#

On error sums formed by rational approximations with split denominators

Thomas Baruchel and Carsten Elsner Lycée naval, Centre d’instruction naval,
Brest, France, e-mail<EMAIL_ADDRESS>
Fachhochschule für die Wirtschaft, University of Applied Sciences, Freundallee
15, D-30173 Hannover, Germany, e-mail<EMAIL_ADDRESS>
###### Abstract

In this paper we consider error sums of the form

$\sum_{m=0}^{\infty}\varepsilon_{m}\Big{(}\,b_{m}\alpha-\frac{a_{m}}{c_{m}}\,\Big{)}\,,$

where $\alpha$ is a real number, $a_{m}$, $b_{m}$, $c_{m}$ are integers, and
$\varepsilon_{m}=1$ or $\varepsilon_{m}={(-1)}^{m}$. In particular, we
investigate such sums for

$\alpha\in\big{\\{}\pi,e,e^{1/2},e^{1/3},\dots,\log(1+t),\zeta(2),\zeta(3)\big{\\}}$

and exhibit some connections between rational coefficients occurring in error
sums for Apéry’s continued fraction for $\zeta(2)$ and well-known integer
sequences. The concept of the paper generalizes the theory of ordinary error
sums, which are given by $b_{m}=q_{m}$ and $a_{m}/c_{m}=p_{m}$ with the
convergents $p_{m}/q_{m}$ from the continued fraction expansion of $\alpha$.

Keywords: Errorr sums, continued fractions
AMS Subject Classification: 11J70, 11J04, 33B10, 33C05, 11M06.

##  1 Introduction

Let $\alpha$ be a real number. We assume that there is a sequence
$B:={(b_{n})}_{n\geq 0}$ of integers, a sequence $R:={(r_{n})}_{n\geq 0}$ of
rationals $r_{n}=a_{n}/c_{n}$, say, with $a_{n}\in{\mathbb{Z}}$ and
$c_{n}\in{\mathbb{N}}$, and a real number $\omega>1$ satisfying

$\big{|}b_{n}c_{n}\alpha-
a_{n}\big{|}\,\ll\,\frac{c_{n}}{\omega^{n}}\qquad(n\geq 0)\,.$ (1.1)

This is equivalent with

$\big{|}b_{n}\alpha-
r_{n}\big{|}\,=\,\Big{|}b_{n}\alpha-\frac{a_{n}}{c_{n}}\Big{|}\,\ll\,\frac{1}{\omega^{n}}\qquad(n\geq
0)\,.$ (1.2)

We consider the fraction $a_{n}/b_{n}c_{n}$ as a rational approximation of
$\alpha$ with split denominator $b_{n}c_{n}$. Since $\omega>1$, the error sums

$\displaystyle\mathcal{E}^{*}\big{(}B,R,\alpha\big{)}$ $\displaystyle:=$
$\displaystyle\sum_{m=0}^{\infty}\big{(}b_{m}\alpha-
r_{m}\big{)}\,=\,\sum_{m=0}^{\infty}\Big{(}b_{m}\alpha-\frac{a_{m}}{c_{m}}\Big{)}\,,$
(1.3) $\displaystyle\mathcal{E}\big{(}B,R,\alpha\big{)}$ $\displaystyle:=$
$\displaystyle\sum_{m=0}^{\infty}\big{|}b_{m}\alpha-
r_{m}\big{|}\,=\,\sum_{m=0}^{\infty}\Big{|}b_{m}\alpha-\frac{a_{m}}{c_{m}}\Big{|}$
(1.4)

exist. Let ${(p_{n}/q_{n})}_{n\geq 0}$ be the sequence of convergents of
$\alpha$ defined by $p_{n}/q_{n}=\langle a_{0};a_{1},a_{2},\dots a_{n}\rangle$
from the regular continued fraction expansion

$\alpha\,=\,\langle
a_{0};a_{1},a_{2},\dots\rangle\,=\,a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{a_{2}+\cfrac{1}{\ddots}}}$

of $\alpha$. The error sums of $\alpha$ for $B={(q_{n})}_{n\geq 0}$ and
$R={(p_{n})}_{n\geq 0}$, namely

$\displaystyle\mathcal{E}^{*}(\alpha)$ $\displaystyle:=$
$\displaystyle\mathcal{E}^{*}\big{(}B,R,\alpha\big{)}\,=\,\sum_{m=0}^{\infty}\big{(}q_{m}\alpha-
p_{m}\big{)}\,,$ $\displaystyle\mathcal{E}(\alpha)$ $\displaystyle:=$
$\displaystyle\mathcal{E}\big{(}B,R,\alpha\big{)}\,=\,\sum_{m=0}^{\infty}\big{|}q_{m}\alpha-
p_{m}\big{|}\,,$

were already studied in some papers [6, 7, 8, 9]. We call
$\mathcal{E}^{*}(\alpha)$ and $\mathcal{E}(\alpha)$ ordinary error sums.
Conversely, for $B={(1)}_{n\geq 0}$ and $R={(p_{n}/q_{n})}_{n\geq 0}$, until
now nobody has found any remarkable approach to the error sums

$\displaystyle\mathcal{E}^{*}(B,R,\alpha)$ $\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\Big{(}\alpha-\frac{p_{m}}{q_{m}}\Big{)}\,,$
$\displaystyle\mathcal{E}(B,R,\alpha)$ $\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}\Big{|}\alpha-\frac{p_{m}}{q_{m}}\Big{|}\,.$

In this paper we focus our interest on the series in (1.3) and (1.4) in the
case of particular values of $\alpha$ and well-known rational approximations
of the form

$0\,<\,\Big{|}b_{n}\alpha-\frac{a_{n}}{c_{n}}\Big{|}\,\ll\,\frac{1}{\omega^{n}}\qquad(n\geq
0)\,.$ (1.5)

Among others we are going to study the numbers

$\alpha\,\in\,\Big{\\{}\,\pi,\,e^{1/l},\,\frac{\log\rho}{\sqrt{5}},\,\log(1+t),\,\zeta(2),\,\zeta(3)\,\Big{\\}}\,,$

where $l=1,2,\dots$, $e=\exp(1)$, $\rho=(1+\sqrt{5})/2$, and $-1<t\leq 1$, and
we shall investigate extraordinary properties of corresponding error sums
(1.3) and (1.4).

##  2 Ordinary error sums for values of the exponential function

Ordinary error sums connected with the exponential function are studied in [1,
10]. Here, our goal is to express this usual error sums itselves by a non-
regular continued fraction. For this purpose we express the error integral

$\mbox{erf\,}(z)\,=\,\frac{2}{\sqrt{\pi}}\int_{0}^{z}e^{-t^{2}}\,dt$

by a hypergeometric series, which again can be transformed into a Gauss-type
continued fraction.

######  Theorem 2.1.

Let $l\geq 2$ be an integer, and let $p_{n}/q_{n}$ denote the convergents of
$e^{1/l}$. Then we have

$\displaystyle\mathcal{E}(e^{1/l}\,)$ $\displaystyle=$
$\displaystyle\sum_{n\geq
0}\big{|}e^{1/l}q_{n}-p_{n}\big{|}\,=\,e^{1/l}\sqrt{\frac{\pi}{l}}\mbox{erf\,}\Big{(}\,\frac{1}{\sqrt{l}}\,\Big{)}\,=\,\frac{2e^{1/l}}{\sqrt{l}}\int_{0}^{1/\sqrt{l}}e^{-t^{2}}\,dt$
$\displaystyle=$
$\displaystyle\cfrac{1/l}{1/2-\cfrac{1/2l}{3/2+\cfrac{2/2l}{5/2-\cfrac{3/2l}{7/2+\cfrac{4/2l}{9/2-\cfrac{5/2l}{11/2+\ddots\cfrac{{(-1)}^{m}m/2l}{(2m+1)/2+\ddots}}}}}}}\qquad(m\geq
1)\,.$

Proof: The first identity of the theorem expressing $\mathcal{E}(e^{1/l})$ by
an error integral is already known from [1, 10]. In order to prove the
continued fraction expansion, we set

$f(z)\,:=\,\frac{\sqrt{\pi}}{2}ze^{z^{2}}\mbox{erf\,}(z)\,=\,ze^{z^{2}}\int_{0}^{z}e^{-t^{2}}\,dt\,.$

We express $f(z)$ in terms of a hypergeometric function
${}_{1}F_{1}(\alpha,\beta;z^{2})$.

$\displaystyle f(z)$ $\displaystyle=$ $\displaystyle
ze^{z^{2}}\int_{0}^{z}\sum_{\nu=0}^{\infty}\frac{{(-1)}^{\nu}t^{2\nu}}{\nu!}\,dt\,=\,ze^{z^{2}}\sum_{\nu=0}^{\infty}\frac{{(-1)}^{\nu}z^{2\nu+1}}{(2\nu+1)\nu!}$
$\displaystyle=$ $\displaystyle
z^{2}\left(\sum_{\mu=0}^{\infty}\frac{z^{2\mu}}{\mu!}\right)\left(\sum_{\nu=0}^{\infty}\frac{{(-1)}^{\nu}z^{2\nu}}{(2\nu+1)\nu!}\right)$
$\displaystyle=$ $\displaystyle
z^{2}\sum_{\mu=0}^{\infty}\,\sum_{\nu=0}^{\infty}\frac{{(-1)}^{\nu}z^{2(\nu+\mu)}}{(2\nu+1)\nu!\mu!}\,=\,z^{2}\sum_{k=0}^{\infty}\Big{(}\mathop{\sum_{\mu=0}^{\infty}\,\sum_{\nu=0}^{\infty}}_{\mu+\nu=k}\frac{{(-1)}^{\nu}}{(2\nu+1)\nu!\mu!}\Big{)}z^{2k}$
$\displaystyle=$ $\displaystyle
z^{2}\sum_{k=0}^{\infty}\Big{(}\,\sum_{\nu=0}^{k}\frac{{(-1)}^{\nu}}{(2\nu+1)\nu!(k-\nu)!}\Big{)}z^{2k}\,=\,z^{2}\sum_{k=0}^{\infty}\frac{1}{k!}\Big{(}\,\sum_{\nu=0}^{k}\frac{{(-1)}^{\nu}{k\choose\nu}}{2\nu+1}\Big{)}z^{2k}\,.$

From [15, p. 68], Remark 8.5, we have the following formula (with $k$ replaced
by $\nu$ and $n$ replaced by $k$)

$\frac{1}{d^{k}{[k]}_{k}}\sum_{\nu=0}^{k}\frac{{(-1)}^{\nu}{k\choose\nu}}{c+\nu
d}\,=\,\frac{1}{c(c+d)(c+2d)\cdots(c+kd)}\,,$

where ${[k]}_{k}=k!$. Setting $c=1$ and $d=2$, it follows that

$\frac{1}{k!}\,\sum_{\nu=0}^{k}\frac{{(-1)}^{\nu}{k\choose\nu}}{2\nu+1}\,=\,\frac{2^{k}}{1\cdot
3\cdot 5\cdots(2k+1)}\,=\,\frac{1}{{(3/2)}_{k}}\,.$

This gives

$f(z)\,=\,z^{2}\sum_{k=0}^{\infty}\frac{z^{2k}}{{(3/2)}_{k}}\,=\,z^{2}\sum_{k=0}^{\infty}\frac{{(1)}_{k}}{k!{(3/2)}_{k}}z^{2k}\,=\,z^{2}\,{}_{1}F_{1}\big{(}1,3/2;z^{2}\big{)}\,.$

The function ${}_{1}F_{1}\big{(}1,3/2;z^{2}\big{)}$ can be expressed by a
Gauss-type continued fraction. Using formula (8) on page 123 in [16] with
$\gamma=3/2$ and $x=z^{2}$, we have

${}_{1}F_{1}\big{(}1,3/2;z^{2}\big{)}\,=\,\cfrac{1/2}{1/2-\cfrac{z^{2}/2}{3/2+\cfrac{2z^{2}/2}{5/2-\cfrac{3z^{2}/2}{7/2+\cfrac{4z^{2}/2}{9/2-\cfrac{5z^{2}/2}{11/2+\ddots\cfrac{{(-1)}^{m}mz^{2}/2}{(2m+1)/2+\ddots}}}}}}}\qquad(m\geq
1)$

Hence the continued fraction expansion given by the theorem follows from

$\sum_{n\geq
0}\big{|}e^{1/l}q_{n}-p_{n}\big{|}\,=\,2\cdot\frac{\sqrt{\pi}}{2}\cdot\frac{e^{1/l}}{\sqrt{l}}\cdot\mbox{erf\,}\big{(}1/\sqrt{l}\,\big{)}\,=\,2f\big{(}1/\sqrt{l}\,\big{)}\,=\,\frac{2}{l}\,{}_{1}F_{1}\big{(}1,3/2;1/l\big{)}\,.$

∎
We point out the particular case $z=1$.

######  Corollary 2.1.

We have

$\displaystyle{}_{1}F_{1}\big{(}1,3/2;1\big{)}$ $\displaystyle=$
$\displaystyle e\int_{0}^{1}e^{-t^{2}}\,dt$ $\displaystyle=$
$\displaystyle\mathcal{E}_{MC}(e)\,=\,e-2+\sum_{n=1}^{\infty}\,\sum_{b=1}^{a_{n}}\big{|}(bq_{n-1}+q_{n-2})e-(bp_{n-1}+p_{n-2})\big{|}$
$\displaystyle=$
$\displaystyle\cfrac{1/2}{1/2-\cfrac{1/2}{3/2+\cfrac{1}{5/2-\cfrac{3/2}{7/2+\cfrac{2}{9/2-\cfrac{5/2}{11/2+\ddots\cfrac{{(-1)}^{m}m/2}{(2m+1)/2+\ddots}}}}}}}\qquad(m\geq
1)\,,$

where $\mathcal{E}_{MC}(e)$ is the error sum of $e$ taking into account all
the minor convergents of

$e\,=\,\langle 2;1,2,1,1,4,1,\dots\rangle\,=\,\langle
2;a_{1},a_{2},a_{3},\dots\rangle\,.$ (2.1)

Proof: The formula

$\mathcal{E}_{MC}(e)\,=\,e\int_{0}^{1}e^{-t^{2}}\,dt$

follows by (2.1) using

$\mathcal{E}_{MC}(e)\,=\,e-1+\sum_{\nu=0}^{\infty}{(-1)}^{\nu+1}\big{(}q_{\nu}e-p_{\nu}\big{)}\Big{(}\,\frac{1}{2}(1+a_{\nu+1})a_{\nu+1}-a_{\nu+2}\,\Big{)}$

and the formulas

$\displaystyle q_{3m-1}e-p_{3m-1}$ $\displaystyle=$
$\displaystyle-\int_{0}^{1}\frac{x^{m+1}{(x-1)}^{m}}{m!}e^{x}\,dx\,,$
$\displaystyle q_{3m}e-p_{3m}$ $\displaystyle=$
$\displaystyle-\int_{0}^{1}\frac{x^{m}{(x-1)}^{m+1}}{m!}e^{x}\,dx\,,$
$\displaystyle q_{3m+1}e-p_{3m+1}$ $\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{x^{m+1}{(x-1)}^{m+1}}{(m+1)!}e^{x}\,dx$

due to H. Cohn [5]. ∎
Let $l\geq 1$. Then we know from [18, p. 193] that the numbers $e^{1/l}$ and
$\int_{0}^{1/\sqrt{1/l}}e^{-t^{2}}\,dt$ are algebraically independent over
${\mathbb{Q}}$. This proves

######  Corollary 2.2.

Let $l\geq 2$ be an integer. Then the numbers $\mathcal{E}(e^{1/l}\,)$ and
$\mathcal{E}_{MC}(e)$ are transcendental.

##  3 Error sums for $\pi$ and $(\log\rho)/\sqrt{5}$

In [11], A.Klauke and the second-named author have found new continued
fractions for $1/\pi$ and $(\log\rho)/\sqrt{5}$. In this section we are going
to apply these results to compute the corresponding error sums and to decide
on their algebraic character. We start with the continued fraction for
$1/\pi$.
1.) From Theorem 8 in [11] and its proof we have the following results.

