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\title {Rationality Theorems for Hecke Operators on $GL_n$ } 

\author {John A. Rhodes and Thomas R. Shemanske}

\date{\today}
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\evenhead{John A. Rhodes and Thomas R. Shemanske}     %Running Heads on Even Pages
\oddhead{Rationality Theorems for Hecke Operators on $GL_{n}$}       %Running Heads on Odd Pages

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%\dedication{This is dedicated to the one I love}

\subjclass {Primary 11F25, 20E42; Secondary 11F60}

\keywords{Hecke Operators, Rationality, Generating Series}


\begin{abstract}
  We define $n$ families of Hecke operators $T_k^n(p^\ell)$ for $GL_n$
  whose generating series $\sum T_k^n(p^\ell) u^\ell$ are rational
  functions of the form $q_k(u)^{-1}$ where $q_k$ is a polynomial of
  degree $\binom{n}{k}$, and whose form is that of the $k$th exterior
  product.  This work can be viewed as a refinement of work of
  Andrianov \cite{Andrianov70}, in which he defined Hecke operators
  the sum of whose generating series was a rational function
  with nontrivial numerator and whose denominator was essentially
  $\prod_k q_k$.

  
  By a careful analysis of the Satake map which defines an isomorphism
  between a local Hecke algebra and a ring of symmetric polynomials,
  we define $n$ families of (polynomial) Hecke operators and
  characterize their generating series as rational functions.  We then
  give an explicit means by which to locally invert the Satake
  isomorphism, and show how to translate these polynomial operators
  back to the classical double coset setting. The classical Hecke
  operators have generating series of exactly the same form as their
  polynomial counterparts, and hence are of number-theoretic interest.
  We give explicit examples for $GL_3$ and $GL_4$.
\end{abstract}


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\section{Introduction}\label{sec:introduction}
While Hecke theory for automorphic forms on $GL_2$ is fairly robust,
far less is known for $GL_n$ when $n > 2$, especially as it relates to
the arithmetic theory of modular forms.  It is well-known (see e.g.,
Cartier \cite{Cartier-Corvallis}, Theorem 4.1) that the Satake map
shows that the $p$-part of the Hecke algebra associated to certain
$p$-adic reductive groups is isomorphic to a polynomial ring invariant
under an associated Weyl group.  In \cite{AZbook}, Andrianov and Zhuravlev
refer to this isomorphism as the spherical map, and give a description
of it for the general linear and symplectic groups in terms of right
cosets of the double cosets which generate the Hecke algebra.

An important aspect of this isomorphism is that it allows one to think
about Hecke operators in the polynomial algebra, a setting in which
the multiplication is more straightforward than the usual product of
double cosets.  Our goals in this paper are to define families of
Hecke operators in the polynomial algebra which have generating series
whose sum is an extremely nice rational function, and then to invert
the Satake isomorphism, to produce ``classical'' Hecke operators
having generating series of exactly the same form, and hence of
number-theoretic interest.

 In \cite{Andrianov70},
Andrianov considers classical Hecke operators for $GL_n$, and studies
their generating series.  He shows (Theorem 3) that the image of their
sum under the spherical map is a rational function (in $\Q(x_1, \dots,
x_n, u)$) whose denominator
is a product essentially of the form:
\[ 
\prod_{k = 1}^n \ \ \prod_{1\le i_1<i_2<\cdots <i_k\le n} 
(1 - x_{i_1} x_{i_2} \cdots x_{i_k} u)
\]
Notice that the monomials which occur in the inner product, $x_{i_1}
x_{i_2} \cdots x_{i_k}$, are $k$th exterior powers of the variables
$x_1, \dots, x_n$.

In this paper, we take as a starting point the polynomial ring, and
define families of symmetric polynomials, denoted $t_k^n(p^\ell)$ ($k
= 1, \dots, n$) and which we will call polynomial Hecke operators,
whose associated generating series are rational functions.  In
particular (see Proposition~\ref{prop:generaleuler})
\[
\sum_{\ell \ge 0} t_k^n(p^\ell) u^\ell = \prod_{1\le i_1<i_2<\cdots <i_k\le n} 
(1 - x_{i_1} x_{i_2} \cdots x_{i_k} u)^{-1}.
\]
Thus the operators $t_k(p^\ell)$ give a significant refinement to the
operators investigated by Andrianov.  Moreover, we shall see that one
family, the $t_1^n(p^\ell)$, are essentially formal analogs of the
classical Hecke operators $T(p^\ell)$.  The more difficult task is to
determine to what combination of double cosets these polynomial
operators correspond.  To accomplish this one must invert the
spherical map.

By studying the restriction of the spherical map to those double
cosets of a fixed determinant, it is possible to (locally) invert the
spherical map, the matrix of the restricted map being upper
triangular.  At the heart of the computations to determine the matrix
of the linear map is the question of determining the number of right
cosets of a specified form which occur in the decomposition of a given
double coset.  In general, this is quite onerous, however we give a
number of ways in which the computation can be significantly
simplified.

Intertwined with the general theory, we provide examples which
explicitly compute families ($T_k^4(p^\ell)$) of classical Hecke
operators for $GL_4$.  The interest in $GL_4$ is that $n = 4$ is the
first case in which we associate ``Euler factors'' of degree greater
than $n$ with Hecke operators for $GL_n$.  In particular,
Proposition~\ref{prop:generaleuler} shows that the denominator of the
rational function associated to $\sum_{\ell \ge 0} t_k^n(p^\ell)
u^\ell$ is the product of $\binom{n}{k}$ linear factors, so have
produced Hecke operators for $GL_n$ with associated Euler factors of
degree $\binom{n}{k}$.

\section{The Hecke Algebras}\label{sec:preliminaries}

%\subsection{}




The global Hecke algebra described by Shimura \cite{Shimurabook} is
generated as a rational vector space by double cosets $\Lambda \xi
\Lambda$ with respect to the discrete group $\Lambda = SL_n(\Z)$ and
$\xi \in M_n^+(\Z)$ (integer matrices with positive determinant),
while the algebra described by Andrianov and Zhuravlev \cite{AZbook}
is generated by double cosets $\G \xi \G$ with $\G = GL_n(\Z)$ and
$\xi \in GL_n(\Q)$.  If we denote by $H(K,S)$ the rational Hecke
algebra generated by double cosets $K \xi K$ for a subgroup $K$ of $G$
and a semigroup $S \supset K$ in the commensurator of $K$, then we
have natural maps

\[H(SL_n(\Z),M_n^+(\Z)) \to H(SL_n(\Z),GL_n^+(\Q)) \to
H(GL_n(\Z),GL_n(\Q)),\] where the map on the left is an injection, and
that on the right an isomorphism.  Moreover, a straightforward
argument shows that


\begin{prop} 
  $H(GL_n(\Z),GL_n(\Q))$ is generated, as a ring, by
  $H(SL_n(\Z),M_n^+(\Z))$ together with the elements $GL_n(\Z)
  (p^{-1}I_n)GL_n(\Z)$ for all primes $p$. ($I_n$ denotes the $n\times n$
  identity matrix.)
\end{prop}

Thus for investigating Hecke theory, it makes little difference which
algebra we choose.  We adopt the notation of \cite{AZbook}, and
set the global Hecke algebra $H^n=H(GL_n(\Z), GL_n(\Q))$.  By Theorems
2.3 and 2.8 of \cite{AZbook}, $H^n$ is
commutative, and generated by its local
subrings, $H_p^n$ one for each prime
$p$, defined by
\[
H_p^n = H(GL_n(\Z),GL_n(\Z [ p^{-1} ] )) \subset H^n.
\]
The integral subring $\underline H_p^n$ of $H_p^n$ is
\begin{align*}
  \underline H_p^n &= \langle \G \xi\G \mid \xi \in M_n(\Z), \det(\xi)
  =
  \pm p^\lambda \rangle\\
  &= \langle \G \diag(p^{i_1}, \ldots, p^{i_n})\G \mid 0\le i_1 \le
  \cdots
  \le i_n \rangle \\
\end{align*}
where the angle brackets enclose generators as a $\Q$-vector space.
As a ring, $H_p^n$ is generated by $\underline H_p^n$ together with
the single element $\G (p^{-1} I_n)\G$.  Thus the study of the the
global Hecke ring $H^n$ reduces to study of the local Hecke rings
$H_p^n$, which in turn can be understood through their integral
subrings $\underline H_p^n$.


Equivalently, we could focus on the $p$-adic Hecke ring.  Letting $\Gp
= GL_n(\Zp)$, the $p$-adic Hecke ring is
$$\cH_p^n=H(GL_n(\Zp),GL_n(\Qp)),$$
and its integral subring
$$\underline\cH_p^n = \langle \Gp (\diag(p^{i_1}, \ldots, p^{i_n}))\Gp \mid
0\le i_1 \le \cdots \le i_n \rangle ,$$
which together with the single
element $\Gp (p^{-1} I_n) \Gp$ generates $\cH_p^n$.  Since $\cH_p^n$
(resp.  $\underline\cH_p^n$) is canonically isomorphic to $H_p^n$
(resp.  $\underline H_p^n$), we could work in the $p$-adic setting (as
in studying $p$-adic reductive groups and buildings), but for our
purposes here, we shall simply keep things local and not $p$-adic.


