# Two Families of Unravelled Abstract Regular Polytopes in $\mathrm{B}_{n}$

Robert Nicolaides and Peter Rowley

(2020)

###### Abstract

This paper exhibits two families of unravelled abstract regular polytopes in
Coxeter groups of type $\mathrm{B}_{n}$. For one family they have rank 4 while
the other family has arbitrarily large rank.

##  1 Introduction

Unravelled abstract regular polytopes were first introduced and studied in
[4]. In that paper, among other things, certain unravelled polytopes for
$\mathrm{SL}_{3}(q)\rtimes\langle t\rangle,$ $t$ being the transpose inverse
automorphism of $\mathrm{SL}_{3}(q)$, were analyzed. Here we continue
investigating unravelled polytopes this time in Coxeter groups of type
$\mathrm{B}_{n}$.

The study of abstract regular polytopes may be carried out entirely in a group
theoretic environment. The parallel group theoretic concept is that of a
C-string. See [3] for details of this reformulation. For a finite group $G$
and $\\{t_{1},\ldots,t_{n}\\}$ a set of involutions is $G$, set
$I=\\{1,\ldots,n\\}$ and for $J\subseteq I$ $G_{J}=\langle t_{j}|j\in
J\rangle$, with $G_{\emptyset}=1$ by convention. If
$J=\\{j_{1},\ldots,j_{k}\\}$, then $G_{J}$ is sometimes written as
$G_{j_{1}\ldots j_{k}}$. A C-string of rank $n$ for $G$ is a set of
involutions $\\{t_{1},\ldots,t_{n}\\}$ in $G$ such that

  1. (i)

$G=\langle t_{1},\ldots t_{n}\rangle$

  2. (ii)

$t_{i}$ and $t_{j}$ commute whenever $|i-j|\geq 2;$ and

  3. (iii)

for $J,K\subseteq I,$ $G_{J}\cap G_{K}=G_{J\cap K}.$

We can now define unravelled C-strings and unravelled abstract regular
polytopes.

######  Definition 1.1.

Suppose $G$ is a finite group with $\\{t_{1},\ldots,t_{n}\\}$ a C-string for
$G$ of rank $n$.

  1. $(i)$

For $N\trianglelefteq G,$ set $\overline{G}=G/N$ and use $\overline{g}$ for
the image of $g$ in $\overline{G}$, $g\in G$. If either
$|\\{\overline{t_{1}},\ldots,\overline{t_{n}}\\}|<n$ or,
$|\\{\overline{t_{1}},\ldots,\overline{t_{n}}\\}|=n$ and
$\\{\overline{t_{1}},\ldots,\overline{t_{n}}\\}$ is not a C-string for
$\overline{G},$ we say that $\\{t_{1},\ldots,t_{n}\\}$ is an $N$-unravelled
C-string for $G$.

  2. $(ii)$

If $\\{t_{1},\ldots,t_{n}\\}$ is an $N$-unravelled C-string for all non-
trivial proper normal subgroups $N$ of $G$, we call $\\{t_{1},\ldots,t_{n}\\}$
an unravelled C-string for $G$, and refer to the associated abstract regular
polytope as being unravelled.

Our first theorem, just as for those on $\mathrm{SL}_{3}(q)\rtimes\langle
t\rangle,$ yields unravelled C-strings of rank 4.

######  Theorem 1.2.

Suppose that $G=\mathrm{B}_{n}$ where $n$ is odd and $n\geq 5$. Then $G$ has a
rank 4 C-string $\\{t_{1},t_{2},t_{3},t_{4}\\}$ with Schlafli symbol

$2n-4$$6$$4$$t_{1}$$t_{2}$$t_{3}$$t_{4}$

,

which is unravelled when $n>5.$ Further $G_{123}\cong\mathrm{Sym}({n})$ and
$G_{234}\cong\mathbb{Z}_{2}\times\mathrm{Sym}({5}).$

Our second theorem, by contrast, gives examples in $\mathrm{B}_{n}$ of
arbitrary large rank.

######  Theorem 1.3.

Suppose that $G=\mathrm{B}_{n}$ where $n\geq 8,$ and set $m=n-4.$ Then $G$ has
a rank $m$ C-string $\\{t_{1},t_{2},\ldots,t_{m}\\}$ with Schlafli symbol

$12$$12$$6$$3$$3$$t_{1}$$t_{2}$$t_{3}$$t_{4}$$t_{5}$$t_{m-1}$$t_{m}$

.

Further, when $n$ is even, this C-string is unravelled.

The proofs of Theorems 1.2 and 1.3 occupy Sections 3 and 4 respectively, while
in Section 2 we cover preliminary results for these proofs.

##  2 Preliminary Results

This short section contains the results we need in the two following sections.
First of these results is one which identifies the Coxeter groups of type
$\mathrm{B}_{n}$. For a set $\Omega=\\{1,\ldots,n\\},$
$\mathrm{Sym}({n})=\mathrm{Sym}({\Omega})$ denotes the symmetric group of
degree $n$ defined on $\Omega$.

######  Lemma 2.1.

Suppose that $\Omega=\\{1,2\ldots,n,n+1,\ldots,2n\\}.$ Let $\beta_{0}=(1,n+1)$
and $\beta_{i}=(i,i+1)(n+i,n+i+1)$ for $1\leq i<n.$ Then
$\langle\beta_{0},\beta_{1},\ldots,\beta_{n-1}\rangle$ is isomorphic to
$\mathrm{B}_{n}$.

###### Proof.

See (2.10) of [2]. ∎

In a similar vein to Lemma 2.1, we have the well-known characterization of
$\mathrm{Sym}({n})$.

######  Lemma 2.2.

Suppose that $H$ is a group with presentation $\langle
r_{1},\ldots,r_{n-1}|(r_{i}r_{j})^{m_{ij}})\rangle$. If $m_{ii}=1$ for
$i=1,\ldots,n-1,m_{ij}=3$ if $|i-j|=1$ and $m_{ij}=2$ if $|i-j|>1,$ then
$H\cong\mathrm{Sym}({n})$.

###### Proof.

See (6.4) of [2]. ∎

We make frequent use of the next lemma to verify that particular sets of
involutions are C-strings.

######  Lemma 2.3.

Let $G=\langle t_{1},\ldots t_{n}\rangle$ where for all $1\leq i,j\leq n$ we
have $|i-j|\geq 2$ implies that $(t_{i}t_{j})^{2}=1$ and each $t_{i}$ an
involution. If $\\{t_{1},\ldots,t_{n-1}\\}$ and $\\{t_{2},\ldots,t_{n}\\}$ are
C-strings for $\langle t_{1},\ldots,t_{n-1}\rangle$ and $\langle
t_{2},\ldots,t_{n}\rangle$, then $\\{t_{1},\ldots,t_{n}\\}$ is a C-string for
G if

$\langle t_{1},\ldots,t_{n-1}\rangle\cap\langle
t_{2},\ldots,t_{n}\rangle=\langle t_{2},\ldots,t_{n-1}\rangle.$

###### Proof.

See 2E16(a) in [3]. ∎

Our group theoretic notation is standard, as given for example in [5], with
the addition that $\mathrm{Dih}({n})$ denotes the dihedral group of order $n$.

##  3 Rank 4 Unravelled C-strings

Here we establish Theorem 1.2. So we are assuming that $n$ is odd and $n\geq
5$.We shall construct the C-string for $\mathrm{B}_{n}$ working in
$\mathrm{Sym}({2n}).$ First we define the involutions $t_{i}$, $i=1,2,3,4$ in
$\mathrm{Sym}({2n})=\mathrm{Sym}({\Omega})$, where $\Omega=\\{1,\ldots,2n\\}.$

######  $\boldsymbol{(}\mkern-3.0mu$ 3.1.

$\boldsymbol{\mkern-7.0mu)}$

$\displaystyle t_{1}$
$\displaystyle=\prod_{i=0}^{\lfloor\frac{n-2}{2}\rfloor}(1+2i,2+2i)(n+1+2i,n+2+2i)$
$\displaystyle t_{2}$
$\displaystyle=\prod_{i=1}^{\lfloor\frac{n-3}{2}\rfloor}(2+2i,3+2i)(n+2+2i,n+3+2i)$
$\displaystyle t_{3}$ $\displaystyle=(1,3)(2,4)(n+1,n+3)(n+2,n+4)$
$\displaystyle t_{4}$
$\displaystyle=(1,2)(n+1,n+2)\prod_{i=1}^{n-2}(2+i,n+2+i)$

Observe that $t_{1}t_{2}$ when written as a product of pairwise disjoint
cycles has two of length 2 and two of length $n-2$. Hence, as $n-2$ is odd,
$t_{1}t_{2}$ has order $2n-4$. It is easy to check that $t_{2}t_{3}$ has order
6 and $t_{3}t_{4}$ has order 4. Also we see that $t_{1}t_{3}=t_{3}t_{1}$ and
$t_{2}t_{4}=t_{4}t_{2}$.

Put $G=\langle t_{1},t_{2},t_{3},t_{4}\rangle$. We will show in (3.10) that
$G\cong\mathrm{B}_{n}$, after we have first investigated the subgroups
$G_{123}=\langle t_{1},t_{2},t_{3}\rangle$ and $G_{234}=\langle
t_{2},t_{3},t_{4}\rangle$.

Beginning with $G_{234}$ and setting

$\displaystyle\Delta_{1}=\\{1,2,3,4,5,n+1,n+2,n+3,n+4,n+5\\},$
$\displaystyle\Delta_{6}=\\{6,7,n+6,n+7\\},$
$\displaystyle\Delta_{8}=\\{8,9,n+8,n+9\\},$ $\displaystyle\,\,\vdots$
$\displaystyle\Delta_{n-1}=\\{n-1,n,2n-1,2n\\},$

we note that

######  $\boldsymbol{(}\mkern-3.0mu$ 3.2.

$\boldsymbol{\mkern-7.0mu)}$ the $G_{234}$ orbits of the $\Omega$ are
$\Delta_{1},\Delta_{6},\Delta_{8},\ldots,\Delta_{n-1}$.

We also note

######  $\boldsymbol{(}\mkern-3.0mu$ 3.3.

$\boldsymbol{\mkern-7.0mu)}$ the induced action of $\langle
t_{2},t_{4}\rangle$ on each of $\Delta_{6},\Delta_{8},\ldots,\Delta_{n-1}$ is
identical to its action on $\Delta_{4}=\\{4,5,n+4,n+5\\}$.

Set $s_{2}=(4,5)(9,10),s_{3}=(1,3)(2,4)(6,8)(7,9)$ and
$s_{4}=(1,2)(3,8)(4,9)(5,10)(6,7)$ (these are $t_{2},t_{3},t_{4}$ for the case
$n=5$), and $H=\langle s_{2},s_{3},s_{4}\rangle$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.4.

$\boldsymbol{\mkern-7.0mu)}$ $G_{234}\cong
H\cong\mathbb{Z}_{2}\times\mathrm{Sym}({5})$ with $\\{t_{2},t_{3},t_{4}\\}$ a
C-string for $G_{234}$. Further
$\langle(t_{2}t_{3}t_{4})^{5}\rangle=Z(G_{234})$.

Restricting $G_{234}$ to $\Delta_{1}$ yields a homomorphism from $G_{234}$ to
$H$, and then (3.3) implies $G_{234}\cong H$. Employing Magma[1] quickly
reveals the structure of $H$ and that $\\{s_{2},s_{3},s_{4}\\}$ is a
$C$-string for $H$. This proves (3.4).

We now turn our attention to $G_{123}$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.5.

$\boldsymbol{\mkern-7.0mu)}$

  1. (i).

