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%% Submission to Seminaire Lotharingien de Combinatoire.
%% Extension to
%% Submission to proceedings of 4th international colloquium "Quantum
%% Groups and Integrable Systems".
%% Author: T. A. Welsh.
%% Title: TWO-ROWED A-TYPE HECKE ALGEBRA REPRESENTATIONS AT ROOTS OF UNITY
%% Pages: 20
%% Version: 6 (slight ammendment to 5).
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\documentstyle[12pt,twoside]{article}
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\markboth{T. A. Welsh}{Non-generic two-rowed $H_n(q)$-modules}
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% Or preferably ...

% \newfam\msymfam
% \font\tensym=msym10
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%  \def\C{{\fam\msymfam C}}

\def\beq{\begin{equation}}
\def\eeq{\end{equation}}
\def\bea{\begin{eqnarray}}
\def\eea{\end{eqnarray}}
\def\nn{\nonumber}
\def\ba{\begin{array}}
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\def\titem#1{\par\hang\indent\hbox to 0pt{\hss\hbox to \parindent{
\enspace\hss#1}}\ignorespaces}
\def\newline{\par\noindent}

\def\mod{\mathop{\rm mod}}
\def\sds{\mathbin{\rlap{\raise0.9pt\hbox{$\scriptstyle+$}}\!\!\supset}}

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% Macros for tableaux: with or without boxes. Normal size or scriptsize.
% Written by Trevor Welsh (1992ish - modified 26/1/96).
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\def\makestrut#1#2{{\dimen12=#2
\divide\dimen12 by 4\dimen11=\dimen12\multiply\dimen11 by 3
\global\setbox#1=\hbox{\vrule height\dimen11 depth\dimen12 width0pt}}}

\newdimen\tadhdimen \newdimen\tabhdimen \newdimen\vdimen
\newdimen\smtadhdimen \newdimen\smtabhdimen
\newcount\splurt
\newbox\tadstrut \newbox\tabstrut
\newbox\smtadstrut \newbox\smtabstrut

% the following two enable the size of the boxes to be set and reset:
% they are for the normal size and the scriptsize of entries respectively.

\def\setyoungsize#1#2{          % Set box sizes for all regular tableaux
  \tadhdimen=#1\tabhdimen=#1\advance\tabhdimen by -0.4truept%
  \vdimen=#2%
  \makestrut\tadstrut\vdimen
  \advance\vdimen by -0.4pt%
  \makestrut\tabstrut\vdimen}

\def\setsmyoungsize#1#2{        % Set box sizes for all script tableaux
  \smtadhdimen=#1\smtabhdimen=#1\advance\smtabhdimen by -0.4truept%
  \vdimen=#2%
  \makestrut\smtadstrut\vdimen
  \advance\vdimen by -0.4pt%
  \makestrut\smtabstrut\vdimen}

% these are now used to set default sizes of tableaux.

\setyoungsize{13.5pt}{12pt}
\setsmyoungsize{7pt}{7pt}

% The following provide tableaux with the entries the current size:
% youngt does not provide boxes, youngd does (different formats).

\def\youngt#1{%
  \vcenter{\offinterlineskip
  \halign{&\copy\tadstrut\hbox to \tadhdimen{\hss$##$\hss}\cr #1}}}
\def\youngd#1{%
  \vcenter{\offinterlineskip
  \halign{&\vrule##&\copy\tabstrut\hbox to \tabhdimen{\hss$##$\hss}\cr #1}}}

% The following provide tableaux with the entries the current scriptsize:
% smyoungt does not provide boxes, smyoungd does (different formats).

\def\smyoungt#1{{\vcenter{\offinterlineskip
  \halign{&\copy\smtadstrut
  \hbox to \smtadhdimen{\hss$\scriptstyle ##$\hss}\cr #1}}}}
\def\smyoungd#1{{\vcenter{\offinterlineskip
  \halign{&\vrule##&\copy\smtabstrut
  \hbox to \smtabhdimen{\hss$\scriptstyle ##$\hss}\cr #1}}}}

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\begin{document}
\renewcommand{\thefootnote}{\fnsymbol{footnote}}
\setcounter{footnote}{1}

\begin{center}
{\bf TWO-ROWED A-TYPE HECKE ALGEBRA REPRESENTATIONS AT ROOTS OF UNITY}%
\footnote{An extended version of that originally presented at the 4th
International Colloquium ``Quantum Groups and Integrable Systems,''
Prague, 22-24 June 1995; and appearing in 
{\it Czech J.~Phys. \bf45}, (1996), 283--291.}

\vspace{1cm}

{\sc Trevor Alan Welsh}\footnote{
E-mail~: trevor@btm2x2.mat.uni-bayreuth.de}\\[2mm]
{\sl Faculty of Mathematical Studies,} \\
{\sl University of Southampton, Southampton, SO17 1BJ, U.K.}\\[2mm]
{\sl Lehrstuhl II f\"ur Mathematik,} \\
{\sl Universit\"at Bayreuth, 95440 Bayreuth, Germany.}
\end{center}

\vspace{1cm}

{\small In this paper, we describe a study into the explicit
construction of irreducible representations of the Hecke algebra
$H_n(q)$ of type $A_{n-1}$ in the non-generic case where $q$ is
a root of unity. The approach is via the Specht modules of $H_n(q)$
which are irreducible in the generic case, and possess a natural
basis indexed by Young tableaux. 
The general framework in which the irreducible non-generic
$H_n(q)$-modules are to be constructed is
set up and exploited in the case of two-part partitions.
For such partitions, we obtain the composition series
of the Specht modules, describe a basis for each irreducible
module in terms of a subset of the set of standard tableaux,
and detail an algorithm by which their explicit matrix
representations may be calculated.
Plentiful examples are given. Full proofs will be given elsewhere.}


\vspace{5mm}
\begin{center}
{\bf 1\ \ Introduction and notation}
\end{center}

The Hecke algebra $H_n(q)$ (of type $A_{n-1}$) is the unital associative
algebra over $\C$, generated by $h_i$, $i=1,2,\ldots,n-1$, subject
to the relations:
\beq
\ba{l}
h_ih_{i+1}h_i=h_{i+1}h_ih_{i+1};\\
h_ih_j=h_jh_i\qquad\hbox{for }\vert i-j\vert>1;\\
h_i^2=(q-1)h_i+q.
\label{hoopy}
\ea\eeq

The parameter $q\in\C$ will be permitted to take any non-zero value.
It is said to be generic if $q=1$ or $q^p\ne 1$ for $p=2,3,4,\ldots$.
Otherwise, if $q$ is a primitive $p$th root of unity for $p\ge2$,
it is said to be non-generic. In this latter case, $[p]_q=0$
where we define $[x]_q=1+q+q^2+\cdots+q^{x-1}$. In addition,
define $[x]_q!=[x]_q[x-1]_q\cdots [2]_q$.

When $q=1$, $H_n(q)$ may be identified with the group algebra
$\C S_n$ of the symmetric group on $n$ symbols, through identifying
each $h_i$ with the simple transposition $s_i=(i,i+1)\in S_n$.