The integral Hecke algebra $\underline H_p$ is generated by the $n$
Hecke operators
\begin{equation*}
  \pi_k^n(p) = \G
  \diag(\underbrace{1,\ldots,1}_{n-k},\underbrace{p,\ldots,p}_k)\G,\
  \ k = 1, \dots,n
\end{equation*}
and the local Hecke algebra $H_p$ is generated by the $n$ elements
above, plus the element $\pi_n^n(p)^{-1}$ (see \cite{AZbook}).  The
Satake map is an isomorphism between the ($p$-adically defined) local
Hecke algebra and a ring of symmetric polynomials.  In \cite{AZbook},
it is called the spherical map, and we will examine it carefully in
\S4.  For now, simply note that if we let $S_n$ denote the symmetric
group on $n$ letters and $\Q[x_1^{\pm1}, \dots, x_n^{\pm 1}]^{S_n}$
the ring of symmetric polynomials in the variables $x_1^{\pm1}, \dots,
x_n^{\pm1}$ (i.e.  those polynomials invariant under the natural
action of the symmetric group on the variables), then the spherical
map is denoted $\omega: H_p \to \Q[x_1^{\pm1}, \dots, x_n^{\pm
  1}]^{S_n}$, and takes $\underline H_p$ isomorphically onto
$\Q[x_1,\dots,x_n]^{S_n}$ (the usual ring of symmetric polynomials).
In particular, it maps $\pi_k^n(p)$ to a power of $p$ times the $k$th
elementary symmetric polynomial (see \cite{AZbook}).


\section{Polynomial Hecke Operators}

The Satake isomorphism is not only a very esthetic result, it is also
a very useful one.  It allows one to study the Hecke algebra in the
context of a polynomial ring, with very familiar ring operations,
instead of its defined setting where the multiplication of double
cosets is cumbersome.

On the other hand, certain families of Hecke operators are naturally
defined in the context of double cosets, and generating series for
these families have been studied by many.  In \cite{Andrianov70},
Andrianov considers classical Hecke operators for $GL_n$, and studies
their generating series.  He shows (Theorem 3) that the image of their
sum under the spherical map is a rational function whose denominator
is a product essentially of the form:

\[ 
\prod_{k = 1}^n \ \ \prod_{1\le i_1<i_2<\cdots <i_k\le n} 
(1 - x_{i_1} x_{i_2} \cdots x_{i_k} u)
\]
Notice that the monomials which occur in the inner product, $x_{i_1}
x_{i_2} \cdots x_{i_k}$, are $k$th exterior powers of the variables
$x_1, \dots, x_n$.

In this paper, we take as a starting point the polynomial ring, and
define families of symmetric polynomials, denoted $t_k^n(p^\ell)$ ($k
= 1, \dots, n$) and which we will call polynomial Hecke operators,
whose associated generating series are rational functions.  In
particular (see Proposition~\ref{prop:generaleuler})
\[
\sum_{\ell \ge 0} t_k^n(p^\ell) u^\ell = \prod_{1\le i_1<i_2<\cdots <i_k\le n} 
(1 - x_{i_1} x_{i_2} \cdots x_{i_k} u)^{-1}.
\]

Thus the operators $t_k(p^\ell)$ give a significant refinement to the
operators investigated by Andrianov.  Moreover, we shall see that one
family, the $t_1^n(p^\ell)$, are essentially formal analogs of the
classical Hecke operators $T(p^\ell)$.  In later sections, we address
the question of how to determine the classical analogs, in terms of
double cosets, of these polynomial operators.

To begin our investigation, note that the symmetric group $S_n$ acts
naturally on polynomials in $n$ variables.  Thus for a polynomial $p$
it makes sense to refer to the stabilizer $\Stab(p)$ in $S_n$ of $p$.
For a polynomial $p$ in $n$ variables, define the {\it symmetrized
  polynomial} associated to $p$, $\Sym_n(p)$, by

\begin{equation*}
\Sym_n(p) = \sum_{\sigma \in S_n / \Stab(p)} \sigma(p)
\end{equation*}
We understand that if $p$ is a constant, that $\Sym_n(p) = p$.


\begin{example}
  In $\Q[z_1, \dots, z_n]$, $\ds \Sym_n(z_1\cdots z_k) = \sum_{1\le
    i_1< \cdots < i_k\le n} z_{i_1} \cdots z_{i_k}$.
\end{example}

\dr{   %Begin draft remark
\begin{proof}
  We observe the orbit of $z_1 z_2\cdots z_k$ is the set of distinct
  monomials in $k$ different variables which one can form with $n$
  variables $z_i$.  The number of summands in the sum, $[S_n :
  \Stab(z_1 z_2\cdots z_k)]$, is the size of the orbit which is
  clearly $\binom{n}{k}$, the binomial coefficient.
\end{proof}
}    %End of Draft Remark

\dr{   %Begin draft remark
  For our later convenience, we observe the following.

\begin{lemma}
  Let $M$ be a monomial, and $F = \Sym_n(M)$.  If $M'$ is another
  monomial occurring in $F $, then $F = \Sym_n(M')$.
\end{lemma}


\begin{proof}
  Let $H = \Stab(M)$ in $S_n$.  Then $F = \sum_{\sigma \in S_n / H}
  \sigma(M)$.  If $M'$ i s another monomial in $F$, then $M' = \tau M$
  for some $\tau \in S_n$.  Since $\Stab(M') = \tau \Stab(M)
  \tau^{-1}$, by definition, we have
\begin{equation*}
\Sym_n(M') = \sum_{\varphi \in S_n / \tau H \tau^{-1}} \varphi(M') =
\sum_{\varphi \in S_n / \tau H \tau^{-1}} \varphi \tau (M)
\end{equation*}
We need only establish that as $\varphi$ runs through a set of coset
representatives of $S_n/\tau H \tau^{-1}$, $\varphi\tau$ runs through
a set of coset representatives for $S _n/H$.  We have that
$\varphi\tau H = \varphi'\tau H$ if and only if
$\tau^{-1}\varphi^{-1}\varphi'\tau \in H$ from which we see the
correspondence is $1-1$.  As there are the same number of cosets in
both cases, we are done.
\end{proof}
}    %End of Draft Remark

Let $m$ and $n$ be positive integers and denote by $P_n(m)$ the set of
partitions of $m$ into $n$ pieces such that a given partition
satisfies: $m \ge i_1 \ge \cdots \ge i_n \ge 0$ (and $\sum i_k = m$).
Introduce the lexicographic ordering on $P_n(m)$, and let $\bi =
(i_1,i_2, \ldots,i_n) \in P_n(m)$. For indeterminates $z_1, z_2,
\ldots,z_n$ define

\begin{align*}
  &hp(0,\ldots,0)  = 1,\\
  & hp(\bi) = hp(i_1,i_2,\ldots,i_n) = \sum_{\substack{\bj \le \bi\\
      \bj \in P_n(m)}}
  \Sym_n(z_1^{j_1}z_2^{j_2} \cdots z_n^{j_n}), \text{ and}\\
  & h^n(\ell) = hp(\underbrace{\ell,0,\ldots,0}_n) = \sum_{\bj \in
    P_n(\ell)} \Sym_n(z_1^{j_1} z_2^{j_2} \cdots z_n^{j_n})=
  \sum_{\substack{\sum j_k = \ell\\j_k \ge 0}}z_1^{j_1}z_2^{j_2}
  \cdots z_n^{j_n}
\end{align*}

Note that the $hp(\bi)$ form a basis for the $\Q$-algebra of symmetric
polynomials. However, our current interest is in the polynomials
$h^n(\ell)$ which have a generating series with a particularly simple
rational expression.


\begin{prop}
\label{prop:basiceuler}
The generating series associated to the $h^r(\ell)$ satisfies
\[
\sum_{\ell \ge 0} h^r(\ell) u^\ell = \left[(1 - uz_1) \cdots
  (1-uz_r)\right]^{-1}
\]
\end{prop}

\begin{proof}
  This is essentially obvious:

\begin{align*}
  \left[(1 - uz_1) \cdots (1-uz_r)\right]^{-1} & = \left(\sum_{a_1 \ge
      0}(uz_1)^{a_1}\right) \cdots
  \left(\sum_{a_r \ge 0}(uz_r)^{a_r}\right)\\
  &= \sum_{\ell \ge 0} u^\ell\cdot \big[\sum_{\substack{\sum a_i =
      \ell\\a_i \ge 0}} z_1^{a_1} \cdots z_r^{a_r}\big]
\end{align*}
It is clear from the definitions above that the coefficient of
$u^\ell$ in the given expression is $h^r(\ell)$.
\end{proof}






Let $x_1, x_2, \ldots, x_n$ be indeterminates.  We now define our
family of Hecke operators as symmetric polynomials in $\Q[x_1^{\pm1},
x_2^{\pm1}, \ldots, x_n^{\pm1}]^{S_n}$: For $1 \le k \le n$, define

\[
t_k^n(p^\ell) = h^{\binom{n}{k}}(\ell)\Big\vert_ {\substack{z_i
    \mapsto \sigma_i(x_1 x_2\cdots x_k)
    \\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)}}
\]
This definition is not as mysterious as it first appears.  We start
with $h^r(\ell)$ where $r = \binom{n}{k}$ is the size of the orbit,
and substitute for the variables $z_i$ the elements in the orbit.
Thus there is one element $\sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)$
to correspond to each of the variables $z_1, \ldots,
z_{\binom{n}{k}}$.