$\displaystyle\frac{1}{\pi}$ $\displaystyle=\frac{3}{10}\begin{array}[]{c}\\\
-\end{array}\frac{14}{25}\begin{array}[]{c}\\\
-\end{array}\frac{110}{171}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{\frac{1}{9}m(m-1)(2m-1)(2m+1)(4m-5)(4m+3)}{(4m+1)(4m^{2}+2m-1)}\begin{array}[]{c}\\\
-\ldots\end{array}$ $\displaystyle=\frac{p_{0}}{q_{0}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{1}}{q_{1}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{2}}{q_{2}}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{p_{m}}{q_{m}}\begin{array}[]{c}\\\
-\ldots\end{array}\qquad(m\geq 2)\,.$

Let $n=0,1,2,\dots$. Set

$\displaystyle B_{n}$ $\displaystyle:=$ $\displaystyle\frac{2\cdot
4^{n+1}}{n!}\sum_{k=0}^{n}{n\choose k}(2k+3){\big{(}k+5/2\big{)}}_{n}\,,$
$\displaystyle A_{n}$ $\displaystyle:=$ $\displaystyle\frac{2\cdot
4^{n+1}}{n!}\sum_{k=0}^{n}\sum_{\nu=0}^{k}{(-1)}^{k+\nu}{n\choose
k}\cfrac{(2k+3){\big{(}k+5/2\big{)}}_{n}}{2k-2\nu+1}+{(-4)}^{n+1}\,.$

Here,

${\big{(}k+5/2\big{)}}_{n}\,=\,(k+5/2)(k+7/2)(k+9/2)\cdots(k+n+3/2)\,.$

Note that $A_{n}$ is a rational number, but no integer, while $B_{n}/4$ is an
integer. Then, for $n\geq 0$, one has

$\begin{aligned} \frac{p_{0}}{q_{0}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{1}}{q_{1}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{2}}{q_{2}}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{p_{n}}{q_{n}}\end{aligned}\,=\,\frac{B_{n}}{4A_{n}}\,,$

and

$0\,<\,A_{n}-\frac{\pi
B_{n}}{4}\,=\,4(n+3/2)\int_{0}^{1}\frac{t\sqrt{1-t}}{2-t}{\Big{(}\,\frac{4t(1-t)}{2-t}\,\Big{)}}^{n}\,dt\,.$

For $0\leq t\leq 1$ the rational function $4t(1-t)/(2-t)$ takes its maximum
$2(6-4\sqrt{2})$ at the point $2-\sqrt{2}$. Therefore, it follows that

$0\,<\,A_{n}-\frac{\pi
B_{n}}{4}\,<\,8(n+3/2){(6-4\sqrt{2})}^{n}\int_{0}^{1}\frac{t\sqrt{1-t}}{2-t}\,dt\,=\,8(10/3-\pi)(n+3/2){(6-4\sqrt{2})}^{n}\,.$

The integral on the right-hand side is a Pochhammer integral of a certain
hypergeometric function. We show the analogous details below in part 2.) which
is devoted to the number $(\log\rho)/\sqrt{5}$.

For (1.3) and (1.4) we define the sequences $B:={(b_{n})}_{n\geq 0}$ and
$R:={(r_{n})}_{n\geq 0}$ by $b_{n}:=-B_{n}/4$ and $r_{n}=-A_{n}$. Then we have
the error sums

$\displaystyle\mathcal{E}^{*}(B,R,\pi)$ $\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}(b_{m}\pi-r_{m})\,=\,-\mathcal{E}(B,R,\pi)$
$\displaystyle=$
$\displaystyle-4\int_{0}^{1}\frac{t\sqrt{1-t}}{2-t}\sum_{m=0}^{\infty}(m+3/2){\Big{(}\,\frac{4t(1-t)}{2-t}\,\Big{)}}^{m}\,dt$
$\displaystyle=$
$\displaystyle-4\int_{0}^{1}\frac{t\sqrt{1-t}}{2-t}\cdot\frac{(2-t)(4t^{2}-7t+6)}{2{(4t^{2}-5t+2)}^{2}}\,dt$
$\displaystyle=$
$\displaystyle-4\int_{0}^{1}\frac{u^{2}(1-u^{2})(4u^{4}-u^{2}+3)}{{(4u^{4}-3u^{2}+1)}^{2}}\,du\,.$

Here we have introduced the new variable $u:=\sqrt{1-t}$. Computing this
integral, we have the following theorem.

######  Theorem 3.1.

For the sequences $B:={(b_{n})}_{n\geq 0}$ and $R:={(r_{n})}_{n\geq 0}$
defined by $b_{n}:=-B_{n}/4$ and $r_{n}=-A_{n}$ we have

$\mathcal{E}^{*}(B,R,\pi)\,=\,-\mathcal{E}(B,R,\pi)\,=\,\frac{\sqrt{7}}{49}\log\Big{(}\,\frac{3-\sqrt{7}}{3+\sqrt{7}}\,\Big{)}-\frac{3\pi}{2}-\frac{4}{7}\,=\,-5.4333111067784\dots.$

Expressing $\pi$ by $\pi=\frac{2\log i}{i}$, we see that
$\mathcal{E}(B,R,\pi)$ is a nonvanishing linear form in logarithms with
algebraic arguments and algebraic coefficients. Then, by Theorem 2.2 in [2],
we have the following corollary.

######  Corollary 3.1.

For the sequences $B:={(b_{n})}_{n\geq 0}$ and $R:={(r_{n})}_{n\geq 0}$
defined by $b_{n}:=-B_{n}/4$ and $r_{n}=-A_{n}$ the error sum
$\mathcal{E}(B,R,\pi)$ is transcendental, and so is the error sum
$\mathcal{E}^{*}(B,R,\pi)$.

2.) From Theorem 6 in [11] and its proof we have the following results.

$\displaystyle\frac{\sqrt{5}}{\log\rho}$
$\displaystyle=\frac{60}{13}\begin{array}[]{c}\\\
-\end{array}\frac{7}{80}\begin{array}[]{c}\\\
-\end{array}\frac{110}{522}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{\frac{1}{9}m(m-1)(2m-1)(2m+1)(4m-5)(4m+3)}{2(4m+1)(6m^{2}+3m-1)}\begin{array}[]{c}\\\
-\ldots\end{array}$ $\displaystyle=\frac{p_{0}}{q_{0}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{1}}{q_{1}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{2}}{q_{2}}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{p_{m}}{q_{m}}\begin{array}[]{c}\\\
-\ldots\end{array}\qquad(m\geq 2)\,.$

Let $n=0,1,2,\dots$. Set (cf. (25) in [11] with $c=d=1$)

$\displaystyle D_{n}$ $\displaystyle:=$ $\displaystyle\frac{5\cdot
4^{n+1}}{n!}\sum_{k=0}^{n}{(-1)}^{n+k}{n\choose
k}(2k+3){\big{(}k+5/2\big{)}}_{n}5^{k}\,,$ $\displaystyle C_{n}$
$\displaystyle:=$ $\displaystyle
4^{n}+\frac{4^{n+1}}{n!}\sum_{k=0}^{n}\sum_{\nu=0}^{k}{(-1)}^{n+k}{n\choose
k}\cfrac{(2k+3){\big{(}k+5/2\big{)}}_{n}5^{\nu}}{2k-2\nu+1}\,.$

Applying Lemma 6 in [11] with $x=\tau=1$, we find that

$\begin{aligned} \frac{p_{0}}{q_{0}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{1}}{q_{1}}\begin{array}[]{c}\\\
-\end{array}\frac{p_{2}}{q_{2}}\begin{array}[]{c}\\\
-\ldots-\end{array}\frac{p_{n}}{q_{n}}\end{aligned}\,=\,\frac{D_{n}}{C_{n}}\,,$

and

$0\,<\,C_{n}-\frac{D_{n}\log\rho}{\sqrt{5}}\,=\,\frac{{(5/2)}_{n}(n+1)!}{4{(5/2)}_{2n+1}}\,{}_{2}F_{1}\left(\begin{array}[]{c}n+1\quad
n+2\\\ 2n+7/2\end{array}\Big{|}-\frac{1}{4}\right)\,.$

We define the sequences $B:={(b_{n})}_{n\geq 0}$ and $R:={(r_{n})}_{n\geq 0}$
by $b_{n}:=-D_{n}$ and $r_{n}=-C_{n}$. Then we have the error sum

$\mathcal{E}^{*}\Big{(}\,B,R,\frac{\log\rho}{\sqrt{5}}\,\Big{)}\,=\,\sum_{m=0}^{\infty}\Big{(}\,b_{m}\frac{\log\rho}{\sqrt{5}}-r_{m}\,\Big{)}\\\
=-\frac{1}{4}\sum_{m=0}^{\infty}\frac{{(5/2)}_{m}(m+1)!}{4{(5/2)}_{2m+1}}\,{}_{2}F_{1}\left(\begin{array}[]{c}m+1\quad
m+2\\\ 2m+7/2\end{array}\Big{|}-\frac{1}{4}\right)\,.$

To compute this error sum, the method is the same as used above for the error
sum of $\mathcal{E}(\pi)$. First we express the hypergeometric function by
Pochhammer’s integral. Let $a,b,c,z$ be complex numbers satisfying $|z|<1$,
$\Re(c-b)>0$, and $\Re(b)>0$. Then we have the identity

${}_{2}F_{1}\left(\begin{array}[]{c}a\quad b\\\
c\end{array}\Big{|}z\right)\,=\,\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_{0}^{1}t^{b-1}{(1-t)}^{c-b-1}{(1-zt)}^{-a}\,dt\,,$

cf. [19, p. 20]. The conditions are fulfilled for $a=n+1$, $b=n+2$,
$c=2n+7/2$, and $z=-1/4$, where $n=0,1,2,\dots$. Hence, it follows that

${}_{2}F_{1}\left(\begin{array}[]{c}n+1\quad n+2\\\
2n+7/2\end{array}\Big{|}-\frac{1}{4}\right)\,=\,\frac{\Gamma(2n+7/2)}{\Gamma(n+2)\Gamma(n+3/2)}\int_{0}^{1}t^{n+1}{(1-t)}^{n+1/2}{\Big{(}\,1+\frac{t}{4}\,\Big{)}}^{-n-1}\,dt\,.$

In order to simplify the above expressions we need two identities involving
Pochhammer’s symbol ([19, p. 239]).

$\displaystyle\frac{{(5/2)}_{n}}{{(5/2)}_{2n+1}}$ $\displaystyle=$
$\displaystyle\frac{1}{{(n+5/2)}_{n+1}}\,,$
$\displaystyle\frac{\Gamma(2n+7/2)}{\Gamma(n+3/2)}$ $\displaystyle=$
$\displaystyle(n+3/2){(n+5/2)}_{n+1}\,.$

Collecting together all the above results, it follows that

$\displaystyle\mathcal{E}^{*}\Big{(}\,B,R,\frac{\log\rho}{\sqrt{5}}\,\Big{)}$
$\displaystyle=$
$\displaystyle-\frac{1}{4}\sum_{m=0}^{\infty}(m+3/2)\int_{0}^{1}\frac{t^{m+1}{(1-t)}^{m+1/2}}{{(1+t/4)}^{m+1}}\,dt$
$\displaystyle=$
$\displaystyle-\frac{1}{4}\int_{0}^{1}\frac{4t\sqrt{1-t}}{4+t}\sum_{m=0}^{\infty}(m+3/2){\Big{(}\,\frac{4t(1-t)}{4+t}\,\Big{)}}^{m}\,dt$
$\displaystyle=$
$\displaystyle-\frac{1}{2}\int_{0}^{1}\frac{t(4t^{2}-t+12)\sqrt{1-t}}{{(4t^{2}-3t+4)}^{2}}\,dt$
$\displaystyle=$
$\displaystyle\int_{0}^{1}\frac{u^{2}(1-u^{2})(4u^{4}-7u^{2}+15)}{{(4u^{4}-5u^{2}+5)}^{2}}\,du\,,$

where $u=\sqrt{1-t}$. This proves

######  Theorem 3.2.

For the sequences $B:={(b_{n})}_{n\geq 0}$ and $R:={(r_{n})}_{n\geq 0}$
defined by $b_{n}:=-D_{n}$ and $r_{n}=-C_{n}$ we have

$\displaystyle\mathcal{E}^{*}\Big{(}\,B,R,\frac{\log\rho}{\sqrt{5}}\,\Big{)}$
$\displaystyle=$
$\displaystyle\cfrac{\sqrt{124\sqrt{5}-265}\log\Big{(}\,1+\cfrac{\sqrt{5}}{2}-\cfrac{\sqrt{4\sqrt{5}+5}}{2}\,\Big{)}-\sqrt{124\sqrt{5}+265}\arccos\Big{(}\,\cfrac{\sqrt{5}}{2}-1\,\Big{)}}{55\sqrt{11}}+\frac{1}{11}$
$\displaystyle=$ $\displaystyle-0.1210649459927\dots\,.$

Using $\arccos z=\frac{1}{i}\log(z+\sqrt{z^{2}-1})$, we obtain by Theorem 2.2
in [2] the following result.

######  Corollary 3.2.

For the sequences $B:={(b_{n})}_{n\geq 0}$ and $R:={(r_{n})}_{n\geq 0}$
defined by $b_{n}:=-D_{n}$ and $r_{n}=-C_{n}$ the error sum
$\mathcal{E}(B,R,\log\rho/\sqrt{5})$ is transcendental.

##  4 An error sum for $\log(1+t)$

In this section we generalize a concept from the proof of Theorem 3 in [12],
where a nonregular continued fraction for $\log 2$ is established. First we
shall prove a continued fraction expansion for $\log(1+t)$ with $-1<t\leq 1$,
namely

$\log(1+t)\,=\,\frac{2t}{2+t}\begin{array}[]{c}\\\
-\end{array}\frac{1^{2}t^{2}}{3(2+t)}\begin{array}[]{c}\\\
-\end{array}\frac{2^{2}t^{2}}{5(2+t)}\begin{array}[]{c}\\\
-\end{array}\frac{3^{2}t^{2}}{7(2+t)}\begin{array}[]{c}\\\
-\dots-\end{array}\frac{m^{2}t^{2}}{(2m+1)(2+t)}\begin{array}[]{c}\\\
-\dots\end{array}\,,$ (4.1)

where $m=1,2,\dots$. O.Perron [16, p. 152] cites by equation (7) the continued
fraction

$\log(1+t)\,=\,\frac{t}{1}\begin{array}[]{c}\\\
+\end{array}\frac{1^{2}t}{2}\begin{array}[]{c}\\\
+\end{array}\frac{1^{2}t}{3}\begin{array}[]{c}\\\
+\end{array}\frac{2^{2}t}{4}\begin{array}[]{c}\\\
+\end{array}\frac{2^{2}t}{5}\begin{array}[]{c}\\\
+\end{array}\frac{3^{2}t}{6}\begin{array}[]{c}\\\
+\end{array}\frac{3^{2}t}{7}\begin{array}[]{c}\\\ +\dots\end{array}\,.$

Here we shall give full details of the proof, since a new argument is needed
in H.Cohen’s method [4] established for Apéry’s irregular continued fractions
of $\zeta(2)$ and $\zeta(3)$, and we need the details in order to compute the
error sum. Similar to Apéry’s approach we have to handle with combinatorial
series.
In the sequel we fix a real number $t$ with $-1<t\leq 1$. Let $n\geq 0$ be an
integer. We define two combinatorial series by

$\displaystyle B_{n}$ $\displaystyle:=$ $\displaystyle\sum_{k=0}^{n}{n\choose
k}{n+k\choose k}t^{n-k}\,,$ $\displaystyle A_{n}$ $\displaystyle:=$
$\displaystyle\sum_{k=0}^{n}{n\choose k}{n+k\choose k}t^{n-k}c_{k}\,,$

where

$c_{k}\,:=\,\sum_{m=1}^{k}\frac{{(-1)}^{m-1}t^{m}}{m}\,.$

By applying Zeilberger’s algorithm [14, Ch. 7] (Algorithm 7.1) on a computer
algebra system, it turns out that the numbers $B_{n}$ satisfy the linear
three-term recurrence formula

$(n+1)X_{n+1}-(2n+1)(2+t)X_{n}+nt^{2}X_{n-1}\,=\,0\qquad(n\geq 1)\,.$ (4.2)

In the sequel we prove this formula for $X_{n}=B_{n}$ without using a
computer, since we need the details to show that even $X_{n}=A_{n}$ satisfy
the recurrence. Let $k,n$ denote integers. Set

$\displaystyle\lambda_{n,k}$ $\displaystyle:=$ $\displaystyle{n\choose
k}{n+k\choose k}t^{n-k}\,,$ $\displaystyle B_{n,k}$ $\displaystyle:=$
$\displaystyle-(4n+2)\lambda_{n,k}\,,$ $\displaystyle A_{n,k}$
$\displaystyle:=$ $\displaystyle B_{n,k}c_{k}\,,$ $\displaystyle S_{n,k}$
$\displaystyle:=$
$\displaystyle(n+1)\lambda_{n+1,k}c_{k}-(2n+1)(2+t)\lambda_{n,k}c_{k}+nt^{2}\lambda_{n-1,k}c_{k}\,.$

Note that ${n\choose k}=0$ for $k<0$ or $k>n$, which implies that
$A_{n,n+1}=B_{n,n+1}=A_{n,-1}=B_{n,-1}=0$. One easily verfies the
identities111We should like to point out that there is a misprint in the
formula for $\lambda_{n+1,k}/\lambda_{n,k}$ in [12].

