Why have your resume in MyKad?
TO make it more convenient for job seekers, the Labour Department has proposed that soft copies of resumes be stored in the MyKad.
According to a report two Sundays ago, the department director-general Datuk Sheikh Yahya Sheikh Mohamed said the proposal had been made to the Human Resources Ministry, which would study the available technology and costs. The move is to relieve the burden of job seekers, especially school leavers.
While the idea is good, I wonder if it is practical, considering the fact that resumes need to be updated each time one changes jobs, acquires new skills or gains new paper qualifications.
If a self-updating feature is available, and the user is allowed to freely access the MyKad, then the authorities will have security issues to contend with.
Is the technology worth the investment when there are so many online networking sites that offer resume templates, and which can be easily accessed via any mobile or desktop web browser?
Some of these sites are not only created by professionals, but are as easy to use as the email.
A good example is Linked-In (www.linkedin.com), created nine years ago. It has 90 million users worldwide and an infrastructure few can equal. Not only does it provide resume templates, it also has a professional contact networking feature. It can be used to find jobs, people or business opportunities. You can even list jobs and seek potential candidates. Job seekers can check out the profile of the potential employer and see which of their contacts can introduce them to the vacancy.
Moreover, Linked-In is not the only site with all the bells and whistles that young job seekers today expect.
So, how relevant will the proposed resume feature in the MyKad be to the average modern job seeker, who is well-connected and technology-savvy?
That aside, there is little sense in using sophisticated -- and probably costly -- technology to perform a job as simple as the storing of a resume.
The MyKad should instead contain vital information such as its holder's medical history and allergies (like the Medic Alert band).
It could even have a geo-location of the holder's home or workplace, with widespread use of global positioning system (GPS) devices. The phone numbers of the card holder's next-of-kin would inarguably be useful in an emergency.
For resumes, just use a flash drive -- it's easier to carry and cheaper, too.
HEARD the one about a country bumpkin who was trying to cross a busy street downtown? Seeing the poor man in his predicament, a traffic policeman walked up to him and said: "Sir, there is a zebra crossing down the street."
The chap looked at the policeman, looked annoyed, and exclaimed: "Oh, yeah? I hope he's got better luck than me."
If you are smiling because you know what a zebra crossing is, I wonder if you recognise it when you see one.
Plentiful in the 1970s, zebra crossings are black and white (now yellow) stripes painted across roads with flashing amber lights stationed at either end.
They have no standard traffic lights facing oncoming traffic, but motorists are obliged to stop when someone is spotted on the crossing, or at the road shoulder waiting to cross.
The pedestrian crossings you see around the city today, including those at traffic light junctions, are not zebra crossings in the strict sense of the word.
In the United Kingdom, they are called pelican crossings -- pelican being a modified acronym for pedestrian light controlled (or pelicon). These have standard traffic lights mounted instead of flashing amber lights.
There is a button for you to press if you need to cross, but you have to wait for the light on your side to turn green before you should do so.
Some of these crossings are disabled-friendly, with chimes to alert the blind when it is safe to cross. Many have countdowns to let you know how much time you have to cross before the light for oncoming traffic goes green.
Compared with viaducts or overhead bridges, zebra or pedestrian crossings are more viable alternatives to connect walking pathways.
They can be built at a fraction of the cost of a bridge and in much shorter time.
They are also easier to maintain and more likely to be used by everyone than the pedestrian bridges or underpasses, many of which have become crime havens.
The only problem with pedestrian or zebra crossings is that many city motorists are like country bumpkins.
They do not recognise the crossings when they see one -- perhaps out of ignorance or apathy.
Take the zebra crossing at Sentul Pasar for instance.
Each morning, while driving to work, I see market-goers making their daily dash to cross the busy road despite using the crossing as they lug their baskets or with children in tow.
Motorists zip pass without batting an eyelid, let alone stop.
There have been many near-misses and I have watched startled pedestrians hurling verbal abuses at inconsiderate motorists.
Fortunately, I have not seen anyone hurt.
I have not seen any camera mounted at this zebra crossing or any pedestrian crossing around the city for that matter to record the misdeeds of errant motorists.
Because of this, brazen motorists continue to give scant regard to the rights of pedestrians on crossings.
In one of my trips to Johor, I saw a pedestrian crossing that can perhaps be an eye-opener for traffic authorities in the Klang Valley.
This crossing did not have flashing amber lights or signalised traffic lights. The only indicators of the crossing's location were two signboards placed several metres up and downstream of it.
But motorists obediently slow down when approaching the crossing even when no one was using it.
Humps high enough to slow down the most reckless speed demon have been put up on both sides of the crossing.
Someone must have been irked by speeding motorists in the area to have come up with the bumpy ride idea to teach errant motorists to slow down.
Perhaps the traffic managers in the city can learn a thing or two from this so that errant motorists learn to respect the rights of pedestrians at crossings.


Do you support Catholic Schools and the right of Catholics to send their children to them? Then we need you to make your voice heard.
The Government is making a critical decision on whether to overturn the admissions cap which prevents Catholic schools from allowing all Catholic pupils to attend. They have acknowledged that this policy discriminates against Catholics and promised to abandon it. Now we need you to tell them not to u-turn on their promise to Catholics. Please visit the Bishops' Conference website to find an online petition to the Government to urge them to drop their policy which bans new Catholic schools.


text:
Million Dollar Arm
20th Century Fox, A Million Ways to Die in the West, Blended, cameron diaz, Captain America: The Wnter Sldier, Disney Entertainment, Godzilla, Jon Hamm, Kate Upton, Leslie Mann, Million Dollar Arm, Neighbors, Rose Byrne, Seth Rogen, Sony Pictures, The Amazing Spider-Man 2, The Other Woman, Universal Pictures, Warner Bros. Pictures, X-Men: Days of Future Past, Zac Efron
Godzilla Climbs to Box Office Top Warner Bros. Pictures' Godzilla was off to a huge start last weekend (May 16-18) with a first place debut of $93.19 million. The highly anticipated and well received sci-fi action remake outpaced expectations, which ranged from $65-$80 million,
"Million Dollar Arm" Movie Review
Alan Arkin, Asif Maandvi, Bill Paxton, Britain's Got Talent, Craig Gillespie, Dinesh Patel, I Dreamed a Dream, J.B. Bernstein, Jerry Maguire, Jon Hamm, Lake Bell, Life of Pi, Madhur Mittal, Michael Mandt, Million Dollar Arm, Neil Mandt, Rinku Singh, Slumdog Millionaire, Suraj Sharma, Susan Boyle, The Million Dollar Arm, Tom House, Tom McCarthy
Million Dollar Arm Hits a Home Run In a last ditch effort to save his career as a sports agent, J.B. Bernstein concocts a scheme to find baseball's next great pitching ace. Hoping to find a young cricket pitcher he can turn into a major league baseball star, Bernstein

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}

The Yang-Baxter equation in its spectral parameter-dependent form has  well-known
applications in a variety of areas of mathematical physics, including exactly
solvable classical models of two-dimensional statistical mechanics and
integrable quantum systems.
The origins of the field trace back to the influential works by McGuire concerning
many-body systems with delta function interactions \cite{jim}, Yang's solution for interacting fermions \cite{cny}, and Baxter's solution of the eight-vertex model \cite{rodney}.
Since then numerous solutions have been obtained including those leading to  the Perk-Schultz model \cite{perkschultz}, the Andrews-Baxter-Forrester model \cite{abf}, the chiral Potts model \cite{cp1,cp2} and the Hubbard model \cite{shastry,usw} to name a few.

One systematic method for solving the
Yang-Baxter equation was found with the advent of affine quantum algebras, which
are deformations of the universal enveloping algebras of affine classical Lie
algebras \cite{Drinfeld1986,jimbo86}.
Through the Drinfeld double construction \cite{Drinfeld1986}, the affine quantum algebras can be seen to belong to a larger class of algebras which
are referred to as {\it quasi-triangular Hopf} algebras. The double construction provides a
universal prescription for constructing a quasi-triangular Hopf algebra,
denoted $D(H)$, from any Hopf algebra $H$ and its dual algebra $H^*$. Via this
construction there exists a canonical element in the tensor product algebra of
the Drinfeld double, $R\in D(H)\otimes D(H)$, known as the universal $R$-matrix.
This element provides an algebraic solution of the Yang-Baxter equation.
Thus for each matrix representation of $D(H)$, a matrix solution of the
Yang-Baxter equation is automatically obtained. Solutions of the Yang-Baxter
equation with spectral parameter arise whenever matrix representations of $D(H)$
can be parameterised by one or more continuous variables. In the context of affine
quantum algebras, the spectral parameter arises naturally by consideration of an
evaluation homomorphism from the affine quantum algebra to its non-affine
counterpart. In this manner the representations of the affine quantum algebras
which give solutions with spectral parameter are the loop
representations. In principle the evaluation homomorphism provides a route to
algebraic solutions of the Yang-Baxter equation with spectral parameter. However
it is a technically challenging task to undertake and has only been implemented
explicitly for some low rank cases \cite{zg94,bgz95}.

Often, a more tractable approach is to start with a constant matrix solution of the
Yang-Baxter equation and then ask the question of whether a spectral
parameter-dependent generalisation exists, a procedure which is commonly referred to as
{\it Baxterisation}. For example, in cases where the solution of the
Yang-Baxter equation without spectral parameter affords a representation of the
Hecke algebra, there exists a procedure for constructing the
associated spectral parameter-dependent solution \cite{j90}. This also applies in
cases where the constant solution gives rise to a representation of the
Birman-Wenzl-Murakami algebra \cite{cgx91}. For solutions associated with representations
of quantum algebras, a general prescription for Baxterising constant solutions of
the Yang-Baxter equation (subject to satisfying certain conditions) was developed
in \cite{zgb91,dgz94,dgz96} using the notion of the {\it tensor product graph} method.

An alternative route to
building solutions of the Yang-Baxter equation is in the framework of descendants.
The most notable instances were found by Bazhanov and Stroganov
\cite{BazhStrog1990}, who showed that the chiral Potts model is a descendant of the
six-vertex model, and Hasegawa and Yamada \cite{HaseYamada1990}, who showed that the
Kashiwara-Miwa model is a descendant of the zero-field eight-vertex model. In both
cases the connection was found by constructing an appropriate algebraic structure
through an $L$-operator. In \cite{BazhStrog1990} the algebraic structure is
closely related to that of the quantum algebra $U_q(sl(2))$ with $q$ a root of unity, and in
\cite{HaseYamada1990} the $L$-operator is expressed in terms of the Sklyanin algebra
\cite{sklyanin}. In this approach use of the $L$-operator allows for the construction of
descendants through higher dimensional representations of the algebra. It is known
\cite{Baxter2004} that the six-vertex and the zero-field eight-vertex models intersect
at the zero-field six-vertex model whose descendant is the Fateev-Zamolodchikov model
\cite{FateevZam1982b}, which is precisely the intersection between the chiral Potts and
 Kashiwara-Miwa models.

Another principal class of quasi-triangular
Hopf algebras is the class of Drinfeld doubles of finite group algebras.  These algebras have recently received attention in relation
to the description of anyonic symmetries in quantum systems,
specifically in terms of the braiding properties of anyonic quasiparticle excitations
which are described by constant solutions of the Yang-Baxter equation \cite{wildb98,k03,m03,nssfd08}. However these algebras
these have not been investigated in the context of solving the Yang-Baxter equation
with spectral parameter, apart from some preliminary investigations reported in \cite{DIL2006,dl09}.
Arguably the simplest family of Drinfeld double algebras for finite groups is that
associated with the dihedral groups, denoted $D_n$. The Drinfeld doubles of these
algebras, denoted $D(D_n)$, are finite-dimensional quasi-triangular Hopf algebras with
a finite number of irreducible representations. For $D(D_n)$
 all the irreducible representations are known \cite{DIL2006}. In the
case where $n$ is odd, the irreducible representations can only have dimensions 1, 2 or
$n$, whereas for $n$ even the dimensions are 1, 2 or $n/2$. The constant $R$-matrices
associated with the 2-dimensional representations can be Baxterised to reproduce the
six-vertex model at roots of unity \cite{DIL2006}.

Our goal below is to obtain spectral
parameter-dependent solutions of the Yang-Baxter equation associated with the
$n$-dimensional ($n$ odd) and $n/2$-dimensional ($n$ even) representations. The general
strategy we follow is to view these solutions as descendants of the six-vertex model,
by firstly determining the appropriate $L$-operators. Once this is achieved, the technical
aspects for determining the explicit form of the descendants is accommodated by adapting
the tensor product graph method developed for quantum algebras \cite{zgb91,dgz94,dgz96}
to the present case. It turns out that solutions we obtain for $n$ odd are limiting cases of the Fateev-Zamolodchikov models \cite{BazhPerk2009}. However we stress that the anticipated $U_q(sl(2))$ symmetry with $q$ a root of unity degenerates in this limit, with the $D(D_n)$ symmetry emerging.

\section{Preliminaries}

\noindent
We first define some notation which is used extensively throughout this article.
We often work
modulo $n \in \mathbb{N}$, and hence define the following map: given $a \in \mathbb{Z}$ then $0 \leq \overline{a} \leq n-1$ is the integer
which satisfies $a \equiv \overline{a} \modn$. We also use the following two delta functions:

\begin{equation*}
\delta^j_i =
	\begin{cases}
		1, & ~~ i = j, \\
		0, & ~~ i \neq j,
	\end{cases}
~~~~
\bar{\delta}_i^j =
	\begin{cases}
		1, & ~~ i \equiv j \modn, \\
		0, & ~~ i \not\equiv j \modn.
	\end{cases}
\end{equation*}

\noindent
We use $e_{i,j}$ to denote an elementary matrix in
$M_{n \times n}(\mathbb{C})$ whose indices are considered modulo $n$. We adopt the convention that $e_{0,0}$ corresponds to the matrix with an entry in the $n$th row and $n$th column.
These matrices obey
the relation
$$ e_{i,j}e_{k,l} = \bar{\delta}_{j}^{k}e_{i,l}. $$

\noindent
We use the following convention for the product symbol:
$$ \prod_{i=j}^{k} a_{i} = \left\{ \begin{array}{cc} 1, & k < j, \\ a_{j}a_{j+1}...a_{k}, & k \geq j. \end{array} \right. $$

\subsection{The Yang-Baxter equation with spectral parameter}
We will use three variants of the Yang-Baxter equation (YBE), which is a non-linear matrix equation in $\mbox{End }(V \tp V \tp V)$ for some vector space $V$. The first
of these forms we call the constant YBE, given by
$$ R_{12}R_{13}R_{23} = R_{23}R_{13}R_{12}, $$
where the subscripts of $R$ refer to which vector spaces the operator is
acting upon. That is, given
$$ R = \sum_{i} a_{i} \tp b_{i} \in \mbox{End } (V \tp V), $$
we have
$$ R_{12} = \sum_{i} a_{i} \tp b_{i} \tp I, \hspace{1cm} R_{13} = \sum_{i} a_{i} \tp I \tp b_{i}, \hspace{1cm} \mbox{etc.} $$
Here $I$ denotes the identity matrix. The second form is known as the spectral parameter-dependent YBE and is given by

\begin{equation} \label{YBE1}
 R_{12}(x)R_{13}(xy)R_{23}(y) = R_{23}(y)R_{13}(xy)R_{12}(x),
\end{equation}

\noindent
where $x,y \in \mathbb{C}$.
