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\title{\bf A Plethysm formula on the characteristic\\ map
of induced linear characters\\ from $U_n(\mathbb F_q)$ to
$GL_n(\mathbb F_q)$}

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\author{Zhi Chen\\
\small Department of Mathematics and Statistics\\[-0.8ex]
\small  York University\\[-0.8ex]
\small To\-ron\-to, ON, Canada\\
\small\tt czhi@mathstat.yorku.ca}

% \date{\dateline{submission date}{acceptance date}\\
% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{May 19, 2012}{Aug 2, 2013}{Aug 9, 2013} \\
\small Mathematics Subject Classifications: 05E10, 20C33}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
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% principal new results that are to be found in the paper. The abstract
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\begin{abstract} This paper gives a plethysm formula on the
characteristic map of the induced linear characters from the
unipotent upper-triangular matrices $U_n(\mathbb F_q)$ to
$GL_n(\mathbb F_q)$, the general linear group over finite field
$\mathbb F_q$. The result turns out to be a multiple of a twisted
version of the Hall-Littlewood symmetric functions
$\tilde{P}_n[Y;q]$. A recurrence relation is also given which makes
it easy to carry out the computation.


% keywords are optional
\bigskip\noindent \textbf{Keywords:} representation theory; induced characters;
symmetric functions; supercharacter theory
\end{abstract}

\section{Introduction}

Let $\mathbb F_q$ be a fixed finite field and $GL_n(\mathbb F_q)$
the finite general linear group over $\mathbb F_q$. The
representation theory of $GL_n(\mathbb F_q)$ over $\mathbb{C}$ was
presented by J.A.Green~\cite{Green}. He also constructed the
characteristic map which builds a connection between the character
spaces of $GL_n(\mathbb F_q)$ for $n\geq 0$ and the Cartesian
product over infinitely indexed sets of rings of symmetric
functions. In character theory, the study of induced linear
characters from subgroups is very useful in order to understand the
character ring of the larger group.

In this paper, we consider certain induced linear characters from
the group of unipotent upper-triangular matrices $U_n(\mathbb F_q)$
to $GL_n(\mathbb F_q)$. The representations of these induced linear
characters are known as Gelfand-Graev modules, which play an
important role in the representation theory of finite groups of Lie
type~(\cite{Gelgra}, \cite{Stein}). The formula for the
characteristic map of the induced linear characters is given by
Thiem~\cite{Nat}. We then apply a homomorphism $\rho$ (see
Definition~\ref{def:hom}) on the image of the characteristic map. We
refer to the result as a plethysm formula because it involves a
composition of symmetric functions using the plethysm operation.
There are two advantages in doing so: to get a simpler formula and
to express the result as a multiple of a twisted version of the
Hall-Littlewood symmetric functions $\tilde{P}_n[Y;q]$. We hope this
method could contribute to the study of the irreducible
decomposition of the induced characters from $U_n(\mathbb F_q)$ to
$GL_n(\mathbb F_q)$.

In section $2$ we give some background knowledge on symmetric
functions and representation theory of $GL_n(\mathbb F_q)$ and
$U_n(\mathbb F_q)$. Since the character theory of $U_n(\mathbb F_q)$
is known as a wild problem, supercharacter theory is built up as an
approximation of the ordinary character theory. The linear
characters of $U_n(\mathbb F_q)$ that we are considering are part of
the category of supercharcters of $U_n(\mathbb F_q)$.  We introduce
further questions about the induction of all supercharacters in
Section $4$. In Section $3$ we give our main result about the
plethysm formula. A natural recurrence relation is obtained so that
we can carry out the computation of the homomorphism $\rho$ on the
characteristic map of the induced linear characters more easily. We
also give a relation between the characteristic map of the induced
characters from $U_n(\mathbb F_q)$ to $GL_n(\mathbb F_q)$, and the
homomorphism $\rho$ on those characteristics. This is depicted in
the following diagram \xyoption{all}
\begin{displaymath}
\xymatrix{
  \otimes_{\varphi\in\Theta} \Lambda_{\mathbb C}(Y^{\varphi}) \ar[d]_{\rho} \ar @{=}^{Id} [r]
                & \otimes_{f\in\Phi} \Lambda_{\mathbb C}(X_f) \ar[d]^{\Pi |_{\Lambda_{\mathbb C}(X_{f=x-1})}}  \\
 \Lambda_{\mathbb C}(Y)           \ar[r]_{\tau\circ\omega}
                & \Lambda_{\mathbb C}(X_{x-1})}
\end{displaymath}
where the notation is explained in Theorem~\ref{thm:comm}. From the
above commutative diagram we show that our simplified plethysm
formula does not lose any information on the characteristic map of
the induced characters from $U_n(\mathbb F_q)$ to $GL_n(\mathbb
F_q)$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Background}
\subsection{Symmetric functions}
The notation in this paper follows closely the book of
Macdonald~\cite{Mac} and Thiem~\cite{Nat2}.
\begin{definition}
A {\sl partition} $\lambda$ of $n\in  \textbf{N}$, is a sequence
$\lambda=(\lambda_1, \lambda_2, \ldots, \lambda_l)$ of positive
integers in weakly decreasing order:
$\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_l$, such that
$\lambda_1+\lambda_2+\cdots+\lambda_l=n$. We denote this by
$\lambda\vdash n$. Here, each $\lambda_i\ (1\leq i\leq l)$ is called
a {\sl part} of $\lambda$. We say the {\sl length} of the partition
$\lambda$ is $l=l(\lambda)$, which is the number of parts of
$\lambda$. We use $|\lambda|$ to denote the sum of all parts
$\lambda_1+\lambda_2+\cdots+\lambda_l$, and we call $|\lambda|$ the
{\sl size} of the partition. Sometimes we also use the notation:
$$\lambda = (1^{m_1},2^{m_2},\ldots,n^{m_n},\ldots),$$
where each $m_i$ means there are $m_i$ parts in $\lambda$ equal to
$i$.
\end{definition}
Let $\Lambda_{\mathbb C}(Y)$ denote the ring of symmetric functions
with complex coefficients in the variables $Y=\{y_1,y_2,\ldots\}$.
We denote the complete symmetric functions, elementary symmetric
functions, monomial symmetric functions, power-sum symmetric
functions, and Schur symmetric functions by $h_{\lambda}[Y]$,
$e_{\lambda}[Y]$, $m_{\lambda}[Y]$, $p_{\lambda}[Y]$, and
$s_{\lambda}[Y]$ respectively.

Following other references (e.g. Garsia and Tesler~\cite{GarTes}),
we use the notation $\Omega$ for the basic symmetric function
kernel. Then the generating function for $h_n[Y]$ is
$$\Omega[tY] = \sum_{n\geq 0} h_n[Y] t^n = \prod_{j\geq 1} (1-y_j t)^{-1}.$$
It is also well known that
$$\Omega[X]\Omega[Y]=\Omega[X+Y], \rm{and}\ \Omega[X]/\Omega[Y]=\Omega[X-Y].$$

Let $X=\{x_1,x_2,\ldots\}$ be another set of finite or infinite
variables. We have the following identity:
\begin{equation}\label{eq:omg}
\Omega[XY]:=\prod_{i,j}(1-x_iy_j)^{-1} =
\sum_{\lambda}m_{\lambda}[X]h_{\lambda}[Y]
\end{equation}
summed over all partitions $\lambda$.

There is a scalar product defined on $\Lambda_{\mathbb C}(Y)$, which
makes $(m_{\lambda})$ and $(h_{\lambda})$ dual to each other:
$$\langle h_{\lambda}, m_{\mu} \rangle = \delta_{\lambda\mu}$$
for all partitions $\lambda, \mu$, where $\delta_{\lambda\mu}$ is
the Kronecker delta.

We use $P_{\lambda}[Y;t]$ to denote the Hall-Littlewood symmetric
functions, as defined in \cite{Mac} (see page 209). If we define
\begin{align*}
\bm{q}_r &= \bm{q}_r[Y;t] = (1-t)P_{(r)}[Y; t]\ \ \text{for}\ r\geq 1 ,\\
\bm{q}_0 &= \bm{q}_0[Y;t] = 1 ,
\end{align*}
then the generating function for $\bm{q}_r[Y;t]$ is
\begin{equation}\label{eq:genqr}
Q(u) = \sum_{r\geq 0} \bm{q}_r[Y; t] u^r = \prod_i\frac{1-y_i t
u}{1-y_i u}=\Omega[(1-t)uY]\, \lower 6pt\hbox{.}
\end{equation}
For each partition $\lambda$, let $n(\lambda) = \sum_{i\geq 1}
(i-1)\lambda_i $. Define
\begin{equation*}
\tilde{P}_{\lambda}[Y;q] = q^{-n(\lambda)} P_{\lambda}[Y;q^{-1}],
\end{equation*}
and we call $\tilde{P}_{\lambda}[Y;q]$ the twisted Hall-Littlewood
symmetric functions.