$\displaystyle\frac{\lambda_{n,k-1}}{\lambda_{n,k}}$ $\displaystyle=$
$\displaystyle\frac{k^{2}t}{(n+k)(n-k+1)}\,,$
$\displaystyle\frac{\lambda_{n+1,k}}{\lambda_{n,k}}$ $\displaystyle=$
$\displaystyle\frac{(n+k+1)t}{n-k+1}\,,$
$\displaystyle\frac{\lambda_{n-1,k}}{\lambda_{n,k}}$ $\displaystyle=$
$\displaystyle\frac{n-k}{(n+k)t}\,,$

which can be applied to prove the identity

$B_{n,k}-B_{n,k-1}\,=\,(n+1)\lambda_{n+1,k}-(2n+1)(2+t)\lambda_{n,k}+nt^{2}\lambda_{n-1,k}\,.$
(4.3)

Summing up on both sides of (4.3) from $k=0$ to $k=n+1$, we obtain

$0\,=\,B_{n,n+1}-B_{n,-1}\,=\,\sum_{k=0}^{n+1}\big{(}B_{n,k}-B_{n,k-1}\big{)}\,=\,(n+1)B_{n+1}-(2n+1)(2+t)B_{n}+nt^{2}B_{n-1}\,,$

which proves (4.2) for $X_{n}=B_{n}$.
Multiplying (4.3) by $c_{k}$, we obtain $S_{n,k}=(B_{n,k}-B_{n,k-1})c_{k}$.
Hence,

$\displaystyle A_{n,k}-A_{n,k-1}$ $\displaystyle=$ $\displaystyle
B_{n,k}c_{k}-B_{n,k-1}c_{k-1}\,=\,\big{(}B_{n,k}-B_{n,k-1}\big{)}c_{k}+B_{n,k-1}\big{(}c_{k}-c_{k-1}\big{)}$
$\displaystyle=$ $\displaystyle
S_{n,k}+B_{n,k-1}\frac{{(-1)}^{k-1}t^{k}}{k}\,.$

Again, we sum up from $k=0$ to $k=n+1$. This gives

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
A_{n,n+1}-A_{n,-1}\,=\,\sum_{k=0}^{n+1}\big{(}A_{n,k}-A_{n,k-1}\big{)}$
$\displaystyle=$
$\displaystyle\sum_{k=0}^{n+1}S_{n,k}-(4n+2)\sum_{k=1}^{n+1}{n\choose
k-1}{n+k-1\choose k-1}t^{n-k+1}\frac{{(-1)}^{k-1}t^{k}}{k}$ $\displaystyle=$
$\displaystyle(n+1)A_{n+1}-(2n+1)(2+t)A_{n}+nt^{2}A_{n-1}-(4n+2)t^{n+1}\sum_{k=0}^{n}{n\choose
k}{n+k\choose k}\frac{{(-1)}^{k}}{k+1}\,.$

Finally Vandermonde’s theorem for the hypergeometric series
${}_{2}F_{1}(n+1,-n,2;1)$ ([19, eq. (1.7.7)]) completes our proof of (4.2) for
$X_{n}=A_{n}$ by

$\sum_{k=0}^{n}{n\choose k}{n+k\choose
k}\frac{{(-1)}^{k}}{k+1}\,=\,{}_{2}F_{1}(n+1,-n,2;1)\,=\,\frac{{(1-n)}_{n}}{{(2)}_{n}}\,=\,0$

for $n\geq 1$. In the next step we prove that

$\lim_{n\to\infty}\frac{A_{n}}{B_{n}}\,=\,\log(1+t)\,.$ (4.4)

For this purpose we shall prove that for every fixed integer $\nu\geq 0$ we
have the limit

$\lim_{n\to\infty}\frac{\displaystyle{n\choose\nu}{n+\nu\choose\nu}t^{n-\nu}}{\displaystyle\sum_{k=0}^{n}{n\choose
k}{n+k\choose k}t^{n-k}}\,=\,0\,.$ (4.5)

Then, (4.5) implies (4.4) by a theorem of O.Toeplitz ([17, p. 10, no. 66]),
since

$\lim_{n\to\infty}c_{n}\,=\,\sum_{m=1}^{\infty}\frac{{(-1)}^{m-1}t^{m}}{m}\,=\,\log(1+t)\,.$

There is nothing to show for $t=0$, because $A_{n}=0$ and $B_{n}={2n\choose
n}\not=0$. Therefore, keep $\nu\in{\mathbb{N}}_{0}$ and
$t\in(-1,1]\setminus\\{0\\}$ fixed. We substitute $X_{n}=B_{n}$ into (4.2) and
divide the equation by $(n+1)B_{n}$. Then we obtain

$\frac{B_{n+1}}{B_{n}}-\frac{(2n+1)(2+t)}{n+1}+\frac{nt^{2}}{n+1}\cdot\frac{1}{B_{n}/B_{n-1}}\,=\,0\,.$

Let $\alpha:=\lim_{n\to\infty}B_{n+1}/B_{n}$. By taking the limit
$n\to\infty$, it follows that $\alpha$ satisfies the quadratic equation

$\alpha-2(2+t)+\frac{t^{2}}{\alpha}\,=\,0\,,$

which yields

$\alpha\,=\,2+t+2\sqrt{1+t}\,>\,|t|\qquad(-1<t\leq 1)\,.$

Put $\beta:=(\alpha+|t|)/2$. Then,

$0\,<\,|t|\,<\,\beta\,<\,\alpha\,.$ (4.6)

There is an integer $n_{0}=n_{0}(t)$ satisfying

$\frac{B_{m}}{B_{m-1}}\,>\,\beta\,>\,0\qquad(m\geq n_{0})\,,$

or

$|B_{m}|\,>\,\beta|B_{m-1}|\qquad(m\geq n_{0})\,.$

Then, for $n\geq 2n_{0}$ and $k:=n-n_{0}\geq n_{0}$, we have

$|B_{n}|\,>\,\beta|B_{n-1}|\,>\,\beta^{2}|B_{n-2}|\,>\dots>\beta^{k}|B_{n-k}|\,=\,\beta^{n-n_{0}}|B_{n_{0}}|\,.$

Consequently, we obtain

$\displaystyle\lim_{n\to\infty}\left|\,\frac{\displaystyle{n\choose\nu}{n+\nu\choose\nu}t^{n-\nu}}{\displaystyle\sum_{k=0}^{n}{n\choose
k}{n+k\choose
k}t^{n-k}}\,\right|\,=\,\lim_{n\to\infty}\frac{\displaystyle{n\choose\nu}{n+\nu\choose\nu}{|t|}^{n-\nu}}{|B_{n}|}$
$\displaystyle\leq$
$\displaystyle\lim_{n\to\infty}\frac{\displaystyle{n\choose\nu}{n+\nu\choose\nu}{|t|}^{n-\nu}}{\beta^{n-n_{0}}|B_{n_{0}}|}\,=\,\frac{\beta^{n_{0}}}{{|t|}^{\nu}|B_{n_{0}}|}\lim_{n\to\infty}{n\choose\nu}{n+\nu\choose\nu}{\Big{(}\,\frac{|t|}{\beta}\,\Big{)}}^{n}$
$\displaystyle=$ $\displaystyle 0\,,$

since $0<|t|{\beta}^{-1}<1$ by (4.6), and ${n\choose\nu}{n+\nu\choose\nu}$ is
a polynomial in $n$ of degree $2\nu$. This completes the proof of (4.5) and,
consequently, of (4.4).
We rewrite the recurrence formula (4.2) as

$P(n+1)X_{n+1}-Q(n+1)X_{n}-R(n+1)X_{n-1}\,=\,0\,,$

where

$\displaystyle P(n+1)$ $\displaystyle:=$ $\displaystyle n+1\,,$ $\displaystyle
Q(n+1)$ $\displaystyle:=$ $\displaystyle(2n+1)(2+t)\,,$ $\displaystyle R(n+1)$
$\displaystyle:=$ $\displaystyle-nt^{2}\,.$

Then, we obtain

$\log(1+t)\,=\,b_{0}+\frac{a_{1}}{b_{1}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{2}}{b_{2}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{3}}{b_{3}}\begin{array}[]{c}\\\ +\dots\end{array}$

with

$\begin{array}[]{lcrlrllc}b_{0}&=&0\,,\quad b_{1}&=&2+t\,,\quad
b_{n+1}&=&\displaystyle\frac{Q(n+1)}{P(n+1)}\,=\,\frac{(2n+1)(2+t)}{n+1}\,,\\\
\\\ &&a_{1}&=&2t\,,\quad
a_{n+1}&=&\displaystyle\frac{R(n+1)}{P(n+1)}\,=\,-\frac{nt^{2}}{n+1}\,.\end{array}$

This gives the continued fraction

$\log(1+t)\,=\,\frac{2t}{2+t}\begin{array}[]{c}\\\
-\end{array}\frac{t^{2}/2}{3(2+t)/2}\begin{array}[]{c}\\\
-\end{array}\frac{2t^{2}/3}{5(2+t)/3}\begin{array}[]{c}\\\
-\end{array}\frac{3t^{2}/4}{7(2+t)/4}\begin{array}[]{c}\\\
-\dots\end{array}\,,$

which is equivalent with (4.1).
Next, we compute the error sum
$\mathcal{E}^{*}(B,R,\log(1+t))=\sum_{m=0}^{\infty}(B_{m}\log(1+t)-A_{m})$ for
$B:={(B_{n})}_{n\geq 0}$ and $R:={(A_{n})}_{n\geq 0}$.

######  Lemma 4.1.

Let $-1<t\leq 1$. For every integer $n\geq 0$ we have

$B_{n}\log(1+t)-A_{n}\,=\,t^{2n+1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}}{{(1+tx)}^{n+1}}\,dx\,.$

Proof. For $A_{n}$ and $B_{n}$ defined above, we obtain

$\displaystyle B_{n}\log(1+t)-A_{n}$ $\displaystyle=$
$\displaystyle\sum_{k=0}^{n}{n\choose k}{n+k\choose
k}t^{n-k}\sum_{m=k+1}^{\infty}\frac{{(-1)}^{m-1}t^{m}}{m}$ $\displaystyle=$
$\displaystyle\sum_{m=0}^{\infty}{(-1)}^{m}t^{m+1}\sum_{k=0}^{n}{(-1)}^{k}{n\choose
k}{n+k\choose k}\frac{t^{n-k}t^{k}}{m+k+1}$ $\displaystyle=$ $\displaystyle
t^{n+1}\int_{0}^{1}\sum_{m=0}^{\infty}{(-1)}^{m}{(tx)}^{m}\sum_{k=0}^{n}{(-1)}^{k}{n\choose
k}{n+k\choose k}x^{k}\,dx$ $\displaystyle=$ $\displaystyle
t^{n+1}\int_{0}^{1}\sum_{m=0}^{\infty}{(-tx)}^{m}\frac{d^{n}}{dx^{n}}\Big{(}\,\frac{x^{n}{(1-x)}^{n}}{n!}\,\Big{)}\,dx$
$\displaystyle=$ $\displaystyle
t^{n+1}\int_{0}^{1}\frac{1}{1+tx}\cdot\frac{d^{n}}{dx^{n}}\Big{(}\,\frac{x^{n}{(1-x)}^{n}}{n!}\,\Big{)}\,dx$
$\displaystyle=$
$\displaystyle{(-1)}^{n}t^{n+1}\int_{0}^{1}\frac{d^{n}}{dx^{n}}\Big{(}\,\frac{1}{1+tx}\,\Big{)}\cdot\frac{x^{n}{(1-x)}^{n}}{n!}\,dx$
$\displaystyle=$ $\displaystyle
t^{2n+1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}}{{(1+tx)}^{n+1}}\,dx\,.$

The antiderivative in the last but one line was obtained using $n$-fold
integration by parts. The lemma is proven. ∎

A first consequence of Lemma 4.1 is an explicit formula for the error sum of
$\log(1+t)$.

######  Corollary 4.1.

Let $-1<t\leq 1$. For the sequences $B:={(B_{n})}_{n\geq 0}$ and
$R:={(A_{n})}_{n\geq 0}$ we have

$\mathcal{E}^{*}\big{(}B,R,\log(1+t)\big{)}\,=\,\frac{2}{\sqrt{3+2t-t^{2}}}\left(\arctan\Big{(}\,\frac{1+t}{\sqrt{3+2t-t^{2}}}-\arctan\Big{(}\,\frac{1-t}{\sqrt{3+2t-t^{2}}}\,\Big{)}\right)\,.$

In particular, $\mathcal{E}^{*}(B,R,\log 2)=\pi/4$.

Proof: From Lemma 4.1 we obtain

$\displaystyle\mathcal{E}^{*}\big{(}B,R,\log(1+t)\big{)}$ $\displaystyle=$
$\displaystyle
t\int_{0}^{1}\sum_{m=0}^{\infty}\frac{t^{2m}x^{m}{(1-x)}^{m}}{{(1+tx)}^{m+1}}\,dx$
$\displaystyle=$ $\displaystyle
t\int_{0}^{1}\frac{1}{1+tx}\sum_{m=0}^{\infty}{\Big{(}\,\frac{t^{2}x(1-x)}{1+tx}\,\Big{)}}^{m}\,dx$
$\displaystyle=$ $\displaystyle t\int_{0}^{1}\frac{dx}{1+t(1-t)x+t^{2}x^{2}}$
$\displaystyle=$
$\displaystyle\frac{2}{\sqrt{3+2t-t^{2}}}\left(\arctan\Big{(}\,\frac{1+t}{\sqrt{3+2t-t^{2}}}-\arctan\Big{(}\,\frac{1-t}{\sqrt{3+2t-t^{2}}}\,\Big{)}\right)\,.$

This proves the corollary. ∎
By straightforward computations it can be seen that the error sum from
Corollary 4.1 satisfies a linear first order differential equation.

######  Corollary 4.2.

Let $-1<t\leq 1$. For the sequences $B:={(B_{n})}_{n\geq 0}$ and
$R:={(A_{n})}_{n\geq 0}$ the function $f(t):=\mathcal{E}^{*}(B,R,\log(1+t))$
satisfies the differential equation

$(3+2t-t^{2})f^{\prime}+(1-t)f-3\,=\,0\,,$

where $f^{\prime}=df/dt$.

A second consequence of Lemma 4.1 is

$\mathcal{E}^{*}\big{(}B,R,\log(1+t)\big{)}\,=\,\mbox{sign}(t)\mathcal{E}\big{(}B,R,\log(1+t)\big{)}\,.$

Finally, the continued fraction (4.1) and Lemma 4.1 allow to prove the
irrationality of $\log(1+t)$ for certain rationals $t:=a/b$.

######  Corollary 4.3.

Let $0<a/b\leq 1$ be a rational number with $ea^{2}<4b$. Then the number
$\log(1+a/b)$ is irrational. In particular, for every integer $k\geq 1$ the
number $\log(1+1/k)$ is irrational.

