_r\Delta^r+A_{r-1}\Delta^{r-1}+\cdots+A_1\Delta+A_0,$$
with $A_i\in K$ for all $i$, and $A_r,A_0\neq0$. Then the dual of $K[\Delta,\Delta^{-1}]/(L)$ is $\Delta$-isomorphic
to $K[\Delta,\Delta^{-1}]/(L^*)$ where
$$L^*=\Delta^{r-1}(A_0)\Delta^r+\Delta^{r-2}(A_1)\Delta^{r-1}+
\cdots+A_{r-1}\Delta+\Delta^{-1}(A_r).$$
\end{theorem}

\begin{proof}Consider the element
\begin{multline*}
\Omega=-\Delta^{-1}(A_r)(1\otimes\Delta^{r-1})+A_{r-2}(\Delta\otimes\Delta^{r-2})\\+(1+\Delta)(A_{r-3}\Delta\otimes\Delta^{r-3})
+\dots+(1+\Delta+\cdots+\Delta^{r-2})(A_0\Delta\otimes1).
\end{multline*}
in $(K[\Delta,\Delta^{-1}]/(L^*))\otimes(K[\Delta,\Delta^{-1}]/(L))$.
Thanks to Proposition~\ref{proposition:deltaisomorphism},
it suffices to prove that $\Delta(\Omega)=\Omega$
and that $\Omega$ is non-degenerate. We compute
\begin{eqnarray*}
(1-\Delta)\Omega&=&-\Delta^{-1}(A_r)(1\otimes\Delta^{r-1})+A_r(\Delta\otimes\Delta^r)\\
&&+(1-\Delta)(A_{r-2}\Delta\otimes\Delta^{r-2})+(1-\Delta^2)(A_{r-3}\Delta\otimes\Delta^{r-3})\\
&&+\dots+(1-\Delta^{r-1})(A_0\Delta\otimes1)\\
&=&A_r\Delta\otimes\Delta^r+A_{r-2}\Delta\otimes
\Delta^{r-2}+\cdots+A_0\Delta\otimes1\\
&&-\Delta^{-1}(A_r)(1\otimes\Delta^{r-1})\\
&&-\Delta(A_{r-2})(\Delta^2\otimes\Delta^{r-1})-\cdots-(\Delta^{r-1}A_0)(\Delta^r\otimes\Delta^{r-1})\\
&=&\Delta\otimes L-L^*\otimes\Delta^{r-1}=0-0=0.
\end{eqnarray*}
Non-degeneracy of our $\Omega$ follows from the invertibility of the matrix $\Psi_q$ defined below, and is equivalent to $A_0,A_r\neq0$.
\end{proof}

\begin{remark}
Our choice of $L^*$ is by no means unique. As M.~van~der~Put pointed out to us,
$$\Delta\circ L^*\circ\Delta^{-1}=\Delta^r\circ A_0+\Delta^{r-1}\circ A_1+\cdots+\Delta\circ A_{r-1}+A_r$$
or rather,
$$\Delta^r\left(\Delta^{-r}\circ A_r+\Delta^{-(r-1)}\circ A_{r-1}+\cdots+\Delta^{-1}\circ A_1+A_0\right)$$
might be a more natural candidate. The subsequent arguments in this paper would also be simpler.
However, for some reason, this operator produces
a version of Theorem~\ref{qduality} with a matrix $(M_{kl}(q))_{0\leq k,l\leq r-1}$ having fewer zeros than the present matrix.
For this reason we have given preference to the more complicated $L^*$ defined above.
\end{remark}

We get the following immediate consequence.
\begin{corollary}\label{corollary:qomegafixed}
Let ${\mathcal K}$ be a $\Delta$-extension of $K$ and suppose that
$f,g\in {\mathcal K}$ satisfy the equations $L(f)=0$ and $L^*(g)=0$, where $L$ and $L^*$ are defined in Theorem~\ref{theorem:qdual}. Then
the element
\begin{eqnarray*}
\Omega(g,f)&:=&-\Delta^{-1}(A_r)g\Delta^{r-1}(f)\\
&&+A_{r-2}\Delta(g)\Delta^{r-2}(f)\\
&&+A_{r-3}\Delta(g)\Delta^{r-3}(f)+\Delta(A_{r-2})\Delta^2(g)\Delta^{r-2}(f)\\
&&\vdots\\
&&+A_0\Delta(g)f+\Delta(A_0)\Delta^2(g)\Delta(f)+\cdots+\Delta^{r-2}(A_0)\Delta^{r-1}(g)\Delta^{r-2}(f)
\end{eqnarray*}
belongs to $K_0$, the subfield of elements of ${\mathcal K}$ fixed under $\Delta$.
\end{corollary}

The element $\Omega(g,f)$ can be written in matrix form as follows:
$$(g,\Delta (g),\ldots,\Delta^{r-1}(g))
\left(\begin{matrix}0 & 0 & \cdots      & 0 &-\Delta^{-1}(A_r)\\
A_0 & A_1 & \cdots              &  A_{r-2} & 0\\
0 & \Delta (A_0) &   \cdots          & \Delta (A_{r-3}) & 0\\
\vdots &\vdots  & &\vdots &\vdots\\
0 & 0 &     \cdots                &  \Delta^{r-3}(A_1) & 0\\
0 & 0 & \cdots               & \Delta^{r-2}(A_0) & 0\end{matrix}\right)
\left(\begin{matrix}f\\ \Delta (f)\\ \vdots\\ \Delta^{r-1}(f)\end{matrix}\right).$$
Denote the middle $r\times r$-matrix, whose entries are in $K$, by $\Psi_q$.

Suppose we have a basis of solutions $f_1,\ldots,f_r$ of $L(f)=0$ and a basis
of solutions $g_1,\ldots,g_r$ of $L^*(g)=0$. Then, as $A_0,A_r\neq0$, the matrix $\Psi_q$ is invertible. Let us write $C_{ij}=\Omega(g_i,f_j)$
for all $i,j$ in $\{1,2,\ldots,r\}$ and denote the $r\times r$-matrix with these
entries by $C$. Letting $\bbbf:=W(f_1,\ldots,f_r)$ and $\bbbg:=W(g_1,\ldots,g_r)$, the previous matrix relation can be rewritten as
\begin{equation}\label{prematrixqduality}
\bbbg^t\Psi_q\bbbf=C.
\end{equation}
As a consequence of Lemma~\ref{lemma:wronskian} the matrices $\bbbf$ and $\bbbg$ are invertible and we get
\begin{equation}\label{matrixqduality}
\bbbf C^{-1}\bbbg^t=\Psi_q^{-1}.
\end{equation}
Notice that $\Psi_q^{-1}$ has zeros on and below the diagonal, except for the
element on place $r,1$. Moreover, by appropriately choosing the basis $g_1,\ldots,g_r$, we can assume that $C$ is the identity matrix, therefore yielding the following immediate consequence of~\eqref{matrixqduality}.
\begin{corollary}\label{qdiagonalidentity}
Let $f_1,\ldots,f_r$ be a basis of solutions of $L(f)=0$, where $L$ is defined in Theorem~\ref{theorem:qdual}. Then there
exists a solution basis $g_1,\ldots,g_r$ of $L^*(g)=0$ such that
$$\sum_{i=1}^r\Delta^k(g_i)\Delta^l(f_i)\in K$$
for all $k,l\ge0$. In particular we have $\sum_{i=1}^rg_i\Delta^l(f_i)=0$
for $l=0,1,\ldots,r-2$.
\end{corollary}

