text:
\begin{document}

\title[Irreducible skew polynomials over domains]
{Irreducible skew polynomials over domains}

\begin{abstract}
Let $S$ be a domain and $S[t;\sigma,\delta]$ a
skew polynomial ring, where $\sigma$ is an injective endomorphism of $S$ and $\delta$ a left $\sigma$-derivation.
 We give criteria for skew polynomials  $f\in S[t;\sigma,\delta]$ of degree less or equal to four to be irreducible
 and apply them to skew polynomials over polynomial rings, for instance quantized Weyl algebras and quantum planes.
 We also consider the skew polynomial $f(t)=t^m-a$.
\end{abstract}

\maketitle

\section*{Introduction}

Most of the results  obtained so far on the irreducibility of skew polynomials in $R=S[t;\sigma,\delta]$, where $\sigma$ is an injective endomorphism  of $S$ and $\delta$ a left
$\sigma$-derivation, assume that $S$ is a division algebra:
A sequence of well-known papers by Lam, Leroy and others greatly contributed to our understanding
 of skew polynomials
over division rings and their factorization behaviour, starting with \cite{lam1988vandermonde}.
Some earlier results are contained in \cite{B, C, P66}. More recently,
 two general algorithms for computing
 the bound of a skew polynomial over a skew field were given in \cite{GLN}. As an application,
 a criterion for deciding whether a bounded skew polynomial is irreducible was developed.
 The computational method presented there heavily relies on being able to find the zero divisors in certain
 central simple algebras and is only
  applicable for certain set-ups, that is when the input data $S$, $\sigma$ and $\delta$ are effective and computable.
  It works for bounded skew polynomials. Recall that  $f \in R$ is \emph{bounded} if there exists $0 \neq f^* \in
R$ such that $Rf^* = f^* R$ is the largest two-sided ideal of $R$
contained in $Rf$. The element $f^*$ is determined by $f$ up to
multiplication on the left by elements of $S^{\times}$.

 Some first results on factoring skew polynomials of degree two in
 quantum planes and quantized Weyl algebras were collected in \cite{CP, HP, CookP}, but
 the main interest in these papers were what the authors called quadratic forms, where additional restrictions are placed
 on the type of degree two skew polynomial considered.

 We only assume that $S$ is a domain, i.e., a unital ring without non-trivial zero divisors. We look at some skew polynomials of low degree in $S[t;\sigma,\delta]$ and the skew polynomial  $f(t)=t^m-a$
 and give  criteria for them to be irreducible.

 The paper is organized as follows: In Section \ref{sec:irreduciblebenodomain}, we establish
  some necessary and sufficient criteria for skew polynomials in
 $S[t;\sigma,\delta]$  of degree less than or equal to four to be irreducible.  The situation is easiest when
  $S$ is a right Ore domain, then
we can take a skew polynomial $f\in S[t;\sigma,\delta]$ and check if it is irreducible over  $D[t;\sigma,\delta]$,
where $D$ is the right ring of fractions of $S$, in order to obtain necessary conditions of $f$ to be irreducible. We employ this for the polynomial $f(t)=t^m-a\in S[t;\sigma,\delta]$.
 In Section \ref{subsec:irred},
 we look at irreducible polynomials of degree two and three
 in skew polynomial rings over a polynomial ring $K[y][t;\sigma]$.
 Following \cite{BI, BII, BIII}, we define $A_h=K[y][t; \delta]$ with $\delta(r) = r'h$ for some $h(y)
\in K[y]$, where $r'$ is the usual derivation of $r$ with respect to $y$. For instance, $A_y$ is the universal enveloping algebra of the two-dimensional non-abelian Lie algebra, and the algebra
$A_{y^2}$ is also known as the Jordan plane which appears in noncommutative algebraic geometry.
Irreducible polynomials of degree two and three as well as $f(t)=t^m-a$ in the quantum plane,
 and irreducible skew polynomials
of degree two in a quantized Weyl algebra and in $A_h=K[y][t; \delta]$  are considered in Section \ref{sec:Weyletc}.

In particular, we show that a monic polynomial $f(t) \in K[t]$ of degree two or three is irreducible in the quantum plane
$K[y][t;\sigma]$ if and only if it is irreducible in $K[t]$ (Corollary \ref{thm:irredquantum}), and that
a degree two polynomial  $f(t) \in K[t]$ is irreducible in the quantized Weyl algebra
$K[y][t;\sigma,\delta]$ if and only if it is irreducible in $K[t]$ (Corollary \ref{thm:irredWeyl}).

\section{Irreducibility criteria for polynomials of low degree in $S[t;\sigma,\delta]$ where $S$ is a domain}
\label{sec:irreduciblebenodomain}

Let $S$ be a domain, $\sigma$  an injective endomorphism of $S$ and $\delta$  a left $\sigma$-derivation of $S$,
i.e. an additive map such that $\delta(ab)=\sigma(a)\delta(b)+\delta(a)b$ for all $a,b\in S$.
The \emph{skew polynomial ring} $R=S[t;\sigma,\delta]$ is the
set of skew polynomials $g(t)=a_0+a_1t+\dots +a_nt^n$ with $a_i\in
S$, with term-wise addition and where the multiplication is defined
via $ta=\sigma(a)t+\delta(a)$ for all $a\in S$ \cite{O1}. That means,
$$at^nbt^m=\sum_{j=0}^n a(\Delta_{n,j}\,b)t^{m+j}$$ for all $a,b\in
S$, where the map $\Delta_{n,j}$ is defined recursively via
$$\Delta_{n,j}=\delta(S_{n-1,j})+\sigma (\Delta_{n-1,j-1}),$$ with
$\Delta_{0,0}=id_S$, $\Delta_{1,0}=\delta$, $\Delta_{1,1}=\sigma $.
Thus $\Delta_{n,j}$ is the sum of all polynomials in $\sigma$
and $\delta$ of degree $j$ in $\sigma$ and degree $n-j$ in $\delta$
\cite[p.~2]{J96}. If $\delta=0$, then $\Delta_{n,n}=\sigma^n$.
For $\delta=0$, we obtain the ring of \emph{twisted polynomials} $S[t;\sigma]=S[t;\sigma,0]$.

 For $f(t)=a_0+a_1t+\dots +a_nt^n\in R$ with $a_n\not=0$ define ${\rm deg}(f)=n$ and ${\rm deg}(0)=-\infty$.
 An element $f\in R$ is \emph{irreducible} in $R$ if it is not a unit and  it has no proper factors, i.e if there do not exist $g,h\in R$ with
 ${\rm deg}(g),{\rm deg} (h)<{\rm deg}(f)$ such that $f=gh$.

\subsection{}
Since $S$ is a domain, we have $\mathrm{deg}(g(t)h(t))
=\mathrm{deg}(g(t)) + \mathrm{deg}(h(t))$ for all $g(t), h(t) \in S[t;\sigma,\delta]$ which implies that
the proof of the results in this subsection are analogous to those mentioned in the literature cited in the following,
where, however, $S$ is assumed to be a division algebra:

\begin{theorem} \label{thm:irreducibility criteria when S is a ring without zero divisors}
 (i) $f(t) = t^2 - a_1 t - a_0 \in S[t;\sigma]$ is irreducible if and only if
$\sigma(b)b - a_1 b - a_0 \neq 0$
for all $b \in S$.
\\ (ii) Suppose $\sigma$ is an automorphism of $S$. Then $f(t) = t^3 - a_2 t^2 - a_1 t - a_0 \in S[t;\sigma]$ is irreducible if and only if
\begin{equation*}
\sigma^2 (b) \sigma(b) b - \sigma^2 (b)\sigma(b) a_2 - \sigma^2 (b)
\sigma(a_1) - \sigma^2 (a_0) \neq 0,
\end{equation*}
and
\begin{equation*}
\sigma^2 (b) \sigma(b) b - a_2 \sigma(b) b - a_1 b - a_0 \neq 0
\end{equation*}
for all $b \in S.$
\\ (iii) Suppose $\sigma$ is an automorphism of $S$, then $f(t) = t^3 - a \in S[t;\sigma]$ is irreducible if and only if $\sigma^2 (b) \sigma(b) b \neq a$ for all $b \in S$.
\\ (iv) Suppose $\sigma$ is an automorphism of $S$, then $f(t) = t^4 - a_3 t^3 - a_2 t^2 - a_1 t - a_0 \in S[t;\sigma]$ is irreducible if and only if
\begin{equation} \label{eqn:degree 4 irreducible 1}
\sigma^3 (b) \sigma^2 (b) \sigma(b) b + a_3 \sigma^2 (b) \sigma(b) b
+ a_2 \sigma(b) b + a_1 b + a_0 \neq 0,
\end{equation}
and
\begin{equation} \label{eqn:degree 4 irreducible 2}
\begin{split}
&\sigma^3 (b) \sigma^2 (b) \sigma (b) b + \sigma^3 (b) \sigma^2 (b)
\sigma (b) a_3 \\ &+ \sigma^3 (b) \sigma^2 (b) \sigma (a_2) +
\sigma^3 (b) \sigma^2 (a_1) + \sigma^3 (a_0) \neq 0,
\end{split}
\end{equation}
for all $b \in S$, and for every $ c , d \in S$, we have either
\begin{equation} \label{eqn:degree 4 irreducible 3}
\sigma^2 (c) \sigma(c)c + \sigma^2 (d)c + \sigma^2 (c) \sigma(d) +
a_3 (\sigma(d) + \sigma(c)c) + a_2 c + a_1 \neq 0,
\end{equation}
 or
\begin{equation} \label{eqn:degree 4 irreducible 4}
\sigma^2 (d)d + \sigma^2 (c) \sigma(c) d + a_3 \sigma(c)d + a_2 d +
a_0 \neq 0.
\end{equation}

\end{theorem}

That means in Theorem \ref{thm:irreducibility criteria when S is a ring without zero divisors} (iv),
the skew polynomial $f(t)$ is irreducible if and only if \eqref{eqn:degree 4
irreducible 1} and \eqref{eqn:degree 4 irreducible 2} and
(\eqref{eqn:degree 4 irreducible 3} or \eqref{eqn:degree 4
irreducible 4}) holds. The proofs are the same as the ones for \cite[Theorem 34, Theorem 41]{CP18}
where $S$ is assumed to be a division algebra.

\begin{corollary}
 Suppose $\sigma$ is an automorphism of $S$. Then
 $f(t) = t^4 - a \in S[t;\sigma]$ is reducible if and only if
$$\sigma^2 (c) \sigma(c)c + \sigma^2 (d)c + \sigma^2 (c) \sigma(d) =
0 \quad \text{and} \quad \sigma^2 (d)d + \sigma^2 (c) \sigma(c) d =a,$$ for some $c, d \in S$.
\end{corollary}

This is proved analogously as \cite[Corollary 43]{CP18}.

We recursively define  two sequences of maps: $N_i : S \rightarrow S$,
and $M_i:S \rightarrow S$, $i \geq 0$.
The maps $N_i$ are given via
$$N_0(b) = 1, \ N_{i+1}(b) = \sigma(N_i(b))b + \delta(N_i(b)), $$ i.e.,
$ N_1(b) = b, \ N_2(b) = \sigma(b)b + \delta(b),\ldots$
For the definition of the maps $M_i$ we
assume that $\sigma$ is an automorphism of $S$, and define
$$ M_0(b) = 1, \ M_{i+1}(b) = b \sigma^{-1}(M_i(b)) - \delta(\sigma^{-1}(M_i(b))),$$
i.e., $M_1(b) = b$, \ $M_2(b) = b \sigma^{-1}(b) - \delta(\sigma^{-1}(b)), \ldots$
\cite{lam1988vandermonde}.

Let $f(t) = t^m - \sum_{i=0}^{m-1} a_i t^i \in S[t;\sigma,\delta]$. Then $(t-b) \vert_r f(t)$ is equivalent to
$N_m(b) - \sum_{i=0}^{m-1} a_i N_i (b) = 0$
which is proved analogously as \cite[Lemma 2.4]{lam1988vandermonde}.

If $\sigma$ is an automorphism of $S$, we can also view $R=S[t;\sigma,\delta]$ as a right
polynomial ring and write $f(t) = t^m -\sum_{i=0}^{m-1} a_i t^i \in R$ in the form $f(t) = t^m -
\sum_{i=0}^{m-1} t^i a_i'$ for some uniquely determined $a_i' \in S$.
 The remainder after dividing $f(t)$ on the left by $t-b$ is then given by  $M_m(b) - \sum_{i=0}^{m-1}
M_i(b) a_i'$ which is proved analogously as \cite[Proposition 49]{CP18}.

The following result is then again proven analogously to  \cite[Theorem 50]{CP18}:

\begin{theorem}  \label{prop:irreducibility criteria when S is a ring without zero divisors delta not 0}
 (i) $f(t) = t^2 - a_1 t - a_0 \in S[t;\sigma,\delta]$ is irreducible if and only if
 $\sigma(b)b + \delta(b) - a_1 b - a_0 \neq 0$  for all $b \in S$.
\\ (ii) Suppose $\sigma$ is an automorphism of $S$ and $f(t) = t^3 - a_2 t^2 - a_1 t - a_0 \in S[t;\sigma,\delta]$.
Write $f(t) = t^3 - t^2 a_2' - t a_1' - a_0'$ for some unique $a_0', a_1', a_2' \in S$. Then $f(t)$ is irreducible if and
only if
\begin{equation} \label{eqn:irredcriteriadelta 1}
N_3(b) - \sum_{i=0}^{2}a_i N_i(b) \neq 0,
\end{equation}
and
\begin{equation} \label{eqn:irredcriteriadelta 2}
M_3(b) - \sum_{i=0}^{2} M_i(b) a_i' \neq 0,
\end{equation}
for all $b \in S$.
\end{theorem}

\subsection{Right Ore domains} \label{subsec:2}

Recall that a domain $S$ is a \emph{right Ore domain} if $aS \cap bS\neq \{ 0 \}$ for all $0 \neq a, b \in S$.
The \emph{ring of right
fractions} of $S$ is a division ring $D$ containing $S$, such that every element of $D$ is of the form $rs^{-1}$ for some $r \in S$ and $0 \neq s \in S$.
Moreover, $\sigma$ and $\delta$ extend uniquely to $D$ via
\begin{equation} \label{eqn:extend sigma delta to right ring of fractions}
\sigma(rs^{-1}) = \sigma(r)\sigma(s)^{-1} \text{ and }
\delta(rs^{-1}) = \delta(r)s^{-1} - \sigma(rs^{-1}) \delta(s) s^{-1},
\end{equation}
for all $r \in S, \ 0 \neq s \in S$ by \cite[Lemma 1.3]{Goodearl}.
Note that any integral domain is a right Ore domain; its  ring of right fractions is equal to its quotient field.
In this subsection, we assume that $S$ is a right Ore domain with ring of right fractions $D$, $\sigma$ is
 an injective endomorphism of $S$ and $\delta$ a
$\sigma$-derivation of $S$. Let $C(D)$ denote the center of $D$.

