text:
90sComedyCult LegendsHard to DieHorrorReviewRevTerry
Hard to Die (1990) Review by RevTerry
Jim Wynorski is one of the unsung kings of trash, or at the very least a prince. A student of Roger Corman's school of cheap blood and boobs, Wynorski has been making bargain bin classics, mockbusters and schlocky sequels for over thirty years. Though he may not be a household name in most circles, everyone has no doubt seen the cover of one of his over 100 films in a 4 pack of movies at Wal-Mart, on netflix or a title listed on their digital cable TV guide during a late night. He makes a lot of shit and it is shit, but fun shit and I fucking love even some of his worst. But in my opinion the apex of his art is the sorority-less "Massacre" film Hard to Die (aka Tower of Terror aka Sorority House Massacre 3:Hard to Die). He has better films (see Chopping Mall, Not of This Earth) and more standard-fare genre films (see Deathstalker II) but still something just feels so right about this obviously rushed Frankenstein's monster of a movie, stitched together of sleaze and recycled parts.
After finishing Sorority House Massacre 2: Nighty Nightmare (1990) for Julie Corman using the pseudonym Arch Stanton, Wynorski was tasked with "remaking" the same film (cast and all) using the sets pieces from the recently wrapped Corporate Affairs (1990) by her husband Roger . Corman brought Mark Mcgee and J.B. Rogers to work a rushed script and the team pumped out the film for the VHS market within the same year as the films cannibalized in its inception. The film was written and shot in about 15 days according to MacGee and Wynorski. It retained most of the cast, plot and wardrobe as Sorority House Massacre 2 and is considered the unofficial third in the series, even going under the title Sorority House Massacre 3 (among other names). Plot elements had to be reworked for the sky-rise setting and the writer struggled to meet Cormans requirements (boobs mostly). McGee remarked "[The sequence] ...is as ridiculous and infantile as anything you can imagine. But half-naked women is just about all that a film like this has to offer" The film lived two lives getting released once again, this time in theaters, under the name Tower Of Terror. Look for cameos from Kelli Maroney, Forrest J. Ackerman and Toni Naples. Gail Harris, Karen Mayo-Chandler,Deborah Dutch,Melissa Moore and Bridget Carney make up the gang of scream queens, some playing what can only be lost relatives of their supposedly dead characters from SHM2 .
The film follows the already loose plot of SHM2, while shoehorning it into the setting of a high-rise. Five very early 90's bombshells are (locked?) in a building over night for a lingerie innovatory, for which they have been hired as temps. Orville Ketchem from SHM 2 is somehow alive, now the janitor, and for some reason recollects the events of Sorority House Massacre (1986), the first one, even though he isn't in that one. I guess maybe he could have been lurking in the background in 1986 during the events of the first film, cleaning up a mess or something. Who gives a fuck? Orville is the shit and I'm glad he's back. Anyway before any real work can get done of course everyone must take a shower, ...everyone. The movie may be more than 1/3 shower scenes, complete with very un-sexy cartoon squeaky sound effects. It's great. Then, before trying on lingerie, they release the spirit of a once imprisoned serial killer that causes bloody, mostly off screen, mayhem. The girls fight for their lives with the help of Orville, who acts as human meat shield, suffering through perils that would kill any less a pure man.
The movie is nothing but a cheese sandwich, in-between more cheese and with a delicious E-Z-Cheeze spread. When you know exactly what you're getting these kind of films can be a lot fun and if you have seen SHM2 and can picture an office building, then you know what you're are getting in to. Even as a make-it-fit clone of SHM2, the movie is the perfect amount of trashy, brainless crap for a good watch or background noise. It is high on kills and has some gore but moves most of the action off screen, likely due to budget, and tries to make up for it in nudity. The dialog is silly as shit "...don't worry, I'm no hero." and the actresses (although legends in their own right) haven't taken any classes in the weeks since SHM2. Sound quality and lighting carries over from SHM2 in terms of quality along with Wynorski's "eye". I'm never bored, although with the lack of story and strange jumps from scenes I have never quite understand why this is all happening despite watching several times. Somehow, altogether combined, the corny mess works on every note it haphazardly hits.
Wynorski, like David DeCoteau and Fred Olen Ray, may never receive the accolades deserved for his many rainy day classics and I probably would have even less social contact if everyone knew how many times I had actually watched this fucking shitty movie, because the world sucks that way. This heartless globe doesn't deserve the majestic, squeaky rubdown that is Hard to Die. Now sign my petition to have Wynorski make a SHM4 where Orville Ketchum gets his much deserved happy ending. Maybe a wife, some kids or at least a job where naked people are not getting killed all over the place.
NC-17 | 1h 17min | 1990/1993
Director: Jim Wynorski (as Arch Stanton)
Writers: Mark Thomas McGee (as Mark McGee), J.B. Rogers (as James B. Rogers)
IMBD
VHS on Amazon
Tags: 90s , Comedy , Cult Legends , Hard to Die , Horror , Review , RevTerry


text:
HomeSportsTennessee Vols play #18 Purdue at Bad Boy Mowers Battle 4 Atlantis,...
Tennessee Vols play #18 Purdue at Bad Boy Mowers Battle 4 Atlantis, Wednesday
Tennessee Vols head to the Bahamas to take on #18 Purdue at the Bad Boy Mowers Battle 4 Atlantis Wednesday at 11:00am CT. (Tennessee Athletics)
Tennessee vs. #18 Purdue
Wednesday, November 22nd, 2017 | 11:00am CT
Paradise Island, Bahamas | Atlantis Resort | TV: ESPN2
Knoxville, TN – Tennessee opens play at the Bad Boy Mowers Battle 4 Atlantis against one of the early season's top scoring teams in 18th-ranked Purdue. The Boilermakers enter Wednesday's contest (11:00am CT, ESPN2) boasting a scoring average of 102.0 points per game through the first four games of the season.
In fact, Purdue has eclipsed the century mark in three of its four games, including a 106-64 romp over Fairfield on Saturday in which it buried a program-record 19 3-pointers.
Purdue's quick start—aided by its 11-game international schedule over the summer—has captured the attention of Tennessee head coach Rick Barnes, who believes his team is preparing to face a Boilermaker squad already nearing midseason form.
"This will be good for us," Barnes said. "I'm really impressed with Purdue, having watched them play. I think they have benefitted greatly from their trip to the World University Games this summer. They are good, and there is no doubt about it. For our team, this is the time of year to get going, and the competition we will face [at the Battle 4 Atlantis] is going to be good."
Purdue is only one member of an impressive field set to the take the court this week on Paradise Island, Bahamas. On Day Two, the Vols will meet either No. 5 Villanova—last season's No. 1 overall seed in the NCAA Tournament—or Western Kentucky. Second-ranked Arizona, NC State, Northern Iowa and SMU round out the field. Four of the tournament's eight teams (Arizona, Purdue, SMU, Villanova) were NCAA Tournament teams a year ago.
"I can't imagine there is any better tournament around the country than this one," added Barnes. "When you look at the teams on the side we are on, it's pretty interesting. These tournaments are good for everyone. I think it gives you and early gauge of where you are and what you need to do."
Schofield Learning and Improving at the Wing
One of the key storylines heading into this season was Admiral Schofield's transition to playing the wing for the Vols. Though Schofield is still adapting to the new role, Barnes has noticed some positive progress from the 6-5, 238-pound junior through UT's first two games.
"He is still learning to play the position," Barnes said. "Since he's been here, he has been able to play on the perimeter some, but it's a different position. You can also drive inside and score, which he can do. I think he will continue getting better and better each game. He will see different things that people will do. People are going to try and exploit him in some ways, but he will have to guard on the other end as well. He's working to try and become a better perimeter defender. I think he will get better every game."
Through two games, Schofield is averaging 18 points and 5.5 rebounds per contest, while shooting 63 percent (12-19) from the field and 57 percent (4-7) from beyond the arc. Against Presbyterian, he posted 22 points on 8-of-12 shooting, grabbed five boards and had a pair of steals for an all-around performance.
With every game, he is becoming more settled into Barnes' system and embracing his role as a leader on the team.
"I've gotten more comfortable each game," Schofield said. "I'm starting to understand how I fit in with the flow of things. I just have to understand the details and get better every week. I think I've done a pretty good job. This is another opportunity for myself and my teammates to go out and display what kind of talent we have as a team and individuals."
Bad Boy Mowers Battle 4 Atlantis
Admiral Schofield
Paradise Island Bahamas
UT Basketball
UT Men's Basketball
UT Vols
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text:
\section{Introduction and Statement of Results}

In \cite{HandT} Terwilliger and the present author gave a
presentation of the three-point $\mathfrak{sl}_2$ loop algebra via
generators and relations.  To obtain this presentation we defined
a Lie algebra $\boxtimes$ by generators and relations and
displayed an isomorphism from $\boxtimes$ to the three-point
$\mathfrak{sl}_2$ loop algebra.

In this paper we classify the finite-dimensional
irreducible $\boxtimes$-modules.  To obtain this classification, we
exploit a connection between $\boxtimes$ and a certain
infinite-dimensional Lie algebra which appears in the physics literature,
called the Onsager algebra \cite{A}, \cite {DG}, \cite{GR}, \cite{ons1}.
Before we explain our results we first give some
background on $\boxtimes$ and the Onsager algebra.  We start with their
definitions.

Throughout the paper $\Bbb K$ denotes an algebraically closed field
of characteristic $0$.

\begin{definition}
\label{def:main} \rm \cite{HandT}
Let $\boxtimes$ denote the Lie algebra over $\Bbb K$ with generators
$$ \{ X_{rs} | r,s \in \mathbb I, r \not=s \} \qquad \qquad \mathbb I = \{0,1,2,3 \}
$$
and the following relations.
\begin{enumerate}
\item
For all distinct $r,s \in \mathbb I$,
\begin{eqnarray}
X_{rs}+ X_{sr} =0. \label{eq:1}
\end{eqnarray}
\item
For all mutually distinct $r,s,t \in \mathbb I$,
\begin{eqnarray}
[X_{rs},X_{st}]=2X_{rs} + 2X_{st}. \label{eq:2}
\end{eqnarray}
\item
For all mutually distinct $r,s,t,u \in \mathbb I$,
\begin{eqnarray}
[X_{rs},[X_{rs},[X_{rs},X_{tu}]]] = 4[X_{rs},X_{tu}]. \label{eq:3}
\end{eqnarray}
\end{enumerate}
We call $\boxtimes$ the {\it tetrahedron algebra}.
\end{definition}

\begin{definition}
\rm \cite{ons1}, \cite{P} \label{def:oalg} Let $\mathcal{O}$ denote the Lie
algebra over $\Bbb K$ with generators $X,Y$ and relations
\begin{eqnarray}
[X,[X,[X,Y]]] = 4 [X,Y], \label{eq:fo1} \\
\left[ Y,[Y,[Y,X]] \right] = 4 [Y,X]. \label{eq:fo2}
\end{eqnarray}
We call $\cal O$ the {\it Onsager algebra}. We call $X,Y$ the {\it
standard generators} for $\mathcal{O}$.
\end{definition}

We recall the connection between $\boxtimes$ and $\mathcal{O}$.  For mutually
distinct $r,s,t,u \in \mathbb I$ there exists a Lie algebra
injection from $\mathcal{O}$ into $\boxtimes$ that sends \begin{eqnarray*}
X & \mapsto & X_{rs} \\
Y & \mapsto & X_{tu} \end{eqnarray*} \cite[Corollary 12.2]{HandT}.  We
call the image of this injection an {\it Onsager subalgebra} of
$\boxtimes$.  Observe that $\boxtimes$ has three Onsager subalgebras.
By \cite[Theorem 11.6]{HandT} the $\Bbb K$-vector space $\boxtimes$ is the
direct sum of its three Onsager subalgebras.

We now summarize the classification of finite-dimensional irreducible
$\mathcal{O}$-modules \cite{Dav1}, \cite{Dav2}.  We begin
with a comment.  Let $V$ denote a finite-dimensional irreducible
$\mathcal{O}$-module.  As we will see in Section \ref{sect:2}, the standard
generators $X,Y$ are diagonalizable on $V$.  Furthermore there
exist an integer $d \geq 0$ and scalars $\alpha, \alpha ^* \in
\Bbb K$ such that the set of distinct eigenvalues of $X$ (resp.
$Y$) on $V$ is $\{d-2i+\alpha | 0 \leq i \leq d \}$ (resp.
$\{d-2i+\alpha^* | 0 \leq i \leq d \}$).  We call the ordered pair
$(\alpha , \alpha^*)$ the {\it type} of $V$.  Replacing $X,Y$ by
$X-\alpha I, Y-\alpha ^* I$ the type becomes $(0,0)$.  Therefore it
suffices to classify the finite-dimensional irreducible $\mathcal{O}$-modules
of type $(0,0)$.