From \cite{Mac}, it is well known that the {\sl plethysm} can be
defined by
\begin{equation}\label{eq:defplethysm}
h_a[ p_b [Y]] = h_a[y_1^b,y_2^b,\ldots],
\end{equation}
which is the coefficient of $t^{ab}$ in $\prod_{j\geq 1} (1-y_j^b
t^b)^{-1}.$

\subsection{ Representation theory of $GL_n(\mathbb F_q)$}
The representation theory of the finite general linear group
$G_n=GL_n(\mathbb F_q)$ over $\mathbb{C}$ can be found in
J.A.Green~\cite{Green}, Macdonald~\cite{Mac} and Thiem~\cite{Nat}.
Here we give a short description of the characteristic map
constructed by J.A.Green.

Let $\bar{\mathbb F}_q$ denote the algebraic closure of the finite
field $\mathbb F_q$. The multiplicative group of $\bar{\mathbb F}_q$
is denoted by $\bar{\mathbb F}_q^{\times}$. The Frobenius
automorphism of $\bar{\mathbb F}_q$ over $\mathbb F_q$ is given by
$$F: x \rightarrow x^q, \text{where}\ x\in \bar{\mathbb F}_q.$$
For each $n\geq 1$, we use $\bar{\mathbb F}_{q,n}^{\times}$ to
denote the fixed points of $F^n$ in $\bar{\mathbb F}_q^{\times}$.

Let $\bar{\mathbb F}_q^{\ast}=\{\phi: \bar{\mathbb F}_q^{\times}
\rightarrow \mathbb C^{\times}\}$ be the group of complex-valued
multiplicative characters of $\bar{\mathbb F}_q^{\times}$. The
Frobenius automorphism on $\bar{\mathbb F}_q^{\ast}$ is
$$F: \xi \rightarrow \xi^q, \text{where}\ \xi\in \bar{\mathbb F}_q^{\ast}.$$
For each $n\geq 1$, let $\bar{\mathbb F}_{q,n}^{\ast}$ be the group
of elements fixed by $F^n$. We also define a pairing of
$\bar{\mathbb F}_q^{\ast}$ with $\bar{\mathbb F}_q^{\times}$ by
$$\langle\xi,x\rangle_n=\xi(x)$$
for $\xi\in\bar{\mathbb F}_{q,n}^{\ast}$ and $x\in\bar{\mathbb
F}_{q,n}^{\times}$.

We then define
$$\Phi=\{F\text{-orbits\ of}\ \bar{\mathbb F}_q^{\times}\}\ \text{and}\
\Theta=\{F\text{-orbits\ of}\ \bar{\mathbb F}_q^{\ast}\}.$$ Since
each $F\text{-orbit\ of}\ \bar{\mathbb F}_q^{\times}$ is in
one-to-one correspondence with an irreducible polynomial over
$\mathbb F_q$, we can also use $f$ to denote each $F\text{-orbit}$
in $\Phi$. A partition-valued function $\bm{\mu}$ on $\Phi$ is a
function which maps each $f\in\Phi$ to a partition $\bm{\mu}(f)$.
The size of $\bm{\mu}$ is
$$\|\bm{\mu}\|=\sum_{f\in\Phi} d(f)|\bm{\mu}(f)|,$$
where $d(f)$ is equal to the degree of $f\in\Phi$.

Let $\mathbb P$ denote the set of all partitions and
\begin{equation*}
\mathcal{P}^{\Phi}=\bigcup_{n\geq 0}\mathcal{P}_n^{\Phi},\
\text{where}\ \mathcal{P}_n^{\Phi}=\{\bm{\mu}: \Phi \rightarrow
\mathbb{P}\,;\ \|\mbox{\boldmath{$\mu$}}\|=n \}.
\end{equation*}
We use $K^{\bm{\mu}}$ to denote the conjugacy classes in $G_n$
parameterized by $\bm{\mu}\in\mathcal{P}_n^{\Phi}$~\cite{Mac}. The
characteristic function of the conjugacy class $K^{\bm{\mu}}$ is
denoted by $\pi_{\bm{\mu}}$.

Similarly, for each partition-valued function
$\mbox{\boldmath{$\lambda$}}: \Theta \rightarrow \mathbb{P}$, the
size of $\bm{\lambda}$ is
$$\|\bm{\lambda}\|=\sum_{\varphi\in\Theta} d(\varphi)|\bm{\lambda}(\varphi)|,$$
where $d(\varphi)$ is equal to the number of elements in $\varphi$.
Let
\begin{equation*}
\mathcal{P}^{\Theta}=\bigcup_{n\geq 0}\mathcal{P}_n^{\Theta},\
\text{where}\ \mathcal{P}_n^{\Theta}=\{\bm{\lambda}: \Theta
\rightarrow \mathbb{P}\,;\ \|\bm{\lambda}\|=n \}.
\end{equation*}
We use $G_n^{\bm{\lambda}}$ to denote the irreducible $G_n$-modules
indexed by $\bm{\lambda}\in\mathcal{P}_n^{\Theta}$ \cite{Mac}. The
character of the irreducible $G_n$-module $G_n^{\bm{\lambda}}$ is
denoted by $\chi^{\bm{\lambda}}$.

For every $f\in\Phi$, let $X_f:=\{X_{1,f}, X_{2,f},\ldots\}$ be a
set of infinitely many variables. Each $X_{i,f}$ has degree $d(f)$.

Let
\begin{equation*}
\tilde P_{\eta}[f] =  \tilde{P}_{\eta}[X_{f}; q^{d(f)}] =
q^{-d(f)n(\eta)} P_{\eta}[X_{f}; q^{-d(f)}],
\end{equation*}
where $\tilde{P}_{\eta}[X_{f}; q^{d(f)}]$ is the twisted
Hall-Littlewood symmetric function. Define
\begin{equation*}
\tilde P_{\bm{\mu}} = \prod_{f\in\Phi} \tilde P_{\bm{\mu}(f)}[f].
\end{equation*}

For every $\varphi\in\Theta$, let $Y^{\varphi}:=\{Y_{1}^{\varphi},
Y_{2}^{\varphi},\ldots\}$ be a set of infinitely many variables.
Each $Y_{i}^{\varphi}$ has degree $d(\varphi)$. Define
\begin{equation*}
S_{\bm{\lambda}} =
\prod_{\varphi\in\Theta}s_{\bm{\lambda}(\varphi)}[Y^{\varphi}],
\end{equation*}
where $s_{\bm{\lambda}(\varphi)}[Y^{\varphi}]$ is the Schur
symmetric function.

We define the transformation between the symmetric functions in the
variables $\{X_{f}: f\in \Phi\}$ and those in the variables
$\{Y^{\varphi}: \varphi\in\Theta\}$ by the following identity:
\begin{equation}\label{eq:transform}
p_k[Y^{\varphi}]=(-1)^{n-1}\sum_{x\in \bar{\mathbb
F}_{q,n}^{\times}}\xi (x)p_{n/d(f_x)}[X_{f_x}]\,,
\end{equation}
where $n= k\times d(\varphi)$, $\xi$ is any element in $\varphi$,
and $f_x$ is the irreducible polynomial that contains $x$ as a root.
Since $f_x=f_y$ if $x, y$ are in the same $F\text{-orbit}$ in
$\bar{\mathbb F}_{q,n}^{\times}$, equation~(\ref{eq:transform}) is
well defined for any $\xi\in\varphi$. Also here $p_k[Y^{\varphi}]$
and $p_{n/d(f_x)}[X_{f_x}]$ are power sums in different sets of
variables. Equation~(\ref{eq:transform}) is an isomorphism and its
inverse can be found in~\cite{Mac}. For interested readers willing
to know more details, please refer to~\cite{Mac} and ~\cite{NatVin}.

Since the power-sum symmetric functions form a basis of the ring of
symmetric functions, with the basis transformation given by
equation~(\ref{eq:transform}), we let
\begin{equation*}
\Lambda_{\mathbb C} = \otimes_{f\in\Phi} \Lambda_{\mathbb C}(X_f)=
\otimes_{\varphi\in\Theta} \Lambda_{\mathbb C}(Y^{\varphi}),
\end{equation*}
where $\Lambda_{\mathbb C}(X_f)$ is the ring of symmetric functions
in $X_{f}$, and $\Lambda_{\mathbb C}(Y^{\varphi})$ is the ring of
symmetric functions in $Y^{\varphi}$. As a graded ring, we have
\begin{align*}
\Lambda_{\mathbb C} &={\mathbb C}\text{-span}\{\tilde
P_{\bm{\mu}} | \bm{\mu}\in\mathcal{P}^{\Phi} \} \\
&={\mathbb C}\text{-span}\{S_{\bm{\lambda}} |
\mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \}.
\end{align*}

From \cite{Mac} we know that the conjugacy classes $K^{\bm{\mu}}$ of
$G_n$ are parameterized by $\bm{\mu}\in\mathcal{P}_n^{\Phi}$, and
the irreducible characters $\chi^{\bm{\lambda}}$ of $G_n$ are
indexed by $\bm{\lambda}\in\mathcal{P}_n^{\Theta}$. The following
theorem gives the characteristic map of $G_n$.