Given a parameter-dependent solution, $R(z)$, constant solutions are
recovered when we take the limits $z \rightarrow 0,1$ and $\infty$. We
refer to invertible solutions of either form as {\it $R$-matrices}.
As this article deals primarily with
the parameter-dependent YBE we shall henceforth refer to it simply as the
Yang-Baxter equation.

\

\noindent
Now suppose $r(z)\in \mbox{End }(V \tp V)$ and $R(z)\in \mbox{End }(W \tp W)$ are $R$-matrices
with $\dim V < \dim W$. We describe $R(z)$ as a {\it descendant} of $r(z)$ provided
that there exists a non-trivial invertible operator $L(z) \in \mbox{End }(V \tp W)$ (referred to as an {\it $L$-operator})
which satisfies
\begin{equation}
	\label{eqnrLL}
	r_{12}(xy^{-1})L_{13}(x)L_{23}(y) = L_{23}(y)L_{13}(x)r_{12}(xy^{-1}), \quad \forall x,y \in \mathbb{C},
\end{equation}
and
\begin{equation}
  \label{eqnLLR}
  L_{12}(x)L_{13}(y)R_{23}(x^{-1}y) = R_{23}(x^{-1}y)L_{13}(y)L_{12}(x), \quad \forall x,y \in \mathbb{C}.
 \end{equation}

In this article we start with an $R$-matrix associated with the zero-field
six-vertex model and use the Drinfeld doubles of dihedral groups to generate
$L$-operators and descendants.

The third variant of the YBE used in this paper is
$$ \check{R}_{12}(x)\check{R}_{23}(xy)\check{R}_{12}(y) = \check{R}_{23}(y)\check{R}_{12}(xy)\check{R}_{23}(x). $$
Given a matrix solution $R(z)$ to Equation \eqref{YBE1}, a solution to this form of the YBE is given by
$$
\check{R}(z) = P\, R(z),
$$
where
$P$ is the permutation operator, i.e.
$$ P = \sum_{i,j} e_{i,j} \tp e_{j,i}. $$

\subsection{Drinfeld doubles of finite groups}

Next we recall relevant results from the Drinfeld double construction applied to finite group algebras, following \cite{DPR1990,Gould1993}. Given a group $G$, with identity $e$, we consider the algebra $\mathbb{C} G$, whose basis
vectors are the elements of the group. The multiplication and unit of $\mathbb{C} G$
are inherited from the group in the natural way. We equip $\mathbb{C} G$ with a coproduct, counit
and antipode defined respectively by
$$ \Delta(g) = g \tp g, \hspace{1cm} \epsilon(g) = 1 \hspace{0.7cm} \mbox{and} \hspace{0.7cm} \gamma(g) = g^{-1}, \hspace{1cm} \forall g \in G. $$
With these maps $\mathbb{C} G$ becomes a Hopf algebra. We next consider the dual
space of $\mathbb{C} G$,
$$ (\mathbb{C} G)_{o} = \mathbb{C} \{g^{*} | g \in G \}. $$
The multiplication and unit are given, respectively, by
$$ g^{*}h^{*} = \delta_{g}^{h} \hspace{1cm} \mbox{and} \hspace{1cm} u(1) = \sum_{g\in G}g^{*}. $$
The costructure and antipode are defined by
$$ \Delta(g^{*}) = \sum_{h\in G} (hg)^{*} \tp (h^{-1})^{*}, \hspace{1cm} \epsilon(g^{*}) = \delta_{g}^{e} \hspace{0.7cm} \mbox{and} \hspace{0.7cm} \gamma(g^{*}) = (g^{-1})^{*}, \hspace{1cm} \forall g \in G. $$
Under these maps $(\mathbb{C} G)_{o}$ is also a Hopf algebra. Using the dual and the
original algebra we can construct the Drinfeld double,
$$ D(G) = \mathbb{C} \{gh^{*} | g,h \in G \}. $$
We impose the relation
$$ h^{*}g = g(g^{-1}hg)^{*} $$
and adopt the required remaining structure from $\mathbb{C} G$ and $(\mathbb{C} G)_{o}$.
Furthermore it is known that $D(G)$ is a quasi-triangular Hopf algebra, containing the canonical element
$$ \mathcal{R} = \sum_{g \in G} g \tp g^{*}. $$
This element satisfies the following relations:
$$ \begin{array}{rcll}
  \mathcal{R} \Delta(a) & = & \Delta^{T}(a) \mathcal{R}, & \forall a \in D(G), \\
  (\Delta \tp \mbox{id}) \mathcal{R} & = &  \mathcal{R}_{13}\mathcal{R}_{23}, \\
  (\mbox{id} \tp \Delta) \mathcal{R} & = &  \mathcal{R}_{13}\mathcal{R}_{12}, \\
\end{array} $$
where $\Delta^T$ denotes the opposite coproduct.  Consequently matrix representations of $\mathcal{R}$ provide solutions of the constant Yang-Baxter equation.

In this paper we use the dihedral group $D_n$, which is the symmetry group of a regular polygon with $n$ vertices.  That is,
$$ D_{n} = \{\sigma, \tau | \sigma^{n} = \tau^{2} = e, \sigma \tau \sigma = \tau \}, $$
and has order $2n$. For ease of calculation we divide the description of the representation theory into different cases.

\subsection{Representations of $D(D_{n})$} \label{r

\subsubsection{The case when $n$ is odd}
We first consider $D(D_{n})$ for the case where $n$ is odd. As stated in \cite{DIL2006}
the representations of the double of a group are naturally partitioned by the
conjugacy classes of the group. For these representations we consider $w \in \mathbb{C}$
to be a primitive $n$th root of unity.
Then the irreducible representations (irreps) are given in Table \ref{oddreps} below:

\begin{table}[ht]

$$
\begin{array}{|c|c|c|c|c|} \hline
 \mbox{Irrep }\pi & \mbox{Constraints} & \pi(\sigma) & \pi(\tau) & \pi(g^*),\; g \in D_n \\ \hline
	\pi_1^\pm & & 1 & \pm 1 & \delta^e_g \\[2pt]
	\pi_2^{(0,k)} & 1 \leq k \leq \frac{n-1}{2} & \begin{pmatrix} w^k & 0 \\ 0 & w^{-k} \end{pmatrix}
		& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & \begin{pmatrix} \delta^e_g  & 0 \\ 0 & \delta^e_g  \end{pmatrix} \\
		[12pt]
	\pi_2^{(l,k)} & \begin{array}{c} 1 \leq l \leq \frac{n-1}{2}, \\ 0 \leq k \leq n-1 \end{array} &
		\begin{pmatrix} w^k & 0 \\ 0 & w^{-k} \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} &
		\begin{pmatrix} \delta^{\sigma^l}_g & 0 \\ 0 & \delta^{\sigma^{-l}}_g \end{pmatrix} \\ [12pt]
	\pi_n^\pm & & \sum_{i=1}^n e_{i+1, i} & \pm \sum_{i=1}^n e_{i, 2-i} & \delta^{\sigma^{2j} \tau}_g e_{j+1, j+1} \\
[2pt] \hline
\end{array}$$ \vspace{-6mm}
\caption{\label{oddreps} The irreps of $D(D_n)$ when $n$ is odd.}
\end{table}

\noindent Here $\pi_1^\pm, \pi_2^{(a,b)}$ and $\pi_n^\pm$ have dimensions 1, 2 and $n$ respectively,
and their associated modules are denoted $V_1^\pm, V_2^{(a,b)}$ and $V_n^\pm$.

Also associated with these representations are the solutions of the Yang-Baxter
equation arising from the canonical element. If we apply a two-dimensional irrep
we find the $R$-matrix
$$ (\pi_{2}^{(l,k)} \tp \pi_{2}^{(l,k)}) \mathcal{R} =
 \left(
	\begin{array}{cccc}
		w^{kl} & 0 & 0 & 0\\
		0 & w^{-kl} & 0 & 0\\
		0 & 0 & w^{-kl} & 0\\
		0 & 0 & 0 & w^{kl}
	\end{array}
\right). $$
It was shown in \cite{DIL2006} that this constant solution leads to the
parameter-dependent solution
\begin{equation}
r(z) =
	\left(
	\begin{array}{cccc}
		w^{2kl}z^{-1} - w^{-2kl}z & 0 & 0 & 0\\
		0 & z^{-1} - z & w^{2kl} - w^{-2kl} & 0\\
		0 & w^{2kl} - w^{-2kl} & z^{-1} - z & 0\\
		0 & 0 & 0 & w^{2kl}z^{-1} - w^{-2kl}z
	\end{array}
\right). \label{rznodd}
\end{equation}
This $R$-matrix corresponds to the six-vertex model with zero-field. Also of interest
is the representation of the canonical element using the $n$-dimensional irreps,
$$ (\pi_{n}^{\pm} \tp \pi_{n}^{\pm}) \mathcal{R} = \pm \sum_{i,j=0}^{n-1}  e_{i+j,i-j} \tp e_{i,i}. $$

In our construction of the descendant of $r(z)$ we will use a linear combination
of projection operators. It is known \cite{Gould1993} that there exist operators
which project $D(D_{n})$ onto its ideals. These projection operators are
defined by
$$E_{\alpha} = \frac{d[\alpha]}{|G|} \sum_{g,h \in G}
\chi_{\alpha}(h^{*}g^{-1})gh^{*}, $$
where $\alpha$ is an irrep, $d[\alpha]$ its dimension and $\chi_{\alpha}$ is the
group character defined by
$$
\chi_\alpha(a) = \mbox{tr } \pi_\alpha(a),\ \ \forall a\in D(D_n).
$$
We also consider the projection operators
\begin{equation}
	p_n^{\alpha} = (\pi_{n}^\pm \tp \pi_{n}^\pm) \Delta (E_{\alpha}).
\label{eqnalgebraicprojection2}
\end{equation}
These operators project from $\pi_{n}^\pm \tp
\pi_{n}^\pm$ onto copies of the irrep associated with $\alpha$ in the
decomposition of $\pi_{n}^\pm \tp \pi_{n}^\pm$.
Note that we consider $\pi_n^+\tp\pi_n^+$ and $\pi_n^-\tp\pi_n^-$ together
since
$$ (\pi_{n}^{+} \tp \pi_{n}^{+})\Delta(a) = (\pi_{n}^{-} \tp \pi_{n}^{-})\Delta(a), \hspace{1cm} \forall a \in D(D_{n}). $$
We calculate $p_n^\alpha$ explicitly, using the expression given in Equation (\ref{eqnalgebraicprojection2}), obtaining
\begin{equation*}
p_n^{\alpha}
 =  \frac{d[\alpha]}{2n} \sum_{g \in D_{n}}
\sum_{i,j = 0}^{n-1}\chi_{\alpha}((\sigma^{2j})^{*}g^{-1})
\pi_{n}^\pm(g) e_{i-j,i-j} \tp \pi_{n}^\pm(g) e_{i,i}.
\end{equation*}
From this we can see that the only irreps with non-zero projection
operators will be associated with the conjugacy classes $\{e\}$ and
$\{\sigma^{i}, \sigma^{-i} \}$ for $ 1 \leq i \leq \frac{n-1}{2}$.
This implies that $\pi_{n}^\pm \tp \pi_{n}^\pm$ decomposes only into one
and two-dimensional irreps.

For convenience, we slightly modify our notation for the irreps. Instead of
using $\pi_1^+$ and $\pi_2^{(l,k)}$, we use ordered pairs corresponding only to
irreps that appear in the direct sum decomposition of $\pi_n^\pm\tp\pi_n^\pm$.
The correspondence is summarised in Table \ref{alphaodd}:

\begin{table}[ht]
$$ \begin{array}{|c|c|c|c|} \hline
  \alpha & \mbox{Irrep} & \mbox{Constraint on } a & \mbox{Constraint on } b \\ \hline
  (0,0) & \pi_{1}^{+} & & \\
  (0,b) & \pi_{2}^{(0,2b)} & & 1 \leq b \leq \floor{\frac{n-1}{4}} \\
  (0,b) & \pi_{2}^{(0,n-2b)} & & \floor{\frac{n+3}{4}} \leq b \leq \frac{n-1}{2} \\
  (a,b) & \pi_{2}^{(2a,\overline{2b})} & 1 \leq a \leq \floor{\frac{n-1}{4}} & 0 \leq b \leq n-1 \\
  (a,b) & \pi_{2}^{(n-2a,\overline{n-2b})} & \floor{\frac{n+3}{4}} \leq a \leq \frac{n-1}{2} & 0 \leq b \leq n-1 \\ \hline \end{array} $$ \vspace{-6mm}
\caption{\label{alphaodd} The ordered pairs labelling the irreps for $D(D_n)$, $n$ odd.}
\end{table}
\noindent Here $\floor{a}$ denotes the floor of $a$.
Then the projection operator for irrep $\alpha = (a,b)$ is given by
$$ p_n^{\alpha} = \frac{c^\alpha}{n} \sum_{i,j = 0}^{n-1}[ w^{2bj} e_{i+a+j,i+a} \tp e_{i+j,i} + w^{-2bj} e_{i-a+j,i-a} \tp e_{i+j,i}]$$

\noindent where

$$c^\alpha = \begin{cases}
\frac{1}{2}, &\quad \alpha = (0,0), \\
1, & \quad \alpha \neq (0,0).
\end{cases}$$

We have calculated non-zero projection operators for $(n^2-1)/2$ two-dimensional irreps and one 1-dimensional irrep.
Using a counting argument, it is clear that these provide a complete decomposition of $\pi_n^\pm \tp \pi_n^\pm$ into irreps for $n$ odd.

\subsubsection{The case when $n$ is even \label{sub2-3-2}}
We similarly catalogue the irreps of $D(D_{n})$ for the case when $n$ is even.
We set $n=2m$ and let $w \in \mathbb{C}$ be a primitive $2m$th
root of unity.