We begin with a special case.  Observe that up to isomorphism, there
exists a unique irreducible $\mathcal{O}$-module of dimension $1$ and type
$(0,0)$.  We call this the {\it trivial} $\mathcal{O}$-module.

Let $\mathfrak{sl}_2$ denote the Lie algebra over $\Bbb K$ with basis
$e,f,h$ and Lie bracket $$ \left[ e,f \right] = h, \qquad \left[ h,e
\right] = 2e, \qquad
\left[ h,f \right] = -2f. $$ Let $t$ denote an indeterminate and let
$L(\mathfrak{sl}_2)$ denote the loop algebra $\mathfrak{sl}_2 \otimes
\Bbb K [t,t^{-1}]$, where $\otimes$ means $\otimes_{\Bbb K}$.  By
\cite[Proposition 1]{roan} there exists a Lie algebra injection from
$\mathcal{O}$ into $L(\mathfrak{sl}_2)$ that sends \begin{eqnarray*}
X & \mapsto & e \otimes 1 + f \otimes 1 \\
Y & \mapsto & e \otimes t + f \otimes t^{-1}. \end{eqnarray*} For the moment
we identify $\mathcal{O}$ with its image under the above injection.  For
nonzero $a \in \Bbb K$ we define the Lie algebra homomorphism
$EV_a: L(\mathfrak{sl}_2) \rightarrow \mathfrak{sl}_2$ by $EV_a(u
\otimes g(t)) = g(a)u$ for all $u \in \mathfrak{sl}_2$ and $g(t) \in \Bbb
K [t,t^{-1}]$.  Let $ev_a$ denote the restriction of
$EV_a$ to $\mathcal{O}$.  Then $ev_a:\mathcal{O} \rightarrow \mathfrak{sl}_2$ is a Lie
algebra homomorphism which we call the {\it evaluation homomorphism}
for $a$. Let $V$ denote an irreducible $\mathfrak{sl}_2$-module with
finite dimension at least $2$.  We pull back by $ev_a$ to get
an $\mathcal{O}$-module structure on $V$.  We call this an {\it evaluation
module} for $\mathcal{O}$ and denote it by $V(a)$.  The $\mathcal{O}$-module $V(a)$
is irreducible if and only if $a \not= \pm 1$ \cite[Lemma 4]{DR}.
Irreducible $\mathcal{O}$-modules $V(a)$ and $V(b)$ are isomorphic if and only
if $a = b^{\pm 1}$ \cite[Proposition 5]{roan}.

Let $U,V$ denote $\mathcal{O}$-modules.  Then $U \otimes V$
has an $\mathcal{O}$-module structure given by
$$ x.(u \otimes v) = (x.u) \otimes v + u \otimes (x.v) \qquad \quad
x \in \mathcal{O}, \quad u \in U, \quad v \in V.$$ Let $V$ denote an
$\mathcal{O}$-module that is the tensor product of finitely many evaluation
modules.  If $V$ is irreducible then it is type $(0,0)$ \cite[p.
3281]{DR}.

The classification of finite-dimensional irreducible $\mathcal{O}$-modules
of type $(0,0)$ is given in the following three theorems.

\begin{theorem} \cite[Theorem 6]{DR}
Every nontrivial finite-dimensional irreducible $\mathcal{O}$-module of type
$(0,0)$ is isomorphic to a tensor product of evaluation modules.
\end{theorem}

\begin{theorem} \cite[Proposition 5]{DR}
Let $V_1(a_1), \ldots, V_n(a_n)$ denote a finite sequence of
evaluation modules for $\mathcal{O}$, and consider the $\mathcal{O}$-module $V_{1}(a_1)
\otimes \cdots \otimes V_{n}(a_n)$.  This module is irreducible if and
only if $a_1, a_1^{-1}, \ldots, a_n,a_n^{-1}$ are
mutually distinct.
\end{theorem}

\begin{definition} \rm
Let $V_1(a_1), \ldots, V_n(a_n)$ denote a finite sequence of
evaluation modules for $\mathcal{O}$.  Let $V$ denote the $\mathcal{O}$-module
$V_{1}(a_1) \otimes \cdots \otimes V_{n}(a_n)$.  Any tensor product of
evaluation modules that can be obtained from $V$ by permuting the
order of the factors and replacing any number of the $a_i$'s with
their multiplicative inverses will be called {\it equivalent} to $V$.
\end{definition}

\begin{theorem}  \cite[Proposition 5]{DR}
Let $U$ and $V$ denote tensor products of finitely many evaluation modules
for $\mathcal{O}$.  Assume each of $U,V$ is irreducible as an $\mathcal{O}$-module.
Then the $\mathcal{O}$-modules $U$ and $V$ are isomorphic if and only if they
are equivalent.
\end{theorem}

This completes the classification of finite-dimensional irreducible
$\mathcal{O}$-modules.

We now state our main results.  They are contained in the
following two theorems and subsequent remark.

\begin{theorem} \label{th:H1}  Let $V$ denote a finite-dimensional
irreducible $\boxtimes$-module.  Then there exists a unique
$\mathcal{O}$-module structure on $V$ such that the standard generators $X,Y$ act
on $V$ as $X_{01},X_{23}$ respectively.  This $\mathcal{O}$-module structure is
irreducible and has type $(0,0)$.
\end{theorem}

\begin{theorem}  \label{th:H2}  Let $V$ denote a
finite-dimensional irreducible $\mathcal{O}$-module of type $(0,0)$. Then
there exists a unique $\boxtimes$-module structure on $V$ such that
the standard generators $X,Y$ act on $V$ as $X_{01},X_{23}$
respectively. This $\boxtimes$-module structure is irreducible.
\end{theorem}

\begin{remark}
Combining the previous two theorems we obtain a bijection between
the following two sets:
\begin{enumerate}
\item
The isomorphism classes of finite-dimensional irreducible
$\mathcal{O}$-modules of type $(0,0)$.
\item
The isomorphism classes of finite-dimensional irreducible
$\boxtimes$-modules.
\end{enumerate}
\end{remark}

\section{$\mathcal{O}$-Modules and Tridiagonal Pairs \label{sect:2}}

In order to prove Theorem \ref{th:H1} and Theorem \ref{th:H2} it will
be useful to consider how finite-dimensional irreducible $\mathcal{O}$-modules
are related to tridiagonal pairs.  To explain this relationship we
use the following concepts.  Let $V$ denote a vector space
over $\Bbb K$ with positive finite dimension.  By a {\it linear
transformation on $V$} we mean a $\Bbb K$-linear map from $V$ to
$V$.  Let $A$ denote a linear transformation on $V$.  For $\lambda
\in \Bbb K$ define
\begin{eqnarray}
V_A(\lambda) := \{ v \in V | A v = \lambda v \}.  \label{eq:espace}
\end{eqnarray}
Observe that $\lambda$ is an eigenvalue for $A$ if and only if
$V_A(\lambda) \not= 0$, and in this case $V_A(\lambda)$ is the
corresponding eigenspace of $A$.
We recall that the sum $\sum_{\lambda \in \Bbb
K}V_A(\lambda)$ is direct.  We call $A$ {\it diagonalizable} whenever
$V = \sum_{\lambda \in \Bbb K}V_A(\lambda)$.

\begin{lemma}
\label{lem:conf}
Let $V$ denote a vector space over $\Bbb K$ with positive finite
dimension.  Let $A$ and $A^*$ denote linear transformations on $V$.
Then for $\lambda \in \Bbb K$ the following $(i),(ii)$ are equivalent.
\begin{enumerate}
\item
The expression $[A,[A,[A,A^*]]] - 4[A,A^*]$ vanishes on
$V_A(\lambda)$.
\item
\begin{eqnarray}
\label{eq:un2}
A^* V_A( \lambda) \subseteq V_A( \lambda+2)+V_A( \lambda)+V_A( \lambda-2).
\end{eqnarray}
\end{enumerate}
\end{lemma}

{\it Proof.}
Let $\Phi$ denote the expression in $(i)$ and observe
\begin{eqnarray*}
\Phi = A^3 A^* - 3A^2 A^* A + 3A A^* A^2 - A^* A^3 - 4A
A^* + 4 A^* A.
\end{eqnarray*}
For $v \in V_A(\lambda)$ we evaluate $\Phi v$ using $Av = \lambda v$
to find
\begin{eqnarray*}
\Phi v &=& (A^3 A^* - 3 \lambda A^2 A^* + 3 \lambda^2 A A^* -
\lambda^3 A^* - 4A A^* + 4 \lambda A^*)v \\
&=& (A - (\lambda+2)I)(A - \lambda I)(A - (\lambda-2)I)A^*v.
\end{eqnarray*} The scalars $\lambda+2,\lambda,\lambda-2$ are mutually
distinct since ${\rm Char} (\Bbb K) = 0$.  The result follows.
{~~~\vrule height .75em width .4em depth .2em}

\medskip

We now recall the concept of a {\it tridiagonal pair}
\cite{bandt}, \cite{iandt}, \cite{ITT}, \cite{ito}.

\begin{definition}
\label{def:tp} \rm \cite{ITT} Let $V$ denote a vector space over
$\Bbb K$ with positive finite dimension.  By a {\it tridiagonal
pair} on $V$, we mean an ordered pair $A,A^*$ where $A$ and $A^*$
are linear transformations on $V$ that satisfy the following four
conditions.
\begin{enumerate}
\item Each of $A,A^*$ is diagonalizable.
\item There exists an ordering $V_0,V_1, \ldots, V_d$ of the
eigenspaces of $A$ such that
\begin{eqnarray}
\label{eq:un1}
A^* V_i \subseteq V_{i+1} + V_i + V_{i-1} \qquad \qquad 0 \leq i \leq d,
\end{eqnarray}
where $V_{-1}=0$, $V_{d+1} =0$.
\item  There exists an ordering $V_0^*,V_1^*, \ldots, V_{\delta}^*$ of the
eigenspaces of $A^*$ such that
$$A V_i^* \subseteq V_{i+1}^* + V_i^* + V_{i-1}^* \qquad \qquad 0 \leq i
\leq \delta,$$
where $V_{-1}^*=0$, $V_{\delta+1}^* =0$.
\item  There does not exist a subspace $W$ of $V$ such that $AW
\subseteq W$, $A^*W \subseteq W$, $W \not= 0$, and $W \not= V$.
\end{enumerate}
\end{definition}

We comment on Definition \ref{def:tp}.  Let $A,A^*$ denote a
tridiagonal pair on $V$ and let $d$ and $\delta$ be as in
Definition \ref{def:tp}.  From \cite[Lemma 4.5]{ITT} we find $d=
\delta$.  We call this common value the {\it diameter} of $A,A^*$.
An ordering of the eigenspaces of $A$ (resp. $A^*$) will be called
{\it standard} whenever it satisfies Definition \ref{def:tp}(ii)
(resp. Definition \ref{def:tp}(iii)).   Let $V_0, \ldots, V_d$
denote a standard ordering of the eigenspaces of $A$.  By \cite[p.
388]{ter2}, the ordering $V_d, \ldots, V_0$ is standard
and no other ordering is standard.  A similar result holds for
$A^*$.  By an {\it eigenvalue sequence} (resp. {\it dual
eigenvalue sequence}) of $A,A^*$ we mean an ordering of the
eigenvalues for $A$ (resp. $A^*$) for which the corresponding
ordering of the eigenspaces is standard.

\begin{definition}
\rm Let $\theta_0, \theta_1, \ldots, \theta_d$ denote a finite
sequence of scalars in $\Bbb K$.  We say this sequence is {\it
arithmetic with common difference $2$} whenever $\theta_{i-1} -
\theta_{i}= 2$ for $1 \leq i \leq d$.
\end{definition}

\begin{theorem}
\label{th:p1} Let $V$ denote a vector space over $\Bbb K$ with
positive finite dimension.  Let $A$ and $A^*$ denote linear
transformations on $V$.  Then the following $(i),(ii)$ are
equivalent.
\begin{enumerate}
\item
$A,A^*$ is a tridiagonal pair on $V$ whose eiqenvalue
sequence and dual eigenvalue sequence are both
arithmetic with common difference $2$.
\item
There exists an irreducible $\mathcal{O}$-module structure on $V$ such that
the standard generators $X,Y$ act on $V$ as $A,A^*$ respectively.
\end{enumerate}
\end{theorem}

{\it Proof.} $(i) \Rightarrow (ii):$  We
first show that $A,A^*$ satisfy
\begin{eqnarray}
\label{eq:o1} [A,[A,[A,A^*]]] = 4[A,A^*].
\end{eqnarray}
By assumption the tridiagonal pair $A,A^*$ has an eigenvalue
sequence that is arithmetic with common difference $2$. Let
$V_0, V_1, \ldots, V_d$ denote the corresponding ordering of the
eigenspaces of $A$.  This ordering is standard by construction so
it satisfies (\ref{eq:un1}).  By these comments we find that $A^*$
satisfies (\ref{eq:un2}) for all $\lambda \in \Bbb K$.  By Lemma
\ref{lem:conf} and since $A$ is diagonalizable on $V$ we find $A,A^*$
satisfy (\ref{eq:o1}).  Reversing the roles of $A$ and $A^*$ in
the above argument we find $A,A^*$ satisfy
\begin{eqnarray}
\label{eq:o2} [A^*,[A^*,[A^*,A]]] = 4[A^*,A].
\end{eqnarray}
By (\ref{eq:o1}) and (\ref{eq:o2}) there exists an $\mathcal{O}$-module
structure on $V$ such that $X,Y$ act as $A,A^*$ respectively.
This $\mathcal{O}$-modules structure is irreducible by Definition
\ref{def:tp}(iv).