Proof: Let $d_{n}:=\mbox{l.c.m.}(1,2,3,\dots,n)$ denote the least common
multiple of the integers $1,2,3,\dots,n$. One knows by the prime number
theorem that

$\log d_{n}\,=\,\sum_{p\leq n}\Big{[}\,\frac{\log n}{\log p}\,\Big{]}\log
p\,\sim\,n\,,$

where $p$ runs through all primes less than or equal to $n$ ([13, Theorem
434]). By the hypothesis $ea^{2}<4b$ there is a positive real number
$\varepsilon$ such that $e^{1+\varepsilon}a^{2}<4^{1-\varepsilon}b$. Hence,
for all sufficiently large numbers $n$, it follows that

$d_{n}\,<\,e^{(1+\varepsilon)n}\,.$

Let $t=a/b$. With $b^{n}d_{n}A_{n}\in{\mathbb{Z}}$ and
$b^{n}d_{n}B_{n}\in{\mathbb{Z}}$ we know by Lemma 4.1 that

$\displaystyle 0$ $\displaystyle<$
$\displaystyle\big{|}b^{n}d_{n}B_{n}\log(1+t)-b^{n}d_{n}A_{n}\big{|}$
$\displaystyle=$
$\displaystyle\frac{a}{b}b^{n}d_{n}{\Big{(}\,\frac{a}{b}\,\Big{)}}^{2n}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}}{{(1+ax/b)}^{n+1}}\,dx$
$\displaystyle<$
$\displaystyle\frac{a}{b}\cdot\frac{e^{(1+\varepsilon)n}a^{2n}}{b^{n}}\int_{0}^{1}x^{n}{(1-x)}^{n}\,dx$
$\displaystyle<$ $\displaystyle\frac{a}{b}\cdot
4^{(1-\varepsilon)n}\int_{0}^{1}\frac{dx}{4^{n}}\,dx$ $\displaystyle=$
$\displaystyle\frac{t}{4^{\varepsilon n}}\,\to\,0$

for $n$ tending to infinity. This completes the proof of Corollary 4.3. ∎

##  5 On error sums formed by Apéry’s continued fractions for $\zeta(2)$ and
$\zeta(3)$

Computing the error sums formed by the linear three term recurrences and
continued fractions of $\zeta(2)$, $\zeta(3)$ introduced by R. Apéry, this
leads unexpectedly into a wide field of connections between famous sequences
of integers. For the needed results we refer to [4] and [3].
1.) Error sums for $\zeta(2)$. We have

$\displaystyle\zeta(2)$ $\displaystyle=$
$\displaystyle\frac{\pi^{2}}{6}\,=\,\frac{5}{3}\begin{array}[]{c}\\\
+\end{array}\frac{1^{4}}{25}\begin{array}[]{c}\\\
+\end{array}\frac{2^{4}}{69}\begin{array}[]{c}\\\
+\dots+\end{array}\frac{n^{4}}{11n^{2}+11n+3}\begin{array}[]{c}\\\
+\dots\end{array}$ $\displaystyle=$ $\displaystyle
b_{0}+\frac{a_{1}}{b_{1}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{2}}{b_{2}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{3}}{b_{3}}\begin{array}[]{c}\\\ +\dots\end{array}$

with

$b_{0}=0\,,\quad b_{1}=3\,,\quad b_{n+1}=11n^{2}+11n+3\quad(n\geq 1)\,,$
$a_{1}=5\,,\quad a_{n+1}=n^{4}\quad(n\geq 1)\,.$

A recurrence formula for both sequences

$\displaystyle B_{n}$ $\displaystyle:=$ $\displaystyle\sum_{k=0}^{n}{n\choose
k}^{2}{n+k\choose k}\,,$ $\displaystyle A_{n}$ $\displaystyle:=$
$\displaystyle\sum_{k=0}^{n}{n\choose k}^{2}{n+k\choose
k}\left(2\sum_{m=1}^{n}\frac{{(-1)}^{m-1}}{m^{2}}+\sum_{m=1}^{k}\cfrac{{(-1)}^{n+m-1}}{m^{2}{n\choose
m}{n+m\choose m}}\right)\,,$

is

$0\,=\,{(n+1)}^{2}X_{n+1}-(11n^{2}+11n+3)X_{n}-n^{2}X_{n-1}\,.$

Then,

$b_{0}+\frac{a_{1}}{b_{1}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{2}}{b_{2}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{3}}{b_{3}}\begin{array}[]{c}\\\
+\dots+\end{array}\frac{a_{n}}{b_{n}}\\\ \,=\,\frac{A_{n}}{B_{n}}\,.$

We obtain from [3, eq. (5)] for the sequences $B_{2}:={(B_{n})}_{n\geq 0}$ and
$R_{2}:={(A_{n})}_{n\geq 0}$,

$\displaystyle\mathcal{E}^{*}\big{(}B_{2},R_{2},\zeta(2)\big{)}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\Big{(}\,B_{n}\zeta(2)-A_{n}\,\Big{)}\,=\,\sum_{n=0}^{\infty}{(-1)}^{n}\int_{0}^{1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}y^{n}{(1-y)}^{n}}{{(1-xy)}^{n+1}}\,dx\,dy$
(5.3) $\displaystyle=$
$\displaystyle\int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1+x^{2}y^{2}-xy^{2}-yx^{2}}\,=\,1.5832522167\dots\,.$

Similarly, one has

$\displaystyle\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\Big{|}\,B_{n}\zeta(2)-A_{n}\,\Big{|}\,=\,\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}y^{n}{(1-y)}^{n}}{{(1-xy)}^{n+1}}\,dx\,dy$
(5.4) $\displaystyle=$
$\displaystyle\int_{0}^{1}\int_{0}^{1}\frac{dx\,dy}{1-x^{2}y^{2}-2xy+xy^{2}+yx^{2}}\,=\,1.7141459142\dots\,.$

2.) Error sums for $\zeta(3)$. Here,

$\displaystyle\zeta(3)$ $\displaystyle=$
$\displaystyle\frac{6}{5}\begin{array}[]{c}\\\
-\end{array}\frac{1^{6}}{117}\begin{array}[]{c}\\\
-\end{array}\frac{2^{6}}{535}\begin{array}[]{c}\\\
-\dots-\end{array}\frac{n^{6}}{34n^{3}+51n^{2}+27n+5}\begin{array}[]{c}\\\
-\dots\end{array}$ $\displaystyle=$ $\displaystyle
b_{0}+\frac{a_{1}}{b_{1}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{2}}{b_{2}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{3}}{b_{3}}\begin{array}[]{c}\\\ +\dots\end{array}$

with

$b_{0}=0\,,\quad b_{1}=5\,,\quad b_{n+1}=34n^{3}+51n^{2}+27n+5\quad(n\geq
1)\,,$ $a_{1}=6\,,\quad a_{n+1}=-n^{6}\quad(n\geq 1)\,.$

A recurrence formula for both sequences,

$\displaystyle D_{n}$ $\displaystyle:=$ $\displaystyle\sum_{k=0}^{n}{n\choose
k}^{2}{n+k\choose k}^{2}\,,$ $\displaystyle C_{n}$ $\displaystyle:=$
$\displaystyle\sum_{k=0}^{n}{n\choose k}^{2}{n+k\choose
k}^{2}\left(\sum_{m=1}^{n}\frac{1}{m^{3}}+\sum_{m=1}^{k}\cfrac{{(-1)}^{m-1}}{2m^{3}{n\choose
m}{n+m\choose m}}\right)\,,$

is

$\displaystyle 0$ $\displaystyle=$ $\displaystyle
P(n+1)X_{n+1}-Q(n+1)X_{n}-R(n+1)X_{n-1}$ $\displaystyle=$
$\displaystyle{(n+1)}^{3}X_{n+1}-(34n^{3}+51n^{2}+27n+5)X_{n}+n^{3}X_{n-1}\,.$

The construction of $C_{n}$ and $D_{n}$ leads to the identity

$b_{0}+\frac{a_{1}}{b_{1}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{2}}{b_{2}}\begin{array}[]{c}\\\
+\end{array}\frac{a_{3}}{b_{3}}\begin{array}[]{c}\\\
+\dots+\end{array}\frac{a_{n}}{b_{n}}\\\ \,=\,\frac{C_{n}}{D_{n}}\,.$

We obtain from [3, eq. (7)] for the sequences $B_{3}:={(D_{n})}_{n\geq 0}$ and
$R_{3}:={(C_{n})}_{n\geq 0}$,

$\displaystyle\mathcal{E}^{*}\big{(}B_{3},R_{3},\zeta(3)\big{)}$ (5.7)
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\Big{(}\,D_{n}\zeta(3)-C_{n}\,\Big{)}\,=\,\sum_{n=0}^{\infty}\Big{|}\,D_{n}\zeta(3)-C_{n}\,\Big{|}\,=\,\mathcal{E}\big{(}B_{3},R_{3},\zeta(3)\big{)}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}y^{n}{(1-y)}^{n}w^{n}{(1-w)}^{n}}{{\big{(}1-(1-xy)w\big{)}}^{n+1}}\,dx\,dy\,dw$
$\displaystyle=$
$\displaystyle\frac{1}{2}\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{dx\,dy\,dw}{1+x^{2}y^{2}w^{2}-xy^{2}w^{2}-x^{2}yw^{2}-x^{2}y^{2}w+xyw^{2}+xy^{2}w+x^{2}yw-w}$
$\displaystyle=$ $\displaystyle 1.2124982529\dots\,.$

3.) Now we focus our interest on various methods in order to express the
multiple integrals in (5.3), (5.4), and (5.7), by series with rational terms.
A first approach to this subject involves the hypergeometric function.

######  Theorem 5.1.

For the sequences $B_{i},R_{i}$ $(i=2,3)$ defined above for $\zeta(2)$ and
$\zeta(3)$, respectively, we have

$\displaystyle\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\cfrac{\displaystyle{n+k\choose
n}}{\displaystyle(2n+k+1)^{2}{2n+k\choose n}^{2}}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\frac{\displaystyle{}_{3}F_{2}\left(\begin{array}[]{c}n+1\quad
n+1\quad n+1\\\ 2n+2\quad
2n+2\end{array}\,\Big{|}1\right)}{\displaystyle{(2n+1)}^{2}{2n\choose
n}^{2}}\,,$ $\displaystyle\mathcal{E}^{*}\big{(}B_{2},R_{2},\zeta(2)\big{)}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\cfrac{\displaystyle{(-1)}^{n}{n+k\choose
n}}{\displaystyle(2n+k+1)^{2}{2n+k\choose n}^{2}}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}{(-1)}^{n}\frac{\displaystyle{}_{3}F_{2}\left(\begin{array}[]{c}n+1\quad
n+1\quad n+1\\\ 2n+2\quad
2n+2\end{array}\,\Big{|}1\right)}{\displaystyle{(2n+1)}^{2}{2n\choose
n}^{2}}\,,$ $\displaystyle\mathcal{E}\big{(}B_{3},R_{3},\zeta(3)\big{)}$
$\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\,\,\sum_{l=0}^{k}\cfrac{\displaystyle{(-1)}^{l}{k\choose
l}{n+l\choose n}}{\displaystyle(2n+k+1)(2n+l+1)^{2}{2n+k\choose n}{2n+l\choose
n}^{2}}$ $\displaystyle=$
$\displaystyle\frac{1}{2}\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\frac{\displaystyle{}_{4}F_{3}\left(\begin{array}[]{c}n+1\quad
n+1\quad n+1\quad-k\\\ 2n+2\quad 2n+2\quad
1\end{array}\,\Big{|}1\right)}{\displaystyle{(2n+1)}^{2}(2n+k+1){2n\choose
n}^{2}{2n+k\choose n}}\,.$

Note that the hypergeometric function

${}_{4}F_{3}\left(\begin{array}[]{c}n+1\quad n+1\quad n+1\quad-k\\\ 2n+2\quad
2n+2\quad 1\end{array}\,\Big{|}1\right)$

takes rational values for all $0\leq k,n<\infty$.
Proof: It suffices to prove the identities for
$\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$, since the arguments are the
same for the remaining error sums. The basic idea is to use the expansion

$\frac{1}{{(1-t)}^{n+1}}\,=\,\sum_{k=0}^{\infty}\frac{{(n+1)}_{k}}{k!}t^{k}\,.$

Then, (5.4) gives

$\displaystyle\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$ $\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\frac{x^{n}{(1-x)}^{n}y^{n}{(1-y)}^{n}}{{(1-xy)}^{n+1}}\,dx\,dy$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\int_{0}^{1}\int_{0}^{1}\sum_{k=0}^{\infty}\frac{{(n+1)}_{k}}{k!}{(xy)}^{k}x^{n}{(1-x)}^{n}y^{n}{(1-y)}^{n}\,dx\,dy$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\frac{{(n+1)}_{k}}{k!}\int_{0}^{1}x^{n+k}{(1-x)}^{n}\,dx\int_{0}^{1}y^{n+k}{(1-y)}^{n}\,dy$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\frac{{(n+1)}_{k}}{k!}{\Big{(}\,\frac{\Gamma(n+1)\Gamma(n+k+1)}{\Gamma(2n+k+2)}\,\Big{)}}^{2}$
$\displaystyle=$
$\displaystyle\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\frac{n!{(n+k)!}^{3}}{k!{(2n+k)!}^{2}{(2n+k+1)}^{2}}\,=\,\sum_{n=0}^{\infty}\,\,\sum_{k=0}^{\infty}\cfrac{\displaystyle{n+k\choose
n}}{\displaystyle(2n+k+1)^{2}{2n+k\choose n}^{2}}\,.$

The second identity for $\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$ in
Theorem 5.1 follows from (5) and from

$\frac{\Gamma^{2}(n+1)\Gamma^{2}(n+k+1)}{\Gamma^{2}(2n+k+2)}\,=\,\frac{1}{{(2n+1)}^{2}\displaystyle{2n\choose
n}^{2}}\cdot\frac{{(n+1)}_{k}{(n+1)}_{k}}{{(2n+2)}_{k}{(2n+2)}_{k}}\,,$

which can be verified by straightforward computations. ∎

Next, we define recursively a sequence $p_{\nu}(t)$ $(\nu=1,2,\dots)$ of
polynomias in one variable $t$, namely

$\displaystyle p_{1}(t)$ $\displaystyle=$ $\displaystyle t^{2}\,,$ (5.12)
$\displaystyle p_{2}(t)$ $\displaystyle=$ $\displaystyle t^{4}-t^{2}+t\,,$
(5.13) $\displaystyle p_{\nu}(t)$ $\displaystyle=$ $\displaystyle
t^{2}p_{\nu-1}(t)+t(1-t)p_{\nu-2}(t)\qquad(\nu=3,4,\dots).$ (5.14)

It is clear that $\deg p_{\nu}=2\nu$, which follows easily by induction for
$\nu$ with $\deg p_{1}=2$ and $\deg p_{2}=4$. The leading coefficient of
$p_{\nu}$ is 1 for $\nu=1,2,\dots$. Let

$p_{\nu}(t)\,=\,\sum_{\mu=0}^{2\nu}a_{\nu,\mu}t^{\mu}\,.$

######  Lemma 5.1.

For $\nu\geq 3$ we have

$p_{\nu}(t)\,=\,t^{2\nu}+(1-\nu)t^{2\nu-2}+\sum_{\mu=2}^{2\nu-3}\big{(}a_{\nu-1,\mu-2}+a_{\nu-2,\mu-1}-a_{\nu-2,\mu-2}\big{)}t^{\mu}$

with

$a_{\nu,\mu}\,=\,a_{\nu-1,\mu-2}+a_{\nu-2,\mu-1}-a_{\nu-2,\mu-2}\qquad(2\leq\mu\leq
2\nu-3)\,.$

Proof: Using the definition of $p_{\nu}(t)$ from (5.12) to (5.14) with
$\nu\geq 3$, we obtain