If $w=s_{i_1}s_{i_2}\cdots s_{i_k}$ and $w\in S_n$ cannot be
expressed as a shorter product of the generators $s_i$, then
$s_{i_1}s_{i_2}\cdots s_{i_k}$ is said to be a reduced expression
for $w$ and the value of $k$ is the length $l(w)$ of $w$.
Thereupon, the relations (\ref{hoopy}) imply that the map
$h:\C S_n\rightarrow H_n(q)$ for which $h(s_i)=h_i$ and
$h(ww^\prime)=h(w)h(w^\prime)$ for $w,w^\prime\in S_n$ satisfying
$l(ww^\prime)=l(w)+l(w^\prime)$, and extended linearly, is well defined.
It follows that if $l(w)=k$ and $w=s_{i_1}s_{i_2}\cdots s_{i_k}$,
then $h(w)=h_{i_1}h_{i_2}\cdots h_{i_k}$. Furthermore, a basis of
$H_n(q)$ is provided by $\{h(w):w\in S_n\}$.
\par
It may be shown that if $q$ is generic then $H_n(q)$
is isomorphic to $\C S_n$ \cite{DJ86,Wn88} and
the representation theory of $H_n(q)$ is much the
same as that of $S_n$. In particular, the inequivalent irreducible
representations of $H_n(q)$ are indexed by partitions $\lambda$ of $n$.
That is, by finite integer sequences
$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_r)$ for which
$\lambda_1+\lambda_2+\cdots+\lambda_r=n$ and
$\lambda_1\ge\lambda_2\ge\cdots\ge\lambda_r>0$.
A partition for which no part $\lambda_i$
is repeated more than $p-1$ times is said to be $p$-regular.
In Section 2, an explicit construction of
the irreducible modules of $H_n(q)$ with $q$ generic will be described.
This generalisation of the well known Specht module
construction (see \cite{JK81}) was first described in \cite{KWy92},
and is based on the use of Young diagrams, Young tableaux
and $q$-analogues of Young symmetrisers.
The Young diagram $F^\lambda$ associated with the partition
$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_r)$ is a left-adjusted,
top-adjusted array of square boxes such that the $i$th row
(counting from the top) contains $\lambda_i$ boxes.
For instance, if $\lambda=(5,3,2,2)$, then
\beq
\setsmyoungsize{9pt}{9pt}
F^\lambda=\smyoungd{
\multispan{11}\hrulefill\cr
&&&&&&&&&&\cr \multispan{11}\hrulefill\cr
&&&&&&\cr \multispan{7}\hrulefill\cr
&&&&\cr \multispan{5}\hrulefill\cr
&&&&\cr \multispan{5}\hrulefill\cr}.
\eeq
Filling (or replacing) each of the $n$ boxes of $F^\lambda$ with 
elements of $\{1,2,\ldots,n\}$ so that no element appears more than once,
yields what is known as a Young tableau.
%(Tableaux containing repeated entries will not be considered here.)
Of the possible $n!$ tableaux of a given shape,
those for which the entries are increasing across each row and down
each column are
known as {\em standard\/} tableaux.
Examples may be found at (\ref{std1}), (\ref{std2}) and (\ref{std3}).
%For example,
%\beq
%T^{(5,3,2,2)}=\youngt{1&3&4&10&12\cr 2&6&9\cr 5&7\cr 8&11\cr}.
%\eeq
That particular standard tableau of shape $\lambda$ for which the entries
increase down first the leftmost column and then down successive columns
taken left to right is denoted $t^\lambda_-$.
For example,
\beq
\setsmyoungsize{9pt}{9pt}
t^{(5,3,2,2)}_-=
\smyoungt{1&5&9&11&12\cr 2&6&10\cr 3&7\cr 4&8\cr}.
\eeq
The total number of standard tableaux of shape $\lambda$ is equal to
$f^\lambda$, the dimension of the irreducible representation of $S_n$
(and $H_n(q)$ with $q$ generic) labelled by $\lambda$ (see \cite{JK81}).
In fact, the Specht module construction enables a basis to be
identified naturally with the set of standard tableaux.

\begin{center}
{\bf 2\ \ The Specht modules}
\end{center}

If $\lambda$ is a partition of $n$, the Specht module
$S^\lambda$ of $H_n(q)$ is defined to be the linear span of the vectors
$v_{t^\lambda}$, indexed by Young tableaux $t^\lambda$
and subject to certain relations (which will be defined below).
The {\em natural\/} action of $H_n(q)$ on these vectors is defined in
the following way.
We say that the entry $i$ precedes $j$ in $t^\lambda$ if $i$ occurs
before $j$ on reading the entries of $t^\lambda$
down the first and then successive columns.
If $x^\lambda$ is identical to $t^\lambda$ apart from the transposition
of $i$ and $i+1$, then $h_i$ acts on $v_{t^\lambda}$ as follows:
\beq
h_i v_{t^\lambda} =
\cases{
v_{x^\lambda}
&if $i$ precedes $i+1$ in $t^\lambda$;\cr
q v_{x^\lambda} + (q-1) v_{t^\lambda}
&if $i+1$ precedes $i$ in $t^\lambda$.\cr}
\label{action}
\eeq
\par
It is possible to express every $v_{z^\lambda}$ in terms of standard
tableaux, by means of the following two types of relation:
\smallskip\begingroup\parindent=22pt
\titem{1. } {\bf Column relations.} Entries within a column may be
transposed, if the corresponding vector is multiplied by $-1$.
Thus if $x^\lambda$ differs from $z^\lambda$ only in that a single pair of
entries within a column are transposed then:
\beq
v_{z^\lambda}=-v_{x^\lambda}.
\label{colrel}
\eeq
For example (denoting $v_{t^\lambda}$ by $t^\lambda$ for
typographical reasons),
\beq
\youngt{1&8&5&10&4&12\cr 6&11&3\cr 9&2&7\cr 13\cr}
=-\youngt{1&8&5&10&4&12\cr 6&2&3\cr 9&11&7\cr 13\cr}
=-\youngt{1&2&3&10&4&12\cr 6&8&5\cr 9&11&7\cr 13\cr}.
\eeq
\titem{2. } {\bf Garnir relations.} Assume that $z^\lambda$ is such that
its entries increase down each column.
If $z^\lambda$ is not standard then an adjacent pair of entries
exists with that on
the left greater than that on the right. Consider these two entries
together with all those below the left one and all those above the right one.
For example, we could consider the highlighted entries in:
\beq
\youngt{1&2&\bf3&10&4&12\cr 6&\bf8&\bf5\cr 9&\bf11&7\cr 13\cr}.
\eeq
Now form all possible tableaux $t^\lambda$ by permuting these entries
in all ways such that the permuted entries are increasing
down the portions of each of the two columns being considered.
The Garnir relation is then the following expression in which the sum
is over all such tableaux:
\beq
(-q)^{l(w_{z^\lambda})}
\sum_{t^\lambda} (-q)^{-l(w_{t^\lambda})}
v_{t^\lambda}=0,
\label{garnirrel}
\eeq
where $w_{t^\lambda}\in S_n$ maps $t^\lambda_-$ to $t^\lambda$.
The above example gives the
%six term ($4!/2!2!=6$)
Garnir relation:
\beq
\ba{l}
\!\!\youngt{1&2&\bf3&10&4&12\cr 6&\bf8&\bf5\cr 9&\bf11&7\cr 13\cr}
-q\youngt{1&2&\bf3&10&4&12\cr 6&\bf5&\bf8\cr 9&\bf11&7\cr 13\cr}
+q^2\youngt{1&2&\bf5&10&4&12\cr 6&\bf3&\bf8\cr 9&\bf11&7\cr 13\cr}\\
+q^2\youngt{1&2&\bf3&10&4&12\cr 6&\bf5&\bf11\cr 9&\bf8&7\cr 13\cr}
-q^3\youngt{1&2&\bf5&10&4&12\cr 6&\bf3&\bf11\cr 9&\bf8&7\cr 13\cr}
+q^4\youngt{1&2&\bf8&10&4&12\cr 6&\bf3&\bf11\cr 9&\bf5&7\cr 13\cr}=0.\\
\ea
\label{garnireg}
\eeq
\endgroup
\par\noindent
As in the example above, these relations do not necessarily immediately
express an arbitrary $v_{t^\lambda}$ in terms of standard tableaux.
However, it may be shown through employing a suitable order on the set
of tableaux \cite{JK81}, that repeated application of the column
and Garnir relations enables any term to be rendered in terms of
standard tableaux in a finite number of steps.
This completes the construction of the irreducible Specht module
$S^\lambda$ of $H_n(q)$ since the number of standard tableaux is equal
to the dimension of the representation of $H_n(q)$ indexed by $\lambda$
and consequently,
\beq
\{v_{t^\lambda}:t^\lambda {\rm\ is\ standard}\}
\eeq
is a basis for $S^\lambda$.