\begin{theorem}{\rm(Green~\cite{Green}, Macdonald~\cite{Mac},
Zelevinski~\cite{Zele})} Let $A_n$ denote the space of
complex-valued class functions on $G_n$ and $A = \oplus_{n\geq 0}
A_n$. The linear map
\begin{align*}
ch : A &\longrightarrow \Lambda_{\mathbb C}\\
    \chi^{\bm{\lambda}} &\mapsto S_{\bm{\lambda}}, \ \ \ \text{for}\ \bm{\lambda} \in
\mathcal{P}^{\Theta},\\
\pi_{\bm{\mu}} &\mapsto \tilde{P}_{\bm{\mu}}, \ \ \ \text{for}\
\bm{\mu}\in \mathcal{P}^{\Phi},
\end{align*}
is a Hopf algebra isomorphism.
\end{theorem}


\subsection{ Supercharacter theory}
Let $U_n$ be the group of unipotent upper-triangular matrices with
entries in the finite field $\mathbb F_q$, and ones on the diagonal.
This group is the subgroup of the finite general linear group $G_n$.
Although the character theory on $U_n$ is a wild problem, people
came up with a slightly coarser version called superclass and
supercharacter theory (Andr\'e~\cite{An95}, Yan~\cite{Yan}). This
approximation of the character theory of $U_n$ is relatively easier
to study and compute. Superclasses are certain unions of conjugacy
classes and supercharacters are sums of irreducible characters. They
are compatible in the sense that supercharacters are constant on
superclasses. The supercharacter theory has a rich combinatorial
structure (ref.~\cite{Nat2}) and connects to some other algebraic
structures as well (ref.~\cite{MulAu}).

The superclasses of $U_n$ can be indexed by the $\mathbb
F_q^{\times}$-labeled set partitions, and a supercharacter becomes
an irreducible character if the corresponding indexed $\mathbb
F_q^{\times}$-labeled set partition has no crossing arcs. For the
strict definitions and more details on supercharacters please see
~\cite{Nat2} or ~\cite{MulAu}.

In this paper we consider the linear supercharacters of $U_n$
indexed by
\begin{align*}
\begin{tikzpicture}
\filldraw [black] (0,0) circle (1.5pt) (.5,0) circle (1.5pt) (1,0)
circle (1.5pt) (1.7,0) circle (0.5pt) (2,0) circle (0.5pt) (2.3,0)
circle (0.5pt) (3,0) circle (1.5pt); \node at (0, -.2) {$\scscs 1$};
\node at (0.25, .35) {$\scs q_1$};\node at (0.5, -.2) {$\scscs 2$};
\node at (0.75, .35) {$\scs q_2$}; \node at (1, -.2) {$\scscs 3$};
\node at (1.25, .35) {$\scs q_3$}; \node at (3, -.2) {$\scscs n$};
\node at (2.75, .35) {$\scs q_{n-1}$}; \draw (0,0) .. controls
(.25,.25) .. (.5,0); \draw (.5,0) .. controls (.75,.25) .. (1,0);
\draw (1,0) .. controls (1.25,.25) .. (1.5,0); \draw (2.5,0) ..
controls (2.75,.25) .. (3,0);
\end{tikzpicture}
\end{align*}
where $q_1,\ldots,q_{n-1}\in\mathbb F_q^{\times}$ (see
Thiem~\cite{Nat2} for this notation). Let
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}$ denote the above character. We
induce $\chi^{(n)}_{(q_1,\ldots,q_{n-1})}$ from $U_n$ to $G_n$ by
the formula
\begin{equation}\label{eq:inform}
\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}(g) =
\frac{1}{|U_n|}\sum_{h\in G_n}
\bar{\chi}^{(n)}_{(q_1,\ldots,q_{n-1})} (h g h^{-1}),
\end{equation}
where $\bar{\chi}(s)=\chi(s)$ if $s\in U_n$, and $\bar{\chi}(s)=0$
if $s\not\in U_n$.

The induced character
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$ is a
character of $G_n$, which is known as the character of the
Gelfand-Greav module (\cite{Gelgra}, \cite{Nat}). Applying the
homomorphism $\rho$ defined in Section~\ref{sec:Plethysm} to the
characteristic map of
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$, we get a
multiple of the twisted Hall-Littlewood symmetric function
$\tilde{P}_n$. An explicit formula for this result together with a
recurrence relation is also given in Section~\ref{sec:Plethysm}.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Plethysm Formula for the induced character}\label{sec:Plethysm}

We start from the formula of the characteristic map of
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$, which is
given by Thiem~\cite{Nat}.

\begin{theorem} {\rm (Thiem~\cite{Nat})}
\begin{equation}\label{eq:chamap}
ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}) =
\sum_{\bm{\lambda} \in \mathcal{P}_n^{\Theta}\atop
ht(\bm{\lambda})=1} S_{\bm{\lambda}},
\end{equation}
where $ht(\bm{\lambda})=max\{l(\bm{\lambda} (\varphi)) | \varphi \in
\Theta\}$.
\end{theorem}

Notice that $ht(\bm{\lambda})=1$ implies for every
$\varphi\in\Theta$ we have $l(\bm{\lambda} (\varphi))\leq 1$, which
means $\bm{\lambda} (\varphi)$ contains at most one part. Since for
each $a\in\mathbb{N}$, $s_a[Y^{\varphi}]=h_a[Y^{\varphi}]$. From the
definition of $S_{\bm{\lambda}}$, we can write (\ref{eq:chamap}) as
\begin{align}\label{eq:char}
ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}) & =
\sum_{\bm{\lambda} \in \mathcal{P}_n^{\Theta}\atop
ht(\bm{\lambda})=1} \prod_{\varphi\in\Theta}
s_{\bm{\lambda}(\varphi)}[Y^{\varphi}]= \sum_{\bm{\lambda} \in
\mathcal{P}_n^{\Theta}\atop ht(\bm{\lambda})=1}
\prod_{\varphi\in\Theta}
h_{\bm{\lambda}(\varphi)}[Y^{\varphi}]\nonumber\\
& =\sum_{a_1b_1+\cdots+a_kb_k=n\atop
\bm{\lambda}(\Theta)=\{a_1,\ldots,a_k\}\in
\mathcal{P}_n^{\Theta}}\sum_{deg(\varphi_i)=b_i\atop
\varphi_1,\ldots,\varphi_k \text{ distinct}}
h_{a_1}[Y^{\varphi_1}]h_{a_2}[Y^{\varphi_2}]\cdots
h_{a_k}[Y^{\varphi_k}] ,
\end{align}
where $Y^{\varphi_1}, Y^{\varphi_2},\ldots, Y^{\varphi_k}$ are
different sets of variables. For $i$ from $1$ to $k$, each variable
in the set $Y^{\varphi_i}=\{Y^{\varphi_i}_1,Y^{\varphi_i}_2,\ldots
\}$ has degree $b_i$.

We give an example to better understand formulae~(\ref{eq:chamap})
and (\ref{eq:char}).
\begin{example}
For $n=3$, we have
\begin{align*}
ch(\chi^{(3)}_{(q_1,q_{2})}\uparrow_{U_3}^{G_3}) &=
\sum_{\varphi_1,\varphi_2,\varphi_3 \text{distinct}\atop
deg(\varphi_i)=1}
h_1[Y^{\varphi_1}]h_1[Y^{\varphi_2}]h_1[Y^{\varphi_3}] \\
& + \sum_{\psi_1,\psi_2 \text{distinct}\atop deg(\psi_i)=1}
h_2[Y^{\psi_1}]h_1[Y^{\psi_2}] + \sum_{deg(\bar{\varphi}_1)=2\atop
deg(\bar{\varphi}_2)=1} h_1[Y^{\bar{\varphi}_1}]h_1[Y^{\bar{\varphi}_2}]\\
& + \sum_{deg(\varphi)=1} h_3[Y^{\varphi}] + \sum_{deg(\psi)=3}
h_1[Y^{\psi}].
\end{align*}
\end{example}
From the above example we see that the expansion on the right-hand
side of~(\ref{eq:char}) becomes more complicated as $n$ increases.
This inspires us to come up with the idea of using a homomorphism,
as we will see later in Definition~\ref{def:hom}, to simplify the
computation.

For each term in equation~(\ref{eq:char}), we have a {\sl two row}
array $\left(\begin{array}{cccc}
b_1 & b_2 & \cdots & b_k \\
a_1 & a_2 & \cdots & a_k
\end{array}\right)$ where $b_i=d(\varphi_i)$ and it satisfies the condition $\sum_{i=1}^{k} a_i b_i =
n$. We arrange the pairs $(b_i,a_i)$ such that:

(1) $b_1\leq b_2\leq\ldots\leq b_k$,

(2) $a_j\leq a_{j+1}$ if $b_j=b_{j+1}$ for $1\leq j<k$.