\begin{table}[ht]
$$
\begin{array}{|c|c|c|c|c|} \hline
 \mbox{Irrep }\pi & \mbox{Constraints} & \pi(\sigma) & \pi(\tau) & \pi(g^*),\; g \in D_{2m} \\ \hline
	\pi_{1,e}^{(a,b)} & a,b \in \{0,1\} & (-1)^b & (-1)^a & \delta^e_g \\[5pt]
	\pi_{1,\sigma^m}^{(a,b)} & a, b \in \{0,1\} & (-1)^b & (-1)^a & \delta^{\sigma^m}_g \\[5pt]
	\pi_2^{(0,k)} & 1 \leq k < m & \begin{pmatrix} w^k & 0 \\ 0 & w^{-k} \end{pmatrix}
		& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} & \begin{pmatrix} \delta^e_g  & 0 \\ 0 & \delta^e_g  \end{pmatrix} \\
		[12pt]
	\pi_2^{(m,k)} & 1 \leq k < m & \begin{pmatrix} w^k & 0 \\ 0 & w^{-k} \end{pmatrix}
		& \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} &
		\begin{pmatrix} \delta^{\sigma^m}_g  & 0 \\ 0 & \delta^{\sigma^m}_g  \end{pmatrix} \\[12pt]
	\pi_2^{(l,k)} & \begin{array}{c} 1 \leq l \leq m-1, \\ 0 \leq k \leq 2m-1 \end{array} &
		\begin{pmatrix} w^k & 0 \\ 0 & w^{-k} \end{pmatrix} & \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} &
		\begin{pmatrix} \delta^{\sigma^l}_g & 0 \\ 0 & \delta^{\sigma^{-l}}_g \end{pmatrix} \\ [12pt]
	\pi_{m,\tau}^{(a,b)} & a, b \in \{0,1\} & \sum_{i=1}^m (-1)^{a\delta_{i}^{1}}e_{i,i-1} & \sum_{i=1}^m
		(-1)^{a\delta_{i}^{1}+b}e_{i,2-i} &	\delta^{\sigma^{2k}\tau}_g e_{k+1, k+1} \\ [5pt]
	\pi_{m, \sigma \tau}^{(a,b)} & a, b \in \{0,1\} & \sum_{i=1}^m (-1)^{a\delta_{i}^{1}}e_{i,i-1} & (-1)^b \sum_{i=1}^m e_{i,1-i} &
		\delta^{\sigma^{2k+1}\tau}_g e_{k+1, k+1}  \\
[2pt] \hline
\end{array}$$ \vspace{-6mm}
\caption{The irreps of $D(D_{2m})$.}
\end{table}

We again concern ourselves with the two-dimensional representations applied to
the canonical element. We find the constant solution
$$ (\pi_{2}^{(l,k)} \tp \pi_{2}^{(l,k)}) \mathcal{R} =
 \left(
	\begin{array}{cccc}
		w^{kl} & 0 & 0 & 0\\
		0 & w^{-kl} & 0 & 0\\
		0 & 0 & w^{-kl} & 0\\
		0 & 0 & 0 & w^{kl},
	\end{array}
\right), $$
which leads to the parameter-dependent solution
$$ r(z) =
	\left(
	\begin{array}{cccc}
		w^{2kl}z^{-1} - w^{-2kl}z & 0 & 0 & 0\\
		0 & z^{-1} - z & w^{2kl} - w^{-2kl} & 0\\
		0 & w^{2kl} - w^{-2kl} & z^{-1} - z & 0\\
		0 & 0 & 0 & w^{2kl}z^{-1} - w^{-2kl}z
	\end{array}
\right). $$
This result only differs to that of $D(D_{n})$ where $n$ is odd by the possible
choices of the root of unity.
Note also the similarity in the representation of the canonical element, given by
$$ (\pi_{m} \tp \pi_{m})\mathcal{R} = (-1)^{b} \sum_{i,j=0}^{m-1} e_{i+j,i-j} \tp e_{i,i}, $$
when $\pi_{m}$ is either $\pi_{m,\tau}^{(0,b)}$ or $\pi_{m,\sigma\tau}^{(0,b)}$ for $b\in \{0,1\}$.

We again use Equation $(\ref{eqnalgebraicprojection2})$ to derive our projection
operators. We first look at the case when $m=n/2$ is odd. We find that
$$
p_m^{\alpha} = \frac{d[\alpha]}{4m} \sum_{g \in G} \sum_{i,j=0}^{m-1}
\chi_{\alpha}((\sigma^{2j})^{*}g^{-1}) \pi_{m}(g) e_{i-j,i-j} \tp \pi_{m}(g)
e_{i,i},
$$
where $\pi_{m}$ is any one of the $m$-dimensional irreps.
As in the case of $D(D_n)$ for odd $n$, we introduce an ordered pair notation for the irreps
that appear in the direct sum decomposition of $\pi_{m,\tau}^{(0,b)} \tp \pi_{m,\tau}^{(0,b)}$ and $\pi_{m,\sigma\tau}^{(0,b)} \tp \pi_{m,\sigma\tau}^{(0,b)}$ for $b\in \{0,1\}$.
The correspondence is summarised in Table \ref{alpha2odd}:
\begin{table}[ht]
$$ \begin{array}{|c|c|c|c|} \hline
  \alpha & \mbox{Irrep} & \mbox{Constraint on } a & \mbox{Constraint on } b \\ \hline
  (0,0) & \pi_{1,e}^{+} & & \\
  (0,b) & \pi_{2}^{(0,2b)} & & 1 \leq b \leq \frac{m-1}{2} \\
  (a,b) & \pi_{2}^{(2a,2b)} & 1 \leq a \leq \frac{m-1}{2} & 0 \leq b \leq m-1 \\ \hline
\end{array} $$ \vspace{-6mm}
\caption{\label{alpha2odd}The ordered pairs labelling the irreps for $D(D_{2m})$, $m$ odd.}
\end{table}

Then the projection operator associated with irrep $\alpha = (a,b)$ is given by
$$ p_m^{\alpha} = \frac{c^\alpha}{m} \sum_{i,j=0}^{m-1}
[w^{2bj} e_{i+a+j,i+a} \tp e_{i+j,i} +  w^{-2bj} e_{i-a+j,i-a} \tp e_{i+j,i}]$$

\noindent where $$c^\alpha = \begin{cases}
\frac{1}{2}, &\quad \alpha = (0,0), \\
1, & \quad \alpha \neq (0,0).
\end{cases}$$

We note that these projection operators here match those arising in the case of $D(D_{n})$ where is $n$ odd.

Similarly, when $m=n/2$ is even we have the projection operators
$$ p_m^{\alpha} = \frac{d[\alpha]}{4m} \sum_{g \in G} \sum_{i,j=0}^{m-1}
\chi_{\alpha}((\sigma^{2j})^{*}g^{-1}) \pi_{m}(g) e_{i-j,i-j} \tp \pi_{m}(g)
e_{i,i}, $$
for any $m$-dimensional irrep $\pi_m$.
The ordered pair notation for the irreps occuring in the decomposition of
$\pi_{m,\tau}^{(0,b)} \tp \pi_{m,\tau}^{(0,b)}$ for $b\in \{0,1\}$ is given in Table \ref{case i}, whereas
those occuring in the decomposition of $\pi_{m,\sigma\tau}^{(0,b)} \tp \pi_{m,\sigma\tau}^{(0,b)}$ for $b\in \{0,1\}$ are given in
Table \ref{case ii}.
\begin{table}[ht]
$$ \begin{array}{|c|c|c|c|} \hline
  \alpha & \mbox{Irrep} & \mbox{Constraint on } a & \mbox{Constraint on } b \\ \hline
  (0,b\frac{m}{2}) & \pi_{1,e}^{(0,b)} & & b \in \{0,1\} \\
  (\frac{m}{2},b\frac{m}{2}) & \pi_{1,\sigma^{m}}^{(0,b)} & & b \in \{0,1\} \\
  (0,b) & \pi_{2}^{(0,2b)} & & 1 \leq b \leq \frac{m}{2} - 1 \\
  (\frac{m}{2},b) & \pi_{2}^{(m,2b)} & & 1 \leq b \leq \frac{m}{2} - 1 \\
  (a,b) & \pi_{2}^{(2a,2b)} & 1 \leq a \leq \frac{m}{2} - 1 & 0 \leq b \leq m-1 \\ \hline \end{array} $$
\vspace{-6mm}
\caption{The ordered pairs labelling the irreps for $D(D_{2m})$, $m$ even, case $(i)$. \label{case i}}
\end{table}

\begin{table}[ht]
$$ \begin{array}{|c|c|c|c|} \hline
  \alpha & \mbox{Irrep} & \mbox{Constraint on } a & \mbox{Constraint on } b \\ \hline
  (0,b\frac{m}{2}) & \pi_{1,e}^{(b,b)} & & b \in \{0,1\} \\
  (\frac{m}{2},b\frac{m}{2}) & \pi_{1,\sigma^{m}}^{(b,b)} & & b \in \{0,1\} \\
  (0,b) & \pi_{2}^{(0,2b)} & & 1 \leq b \leq \frac{m}{2} - 1 \\
  (\frac{m}{2},b) & \pi_{2}^{(m,2b)} & & 1 \leq b \leq \frac{m}{2} - 1 \\
  (a,b) & \pi_{2}^{(2a,2b)} & 1 \leq a \leq \frac{m}{2} - 1 & 0 \leq b \leq m-1 \\ \hline \end{array} $$
\vspace{-6mm}
\caption{The ordered pairs labelling the irreps for $D(D_{2m})$, $m$ even, case $(ii)$. \label{case ii}}
\end{table}

We obtain the same projection operators irrespective of which of the two cases we consider.  In particular, the
projection operator corresponding to the irrep $\alpha = (a,b)$ is

$$ p_m^{\alpha} = \frac{c^\alpha}{m} \sum_{i,j=0}^{m-1}
[w^{2bj} e_{i+a+j,i+a} \tp e_{i+j,i} + w^{-2bj} e_{i-a+j,i-a} \tp  e_{i+j,i}] $$

\noindent where $$c^\alpha = \begin{cases}
\frac{1}{2}, &\quad a \in \{0,\frac{m}{2}\} \mbox{ and } b \in \{0, \frac{m}{2}\}, \\
1, & \quad \mbox{otherwise}.
\end{cases}$$

Again, a counting argument verifies that these are the only non-zero projection operators.
As the projection operators are the same for the different $m$-dimensional irreps, we can
without loss of generality consider only
$\pi_{m,\tau}^{(0,0)}.$

\section{Descendants associated with $D(D_{n})$ - $n$ odd}
In this section we construct a family of solutions of the Yang-Baxter equation
using $D(D_{n})$, where $n$ is odd and $n>2$.

\subsection{Construction of the $L$-operator\label{ssecLopDnodd}}
We begin by constructing an operator $L(z) \in {\rm End}(V_{2}^{(l,k)} \tp V_{n}^\pm)$. We observe that
$V_{2}^{(l,k)} \tp V_n^\pm \cong V_n^+ \oplus V_n^-$, suggesting that the $L$-operator will be a linear
combination of two intertwining operators with one parameter-dependent coefficient (after scaling). Hence we adopt the following
ansatz for the $L$-operator:

$$ L(z) = (\pi_{2}^{(l,k)} \tp \pi_{n}^{\pm}) [\mathcal{R} + h(z) (\mathcal{R}^{T})^{-1}] $$

\noindent where $(\mathcal{R}^{T})^{-1} = \underset{g \in G}{\sum} g^* \tp g^{-1}$. Applying the representations leads to the following expression for $L(z)$:
\begin{equation*}
L(z) = \sum_{i=0}^{n-1}\left \{ (w^{2(i-1)k}e_{1,2}
+ w^{-2(i-1)k}e_{2,1}) \tp e_{i,i} + h(z) \left[ e_{1,1} \tp e_{i-l,i}
 + e_{2,2} \tp e_{i+l,i} \right] \right\}.
\end{equation*}
Performing a basis transformation which leaves $r(z)$ invariant, we obtain
$$ L(z) = \sum_{i=0}^{n-1}\left \{ (w^{2ik}e_{1,2} + w^{-2ik}e_{2,1}) \tp e_{i,i}
+ h(z) \left[ e_{1,1} \tp e_{i-l,i} + e_{2,2} \tp e_{i+l,i} \right] \right\}. $$

Substituting $r(z)$ and $L(z)$ into Equation $(\ref{eqnrLL})$,
we find only one constraint on $h(z)$, namely
$$ h(x)y - h(y)x = 0, $$
which has the solution
$$ h(z) = Cz, $$
for any $C \in \mathbb{C}$. We are free to rescale the parameter $z$ without affecting the descendants given in the next section, so without
loss of generality we can choose $C = 1$. We have therefore shown the following:
\begin{prop}
The $L$-operator given explicitly by
\begin{eqnarray}
 L(z) = \sum_{i=0}^{n-1}\left \{ (w^{2ik}e_{1,2} + w^{-2ik}e_{2,1}) \tp e_{i,i}
+ z \left[ e_{1,1} \tp e_{i-l,i} + e_{2,2} \tp e_{i+l,i} \right] \right\},
\label{lop}
\end{eqnarray}
and the $r(z)$ given in Equation (\ref{rznodd}) together satisfy Equation (\ref{eqnrLL}).
\end{prop}
It is important to comment that, up to a basis transformation, the $L$-operator (\ref{lop}) is a particular limit
of the general $L$-operator discussed in \cite{BazhStrog1990} in relation to the chiral Potts model. However, the
associated $U_q(sl(2))$ structure with $q^3=1$ described in \cite{BazhStrog1990} is lost in this limit.
Specifically,  Equation (2.9) in \cite{BazhStrog1990} does not hold. Amongst the defining relations of the generalised
$U_q(sl(2))$ algebra this means, in particular, that
$$\left[e,\,f\right]=0 $$
where $e$ and $f$ are the raising and lowering generators respectively. The above relation is indicative of the fact
that the $U_q(sl(2))$ structure degenerates in this limit. Instead we find that the symmetry algebra of  $D(D_n)$
emerges when the $L$-operator is given by Equation (\ref{lop}).

\subsection{Construction of the descendants\label{props}}
Here we put forth a predicted form of a descendant, $R(z)$, and determine the
constraints on this form. We then impose additional constraints enforcing that
the descendant inherits certain properties from representations of the canonical element
from $D(D_{n})$. For convenience we shall use an alternate form of Equation $(\ref{eqnLLR})$:
\begin{equation}
	\check{R}_{23}(xy^{-1})L_{13}(x)L_{12}(y)
= L_{13}(y)L_{12}(x)\check{R}_{23}(xy^{-1}), \label{eqnLLRc}
\end{equation}
where
$$ R(z) = P\check{R}(z)$$
and $L(z)$ is given by Equation \eqref{lop}.  As the descendants $\check{R}(z)$ commute with the action of the coproduct,
we assume they are of the form
\begin{equation}
\check{R}(z) = \sum_{\alpha \in S} f_{\alpha}(z) p^{\alpha}, \label{Rcheck}
\end{equation}
where $p^{\alpha}$ are the projection operators previously calculated,
$f_{\alpha}(z)$ are continuous functions and $S$ is the set of ordered pairs
which correspond to non-zero projection operators. For convenience we rescale our projection operators, henceforth using
\begin{equation*}
\tilde{p}^{\alpha}
 = \sum_{i,j = 0}^{n-1}[ w^{2jb} e_{i+a+j,i+a} \tp e_{i+j,i} + w^{-2bj} e_{i-a+j,i-a} \tp e_{i+j,i}], \quad \alpha = (a,b) \in S
\end{equation*}
Applying
$\tilde{p}^{\alpha}_{23}$ and $\tilde{p}^{\beta}_{23}$ to the left- and right-hand sides
respectively of Equation $(\ref{eqnLLRc})$, we find
\begin{equation}
f_{\alpha}(xy^{-1}) \tilde{p}^{\alpha}_{23} L_{13}(x)L_{12}(y) \tilde{p}_{23}^{\beta}
= f_{\beta}(xy^{-1}) \tilde{p}^{\alpha}_{23} L_{13}(y)L_{12}(x) \tilde{p}_{23}^{\beta}.