The above definition is sufficiently complex to warrant an example,
however we note before giving detail that $t_k^n(p)$ is nothing more
than the $k^{\text{th}}$ elementary symmetric polynomial in the
variables $x_1, x_2, \ldots, x_n$.

\begin{ex}  We begin by computing

\[h^r(1) = hp(\underbrace{1,0,\ldots,0}_r) = \Sym_r(z_1) = 
z_1 + z_2 + \cdots + z_r.
\]
Then for example,
\[
t^n_1(p) = h^n(1)\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1)\\
    \sigma_i \in S_n/\Stab(x_1)}} = x_1 + x_2 + \cdots + x_n =
s_1(x_1, x_2,\dots, x_n),
\]
the first elementary symmetric polynomial, and in general
\[t^n_k(p) = h^{\binom{n}{k}}(1)\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1 x_2\cdots x_k)\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)}} = \sum\limits_{1\le
  i_1< \cdots <i_k\le n} x_{i_1} \cdots x_{i_k} = s_k(x_1, x_2,
\ldots, x_n),
\]
the $k^{\text{th}}$ elementary symmetric polynomial.
\end{ex}

\begin{remark} First recall that under the Satake isomorphism, the
  full local Hecke algebra $\H_p$ is isomorphic to
  $\Q[x_1^{\pm1},\ldots,x_n^{\pm1}]^{S_n} =
  \Q[x_1,\ldots,x_n]^{S_n}[(x_1\cdots x_n)^{-1}]$, while the integral
  local Hecke algebra $\underline\H_p \cong \Q[x_1,\ldots,x_n]^{S_n}$.
  Then note that the polynomials $t_k^n(p^\ell)$ are symmetric
  polynomials in the variables $x_1, x_2, \ldots, x_n$, so our
  operators $t_k^n(p^\ell)$ lie in the integral local Hecke algebra.
\end{remark}

Turning to generating series, we find

\begin{prop}
\label{prop:generaleuler}
For $k = 1, 2, \ldots, n$, the generating series for the Hecke
operators $t_k^n(p^\ell)$ is a rational function:

\[
\sum_{\ell \ge 0} t_k^n(p^\ell) u^\ell = \Big[\prod_{\sigma \in
  S_n/\Stab(x_1 x_2\cdots x_k)} (1 - u\sigma(x_1 x_2\cdots
x_k))\Big]^{-1}
\]
\end{prop}

\begin{proof}
  From Proposition~\ref{prop:basiceuler}, we know that

\[
\sum_{\ell \ge 0} h^{\binom{n}{k}}(\ell) u^\ell =
\Big[\prod_{i=1}^{\binom{n}{k}}(1 - uz_i)\Big]^{-1}.
\]
But
\[
t_k^n(p^\ell) = h^{\binom{n}{k}}(\ell)\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1 x_2\cdots x_k)\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)}}
\]
which completes the proof.
\end{proof}

\begin{remark}
  As we said earlier, the Euler factors which appear here also
  appeared in work of Andrianov \cite{Andrianov70}.  He defined a zeta
  function which encapsulated the algebra structure of the full local
  Hecke algebra and whose rational expression had denominator equal to
  the product of all of the denominators above, i.e.,

\[
\prod_{k=1}^n \prod_{\sigma \in S_n/\Stab(x_1 x_2\cdots x_k)} (1 -
u\sigma(x_1 x_2\cdots x_k)).
\]
However, the numerator of that rational expression was non-trivial and
not explicitly given. In this paper, we are producing a family of
operators whose generating series are rational functions whose
denominators are the individual factors of this product, with
numerators 1.
\end{remark}




\begin{ex}\label{ex:GL4}
  Consider the case of $n = 4$.  We have the four Hecke series:
\begin{align*}
  \sum_{\ell \ge 0} t_1^4(p^\ell) u^\ell &=
  [(1 - ux_1)(1 - ux_2)(1 - ux_3)(1-ux_4)]^{-1}\\
  \sum_{\ell \ge 0} t_2^4(p^\ell) u^\ell &= [(1 - ux_1x_2)(1 -
  ux_1x_3)(1 - ux_1x_4)(1-ux_2x_3)
  (1-ux_2x_4)(1-ux_3x_4)]^{-1}\\
  \sum_{\ell \ge 0} t_3^4(p^\ell) u^\ell &=
  [(1 - ux_1x_2x_3)(1 - ux_1x_2x_4)(1 - ux_1x_3x_4)(1-ux_2x_3x_4)]^{-1}\\
  \sum_{\ell \ge 0} t_4^4(p^\ell) u^\ell &= [(1 - ux_1x_2x_3x_4)]^{-1}
\end{align*}
\end{ex}

\begin{remark}
  Later in this paper we establish a connection between these $t_k^n$
  and classical Hecke operators.  Accepting such a connection for now,
  note that the Euler factor corresponding to $t_k^n$ has degree
  $\binom{n}{k}$ and the monomials are the $k$th exterior powers of
  the $x_i$'s.  Then, observing that the Euler factor corresponding to
  $\sum t_1^n(p^\ell) u^\ell$ seems most basic to $GL_n$ and that the
  Euler factor corresponding to $\sum t^n_k(p^\ell) u^\ell$ has degree
  $\binom{n}{k}$ perhaps suggests a correspondence between forms on
  $GL_n$ and forms on $GL_{\binom{n}{k}}$ via the exterior power
  $L$-series.
\end{remark}

\section{The Spherical Map}

Now that we have defined interesting Hecke operators in the polynomial
ring, we need to invert the spherical map to characterize their
``classical'' counterparts.

Let $\G = GL_n(\Z)$ and $G = GL_n(\Z[p^{-1}])$.  Andrianov
\cite{AZbook} defines the spherical map \mbox{$\omega: \G\backslash G
  \to \Q[x_1^{\pm1},\dots, x_n^{\pm1}]$}, by observing that every
right coset has a unique representative in upper-triangular form with
powers of $p$ on the diagonal, and entries in any given column taken
modulo the corresponding diagonal entry (analogous to Hermite normal
form). On such representatives he defines the map (extending linearly)
by:

\[
\omega\left(\G
  \begin{pmatrix}
    p^{b_1}&* & *\\ \vdots & \ddots& *\\ 0&\dots&p^{b_n}
  \end{pmatrix}
\right)= p^{-m}x_1^{b_1}x_2^{b_2}\cdots x_n^{b_n}
\]
where $\sum i b_i = m$ and the $b_i$ are integers. He later proves
that the restriction of $\omega$ to $H_p$ (viewing the double coset as
a sum of right cosets) gives an isomorphism onto the ring of symmetric
polynomials $\Q[x_1^{\pm1},\dots, x_n^{\pm1}]^{S_n}$.

\subsection{Inverting the spherical map}

As we mentioned earlier, we may assume we are working with the
integral Hecke ring $\underline H_p$, so we need to analyze the image
of the spherical map on double cosets $\G \xi \G$ with $\xi \in
GL_n(\Z[p^{-1}]) \cap M_n(\Z)$, $\det(\xi) = p^n$.  Restricted to such
double cosets, the spherical map has image equal to the space spanned
by symmetric polynomials of total degree $n$.  In fact, with respect
to suitable bases, the restriction of the spherical map to the space
of cosets spanned by double cosets $\G \xi \G$ with $\det(\xi) = p^n$
has a matrix which is upper triangular.  Given this fact, the
spherical map is (at least in principle) easy to invert.  That the
spherical map has this property could no doubt be deduced from the
analogous result regarding representations of $p$-adic groups
(\cite{Satake},\cite{Cartier-Corvallis}), but our method of proof is
also computationally valuable, so we include it here, especially since
we need to find some explicit inverse images.



Let $m$ and $n$ be positive integers and as above, denote by $P_n(m)$
the set of partitions of $m$ into $n$ pieces with a lexicographic
ordering on $P_n(m)$, Note that in particular, the elements of
$P_n(m)$ are linearly ordered.

By standard elementary divisor theory, for any $g \in GL_n(\Z[p^{-1}])
\cap M_n(\Z)$, with $\det g = p^m$, $\G g \G = \G\diag(p^{b_1},
p^{b_2}, \dots, p^{b_n}) \G$ for precisely one $\bb =
(b_1,b_2,\ldots,b_n) \in P_n(m)$.  For simplicity, write $\G p^\bb \G$
for $\G\diag(p^{b_1}, p^{b_2}, \dots, p^{b_n}) \G$, and write
$\Sym_n(x^\bb)$ for $\Sym_n(x_1^{b_1}\cdots x_n^{b_n})$, the
symmetrized polynomial under the action of the symmetric group.