\section{Basic hypergeometric equation of order $r$}\label{qhypergeometric}

As an application of the previous section,  consider $\Delta=\sigma_q$ and the basic hypergeometric equation~\eqref{eq:qhyperg}, which we rewrite below:
$$(1-b_1\Delta/q)\cdots(1-b_{r}\Delta/q)f=z(1-a_1\Delta)\cdots(1-a_r\Delta)f.$$
By Theorem~\ref{theorem:qdual}, the dual equation reads
$$(\Delta-b_1/q)\cdots(\Delta-b_{r}/q)g=(\Delta-a_1)\cdots(\Delta-a_r)(z/q)g.$$
After rearranging factors we obtain
$$(1-q\Delta/b_1)\cdots(1-q\Delta/b_{r})g=\frac{a_1\cdots a_r}{b_1\cdots b_{r-1}}q^{r-2}z
(1-q\Delta/a_1)\cdots(1-q\Delta/a_r)g.$$
Note that this way of writing the dual general hypergeometric equation  was already  given in~\cite[Proposition~5]{Ro11}.
So the dual equation is again a hypergeometric equation with parameters
$a'_i=q/a_i,b'_i=q^2/b_i$ for $i=1,\ldots,r$ and $a_1\cdots a_rq^{r-2}z/(b_1\cdots b_{r-1})$
instead of $z$.

Suppose that none of the ratios $b_i/b_j$ with $i\ne j$ is an integer power of $q$.
Recall from the introduction that a basis of solutions of~\eqref{eq:qhyperg} reads
$$f_i(q;z)=z^{1-\beta_i}\,{}_r\phi_{r-1}\!\left[\begin{matrix}qa_1/b_i,\ldots,qa_r/b_i\\qb_1/b_i,\ldots,\vee,
\ldots,qb_r/b_i\end{matrix};q,z\right],\qquad 1\leq i\leq r.$$
Therefore a basis of solutions for the dual equation is given by
$$g_i(q;z)=z^{\beta_i-1}\,{}_r\phi_{r-1}\!\left[\begin{matrix}b_i/a_1,\dots,b_i/a_r\\qb_i/b_1,\dots,\lor,\dots,qb_i/b_r\end{matrix};q;\frac{a_1\dots a_rzq^{r-2}}{b_1\dots b_{r-1}}\right],$$
 as was defined in the introduction.

In a formal setup the ground field is now the field of rational functions $H_q(z)$ where $H_q$ is
the field $\bbbq$ extended with $q$ and the $a_i,b_j$. For the field ${\mathcal K}$ containing the solutions of the
difference equation we can take the field $H_q((z))$ of Laurent series with coefficients in  $H_q$ extended with the functions $z^{1-\beta_i}$,
which are defined as non-trivial solution of the difference equation $\Delta(f)=(q/b_i) f$. So loosely
speaking, $\beta_i=\log(b_i)/\log(q)$.

\begin{proposition}\label{dualpairingheine}
Let $\Psi_q$ and $C$ be the $r\times r$-matrices defined in Section~\ref{deltamodules},
but for the case of the q-hypergeometric equation of order $r$. For the solution bases of the equation
and its dual we take the bases just defined. Then $C$ is a diagonal matrix. The $i$-th diagonal
entry reads
$$C_{ii}={1\over qb_i^{r-2}}\prod_{j=1\atop j\neq i}^r(b_i-b_j),$$
where $b_r=q$.
\end{proposition}

\begin{proof} Let $H_q$ be the field generated over $\bbbq$ by the $a_i$'s, the $b_j$'s and $q$.
This is contained in the field $K_0$ of
constants. The elements of $C$ belong to $H_q$. In particular, $C_{ij}\in H_q$ for all $i\ne j$.
However, $C_{ij}$ is also of the form $z^{-\beta_i+\beta_j}$ times a power series in $z$.
Since $\beta_i\ne\beta_j$ if $i\ne j$ we can only conclude that $C_{ij}=0$ for all
distinct $i,j$. For the computation of $C_{ii}$ it suffices to look only at
constant terms after cancellation of the powers $z^{1-\beta_i}$ and $z^{\beta_i-1}$
in the evaluation of equation~\eqref{prematrixqduality}.
This means $C_{ii}$ equals
$$\left(1,b_i/q,(b_i/q)^2,\ldots,(b_i/q)^{r-1}\right)
\left(\begin{matrix}0 & 0 & \cdots     & 0 &-B_r\\
B_0 & B_1 &  \cdots              & B_{r-2} & 0\\
0 & B_0 &    \cdots          &B_{r-3} & 0\\
\vdots &\vdots & &\vdots  &\vdots\\
0 & 0 &     \cdots                 &  B_1 & 0\\
0 & 0 & \cdots              & B_0 & 0\end{matrix}\right)
\left(\begin{matrix}1\\ q/b_i\\ (q/b_i)^2\\ \vdots\\ (q/b_i)^{r-1}\end{matrix}\right),
$$
where the $B_j$'s are the constants in the coefficients of the q-hypergeometric
operator. That is,
$$B_0+B_1\Delta+\cdots+B_r\Delta^r=(1-b_i\Delta/q)\cdots(1-b_{r}\Delta/q).$$
Rewrite this as
$$B_0x^r+B_1x^{r-1}+\cdots+B_r=(x-b_1/q)(x-b_2/q)\cdots(x-b_r/q),$$
where $b_r=q$ and $B_0=1$. Note that this implies for any $i$ that
$$B_0(b_i/q)^r+B_1(b_i/q)^{r-1}+\cdots+B_{r-1}(b_i/q)+B_r=0,$$
and after differentiation of the polynomial,
$$rB_0(b_i/q)^r+(r-1)B_1(b_i/q)^{r-1}+\cdots+B_{r-1}=b_i(b_i-b_1)\cdots\vee\cdots(b_i-b_r)/q^r.$$
Subtraction of the first identity yields
\begin{multline*}
(r-1)B_0(b_i/q)^r+(r-2)B_1(b_i/q)^{r-1}\\
+\cdots+B_{r-2}(b_i/q)^2-B_r
=
b_i(b_i-b_1)\cdots\vee\cdots(b_i-b_r)/q^r.
\end{multline*}
But evaluation of $C_{ii}$ gives
$$C_{ii}=(q/b_i)^{r-1}(-B_r+(r-1)B_0(b_i/q)^r+\dots+B_{r-2}(b_i/q)^2).$$
In view of the previous calculation this implies our assertion.
\end{proof}