 If $S$ is a right Ore domain,
we can take a skew polynomial $f\in S[t;\sigma,\delta]$ and check if it is irreducible over  $D[t;\sigma,\delta]$, in which case it
will be irreducible over $ S[t;\sigma,\delta]$ as well. We thus obtain the following results as elementary
corollaries of
the corresponding results  \cite[Theorems 39 and 51, Corollary 52]{CP18}:

\begin{theorem} \label{thm:Petit(19)}
Suppose $m$ is prime and $C(D) \cap \mathrm{Fix}(\sigma)$
contains a primitive $m$th root of unity.
\\ (i)  $f(t) = t^m - a \in S[t;\sigma]$ is irreducible if
$$a \neq \sigma^{m-1} (b) \cdots\sigma(b) b$$
 for all $b \in D$.
 \\ (ii) If $\mathrm{char}(D) \neq m$ then $f(t) = t^m - a \in
S[t;\sigma,\delta]$ is irreducible if $N_m(b) \neq a$ for
all $b \in D$.
 \\ (iii) Suppose $m=3$ and $\mathrm{char}(D) \neq 3$. Then $f(t) = t^3-
a \in S[t; \delta]$ is irreducible if $$N_3(b)=b^3 + 2
\delta(b)b + b\delta(b) + \delta^2(b) \neq a,$$ for all $b \in D$.
\end{theorem}

\section{Irreducibility criteria for some polynomials of degree two and three in $R=K[y][t;\sigma]$} \label{subsec:irred}

Let $R = K[y][t;\sigma]$ where $K$ is a field of characteristic not $2$, $y$ is an indeterminate, and
$\sigma$ an automorphism of the integral domain $K[y]$. We know that $\sigma|_{K} =id$ and $\sigma(y) = \alpha y + \beta$ for some $\alpha, \beta
\in K, \alpha \neq 0$.

This implies that
 $$\sigma^2(y)=\alpha^2y+(\alpha\beta+\beta),\dots,\sigma^l(y)=\alpha^ly+\sum_{i=0}^{l}\alpha^{l-1}\beta.$$
Let $z=z(y)=\sum_{i=0}^{n}z_iy^i\in K[y]$,  then
$$\sigma(z)=\sum_{i=0}^{n}z_i(\alpha x+\beta)^i=z_n \alpha^n y^n+\dots+\sum_{j=0}^{n}z_j \beta^j$$
 has constant term
$z(\beta)=\sum_{j=0}^{n}z_j \beta^j$ and
$$\sigma^2(z)=\sum_{i=0}^{n}a_i(a^2y+\alpha\beta+\beta)^i=a_n \alpha^{2n}
y^{n}+\dots+\sum_{j=0}^{n}a_j (\alpha \beta+\beta)^j$$ has constant term
$z(\alpha \beta+\beta)=\sum_{j=0}^{n}a_j (\alpha \beta+\beta)^j=\sum_{j=0}^{n}a_j (\alpha+1)^j\beta^j$. Continuing in this manner,
we see that
\begin{equation} \label{eqn:general}
\sigma^l(z)=\sum_{i=0}^{n}a_i(\alpha^lx+\sum_{j=0}^{l}\alpha^{l-1}\beta^j)^i=a_n
\alpha^{ln} y^{n}+\dots+ \sum_{j=0}^{n}a_j (\sum_{i=0}^{l}\alpha^{l-1}\beta)^j
\end{equation}
has constant term $z(\sum_{i=0}^{l}\alpha^{l-1}\beta)=\sum_{j=0}^{n}a_j
(\sum_{i=0}^{l}\alpha^{l-1}\beta)^j$ for  $l\geq1$ an integer. In particular,
$$\sigma(z)z=(a_n \alpha^n y^n+\dots+\sum_{j=0}^{n}a_j \beta^j)(a_n y^n+\dots
a_0)=a_n^2 \alpha^n y^{2n}+\dots+a_0z(\beta),$$
$$\sigma^2(z)\sigma(z)z=(a_n \alpha^{2n} x^{n}+\dots+\sum_{j=0}^{n}a_j (\alpha\beta+\beta)^j)(a_n^2 \alpha^n y^{2n}+\dots+a_0z(\beta))$$
$$=a_n^3 \alpha^{3n}
y^{3n}+\dots+a_0z(\beta)z(\alpha\beta+\beta),$$
and
$$\sigma^3(z)\sigma^2(z)\sigma(z)z=(z_n \alpha^{3n}y^{n}+\dots+z(\alpha^2\beta+\alpha\beta+\beta)) (z_n^3 \alpha^{3n}
y^{3n}+\dots+z_0z(\beta)z(\alpha \beta+\beta))$$
$$=z_n^4 \alpha^{6n} y^{4n}+\dots+z_0z(\beta)z(\alpha \beta+\beta)z
(\alpha^2\beta+\alpha \beta+\beta).$$
Note that if $\beta=0$ then the constant
term of $\sigma^l(z)$ simplifies to $a_0$ and thus the constant term
of $\sigma(z)z$ to $a_0^2$,  the constant term of
$\sigma^2(z)\sigma(z)z$ to $a_0^2$, and the one of $\sigma^3(z)\sigma^2(z)\sigma(z)z$ to $a_0^4$.

Applying Theorem \ref{thm:irreducibility criteria when S is a ring without zero divisors} this means for instance:

\begin{theorem}
Let $a=a(y)=\sum_{j=0}^sd_jy^j\in K[y]$ and $f(t)=t^2-a\in K[y][t;\sigma]$. Then $f(t)$ is irreducible in $K[y][t;\sigma]$
in the following cases:
\\ (i) $a(y)\in K[y]$ has odd degree,
\\ (ii) $a(y)\in K^\times\setminus K^{\times2}$,
\\ (iii) $\alpha=1$, $a(y)$ has even degree with leading coefficient not a square,
\\ (iv)  $\alpha\not=1$, $a(y)$ has even degree and leading coefficient
  not of the form $c^2 \alpha^s$ for some $c\in K^\times$, and any integer $s\geq0$,
\\ (v) $\beta=0$, $a(y)$ has even degree and $d_0\not\in K^{\times 2}$.
\end{theorem}

\begin{proof}
 $f(t)=t^2-a$ is irreducible in $K[y][t;\sigma]$ if and only if $\sigma(z)z\not=a$ for all $z\in K[y]$.
Now $\sigma(z)z=z_n^2 \alpha^n y^{2n}+\dots+z_0z(\beta)$ is either constant or a polynomial of even degree.
Thus if $a=a(y)\in K[y]$ has odd degree,
we know that $\sigma(z)z\not=a$ for all $z\in K[y]$ which shows (i).
\\
If $a\in K^\times$, we know that $\sigma(z)z=a$ is only possible for $z=a_0\in
K$ in which case we have $\sigma(z)z=z_0^2$. Thus if $a$ is not a
square in $K$, $\sigma(z)z\not=a$ for all $z\in K[y]$ which proves (ii).
\\ Suppose next
that $a=\sum_{j=0}^sd_jy^j$ has even degree. Then $\sigma(z)z=a$ for
some $z\in K[y]$ is equivalent to $z_n^2 \alpha^n y^{2n}+\dots+z_0z(\beta)=a$
for some $n$, $z_i\in K$, $z_n\not=0$. This means $d_s=z_n^2 \alpha^n$ and $d_0=z_0z(\beta)$.

If $d=1$ then this implies that for all $a$ of even degree with leading coefficient not a square, $f(t)$ is irreducible,
proving (iii).
\\ If $d\not=1$ then this implies that for all $a$ of even degree $s=2n$ with leading coefficient
  not of the form $c^2 \alpha^n$ for some $c\in K^\times$, $f(t)$ is irreducible, which shows (iv).

Moreover, since $\sigma(z)z=d$ for some $z\in K[y]$ also means $d_0=a_0z(\beta)$, we know that if
$b=0$ and $d_0\not\in K^{\times 2}$ then $f(t)$ is irreducible. This is (v).
\end{proof}

\begin{theorem}
Let $a=a(y)=\sum_{j=0}^sd_jy^j\in K[y]$ and $f(t)=t^3-a\in K[y][t;\sigma]$. Then $f(t)$ is irreducible in $K[y][t;\sigma]$ in the
following cases:
\\ (i) $a(y)\in K[y]$ has degree not divisible by 3,
\\ (ii) $a(y)\in K^\times\setminus K^{\times 3}$,
\\ (iii) $\alpha=1$, $a(y)$ has degree divisible by 3 with leading coefficient not a cube,
\\ (iv)  $\alpha\not=1$, $a(y)$ has degree divisible by 3 and leading coefficient
  not of the form $c^3 \alpha^s$ for some $c\in K^\times$, and any integer $s\geq0$,
\\ (v)  $\beta=0$, $a(y)$ has degree divisible by 3 and $d_0\not\in K^{\times 3}$.
\end{theorem}

\begin{proof}
 $f(t)=t^3-a$ is irreducible in $K[y][t;\sigma]$ if and only if
$a\not=\sigma^2(z)\sigma(z)z$ for all $z\in K[y]$ by Theorem
\ref{thm:irreducibility criteria when S is a ring without zero divisors} (iii). Now
$\sigma^2(z)\sigma(z)z=z_n^3 \alpha^{3n} y^{3n}+\dots+z_0z(\beta)z(\alpha\beta+\beta)$ is either constant or a polynomial of degree divisible by 3.
 Thus if $a=a(y)\in K[y]$ has degree not divisible by 3, we know that
$\sigma^2(z)\sigma(z)z\not=a$ for all $z\in K[y]$ proving (i).
\\ If $a\in K^\times$, we know that $\sigma^2(z)\sigma(z)z=a$
is only possible for $z=z_0\in K$ in which case we have $\sigma(z)z=z_0^3$. Thus if $a\in K^\times\setminus K^{\times 3}$,
$\sigma^2(z)\sigma(z)z\not=a$
for all $z\in K[y]$ which shows (ii).
\\ Suppose next that $a=\sum_{j=0}^sd_jy^j$ has
degree divisible by 3. Then $\sigma^2(z)\sigma(z)z=a$ for some $z\in
K[y]$ is equivalent to  $z_n^3 \alpha^{3n} y^{3n}+\dots+z_0z(\beta)z(\alpha\beta+\beta)=a$
 for some $n$, $z_i\in K$, $z_n\not=0$. This means $d_s=z_n^3 \alpha^{3n}$.

If $\alpha=1$ then this implies that for all $a$ of degree divisible by 3 with leading coefficient not a cube, $f(t)$
is irreducible  and we have proved (iii).
\\
 If $\alpha\not=1$ then this implies that for all $a$ of degree $s=3n$ with leading coefficient
not of the form $c^3 \alpha^n$ for some $c\in K^\times$, $f(t)$ is irreducible and we got (iv).
\\ Moreover, since $\sigma^2(z)\sigma(z)z=a$ for some $z\in K[y]$ also means $d_0=z_0z(\beta)z(\alpha\beta+\beta)$, we know
that if $\beta=0$ and $d_0\not\in K^{\times 3}$ then $f(t)$ is irreducible. This shows (v).
\end{proof}

If desired, similar results can be obtained for higher degree skew polynomials, the calculations become more tedious but follow the same
pattern.

\section{Irreducible polynomials of low degree in a quantum plane, a quantized Weyl algebra, and in $A_h$}
\label{sec:Weyletc}

Let $R = K[y][t;\sigma,\delta]$ where $K$ is a field of characteristic not $2$, $y$ is an indeterminate,
$\sigma$ an automorphism of the domain $K[y]$, i.e. $\sigma|_K=id_K$, $\sigma(y) = \alpha y + \beta$ for some $\alpha, \beta
\in K, \alpha \neq 0$, and $\delta$  a left
$\sigma$-derivation. We know that $R$ is isomorphic to either a quantum plane, a quantized Weyl algebra, or  the infinite-dimensional
unital associative algebra $A_h=K[y][t;{\rm id}_{K[y]},\delta]$ studied in  \cite{BI, BII, BIII}, where
$\delta: K[y] \rightarrow K[y]$ is the $K$-linear
derivation $\delta(r) = r'h$ for some $h \in K[y]$ and $r'$ denotes
the usual derivative of $r$ with respect to $y$.
Given $a(y) \in K[y]$, we denote  the degree of $a(y)$ as a polynomial in $y$ by $\mathrm{deg}_y(a(y))$.

\subsection{Irreducible polynomials of low degree and $f(t)=t^m-a$ in the quantum plane}

Let
 $\sigma$ be the automorphism of $K[y]$ such that $\sigma = id_K$ on
$K$ and $\sigma(y) = qy$ for some $1\not=q \in K^{\times}$.  Then $K[y][t;
\sigma]$ is called the \emph{quantum plane}.

\begin{lemma} \label{quantum plane sigma(b(y))=b(qy)}
$\sigma^j(b(y)) = b(q^j y)$ for all $j \in \mathbb{N}$ and all $b(y)\in K[y]$.
\end{lemma}

\begin{proof}
If $b(y) = b_0 + b_1y + \ldots + b_l y^l \in K[y]$, then
\begin{align*}
\sigma^j(b(y)) &= \sigma^j(b_0) + \sigma^j(b_1 y) + \ldots +
\sigma^j(b_l y^l) = b_0 + b_1 \sigma^j(y) + \ldots + b_l
\sigma^j(y^l) \\ &= b_0 + b_1 q^j y + \ldots + b_l (q^j y)^l = b(q^j
y)
\end{align*}
as in Equation \eqref{eqn:general}.
\end{proof}

Lemma \ref{quantum plane sigma(b(y))=b(qy)} and Theorem
\ref{thm:irreducibility criteria when S is a ring without zero
divisors} immediately yield:

\begin{proposition} \label{Irreducibility_criteria_quantum_plane}
 (i) $f(t) = t^2 - a_1(y) t - a_0(y) \in K[y][t;\sigma]$ is irreducible if and only if
$$b(qy)b(y) - a_1(y)b(y) - a_0(y) \neq 0$$ for all $b(y) \in K[y]$.
\\ (ii) $f(t) = t^3 - a_2(y) t^2 - a_1(y) t - a_0(y) \in K[y][t;\sigma]$ is irreducible if and only if
$$b(q^2y)b(qy)b(y) - b(q^2 y)b(qy)a_2(y) - b(q^2y) a_1(qy) -
a_0(q^2y) \neq 0$$ and $$b(q^2y)b(qy)b(y) - b(qy)b(y) a_2(y) - b(y)
a_1(y) - a_0(y) \neq 0$$ for all $b(y) \in K[y]$.
\end{proposition}

\begin{proposition}  \label{prop:I}
Let $f(t) = t^2 - a_1(y)t - a_0(y) \in K[y][t;\sigma]$ and let $a_{i,0}$ denote the constant terms of
$a_i(y)$, $0\leq i\leq 1$.
\\ (i) If  ${\rm deg}(d(y)) > {\rm
deg}(a(y))$, then $f(t)$ is irreducible.
\\ (ii) If $ t^2-a_{1,0}t-a_{0,0}\in K[t]$ is irreducible, then $f(t)$ is irreducible.
\end{proposition}

\begin{proof}
$f(t)$ is irreducible if and only if $$b(qy)b(y) - d(y)b(y) \neq
a(y)$$ for all $b(y) \in K[y]$ by Proposition
\ref{Irreducibility_criteria_quantum_plane}.
\\ (i) Suppose that
$$a(y) = b(qy)b(y) - d(y)b(y) = (b(qy) - d(y)) b(y)$$
for some $b(y) \in K[y]$ and notice that $b(y) \neq 0$ since $a(y)
\neq 0$. Then $${\rm deg}(b(qy) - d(y)), \ {\rm deg}(b(y)) =
{\rm deg}(b(qy)) \leq {\rm deg}(a(y)).$$ Hence ${\rm deg}(d(y)) \leq {\rm deg}(a(y))$.
\\ (ii) A look at the constant terms of the above equation yields the assertion.
\end{proof}

\begin{proposition} \label{prop:II}
Let  $f(t)=t^3 - a_2(y) t^2 - a_1(y) t - a_0(y) \in K[y][t;\sigma]$ and let $a_{i,0}$ denote the constant terms of
$a_i(y)$, $0\leq i\leq 2$.
 If $$ t^3-a_{2,0}t^2-a_{1,0}t-a_{0,0}\in K[t]$$
  is irreducible, then
 $f(t) = t^3 - a_2(y) t^2 - a_1(y) t - a_0(y) \in K[y][t;\sigma]$
  is irreducible.
\end{proposition}

\begin{proof}
 Comparing constants in the two equations of Proposition \ref{Irreducibility_criteria_quantum_plane} (ii)
immediately yields the assertion.
\end{proof}