As an example, consider representing $h_1\in H_5(q)$ in the Specht
module $S^{(3,2)}$, by acting on each $v_{t^{(3,2)}}$ for which $t^{(3,2)}$
is standard (once more $v_{t^\lambda}$ is written as $t^\lambda$):
$$
\setyoungsize{10pt}{12pt}
\ba{l}
{ h_1} \youngt{1&3&5\cr 2&4\cr}
= \youngt{2&3&5\cr 1&4\cr}
= -\youngt{1&3&5\cr 2&4\cr},\\[3mm]
{ h_1} \youngt{1&2&5\cr 3&4\cr}
= \youngt{2&1&5\cr 3&4\cr}
= q\youngt{1&2&5\cr 3&4\cr}-q^2\youngt{1&3&5\cr 2&4\cr},\\[3mm]
{ h_1} \youngt{1&3&4\cr 2&5\cr}
= \youngt{2&3&4\cr 1&5\cr}
= -\youngt{1&3&4\cr 2&5\cr},\\[3mm]
{ h_1} \youngt{1&2&4\cr 3&5\cr}
= \youngt{2&1&4\cr 3&5\cr}
= q\youngt{1&2&4\cr 3&5\cr}-q^2\youngt{1&3&4\cr 2&5\cr},\\[3mm]
{ h_1} \youngt{1&2&3\cr 4&5\cr}
= \youngt{2&1&3\cr 4&5\cr}
= q\youngt{1&2&3\cr 4&5\cr}-q^2\youngt{1&4&3\cr 2&5\cr}
%\qquad\qquad\qquad
=q \youngt{1&2&3\cr 4&5\cr}
-q^3 \youngt{1&3&4\cr 2&5\cr}
+q^4 \youngt{1&3&5\cr 2&4\cr}.\\
\ea
$$
Here, column relations have been used in the first and third calculations,
and Garnir relations have been used in the second, fourth and last (twice),
to express each result in terms of the standard tableaux.
Consequently, in the representation labelled by the partition $(3,2)$,
$h_1$ is mapped to the matrix (where zeros are denoted by dots):
\beq
\pmatrix{
-1&-q^2&.&.&q^4\cr
.&q&.&.&.\cr
.&.&-1&-q^2&-q^3\cr
.&.&.&q&.\cr
.&.&.&.&q\cr}.
\eeq
The matrices representing the generators $h_i$ of $H_n(q)$ in each
irreducible representation for $n\le5$ given
in \cite{KWy92} have been produced in a similar way.

\begin{center}
{\bf 3\ \ The Young symmetriser and its annihilators}
\end{center}

For each entry $a$ of $t^\lambda_-$ which is not at the bottom of
a column, define the {\em column element}:
\beq
C^\lambda_a=1+h_a.
\label{coldef}
\eeq
Its action on $v_{t^\lambda_-}$ gives rise to a Column relation
(cf.~(\ref{colrel})):
\beq
C^\lambda_a v_{t^\lambda_-}=0.
\label{annih1}
\eeq

The {\em Garnir element\/} $G^\lambda_a$ is defined for each $a$ which is
not at the end of a row of $t^\lambda_-$, through first letting $d$ be the
entry to the right of $a$, $b$ be the entry at the bottom of the column
containing $a$, and $c$ (${}=b+1$) the entry at the top of the column
containing $d$ in $t^\lambda_-$. 
With $W_{ij}$ the subgroup of $S_n$ permuting only $\{i,i+1,\ldots,j\}$,
let ${\cal G}^\lambda_a$ be a set of
left coset representatives for $W_{ab}\times W_{cd}$ in $W_{ad}$
chosen so that each representative is of minimal length in its coset
(it is unique). Then let \cite{KWy92}:
\beq
G^\lambda_a=q^l\sum_{d\in{\cal G}^\lambda_a} (-q)^{-l(d)} h(d),
\label{garnirdef}
\eeq
where $l$ is the length of the longest element in ${\cal G}^\lambda_a$.
Its action on $v_{t^\lambda_-}$ gives rise to a Garnir relation
(cf.~(\ref{garnirrel})):
\beq
G^\lambda_a v_{t^\lambda_-}=0.
\label{annih2}
\eeq

It is easily shown that the general column and Garnir relations
of Section 2 are a consequence of (\ref{annih1}) and (\ref{annih2}).
These properties themselves arise by identifying
$v_{t^\lambda_{\smash{-}}}$
with the $q$-analogue $Y^\lambda(q)$ of the Young symmetriser.
$Y^\lambda(q)$ was originally defined in \cite{DJ86,Gy86} and cast
in a form suitable for the current purposes in \cite{KWy92,BKW93}.
However, as is seen, only its $2n-r-\lambda_1$ annihilators $C_a$ and $G_a$
are required in the construction of the Specht module $S^\lambda$.
Thus $S^\lambda$ may be defined as the free module generated by a non-zero
vector
(say $v_{t^\lambda_{\smash{-}}}$)
subject to (\ref{annih1}) and (\ref{annih2}).
This viewpoint of $S^\lambda$ will be utilised in what follows.
\par
To illustrate it, consider $\lambda=(6,3,3,1)$, for which:
\beq
t^\lambda_-=
\youngt{1&5&8&11&12&13\cr 2&6&9\cr 3&7&10\cr 4\cr}.
\label{std1}
\eeq
Here we have the seven column elements $1+h_1$, $1+h_2$, $1+h_3$,
$1+h_5$, $1+h_6$, $1+h_8$ and $1+h_9$, each of which annihilates
$v_{t^\lambda}$. There are nine Garnir elements $G^\lambda_a$ for
$a=1,2,3,5,6,7,8,11,12$, each of which annihilates $v_{t^\lambda}$.
Typically:
\beq
\ba{l}
G^\lambda_6=
q^4-q^3h_7+q^2h_6h_7+q^2h_8h_7-qh_6h_8h_7+h_7h_6h_8h_7;\\
G^\lambda_8=
q^3-q^2h_{10}+qh_9h_{10}-h_8h_9h_{10};\\
G^\lambda_{11}=
q-h_{11}.\\
\ea\eeq
In fact $G^\lambda_6 v_{t^\lambda_-}=0$ gives rise to (\ref{garnireg}).
\goodbreak
\vfill\eject

\begin{center}
{\bf 4\ \ Decomposing \boldmath $S^\lambda$ at roots of unity}
\end{center}