Once the array is sorted, we can denote it as follows:
\begin{equation*}
\left(\begin{array}{ccccccccccccc}
   &  & 1^{m_1} &  &  &  & 2^{m_2} &  & \cdots  &  &  & n^{m_n} &  \\
  1^{m_{1,1}} & 2^{m_{1,2}} & \cdots & n^{m_{1,n}} & 1^{m_{2,1}} & 2^{m_{2,2}} & \cdots & n^{m_{2,n}} & \cdots & 1^{m_{n,1}} & 2^{m_{n,2}} & \cdots & n^{m_{n,n}}
\end{array}\right)
\end{equation*}
where $\sum_{i,j=1}^{n} (m_{i,j}\times j\times i)=n$ and
$m_{i,1}+m_{i,2}+\ldots+m_{i,n}=m_i$ for $1\leq i \leq n$. Each
$m_i$ counts the number of different sets of variables appearing in
the term with the same degree $i$. Each $m_{i,j}$ counts the number
of complete symmetric functions $h_{j}$ in variables with degree
$i$.

For a given $i$, let $l_q(i)$ denote the number of all different
sets of variables with the same degree $i$. We know that $l_q(i)$ is
equal to the number of irreducible polynomials $f$ over finite field
$\mathbb F_q$ with degree $i$ and satisfying $f(0)\neq 0$. The
number of irreducible polynomials of degree $i$ over $\mathbb F_q$
is given by the formula
$$L_q(i)=\frac{1}{i}\sum_{d|i}\mu(d)q^{\frac{i}{d}},$$
where $\mu$ is the M\"{o}bius function. Then we have
\begin{equation}\label{eq:numirr}
l_q(i)=\left\{
  \begin{array}{ll}
    L_q(1)-1, & \text{for}\ i=1; \\
    L_q(i), & \text{for}\ i\geq 2.
  \end{array}
\right.
\end{equation}
Thus for a given $i$ and a list of numbers $[m_{i,1}, m_{i,2},
\ldots, m_{i,n}]$ where $m_{i,1}+m_{i,2}+\ldots + m_{i,n}=m_i$, the
number of products of the form
\begin{align}\label{eq:prod}
& h_{1}[Y^{\varphi_{i,1}}]h_{1}[Y^{\varphi_{i,2}}]\cdots
h_{1}[Y^{\varphi_{i,m_{i,1}}}] \nonumber\\
& \times
h_{2}[Y^{\varphi_{i,m_{i,1}+1}}]h_{2}[Y^{\varphi_{i,m_{i,1}+2}}]\cdots
h_{2}[Y^{\varphi_{i,m_{i,1}+m_{i,2}}}]\nonumber\\
& \times \cdots \nonumber\\
& \times h_{n}[Y^{\varphi_{i,m_{i,1}+\cdots +
m_{i,n-1}+1}}]h_{n}[Y^{\varphi_{i,m_{i,1}+\cdots +
m_{i,n-1}+2}}]\cdots h_{n}[Y^{\varphi_{i,m_{i}}}]
\end{align}
is equal to
$$\frac{l_q(i)(l_q(i)-1)\cdots(l_q(i)-m_{i}+1)}{m_{i,1}!m_{i,2}!\cdots
m_{i,n}!} , $$ where $Y^{\varphi_{i,1}}, Y^{\varphi_{i,2}}, \ldots,
Y^{\varphi_{i,m_{i}}}$ are $m_i$ different sets of variables with
the same degree $i$. Notice that when $n$ increases, we get more
terms on the right-hand side of equation~(\ref{eq:char}).

In order to simplify the computation, we send each set of variables
$Y^{\varphi_i}$ to $\{y^{b_i}_1, y^{b_i}_2, \ldots\}$. Using
plethysm defined as equation~(\ref{eq:defplethysm}), we give the
following definition.
\begin{definition}\label{def:hom}
Define a {\sl homomorphism} $\rho: {\mathbb
C}\text{-span}\{S_{\bm{\lambda}} |
\mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \} \rightarrow
\Lambda_{\mathbb C}(Y) $ as follows:
$$\rho(h_{a}[Y^{\varphi}]) = h_{a}[p_{b}[Y]], \ \forall \varphi\in\Theta, b=deg(\varphi).$$
\end{definition}
The image of the homomorphism doesn't differentiate the sets of
variables and therefore simplifies the computation. Applying the
homomorphism does lose information on the characteristics of
characters in $G_n$. However, as we will see later on in
Theorem~\ref{thm:comm} and Corollary~\ref{col:equi}, if we consider
the induced characters from $U_n$ to $G_n$, then the homomorphism
carries all the information and this will be discussed later in this
section.

Since $ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})$ is
independent of $q_1,\ldots,q_{n-1}$, and only depends on $n$ for
$n\geq 1$, we can simply denote
$\rho(ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}))$ by
$\rho_n$ and set $\rho_0=1$. We also use $\rho_{[m_{i,1},\ldots,
m_{i,n}]}$ to denote the results of applying $\rho$ on the sum of
all different products in the form of (\ref{eq:prod}) for the same
index list $[m_{i,1},\ldots, m_{i,n}]$, i.e.
$$\rho_{[m_{i,1},\ldots, m_{i,n}]}:=\frac{l_q(i)(l_q(i)-1)\cdots(l_q(i)-m_{i}+1)}{m_{i,1}!m_{i,2}!\cdots
m_{i,n}!}(h_{1}[p_{i}])^{m_{i,1}}\cdots
(h_{n}[p_{i}])^{m_{i,n}}\,\lower 4pt \hbox{.}$$ Applying $\rho$ to
both sides of (\ref{eq:char}) we get
\begin{equation}\label{eq:plechar}
\rho_n = \sum_{\sum_{i,j=1}^{n} m_{i,j}\times j\times i=n}
\rho_{[m_{1,1},\ldots, m_{1,n}]}\cdots \rho_{[m_{n,1},\ldots,
m_{n,n}]}\,\lower 4pt \hbox{.}
\end{equation}

The following theorem follows naturally.
\begin{theorem}\label{thm:lm1}
Let $CH_q(t)$ denote the generating function for $\rho_n$:
$$CH_q(t) = 1+ \rho_1 t+\rho_2 t^2+\cdots = \sum_{n\geq 0} \rho_n t^n . $$
Then we have
$$CH_q(t) = \prod_{i\geq 1}\left(\prod_{j\geq 1}(1-y_j^{i}t^{i})^{-1}\right)^{l_q(i)} = \prod_{i\geq 1}\prod_{j\geq 1}(1-y_j^{i}t^{i})^{-l_q(i)}.$$
\end{theorem}
\begin{proof}
Since for every $i\geq 1$,
\begin{align*}
\prod_{j\geq 1}(1-y_j^{i}t^{i})^{-1} & = \sum_{a\geq 0} h_a[y_1^i,
y_2^i, \ldots]t^{a\cdot i}\\
& = 1+ (h_{1}[p_{i}])\cdot t^{i}+(h_{2}[p_{i}])\cdot t^{2i} + \cdots
.
\end{align*}
We have
\begin{align*}
&\left(\prod_{j\geq 1} (1-y_j^{i}t^{i})^{-1}\right)^{l_q(i)} \\
&\ \ \ \  = \left( 1+ (h_{1}[p_{i}]) t^{i}+(h_{2}[p_{i}]) t^{2i} +
\cdots\right)^{l_q(i)}\\
&\ \ \ \  = \sum_{m_{i,1}+m_{i,2}+\cdots+m_{i,n}=m_{i}\atop 0\leq
m_i \leq l_q(i)} {l_q(i)\choose
m_{i}}{m_{i}\choose m_{i,1} m_{i,2} \cdots m_{i,n}}\\
&\ \ \ \ \qquad\qquad\quad \times
(h_{1}[p_{i}])^{m_{i,1}}(h_{2}[p_{i}])^{m_{i,2}}\cdots(h_{n}[p_{i}])^{m_{i,n}}\cdot
t^{(m_{i,1}+2m_{i,2}+\cdots+n\cdot
m_{i,n})\cdot i}\\
&\ \ \ \ = \sum_{m_{i,1}+m_{i,2}+\cdots+m_{i,n}=m_{i}\atop 0\leq m_i
\leq l_q(i)} \rho_{[m_{i,1},\ldots, m_{i,n}]}\cdot
t^{(m_{i,1}+2m_{i,2}+\cdots+n\cdot m_{i,n})\cdot i}\,\lower 5pt
\hbox{.}
\end{align*}
From (\ref{eq:plechar}) we see that the coefficient of $t^n$ in the
product $\prod_{i\geq 1}\left(\prod_{j\geq
1}(1-y_j^{i}t^{i})^{-1}\right)^{l_q(i)}$ is exactly equal to
$\rho_n$ for $n\geq 1$. Thus we get the Theorem.
\end{proof}

\begin{theorem}\label{thm:lm2}
\begin{equation}\label{eq:lqt}
\prod_{i\geq 1}\prod_{j\geq 1}(1-y_j^{i}t^{i})^{-l_q(i)} =
\frac{\prod_{j\geq 1}(1-y_j t)}{\prod_{j\geq 1}(1-y_j q t)} \lower
13pt\hbox{.}
\end{equation}
\end{theorem}
\begin{proof}From equation~(\ref{eq:numirr}) we know that when $i=1$,
\begin{equation*}
\prod_{j\geq 1}(1-y_jt)^{-l_q(1)}\prod_{j\geq
1}(1-y_jt)^{-1}=\prod_{j\geq 1}(1-y_jt)^{-L_q(1)}\,\lower 4pt
\hbox{,}
\end{equation*}
and when $i\geq 2$,
\begin{equation*}
\prod_{j\geq 1}(1-y_j^{i}t^{i})^{-l_q(i)}=\prod_{j\geq
1}(1-y_jt)^{-L_q(i)}\lower 5pt \hbox{.}
\end{equation*}
Thus the above identity is equivalent to the identity
\begin{equation}\label{eq:qt}
\prod_{i\geq 1}\prod_{j\geq 1}(1-y_j^{i}t^{i})^{L_q(i)} =
\prod_{j\geq 1}(1-y_j q t),
\end{equation}
where $L_q(i)$ denotes the number of irreducible polynomials over
$\mathbb F_q$ for $i\geq 1$, as we stated before.