Note that $K[t] \subset K[y][t;\sigma]$
and so if $f(t) \in K[t]$ is irreducible in $K[y][t;\sigma]$ then $f(t)$ is irreducible in $K[t]$.
Thus Propositions \ref{prop:I} and \ref{prop:II} yield:

\begin{corollary} \label{thm:irredquantum}
Let $f(t) \in K[t] \subset K[y][t;\sigma]$.
\\ (i) $f(t) = t^2 - a_1 t - a_0$ is irreducible in $K[y][t;\sigma]$ if and only if it is irreducible in $K[t]$
 if and only if $a_1^2 + 4 a_0$ is not a square in $K$.
\\ (ii) $f(t) = t^3 - a_2 t^2 - a_1 t - a_0$ is irreducible in $K[y][t;\sigma]$ if and
 it is irreducible in $K[t]$.
\end{corollary}

The following result yields a large class of irreducible polynomials of degree $m$:

\begin{theorem}
Let $f(t) = t^m - a(y) \in K[y][t;\sigma]$, $ a(y)=\sum_{j=0}^sa_jy^j \not=0,$ where $m$ is prime and $K$
contains a primitive $m$th root of unity.
\\ (i) If $m \nmid {\rm deg} (a(y))$, then $f(t)$ is irreducible over $K[y][t;\sigma]$.
\\ (ii) If $a_0\not\in K^\times \setminus K^{\times m},$  then $f(t)$ is irreducible over $K[y][t;\sigma]$.
\\ (iii) If $a_s\not\in K^{\times m}q^e$ for any $e\geq 0,$ then $f(t)$ is irreducible over $K[y][t;\sigma]$.
\end{theorem}

\begin{proof}
 Extend $\sigma$
to an automorphism of the field of fractions $K(y)$ of $K[y]$ as in
\eqref{eqn:extend sigma delta to right ring of fractions}, i.e.
$$\sigma(\frac{c}{d})=\frac{\sigma(c)}{\sigma(d)}$$
for all $c,d\in K[y]$, $d\not=0$.
By Theorem \ref{thm:Petit(19)}, $f(t)$ is irreducible  over
$K(y)[t;\sigma]$ and hence over $K[y][t;\sigma]$, if
$$N_m(b(y)) = \sigma^{m-1}(b(y)) \cdots \sigma(b(y)) b(y) \neq a(y)$$
for all $b(y) \in K(y)$. Write $b(y) =
c(y)/d(y)$ for some $c(y), d(y) \in K[y]$ with $d(y) \neq 0$, then
\begin{align*}
N_m(b(y)) &= \frac{c(q^{m-1}y) \cdots c(qy) c(y)}{d(q^{m-1}y) \cdots
d(qy) d(y)}
\end{align*}
by Lemma \ref{quantum plane sigma(b(y))=b(qy)}. If $N_m(b(y)) \notin
K[y]$, we immediately conclude $N_m(b(y)) \neq a(y)$ because $a(y)
\in K[y]$. So suppose that $N_m(b(y)) \in K[y]$  and  $N_m(b(y)) = a(y)$ for some $b(y) =
c(y)/d(y)$ with $c(y)=\sum_{j=0}^nc_jy^j \not=0$, $d(y)=\sum_{j=0}^ld_jy^j \not=0$.
\\ (i) If $N_m(b(y)) \in K[y]$, then
$$\mathrm{deg}(N_m(b(y))) = m \mathrm{deg}(c(y)) - m
\mathrm{deg}(d(y))$$ for all $0 \neq b(y) \in K(y)$. Therefore
$\mathrm{deg}(N_m(b(y)))$ is a multiple of $m$, thus $N_m(b(y))
\neq a(y)$ for all $b(y) \in K(y)$, and so $f(t)$ is irreducible.
\\ (ii) Comparing constants in the equation $N_m(b(y)) = a(y)$ yields
$c_0^m=a_0d_0^m$, hence $a_0\in K^\times \setminus K^{\times m}.$ Thus if $a_0\not\in K^\times \setminus K^{\times m},$
$f(t)$ is irreducible.
\\ (iii) Comparing highest terms in the equation $N_m(b(y)) = a(y)$ implies that $a_s\in  K^{\times m}q^e$
for some $e\geq 0$.
\end{proof}

\subsection{Irreducible polynomials of degree two in the quantized Weyl algebra}

Let $K$ be a field of characteristic not 2, $y$ be an indeterminate
and $\sigma$ be the automorphism of $K[y]$ such that $\sigma = id$ on
$K$ and $\sigma(y) = qy$ for $q \in K^{\times}, q \neq 1$. Define  $$\delta(g) = \frac{g(qy)
- g(y)}{qy - y}$$ for all $g \in K[y]$. The algebra $K[y][t; \sigma,
\delta]$ is called a \emph{quantized Weyl algebra}.

\begin{proposition} \label{Irreducibility_criteria_quantized_Weyl_Algebra}
 $f(t) = t^2 - a_1(y) t - a_0(y) \in K[y][t;\sigma,\delta]$ is
irreducible if and only if $$b(qy)b(y) + \frac{b(qy) - b(y)}{qy - y}
- a_1(y) b(y) - a_0(y) \neq 0$$ for all $b(y) \in K[y]$.
\end{proposition}

\begin{proof}
 By Lemma \ref{quantum plane
sigma(b(y))=b(qy)}, we have $\sigma(b(y)) = b(qy)$. Theorem \ref{prop:irreducibility criteria
when S is a ring without zero divisors delta not 0} then yields the
result.
\end{proof}

\begin{corollary} \label{thm:irredWeyl}
Let $f(t) = t^2 - a_1(y)t - a_0(y) \in K[y][t;\sigma,\delta]$ where
$a_0(y) \neq 0$.
\\ (i)
 Suppose $f(t) \in K[t]$. Then $f(t)$ is irreducible in $K[y][t;\sigma,\delta]$ if and only if $a_1^2 + 4 a_0$ is
 not a square in $K$ if and only if $f(t)$ is irreducible in $K[t]$.
\\ (ii) Suppose $a_0(y), a_1(y)$ are such that $2 {\rm deg}(a_1(y)) < {\rm deg}(a_0(y))$ and
${\rm deg}(a_0(y))$ is odd. Then $f(t)$ is irreducible in $K[y][t;\sigma,\delta]$. In particular, if $a_1 \in K$ and
${\rm deg}(a_0(y))$ is odd then $f(t)$ is irreducible.
\end{corollary}

\begin{proof}
By Proposition
\ref{Irreducibility_criteria_quantized_Weyl_Algebra}, $f(t)$ is
irreducible in $K[y][t;\sigma,\delta]$ if and only if $$b(qy)b(y) +
\frac{b(qy) - b(y)}{qy - y} - a_1(y)b(y) \neq a_0(y)$$ for all $b(y)
\in K[y]$, if and only if
\begin{equation} \label{eqn:quantised weyl degree1}
b(qy)b(y)(qy - y) + b(qy) - b(y) - a_1(y)b(y)
\neq a_0(y) (qy - y)
\end{equation}
for all $b(y) \in K[y]$.

Note that \eqref{eqn:quantised weyl degree1} is trivially
satisfied if $b(y) = 0$. If $b(y) = b \in K^{\times}$ then \eqref{eqn:quantised
weyl degree1} simplifies to
\begin{equation} \label{eqn:quantised weyl degree2}
b^2 - a_1(y)b \neq a_0(y).
\end{equation}
 (i) Suppose $a_0, a_1 \in K$. If $l={\rm deg}(b)  > 0$ then
$${\rm deg} \Big( b(qy)b(y)(qy - y) + b(qy) - b(y) -
a_1(y)b(y)(qy - y) \Big) = (2l + 1) > 1$$ and so
$$b(qy)b(y)(qy - y) + b(qy) - b(y) - a_1b(y)(qy - y)\neq a_0 (qy - y)$$
for all $b(y) \in K[y]$. Therefore $f(t)$ is irreducible in
$K[y][t;\sigma,\delta]$ if and only if \eqref{eqn:quantised weyl
degree2} is satisfied for all $b \in K$,  which is equivalent to $f(t)$ being irreducible in
$K[t]$, and this holds in turn if and only if $a_1^2 + 4
a_0$ is not a square in $K$.
\\ (ii) Suppose now that $2 {\rm deg}(a_1(y)) < {\rm deg}(a_0(y))$ and ${\rm deg}(a_0(y))$ is odd. Then for all
$b \in K^{\times}$ we have
$${\rm deg}(b^2 - a_1(y)b) = {\rm deg}(a_1(y)) < {\rm deg}(a_0(y))$$ unless $b = a_1(y) \in K$ in which case $b^2 -
a_1(y)b \in K$. In either case \eqref{eqn:quantised weyl degree2} is
satisfied.

Put $l={\rm deg}(b(y))  > 0$ then we have to consider 3 cases:
\\ If $l = {\rm deg}(a_1(y))$ then
$${\rm deg}(b(qy)b(y)(qy-y)) = 2l+1 = {\rm deg}(a_1(y)b(y)(qy-y))$$ which implies
\begin{equation*}
\begin{split}
{\rm deg} \Big( & \frac{b(qy)b(y)(qy - y) + b(qy) - b(y) -
a_1(y)b(y)(qy - y)}{qy - y} \Big) \\ & \qquad \leq (2l + 1)-1 = 2
{\rm deg}(a_1(y)) < {\rm deg}(a_0(y)).
\end{split}
\end{equation*}
\\ If $l < {\rm deg}(a_1(y))$ then
$${\rm deg}(a_1(y)b(y)(qy-y)) > {\rm deg}(b(qy)b(y)(qy-y)), \
{\rm deg}(b(qy)), \ {\rm deg}(b(y))$$ which implies
\begin{equation*}
\begin{split}
{\rm deg} \Big( & \frac{b(qy)b(y)(qy - y) + b(qy) - b(y) -
a_1(y)b(y)(qy - y)}{qy - y} \Big) \\ &= {\rm deg}(a_1(y)) + l + 1
-1 \\&=  {\rm deg}(a_1(y)) + l < 2 {\rm deg}(a_1(y)) < {\rm deg}(a_0(y)).
\end{split}
\end{equation*}
\\ Finally, if $l > {\rm deg}(a_1(y))$ then
$${\rm deg}(b(qy)b(y)(qy-y)) > {\rm deg}(a_1(y)b(y)(qy-y)), \
{\rm deg}(b(qy)), \ {\rm deg}(b(y))$$ which implies
\begin{equation*}
\begin{split}
{\rm deg} \Big( & \frac{b(qy)b(y)(qy - y) + b(qy) - b(y) -
a_1(y)b(y)(qy - y)}{qy - y} \Big) = (2l + 1)-1
\end{split}
\end{equation*}
is even.

In all cases we have $$b(qy)b(y)(qy - y) + b(qy) - b(y)-a_1(y)b(y)(qy - y)\neq a_0(y)(qy - y),$$
therefore $f(t)$ is irreducible.
\end{proof}

\subsection{Irreducible polynomials of degree two in $A_h$}

Recall that $A_h=K[y][t; \delta]$ with $\delta(r) = r'h$ for some $h(y)
\in K[y]$, where $r'$ is the usual derivation of $r$ with respect to $y$.
These algebras were comprehensively studied in \cite{BI, BII, BIII}. $A_h$ is simple if and only if $F$ has characteristic 0
and $h\in F^\times$ \cite[Corollary 7.5]{BI}.
 If $F$ has characteristic 0 then $A_h$ is a unique factorization domain \cite[Lemma 7.6]{BI}.

The irreducibility of a polynomial in $A_h$ clearly depends on the choice of $h$:

\begin{proposition} \label{Irreducibility_criteria_Ah_degree_2}
 $f(t) = t^2 - a_1 (y)t - a_0(y) \in K[y][t; \delta]$ is
irreducible if and only if
\begin{equation*}
b(y)^2 + b'(y)h(y) - a_1(y) b(y) - a_0(y) \neq 0
\end{equation*}
for all $b(y) \in K[y]$.
\end{proposition}

\begin{proof}
We have $$\sigma(b(y))b(y) + \delta(b(y)) - a_1(y)b(y) - a_0(y) =
b(y)^2 + b'(y)h(y) - a_1(y) b(y) - a_0(y),$$ hence the result follows
from Theorem \ref{prop:irreducibility criteria when S is a ring
without zero divisors delta not 0}.
\end{proof}

\begin{corollary} \label{A_h degree irreducible t^2-a_1(y)t-a_0(y)}
Let $f(t) = t^2 - a_1(y)t - a_0(y) \in K[y][t;\delta]$ and let $a_{i,0}$ denote the constant terms of
$a_i(y)$, $0\leq i\leq 1$.
\\ (i) If
$${\rm deg}(a_0(y)) > 2 {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y))-1 \}$$ and ${\rm deg}(a_0(y))$ is odd, then $f(t)$ is
irreducible in $K[y][t;\delta]$.
\\ (ii) If $h(y)$ has zero constant term and $g(t)=t^2-a_{1,0}t-a_{0,0}$ is irreducible in $K[t]$, then
 then $f(t)$ is irreducible in $K[y][t;\delta]$.
\end{corollary}

\begin{proof}
$f(t)$ is irreducible in $K[y][t;\delta]$ if and only if
\begin{equation} \label{eqn:A_h degree irreducible t^2-a_1(y)t-a_0(y)}
b(y)^2 + b'(y)h(y) - a_1(y)b(y) \neq a_0(y)
\end{equation}
for all $b(y) \in K[y]$ by Proposition
\ref{Irreducibility_criteria_Ah_degree_2}.
\\ (i) If $b(y) = 0$
then \eqref{eqn:A_h degree irreducible t^2-a_1(y)t-a_0(y)} is
satisfied since $a_0(y) \neq 0$. If $b(y) \in K^{\times}$ then
$b'(y)h(y)=0$ which implies $${\rm deg} \big( b(y)^2 + b'(y)h(y) -
a_1(y)b(y) \big) = {\rm deg}(a_1(y)) < {\rm deg}(a_0(y)),$$
unless $a_1(y) \in K$ in which case $b(y)^2 + b'(y)h(y) - a_1(y)b(y)
\in K$. In either case \eqref{eqn:A_h degree irreducible
t^2-a_1(y)t-a_0(y)} is satisfied.