$\displaystyle p_{\nu}(t)$ $\displaystyle=$
$\displaystyle\sum_{\mu=0}^{2\nu}a_{\nu,\mu}t^{\mu}\,=\,\sum_{\mu=0}^{2\nu-2}a_{\nu-1,\mu}t^{\mu+2}+\sum_{\mu=0}^{2\nu-4}a_{\nu-2,\mu}t^{\mu+1}-\sum_{\mu=0}^{2\nu-4}a_{\nu-2,\mu}t^{\mu+2}$
$\displaystyle=$
$\displaystyle\sum_{\mu=2}^{2\nu}a_{\nu-1,\mu-2}t^{\mu}+\sum_{\mu=1}^{2\nu-3}a_{\nu-2,\mu-1}t^{\mu}-\sum_{\mu=2}^{2\nu-2}a_{\nu-2,\mu-2}t^{\mu}$
$\displaystyle=$ $\displaystyle
a_{\nu-1,2\nu-2}t^{2\nu}+a_{\nu-1,2\nu-3}t^{2\nu-1}+a_{\nu-1,2\nu-4}t^{2\nu-2}+a_{\nu-2,0}t-a_{\nu-2,2\nu-4}t^{2\nu-2}$
$\displaystyle+\sum_{\mu=2}^{2\nu-3}\big{(}a_{\nu-1,\mu-2}+a_{\nu-2,\mu-1}-a_{\nu-2,\mu-2}\big{)}t^{\mu}$
$\displaystyle=$ $\displaystyle
t^{2\nu}+(1-\nu)t^{2\nu-2}+\sum_{\mu=2}^{2\nu-3}\big{(}a_{\nu-1,\mu-2}+a_{\nu-2,\mu-1}-a_{\nu-2,\mu-2}\big{)}t^{\mu}\,,$

since the four identities

$\displaystyle a_{\nu-1,2\nu-2}$ $\displaystyle=$ $\displaystyle 1\,,$
$\displaystyle a_{\nu-1,2\nu-3}$ $\displaystyle=$ $\displaystyle 0\,,$
$\displaystyle a_{\nu-1,2\nu-4}-a_{\nu-2,2\nu-4}$ $\displaystyle=$
$\displaystyle 1-\nu\,,$ $\displaystyle a_{\nu-2,0}$ $\displaystyle=$
$\displaystyle 0$

follow easily from (5.12) to (5.14) by $a_{\nu,2\nu}=1$, $a_{\nu,2\nu-1}=0$,
$a_{\nu,2\nu-2}=1-\nu$, and $a_{\nu,0}=0$ for $\nu\geq 1$. The lemma is
proven. ∎

######  Theorem 5.2.

For the sequences $B_{2},R_{2}$ defined above for $\zeta(2)$ we have

$\mathcal{E}^{*}\big{(}B_{2},R_{2},\zeta(2)\big{)}\,=\,1+\sum_{\nu=1}^{\infty}\,\sum_{\mu=0}^{2\nu}\frac{a_{\nu,\mu}}{(\nu+1)(\mu+1)}\,.$

Proof: With (5.12) to (5.14) we obtain

$\displaystyle(1+x^{2}y^{2}-xy^{2}-yx^{2})\Big{(}\,1+\sum_{\nu=1}^{\infty}p_{\nu}(y)x^{\nu}\,\Big{)}$
$\displaystyle=$ $\displaystyle
1+x^{2}y^{2}-xy^{2}-yx^{2}+\sum_{\nu=1}^{\infty}p_{\nu}(y)x^{\nu}+\sum_{\nu=1}^{\infty}y^{2}p_{\nu}(y)x^{\nu+2}-\sum_{\nu=1}^{\infty}y^{2}p_{\nu}(y)x^{\nu+1}$
$\displaystyle-\,\sum_{\nu=1}^{\infty}yp_{\nu}(y)x^{\nu+2}$ $\displaystyle=$
$\displaystyle
1+x^{2}y^{2}-xy^{2}-yx^{2}+\sum_{\nu=1}^{\infty}p_{\nu}(y)x^{\nu}-\sum_{\nu=3}^{\infty}y(1-y)p_{\nu-2}(y)x^{\nu}-\sum_{\nu=2}^{\infty}y^{2}p_{\nu-1}(y)x^{\nu}$
$\displaystyle=$ $\displaystyle
1-p_{1}(y)x-p_{2}(y)x^{2}+y^{2}p_{1}(y)x^{2}+\sum_{\nu=1}^{\infty}p_{\nu}(y)x^{\nu}-\sum_{\nu=3}^{\infty}y(1-y)p_{\nu-2}(y)x^{\nu}-\sum_{\nu=2}^{\infty}y^{2}p_{\nu-1}(y)x^{\nu}$
$\displaystyle=$ $\displaystyle
1+\sum_{\nu=3}^{\infty}p_{\nu}(y)x^{\nu}-\sum_{\nu=3}^{\infty}y(1-y)p_{\nu-2}(y)x^{\nu}-\sum_{\nu=3}^{\infty}y^{2}p_{\nu-1}(y)x^{\nu}$
$\displaystyle=$ $\displaystyle
1+\sum_{\nu=3}^{\infty}\Big{[}\,p_{\nu}(y)-\big{(}y^{2}p_{\nu-1}(y)+y(1-y)p_{\nu-2}(y)\big{)}\,\Big{]}x^{\nu}$
$\displaystyle=$ $\displaystyle 1\,.$

Hence,

$\frac{1}{1+x^{2}y^{2}-xy^{2}-yx^{2}}\,=\,1+\sum_{\nu=1}^{\infty}p_{\nu}(y)x^{\nu}\,=\,1+\sum_{\nu=1}^{\infty}\,\Big{(}\,\sum_{\mu=0}^{2\nu}a_{\nu,\mu}y^{\mu}\,\Big{)}x^{\nu}.$

Now the theorem follows from (5.3) by two-fold integration with respect to $x$
and $y$. ∎

We can proceed similarly in order to obtain similar results for
$\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}$ and for
$\mathcal{E}\big{(}B_{3},R_{3},\zeta(3)\big{)}$. Therefore, we state them
without proofs. Again we define recursively a sequence $q_{\nu}(t)$
$(\nu=1,2,\dots)$ of integer polynomias in one variable $t$,

$\displaystyle q_{1}(t)$ $\displaystyle=$ $\displaystyle 2t-t^{2}\,,$
$\displaystyle q_{2}(t)$ $\displaystyle=$ $\displaystyle
t^{4}-4t^{3}+5t^{2}-t\,,$ $\displaystyle q_{\nu}(t)$ $\displaystyle=$
$\displaystyle t(2-t)q_{\nu-1}(t)+t(t-1)q_{\nu-2}(t)\qquad(\nu=3,4,\dots).$

Let

$q_{\nu}(t)\,=\,\sum_{\mu=0}^{2\nu}b_{\nu,\mu}t^{\mu}\,.$

Here, we have

$\frac{1}{1-x^{2}y^{2}-2xy+xy^{2}+yx^{2}}\,=\,\sum_{\nu=0}^{\infty}q_{\nu}(y)x^{\nu}\,=\,\sum_{\nu=0}^{\infty}\,\Big{(}\,\sum_{\mu=0}^{2\nu}b_{\nu,\mu}y^{\mu}\,\Big{)}x^{\nu}.$

######  Lemma 5.2.

For $\nu\geq 3$ we have

$\displaystyle q_{\nu}(t)$ $\displaystyle=$
$\displaystyle{(-1)}^{\nu}t^{2\nu}+2{(-1)}^{\nu+1}\nu
t^{2\nu-1}+{(-1)}^{\nu}(2\nu^{2}-\nu-1)t^{2\nu-2}$
$\displaystyle+\,\sum_{\mu=2}^{2\nu-3}\big{(}-b_{\nu-1,\mu-2}+2b_{\nu-1,\mu-1}+b_{\nu-2,\mu-2}-b_{\nu-2,\mu-1}\big{)}t^{\mu}$

with

$b_{\nu,\mu}\,=\,-b_{\nu-1,\mu-2}+2b_{\nu-1,\mu-1}+b_{\nu-2,\mu-2}-b_{\nu-2,\mu-1}\qquad(2\leq\mu\leq
2\nu-3)\,.$

######  Theorem 5.3.

For the sequences $B_{2},R_{2}$ defined above for $\zeta(2)$ we have

$\mathcal{E}\big{(}B_{2},R_{2},\zeta(2)\big{)}\,=\,1+\sum_{\nu=1}^{\infty}\,\sum_{\mu=0}^{2\nu}\frac{b_{\nu,\mu}}{(\nu+1)(\mu+1)}\,.$

The above method can be generalized such that it also works for
$\mathcal{E}\big{(}B_{3},R_{3},\zeta(3)\big{)}$. Let

$\displaystyle r_{0}(x,y)$ $\displaystyle=$ $\displaystyle 1\,,$
$\displaystyle r_{1}(x,y)$ $\displaystyle=$ $\displaystyle
x^{2}y^{2}-xy^{2}-x^{2}y+1\,,$ $\displaystyle r_{2}(x,y)$ $\displaystyle=$
$\displaystyle
x^{4}y^{4}-2x^{3}y^{4}-2x^{4}y^{3}+x^{2}y^{4}+x^{4}y^{2}+2x^{3}y^{3}+x^{2}y^{2}-xy^{2}-x^{2}y-xy+1\,,$
$\displaystyle r_{\nu}(x,y)$ $\displaystyle=$
$\displaystyle\big{(}x^{2}y^{2}-xy^{2}-x^{2}y+1\big{)}r_{\nu-1}(x,y)-\big{(}x^{2}y^{2}-xy^{2}-x^{2}y+xy\big{)}r_{\nu-2}(x,y)\,,$

where $\nu\geq 3$. Setting

$r_{\nu}(x,y)\,=\,\sum_{\mu_{1}=0}^{2\nu}\,\sum_{\mu_{2}=0}^{2\nu}c_{\nu,\mu_{1},\mu_{2}}x^{\mu_{1}}y^{\mu_{2}}\,,$

it turns out that

$\frac{1}{1+x^{2}y^{2}w^{2}-xy^{2}w^{2}-x^{2}yw^{2}-x^{2}y^{2}w+xyw^{2}+xy^{2}w+x^{2}yw-w}\,=\,\sum_{\nu=0}^{\infty}r_{\nu}(x,y)\cdot
w^{\nu}\,.$

Then, (5.7) underlies the following result.

######  Theorem 5.4.

For the sequences $B_{3},R_{3}$ defined above for $\zeta(3)$ we have

$\mathcal{E}\big{(}B_{3},R_{3},\zeta(3)\big{)}\,=\,\frac{1}{2}+\frac{1}{2}\sum_{\nu=1}^{\infty}\,\sum_{\mu_{1}=0}^{2\nu}\,\sum_{\mu_{2}=0}^{2\nu}\frac{c_{\nu,\mu_{1},\mu_{2}}}{(\nu+1)(\mu_{1}+1)(\mu_{2}+1)}\,.$

4.) As mentionned at the beginning of Section 5, some connections between
rational coefficients involved in computing the error sums for Apéry’s
continued fraction and well-known integer sequences may be noticed.
Furthermore some unproved identities have been empirically found for such
coefficients.

The Theorems 5.4 and 5.7 rely on two triangles of integer coefficients, namely
$a_{\nu,\mu}$ for the former and $b_{\nu,\mu}$ for the latter. Both can be
expressed by binomial sums as follows.

$\begin{array}[]{lcl}a_{\nu,\mu}&=&\displaystyle\sum_{k=0}^{\nu}\sum_{i=0}^{k}\left(-1\right)^{\nu+k}{{\mu-k}\choose{2\mu-\nu-
k-i}}{{\mu-k}\choose i}{{\mu-i}\choose{k-i}}\,,\\\\[20.0pt]
b_{\nu,\mu}&=&\displaystyle\sum_{k=0}^{\nu}\sum_{i=0}^{k}\left(-1\right)^{\nu+\mu}{{\mu-k}\choose{2\mu-\nu-
k-i}}{{\mu-k}\choose i}{{\mu-i}\choose{k-i}}\,,\end{array}$

which both lead to non-recurrent formulas for the error sums as quadruple
sums.

Several basic properties concerning the coefficients $a_{\nu,\mu}$ and
$b_{\nu,\mu}$ can be noticed, including

$\displaystyle\sum_{\mu=0}^{2\nu}a_{\mu,\nu}=1\qquad\textrm{and}\qquad\displaystyle\sum_{\mu=0}^{2\nu}b_{\mu,\nu}=1\qquad\qquad\left(\nu\in\mathbb{N}\right)$

and

$a_{\nu,\mu}=a_{\mu,\nu}\qquad\textrm{and}\qquad b_{\nu,\mu}=b_{\mu,\nu}\,.$

More unproved identities come from the theory of generating functions. Both
coefficients $a_{\nu,\mu}$ and $b_{\nu,\mu}$ seem to be the coefficients of
degree $2\nu-\mu$ in the MacLaurin series expansion of

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\frac{\left(\displaystyle\frac{x^{2}+1-\sqrt{x^{4}-4x^{3}+2x^{2}+1}}{2x^{3}}\right)^{\mu-\nu}}{\sqrt{x^{4}-4x^{3}+2x^{2}+1}}&\textrm{for
$a_{\nu,\mu}$}\\\\[20.0pt]
\displaystyle\frac{\left(\displaystyle\frac{x^{2}+2x-1+\sqrt{x^{4}+2x^{2}-4x+1}}{2x^{3}}\right)^{\mu-\nu}}{\sqrt{x^{4}+2x^{2}-4x+1}}&\textrm{for
$b_{\nu,\mu}$}\end{array}\right.\quad.$

These generating functions actually allow to build the triangles of
coefficients $a_{\nu,\mu}$ and $b_{\nu,\mu}$ by diagonals rather than by rows.

Summing these coefficients by rows according to Theorems 5.4 and 5.7, the
results can be easely achieved by applying the following unproved recursive
identities.

$\left\\{\begin{array}[]{l}\alpha_{0}=1,\qquad\alpha_{1}=1/3,\qquad\alpha_{2}=11/30,\qquad\alpha_{3}=17/70\\\\[12.0pt]
\begin{array}[]{lclcl}\alpha_{n}&=&\displaystyle\frac{4n-1}{2n+1}\alpha_{n-1}&-&\displaystyle\frac{2n-2}{2n+1}\alpha_{n-2}\\\\[12.0pt]
&-&\displaystyle\frac{n-1}{4n+2}\alpha_{n-3}&+&\displaystyle\frac{n-2}{4n+2}\alpha_{n-4}\end{array}\end{array}\right.\qquad$

and

$\left\\{\begin{array}[]{l}\beta_{0}=1,\qquad\beta_{1}=2/3,\qquad\beta_{2}=11/30,\qquad\beta_{3}=47/210\\\\[12.0pt]
\begin{array}[]{lclcl}\beta_{n}&=&\displaystyle\frac{6n-1}{2n+1}\beta_{n-1}&-&\displaystyle\frac{6n-5}{2n+1}\beta_{n-2}\\\\[12.0pt]
&+&\displaystyle\frac{5n-7}{4n+2}\beta_{n-3}&-&\displaystyle\frac{n-2}{4n+2}\beta_{n-4}\end{array}\end{array}\right.\quad,$

where

$\alpha_{\nu}=\displaystyle\sum_{\mu=0}^{2\nu}\displaystyle\frac{a_{\nu,\mu}}{\mu+1}\qquad\textrm{and}\qquad\beta_{\nu}=\displaystyle\sum_{\mu=0}^{2\nu}\displaystyle\frac{b_{\nu,\mu}}{\mu+1}\,.$

The special case $b_{n,n}$, which may be called the main diagonal in the
triangle of coefficients $b_{\nu,\mu}$, leads to the following
simplifications. We have

$b_{n,n}=\displaystyle\sum_{k=0}^{n}\sum_{i=0}^{k}\displaystyle{{n-k}\choose
i}^{2}{{n-i}\choose{k-i}}\,,$

where the generating function of the $b_{n,n}$ is given by
$1/\sqrt{x^{4}+2x^{2}-4x+1}$. This is the sequence A108626 from the On-Line
Encyclopedia of Integer Sequences. This sequence gives the antidiagonal sums
of the square array A108625 itself known to be highly related to the constant
$\zeta(2)$.