Now fix the integer $m$, and list the elements of $P_n(m)$ in order:
$P_n(m) = \{ \ba_1 < \ba_2 < \cdots < \ba_t\}$.  Consider the double
cosets $\{ \G p^{\ba_i} \G\}$ and the symmetric polynomials $\{
\Sym_n(x^{\ba_i}) \}$ with an ordering induced from the one on
$P_n(m)$.  Each set is a linearly independent subset of the
appropriate vector space.  Consider the restriction of Andrianov's
spherical map $\omega: H_p \to \Q[x_1^{\pm1}, \dots, x_n^{\pm1}]$, to
the subspace generated by the $\{ \G p^{\ba_i} \G\}$.  The image of
the spherical map restricted to this domain is spanned by the $\{
\Sym_n(x^{\ba_i}) \}$, and we show below that the matrix of this
linear map is upper triangular.  We begin with a proposition of use in
its own right.

\begin{prop}\label{prop:doublecosetreps}
  Let $n\ge 2$ and $m \ge 1$ be integers, and let $\bb \in P_n(m)$.
  Then
\[
\G p^\bb \G = \bigcup\ \G
  \begin{pmatrix}
    p^{a_1}&* & *\\ \vdots & \ddots& *\\ 0&\dots&p^{a_n} 
  \end{pmatrix} \textrm{ (disjoint)}
\]
where, in any given column, the entries are taken modulo the
corresponding diagonal entry.  Moreover, for some $\sigma \in S_n$,
$\ba = (a_{\sigma(1)}, \dots, a_{\sigma(n)}) \in P_n(m)$ and $\ba \le
\bb$.
\end{prop}

\begin{proof} All but the last statement is established in Lemma 2.7
  of \cite{AZbook}.  Since all diagonal entries are non-negative powers of
  $p$ and any right coset representative must have the same
  determinant as $p^\bb$, it follows that there is a $\sigma \in S_n$
  with $\ba = (a_{\sigma(1)}, \dots, a_{\sigma(n)}) \in P_n(m)$.  For
  any right coset occurring in the decomposition of the double coset
  $\G p^\bb \G$, we must have that the corresponding double cosets
  agree, that is $\G
  \begin{pmatrix}
    p^{a_1}&* & *\\ \vdots & \ddots& *\\ 0&\dots&p^{a_n}
  \end{pmatrix}\G = \G p^\bb\G$. From Theorem II.10 of
  \cite{Newman}, this is true if and only if the two matrices,
  $\begin{pmatrix} p^{a_1}&* & *\\ \vdots & \ddots& *\\ 
    0&\dots&p^{a_n}
  \end{pmatrix}$ and $\diag(p^{b_1}, p^{b_2}, \dots, p^{b_n})$,
  have the same determinantal divisors. Recall that the $k$th
  determinantal divisor is simply the gcd of the determinants of all
  $k\times k$ sub-matrices of the matrix, and that the quotients of
  successive determinantal divisors yield the elementary divisors.  By
  a $k\times k$ sub-matrix we mean the matrix obtained by fixing $k$
  row and $k$ column indices.  To proceed we need a small

\begin{lemma}
  Let $\ba = (a_1, \dots, a_n)$, $\bb = (b_1, \dots, b_n) \in P_n(m)$
  with $\bb < \ba$.  There is a smallest index $k$ for which $b_k >
  a_k$. For this $k$, $\ds \sum_k^n b_i > \sum_k^n a_i$.
\end{lemma}

\begin{proof}
  Since $\bb < \ba$ there is an index $i_0 \ge 1$ such that $b_i =
  a_i$ for $i < i_0$, and $b_{i_0} < a_{i_0}$.  Thus $\ds \sum_1^{i_0}
  b_i < \sum_1^{i_0} a_i$.  But since $\ba, \bb \in P_n(m)$,
  $\ds\sum_1^n a_i = \sum_1^n b_i$, so there must exist a smallest
  index $k$ (clearly $> i_0$) for which $b_k > a_k$.  It follows that
  $\ds \sum_1^{k-1}b_i < \sum_1^{k-1} a_i$, so that $\ds \sum_k^n b_i
  > \sum_k^n a_i$, as required.
\end{proof}

To complete the proof of the proposition, we have that
$\G\begin{pmatrix} p^{a_1}&* & *\\ \vdots & \ddots& *\\
  0&\dots&p^{a_n}
  \end{pmatrix}$ is a right coset representative of $\G p^\bb \G$, and
  that $\ba = (a_{\sigma(1)}, \dots, a_{\sigma(n)}) \in P_n(m)$ for
  some $\sigma \in S_n$.  Suppose to the contrary that $\ba > \bb$.
  Recall we have $m \ge b_1 \ge \cdots \ge b_n \ge 0$ and $m \ge
  a_{\sigma(1)} \ge \cdots \ge a_{\sigma(n)} \ge 0$.  Denote by
  $d_k^\bb$ and $d_k^\ba$ the $k$th determinantal divisor of
  $\diag(p^{b_1}, \dots, p^{b_n})$ and
  $\begin{pmatrix} p^{a_1}&* & *\\ \vdots & \ddots& *\\
    0&\dots&p^{a_n}
  \end{pmatrix}$ respectively.
  
  The determinantal divisors of $\diag(p^{b_1}, \dots, p^{b_n})$ are
  clear: $d_1^\bb = p^{b_n}$, $d_2^\bb = p^{b_n + b_{n-1}}$, \dots,
  $d_n^\bb = p^{b_n + \cdots + b_1}$.  Since we are assuming that $\ba
  > \bb$, we have by the lemma a smallest index $k$ for which $b_k >
  a_{\sigma(k)}$, and hence for which $\ds d^\bb_{n+1-k} = p^{\sum_k^n
    b_i} > p^{\sum_k^n a_{\sigma(i)}}$.  However $d_{n+1-k}^\ba$ is
  the gcd of all possible determinants of $(n+1-k) \times (n+1-k)$
  sub-matrices, and choosing rows and columns $\sigma(k), \dots ,
  \sigma(n)$ for one such matrix produces a determinant smaller than
  $d_{n+1-k}^\bb$, thus showing that $d_{n+1-k}^\ba < d_{n+1-k}^\bb$, a
  contradiction.
\end{proof}
 
We now have a characterization of the spherical map.

\begin{thm}\label{thm:uppertriangular}
  The matrix of the spherical map, restricted as above, is nonsingular
  and upper triangular.  In particular, if $\bb \in P_n(m)$, then
  \[
  \omega(\G p^\bb \G) = \sum_{\ba \in P_n(m),\ \ba \le \bb} c(\ba)
  \Sym_n(x^\ba)
\]
for constants $c(\ba)$ with $c(\bb) \ne 0$.
\end{thm}

\begin{proof}
  It is enough to consider $\omega$ restricted to the integral subring
  $\underline H_p^n$. We compute the image of $\omega$ on the double
  coset $\G \diag(p^{b_1},\dots,p^{b_n})\G$ for $\bb = (b_1, \dots,
  b_n) \in P_n(m)$.  From Proposition~\ref{prop:doublecosetreps}
  above, the double coset is a
  union of right cosets $\G\begin{pmatrix} p^{a_{\sigma(1)}}&* & *\\
    \vdots & \ddots& *\\ 0&\dots&p^{a_{\sigma(n)}}
  \end{pmatrix}$ where $\ba=(a_1, \dots, a_n) \in P_n(m)$, $\ba \le \bb$, and $\sigma
  \in S_n$. By definition of the spherical map \cite{AZbook},
\[
\omega\left(
  \G\begin{pmatrix} p^{a_{\sigma(1)}}&* & *\\ \vdots & \ddots& *\\
    0&\dots&p^{a_{\sigma(n)}}\end{pmatrix} \right) = p^{-\sum i
  a_{\sigma(i)}}x_1^{a_{\sigma(1)}} \cdots x_n^{a_{\sigma(n)}},
\]
so to understand the image of a double coset, we obviously need to
count the number of such right cosets occur in $\G p^\bb \G$.  Denote
this number by $\psi^\sigma_{\ba,\bb}$.

Then
\[\omega(\G p^\bb \G) = \sum_{\ba\le\bb,\ \ba \in P_n(m)}\  
\sum_{\sigma^{-1} \in S_n/\Stab(x^\ba)} \psi^\sigma_{\ba,\bb}\,
p^{-\sum i a_{\sigma(i)}} \, x_1^{a_{\sigma(1)}} \cdots
x_n^{a_{\sigma(n)}}
\]
Note that the inner sum is over $\sigma^{-1}$ and not $\sigma$ since
if $\sigma(i)=j$, then $x_i^{a_{\sigma(i)}} =
x_{\sigma^{-1}(j)}^{a_j}$.  Moreover, the sum over $\sigma^{-1}$ is
not generally the same as the sum over $\sigma$ unless $\Stab(x^\ba)$
is a normal subgroup of $S_n$.


On the other hand, $\omega(\G p^\bb \G)$ is
a symmetric polynomial, so that $\psi^\sigma_{\ba,\bb}\, p^{-\sum i
  a_{\sigma(i)}} = \psi^\tau_{\ba,\bb}\, p^{-\sum i a_{\tau(i)}}$ for
any two $\sigma, \tau \in S_n$.  In particular, $\psi_{\ba,\bb}^\sigma
= p^{\sum i(a_{\sigma(i)} - a_i)} \psi_{\ba,\bb}^1$, where $1$ denotes
the identity permutation, and consequently

\begin{equation}\label{eq:sphercial}\omega(\G p^\bb \G) = \sum_{\ba\le\bb,\ \ba \in P_n(m)} 
\psi^1_{\ba,\bb}\, p^{-\sum i a_i}
\, \Sym_n(x^\ba).
\end{equation}
Thus $c(\ba) = \psi^1_{\ba,\bb}\, p^{-\sum i a_i}$, and it is clear
that $c(\bb) \ne 0$ since one of the right coset representatives of
the double coset is the diagonal matrix
$\diag(p^{b_1},\dots,p^{b_n})$.
\end{proof}

\begin{remark}
  Of course the coefficients $c(\ba) = \psi^\sigma_{\ba,\bb}\,
  p^{-\sum i a_{\sigma(i)}}$ for any $\sigma \in S_n$, however
  $\psi^1_{\ba,\bb}$ is the easiest to compute since we are assuming
  the right coset has the form $\begin{pmatrix} p^{a_1}&* & *\\ \vdots
    & \ddots& *\\ 0&\dots&p^{a_n}
  \end{pmatrix}$
  with $a_1 \ge a_2 \ge \cdots \ge a_n$, and where the entries in the
  $i$th column can be taken modulo $p^{a_i}$, and hence this case
  presents the fewest right cosets to analyze.
  