Now suppose $l={\rm deg}(b(y)) > 0$. We  consider the following two cases:
\\ If $l > {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y))-1 \}$ then
\begin{equation*}
\begin{split}
{\rm deg} \big( b(y)^2 + & b'(y)h(y) - a_1(y)b(y) \big) = {\rm deg}(b(y)^2) = 2l \\ &> 2 {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y))-1 \}
\end{split}
\end{equation*}
and is even.
\\ If $l \leq {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y))-1 \}$ then
\begin{align*}
{\rm deg} \big( b(y)^2 + b'(y)h(y) - a_1(y)b(y) \big) & \leq {\rm
max} \{ {\rm deg}(a_1(y)) + l, {\rm deg}(h(y)) - 1 + l \} \\ &= l
+ {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y)) - 1 \} \\ &
\leq 2 {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y)) - 1 \}.
\end{align*}
Therefore if ${\rm deg}(a_0(y)) > 2 {\rm max} \{ {\rm deg}(a_1(y)), {\rm deg}(h(y))-1 \}$ and ${\rm deg}(a_0(y))$ is
odd then \eqref{eqn:A_h degree irreducible t^2-a_1(y)t-a_0(y)} is
satisfied which implies $f(t)$ is irreducible.
\\ (ii) Suppose there exists some $b(y)$ such that $b(y)^2 + b'(y)h(y) - a_1(y)b(y) = a_0(y)$. Let $b(y)=\sum_{i=0}^sb_iy^i$.
Looking at the constant terms the equation then yields $b_0^2 - a_{1,0}b_0 = a_{0,0}$.
Thus if  $g(t)=t^2-a_{1,0}t-a_{0,0}$ is irreducible in $K[t]$ there is no such $b(y)$ and the assertion follows.
\end{proof}

\begin{corollary} \label{A_h t^2-a degree irreducible}
Let $f(t) = t^2 - a(y) \in K[y][t;\delta]$ and $h(y) \in K[y]$.
\\ (i) Suppose ${\rm deg} (h(y))\in\{0,1\}$. If ${\rm deg}(a(y))$ is odd then $f(t)$ is irreducible.
\\ (ii) Suppose ${\rm deg}(h(y)) = n \geq 2$ and ${\rm deg}(a(y)) \geq 2n - 2$ is odd. Then $f(t)$ is irreducible.
\end{corollary}

\begin{proof}
Let $b(y) \in K[y]$. If $b(y) \in K$ then $b(y)^2 + b'(y)h(y) =
b(y)^2 \in K$ so is not equal to $a(y)$. Now suppose ${\rm deg}(b(y)) \geq 1$.
\\ (i) We see that
 $${\rm deg}(b(y)^2 + b'(y)h(y)) = {\rm deg}(b(y)^2) = 2 {\rm deg}(b(y))$$
is even. Thus if ${\rm deg}(a(y))$ is odd, $b(y)^2 + b'(y)h(y) \neq
a(y)$ and so $f(t)$ is irreducible by Proposition
\ref{Irreducibility_criteria_Ah_degree_2} implying (i).
\\ (ii)
We have to consider the following cases:
\\ If ${\rm deg}(b(y)) < n - 1$ then
$ \label{eqn:A_h t^2-a degree irreducible1}
{\rm deg}(b(y)^2 + b'(y)h(y)) = {\rm deg}(b(y)) - 1 + n < 2n - 2.
$
\\ If ${\rm deg}(b(y)) > n - 1$ then
${\rm deg}(b(y)^2 + b'(y)h(y)) = 2 {\rm deg}(b(y)) > 2n - 2$
is even.
\\ If ${\rm deg}(b(y)) = n - 1$ then ${\rm deg}(b(y)^2) = {\rm deg}(b'(y)h(y))$ and so
${\rm deg}(b(y)^2 + b'(y)h(y)) \leq 2n-2.$
(This may be a $<$ if the leading coefficient of $b(y)^2$ and
$-b'(y)h(y)$ are equal.)
\\
Therefore, if ${\rm deg}(a(y)) > 2n - 2$ is odd then $b(y)^2 + b'(y)h(y) \neq a(y)$ which implies $f(t)$ is irreducible by
Proposition \ref{Irreducibility_criteria_Ah_degree_2}.
\end{proof}

config: arxiv


TITLE: How to know when can I use proof by contradiction to prove operations with irrational/rational numbers?
QUESTION [0 upvotes]: I know that the proof by contradiction is useful when we know that there is only one answer and we can pick from two exclusive options. Therefore, if one doesn't work, we know the other is the correct answer. It seems to me that to use the proof I require, beforehand, the knowledge that the result can be only one option or the other.
When proving, that the sum of an irrational number and a rational number is irrational, we can use proof by contradiction to demonstrate that it can't be rational. Doesn't this assume that the sum can't be sometimes rational and other times irrational?
However, with the sum of two irrational numbers, the result can be either rational or irrational based on the chosen numbers. How can I deduce, beforehand, that the proof by contradiction won't be useful in this case?

REPLY [0 votes]: When we prove that the sum of a rational and an irrational number cannot be rational, we are certainly not assuming that the sum cannot be rational in some cases and irrational in others: we are proving that this is the case. This really has nothing to do specifically with proofs by contradiction; it just happens that in this case we prove that something is true by proving that its negation is false. And in fact it’s entirely possible to prove the result without using proof by contradiction.

Proposition: If $x,y\in\Bbb R$, and $x$ is rational, then $x+y$ is rational if and only if $y$ is rational.

Proof: If $y$ is rational, then $x+y$ is the sum of two rational numbers and is therefore rational. If $x+y$ is rational, then $y=(x+y)+(-x)$ is the sum of two rational numbers, so $y$ is rational. $\dashv$

Knowing when a proof by contradiction or a proof of the contrapositive is likely to be a good way to proceed is in large part a matter of experience, but there is at least one moderately useful guide: if the hypotheses available for a proof of the contrapositive or a proof by contradiction seem to give you more to work with — more specific details, for instance — than the hypotheses available for a direct proof, then it’s definitely worth your while to consider an indirect proof.


TITLE: Prove that $\sum_{n=1}^\infty\frac{\mu(n)}{n}H_n\sum_{k=n+1}^\infty\frac{\mu(k)}{k^2}$ is convergent
QUESTION [2 upvotes]: For integers $n\geq 1$ let $$H_n=1+\frac{1}{2}+\ldots+\frac{1}{n}$$ the $nth$ harmonic number, and $\mu(n)$ the Möbius function. See, if you need it, this Wikipedia to know the definition of Möbius function.
After I've read a problem due to Furdui from [1], I would like to ask about the convergence of a similar series:

Question. What is a good way to deduce if the series $$\sum_{n=1}^\infty\frac{\mu(n)}{n}H_n\sum_{k=n+1}^\infty\frac{\mu(k)}{k^2}$$ does converge? Thanks in advance.

References:
[1] Furdui, PROBLEMA 94 from La Gaceta de la Real Sociedad Matemática Española Vol. 11, No. 4 (2008).

REPLY [2 votes]: With a very crude estimation
$$\left|\sum_{k\geq n+1}\frac{\mu(k)}{k^2}\right|\leq \sum_{k\geq n+1}\frac{1}{k^2}\leq \int_{n}^{+\infty}\frac{dx}{x^2}=\frac{1}{n} $$
since $\left|\mu(k)\right|\leq 1$. Since
$$\sum_{n\geq 1}\frac{H_n}{n^2}=2\,\zeta(3) $$ is a convergent series, the given series is convergent too.


TITLE: Why doesn't a Gaussian beam converge to a point?
QUESTION [1 upvotes]: No matter what lens is put in the beam path of a Gaussian beam, it will always go through a waist of non-zero width.
Why not just a point?
I know the maths, I'm wondering whether there is any physics that prevents it.

REPLY [1 votes]: While I concur that you may use the uncertainty principle to understand this, it isn't necessary.  If you have a classical EM field that's governed by the nice wave equation derived from Maxwell's equations, then you can compute a diffraction integral that tells you that you must have a finite waist, even if the far-field divergence is very large.


TITLE: Decomposing adelic points using torsors
QUESTION [4 upvotes]: Let $k$ be a number field and $X$ be a $k$-scheme. Let $G$ be a linear algebraic group over $k$ and let $f: Z \to X$ be a $G_X$-torsor ($G_X = G \times_k X)$. We can twist the torsor $f$ by 1-cocycles in $Z^1(k,G)$ (see Skorobogatov's "Torsors and rational points" for more details), and we denote the twist of $f$ by $[\tau] \in H^1(k,G)$ by $f^\tau : Z^\tau \to X$, which is a $G_X^\tau$-torsor. Now to my question.

Is it true that $X(\mathbb{A}_k) = \bigcup_{[\tau] \in H^1(k,G)} f^\tau(Z^\tau(\mathbb{A}_k))$?

I remark the level of generality of $X$ in the above.

REPLY [11 votes]: This question for connected $G$ is exactly asking for a "strong approximation" property for the construction of torsors, and that generally fails for both tori and connected semisimple groups beyond the simply connected case (due to various phenomena explained by class field theory), and much deeper is the affirmative answer when $G$ is a connected semisimple group that is simply connected.
Michael Stoll's counterexample for $G = \mu_2$ over $K = \mathbf{Q}$ generalizes to give a negative answer in the disconnected case over every number field for $G = \mu_n$ and $f$ the $n$th-power endomorphism of $\mathbf{G}_m$ with any $n > 1$, again explained by class field theory.
To see what is going on, we first translate the problem into more hands-on terms in a more general setting without connectedness conditions.
Let $K$ be a global field and $G$ a smooth $K$-group of finite type. (I want to incorporate the case of abelian varieties into the preliminary discussion.)  The question for $G$ concerns the pullback of $f$ along $K$-morphisms $x: {\rm{Spec}}(\mathbf{A}_K) \rightarrow X$, which is to say that one is given a $G$-torsor $E$ over ${\rm{Spec}}(\mathbf{A}_K)$ and is asking if it is split by twisting along a $G$-torsor over $K$. Since you emphasize that you allow $X$ to be an arbitrary $K$-scheme (so not even assuming it to be of finite type over $K$), what you are asking is equivalent to the question of whether or not ${\rm{H}}^1(K,G) \rightarrow {\rm{H}}^1(\mathbf{A}_K,G)$ is surjective, where we are working with the etale topology (since $G$ is $K$-smooth).
The adelic H$^1$ enforces an "almost everywhere integral" condition that is not present when considering $\prod_v {\rm{H}}^1(K_v,G)$. To explain this in precise terms (and then to address the surjectivity question), first observe that by "spreading out" of the connected-etale sequence for $G$, we may choose a finite set $S_0$ of places of $K$ large enough so that $G$ extends to a smooth separated $O_{K,S_0}$-group $\mathscr{G}$ of finite type (even affine if $G$ is affine) which fits into a short exact sequence
$$1 \rightarrow \mathscr{G}^0 \rightarrow \mathscr{G} \rightarrow \Gamma \rightarrow 1$$
where $\mathscr{G}^0$ is a smooth closed $O_{K,S_0}$-subgroup with connected fibers and $\Gamma$ is a finite etale $O_{K,S_0}$-group. Since $\mathbf{A}_K$ is the direct limit of its subrings $\mathbf{A}_{K,S} = (\prod_{v \in S} K_v) \times \prod_{v \not\in S} O_{K,v}$ as $S$ varies through the finite sets of places of $K$ containing $S_0$, by standard limit arguments we have ${\rm{H}}^1(\mathbf{A}_K,G) = \varinjlim {\rm{H}}^1(\mathbf{A}_{K,S},\mathscr{G})$. But $\mathbf{A}_{K,S}$ is a direct product of local henselian rings, so a cofinal system of etale covers of its spectrum is given by precisely the spectra of the product rings $R = \prod_v R_v$ where $R_v$ is a local finite etale algebra over $K_v$ for $v \in S$ and over $O_{K,v}$ for $v\not\in S$ with these local algebras of bounded degree  as $v$ varies (see Lemma 7.5.5 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148).  This provides a natural injection
$${\rm{H}}^1(\mathbf{A}_{K,S},\mathscr{G}) = \varinjlim_R {\rm{H}}^1(R/\mathbf{A}^1_{K,S}, \mathscr{G}) \rightarrow \prod_{v \in S} {\rm{H}}^1(K_v,G) \times \prod_{v \not\in S} {\rm{H}}^1(O_{K,v},\mathscr{G})$$
whose image consists of precisely the collections $(E_v)_v$ of classes $E_v$ split by a finite etale cover of bounded degree over the local $v$-factor ring for all $v$.
This uniform boundedness condition is automatic (so the degree-1 ${\mathbf{A}}_{K,S}$-cohomology set is identified with the indicated direct product).  To prove this, first recall that for $v\not\in S$, Lang's theorem over the finite residue field of $O_{K,v}$ implies the triviality of torsors for smooth $O_{K,v}$-groups with connected fibers. We may apply this to the $O_{K,v}$-groups arising by "twisting" $\mathscr{G}^0_{O_{K,v}}$ by its torsors over $O_{K,v}$ (i.e., automorphism schemes of such torsors). Hence, the twisting method for torsors (for the etale topology) gives that for $v \not\in S$ the natural map ${\rm{H}}^1(O_{K,v},\mathscr{G}) \rightarrow {\rm{H}}^1(O_{K,v},\Gamma)$ is injective.  Since $\Gamma$ is finite etale (so likewise for its torsors over $O_{K,v}$), it follows via the valuative criterion for properness (in the easy case of finite morphisms) that ${\rm{H}}^1(O_{K,v},\Gamma)$ restricts isomorphically onto the set ${\rm{H}}^1(K_v^{\rm{un}}/K_v,\Gamma)$ of classes in ${\rm{H}}^1(K_v,\Gamma)$ split by a finite unramified extension of $K_v$; note also that $\Gamma_{K_v}$ is itself split by an unramified extension of $K_v$.  But every class in ${\rm{H}}^1(K_v^{\rm{un}}/K_v,\Gamma)$ is an "unramified" finite etale $K_v$-scheme of degree equal to the order of $\Gamma$ and hence is split by an unramified extension of degree bounded solely in terms of the order of $\Gamma$.  This provides the desired uniform boundedness of local splitting degrees as $v$ varies.
To summarize, $${\rm{H}}^1(\mathbf{A}_K,G) = \varinjlim_{S \supset S_0} \prod_{v \in S} {\rm{H}}^1(K_v,G) \times \prod_{v \not\in S} {\rm{H}}^1(O_{K,v},\mathscr{G})$$ and via restriction we have
${\rm{H}}^1(O_{K,v},\mathscr{G}) \subset {\rm{H}}^1(K_v^{\rm{un}}/K_v,G)$
for all $v\not\in S_0$.  This latter injective restriction map is bijective when $\mathscr{G}$ is $O_{K,v}$-proper (as can be arranged to happen when $G$ is $K$-proper, such as $K$-finite), but it is generally not an equality; e.g., if $\mathscr{G}_{O_{K,v}}$ has connected fibers (as happens when $G$ is connected) then the $O_{K,v}$-cohomology set vanishes by Lang's theorem but the unramified cohomology set is generally non-trivial (e.g.,for $G = {\rm{PGL}}_n$ is recovers the $n$-torsion in ${\rm{Br}}(K_v)$).
Hence, we always have ${\rm{H}}^1(\mathbf{A}_K,G) \subset \prod_v {\rm{H}}^1(K_v,G)$ and this lies inside the set of classes $(c_v)_v$ that are unramified at all but finitely many places but is generally much smaller than that: it is characterized by an "almost everywhere integral" condition relative to a fixed choice of smooth separated finite type $S_0$-integral model $\mathscr{G}$ as above.
In the special case $G = \mathbf{G}_m$ over $K$ we may take $\mathscr{G} = \mathbf{G}_m$, so it follows that ${\rm{Pic}}(\mathbf{A}_K) = 1$.