$b_{\nu,\mu}$ is defined recursively by

$b_{\nu,\mu}=2b_{\nu-1,\mu-1}-b_{\nu-1,\mu-2}+b_{\nu-2,\mu-2}-b_{\nu-2,\mu-1}\,.$

Assuming $b_{n,n+1}=b_{n+1,n}$ (unproved), a new recursive identity can be
given concerning A108626:

$\begin{split}\texttt{A108626}\left(n+2\right)-2\times\texttt{A108626}\left(n+1\right)-\texttt{A108626}\left(n\right)\\\
=2\displaystyle\sum_{k=0}^{n}\sum_{i=0}^{k}{{n-k+1}\choose{i-1}}{{n-k+1}\choose
i}{{n-i+1}\choose{k-i}}\,.\end{split}$

The previous relation actually happens to be the simplest case from a more
general sequence of recurrence relations of order $2d$ given by:

$\displaystyle\sum_{k=0}^{2d}c_{k}\texttt{A108626}\left(n+k\right)=\left(-1\right)^{d}\displaystyle\sum_{k=0}^{n}\sum_{i=0}^{k}{{n-k}\choose{d+i}}{{n-k}\choose
i}{{n-i}\choose{k-i}}\,,$

where the numbers $c_{k}$ are coefficients of order $2d-k$ in the
characteristic polynomial

$\displaystyle\frac{1}{2^{d}}\sum_{i=0}^{\left\lfloor\frac{d}{2}\right\rfloor}{d\choose{2i}}\left(x^{4}+2x^{2}-4x+1\right)^{i}\left(x^{2}+2x-1\right)^{d-2i}\,.$

These recurrence relations, as well as similar ones related to the
coefficients $a_{\nu,\mu}$, can be written as new generating functions, the
diagonal of order $d$ being made from the coefficients of terms with positive
powers in

$\left\\{\begin{array}[]{l@{\qquad}l}\displaystyle\frac{\displaystyle\sum_{k=0}^{\left\lfloor\frac{d}{2}\right\rfloor}{d\choose{2k}}\left(x^{4}-4x^{3}+2x^{2}+1\right)^{k}\left(x^{2}+1\right)^{d-2k}}{\left(2x^{3}\right)^{d}\sqrt{x^{4}-4x^{3}+2x^{2}+1}}&\textrm{for
$a_{n,n+d}$}\\\\[20.0pt]
\displaystyle\frac{\displaystyle\sum_{k=0}^{\left\lfloor\frac{d}{2}\right\rfloor}{d\choose{2k}}\left(x^{4}+2x^{2}-4x+1\right)^{k}\left(x^{2}+2x-1\right)^{d-2k}}{\left(2x^{3}\right)^{d}\sqrt{x^{4}+2x^{2}-4x+1}}&\textrm{for
$b_{n,n+d}$}\end{array}\right.\quad.$

## References

  * [1] J. P. Allouche and T. Baruchel, Variations on an error sum function for the convergents of some powers of _e_ , http://arxiv.org/abs/1408.2206
  * [2] A. Baker, Transcendental Number Theory, Cambridge University Press, 1975.
  * [3] F. Beukers, A note on the irrationality of $\zeta(2)$ and $\zeta(3)$, Bull. London Math. Soc. 11 (1979), 268-272.
  * [4] H. Cohen, Demonstration de l’irrationalite de $\zeta(3)$ (d’aprés R.Apéry), Séminaire de Théorie des Nombres, 5 octobre 1978, Grenoble, VI.1 - VI.9.
  * [5] H. Cohn, A short proof of the simple continued fraction expansion of $e$, Amer. Math. Monthly 113 (2006), 57-62.
  * [6] C. Elsner, Series of error terms for rational approximations of irrational numbers, Journal of Integer Sequences 14 (2011), Article 11.1.4;
http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Elsner/elsner9.html

  * [7] C. Elsner and M. Stein, On error sum functions formed by convergents of real numbers, Journal of Integer Sequences 14 (2011), Article 11.8.6;
http://www.cs.uwaterloo.ca/journals/JIS/VOL14/Elsner2/elsner10.html

  * [8] C. Elsner and M. Stein, On the value distribution of Error Sums for approximations with rational numbers, Integers 12 (2012), A66, 1–28.
  * [9] C. Elsner, On error sums for square roots of positive integers with applications to Lucas and Pell numbers, Journal of Integer Sequences, 17 (2014), Article 14.4.4 .
https://cs.uwaterloo.ca/journals/JIS/VOL17/Elsner/elsner15.html

  * [10] C. Elsner and A. Klauke, Errorsums for the values of the exponential function, Forschungsberichte der FHDW Hannover, Bericht Nr. 02014/01, 1 - 19; RS 8153 (2014,1)
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  * [12] C. Elsner, On prime-detecting sequences from Apéry’s recurrence formulae for $\zeta(3)$ and $\zeta(2)$, Journal of Integer Sequences 11 (2008), Article 08.5.1.
  * [13] G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, fifth ed., Clarendon Press, 1979.
  * [14] W. Koepf, Hypergeometric Summation, Vieweg, 1998.
  * [15] M. E. Larsen, Summa Summarum, CMS Treatises in Mathematics, A K Peters, Ltd., 2007.
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List of OEIS sequence numbers.

A001850,
A003417,
A005258,
A005259,
A051451,
A108626,
A108626.


††institutetext: Indian Institute of Technology Kanpur, Kalyanpur, Kanpur
208016. INDIA.

# BRST Symmetry of Non-Lorentzian Yang-Mills Theory

Minhajul Islam
<EMAIL_ADDRESS>
###### Abstract

We explore the realization of BRST symmetry in the non-Lorentzian Yang-Mills
Lagrangian within the context of Galilean and Carrollian Yang-Mills theory.
Firstly we demonstrate the nilpotent property of classical BRST
transformations and construct corresponding conserve charges for both cases.
Then we analyze the algebra of these charges and observe the nilpotent
properties at the algebraic level. The findings of this study contribute to a
deeper understanding of BRST symmetry in non-Lorentzian Yang-Mills Lagrangians
and provide insights into the algebraic properties of related conserve
charges.

##  1 Introduction

Yang-Mills theories, named after Chen Ning Yang and Robert Mills Yang:1954vj ,
are quantum field theories (QFT) that describe the behavior of elementary
gauge bosons. This is the main ingredient of the Standard Model of particle
physics, which is our best current understanding of the behavior of subatomic
particles.

Our current comprehension of the physical world thus heavily relies on the
framework of QFT. The textbook formulation of QFT is closely tied to the
principles of relativistic physics, including Lorentz and Poincaré symmetry.
However, when examining real-life systems, it is often necessary to consider
approximations and limits of the fundamental theory. In this paper, we will be
interested in QFTs where Poincaré symmetry is replaced by Galilean and
Carrollian symmetries which constitute the the low and high energy sectors of
relativistic QFTs.

To understand Galilean and Carrollian QFTs, we will adopt a group-theoretic
approach starting from the Poincaré algebra and taking the limits of large c
(speed of light) and small c. These limits yield two different symmetry
algebras: the familiar Galilean algebra and the less familiar Carrollian
algebra. In both limits, several counter-intuitive concepts emerge. The
spacetime metrics degenerate, light-cones open up for non-relativistic theory,
and close up for Carrollian theory. Moreover, symmetry algebra gets enhanced
in both cases.

Galilean theories, which correspond to the limit of $c\rightarrow\infty$
(where c is the speed of light), are important for various areas of physics
such as condensed matter physics, non-AdS holography, and hydrodynamics. In
this limit, the metric of spacetime degenerates, and the structure of
spacetime changes from the usual Riemannian structure to a new one called
Newton-Cartan spacetime Duval:1984cj ; Duval:2009vt ; Bleeken:2015ykr ;
Bergshoeff:2015sic ; Hansen:2020pqs . For developing non-relativistic physics
involves starting from a Poincaré-invariant theory and expanding it using a
large c-expansion provides many insights into non-relativistic physics, such
as enhanced symmetry algebra and actions at each order Hansen:2019svu ;
Hansen:2020wqw ; Hansen:2020pqs ; Ergen:2020yop .

The Carrollian limit is the opposite of the limit mentioned earlier,
corresponding to $c\rightarrow 0$. The Carroll algebra was first discussed in
LevyLeblond ; NDS . It has recently become important in various applications,
particularly in the understanding of flat space holography Susskind:1998vk .
AdS/CFT duality is one of the most promising tools for understanding quantum
gravity. When the radius of curvature is infinite, AdS spacetime becomes flat
spacetime. Correspondingly, on the dual side, sending the speed of light to
zero results in a Carrollian conformal field theory Bagchi:2012cy . Some
important references for holography for asymptotically flat spacetime are
Susskind:1998vk ; Barnich:2010eb ; Bagchi:2010eg ; Bagchi:2012cy ;
Bagchi:2014iea ; Bagchi:2016bcd ; Barnich:2012aw ; Barnich:2012xq . The
understanding of flat space holography recently has taken two different
directions, viz. Celestial holography and Carrollian holography. Celestial
holography relates gravity in 4d asymptotically flat spacetimes to a 2d CFT
living on the celestial sphere Raclariu:2021zjz ; Pasterski:2021rjz . On the
other hand, Carrollian holography relates 4d asymptotically flat gravity to 3d
Carrollian CFTs living on the entire null boundary of 4d bulk spacetime
Bagchi:2016bcd ; Bagchi:2019xfx ; Bagchi:2019clu ; Duval:2014uva ;
Duval:2014lpa ; Dutta:2022vkg ; Bagchi:2023fbj ; Nguyen:2023vfz ;
Salzer:2023jqv ; Saha:2023hsl . Some works Bagchi:2022emh ; Donnay:2022aba ;
Donnay:2022wvx  connect both formalisms.

Carrollian physics appears on any null hypersurface, including the horizon of
a black hole Donnay:2019jiz ; Bicak:2023vxs . Carrollian gravity may provide a
tractable version of general relativity and may be useful in various physical
contexts Hartong:2015xda ; Bergshoeff:2017btm . The theory of Carroll is also
important in cosmology, inflation deBoer:2021jej , for fluids flowing at very
high velocities Bagchi:2023ysc , fractons Bidussi:2021nmp ; Perez:2022kax ;
Figueroa-OFarrill:2023vbj , and the study of flat physics in condensed matter
systems Bagchi:2022eui .The Carrollian limit of the string theory worldsheet
leads to the very high energy tensionless regime of strings Bagchi:2013bga ;
Bagchi:2015nca ; Bagchi:2020fpr .

Before we begin our investigations of aspects of non-Lorentzian QFTs, we will
quickly summarize summarize previous research on Galilean and Carrollian gauge
theories. Galilean electrodynamics was first studied a long time ago in LBLL .
Later, in Bagchi:2014ysa ; Bagchi:2015qcw ; Bagchi:2017yvj , authors
discovered infinite-dimensional Galilean conformal symmetry in both Galilean
abelian and Galilean Yang-Mills theory at the level of equations of motion.
More detailed work has been done on constructing actions for both Galilean
abelian Festuccia:2016caf ; Banerjee:2019axy  and Yang-Mills theory
Bagchi:2022twx . The quantum properties of Galilean scalar electrodynamics
were examined in Chapman:2020vtn , Galilean QED, Scalar QED, non-linear
electrodynamics in Banerjee:2022uqj ; Baiguera:2022cbp ; Sharma:2023chs ;
Mehra:2023rmm ; Banerjee:2022sza .

In the Carroll case, conformal structures at the level of equations of motion
in Bagchi:2016bcd ; Bagchi:2019clu ; Bagchi:2019xfx . Basu:2018dub  presented
the Carrollian action for the so-called electric abelian theory, which is an
interacting field theory with a scalar field Banerjee:2020qjj ; deBoer:2021jej
. Using the small c-expansion, the magnetic sector of Carrollian abelian
theory has been recently constructed in deBoer:2021jej , and the conformal
structure of this magnetic action was analyzed. In Henneaux:2021yzg , the
authors constructed the off-shell Carrollian Yang-Mills theory in the
Hamiltonian formulation. Finally, in Islam:2023rnc , the action formulation
for Carrollian Yang-Mills theory was constructed. In Chen:2023pqf  the authors
constructed Carrollian field theory using null reduction techniques.

BRST symmetry, first introduced by Becchi, Rouet, Stora, and Tyutin in the
1970s Becchi:1975nq ; Becchi:1974md , plays a crucial role in in the study of
gauge theories. BRST symmetry is a type of ghost symmetry, which means that it
involves the introduction of additional degrees of freedom that are not
physical, but help to resolve the ambiguity in the system. BRST symmetry of
$SU(N)$ Yang-Mills theory is defined using a set of operators known as BRST
operators, which act on the fields of the system and generate BRST symmetry
transformations. BRST operators have the property that they are nilpotent,
which means that they square to zero. This property is crucial in the
construction of the BRST symmetry, as it ensures that the BRST transformations
form a closed algebra. BRST symmetry of $SU(N)$ Yang-Mills theory is important
for the computation of physical observables in the system.

In this paper, we will construct BRST symmetry for Galilean and Carrollian
Yang-Mills theories. We incorporate important results and notation from our
recent previous works, namely Bagchi:2022twx  and Islam:2023rnc . We first
provide a brief review of these theories in Section 2. We then proceed to
explore BRST symmetry for Galilean field theories in Section 3.1, and in
Section 3.3 we do the same for Carrollian field theories. Finally, in Section
4, we conclud the paper by summarizing our findings and discussing the
implications of our results. Overall, our analysis sheds light on BRST
symmetry in non-Lorentzian field theories, and provides a foundation for
future investigations in this area.

##  2 Brief review of Non-Lorentzian Yang-Mills

###  2.1 Galilean Yang-Mills

In Bagchi:2022twx , we constructed the action for Galilean Yang-Mills theory
by using the null reduction procedure. We will give a brief review on that. We
write the Lagrangian density of Yang-Mills theory in $(d+1)$ dimensions in
null coordinates:

$\displaystyle\mathcal{L}_{YM}=-\frac{1}{4}\eta^{\tilde{\mu}\tilde{\rho}}\eta^{\tilde{\nu}\tilde{\sigma}}F_{\tilde{\mu}\tilde{\nu}}^{a}F_{\tilde{\rho}\tilde{\sigma}}^{a}=-\frac{1}{4}\Big{[}2F^{a}_{ut}F^{a}_{tu}+F^{ija}F^{a}_{ij}+4F^{a}_{ui}F^{ia}_{t}\Big{]},$
(1)

and perform null reduction along the null direction parametrized by the
coordinate $u$. We take the gauge field to be independent of the
$u$-coordinate, _i.e._ $\partial_{u}A_{\tilde{\mu}}^{a}=0$, and decompose its
components as

$\displaystyle A_{u}^{a}=\phi^{a},\quad A_{t}^{a}=a_{t}^{a},\quad
A_{i}^{a}=a_{i}^{a}.$ (2)

Then the null reduction gives the Lagrangian density in $d$ spacetime
dimensions as

$\displaystyle\mathcal{L}_{GYM}=\Big{[}\frac{1}{2}(\partial_{t}\phi^{a}-gf^{abc}\phi^{b}a_{t}^{c})(\partial_{t}\phi^{a}-gf^{ade}\phi^{d}a_{t}^{e})$
$\displaystyle\hskip
42.67912pt-\frac{1}{4}(\partial^{i}a^{j}-\partial^{j}a^{i}+gf^{ade}a^{id}a^{je})(\partial_{i}a_{j}-\partial_{j}a_{i}+gf^{abc}a_{i}^{b}a_{j}^{c})$
$\displaystyle\hskip
42.67912pt+(\partial_{i}\phi^{a}-gf^{abc}\phi^{b}a_{i}^{c})(\partial_{t}a^{ia}-\partial^{i}a_{t}^{a}+gf^{abc}a_{t}^{b}{a}^{ic})\Big{]},$
(3)

where the subscript GYM stands for Galilean Yang-Mills. It can also be written
in a compact form given by

$\displaystyle\mathcal{L}_{GYM}=\frac{1}{2}D_{t}\phi^{a}D_{t}\phi^{a}+D_{i}\phi^{a}E^{ia}-\frac{1}{4}W^{ija}W_{ij}^{a},$
(4)

where $D_{t}$, $D_{i}$ are gauge-covariant derivatives and $E^{ia}$,
$W^{a}_{ij}$ are field strength variables defined as

$\displaystyle
D_{t}\phi^{a}=\partial_{t}\phi^{a}-gf^{abc}\phi^{b}a_{t}^{c},\quad
D_{i}\phi^{a}=\partial_{i}\phi^{a}-gf^{abc}\phi^{b}a_{i}^{c},$ (5a)
$\displaystyle
E^{ia}=\partial_{t}a^{ia}-\partial^{i}a_{t}^{a}+gf^{abc}a_{t}^{b}{a}^{ic},\quad
W_{ij}^{a}=\partial_{i}a_{j}-\partial_{j}a_{i}+gf^{abc}a_{i}^{b}a_{j}^{c}.$
(5b)

The EOM for the Lagrangian (4) are given by

$\displaystyle D_{t}D_{t}\phi^{a}+D_{i}E^{ia}=0,$ (6a) $\displaystyle
D_{i}D_{i}\phi^{a}+gf^{abc}\phi^{b}D_{t}\phi^{c}=0,$ (6b) $\displaystyle
D_{t}D_{i}\phi^{a}-D_{j}W_{ji}^{a}-gf^{abc}\phi^{b}E^{ic}=0.$ (6c)

We can also find these equations by doing the procedure of null reduction on
relativistic equations.111The Lagrangian was also introduced in Gomis:2020fui
to derive the EOM for Galilean Yang-Mills theory with $U(N)$ gauge group
obtained as an effective theory from non-relativistic open string theory.