As an immediate consequence of Corollary~\ref{qdiagonalidentity} and Proposition~\ref{dualpairingheine}, we get the following result.
\begin{corollary}\label{finalheine}
Let $(f_i)_{1\leq i\leq r}$ and $(g_j)_{1\leq j\leq r}$ be the basis of solutions of the q-hypergeometric equation and the
dual equation, as given above. Let $H_q$ be the field generated over $\bbbq$ by the $a_i,b_j$ and $q$. Then, with
$C_{ii}$ as in the previous theorem we get
$$\sum_{i=1}^r{1\over C_{ii}}\Delta^k(f_i)\Delta^l(g_i)\in H_q(z),$$
for all integers $k,l\ge0$. In particular we have $\sum_{i=1}^r{1\over C_{ii}}\Delta^k(f_i)g_i=0$
for $k=0,1,\ldots,r-2$.
\end{corollary}
Finally, to prove the cases $k=r-1,l=0$ and $l=k+1$ of Theorem~\ref{qduality}, it remains to explicitly invert the matrix $\Psi_q$, and to use~\eqref{matrixqduality} and Proposition~\ref{dualpairingheine}.

\ifx
Directly from the definition of $\Psi$, we have
\begin{equation*}\label{inverse}
\Psi^{-1}=\left(
\begin{matrix}
0&x_0&x_1&\dots&x_{r-2}\\
&&&&\\
0&0&\Delta x_0&\dots&\Delta x_{r-3}\\
\vdots&\vdots&\vdots&&\vdots\\
0&0&0&\dots&\Delta^{r-2} x_{0}\\
&&&&\\
-\Delta^{-1}(1/A_r)&0&0&\dots&0
\end{matrix}\right),\end{equation*}
where $x_0=1/A_0=1/(1-z)$ and for all $r\geq3$, $\sum_{j=0}^{r-2}x_j\Delta^j(A_{r-2-j})=0$. As $\Delta^j(A_{r-2-j})\in h(z)$, this gives the desired recursive definition of the $x_j$'s.\\

Thanks to the previous expression of $\Psi^{-1}$, note that there are only $2r-1$ different duality relations encoded in~\eqref{matrixduality}, all other $r^2-2r+1$ identities being consequences of these ones.
\fi

config: arxiv


TITLE: Questions about the proof of generalized Poisson summation formula.
QUESTION [4 upvotes]: The generalized Poisson summation formula is
$$
\sum_{\gamma \in \Gamma} f(\gamma) = \sum_{ \pi \in \widehat{\Gamma \backslash G}} \hat{f}(\pi),
$$
where $G$ is a locally compact Abelian group, $\Gamma$ is a discrete subgroup of $G$, $f$ is some function on $G$, $\widehat{\Gamma \backslash G}$ is the set of all unitary character of $\Gamma \backslash G$,
$$
\hat{f}(\pi) = \int_G f(x)\pi^{-1}(x)dx
$$
is the Fourier transform. If $G=\mathbb{R}$, $\Gamma=\mathbb{Z}$, then we obtain the usual Poisson summation formula.
The following is the proof of the generalized Poisson summation formula. Let
$$
F(x)=\sum_{\gamma \in \Gamma} f(\gamma x).
$$
Then we have Fourier expansion
$$
F(x) = \sum_{\pi \in \widehat{\Gamma \backslash G}} c(\pi)\pi(x),
$$
$c(\pi)$ are Fourier coefficients.
We have
$$
c(\pi) = \int_{\Gamma \backslash G} F(x)\pi^{-1}(x)dx \\
= \int_G f(x)\pi^{-1}(x)dx \\
= \hat{f}(\pi)
$$
Therefore
$$
\sum_{\gamma \in \Gamma}f(\gamma x) = F(x) = \sum_{\pi \in \widehat{\Gamma \backslash G}} \hat{f}(\pi)\pi(x).
$$
The generalized Poisson formula is proved.
My question is: how to prove the equality
$$
\int_{\Gamma \backslash G} F(x)\pi^{-1}(x)dx = \int_G f(x)\pi^{-1}(x)dx?
$$
I think that
$$
\int_{\Gamma \backslash G} F(x)\pi^{-1}(x)dx \\
= \int_{\Gamma \backslash G} \sum_{\gamma \in \Gamma} f(\gamma x)\pi^{-1}(x) dx \\
= \int_{\Gamma \backslash G} \sum_{\gamma \in \Gamma} f( x)\pi^{-1}( \gamma^{-1} x) d(\gamma^{-1}x).
$$
Do we have $$
\int_{\Gamma \backslash G} \sum_{\gamma \in \Gamma} f( x)\pi^{-1}( \gamma^{-1} x) d(\gamma^{-1}x) = \int_{\Gamma \backslash G} \sum_{\gamma \in \Gamma} f( x)\pi^{-1}(   x) dx = \int_{\Gamma \backslash G}  f( x)\pi^{-1}(   x) dx ?
$$
Thank you very much.

REPLY [4 votes]: Your assertion is correct, but you are missing two ingredients in the proof
1) The dual group of $\Gamma/G$ is the group of characters that map each element of $\Gamma$ to $1$.
2) It helps if you distinguish cosets [x] from group elements x. You need the fact that if $H$ is a closed subgroup of $G$, then given Haar measures on $G$ and $H$ there exists a Haar measure on $H/G$ so that
$$ \int_{G}f(g)dg = \int_{H/G}\int_{H}f(gh)dhd[g].$$
In your case, $H = \Gamma$ and Haar measure on it is just counting measure.


TITLE: Why does the sum of the residuals not equal to 0 when there is no intercept of SLR
QUESTION [3 upvotes]: As we all know that the residuals sum to 0 when there is an intercept. But why the residuals do not sum to 0 when we don't have the intercept in simple linear regression?