\label{eqnpLLp}
\end{equation}
We calculate that
$$
L_{13}(x)L_{12}(y) = e_{11} \tp A(x,y) + e_{12} \tp B(x,y) + e_{21} \tp C(x,y) + e_{22} \tp D(x,y),
$$
where
\begin{eqnarray}
	A(x,y) & = & \sum_{i,j=0}^{n-1} w^{2(j-i)k}  e_{i,i} \tp e_{j,j} + xy\ e_{i-l,i} \tp e_{j-l,j}, \nonumber\\
	B(x,y) & = & \sum_{i,j=0}^{n-1} x w^{2ik} e_{i,i} \tp e_{j-l,j} + y w^{2jk} e_{i+l,i} \tp e_{j,j}, \nonumber\\
	C(x,y) & = & \sum_{i,j=0}^{n-1} y w^{-2jk} e_{i-l,i} \tp e_{j,j} + x w^{-2ik} e_{i,i} \tp e_{j+l,j}, \nonumber\\
	D(x,y) & = & \sum_{i,j=0}^{n-1} w^{2(i-j)k} e_{i,i} \tp e_{j,j} + xy\ e_{i+l,i} \tp e_{j+l,j}. \label{ABCD}
\end{eqnarray}
Substituting these functions into Equation $(\ref{eqnpLLp})$, we find the following constraints on $f_{\alpha}(xy^{-1})$:
\begin{eqnarray}
f_{\alpha}(xy^{-1}) \tilde{p}^{\alpha} A(x,y) \tilde{p}^{\beta}
&=&	f_{\beta}(xy^{-1}) \tilde{p}^{\alpha} A(x,y) \tilde{p}^{\beta},\label{Aconstraint} \\
f_{\alpha}(xy^{-1}) \tilde{p}^{\alpha} B(x,y) \tilde{p}^{\beta}
&=&	f_{\beta}(xy^{-1}) \tilde{p}^{\alpha} B(y,x) \tilde{p}^{\beta},\label{Bconstraint} \\
f_{\alpha}(xy^{-1}) \tilde{p}^{\alpha} C(x,y) \tilde{p}^{\beta}
&=&	f_{\beta}(xy^{-1}) \tilde{p}^{\alpha} C(y,x) \tilde{p}^{\beta},\label{Cconstraint} \\
f_{\alpha}(xy^{-1}) \tilde{p}^{\alpha} D(x,y) \tilde{p}^{\beta}
&=&	f_{\beta}(xy^{-1}) \tilde{p}^{\alpha} D(x,y) \tilde{p}^{\beta}. \label{Dconstraint}
\end{eqnarray}
Note that Equations (\ref{Aconstraint},\ref{Dconstraint}) are automatically satisfied.

The above relations provide a tensor product graph \cite{zgb91,dgz94,dgz96}. Assigning each irrep to a vertex of a graph, we connect the
vertices labelled $\alpha$ and $\beta$ by an edge if
$$\tilde{p}^{\alpha} \chi(x,y) \tilde{p}^{\beta}\neq 0$$
for either $\chi = B$ or $\chi = C$. The tensor product graph for the case $l=k=1$ is depicted in Figure \ref{FigTensorOdd}.
An edge connecting vertices $\alpha$ and $\beta$ signifies that the functions $f_\alpha(x)$ and $f_\beta(x)$ are constrained by
Equations (\ref{Bconstraint},\ref{Cconstraint}).

\begin{figure}[ht]
\begin{center}
\input{TensorDiagramOdd.pdf_t}
\end{center}
\caption{Tensor product graph for odd $n$ when $l = k = 1$. \label{FigTensorOdd}}
\end{figure}

We now give a series of propositions and lemmas which result in a solution to Equation \eqref{eqnLLRc}.
\begin{prop} \label{prop3-4}
For $\check{R}(z)$ defined by Equation (\ref{Rcheck}) to be a solution of
Equation (\ref{eqnLLRc}), the following four constraints must be satisfied:
\begin{eqnarray*}
	0 & = & \bar{\delta}_{k+b}^{d} \bar{\delta}_{a+l}^{c} \left[f_{(a,b)}(z)(zw^{2((-l-a)k-bl)} + 1) - f_{(c,d)}(z)(w^{2((-l-a)k-bl)} + z) \right], \\
	0 & = & \bar{\delta}_{k-b}^{d} \bar{\delta}_{c}^{l-a} \left[f_{(a,b)}(z)(zw^{2((-l+a)k+bl)} + 1) - f_{(c,d)}(z)(w^{2((-l+a)k+bl)} + z) \right], \\
	0 & = & \bar{\delta}_{k+b}^{-d} \bar{\delta}_{-c}^{l+a} \left[f_{(a,b)}(z)(zw^{2((-l-a)k-bl)} + 1) - f_{(c,d)}(z)(w^{2((-l-a)k-bl)} + z) \right], \\
	0 & = & \bar{\delta}_{k-b}^{-d} \bar{\delta}_{c}^{a-l} \left[f_{(a,b)}(z)(zw^{2((-l+a)k+bl)} + 1) - f_{(c,d)}(z)(w^{2((-l+a)k+bl)} + z) \right],
\end{eqnarray*}
where the pairs $(a,b)$ and $(c,d)$ belong to $S$.
\end{prop}
\proof{
 We consider constraint Equation \eqref{Bconstraint}.
Given the operators
$$ B(x,y) = \sum_{i,j=0}^{n-1} x w^{2ik} e_{i,i} \tp e_{j-l,j} + y w^{2jk} e_{i+l,i} \tp e_{j,j}, $$
and
$$ \tilde{p}^{\alpha} = \sum_{i,j = 0}^{n-1}[ w^{2jb} e_{i+a+j,i+a} \tp e_{i+j,i}
+ w^{-2bj} e_{i-a+j,i-a} \tp e_{i+j,i}], $$
we first calculate
\begin{align*}
	\tilde{p}^{(a,b)}& B(x,y)\\
	& = \sum_{i,j,s,t=0}^{n-1} [ w^{-2bj} e_{i-a+j,i-a} \tp  e_{i+j,i} + w^{2bj} e_{i+a+j,i+a} \tp e_{i+j,i}]
		[ x w^{2sk}e_{s,s} \tp e_{t-l,t} + yw^{2tk} e_{s+l,s} \tp e_{t,t}] \\
	& = \sum_{j,t=0}^{n-1}
		[ xw^{2((t-l-a)k-bj)} e_{t-l-a+j,t-l-a} \tp e_{t-l+j,t}
		+ xw^{2((t-l+a)k+bj)} e_{t-l+a+j,t-l+a} \tp e_{t-l+j,t}\\
	&  \hspace{12mm} + yw^{2(tk-bj)} e_{t-a+j,t-a-l} \tp e_{t+j,t}
		+ yw^{2(tk+bj)} e_{t+a+j,t+a-l} \tp e_{t+j,t}].
\end{align*}
This leads to
\begin{eqnarray*}
	&&\tilde{p}^{(a,b)} B(x,y) \tilde{p}^{(c,d)} \\
	& = & \sum_{j,t=0}^{n-1}
		[ xw^{2((t-l-a)k-bj)} e_{t-l-a+j,t-l-a} \tp e_{t-l+j,t} +
		xw^{2((t-l+a)k+bj)} e_{t-l+a+j,t-l+a} \tp e_{t-l+j,t}\\
	& & \hspace{12mm} + yw^{2(tk-bj)} e_{t-a+j,t-a-l} \tp e_{t+j,t}
		+ yw^{2(tk+bj)} e_{t+a+j,t+a-l} \tp e_{t+j,t}]\tilde{p}^{(c,d)}\\
	& = & \sum_{j,t,v=0}^{n-1}
	[     xw^{2((t-l-a)k-bj-dv)} \bar{\delta}_{a+l}^{c} e_{t-l-a+j,t-c-v} \tp e_{t-l+j,t-v} \\
	& & \hspace{12mm} + xw^{2((t-l+a)k+bj-dv)} \bar{\delta}_{c}^{l-a} e_{t-l+a+j,t-c-v} \tp e_{t-l+j,t-v} \\
	& & \hspace{12mm} + yw^{2(tk-bj-dv)} \bar{\delta}_{c}^{a+l} e_{t-a+j,t-c-v} \tp e_{t+j,t-v}
		+ yw^{2(tk+bj-dv)} \bar{\delta}_{c}^{l-a} e_{t+a+j,t-c-v} \tp e_{t+j,t-v} \\
	& & \hspace{12mm} + xw^{2((t-l-a)k-bj+dv)} \bar{\delta}_{-c}^{l+a} e_{t-l-a+j,t+c-v} \tp e_{t-l+j,t-v} \\
	& &	\hspace{12mm} + xw^{2((t-l+a)k+bj+dv)} \bar{\delta}_{c}^{a-l} e_{t-l+a+j,t+c-v} \tp e_{t-l+j,t-v}\\
	& & \hspace{12mm} + yw^{2(tk-bj+dv)} \bar{\delta}_{-c}^{a+l} e_{t-a+j,t+c-v} \tp e_{t+j,t-v}
		+ yw^{2(tk+bj+dv)} \bar{\delta}_{c}^{a-l} e_{t+a+j,t+c-v} \tp e_{t+j,t-v}]\\
	& = & n\sum_{j,v=0}^{n-1}
	\biggl\{     w^{2(dv-bj)} \left[xw^{2((-l-a)k-bl)} + y\right] \bar{\delta}_{k+b}^{d} \bar{\delta}_{a+l}^{c} e_{j-a,v-c} \tp e_{j,v} \\
	& & \hspace{12mm}+ w^{2(dv+bj)}\left[xw^{2((-l+a)k+bl)} + y\right] \bar{\delta}_{k-b}^{d} \bar{\delta}_{c}^{l-a} e_{j+a,v-c} \tp e_{j,v} \\
	& & \hspace{12mm}+ w^{-(2dv+bj)}\left[xw^{2((-l-a)k-bl)} + y\right] \bar{\delta}_{k+b}^{-d} \bar{\delta}_{-c}^{l+a} e_{j-a,v+c} \tp e_{j,v} \\
	& & \hspace{12mm}+ w^{-(2dv-bj)}\left[xw^{2((-l+a)k+bl)} + y\right] \bar{\delta}_{k-b}^{-d} \bar{\delta}_{c}^{a-l}
e_{j+a,v+c} \tp e_{j,v} \biggr\}.
\end{eqnarray*}
We now substitute the above result into Equation (\ref{Bconstraint}),
$$p^{\alpha}[f_{\alpha}(xy^{-1})B(x,y) - f_{\beta}(xy^{-1}) B(y,x)]p^{\beta} = 0,$$
and finally arrive at the desired four constraints.

Very similar calculations for Equation \eqref{Cconstraint} leads to the same four constraint equations.}

\begin{lem} \label{lem1}
Let $w^2$ be a primitive $n$th root of unity and $l, k$ be integers such that $gcd(l,n)=gcd(k,n)=1$. Then
$$ \prod_{j=1}^{a} \left( \frac{z + w^{2l((2j-1)k + b)}}{1 + zw^{2l((2j-1)k + b)}} \right)
= \prod_{j=1}^{\overline{a}} \left( \frac{z + w^{2l((2j-1)k + b)}}{1 + zw^{2l((2j-1)k + b)}} \right), $$
for all $a\in \mathbb{N}$, $b\in \mathbb{Z}.$
\end{lem}
The proof is omitted.  Henceforth we choose $l,k$ satisfying $gcd(l,n)=gcd(k,n)=1$ so that we may use the above lemma.
\begin{prop} \label{S'}
Let $S' = \{ (a,b) | a \geq 0, b \in \mathbb{Z} \}$, and consider a set of functions $\{f_{(a,b)}| (a,b) \in S'\}$.
If the functions satisfy the relations
\begin{eqnarray*}
f_{(a+l,b+k)}(z) & = \es & \left(
\frac{z + w^{2((a+l)k+bl)}}{1 + zw^{2((a+l)k+bl)}} \right) f_{(a,b)}(z), \\
f_{(0,b)}(z) &=  \es & f_{(0,-b)}(z), \\
f_{(a,b)}(z) & = \es & f_{(a,b+n)}(z), \\
f_{(a,b)}(z) & = \es & f_{(a+n,b)}(z),  \\
f_{(\overline{a},b)}(z) & = \es & f_{(n-\overline{a},-b)}(z), \  \ (a,b) \in S',
\end{eqnarray*}

\noindent then the functions also satisfy the constraints given in Proposition \ref{prop3-4} for all
$(a,b) \in S$.  Moreover, every set of functions satisfying the conditions of Proposition \ref{prop3-4} can
be extended in a unique way to a set of functions defined on $S'$ satisfying the above conditions.  Hence
the two sets of constraints are equivalent.
\end{prop}
The proof is straightforward and omitted.
\begin{lem} \label{lem2}
The set of functions
$$
f_{(a,b)}(z) = \prod_{j=1}^{\overline{al^{-1}}}
\left( \frac{z + w^{2l((2j-1)k + b-akl^{-1})}}{1 + zw^{2l((2j-1)k + b-akl^{-1})}} \right)
f_{(0,b-akl^{-1})}(z)
$$
satisfies the conditions in Proposition \ref{S'} given
$$ f_{(0,b)}(z) = f_{(0,-b)}(z) = f_{(0,b+n)}, $$
for $b \in \mathbb{Z}$.
\end{lem}
The proof follows from Lemma \ref{lem1}.

Substituting the functions given in Lemma \ref{lem2} and their associated projection operators into our form for $\check{R}(z)$ gives the operator
$$ \check{R}(z) = \frac{1}{n}\sum_{i,j,a,b=0}^{n-1} w^{2bj} \prod_{p=1}^{\overline{al^{-1}}}
\left( \frac{z + w^{2l((2p-1)k+b-akl^{-1})}}{1 + zw^{2l((2p-1)k+b-akl^{-1})}} \right)
f_{(0,b-akl^{-1})}(z) e_{i+a+j,i+a} \tp e_{i+j,i}, $$
which satisfies Equation $(\ref{eqnLLRc})$.
Note that in these calculations the only property that we have used is that $w^{2}$
is a primitive $n$th root of unity, irrespective of whether $n$ is odd or even.