That's enough of generalities, so we now return to the question posed.  In the special case where $f:Z \rightarrow X$ is the $n$th-power endomorphism of $\mathbf{G}_m$ over a number field $K$, the question is asking about surjectivity of ${\rm{H}}^1(K,\mu_n) \rightarrow {\rm{H}}^1(\mathbf{A}_K,\mu_n)$, which is to say surjectivity of $K^{\times}/(K^{\times})^n \rightarrow \mathbf{A}_K^{\times}/(\mathbf{A}_K^{\times})^n$. (Here we have computed the $\mu_n$-cohomology with the Kummer sequence, using the Pic vanishes for $\mathbf{A}_K$ too.) In other words, is the $n$th-power endomorphism of the idele class group $\mathbf{A}_K^{\times}/K^{\times}$ surjective?  Specializing to $K = \mathbf{Q}$ and $n=2$ we recover the counterexample provided by Stoll, and more generally for every number field $K$ this fails for every $n > 1$ since the quotient of $\mathbf{A}_K^{\times}/K^{\times}$ modulo its subgroup of infinitely divisible elements is the abelianized Galois group of $K$ (by class field theory) and that quotient is not $n$-divisible for any $n > 1$ due to the existence of finite abelian extensions of prime degree dividing $n$ (e.g., using cyclotomic extensions).
Now we assume $G$ is connected and have to address if ${\rm{H}}^1(K,G) \rightarrow {\rm{H}}^1(\mathbf{A}_K,G)$ is surjective.  As above, we may build a smooth separated finite type $O_{K,S_0}$-group $\mathscr{G}$ with generic fiber $G$ and with all fibers connected, so we have seen that ${\rm{H}}^1(\mathbf{A}_K,G)$ consists of exactly the set $\coprod {\rm{H}}^1(K_v,G)$ of collections $(c_v)_v$ of local classes that are trivial for all but finitely many $v$. In other words, this is precisely the question of whether a collection of $G$-torsors $E_v$ over $K_v$ which are trivial for all but finitely many $v$ can be assembled into a global torsor $E$ over $K$.  One may regard this as a "strong approximation" property for the construction of torsors.
Now assume $K$ is a number field and $G$ is affine, as in the question posed.
By Mostow we have $G = H \ltimes U$ for reductive $H$ and unipotent $U$, and an elementary Galois-twisting argument shows that ${\rm{H}}^1(K,H) \rightarrow {\rm{H}}^1(K,G)$ is bijective (since any $K$-form of $U$ is unipotent and smooth connected unipotent $K$-groups have vanishing degree-1 cohomology since ${\rm{char}}(K)=0$). Thus, the problem for connected affine $G$ is equivalent to that for its maximal reductive quotient, so now assume $G$ is reductive.
For tori we get counterexamples via consideration of norms.  Namely, let $T$ be the norm-1 torus $$T = \ker({\rm{R}}_{K'/K}(\mathbf{G}_m) \twoheadrightarrow \mathbf{G}_m)$$ for finite separable $K'/K$.  Then by Hilbert 90 we see that the problem for $T$ is exactly the question of whether or not the norm map
$$\mathbf{A}_{K'}^{\times}/{K'}^{\times} \rightarrow \mathbf{A}_K^{\times}/K^{\times}$$
is surjective. Infinitely divisible elements in the target are certainly hit, so to make a counterexample it suffices for surjectivity to fail modulo the subgroups of infinitely divisible elements.  By class field theory that induced map is exactly the natural map $\Gamma_{K'}^{\rm{ab}} \rightarrow \Gamma_K^{\rm{ab}}$ between abelianized Galois groups.  This is surjective if and only if every finite abelian extension $F/K$ is linearly disjoint from $K'$ over $K$, and that fails precisely when $K'$ contains a nontrivial finite abelian extension of $K$.
Hence, the "last chance" for an affirmative answer is in the connected semisimple case.  Here one encounters counterexamples when $G$ is not simply connected.  For example, if $G = {\rm{PGL}}_n$ with $n > 1$ then ${\rm{H}}^1(K,G) = {\rm{Br}}(K)[n]$ and likewise for $K_v$, and by class field theory the natural map ${\rm{Br}}(K)[n] \rightarrow \oplus_v {\rm{Br}}(K_v)[n]$ is never surjective for $n > 1$.
Finally, suppose $G$ is simply connected too.  Then we get a positive answer!  Indeed, by the deep Kneser-Bruhat-Tits theorem we have ${\rm{H}}^1(K_v,G)=1$ for all non-archimedean $v$, so the problem is surjectivity of ${\rm{H}}^1(K,G) \rightarrow \prod_{v|\infty} {\rm{H}}^1(K_v,G)$ for such $G$.  This latter map is actually surjective for all connected linear algebraic groups over number fields (see Proposition 6.17 in the book of Platanov-Rapinchuk, which rests on hard work with tori).


text:
\begin{document}

\title{A note on tensor categories of Lie type $E_9$}

\thanks{\textit{Department of Mathematics, Indiana University, Bloomington, IN 47401 USA}\\
email: \texttt{errowell@indiana.edu}}
\author{Eric C. Rowell}

\maketitle
\begin{abstract}
We consider the problem of decomposing tensor powers of the fundamental
level 1 highest weight representation $V$ of the affine Kac-Moody
algebra $\g(E_9)$.  We describe an elementary algorithm for
determining the decomposition of the submodule of $\Vn$ whose
irreducible direct summands have highest weights which are maximal
with respect to the null-root. This decomposition is based on
Littelmann's path algorithm and conforms with the uniform
combinatorial behavior recently discovered by H. Wenzl for the
series $E_N$, $N\not=9$.
\end{abstract}
\section{Introduction}

While a description of the tensor product decompositions for
irreducible highest weight modules over affine algebras can be
found in the literature (see e.g. \cite{Gould}), effective
algorithms for computing explicit tensor product multiplicities
are scarce.  Some partial results in this direction have been
obtained by computing
characters (see e.g. \cite{King1}) and  by employing crystal bases (see
e.g. \cite{Shimo}) or the equivalent technique of Littelmann
paths. In this note we look at the particular case of the affine
Kac-Moody algebra associated to the Dynkin diagram $E_9$, with any
eye towards extending the results of \cite{Wenzl3}.

Let $V$ be the irreducible highest weight representation of the
$\g(E_N) \ N\geq 6$ with highest weight $\La_1$ corresponding to
the vertex in the Dynkin diagram furthest from the triple point.
For $N\not=9$ H. Wenzl \cite{Wenzl3} has found uniform
combinatorial behavior for decomposing a certain submodule
$\Vn_{new}$ of $\Vn$ using Littelmann paths \cite{Li}.
These submodules have the property that each irreducible
summand of $\Vn_{new}$ appears in $\Vn$ for the first time (for $N\leq
8$) or last time (for $N\geq 10$).  The degeneracy of the
invariant form was an obstacle to including the
affine, $N=9$ case.

We extend Wenzl's combinatorial description to the case $N=9$ by
finding submodules $\M_n$ analogous to his $\Vn_{new}$.
Specifically, we look at the (full multiplicity) direct sum of
those submodules of $\Vn$ whose highest weights have maximal
null-root coefficient. Not surprisingly, these summands appear
\emph{only} in $\Vn$. The particular
utility of considering this submodule is that whereas decomposing
the full tensor power $\Vn$ into its simple constituents would
require an infinite path basis, only a finite sub-basis
(consisting of 200 straight paths) is needed to determine the
decomposition of $\M_n$.  Although this note was inspired by the
results of \cite{Wenzl3}, the module $\M_n$ appears so naturally
that this case may shed some light on the combinatorial behavior
described by Wenzl.

This paper is organized in the following way.  In Section \ref{defs} we
give the data and standard definitions for the Kac-Moody algebra
$\g(E_9)$.  Section \ref{paths} is dedicated to summarizing the general
technique of Littelmann paths, while in Section \ref{e9} we apply this
technique to the present case and present some new definitions.
Table \ref{notation} gives a glossary of notation for the
reader's convenience. All the lemmas we prove are
contained in Section \ref{lemmas}, and the main theorem and algorithm they lead to is
described and illustrated in Section \ref{main}.  We briefly mention a possible
application and a generalization in Section \ref{future}, as well as connections to Wenzl's
results.

I would like to thank H. Wenzl for bringing this problem to my
attention and for many useful discussions.
 \section{Notation and Definitions}\label{defs}
We begin by fixing a realization of the generalized Cartan matrix
of $\g(E_9)$ sometimes denoted in the literature by $\g(E^{(1)}_8)$.
Observe that our realization is different than that of Kac
\cite{Kac}. In particular the vertex Kac labels with a $0$ we
label with a $1$ in our Dynkin diagram (Figure \ref{Dynkin}). This
is done to conform with the notation of \cite{Wenzl3}.

\begin{figure}\label{Dynkin}\begin{picture}(300,80)(50,30)
\put(118,50){\circle{4}} \put(120,50){\line(1,0){18}}
\put(140,50){\circle{4}} \put(142,50){\line(1,0){18}}
\put(162,50){\circle{4}} \put(164,50){\line(1,0){18}}
\put(184,50){\circle{4}} \put(186,50){\line(1,0){18}}
\put(206,50){\circle{4}} \put(208,50){\line(1,0){18}}
\put(228,50){\circle{4}} \put(228,52){\line(0,0){18}}
\put(228,72){\circle{4}} \put(230,50){\line(1,0){18}}
\put(250,50){\circle{4}} \put(252,50){\line(1,0){18}}
\put(272,50){\circle{4}} \put(115,35){1} \put(137,35){2}
\put(159,35){3} \put(181,35){4} \put(203,35){5} \put(225,35){6}
\put(247,35){7} \put(232,67){8} \put(269,35){0}
\end{picture}\caption{Dynkin Diagram of $E_9$}
\end{figure}

\begin{definition}
Let $\{\ep_0,\delta,\ep_1,\ldots,\ep_8\}$
 be an ordered basis for $\R^{10}$, with symmetric bilinear form
$\lan\  ,\  \ra$ such that $\lan\ep_i,\ep_j\ra=\delta_{ij}$
 for $1\leq i,j\leq 8$
and $\lan\delta,\ep_0\ra=1$ with all other pairings $0$.
The simple
roots of $\mathfrak{g}(E_9)$ are defined by:
\begin{equation*}
\alpha_i=
\begin{cases} \ep_i-\ep_{i+1} & \text{if $1\leq i\leq 7$,}\\
\ep_7+\ep_8& \text{if $i=8$,}\\
\frac{1}{2}(\delta+\ep_8-\sum_{i=1}^7\ep_i)& \text{if $i=0$}
\end{cases}
\end{equation*}
\end{definition}
The simple roots generate the root lattice $Q:=\rm{span}_\Z\{\al_i\}$, and we define
coroots $\check\al_i:=\frac{2}{\lan \al_i,\al_i \ra}\al_i$.  As $\g(E_9)$
is simply-laced we abuse notation and identify each coroot with the
corresponding root.  Since we are only concerned with the combinatorics,
we refer the reader to Chapter 6 of the book \cite{Kac} for the full description of
the Kac-Moody algebra $\g(E_9)$.
\begin{definition}
We define the fundamental weights by:
\begin{equation*}
\La_i=
\begin{cases} (i,0;1,\ldots, 1,0,\ldots,0)\  \text{$i$ ones}
 & \text{if $1\leq i\leq 6$,}\\
\frac{1}{2}(8,0;1,\ldots,1,-1)& \text{if $i=7$,}\\
\frac{1}{2}(6,0;1,\ldots,1)& \text{if $i=8$,}\\
(2,0;0,\ldots,0)& \text{if $i=0$}
\end{cases}
\end{equation*}

Note that $\lan \alpha_i,\La_j\ra=\delta_{ij}$, and the
fundamental weights of $\mathfrak{g}(E_9)$ are determined up to a
multiple of $\delta$ by this relation. The set of dominant weights $P_+$ is
the $\N$-span of the fundamental weights plus $\C\delta$, and the $\Z$-span
$P$ is called the weight lattice.  It will be useful to denote by $\hat{P}_+$ those
dominant weights whose second coordinate is $0$.
\end{definition}
\begin{definition}
We define the Weyl group in the usual way: $W=\lan
s_i:i=0,\ldots,8\ra$ where $s_i(\nu)=\nu-\lan
\nu,\alpha_i\ra\alpha_i$ for $\nu \in \R^{10}$.
\end{definition}
The simple reflections $\{s_i\}_{i=1}^8$ generate a finite
subgroup $\overline{W}$ acting on the last eight coordinates by
permutations and an even number of sign changes. For any $\la$
with $\lan \la,\alpha_0\ra>0$ the vector $s_0(\la)$ has a strictly
smaller $\delta$-coefficient, and by applying elements of
$\overline{W}$ to arrange $\lan \la,\alpha_0\ra>0$ one can
construct an infinite sequence of vectors with strictly decreasing
$\delta$-coefficients.  Thus one sees that $W$ is an infinite
group.

\section{Littelmann Paths}\label{paths}
 To decompose the tensor powers of
$V$ we use the Littelmann path formalism (see \cite{Li}).
For this section we consider general Kac-Moody algebras
$\g$.

Littelmann considers the space of piecewise linear paths $\pi:
[0,1]\rightarrow P_\Q$ beginning at $0$ and ending at some point
in the weight lattice $P$.  He defines \emph{root operators} on
the space of all such paths $e_{i}$ and $f_{i}$ for each simple
root $\al_i$, which, when applied repeatedly to the straight path
$\pi_\la$ from $0$ to a dominant weight $\la$ give a \emph{path
basis} $\mathcal{B}_\la$ for the corresponding irreducible highest
weight module $V_\la$. The operators $f_i$ are defined on paths
$\pi$ as follows (see \cite{Li} for full details): let
$h_i(t)=\lan \pi(t),\al_i\ra$, and $m_i=\min(h_i(t))$.  If
$h_i(1)-m_i\geq 1$, split the interval $[0,1]$ into three pieces:
$[0,t_0]\cup [t_0,t_1]\cup [t_1,1]$ where $t_0$ is the maximal $t$
such that $h_i(t)=m_i$ and $t_1$ is the minimal $t$ such that
$h_i(t)=m_i+1$.  Then $$f_i\pi=\begin{cases}\pi(t) & \text{on
$[0,t_0]$}\\
s_i(\pi(t)) & \text{on $[t_0,t_1]$}\\
\pi(t)-\al_i & \text{on $[t_1,1]$}\end{cases}$$
If $h_i(1)-m_i<1$ then $f_i\pi=0$.  The operators
$e_i$ are defined similarly.  Since all paths begin at the weight
$0$, we may concatenate paths in the usual way.  For any $\la\in P$
define the path $\pi_\la: [0,1]\rightarrow P_\Q$ by
$t\rightarrow t\la$.  We denote concatenation by $*$, i.e.
$\pi_\la*\pi_\mu$ passes through $\la$ and terminates at $\la+\mu$.

Let $\la$ and $\mu$ be dominant weights of a Kac-Moody algebra
and $V_\la$, $V_\mu$ the corresponding irreducible highest weight
modules. We collect together those of Littelmann's results that we will need in:
\begin{proposition}
\begin{enumerate}
\item[(a)] $\B_\la\subset\{f_{1_j}f_{2_j}\cdots f_{s_j}\pi_\la\}$, that is, every
path in the basis $\B_\la$ is obtained from $\pi_\la$ by applying a finite sequence of
the root operators $f_i$.
\item[(b)]
The decomposition rules for the tensor product is given as
follows:
\[ V_\mu\otimes V_\la \cong \bigoplus_{\pi}V_{\pi(1)}
\] where $\pi=\pi_\mu *\pi_i$ with $\pi_i\in \mathcal{B}_\la$ and the image of $\pi$
contained in the closure of the dominant Weyl chamber.
\item[(c)] The multiplicity of $V_\nu$ in $\Vn_\la$ is
equal to the number of paths whose image is
contained in the closure of the
dominant Weyl chamber that terminated at $\nu$ that
are obtained by concatenating $n$ basis paths.
\end{enumerate}
\end{proposition}
\section{Lie type $E_9$}\label{e9}
Now we consider
the set $P(\La_1)$ of weights of $V$.  Following Kac \cite{Kac}
we call the weights in the Weyl group orbit $W\cdot\La_1$
\emph{maximal} and note that
$$P(\La_1)=\bigcup_{\om \in W\cdot\La_1}\{\om-t\delta : t \in \N\}.$$
Any $V_{\la-s\delta}$ that appears in some $\Vn$ must be of the
form:
$$\la-s\delta=\sum_{\om_i\in P(\La_1)}\om_i$$ with
$s\in\half\N$.  It is well known that the maximal weights appear in the
multiset $P(\La_1)$ with
multiplicity one (for example, see \cite{Gould}).