REPLY [2 votes]: Recall in the case with an intercept, we get the least squares regressor by minimizing $$ L=\sum_i(y_i-(b_1x_i+b_0))^2$$ in terms of $b_1 $ and $b_0.$ We can do this in the usual way by taking partial derivatives with respect to $b_0$ and $b_1,$ setting to zero and solving.
The fact that the residuals sum to zero comes directly from setting the partial derivative with respect to the interception coefficient $b_0$ to zero. Taking the derivative with respect to $b_0$ gives $$\frac{\partial L}{\partial b_0}=-2\sum_i(y_i- (b_1x_i+b_0)).$$ But the residuals are exactly $$\hat \epsilon_i = (y_i-(\hat b_1 x_i+\hat b_0)) $$ so setting the derivative to zero gives the condition $$\sum_i \hat \epsilon_i=0.$$
Notice that when you take the derivative with respect to $b_1$ you get a factor of $x_i$ out front, so it's a different kind of condition and not one that guarantees the residuals sum to zero.
Also, there was nothing specific to simple linear regression here, this generalizes immediately to the case where there are multiple regressors. The only important thing is that there is a constant term whose minimization equation gives us the condition. If there is no constant term, there is no such condition and thus no guarantee that the residuals sum to zero.


TITLE: Algebraic proof for the identity $\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$
QUESTION [0 upvotes]: I can't figure out an algebraic proof for the following identity, (and I don't know if I can use the binomial theorem for this one):
$$\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$$
Thank you, for your help in advance.

REPLY [0 votes]: Notice that $$(-1)^{m-k}=\binom{-1}{m-k},$$ so you want to prove
$$\sum_{k=0}^m\binom{-1}{m-k}{\binom{n}{k}}= {\binom{n-1}{m}}.$$ But that's just a special case of the Vandermonde identity.
Using generating functions, the identity follows from the evaluation of the coefficient at $x^m$ on both sides of the identity $$\frac{1}{1+x}\cdot(1+x)^n=(1+x)^{n-1}.$$


TITLE: Decide $\overline{A} $ U $\overline{B}$
QUESTION [0 upvotes]: $U = \left \{ 1,2,3,4,5,6,7,8 \right \}$
$ A = \left \{ 1,2,3 \right \}$
$ B = \left \{ 3,4,5,6 \right \} $
a) Decide $ \overline{A}  $ U $\overline{B} $
Correct answer:
\begin{Bmatrix}
 1,2,4,5,6,7,8
\end{Bmatrix}
I don't understand that answer. What does  $\overline{A} $ and  $\overline{B}$ mean?

REPLY [0 votes]: You can also use $\overline{A}\cup\overline{B}=\overline{A\cap B}=\overline{\{3\}}=U\setminus\{3\}=\{1,2,4,5,6,7,8\}$

When the context in unambiguous $\overline{X}$ designates $U\setminus X$.
When the context in ambiguous (i.e when $\overline{X}$ designates the closure of $X$ and $\mathring X$ its interior) then $X^\complement$ is prefered.


text:
\begin{document}
\begin{abstract}
Among the class of functions with Fourier modes up to degree 30, constant functions are the unique real-valued maximizers for the endpoint Tomas--Stein inequality on the circle.
\end{abstract}

\maketitle

\section{Introduction}

We are interested in the sharp constant of the endpoint Tomas--Stein adjoint restriction inequality \cite{To75} on the circle $\mathbb{S}^1=\{\omega\in\R^2:|\omega|=1\}$. More precisely, we seek
a maximizer for the functional $\Phi$ defined on nonzero $f\in L^2(\mathbb{S}^1)$ by
$$\Phi(f) := \|\widehat{f\sigma}\|^6_{L^6(\R^2)}\|f\|_{L^2(\mathbb{S}^1)}^{-6}.$$
We have written $\sigma$ for the arc length measure on the circle $\mathbb{S}^1$, and we have used the Fourier transform
\begin{equation*}
\widehat{f\sigma}(x):=\int_{\mathbb{S}^1} f(\omega)\,e^{-i x\cdot\omega}\, \d\sigma_\omega,\;\;\;(x\in\R^2).
\end{equation*}
It is known that maximizers of $\Phi$ exist \cite{Sh16} and are smooth \cite{Sh16b}, and that the constant function ${\bf 1}$ is a local maximizer of $\Phi$, see \cite[Theorem 1.1]{CFOST17}.
Moreover, real-valued maximizers of $\Phi$ are known to be nonnegative, antipodally symmetric functions, that is
$$f(\omega)\ge 0,\ f(\omega)=f(-\omega),$$
for every $\omega\in \mathbb{S}^1$.
It is natural to conjecture that constant functions are global maximizers of $\Phi$, in which case a complete characterization of complex-valued maximizers is given by \cite[\S1, Step 6]{CFOST17}.

In this paper, we report on numerical verification of a finite dimensional variant of this conjecture:

\begin{theorem}\label{thm:main}
Let $f\in L^2(\mathbb{S}^1)$ be non-negative and antipodally symmetric.
Assume that $\widehat{f}(n)=0$ if $|n|>30$. Then
$$\Phi(f)\leq\Phi({\bf 1}),$$
with equality if and only if $f$ is constant.
\end{theorem}

A numerical difficulty in this problem is that there are close competitors for maximizers, namely functions that concentrate in the vicinity of two antipodal points.
Heisenberg uncertainty allows for functions with Fourier modes up to degree $30$ to localize roughly in a $\frac{2\pi}{30}$-neighborhood of these points.

This paper continues efforts to implement Foschi's program \cite{Fo15}
for the 2-sphere in the case of the circle, see also \cite{CFOST17}. The
approach works through positive semi-definiteness of a certain quadratic form on the relevant finite dimensional space. It would be nice to establish positive semi-definiteness for the full space. For recent similar work on the paraboloid, \mbox{see \cite{Go17}.}

\section{Proof of Theorem \ref{thm:main}}

 With  $f$ as in  Theorem \ref{thm:main}, we compute
	$$\|\widehat{f\sigma}\|_{L^6(\R^2)}^6
	= (2\pi)^2 \int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j }   \Big(\prod_{j=1}^3 f_j(\omega_j)\d\sigma_{\omega_j}\Big)\Big(\prod_{j=4}^6 f_j(-\omega_j)\d\sigma_{\omega_j}\Big) $$
 	$$= 5 \pi^2 \int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j }  (|\omega_4+\omega_5+\omega_6|^2-1)
 	\Big(\prod_{j=1}^3 f_j(\omega_j)\d\sigma_{\omega_j}\Big)
 	\Big(\prod_{j=4}^6 f_j(-\omega_j)\d\sigma_{\omega_j}\Big)
    $$
  	$$\le 5\pi ^2 \int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j}  (|\omega_4+\omega_5+\omega_6|^2-1) \Big(\prod_{j=1}^3 f_j(\omega_j)^2\Big)
  	\prod_{j=1}^6 \d\sigma_{\omega_j} $$
  	$$\le  5 \pi^2 \frac{\|f\|_{L^2(\mathbb{S}^1)}^6}{\|{\bf 1}\|_{L^2(\mathbb{S}^1)}^6}
	\int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j}  (|\omega_4+\omega_5+\omega_6|^2-1)
\prod_{j=1}^6 \d\sigma_{\omega_j}= \Phi({\bf 1}) \|f\|_{L^2(\mathbb{S}^1)}^6.$$