$(ii) \Rightarrow (i):$  \cite[Example 1.6]{ITT} Since $\Bbb K$ is
algebraically closed
all the eigenvalues for $A$ are contained in $\Bbb K$.  Since $V$
has positive finite dimension, $A$ has at least one eigenvalue
$\lambda$.  Since ${\rm Char} (\Bbb K) = 0$, the scalars $\lambda,
\lambda+2, \ldots$ are mutually distinct and therefore
cannot all be eigenvalues for $A$.  Hence there exists an
eigenvalue $\theta$ for $A$ such that $\theta +2$ is not an
eigenvalue for $A$.  The scalars $\theta, \theta-2,
\ldots$ are mutually distinct and therefore cannot all be
eigenvalues for $A$.  Hence there exists a nonnegative integer $d$
such that $\theta - 2i$ is an eigenvalue of $A$ for $0 \leq i
\leq d$ but is not an eigenvalue of $A$ for $i = d+1$. Abbreviate
$V_A(\theta -2i)$ by $V_i$ for $0 \leq i \leq d$.  By construction
$\sum_{i=0}^{d} V_i$ is $A$-invariant. By Lemma \ref{lem:conf} we
find $$A^* V_i \subseteq V_{i+1} + V_i + V_{i-1} \qquad (0 \leq i
\leq d),$$ where $V_{-1}=0$, $V_{d+1} =0$. Therefore
$\sum_{i=0}^{d} V_i$ is $A^*$-invariant.  Since the $\mathcal{O}$-module
structure is irreducible and since $\sum_{i=0}^{d} V_i \not= 0$ we
find $\sum_{i=0}^{d} V_i = V$.  We have now shown that $A$ is
diagonalizable and Definition \ref{def:tp}(ii) holds. Reversing
the roles of $A$ and $A^*$ in the above argument we find $A^*$ is
diagonalizable and Definition \ref{def:tp}(iii) holds. Definition
\ref{def:tp}(iv) is immediate since the $\mathcal{O}$-module $V$ is
irreducible. We have now shown that $A,A^*$ is a tridiagonal pair
on $V$. Recall that $\theta -2i$ is the eigenvalue for $A$
associated with $V_i$ for $0 \leq i \leq d$. The ordering
$V_0, \ldots, V_d$ is standard so the sequence $\theta -2i$
$(0 \leq i \leq d)$ is an eigenvalue sequence for $A,A^*$. This
sequence is arithmetic with common difference $2$.  We have now
shown $A,A^*$ has an arithmetic eigenvalue sequence with common
difference $2$. Reversing the roles of $A$ and $A^*$ in the above
argument we find that $A,A^*$ has an arithmetic dual eigenvalue
sequence with common difference $2$. {~~~\vrule height .75em width .4em depth .2em}

\medskip

\begin{definition}
\rm Let $V$ denote a finite-dimensional irreducible $\mathcal{O}$-module.
By Theorem \ref{th:p1} there exist an integer $d \geq 0$ and
scalars $\alpha, \alpha ^* \in \Bbb K$ such that the set of
distinct eigenvalues of $X$ (resp. $Y$) on $V$ is $\{d-2i+\alpha |
0 \leq i \leq d \}$ (resp. $\{d-2i+\alpha^* | 0 \leq i \leq d
\}$). We call $d$ the {\it diameter} of $V$.  We call the ordered pair
$(\alpha, \alpha^*)$ the {\it
type} of $V$.
\end{definition}

\begin{note}
\rm Let $V$ denote a finite-dimensional irreducible $\mathcal{O}$-module of
type $(\alpha, \alpha^*)$.  Replacing $X,Y$ by $X-\alpha I,
Y-\alpha ^* I$ the type becomes $(0,0)$.
\end{note}

Restricting Theorem \ref{th:p1} to type $(0,0)$ we get the
following corollary.

\begin{corollary}
\label{cor:t0o} Let $V$ denote a vector space over $\Bbb K$ with
positive finite dimension.  Let $A$ and $A^*$ denote linear
transformations on $V$.  Then the following $(i),(ii)$ are
equivalent.
\begin{enumerate}
\item
$A,A^*$ is a tridiagonal pair on $V$ and $d-2i$ $(0 \leq i \leq
d)$ is both an eigenvalue sequence and dual eigenvalue sequence
for $A,A^*$, where $d$ denotes the diameter.
\item
There exists an irreducible $\mathcal{O}$-module structure on $V$ of type
$(0,0)$ such that the standard generators $X,Y$ act on $V$ as $A,A^*$
respectively.
\end{enumerate}
\end{corollary}

{\it Proof.}  Immediate from Theorem \ref{th:p1} and the definition of
type.

\medskip

\section{Finite-Dimensional Irreducible $\boxtimes$-Modules}

Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module. In this section we obtain the following
description of $V$. We show that each generator $X_{rs}$ of
$\boxtimes$ is diagonalizable on $V$.  We show that there exists
an integer $d \geq 0$ such that for each $X_{rs}$ the eigenvalues
for $X_{rs}$ on $V$ are $d, d-2, \ldots, -d$.  We give the action
of each $X_{rs}$ on the eigenspaces of the other generators.  In
our investigation we use the following lemma.

\begin{lemma}
\label{lem:conf2}
Let $V$ denote a vector space over $\Bbb K$ with positive finite
dimension.  Let $A$ and $B$ denote linear transformations on $V$.
Then for $\lambda \in \Bbb K$ the following $(i),(ii)$ are
equivalent.
\begin{enumerate}
\item
The expression $[A,B] - 2A - 2B$ vanishes on $V_A(\lambda)$.
\item
$(B + \lambda I) V_A(\lambda) \subseteq V_A(\lambda+2).$
\end{enumerate}
\end{lemma}

{\it Proof.}
Let $\Psi$ denote the expression in $(i)$ and observe
\begin{eqnarray*}
\Psi = AB-BA-2A-2B.
\end{eqnarray*}
For $v \in V_A(\lambda)$ we evaluate $\Psi v$ using $Av= \lambda v$
and find
\begin{eqnarray*}
\Psi v &=& (A - (\lambda + 2)I)(B + \lambda I)v.
\end{eqnarray*}
The result follows.
{~~~\vrule height .75em width .4em depth .2em}

\medskip

We refine notation (\ref{eq:espace}) as follows.

\begin{notation}
\rm \label{note:u} With reference to Definition \ref{def:main} let
$V$ denote a finite-dimensional irreducible $\boxtimes$-module.
For distinct $r,s \in \mathbb I$ and for all $\lambda \in \Bbb K$ we
define
\begin{eqnarray} \label{eq:es2} V_{rs} (\lambda) = \{ v \in V |
X_{rs} v = \lambda v \}.
\end{eqnarray}
\end{notation}

\begin{theorem}
\label{lem:1} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  For all $r,s,t,u \in \mathbb I$ ($r \not= s, t \not=
u$) and for all $\lambda \in \Bbb K$ the action of $X_{tu}$ on
$V_{rs}(\lambda)$ is given as follows.

\medskip

\begin{tabular}{c|c}
Case & Action of $X_{tu}$ on $V_{rs}(\lambda)$ \\
\hline
$t=r, \qquad u=s$ & $(X_{tu} - \lambda I) V_{rs}(\lambda ) = 0$ \\
$t=s, \qquad u=r$ & $(X_{tu} + \lambda I) V_{rs}(\lambda ) = 0$ \\
$t=s, \qquad u \not= r$ & $(X_{tu} + \lambda I) V_{rs}(\lambda ) \subseteq V_{rs}(\lambda + 2)$ \\
$t \not= r, \qquad u=s$ & $(X_{tu} - \lambda I) V_{rs}(\lambda )
\subseteq V_{rs}(\lambda + 2)$ \\
$t=r, \qquad u \not= s$ & $(X_{tu} - \lambda I) V_{rs}(\lambda )
\subseteq V_{rs}(\lambda - 2)$ \\
$t \not= s, \qquad u=r$ & $(X_{tu} + \lambda I) V_{rs}(\lambda )
\subseteq V_{rs}(\lambda - 2)$ \\
$r,s,t,u$ distinct & $X_{tu} V_{rs}(\lambda ) \subseteq
V_{rs}(\lambda + 2) + V_{rs}(\lambda ) + V_{rs}(\lambda - 2) $
\end{tabular}

\medskip

\noindent
 We are using Notation \ref{note:u}.
\end{theorem}

{\it Proof.}  We consider each row in the above table.

$t=r, u=s$: Immediate from (\ref{eq:es2}).

$t=s, u=r$: Immediate from (\ref{eq:es2}) and since $X_{rs}
= - X_{tu}$ by Definition \ref{def:main}(i).

$t=s, u \not= r$: Combine Definition \ref{def:main}(ii) and
Lemma \ref{lem:conf2}.

$t \not= r, u=s$: By case $t=s, u \not= r$ above and since
$X_{ut} = -X_{tu}$.

$t=r, u \not= s$: By case $t=s,u \not= r$ above and since
$V_{rs}(\lambda) = V_{sr}(-\lambda)$.

$t \not= s, u=r$: By case $t=r, u \not= s$ above and since $X_{ut}
= -X_{tu}$.

$r,s,t,u$ distinct: Combine Definition \ref{def:main}(iii) and
Lemma \ref{lem:conf}. {~~~\vrule height .75em width .4em depth .2em}

\begin{corollary}
\label{cor:act} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  Then for all $r,s,t,u \in \mathbb I$ ($r \not= s, t
\not= u$) and for all $\lambda \in \Bbb K$ we have
$$X_{tu} V_{rs}(\lambda) \subseteq V_{rs}(\lambda+2) + V_{rs}(\lambda) +
V_{rs}(\lambda-2).$$ We are using Notation \ref{note:u}.
\end{corollary}

{\it Proof.}
The above inclusion holds for each row of the table in Theorem
\ref{lem:1}.
{~~~\vrule height .75em width .4em depth .2em}

\begin{theorem}
\label{cor:1} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  Let $r,s,t$ denote mutually distinct elements
in $\mathbb I$. Then for all $\lambda \in \Bbb K$ we have
\begin{eqnarray}
V_{rs}(\lambda) + V_{rs}(\lambda-2) + \cdots = V_{rt}(\lambda) +
V_{rt}(\lambda-2) + \cdots . \label{eq:eq}
\end{eqnarray}
We are using Notation \ref{note:u}.
\end{theorem}

{\it Proof.} Let $Y$ denote the left-hand side of (\ref{eq:eq})
and let $Z$ denote the right-hand side of (\ref{eq:eq}).  We show
$Y = Z$.  We first show $Y \subseteq Z$. Since the dimension of
$V$ is finite, there exists a nonnegative integer $j$ such that
$V_{rs}(\lambda-2i) = 0$ and $V_{rt}(\lambda-2i) = 0$ for all
integers $i
> j$. Observe that $Y= \sum_{i=0}^j V_{rs}(\lambda -2i)$ and $Z=
\sum_{i=0}^{j} V_{rt}(\lambda -2i)$.  By (\ref{eq:es2}) we find
$Z$ is the set of vectors in $V$ on which
\begin{eqnarray}
\prod_{i=0}^{j} (X_{rt} - (\lambda-2i)I) \label{eq:in}
\end{eqnarray}
vanishes.  Using the table in Theorem \ref{lem:1} (row $t=r,u \not=
s$) we find that (\ref{eq:in}) vanishes on
$V_{rs}(\lambda-2i)$ for $0 \leq i \leq j$.  Therefore
$V_{rs}(\lambda-2i) \subseteq Z$ for $0 \leq i \leq j$ so $Y
\subseteq Z$.