This allows us to use these calculations later on without alteration.

For $\check{R}(z)$ to be a descendent, it must also satisfy the Yang--Baxter equation.
For the moment we will set $l=k=1$, and consider the more general case in Section \ref{other reps}. Then the functions become
\begin{equation}
f_{(a,a + b)}(z) = \prod_{j=1}^{a}
\left( \frac{z + w^{2(2j-1+b)}}{1 + zw^{2(2j-1+b)}} \right) f_{(0,b)}(z).
\label{ph1}
\end{equation}
This gives the operator
\begin{eqnarray*}
\check{R}(z)
&= & f_{(0,0)}(z)\left \{ \sum_{a=0}^{\frac{n-1}{2}} \prod_{j=1}^{a}
\left( \frac{z + w^{2(2j-1)}}{1 + zw^{2(2j-1)}} \right) p^{(a,a)} \right \} \\
&  & + \sum_{b=1}^{\frac{n-1}{2}} f_{(0,b)}(z) \left\{p^{(0,b)}
+ \sum_{a=1}^{\frac{n}{2}-1}\left[\prod_{j=1}^{a}
\left( \frac{z + w^{2(2j-1 + b)}}{1 + zw^{2(2j-1+ b)}} \right)
p^{(a,\overline{a+b})} \right. \right. \\
&  & \left. \left. + \prod_{j=1}^{a}
\left( \frac{z + w^{2(2j-1 - b)}}{1 + zw^{2(2j-1-b)}} \right)
p^{(a,\overline{a-b})} \right] \right\},
\end{eqnarray*}
or equivalently
\begin{equation}
\check{R}(z) = \sum_{i,j,a=0}^{n-1}
\left[\frac{1}{n} \sum_{b=0}^{n-1} w^{2bj} \prod_{p=1}^{a}
\left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+b-a)}} \right)
f_{(0,b-a)}(z) \right] e_{i+a+j,i+a} \tp e_{i+j,i}.
\label{ph2}
\end{equation}
In $\check{R}(z)$ there remain $\frac{n+1}{2}$ arbitrary functions, one for each disconnected component in
the tensor product graph of Figure \ref{FigTensorOdd}. This is in some contrast to the examples studied in \cite{zgb91,dgz94,dgz96} for which the tensor product graphs are connected.
To ensure $R(z)$ inherits properties from the representation of the canonical element,
we enforce additional conditions. Firstly, we impose that $R(z)$ is self-adjoint,
which is equivalent to
$$ \check{R}_{12}(z) = \check{R}_{21}^{\dagger}(z), \hspace{1cm} z \in \mathbb{R},$$
where $\dagger$ is the adjoint operator.

\begin{prop} \label{propad}
The following three statements about the matrix $\check{R}(z)$ of Equation (\ref{ph2})
and coefficient functions of Equation (\ref{ph1}) are equivalent:
\begin{itemize}
\item[(i)]
$\check{R}_{12}(z) = \check{R}_{21}^{\dagger}(z).$
\item[(ii)]
$ (f_{(a,b)}(z))^{*} = f_{(a,-b)}(z), \quad \forall (a,b) \in S'.$
\item[(iii)]
$(f_{(0,b)}(z))^{*} = f_{(0,b)}(z) = f_{(0,b+2c)}(z), \quad \forall z \in \mathbb{R}$ and $b,c \in \mathbb{Z}$.
\end{itemize}
Here $^{*}$ denotes complex conjugation.
\end{prop}
\proof{
Assume $(i)$ holds. Looking at the general form of the projection operators we find that
$$ (\tilde{p}^{(a,b)}_{21})^{\dagger} = (\tilde{p}^{(a,n-b)}_{12}). $$
Recalling the form of $\check{R}(z)$ given in Equation
(\ref{Rcheck}) and using the linear independence of the projection operators,
this implies that
$$ (f_{(a,b)}(z))^{*} = f_{(a,-b)}(z), \quad \forall (a,b) \in S'. $$

Conversely, $(i)$ follows directly from $(ii)$, and hence statements $(i)$ and $(ii)$ are equivalent.

Now assume $(ii)$ holds. Equation (\ref{ph1}) gives us
$$ f_{(1,1+b)}(z) = \left( \frac{z + w^{2(b+1)}}{1 + zw^{2(b+1)}} \right) f_{(0,b)}(z), $$
and
$$ f_{(1,-1-b)}(z) = \left( \frac{z + w^{-2(b+1)}}{1 + zw^{-2(b+1)}} \right) f_{(0,-b-2)}(z). $$
We see that
$$ (f_{(1,-1-b)}(z))^{*} = \left( \frac{z + w^{2(b+1)}}{1 + zw^{2(b+1)}} \right)
(f_{(0,-b-2)}(z))^{*} = \left( \frac{z + w^{2(b+1)}}{1 + zw^{2(b+1)}} \right)
f_{(0,b+2)}(z), $$
which implies that
$$ f_{(0,b)}(z) = f_{(0,b+2)}(z). $$
Combining this with the constraint of $f_{(0,b)}(z)$ we find
$$ (f_{(0,b)}(z))^{*} = f_{(0,b)}(z) = f_{(0,b+2c)}(z), $$
for $b,c \in \mathbb{Z}$.

Conversely, suppose $(iii)$ holds.
It follows that
$$
f_{(a,b)}(z) = \prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1+b-a)}}{1 +
zw^{2(2j-1+b-a)}} \right) f_{(0,b-a)}(z).
$$
Hence
\begin{eqnarray*}
  (f_{(a,b)}(z))^{*}
  & = & \left(\prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1+b-a)}}{1 + zw^{2(2j-1+b-a)}} \right) f_{(0,b-a)}(z)\right)^{*} \\
  & = & \prod_{j=1}^{a} \left( \frac{z + w^{-2(2j-1+b-a)}}{1 + zw^{-2(2j-1+b-a)}} \right) \left(f_{(0,b-a)}(z)\right)^{*} \\
  & = & \prod_{j=1}^{a} \left( \frac{z + w^{2(2(-j)+1-b+a)}}{1 + zw^{2(2(-j)+1-b+a)}} \right) f_{(0,0-b-a)}(z) \\
  & = & \prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1-b-a)}}{1 + zw^{2(2j-1-b-a)}} \right) f_{(0,-b-a)}(z) \\
  & = & f_{(a,-b)}(z).
\end{eqnarray*}

This completes the proof.
}

As the second index of the function can be considered modulo $n$ and $n$ is
odd, imposing self-adjointness implies that there is only one arbitrary function left. Without loss
of generality we consider it to be $f_{(0,0)}(z)$. This can be seen as an
overall scalar of our operator thus we are able to set it to a constant. Now we impose the limiting condition
\begin{equation} \label{limit}
\lim_{z \rightarrow 0}{R}(z) = \pm (\pi_n^{\pm} \tp \pi_n^{\pm}) \mathcal{R}.
\end{equation}

This sets the scalar to be $f_{(0,0)}(z) = 1$, so the functions become
$$
f_{(a,b)}(z)
 = \prod_{j=1}^{a}
\left( \frac{z + w^{2(2j-1+b-a)}}{1 + zw^{2(2j-1+b-a)}} \right).
$$
Hence
$$
\check{R}(z)
 = \sum_{b=0}^{\frac{n-1}{2}} p^{(0,b)} + \sum_{a=1}^{\frac{n-1}{2}}\sum_{b=0}^{n-1}
\prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1+b-a)}}{1 + zw^{2(2j-1+b-a)}} \right) p^{(a,b)},
$$
or equivalently
$$
\check{R}(z)
= \sum_{i,j,a=0}^{n-1} \left[\frac{1}{n} \sum_{b=0}^{n-1} w^{2bj}
\prod_{p=1}^{a} \left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+b-a)}} \right) \right]
e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
To simplify this we define the functions
$$
g_{(a,j)}(z) = \frac{1}{n} \sum_{b=0}^{n-1} w^{2bj} f_{(a,b)}(z)
 = \frac{1}{n} \sum_{b=0}^{n-1} w^{2bj} \prod_{p=1}^{a}
\left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+b-a)}} \right) ,
$$
where $a \in \mathbb{N}$ and $j \in \mathbb{Z}$. In terms of these new functions $g_{(a,j)}(z)$, our
operator becomes
\begin{equation*}
\check{R}(z) = \sum_{i,j,a=0}^{n-1} g_{(a,j)}(z) e_{i+a+j,i+a} \tp e_{i+j,i},
\end{equation*}

\noindent or

\begin{equation} \label{newR}
R(z) = \sum_{i,j,a=0}^{n-1} g_{(a,j)}(z) e_{i+j,i+a} \tp e_{i+a+j,i}.
\end{equation}

\begin{rmk}
The operator $R(z)$ of Equation (\ref{newR}) satisfies the following properties:

\begin{itemize}
\item[(i)] $R(z)^{*} = R(z), \; \forall z \in \mathbb{R}$,
\item[(ii)] $R^{t}(z) = R(z), \; \forall z \in \mathbb{C}$,
\item[(iii)] $R^{-1}(z) = R(z), \; \forall z \in \mathbb{C}$,
\item[(iv)] $R_{12}(z) R_{21}(z^{-1}) = I \tp I, \; \forall z \in \mathbb{C}$,
\item[(v)] $\underset{z \rightarrow 0}{\lim} R(z) = \pm(\pi_n^\pm \tp \pi_n^\pm) \mathcal{R}$,
\item[(vi)] $\underset{z \rightarrow 1}{\lim} R(z) = P$.

\end{itemize}
\end{rmk}

\begin{prop} \label{grel}
The operator $R(z)$ of Equation (\ref{newR}) is a descendant if and only if
$$
\sum_{k=0}^{n-1} g_{(a,k-d)}(x)g_{(k,a-b)}(xy)g_{(b,c-k)}(y)
= \sum_{k=0}^{n-1}  g_{(c,k-b)}(x)g_{(k,c-d)}(xy)g_{(d,a-k)}(y),
$$
for $0 \leq a,b,c,d \leq n-1$.
\end{prop}
This follows directly from the Yang--Baxter equation.

The matrix $R(z)$ of Equation (\ref{newR}) agrees with the $R$-matrix obtained in \cite{DIL2006} when $n=3$.  It has also been verified computationally for
all odd $n \leq 17$ that the functions $g(z)$ satisfy the equation given in Proposition \ref{grel} above, and
thus that $R(z)$ is a solution to the Yang--Baxter equation.  The computations were performed with
$z$ treated as an arbitrary complex number and $w$ as an arbitrary primitive $n$th root of unity.

\begin{conj}
The matrix $R(z)$ given in Equation \eqref{newR} is a descendant of the zero-field six-vertex model with $D(D_n)$ symmetry for all odd $n$.
\end{conj}

\section{Descendants associated with $D(D_{2m})$}
In this section we construct a family of solutions of the Yang-Baxter equation
using $D(D_{2m})$, treating the cases where $m$ is even and
odd separately. It will be shown that although in both cases the
representations have the same general form, the case where $m$ is odd has more in common
with the solutions from the previous section associated with $D(D_n)$ where $n$ is
odd.

Throughout this entire section, we consider $w$ to be
a primitive $2m$th root of unity and we use the $r(z)$ stated previously and
associated with $\pi_{2} = \pi_{2}^{(k,l)}$.
The $L$-operator arising from the cases of $m$ odd or even can be considered at
once, since the approach is similar to that of Section \ref{ssecLopDnodd}.

\subsection{Construction of the $L$-operator}
In this subsection we construct an operator $L(z) \in {\rm End}(V_{2} \tp
V_{m})$
using a similar approach as before in Section \ref{ssecLopDnodd}.
Again, we assume that the $L$-operator is of the form
$$ L(z) = (\pi_{2} \tp \pi_{m}) [R + h(z) (R^{T})^{-1}], $$
where $\pi_{m}$ is either $\pi_{m,\tau}^{(0,b)}$ or $\pi_{m,\sigma\tau}^{(0,b)}$ with $b\in \{0,1\}$.
Applying this representation and an appropriate basis transformation on the two-dimensional space we find
$$
L(z) = \sum_{i=0}^{m-1}
\left \{ (w^{2ik}e_{1,2} + w^{-2ik}e_{2,1}) \tp e_{i,i} + h(z)
\left[ e_{1,1} \tp e_{i-l,i} + e_{2,2} \tp e_{i+l,i} \right] \right\}.
$$
We note that the basis transformation is of the same general form used
in Section $\ref{ssecLopDnodd}$, as are the operators $r(z)$ and $L(z)$. Thus it follows that
our $L$-operator is of the form
$$
L(z)
= \sum_{i=0}^{m-1}
\left \{ (w^{2ik}e_{1,2} + w^{-2ik}e_{2,1}) \tp e_{i,i} + z
\left[ e_{1,1} \tp e_{i-l,i} + e_{2,2} \tp e_{i+l,i} \right] \right\}.
$$

\subsection{Construction of the descendants when $m$ is even}
Here we consider the case of $D(D_{2m})$ where $m$ is even. We use $r(z)$
associated with $\pi_{2}^{(l,k)}$ with $w$ a primitive $2m$th root
of unity.
As before, we look for descendants of the form
$$
P R(z) = \check{R}(z) = \sum_{\alpha \in S} f_{\alpha}(z) p^{\alpha},
$$
where $p^{\alpha}$ are the projection operators previously calculated,
$f_{\alpha}(z)$ are continuous functions and $S$ is the set of ordered pairs
which correspond to non-zero projection operators. We require that $\check{R}(z)$
satisfies Equation $(\ref{eqnLLRc})$. As before, we use rescaled projection
operators:
$$
\tilde{p}^{\alpha}
= \sum_{i,j = 1}^{m}[ w^{2jb} e_{i+a+j,i+a} \tp e_{i+j,i} + w^{-2bj} e_{i-a+j,i-a} \tp e_{i+j,i}], \quad \alpha = (a,b) \in S.
$$
As these operators
and the $L$-operator are equivalent to those found earlier, we can use the previous calculations to arrive at the tensor product graph shown in
Figure $\ref{FigTensorEven}$.

\begin{figure}[ht]
\begin{center}
\input{TensorDiagramEven.pdf_t}
\end{center}
\caption{Tensor product graph for even $n$\label{FigTensorEven} when $l=k=1$.}
\end{figure}

We again define the set
$$ S' = \{(a,b) | a \in \mathbb{N}, b \in \mathbb{Z} \},$$
and extend the functions $f_\alpha(z)$ to that set in a way analogous to that of Proposition \ref{S'}.  We restrict ourselves
to $k,l$ satisfying gcd$(l,m) = $gcd$(k,m) = 1$. Following the calculations of the previous section, we determine that if
 $R(z)$ is a descendant then the functions $f_{(a,b)}(z)$ must satisfy
\begin{align*}
f_{(a,akl^{-1} + b)}(z)
&= \prod_{j=1}^{\overline{al^{-1}}}
\left( \frac{z + w^{2l((2j-1)k + b)}}{1 + zw^{2l((2j-1)k + b)}} \right) f_{(0,b)}(z), \\
	f_{(0,b)}(z) & =  f_{(0,b+m)}(z), \\
	f_{(0,b)}(z) & =  f_{(0,-b)}(z), \qquad \forall (a,b) \in S'.