Equation~(\ref{eq:qt}) is a classical identity which could be found,
e.g. in~\cite{ChrisReu} page 171. Here we give a direct
computational proof by taking the logarithm on both sides of
(\ref{eq:qt}) and showing they are equal.
\begin{align*}
\ln\left(\prod_{i\geq 1}\prod_{j\geq
1}(1-y_j^{i}t^{i})^{L_q(i)}\right) &= \sum_{j\geq
1}\left(\sum_{i\geq 1}
L_q(i)\ln(1-y_j^{i}t^{i})\right)\\
&= \sum_{j\geq 1}\left(\sum_{i\geq 1} L_q(i)\left(\sum_{r\geq 1}
\frac{(y_j^i t^i)^{r}}{r} \right)\right)\\
&= \sum_{j\geq 1}\left(\sum_{i\geq 1} \sum_{r\geq 1}
L_q\left(\frac{i\cdot r}{r}\right)\cdot \frac{i\cdot r}{r}\cdot
\frac{y_j^{(i\cdot r)}\cdot t^{(i\cdot r)}}{i\cdot r} \right)\\
&=\sum_{j\geq 1}\left(\sum_{N\geq 1\atop N=i\cdot r}\frac{
y_j^{N}\cdot t^{N}}{N}\left( \sum_{r|N}
L_q\left(\frac{N}{r}\right)\cdot \frac{N}{r} \right)\right)\\
&=\sum_{j\geq 1}\left(\sum_{N\geq 1\atop N=i\cdot r}\frac{
y_j^{N}\cdot t^{N}}{N}\cdot q^{N}\right)\\
&=\sum_{j\geq 1}\left(\ln(1-y_j q t)\right) = \ln\left(\prod_{j\geq
1}(1-y_j q t)\right)\!\lower 8pt \hbox{.}
\end{align*}
Thus we get (\ref{eq:qt}).
\end{proof}
Theorem~(\ref{thm:lm1}) and Theorem~(\ref{thm:lm2}) together yield
the formula for the generating function of $\rho_n$ as follows:
\begin{equation}\label{eq:gen}
CH_q(t) = \prod_{j\geq 1}\frac{1-y_j t}{1-y_j q t}\, \lower
8pt\hbox{.}
\end{equation}
Before we link it to Hall-Littlewood polynomials, we give a
recurrence relation for $\rho_n$ using formula~(\ref{eq:gen}).

\begin{corollary}\label{col:thmrec}
For every $n\geq 1$, we have
\begin{equation}\label{eq:rec}
\rho_n = (q^n-1) h_{n} - \rho_{n-1} h_{1} - \rho_{n-2} h_{2}- \cdots
- \rho_1 h_{n-1} .
\end{equation}
\end{corollary}
\begin{proof}
From~(\ref{eq:gen}) we have
\begin{equation*}
CH_q(t)\times \Omega[tY] = \prod_{j\geq 1}(1-y_j q t)^{-1} .
\end{equation*}
Comparing the coefficients of $t^n$ on both sides, we get
\begin{equation*}
\rho_0 h_{n}+\rho_1 h_{n-1}+\cdots +\rho_n h_{0} = q^n h_{n},
\end{equation*}
which yields the theorem.
\end{proof}

\begin{example}
\begin{align*}
\rho_1 &= (q-1) h_1 ;\\
\rho_2 &= (q^2-1) h_2 - \rho_1 h_{1} \\
        &= (q^2-1) h_2 - (q-1) h_{1,1}\\
        &= (q-1)[(q+1) h_2 - h_{1,1}] ;\\
\rho_3 &= (q^3-1) h_3 - \rho_1 h_{2} - \rho_2 h_{1}\\
        &= (q^3-1) h_3 - (q-1) h_{2,1} - (q^2-1) h_{2,1} + (q-1) h_{1,1,1}\\
        &= (q-1)[(q^2+q+1) h_3 -(q+2) h_{2,1} + h_{1,1,1}] .\\
\end{align*}
\end{example}
From the above examples we notice that the coefficients of
$h_{\lambda}$ are in $\pm\mathbb N[q]\times (q-1)$.  Let
$[h_{\lambda}]\rho_n$ denote the coefficients of $h_{\lambda}$ in
the expansion of $\rho_n$. In particular, we have $[h_{n}]\rho_n =
q^n-1$ for all $n\geq 1$. The following corollary gives the
recurrence relation on the coefficients.

\begin{corollary}\label{col:rec}
For any $\lambda = (a_1^{l_1}, a_2^{l_2}, \ldots, a_k^{l_k}) \vdash
n$ with $l_i \geq 1$ for all $1\leq i \leq k$ and $l(\lambda)\geq
2$, we have \vskip -0.2cm
\begin{equation}\label{eq:reccoeff}
[h_{\lambda}]\rho_n = -[h_{(a_1^{l_1-1}, a_2^{l_2}, \ldots,
a_k^{l_k})}]\rho_{n-a_1} -[h_{(a_1^{l_1}, a_2^{l_2-1}, \ldots,
a_k^{l_k})}]\rho_{n-a_2}- \cdots -[h_{(a_1^{l_1}, a_2^{l_2}, \ldots,
a_k^{l_k-1})}]\rho_{n-a_k}.
\end{equation}
Here, if $l_i=1$ for some $1\leq i\leq k$, then we set
$$(a_1^{l_1}, \ldots, a_i^{l_i-1}, \ldots, a_k^{l_k}):= (a_1^{l_1}, \ldots, \hat{a_i}, \ldots, a_k^{l_k}) , $$
where $\hat{a_i}$ means simply remove $a_i$ from the partition
$\lambda$. In particular, $[h_{\lambda}]\rho_n\in \pm\mathbb
N[q]\times(q-1)$, while the sign is given by $(-1)^{l(\lambda)-1}$.
\end{corollary}
\begin{proof}
Equation~(\ref{eq:reccoeff}) follows directly from
Corollary~\ref{col:thmrec} by comparing the coefficients of\,
$h_{\lambda}$\, from two sides. The claim that $[h_{\lambda}]\rho_n$
is in $\pm\mathbb N[q]\times(q-1)$ together with the sign property
can be proved easily by using induction on
equation~(\ref{eq:reccoeff}).
\end{proof}

\begin{remark}
Corollary~\ref{col:thmrec} and Corollary~\ref{col:rec} give an easy
way of computing $\rho_n$ for every $n\geq 1$ simply by knowing
$[h_i]\rho_i=q^i-1$ for every $i\geq 1$.
\end{remark}

\begin{example}
\begin{align*}
[h_{2,1}]\rho_3& = -[h_{1}]\rho_1 - [h_{2}]\rho_2\\
                & = - (q-1) - (q^2-1) \\
                & = -(q-1)(q+2).
\end{align*}
\begin{align*}
[h_{1,1,1}]\rho_3& = -[h_{1,1}]\rho_2 =[h_{1}]\rho_1\\
                  & = q-1.
\end{align*}
\end{example}

Now back to our formula~(\ref{eq:gen}). We rewrite it in the
following form so that we can easily use the generating function for
$\bm{q}_r$ as in equation~(\ref{eq:genqr}).
\begin{align*}
CH_q(t) &= \prod_{j\geq 1}\frac{1-y_j t}{1-y_j q t} = \prod_{j\geq
1}\frac{1-y_j\cdot \frac{1}{q}\cdot (q t)}{1-y_j\cdot (q t)} \\
      &=\sum_{r\geq 0} \bm{q}_r[Y;q^{-1}] q^{r} t^{r}  \lower
5pt\hbox{,}
\end{align*}
where $Y=\{y_1, y_2, \ldots\}$. Comparing the coefficients from two
sides we get the following corollary.
\begin{corollary}\label{col:halli}
\begin{align}\label{eq:HL}
\rho_n &=  \bm{q}_n[Y;q^{-1}] q^{n} = (1-q^{-1})P_n\left[Y;q^{-1}\right] q^{n}\nonumber\\
        &= q^{n-1}(q-1) P_n\left[Y;q^{-1}\right]= q^{n-1}(q-1) \tilde{P}_n\left[Y;q\right].
\end{align}
\end{corollary}
Corollary~\ref{col:halli} gives the connection between the image of
the homomorphism $\rho$ on the characteristic map of
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$ and the
Hall-Littlewood symmetric functions. As we mentioned in the
introduction, we refer to this result as a plethysm formula on
$ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})$.