\nobreak
In the generic case when $q$ is not a root of unity, each Specht module
$S^\lambda$ of $H_n(q)$ is irreducible.
However, this is no longer so if
$q$ is a root of unity, although $S^\lambda$ remains well-defined.
For $q$ a primitive $p$th root of unity,
let $D^\lambda_p$ be the irreducible $H_n(q)$-module
obtained by factoring out the maximal proper submodule from $S^\lambda$.
It is shown in \cite{DJ86} that in this case,
\beq
\{D^\lambda_p: \lambda {\rm\ is\ } p {\rm -regular}\}
\label{irreps}
\eeq
is a complete and irredundant set of irreducible $H_n(q)$-modules.
Very little is known about the $D^\lambda_p$ or the composition
series of $S^\lambda$ in terms of the $D^\lambda_p$ except in a few
specific cases (see \cite{Jm90} for $n\le10$, \cite{CK92} for $n\le5$,
and \cite{JM95} for various results when $p=2$).
\par
The viewpoint developed in the previous section provides a means of
tackling these questions in a quite general way.
It relies on the fact that within the set (\ref{irreps}),
the module $D^\mu_p$ is characterised by the
presence of a non-zero vector
$v_{t^\mu_{\smash{-}}}$
which is annihilated by the set of column and Garnir elements,
$C^\mu_a$ and $G^\mu_a$ defined above.
This follows because, via (\ref{action}),
$v_{t^\mu_{\smash{-}}}$
generates the whole of $S^\mu$, and hence
$v_{t^\mu_{\smash{-}}}$
cannot be present in any proper submodule of $S^\mu$.
Thus, to determine whether $S^\lambda$ is reducible, it is sufficient
to prove the existence of a non-zero $v^\mu\in S^\lambda$ having
the same set of annihilators as
$v_{t^\mu_{\smash{-}}}\in S^\mu$
for some $p$-regular partition $\mu\ne\lambda$ of $n$.
Conversely, the absence of all such
$v_{t^\mu_{\smash{-}}}$
would prove $S^\lambda$ to be irreducible.
(In fact, the results of \cite{DJ86} and \cite{DJ87} considerably restrict
the set of $\mu$ for which $D^\mu_p$ may occur as a composition factor
of $S^\lambda$.)
\par
As an example, consider $\lambda=(3,2)$.
We will show that if $p=3$ then
\beq
\setyoungsize{10pt}{12pt}
v^\mu=(1+h_4)v_{t^{(3,2)}_-}=
\youngt{1&3&5\cr 2&4\cr}+\youngt{1&3&4\cr 2&5\cr}
\label{anniheg1}
\eeq
is annihilated by the column and Garnir elements of $\mu=(4,1)$, and hence
that $S^{(3,2)}$ has a submodule $D^{(4,1)}_3$.
Since
$t^{(4,1)}_-=\smyoungt{1&3&4&5\cr 2\cr}$,
the column and Garnir elements of $\mu=(4,1)$, are:
\beq
\ba{ll}
\romannumeral1) & (1+h_1); \\
\romannumeral2) & (q^2-qh_2+h_1h_2); \\
\romannumeral3) & (q-h_3); \\
\romannumeral4) & (q-h_4). \\
\ea\eeq
Acting on (\ref{anniheg1}) with each of these, using (\ref{action}) gives:
$$
\setyoungsize{9pt}{12pt}
\ba{ll}
\romannumeral1) & (1+h_1)v^\mu=
\youngt{1&3&5\cr 2&4\cr}+\youngt{2&3&5\cr 1&4\cr}
+\youngt{1&3&4\cr 2&5\cr}+\youngt{2&3&4\cr 1&5\cr}
=0;\\
\romannumeral2) & (q^2-qh_2+h_1h_2) v^\mu=
q^2\youngt{1&3&5\cr 2&4\cr}-q\youngt{1&2&5\cr 3&4\cr}+\youngt{2&1&5\cr 3&4\cr}
\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
+q^2\youngt{1&3&4\cr 2&5\cr}-q\youngt{1&2&4\cr 3&5\cr}+\youngt{2&1&4\cr 3&5\cr}
=0;\\
\romannumeral4) & (q-h_4)v^\mu=
q\youngt{1&3&5\cr 2&4\cr}-\youngt{1&3&4\cr 2&5\cr}
+q\youngt{1&3&4\cr 2&5\cr}-q\youngt{1&3&5\cr 2&4\cr}
-(q-1)\youngt{1&3&4\cr 2&5\cr}
=0;\\
\romannumeral3) & (q-h_3)v^\mu=
q\youngt{1&3&5\cr 2&4\cr}-\youngt{1&4&5\cr 2&3\cr}
+q\youngt{1&3&4\cr 2&5\cr}-\youngt{1&4&3\cr 2&5\cr}
\\
&\qquad
=(1+q)\youngt{1&3&5\cr 2&4\cr}+q\youngt{1&3&4\cr 2&5\cr}
-q\youngt{1&3&4\cr 2&5\cr}+q^2\youngt{1&3&5\cr 2&4\cr}
%\\
%&\qquad
=(1+q+q^2)\youngt{1&3&5\cr 2&4\cr}=0,
\\
\ea
$$
since if $p=3$, then $1+q+q^2=0$.
Therefore, $D^{(4,1)}_3$ is a submodule of $S^{(3,2)}$.
It may be shown that the 4-dimensional $S^{(4,1)}$ is irreducible when
$p=3$, so that $D^{(4,1)}_3\equiv S^{(4,1)}$.
Hence $D^{(3,2)}_3$ is of dimension $5-4=1$.
It is spanned by $v_{t^{(3,2)}_-}$.

In order to generalise the $v^\mu$ of the previous example,
let $\lambda=(\lambda_1,\lambda_2)$ and for $1\le x\le\lambda_2$,
define the standard tableau $t^\lambda_{x-}$ as follows.
After ignoring the rightmost $x$ boxes of the bottom row and
the rightmost $\lambda_1-\lambda_2$ boxes of the top row,
fill the diagram as for $t^{(\lambda_2,\lambda_2-x)}_-$.
Then put the entries $\{2\lambda_2-x+1,\ldots,n\}$ in increasing
order across first the remaining $x$ boxes of the bottom row and
then the remaining $\lambda_1-\lambda_2$ boxes of the top row.
For example:
\beq
t^{(4,2)}_{2-}=
\youngt{1&2&\bf5&\bf6\cr \bf3&\bf4\cr},
\quad
t^{(7,4)}_{3-}=
\youngt{1&3&4&5&\bf9&\bf10&\bf11\cr 2&\bf6&\bf7&\bf8\cr},
\quad
t^{(7,4)}_{1-}=
\youngt{1&3&5&7&\bf9&\bf10&\bf11\cr 2&4&6&\bf8\cr},
\quad
\label{exxminus}
\eeq
where the latter set of entries have been highlighted.
In addition, define the standard tableau $t^\lambda_{x+}$
which also has the entries $\{1,2,\ldots,2\lambda_2-x\}$
placed exactly as for $t^{(\lambda_2,\lambda_2-x)}_{-}$.
The entries $\{2\lambda-x+1,\ldots,n\}$ are then
placed in increasing order across first the remaining
$\lambda_1-\lambda_2$ boxes
of the top row and then the remaining $x$ boxes of the bottom row.
For example:
\beq
t^{(4,2)}_{2+}=
\youngt{1&2&\bf3&\bf4\cr \bf5&\bf6\cr},
\quad
t^{(7,4)}_{3+}=
\youngt{1&3&4&5&\bf6&\bf7&\bf8\cr 2&\bf9&\bf10&\bf11\cr},
\quad
t^{(7,4)}_{1+}=
\youngt{1&3&5&7&\bf8&\bf9&\bf10\cr 2&4&6&\bf11\cr}.
\quad
\label{exxplus}
\eeq

For $a\le b\le m$, let ${\cal D}^b_{a,m}$ be the set of left coset
representatives of $W_{b,m}$ in $W_{a,m}$ chosen so that
each representative is of minimal length in its coset
(once more, it is unique). Then define:
\beq
v^{(\lambda_1+x,\lambda_2-x)}=
\sum_{d\in{\cal D}^{2\lambda_2+1}_{\lambda_2+i,n}}
v_{dt^\lambda_{x-}}.
\label{symterm}
\eeq
(For each of the tableaux in (\ref{exxminus}), this sum is over all
tableaux in which the highlighted entries have been permuted
such that those in the top row are in increasing order.)

\proclaim Theorem 1. For $\lambda=(\lambda_1,\lambda_2)$ and
$q$ a primitive $p$th root of unity,
let $x=p-1-(\lambda_1-\lambda_2)\mod p$ and
$\mu=(\lambda_1+x,\lambda_2-x)$.
If $0<x\le\lambda_2$ then $v^{\mu}\in S^\lambda$ satisfies:
\parindent=30pt
\titem{1. } $C^{\mu}_a v^{\mu}=0$ for all $a$ not at the bottom of
a column of $t^\mu_-$;
\titem{2. } $G^{\mu}_a v^{\mu}=0$ for all $a$ not at the end of
a row of $t^\mu_-$.
\newline
Furthermore, if $v^{\mu}$ is written in
terms of standard tableaux then, subject to $[p]_q=0$,
each polynomial coefficient has a polynomial factor $[x]_q!$.
Moreover the coefficient of $v_{t^\lambda_{x+}}$ in
$v^{\mu}/[x]_q!$ is 1.

\noindent
This theorem therefore has the consequence that $S^{(\lambda_1,\lambda_2)}$
has a submodule $D^{\mu}_p$ where $\mu$ is determined by $p$ as in
the statement of Theorem 1.
It may be further shown that all submodules of $S^{(\lambda_1,\lambda_2)}$
arise in this way, and this enables the full composition of $S^\lambda$
in the case of two-part partitions $\lambda$ to be determined.
The result is best expressed using the notion of a boundary strip
(sometimes called a rim hook \cite{JK81}) of a Young diagram $F^\lambda$.
It is a continuous strip of boxes obtained by starting at the
rightmost end of a row of $F^\lambda$ and, for a number of steps,
recursively passing to the box below if one exists, otherwise passing
to the box to the left. The strip ends at the bottom of any column
to the left of, or in, the column in which it started.
%It must be such that if removed from $F^\lambda$, a valid Young
%diagram remains.
The length of the boundary strip is the number of boxes that it comprises.