\end{align*}

We now choose to set $l=k=1$; this gives the functions
$$
f_{(a,a + b)}(z)
= \prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1+b)}}{1 + zw^{2(2j-1+b)}} \right) f_{(0,b)}(z).
$$
This produces the operator
\begin{eqnarray*}
\check{R}(z)
&=& f_{(0,0)}(z)\left \{ \sum_{a=0}^{\frac{m}{2}}
    \prod_{j=1}^{a} \left( \frac{z + w^{2(2j-1)}}{1 + zw^{2(2j-1)}} \right) p^{(a,a)} \right \} \\
& & + f_{(0,\frac{m}{2})}(z)\left \{ \sum_{a=0}^{\frac{m}{2}} \prod_{j=1}^{a}
    \left( \frac{z + w^{2(2j-1 + \frac{m}{2})}}{1 + zw^{2(2j-1 + \frac{m}{2})}} \right) p^{(a,\overline{\frac{m}{2} + a})} \right\} \\
& & + \sum_{b=1}^{\frac{m}{2}-1} f_{(0,b)}(z)
    \left\{p^{(0,b)} + \sum_{a=1}^{\frac{m}{2}-1}\left[\prod_{j=1}^{a}
    \left( \frac{z + w^{2(2j-1+b)}}{1 + zw^{2(2j-1 + b)}} \right) p^{(a,\overline{a+b})} \right. \right. \\
& & \left. \left. + \prod_{j=1}^{a}
    \left( \frac{z + w^{2(2j-1 - b)}}{1 + zw^{2(2j-1 - b)}} \right)
    p^{(a,\overline{a-b})} \right] + \prod_{j=1}^{a}
    \left( \frac{z + w^{2(2j-1+ \frac{m}{2})}}{1 + zw^{2(2j-1 + \frac{m}{2})}} \right) p^{(a,\overline{a+\frac{m}{2}})} \right\} \\
\end{eqnarray*}
or equivalently
$$
R(z) = \sum_{i,j,a=0}^{m-1}
\left[\frac{1}{n}\sum_{b=0}^{m-1} w^{2bj} \prod_{p=1}^{a}
\left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+ b-a)}} \right)
f_{(0,b-a)}(z) \right] e_{i+j,i+a} \tp e_{i+a+j,i}.
$$
Here we again enforce that our operator is self-adjoint. We are able to use the
previous calculations and recall that
$$
(f_{(0,b)}(z))^{*} = f_{(0,b)}(z) = f_{(0,b+2c)}(z),
$$
$\forall z \in \mathbb{R}$ and $b,c \in \mathbb{N}$. The second index of the function can be
considered modulo $m$ and as $m$ is even the functions are partitioned into
two sets. We see that we have only two functions in which we have any freedom
left; without loss of generality we consider them to be $f_{(0,0)}(z)$ and
$f_{(0,1)}(z)$. As there are two functions we find that we cannot consider them
an overall scalar, so we need to impose additional conditions. As the $m$-dimensional
representations of the canonical element are self-adjoint, we enforce that
$R(z)$ is unitary . From this it follows that
$$
f_{(0,0)}(z) = \pm 1 \mbox{ and } f_{(0,1)}(z) = \pm 1.
$$
Imposing the limiting condition given in Equation (\ref{limit}) sets
$$
f_{(0,0)}(z) = f_{(0,1)}(z) = 1.
$$
This yields the operator
\begin{equation}
\check{R}(z) = \sum_{i,j,a=0}^{m-1} \left[\frac{1}{m} \sum_{b=0}^{m-1} w^{2bj}
\prod_{p=1}^{a} \left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+b-a)}} \right) \right]
e_{i+a+j,i+a} \tp e_{i+j,i}.
\label{rcheckeven}
\end{equation}
To simplify this we define the functions
$$ g_{(a,j)}(z) = \frac{1}{m} \sum_{b=0}^{m-1} w^{2bj} f_{(a,b)}(z)
= \frac{1}{m} \sum_{b=0}^{m-1} w^{2bj} \prod_{k=1}^{a}
\left( \frac{1 + zw^{2(a-b+1-2k)}}{z + w^{2(a-b+1-2k)}} \right) ,
$$
where $a \in \mathbb{N}$ and $j \in \mathbb{Z}$. With the use of these new functions our operator
becomes
\begin{equation}
R(z) = \sum_{i,j,a=0}^{m-1} g_{(a,j)}(z) e_{i+j,i+a} \tp e_{i+a+j,i}.
\label{Reven}
\end{equation}

\begin{rmk} The operator $R(z)$ of Equation (\ref{Reven}) satisfies the following properties:

\begin{itemize}
\item[(i)] $R(z)^{*} = R(z), \; \forall z \in \mathbb{R}$,
\item[(ii)] $R^{t}(z) = R(z), \; \forall z \in \mathbb{C}$,
\item[(iii)] $R^{-1}(z) = R(z), \; \forall z \in \mathbb{C}$,
\item[(iv)] $R_{12}(z) R_{21}(z^{-1}) = I \tp I, \; \forall z \in \mathbb{C}$,
\item[(iv)] $\underset{z \rightarrow 0}{\lim} R(z) = \pm(\pi_n^\pm \tp \pi_n^\pm) \mathcal{R}$,
\item[(v)] $\underset{z \rightarrow 1}{\lim} R(z) \neq P$.

\end{itemize}
\end{rmk}

Note that the last property shows that $R(z)$ does not satisfy regularity, unlike the $R(z)$ constructed from
$D(D_n)$ when $n$ is odd.
We do, however, have the following analogue of Proposition \ref{grel}:
\begin{prop} \label{geven}
The operator of Equation (\ref{Reven}) is a descendant if and only if the following constraint
is satisfied:
$$
\sum_{k=0}^{m-1} g_{(a,k-d)}(x)g_{(k,a-b)}(xy)g_{(b,c-k)}(y)
= \sum_{k=0}^{m-1}  g_{(c,k-b)}(x)g_{(k,c-d)}(xy)g_{(d,a-k)}(y),
$$
for $0 \leq a,b,c,d \leq m-1$.
\end{prop}
It has been computationally verified that the functions $g(z)$ satisfy the conditions in
Proposition \ref{geven} above for all even $m \leq 16$, and hence that $R(z)$ is a solution
to the Yang--Baxter equation.

\begin{conj}
The matrix $R(z)$ given by Equation \eqref{Reven} is a descendent of the zero-field six-vertex
model with $D(D_{2m})$ symmetry for all even $m$.
\end{conj}

\subsection{Construction of the descendants when $m$ is odd}
For completeness, we include the descendants associated with $D(D_{2m})$ for odd
$m$. In our construction of the descendant of $r(z)$ we use a linear
combination of projection operators found in Subsection \ref{sub2-3-2}.
Using these projections, for a self-adjoint descendant of $r(z)$ which limits to
a representation of the canonical element, we obtain the operator
$$
R(z) = \sum_{i,j,a=0}^{m-1} \left[\frac{1}{m}\sum_{b=0}^{m-1} w^{bj}
\prod_{p=1}^{a} \left( \frac{z + w^{2(2p-1+b-a)}}{1 + zw^{2(2p-1+ b-a)}}
\right) \right] e_{i+j,i+a} \tp e_{i+a+j,i}.
$$
This is equivalent to the operator $R(z)$ found using $D(D_{m})$ where $m$ is odd.  Hence it
obeys the properties stated in the remark in Section \ref{props}. Moreover, $R(z)$ is a descendant
of the zero-field six-vertex model with $D(D_{2m})$ symmetry for odd $m \leq 17$ and conjectured to be a descendant for all odd $m \geq 3$.

\section{Generalised descendants}
In this section we investigate some of the choices we made in order to obtain
descendants. Specifically, we investigate the effects of choosing a different initial 2-dimensional
irrep and of imposing fewer conditions on the descendant.

\subsection{Dependency on choice of irreducible representations}\label{other reps}
In earlier sections we constructed descendants by starting from the $R$-matrix associated with
the two-dimensional irrep $\pi_2^{(1,1)}$. Here we explore the consequences of starting with a different
two-dimensional irrep, namely any $\pi_2^{(l,k)}$ where $\gcd(l,n) = \gcd(k,n)=1.$  We consider
$n$-dimensional descendants for any $n \geq 3$,
with $w$ a primitive $n$th root of unity. We recall that
$$ r(z) =
	\left(
	\begin{array}{cccc}
		w^{kl}z^{-1} - w^{-kl}z & 0 & 0 & 0\\
		0 & z^{-1} - z & w^{kl} - w^{-kl} & 0\\
		0 & w^{kl} - w^{-kl} & z^{-1} - z & 0\\
		0 & 0 & 0 & w^{kl}z^{-1} - w^{-kl}z
	\end{array}
\right)$$
and
$$
L(z) = \sum_{i=0}^{n-1}\left \{ (w^{ik}e_{1,2} + w^{-ik}e_{2,1}) \tp e_{i,i}
+ z \left[ e_{1,1} \tp e_{i-l,i} + e_{2,2} \tp e_{i+l,i} \right] \right\}
$$
where $0 \leq l \leq \frac{n}{2}$ and  $0 \leq k \leq n-1$, although here $w$ differs slightly
from that used in Section \ref{r}.  We now repeat our earlier calculations for
our more general $l$ and $k$, obtaining
$$
\check{R}(z) = \frac{1}{n}\sum_{i,j,a,b=0}^{n-1} w^{bjk}
\prod_{p=1}^{\overline{al^{-1}}}
\left( \frac{z + w^{lk(2p-1+b-al^{-1})}}{1 + zw^{lk(2p-1+b-al^{-1})}} \right)
f_{(0,bk-akl^{-1})}(z) e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
We consider the change of basis on the $n$-dimensional space which yields
$$ e_{i,j} \rightarrow e_{il^{-1},jl^{-1}}, \hspace{1cm} i \in \mathbb{Z}. $$
Under this change of basis we find that our $L$-operator becomes
$$
L(z) = \sum_{i=0}^{n-1}\left \{ (w^{ilk}e_{1,2} + w^{-ilk}e_{2,1}) \tp e_{i,i}
+ z \left[ e_{1,1} \tp e_{i-1,i} + e_{2,2} \tp e_{i+1,i} \right] \right\},
$$
while $\check{R}(z)$ becomes
$$
\check{R}(z) = \frac{1}{n}\sum_{i,j,a,b=0}^{n-1} w^{bjlk}
\prod_{p=1}^{a} \left( \frac{z + w^{lk(2p-1+b-a)}}{1 + zw^{lk(2p-1+b-a)}} \right)
f_{(0,(b-a)k)}(z) e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
From this we see that these different choices of initial two-dimensional representation
provide equivalent descendants. The different choices yield different basis
transformations, a permutation on the arbitrary functions and a change of the
root of unity, which must remain primitive. Thus any 2-dimensional
irrep satisfying $\mbox{gcd}(l,n) = \mbox{gcd}(k,n)=1$ results in equivalent descendants.
This is unsurprising as $\check{R}(z)$ is real whenever $z \in \mathbb{R}$, and choosing different
$l,k$ effectively just changes the root of unity being used.

\subsection{Descendants with an extra parameter associated with $D(D_{2m})$ when $m$ is even. \label{secOtherEvenDes}}
Here we return to $D(D_{2m})$ where $m$ is even and we find more descendants by
imposing fewer constraints. We use the general form
$$ \check{R}(z) = \sum_{i,j,a=0}^{m-1} \left[\frac{1}{m} \sum_{b=0}^{m-1} w^{bj}
\prod_{p=1}^{a} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right)
f_{(0,b-a)}(z) \right] e_{i+a+j,i+a} \tp e_{i+j,i} $$
where $w$ is now a primitive $m$th root of unity.  We recall that previously we imposed that $R(z)$ is self-adjoint, unitary and obeys a limiting
condition. If we ignore the limiting condition but enforce the other two conditions
we find that
$$ f_{(0,2b)}(z) = f_{(0,0)}(z) = \pm 1 \hspace{0.7cm} \mbox{and} \hspace{0.7cm}
f_{(0,2b+1)}(z) = f_{(0,1)}(z) = \pm 1, $$
for all $b \in \mathbb{Z}$. Without loss of generality we can set
$$ f_{(0,0)}(z) = 1. $$
This gives us two possibilities; the first choice is $f_{(0,1)}(z) = 1$,
which yields
\begin{equation}
\check{R}^{+}(z) = \sum_{i,j,a=0}^{m-1} \left[\frac{1}{m} \sum_{b=0}^{m-1} w^{bj}
\prod_{p=1}^{a} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right) \right]
e_{i+a+j,i+a} \tp e_{i+j,i}.
\label{r++}
\end{equation}
This operator corresponds to the conjectured descendant given in Equation \ref{rcheckeven}. The second
option is $f_{(0,1)}(z) = -1$, which gives the operator
\begin{equation}
\check{R}^{-}(z) = \sum_{i,j,a=0}^{m-1} \left[\frac{1}{m}
\sum_{b=0}^{m-1} (-1)^{a+b} w^{bj} \prod_{p=1}^{a}
\left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right) \right]
e_{i+a+j,i+a} \tp e_{i+j,i}.
\label{r--}
\end{equation}
The matrices $\check{R}^+(z)$ and $\check{R}^{-}(z)$ are different; nonetheless they share many properties,
including that they square to the identity and obey the unitarity condition. Moreover, we have the following
proposition:
\begin{prop} \label{propr+r-}
The descendant $\check{R}^{-}(z)$ of Equation (\ref{r--}) satisfies the Yang--Baxter
equation if and only if $\check{R}^{+}(z)$ given by Equation (\ref{r++}) does.
\end{prop}
\proof{
We use the identity
\begin{equation*}
\prod_{p=c+1}^{c+\frac{m}{2}} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right)
=(-1)^{a+b+\frac{m}{2}}.