 The second coordinate (essentially determined by the number of times $s_0$ occurs in a
 minimal expression)
 provides a gradation on
$W\cdot\La_1$ which motivates the following lemma, the proof of
which is a computation.
\begin{lemma}\label{forms} Every $\om\in W\cdot\La_1$ is of one of the
following 4 forms:
\begin{enumerate}
\item[(I)] Type I: $(1,0;\pm\ep_i)$
\item[(II)] Type II: $\frac{1}{2}(2,-1;\pm1,\ldots,\pm1)$
with an even number of minuses among the last eight coordinates.
\item[(III)] Type III: $(1,-1;w(1,1,1,0,\ldots,0))$
where $w \in S_8$, the group of permutations on $8$ symbols.
\item[(IV)] Type IV: all others, i.e. $(1,-j;\nu)$ where
$j\geq 1$ and if $j=1$, $\nu\not\in S_8\{(1,1,1,0,\ldots,0)\}$.
\end{enumerate}
\end{lemma}
The weights of types I, II and III will be particularly useful and
we will call them \emph{straight weights} and denote the set of
straight weights by $\Omega$. It is a simple but tedious
computation
 to show that for any $\omega\in\Omega$, the straight line path
 $\pi_\omega$ from $0$
 to $\omega$ is in the path basis $\mathcal{B}_{\La_1}$.
The idea of the computation is to start with the path $\pi_{\La_1}$
and inductively apply only those operators $f_i$ for which the height function
$h_i(t)=t$ so that two of the three intervals in the definition of the operator $f_i$
are degenerate, and the image of the paths remain straight lines.
Type I and II weights are in fact \emph{all} maximal weights with
second coordinate $0$ or $-\half$, while there are maximal weights
with second coordinate $-1$ besides those of type III.  Observe that since $\La_1$ is
the unique level one dominant weight (modulo $\delta$),
all paths obtained from concatenation of basis paths whose image lies in
the dominant Weyl chamber must pass through $\La_1$.
\begin{definition}\label{leveldef}
The \emph{level} $n(\la)$ of a weight $\la$ is the $\ep_0$
coordinate. Note that all weights in $P(\La_1)$ are level $1$,
thus $\la$ has level $n$ iff $\la-t\delta$ appears in $V^{\otimes
n}$ for some $t$ since $\la$ will be a sum of weights in
$P(\La_1)$.  Denote by $\hat{P}_+(n)$ the set of level $n$
dominant weights modulo $\delta$.
\end{definition}
The following definition appears in \cite{Wenzl3} and is critical
in the sequel.
\begin{definition}
We define the function $k: Q\rar\Z$ in one of the following
equivalent ways:
\begin{enumerate}
\item[(a)]  $k(\om)= -\lan\om,2\hat{\al}_0\ra$, where
 $\hat{\al}_0=\al_0-\ep_{8}$,
\item[(b)] If $\om=\sum_{i=0}^8 M_i\La_i$,
then $k(\om)=M_{8}-M_{7}-2M_0$.
\end{enumerate}
We will also need the quantity $[(\la)]_3$ defined to be the
remainder of $k(\la)$ upon division by $3$.
\end{definition}
We compute these values for the maximal weights and record them in
the following:
\begin{lemma}\label{omkvalues}
The values of the function $k$ for the maximal weights $\om$ of types I, II,
III and IV (as in Lemma \ref{forms}) satisfy:
\begin{enumerate}
\item[(I)] Type I: $k(\omega)\in\{0,-2\}$
\item[(II)] Type II: $k(\omega)\in\{3,1,-1,-3,-5\}$
\item[(III)] Type III: $k(\omega)=2$
\item[(IV)] Type IV: $k(\omega)\leq(6j-6)$ where $\om=(1,-j;\nu)$ and $1\leq j\in \half\Z$.
\end{enumerate}
\end{lemma}

The dominant weights are only defined up to a multiple
of $\delta$, but we are interested in those that appear in $P_+(\Vn)$ which motivates:
\begin{definition}
A level $n$ dominant weight $\la -m_\la\delta$ is called
\emph{initial} if $m_\la$ is minimal such that $V_{\la
-m_\la\delta}$ appears in $\Vn$.
\end{definition}
\begin{remark}  It is easy to see that there are
finitely many initial weights of a fixed level $n$, since there is
a one-to-one correspondence between the finite set $\hat{P}_+(n)$ and initial weights.
The term \emph{initial} comes from the fact that if $\la-m_\la\delta\in\hat{P}_+(n)$
so is $\la-(m_\la+1)\delta$.
Moreover, it is clear that $m_\la$ is always a non-negative
half-integer, since the coefficient of $\delta$ for any weight
$\om\in P(\La_1)$ is a non-positive half integer.
\end{remark}

We will eventually show that the $m_\la$ is computed from
the value of $k(\la)$ via the function:
\begin{definition}Let $\la\in\hat{P}_+$:
\begin{equation}
\Delta(\la)=
\begin{cases}
0 & \text{if $k(\la)\leq 0$ and even,}\\
\frac{1}{2} & \text{if $k(\la)\leq 1$ and odd,}\\
\frac{1}{6}(k(\la)+2[\la]_3)& \text{if $k(\la)\geq 1$}
\end{cases}
\end{equation}

Observe that when $k(\la)=1$ we have
$\frac{1}{6}(k(\la)+2[\la]_3)=\half$ so $\Delta$ is well defined.
\end{definition}
\begin{definition}
Define $\M_n$ to be the largest submodule of $V^{\otimes n}$ such
that all irreducible direct summands have highest weights of the form
$\la-m_\la\delta$ (i.e. initial weights).
\end{definition}
We illustrate this definition with an example:
\begin{example} The highest weight module $V_{\La_8}$ does not appear in
$V^{\otimes 3}$ as it is not a sum of $3$ type I weights. However,
$V_{\La_8-\frac{\delta}{2}}$ does appear in  $V^{\otimes 3}$ as
$$\La_8-\frac{\delta}{2}=\La_0+\frac{1}{2}(2,-1;1,\ldots,1)=
(1,0;\ep_1)+(1,0;-\ep_1)+\frac{1}{2}(2,-1;1,\ldots,1).$$  Notice
also that $V_{\La_8-\frac{t}{2}\delta}$ will also appear in
$V^{\otimes 3}$ for any $t\geq 1$, but only
$V_{\La_8-\frac{\delta}{2}}$ will appear in $\M_3$.\end{example}
The complete reducibility of $\Vn$ (see e.g. \cite{Gould}) allows
us to write:
$$\Vn\cong\M_n\oplus Z_n$$ where $Z_n$ consists of those simple submodules whose
highest weights are not initial.
\begin{table}\label{notation}
\caption{Notation}
\begin{tabular}{*{2}{lr}}
$\al_i$ & $i$th simple root\\
$Q$ & root lattice\\
$\La_i$ & $i$th fundamental weight\\
$P$ & weight lattice\\
$P_+$ & dominant weights \\
$\hat{P}_+$ & dominant weights $\pmod\delta$  \\
$\hat{P}_+(n)$ & level $n$ dominant weights $\pmod\delta$\\
$n(\la)$ & level of $\la$\\
$P(\La_1)$ & weights of $V$\\
$W\cdot\La_1$ & maximal weights of $V_{\La_1}$\\
$\Omega$ & set of straight weights\\
$P_+(\Vn)$ & dominant weights of $\Vn$\\
$[\la]_3$ & least residue of $k(\la)\pmod 3$\\
$\pi_\la$ &  path $t\rightarrow t\la$\\
$W$ & (affine) Weyl group\\
$\S(k)$ & level $k$ initial weights\\
$\la\rightarrow\mu$ & straight weight path\\
\end{tabular}
\end{table}
\section{Lemmas}\label{lemmas}
In this section we describe the combinatorial rules for
decomposing the modules $\M_n$.  The first two lemmas show that
$m_\la=\Delta(\la)$, while the two that follow show that one
may determine $\M_{n+1}$ from $\M_n$ and the (finitely many)
straight weights.
\begin{lemma}\label{exist}
Let $\la\in\hat{P}_+$ so that $\la-\Delta(\la)\delta$ is a level
$n$ dominant weight.  Then $V_{\la-\Delta(\la)\delta}$ appears in
$\Vn$. Moreover, there is a straight weight path from $0$ to $\la$
passing through only weights of the form $\mu -\Delta(\mu)\delta$
with $\mu\in\hat{P}_+$.
\end{lemma}
$Proof.$ Since we are not concerned with computing multiplicities,
it suffices to construct a piecewise linear straight weight path
from $0$ to $\la-\Delta(\la)\delta$ contained entirely within the
dominant Weyl chamber. Assume $\la=\sum_{i=0}^8 M_i\La_i$. We will
construct the required path in reverse by starting from the weight
$\la-\Delta(\la)\delta$ and removing path segments until we reach
the weight $0$. By concatenating the paths we remove in reverse
order we obtain the desired path.  The first set of useful paths
are the sub-paths of:
\[
\pi_1: 0\rar\La_1\rar\La_2\rar\La_3\rar\La_4\rar\La_5\rar\La_6\rar
(\La_7+\La_8)\rar(\La_0+2\La_8)
\] constructed by concatenating straight paths $\pi_\om$
terminating at $\om=(1,0;\ep_i)$, $i=1,\ldots,8$.  We denote by
$\pi_1^{(i)}$ the $i$th sub-path of $\pi_1$. In a similar fashion
we construct
\[ \pi_2:  0\rar\La_1\rar\La_0 \quad
\text{and} \quad \pi_3:0\rar\La_1\cdots\rar(\La_7+\La_8)\rar 2\La_7
\]
again using only paths with type I straight weight segments.  The
affect of removing these path segments on the value of $k(\la)$ is
as follows (where the value of a path at $1$ is $\la$):
\begin{enumerate}
\item $k(\pi*\pi_1^{(i)}(1))=k(\pi(1))$ i.e. deleting sub-paths of $\pi_1$ has
no affect on the value of $k(\la)$.
\item $k(\pi*\pi_2(1))=k(\pi(1))-2$ so deleting the path $\pi_2$ increases the value of
$k(\la)$ by 2.
\item $k(\pi*\pi_3(1))=k(\pi(1))-2$ so deleting the path $\pi_3$
increases the value of $k(\la)$ by 2.
\end{enumerate}
\subsection*{Case I} $k(\la)=M_8-M_7-2M_0\leq 0$ is even.\\
In this case $\Delta(\la)=0$.  Since $k(\la)$ does not
depend on $M_i$ $1\leq i\leq 6$ we can reduce the the case
$M_i=0$, $1\leq i\leq 6$ using sub-paths of $\pi_1$.  For $\la$
with $M_8=M_7=M_0=0$ we are done.  If not, we observe that $M_7$
and $M_8$ have the same parity. Again using the sub-path of
$\pi_1$ terminating at $\La_7+\La_8$ as many times as is
necessary, we may assume either $M_7=0$ or $M_8=0$.

\subsubsection*{Case I.1} $M_7=0$.\\
In this case we have $M_8\leq 2M_0$ and $M_8$ even. So by removing
path segments $\pi_1$ as many times is as necessary we can reduce
to $M_8=0$ with $k(\la)=-2M_0$ unchanged.  At this point we are
left with the case $\la=M_0\La_0$, to which we remove the path
segments $\pi_2$ as many times as necessary to reduce to $0$.

\subsubsection*{Case I.2} $M_8=0$.\\
Here we have that $M_7\geq -2M_0$ and $M_7$ is even, so we reduce by $\pi_3$
until $M_7=0$ and then reduce by the path $\pi_2$ until $M_0=0$ and we are left with
the weight $0$.  Observe that $k(\la)=-M_7-2M_0$ in this case so while deleting path segments
$\pi_2$ or $\pi_3$ result in a raised $k$-value,
it will always be non-positive and even, regardless.

\subsection*{Case II} $k(\la)=M_8-M_7-2M_0\leq 1$ is odd.\\
Here $M_8$ and $M_7$ have opposite parity; so, as in Case I, we
reduce by sub-paths of $\pi_1$ until either $M_7=0$ or $M_8=0$.  Then we reduce by
paths as in Case I
until we are left with two cases: $\la=\La_7-\frac{\delta}{2}$ and
$\la=\La_8-\frac{\delta}{2}$.  These are achieved by the paths:
$$0\rar\La_1\rar\La_2\rar\La_3\rar (\La_7-\frac{\delta}{2}) \quad
\text{and} \quad 0\rar\La_2\rar (\La_8-\frac{\delta}{2})$$ using
the straight weights $\half(2,-1;-1,-1,-1,1,\ldots,1,-1)$ and
$\half(2,-1;-1,-1,1,\ldots,1)$ respectively.