\goodbreak
\proclaim Theorem 2. If $\lambda=(\lambda_1,\lambda_2)$ and $q$
is a primitive $p$th root of unity then
$S^\lambda$ is reducible if and only if for some integer $k>0$,
$F^\lambda$ has a boundary strip of length $kp$ having at least one,
but not more than $p-1$ boxes in the second row
(or equivalently, if there exists an integer $k>0$ such that
$\lambda_1-\lambda_2+2\le kp\le\min\{\lambda_1+1,\lambda_1-\lambda_2+p\}$).
If so, $S^\lambda$ has an irreducible submodule corresponding to the diagram
obtained by moving all the boxes of the boundary strip into the top row.
The corresponding quotient module is irreducible.
That is:
\beq
S^{(\lambda_1,\lambda_2)}=
D^{(\mu_1,\mu_2)}_p
\sds D^{(\lambda_1,\lambda_2)}_p,
\label{reduction}
\eeq
where
\beq
(\mu_1,\mu_2)=(\lambda_1+p-1-(\lambda_1-\lambda_2)\mod p,
\lambda_2-p+1+(\lambda_1-\lambda_2)\mod p)
\eeq
\par\noindent
This theorem is illustrated by the following table, which for various
$\lambda=(\lambda_1,\lambda_2)$ and $p$, displays the Young diagram $F^\lambda$
with the appropriate boundary strip indicated, says whether $S^\lambda$
is reducible, and shows its composition series.
$$
\vcenter{\openup\jot
\halign{\enspace$#$\quad\hfill&$#$\qquad\hfill&$#$\qquad\hfill&#\quad\hfill
&$#$\enspace\hfill\cr
\enspace\lambda&p&F^\lambda&$S^\lambda$&\rm{Composition}\cr
%
(5,4)&3&
\smyoungd{\multispan{11}\hrulefill\cr
 &&&&&&&\bullet&&\bullet&\cr \multispan{11}\hrulefill\cr
 &&&&&&&\bullet&\cr \multispan{9}\hrulefill\cr}
&reducible&S^{(5,4)}=D^{(6,3)}_3\sds D^{(5,4)}_3\cr
%
(6,4)&3&
\smyoungd{\multispan{13}\hrulefill\cr
 &&&&&&&&&&&&\cr \multispan{13}\hrulefill\cr
 &&&&&&&&\cr \multispan{9}\hrulefill\cr}
&irreducible&S^{(6,4)}=D^{(6,4)}_3\cr
%
(7,4)&3&
\smyoungd{\multispan{15}\hrulefill\cr
 &&&&&&&\bullet&&\bullet&&\bullet&&\bullet&\cr \multispan{15}\hrulefill\cr
 &&&&&\bullet&&\bullet&\cr \multispan{9}\hrulefill\cr}
&reducible&S^{(7,4)}=D^{(9,2)}_3\sds D^{(7,4)}_3\cr
%
(8,3)&3&
\smyoungd{\multispan{17}\hrulefill\cr
 &&&&&&&&&&&&&&&&\cr \multispan{17}\hrulefill\cr
 &&&&&&\cr \multispan{7}\hrulefill\cr}
&irreducible&S^{(8,3)}=D^{(8,3)}_3\cr
%
(8,2)&5&
\smyoungd{\multispan{17}\hrulefill\cr
 &&&&&&&&&&&&&&&&\cr \multispan{17}\hrulefill\cr
 &&&&\cr \multispan{5}\hrulefill\cr}
&irreducible&S^{(8,2)}=D^{(8,2)}_5\cr
%
(9,3)&5&
\smyoungd{\multispan{19}\hrulefill\cr
 &&&&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet&&\bullet&\cr
 \multispan{19}\hrulefill\cr
 &\bullet&&\bullet&&\bullet&\cr \multispan{7}\hrulefill\cr}
&reducible&S^{(9,3)}=D^{(12)}_5\sds D^{(9,3)}_5\cr }}
$$
\par
Theorem 2 has the consequence that the character $\tilde\chi^\lambda_p$
of the irreducible representation corresponding to $D^\lambda_p$
may be expressed as a finite sum over the characters
$\chi^\lambda(q)$ of the generic representations of $H_n(q)$
(which themselves may be calculated using the methods and formulae of
\cite{KWy90,KWy92,Rm91,Vj91}).

\proclaim Theorem 3. If $S^{(\lambda_1,\lambda_2)}$ is reducible then,
using the notation of Theorem 2,
\beq
\tilde\chi^\lambda_p=
\sum_{j=0}^{[\lambda_2/p]} \chi^{(\lambda_1+jp,\lambda_2-jp)}(q)
-\sum_{j=0}^{[\mu_2/p]} \chi^{(\mu_1+jp,\mu_2-jp)}(q),
\eeq
where $[x]$ is the largest integer less than or equal to $x$.
\par\noindent
Of course, this Theorem may be used to give the dimension of
$D^\lambda_p$ in terms of the dimensions $f^\nu$ of the
irreducible representations of $S_n$.
For example,
\beq\ba{l}
\dim D^{(4,2)}_4=f^{(4,2)}-f^{(5,1)}=9-5=4;\\
\dim D^{(4,2)}_2=f^{(4,2)}+f^{(6,0)}-f^{(5,1)}=9+1-5=5;\\
\dim D^{(6,5)}_3=f^{(6,5)}+f^{(9,2)}-f^{(7,4)}-f^{(10,1)}
=132+44-165-10=1.
\label{exdims}
\ea\eeq
In fact, $D^{(6,5)}_3\cong S^{(1^{11})}$ (note that $(1^{11})$ is
not a $3$-regular partition).
%\vfill\eject

\begin{center}
{\bf 5\ \ The root-standard basis}
\end{center}

\nobreak
When $q$ is a primitive $p$th root of unity, the irreducible
$H_n(q)$-module $D^{(\lambda_1,\lambda_2)}_p$ may be constructed
along lines similar to the construction of the Specht modules.
A basis for $D^{(\lambda_1,\lambda_2)}_p$ may be defined in terms
of a certain subset of the set of standard tableaux.
In order to specify this set, let
\beq
T^{(\lambda_1,\lambda_2)}=
\youngt{a_1&a_2&a_3&\cdot&\cdot&\cdot&\cdot&\cdot&a_{\lambda_1}\cr
b_1&b_2&b_3&\cdot&\cdot&b_{\lambda_2}\cr},
\eeq
and say that $T^\lambda$ is $s$-strip standard at the $i$th position
if:
\beq
b_i<a_{i+s-2}
\eeq
(or if $i>\lambda_1-s+2$, when of course $a_{i+s-2}$ is undefined).

\proclaim Definition 1. If $\lambda=(\lambda_1,\lambda_2)$ and the
positive integers $p$ and $k$ are such that
$\lambda_1-\lambda_2+2\le kp\le\min\{\lambda_1+1,\lambda_1-\lambda_2+p\}$,
then $T^\lambda$ is said to be {\em $p$-root standard} if $T^\lambda$
is standard and either:
\parindent=30pt
\titem{1. } $T^\lambda$ is $kp$-strip standard at positions
$1,2,\ldots,\lambda_2$;
\titem{or 2. } to the right of the rightmost position of
a non-standard $kp$-strip, there is a position at which
$T^\lambda$ is $((k-1)p+2)$-strip standard.

\noindent
Note that in the important case of $k=1$, the second condition here
can never be satisfied because standardness denies 2-strip standardness.
In this case, the tableaux are identical to those defined in
\cite{Wn88} for the corresponding representations.

As an example, consider $\lambda=(7,4)$ and $p=3$ (so that $k=2$).
In this case, the following are $3$-root standard:
\beq
\youngt{1&2&3&4&6&8&10\cr 5&7&9&11\cr},
\qquad
\youngt{1&3&4&5&6&7&11\cr 2&8&9&10\cr},
\qquad
\youngt{1&2&3&4&5&9&10\cr 6&7&8&11\cr};
\qquad
\label{std2}
\eeq
whereas the following are not $3$-root standard:
\beq
\youngt{1&3&5&6&7&8&\bf9\cr 2&4&\bf10&11\cr},
\qquad
\youngt{1&3&4&5&6&\bf7&10\cr 2&\bf8&9&11\cr},
\qquad
\youngt{1&2&3&4&\bf5&8&10\cr \bf6&7&9&11\cr};
\qquad
\label{std3}
\eeq
with the highlighted entries indicating the offending
$6$-strip.

\proclaim Theorem 4. If $\lambda=(\lambda_1,\lambda_2)$,
the dimension of $D^\lambda_p$ is equal to the number of
$p$-root standard tableaux of shape $\lambda$.