\end{equation*}
This implies that
$$
\check{R}^{-}(z) = (-1)^{\frac{m}{2}}\frac{1}{m} \sum_{i,j,a,b=0}^{m-1}
\left[ w^{bj} \prod_{p=1}^{a+\frac{m}{2}}
\left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right)
\right] e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
We now consider a basis transformation. Given any $\lambda \in \mathbb{C}$,
there exists a basis transformation which yields
$$
e_{i,j} \rightarrow  \lambda^{\overline{i} - \overline{j}} e_{i,j}.
$$
We choose $\lambda$ such that our operator becomes (after scaling)
$$
\check{R}^{-}(z) = \frac{1}{m} \sum_{i,j,a,b=0}^{m-1} \left[ w^{(b+\frac{m}{2})j}
\prod_{p=1}^{a+\frac{m}{2}} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right)
\right]  e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
We define the functions
$$
g_{(a,j)}(z) = \sum_{b=0}^{m-1} w^{bj} \prod_{p=1}^{a}
\left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right),
$$
which allow to write the operators
$$
\check{R}^{+}(z) = \sum_{i,j,a=0}^{m-1} g_{(a,j)}(z) e_{i+a+j,i+a} \tp e_{i+j,i}
$$
and
$$
\check{R}^{-}(z) = \sum_{i,j,a=0}^{m-1} g_{(a+\frac{m}{2},j)}(z) e_{i+a+j,i+a} \tp e_{i+j,i}.
$$
Hence $\check{R}^{-}(z)$ differs from $\check{R}^{+}(z)$ only by a
basis transformation and a permutation of the entries. We calculated previously
that $\check{R}^{+}(z)$ satisfies the Yang--Baxter equation if and only if
$$
\sum_{k=0}^{m-1} \left[g_{(a,k-d)}(x) g_{(k,a-b)}(xy) g_{(b,c-k)}(y)
-  g_{(c,k-b)}(x)g_{(k,c-d)}(xy)g_{(d,a-k)}(y)\right] =0,
$$
$0 \leq a,b,c,d \leq m-1$. Similarly we have that $\check{R}^{-}(z)$ satisfies
the Yang--Baxter equation if and only if
\fontsize{10}{12} $$ \sum_{k=0}^{m-1}
\left[g_{(a+\frac{m}{2},k-d)}(x) g_{(k+\frac{m}{2},a-b)}(xy) g_{(b+\frac{m}{2},c-k)}(y)
-  g_{(c+\frac{m}{2},k-b)}(x)g_{(k+\frac{m}{2},c-d)}(xy)g_{(d+\frac{m}{2},a-k)}(y)\right]
= 0 ,
$$ \fontsize{12}{14}
$0 \leq a,b,c,d \leq m-1$. As we can consider each of the indices of the functions
modulo $m$ we find that the two above constraints are equivalent,
hence the result.
}
Thus we have found another family of conjectured descendants. We explain the
existence of $\check{R}^{-}(z)$ by considering the values of functions associated with the irreps.
To obtain $\check{R}^{+}(z)$ we simply set every function
associated with irreps from the conjugacy class $\{ e \}$ equal to one, i.e. $f_{(0,b)}=1$ for $b \in \mathbb{Z}$. Conversely,
$\check{R}^{-}(z)$ is obtained by setting every function associated with irreps from the
conjugacy class $\{ \sigma^{m} \}$ equal to one, i.e. $f_{(\frac{m}{2},b)}(1)=1$ for $b \in \mathbb{Z}$. We recall that $e$ and $\sigma^{m}$
are the central elements in $D_{2m}$ when $m$ is even. It is in part due to this enlarged centre that we obtain more general solutions in
this case.

It is also possible for us to ignore the unitary condition and only impose that $R(z)$ is self-adjoint.
This leads to the constraints
$$
f_{(0,2b)}(z) = f_{(0,0)}(z) = (f_{(0,0)}(z))^{*} \hspace{0.7cm} \mbox{and}
\hspace{0.7cm} f_{(0,2b+1)}(z) = f_{(0,1)}(z) = (f_{(0,1)}(z))^{*},
$$
for all $b \in \mathbb{Z}$ and $z \in \mathbb{R}$. Using these constraints we set
$$ f_{(0,0)}(z) = 1 + f(z) \hspace{0.7cm} \mbox{and} \hspace{0.7cm}
f_{(0,1)}(z) = 1 - f(z), $$
where $f(z)$ is an arbitrary real function. This yields the operator
\begin{equation}
\check{R}(z) = \check{R}^{+}(z) + f(z) \check{R}^{-}(z).
\label{Rcheckz}
\end{equation}
This is invertible provided $$ f(z) \neq \pm 1. $$
The function $f(z)$ is equivalent to a second
parameter. That is, the operator
\begin{equation}
\check{R}(z,\mu) = \check{R}^{+}(z) + \mu \check{R}^{-}(z),
\label{Rcheckzmu}
\end{equation}
is invertible for $\mu\neq \pm 1$. Furthermore, we have the following:
\begin{prop}
The operator $\check{R}(z,\mu)$ given by Equation (\ref{Rcheckzmu})
satisfies
\begin{equation}
\check{R}_{12}(x,\lambda)\check{R}_{23}(xy, \mu)\check{R}_{12}(y, \nu)
= \check{R}_{23}(y, \nu)\check{R}_{12}(xy, \mu)\check{R}_{23}(x,\lambda), \label{eqnYBE2par}
\end{equation}
if and only if $\check{R}(z)$ given by Equation (\ref{Rcheckz}) is a solution to
the Yang--Baxter equation with no constraints on $f(z)$.
\end{prop}
\proof{
Consider a solution of the Yang-Baxter equation $R(z) \in \mbox{End}(V_{n} \tp V_{n})$
which contains an arbitrary function $f(z)$. Furthermore suppose every entry of
$R(z)$ can be written as a polynomial in terms of $z$ and $f(z)$. Let
$$ \Omega = R_{12}(x)R_{13}(xy)R_{23}(y) - R_{23}(y)R_{13}(xy)R_{12}(x). $$
We observe that every entry of $\Omega$ is expressible as
$$ \sum_{i,j,k = 0}^{\infty} h_{ijk}^{l}(x,y) f^{i}(x)f^{j}(xy)f^{k}(y), $$
where $h_{ijk}^{l}(x,y)$ are polynomials in $x$ and $y$
and $l$ indexes the entry of $\Omega$. As $R(z)$ is a solution of the YBE, i.e. $\Omega = 0$, and
$f(z)$ is an arbitrary function, we deduce that.
$$ h_{ijk}^{l}(x,y) = 0, \hspace{1cm} \forall x,y \in \mathbb{C} / \{ 0,1, \infty \}
\hspace{0.1cm} \mbox{s.t.} \hspace{0.1cm} x \neq y. $$
Let $y_0 \in \mathbb{C} / \{ 0,1, \infty \}$. There are at most we have four values of $x$ for which
$h_{ijk}^{l}(x,y_0)$ can be non-zero, but $h_{ijk}^{l}(x,y_0)$ is continuous in $x$.
This means that
$$
h_{ijk}^{l}(x,y_0) = 0 \hspace{1cm} \forall x \in \mathbb{C}.
$$
By symmetry
$$
h_{ijk}^{l}(x,y) = 0, \hspace{1cm} \forall (x,y)
\in (\mathbb{C} \times \mathbb{C}) / (\{ 0,1, \infty \} \times \{ 0,1, \infty \}).
$$
Thus there are at most 9 points in which $h_{ijk}^{l}(x,y)$ can be non-zero; however
$h_{ijk}^{l}(x,y)$ is continuous in $x$ and $y$. Hence
$$
h_{ijk}^{l}(x,y) = 0, \hspace{1cm} \forall x,y \in \mathbb{C}.
$$
Let $R(z,\mu)$ be the operator derived from $R(z)$ in which we have
replaced the free function $f(z)$ with $\mu$. Every entry of $R(z)$ must
be expressible as a polynomial in terms of $z$ and $\mu$. If we let
$$ \Omega = R_{12}(x,\lambda)R_{13}(xy, \mu)R_{23}(y, \nu)
- R_{23}(y, \nu)R_{13}(xy, \mu)R_{12}(x,\lambda), $$
$x,y,\lambda,\mu,\nu \in \mathbb{C}$, then we find every entry of $\Omega$ can be
written
$$
\sum_{i,j,k = 0}^{\infty} h_{ijk}^{l}(x,y) \lambda^{i}\mu^{j}\nu^{k}.
$$
As shown previously
$$ h_{ijk}^{l}(x,y) = 0, \hspace{1cm} \forall x,y \in \mathbb{C}. $$
Thus $\Omega = 0$ and $R(z,\mu)$ satisfies
$$ R_{12}(x,\lambda)R_{13}(xy, \mu)R_{23}(y, \nu)
= R_{23}(y, \nu)R_{13, \mu}(xy)R_{12}(x,\lambda). $$

Conversely, suppose we have $R(z,\mu)$ which satisfies
$$
R_{12}(x,\lambda)R_{13}(xy, \mu)R_{23}(y, \nu)
= R_{23}(y, \nu)R_{13}(xy, \mu)R_{12}(x,\lambda).
$$
If we consider $R(z) = R(z,f(z))$ where $f(z)$ is an arbitrary function then
$R(z)$ must satisfy the YBE.
To recover the result, we use the fact that $\check{R}(z)=P\ R(z)$ and
$\check{R}(z,\mu)=P\ R(z,\mu)$.
This proof works if the entries of $R(z)$ can be written as a
polynomial $f(z)$ whose coefficients are rational functions of $z$. We are able
to scale $R(z)$ by the product of all the denominators of the coefficients of
$f(z)$, hence turning it into a polynomial.
}
We are able to verify using Maple that $\check{R}(z)$ given by Equation (\ref{Rcheckz})
satisfies the Yang--Baxter equation for even $n$ up to 12, and hence that
$\check{R}(z,\mu)$ satisfies Equation (\ref{eqnYBE2par}). We conjecture that this holds
true for all even $n$, and have shown it to hold in the limit $x=y=0$.

\section{Connection to the Fateev--Zamolodchikov model}
Closely associated with the Yang-Baxter equation is the star-triangle relation (STR), given by
$$ \sum_{d=0}^{N-1} \bar{W}(x|a-d) W(xy|d,c) \bar{W}(y|d-b) = W(x|b-c) \bar{W}(xy|a-b) W(y|a-c),$$
for $0 \leq a,b,c \leq N-1$. One well-known solution of the STR is the $N$-state Fateev--Zamolodchikov model \cite{FateevZam1982b}, which has weights
$$ W(z|l) = \prod_{j=1}^{l} \frac{\lambda^{2j-1}z-1}{\lambda^{2j-1}-z} \hspace{0.7cm} \mbox{and} \hspace{0.7cm} \bar{W}(z|l) = \prod_{j=1}^{l} \frac{\lambda^{2j-1} - \lambda z}{\lambda^{2j}z  - 1}, $$
for $0 \leq l \leq N-1$, where $\lambda$ is a primitive $2N$th root of unity. These weights are extended by the relations
$$ W(z|l) = W(z|N+l) \hspace{0.7cm} \mbox{and} \hspace{0.7cm} \bar{W}(z|l) = \bar{W}(z|N+l),  $$
while also satisfying
$$ W(z|l) = W(z|N-l) \hspace{0.7cm} \mbox{and} \hspace{0.7cm} \bar{W}(z|l) = \bar{W}(z|N-l), $$
for $l \in \mathbb{Z}$. These weights satisfy the STR and lead to the $R$-matrix defined by
$$ R(\tilde{x},\tilde{y}) = \sum_{a_{1},a_{2},b_{1},b_{2}=1}^{N} R_{a_{1}a_{2}}^{b_{1}b_{2}}(\tilde{x},\tilde{y}) e_{b_{1},a_{1}} \tp e_{b_{2},a_{2}}, $$
where
$$ \tilde{x} ={x_{1}\choose x_{2}}, \hspace{1cm} \tilde{y} = {y_{1}\choose y_{2}} $$
and
$$ R_{a_{1}a_{2}}^{b_{1}b_{2}}(\tilde{x},\tilde{y}) = \bar{W}(x_{1}y_{1}^{-1}|a_{1}-b_{2}) W(x_{2}y_{1}^{-1}|a_{1}-a_{2}) \bar{W}(x_{2}y_{2}^{-1}|a_{2}-b_{1}) W(x_{1}y_{2}^{-1}|b_{2}-b_{1}). $$

Through a private communication with V. Bazhanov and J. Perk \cite{BazhPerk2009} we learnt of a connection between the $D(D_{n})$ solution and the Fateev--Zamolodchikov model. Specifically in the private communication a limiting case of the 3-state Fateev--Zamolodchikov model was shown to reduce to the $D(D_{3})$ solution. Using the ideas presented in \cite{BazhPerk2009} we are able to establish a connection between the $N$-state Fateev--Zamolodchikov model and $D(D_{n})$ (or equivalently $D(D_{2n})$) solution in the case where $N=n$ and $n$ is odd.

To investigate which limit of the Fateev--Zamolodchikov model leads to the $D(D_{n})$ model, we determine when the Fateev--Zamolodchikov $R$-matrix squares to the identity. Using certain properties of the weights it is possible to show that
$$ R(\tilde{x},\tilde{y})R(\tilde{x}^{-T},\tilde{y}^{-T}) \propto I \tp I, $$
where
$$ \tilde{x} = {x_{1} \choose x_{2}
\hspace{0.5cm} \mbox{and}\hspace{0.5cm} \tilde{x}^{-T} = {x_{2}^{-1} \choose x_{1}^{-1}}. $$
Thus the inverse of the $R$-matrix is known up to a scalar multiple. Furthermore we are able to show that if the $R$-matrix squares to the identity then it is equivalent to one which satisfies the  constraint
$$ \tilde{x} = \tilde{x}^{-T} \hspace{0.5cm} \mbox{and}\hspace{0.5cm} \tilde{y} = \tilde{y}^{-T}. $$

This implies that the $R$-matrix can be reduced to
$$ R(x,y) = \sum_{a_{1},a_{2},b_{1},b_{2}=1}^{N} R_{a_{1}a_{2}}^{b_{1}b_{2}}(x,y) e_{b_{1},a_{1}} \tp e_{b_{2},a_{2}}, $$
where
$$ R_{a_{1}a_{2}}^{b_{1}b_{2}}(x,y) = \bar{W}(xy^{-1}|a_{1}-b_{2}) W(x^{-1}y^{-1}|a_{1}-a_{2}) \bar{W}(x^{-1}y|a_{2}-b_{1}) W(xy|b_{2}-b_{1}). $$
To obtain the \textit{difference property} we set
\begin{eqnarray}
 R(z) = \lim_{x,y\rightarrow \infty} R(x,y),
 \label{fzz}
 \end{eqnarray}
where $z= \frac{x}{y}$. For odd $n \leq 11$ we are able to computationally verify that this $R$-matrix is equivalent (up to a basis transformation) to the $D(D_{n})$ $R$-matrix while setting $\lambda = - w^{-1}$. For odd $n>11$ we can verify that $R(0)$ is indeed equivalent to $(\pi_{n}^{+} \tp \pi_{n}^{+})\mathcal{R}$.

We now briefly comment on the descendants obtained from $D(D_{2n}),$ $n$ even.  There are two distinct $R$-matrices, $R^{+}(z)$ and $R^{-}(z)$, which both square to the identity.
The multiplicities of the eigenvalues of $R^{+}(z)$ and $R^{-}(z)$ differ; furthermore neither eigenvalue spectrum matches that of $R(z)$ as defined by Equation (\ref{fzz}).