Here the first line is simply Plancherel's identity. The second line
is Foschi's idea to improve the situation by artificially
inserting a weight, which after symmetrization over the
indices $j$ reverts to a constant, see the computation following \cite[Lemma 1.3]{CFOST17}.
The third line is the crucial inequality. We defer its proof for the moment.
The inequality in the fourth line is an application of the main result proved in  \cite[Theorem 1.2]{CFOST17}. Equality is attained if and only
if $f$ is constant. Identification of the constant in the fourth line
is then easy and was also observed in \cite{CFOST17}.

This proves Theorem \ref{thm:main}, safe for verification of the crucial inequality in the third line. Note that this inequality would follow from
$$|\omega_1+\omega_2+\omega_3|=|\omega_4+\omega_5+\omega_6|$$
and the inequality between the arithmetic mean and the geometric mean,
$$\prod_{j=1}^3 f_j(\omega_j)\prod_{j=4}^6 f_j(-\omega_j)
\le \frac 12\Big(\prod_{j=1}^3 f_j(\omega_j)^2
+\prod_{j=4}^6 f_j(-\omega_j)^2\Big),$$
if the weight were positive. Unfortunately, the weight is not positive.
One reason to believe that the inequality  still holds as stated is that the negative part of the weight is small, and via antipodal symmetry the values of the function on the negative part of the weight have a strong correlation with the values on the positive part.
However, the support of the measure	 $$\ddirac{\sum_{j=1}^6 \omega_j}$$
is not preserved under antipodal symmetry, which makes it difficult to exploit this correlation. We resort to numerical verification
of the crucial inequality in the given finite dimensional space of functions.

Consider the index set
$$Z=\{k\in 2\mathbb{Z}, |k|\le 30\},$$
and expand the band-limited function $f$ into a Fourier series
$$f(\omega)=\sum_{k\in Z} a_k \omega^k,$$
where we identify $\mathbb{R}^2$ with the complex plane and correspondingly define products and powers of elements in $\mathbb{R}^2$.
Note that
\begin{equation}
\label{realsym}a_{-k}=\overline{a_k}
\end{equation}
for every $k\in Z$.
We write a constant multiple of the left-hand side of the crucial \mbox{inequality as}
$$ \sum_{{\bf k}\in \Z^6}
 L_{\bf k} \Big(\prod_{j=1}^3 a_{k_j}\Big)\Big(\prod_{j=4}^6 a_{-k_j}\Big),
$$
$$L_{{\bf k}}:= (2\pi)^{-5}\int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j}   (|\omega_4+\omega_5+\omega_6|^2-1) \Big(\prod_{j=1}^3 \omega_j^{k_j}\Big)\Big(\prod_{j=4}^6 \omega_{j}^{-k_j}\Big)
\prod_{j=1}^6 \d\sigma_{\omega_j},$$
and the same multiple of the right-hand side as

$$ \sum_{{\bf k}\in \Z^6}
R_{\bf k} \Big(\prod_{j=1}^3 a_{k_j}\Big)\Big(\prod_{j=4}^6 a_{-k_j}\Big),$$
$$R_{{\bf k}}:= (2\pi)^{-5}\int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j}   (|\omega_4+\omega_5+\omega_6|^2-1) \Big(\prod_{j=1}^3 \omega_j^{k_j-k_{j+3}}\Big)
\prod_{j=1}^6 \d\sigma_{\omega_j}.$$

Define for ${\bf m}\in Z^3$
$$s_{\bf m}:=a_{m_1}a_{m_2}a_{m_3},$$
and note that $(s_{\bf m})_{{\bf m}\in Z^3}$ is an element of $Sym(Z^3)$, the vector space of functions on $Z^3$ symmetric under permutation of the three indices.
At this point we do not require the symmetry \eqref{realsym},
instead we pass to a larger space allowing for a convenient orthogonal splitting later.

The crucial inequality then follows from positive semi-definiteness
of the quadratic form
$$\sum_{{\bf m},{\bf n}\in Z^3}
Q_{{\bf m},{\bf n}}s_{\bf m} \overline{s_{ \bf n}}:=
\sum_{{\bf m},{\bf n}\in Z^3} \frac 16 \sum_{\sigma\in S_3}
(R_{{\bf m},{\bf n_\sigma}}-L_{{\bf m},{\bf n_\sigma }})s_{\bf m} \overline{s_{ \bf n}}$$
on $Sym(Z^3)$,
where we write $S_3$ for the group of permutations of three elements and
$${\bf n}_\sigma=(n_{\sigma(1)},n_{\sigma(2)},n_{\sigma(3)}).$$
Note that the symmetrization over $S_3$ does not change the value of the quadratic form whenever the coefficients $s_{\bf n}$ are symmetric. It merely symmetrizes the coefficients of the quadratic form, and allows to reduce the dimension of the matrix by identifying equivalent
tuples. Letting $\tilde{X}$ be the space of tuples in $Z^3$
satisfying $m_1\le m_2\le m_3$, it therefore suffices to prove
positive definiteness of the quadratic form
$$Q(s,s)= \sum_{{\bf m},{\bf n}\in \tilde{X}} Q_{{\bf m},{\bf n}} s_{\bf m} \overline{s_{\bf m}} .$$

Note that the matrix
 $(Q_{{\bf m},{\bf n}})_{{\bf m},{\bf n}\in \tilde{X}}$ is Hermitian.
Moreover, for all ${\bf m}\in Z^3$ we have
$$R_{{\bf m},(0,0,0)}=L_{{\bf m},(0,0,0)}$$
and hence the Dirac delta vector $\delta_{(0,0,0)}$ corresponding to constant functions on the circle is in the kernel
of the matrix $(Q_{{\bf m},{\bf n}})_{{\bf m},{\bf n}\in \tilde{X}}$.
Therefore we replace $\tilde{X}$ by $X = \tilde{X} \setminus \{(0,0,0)\}$.