\noindent
Let $d^\lambda_p$ denote the dimension of $D^\lambda_p$.
In the case $p=2$, we immediately deduce from Theorem 1 that if 
$n=\lambda_1+\lambda_2$ is odd then $d^\lambda_2=f^\lambda$.
Furthermore, if $n$ is even
(so that $\lambda_1-\lambda_2$ is even and positive), then
Definition 1 forces the entry $n$ to be in the final box of
the top row. The other entries are then restricted only to be
in a standard configuration. Thus if $\lambda$ is 2-regular:
\beq
d^{(\lambda_1,\lambda_2)}_2 =
\cases{
f^{(\lambda_1,\lambda_2)}
&if $n$ is odd;\cr
f^{(\lambda_1-1,\lambda_2)}
&if $n$ is even.\cr}
\label{dimp2}
\eeq
This result was obtained in \cite{JM95} by considering the restriction
of the $H_n(-1)$-module $D^\lambda_2$ to $H_{n-1}(-1)$.
In fact, either of these arguments may be generalised to yield:
\beq
d^{(\lambda_1,\lambda_2)}_p=
d^{(\lambda_1-1,\lambda_2)}_p
\quad\hbox{ if }
(\lambda_1-\lambda_2+2)\mod p\equiv0.
\eeq

\begin{center}
{\bf 6\ \ Explicit \boldmath $D^{(\lambda_1,\lambda_2)}_p$}
\end{center}

As indicated above, the non-generic $H_n(q)$-module $D^\lambda_p$
may be explicitly constructed with basis:
\beq
\{v_{t^\lambda}: t^\lambda {\rm\ is\ } p {\rm-root\ standard}\}.
\eeq
As for the Specht module, the explicit construction process
relies on being able to express terms indexed by arbitrary
tableaux in terms of those that are in the basis.
For the explicit construction of $D^\lambda_p$, 
the column and Garnir relations are retained and are
supplemented by additional relations. These relations are
described here.

By using the column and Garnir relations, any term may be
expressed in terms of standard tableaux. The rewriting of
$v_{t^\lambda}$ with $t^\lambda$ standard in terms of $p$-root
standard tableaux involves $v^{\mu}$
(as defined in (\ref{symterm})) and similar expressions.
So assume that $t^\lambda$ is standard and that $i$ is the largest
number such that $t^\lambda$ is not $kp$-strip standard at
the $i$th position. Three cases need to be considered.

\setyoungsize{10pt}{12pt}
{\bf Case 1. $i>\lambda_2-p$.} Let $x=\lambda_2-i+1$ which is the number
of boxes in the second row to the right of, and including, 
the non strip-standard position.
Now let:
\beq
v_0=
{1\over [x]_q!}
\sum_{d\in{\cal D}^{2\lambda_2+1}_{\lambda_2+i,\lambda_2+i+kp-2}}
v_{dt^\lambda_{x-}}
=\sum_{{\rm standard\ } z^\lambda} c(z^\lambda) v_{z^\lambda},
\label{symcase1}
\eeq
where the column and Garnir relations have been used to obtain
the sum over standard tableaux, and where each $c(t^\lambda)$
is a polynomial in $q$. From Theorem 1, it can be seen that
$c(t^\lambda_{*+})=1$ here where $t^\lambda_{*+}$ has
$\{\lambda_2+i+kp-1,\ldots,n\}$ in the last 
$\lambda_1-i-kp+2$ boxes of the top
row but is otherwise identical to $t^{(i+kp-2,\lambda_2)}_{x+}$. 
Then, as may be shown, the quotienting out of the submodule
implies that $v_0=0$.
Thereby, an expression for $v_{t^\lambda_{*+}}$
is obtained in terms of other tableaux.
Acting on the tableaux that index the vectors in this expression with
$w\in S_n$ defined such that $t^\lambda=wt^\lambda_{*+}$, may be
shown to yield an expression for $v_{t^\lambda}\in D^\lambda_p$:
\beq
v_{t^\lambda}
=-\sum_{{\rm standard\ } z^\lambda\ne t^\lambda_{*+}}
c(z^\lambda) v_{wz^\lambda}.
\eeq
For $\lambda=(4,2)$ and $p=4$ (so that $k=1$), we will consider 
a number of examples.
First let
\beq
t^\lambda_1=
\youngt{1&3&4&5\cr 2&6\cr},
\eeq
for which $i=2$ and $x=1$, whereupon
\beq
t^\lambda_{x-}=
\youngt{1&3&\bf5&\bf6\cr 2&\bf4\cr},
\eeq
$t^\lambda_{*+}=t^\lambda_1$, $w=1$ and we require
${\cal D}^5_{4,6}=\{1,s_4,s_5s_4\}$ which permutes the
highlighted entries.
Then on setting this particular case of (\ref{symcase1}) to zero yields:
\beq
\youngt{1&3&4&5\cr 2&6\cr}
=-\youngt{1&3&4&6\cr 2&5\cr}
-\youngt{1&3&5&6\cr 2&4\cr},
\label{ex1}
\eeq
an expression which immediately gives $v_{t^\lambda_1}$ in terms
of $4$-root standard tableaux.

For the tableau
\beq
t^\lambda_2=
\youngt{1&2&3&4\cr 5&6\cr},
\eeq
again $i=2$, and $x=1$, so that
$t^\lambda_{x-}$ and $t^\lambda_{*+}$ are as above, but now $w=s_4s_3s_2$.
Thus, it is required to act on the tableaux in the expression (\ref{ex1})
with $s_4s_3s_2$, which gives:
\beq
\youngt{1&2&3&4\cr 5&6\cr}
=-\youngt{1&2&3&6\cr 5&4\cr}
-\youngt{1&2&4&6\cr 5&3\cr}.
\label{ex2}
\eeq
Here the resulting terms are not $4$-root standard and further processing
would be required to render $v_{t^\lambda_2}$ in terms of 4-root
standard tableaux.

For the tableau
\beq
t^\lambda_3=
\youngt{1&2&3&6\cr 4&5\cr},
\eeq
$i=1$ and $x=2$, whereupon
\beq
t^\lambda_{x-}=
\youngt{1&2&\bf5&6\cr \bf3&\bf4\cr},
\eeq
$t^\lambda_{*+}=t^\lambda_3$, $w=1$ and we require
${\cal D}^5_{3,5}=\{1,s_3,s_4,s_4s_3,s_3s_4,s_3s_4s_3\}$,
Thus, in this case, (\ref{symcase1}) yields:
\beq
v_0=
\youngt{1&2&5&6\cr 3&4\cr}
+\youngt{1&2&5&6\cr 4&3\cr}
+\youngt{1&2&4&6\cr 3&5\cr}
+\youngt{1&2&4&6\cr 5&3\cr}
+\youngt{1&2&3&6\cr 4&5\cr}
+\youngt{1&2&3&6\cr 5&4\cr},
\eeq
which, on standardisation using the column and Garnir relations,
results in:
\beq
v_0=
(1+q)
\left(
(q^3\!-\!q^2)
\youngt{1&3&5&6\cr 2&4\cr}
-q^2
\youngt{1&3&4&6\cr 2&5\cr}
+\youngt{1&2&5&6\cr 3&4\cr}
+\youngt{1&2&4&6\cr 3&5\cr}
+\youngt{1&2&3&6\cr 4&5\cr}
\right),
\eeq
where, in this particular case, it has not been necessary to use
$[4]_q=0$ to enable the factor $[x]_q=(1+q)$ to be extracted.
Setting this expression to zero yields the requisite expression
for $v_{t^\lambda_3}$:
\beq
\youngt{1&2&3&6\cr 4&5\cr}
=(q-1)
\youngt{1&3&5&6\cr 2&4\cr}
-\youngt{1&3&4&6\cr 2&5\cr}
-\youngt{1&2&5&6\cr 3&4\cr}
-\youngt{1&2&4&6\cr 3&5\cr},
\label{resultcase1}
\eeq
where use has been made of $q^2=-1$.