To get $Z \subseteq Y$ interchange the roles of $s,t$ in the
argument so far.  We conclude that $Y=Z$ and the result follows.
{~~~\vrule height .75em width .4em depth .2em}

\begin{corollary}
\label{cor:2} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module and fix $\lambda \in \Bbb K$.  Then for
distinct $ r,s \in  \mathbb I$ the dimension of $V_{rs}(\lambda)$ is
independent of $r,s$. We are using Notation \ref{note:u}.
\end{corollary}

{\it Proof.}  We define a binary relation on the generators of
$\boxtimes$ called {\it adjacent}.  By definition two distinct
generators $X_{rs},X_{tu}$ are adjacent whenever $r=t$ or $s=u$.
We observe the adjacency relation is symmetric.  We further
observe that the generators of $\boxtimes$ are connected with
respect to adjacency. By these comments it suffices to show
$V_{rs}(\lambda), V_{tu}(\lambda)$ have the same dimension for
adjacent generators $X_{rs},X_{tu}$. First assume $r=t$.  By
Theorem \ref{cor:1} we find
\begin{eqnarray}
V_{rs}(\lambda) + V_{rs}(\lambda-2) + \cdots = V_{tu}(\lambda) +
V_{tu}(\lambda-2) + \cdots . \label{eq:star2}
\end{eqnarray}
Applying Theorem \ref{cor:1} (with
$\lambda$ replaced by $\lambda -2$), we find
\begin{eqnarray}
V_{rs}(\lambda-2) + V_{rs}(\lambda-4) + \cdots =  V_{tu}(\lambda-2)
+V_{tu}(\lambda-4) + \cdots .  \label{eq:star}
\end{eqnarray}
Let $H$ denote the sum on either side of (\ref{eq:star}).  Comparing
(\ref{eq:star2}) and (\ref{eq:star}) we find
\begin{eqnarray}
V_{rs}(\lambda) + H = V_{tu}(\lambda) + H. \label{eq:star1}
\end{eqnarray}
The sum on either side of (\ref{eq:star1}) is direct, so
$V_{rs}(\lambda)$ and $V_{tu}(\lambda)$ have the same dimension.

Next we assume $s=u$.  Applying what we have done so
far with $(r,s,t,u,\lambda)$ replaced by $(s,r,u,t, -\lambda)$ we find
$V_{sr}(-\lambda)$
and $V_{ut}(-\lambda)$ have the same dimension.  Recall that $X_{sr} =
-X_{rs}$ and $X_{ut} = -X_{tu}$ so $V_{sr}(-\lambda) =
V_{rs}(\lambda)$ and $V_{ut}(-\lambda) =
V_{tu}(\lambda)$.  By these comments $V_{rs}(\lambda)$ and $V_{tu}(\lambda)$
have the same dimension.  We have now shown $V_{rs}(\lambda),
V_{tu}(\lambda)$ have the
same dimension for adjacent generators $X_{rs},X_{tu}$ and the result
follows.
{~~~\vrule height .75em width .4em depth .2em}

\begin{corollary}
\label{cor:samedim} Let $V$ denote a finite-dimensional
irreducible $\boxtimes$-module and fix $\lambda \in \Bbb K$.  Then
for distinct $r,s \in \mathbb I$ the spaces $V_{rs}(\lambda),
V_{rs}(-\lambda)$ have the same dimension.  We are using Notation
\ref{note:u}.
\end{corollary}

{\it Proof.}  By Corollary \ref{cor:2} we find
$V_{rs}(\lambda),V_{sr}(\lambda)$ have the same dimension.  But
$X_{rs} = -X_{sr}$ so $V_{sr}(\lambda) = V_{rs}(-\lambda)$. {~~~\vrule height .75em width .4em depth .2em}

\begin{theorem}
\label{lem:2facts}
Let $V$
denote a finite-dimensional irreducible $\boxtimes$-module.  Then
the following $(i),(ii)$ hold.
\begin{enumerate}
\item
For distinct $r,s \in \mathbb I$ the generator $X_{rs}$ is diagonalizable
on $V$.
\item
There exists an integer $d \geq 0$ such that for all distinct $r,s
\in \mathbb I$ the eigenvalues for $X_{rs}$ on $V$ are $d, d-2, \ldots,
-d$.
\end{enumerate}
\end{theorem}

{\it Proof.} Throughout this proof we will use Notation
\ref{note:u}. Let $r,s$ denote distinct elements of $\mathbb I$.  Since
$\Bbb K$ is algebraically closed, all the eigenvalues for $X_{rs}$
on $V$ are contained in $\Bbb K$.  Since $V$ has positive finite
dimension, the action of $X_{rs}$ on $V$ has at least one
eigenvalue $\lambda$.  Since ${\rm Char} (\Bbb K) = 0$, the
scalars $\lambda, \lambda +2, \ldots$ are mutually
distinct, so they cannot all be eigenvalues for $X_{rs}$ on $V$;
consequently there exists $\theta \in \Bbb K$ such that
$V_{rs}(\theta) \not= 0$ but $V_{rs}(\theta+2) = 0$.  Similarly
the scalars  $\theta, \theta -2, \ldots$ are mutually
distinct, so they cannot all be eigenvalues for $X_{rs}$ on $V$;
consequently there exists an integer $d \geq 0$ such that
$V_{rs}(\theta - 2i)$ is nonzero for $0 \leq i \leq d$ and zero
for $i=d+1$.  We will now show that
\begin{eqnarray}
V_{rs}(\theta) + \cdots + V_{rs}(\theta -
2d) \label{eq:vis}
\end{eqnarray}
is equal to $V$.

By Corollary \ref{cor:act} and since $V_{rs}(\theta+2) = 0$,
$V_{rs}(\theta - 2d - 2) = 0$ we find that (\ref{eq:vis}) is
$X_{tu}$-invariant for all distinct $t,u \in \mathbb I$.  Therefore
(\ref{eq:vis}) is $\boxtimes$-invariant.  Recall the
$\boxtimes$-module $V$ is irreducible so (\ref{eq:vis}) is equal
to either $0$ or $V$. By construction each term in (\ref{eq:vis})
is nonzero and there is at least one term, so (\ref{eq:vis}) is
nonzero.  Therefore (\ref{eq:vis}) is equal to $V$.  This shows
that the action of $X_{rs}$ on $V$ is diagonalizable with
eigenvalues $\Delta = \{ \theta -2i | 0 \leq i \leq d \} $.

It remains to show $\theta = d$.  By Corollary \ref{cor:samedim} we
find $\Delta = - \Delta$.  It follows that $\theta = -\theta + 2d$ so
$\theta = d$.
{~~~\vrule height .75em width .4em depth .2em}

\begin{definition}
\rm Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module. By the {\it diameter} of $V$ we mean the
nonnegative integer $d$ from Theorem \ref{lem:2facts}$(ii)$.
\end{definition}

Our discussion of $\boxtimes$-modules will continue after we recall
the notion of a flag.

\section{Flags}
\label{sect:flag}

\begin{definition}
\rm Let $V$ denote a vector space over $\Bbb K$ with positive
finite dimension.  Let $d$ denote a nonnegative integer.  By a
{\it flag on $V$ of diameter $d$}, we mean a sequence $F_0,F_1,
\ldots, F_d$ consisting of mutually distinct subspaces of $V$ such
that $F_0 \not= 0$, $F_{i-1} \subseteq F_i$ for $1 \leq i \leq d$,
and $F_d = V$.  We call $F_i$ the $i^{th}$ {\it component} of the
flag.
\end{definition}

Let $V$ denote a vector space over $\Bbb K$
with positive finite dimension.  Let $d$ denote a
nonnegative integer.  By a {\it decomposition of $V$ of diameter $d$}, we
mean a sequence $V_0,V_1, \ldots, V_d$ consisting of nonzero subspaces
of $V$ such that
$$V = V_0 + \cdots + V_d \qquad \mbox{(direct sum).}$$
We do not assume each of $V_0, \ldots, V_d$ has dimension $1$.
For $0 \leq i \leq d$ we call $V_i$ the {\it $i^{th}$ subspace} of the
decomposition.

The following construction yields a flag on $V$.  Let $V_0,
\ldots, V_d$ denote a decomposition of $V$.  Set
$$F_i = V_0 + \cdots + V_i$$
for $0 \leq i \leq d$.  Then the sequence $F_0, \ldots, F_d$ is a
flag on $V$.  We say this flag is {\it induced} by $V_0, \ldots, V_d$.

Let $V_0, \ldots, V_d$ denote a decomposition of $V$.  By the {\it
inversion} of this decomposition, we mean the decomposition $V_d,
\ldots, V_0$ of $V$.

We now discuss the notion of {\it opposite} flags.  Let $F$ and $G$
denote flags on $V$.  These flags are said to be {\it opposite} whenever
there exists a decomposition $V_0, \ldots, V_d$ of $V$ such that $F$ is
induced by $V_0, \ldots, V_d$ and $G$ is induced by $V_d,
\ldots, V_0$. In this case
\begin{eqnarray}
\label{eq:note1}
F_i \cap G_j = 0 \qquad \mbox{if } i+j < d
\end{eqnarray}
and
\begin{eqnarray}
\label{eq:note2}
V_i = F_i \cap G_{d-i} \qquad \mbox{for } 0 \leq i \leq d.
\end{eqnarray}
In particular $V_0, \ldots, V_d$ is uniquely determined by the
ordered pair $F,G$.  We say $V_0, \ldots, V_d$ is {\it
induced} by $F,G$.

\section{Finite-Dimensional Irreducible $\boxtimes$-Modules,
  Revisited}

We return our attention to a finite-dimensional irreducible
$\boxtimes$-module $V$.  Recall the set $\mathbb I$ from Definition
\ref{def:main} has four elements.  With each element of $\mathbb I$ we
associate a flag on $V$.  We start with a definition.

\begin{definition}
\rm Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  Let $(r,s)$ denote an ordered pair of
distinct elements of $\mathbb I$.  By Theorem \ref{lem:2facts} the
sequence $V_{rs}(-d),V_{rs}(2-d), \ldots, V_{rs}(d)$ is a
decomposition of $V$, where $d$ denotes the diameter.  We will
call this the {\it decomposition of $V$ associated with $(r,s)$}.
\end{definition}

\begin{note}
\label{note:29}
\rm For distinct $r,s \in \mathbb I$ the decomposition of $V$ associated
with $(s,r)$ is the inversion of the decomposition of $V$
associated with $(r,s)$.
\end{note}

\begin{lemma}
\label{lem:nwaa} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module. Let $r,s$ denote distinct elements of $\mathbb I$ and
consider the decomposition of $V$ associated with $(r,s)$.  Then
the flag induced by this decomposition is independent of $s$.
\end{lemma}

{\it Proof.}  Let $t$ denote an element of $\mathbb I$ such that $r,s,t$
are mutually distinct.  It suffices to show that the flag induced
by the decomposition of $V$ associated with $(r,s)$ equals the
flag induced by the decomposition of $V$ associated with $(r,t)$.
Let $d$ denote the diameter of $V$.  For $0 \leq i \leq d$ the
$i^{th}$ component of the first flag is $V_{rs}(-d)
+ \cdots + V_{rs}(2i-d)$ and the $i^{th}$ component of the second
flag is $V_{rt}(-d) + \cdots + V_{rt}(2i-d)$.  These
components are equal by Theorem \ref{cor:1} (with $\lambda =
2i-d$). Therefore the flags are equal. {~~~\vrule height .75em width .4em depth .2em}

\medskip

\begin{definition}
\label{def:nwaa} \rm Let $V$ denote a finite-dimensional
irreducible $\boxtimes$-module. For $r \in \mathbb I$, by the {\it flag on
$V$ associated with $r$} we mean the flag discussed in Lemma
\ref{lem:nwaa}.
\end{definition}

\medskip

The next corollary restates Lemma \ref{lem:nwaa} in light of
Definition \ref{def:nwaa}.