For any linear supercharacter~\cite{Nat2, Nat, MulAu} of $U_n$,
there is a unique way to decompose the indexed set partition into
connected components. We denote a linear supercharacter with $k$
connected components by $\chi^{n_1 | n_2 | \ldots |
n_k}_{\vec{q_1},\ldots,\vec{q_{k}}}$, where for $i$ from $1$ to $k$,
each $n_i$ counts the size of the $i^{th}$ connected component, and
$\vec{q_i}=(q_{i,1}, \ldots, q_{i,n_i-1})\in (\mathbb
F_q^{\times})^{n_i-1}$ denotes the labels of the arcs for the
$i^{th}$ connected component. The following corollary follows from
the property of the linear supercharacters~\cite{Nat2, Nat, MulAu}.
\begin{corollary}\label{col:multilichar}
$$\rho(ch(\chi^{n_1 | n_2 | \ldots |
n_k}_{\vec{q_1},\ldots,\vec{q_{k}}}\uparrow_{U_n}^{G_n})) =
\prod_{i=1}^k \rho_{n_i}.$$
\end{corollary}

\begin{example} For the following linear supercharacter of $U_6$
\begin{align*}
\raise 2pt \hbox{$\chi^{1 | 2 |
3}_{\vec{q_1},\vec{q_2},\vec{q_{3}}}$ = } \chi^{\begin{tikzpicture}
\filldraw [black] (0,0) circle (1.5pt) (.5,0) circle (1.5pt) (1,0)
circle (1.5pt) (1.5,0) circle (1.5pt) (2,0) circle (1.5pt) (2.5,0)
circle (1.5pt); \node at (0, -.2) {$\scscs 1$}; \node at (0.5, -.2)
{$\scscs 2$}; \node at (0.75, .35) {$\scs q_{2,1}$}; \node at (1,
-.2) {$\scscs 3$}; \node at (1.5, -.2) {$\scscs 4$}; \node at (1.75,
.35) {$\scs q_{3,1}$};\node at (2, -.2) {$\scscs 5$}; \node at
(2.25, .35) {$\scs q_{3,2}$};\node at (2, -.2) {$\scscs 5$}; \node
at (2.5, -.2) {$\scscs 6$}; \draw (0.5,0) .. controls (.75,.25) ..
(1,0); \draw (1.5,0) .. controls (1.75,.25) .. (2,0); \draw (2,0) ..
controls (2.25,.25) .. (2.5,0);
\end{tikzpicture}}
\end{align*}
where $\vec{q_1}= ( ), \vec{q_2}=(q_{2,1}),
\vec{q_{3}}=(q_{3,1},q_{3,2})$ and $q_{2,1}, q_{3,1},
q_{3,2}\in\mathbb F_q^{\times}$, we have
$$\rho\circ ch(\chi^{1 | 2 |
3}_{\vec{q_1},\vec{q_2},\vec{q_{3}}}\uparrow_{U_6}^{G_6}) =
\rho_1\rho_2\rho_3.$$
\end{example}

Let the transition matrix between $\{m_{\lambda}[X]\}_{\lambda\vdash
n} $ and  $\{p_{\mu}[X]\}_{\mu\vdash n} $ be $C_{\lambda, \mu}$,
i.e.
\begin{equation*}
m_{\lambda}[X]=\sum_{\mu} C_{\lambda, \mu} p_{\mu}[X].
\end{equation*}
Define $m_{\lambda}[q-1]$ by the following equation
\begin{equation*}
m_{\lambda}[q-1]=\sum_{\mu} C_{\lambda, \mu} p_{\mu}[q-1],
\end{equation*}
where $p_{n}[q-1] = q^n-1$ for every $n\geq 1$, and $p_{\mu}[q-1] =
p_{\mu_1}[q-1]\cdots p_{\mu_l}[q-1]$ for $\mu=(\mu_1,\ldots,\mu_l)$.

\begin{remark}
Using the orthogonal relation between the bases $\{m_{\lambda}\}$
and $\{h_{\mu}\}$, we give another expression for $\rho_n$ as
follows:
\begin{equation*}
\rho_n = \sum_{\lambda\vdash n}m_{\lambda}[q-1]\times
h_{\lambda}[Y].
\end{equation*}
\end{remark}
\begin{proof}
Using the notation in Section~\ref{eq:omg}, we have
\begin{equation*}
\Omega[qtY] = \prod_{j\geq 1}\frac{1}{1-y_j q t} \,\lower 8pt
\hbox{,} \ \  \Omega[-tY] = \prod_{j\geq 1}(1-y_j t)\lower 2pt
\hbox{.}
\end{equation*}
\begin{align*}
CH_q(t) &= \prod_{j\geq 1}\frac{1-y_j t}{1-y_j q t} \\
      &= \Omega [(q-1)tY]\\
      &= \sum_{n\geq 0}\left(\sum_{\lambda\vdash n}m_{\lambda}[q-1]\cdot
      h_{\lambda}[Y]
\right) t^n \,\lower 8pt\hbox{.}
\end{align*}
\end{proof}

As we mentioned above, it seems that we lose information by applying
$\rho$ to the characteristic map of
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$. However,
since $\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$ is a
character induced from $U_n$, this allows us to express
$ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})$ in basis
$\{\tilde P_{\bm{\mu}} | \bm{\mu}\in\mathcal{P}^{\Phi} \}$ from the
result of applying $\rho$. To show this fact, we first introduce the
following homomorphism on symmetric function ring defined
in~\cite{Mac}:
$$\omega: \Lambda_{\mathbb C}(Y) \rightarrow \Lambda_{\mathbb C}(Y)$$
by
$$\omega(e_r[Y])=h_r[Y],\ \text{for\ all}\ r\geq 0.$$

The map $\omega$ is a well known involution and automorphism on
$\Lambda_{\mathbb C}(Y)$. Also, we have
$$\omega(p_r[Y])=(-1)^{r-1}p_r[Y],\ \text{for\ all}\ r\geq 0.$$

The following theorem illustrates the relation between applying the
homomorphism $\rho$ to the characteristic map in basis
$\{S_{\bm{\lambda}} | \mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta}
\}$ and the projection on the characteristic map in basis $\{\tilde
P_{\bm{\mu}} | \bm{\mu}\in\mathcal{P}^{\Phi} \}$ to the space
$\Lambda_{\mathbb C}(X_{x-1})$.
\begin{theorem}\label{thm:comm}
The following diagram commutes:
\input xy
\xyoption{all}
\begin{displaymath}
\xymatrix{
  \otimes_{\varphi\in\Theta} \Lambda_{\mathbb C}(Y^{\varphi}) \ar[d]_{\rho} \ar @{=}^{Id} [r]
                & \otimes_{f\in\Phi} \Lambda_{\mathbb C}(X_f) \ar[d]^{\Pi |_{\Lambda_{\mathbb C}(X_{f=x-1})}}  \\
 \Lambda_{\mathbb C}(Y)           \ar[r]_{\tau\circ\omega}
                & \Lambda_{\mathbb C}(X_{x-1})}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\lower 43pt \hbox{,}
\end{displaymath}
where $\tau$ is the map of changing variables $y_i$ into $X_{i,x-1}$
for $i=1,2,\ldots$, and\, $\Pi |_{\Lambda_{\mathbb C}(X_{f=x-1})}$
is the projection to the space $\Lambda_{\mathbb C}(X_{x-1})$.
\end{theorem}
\begin{proof}
Recall equation~(\ref{eq:transform})
\begin{equation*}
p_k[Y^{\varphi}]=(-1)^{n-1}\sum_{x\in M_{n}}\xi
(x)p_{n/d(f_x)}[X_{f_x}]\,,
\end{equation*}
where $\xi\in\varphi$, $x\in f_x$ and $n= k\times d(\varphi)$. If we
apply $\rho$ to $p_k[Y^{\varphi}]$ we get $p_n[Y]$. Applying the
projection map $\Pi |_{\Lambda_{\mathbb C}(X_{f=x-1})}$ on the
right-hand side of equation~(\ref{eq:transform}) yields
$(-1)^{n-1}p_n[X_{x-1}]$. Since $\{p_n: n=1,2,\ldots\}$ are
algebraically independent over $\mathbb C$ and $\{p_{\lambda}:
\lambda\ \text{a\ partition}\}$ form a basis of the symmetric
function ring, we get the theorem.
\end{proof}
Since $\Lambda_{\mathbb C}$ has two bases $\{S_{\bm{\lambda}} |
\mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \}$ and $\{\tilde
P_{\bm{\mu}} | \bm{\mu}\in\mathcal{P}^{\Phi} \}$, we use
$\nobreak{ch(\chi\uparrow_{U_n}^{G_n})}\nobreak{(X_f: f\in \Phi)}$
to denote the expression in terms of basis  $\{\tilde P_{\bm{\mu}} |
\bm{\mu}\in\mathcal{P}^{\Phi}\}$, and
$ch(\chi\uparrow_{U_n}^{G_n})(Y^{\varphi}:\varphi\in\Theta)$ the
expression in terms of basis $\{S_{\bm{\lambda}} |
\mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \}$. We then have the
following identity.
\begin{corollary}\label{col:equi}
\begin{align*}
ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})(X_f: f\in
\Phi)&=\tau\circ\omega (\rho_n)\\
&=q^{n-1}(q-1) \omega(\tilde{P}_n\left[X_{x-1};q\right]).
\end{align*}
\end{corollary}
\begin{proof}
From the definition of the induced character by
equation~(\ref{eq:inform}) we know that
$$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}(g)=0$$ for
all $g\in G_n$ which are not similar to any unipotent
upper-triangular matrices. Notice that the characteristic polynomial
for all matrices in $U_n$ is $(x-1)^n$. Since similar matrices have
the same characteristic polynomial,
$\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}$ could
possibly take nonzero values only on those matrices in $G_n$ with
characteristic polynomials equal to $(x-1)^n$. We then have
$$ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})(X_f: f\in
\Phi) \in \Lambda_{\mathbb C}(X_{x-1}),$$ and so
$$\Pi |_{\Lambda_{\mathbb C}(X_{f=x-1})}[ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})]
=ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}).$$ By
theorem~\ref{thm:comm} we obtain the corollary.
\end{proof}
\begin{remark}\label{rm:equi}
From the proof of Corollary~\ref{col:equi} we conclude that for any
character $\chi$ of $U_n$, if we induce $\chi$ from $U_n$ to $G_n$,
then we have
\begin{equation*}
ch(\chi\uparrow_{U_n}^{G_n})(X_f: f\in \Phi) =
\tau\circ\omega\circ\rho
(ch(\chi\uparrow_{U_n}^{G_n})(Y^{\varphi}:\varphi\in\Theta)).
\end{equation*}
\end{remark}