{\bf Case 2. $i\le\lambda_2-p$ and $k=1$.} The relations obtained
in this case are similar to those obtained in Case 1. They may
be viewed as those in that case having been moved leftward through
the tableaux.
Here let $x=p-1$ which is again the number of boxes in the second
row over which a symmetrisation process takes place.
Now define $t^\lambda_{*}$ to be identical to $t^\lambda_-$ in the
first $i-1$ columns and also from columns $i+p-1$ to the last.
The remaining entries $\{2i-1,\ldots,2(i+p)-4\}$ are placed
in the remaining positions, across first the top row and then across
the bottom row (so that in its first $i+p-2$ columns,
$t^\lambda_{*}$ is identical to $t^{(i+p-2,i+p-2)}_{x-}$).
For example if $\lambda=(12,9)$, $p=5$ and $i=4$ then
\beq
\setyoungsize{15pt}{12pt}
t^{(12,9)}_{*}=
\youngd{\multispan{25}\hrulefill\cr
&1&\omit&3&\omit&5&&7&\omit&8&\omit&9&\omit&10&&15&\omit
&17&\omit&19&\omit&20&\omit&21&\cr
\multispan{25}\hrulefill\cr
&2&\omit&4&\omit&6&&\bf11&\omit&\bf12&\omit&\bf13&\omit
&\bf14&&16&\omit&18&\cr
\multispan{19}\hrulefill\cr},
\eeq
where the partitioning gives an indication as to how the entries
have been entered.
Now, in analogy with (\ref{symcase1}), let:
\beq
v_0=
{1\over [x]_q!} \sum_{d\in{W}_{2i+p-2,2(i+p-2)}} v_{dt^\lambda_{*}}
=\sum_{{\rm standard\ } z^\lambda} c(z^\lambda) v_{z^\lambda},
\label{symcase2}
\eeq
where again the column and Garnir relations have been used to
obtain the sum over standard tableaux. As before,
$c(t^\lambda_{*})=1$. This $v_0$ also lies in the submodule
and thus setting $v_0=0$ results in an expression for
$v_{t^\lambda_{*}}\in D^\lambda_p$.
To obtain an expression for $v_{t^\lambda}$, it is again valid 
to act on each standard tableau with $w\in S_n$ for which
$wt^\lambda_{*}=t^\lambda$:
\beq
v_{t^\lambda}
=-\sum_{{\rm standard\ } z^\lambda\ne t^\lambda_{*}}
c(z^\lambda) v_{wz^\lambda}.
\eeq
To illustrate this case, let $\lambda=(4,4)$, $p=4$ (so that $k=1$)
and
\beq
t^\lambda_4=
\youngt{1&2&3&6\cr 4&5&7&8\cr},
\eeq
for which $i=1$ and $x=3$, whereupon
\beq
t^\lambda_{*}=
\youngt{1&2&3&7\cr \bf4&\bf5&\bf6&8\cr},
\eeq
and $w=s_6$.  Then using ${W}_{4,6}=\{1,s_4,s_5,s_5s_4,s_4s_5,s_4s_5s_4\}$, 
in accordance with (\ref{symcase2}),
\beq\ba{rl}
[3]_q!v_0=&%
\youngt{1&2&3&7\cr 4&5&6&8\cr}
+\youngt{1&2&3&7\cr 5&4&6&8\cr}
+\youngt{1&2&3&7\cr 4&6&5&8\cr}
+\youngt{1&2&3&7\cr 6&4&5&8\cr}
+\youngt{1&2&3&7\cr 5&6&4&8\cr}
+\youngt{1&2&3&7\cr 6&5&4&8\cr}\\[3mm]
=&
q^4(1+q)^2 \youngt{1&3&5&7\cr 2&4&6&8\cr}
-q^3(1+q)\left(
\youngt{1&3&4&7\cr 2&5&6&8\cr}
+\youngt{1&2&5&7\cr 3&4&6&8\cr}
+\youngt{1&2&4&7\cr 3&5&6&8\cr}
\right)\\
&
+(1+q)(1+q+q^2)
\youngt{1&2&3&7\cr 4&5&6&8\cr},\\
\ea\eeq
for which the $[3]_q!$ factor on the right is made manifest on
using $[4]_q=0$ in the form $q^3=-1-q-q^2$. This results in:
\beq
v_0=
-q(1+q)
\youngt{1&3&5&7\cr 2&4&6&8\cr}
+\youngt{1&3&4&7\cr 2&5&6&8\cr}
+\youngt{1&2&5&7\cr 3&4&6&8\cr}
+\youngt{1&2&4&7\cr 3&5&6&8\cr}
+\youngt{1&2&3&7\cr 4&5&6&8\cr},
\eeq
whence, on setting $v_0=0$, $q^2=-1$, and acting on each tableau
with $w=s_6$, we get the requisite expression for $v_{t^\lambda_4}$:
\beq
\youngt{1&2&3&6\cr 4&5&7&8\cr}=
(q-1)
\youngt{1&3&5&6\cr 2&4&7&8\cr}
-\youngt{1&3&4&6\cr 2&5&7&8\cr}
-\youngt{1&2&5&6\cr 3&4&7&8\cr}
-\youngt{1&2&4&6\cr 3&5&7&8\cr}.
\eeq
Note the similarity to (\ref{resultcase1}). This is because essentially
the same symmetrisation process has taken place.

\setyoungsize{14pt}{12pt}
{\bf Case 3. $i\le\lambda_2-p$ and $k>1$.} The final case may
also be viewed as the symmetrisation process moved leftward.
However, the situation here is not so straightforward in
that the entries to the right of the symmetrised section are not
constant across the analogue of (\ref{symcase1}) and (\ref{symcase2}).
For the first part of the following algorithm, these entries are ignored.
Again let $x=p-1$ and consider just $t^\nu_{x-}$ where
$\nu=(kp+i-2,p+i-2)$. This tableau is used exactly as in
(\ref{symcase1}) to give:
\beq
v^{\rm ig}_0=
{1\over [x]_q!}
\sum_{d\in{\cal D}^{2\nu_2+1}_{\nu_2+i,\nu_2+i+kp-2}}
v_{dt^\nu_{x-}}
=\sum_{{\rm standard\ } z^\nu} c(z^\nu) v_{z^\nu},
\label{symcase3}
\eeq
the Garnir and column relations having been used as for tableaux
of shape $\nu$.
Now for each standard tableau $z^\nu$ in the sum here, 
form the tableau $z^\lambda_{\rm aug}$ by first appending the
entries $\{2i+(k+1)p-3,\ldots,2\lambda_2+(k-1)p\}$ one at
a time, alternately onto the bottom row and onto the top row.
The remaining entries $\{2\lambda_2+(k-1)p+1,\ldots,n\}$
(if any), are used to complete the top row.
For example, if $\lambda=(17,9)$, $p=5$, (so that $k=2$)
and $i=3$, then each $z^\lambda_{\rm aug}$ will be a standard
tableau of the form:
\beq
z^{(17,9)}_{\rm aug}=
\youngd{\multispan{35}\hrulefill\cr
&1&\omit&3&
&\bullet&\omit&\bullet&\omit&\bullet&\omit&\bullet&\omit
&\bullet&\omit&\bullet&\omit&\bullet&\omit&\bullet&\omit
&\bullet&&19&\omit&21&\omit&23&&24&\omit&25&\omit&26&\cr
\multispan{35}\hrulefill\cr
&2&\omit&4&
&\bullet&\omit&\bullet&\omit&\bullet&\omit&\bullet
&&18&\omit&20&\omit&22&\cr
\multispan{19}\hrulefill\cr},
\eeq
where each $\bullet$ represents an entry from $\{5,6,\ldots,17\}$.
Now from $v^{\rm ig}_0$, form the sum:
\beq
v^{\rm aug}_0=
{1\over [x]_q!} 
\prod_{i\in{\cal E}_{2i+(k+1)p-3}^{2\lambda_2+(k-1)p-1}} (h_i-q)
\sum_{{\rm standard\ } z^\nu} c(z^\nu) v_{z^\lambda_{\rm aug}},
\label{symcase4}
\eeq
where ${\cal E}_{a}^{b}=\{a,a+2,a+4,\ldots,b\}$.
In fact, the action of each $h_i$ here may be accomplished
(as may be seen from (\ref{action})) directly through the
action of $s_i$ on the tableau:
$h_i v_{z^\lambda_{\rm aug}}=v_{s_iz^\lambda_{\rm aug}}$.
In this sum, the term $v_{t^\lambda_*}$ has coefficient 1
where $t^\lambda_*$ is defined as $t^\nu_{x+}$
augmented with the entries 
$\{2i+(k+1)p-3,\ldots,2\lambda_2+(k-1)p\}$ placed one at a time,
alternately onto the {\it top} row and onto the {\it bottom} row.
Again, the remaining entries $\{2\lambda_2+(k-1)p+1,\ldots,n\}$
(if any), are used to complete the top row. For the example
$\lambda=(17,9)$, $p=5$, $i=3$ considered above:
\beq
t^{(17,9)}_{*}=
\youngd{\multispan{35}\hrulefill\cr
&1&\omit&3&&5&\omit&6&\omit&7&\omit&8&\omit&9&\omit&10&\omit
&11&\omit&12&\omit&13&&18&\omit&20&\omit&22&&24&\omit&25&\omit&26&\cr
\multispan{35}\hrulefill\cr
&2&\omit&4&&14&\omit&15&\omit&16&\omit&17&&19&\omit&21&\omit&23&\cr
\multispan{19}\hrulefill\cr}.
\eeq
Now let $w$ be such that $t^\lambda=wt^\lambda_*$.
Unfortunately, in this case, the action of $w$ on $v^{\rm aug}_0$,
given by (\ref{symcase4}), must take place
through the Hecke algebra action of $h(w)$.
Nonetheless, in the standardised expression for $h(w)v^{\rm aug}_0$,
the term $v_{t^\lambda}$ appears with a coefficient of 1,
whereupon, on setting $h(w)v^{\rm aug}_0$ to zero
(it also lies in the submodule),
an expression for $v_{t^\lambda}\in D^\lambda_p$ is obtained.