\subsection*{Case III} $k(\la)=M_8-M_7-2M_0\geq 2$.\\
In this case $M_8\geq 2+M_7+2M_0$, so we can use sub-paths of
$\pi_1$ to reduce to $M_7=0$ and then $M_0=0$ without changing the
value of $k(\la)$ and we are left with the task of constructing a
path terminating at $\la=M_8\La_8-\Delta(M_8\La_8)$, where
$k(\la)=M_8\geq 2$. The weight $3\La_8-\frac{\delta}{2}$ is of the
form $\mu-\Delta(\mu)\delta$ with $\mu\in\hat{P}_+$ and the path:
$$\pi_1*[(2\La_8+\La_0)\rar (3\La_8-\frac{\delta}{2})]$$
 allows us to reduce to $M_8\leq 2$ since:
\begin{equation}\label{three}
\Delta((M_8-3\ell)\La_8)=\frac{1}{6}(M_8-3\ell+2[(M_8-3\ell)\La_8]_3)=
\Delta(M_8\La_8)-\frac{\ell}{2}
\end{equation}
 so that
$$M_8\La_8-\Delta(M_8\La_8)\delta-\ell(3\La_8-\frac{\delta}{2}))=
(M_8-3\ell)\La_8-\Delta((M_8-3\ell)\La_8)\delta.$$  Now the cases
$M_8=0,1$ were covered in cases I and II respectively, so we need
only construct a path to $2\La_8-\delta$.  But this is nothing
more than a doubling of the path $$0\rar\La_2\rar
(\La_8-\frac{\delta}{2})$$ constructed above. This completes the
proof.   $\Box$

\begin{lemma}\label{minimal}
$\Delta(\la)=m_\la$ for all $\la\in\hat{P}_+$.
\end{lemma}
$Proof.$  By Lemma \ref{exist} it is sufficient to show that
$m_\la\geq\Delta(\la)$ since $\la-\Delta(\la)\delta$ appears in $\Vn$ hence
$m_\la\leq\Delta(\la)$.
Again we consider cases.
\subsection*{Case I} $k(\la)\leq 0$ and even.\\
Since $m_\la\geq 0=\Delta(\la)$ there is nothing to prove.
\subsection*{Case II} $k(\la)\leq 1$ and odd.\\
We need only show that $m_\la\not=0$.  The only way that this can
occur is if $\la$ can be expressed as a sum of type I weights. But
$k(\om)=0$ or $-2$ for $\om$ of type I, so if $\la$ were a sum of
type I weights $k(\la)$ would be even.
\subsection*{Case III} $k(\la)\geq 2$.\\  In this case we will
reduce to the case where $\la=M_8\La_8-t\delta$ using sub-paths of
$\pi_1$ as in the proof of Lemma \ref{exist}.  Suppose
$$\la=\sum_{i=0}^8M_i\La_i-t\delta$$ with $t$ minimal and
$t\in\half\N$. We compute $k(\la)=M_8-M_7-2M_0\geq 2$, so that
$M_8>M_7+2M_0$.  We can reduce to the case where
$M_1=M_2=\cdots=M_6=M_7=0$ using sub-paths of $\pi_1$ and
observing that
$$\la^\prime=M_0\La_0+(M_8-M_7)\La_8-t\delta$$ has
$k(\la^\prime)=k(\la)$ and $t$ minimal for $\la^\prime$ if and
only if $t$ is minimal for $\la$.  Setting $M_0^\prime=M_0$ and
$M_8^\prime=(M_8-M_7)$ we have
$k(\la^\prime)=M_8^\prime-2M_0^\prime\geq 2$.  Reducing by the
path $\pi_1$ and setting
$M_8^{\prime\prime}=M_8^\prime-2M_0^\prime$ we see that
$$\la^{\prime\prime}=M_8^{\prime\prime}\La_8-t\delta$$
has $t$ minimal if and only if $t$ is minimal for $\la^\prime$. So
we are left with showing that $m_\la\geq\Delta(\la)$ for
$\la=M_8\La_8$.  This will follow by an induction argument once we
show it for the cases $M_8=1,2$ and $3$.
\subsubsection*{$M_8=1$} This case was already covered in Case II
above.
\subsubsection*{$M_8=2$} If $2\La_8$ were a sum of type I
weights we would have $k(2\La_8)\leq 0$ so we must have at least
one weight of type II, III or IV. By considering the values of $k$
on these weights we see that that $t=1$ is minimal.
\subsubsection*{$M_8=3$} Again considering the values of $k$ we
see that type I weights are not sufficient and that $t=\half$ is
minimal.

Observing that
$\Delta((M_8-3\ell)\La_8)+\frac{\ell}{2}=\Delta(M_8\La_8)$ (see
Equation \ref{three}) the case $M_8>3$ follows by induction and we
are done.
   $\Box$

\begin{remark}
We may now redefine \emph{initial weight} to be any dominant
weight of the form $\la-\Delta(\la)\delta$, and we denote the set
of initial weights of level $n$ by
$\S(n)=\{\la-\Delta(\la)\delta:\la\in\hat{P}_+(n)\}$.
\end{remark}
\begin{lemma}If $\la-\Delta(\la)\delta$ is an initial weight
and $\om\in\Omega$ then $\la-\Delta(\la)\delta -\om$
is either an initial weight or not in the dominant Weyl chamber.
\end{lemma}
$Proof.$  Let $\mu-t\delta=\la-\Delta(\la)\delta-\om$ where
$\mu\in\hat{P}_+$.  Assume that $\mu$ is
in the dominant Weyl chamber. We must demonstrate that $t=\Delta(\mu)$. By
Lemma \ref{minimal}, we have that $t\geq\Delta(\mu)$ as $t\leq\Delta(\mu)$
would contradict the minimality of $\Delta(\mu)$. Using Lemma \ref{omkvalues}
we have:
\begin{equation}
t=
\begin{cases}
\Delta(\la) & \text{if $\om$ is type I,}\\
\Delta(\la)-\half & \text{if $\om$ is of type II,}\\
\Delta(\la)-1 & \text{if $\om$ is of type III}
\end{cases}
\end{equation}
It is sufficient to show that $\Delta(\mu)\geq t$ for all of these
cases.  We organize them by considering the value of $k(\la)$ as
follows:
\subsection*{Case I} $k(\la)\leq 0$ and even.\\
Here we have $\Delta(\la)=0$.  Since $\Delta(\mu)\geq 0$ the only
possibility is that $\om$ is of type I, for which it is clear.
\subsection*{Case II} $k(\la)\leq 1$ and odd.\\
The only possibilities are $\om$ of type I or II, since
$\Delta(\la)=\half$ in this case and $\Delta(\mu)\geq 0$.  If
$\om$ is of type II, then $t=0$, hence $\Delta(\mu)\geq t$ is
obvious.  If $\om$ is of type I, then $t=\half$ and Lemma
\ref{omkvalues} implies $k(\mu)=k(\la)-k(\om)\leq 3$ and odd.  If
$k(\mu)\leq 1$ and odd then $\Delta(\mu)=\half$ and we are done.
Otherwise $k(\mu)=3$, and we compute $\Delta(\mu)=\half$ as
required.
\subsection*{Case III} $k(\la)\geq 2$.\\
Here there are 3 cases depending on the type of $\om$.  The
computations are somewhat tedious, but straightforward.
\subsubsection*{Case III.1} $\om$ is of type I.\\
If $k(\om)=0$ then $k(\mu)=k(\la)$ hence $\Delta(\mu)=\Delta(\la)$
and we are done.  If $k(\om)=-2$, then $k(\mu)=k(\la)+2$ and we
must check the three 3 cases corresponding to the values of
$[\mu]_3$ (depending on $[\la]_3$) by evaluating
$\Delta(\mu)=\frac{1}{6}(k(\mu)+2[\mu]_3)$.

\subsubsection*{Case III.2} $\om$ is of type II.\\
We must show that $\Delta(\mu)\geq \Delta(\la)-\half$.  This is
the most involved case as $k(\om)\in\{3,1,-1,-3,-5\}$ and we must
check a total of 15 subcases corresponding to the 3 values of
$[\la]_3$ and 5 values of $k(\om)$. As an example of what is
involved we work out the cases where $[\la]_3=1$ and $k(\om)=-1$.
Then $k(\mu)=k(\la)+1$, $[\mu]_3=2$ and
$$\Delta(\mu)=\frac{1}{6}(k(\la)+1+2[\mu]_3)=
\frac{1}{6}(k(\la)+1+4)\geq\frac{1}{6}(k(\la)+2)=\Delta(\la)>\Delta(\la)-\half.$$
Notice that in this case $\mu$ is not dominant. The remaining
cases are handled similarly.
\subsubsection*{Case III.3} $\om$ is of type III.\\
We must show that $\Delta(\mu)\geq \Delta(\la)-1$.  Here
$k(\om)=2$ so $k(\mu)=k(\la)-2$ and we must again check cases by
evaluating $\Delta(\mu)$.   $\Box$

\begin{lemma}If $\la-\Delta(\la)\delta$ is an initial weight and
$\la-\Delta(\la)\delta +\om$ is also initial, then $\om$ is a
straight weight.
\end{lemma}
Before giving a proof, we mention a \emph{caveat}: the requirement
that $\la-\Delta(\la)\delta +\om$ is initial is not superfluous.
For example, $2\La_8-\delta$ is an initial weight and
$(1,0;-\ep_8)$ is a straight weight, but
$2\La_8-\delta+(1,0;-\ep_8)=\La_7+\La_8-\delta$ is not initial.\\
$Proof.$ It is enough to show that $\la-\Delta(\la)\delta+\om$ is
not initial if $\om$ is not straight.  The key fact here is from
Lemma \ref{omkvalues}: $k(\om)\leq 6j-6$ for the type IV weight
$\om=(1,-j;\nu)$ where $j\geq 1$ is a half-integer. Let
$\mu-s\delta=\la-\Delta(\la)\delta+\om$ for such a weight $\om$.
Observing that $s=\Delta(\la)+j$ we will show that
$s\not=\Delta(\mu)$.
\subsection*{Case I} $k(\mu)\leq 1$.\\
Since $s\geq j\geq 1$ and $\Delta(\mu)\leq \half$, it is clear
that $s\geq\Delta(\mu)$.
\subsection*{Case II} $k(\la)\leq 1$.\\
Here we have that
$$\Delta(\mu)=\frac{1}{6}(k(\la)+k(\om)+2[k(\la)+k(\om)]_3)\leq
\frac{1}{6}(1+6j-6+4)=\frac{6j-1}{6}<j\leq s$$ so once again
$\Delta(\mu)\not=s$.
\subsection*{Case III} $k(\la)\geq 1$ and $k(\mu)\geq 1$.\\
Computing as above we have
$$\Delta(\mu)=\frac{1}{6}(k(\la)+k(\om)+2[k(\la)+k(\om)]_3)\leq
\frac{1}{6}(k(\la)+6j-6+4)\leq\frac{1}{6}(k(\la))+\frac{6j-2}{6}<
\Delta(\la)+j=s$$
So we see that $\Delta(\mu)\not=s$ in all cases and we are done.
  $\Box$

\section{The Main Theorem and an Algorithm}\label{main}
The following theorem is a immediate corollary of the
lemmas in the previous section:
\begin{theorem}\label{summarized}
If $\la-\Delta(\la)\delta\in\S(n)$ then
any straight weight path from $0$ to $\la-\Delta(\la)\delta$
passes through only initial weights.  Thus
$$\M_n\cong\bigoplus_{\la\in\hat{P}_+(n)}c_\la V_{\la-\Delta(\la)\delta}$$
where the multiplicities $c_\la$ are determined by counting the
straight weight paths terminating at $\la-\Delta(\la)\delta$.
\end{theorem}
Applying the results we have the following simple inductive
algorithm for decomposing $\M_n$ as a sum of simple highest weight
modules:
\begin{enumerate}
\item[{\bf Step 1}] Initialize with $\M_1\cong V_{\La_1}$.
\item[{\bf Step 2}] Having determined the multiplicities $c_\la$
so that
$$\M_n\cong\bigoplus_{\la\in\hat{P}_+(n)}c_\la V_{\la-\Delta(\la)\delta}$$ compute the
set $A_\la=\{\la-\Delta(\la)\delta +\om: \om \in\Omega\}$ for each
$\la\in\hat{P}_+(n)$.
\item[{\bf Step 3}] Compute the set $\S(n+1)$.  The size of $\S(k)$
is computed from the generating
function:
\begin{equation}\label{genfun}
\prod_{0\leq i\leq 8}\frac{1}{1-x^{n(\La_i)}}=
1+x+3x^2+5x^3+10x^4+15x^5+27x^6+39x^7+63x^8+O[x^9]\end{equation}
\item[{\bf Step 4}] For each $\mu-\Delta(\mu)\delta\in\S(n+1)$, let
$B_\mu=\{\la\in\hat{P}_+(n): \mu-\Delta(\mu)\delta\in
A_\la\}$. Then
$$c_\mu=\sum_{\la\in B_\mu}c_\la.$$
\begin{remark} The formula in Step 3 is valid since
the level of a dominant weight $\la$ is determined by the
decomposition $\la=\sum_i M_i\La_i$ and the levels $n(\La_i)$ of
the fundamental weights $\La_i$ (see Definition \ref{leveldef}).
One identifies a level $n$
dominant weight with a partition of $n$ into parts whose sizes are
in the multi-set $\{n(\La_i)\}$, and standard combinatorics lead to Eq. \ref{genfun}.
For arbitrary $N$ the highest weight
module $V_\la$ appears in $V^{\otimes n(\la)}$ where the formula
for $n(\la)$ is given in \cite{Wenzl3}, Eq. (3.1) in case $N\not=9$.
However, his formula
breaks into three cases which depend on $k(\la)$ in a way that makes the
problem of constructing a generating function valid for all $N$ rather
complicated combinatorially.
\end{remark}
\end{enumerate}
As an application we compute the decompositions of the first few
$\M_n$:
\begin{eqnarray*}
 &\M_2\cong&V_{\La_0}\oplus V_{\La_2}\oplus
 V_{2\La_1}\\
&\M_3\cong&V_{3\La_1}\oplus 2 V_{\La_1+\La_2}\oplus 3
V_{\La_0+\La_1}\oplus V_{\La_3}\oplus 2
V_{\La_8-\frac{\delta}{2}}\\
&\M_4\cong&V_{4\La_1}\oplus V_{\La_4}\oplus
6V_{\La_0+2\La_1}\oplus 3V_{\La_2+2\La_1}\oplus
6V_{\La_7-\frac{\delta}{2}}\oplus \\
& &6V_{\La_0+\La_2}\oplus 3V_{2\La_0}\oplus
8V_{\La_1+\La_8-\frac{\delta}{2}}\oplus 3V_{\La_1+\La_3}\oplus
2V_{2\La_2}
\end{eqnarray*}

\section{Connections and Further Directions}\label{future}
\subsection{$E_N$ Series}
Wenzl introduces a generic labeling set, $\Gamma$, for
the dominant integral weights of $\g(E_N)$, $N\not=9$ consisting
of triples $(n,\mu,i)$ where $n\in\N$, $\mu$ a Young diagram with
$|\mu|\leq n$ and $i\in\{0,1,2\}$, subject to some further
conditions (see \cite{Wenzl3}, Section 2).  The labeling is
realized via a map $\Phi$ assigning an element of $\Gamma$ to each
integral dominant weight.  The ambiguity in the dominant weights
due to the null-root precludes extending $\Phi$ directly to the
excluded case; however, the set of integral dominant weights of
$\g(E_9)$ whose image under $\Phi$ is in $\Gamma$ is precisely the
set of initial weights!  Thus one sees that our submodule $\M_n$
must be the ``missing link'' replacing $V_{new}^{\otimes n}$
required to extend Wenzl's main combinatorial result
(\cite{Wenzl3}, Proposition 3.10) for the $E_N$ $N\geq 6$ series
to the $N=9$ case:
\begin{proposition}
Assume $N>n$.  Then the branching rules for $V_{new}\subset
V^{\otimes 2}_{new}\subset\cdots\subset V^{\otimes n}_{new}$ do
not depend on $N$.
\end{proposition}
This proposition implies that when $k<9$ and $N>k$ the
combinatorial formula given is Step 3 of the algorithm holds.

\subsection{Braid Representations}
For generic $q$, the tensor product rules for the quantum group
$U_q\g(E_N)$ are the same as those of the Kac-Moody algebra
$\g(E_N)$. Wenzl was also able to show that, for $N\not=9$, the
centralizer algebra of the corresponding $U_q\g(E_N)$-module
$\Vn_{new}$ is generated by the image of the braid group $B_n$
(acting by $R$-matrices) and one more operator called the
\emph{quasi-Pfaffian}.  It should be possible to extend this
result to the $N=9$ case using the quantum group version of the
modules $\M_n$ together with the specific knowledge of the
decomposition rules.

\subsection{Other Lie Types}
It may be possible to use the same approach to derive a similar
algorithm for decomposing the tensor powers of low-level highest
weight modules for any affine Kac-Moody algebra.  By defining the
submodules analogous to $\M_n$ one would just need to determine
the subset of maximal weights corresponding to the set $\Omega$ of
straight weights.


TITLE: Taylor's formula
QUESTION [1 upvotes]: Taylor's Formula
Write taylor's formula for $F(x,y)= \sin(x)\sin(y)$ using $a=0$, $b=0$, and $n=2$.
$$\sin(h)\sin(k)=hk−\frac 16h(h^2+3k^2)\cos\theta h\sin\theta k−\frac 16 k(3h^2+k^2)\sin\theta h\cos\theta k$$
That's the answer, I don't get how to get it though.
I don't understand how to use the taylor's formula to find this answer. I've tried and failed, I have no idea where $h$,$k$, and $\theta$ comes from. Is anyone able to help me out?