  In any real computation, this simplification represents an enormous
  saving of time, and to some extent complexity since determining
  which right cosets belong to a given double coset involves computing
  determinantal divisors.  We note that the complexity of the problem
  depends not so much on $n$ or $m$, but on the cardinality of
  $P_n(m)$ which grows extremely fast with $m$ even for small $n$.
\end{remark}

\begin{cor}
  The matrix of the inverse of the spherical map, restricted as above,
  is upper triangular.  In particular, if $\bb \in P_n(m)$, then
  \[
  \Sym(x^\bb) = \sum_{\ba \in P_n(m),\ \ba \le \bb} d(\ba)\omega (\G
  p^\ba \G)
\]
for constants $d(\ba)$.
\end{cor}


\begin{remark}\label{rmk:simplifications}
  We point out three further observations which can be useful for
  simplifying such computations.  Let $P_n(m) = \{\bl_1, \dots,
  \bl_r\}$ and assume that $\bl_1 < \cdots < \bl_r$.  If
  $\bl_i=(\lambda_{i,1}, \dots, \lambda_{i,n})$, let $\chi(\bl_i,
  \bl_j)$ denote the number of right cosets of the form
  $\G\begin{pmatrix} p^{\lambda_{i,\sigma(1)}}&* & *\\ \vdots &
    \ddots& *\\ 0&\dots&p^{\lambda_{i,\sigma(n)}}
  \end{pmatrix}$
  which occur in the double coset $\G p^{\bl_j}\G$, as $\sigma$ ranges
  over $S_n$.  From the discussion above, we know $\chi(\bl_i,\bl_j) =
  0$ whenever $\bl_i > \bl_j$.
  
  Certainly, we already know that for any $\ba, \bb \in P_n(m)$
  ($\ba \le \bb$)
\begin{equation}\label{eq:chi-psi}
\chi(\ba,\bb) = \sum_{\sigma^{-1} \in S_n/\Stab(x^\ba)}
\psi^\sigma_{\ba,\bb} = \sum_{\sigma^{-1} \in S_n/\Stab(x^\ba)}
 p^{\sum
  i(a_{\sigma(i)} - a_i)} \psi_{\ba,\bb}^1,
\end{equation}
but we also have the following relations:
\begin{enumerate}
\item $\mu(\bl_j) = \chi(\bl_1, \bl_j) + \cdots + \chi(\bl_j, \bl_j)$ is the
degree (i.e., the total number of right cosets in the double coset) of
the double coset $\G p^{\bl_j} \G$ which is explicitly computed in
Lemma 6 of \cite{Andrianov70}.

\item $\chi(\bl_j, \bl_j)
+ \cdots + \chi(\bl_j, \bl_r)$ is the total number of possible right
cosets with $p^{\lambda_{j,k}}$'s ($j$ fixed) on the diagonal.  This
is trivial to compute since for any given upper triangular matrix, all
the entries above are taken modulo the diagonal entry in that column.

\item Refining the previous item, we have that
\[
\psi^1_{\bl_j, \bl_j} + \cdots + \psi^1_{\bl_j, \bl_r} = p^\ell,\quad
 \ell = {\sum_{k=1}^r \lambda_{j,k} (k-1)}
\]
\end{enumerate}

 

The first two observations produce $2r$ linear relations among the
$r(r+1)/2$ variables $\chi(\bl_i, \bl_j)$.  These together with the
third observation will be used in the example below to further reduce
the computational load of inverting the spherical map.

\end{remark}




\subsection{Some Examples}
In this section we provide two examples one trivial, one not, which
will be used to define our new families of Hecke operators for $GL_4$,
and which demonstrate the observations given in the previous section.
We shall compute the inverse image (under the spherical map) of
$t_1^4(p^2)$ and $t_2^4(p^2)$.


We consider first the almost trivial case of $t_1^4(p^2)$ which
requires we analyze matrices of determinant $p^2$. $P_4(2)=\{\bl_1,
\bl_2\}$ where $\bl_1 = (1,1,0,0)< \bl_2=(2,0,0,0)$.  To ease the
notation, we write $\psi_{ij}$ for $\psi^1_{\bl_i,\bl_j}$ in both
examples below, and denote the $k$th determinantal divisor
$d_k^{\bl_i}$ as $d_k^i$.

There is only one right coset with diagonal $(p^2,1,1,1)$ since all the
upper triangular entries are read modulo the diagonal entry, so it is
clear that $\psi_{2,2} = 1$.  The right cosets of the form
$\G\left(\begin{smallmatrix}p&a&0&0\\&p&0&0\\&&1&0\\&&&1\end{smallmatrix}\right)$
belong either to the double coset $\G\diag(p,p,1,1)\G$ or to
 $\G\diag(p^2,1,1,1)\G$, and the third determinantal divisor easily
determines which for us, with $d_3^1 = p$ while $d_3^2 = 1$.  For a
typical coset we see that the third determinantal divisor is $d_3 =
\gcd(p, a)$.  Now $a$ runs mod $p$, so we immediately see that
$\psi_{11} = 1$ and $\psi_{12}=p-1$.

Recall from equation~\ref{eq:sphercial}, $\omega(\G p^\bb \G) =
\sum_{\ba\le\bb,\ \ba \in P_n(m)} \psi^1_{\ba,\bb}\, p^{-\sum i a_i}
\, \Sym(x^\ba)$.  Let $A = (a_{ij})$ denote the (upper triangular)
matrix of the (restricted) spherical map, so $\omega(\G p^{\bl_j} \G)
= \sum_{i\le j} a_{ij} \Sym(x^{\bl_i})$.  If $\bl_i = (\lambda_{i,1},
\dots, \lambda_{i,n})$, then $a_{ij} = p^{-\sum k
  \lambda_{i,k}}\psi_{ij}$, thus $A= \left(
\begin{smallmatrix}
  p^{-3}&p^{-3}(p-1)\\0&p^{-2}
\end{smallmatrix}\right)$.

It's inverse is $A^{-1} = \left(\begin{smallmatrix}
    p^{3}&-p^2(p-1)\\0&p^{2}
\end{smallmatrix}\right)$, that is
\begin{align*}
  \omega^{-1}(\Sym_4(x^{\bl_1})) &= \omega^{-1}(\Sym_4(x_1x_2)) =
  p^3\G\diag(p,p,1,1)\G\textrm{ and}\\
  \omega^{-1}(\Sym_4(x^{\bl_2})) &= \omega^{-1}(\Sym_4(x_1^2)) =
  p^2\G\diag(p^2,1,1,1)\G - p^2(p-1)\G\diag(p,p,1,1)\G
\end{align*}

Since $t_1^4(p^2) = \Sym_4(x_1^2) + \Sym_4(x_1x_2)$, it follows that
\[
\omega^{-1}(t_1^4(p^2)) = p^2\G\diag(p^2,1,1,1)\G +
p^2 \G\diag(p,p,1,1)\G.
\]

\medbreak Next we consider the harder example of $t_2^4(p^2)$.  This
requires that we study matrices of determinant $p^4$.  $P_4(4) =
\{\bl_1, \dots,\bl_5\}$, where $\bl_1=(1,1,1,1) < \bl_2=(2,1,1,0) <
\bl_3=(2,2,0,0) < \bl_4=(3,1,0,0) < \bl_5=(4,0,0,0)$.  We need to
compute $\psi_{ij}$ for $1\le i\le j \le 5$.  For $2\le i \le 5$, the
$\psi_{ij}$ are computed by means of analyzing determinantal divisors.
For example for $i=2$, the task is to determine how many right cosets
$\G\left(
\begin{smallmatrix}p^2&a&b&0\\&p&c&0\\&&p&0\\&&&1
\end{smallmatrix}\right)$ (with $a, b, c\pmod p$) belong to the double
cosets $\G\diag(p^2,p,p,1)\G$, $\G\diag(p^2,p^2,1,1)\G$,
$\G\diag(p^3,p,1,1)\G$, and $\G\diag(p^4,1,1,1)\G$.  These numbers
will be (respectively) $\psi_{22}, \psi_{23},\psi_{24}, \psi_{25}$.
If we put $\Psi = (\psi_{ij})$, then careful counting with
determinantal divisors gives us all of $\Psi$ except the first row:

\[
\Psi =
\begin{pmatrix}
  *&*&*&*&*\\&1&p-1&2p(p-1)&p(p-1)^2\\&&1&p-1&p(p-1)\\&&&1&p-1\\&&&&1
\end{pmatrix}
\]

Note that as a reality check, the row sums are consistent with
observation 3 in Remark~\ref{rmk:simplifications}.  In principle, the
first row could also be computed in this fashion although a quick
inspection of the problem reveals a myriad of cases which must be
analyzed to determine the determinantal divisors.  So instead, we take
advantage of the other observations we made in the previous section.