The $G_{123}$\- orbits of $\Omega$ are $\Lambda_{1}=\\{1,..,n\\}$ and
$\Lambda_{n+1}=\\{n+1,\ldots,2n\\}.$

  2. (ii).

For $j\in\Lambda_{1}$ and $g\in G_{123}$, $\,(j)g=k$ if and only if
$(j+n)g=k+n$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.6.

$\boldsymbol{\mkern-7.0mu)}$ For $1\leq i<n$, we have $(i,i+1)(n+i,n+i+1)\in
G_{123}.$

In view of (3.5)($ii$) it will be sufficient to look at the action of elements
of $G_{123}$ on $\Lambda_{1}$. So, for $i=1,2,3,$ let $\hat{t}_{i}$ denote the
induced action of $t_{i}$ on $\Lambda_{1}$. Hence

$\displaystyle\hat{t}_{1}=\prod_{i=0}^{\lfloor\frac{n-2}{2}\rfloor}(1+2i,2+2i),$
$\displaystyle\hat{t}_{2}=\prod_{i=1}^{\lfloor\frac{n-3}{2}\rfloor}(2+2i,3+2i)\,\text{
and }$ $\displaystyle\hat{t}_{3}=(1,3)(2,4).$

Consequently

$\hat{t}_{1}\hat{t}_{2}=(1,2)(3,5,7,\ldots,n,n-1,n-3,\ldots,6,4).$

Since $n-2$ is odd, $(\hat{t}_{1}\hat{t}_{2})^{n-2}=(1,2)$. Therefore,
$(1,2)(n+1,n+2)\in G_{123}$. Also

$\displaystyle\hat{t}_{3}\hat{t}_{1}\hat{t}_{2}$
$\displaystyle=(1,3)(2,4)(1,2)(3,5,7,\ldots,n,n-1,n-3,\ldots,6,4)$
$\displaystyle=(1,5,7,\ldots,n,n-1,n-3,\ldots,6,4)(2,3)$

which is in disjoint cycle form. Again, since $n-2$ is odd, we have
$(\hat{t}_{3}\hat{t}_{1}\hat{t}_{2})^{n-2}=(2,3)$. Hence $(2,3)(n+2,n+3)\in
G_{123}.$ Now $(3,4)=(1,2)^{\hat{t}_{3}}$, so $(3,4)(n+3,n+4)\in G_{123}.$

Now we recursively construct the remaining $(i,i+1)(n+i,n+i+1)$ for all $i$,
with $3<i<n$. Supposing we have $(i,i+1)(n+i,n+i+1)\in G_{123}$ for all $3\leq
i\leq k$. We show that $(k+1,k+2)(n+k+1,n+k+2)\in G_{123}$. If $k$ is even,
then

$\displaystyle(k,k+1)(n+k,n+k+1)^{t_{1}}$
$\displaystyle=(k-1,k+2)(n+k-1,n+k+2).$

Since

$(k-1,k+2)(n+k-1,n+k+2)^{(k-1,k+1)(n+k-1,n+k+1)}=(k+1,k+2)(n+k+1,n+k+2)$

and

$(k-1,k+1)(n+k-1,n+k+1)=(k-1,k)(n+k-1,n+k)^{(k,k+1)(n+k,n+k+1)}\in G_{123},$

we deduce that $(k+1,k+2)(n+k+1,n+k+2)\in G_{123}$. When $k$ is odd, a similar
calculation using $t_{2}$ in place of $t_{1}$, also yields the same
conclusion, so proving (3.6).

######  $\boldsymbol{(}\mkern-3.0mu$ 3.7.

$\boldsymbol{\mkern-7.0mu)}$ $G_{123}\cong\mathrm{Sym}({n})$.

From (3.6)

$K=\langle\,\,(i,i+1)(n+i,n+i+1)\,\,|\,\,1\leq i<n\,\,\rangle\leq G_{123},$

with the generators of $K$ satisfying the Coxeter relations for
$\mathrm{Sym}({n})$. Thus, by Lemma 2.2, $K$ is isomorphic to a quotient of
$\mathrm{Sym}({n})$ and hence $K\cong\mathrm{Sym}({n})$. The action of
$G_{123}$ on $\\{\\{i,n+i\\}|i\in\Lambda_{1}\\}$ forces $K=G_{123}$, so giving
(3.7).

######  $\boldsymbol{(}\mkern-3.0mu$ 3.8.

$\boldsymbol{\mkern-7.0mu)}$ $\\{t_{1},t_{2},t_{3}\\}$ is a C-string for
$G_{123}$.

We only need check $\langle t_{1},t_{2}\rangle\cap\langle
t_{2},t_{3}\rangle=\langle t_{2}\rangle$, the other intersections being clear.
Now $\langle t_{2}t_{3}\rangle$ has $\\{1,3\\}$ and $\\{2,4,5\\}$ as an orbits
on $\Omega$ and so $\langle(t_{2}t_{3})^{2}\rangle$ has $\\{2,4,5\\}$ as an
orbit and $\langle(t_{2}t_{3})^{3}\rangle$ has $\\{1,3\\}$. Since $\langle
t_{1}t_{2}\rangle$ has $\\{1,2\\}$ as an orbit and $t_{2}t_{3}$ has order 6,
we conclude that $\langle t_{1},t_{2}\rangle\cap\langle
t_{2},t_{3}\rangle=\langle t_{2}\rangle.$ So (3.8) holds.

Set $\omega_{0}=\prod_{i=1}^{n}(i,n+i)$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.9.

$\boldsymbol{\mkern-7.0mu)}$ $\omega_{0}=(t_{1}t_{2}t_{3}t_{4})^{n}.$

We calculate that

$\displaystyle t_{1}t_{2}$
$\displaystyle=(1,2)(3,5,7,\ldots,n,n-1,\ldots,6,4)(n+1,n+2)(n+3,\ldots,2n,2n-1,\ldots
n+4),$ $\displaystyle t_{1}t_{2}t_{3}$
$\displaystyle=t_{1}t_{2}(1,3)(2,4)(n+1,n+3)(n+2,n+4)$
$\displaystyle=(1,4)(3,5,\ldots,n,n-1,\ldots,8,6,2)(n+1,n+4)(n+3,\ldots,2n,2n-1,\ldots,n+4),$
$\displaystyle t_{1}t_{2}t_{3}t_{4}$
$\displaystyle=t_{1}t_{2}t_{3}(1,2)(n+1,n+2)\prod_{i=1}^{n-2}(2+i,(n+2)+i)$
$\displaystyle=(\underbrace{1,n+4,n+2,3,n+5,7,\ldots,n,2n-1,n-3,2n-5,\ldots,n+6,n+1}_{n+1,}$
$\displaystyle\underbrace{4,2,n+3,\ldots,6}_{n-1})$ $\displaystyle\text{when
}n\equiv 1\mod 3\text{ and }$
$\displaystyle=(\underbrace{1,n+4,n+2,3,n+5,7,\ldots,2n,n-1,2n-3,n-5,\ldots,n+6,n+1}_{n+1},$
$\displaystyle\underbrace{4,2,n+3,\ldots,6}_{n-1})$ $\displaystyle\text{when
}n\equiv 3\mod 3.$

Therefore $t_{1}t_{2}t_{3}t_{4}=\prod_{i=1}^{n}(i,n+i)$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.10.

$\boldsymbol{\mkern-7.0mu)}$ $G\cong\mathrm{B}_{n}$.

Set $\beta_{0}=(1,n+1)$ and, for $1\leq i<n$, $\beta_{i}=(i,i+1)(n+i,n+i+1)$.
Put $L=\langle\beta_{0},\beta_{1},\ldots,\beta_{n-1}\rangle$. By Lemma 2.2
$L\cong\mathrm{B}_{n}$.

Directly from their definitions, we have

$\displaystyle t_{1}$
$\displaystyle=\prod_{i=0}^{\lfloor\frac{n-2}{2}\rfloor}\beta_{1+2i}\text{ and
}$ $\displaystyle t_{2}$
$\displaystyle=\prod_{i=0}^{\lfloor\frac{n-3}{2}\rfloor}\beta_{2+2i}.$

An easy check shows $t_{3}={\beta_{1}}^{\beta_{2}}{\beta_{2}}^{\beta_{3}}$.
Since

$\displaystyle t_{4}$
$\displaystyle=(1,2)(n+1,n+2)\prod_{i=1}^{n-2}(2+i,n+2+i)$
$\displaystyle=\beta_{1}\prod_{i=2}^{n}\beta_{0}^{\prod_{j=i}^{i}\beta_{j}},$

we infer that $G\leq L$.

By (3.6) $\beta_{i}\in G$ for $1\leq i<n.$ Thus to complete the proof of
(3.10) we need to demonstrate that $\beta_{0}\in G$. Now

$\displaystyle{t_{2}}^{t_{3}}$
$\displaystyle=(\prod_{i=1}^{\lfloor\frac{n-3}{2}\rfloor}(2+2i,3+2i)(n+2+2i,n+3+2i))^{(1,3)(2,4)(n+1,n+3)(n+2,n+4)}$
$\displaystyle=(2,5)(n+2,n+5)t_{2}(4,5)(n+4,n+5)$ and
$\displaystyle{t_{2}}^{t_{3}t_{4}}$
$\displaystyle=(1,n+5)(n+1,5)t_{2}(4,5)(n+4,n+5).$ Hence
$\displaystyle{t_{2}}^{t_{3}}{t_{2}}^{t_{3}t_{4}}$
$\displaystyle=(2,5)(n+2,n+5)t_{2}(4,5)(n+4,n+5)(1,n+5)(n+1,5)t_{2}(4,5)(n+4,n+5)$
$\displaystyle=(2,5)(n+2,n+5)(1,n+5)(n+1,5)$
$\displaystyle=(2,n+1,5)(1,n+5,n+2).$ Therefore $\displaystyle
t_{1}{t_{2}}^{t_{3}}{t_{2}}^{t_{3}t_{4}}$
$\displaystyle=t_{1}t_{2}(2,n+1,5)(1,n+5,n+2)$
$\displaystyle=(1,n+1)(2,n+5,n+7,\ldots,2n,2n-1,\ldots,n+4,n+3,$
$\displaystyle\quad\quad n+2,5,7,9,\ldots,n,n-1,\ldots,6,4,3),$ and so
$\displaystyle(t_{1}{t_{2}}^{t_{3}}{t_{2}}^{t_{3}t_{4}})^{n-1}$
$\displaystyle=\prod_{i=2}^{n}(i,n+i),$

Hence, using (3.9),

$\beta_{0}=\prod^{n}_{i=1}(i,n+i)(t_{1}{t_{2}}^{t_{3}}{t_{2}}^{t_{3}t_{4}})^{-n+1}\in
G,$

which proves $(\ref{(3.10)})$.

######  $\boldsymbol{(}\mkern-3.0mu$ 3.11.

$\boldsymbol{\mkern-7.0mu)}$ $\\{t_{1},t_{2},t_{3},t_{4}\\}$ is a string
C-string for $G$.

From (3.4) $G_{234}\cong\mathbb{Z}_{2}\times\mathrm{Sym}({5})$ and
$\langle(t_{2}t_{3}t_{4})^{5}\rangle=Z(G_{234})$. If $G_{123}\cap
G_{234}>G_{23}$, then, as $G_{23}\cong\mathrm{Dih}({12})$, we must have either
$(t_{2}t_{3}t_{4})^{5}\in G_{123}\cap G_{234}$ or $G_{123}\cap G_{234}$ has
index at most $2$ in $G_{234}$ (and so $t_{3}t_{4}\in G_{123}$). Either of
these possibilities would contradict (3.5)($i$) as
$(t_{2}t_{3}t_{4})^{5}:1\rightarrow n+1$ and $t_{3}t_{4}:5\rightarrow n+5$.
Thus $G_{123}\cap G_{234}=G_{23}$. Using (3.4) and (3.8) we now obtain (3.11).