Hence unlike the case when $n$ is odd, the $R$-matrix \eqref{Rcheckz} is not equivalent up to basis transformation of a limit of the Fateev--Zamolodchikov $R$-matrix.

\section{Summary}
In this paper we used the framework of descendants to construct $R$-matrices from the Drinfeld
doubles of dihedral groups.
For $3 \leq n \leq 17 $ and $w$ a primitive $n$th
root of unity,
$$
R(z) = \sum_{i,j,a=0}^{n-1} \left[\frac{1}{n}\sum_{b=0}^{n-1} w^{bj}
\prod_{p=1}^{a} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+ b-a)}}
\right) \right] e_{i+j,i+a} \tp e_{i+a+j,i}.
$$
is a descendant of the six-vertex model
$$
	r(z) =
	\left(
	\begin{array}{cccc}
		wz^{-1} - w^{-1}z & 0 & 0 & 0\\
		0 & z^{-1} - z & w - w^{-1} & 0\\
		0 & w - w^{-1} & z^{-1} - z & 0\\
		0 & 0 & 0 & wz^{-1} - w^{-1}z
	\end{array}
	\right),
$$
with corresponding $L$-operator
$$
L(z) = \sum_{i=0}^{n-1}\left \{ (w^{i}e_{1,2} + w^{-i}e_{2,1}) \tp e_{i,i}
+ z \left[ e_{1,1} \tp e_{i-1,i} + e_{2,2} \tp e_{i+1,i} \right] \right\}.
$$
We conjecture that this holds true for all $n \geq 2$.

We also showed that when $n$ is even we obtain $R$-matrices with a second, non-spectral parameter.
Specifically, let $n>2$ be an even integer and $w$ a primitive $n$th root of unity. Given
$$ R^{+}(z) = \sum_{i,j,a=0}^{n-1} \left[\frac{1}{n} \sum_{b=0}^{n-1} w^{bj}
\prod_{p=1}^{a} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}} \right)
\right] e_{i+j,i+a} \tp e_{i+a+j,i} $$
and
$$ R^{-}(z) = \sum_{i,j,a=0}^{n-1} \left[\frac{1}{n} \sum_{b=0}^{n-1}
(-1)^{b-a} w^{bj} \prod_{p=1}^{a} \left( \frac{z + w^{(2p-1+b-a)}}{1 + zw^{(2p-1+b-a)}}
\right) \right] e_{i+j,i+a} \tp e_{i+a+j,i}, $$
then
$$ R(z, \mu) = R^{+}(z) + \mu R^{-}(z) $$
satisfies Equation $(\ref{eqnYBE2par})$ for $n \leq 12$ and is conjectured to satisfy it for larger $n$. Moreover,
$R(z, \mu)$ and $L(z)$ together satisfy
$$
L_{12}(x)L_{13}(y)R_{23}(x^{-1}y, \mu) = R_{23}(x^{-1}y, \mu)L_{13}(y)L_{12}(x).
$$

{\bf Acknowledgements} The authors thank H. Au-Yang, V. Bazhanov, V. Mangazeev, and J. Perk for their insightful comments.
P.S.I. acknowledges the support of an Early Career Researcher Grant from The University of Queensland.  K.A.D. acknowledges
the support of the Australian Research Council under Discovery Project DP1092513.

redpajama_set_name: RedPajamaArXiv


Contracting PLUS raises vital funds for the Peter McVerry Trust!
Contracting PLUS is delighted to announce that we recently raised and donated 2,100 to the Peter McVerry Trust following our annual Agency Charity Table Quiz!
As part of our corporate and social responsibility commitments for 2016, we partnered with our friends and local business networks to support a causes that is close to all our hearts, to try and make a real difference in the lives of people who need it most.
Commenting upon receiving the cheque at Contracting PLUS HQ, Nell Ward, Corporate Fundraising Manager of Peter McVerry Trust said: "Corporate support provides an invaluable contribution to the work of the Peter McVerry Trust. It's with the support of companies like Contracting PLUS, their clients, networks and friends that we can continue our work providing vital housing, homeless, prevention and support services to those on the margins of our society".
We appreciate the support of all of recruitment community who took park and of the below corporate sponsors who came together to truly make a difference in raising vital funds to support the Peter McVerry Trust.
Thank you so much to all who helped make our night such a success! We hope you enjoyed the night and please remember to thank and acknowledge our wonderful sponsors without whom, the night wouldn't have been so prosperous!


Tasting of wines from The Hilt at Mayfair Fine Wines.
2010 The Hilt The Old Guard Chardonnay - alcoholic, pretty ripe, really sharp on the nose, oaky, a little pungent, with vanilla notes.
2010 The Hilt The Vanguard Pinot Noir - sweet and ripe, almost jammy. A little pungent with sulfur. Later some forest notes when it opened up more.
2009 The Hilt The Vanguard Pinot Noir - pretty sweet, less pungent. Still lots of oak here, and some forest notes.
2010 The Hilt The Old Guard Pinot Noir - sweet fruit and coffee notes.
2011 Château Alcée - pretty pleasant and easy drinking. A little young now. Sweet fruit with dried herbs.


Chicago Processing, a member of the Plymouth Tube company, offers high quality Edging, Slitting and Processing of stainless steel and various alloys.
We've got the edge on tough applications.
Chicago Processing offers a variety of value-added services and capabilities to better serve the needs of our customers. We do more than just tubes and shapes. We provides steel coil slitting and edge in a wide variety of sizes and materials for customers in a number of industries. Our Chicago location, in operation since 1983, can address the unique needs of our customers with speed and clarity.


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As Christians, we know that prayer is the most powerful way for us to experience God. But most Christians would agree that their prayer life is not what it should be. I've read many books about prayer in an effort to improve my prayer life. Here are 10 of those books that I would recommend.
Many Christians want to know the way in which to pray, but they are just not sure how to get started. There are a number of models that people follow, such as what is known as "The Lord's Prayer" (Matthew 6: 9-13). A new book I would recommend that discusses this model is Albert Mohler's The Prayer That Turns the World Upside Down: The Lord's Prayer as a Manifesto for Revolution. David's prayer in Psalm 51 is a good model for contrition. Paul Miller in his excellent book A Praying Life, talks about the use of prayer cards (one for each person you are praying for), while a friend of mine uses a prayer jar. I have tended to use a prayer list to capture my prayer requests. In addition, in his book Prayer: Experiencing Awe and Intimacy with God, the best book on prayer that I have read, Tim Keller writes "From earliest times, the Christian church adopted the Psalms of the Old Testament to be its prayer book".
Some people use a daily "Quiet Time" for their prayer time, while others, including my wife, pray throughout the day.
Rejoice always, pray without ceasing, give thanks in all circumstances; for this is the will of God in Christ Jesus for you. 1 Thessalonians 5:16-18.
A – Adoration. Far too often, we begin our prayers to God with what we are asking from Him, which in this model is called "Supplication". I like this acrostic as it begins with adoration, or praise. Beginning with a time of praise for who God is and what He has done, is a wonderful way to start your time of prayer.
C – Confession. After beginning with a time of adoration, we move to a time of confession. This is a time to pray for sins that we have committed against God and others and ask for forgiveness from God.
T – Thanksgiving. The "T" in the acrostic is a wonderful time to thank our Heavenly Father for all of the blessings that you have been given. Far too often we just take the blessings we have been given for granted.
S – Supplication. Supplication is not a word we use often. Think of the "S" in the acrostic as the requests you bring before your Heavenly Father – for your loved ones, your church, your workplace, national and world issues, yourself, etc. As I mentioned earlier, this is where many of us start (and end) our prayers.
The A.C.T.S. model of prayer helps me stay focused and organized in my prayers, beginning with adoration or praise. What model or method do you follow for your prayers?
Categories: Articles ~ Personal | Tags: A.C.T.S., model for prayer, prayer | Permalink.
I have found Luther's Morning/Evening Prayers are a good model for me, with some contemporizing (the lingo of the 1500's isn't quite my style!).


The island of Sao Miguel in the Mid-Atlantic ocean is hundreds of miles away from mainland Europe with the other 8 islands all within a short flight away. Sao Miguel is the largest island of the Portuguese islands known as the Azores. Discovered in the early 15th century the islands location became one of the first major commercial posts between Europe and North America in the early days of exploration. The island is filled with unique, nature based activities while visiting this beautiful green island. When arriving off the coast of the Azores, just think you are arriving like the majority of historical figures throughout history. The island of Sao Miguel looks like a grain of sand in the distance, until you approach closer and witness the amazing flora and fauna if this island paradise. It was through these waters that boats sailed from Europe to the new world, being the last land most navigators would see for months during the years of early exploration. The island is filled with natural wonders, museums, great beaches and unique restaurants. Make the most of your trip and hire a local guide to make the most of your island experience. Hailed, by National Geographic Traveller for its unique Eco-friendly existence, here the waterfalls, hot springs and jaw dropping gardens are all at your fingertips. The marina, which is located on the islands capital city of Ponta Delgada, right in the heart of a very beautiful, dynamic small city is the perfect starting point to begin your "Day in the Azores". It is recommended to try the local cuisine while visiting the island as the Azores are recognized for their high quality meat and dairy industry with the seafood industry being a driving force for the economy, due to the abundant, high quality fish that are caught of the Azorean coastlines. Make sure that if you are looking for authentic gifts from the island that it states that it was made in the Azores. A must is to buy the green or black tea grown here on the island of Sao Miguel since 1883. The famous Cha Gorreana brand in Portugal produces organic teas at their family owned and operated estate where they are open daily from 8am-8pm free for tourists to come explore the vast tea plantation and unique 18th century machinery.


Caren Helene Rudman works with mixed media, photography, and writing. She has exhibited locally in Chicago and Nationally including shows in NYC. She received her MA from NYU in a program affiliated with the International Center of Photography. Since 2010 she has curated an biennial exhibition, Voices and Visions; Standing on the Bridge Between Health and Disease, at The Art Center Highland Park, which traveled nationally. Caren's work examines the dichotomies between self and other, inside and outside, mind and body. She delves into the notion of genetics, linking past and future generations. She is currently the curator at The Art Center Highland Park.
Redefining Beauty Project is an ongoing collaboration with women across the world who share images of themselves after enduring surgeries, biopsies, or radiation. The mixed media images are not shocking or disturbing, but rather they help all of us see the beauty of what lies beneath the surface. Breast cancer is uniquely linked to external archetypes of femininity and when bodies are altered, identity is often affected. Like the clothes and bras that cover the scars to make us feel "normal", and "like ourselves again", the series reveals the layers that reflect the complexities of accepting our imperfect bodies.


text:
Io e il sig. Rosso è il quarto album in studio del cantautore italiano Stefano Rosso, pubblicato nel 1980.

Il disco

Dopo l'abbandono della RCA Italiana, Stefano Rosso passa ad una nuova etichetta, la Ciao Records, nata appena due anni prima a Milano ed il cui direttore artistico è Osvaldo Bernasconi, ex batterista dei Profeti.

L'etichetta decide di far partecipare il cantante al Festival di Sanremo 1980, con una delle canzoni registrate per l'album in uscita, L'italiano, canzone che ironizza sull'italiano medio, che si classifica all'ottavo posto e che viene pubblicata su 45 giri, L'italiano/Quarant'anni, riscuotendo un buon successo di vendite.

L'album segue il singolo di qualche settimana, ed è particolarmente importante nella discografia del cantautore perché è il primo in cui compaiono due brani interamente strumentali (Banjorosso e Metro's rag) in cui Stefano Rosso può mettere in evidenza le sue doti di chitarrista e banjoista, decidendo quindi di far risaltare in maniera più decisa l'aspetto musicale.

La produzione e gli arrangiamenti sono ancora curati da Gianni Marchetti, mentre le registrazioni vengono effettuate allo studio Trafalgar di Roma con il tecnico del suono Giorgio Agazzi, allo Studio Bach di Milano con il tecnico del suono Nino Iorio e soprattutto negli innovativi Stone Castle Studios di Carimate, con Paolo Tofani, chitarrista degli Area come tecnico del suono.

Tutte le canzoni sono scritte dallo stesso Rosso, che per la prima volta non si firma con il suo vero nome, Stefano Rossi, ma con lo pseudonimo, e sono edite dalla Edizioni Musicali Itaca.

Oltre ai due strumentali ed alle canzoni del 45 giri, altri brani degni di nota sono l'allegra Quando partì Noè, che viene ripresa anche nel finale, e che in origine era stata scritta da Stefano Rosso per uno spettacolo del gruppo di cabaret La Smorfia (il trio formato da Massimo Troisi, Lello Arena e Enzo De Caro), Ma dove andiamo, brano con influenze musicali popolari, Quarant'anni, bilancio di un'età che, comunque, il cantautore doveva ancora raggiungere (avendo all'epoca 32 anni), Milano (ispirata da un suo soggiorno in questa città nel 1975 per partecipare come chitarrista fisso alla trasmissione Alle cinque della sera, condotta da Gianni Morandi ed Elisabetta Viviani ed in onda su Rai 2; ma i due brani più belli sono Quello che mi resta e Per male che gli vada, due ballate malinconiche in cui il cantautore, accompagnandosi con la chitarra acustica, nella prima fa un'analisi di quello che è stata la sua vita fino a quel momento (« Ecco quello che mi resta: / tanta voglia dentro i panni, / tanta confusione in testa ») mentre nella seconda allarga lo sguardo al mondo, rilevandone i problemi e le inadempienze dei potenti.

La copertina del disco, ideata da Giorgio Pertici e realizzata da Gianni Cicarelli e Luca Coelli, raffigura una carta da gioco, il fante di cuori, con il volto del cantante; è apribile e all'interno sono riportati i testi delle canzoni, ma non i nomi dei musicisti che hanno partecipato alle sessioni di registrazione.

Il 33 giri, pur entrando in classifica, rimane in posizioni molto basse, non superando la trentesima posizione rimanendo in hit-parade per sole 4 settimane, al contrario del 45 giri che riscuoterà un successo maggiore.

Io e il sig. Rosso è l'unico album di Stefano Rosso pubblicato dalla Ciao Records, il cantautore infatti l'anno successivo passerà alla Lupus.

Tracce
LATO A
Quando partì Noè - 3:33
Ma dove andiamo - 3:12
Banjorosso - 1:34
L'italiano - 3:35
Per male che gli vada - 3:19

LATO B
Grazie a Dio - 4:08
Milano - 3:19
Quarant'anni - 3:49
Quello che mi resta - 3:05
Metro's rag - 1:40
Quando partì Noè (finale) - 0:42

Note

Bibliografia
 Maurizio Becker, C'era una volta la RCA. Conversazioni con Lilli Greco, Coniglio Editore, Roma, 2007
 Stefania Rosso & Mario Bonanno, Che mi dici di Stefano Rosso?, edizioni Stampa Alternativa, 2011.

Collegamenti esterni

redpajama_set_name: RedPajamaWikipedia