If we put $\phi^{a}=0$, the equations become

$\displaystyle D_{i}E^{ia}=0,~{}~{}D_{j}W_{ji}^{a}=0.$ (7)

An interesting point to re-emphasise is that the EOM even with the fields
$\phi^{a}$ turned off are different from the ones obtained by taking limits as
was described earlier in the section. It is possible that one needs to
consider scalings of the gauge coupling $g$ in order to obtain these
results222A similar phenomenon was observed when constructing the action of
Carrollian scalar electrodynamics in Bagchi:2019clu .

###  2.2 Carrollian Yang-Mills

In Islam:2023rnc , we analyzed the Carrollian limit of Yang-Mills theory and
obtained electric and magnetic sectors, where one subsector of each contained
non-abelian or self-interaction terms while the other subsector contained
copies of the Carrollian abelian theory. In the above mentioned paper, we
obtained Carrollian Yang-Mills actions by taking a small $c$-expansion of the
Poincaré invariant Yang-Mills action, where different values of the parameter
$\delta$ lead to different sectors for Carrollian Yang-Mills theory.

All four sectors were found to be invariant under infinite Carrollian
conformal algebra in $4$ dimensions. Below we provide a brief review of the
two non-trivial sectors, electric and magnetic, which are relevant to our
current purpose.

#### Electric Action

The electric sector action which has a non-abelian term, can be written in
compact form:

$\displaystyle\mathcal{L}_{0}=\frac{1}{2}\bigg{(}(\partial_{t}a_{i}^{a(0)}-\partial_{i}a_{t}^{a(0)})(\partial_{t}a_{i}^{a(0)}-\partial_{i}a_{t}^{a(0)})+2gf^{abc}(\partial_{t}a_{i}^{a{(0)}}-\partial_{i}a_{t}^{a(0)})a_{t}^{b(0)}a_{i}^{c(0)}$
$\displaystyle\hskip
142.26378pt+g^{2}f^{abc}f^{ade}a_{t}^{b(0)}a_{i}^{c(0)}a_{t}^{d(0)}a_{i}^{e(0)}\bigg{)}=\frac{1}{2}E_{i}^{a(0)}E_{i}^{a(0)},$
(8)

where
$E_{i}^{a(0)}=\partial_{t}a_{i}^{a(0)}-\partial_{i}a_{t}^{a(0)}+gf^{abc}a_{t}^{a(0)}a_{i}^{a(0)}$.
The equations of motion following from the action are given by

$\displaystyle\partial_{i}E_{i}^{a(0)}+gf^{abc}a_{i}^{b(0)}E_{i}^{c(0)}=D_{i}^{(0)}E_{i}^{a(0)}=0,$
(9a)
$\displaystyle\partial_{t}E_{i}^{a(0)}+gf^{abc}a_{t}^{b(0)}E_{i}^{c(0)}=D_{t}^{(0)}E_{i}^{a(0)}=0,$
(9b)

where
$D_{i}\mathcal{O}^{a}=\partial_{i}\mathcal{O}^{a}+gf^{abc}a_{i}^{b(0)}\mathcal{O}^{c},\,D_{t}\mathcal{O}^{a}=\partial_{t}\mathcal{O}^{a}+gf^{abc}a_{t}^{b(0)}\mathcal{O}^{c}$.

The gauge transformations under which the action (2.2) is invariant are given
by

$\displaystyle a_{t}^{a(0)}\rightarrow
a_{t}^{a(0)^{\prime}}=a_{t}^{a(0)}+\frac{1}{g}\partial_{t}\alpha^{a}+f^{abc}a_{t}^{b(0)}\alpha^{c},$
(10a) $\displaystyle a_{i}^{a(0)}\rightarrow
a_{i}^{a(0)^{\prime}}=a_{i}^{a(0)}+\frac{1}{g}\partial_{i}\alpha^{a}+f^{abc}a_{i}^{b(0)}\alpha^{c}.$
(10b)

This gauge transformation is the same as parent theory, but now we cannot
write it in covariant form like relativistic theory. Because, like the non-
relativistic theory, the metrics in Carrollian theory are degenerate, and time
and space are not on the same footing.

#### Magnetic Action

Now, we will discuss on the magnetic sector. The next to leading order( NLO)
Lagrangian take from Islam:2023rnc  contains leading order and NLO fields.
From the expansion of action section, we have the NLO Lagrangian (coefficient
of $c^{0}$) as

$\displaystyle\mathcal{L}^{(1)}=\big{(}D_{t}^{(0)}a_{i}^{a(1)}\big{)}E_{i}^{a(0)}-\big{(}D_{i}^{(0)}a_{t}^{a(1)}\big{)}E_{i}^{a(0)}-\frac{1}{4}f^{ija(0)}f_{ij}^{a(0)}.$
(11)

If we take the variation of the Lagrangian with respect to next to leading
order fields $a_{t}^{a(1)},a_{i}^{a(1)}$ we will get Eq.(9), leading order
equations of motion as a property of this formalism. If we take variation with
respect to leading order fields ($a_{t}^{a(0)},a_{i}^{a(0)}$), equations of
motion are

$\displaystyle
D_{i}^{(0)}D_{i}^{(0)}a_{t}^{a(1)}-D_{i}^{(0)}D_{t}^{(0)}a_{i}^{a(1)}-gf^{abc}a_{i}^{b(1)}E_{i}^{c(0)}=0,$
(12a) $\displaystyle
D_{t}^{(0)}D_{t}^{(0)}a_{i}^{a(1)}-D_{t}^{(0)}D_{i}^{(0)}a_{t}^{a(1)}-gf^{abc}a_{t}^{b(1)}E_{i}^{c(0)}-D_{k}^{(0)}f_{ki}^{a(0)}=0,$
(12b)

where
$D_{k}^{(0)}f_{ki}^{a(0)}=\partial_{k}f_{ki}^{a(0)}+gf^{abc}a_{k}^{b(0)}f_{ki}^{c(0)}$.
Although the action and the equations of motion look nice in compact form,
these are not Carroll invariant. To make Carroll invariant, we have to take
the constraint $E_{i}^{a(0)}=0$ at the level of action Eq.(11). Then action
will become $-\frac{1}{4}f^{ija(0)}f_{ij}^{a(0)}$ and equations of motion will
be $D_{k}^{(0)}f_{ki}^{a(0)}=0$.

We can derive the Carroll invariant magnetic sector from the relativistic
Yang-Mills action if we consider a Lagrange multiplier in relativistic
Lagrangian and then take speed of light to zero limit. The relativistic
Lagrangian with Lagrange multiplier $\xi_{i}^{a}$ and explicit $c$ factor is
given by

$\displaystyle\mathcal{L}=-\frac{c^{2}}{2}\xi_{i}^{a}\xi_{i}^{a}+\xi_{i}^{a}F_{0i}^{a}-\frac{1}{4}F_{ij}^{a}F_{ij}^{a}.$
(13)

From here, we can get back to the usual Yang-Mills action if we integrate out
$\xi_{i}$ fields. Now we can see if we take the small $c$ limit here, we will
get

$\displaystyle\mathcal{L}^{NLO}=\xi_{i}^{a}(\partial_{t}a_{i}^{a(0)}-\partial_{i}a_{t}^{a(0)})-\frac{1}{4}(\partial_{i}a_{j}^{a}-\partial_{j}a_{i}^{a})(\partial_{i}a_{j}^{a}-\partial_{j}a_{i}^{a})+gf^{abc}a_{t}^{b}a_{i}^{c}\xi_{i}^{a}$
$\displaystyle\hskip 56.9055pt-
gf^{abc}a_{i}^{b}a_{j}^{c}\partial_{i}a_{j}^{a}-\frac{1}{4}g^{2}f^{abc}f^{ade}a_{i}^{b}a_{j}^{c}a_{i}^{d}a_{j}^{e}=\xi_{i}^{a}E_{i}^{a}-\frac{1}{4}f_{ij}^{a}f_{ij}^{a}.$
(14)

The Lagrangian contains non-trivial self-interaction terms or non-abelian
terms. The equations of motion of this action are

$\displaystyle E_{i}^{a}=0,\quad D_{i}\xi_{i}^{a}=0,\quad
D_{t}\xi_{i}-D_{j}f_{ji}=0.$ (15)

Here we are getting the constraints $E_{i}^{a(0)}=0$ as an equations of motion
for the Lagrange($\xi_{i}^{a}$). Below we will see the full spacetime symmetry
of this action.

The action Eq.(2.2) is invariant under the gauge transformation

$\displaystyle a_{t}^{a}\rightarrow
a_{t}^{{}^{\prime}a}=a_{t}^{a}+\frac{1}{g}\partial_{t}\alpha^{a}+f^{abc}a_{t}^{b}\alpha^{c},$
(16a) $\displaystyle a_{i}^{a}\rightarrow
a_{i}^{{}^{\prime}a}=a_{i}^{a}+\frac{1}{g}\partial_{i}\alpha^{a}+f^{abc}a_{i}^{b}\alpha^{c},$
(16b)
$\displaystyle\xi_{i}^{a}\rightarrow\xi_{i}^{{}^{\prime}a}=\xi_{i}^{a}+f^{abc}\xi_{i}^{b}\alpha^{c}.$
(16c)

The temporal and spatial component of the gauge field is transformed in the
same way as the electric sector. The Lagrange multiplier $\xi_{i}^{a}$
transforms as a scalar in the adjoint representation of the underlying gauge
group.

##  3 BRST Symmetry

###  3.1 Yang-Mills theory

One of the most important techniques for quantization of gauge theories is
BRST quantization. The canonical quantization of the modified action obtained
after path integral formulation is important to define physical states and to
eliminate the gauge symmetry when restricted to the subspace of the physical
Hilbert space.

The total gauge-fixed Lagrangian of the relativistic Yang-Mills theory
including the gauge fixing and ghost terms is

$\displaystyle\mathcal{L}=-\frac{1}{4}F^{\mu\nu
a}F^{a}_{\mu\nu}-\frac{1}{2\xi}\big{(}\partial^{\mu}A_{\mu}\big{)}^{2}-\partial^{\mu}\bar{c}^{a}D_{\mu}c^{a},$
(17)

where $D_{\mu}c^{a}=\partial_{\mu}c^{a}-gf^{abc}A_{\mu}^{b}c^{c}$. This gauge
fixed Lagrangian is not invariant under the gauge symmetry but invariant under
a global symmetry, _i.e._ BRST symmetry. The BRST transformations are given by

$\displaystyle\delta A_{\mu}^{a}=\frac{\omega}{g}D_{\mu}c^{a},\quad\delta
c^{a}=-\frac{\omega}{2}f^{abc}c^{b}c^{c},\quad\delta\bar{c}^{a}=\frac{\omega}{g\xi}\big{(}\partial_{\mu}A^{\mu
a}\big{)}^{2},$ (18)

where $\omega$ is an anti-commutating constant parameter. The BRST invariance
of the Lagrangian leads to many interesting consequences. It helps us in the
covariant quantization of the Yang-Mills theory, to prove unitarity of the
S-matrix, with gauge fixing and ghost term of the Lagrangian. The BRST
invariance also leads to Ward-Takahashi identities; in Yang-Mills theory these
are called Slavnov-Taylor identities.

###  3.2 Galilean Yang-Mills theory

Performing the null reduction of the full Lagrangian (17), the NR Yang-Mills
Lagrangian with gauge fixing term and ghost term is

$\displaystyle\mathcal{L}_{Gfull}=\frac{1}{2}D_{t}\phi^{a}D_{t}\phi^{a}+D_{i}\phi^{a}E^{ia}-\frac{1}{4}f^{ija}f_{ij}^{a}-\frac{1}{2\xi}(\partial_{t}\phi^{a}+\partial^{i}a_{i}^{a})^{2}-gf^{abc}\partial_{t}\bar{c}^{a}c^{b}\phi^{c}$
$\displaystyle-\partial_{i}\bar{c}^{a}(\partial_{i}c^{a}-gf^{abc}c^{b}a_{i}^{c}).$
(19)

This Galilean Yang-Mills Lagrangian is invariant under the transformations

$\displaystyle\delta\phi^{a}=\omega f^{abc}\phi^{b}c^{c},\,\delta
a_{t}^{a}=\frac{\omega}{g}D_{t}c^{a},\,\delta
a_{i}^{a}=\frac{\omega}{g}(D_{i}c^{a}),$ $\displaystyle\delta
c^{a}=-\frac{\omega}{2}f^{abc}c^{b}c^{c},\,\delta\bar{c}^{a}=\frac{\omega}{g\xi}(\partial_{t}\phi^{a}+\partial_{i}a_{i}^{a}).$
(20)

Here $c^{a}$ is a Grassmannian variable and represents the ghost field. The
transformations of gauge fields contain ghost field. We can see that the
transformations of the gauge fields are actually same as the gauge
transformation but now the gauge parameter is replaced by the ghost field
$c^{a}$. Thus the Yang-Mills piced of the total Lagrangian is naturally
invariant under the above transformations. Under these transformations the
gauge-fixing and the ghost terms transform as

$\displaystyle\delta\big{(}\mathcal{L_{GF+GH}}\big{)}=gf^{abc}\partial_{t}\big{[}\frac{\omega}{g\xi}\big{(}\partial_{t}\phi^{a}+\partial_{i}a_{i}^{a}\big{)}c^{b}\phi^{c}\big{]}-\partial_{i}\big{[}\frac{\omega}{g\xi}\big{(}\partial_{t}\phi^{a}+\partial_{k}a_{k}^{a}\big{)}D_{i}c^{a}\big{]},$
(21)

which is a total divergence, thus showing invariance of the gauge-fixing and
ghost terms.

In relativistic Yang-Mills theories, the BRST transformations are nilpotent,
_i.e._ $\delta_{1}\delta_{2}\Phi^{a}=0$, where $\Phi^{a}$ denotes all the
fields in the theory. The nilpotency of the BRST operator is very important to
introduce physical states.