######  $\boldsymbol{(}\mkern-3.0mu$ 3.12.

$\boldsymbol{\mkern-7.0mu)}$ When $n\geq 7$, $\\{t_{1},t_{2},t_{3},t_{4}\\}$
is an unravelled C-string for $G$.

Let $M_{1}=\langle\,\,(i,n+i)\,\,|\,\,1\leq i\leq
n\,\,\rangle(=\mathrm{O}_{2}(G))$ and
$M_{2}=\langle\,\,(1,n+1)(i,n+i)\,\,|\,\,1<i\leq n\,\,\rangle$. From (3.10)
$G\cong\mathrm{B}_{n}$ and hence for $N\trianglelefteq G$, $1\neq N\neq G$, we
either have $[G:N]\leq 4$ or $N=\langle\omega_{0}\rangle,$ $M_{1}$ or $M_{2}$.
Set $\overline{G}=G/N$. Since $\\{t_{1},t_{2},t_{3},t_{4}\\}$ has rank 4, we
are only required to check that
$\\{\overline{t}_{1},\overline{t}_{2},\overline{t}_{3},\overline{t}_{4}\\}$ is
not a C-string for $N=\langle\omega_{0}\rangle,M_{1}$ and $M_{2}$.

Suppose $N=M_{1}$ or $M_{2}$. Then

$g=(t_{3}t_{4})^{2}=(1,n+2)(2,n+1)(3,n+4)(4,n+3)\in G_{234}.$

Also, using (3.6),

$h=(1,2)(3,4)(n+1,n+2)(n+3,n+4)\in G_{123}.$

Since

$gh^{-1}=(1,n+1)(2,n+2)(3,n+3)(4,n+4)\in M_{2}\leq M_{1},$

we get $\overline{g}=\overline{h}\in\overline{G}_{123}\cap\overline{G}_{234}.$
The non-trivial elements of $M_{1}$ either fix an element of $\Lambda_{1}$ or
maps it to an element of $\Lambda_{2}$. Hence, as $\\{1,3\\}$ is a
$G_{23}$-orbit, $\overline{h}\notin\overline{G}_{23}$. Thus
$\overline{G}_{123}\cap\overline{G}_{234}>\overline{G}_{23}$ when $N=M_{1}$ or
$M_{2}$. Now suppose $N=\langle\omega_{0}\rangle$. This time we take

$g=\prod^{5}_{i=1}(i,n+i)\prod_{j=1}^{\lfloor\frac{n-5}{2}\rfloor}(4+2j,n+5+2j)(5+2j,n+4+2j)\text{
and }$
$h=\prod_{j=1}^{\lfloor\frac{n-5}{2}\rfloor}(4+2j,5+2j)(n+4+2j,n+5+2j).$

Then $g\in G_{234}$, $h\in G_{123}$ and $gh^{-1}=\omega_{0}.$ Therefore
$\overline{g}=\overline{h}\in\overline{G}_{123}\cap\overline{G}_{234}.$ It is
straightforward to also see that $\overline{g}\notin\overline{G}_{23},$ and
consequently (3.12) is proven.

Combining (3.10),(3.11) and (3.12) completes the proof of Theorem 1.2.

##  4 Rank $n-4$ Unravelled C-Strings

This final section is devoted to the proof of Theorem 1.3. Thus we assume
$n\geq 8$ and we set $m=n-4$.

Just as in the proof of Theorem 1.2 we construct
$\\{t_{1},t_{2},\ldots,t_{m}\\}$ as a subset of $\mathrm{Sym}({2n})$ and then
show that it is a C-string for $\mathrm{B}_{n}$. Finally, when $n$ is even, we
prove that it is an unravelled C-string. So again, let
$\Omega=\\{1,\ldots,2n\\}$ and define the $t_{i}$ as follows.

######  $\boldsymbol{(}\mkern-3.0mu$ 4.1.

$\boldsymbol{\mkern-7.0mu)}$

$\displaystyle t_{1}$
$\displaystyle=(2,3)(n+2,n+3)(4,5)(n+4,n+5)\prod_{i=6}^{n}(i,n+i)$
$\displaystyle t_{2}$
$\displaystyle=(1,2)(n+1,n+2)(3,4)(n+3,n+4)(5,6)(n+5,n+6)\prod_{i=7}^{n}(i,n+i)$
$\displaystyle t_{3}$ $\displaystyle=(2,3)(n+2,n+3)(6,7)(n+6,n+7)$ and for
$k=4,\ldots,m$, $\displaystyle t_{k}$ $\displaystyle=(k+3,k+4)(n+k+3,n+k+4).$

Set $I=\\{1,2,\ldots,m\\}.$

######  $\boldsymbol{(}\mkern-3.0mu$ 4.2.

$\boldsymbol{\mkern-7.0mu)}$ $t_{3}t_{4}=(2,3)(n+2,n+3)(6,8,7)(n+6,n+8,n+7).$

Next we show that

######  $\boldsymbol{(}\mkern-3.0mu$ 4.3.

$\boldsymbol{\mkern-7.0mu)}$ $(t_{1}t_{2}\ldots
t_{m}t_{3}t_{4})^{n}=\prod_{i=1}^{n}(i,n+i).$

We calculate that $\displaystyle t_{4}t_{5}\ldots t_{m}=$
$\displaystyle(7,n,n-1,n-2,\ldots,9,8)(n+7,2n,2n-1,\ldots,n+8)$ and
$\displaystyle t_{1}t_{2}t_{3}=$
$\displaystyle(1,3)(n+1,n+3)(2,4,7,6,n+5,n+2,n+4,n+7,n+6,5).$ Hence
$\displaystyle t_{1}t_{2}\ldots t_{m}=$
$\displaystyle(1,3)(n+1,n+3)(2,4,n,n-1,\ldots,9,8,7,6,n+5,$ $\displaystyle
n+2,n+4,2n,2n-1,\ldots,n+8,n+7,n+6,5).$ Using (4.2) we now get $\displaystyle
t_{1}t_{2}\ldots t_{m}t_{3}t_{4}=$
$\displaystyle(1,2,4,n,n-1,\ldots,9,7,8,6,n+5,$ $\displaystyle
n+3,n+1,n+2,n+4,2n,\ldots,n+9,n+7,n+8,n+6,5,3),$ which yields (4.3).

######  $\boldsymbol{(}\mkern-3.0mu$ 4.4.

$\boldsymbol{\mkern-7.0mu)}$ $(t_{1}t_{2}\ldots
t_{m-1}t_{3}t_{4})^{n-1}=\prod_{i=1}^{n-1}(i,n+i).$

First we have $\displaystyle t_{1}t_{2}\ldots t_{m-1}=$
$\displaystyle(1,3)(n+1,n+3)(2,4,n-1,n-2,\ldots,6,$ $\displaystyle
n+5,n+2,n+4,2n-1,\ldots,n+7,n+6,5),$ then, using (4.2), $\displaystyle
t_{1}t_{2}\ldots t_{m-1}t_{3}t_{4}=$
$\displaystyle(1,2,4,n-1,\ldots,9,7,8,6,n+5,n+3,$ $\displaystyle
n+1,n+2,n+4,2n-1,\ldots,n+9,n+7,n+8,n+6,5,3).$ This gives the desired
expression for $(t_{1}t_{2}\ldots t_{m-1}t_{3}t_{4})^{n-1}$.

Combining (4.3) and (4.4) we observe that

######  $\boldsymbol{(}\mkern-3.0mu$ 4.5.

$\boldsymbol{\mkern-7.0mu)}$ $(t_{1}t_{2}\ldots
t_{m-1}t_{3}t_{4})^{n-1}(t_{1}t_{2}\ldots t_{m}t_{3}t_{4})^{n}=(n,2n)$.

######  $\boldsymbol{(}\mkern-3.0mu$ 4.6.

$\boldsymbol{\mkern-7.0mu)}$ For $i,j\in\\{1,2,\ldots,m\\}$, the order of

$t_{i}t_{j}\text{ is }\begin{dcases}1&\text{ if }i=j\\\ 2&\text{ if }|i-j|\geq
2\\\ 12&\text{ if }i=1,j=2\text{ or }i=2,j=3\\\ 6&\text{ if }i=3,j=4\\\
3&\text{ otherwise.}\end{dcases}$

It is evident that each $t_{i}$ is an involution as they are defined as the
products of pairwise disjoint transpositions. Since

$\displaystyle t_{1}t_{2}=$
$\displaystyle(1,2,4,6,n+5,n+3,n+1,n+2,n+4,n+6,5,3),$ $t_{1}t_{2}$ has order
12. Similarly we have $\displaystyle t_{2}t_{3}=$
$\displaystyle(1,3,4,2)(n+1,n+3,n+4,n+2)(5,7,n+6,n+5,n+7,6)\prod_{i=8}^{n}(i,n+i),$
and so $t_{2}t_{3}$ also has the order 12. From (4.2) we see that order of
$t_{3}t_{4}$ is 6. If, $|i-j|=1$ and $4\leq i<j\leq m$, then $\displaystyle
t_{i}t_{j}=$ $\displaystyle(3+i,5+i,4+i)(n+3+i,n+5+i,n+4+i)$

has order 3. That $t_{i}$ and $t_{j}$ commute when $|i-j|\geq 2$ is readily
checked, so verifying (4.6).

Put $G=\langle t_{1},t_{2},\ldots,t_{m}\rangle.$

######  $\boldsymbol{(}\mkern-3.0mu$ 4.7.

$\boldsymbol{\mkern-7.0mu)}$ $G\cong\mathrm{B}_{n}$.

We again employ Lemma 2.1 to identify $G$. So set $\beta_{0}=(1,n+1),$
$\beta_{i}=(i,i+1)(n+i,n+i+1)$, for $1\leq i<n,$ and
$L=\langle\beta_{0},\beta_{1},\ldots,\beta_{n-1}\rangle\leq\mathrm{Sym}({2n})$.
Also set $\eta_{i}=(i,n+i)$ for $1\leq i\leq n$. Note that
$\eta_{1}=\beta_{0}$ and $\eta_{i}={\beta_{0}}^{\prod_{j=1}^{i-1}\beta_{j}}$
for $i=2,\ldots,n.$ Therefore $\eta_{i}\in L$ for $i=1,\ldots,n.$ Because

$\displaystyle t_{1}=$
$\displaystyle\beta_{2}\beta_{4}\prod_{i=6}^{n}\eta_{i},$ $\displaystyle
t_{2}=$ $\displaystyle\beta_{1}\beta_{3}\beta_{5}\prod_{i=7}^{n}\eta_{i},$
$\displaystyle t_{3}=$ $\displaystyle\beta_{2}\beta_{6}\,\,\text{ and, for
$4\leq i\leq m$},$ $\displaystyle t_{i}=$ $\displaystyle\beta_{i+3}$

we conclude that $G\leq L.$

From (4.5), $\eta_{n}=(n,2n)\in G$. Now let $g=t_{m}t_{m-1}\ldots
t_{4}t_{3}t_{2}t_{1}t_{2}t_{1}t_{2}\in G.$ Then we see that
${\eta_{n}}^{g}=\eta_{1}=\beta_{0},$ whence $\beta_{0}\in G.$ Since
$\beta_{i}=t_{i-3}$ for $i=7,\ldots,n-1,$ it remains to show that
$\beta_{1},\beta_{2},\beta_{3},\beta_{4},\beta_{5}$ and $\beta_{6}$ are in
$G$.