 A change of variables
$$(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5,\omega_6)\mapsto
(\omega_1\cdot\omega,\omega_2\cdot\omega,\omega_3\cdot\omega,\omega_4\cdot\omega,\omega_5\cdot\omega,\omega_6\cdot\omega)$$
for some arbitrary $\omega$ of modulus one in the expressions for $R_{\bf k}$ and $L_{\bf k}$ shows that
 $$Q_{{\bf m},{\bf n}}=\omega^{d({\bf m})-d({\bf n})}Q_{{\bf m},{\bf n}},$$
where we have denoted
$$d({\bf m})=m_1+m_2+m_3.$$
We conclude
 $$Q_{{\bf m},{\bf n}}=0$$
whenever $d({\bf m})\neq d({\bf n})$. The matrix $(Q_{{\bf m},{\bf n}})_{{\bf m},{\bf n}\in X}$
therefore has the structure of a diagonal block matrix, with blocks enumerated by $D:=d({\bf m})$.
It suffices to prove positive semi-definiteness for each block $(Q_{{\bf m},{\bf n}})_{{\bf m},{\bf n}\in X_{D}}$ separately, where $X_{D} = \{{\bf m}\in X : d({\bf m}) = D\}$.
This will be done in the following section.

\section{Numerical computations}\label{sec:Numerics}
In order to verify that the matrix $(Q_{{\bf m},{\bf n}})_{{\bf m},{\bf n}\in X_{D}}$ is positive definite, we split it into a numerically computed approximation and an error term.
The smallest eigenvalue of the numerical approximation will be larger than the operator norm of the error term.

 \begin{table}[ht]
 \centering
 \begin{tabular}{|c | c || c | c || c |c || c | c |}
 \hline
 $D$&$\lambda_{\min}$&$D$&$\lambda_{\min}$&$D$&$\lambda_{\min}$& $D$ &$\lambda_{\min}$\\[0.5ex]
 \hline
  0&0.00035 & 24&0.00061 & 48&0.00121 & 72&0.00407\\
  2&0.00037 & 26&0.00064 & 50&0.00133 & 74&0.00501\\
  4&0.00038 & 28&0.00067 & 52&0.00144 & 76&0.00596\\
  6&0.00042 & 30&0.00069 & 54&0.00154 & 78&0.00668\\
  8&0.00045 & 32&0.00073 & 56&0.00171 & 80&0.00937\\
 10&0.00049 & 34&0.00077 & 58&0.00188 & 82&0.01258\\
 12&0.00052 & 36&0.00081 & 60&0.00203 & 84&0.01332\\
 14&0.00055 & 38&0.00086 & 62&0.00229 & 86&0.02997\\
 16&0.00057 & 40&0.00092 & 64&0.00255 & 88&0.04400\\
 18&0.00057 & 42&0.00097 & 66&0.00278 & 90&0.20081\\
 20&0.00058 & 44&0.00105 & 68&0.00324  & &\\
 22&0.00059 & 46&0.00113 & 70&0.00369 & &\\[1ex]
 \hline
 \end{tabular}
 \caption{Minimal eigenvalue for the approximation for the block
 	$D\in\{ 0,2,4,\ldots,90\}$, calculated with 5 significant digits of precision.
	In the case of $D=0$, the null block of the vector $\delta_{(0,0,0)}$ has been removed.}
 \label{table:Eigenvalues}
 \end{table}

Numerical approximation of the integrals $L_{{\bf k}}$ and $R_{{\bf k}}$ will rely on the following family of Bessel integrals for  $\sum_{j=1}^6 k_j=0$:
$$I_{{\bf k}}:= (2\pi)^{-5} \int_{(\mathbb{S}^1)^6} 	\ddirac{\sum_{j=1}^6 \omega_j}  \Big(\prod_{j=1}^6 \omega_j^{k_j} \d\sigma_{\omega_j}\Big)$$
$$= (2\pi)^{-1}\int_{\mathbb{R}^2} 	\prod_{j=1}^6
J_{k_j}(|x|) \d x
= \int_0^\infty 	\prod_{j=1}^6 J_{k_j}(r) r\, \d r,$$
where the Bessel function $J_k$ is defined by
$$ \int_{\mathbb{S}^1} \omega^k e^{-ix\cdot\omega} \d\sigma_\omega
= 2\pi (-i)^k J_k(|x|)(x/|x|)^k.$$
Indeed, writing
$$|\omega_4+\omega_5+\omega_6|^2-1=2+\sum_{j,k \in \{4,5,6\}, k\neq j} \omega_j\omega_k^{-1},$$
we obtain
$$L_{{\bf m},{\bf n}}=2I_{{\bf m},{\bf n}}+\sum_{\sigma\in S_3}
I_{{\bf m},{\bf n}+(1,-1,0)_\sigma},$$
$$R_{{\bf m},{\bf n}}=2I_{{\bf m}+{\bf n},(0,0,0)}+\sum_{\sigma\in S_3}
I_{{\bf m}+{\bf n},(1,-1,0)_\sigma}.$$

Using the fact that $J_{-k} = (-1)^{k} J_{k}$ and the above representation we see that $Q_{{\bf m}, {\bf n}} = Q_{-{\bf m}, -{\bf n}}$, so it suffices to consider $D \geq 0$.

To evaluate the integrals $I_{\bf k}$, we follow the
scheme in  \cite{OST17}.
We split the integrals into
\begin{equation}\label{eq:SplitI}
I_{\bf k}=
 \int_0^S 	\prod_{j=1}^6
J_{k_j}(r) r\, \d r+\int_S^R 	\prod_{j=1}^6
J_{k_j}(r) r\, \d r+\int_R^\infty 	\prod_{j=1}^6 J_{k_j}(r) r\, \d r,
\end{equation}
with $S=3600$ and $R=63000$.
The first two integrals are evaluated with a Newton--Cotes
quadrature rule. The step size is $0.003$ for the first integral and
$0.05$ for the second integral.
In practice, this step involved tabulating the numerical values of $61$ Bessel functions at around $3\times 10^6$ points each,  with $20$ digit precision obtained via the software package {\it Mathematica} \cite{WM17}. This high precision lets the rounding errors be absorbed by the error estimates below.

The approximation error for the first integral in \eqref{eq:SplitI} was estimated  in \cite[\S 8]{OST17}, independently of the vector $\k$, by
\[
E_{\k,1} = 1.5\times 10^{-9}.
\]

The approximation error for the second integral in \eqref{eq:SplitI} was also estimated in \cite{OST17} by
\[
E_{\k,2} = C_{2} \prod_{j=1}^6 \Big(1+\frac {k_j^2} S\Big),
\]
where
\[
C_{2}
=
1.01^6(R-S)w^8\frac{6^3}5 \Big(\frac{2}{\pi(S-1)}\Big)^3\cosh^6(1) (R+1)
\]
with $S=3600$, $R=36000$ and $w=0.05$.