For $\lambda=(\lambda_1,\ldots,\lambda_l)$ let
$\rho_{\lambda}=\rho_{\lambda_1}\rho_{\lambda_2}\ldots\rho_{\lambda_l}$.
By Corollary~\ref{col:halli}, since $\rho_n = q^{n-1}(q-1)
P_n[Y;q^{-1}]$, we know that $\{\rho_{\lambda}\}$ forms a basis for
the symmetric function ring $\Lambda_{\mathbb C}(Y)$. Thus
$\rho(ch(\chi\uparrow_{U_n}^{G_n}))$ can be written as
$\rho(ch(\chi\uparrow_{U_n}^{G_n})) = \sum_{\lambda\vdash n}
C_{\lambda}\rho_{\lambda}$, where $C_{\lambda}\in\mathbb C$. We then
define a map as follows.
\begin{definition}
Define $\hat{\rho}: \Lambda_{\mathbb C}(Y) \rightarrow {\mathbb
C}\text{-span}\{S_{\bm{\lambda}} |
\mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \}$ by
\begin{align*}
\hat{\rho}(\rho_n)&:= \sum_{\bm{\lambda} \in
\mathcal{P}_n^{\Theta}\atop ht(\bm{\lambda})=1}
S_{\bm{\lambda}}\\
&=ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n}) ,
\end{align*}
and
\begin{align*}
\hat{\rho}(\rho_{\lambda})=\hat{\rho}(\rho_{\lambda_1})\hat{\rho}(\rho_{\lambda_2})\ldots\hat{\rho}(\rho_{\lambda_l}),
\end{align*}
where $\lambda=(\lambda_1,\ldots,\lambda_l)$.
\end{definition}

\begin{proposition}\label{prop:conjecture2}
For a fixed finite field $\mathbb F_q$ and a character $\chi$ of
$U_n$, we have
\begin{equation*}
(\hat{\rho}\circ\rho) (ch(\chi\uparrow_{U_n}^{G_n})) =
ch(\chi\uparrow_{U_n}^{G_n}).
\end{equation*}
\end{proposition}
\begin{proof}
First of all, we have $\hat{\rho}\circ\rho
(ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})) =
\hat{\rho} (\rho_n) =
ch(\chi^{(n)}_{(q_1,\ldots,q_{n-1})}\uparrow_{U_n}^{G_n})$. From
Theorem~\ref{thm:comm} and Remark~\ref{rm:equi} we know that $\rho$
restricted on the characteristics of induced characters from $U_n$
to $G_n$ is an isomorphism. Since $\{\rho_n\}_{n\in\mathbb{N}}$
forms an algebraic basis and for all $ch(\chi\uparrow_{U_n}^{G_n})$,
$\rho$ must have a unique inverse, we get the proposition.
\end{proof}

Suppose $\rho(ch(\chi\uparrow_{U_n}^{G_n})) = \sum_{\lambda\vdash n}
C_{\lambda}\rho_{\lambda}$, where $C_{\lambda}\in\mathbb C$. From
the definition of $\hat{\rho}$ we get
\begin{align}\label{eq:inverse}
\hat{\rho}\circ\rho(ch(\chi\uparrow_{U_n}^{G_n})) &=
\sum_{\lambda\vdash n} C_{\lambda}(\hat{\rho}(\rho_{\lambda})) \nonumber\\
&=\sum_{\lambda\vdash n}
C_{\lambda}\hat{\rho}(\rho_{\lambda_1})\hat{\rho}(\rho_{\lambda_2})\ldots\hat{\rho}(\rho_{\lambda_l}).
\end{align}
Using Proposition~\ref{prop:conjecture2} we get the following
corollary.
\begin{corollary}\label{col:irrdecomp}
For a fixed finite field $\mathbb F_q$ and a character $\chi$ of
$U_n$, suppose $\nobreak{ch(\chi\uparrow_{U_n}^{G_n})} =
\sum_{\lambda\vdash n} C_{\lambda}\rho_{\lambda}$ where
$C_{\lambda}\in\mathbb C$. We have
\begin{align*}
ch(\chi\uparrow_{U_n}^{G_n})&= \sum_{\lambda\vdash n}
C_{\lambda}\rho_{\lambda}\\
&=\sum_{\lambda\vdash n} C_{\lambda}\left(\sum_{\bm{\lambda}^{(1)}
\in \mathcal{P}_{\lambda_1}^{\Theta}\atop ht(\bm{\lambda}^{(1)})=1}
S_{\bm{\lambda}^{(1)}}\right) \left(\sum_{\bm{\lambda}^{(2)} \in
\mathcal{P}_{\lambda_2}^{\Theta}\atop ht(\bm{\lambda}^{(2)})=1}
S_{\bm{\lambda}^{(2)}}\right)\cdots\left(\sum_{\bm{\lambda}^{(l)}
\in \mathcal{P}_{\lambda_l}^{\Theta}\atop ht(\bm{\lambda}^{(l)})=1}
S_{\bm{\lambda}^{(l)}}\right) \lower 25pt \hbox{.}
\end{align*}
\end{corollary}

\begin{remark}
It is difficult to get an expression for
$ch(\chi\uparrow_{U_n}^{G_n})$ in terms of basis $\{S_{\bm{\lambda}}
| \mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta} \}$, which gives the
irreducible decomposition of the induced character. However, if we
know the image of $\rho$ on the characteristic map of
$\chi\uparrow_{U_n}^{G_n}$, we may use $\hat{\rho}$ to get the
irreducible decomposition of $ch(\chi\uparrow_{U_n}^{G_n})$. We hope
these results can contribute to research in this problem. We list
some open problems in Section~\ref{sec:probs}.
\end{remark}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Further Questions}\label{sec:probs}
The induced characters that we are studying in this paper are very
special. A natural question to ask is: ``can we give a nice formula
for the characteristics of all the induced supercharacters from
$U_n$ to $G_n$?" Zelevinsky~\cite{Zele} and Thiem and
Vinroot~{\cite{NatVin}} have worked on the case of degenerate
Gelfand-Graev characters. The problem of how the generalized
Gelfand-Graev representations of the finite unitary group decompose
is still open.

The generalized Gelfand-Graev representations, which are defined by
Kawanaka~\cite{Kawa}, are obtained by inducing certain irreducible
representations from a unipotent subgroup~\cite{NatVin}. Here the
supercharacters that we are considering are more general than the
case of the generalized Gelfand-Graev representations. We hope that
the ideas and results developed in this paper could help to work on
this problem.