Employing (4.2) again we have

$\displaystyle\beta_{2}=$ $\displaystyle(t_{3}t_{4})^{3},$
$\displaystyle\beta_{6}=$ $\displaystyle t_{3}(t_{3}t_{4})^{3},$
$\displaystyle\beta_{1}=$
$\displaystyle{\beta_{6}}^{t_{2}t_{3}t_{1}t_{2}t_{1}t_{2}t_{3}}$
$\displaystyle\beta_{3}=$
$\displaystyle{\beta_{6}}^{t_{2}t_{3}t_{1}t_{2}t_{1}t_{3}}$
$\displaystyle\beta_{4}=$
$\displaystyle{\beta_{6}}^{t_{2}t_{1}t_{2}t_{1}t_{3}t_{2}}\text{ and}$
$\displaystyle\beta_{5}=$ $\displaystyle{\beta_{6}}^{t_{2}t_{3}\eta_{6}}.$

Since $\eta_{6}={\eta_{n}}^{h}$ where $h=t_{m}t_{m-1}\ldots t_{4}t_{3}$, we
have now shown that $\beta_{i}\in G$ for $i=0,\ldots,n-1.$ Thus $G=L,$ and
(4.7) is proven.

We now turn our attention to showing that $\\{t_{1},\ldots,t_{m}\\}$ is a
C-string.

For $t_{1},\ldots,t_{m}$ if we wish to highlight that they are permutations in
$\mathrm{Sym}({2n})$ we shall write $t_{1}^{(n)},\ldots,{t_{m}}^{(n)}.$ Set
$G_{1234}^{(n)}=\langle
t_{1}^{(n)},t_{2}^{(n)},t_{3}^{(n)},t_{4}^{(n)}\rangle$,
$G_{123}^{(n)}=\langle t_{1}^{(n)},t_{2}^{(n)},t_{3}^{(n)}\rangle$,
$G_{234}^{(n)}=\langle t_{2}^{(n)},t_{3}^{(n)},t_{4}^{(n)}\rangle$ and
$G_{23}^{(n)}=\langle t_{2}^{(n)},t_{3}^{(n)}\rangle$.

######  $\boldsymbol{(}\mkern-3.0mu$ 4.8.

$\boldsymbol{\mkern-7.0mu)}$ For $n\geq 8$,
$\\{t_{1}^{(n)},t_{2}^{(n)},t_{3}^{(n)},t_{4}^{(n)}\\}$ is a C-string for
$G_{1234}^{(n)}$.

First we may verify (4.8) for $n=8$ using Magma. Then we may define
$\mu\in\mathrm{Sym}({2n})$ by

$\displaystyle\mu$ $\displaystyle:\omega\rightarrow\omega-n+8\text{ for
$\omega\in\Lambda=\\{n+1,n+2,n+3,n+4,n+5,n+6,n+7,n+8\\}$}$
$\displaystyle:\omega\rightarrow\omega\text{ for
$\omega\in\Omega\setminus\Lambda$}.$

For $i=1,2,3,4$ define

$\phi(t_{i}^{(n)})=\widehat{(t_{i}^{(n)})^{\mu}}$

where $\,\widehat{\quad\quad}\,$ denotes the induced action upon the set
$\Phi=\\{1,\ldots,16\\}.$ When we write equalities in this context, it is as a
permutation of $\Phi$. Observe that $\phi$ extends to a homomorphism from
$G_{1234}^{(n)}$ to $\mathrm{Sym}({\Phi})$ with
$\phi(G_{1234}^{(n)}=G_{1234}^{(8)}$.

Because

$\displaystyle\phi(G_{23}^{(n)})\leq\phi(G_{123}^{(n)}\cap
G_{234}^{(n)})\leq\phi(G_{123}^{(n)})\cap\phi(G_{234}^{(n)})=G_{123}^{(n)}\cap
G_{234}^{(n)}$

and (4.8) holds for $n=8$, we have

$\phi(G_{123}^{(n)}\cap G_{234}^{(n)})=G_{23}^{(8)}.$

We now investigate the structure of $H=G_{234}^{(n)}.$ Set
$s_{2}=t_{2}^{(n)},s_{3}=t_{3}^{(n)},s_{4}=t_{4}^{(n)}$ and
$R=\langle(s_{2}s_{3})^{3},(s_{3}s_{4})^{3}\rangle$. Now $(s_{3}s_{4})^{3}$
inverts $(s_{2}s_{3})^{3}$ which has order 4. Therefore
$R\cong\mathrm{Dih}({8}).$ Calculation shows that $s_{2},s_{3}$ and $s_{4}$
normalize $R$ and hence $R\trianglelefteq H$. Set $C=C_{H}(R)$. Further
calculation shows that $s_{2},s_{3},s_{2}s_{3}\notin C$ but
$s_{4},(s_{2}s_{3})^{2}\in C.$ Therefore, as $H=\langle
s_{2},s_{3},s_{4}\rangle$, $H=\langle s_{2},s_{3}\rangle C$ with $H/C\cong
2^{2}.$ Also we have $(s_{2}s_{3})^{6}\in Z(H)$ with $s_{4}(s_{2}s_{3})^{4}$
of order 4. Thus, as $(s_{2}s_{3})^{4}$ has order 3 and $s_{4}$ has order 2,
$\langle(s_{2}s_{3})^{4},s_{4}\rangle\cong\mathrm{Sym}({4}).$ Thus
$S=\langle(s_{2}s_{3})^{2},s_{4}\rangle\cong 2\times\mathrm{Sym}({4})$ with
$S\leq C.$ Since $s_{2}$ and $s_{3}$ normalize $S$, we infer $C=S.$ In
particular, we have shown $G_{234}^{(n)}=H$ has order $2^{6}.3$ for all $n\geq
8.$

Consequently $\phi$ restricted to $G_{234}^{(n)}\rightarrow G_{234}^{(8)}$ is
an isomorphism. So calculation in $G_{234}^{(n)}$ may be performed in
$G_{234}^{(8)}$ and, using $\phi$, we may keep track of the action on
$\Omega$.

Now $G_{234}^{(n)}$ has orbit $\\{5,6,7,8,n+5,n+6,n+7,n+8\\}$ on $\Omega$ and
$G_{123}^{(n)}$ has $\\{8,n+8\\}$ as an orbit. Thus

$G_{23}^{(n)}\leq G_{123}^{(n)}\cap G_{234}^{(n)}\leq
Stab({5,6,7,n+5,n+6,n+7\\}})=T^{(n)}.$

Calculation shows that $T^{(n)}=\langle G_{123}^{(n)},(6,7)(n+6,n+7)\rangle$
with $[T^{(n)}:G_{23}^{(n)}]=2$. If $G_{23}^{(n)}<G_{123}^{(n)}\cap
G_{234}^{(n)},$ then $G_{123}^{(n)}\cap G_{234}^{(n)}=T^{(n)}$ must contain a
normal subgroup of order 2 intersecting $G_{23}^{(n)}$ trivially (the kernel
of $\phi$, restricted to $G_{123}^{(n)}\cap G_{234}^{(n)}$), but it does not.
Thus $G_{23}^{(n)}=G_{123}^{(n)}\cap G_{234}^{(n)}.$ That
$\\{t_{2}^{(n)},t_{3}^{(n)},t_{4}^{(n)}\\}$ is a C-string for $G_{234}^{(n)}$
follows from $G_{234}^{(n)}$ and $G_{234}^{(8)}$ being isomorphic and the fact
that $\phi$ maps generator to generator. Observe that the $G_{23}-$orbits of
$\Omega$ are
$\\{1,2,3,4\\},\\{5,6,7,n+5,n+6,n+7\\},\\{n+1,n+2,n+3,n+4\\},\\{i,n+i\\}(i=8,\ldots,n).$
If $G_{12}\cap G_{23}>G_{2}$ then one of

$\displaystyle(t_{1}t_{2})^{4}$
$\displaystyle=(1,n+5,n+4)(2,n+3,n+6)(4,n+1,5)(6,n+2,3)\text{ and }$
$\displaystyle(t_{2}t_{3})^{6}$
$\displaystyle=(1,n+1)(2,n+2)(3,n+3)(4,n+4)(5,n+5)(6,n+6)$

would be in $G_{23}$. But then $1$ and $n+1$ would be in the same
$G_{23}-$orbit, a contradiction. Therefore $G_{12}\cap G_{23}=G_{2}.$
Appealing to Lemma 2.3 this now proves (4.8).

######  $\boldsymbol{(}\mkern-3.0mu$ 4.9.

$\boldsymbol{\mkern-7.0mu)}$ Let $4\leq k\leq m$ and set
$J_{k}=I_{k}\setminus\\{1,2\\}$. Then

  1. $(i)$

$G_{J_{k}}\cong\mathrm{Sym}({k-2})$ and

  2. $(ii)$

$\\{t_{4},\ldots,t_{k}\\}$ is a C-string for $G_{J_{k}}$.

Since $G\cong\mathrm{B}_{n}$ by (4.7), $\langle
t_{4},\ldots,t_{k}\rangle=\langle\beta_{1},\ldots,\beta_{k+3}\rangle$ is a
standard parabolic subgroup of $G$. Hence (4.9)$(i)$ and $(ii)$ follow.

######  $\boldsymbol{(}\mkern-3.0mu$ 4.10.

$\boldsymbol{\mkern-7.0mu)}$ Let $4\leq k\leq m$ and set
$J_{k}=I_{k}\setminus\\{1,2\\}$. Then

  1. $(i)$

$G_{J_{k}}\equiv\mathbb{Z}_{2}\times\mathrm{Sym}({k-1});$ and

  2. $(ii)$

$\\{t_{3},\ldots,t_{k}\\}$ is a C-string for $G_{J_{k}}$.

Recall that $t_{3}=\beta_{2}\beta_{6}$, and that $t_{i}=\beta_{i+3}$ for
$4\leq i\leq k$. Since $(\beta_{2}\beta_{6}\beta_{7})^{3}=\beta_{2}$, we have

$\displaystyle\langle t_{3},\ldots,t_{k}\rangle=$
$\displaystyle\langle\beta_{2}\beta_{6},\beta_{7}\ldots,\beta_{k+3}\rangle$
$\displaystyle=$
$\displaystyle\langle\beta_{2},\beta_{6},\beta_{7}\ldots,\beta_{k+3}\rangle$
$\displaystyle\cong$ $\displaystyle\mathbb{Z}_{2}\times\mathrm{Sym}({k-1}),$

so giving part $(i)$. Further, as $\langle t_{3},\ldots,t_{k}\rangle$ is a
Coxeter group,

$\displaystyle\langle t_{3},t_{4},\ldots,t_{k-1}\rangle\cap\langle
t_{4},\ldots t_{k}\rangle$
$\displaystyle=\langle\beta_{2},\beta_{6},\beta_{7},\ldots,\beta_{k+2}\rangle\cap\langle\beta_{7},\ldots\beta_{k+3}\rangle$
$\displaystyle=\langle\beta_{7},\ldots,\beta_{k+2}\rangle$
$\displaystyle=\langle t_{4},\ldots,t_{k-1}\rangle.$

To prove $(ii)$ we may argue by induction on $k$, $k=4$ being covered by
(4.8). Thus $\\{t_{3},t_{4},\ldots,t_{k-1}\\}$ is a C-string for
$G_{\\{3,\ldots,k-1\\}}$ and, as $\\{t_{4},\ldots,t_{k}\\}$ is a C-string for
$G_{\\{4,\ldots,k\\}}$, Lemma 2.3 yields $(ii)$.