The third integral in \eqref{eq:SplitI} is approximated by analytic methods.
Since $R=63000$ is large when compared to $n^2\leq 61^2$, we take advantage of the following well-known asymptotic formulae which are a simplified version of \cite[Corollaries 2.6 and 2.7]{OST17}.
\footnote{We record a typo on the first formula in \cite[Corollary 2.7]{OST17}, which should read as follows:
\begin{multline*}
\left|J_{0}^\pm(z)- \left(\frac{2}{\pi z}\right)^{\frac12}\cos(\omega_0)
-\frac 1{8z}  \left(\frac{2}{\pi z}\right)^{\frac12}\sin(\omega_0)\right|\le \\
\le\frac{9}{128|z|^2}\left(\frac{2}{\pi |z|}\right)^{\frac12} \cosh(|\Im (z)|)
\left(\frac{|z|}{|\Re(z)|}\right)^{\frac52}.
\end{multline*}}
\begin{lemma}\label{le:BesselAsymp}
	Let $\omega_n:=z-\frac{\pi}4-\frac{n\pi}2$ and $\hat{n}:=\max\{1,n\}$.
	If $n\geq 0$ and $z>\hat{n}^2$, then
	\begin{equation}\label{eq:BesselAsymp}
	\Big|J_n^{\pm}(z)-\Big(\frac2{\pi z}\Big)^{\frac12}\cos(\omega_n)\Big|
	\leq
	\Big(\frac2{\pi |z|}\Big)^{\frac12}\frac{\hat{n}^2}{|z|},
	\end{equation}
	\begin{multline}\label{eq:BesselAsymp2}
	\Big|J_n^{\pm}(z)-\Big(\frac2{\pi z}\Big)^{\frac12}\cos(\omega_n)+\frac{4n^2-1}{8z}\Big(\frac2{\pi z}\Big)^{\frac12}\sin(\omega_n)\Big|
	\leq
	\frac14\Big(\frac2{\pi |z|}\Big)^{\frac12}\frac{\hat{n}^4}{|z|^2}.
	\end{multline}\end{lemma}
\noindent Here the functions $J_n^\pm$ are obtained by writing $\cos(zt)=(e^{izt}+e^{-izt})/2$ in the Poisson integral representation for $J_n$,
and as such satisfy $J_n=(J_n^++J_n^-)/2$.
Using \eqref{eq:BesselAsymp}, we may split each Bessel function into a main term plus error.
Applying the distributive law yields one main integral of the form
\begin{equation}\label{eq:ExactCosIntegral}
\int_R^\infty \Big(\frac2{\pi r}\Big)^3 \Big(\prod_{j=1}^6  \cos(\omega_{k_j})\Big) r\d r,
\end{equation}
which can be calculated exactly,
plus $2^6-1$ further terms involving one of the two factors $$\left( \frac{2}{\pi r} \right)^{\frac 1 2}\cos(\omega_{k_j}), J_{k_j}(r)-\left( \frac{2}{\pi r} \right)^{\frac 1 2}\cos(\omega_{k_j})$$
for each $j$. We call them cosine factor and error factor.
For the main integral, observe that
\[
\cos(r-\tfrac{\pi}4-\tfrac{k\pi}2)
=
(-1)^{\lfloor\frac{k}2\rfloor} \cdot \begin{cases}
\sin(r-\frac{\pi}4),& \text{if $k$ is odd},\\
\cos(r-\frac{\pi}4),& \text{if $k$ is even,}
\end{cases}
\]
and so \eqref{eq:ExactCosIntegral} equals a multiple of
$$\int_R^\infty \cos^6(r-\tfrac{\pi}4) r^{-2} \d r, \text{ or }\int_R^\infty \cos^4(r-\tfrac{\pi}4)\sin^2(r-\tfrac{\pi}4) r^{-2}\d r,$$
with sign determined by the parity of $\sum_{j=1}^3(\lfloor\frac{m_j}2\rfloor+ \lfloor\frac{n_j}2\rfloor)$.
{\it Mathematica} calculates these expressions with any prescribed accuracy.
For the further terms, consider first those consisting of an integral of a product of five cosine factors and one error factor.

To estimate these six terms, we use the finer information given by \eqref{eq:BesselAsymp2}.
The sine term in \eqref{eq:BesselAsymp2} leads to integrals of the type
$$\frac{4m_1^2-1}{8} \int_R^\infty \Big(\frac{2}{\pi r}\Big)^3 \sin(\omega_{m_1})\cos(\omega_{m_2})\cos(\omega_{m_3})\cos(\omega_{\bf n})\d r$$
and similar terms with a different cosine factor replaced by a sine factor and corresponding prefactor.
The product of the six trigonometric functions is odd about the point $\frac{\pi}4$.
Thus the product integrates to 0 over each period.
On the period $[R+2\pi k, R+2\pi (k+1)]$, we may thus replace the weight $r^{-3}$ by the difference between itself and its mean over that interval, which in turn is bounded by $6\pi r^{-4}$.
Thus the sum of terms arising from the sine term in \eqref{eq:BesselAsymp2} is bounded by
$$E_{\k,3} = 3\pi \sum_{j=1}^6\hat{k}_j^2 \int_R^\infty \Big(\frac2{\pi}\Big)^3 r^{-4}\d r,$$
where $\hat{k}_j:=\max\{1,k_j\}$.
The sum of the six terms arising from the right-hand side of \eqref{eq:BesselAsymp2} can be estimated by
$$E_{\k,4} = \frac14\sum_{j=0}^6 \hat{k}_j^4 \int_R^\infty \Big(\frac2{\pi}\Big)^3 r^{-4}\d r.$$
Next come fifteen terms of the original $2^6-1$ terms which have four cosine factors and two error factors. It suffices to estimate these with
\eqref{eq:BesselAsymp}, since they benefit from an extra integration of a negative power of $r$. Their sum can be estimated by
$$E_{\k,5} = \sum_{i\neq j}\hat{k}_i^2 \hat{k}_j^2\int_R^\infty\Big(\frac2{\pi}\Big)^3 r^{-4}\d r,$$
where the sum is over ${6\choose 2}=15$ choices of distinct indices $i,j\in\{1,2,3,4,5,6\}$.