\begin{corollary}
\label{cor:nwbb} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  For $r \in \mathbb I$ consider the flag on $V$ associated
with $r$.  The components of this flag are described as follows.
Let $d$ denote the diameter of $V$ and pick $s \in \mathbb I$ such that
$r \not= s$. Then for $0 \leq i \leq d$ the $i^{th}$ component of
the flag is
\begin{eqnarray}
\label{eq:nwq} V_{rs}(-d)+V_{rs}(2-d) + \cdots + V_{rs}(2i-d).
\end{eqnarray}
\end{corollary}

\begin{lemma}
\label{lem:nwcc} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module.  Recall from Definition \ref{def:main} that
$\mathbb I$ has four elements and consider the corresponding flags on $V$
from Definition \ref{def:nwaa}.  These four flags are mutually
opposite.
\end{lemma}

{\it Proof.}  For distinct $r,s \in \mathbb I$ let
$F$ denote the flag on $V$ associated with $r$ and let $G$ denote
the flag on $V$ associated with $s$.  We show $F$ and $G$ are
opposite.  By Note \ref{note:29} the decomposition of $V$ associated
with $(r,s)$ is the inversion of the decomposition of $V$ associated
with $(s,r)$.  The first decomposition induces $F$ and the second
decomposition induces $G$.  It follows that $F$ and $G$ are
opposite. {~~~\vrule height .75em width .4em depth .2em}

\begin{lemma}
\label{lem:nwdd} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module of diameter $d$.  Let $r,s$ denote distinct elements
of $\mathbb I$.  For $0 \leq i \leq d$ the subspace $V_{rs}(2i-d)$
is the intersection of the following two sets:
\begin{enumerate}
\item The $i^{th}$ component of the flag on $V$ associated with $r$.
\item The $(d-i)^{th}$ component of the flag on $V$ associated with $s$.
\end{enumerate}
\end{lemma}

{\it Proof.}  By Definition \ref{def:nwaa} the flag on $V$
associated with $r$ is induced by $V_{rs}(-d),
\ldots, V_{rs}(d)$ and the flag on $V$ associated with $s$ is
induced by $V_{rs}(d), \ldots, V_{rs}(-d)$.  The
result follows. {~~~\vrule height .75em width .4em depth .2em}

\medskip

\section{From $\boxtimes$-Modules to $\mathcal{O}$-Modules}

In this section we prove Theorem \ref{th:H1}.  We begin with a
lemma.

\begin{lemma}
\label{lem:ne} Let $V$ denote a finite-dimensional irreducible
$\boxtimes$-module. Let $r,s,t,u$ denote mutually distinct
elements of $\mathbb I$. Let $W$ denote a nonzero subspace of $V$ such
that $X_{rs}W \subseteq W$, $X_{tu}W \subseteq W$.  Then $W = V$.
\end{lemma}

{\it Proof.}  Without loss of generality, we assume $W$ is irreducible
as a module for $X_{rs},X_{tu}$.  To show $W = V$, it suffices to show
that $W$ is $\boxtimes$-invariant.  To this end, we show
\begin{eqnarray}
\label{eq:cl} X_{ab} W \subseteq W
\end{eqnarray}
for all distinct $a,b \in \mathbb I$.  Assume $ab$ is not one of $rs, tu,
sr, ut$; otherwise (\ref{eq:cl}) holds by the construction and
(\ref{eq:1}).  Replacing an appropriate subset of $\{ rs,tu,ab \}$
by the corresponding subset of $\{ sr,ut,ba \}$ we may assume,
without loss of generality, that $a=s$ and $b=t$. Therefore it
suffices to show $X_{st} W \subseteq W$. We define $W' :=\{ w \in
W | X_{st} w \in W \}$ and show $W' = W$. By (\ref{eq:2}) we
routinely find that $X_{rs} W' \subseteq W'$ and $X_{tu} W'
\subseteq W'$. By these comments and the irreducibility of $W$ we
find either $W'=0$ or $W'=W$.

\noindent
{\it Claim.} $W' \not= 0$.

\noindent {\it Proof of claim.}  Let $d$ denote the diameter of $V$.
Define $W_j = W \cap
V_{rs}(2j-d)$ for $0 \leq j \leq d$. The nonzero spaces among $W_0
, \ldots , W_d$ are eigenspaces of $X_{rs}$ on $W$. By Theorem
\ref{lem:2facts} we find $W = \sum_{j=0}^d W_j$.   By this and
since $W \not= 0$ we find $W_0 , \ldots , W_d$ are not all $0$.
Define $ m = \mbox{min} \{ j | 0 \leq j \leq d, W_j \not= 0 \}.$

Define $W_j^* = W \cap V_{tu}(2j-d)$ for $0 \leq j \leq d$.  The
nonzero spaces among $W_0^* , \ldots , W_d^*$ are eigenspaces of
$X_{tu}$ on $W$. By Theorem \ref{lem:2facts} we find $W =
\sum_{j=0}^d W_j^*$.  By this and since $W \not= 0$ we find $W_0^*
, \ldots , W_d^*$ are not all $0$. Define $ n = \mbox{min} \{ j | 0 \leq
j \leq d, W_j^* \not= 0 \}. $

We show $m=n$.  To do this we first assume $m < n$ and get a
contradiction. By construction $W_j^* \subseteq V_{tu}(2j-d)$ for
$0 \leq j \leq d$ and $W = \sum_{j=n}^d W_j^*$ so $W \subseteq
\sum_{j=n}^d V_{tu}(2j-d).$ Therefore $W$ is contained in the
$(d-n)^{th}$ component of the flag on $V$ associated with $u$. By
construction $W_m \subseteq V_{rs}(2m-d)$ so $W_m \subseteq
\sum_{j=0}^m V_{rs}(2j-d)$.  Therefore $W_m$ is contained in the
$m^{th}$ component of the flag on $V$ associated with $r$. Recall
by Lemma \ref{lem:nwcc} that the flag on $V$ associated with $u$
and the flag on $V$ associated with $r$ are opposite. By
(\ref{eq:note1}) the $(d-n)^{th}$ component of the flag on $V$
associated with $u$ and the $m^{th}$ component of the flag on $V$
associated with $r$ have $0$ intersection. Therefore $W \cap W_m =
0$, contradicting the fact that $W_m$ is nonzero in $W$. Therefore
$m \geq n$.

Next we assume $m > n$ and get a contradiction.  By construction
$W_j \subseteq V_{rs}(2j-d)$ for $0 \leq j \leq d$ and $W =
\sum_{j=m}^d W_j$ so $W \subseteq \sum_{j=m}^d V_{rs} (2j-d)$.
Therefore $W$ is contained in the $(d-m)^{th}$ component of the
flag on $V$ associated with $s$. By construction $W_n^* \subseteq
V_{tu}(2n-d)$ so $W_n^* \subseteq \sum_{j=0}^n V_{tu}(2j-d)$.
Therefore $W_n^*$ is contained in the $n^{th}$ component of the
flag on $V$ associated with $t$. Recall by Lemma \ref{lem:nwcc}
that the flag on $V$ associated with $s$ and the flag on $V$
associated with $t$ are opposite.  By (\ref{eq:note1}) the
$(d-m)^{th}$ component of the flag on $V$ associated with $s$ and
the $n^{th}$ component of the flag on $V$ associated with $t$ have
$0$ intersection. Therefore $W \cap W_n^* = 0$, contradicting the
fact that $W_n^*$ is nonzero in $W$.  Therefore $m \leq n$.

Combining the previous two paragraphs we find $m = n$.  By the
previous paragraph we find $W_n^*$ is contained in the
intersection of the $(d-m)^{th}$ component of the flag on $V$
associated with $s$ and the $m^{th}$ component of the flag on $V$
associated with $t$.  By Lemma \ref{lem:nwdd} we find $W_n^*
\subseteq V_{st}(d-2m)$.  Since $V_{st}(d-2m)$ is an eigenspace
for $X_{st}$ we find $X_{st} W_n^* \subseteq W_n^*$. Therefore
$W_n^* \subseteq W'$, so $W' \not= 0$ and the claim is proved.

This shows that $W' = W$ and therefore $X_{st}W \subseteq W$.  By this
and the comments prior to the claim we have now shown that
$X_{ab} W \subseteq W$ for all distinct $a,b \in \mathbb I$.
Therefore $W$ is $\boxtimes$-invariant.  Since the $\boxtimes$-module
$V$ is irreducible we have $W=V$. {~~~\vrule height .75em width .4em depth .2em}

\medskip

We are now ready to prove Theorem \ref{th:H1}.

\medskip

{\it Proof of Theorem \ref{th:H1}.} Since $X_{01},X_{23}$ satisfy
(\ref{eq:fo1}) and (\ref{eq:fo2}) there exists an $\mathcal{O}$-module
structure on $V$ such that $X,Y$ act on $V$ as
$X_{01},X_{23}$ respectively.  This structure is unique since $X,Y$
generate $\mathcal{O}$.  This structure is irreducible by
Lemma \ref{lem:ne}.  This structure has type $(0,0)$ by Theorem
\ref{lem:2facts}$(ii)$. {~~~\vrule height .75em width .4em depth .2em}

\section{From $\mathcal{O}$-Modules to $\boxtimes$-Modules}

In this section we prove Theorem \ref{th:H2}.  We refer to the
following setup.

\begin{definition}
\rm \label{def:asm} Let $V$ denote a finite-dimensional
irreducible $\mathcal{O}$-module of type $(0,0)$.  Let $A,A^*$ denote the
actions on $V$ of the generators $X,Y$ respectively and recall by
Corollary \ref{cor:t0o} that $A,A^*$ is a tridiagonal pair on $V$.
Let $\mathbb I$ denote the set consisting of the four symbols $\{0,1,2,3
\}$. With each element of $\mathbb I$ we will associate a flag on $V$ of
diameter $d$ where $d$ denotes the diameter of $V$.  For $0 \leq i
\leq d$ the $i^{th}$ component of this flag is given as follows.

\medskip

\begin{tabular}{c|c}
Element of $\mathbb I$ & $i^{th}$ component of associated flag \\
\hline
$0$ & $V_A(-d)+ \cdots + V_A(2i-d)$ \\
$1$ & $V_A(d)+ \cdots + V_A(d-2i)$ \\
$2$ &  $V_{A^*}(-d)+  \cdots + V_{A^*}(2i-d)$ \\
$3$ & $V_{A^*}(d)+  \cdots + V_{A^*}(d-2i)$
\end{tabular}
\end{definition}

\begin{lemma}
\label{lem:asm}
\cite[Theorem 7.3]{ter3} The four flags from Definition
\ref{def:asm} are mutually opposite.
\end{lemma}

\begin{definition}
\rm \label{def:12} Adopt the assumptions of Definition
\ref{def:asm}. For distinct $r,s \in \mathbb I$ let $x_{rs}:V \to V$
denote the unique linear transformation satisfying the following
conditions.  By Definition \ref{def:asm} each of $r,s$ is
associated with a flag on $V$ of diameter $d$.  By Lemma \ref{lem:asm}
the flags $r,s$ are opposite; let $ U_0, \ldots, U_d$ denote the
induced decomposition of $V$.  For $0 \leq i \leq d$,
$U_i$ is an eigenspace for $x_{rs}$ with eigenvalue $2i-d$.
\end{definition}

We will be using the following notation.

\begin{notation}
\rm \label{not:11} With reference to Definition \ref{def:12}, for
distinct $r,s \in \mathbb I$ and for all $\lambda \in \Bbb K$ we define
$$V_{rs}(\lambda) = \{ v \in V | x_{rs} v = \lambda v \}.$$
\end{notation}

The next two lemmas illustrate the notation in Definition
\ref{def:12} and will be useful later in the paper.

\begin{lemma}
\label{lem:note} With reference to Definition \ref{def:12}, we
have $x_{01} = A$ and $x_{23} = A^*$.
\end{lemma}

{\it Proof.}  Immediate from Definition \ref{def:12}.
{~~~\vrule height .75em width .4em depth .2em}

\begin{lemma}
\label{lem:go} With reference to Definition \ref{def:12}, for $r
\in \mathbb I$ consider the flag on $V$ associated with $r$. The
components of this flag are described as follows. Pick $s \in \mathbb I$
such that $r \not= s$. Then for $0 \leq i \leq d$ the $i^{th}$
component of the flag is
$$V_{rs}(-d)+ \cdots + V_{rs}(2i-d).$$
We are using Notation \ref{not:11}.
\end{lemma}

{\it Proof.}  By Definition \ref{def:12} the decomposition
$V_{rs}(-d), \ldots, V_{rs}(d)$ of $V$ is induced by
the flag on $V$ associated with $r$ and the flag on $V$ associated
with $s$. By the discussion at the end of Section \ref{sect:flag}
the flag $r$ is induced by $V_{rs}(-d), \ldots,
V_{rs}(d)$. The result follows. {~~~\vrule height .75em width .4em depth .2em}

\medskip

We will need the following three lemmas.

\begin{lemma}
\label{note:minus} With reference to Definition \ref{def:12}, for
distinct $r,s \in \mathbb I$ we have
$$ x_{rs} + x_{sr} =0. $$
\end{lemma}

{\it Proof.}  Let $U_0, \ldots , U_d$ denote the decomposition
induced by the flag on $V$ associated with $r$ and the flag on $V$
associated with $s$.  By Definition \ref{def:12} and with
reference to Notation \ref{not:11}, we find $V_{rs}(2i-d) = U_i =
V_{sr}(d-2i)$ for $0 \leq i \leq d$. The result follows. {~~~\vrule height .75em width .4em depth .2em}

\begin{lemma}
\label{lem:ii} With reference to Definition \ref{def:12}, for mutually
distinct $r,s,t \in \mathbb I$ we have
\begin{eqnarray}
[x_{rs},x_{st}] = 2x_{rs} + 2x_{st}. \label{eq:r}
\end{eqnarray}
\end{lemma}

{\it Proof.}  Throughout this proof we will use Notation \ref{not:11}.