Set $t_{0}={t_{4}}^{(t_{3}t_{4}t_{2}t_{3})}t_{2}.$

######  $\boldsymbol{(}\mkern-3.0mu$ 4.11.

$\boldsymbol{\mkern-7.0mu)}$ For $4\leq k\leq m,$ set
$J_{k}=I_{k}\setminus\\{1\\}.$ Then

$\displaystyle G_{J_{k}}=$ $\displaystyle\langle
t_{0},\beta_{2}\rangle\times\langle\beta_{5},\beta_{6},\ldots,\beta_{k+3}\rangle$
$\displaystyle\cong$ $\displaystyle\mathrm{Dih}({8})\times\mathrm{Sym}({k}).$

Clearly $t_{0}\in G_{J}$ and calculation reveals that

$t_{0}=(1,2)(n+1,n+2)(3,4)(n+3,n+4)\prod_{i=7}^{n}(i,n+i).$

Hence $t_{0}\beta_{5}=t_{2}$.

Set $H=\langle t_{0},\beta_{2},\beta_{5},\beta_{6},\ldots,\beta_{k+3}\rangle$.
Observing that $t_{0}$ and $\beta_{2}$ commute with each of
$\beta_{5},\beta_{6},\ldots,\beta_{k+3},$ we have

$H=\langle
t_{0},\beta_{2}\rangle\times\langle\beta_{5},\beta_{6},\ldots,\beta_{k+3}\rangle.$

We show that $G_{J_{k}}=H.$ Recalling that $\beta_{i+3}=t_{i}$, $4\leq i\leq
m$ and $t_{3}=\beta_{2}\beta_{6}$, $t_{2}=t_{0}\beta_{5}$ implies
$G_{J_{k}}\leq H$. Since $t_{0}=t_{2}\beta_{5}$,
$\beta_{5}=t_{0}t_{2},\beta_{6}=t_{3}(t_{3}t_{4})^{3}$ and
$\beta_{2}=t_{3}\beta_{6}$, we also have $H\leq G_{J_{k}}$. Because
$t_{0}\beta_{2}$ has order 4 and
$\langle\beta_{2},\beta_{6},\ldots,\beta_{k+3}\rangle$ is a standard parabolic
subgroup of $\mathrm{B}_{n}$, we deduce that
$G_{J_{k}}\cong\mathrm{Dih}({8})\times\mathrm{Sym}({k})$.

######  $\boldsymbol{(}\mkern-3.0mu$ 4.12.

$\boldsymbol{\mkern-7.0mu)}$ For $J_{k}=I_{k}\setminus\\{1\\}$ where $4\leq
k\leq m$, $\\{t_{2},\ldots,t_{k}\\}$ is a C-string for $G_{J_{k}}$.

We argue by induction on $k$. By our induction hypothesis we have that
$\\{t_{2},t_{3},\ldots,t_{k-1}\\}$ is a C-string for
$G_{{J_{k}}\setminus\\{k\\}}$. From (4.10) $\\{t_{3},\ldots,t_{k}\\}$ is a
C-string for $G_{{J_{k}}\setminus\\{2\\}}$. Also, from (4.10),

$\displaystyle G_{{J_{k}}\setminus\\{2,k\\}}$
$\displaystyle\cong\mathbb{Z}_{2}\times\mathrm{Sym}({n-6})\,\,\,\text{ and }$
$\displaystyle G_{{J_{k}}\setminus\\{k\\}}$
$\displaystyle\cong\mathbb{Z}_{2}\times\mathrm{Sym}({n-5}).$

Hence $G_{{J_{k}}\setminus\\{2,k\\}}$ is a maximal subgroup of
$G_{{J_{k}}\setminus\\{k\\}}$. So, if $G_{{J_{k}}\setminus\\{k\\}}\cap
G_{{J_{k}}\setminus\\{2\\}}>G_{{J_{k}}\setminus\\{2,k\\}}$, then
$G_{{J_{k}}\setminus\\{k\\}}\leq G_{{J_{k}}\setminus\\{2\\}}$ which means
$t_{2}\in G_{{J_{k}}\setminus\\{2\\}}$. But $G_{{J_{k}}\setminus\\{2\\}}$
fixes 1 whereas $t_{2}$ does not, a contradiction. Therefore
$G_{{J_{k}}\setminus\\{k\\}}\cap
G_{{J_{k}}\setminus\\{2\\}}=G_{{J_{k}}\setminus\\{2,k\\}}$. Thus, using Lemma
2.3, we get (4.12).

######  $\boldsymbol{(}\mkern-3.0mu$ 4.13.

$\boldsymbol{\mkern-7.0mu)}$ For $4\leq k\leq m,$
$\\{t_{1},t_{2},\ldots,t_{k}\\}$ is a C-string for $G_{I_{k}}.$

We again argue by induction on $k$, with (4.8) starting the induction. So
$k>4$ and $\\{t_{1},t_{2},\ldots,t_{k-1}\\}$ is a C-string for $G_{I_{k-1}}$.
By (4.12) we have that $\\{t_{2},\ldots,t_{k-1}\\}$ is a C-string for
$G_{I_{k-1}\setminus\\{1\\}}$. Then, using (4.11) we have that

$\displaystyle G_{{I_{k}}\setminus\\{1\\}}$
$\displaystyle\cong\mathrm{Dih}({8})\times\mathrm{Sym}({k})\text{\quad and}$
$\displaystyle G_{I_{k-1}\setminus\\{1\\}}$
$\displaystyle\cong\mathrm{Dih}({8})\times\mathrm{Sym}({k-1}).$

Hence $G_{I_{k-1}\setminus\\{1\\}}$ is a maximal subgroup of
$G_{{I_{k}}\setminus\\{1\\}}$. So, if $G_{{I_{k}}\setminus\\{1\\}}\cap
G_{{I_{k-1}}}>G_{{I_{k-1}}\setminus\\{1\\}}$, then
$G_{{I_{k}}\setminus\\{1\\}}\leq G_{I_{k-1}}$. But then $t_{k}\in
G_{I_{k-1}},$ which is not the case as $t_{k}$ moves $k+3$ to $k+4$ which are
in different $G_{{I_{k-1}}}$-orbits. Therefore
$G_{{I_{k}}\setminus\\{1\\}}\cap G_{{I_{k-1}}}=G_{{I_{k-1}}\setminus\\{1\\}}$.
Thus, using Lemma 2.3, we get (4.13).

######  $\boldsymbol{(}\mkern-3.0mu$ 4.14.

$\boldsymbol{\mkern-7.0mu)}$ $\\{t_{1},t_{2},\ldots,t_{k}\\}$ is a C-string
for $G.$

Taking $m=k$ in (4.13) gives (4.14).

######  $\boldsymbol{(}\mkern-3.0mu$ 4.15.

$\boldsymbol{\mkern-7.0mu)}$ If $n$ is even, then $\\{t_{1},\ldots,t_{m}\\}$
is an unravelled C-string.

Set $w_{0}=\prod_{i=1}^{n}(i,n+i)$, $g=(t_{3}(t_{3}t_{4}))^{3}(=\beta_{6})$
and

$h=(t_{1}t_{2})^{2}t_{1}(t_{3}t_{2}t_{1})^{3}t_{3}^{t_{2}}(t_{1}t_{3}t_{2})^{4}.$

Then

$h=\prod_{i=1}^{5}(i,n+i)(6,7,n+6,n+7)\prod_{i=8}^{n}(i,n+i).$

Hence $gh=w_{0}$, and $g\in G_{\\{3,4\\}},h\in G_{\\{1,2,3\\}}$.

Put $M_{1}=\langle(i,n+i)|1\leq i\leq n\rangle$ and
$M_{2}=\langle(1,n+1)(i,n+i)|1<i\leq n\rangle.$ Since $n$ is even, $\langle
w_{0}\rangle\leq M_{2}\leq M_{1}$. Let $\overline{G}=G/N$ where $N$ is one of
$\langle w_{0}\rangle,M_{2},M_{1}.$ Then
$\overline{g}\overline{h^{-1}}\in\overline{G}_{\\{1,2,3\\}}\cap\overline{G}_{\\{3,4\\}}$
with $\overline{g}\neq\overline{t_{3}}$ whence
$\overline{G}_{\\{1,2,3\\}}\cap\overline{G}_{\\{3,4\\}}>\overline{G_{\\{3\\}}}$,
which proves (4.14).

Together, (4.14) and (4.15) establish Theorem 1.3.

## References

  * [1] Bosma, Wieb and Cannon, John and Playoust, Catherine, The Magma algebra system. I. The user language, J. Symbolic Comput., 24 (1997), no.3-4 235–265.
  * [2] Humphreys, James E. Reflection groups and Coxeter groups. Cambridge Studies in Advanced Mathematics, 29. Cambridge University Press, Cambridge, 1990.
  * [3] McMullen, Peter. and Schulte, Egon. Abstract regular polytopes. Encyclopedia of Mathematics and its Applications, 92. Cambridge University Press, Cambridge, 2002.
  * [4] Nicolaides, R. and Rowley, P.,  Unravelled Abstract Regular Polytopes. Preprint.
  * [5] Suzuki, M., Group theory v. 1, Grundlehren der mathematischen Wissenschaften, Springer-Verlag, 1982\.


# Representations of $ax+b$ group and Dirichlet Series

Hongyu He 111Key word: $ax+b$ group, unitary representation, smooth vectors,
distributions, Dirichlet series, Mellin transform, generalized matrix
coefficients
Department of Mathematics
Louisiana State University
email<EMAIL_ADDRESS>

###### Abstract

Let $G$ be the $ax+b$ group. There are essentially two irreducible infinite
dimensional unitary representations of $G$, $(\mu,L^{2}(\mathbb{R}^{+}))$ and
$(\mu^{*},L^{2}(\mathbb{R}^{+}))$. In this paper, we give various
characterizations about smooth vectors of $\mu$ and their Mellin transforms.
Let $\mathfrak{d}$ be a linear sum of delta distributions supported on the the
positive integers $\mathbb{Z}^{+}$. We study the Mellin transform of the
matrix coefficients $\mu_{\mathfrak{d},f}(a)$ with $f$ smooth. We express
these Mellin transforms in terms of the Dirichlet series $L(s,\mathfrak{d})$.
We determine a sufficient condition such that the generalized matrix
coefficient $\mu_{\mathfrak{d},f}$ is a locally integrable function and
estimate the $L^{2}$-norms of $\mu_{\mathfrak{d},f}$ over the Siegel set. We
further derive an inequality which may potentially be used to study the
Dirichlet series $L(s,\mathfrak{d})$.

##  1 Introduction

Let $G$ be the $ax+b$ group, the semidirect product
$\mathbb{R}\rtimes\mathbb{R}^{+}$. Here $\mathbb{R}^{+}$ is the multiplicative
group and $\mathbb{R}$ is the additive group. The group $G$ can be realized as
the matrix group consisting of

$\left(\begin{array}[]{clcr}a&0\\\
t&1\end{array}\right)=\left(\begin{array}[]{clcr}a&0\\\
0&1\end{array}\right)\left(\begin{array}[]{clcr}1&0\\\
t&1\end{array}\right),\qquad(a\in\mathbb{R}^{+},t\in\mathbb{R}).$

For simplicity, we denote the first factor by $a$ and second factor by
$b_{t}$. There are two equivalence classes of irreducible infinite dimensional
unitary representations, $\mu$ and its contragredient $\mu^{*}$. The
representation $\mu$ can be modeled on $L^{2}(\mathbb{R}^{+})$ by defining

$\mu(ab_{t})f(x)=a^{-\frac{1}{2}}\exp(-2\pi ita^{-1}x)f(a^{-1}x).$

The unitary representation $(\mu,L^{2}(\mathbb{R}^{+}))$ is well-known ([3]).
Let $\mathcal{H}^{\infty}$ be the Frechet space of smooth vectors in $\mu$.
The first result we proved in this paper is a characterization of smooth
vectors in terms of their Mellin transform $\mathcal{M}$.