The remaining $2^6-1-6-15=42$ terms benefit from an integration of at least the negative fifth power of $r$, and are estimated even more crudely by
$$
E_{\k,6} = \sum_{i,j,\ell} \hat{k}_i^2 \hat{k}_j^2 \hat{k}_\ell^2 \int_R^\infty \Big(\frac2\pi\Big)^3 r^{-5}\d r
+\sum_{i,j,\ell,m} \hat{k}_i^2 \hat{k}_j^2 \hat{k}_\ell^2 \hat{k}_m^2 \int_R^\infty \Big(\frac2\pi\Big)^3 r^{-6}\d r$$
$$+\sum_{i,j,\ell,m,n} \hat{k}_i^2 \hat{k}_j^2 \hat{k}_\ell^2 \hat{k}_m^2 \hat{k}_n^2\int_R^\infty \Big(\frac2\pi\Big)^3 r^{-7}\d r
+\hat{k}_1^2 \hat{k}_2^2 \hat{k}_3^2 \hat{k}_4^2 \hat{k}_5^2 \hat{k}_6^2\int_R^\infty \Big(\frac2\pi\Big)^3 r^{-8}\d r,$$
where the sums are over tuples of distinct indices for a total of ${6\choose 3}=20$, ${6\choose 4}=15$, and ${6\choose 5}=6$ summands, respectively.

Addition of the error bounds $E_{\k,1}+\dotsb+E_{\k,6}$ yields error bounds for $I_{\k}$, which in turn give error bounds for the matrix coefficients $Q_{\mathbf{m},\mathbf{n}}$.
Applying Schur's test to each block with a fixed $D$ individually shows that the matrix consisting of the error bounds has operator norm less than $10^{-5}$.
These steps were again performed via {\it Mathematica}.
Since $10^{-5}$ is smaller than the minimal eigenvalues shown in Table~\ref{table:Eigenvalues}, we can conclude that the matrix $(Q_{\mathbf{m},\mathbf{n}})_{\mathbf{m},\mathbf{n}\in X}$ is positive definite.
This completes the proof of Theorem \ref{thm:main}.

\section{Further remarks}\label{sec:Remarks}

We conclude our discussion with several observations.

Table \ref{table:Eigenvalues} reveals that the smallest eigenvalues of the block $D$ is increasing in the parameter $D\ge 0$.
It might be interesting to find an analytic explanation of this fact.

Zooming into the main block $D=0$,
Figure \ref{fig:EigenvaluesT} shows the non-zero eigenvalues of this block. There is a cluster of very small eigenvalues.
The corresponding eigenvectors seem to suggest that many of these small eigenvalues are related to functions on the circle that are mainly supported in neighborhoods of two antipodally symmetric points.
These functions are close competitors of constants for being maximizers.
A line of attack on this problem, say for larger  or infinite bandwidth, might be to understand how to analytically separate the effect of these antipodal functions.
The remaining eigenvalues may be sufficiently far from zero to allow for crude estimation.

\begin{figure}[htb]
\centering
\includegraphics[height=6cm]{ev_s=0_}
\caption{ Plot of the eigenvalues
  $0<\lambda_{1}\leq\lambda_{2}\leq\ldots\leq\lambda_{127}$
  of the approximation to the block $D=0$.}
\label{fig:EigenvaluesT}
\end{figure}

We calculated the entries of the quadratic form $Q$
numerically. A look at these entries reveals some nice patterns such as circular structures, shown Figures \ref{fig:Circle30v1} and \ref{fig:Circle30v2} below. We do not know how to exactly describe or explain these structures independently of the numerical calculations. These structures merit further investigation.
 Each of the  six figures below shows a row of the block $D=0$.
This corresponds to fixing an index ${\bf m}_0$, and plotting the matrix entries corresponding to $Q_{{\bf m}_0,{\bf n}}$, where ${\bf n}$ ranges over all admissible values. Since $n_1+n_2+n_3=0$, we may
parametrize the entries of the row by $(n_1,n_2)$, which ranges
in a hexagonal region in $\Z^2$, shown in the figures as complement
of the shaded region.

\begin{figure}[htb]
\centering
\includegraphics[height=4.1cm]{C_-24,0,24_30}
\includegraphics[height=4.1cm]{C_-12,0,12_30}
\includegraphics[height=4.1cm]{C_-6,0,6_30}
\caption{${\bf m}_0=(-24,0,24)$, ${\bf m}_0=(-12,0,12)$,  ${\bf m}_0=(-6,0,6)$.}
\label{fig:Circle30v1}
\end{figure}

\begin{figure}[htb]
\centering
\includegraphics[height=4.1cm]{C_-20,8,12_30}
\includegraphics[height=4.1cm]{C_-12,6,6_30}
\includegraphics[height=4.1cm]{C_-10,2,8_30}
\caption{${\bf m}_0=(-20,8,12)$, ${\bf m}_0=(-12,6,6)$, ${\bf m}_0=(-10,2,8)$.}
\label{fig:Circle30v2}
\end{figure}

We close with a remark on the central Bessel integral
$$I_{(0,0,0,0,0,0)}=\int_0^\infty J_0^6(r)r\d r,$$
which up to a factor $(2\pi)^4$ is the conjectured sharp constant
$\Phi({\bf 1})$
in the Tomas--Stein adjoint restriction inequality.
This integral appears in the following related context.
An {\it $n$-step uniform random walk} is a walk in the plane that starts at the origin and consists of $n$ steps of length 1 each taken into a uniformly random direction.

If $p_n$ denotes the radial density of the distance travelled after $n$ steps, then it is a classical result that $p_5(1)=I_{(0,0,0,0,0,0)}$, see e.g. \cite{BSWZ12}. In the same paper, the exact value of the integral
$$p_4(1)=\int_0^\infty J_0^5(r)r\d r=\frac{1}{2\pi^2}\sqrt{\frac{\Gamma(\frac1{15})\Gamma(\frac2{15})\Gamma(\frac4{15})\Gamma(\frac8{15})}{5\Gamma(\frac7{15})\Gamma(\frac{11}{15})\Gamma(\frac{13}{15})\Gamma(\frac{14}{15})}}$$
is determined resorting to  striking modularity properties of the function $p_4$, see \cite[Theorems 4.9 and 5.1]{BSWZ12}.
The corresponding problem with a sixth power remains open.

\section*{Acknowledgements}
We thank Emanuel Carneiro, Damiano Foschi and Felipe Gon\c{c}alves for stimulating discussions.
The software {\it Mathematica} was used to perform the numerical tasks described in \S\ref{sec:Numerics} and \S\ref{sec:Remarks}.
The authors acknowledge support by the Hausdorff Center for Mathematics and the Deutsche Forschungsgemeinschaft through the Collaborative Research Center 1060.