Due to the large number of terms involved in instances of this case,
working a full example is impractical.
So we will outline the example
$\lambda=(9,5)$, $p=4$ (so that $k=2$) and
\beq
t^\lambda_5=
\youngt{1&2&3&4&5&6&7&10&13\cr 8&9&11&12&14\cr},
\eeq
which is not 4-root standard with $i=1$.
So in this case we consider $\nu=(7,3)$ and
\beq
t^\nu_{x-}=
\youngt{1&2&3&\bf7&\bf8&\bf9&\bf10\cr \bf4&\bf5&\bf6\cr}.
\eeq
Then (\ref{symcase3}) produces a sum over all 75 standard tableaux of
shape $\nu$. Augmenting each of these with the entries
$\{11,12,13,14\}$ as indicated, results in a sum over 75 tableaux
each of the form:
\beq
\youngt{
\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&\bullet&12&14\cr
\bullet&\bullet&\bullet&11&13\cr}.
\eeq
The action of $(h_{11}-q)(h_{13}-q)$ then results in a sum over
300 terms. Amongst them appears
\beq
t^\lambda_*=
\youngt{1&2&3&4&5&6&7&11&13\cr 8&9&10&12&14\cr}
\eeq
with a coefficient 1.
Acting on this sum with $h(w)$ where $w=s_{10}$, and setting the
result to zero, then yields the requisite expression for
$v_{t^\lambda_5}$ in terms of 299 other tableaux.

\medskip
In each of the three cases considered above, the resulting
expression for $v_{t^\lambda}$ may include terms that are
not $p$-root standard (indeed this was the situation in the previous
example, as it was also in (\ref{ex2})).
If an expression solely in terms of $p$-root standard tableaux
is required then the above relations together with the column and
Garnir relations would have to recursively applied in order to 
obtain such an expression. This is guaranteed eventually since, 
as in Section 3, the ordering on the tableaux shows that an improvement
takes place with each invocation.

\vfill\eject

\begin{center}
{\bf 7\ \ Final example}
\end{center}
\setyoungsize{10pt}{12pt}

For $\lambda=(4,2)$, the Specht module $S^\lambda$ is 9-dimensional.
When $p=4$, the corresponding irreducible module $D^\lambda_4$ is
4-dimensional, as was given in (\ref{exdims}). The methods of
Section 2 enable the action of each $h_i$ on each of the four vectors 
indexed by the 4-root standard tableaux
\beq
\youngt{1&3&5&6\cr 2&4\cr},
\qquad
\youngt{1&2&5&6\cr 3&4\cr},
\qquad
\youngt{1&3&4&6\cr 2&5\cr},
\qquad
\youngt{1&2&4&6\cr 3&5\cr},
\label{exbasis}
\eeq
to be written in terms of standard tableaux. Those terms which are then
not 4-root standard may be expressed in terms of such using
(\ref{ex1}), (\ref{resultcase1}), and the immediate result of the
action of $s_2$ on the tableaux of (\ref{ex1}).

Consider the action of $h_3$ on the terms indexed by the tableaux
(\ref{exbasis}) above:
$$
\ba{l}
{ h_3} \youngt{1&3&5&6\cr 2&4\cr}
= \youngt{1&4&5&6\cr 2&3\cr}
= -\youngt{1&3&5&6\cr 2&4\cr},\\[3mm]
{ h_3} \youngt{1&2&5&6\cr 3&4\cr}
= \youngt{1&2&5&6\cr 4&3\cr}
= q\youngt{1&2&5&6\cr 3&4\cr}-q^2\youngt{1&3&5&6\cr 2&4\cr},\\[3mm]
{ h_3} \youngt{1&3&4&6\cr 2&5\cr}
= \youngt{1&4&3&6\cr 2&5\cr}
= q\youngt{1&3&4&6\cr 2&5\cr}-q^2\youngt{1&3&5&6\cr 2&4\cr},\\[3mm]
{ h_3} \youngt{1&2&4&6\cr 3&5\cr}
= \youngt{1&2&3&6\cr 4&5\cr}
= (q-1) \youngt{1&3&5&6\cr 2&4\cr}-\youngt{1&3&4&6\cr 2&5\cr}
 -\youngt{1&2&5&6\cr 3&4\cr}-\youngt{1&2&4&6\cr 3&5\cr},
\ea
$$
where the methods of Section 2 were sufficient in all but the last
calculation, where (\ref{resultcase1}) was used.
Thus $D^\lambda_4$ gives rise to the following representation matrix
for $h_3$:
\beq
\pmatrix{
-1&1&1&q-1\cr
.&q&.&-1\cr
.&.&q&-1\cr
.&.&.&-1\cr},
\eeq
where $q^2=-1$ has been used. The representation matrices for the other
generators of $H_6(q)$ may be calculated in a similar manner.

It is interesting to note that for $\lambda=(4,2)$ and $p=2$,
the calculation proceeds in an almost identical manner.
However, an extra basis vector is present since, by (\ref{exdims}),
$D^\lambda_2$ is 5-dimensional. This vector is indexed by the tableau
on the left of (\ref{resultcase1}). That (\ref{resultcase1}) cannot
be used in this $p=2$ case may be traced to the appearance of the
$[2]_q$ factor in its derivation.


\vfill\eject

\begin{center}
{\bf 8\ \ Footnote}
\end{center}

The algorithms which have been described in this paper have been
implemented in the computer algebra package SYMMETRICA.
In the generic and the two-rowed non-generic cases, they
enable representation matrices to be readily obtained for
any element of $H_n(q)$. Further routines can check that the
matrices so produced indeed provide a representation in that
they respect (\ref{hoopy}). These have been used to check the
algorithms presented in this paper in a large number of cases.
In addition, in either the generic or two-rowed non-generic case,
routines are available to generate and
enumerate the appropriate standard tableaux, and to render
an arbitrary tableau in terms of the appropriate standard
tableaux. Calculations in $H_n(q)$ itself may also be undertaken. 

SYMMETRICA has been developed at Bayreuth University, and
is an extensive package of routines concerned with
the symmetric and related groups, together with their
representations and combinatorics. For more information
about SYMMETRICA, access the WorldWideWeb page
{\it http://btm2xd.mat.uni-bayreuth.de} or email
{\it sym@btm2x2.mat.uni-bayreuth.de}.
The full SYMMETRICA package, with documentation,
is available from the above WorldWideWeb site or via FTP from
{\it ftp://btm2x7.mat.uni-bayreuth.de/dist/SYM.tar.Z}.

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% Frobenius formula for the characters of Hecke algebras of
% type $A_{n-1} at roots of unity;
% An algorithm for calculating characters of Hecke algebras
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\end{document}