Let us check the nilpotency of the BRST transformations (20) in Galilean Yang-
Mills theory:

$\displaystyle\delta^{2}\phi^{a}=-\omega\delta(gf^{abc}c^{b}\phi^{c})=0,\quad\text{using
Jacobi identity}$ (22a)
$\displaystyle\delta^{2}a_{t}^{a}=\frac{\omega}{g}\delta(D_{t}c^{a})=0,~{}\delta^{2}a_{i}^{a}=\frac{\omega}{g}\delta(D_{i}c^{a})=0.$
(22b)

To see this we have to use Jacobi identity for the structure constants

$\displaystyle f^{cab}f^{ckl}+f^{cal}f^{cbk}+f^{cak}f^{clb}=0,$ (23)

which is derived from the Lie algebra of the underlying gauge group

$\displaystyle[T^{b},[T^{k},T^{l}]]+[T^{k},[T^{l},T^{b}]]+[T^{l},[T^{b},T^{k}]]=0.$
(24)

For the ghost field, using the Jacobi identity we get

$\displaystyle\delta^{2}c^{a}=\frac{\omega}{2}Q(f^{abc}c^{b}c^{c})=0.$ (25)

However for the anti-ghost field, the transformations are nilpotent only upon
using the equation of motion for the ghost field, _i.e._

$\displaystyle\delta^{2}\bar{c}^{a}=\delta(\partial_{t}\phi^{a}+\partial^{i}a_{i}^{a})=0~{}~{}\text{upon
using}~{}~{}\partial_{i}D^{i}c^{a}-gf^{abc}\partial_{t}(c^{b}\phi^{c})=0.$
(26)

Now we have seen that the Lagrangian is invariant under the BRST
transformations (20) off-shell but to get full nilpotent BRST operator we have
to use the ghost equation of motion. To get the nilpotency off-shell, we
introduce an auxiliary field $F^{a}$. So now relevant part of the modified
Lagrangian is

$\displaystyle\mathcal{L_{GF+GH}}=\frac{\xi}{2}F^{a}F^{a}+\partial^{t}F^{a}\phi^{a}+\partial^{i}F^{a}a_{i}-gf^{abc}\partial_{t}\bar{c}^{a}c^{b}\phi^{c}+\partial_{i}\bar{c}^{a}(\partial_{i}c^{a}-gf^{abc}c^{b}a_{i}^{c}).$
(27)

Now the transformations which keep this Lagrangian invariant are given below.
The transformations of gauge fields and ghost field are same as before,
however the transformation of anti-ghost field is changed and written in terms
$F^{a}$. The transformations are

$\displaystyle\delta\phi^{a}=\omega f^{abc}\phi^{b}c^{c},\,\delta
a_{t}^{a}=\frac{\omega}{g}D_{t}c^{a},\,\delta
a_{i}^{a}=\frac{\omega}{g}(D_{i}c^{a}),$ $\displaystyle\delta
c^{a}=-\frac{\omega}{2}f^{abc}c^{b}c^{c},\,\delta\bar{c}^{a}=\frac{\omega}{g}F^{a},\,\delta
F^{a}=0.$ (28)

Now doing the same analysis as before and using the Jacobi identity of
structure constants, we see that these transformations are nilpotent for all
the fields without using any equations of motion. The Lagrangian is invariant
under these transformations without any total derivetive term. Now let us
calculate the current for the above transformations

$\displaystyle J^{t}=-\omega
f^{abc}\phi^{b}c^{c}D_{t}\phi^{a}+\frac{\omega}{g}D_{i}\phi^{a}D_{i}c^{a}+{\omega}F^{a}f^{abc}c^{b}\phi^{c}$
$\displaystyle\hskip 14.22636pt=-\omega
f^{abc}\phi^{b}c^{c}\Pi_{\phi}^{a}+\frac{\omega}{g}\Pi_{a_{i}}^{a}D_{i}c^{a}-\frac{\omega}{g}F^{a}\Pi_{\bar{c}}^{a}.$
(29)

Then the BRST charge is

$\displaystyle Q=\int d^{3}xJ^{t}.$ (30)

The Lagrangian respect another symmetry called ghost scaling symmetry. In
relativistic Yang-Mills theories, this symmetry is associated with number of
ghost field. The ghost scaling transformation is given by

$\displaystyle\delta c^{a}=\epsilon
c^{a},\,\delta\bar{c}^{a}=-\epsilon\bar{c}^{a},$ (31)

where $\epsilon$ is a constant commuting parameter. The conserved charge
corresponding to this symmetry is

$\displaystyle Q_{c}=\int d^{3}xJ^{t}_{c}=\int
d^{3}x\big{(}gf^{abc}\bar{c}^{a}c^{b}\phi^{c}\big{)}=-\int
d^{3}x\bar{c}^{a}\Pi_{\bar{c}}^{a}.$ (32)

The usual equal time (anti-) commutation relations are

$\displaystyle\big{[}\phi^{a},\Pi_{\phi}^{b}\big{]}=i\delta^{ab}\delta^{3}(x-y),\,\big{[}a_{i}^{a},\Pi_{a_{k}}^{b}\big{]}=i\delta^{ab}\delta_{ik}\delta^{3}(x-y)$
$\displaystyle\big{[}a_{t}^{a},\Pi_{a_{t}}^{b}\big{]}=i\delta^{ab}\delta^{3}(x-y),\,\big{[}F^{a},\Pi_{F}^{b}\big{]}=i\delta^{ab}\delta^{3}(x-y)$
$\displaystyle\big{\\{}c^{a},\Pi_{c}^{b}\big{\\}}=i\delta^{ab}\delta^{3}(x-y),\,\big{\\{}\bar{c}^{a},\Pi_{\bar{c}}^{b}\big{\\}}=i\delta^{ab}\delta^{3}(x-y)$
(33)

Using these we can see that

$\displaystyle\big{\\{}Q,Q\\}=2Q^{2}=0,$ (34)

which shows the nilpotency of the BRST operator. For the ghost operator
$Q_{c}$ we have

$\displaystyle\big{[}Q_{c},Q_{c}\big{]}=0.$ (35)

To understand the relation between $Q$ and $Q_{c}$, we will do the following
analysis:

$\displaystyle(\delta_{c}\delta_{B}-\delta_{B}\delta_{c})\Phi^{a}=\delta_{B}\Phi^{a},$
(36)

where $\Phi^{a}$ is any field
($\phi^{a},\,a_{t}^{a},\,a_{i}^{a},\,c^{a},\,\bar{c}^{a},\,F^{a}$). Also doing
a similar computation in terms of the operators, we get
$\big{[}Q,Q_{c}\big{]}\Phi^{a}=Q\Phi^{a}$ from which we conclude that

$\displaystyle\big{[}Q,Q_{c}\big{]}=Q.$ (37)

With many application there is a very good mathematical foundation of BRST
symmetry. Using BRST symmetry in relativistic QFTs we can define equivalence
class of physical states which constitute the BRST cohomology. In future, we
want to do similar analysis in Galilean field theories.

###  3.3 Carrollian Yang-Mills theory

In this section, we will examine how the BRST symmetry is manifested in the
Carrollian Yang-Mills theory. Specifically, we will first investigate the
realization of the BRST symmetry at the classical level for both the electric
and magnetic sectors of the theory. After that, we will proceed to calculate
the BRST charge.

Additionally, there is an important symmetry, denoted as $U(1)$, that exists
in the ghost part of the Lagrangian, corresponding to the conservation of
ghost particle number. It is important to note that the ghost field is a
Grassmann variable, and therefore the algebra of charge corresponding to this
$U(1)$ symmetry is anti-commuting.

#### Electric Sector

The Lagrangian for the electric sector with gauge fixing term and ghost term
is discussed in Islam:2023rnc . The full Lagrangian for the electric sector is

$\displaystyle\mathcal{L}=\frac{1}{2}E_{i}^{a(0)}E_{i}^{a(0)}-\frac{1}{2\chi}\partial_{t}a_{t}^{a}\partial_{t}a_{t}^{a}++\partial_{t}\bar{c}^{a}D_{t}c^{a}.$
(38)

Similar to the relativistic this Lagrangian also satisfy a global symmetry.
The global transformations under which this Lagrangian is invariant is given
by

$\displaystyle\delta
a_{t}^{a}=\frac{\omega}{g}\big{(}D_{t}c\big{)}^{a}\quad\delta
a_{i}^{a}=\frac{\omega}{g}\big{(}D_{i}c\big{)}^{a}\quad\delta
c^{a}=-\frac{\omega}{2}f^{abc}c^{b}c^{c},\quad\delta\bar{c}=-\frac{\omega}{g\xi}\partial_{t}a_{t}^{a},$
(39)

this is Carrollian BRST transformations for the electric sector. The action is
invariant under these transformation with a total derivative term. One of the
property BRST transformations is that these transformations are nilpotent. To
prove that the above mentioned transformations are nilpotent we have to use
Jacobi identity and equations of motion of ghost
fields($\partial_{t}D_{t}c^{a}=0$). The action under the transformation
changes as

$\displaystyle\delta\mathcal{L}^{electric}=-\partial_{t}\big{(}\frac{\omega}{g\xi}\partial_{t}a_{t}^{a}D_{t}c^{a}\big{)}$
(40)

to derive the above change we have to use ghost equations of motion. So for
nilpotency of the transformations and to see the invariance of the action we
need to use equations of motion for the ghost field. We can also have off-
shell BRST transformations which is nilpotent without using any equations of
motion and the action is invariant without using any equations of motion. To
have BRST symmetry with out using ghost equations of motion we need to
introduce auxiliary field. With the auxilary field full Lagrangian is

$\displaystyle\mathcal{L}^{electric}=\frac{1}{2}E_{i}^{a(0)}E_{i}^{a(0)}+\frac{\xi}{2}F^{a}F^{a}+\partial_{t}F^{a}a_{t}^{a}+\partial_{t}\bar{c}^{a}D_{t}c^{a}$
(41)

using equation of motion of $F^{a}$ we can go back to previous Lagrangian. Now
the transformations are

$\displaystyle\delta
a_{t}^{a}=\frac{\omega}{g}\big{(}D_{t}c\big{)}^{a}\quad\delta
a_{i}^{a}=\frac{\omega}{g}\big{(}D_{i}c\big{)}^{a}\quad\delta
c^{a}=-\frac{\omega}{2}f^{abc}c^{b}c^{c},\quad\delta\bar{c}=-\frac{\omega}{g}F^{a},\quad\delta
F^{a}=0$ (42)

The action is invariant under these transformation without using any equations
of motion. These symmetry of action is off-shell symmetry.

There is another symmetry corresponding to ghost number symmetry, is global
$U(1)$ symmetry. Infinitesimal form of the transformation is

$\displaystyle\delta c^{a}=\epsilon
c^{a},\quad\delta\bar{c}^{a}=-\epsilon\bar{c}$ (43)

The Lagrangian (41) in invariant under this transformation.

Now to go to the Quantum we need calculate charges corresponding to BRST
symmetry and above mention $U(1)$ symmetry. After that we will see relevant
commutation and anti commutation relation. Charges are for the BRST and $U(1)$
respectively are

$\displaystyle Q_{BRST}=\int
d^{3}x\big{[}\frac{\omega}{g}E_{i}^{a}D_{i}c^{a}-\frac{\omega}{g}F^{a}D_{t}c^{a}+\frac{\omega}{2}f^{abc}c^{b}c^{c}\partial_{t}\bar{c}^{a}\big{]}$
(44) $\displaystyle Q_{U(1)}=\int
d^{3}x\big{[}\bar{c}^{a}D_{t}c^{a}+c^{a}\partial_{t}c^{a}\big{]}$ (45)

To calculate the algebra of these charges it will convenient to write it using
conjugate momentum of the fields. Then we will able to usual bracket between
fields and conjugate momentum. Conjugate momentum corresponding to different
fields are

$\displaystyle\Pi_{i}^{a}=\frac{\partial\mathcal{L}}{\partial(\partial_{t}a_{i}^{a})}=E_{i}^{a},\,\Pi^{a}=\frac{\mathcal{L}}{\partial(\partial_{t}F^{a})}=a_{t}^{a},\,\Pi_{c}^{a}=-\partial_{t}\bar{c}^{a},\,\Pi_{\bar{c}}^{a}=D_{t}c^{a}$
(46)

Charges using these definition

$\displaystyle Q_{BRST}=\int
d^{3}x\big{[}\frac{\omega}{g}\Pi_{i}^{a}D_{i}c^{a}-\frac{\omega}{g}F^{a}\Pi_{\bar{c}}^{a}-\frac{\omega}{2}f^{abc}c^{b}c^{c}\Pi_{c}^{a}\big{]}$
(47) $\displaystyle Q_{U(1)}=\int
d^{3}x\big{[}\bar{c}^{a}\Pi^{a}_{\bar{c}}-c^{a}\Pi_{c}^{a}\big{]}$ (48)

Algebra satifying these operator using field Poisson bracket

$\displaystyle\big{[}Q_{BRST},Q_{BRST}\big{]}=0,\quad\\{Q_{U(1)},Q_{U(1)}\\}=0,\quad\big{[}Q_{BRST},Q_{U(1)}\big{]}=iQ_{BRST},$
(49)

We can also see these relations using (42) and (43). From first commutation we
can confirm the nilpotency of BRST symmetry. From second relation we can say
that the ghost particle symmetry is abelian. Lastly from third relation we can
see BRST transformation carries unit ghost particle.

#### Magnetic Sector

The full magnetic sector Lagrangian with gauge fixing term and ghost term
is(details in Islam:2023rnc )

$\displaystyle\mathcal{L}=\xi_{i}^{a}E_{i}^{a}-\frac{1}{4}f_{ij}^{a}f_{ij}^{a}-\frac{1}{2\chi}\partial_{i}a_{i}^{a}\partial_{j}a_{j}^{a}-\partial_{i}\bar{c}^{a}D_{i}c^{a}.$
(50)

. The full Lagrangian which respect the off-shell BRST transformations is

$\displaystyle\mathcal{L}_{magnetic}=\xi_{i}^{a}E_{i}^{a}-\frac{1}{4}f_{ij}^{a}f_{ij}^{a}+\frac{\xi}{2}F^{a}F^{a}+\partial_{i}F^{a}a_{i}^{a}+\partial_{i}\bar{c}^{a}D_{i}c^{a}$
(51)

Off-shell BRST transformations which keep the above Lagrangian invariant is
same as eq.(42) along with transformation of $\xi_{i}^{a}$ as
$\delta\xi_{i}^{a}=\omega f^{abc}\xi_{i}^{b}c^{c}$. For magnetic sector BRST
charge contain only spatial derivative.

$\displaystyle Q_{BRST}=\int d^{3}{x}\big{[}\xi_{i}^{a}D_{i}c^{a}\big{]}$ (52)

The Lagrangian invariant under u(1)symmetry mentioned in previous section. But
charge corresponding that symmetry is identically zero because in the
Lagrangian there is no time derivative of ghost field. The algebra for this
sector is trivially realized.

##  4 Conclusions and Discussion

In this paper, we investigate the BRST symmetry of Galilean and Carrollian
Yang-Mills theories, which is should be crucial to a more detailed study of
gauge theories in the non-Lorentzian regime. We begin by studying the Galilean
Yang-Mills theory, which is a non-relativistic theory that describes the
interaction between gauge fields and matter fields in a Galilean-invariant
framework. We first realized the BRST symmetry for the Galilean Yang-Mills
theory and redefine its Lagrangian to make it more concrete. We then analyze
the classical and quantum level of the BRST symmetry and observe that it is
realized in both cases.

Next, we move on to Carrollian Yang-Mills theory, which is another non-
Lorentzian theory that describes the interaction between gauge fields and
matter fields in a Carrollian-invariant framework. We analyze the non-trivial
sectors of the theory, specifically the electric and magnetic sectors. For the
magnetic sector case, the $U(1)$ charge is zero, and hence the charge algebra
is trivially realized. Again, we observe the BRST symmetry at both the
classical and quantum level.

The study of the BRST symmetry in non-Lorentzian field theories is crucial in
understanding the underlying physical properties of these theories. The BRST
symmetry provides a way to fix gauge ambiguities, and its realization at the
classical and quantum level is an essential aspect in the computation of
physical observables.

In future work, we plan to construct different sectors of the Galilean Yang-
Mills Lagrangian and analyze their BRST symmetry as we discussed for
Carrollian case in this paper. We also aim to extend our analysis to construct
BRST cohomology for non-Lorentzian theories. This analysis will help us gain a
better understanding of the fundamental symmetries of non-Lorentzian field
theories and their physical properties.

Our investigation of the BRST symmetry of Galilean and Carrollian Yang-Mills
theories provides valuable insights into the behavior of non-Lorentzian field
theories. The BRST symmetry is an essential tool in understanding the physical
properties of these theories, and its realization at both the classical and
quantum level is a crucial aspect in the computation of physical observables.

## Acknowledgments

We express our heartfelt gratitude to Arjun Bagchi for fruitful discussions,
insightful suggestions, and valuable comments on this manuscript and our work.
We would also like to thank Nilay Kundu, Rudranil Basu, Kedar Kolekar, Kunal
Pal, and Kuntal Pal for productive discussions.

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