First we observe by equation~\ref{eq:chi-psi}, that $\chi_{1,j} =
\psi_{1j}$ since the stabilizer of $x^{\bl_1} = x_1x_2x_3x_4$ is all
of $S_4$.  Then we note from Remark~\ref{rmk:simplifications} that
$\chi_{1,j} = \mu(\bl_j) - \chi_{2,j} - \chi_{3,j} - \dots -
\chi_{j,j}$, where the $\mu(\bl_j)$ is the degree (i.e., the total
number of right cosets in the double coset) of the double coset $\G
p^{\bl_j} \G$ which is explicitly computed in Lemma 6 of
\cite{Andrianov70}. Unfortunately, we must also compute the
$\chi_{i,j}$ ($i>1$) from our knowledge of the $\psi_{ij}$ using
equation~\ref{eq:chi-psi}.  If we define $\nu(\bl_i)$ by $\chi_{i,j} =
\nu(\bl_i)\psi_{ij}$, we compute that

% \begin{alignat*}{3}
%  \null\quad &\mu(\bl_i)\quad&\hfil\nu(\bl_i)\hfill\\
%   \bl_2\quad&p(p^2+p+1)(p^2+1)(p+1)\quad&1+2p+3p^2+3p^4+2p^5+p^6\\
%   \bl_3\quad&p^4(p^2+p+1)(p^2+1)\quad&1+p^2+2p^4+p^6+p^8\\
%   \bl_4\quad&p^5(p^+p+1)(p^2+1)(p+1)\quad&1+p+2p^2+p^4+2p^5+p^6+2p^8+p^9+p^{10}\\
%   \bl_5\quad&p^9(p^2+1)(p+1)\quad&1+p^4+p^8+p^{12}
% \end{alignat*}

\newcolumntype{Y}{>{\centering\arraybackslash}X}

\renewcommand{\arraystretch}{1.5}
\begin{center}
\begin{tabularx}{6in}
  {|>{\small}Y | >{\small} Y | >{\small}Y |}
  \hline &$\mu(\bl_i)$&$\nu(\bl_i)$\\
  \hline $\bl_2$& $p(p^2+p+1)(p^2+1)(p+1)$ & $1+2p+3p^2+3p^4+2p^5+p^6$\\
  \hline $\bl_3$ &$p^4(p^2+p+1)(p^2+1)$ &$1+p^2+2p^4+p^6+p^8$\\
  \hline $\bl_4$ &$p^5(p^2+p+1)(p^2+1)(p+1)$ &$1+p+2p^2+p^4+2p^5+p^6+2p^8+p^9+p^{10}$\\
  \hline $\bl_5$ &$p^9(p^2+1)(p+1)$ &$1+p^4+p^8+p^{12}$\\
  \hline
\end{tabularx}
\end{center}


Putting this together we conclude:
\begin{align*}
  \psi_{11} &= \chi_{11} = 1\\
  \psi_{12} &= \chi_{12} = 3p^3 - p^2 -p -1\\
  \psi_{13} &= \chi_{13} = 2p^4 - 3p^3 + p\\
  \psi_{14} &= \chi_{14} = 3p^5 - 5p^4 + p^3 + p^2\\
  \psi_{15} &= \chi_{15} = p^6 - 3p^5 + 3p^4 - p^3\\
\end{align*}

Now if $A = (a_{ij})$ once again represents the matrix of the
spherical map with respect to the canonical bases, then $a_{ij} =
p^{-\sum k \lambda_{i,k}}\cdot \psi_{ij}$, and we have

\[
A = \left(
\begin{array}{ccccc}
{\ds \frac {1}{p^{10}}}  & {\ds \frac {3\,p^{
3} - p^{2} - p - 1}{p^{10}}}  & {\ds \frac {2\,p^{4} - 
3\,p^{3} + p}{p^{10}}}  & {\ds \frac {3\,p^{5} - 5\,p^{
4} + p^{3} + p^{2}}{p^{10}}}  & {\ds \frac {p^{6} - 3\,
p^{5} + 3\,p^{4} - p^{3}}{p^{10}}}  \\ [2ex]
0 & {\ds \frac {1}{p^{7}}}  & {\ds \frac {p
 - 1}{p^{7}}}  & {\ds \frac {2p(p-1)}{p^{7}}}  & 
{\ds \frac {p(p - 1)^{2}}{p^{7}}}  \\ [2ex]
0 & 0 & {\ds \frac {1}{p^{6}}}  & {\ds 
\frac {p - 1}{p^{6}}}  & {\ds \frac {p(p - 1)}{p^{6}}}  \\
 [2ex]
0 & 0 & 0 & {\ds \frac {1}{p^{5}}}  & {\ds 
\frac {p - 1}{p^{5}}}  \\ [2ex]
0 & 0 & 0 & 0 & {\ds \frac {1}{p^{4}}} 
\end{array}
\right)
\]

It's inverse $A^{-1}$ is:
\[\small
\left(
\begin{array}{ccccc}
p^{10}& -p^7(3p^3-p^2-p-1)& p^6(p^4-p^3-p+1)&
p^5(2p^5-p^4-2p^3+1)& -p^4(p^6-p^5-p^4+p^2+p-1)\\
0& p^7&-p^6(p-1)& -p^5(p-1)(p+1)& p^4(p-1)^2(p+1)\\
0& 0& p^6&-p^5(p-1)& -p^4(p-1)\\
0& 0& 0& p^5& -p^4(p-1)\\
0& 0& 0& 0&p^4\\
\end{array}
\right)
\]

We compute that $t_2^4(p^2) = \Sym_4(x^{\bl_3}) + \Sym_4(x^{\bl_2}) +
3\Sym_4(x^{\bl_1})$, so
\begin{align*}
  \omega^{-1}(t_2^4(p^2)) &= [(p^4-p^3+1-p)p^6\G p^{\bl_1}\G
  -p^6(p-1)\G
  p^{\bl_2}\G +  p^6\G p^{\bl_3}\G]\\
  &\phantom{xx}+[ -p^7(3p^3-p^2-p-1)\G p^{\bl_1}\G + p^7\G p^{\bl_2}\G
  ]
  + 3[p^{10}\G p^{\bl_1}\G]\\
  &= p^6\left[ (p^4+p^2+1)\G p^{\bl_1}\G + \G p^{\bl_2}\G + \G
    p^{\bl_3}\G\right]\\
  &= p^6\left[ (p^4+p^2+1)\G\diag(p,p,p,p)\G + \G\diag(p^2,p,p,1)\G +
    \G \diag(p^2,p^2,1,1)\G\right]
\end{align*}






\section{New Families of Hecke Operators}

Recall that we have defined families of polynomial Hecke operators
\[
t_k^n(p^\ell) = h^{\binom{n}{k}}(\ell)\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1 x_2\cdots x_k)\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)}}
\]
 where $\ds h^r(\ell)=
\sum_{\bj \in P_r(\ell)} \Sym_r(z_1^{j_1} z_2^{j_2} \cdots z_r^{j_r})=
\sum_{\substack{\sum j_k = \ell\\j_k \ge 0}}z_1^{j_1}z_2^{j_2} \cdots
z_r^{j_r}$, and which satisfy
\begin{equation}\label{eqn:eulerproduct}
\sum_{\ell \ge 0} t_k^n(p^\ell) u^\ell = \Big[\prod_{\sigma \in
  S_n/\Stab(x_1 x_2\cdots x_k)} (1 - u\sigma(x_1 x_2\cdots
x_k))\Big]^{-1},
\end{equation}
and $t_k^n(p)$ is the $k$th elementary symmetric polynomial in the
variables $x_1, \dots, x_n$.

One could attempt simply to apply the inverse of the spherical map to
the elements $t_k^n(p^\ell)$, but this is neither practical, nor
particularly illuminating.  Instead, we realize that even for $GL_2$
the Hecke operators are most clearly defined by defining some base
cases and the recursion relations satisfied by the operators, so this
is how we proceed here.


Equation~\ref{eqn:eulerproduct} implicitly contains the recursion
relations for the $t_k^n(p^\ell)$ via

\begin{equation}\left[\sum_{\ell \ge 0} t_k^n(p^\ell) u^\ell \right] \cdot 
\left[\prod_{\sigma \in
  S_n/\Stab(x_1 x_2\cdots x_k)} (1 - u\sigma(x_1 x_2\cdots
x_k))\right] = 1
\end{equation}

In the following sections, we completely characterize $t_1^n(p^\ell)$
and $t_n^n(p^\ell)$ for any $n$.  Then as we saw hinted at in
Example~\ref{ex:GL4}, we investigate a sort of duality between $t_k^n$ and
$t_{n-k}^n$.  The remaining operators are difficult to characterize in
general, but in the last section, we work out the details for $GL_4$.
This is the first really interesting case as it associates an ``Euler
factor'' of degree $ > n$ to a Hecke operator for $GL_n$.