###### Theorem 1.1

Let $g(z)$ be an analytic function on the right half plane
$\mathbf{H}=\\{\Re(z)>\frac{1}{2}\\}$. Then
$g(z)\in\mathcal{M}(\mathcal{H}^{\infty})$ if and only if the following hold

  1. 1.

For any $n>0$ and any finite closed interval $I$ in $(\frac{1}{2},\infty)$,
$g(\sigma+is_{1})$ is bounded by $C_{I,n}|\sigma+is_{1}|^{-n}$ for all
$s_{1}\in\mathbb{R}$ and $\sigma\in I$;

  2. 2.

$g(\sigma+is_{1})$ has a $L^{2}$ limit $g(\frac{1}{2}+is_{1})$ as
$\sigma\rightarrow\frac{1}{2}$.

  3. 3.

For any $n\geq 1$, ${s_{1}}^{n}g(\sigma+is_{1})$ has a $L^{2}$ limit as
$\sigma\rightarrow\frac{1}{2}$.

Let $(\mathcal{H}^{*})^{-\infty}$ be the dual topological vector space of
$\mathcal{H}^{\infty}$, equipped with the weak star topology. We can extend
the representation $\mu$ to its dual topological vector space
$(\mathcal{H}^{*})^{-\infty}$. We obtain a representation $\mu$ on the
distribution vectors. Since the Schwartz testing function
$\mathbf{D}(\mathbb{R}^{+})$ are dense in $\mathcal{H}^{\infty}$, (Theorem
LABEL:hinfinity), the space $(\mathcal{H}^{*})^{{-\infty}}$ is a linear
subspace of $\mathbf{D}^{\prime}(\mathbb{R})$ ([9]). Hence, all our discussion
can be carried out in the classical sense.

Let $\mathfrak{d}=\sum_{n=1}^{\infty}d_{n}\delta_{n}$ be a distribution in
$\mathbf{D}(\mathbb{R}^{+})^{\prime}$. Here $\delta_{n}$ is the Dirac $\delta$
function supported on $n$ with $n$ a positive integer. Suppose that
$\\{d_{n}\\}$ grow slower than a polynomial. Then
$\mathfrak{d}\in\mathcal{H}^{-\infty}$. In [4], we proved that the classical
definition of matrix coefficients can be extended to a pair of distributions
and the resulting generalized matrix coefficients are distributions on the
group $G$. Hence generalized matrix coefficient $\mu_{\mathfrak{d},f}(ab_{t})$
is defined for any $f\in(\mathcal{H}^{*})^{-\infty}$.

###### Theorem 1.2

Let $f$ be a locally integrable function in $\mathcal{H}^{-\infty}$ such that
for some $K>0$ and $\epsilon>0$

$\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da<\infty.$

Then $\mu_{\mathfrak{d},f}(ab_{t})$ is a well-defined locally integrable
function. In addition, for some positive constant $C_{\epsilon,\mathfrak{d}}$

$|d_{1}|^{2}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{K}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon,\mathfrak{d}}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Our estimate is the $L^{2}$-norm of $\mu_{\mathfrak{d},f}(ab_{t})$ over the
Siegel set.

Since $\\{d_{n}\\}$ is of polynomial growth, the Dirichlet series
$L(s,\mathfrak{d})=\sum_{n=1}^{\infty}d_{n}n^{-s}$ is well-defined in a right
half plane. For $\phi\in\mathcal{(}H^{*})^{\infty}$, the matrix coefficient
$\mu_{\mathfrak{d},\phi}(b_{T}a)$ is a smooth function. It can be expressed in
terms of Mellin transform.

###### Theorem 1.3

Suppose that the sequence $\\{d_{n}\\}$ is bounded by a multiple of
$n^{\epsilon}$ for any $\epsilon>0$. Suppose that $L(s,\mathfrak{d})$ is
holomorphic and bounded by a polynomial on the strip $\\{r\leq\Re(s)\leq
1+\epsilon\\}$ for some $\epsilon>0$. Then for any
$\phi\in\mathcal{H}^{\infty}$, we have

$\mu_{\mathfrak{d},\phi}(b_{T}a)=\langle\mu(b_{T}a)\mathfrak{d},\phi\rangle=\frac{1}{2\pi}\int
L(r+s_{1}i,\mathfrak{d})\mathcal{M}\\{\phi(x)\exp-2\pi
iTx\\}(r+s_{1}i)a^{-\frac{1}{2}-r-s_{1}i}ds_{1}.$

The main result of this paper is the following inequality.

###### Theorem 1.4

Let $\mathfrak{d}=\sum d_{n}\delta_{n}$ with $\\{d_{n}\\}$ bounded by
$C_{\tau}n^{\tau}$ for any $\tau>0$. Let $T_{1}\geq 1$, $\epsilon>0$ and
$C_{\epsilon}=\sum_{n=1}^{\infty}n^{-1-\epsilon}|d_{n}|^{2}$. Let $f\in
L^{2}(\mathbb{R}^{+},(1+x^{\epsilon})dx)$. We have

$\frac{1}{2}T_{1}C_{\epsilon}\int_{1}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{1}^{\infty}\int_{0}^{T_{1}}|\mu_{\mathfrak{d},f}(b_{T}a)|^{2}dTa^{\epsilon}\frac{da}{a}\leq
2T_{1}C_{\epsilon}\int_{1}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Let $\chi_{[1,\infty)}$ be the indicator function of $[1,\infty)$. Then there
exists a constant $C_{\epsilon,T_{1}}$ such that

$\int_{0}^{T_{1}}\|\mathcal{M}^{\frac{\epsilon}{2}}\\{\mu_{\mathfrak{d},f}(b_{T}a)\chi_{[1,\infty)}(a)\\}(s_{1})\|_{L^{2}(\mathbb{R})}^{2}dT\leq
C_{\epsilon,T_{1}}\|\mathcal{M}^{\frac{1+\epsilon}{2}}(f\chi_{[,\infty)})(s_{1})\|_{L^{2}(\mathbb{R})}^{2}.$

Here $\mathcal{M}^{\frac{\epsilon}{2}}\\{*\\}(s_{1})$ is Fourier-Mellin
transform, a variation of $\mathcal{M}\\{*\\}(\frac{\epsilon}{2}+s_{1}i)$.

Notice that Mellin transform
$\mathcal{M}^{\frac{\epsilon}{2}}\\{\mu_{\mathfrak{d},f}(b_{T}a)\chi_{[1,\infty)}(a)\\}(s_{1})$
may be obtained from
$\mathcal{M}^{\frac{\epsilon}{2}}\\{\mu_{\mathfrak{d},f}(b_{T}a)\\}(s_{1})$ by
solving a Riemann-Hilbert problem. By Theorem 1.3, our inequality may then be
expressed in terms of $L(s,\mathfrak{d})$. Furthermore, if we replace the
$ax+b$ group by $SL(2,\mathbb{R})$ and assume that $\\{d_{n}\\}$ are the
Fourier coefficients of an automorphic form, then the factor
$\chi_{[1,\infty)}$ can be removed. We will be able to bound certain
$L^{2}$-norm of the Mellin transform

$\mathcal{M}^{\frac{\epsilon}{2}}\\{\mu_{\mathfrak{d},f}(ga)\\}(s_{1})$

by certain $L^{2}$-norm of the form $M^{\frac{1+\epsilon}{2}}\\{f\\}(s_{1})$
([5]). Tis will shed some lights on the behavior of the Dirichlet series
$L(s,\mathfrak{d})$.

##  2 Representations of $ax+b$ group

Let us define $ax+b$ group to be the semidirect product of
$A\cong\mathbb{R}^{+}$ and $B\cong\mathbb{R}$. More precisely, for
$a\in\mathbb{R}^{+}$ and $b_{t}\in\mathbb{R}$. define
$(a,b_{t})=(a,0)(1,b_{t})=(1,b_{a^{-1}t})(a,0)$. We may simply write $ab_{t}$
for $(a,b_{t})$. Then

$ab_{t}=b_{a^{-1}t}a.$

The product rule is given by

$ab_{t}a^{\prime}b_{t^{\prime}}=aa^{\prime}b_{a^{\prime}t+t^{\prime}}.$

Let $G$ be the $ax+b$ group. The group $G$ can be parametrized by $ab_{t}$ or
$b_{T}a$ with $T=a^{-1}t$. The standard left invariant Haar measure is given
by $\frac{da}{a}dt=dadT$. The standard right invariant measure is given by
$\frac{da}{a}dT=a^{-1}\frac{da}{a}dt$.

The unitary dual of $G$ is easy to describe. There are one dimensional
representations parametrized by $i\lambda\in i\mathbb{R}$:

$\chi_{i\lambda}:ab_{t}\rightarrow a^{i\lambda}.$

There are two infinitely dimensional representations
$(\mu,L^{2}(\mathbb{R}^{+}))$ and $(\mu^{*},L^{2}(\mathbb{R}^{+}))$ defined as
follows

$\mu(a)f(x)=a^{-\frac{1}{2}}f(a^{-1}x),\qquad\mu^{*}(a)f(x)=a^{-\frac{1}{2}}f(a^{-1}x);$
$\mu(b_{t})f(x)=\exp(-2\pi itx)f(x),\qquad\mu^{*}(b_{t})f(x)=\exp(2\pi
ixt)f(x);$

for $f\in L^{2}(\mathbb{R}^{+})$, where $\mathbb{R}^{+}$ is equipped with the
Euclidean measure. See for example [3] for details of the unitary dual of $G$.
Notice that $\mu^{*}$ is simply the contragredient representation of $\mu$.
The representations $\mu$ and $\mu^{*}$ differ by a complex conjugation on the
action of $b_{t}$. We will mainly focus on $\mu$. The support of $\mu|_{B}$ is
the negative half line $\mathbb{R}^{-}$ and the support of $\mu^{*}|_{B}$ is
the positive half line $\mathbb{R}^{+}$.

We start with the following lemma about the action of the Lie algebra
$\mathfrak{g}$.

###### Lemma 2.1

The image of the universal enveloping algebra $\mu(U(\mathfrak{g}))$ is
spanned by

$\\{x^{m}\frac{d^{n}}{dx^{n}},m\geq n\geq 0\\}.$

Proof: Let $H$ be the infinitesimal generator of $A$ and $E$ be the
infinitesimal generator of $B$. Then $\mu(H)=-\frac{1}{2}-x\frac{d}{dx}$ and
$\mu(E)=-2\pi ix$. Since the identity $1$ is contained in the universal
enveloping algebra $U(\mathfrak{g})$, $\mu(U(\mathfrak{g}))$ is generated by
$\\{x,x\frac{d}{dx}\\}$ as a ring. One can then proceed by induction on $n$ to
show that $x^{n}\frac{d^{n}}{dx^{n}}\in\mu(U(\mathfrak{g}))$ for any $n\geq
0$. It follows that

$\\{x^{m}\frac{d^{n}}{dx^{n}},m\geq n\geq 0\\}\subseteq\mu(U(\mathfrak{g})).$

Conversely, $\mu(U(\mathfrak{g}))$ is spanned by $x^{k}(x\frac{d}{dx})^{n}$
since the Lie bracket $[x\frac{d}{dx},x]=x$. But $(x\frac{d}{dx})^{n}$ is a
linear combination of $x^{k}\frac{d^{k}}{dx^{k}}$ with $0\leq k\leq n$. Hence

$\mu(U(\mathfrak{g}))\subseteq\\{x^{m}\frac{d^{n}}{dx^{n}},m\geq n\geq 0\\}.$

$\Box$.