REPLY [0 votes]: It might be a slight overkill to use the general Taylor expansion in two variables for this particular example. In fact, since the function factors in two independent parts, you can proceed as follows:
$$\sin\left(h+\theta h\right)\sin\left(k+\theta k\right)=\left(\sin h\cos\theta h+\sin\theta h\cos h\right)\left(\sin k\cos\theta k+\sin\theta k\cos k\right)$$
$$=\left(\left(h-\frac{h^{3}}{6}+O\left(h^{5}\right)\right)\cos\theta h+\sin\theta h\left(1-\frac{h^{2}}{2}+O\left(h^{4}\right)\right)\right)\left(\left(k-\frac{k^{3}}{6}+O\left(k^{4}\right)\right)\cos\theta k+\sin\theta k\left(1-\frac{h^{2}}{2}+O\left(k^{4}\right)\right)\right)$$
Opening the brackets and gathering terms you should arrive at the same result. If this is not the kind of solution expected from you, it is still a good means of checking the answer


TITLE: How does the weak interaction sense the instability of a nucleus?
QUESTION [1 upvotes]: To reduce the disruptive energy, the weak interaction allows the number of neutrons to exceed that of protons (...) (via nasa.gov).

How did it know it needs to reduce the disruptive energy?
Example: Two electrons (elementary particles) coming together are repelled by the exchange of a photon (mediator). They've come too close that the force-carrier acted. This is easy to grasp.
For the beta decay / weak interaction / W boson, I can't quite understand it in similar intuitive terms. In other words, how does the instability of a nucleus (how far off the beta stability line it is) make it act on a single quark via the very short-ranged W boson?

This question stems from further reading after posting a related question (now deleted). I thought I needed to read more before asking basic questions, but I only found myself more confused at more basic levels. (That question was how does a rare W boson become not rare in elements like francium that beta decay very quickly.)

REPLY [1 votes]: There are basically two things to consider:

isospin, originally, when we did not know what neutrons and protons were made up of, this was the only QM characteristic to distuinguish between them. Now we know the structure of nucleons, but isospin remains. Now it is very important to understand that a neutron and a proton can align their isospin, thus (coming a little closer, where the strong force is even stronger) creating a more stable, lower energy QM system. This means, that the nucleus is trying to equalize the number of neutrons and protons. This is the only way to create a lowest energy state, most stable nucleus.
the strong force is very short ranged, wheres the EM force's range is infinite. When you have a big nucleus, consisting of a lot of nucleons, the strong force will not be able to balance the EM repulsion between protons. Now if you have a proton rich nucleus, that is too big, it will not be stable, the nucleus will try to go to a lower energy state, a more stable energy state by converting protons to neutrons, thus, because of isospin, the size of the whole nucleus will decrease too, and the neighboring nucleons will have a stronger (isospin aligned) bond, with lower total energy, a more stable system.

You are asking how the nucleus knows or feels, that it has to go to a lower energy state. In QM, it is all about probabilities. It is a misconception that the proton is stable. It is only stable if it is free. Inside the nucleus, the proton will decay. Neutrons are unstable, so they will decay too.
Now you might be asking about beta decay.

In nuclear physics, beta decay (β-decay) is a type of radioactive decay in which a beta particle (fast energetic electron or positron) is emitted from an atomic nucleus. For example, beta decay of a neutron transforms it into a proton by the emission of an electron accompanied by an antineutrino, or conversely a proton is converted into a neutron by the emission of a positron (positron emission) with a neutrino, thus changing the nuclide type.

You are saying that two electrons will repel, because they are repelled by a mediator. Now this mediator is a virtual photon. It is a mathematical description of how the EM field of the electron interacts with the other electron. In reality, we do not know how they interact. We have experiments, and the mathematical description of virtual photons fits the data.
Now in beta decay, the whole nucleus's instability does not act directly on the neutron or the proton to make it decay. It is the forces:

strong force attracts
EM force repels

Now the isospin matters too. The nucleus as a whole moves toward the lowest energy state possible. That is if the neutrons and protons number is equalized. And if a nucleus is too big, the strong force can't balance the EM repulsion of the protons. So you:

can't go too big, because big nuclei are unstable (strong vs EM force)
can't have too many protons or too many neutrons, they need to be balanced (isospin)

In QM, it is all about probabilities. The protons and neutrons inside the nucleus are unstable themselves, not just the whole nucleus. They will decay. Thus, they move toward a more stable nucleus.
The answer to your question is that the whole nucleus's unstability does not act on the quarks directly. It is the neutrons and protons themselves that are unstable inside the nucleus and will decay.


TITLE: Doubt in raising a power to a complex number
QUESTION [2 upvotes]: What's the value of $$i^{i^{i^{...}}}$$?
I tried to take log on both sides.
$x=i^x$
$\implies \log x=x \log i$
After this how can I solve this...
I am sorry, that I don't know the methods you are explaining... And more over that the question is in school level only..

REPLY [1 votes]: Such number $x$ statistifies $i^x=x$. This can be solved with Lambert W function. Doing so, gives:
$$x \approx -1.861743075013160441612498412-0.4107999688363923527986399873 i$$
$$x \approx 0.4382829367270321593816295045+0.3605924718713854895879688663 i$$
But we see that they are always positive, so it is the second one.


TITLE: How to show that $\mathbb{Q}$ and $\mathbb{Q}^2$ are elementarily equivalent?
QUESTION [2 upvotes]: I try to prove that $(\mathbb{Q},P)$ and $(\mathbb{Q}^2,P)$ are elementarily equivalent using Ehrenfeucht–Fraïssé games
$P(a,b,c)=True$  when $a+b=c$
$+$ on $\mathbb{Q}^2$ is defined as $(x,x')+(y,y')=(x+y,x'+y')$
$=$ on $\mathbb{Q}^2$ is defined as $(x,x')=(y,y')\iff x=y,x'=y' $
So, if I get a $q\in\mathbb{Q}$ I initially thought about choosing $(q,q)\in\mathbb{Q}^2$.
But as soon as  $(q,q')\in\mathbb{Q}^2$ with $q\neq q'$ is chosen, I lose.
Also the given $(q_1,q_2)$ the function I apply on both of them to get $q\in \mathbb{Q}$ cannot be symmetric, otherwise I won't have an answer for $(q_2,q_1)$.
But there are so many possibilities and nothing I tried is a winning strategy, can you please give me a hint (not a full proof I probably miss some property or function)?
Can this be further generalized for every set of numbers $A$ and $A^2$ or every field at least?

REPLY [2 votes]: First, let us notice that an infinite lasting game cannot have a winning strategy for Duplicator. It is a fact that if Duplicator has a winning strategy in an infinite lasting game between two countable structures, then they are isomorphic. We can easily see that these structures are not isomorphic, e.g. because the operation $+$ is definable in the language (its graph itself is in the language), and $(\mathbb Q,+)$ and $(\mathbb Q^2,+)$ are not isomorphic because they don't have equal dimensions as vector spaces over $\mathbb Q$ (this is also a fact: an additive homomorphism between vector spaces over $\mathbb Q$ is $\mathbb  Q$-linear).
But let us see this directly, because it gives a hint. Assume that Duplicator has a winning strategy in an infinite lasting game. Assume that Spoiler chooses $(1,0)$ in the first step, and Duplicator responds by $a$. Then Spoiler chooses $(0,1)$ in the second step, and Duplicator responds by $b$. Clearly $a,b\neq 0$, so we can write $a/b=m/n$, i.e. $na=mb$, for some integers $m,n$, and let us assume that $m,n>0$ (if not, one can make some obvious adjustments in the argument). In the next $n-1$ steps Spoiler chooses $2a,3a,\ldots,na$ respectively, and we see that in each of them Duplicator is forced to respond by $(2,0),(3,0),\ldots,(n,0)$. In the following $m-1$ steps Spoiler chooses $2b,3b,\ldots,mb$, and as before Duplicator is forced to respond by $(0,2),(0,3),\ldots,(0,m)$. But at this point Duplicator looses because he mapped $na=mb$ to two different elements $(n,0)$ and $(0,m)$. Thus Duplicator cannot have a winning strategy in an infinite lasting game.
But in order to prove elementary equivalence, Duplicator must have a winning strategy in $k$-steps games for every $k<\omega$. As the previous argument shows, the duration of the game, $k$, must play a role in Duplicator's strategy. E.g. in the previous argument if the game has less than $m+n+2$ steps, then we don't obtain the above contradiction (maybe we can obtain some other), so one part of Duplicator's strategy at the begging of the previous game should be to choose $b$ such that at least $m+n+2>k$ (but this is not the only part, of course).
The hint for the general solution would be the following. For a subset $A$ (of either $\mathbb Q$ or $\mathbb Q^2$), denote by $O_n(A)$ by recursively by: $$O_0(A)=\{0\}\cup A,\ \ \ \ \ O_{n+1}(A)=\{a+b,a-b,a/2\mid a,b\in O_n(A)\}.$$
(Here $0$ denotes the zero in the structure we are talking about, as well as the operations $+$,$-$, and $-/2$.) Basically, $O_n(A)$ is the set of elements which we can define by using $P$ in $n$ steps starting from $A$ (and formulae that make sense are $P(a,b,x)$, $P(b,x,a)$, $P(x,x,a)$). So, note that if we start with a finite set $A$, then $O_n(A)$ are also finite.
Assume that we are playing the $k$-steps game. Essentially, Duplicator's strategy is the following: at each step $i\leqslant k$, for already chosen $a_1,\ldots,a_{i-1}\in\mathbb Q$ and corresponding $f(a_1),\ldots,f(a_{i-1})\in\mathbb Q^2$, and by Spoiler given $a_i\in\mathbb Q$ (or $b_i\in\mathbb Q^2$, it doesn't really matter), Duplicator extends the mapping such that it has a unique lifting to an elementary bijection between $O_{k-i}(a_1,\ldots,a_i)$ and $O_{k-i}(f(a_1),\ldots,f(a_i))$.
Now try to prove that this is always possible (I hope it is true, I think that I see how to proceed, but I didn't write everything; finiteness of these sets is a key property).


TITLE: Separate real and imaginary part of $j \cos (z)$
QUESTION [1 upvotes]: Given the following expression
$$w = j \cos \left[ \displaystyle \frac{1}{n} \arccos \left( \frac{j}{\epsilon} \right) + \frac{m \pi}{n} \right] = j \cos (z)$$
(which is related to this question; $n,m \in \mathbf{Z}$, $\epsilon > 0 \in \mathbf{R}$), first of all I would like to express the real and imaginary part of $w$ respectively in terms of $\sinh$ and $\cosh$ of some appropriate angles. The aim is to obtain something like
$$j \cos(z) = A \sinh(u) + j B \cosh(v)$$
1) Is it possible?
I tried
$$j \cos (z) = j \frac{e^{jz} + e^{-jz}}{2} = \frac{e^{j\left( z + \frac{\pi}{2} \right)} + e^{- j\left( z - \frac{\pi}{2} \right)}}{2}$$
but up to now I don't see a connection with $\sinh$ and $\cosh$. Nevertheless:
2) if this could be a suitable way, how to proceed? Or are there other alternative ways?

REPLY [2 votes]: Set $z=x+iy$, where $x,y$ are real. Use the addition formula:
$$ \cos{(x+iy)} = \cos{x}\cos{iy}-\sin{x}\sin{iy}. $$
Now you have
$$ \cos{iy} = \cosh{y}, \quad \sin{iy} = i\sinh{y}, $$
which is easy to prove in the way you suggested. So
$$ i\cos{(x+iy)} = i(\cos{x}\cosh{y}-i\sin{x}\sinh{y}) = i\cos{x}\cosh{y}+\sin{x}\sinh{y}. $$


TITLE: About the series $\sum_{n\geq 0}\frac{1}{(2n+1)^2+k}$ and the digamma function
QUESTION [6 upvotes]: Let we provide a closed form for
$$ S_k = \sum_{n\geq 0}\frac{1}{(2n+1)^2+k} $$
for $k>0$ in terms of elementary functions. It is quite easy to check that $S_k$ can be computed in terms of the digamma function $\psi(x)$, but it is also true that:
$$ \int_{0}^{+\infty}\frac{\sin(mx)}{m}e^{-\sqrt{k}\,x}\,dx =\frac{1}{m^2+k}\tag{1} $$
and that, almost everywhere:
$$ \sum_{n\geq 0}\frac{\sin((2n+1) x)}{2n+1} = \frac{\pi}{4}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor}, \tag{2}$$
hence:
$$\begin{eqnarray*} S_k &=& \frac{\pi}{4}\int_{0}^{+\infty}(-1)^{\left\lfloor\frac{x}{\pi}\right\rfloor} e^{-\sqrt{k}\,x}\,dx = \frac{\pi}{4}\sum_{n\geq 0}(-1)^n \int_{n\pi}^{(n+1)\pi}e^{-\sqrt{k}\,x}\,dx \\&=&\frac{\pi}{4\sqrt{k}}\sum_{n\geq 0}(-1)^n\left(e^{-n\pi\sqrt{k}}-e^{-(n+1)\pi\sqrt{k}}\right)\\&=&\frac{\pi\left(1-e^{-\pi\sqrt{k}}\right)  }{4\sqrt{k}}\sum_{n\geq 0}(-1)^n e^{-n\pi\sqrt{k}}=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}.\tag{3} \end{eqnarray*}$$
Does this identity provide something interesting about special values of the digamma function?

REPLY [6 votes]: Your result just confirms already known special values of the digamma function.
One may recall that the digamma function admits the following series representation
$$\begin{equation}
\psi(x+1) = -\gamma - \sum_{k=1}^{\infty} \left( \frac{1}{n+x} -\frac{1}{n}
\right), \quad \Re x >-1, \tag1
\end{equation}
$$ where $\gamma$ is the Euler-Mascheroni constant. By partial fraction decomposition we have
$$
\begin{align}
\frac{1}{(2n+1)^2+k} &= \frac{i}{2\sqrt{k}}\left(\frac{1}{(2n+1)+i\sqrt{k}}-\frac{1}{(2n+1)-i\sqrt{k}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac{1}{n+\frac{1-i\sqrt{k}}{2}}\right)\\\\
&=\frac{i}{4\sqrt{k}}\left[\left(\frac{1}{n+\frac{1+i\sqrt{k}}{2}}-\frac1n\right)-\left(\frac{1}{n+\frac{1-i\sqrt{k}}{2}}-\frac1n\right)\right]
\end{align}
$$ then summing from $n=1$ to $+\infty$, using $(1)$, we get
$$
\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1-i\sqrt{k}}{2}+1\right)-\psi\left(\frac{1+i\sqrt{k}}{2}+1\right)\right)
$$ using
$$
\psi(x+1)=\psi(x)+\frac1x
$$ it gives the case $n=0$ to obtain
$$
\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2+k} =\frac{i}{4\sqrt{k}}\left(\psi\left(\frac{1}{2}-\frac{i\sqrt{k}}{2}\right)-\psi\left(\frac{1}{2}+\frac{i\sqrt{k}}{2}\right)\right)=\color{red}{\frac{\pi}{4\sqrt{k}}\cdot\frac{e^{\pi\sqrt{k}}-1}{e^{\pi\sqrt{k}}+1}}
$$ where we have used the result (6.3.12):
$$
\Im \psi\left(\frac{1}{2}+iy\right)=\frac{\pi}{2}\tanh (\pi y).
$$