We break our proof into the following three cases:

(i) $st$ is one of $01, 10, 23, 32$.

(ii) $rs$ is one of $01, 10, 23, 32$.

(iii) Neither $rs$ nor $st$ is one of $01,10,23,32$.

Case (i):  We will invoke Lemma \ref{lem:conf2}.  We begin with some
comments.  Recall that
$x_{rs}$ is diagonalizable on $V$ with eigenvalues $d, d-2,
\ldots, -d$. We now show that
\begin{eqnarray}
\label{eq:newe1} (x_{st} + (2i-d)I)V_{rs}(2i-d) \subseteq
V_{rs}(2i-d+2)
\end{eqnarray}
for $0 \leq i \leq d$.  Let $u$ denote the unique element of $\mathbb I$
such that $r,s,t,u$ are mutually distinct. By Lemma \ref{lem:note}
and Lemma \ref{note:minus} we find that $x_{st},x_{ur}$ is either
$\epsilon A, \epsilon^* A^*$ or $\epsilon^* A^*,\epsilon A$ for
some $\epsilon, \epsilon^* \in \{ 1,-1 \}$.  By this and Definition
\ref{def:asm} we find $x_{st},x_{ur}$
is a tridiagonal pair on $V$ of diameter $d$ and $d, d-2, \ldots ,
-d$ is both an eigenvalue sequence and dual eigenvalue sequence of
$x_{st},x_{ur}$. By Lemma \ref{lem:go} and using \cite[Theorem
4.6]{ITT} we find that (\ref{eq:newe1}) holds for $0 \leq i \leq
d$.  By this and Lemma \ref{lem:conf2} we find (\ref{eq:r}) holds.

Case (ii):  Observe that $sr$ is one of $01, 10, 23, 32$.
By case (i) of the current proof we find that
\begin{eqnarray}
\label{eq:fy} [x_{ts},x_{sr}] = 2x_{ts}+2x_{sr}.
\end{eqnarray}
By Lemma \ref{note:minus} we find $x_{ts} = -x_{st}$ and $x_{sr}
= -x_{rs}$.  Substituting this into (\ref{eq:fy}) and simplifying
the result we find that (\ref{eq:r}) holds.

Case (iii):  Similarly to case (i) we will invoke Lemma
\ref{lem:conf2}.  Recall that
$x_{rs}$ is diagonalizable on $V$ with eigenvalues $d, d-2,
\ldots, -d$. We now show that
\begin{eqnarray}
\label{eq:newe2} (x_{st} + (2i-d)I) V_{rs}(2i-d) \subseteq
V_{rs}(2i-d+2)
\end{eqnarray}
for $0 \leq i \leq d$.  Observe $rt$ is one of $01, 10, 23, 32$. By
case (ii) of the current proof we find that $[x_{rt},x_{ts}] = 2x_{rt} +
2x_{ts}$.  By Lemma \ref{lem:conf2} (with $\lambda = d-2i$) and since
$x_{rt} = -x_{tr}$, $x_{ts}= -x_{st}$ we find
\begin{eqnarray}
(x_{st} + (2i-d) I)V_{tr}(2i-d) \subseteq V_{tr}(2i-d -2) \qquad 0
\leq i \leq d. \label{eq:d3}
\end{eqnarray}
For $0 \leq i \leq d$ we have
\begin{eqnarray*} (x_{st} + (2i-d)
I)V_{rs}(2i-d) & \subseteq & (x_{st} + (2i-d) I)
\sum_{n=0}^i V_{rs}(2n-d) \\
& = & (x_{st} + (2i-d) I) \sum_{n=0}^i V_{tr}(d-2n) \qquad \qquad
\mbox{by Lemma \ref{lem:go}} \\
& \subseteq & \sum_{n=0}^{i+1} V_{tr}(d-2n)
\qquad \qquad \qquad \qquad \qquad \quad \mbox{by (\ref{eq:d3})} \\
& = & \sum_{n=0}^{i+1}V_{rs}(2n-d) \qquad \qquad \qquad \qquad \qquad
\quad \mbox{by Lemma
\ref{lem:go}} \end{eqnarray*}
and
\begin{eqnarray*} (x_{st} + (2i-d) I)V_{rs}(2i-d) &
\subseteq & (x_{st} + (2i-d) I)
\sum_{n=i}^dV_{rs}(2n-d) \\
& = & (x_{st} + (2i-d) I) \sum_{n=i}^d  V_{st}(d-2n) \qquad \qquad
\mbox{by Lemma \ref{lem:go}}\\
& = & \sum_{n=i+1}^d  V_{st}(d-2n) \\
& = & \sum_{n=i+1}^d V_{rs}(2n-d) \qquad \qquad \qquad \qquad \qquad
\mbox{by Lemma \ref{lem:go}}. \end{eqnarray*}

Combining these observations we find that (\ref{eq:newe2}) holds
for $0 \leq i \leq d$.  By this and Lemma \ref{lem:conf2} we find
that (\ref{eq:r}) holds. {~~~\vrule height .75em width .4em depth .2em}

\begin{lemma}
\label{lem:iii} With reference to Definition \ref{def:12}, for
mutually distinct $r,s,t,u \in \mathbb I$ we have
\begin{eqnarray}
[x_{rs},[x_{rs},[x_{rs},x_{tu}]]] = 4[x_{rs},x_{tu}]. \label{eq:s}
\end{eqnarray}
\end{lemma}

{\it Proof.}   Throughout this proof we will use Notation \ref{not:11}.

We will invoke Lemma \ref{lem:conf}.  We begin with some comments.
Recall that $x_{rs}$ is diagonalizable on $V$ with eigenvalues $d,
d-2, \ldots, -d$.  We now show that
\begin{eqnarray}
\label{eq:newe3} x_{tu}V_{rs}(2i-d) \subseteq V_{rs}(2i-d+2)+
V_{rs}(2i-d)+ V_{rs}(2i-d-2)
\end{eqnarray}
for $0 \leq i \leq d$.  By Lemma \ref{lem:ii} we find $[x_{st},x_{tu}]
= 2x_{st} + 2x_{tu}$. By this and Lemma \ref{lem:conf2} we find
\begin{eqnarray}
(x_{tu} + (2i-d) I)V_{st}(2i-d) \subseteq  V_{st}(2i-d +2)
\label{eq:e4} \qquad 0 \leq i \leq d.
\end{eqnarray}
By Lemma \ref{lem:ii} we find $[x_{ru},x_{ut}] =
2x_{ru}+2x_{ut}$.  By Lemma \ref{lem:conf2} (with $\lambda = d-2i$)
and since $x_{ru} = -x_{ur}$, $x_{ut} = -x_{tu}$ we find
\begin{eqnarray}
(x_{tu} + (2i-d) I)V_{ur}(2i-d) \subseteq V_{ur}(2i-d -2) \qquad 0
\leq i \leq d. \label{eq:e3}
\end{eqnarray}
For $0 \leq i \leq d$ we have
\begin{eqnarray*}
x_{tu}V_{rs}(2i-d) & \subseteq & x_{tu} \sum_{n=i}^d V_{rs}(2n-d) \\
& = & x_{tu} \sum_{n=i}^d V_{st}(d-2n)
\qquad \qquad \mbox{by Lemma \ref{lem:go}}\\
& \subseteq &  \sum_{n=i-1}^dV_{st}(d-2n)
\qquad \qquad \quad \mbox{by (\ref{eq:e4})} \\
 & = & \sum_{n=i-1}^dV_{rs}(2n-d) \qquad \qquad \quad
 \mbox{by Lemma \ref{lem:go}}
\end{eqnarray*}
and
\begin{eqnarray*}
x_{tu} V_{rs}(2i-d) & \subseteq & x_{tu} \sum_{n=0}^i V_{rs}(2n-d) \\
& = & x_{tu} \sum_{n=0}^i V_{ur}(d-2n)
\qquad \qquad \mbox{by Lemma \ref{lem:go}}\\
& \subseteq & \sum_{n=0}^{i+1}V_{ur}(d-2n)
\qquad \qquad \quad  \mbox{by (\ref{eq:e3})} \\
 & = & \sum_{n=0}^{i+1}  V_{rs}(2n-d) \qquad
 \qquad \quad  \mbox{by Lemma \ref{lem:go}}.
\end{eqnarray*} Combining these observations we find that (\ref{eq:newe3})
holds for all $0 \leq i \leq d$.   By this and Lemma
\ref{lem:conf} we find that (\ref{eq:s}) holds. {~~~\vrule height .75em width .4em depth .2em}

\medskip

\begin{theorem}
\label{th:pth} With reference to Definition \ref{def:12}, there
exists a unique $\boxtimes$-module structure on $V$ such that
the generator $X_{rs}$ acts on $V$ as $x_{rs}$ for all distinct $r,s
\in \mathbb I$. This
$\boxtimes$-module structure is irreducible.
\end{theorem}

{\it Proof.}  Using Lemma \ref{note:minus}, Lemma
\ref{lem:ii}, and Lemma \ref{lem:iii} we find that there exists a
$\boxtimes$-module structure on $V$ such that $X_{rs}$ acts on $V$
as $x_{rs}$ for distinct $r,s \in \mathbb I$.  This module structure is
unique since the set $\{ X_{rs} | r,s \in \mathbb I, r \not= s \}$ is a
generating set for $\boxtimes$.  By Lemma \ref{lem:note} we find
that $A,A^*$ are among the linear transformations $x_{rs}$, $r,s
\in \mathbb I$, $r \not= s$. Recall that $A,A^*$ is a tridiagonal pair on
$V$. Therefore the $\boxtimes$-module $V$ is irreducible by Definition
\ref{def:tp}(iv).
{~~~\vrule height .75em width .4em depth .2em}

\medskip

We are now ready to prove Theorem \ref{th:H2}.

\medskip

{\it Proof of Theorem \ref{th:H2}.}  Consider the $\boxtimes$-module
structure on $V$ from Theorem \ref{th:pth}.  In this module structure
$X_{01},X_{23}$ act on $V$ as $X,Y$ respectively by Definition
\ref{def:asm} and Lemma \ref{lem:note}.  Next we show that this
$\boxtimes$-module structure is unique. Suppose we are given any
$\boxtimes$-module structure on $V$ where $X_{01},X_{23}$ act on $V$
as $X,Y$ respectively.  This module structure is irreducible since the
$\mathcal{O}$-module $V$ is irreducible.  For each generator
$X_{rs}$ of $\boxtimes$ the action on $V$ is determined by the
decomposition of $V$ associated with $(r,s)$.  By Lemma
\ref{lem:nwdd} the decomposition of $V$ associated with $(r,s)$ is
determined by the flag on $V$ associated with $r$ and the flag on
$V$ associated with $s$. Therefore our $\boxtimes$-module
structure on $V$ is determined by the four flags associated with
the four elements of $\mathbb I$.  By Corollary \ref{cor:nwbb} the four
flags associated with the four elements of $\mathbb I$ are determined by
the actions of $X_{01}$ and $X_{23}$ on $V$.  Therefore the given
$\boxtimes$-module structure on $V$ is determined by the actions
of $X_{01}$ and $X_{23}$ on $V$, so the $\boxtimes$-module
structure is unique.  We have now shown there exists a unique
$\boxtimes$-module structure on $V$ where $X_{01},X_{23}$ act on
$V$ as $X,Y$ respectively.  We mentioned earlier that this module
structure is irreducible. {~~~\vrule height .75em width .4em depth .2em}

\bigskip

\noindent
{\bf Acknowledgements:}  This paper was written while the author was
a graduate student at the University of Wisconsin-Madison.  The author
would like to thank his advisor Paul Terwilliger for his many valuable
ideas and suggestions.

redpajama_set_name: RedPajamaArXiv


text:
Il prosciuro (gen. Prosciurus) è un roditore estinto, appartenente agli aplodontidi. Visse tra l'Eocene superiore e l'Oligocene medio (circa 36 - 28 milioni di anni fa) e i suoi resti fossili sono stati ritrovati in Nordamerica e in Asia.