###  2.1 smooth vectors

Let $(\mu,\mathcal{H}=L^{2}(\mathbb{R}^{+}))$ be the irreducible unitary
representation of $G$ defined above. Let $\mathcal{H}^{\infty}$ be the space
of smooth vectors equipped with the canonical Frechet topology defined by
seminorms $\\{\|Df\|;D\in\mu(U(\mathfrak{g}))\\}$. Sometimes the seminorm
$\|Df\|$ is denoted by $\|f\|_{D}$.

###### Lemma 2.2

$\mathcal{H}^{\infty}=\\{f\in C^{\infty}(\mathbb{R}^{+})\mid
x^{m}\frac{d^{n}}{dx^{n}}f\in L^{2}(\mathbb{R}^{+})\,\,\,\forall\,\,m\geq
n\geq 0.\\}.$

Proof: Let $f$ be a smooth vector. Then $Df\in L^{2}(\mathbb{R})$ for any
$D\in\mu(U(\mathfrak{g}))$. The higher order derivatives of $f$ are all
locally $L^{2}$, hence locally $L^{1}$. Therefore any order derivatives of $f$
are continuous. This show that $f\in C^{\infty}(\mathbb{R}^{+})$. $\Box$.

We shall now give a more detailed description of $\mathcal{H}^{\infty}$.

###### Theorem 2.1

$f(x)\in\mathcal{H}^{\infty}$ if and only if $f\in C^{\infty}(\mathbb{R}^{+})$
and the following are satisfied

  1. 1.

For any $n\geq 0$, we have $x^{n}\frac{d^{n}f}{dx^{n}}\in L^{2}(0,\delta)$ for
some $\delta>0$;

  2. 2.

For any $n,m\in\mathbb{N}$ and some $k>0$, there exists a constant $c_{m,n,k}$
such that

$|\frac{d^{n}f(x)}{dx^{n}}|<c_{m,n,k}(1+|x|)^{-m}\qquad(x>k).$

Roughly, the derivatives of $f$ must not behave too badly near zero and must
be rapidly decaying near infinity. This follows from Sobelev inequality.

Proof: To prove ”if” part, assume $(1)$ and $(2)$ and $m\geq n\geq 0$. From
$(1)$, we have $x^{m}\frac{d^{n}f}{dx^{n}}\in L^{2}(0,\delta)$. By the rapidly
decaying condition $(2)$, $x^{m}\frac{d^{n}f}{dx^{n}}\in L^{2}(k,\infty)$.
Finally, since $f\in C^{\infty}(\mathbb{R})$, $x^{m}\frac{d^{n}f}{dx^{n}}$
will be bounded between $\delta$ and $k$, hence in $L^{2}(\delta,k)$.
Therefore, $x^{m}\frac{d^{n}f}{dx^{n}}\in L^{2}(\mathbb{R}^{+})$.

Conversely, let $m\geq n\geq 0$, $(1)$ is obvious. Observe that

$\begin{split}&|a^{m}\frac{d^{n}f(a)}{dx^{n}}-\frac{d^{n}f(1)}{dx^{n}}|=|\int_{1}^{a}(x^{m}\frac{d^{n}f}{dx^{n}})^{\prime}dx|\\\
=&|\int_{1}^{a}mx^{m-1}\frac{d^{n}f}{dx^{n}}+x^{m}\frac{d^{n+1}f}{dx^{n+1}}dx|\\\
\leq&m\int_{1}^{a}|x^{m-1}\frac{d^{n}f}{dx^{n}}|dx+\int_{1}^{a}|x^{m}\frac{d^{n+1}f}{dx^{n+1}}|dx\\\
\leq&m(\int_{1}^{a}|x^{m}\frac{d^{n}f}{dx^{n}}|^{2}dx)^{\frac{1}{2}}(\int_{1}^{a}|x^{-1}|^{2}dx)^{\frac{1}{2}}+(\int_{1}^{a}|x^{m+1}\frac{d^{n+1}f}{dx^{n+1}}|^{2}dx)^{\frac{1}{2}}(\int_{1}^{a}|x^{-1}|^{2}dx)^{\frac{1}{2}}\\\
\leq&C_{n,m}\\\ \end{split}$ (1)

Hence $x^{m}\frac{d^{n}f}{dx^{n}}(x)$ is bounded by a constant. $(2)$ follows.
$\Box$

We shall observe that
$C_{c}^{\infty}(\mathbb{R}^{+})\subseteq\mathcal{H}^{\infty}$.

###  2.2 2nd Characterization of $\mathcal{H}^{\infty}$

For any $x\in\mathbb{R}^{+}$, let $h=\ln x$. Define a unitary operator
$\mathcal{I}:L^{2}(\mathbb{R}^{+})\rightarrow L^{2}(\mathbb{R})$ by

$\mathcal{I}(f)(h)=(\exp\frac{1}{2}h)f(\exp h).$

It is easy to check that
$\mathcal{I}(\mu(H)f)(h)=-\frac{d}{dh}(\mathcal{I}(f))$ and
$\mathcal{I}(\mu(E)f)(h)=(-2\pi i\exp h)\mathcal{I}(f)(h)$. Immediately, we
have

###### Lemma 2.3

$\phi\in\mathcal{I}(\mathcal{H}^{\infty})$ if and only if

  1. 1.

$\phi\in C^{\infty}(\mathbb{R})$;

  2. 2.

for any $n\geq 0$, $\frac{d^{n}\phi(h)}{dh^{n}}\in L^{2}(-\infty,k)$ for some
$k$;

  3. 3.

$\frac{d^{n}\phi(h)}{dh^{n}}$ is bounded by $C_{n,m}(1+\exp h)^{-m}$ for any
$m>0$ and $h>0$.

We need to analyze $(2)$ a little deeper. Since ${\phi}^{\prime}\in
L^{2}(\mathbb{R})$, for $h<-1$,

$|\phi(-1)-\phi(h)|=|\int_{h}^{-1}{\phi}^{\prime}(y)dy|\leq\sqrt{-1-h}\|{\phi}^{\prime}\|_{L^{2}}.$

Hence $\phi(h)$ is bounded by a multiple of $\sqrt{|h|}$ at $-\infty$. If we
work a little harder, we have

###### Theorem 2.2

If $\phi(h)\in\mathcal{I}(\mathcal{H}^{\infty})$, then $\phi(h)$ and all its
higher order derivatives are bounded.

Proof: Consider the Fourier transform

$\mathcal{F}\phi(\xi)=\int\phi(h)\exp(-2\pi ih\xi)dh.$

Since $\phi^{(n)}(h)\in L^{2}(\mathbb{R})$, we have
$\xi^{n}\mathcal{F}\phi(\xi)\in L^{2}(\mathbb{R})$ for any $n\geq 0$. Then

$|\phi(h)|=|\int\mathcal{F}\phi(\xi)\exp(2\pi
ih\xi)d\xi|\leq\int_{-\infty}^{-1}|\mathcal{F}\phi(\xi)|d\xi+\int_{1}^{\infty}|\mathcal{F}\phi(\xi)|d\xi+C.$

The first two terms can all be bounded by the $L^{2}$-norm of
$\xi\mathcal{F}\phi(\xi)$. Hence $\phi(h)$ is bounded. Similarly,
$\frac{d^{n}\phi}{dh^{n}}(h)$ is bounded for all $n\geq 0$. $\Box$

This estimate applies to all $\phi(h)=\exp\frac{1}{2}h\,f(\exp h)$. Returning
to $f$, we have

###### Corollary 2.1

Let $f\in\mathcal{H}^{\infty}$. Then all the derivatives of $f$ will be
bounded by a multiple of $x^{-\frac{1}{2}}$ near $0$.

##  3 Fourier-Mellin transform of $\mathcal{H}^{\infty}$

Let $f$ be a locally integrable function on $\mathbb{R}^{+}$. Narrowly
speaking, Mellin transform

$\mathcal{M}f(s)=\int f(x)x^{s}\frac{dx}{x}$

is required to have a band of convergence. In this paper, we will loosely
define $\mathcal{M}f(s)=\int f(x)x^{s}\frac{dx}{x}$ as long as the integral
converges absolutely. Indeed, if $f(x)x^{\sigma-1}\in L^{1}(\mathbb{R}^{+})$
we can define

$\mathcal{M}^{\sigma}f(s_{1})=\int f(x)x^{\sigma+is_{1}}\frac{dx}{x}.$

Throughout this paper $s_{1}$ will always be real and $s$ will always be
complex with its imaginary part $is_{1}$.

If $x=\exp h$, then

$\mathcal{M}^{\sigma}f(s_{1})=\int f(\exp h)\exp(\sigma+is_{1})hdh.$

Notice that $\mathcal{M}^{\sigma}$ can be extended to $L^{2}$-functions and
beyond, by Fourier transform. We call $\mathcal{M}^{\sigma}$ the Fourier-
Mellin transform. In this context, the Fourier-Mellin inversion for
$L^{2}$-functions will be given by

$f(x)=\frac{1}{2\pi}\int M^{\sigma}f(s)x^{-\sigma-s_{1}i}ds_{1}.$

Unless such interpretation is needed, we will generally retain the usage of
Mellin-transform $\mathcal{M}f(s)$ or Mellin inversion.

###### Theorem 3.1

Suppose that $f\in\mathcal{H}^{\infty}$. Then

  1. 1.

$\mathcal{M}f(s)$ is defined and holomorphic on the open half plane
$\mathbf{H}=\\{\Re s>\frac{1}{2}\\}$;

  2. 2.

$\mathcal{M}f(s)$ decays faster than any $|s|^{-n}$ along any compact vertical
stripe in $\mathbf{H}$;

  3. 3.

$\mathcal{M}^{\sigma}f(s_{1})$ has an $L^{2}$-limit as
$\sigma\rightarrow\frac{1}{2}$. The $L^{2}$-limit is the Fourier-Mellin
transform $\mathcal{M}^{\frac{1}{2}}f(s_{1})$.

Proof: Since $f$ is bounded by a multiple of $x^{-\frac{1}{2}}$ near $0$ and
$x^{-n}$ near $\infty$, $\mathcal{M}f(s)$ is well-defined and analytic on the
open half plane $\mathbf{H}$. $(1)$ is proved.

Observe that

$\mathcal{M}f(\sigma+is_{1})=\int\mathcal{I}(f)(h)\,\exp(\sigma-\frac{1}{2})h\,\exp
is_{1}h\,dh$

in terms of Fourier transform. By Lemma 2.3 and Theorem 2.2, if
$\sigma>\frac{1}{2}$, $\mathcal{I}(f)(h)\exp(\sigma-\frac{1}{2})h$ and all its
derivatives will be in $L^{1}$. It follows that $\mathcal{M}f(\sigma+is_{1})$
decays faster than any $|s_{1}|^{-n}$ as $|s_{1}|\rightarrow\infty$ for
$\sigma$ in a finite closed interval in $(\frac{1}{2},\infty)$. $(2)$ is
proved.

Let $\sigma>\frac{1}{2}$. Observe that
$\mathcal{I}(f)(h)\exp(\sigma-\frac{1}{2})h\rightarrow\mathcal{I}(f)(h)$ in
$L^{2}$ as $\sigma\rightarrow\frac{1}{2}$. On the Fourier-Mellin transform
side, we have
$\mathcal{M}^{\sigma}f(s_{1})\rightarrow\mathcal{M}^{\frac{1}{2}}f(s_{1})$ in
$L^{2}(\mathbb{R})$. $\Box$.

We summarize how the Lie algebra act on the Mellin transform
$\mathcal{M}f(s)$.