\subsection{The operators $t_1^n(p^\ell)$ and $t_n^n(p^\ell)$}
The case $k=1$ recovers the classical Hecke operators (see \S3.2 of
\cite{Shimurabook}).  In particular, we see that if $s_k =
s_k(x_1,\dots,x_n) = t_k^n(p) $ is the $k$th elementary symmetric
polynomial, then
\[\left[\sum_{\ell \ge 0} t_1^n(p^\ell) u^\ell \right]
\cdot \left[1 - s_1u + s_2 u^2 - s_3 u^3 + \dots+ (-1)^n s_n u^n \right] = 1,
\]
or (letting $t_1^n(p^j) = 0$ for $j < 0$),
\begin{align*}
  t_1^n(p^\ell) &= t_1^n(p^{\ell-1})s_1 - t_1(p^{\ell-2})s_2 + \cdots
  + (-1)^{n-1}t_1^n(p^{\ell-n})s_n\\
  &= t_1^n(p^{\ell-1})t_1^n(p) - t_1(p^{\ell-2})t_2^n(p) + \cdots +
  (-1)^{n-1}t_1^n(p^{\ell-n})t_n^n(p)
\end{align*}

At the other extreme are the operators $t_n^n$ which satisfy
$\sum_{\ell \ge 0}t_n^n(p^\ell) u^\ell = [1 - u x_1\cdots x_n]^{-1}$
which is the sum of a geometric series, yielding
\[
t_n^n(p^\ell) = t_n^n(p)^\ell = [s_n(x_1, \dots, x_n)]^\ell
\]









\subsection{Relations between $t_k^n$ and $t_{n-k}^n$}
We note that there is a sort of duality between $t_k^n(p^\ell)$
and $t^n_{n-k}(p^\ell)$.  Since $\binom{n}{k} = \binom{n}{n-k}$, we
see that $\ds t_{n-k}^n(p^\ell) = h^{\binom{n}{n-k}}\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1 x_2\cdots x_{n-k})\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_{n-k})}} =
h^{\binom{n}{k}}\Big\vert_
{\substack{z_i \mapsto x_1x_2\cdots x_n/\sigma_i(x_1 x_2\cdots x_{k})\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_{k})}}$
while $t_k^n(p^\ell) = h^{\binom{n}{k}}(\ell)\Big\vert_
{\substack{z_i \mapsto \sigma_i(x_1 x_2\cdots x_k)\\
    \sigma_i \in S_n/\Stab(x_1 x_2\cdots x_k)}}$.

The correspondence is most easily seen between $t_1$ and $t_{n-1}$. An
elementary computation shows
\[
s_n(x_1, \dots, x_n) \cdot s_k(x_1^{-1}, \dots, x_n^{-1}) =
s_{n-k}(x_1, \dots, x_n)
\]
and hence (letting $s_n = s_n(x_1, \dots, x_n) = x_1\cdots x_n$),
\[s_k(s_n/x_1, \dots, s_n/x_n) = s_n^{k-1}(x_1,\dots,x_n) \cdot
s_{n-k}(x_1, \dots, x_n)
\]


Thus,
\begin{align*}
  \sum_{\ell \ge 0} t_{1}^n(p^\ell) u^\ell &= \left[\prod_{\sigma \in
      S_n/\Stab(x_1)} (1 - u\sigma(x_1))\right]^{-1} \\
  &=\left[1 - u s_1(x_1, \dots, x_n) + u^2 s_2(x_1, \dots, x_n) +
    \cdots + (-1)^n u^n s_n(x_1, \dots, x_n)\right]^{-1}
\end{align*}
while for $t_{n-1}^n$ we have

\begin{equation*}
\begin{split}
  \sum_{\ell \ge 0} t_{n-1}^n(p^\ell) u^\ell &=\Big[\prod_{\sigma \in
    S_n/\Stab(x_1 x_2\cdots x_{n-1})} (1 - u\sigma(x_1 x_2\cdots
  x_{n-1}))\Big]^{-1}\\
  &= \left[\prod_{\sigma \in
      S_n/\Stab(x_1)} (1 - u s_n/\sigma(x_1))\right]^{-1} \\
  &=\left[\prod_{j=1}^n (1 - u s_n/x_j)\right]^{-1}\\
  &=\left[1 - u s_1(s_n/x_1, \dots, s_n/x_n) + \cdots +
    (-1)^n u^n s_n(s_n/x_1, \dots, s_n/x_n)\right]^{-1}\\
  &=[1 - u s_{n-1}(x_1, \dots, x_n) + u^2 s_n(x_1, \dots, x_n)s_{n-2}(x_1, \dots, x_n) \\
  &\phantom{xxxx}-u^3s_n^2(x_1, \dots, x_n)s_{n-3}(x_1, \dots, x_n)+
  \cdots + (-1)^n u^n s_n^{n-1}(x_1, \dots, x_n) ]^{-1},
\end{split}
\end{equation*}
showing the nice symmetry.


\dr{Is there really one for other pairs?  May need to go to $n = 6$ to
  see}


\subsection{The recursion relations for Hecke operators on $GL_2$, $GL_3$, $GL_4$}

Having done what we can for the operators for general $n$, here we
work out all the recursion relations defining the Hecke operators on
$GL_n$ for $n = 2, 3, 4$.  For small $n$ there is little choice in how
these relations are written down, but as we shall see for $n \ge 4$,
it is possible to write down multiple recursion relations which appear
quite distinct. $GL_4$ is the first really interesting case, however for
completeness we write down the recursion relations for $n= 2, 3, 4$.
Throughout, we put $t_k^n(p^j) = 0$ for $j < 0$ (and $t_k^n(1) = 1$ by
definition), and recall $t_n^n(p^\ell) = t_n^n(p)^\ell$.


The relations for the case $n = 2$ simply consists of a base case and
one relation:
\begin{align*}
  &t_k^2(p) = s_k(x_1, x_2), \quad k=1,2\\
  &t_1^2(p^\ell)= t_1^2(p^{\ell-1})t_1^2(p) -
  t_1^2(p^{\ell-2})t_2^2(p)
\end{align*}


For $n=3$, we have a base case and two nontrivial relations:
\begin{align*}
  &t_k^3(p) = s_k(x_1, x_2,x_3), \quad k=1,2,3\\
  &t_1^3(p^\ell) = t_1^3(p^{\ell-1})t_1^3(p) -
  t_1^3(p^{\ell-2})t_2^3(p) + t_1^3(p^{\ell-3})t_3^3(p)\\
  &t_2^3(p^\ell) = t_2^3(p^{\ell-1})t_2^3(p) -
  t_2^3(p^{\ell-2})t_1^3(p)t_3^3(p) + t_2^3(p^{\ell-3})t_3^3(p)^2
\end{align*}


For $n=4$, we analogously have a base case and three nontrivial
relations.  The relations for $t_1^4(p^\ell)$ and $t_3^4(p^\ell)$ are
fairly easy to write down from what we have noted above, but the
relation for $t_2^4(p^\ell)$ is somewhat more involved, and indeed is
much more suggestive of the general case $t_k^n(p^\ell)$ for $n > 4$.
The relations for $t_2^4(p^\ell)$ are deduced from the identity:
\[
[\sum_{\ell \ge 0} t_2^4(p^\ell) u^\ell ]\cdot [(1 -
ux_1x_2)(1-ux_1x_3)(1-ux_1x_4)(1-ux_2x_3)(1-ux_2x_4)(1-ux_3x_4)] = 1
\]

To write things in closed from, we adopt some ad hoc notation to
simplify the exposition.  Let $t(a_1, a_2, a_3, a_4)$ be the symmetric
polynomial $\Sym_4(x_1^{a_1}x_2^{a_2}x_3^{a_3}x_4^{a_4})$.  Then
\begin{equation*}
\begin{split}
  (1 -&ux_1x_2)(1-ux_1x_3)(1-ux_1x_4)(1-ux_2x_3)(1-ux_2x_4)(1-ux_3x_4) =\\
  &1 - u t(1,1,0,0) + u^2(t(2,1,1,0) + 3t(1,1,1,1)) -u^3(t(3,1,1,1) +
  t(2,2,2,0) + 2t(2,2,1,1)) \\
  &+u^4(t(3,2,2,1) + 3t(2,2,2,2)) -u^5 t(3,3,2,2) + u^6 t(3,3,3,3)\\
\end{split}
\end{equation*}

There are often many ways to represent the coefficients above in terms
of the $t_k^4(p^\ell)$.  For example, the coefficient of $u^3$ is


\begin{align*}
  &t(3,1,1,1) + t(2,2,2,0) + 2t(2,2,1,1) =\\
  &t_4^4(p)t_1^4(p^2) + 2t_3^4(p^2) - t_3^4(p)^2 = \\
  &t_4^4(p)(t_1^4(p^2) + t_3^4(p))
\end{align*}
and the shape of the recursion relations we write down depends upon
the choice of the representation. Here we opt for simplicity, and
express
\begin{equation*}
\begin{split}
  (1 -&ux_1x_2)(1-ux_1x_3)(1-ux_1x_4)(1-ux_2x_3)(1-ux_2x_4)(1-ux_3x_4) =\\
  &1 - u t_2^4(p) + u^2 (t_2^4(p)^2 - t_2^4(p^2)) - u^3
  t_4^4(p)(t_1(p^2) + t_3^4(p))\\
  & +u^4 t_4^4(p) (t_2^4(p)^2 - t_2^4(p^2)) - u^5 t_4^4(p)^2 t_2^4(p)
  + u^6 t_4^4(p)^3
\end{split}
\end{equation*}
From this we deduce the recursion relations for $n=4$:
\begin{equation*}
  \begin{split}
    &t_k^4(p) = s_k(x_1, x_2,x_3, x_4), \quad k=1,2,3,4\\
    &t_1^4(p^\ell) = t_1^4(p^{\ell-1})t_1^4(p) -
    t_1^4(p^{\ell-2})t_2^4(p) + t_1^4(p^{\ell-3})t_3^4(p) -
    t_1^4(p^{\ell-4})t_4^4(p)\\
    &t_2^4(p^\ell) = t_2^4(p^{\ell-1})t_2^4(p) -
    t_2^4(p^{\ell-2})(t_2^4(p)^2 -t_2^4(p^2))
    + t_2^4(p^{\ell-3})t_4^4(p)(t_1^4(p^2) + t_3^4(p)) \\
    &\phantom{xxxxx}- t_2^4(p^{\ell-4})t_4^4(p)(t_2^4(p)^2 -
    t_2^4(p^2))
    +t_2^4(p^{\ell-5}) t_2^4(p)t_4^4(p)^2 - t_2^4(p^{\ell-6})t_4^4(p)^3\\
    &t_3^4(p^\ell) = t_3^4(p^{\ell-1})t_3^4(p) -
    t_3^4(p^{\ell-2})t_2^4(p)t_4^4(p) +
    t_3^4(p^{\ell-3})t_1^4(p)t_4^4(p)^2 - t_3^4(p^{\ell-4})t_4(p)^3
  \end{split}
\end{equation*}