Descrizione
È probabile che l'aspetto di questo animale assomigliasse a quello dell'attuale castoro di montagna (Aplodontia rufa), ma le dimensioni erano sicuramente più ridotte e la corporatura doveva essere più snella; la specie tipo, Prosciurus vetustus, doveva essere grande quanto un odierno scoiattolo nordamericano (Sciurus carolinensis). Prosciurus possedeva molari superiori trigonodonti (a tre tubercoli), con il tubercolo interno posteriore appena accennato. La cresta anteriore del trigono era completa, mentre quella posteriore si interrompeva tra metaconulo e protocono. I molari inferiori erano dotati di un grande bacino centrale, dal margine anteriore rettilineo. L'ipolofide era debole, e si inseriva davanti all'ipoconide. L'ipoconulide era ridotto.

Classificazione
Il genere Prosciurus venne descritto per la prima volta da William Diller Matthew nel 1903, sulla base di resti fossili ritrovati in Montana in terreni dell'Oligocene inferiore; la specie tipo è Prosciurus vetustus. A questo genere sono state poi ascritte altre specie, rinvenute in terreni dell'Eocene superiore - Oligocene inferiore degli Stati Uniti (P. relictus, P. magnus, P. albiclivus), ma più di recente sono state descritte anche specie asiatiche: P. pristinus dell'Eocene superiore della Mongolia Interna (Cina), P. mongoliensis dell'Oligocene inferiore della Mongolia e P. ordosicus dell'Oligocene inferiore della Mongolia Interna.

Prosciurus è un rappresentante degli aplodontidi, una famiglia di roditori arcaici attualmente rappresentati dal solo castoro di montagna. Prosciurus è il genere eponimo della sottofamiglia Prosciurinae, ed è possibile che alcune specie (ad esempio la nordamericana P. albiclivus) siano da porre in relazione con la sottofamiglia Meniscomyinae, e in particolare con il genere asiatico Promeniscomys (Korth, 1994).

Bibliografia
W. D. Matthew. 1903. The fauna of the Titanotherium Beds at Pipestone Springs, Montana. Bulletin of the American Museum of Natural History 19(6):197-226
B. Wang. 1987. Discovery of Aplodontidae (Rodentia, Mammalia) from Middle Oligocene of Nei Mongol, China. Vertebrata PalAsiatica 25(1):32-45
W. W. Korth. 1994. A new species of the rodent Prosciurus (Aplodontidae, Prosciurinae) from the Orellan (Oligocene) of North Dakota and Nebraska. Journal of Mammalogy; 752: 478-482
B.-Y. Wang and D. Dashzeveg. 2005. New Oligocene sciurids and aplodontids (Rodentia, Mammalia) from Mongolia. Vertebrata PalAsiatica 43(2):85-99
B.-Y. Wang. 2008. Additional rodent material from Houldjin Formation of Erenhot, Nei Mongol, China. Vertebrata PalAsiatica 46(1):21-30

Collegamenti esterni

Roditori fossili
Taxa classificati da William Diller Matthew

redpajama_set_name: RedPajamaWikipedia


Save the date from 26 to 29 March in China to discover the new Egro Next machine and all other Rancilio products.
Rancilio Group will be in Shanghai from 26 to 29 March for Hotelix, the most important Asian trade fair for the hospitality and catering industries. In recent years, China has recorded a significant growth in its consumption of coffee and, in the wake of this trend, Hotelex has chosen to extend its dedicated exposition area to host the "China Barista Championship" and invite the world's leading operators. In the coffee area alone, there will be more than 300 exhibitors involved, with a guaranteed full market coverage thanks to the presence of the most important buyers.
This will be a special appointment because it marks the Rancilio Group's first ever participation as an exhibitor, with the debut of its "Coffee House" on the Blue river delta, but that's not all. The Group will inaugurate its presence with the official presentation of the new Egro NEXT, the fully automatic coffee machine specifically designed for the Asian market.
Visitors will therefore have an opportunity to discover the potential of the new brewing unit of Egro Next which, in the version with a 22 g chamber, guarantees rapid and high-quality brewing in the preparation of American-style coffee and a Coffee Box of over 2.7 litres. This provides an ideal solution for take-away and self-service venues, which are particularly widespread throughout Asia, enabling them to take advantage of the extremely user-friendly and intuitive touchscreen interface based on the Android operating system. Another new feature of Egro Next, sure to amaze the public of Hotelex, is the Cooled Coffee Module. A cutting-edge system for the preparation of all coffee-based cold beverages, which prevents ice cubes from melting in the glass during the dispensing phase.
Alongside the newly released NEXT machine, at the "Coffee House", Rancilio Group will also be displaying other fully automatic Egro models designed for locations with different requirements and consumptions: Egro One – in the double Touch and Key Pad version - and Egro Zero. The public of Hotelex will also be able to familiarize with the traditional coffee machines branded Rancilio: the top of the line Rancilio Classe 11 with Xcelsius technology for Temperature Profiling and the Classe 9, Classe 7 and Classe 5 lines which, in the TALL configuration, enable the operator to work comfortably even with American coffee cups. There will also be space for Latte Art with test runs and demonstrations to the public aimed at showcasing the steam solutions offered by Rancilio coffee machines, such as the automatic iSteam wand, the C-lever providing full control of the steam tap and the cool touch wands, enabling the operator to work in perfect safety.
As well as coffee machines, visitors may also admire and try out the high-performing features of the KRYO dosing grinders which, both in the "On Demand" and fully automatic versions, guarantee maximum precision when setting the desired particle size.
The Rancilio Group team looks forward to welcoming you to Hotelex Shanghai from 26 to 29 March.


(Gateway Pundit) – President Trump departed Joint Base Andrews Monday afternoon and headed to El Paso, Texas for a MAGA rally.
Well this was embarrassing. Only about 25 people showed up early to see Beto.
I wonder who will win in 2020?


Although we primarily concentrate on Train Simulator, we also love simulation games in general and they cover a huge variety of interests and topics. Whether it be other forms of transport such as buses and planes, the laid back management and upkeep of an expansive farm or being in control of our own city and it's inhabitants, we love them all. Maybe there is something that interests you as well.
Browse our directory featuring games from all corners of the simulation genre. This page is work in progress and is updated regularly.


text:
Konotopî (în ) este un sat în comuna Opilsko din raionul Sokal, regiunea Liov, Ucraina.

Demografie

Conform recensământului din 2001, majoritatea populației localității Konotopî era vorbitoare de ucraineană (%), existând în minoritate și vorbitori de alte limbi.

Note

Konotopî, Sokal, Liov
Konotopî, Sokal, Liov
Konotopî, Sokal, Liov


The medieval city of Brno was founded in the 11th century and in 1641 it became the capital of Moravia. At the beginning of the 20th century, German-speaking residents outnumbered Czechs by a large majority; but today with almost 400,000 inhabitants, Brno is the second largest city in the Czech Republic. Its rich history, culture and nightlife make it a worthwhile weekend getaway and if you've ever wondered what Prague might have been like before the expats moved in and the hordes of tourists arrived, a visit to this charming location might well be the best chance you have left to see for yourself.
Although locals who speak a second language are more likely to speak German than English, Brno is very welcoming. The primary tourist attraction is the fortress of Špilberk, which dates back more than 700 years. Constructed as a residence for the Czech Kings, the highlight of the castle tour is the "casemates", which were originally designed as a military barracks and arsenal, but were later converted into a prison. And what a dark and terrible place it became! During the late 1700s, prisoners convicted of the worst crimes were confined to the darkest and deepest dungeons and shackled to the walls in perpetuity. From the castle itself, you can treat yourself to the enchanting view of the Old Town with its renaissance houses and picturesque churches, as well as the Cathedral of St. Peter and Paul.
In the center of old Brno lies Námestí Svobody (Liberty Square) whose present name dates back to 1918 and the proclamation of an independent Czechoslovakia. The majority of the buildings situated on the square are built on foundations that date back to the Gothic period and many figure prominently in Moravian history and folklore. At n° 18 you can find the Royal House, where Czech Kings resided during their visits to Brno. Just next-door at n° 17 is the House of the Lords of Lipe, a Renaissance palace complete with dungeons and an arcaded interior gallery. It was given to the French colonel, Raduit de Souches, as a reward for successfully defending the city against the siege of the Swedish army.
The Old Town Hall is located on Radnická Street, adjacent to Námestí Svobody. Designed in the High Gothic style, it is best-known for the symbols of the city, the Dragon and the Wheel, that hang over the entry passage. Several conflicting legends attempt to explain the dragon's origin but the only thing they seem to agree on is that the dragon is in fact a crocodile. The wheel is said to have been made by a master wheelwright who in 1638 made a bet with some friends in the nearby village of Lednice that he could cut down a tree at six in the morning, make a wheel, and a roll it to Brno within that same day, before they closed the city gates. He succeeded and presented the wheel to the amazed Mayor of Brno. Since the wheel was a true masterpiece, the Mayor had it hung in the town hall.
Not far from the Town Hall is Zelný trh (the vegetable market) and the Reduta Theater where an 11-year-old Wolfgang Amadeus Mozart once gave a concert. The adjoining Petrská Street leads up to Petrov hill and the Cathedral of St. Peter and Paul, a Gothic edifice that was built during the 14th and 15th centuries, but never completed. During the Swedish siege of 1645 the church was set on fire and partially destroyed. The interior vaults were rebuilt during the 18th and 19th century, but the two towers that symbolize present-day Brno were only completed in 1911.
Legend tells us that the Swedish commander, growing impatient with the time the siege was lasting, told his army that on the morning of August 15, 1645 they would take the town by noon, or leave. At 11 a.m. the walls were breached and the attackers began ransacking the city. An old bell ringer received word from a spy that he must ring the bells so he climbed the damaged tower and succeeded in ringing out midday. The Swedish attackers retreated, Brno was saved and even today the bells ring midday at 11 a.m. in memory of that event.
Brno is also an important cultural center. National Theater companies perform operas, ballets and plays in the Janácek and Mahen Theaters; the city is the location of the Museum of Romanies (Gypsies), the Janácek Museum (devoted to the composer) and the Brno City Museum (located in Špilberk Castle). Brno was also an active participant in the Functionalist architecture movement of the 1920s and 1930s and landmark buildings such as Mies van der Rohe's Tugendhat Villa (a UNESCO World Heritage site) can be visited here.
The city's large student population ensures that things stay lively well into the night. Restaurants, bars and pubs abound on the streets leading off of the central Námestí Svobody and you won't have to search for long to find a suitable restaurant. The excellent Starobrno beer is brewed locally and you shouldn't miss a chance to try it during your visit.
When going out to bars, pubs and nightclubs in Brno be sure to look at the menu or request prices in advance; tourists are still not so very common here and that nice, friendly bartender who chats with you in perfect English might well charge you double what the locals pay and pocket the difference.
If you're looking for a closer glimpse of some of those fabulous local girls, then you should check out Mandarin (Jakubské Nám. 6). Two big rooms offer you the choice of a sterile modern cocktail bar where you can see and be seen or a downstairs music club with a small dance floor playing radio hits and house music. A slightly older crowd ranging from mid 20s to the middle-aged frequents the upscale Tabarin dance club (Divadelní 3), which features a central dance floor surrounded by plush furnishings, while Remix (Brandlova 4) offers a more casual setting. The brick walled club pounds out R&B and hip-hop, while the casually-dressed crowd of students and 20 somethings party the night away. The free Czech-language weekly, METRO, provides listings of the latest clubs, concerts, and events.
Austerlitz ("Slavkov" in Czech) was the site of the Battle of the Three Emperors, Napoleon's famous military victory over the allied forces of Austria and Russia, and town's beautiful Baroque chateau is well worth a visit. It is located just outside Brno and can be reached by public transportation.
The countryside north of Brno is a great place for outdoor activities. Numerous trails serve hikers in the summer and cross-country skiers in winter. Only a little bit farther to the north are the spectacular limestone caves in the Karst region of the Moravský kras. Some of the Czech Republic's best vineyards are located to the south and east of the city and the small towns in this region are great places to participate in the autumn vintage celebrations as well as late winter tasting of the new vintage.
The Grand Hotel Brno, located across from the Main Railway station, is probably the best-known of the four-star hotels (double rooms from 3,500 CZK a night), but the more moderately priced Hotel Slávia (double rooms from 2,250 CZK a night) is a great alternative. This Art Nouveau style hotel features an excellent restaurant and bar and is located right on the edge of the Old Town. The historic two-star Hotel Avion is situated just down the street and, as a notable example of functionalist architecture, it's a real bargain (single rooms from 900 CZK a night). For budget travelers, if it's summer you're in luck; the Traveller's Hostel is only a few hundred meters from the train station and at 290 CZK it's cheap!
Brno is located about 250 km from Prague and both buses and trains run an hourly service to Brno each day.
May - September: 9 a.m. - 6 p.m.
October - April: 9 a.m. - 5 p.m.
July - August: daily including Monday.
Last admission one hour before closing time.