Below we give some examples of computations of $\rho$ acting on the
characteristic map of induced supercharacters.
\begin{example}\label{ex:example}
For $q=2$, we have
\begin{align*}
\lower 5pt \hbox{$\rho\circ ch$} \left(\lower 4pt
\hbox{$\chi^{\begin{tikzpicture} \filldraw [black] (0,0) circle
(1.5pt) (.5,0) circle (1.5pt)  (1,0) circle (1.5pt); \node at (0,
-.2) {$\scscs 1$}; \node at (0.5, -.2) {$\scscs 2$}; \node at (0.5,
.55) {$\scs 1$}; \node at (1, -.2) {$\scscs 3$}; \draw (0,0) ..
controls (.25,.4) and (.75,.4).. (1,0);
\end{tikzpicture}} \uparrow_{U_3}^{G_3}$} \right) = (\rho_{3} +
\rho_2\rho_1)|_{q=2}
\end{align*}
\begin{align*}
\lower 5pt \hbox{$\rho\circ ch$} \left(\lower 4pt
\hbox{$\chi^{\begin{tikzpicture} \filldraw [black] (0,0) circle
(1.5pt) (.5,0) circle (1.5pt)  (1,0) circle (1.5pt) (1.5,0) circle
(1.5pt); \node at (0, -.2) {$\scscs 1$}; \node at (0.5, -.2)
{$\scscs 2$}; \node at (0.75, .75) {$\scs 1$}; \node at (1, -.2)
{$\scscs 3$};  \node at (1.5, -.2) {$\scscs 4$}; \draw (0,0) ..
controls (.35,.6) and (1.15,.6).. (1.5,0);
\end{tikzpicture}} \uparrow_{U_4}^{G_4}$} \right) = (\rho_{4} +
2\rho_3\rho_1+\rho_2\rho_1^2)|_{q=2}
\end{align*}
\begin{align*}
\lower 5pt \hbox{$\rho\circ ch$} \left(\lower 4pt
\hbox{$\chi^{\begin{tikzpicture} \filldraw [black] (0,0) circle
(1.5pt) (.5,0) circle (1.5pt)  (1,0) circle (1.5pt) (1.5,0) circle
(1.5pt); \node at (0, -.2) {$\scscs 1$}; \node at (0.5, -.2)
{$\scscs 2$}; \node at (0.5, .55) {$\scs 1$}; \node at (1, -.2)
{$\scscs 3$};  \node at (1, .55) {$\scs 1$}; \node at (1.5, -.2)
{$\scscs 4$}; \draw (0,0) .. controls (.25,.4) and (.75,.4)..
(1,0);\draw (.5,0) .. controls (.75,.4) and (1.25,.4).. (1.5,0);
\end{tikzpicture}} \uparrow_{U_4}^{G_4}$} \right)
= \lower 5pt \hbox{$\rho\circ ch$} \left(\lower 4pt
\hbox{$\chi^{\begin{tikzpicture} \filldraw [black] (0,0) circle
(1.5pt) (.5,0) circle (1.5pt)  (1,0) circle (1.5pt) (1.5,0) circle
(1.5pt); \node at (0, -.2) {$\scscs 1$}; \node at (0.5, -.2)
{$\scscs 2$}; \node at (0.75, .75) {$\scs 1$};\node at (1, -.2)
{$\scscs 3$}; \node at (0.75, .35) {$\scs 1$};\node at (1.5, -.2)
{$\scscs 4$}; \draw (0,0) .. controls (.35,.7) and (1.15,.7)..
(1.5,0); \draw (.5,0) .. controls (.625,.3) and (.875,.3).. (1,0);
\end{tikzpicture}} \uparrow_{U_4}^{G_4}$} \right) =  (2\rho_{4} +
\rho_2\rho_2+\rho_3\rho_1)|_{q=2}
\end{align*}
\begin{align*}
\lower 5pt \hbox{$\rho\circ ch$} \left(\lower 4pt
\hbox{$\chi^{\begin{tikzpicture} \filldraw [black] (0,0) circle
(1.5pt) (.5,0) circle (1.5pt)  (1,0) circle (1.5pt) (1.5,0) circle
(1.5pt); \node at (0, -.2) {$\scscs 1$}; \node at (0.5, -.2)
{$\scscs 2$}; \node at (0.25, .4) {$\scs 1$}; \node at (1, -.2)
{$\scscs 3$};  \node at (1, .55) {$\scs 1$}; \node at (1.5, -.2)
{$\scscs 4$}; \draw (0,0) .. controls (.125,.2) and (.375,.2)..
(.5,0);\draw (.5,0) .. controls (.75,.4) and (1.25,.4).. (1.5,0);
\end{tikzpicture}} \uparrow_{U_4}^{G_4}$} \right)
= (\rho_{4} + \rho_3\rho_1)|_{q=2}
\end{align*}
\end{example}

Inspired by these results, we propose the following conjecture and
state some questions that remain open.

\begin{conjecture}
For a fixed finite field $\mathbb F_q$ and a supercharacter $\chi$
of $U_n$, we have
\begin{equation*}
\rho(ch(\chi\uparrow_{U_n}^{G_n})) \in \mathbb
N[\rho_1,\ldots,\rho_n].
\end{equation*}
\end{conjecture}
The conjecture is interesting in the sense that not every character
of a larger group can be represented as a non-negative integer
combination of characters induced from linear characters of
subgroups. It is interesting to know which kind of characters have
that property. If the above conjecture is true, then the following
remark is meaningful.
\begin{remark}
For a fixed finite field $\mathbb F_q$ and a character $\chi$ of
$U_n$, suppose $\nobreak{\rho(ch(\chi\uparrow_{U_n}^{G_n}))} =
\sum_{\lambda\vdash n} C_{\lambda}\rho_{\lambda}$ where
$C_{\lambda}\in\mathbb C$. We have
\begin{equation}\label{eq:dim}
\dim(\chi) = \sum_{\lambda\vdash n} C_{\lambda}.
\end{equation}
\end{remark}
\begin{proof}
From Corollary~\ref{col:multilichar} we have
\begin{equation*}
\chi\uparrow_{U_n}^{G_n} = \sum_{\lambda\vdash n} C_{\lambda}
(\chi^{\lambda_1 |\lambda_2 | \ldots |
\lambda_l}_{\vec{q_1},\ldots,\vec{q_{l}}}\uparrow_{U_n}^{G_n}) =
\bigg(\sum_{\lambda\vdash n} C_{\lambda} \chi^{\lambda_1 |\lambda_2
| \ldots |
\lambda_l}_{\vec{q_1},\ldots,\vec{q_{l}}}\bigg)\uparrow_{U_n}^{G_n},
\end{equation*}
where $\vec{q_i}=(q_{i,1},\ldots,q_{i,\lambda_i-1})\in (\mathbb
F_q^{\times})^{\lambda_i-1}$. So we have
\begin{equation*}
\dim(\chi) = \dim\bigg(\sum_{\lambda\vdash n} C_{\lambda}
\chi^{\lambda_1 |\lambda_2 | \ldots |
\lambda_l}_{\vec{q_1},\ldots,\vec{q_{l}}}\bigg) =
\sum_{\lambda\vdash n} C_{\lambda} \dim(\chi^{\lambda_1 |\lambda_2 |
\ldots | \lambda_l}_{\vec{q_1},\ldots,\vec{q_{l}}}).
\end{equation*}
Since $\dim(\chi^{\lambda_1 |\lambda_2 | \ldots |
\lambda_l}_{\vec{q_1},\ldots,\vec{q_{l}}}) = 1$,
equation~(\ref{eq:dim}) follows.
\end{proof}

\begin{question}
For a fixed finite field $\mathbb F_q$ and a supercharacter $\chi$
of $U_n$, find a formula for the image of the homomorphism $\rho$ on
the characteristic map of $\chi\uparrow_{U_n}^{G_n}$,
\begin{equation*}
\rho\circ ch(\chi\uparrow_{U_n}^{G_n})= \sum_{\lambda\vdash n}
C_{\lambda}\rho_{\lambda},
\end{equation*}
where
$\rho_{\lambda}=\rho_{\lambda_1}\rho_{\lambda_1}\ldots\rho_{\lambda_l}$
for $\lambda={\lambda_1,\ldots,\lambda_l}$. It would be nice to give
a combinatorial formula for the coefficient $C_{\lambda}$, since the
examples above suggest a possibility of some rules.
\end{question}
\begin{remark}
If we have the formula of $\rho(ch(\chi\uparrow_{U_n}^{G_n}))$, we
can easily get the expression for the characteristic map of
$\chi\uparrow_{U_n}^{G_n}$ in terms of basis $\{\tilde P_{\bm{\mu}}
| \bm{\mu}\in\mathcal{P}^{\Phi} \}$ by Remark~\ref{rm:equi}. We may
also use $\hat{\rho}$ to get an expression in the basis
$\{S_{\bm{\lambda}} | \mathbf{\bm{\lambda}}\in\mathcal{P}^{\Theta}
\}$ by Corollary~\ref{col:irrdecomp}.
\end{remark}

\begin{question}
Up to now the induced representations that we are considering are in
characteristic zero. Another problem we can think about is what
happens in characteristic $p$ case.
\end{question}

%\parskip=0pt plus 2pt
\parindent=0pt

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
I am grateful to my advisor, Professor Nantel Bergeron, for his
guidance and discussions on this research problem. I also want to
thank the anonymous referee and Professor Mike Zabrocki for many
useful suggestions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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