\section{The Hecke operators in the classical setting}

We finally reach the point where we can define the classical Hecke
operators which correspond to our polynomial ones.  These are the
operators defined in terms of double cosets which should be of
number-theoretic interest because of their well structured generating
series.


Recall our notation for the
standard generators of $H_p$:
\begin{equation*}
  \pi_k^n(p) = \G
  \diag(\underbrace{1,\ldots,1}_{n-k},\underbrace{p,\ldots,p}_k)\G,\
  \ k = 1, \dots,n
\end{equation*}

We let $T_k^n(p^\ell)$ be the inverse image under the spherical map of
$t_k^n(p^\ell)$.  Since the spherical map $\omega$ is an isomorphism,
the $T_k^n(p^\ell)$ will satisfy exactly the same recursion
relations as the $t_k^n(p^\ell)$.  They will also have a generating
series $\sum T_k^n(p^\ell) u^\ell$ which are rational functions of
exactly the same form as the polynomial Hecke operators, in
particular, $\sum T_k^n(p^\ell) u^\ell = [q_k^n(u)]^{-1}$ where
$q_k^n$ is a polynomial of degree $\binom{n}{k}$.


Finally, we note that to define the operators $T_k^n(p^\ell)$, we need
only the base cases and the images of $t_k^n(p^\ell)$ which occur in
the defining recurrence relations.  So for $n = 2$ or 3, we need only
the images of $t_k^n(p)$, while for $n=4$, we need the images of
$t_k^4(p)$ as well as $t_k^4(p^2)$ for $k = 1, 2$ which were the
examples computed in a previous section.


By Lemma 2.21 of \cite{AZbook}, we have that $T_k^n(p) =
p^{k(k+1)/2}\pi_k^n(p)$, and from our earlier examples we obtained:

\begin{align*}
  T_1^4(p^2) &=  p^2(\G\diag(p^2,1,1,1)\G + \G\diag(p,p,1,1)\G\\
  T_2^4(p^2) &= p^6\left[ (p^4+p^2+1)\G\diag(p,p,p,p)\G +
    \G\diag(p^2,p,p,1)\G + \G \diag(p^2,p^2,1,1)\G\right]
\end{align*}
which, together with the recursion relations of the
last section, characterizes the structure of these
new Hecke operators for $GL_4$.

\begin{remark}
  It may occur to the reader that since the $t_k^n(p^\ell)$ are
  symmetric polynomials, they are polynomials in the elementary
  symmetric polynomials ($t_k^n(p)$), and that the inverse image of
  the elementary symmetric polynomials is known, so why go through the
  trouble we have?  The point is that taking this later approach would
  only define our classical Hecke operators as algebraic combinations
  of double cosets instead of linear combinations, and it was the
  cumbersome nature of computing products of double cosets which led
  us to the polynomial setting in the first place.
\end{remark}




\newpage

%-------------------------------------------
%        Begin References
%-------------------------------------------

\small\singlespace \bibliographystyle{amsplain}
\bibliography{shemanske}

%-------------------------------------------
%        Begin Closing Information
%-------------------------------------------
\address{Department of Mathematics, Bates College, Lewiston,
  Maine 04240}\\
\email {jrhodes@bates.edu} \medskip

\address{Department of Mathematics, 6188 Bradley Hall, Dartmouth
  College, Hanover, NH 03755}\\
\fax{(603) 646-1312}\\
\email {thomas.r.shemanske@dartmouth.edu}\\
\URL{http://www.math.dartmouth.edu/\symbol{126}trs/}

%-------------------------------------------
%        End Closing Information
%-------------------------------------------

\end{document}



\begin{example}
  For $GL_3$, I previously sent you a few computations.  They are
  pretty tedious; you more or less need to analyze simultaneously all
  matrices with determinant $p^m$.  Let $\omega$ denote the spherical
  map as in Andrianov. Then

\noindent For $P_3(2)= \{\ba=(1,1,0) < \ba_2=(2,0,0)\}$, we have:
\begin{align*}
  \omega(T(1,p,p)) &=  p^{-3}\Sym(x_1x_2)\\
  \omega(T(1,1,p^2) &=p^{-2}\Sym(x_1^2) + p^{-3}(p-1)\Sym(x_1x_2)
\end{align*}
so
\begin{align*}
  \Sym(x_1x_2)&=\Sym(x^{\ba_1}) = p^3 \omega(T(1,1,p))\\
  \Sym(x_1^2) &= \Sym(x^{\ba_2}) = p^2\omega(T(1,1,p^2) - p^2(p-1)\omega(T(1,p,p))
\end{align*}

\end{example}

\begin{example}
\noindent For $P_3(3)=\{\ba_1=(1,1,1) < \ba_2=(2,1,0) < \ba_3=(3,0,0)\}$, we have:
\begin{align*}
\omega(T(p,p,p)) &=  p^{-6}\Sym(x_1x_2x_3)\\
  \omega(T(1,p,p^2)) &=  p^{-4}\Sym(x_1^2x_2) + p^{-6}(2p+1)(p-1)\Sym(x_1x_2x_3)\\
  \omega(T(1,1,p^3)) &=  p^{-3} \Sym(x_1^3) + p^{-4}(p-1)\Sym(x_1^2x_2) + p^{-5}(p-1)^2
  \Sym(x_1x_2x_3)
\end{align*}
so
\begin{align*}
  \Sym(x_1x_2x_3) &= \Sym(x^{\ba_1}) = p^6 \omega(T(p,p,p))\\
  \Sym(x_1^2x_2) &= \Sym(x^{\ba_2}) = p^4 \omega(T(1,p,p^2) -
  p^4(2p+1)(p-1)\omega(T(p,p,p)\\
  \Sym(x_1^3) &= \Sym(x^{\ba_3}) = p^3\omega(T(1,1,p^3)) - p^3(p-1)
  \omega(T(1,p,p^2)) + p^3(p+1)(p-1)^2\omega(T(p,p,p))
\end{align*}

\end{example}

\begin{example}


\noindent For $P_3(4)=\{\ba_1=(2,1,1) < \ba_2 = (2,2,0) <
  \ba_3=(3,1,0) < \ba_4=(4,0,0)\}$, we have:
\begin{align*}
\omega(T(p,p,p^2)) &=  p^{-7}\Sym(x_1^2x_2x_3)\\
\omega(T(1,p^2,p^2)) &=  p^{-6}\Sym(x_1^2x_2^2) + p^{-7}(p-1) \Sym(x_1^2x_2x_3)\\
\omega(T(1,p,p^3)) &=  p^{-5}\Sym(x_1^3 x_2) + p^{-6}(p-1) \Sym(x_1^2
  x_2^2) + 2p^{-6}(p-1) \Sym(x_1^2x_2 x_3)\\
\omega(T(1,1,p^4)) &= p^{-4} \Sym(x_1^4) + p^{-5}(p-1) \Sym(x_1^3x_2) +
  p^{-5}(p-1) \Sym(x_1^2x_2^2)\\ 
&+  p^{-6}(p-1)^2\Sym(x_1^2x_2x_3)
\end{align*}
so
\begin{align*}
  \Sym(x_1^2x_2x_3) &= \Sym(x^{\ba_1}) = p^7\omega(T(p,p,p^2))\\
  \Sym(x_1^2x_2^2) &= \Sym(x^{\ba_2}) = p^6\omega(T(1,p^2,p^2) -
  p^6(p-1)\omega(T(p,p,p^2))\\
  \Sym(x_1^3x_2) &= \Sym(x^{\ba_3}) = p^5\omega(T(1,p,p^3)) -
  p^5(p-1)\omega(T(1,p^2,p^2)) - p^5(p-1)^2 \omega(T(p,p,p^2))\\
  \Sym(x_1^4) &= \Sym(x^{\ba_4}) = p^4\omega(T(1,1,p^4)) -
  p^4(p-1)\omega(T(1,p,p^3)) - p^4(p-1)\omega(T(1,p^2,p^2)) \\
&\hspace{1in} +  p^4(p-1)^3 \omega(T(p,p,p^2)
\end{align*}
\end{example}