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Charmaine Warren
Dance/Education
Dance/Performances
Yoga/Classes
Yoga/Workshops
Philadanco @ The Joyce - 10/2 - 10/7
What an evening! I also heard that it was quite a week! Philadanco! The Philadelphia Dance Company returned for their annual season at The Joyce last week (10/2 - 10/7) with four works by four acclaimed male
choreographers. There were two new works: Moan (World premiere) by Matthew Rushing, and Wake Up (New York Premiere) by Rennie Harris, plus Ronald K. Brown's Gatekeepers (1999), and the restaging of George
Faison's iconic work Suite Otis (1971). The evening opened with one of the most sincere renditions of Faison's Suite Otis ever! Brown's Gatekeepers, so daringly sparse, yet so contextually full that the audience must take time to think, followed. Moan, a fine example of Rushing finding his voice in dance theatre, then Harris' fantastic Wake Up, in the usual Harris-style, telling a story; opening the eyes about Africans and/or African descended people through his special blend of Hip Hop, closed the second half. We must say thank goodness for repertory companies sometimes, especially those that are able to, in just one evening, present such a broad range of works with marked difference. The dancers, most especially Lindsey Holmes and Tommie Waheed Evans, danced so beautifully on Friday evening. Kudos to the rest of the dancers and congratulations to Founder and Executive Director Joan Myers Brown for continuing such a steadfast journey!
Performance Calendars
What I Saw
I am a performer, historian, consultant and dance writer. I am a Empire State College's online program Center for Distance Learning. I am also a former faculty member at The Ailey School and the Alvin Ailey/Fordham University dance major program, Hunter College, Sarah Lawrence College (Guest), Kean University and The Joffrey Ballet School's Jazz and Contemporary Trainee Program. I write on dance for The Amsterdam News, Dance Magazine and various publications. Click below to read more about me at my home page - "About Me."


The pro pu boxing helmet boxing head guard mma headguard for adult sparring Boxing is manufactured by MagiDeal and was added around June 2017.
This version of the Boxing comes in Color : Blue.
The best price for the Pro PU Boxing Helmet Boxing in India is ₹ 2300 sold at ebay and available with free shipping and delivery within 10 DAYS WITHIN INDIA , 21 days OUTSIDE INDIA.


That looks lovely, thank you for sharing @JP_Finlay.
Let me also extend a very warm welcome to Workshop. It's great to have you join the community. I look forward to seeing more of your posts soon.
Did you make this yourself @JP_Finlay? If so, what else have you made?


Lana's Life: School's Out School's Out!
Now I have these two 24/7 until August 15th!
Oh well, not to worry, there is always plenty to do here on the farm, especially for Tink with FIVE 4-H projects to start and finish by July 9th or something like that.
Here's a look at their Field Day. Such fun, and Tuesday was a beautiful day to be out playing in the sun.
Yay!!! School's out. Aren't field days awesome? Good luck with those 4H projects!!
Looks like everyone had a great time! Enjoy those 4-H projects!
B's last day is today. I have to say, I'm a little sad. I'm excited about the summer, but that means she is a first grader. I'll admit, I teared up a little when she got on the bus this morning!
I hope yall have a great summer! Good luck on all the 4-H projects too!
Wordless Wednesday~ What's Growing on the Farm?
My Rock, "Stayin' Alive" & "On the Radio"


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'Yes, I'm scared': Local respiratory therapist works the front lines
By County Line | Posted March 27th, 2020 | Tweet
By KAREN PARKER
County Line Publisher Emeritus
Kim DeWitt of Ontario is one of those people we have heard much about on the news lately. She is "on the front lines" as a respiratory therapist with the Mayo Clinic Health System.
"When I come home, I throw all my clothes in the washer, take a shower and wash my hair."
She also no longer sees her grandchildren.
Healthcare workers will be consistently exposed to the COVID-19 virus, a respiratory infection that has already killed 20 doctors in Italy and multiple healthcare providers throughout the world.
DeWitt divides her time between Sparta and La Crosse. Sparta has no ICU beds, while La Crosse has 17. According to DeWitt, one therapist can tend to about five ventilators at once. All need to be watched to ensure they are delivering the right levels of oxygen.
DeWitt said Mayo is currently short six therapists and is scrambling to add to the staff. She wasn't certain how well-stocked Mayo was for supplies, but said she calculated that each day, it takes 118 masks to provide lab and other services to ICU patients.
Although COVID-19 patients are not flooding the system yet, DeWitt said this has been a very busy year with flu, respiratory infections and pneumonia.
← Here come the frauds: From bogus vaccine kits to 'Silver Solution,' coronavirus cons begin | Search goes on (sort of) for new N-O-W superintendent →
Two more Covid-19 deaths were reported in Monroe County today, bringing the total to 29 since the start of the pandemic.
Monroe County reported 27 new cases over the weekend and 11 today, bringing its tally to 3,814, of which 166 are active (three are hospitalized), 3,621 have recovered, and 27 have died.
Monroe County today had 28 new Covid-19 cases, bringing the total to 3,776, of which 169 are active (two are hospitalized), 3,580 have recovered, and 27 have died.
Monroe County had 26 new Covid-19 cases, bringing its total to 3,748, of which 173 are active (one is hospitalized), 3,548 have recovered, and 27 have died.
Monroe County had 32 new Covid-19 cases, resulting in a total 3,722 since the start of the pandemic.
Tuesday, Jan. 12: Monroe County has 14 cases; Vernon County, two; and Juneau County, seven and one death
Monroe County reported 14 new Covid-19 cases, resulting in a total of 3,690, of which 143 were active (one was hospitalized), 3,520 had recovered, and 27 had died.
Saturday, Jan. 9, to Monday, Jan. 11: Monroe County has 52 new cases; Vernon County, 12 cases and one death; and Juneau County, 30 cases
Monroe County reported 38 new Covid-19 cases over the weekend and 14 today, resulting in a tally of 3,676, of which 135 are active (one is hospitalized), 3,514 have recovered, and 27 have died.
Friday, Jan. 8: Monroe County reports 20 new cases; Vernon County, 12; and Juneau County, 27
Monroe County had 20 new Covid-19 cases, resulting in a total of 3,624, of which 157 were active (two were hospitalized), 3,440 had recuperated, and 27 had died.
Thursday, Jan. 7: Monroe County reports 23 cases; Vernon County, eight cases and one death; and Juneau County, 14 cases
Monroe County had 23 new Covid-19 cases, bringing its tally to 3,604, of which 151 are active (two are hospitalized), 3,426 have recovered, and 27 have died.
Wednesday, Jan. 6: Monroe County reports 25 new cases; Vernon County, 20; and Juneau County, 33
Monroe County had 25 new Covid-19 cases, bringing the total to 3,581, of which 129 were active (two were hospitalized), 3,425 had recovered, and 27 had died.
Tuesday, Jan. 5: Monroe County reports 43 new cases; Vernon County, 13; and Juneau County, 33
Today, Monroe County had 43 new Covid-19 cases, resulting in a total 3,566, of which 136 are active (three are hospitalized), 3,393 have recovered, and 27 have died.


This section of the site features news from past months and years of this site's existence dating back to June 2001 when it all began here on World of Genesis.com. This is a great place to start if you are more of a passive fan and want to check up on several months (or years) of news quickly and easily in one location. Currently, I've reformatted news from the old World of Genesis site going back all the way to the beginning with the exception of a few months here and there which still need to be reformatted and should resurface in the future (with the exception of June 2008 which appears to have been "lost"). Stay tuned!
WORLD OF GENESIS - CELEBRATING MORE THAN TWELVE YEARS OF FREE ON-LINE SERVICE TO GENESIS FANS EVERYWHERE!
Please note that archive news many contain some rare instances of information that changed over time (such as release dates, etc.) or even a few "dead" links to things that do not exist anymore. This section of the site is really for historical purposes on news that became available at that particular point in time in the world of Genesis. I have chosen not to modify the old text for this reason. Hopefully, you will enjoy reading back over the last nine years of history as much as I am!


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Bailya is een geslacht van weekdieren uit de klasse van de Gastropoda (slakken).

Soorten
 Bailya anomala (Hinds, 1844)
 Bailya cidaris Watters, 2013
 Bailya intricata (Dall, 1884)
 Bailya morgani Watters, 2009
 Bailya parva (C. B. Adams, 1850)
 Bailya sanctorum Watters, 2009

Buccinidae


Leamington Spa 23.59 - Warwick 00.03 - Warwick Parkway 00.06 - Dorridge 00.15 - Solihull 00.20 - Birmingham Moor Street 00.33.
23.07 Marylebone - Birmingham Moor Street is replaced by a bus between Leamington Spa and Birmingham Moor Street.
Bus times: Leamington Spa 00.30 - Warwick 00.40 - Warwick Parkway 00.50 - Dorridge 01.05 - Solihull 01.15 - Birmingham Moor Street 01.40.


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UPDATE 1-HSBC will have to share custody with JPMorgan of $52 bln in gold bars
Peter Hobson
December 1, 2022, 11:53 AM ·2 min read
(Adds detail)
By Peter Hobson
LONDON, Dec 1 (Reuters) - JPMorgan will join HSBC in storing bullion for the world's biggest gold-backed exchange-traded fund (ETF), the fund's operator said on Thursday, ending its rival's sole guardianship of the $52-billion stash of gold.
The change, which begins on Dec. 6, is a boon for JPMorgan, which could rake in millions of dollars of storage fees.
HSBC had been the sole custodian for SPDR Gold Trust, also known as GLD, since it launched in 2004. The bank currently stores about 910 tonnes of gold for GLD in London -- around a quarter of all the gold held for ETFs globally.
"The addition of JPM will change the current, single-custodian and vault operating model, to accommodate the activity of the fund in anticipation of future growth," the World Gold Council (WGC), which runs the fund, said in a statement.
WGC executive Joe Cavatoni said the council wanted to diversify its storage and adding JPMorgan "gives us another commercial entity with a vested interest in supporting the product."
He said the WGC would seek to funnel gold to JPMorgan, for example by sending it new metal added to the fund, and held out the possibility of an eventual even split between the two banks.
"If we get to the point where there's a very equal balance between the relationships, that would be exciting for us," he said.
Typical fees for large clients like GLD are around 0.03-0.04% of the value of the gold stored, a market source said. That means equal division of the fund's 910 tonnes would see JPMorgan take revenue of around $8-10 million a year from HSBC.
Cavatoni said the WGC's agreement with JPMorgan allowed it to store gold in the United States and Switzerland but for the time the fund intended continue storing all its gold in London.
HSBC said: "We're pleased to continue acting as a custodian for the World Gold Council's SPDR Gold Trust."
JPMorgan declined to comment. (Reporting by Peter Hobson Editing by David Goodman and Nick Zieminski)
GC=F
Gold Feb 23
HSBC Holdings plc
Hedge Funds Boost Treasury Shorts to Record on Doubts Over Rally
China stocks open up sharply as holiday rebound lifts sentiment