###### Lemma 3.1

Let $f\in\mathcal{H}^{\infty}$. Then $\mathcal{M}(xf)(s)=\mathcal{M}(f)(s+1)$
and $\mathcal{M}(\mu(H)f)(s)=(s-\frac{1}{2})\mathcal{M}(f)(s)$ for any
$s\in\mathbf{H}$.

Notice that Theorem 3.1 (1)(2) are preserved under the action of Lie algebra.
We can now characterize the Mellin transform of $\mathcal{H}^{\infty}$.

###### Theorem 3.2

Let $g(s)$ be an analytic function on $\mathbf{H}$. Then
$g(s)\in\mathcal{M}(\mathcal{H}^{\infty})$ if and only if the following hold

  1. 1.

For any $n>0$ and any finite closed interval $I$ in $(\frac{1}{2},\infty)$,
$g(\sigma+is_{1})$ is bounded by $C_{I,n}|\sigma+is_{1}|^{-n}$ for all
$s_{1}\in\mathbb{R}$ and $\sigma\in I$;

  2. 2.

$g(\sigma+is_{1})$ has a $L^{2}$ limit $g(\frac{1}{2}+is_{1})$ as
$\sigma\rightarrow\frac{1}{2}$.

  3. 3.

For any $n\geq 1$, ${s_{1}}^{n}g(\sigma+is_{1})$ has a $L^{2}$ limit as
$\sigma\rightarrow\frac{1}{2}$.

Proof: The only if part can be easily seen from Theorem 3.1 and Lemma 3.1. To
show ”if” part, assume $(1)(2)(3)$. We must show that $g(z)=\mathcal{M}f(z)$
for some $f(x)\in\mathcal{H}^{\infty}$. First the $L^{2}$-limit of
${s_{1}}^{n}g(\sigma+is_{1})$ will be ${s_{1}}^{n}g(\frac{1}{2}+is_{1})$ as
$\sigma\rightarrow\frac{1}{2}$ for $s_{1}$ in any compact interval. Hence
${s_{1}}^{n}g(\sigma+is_{1})\rightarrow{s_{1}}^{n}g(\frac{1}{2}+is_{1})$ in
$L^{2}(\mathbb{R})$ as $\sigma\rightarrow\frac{1}{2}$.

Let

$f(x)=\frac{1}{2\pi}\int g(\sigma+s_{1}i)x^{-\sigma-
is_{1}}ds_{1}\qquad(\sigma>\frac{1}{2}).$

By (1), the integral above is independent of $\sigma$. In addition, for any
$n\in\mathbb{N}$,

$\frac{1}{2\pi}\int g(\sigma+s_{1}i)[\prod_{j=0}^{n-1}(-\sigma-
is_{1}-j)]x^{-\sigma-is_{1}-n}ds_{1}$

converges absolutely and defines the higher order derivative
$\frac{d^{n}f}{dx^{n}}(x)$. Hence $f(x)\in C^{\infty}(\mathbb{R}^{+})$. By
letting $\sigma$ arbitrarily large, $\frac{d^{n}f}{dx^{n}}(x)$ is uniformly
bounded by a multiple of $|x|^{-m}$ for $x>1$. This proves $(2)$ of Theorem
2.1. Near $0$, by letting $\sigma\rightarrow\frac{1}{2}$, $f(x)$ is bounded by
a multiple of $x^{-\frac{1}{2}-\delta}$ for arbitrarily small $\delta>0$. It
follows that $\mathcal{M}f(s)$ is well-defined on $\mathbb{H}$. By Mellin
inversion, $\mathcal{M}f(s)=g(s)$ for any $s\in\mathbb{H}$.

To show that $f\in\mathcal{H}^{\infty}$, it suffices to show $(1)$ of Theorem
2.1, namely $x^{n}\frac{d^{n}f}{dx^{n}}\in L^{2}(0,1)$. By Fourier-Mellin
inversion, $x^{\sigma}f(x)\in L^{2}(\mathbb{R}^{+},\frac{dx}{x})$ and
$|x^{\sigma-\frac{1}{2}}f(x)|\in L^{2}(\mathbb{R}^{+})$ for any
$\sigma>\frac{1}{2}$. By $(2)$, $|x^{\sigma-\frac{1}{2}}f(x)|$ has a $L^{2}$
limit in $L^{2}(\mathbb{R}^{+})$ as $\sigma\rightarrow\frac{1}{2}$. This
$L^{2}$-limit must be the pointwise limit $f(x)$. Hence $f(x)\in
L^{2}(\mathbb{R}^{+})$. Similarly, by $(3)$ and Lemma 3.1,
$(x\frac{d}{dx}+\frac{1}{2})^{n}f\in L^{2}(\mathbb{R}^{+})$. By Theorem 2.1,
$f\in\mathcal{H}^{\infty}$. $\Box$

##  4  Dirichlet series and distributions supported on $\mathbb{Z}^{+}$

Recall that $C_{c}^{\infty}(\mathbb{R}^{+})\subseteq\mathcal{H}^{\infty}$. In
addition, for any compact interval $K\subset\mathbb{R}^{+}$, the seminorms of
$\mathcal{H}^{\infty}$ yield the seminorms of $C_{c}^{\infty}(K)$ when
restricted to $C_{c}^{\infty}(K)$. Thus every continuous functional on the
Frechet space $\mathcal{H}^{\infty}$ is a distribution on $\mathbb{R}^{+}$ in
the sense of Schwartz. However a distribution on $\mathbb{R}^{+}$ may not be a
distribution on $\mathcal{H}^{\infty}$. Let $\mathcal{H}^{-\infty}$ be the
space of continuous linear functionals on $\mathcal{H}^{*}$, equipped with the
weak-star topology.

Let $d=(d_{1},d_{2},\ldots,d_{n}\ldots)$ be an infinite sequence of complex
numbers. We say that $d$ is of polynomial growth if there exists a
$k\in\mathbb{N}$ and $C_{k}>0$ such that $d_{n}<C_{k}n^{k}$ for all
$n\in\mathbb{N}$. Let $\delta_{n}$ be the $\delta$ distribution on
$n\in\mathbb{N}$. Define the distribution
$\mathfrak{d}=\sum_{n=1}^{\infty}d_{n}\delta_{n}$. If $d$ is of polynomial
growth, define the Dirichlet series
$L(s,\mathfrak{d})=\sum_{n=1}^{\infty}d_{n}n^{-s}$. $L(s,\mathfrak{d})$ is
well-defined for $\Re(s)>k+1$.

###  4.1 Distributions supported on $\mathbb{N}$

###### Theorem 4.1

If $d$ is of at most polynomial growth, then
$\mathfrak{d}\in\mathcal{H}^{-\infty}$.

Proof: It suffices to show that there exists a $F\in\mathcal{H}$ and $D\in
U(\mathfrak{g})$ such that $\mathfrak{d}=\mu(D)F$. Consider the antiderivative
of $\mathfrak{d}$:

$f(x)=\sum_{n\leq x}d_{n}\qquad(x>0).$

Obviously, $f(x)\leq C_{k}|x|^{k+1}$ for some $C_{k}>0$ and $k\in\mathbb{N}$.
Let $F(x)=x^{-k-2}f(x)$. Then $F(x)\in\mathcal{H}$ and

$\mathfrak{d}=\frac{d}{dx}x^{k+2}F(x)=x^{k+2}\frac{d}{dx}F(x)+(k+2)x^{k+1}F(x)=(x^{k+2}\frac{d}{dx}+(k+2)x^{k+1})F(x).$

Since $x^{k+2}\frac{d}{dx}+(k+2)x^{k+1}\in\mu(U(\mathfrak{g}))$,
$\mathfrak{d}\in\mathcal{H}^{-\infty}$. $\Box$

The Dirichlet series $L(s,\mathfrak{d})$ is well-defined for $s>k+1$ if
$\\{d_{n}\\}$ is bounded by $\\{C_{k}n^{k}\\}$ uniformly. We say that
$L(s,\mathfrak{d})\in\mathcal{R}$ if $\\{d_{n}\\}$ is bounded by
$C_{\epsilon}n^{\epsilon}$ for any $\epsilon>0$. This is the Ramanujan
condition. Clearly, $L(s,\mathfrak{d})$ is an analytic function on the open
half plan $\mathbb{H}_{1}=\\{z\in\mathbb{C}\mid\Re(z)>1\\}$. In addition, if
$L(s,\mathfrak{d})\in\mathcal{R}$ and
$L(s,\mathfrak{d}^{\prime})\in\mathcal{R}$, then
$L(s,\mathfrak{d})L(s,\mathfrak{d}^{\prime})$ is also analytic on
$\mathbb{H}_{1}$. In fact
$L(s,\mathfrak{d})L(s,\mathfrak{d}^{\prime})\in\mathcal{R}$. The following is
well-known.

###### Lemma 4.1

The product $L(s,\mathfrak{d})L(s,\mathfrak{d}^{\prime})=L(s,\mathfrak{j})$
with

$j_{l}=\sum_{mn=l}d_{m}d_{n}^{\prime}.$

If $L(s,\mathfrak{d}),L(s,\mathfrak{d}^{\prime})\in\mathcal{R}$, then
$L(s,\mathfrak{d})L(s,\mathfrak{d}^{\prime})\in\mathcal{R}$.

Hence $\mathcal{R}$ is a ring. We write $j$ as $d\sharp d^{\prime}$ and
$\mathfrak{j}$ as $\mathfrak{d}\sharp\mathfrak{d}^{\prime}$.

###  4.2 $\langle\mathfrak{d},f\rangle$ in terms of Mellin transform

Let $\mathcal{MR}$ be the subring of $\mathcal{R}$ consisting of
$g(z)\in\mathcal{R}$ with meromorphic continuation to $\mathbb{C}$. We shall
see how the pairing $\langle\mathfrak{d},f\rangle$ can be expressed in terms
of $L(s,\mathfrak{d})$.

###### Lemma 4.2

Let $f\in\mathcal{H}^{\infty}$ and $L(s,\mathfrak{d})\in\mathcal{MR}$. Then
for any $r>1$, we have

$\langle\mathfrak{d},f\rangle=\frac{1}{2\pi}\int
L(r+s_{1}i,\mathfrak{d})\mathcal{M}f(r+s_{1}i)ds_{1}.$

Proof: Let $r>1$. By the characterization of Mellin transform of
$\mathcal{H}^{\infty}$ and Mellin inversion, we have

$f(x)=\frac{1}{2\pi}\int\mathcal{M}f(r+is_{1})x^{-r-is_{1}}ds_{1}$

with $\mathcal{M}f(r+s_{1}i)$ fast decaying. Due to absolute convergence of
$L(r+s_{1}i,\mathfrak{d})=\sum_{k=1}^{\infty}\frac{d_{k}}{k^{r+s_{1}i}}$, we
have

$\langle\mathfrak{d},f\rangle=\sum_{k=1}^{\infty}d_{k}f(k)=\frac{1}{2\pi}\sum_{k=1}^{\infty}d_{k}\int\mathcal{M}f(r+is_{1}){k^{-r-s_{1}i}}ds_{1}$
$=\frac{1}{2\pi}\int\mathcal{M}f(r+is_{1})\sum_{k=1}^{\infty}d_{k}k^{-r-s_{1}i}ds_{1}=\frac{1}{2\pi}\int
L(r+s_{1}i,\mathfrak{d})\mathcal{M}f(r+s_{1}i)ds_{1}.$

Now we can move the integral from $r+i\mathbb{R}$ to the left of
$1+i\mathbb{R}$.

###### Theorem 4.2

Let $f\in\mathcal{H}^{\infty}$, $1\geq r>\frac{1}{2}$ and
$L(s,\mathfrak{d})\in\mathcal{MR}$. Suppose that $L(s,\mathfrak{d})$ is
holomorphic and bounded by a polynomial on the strip $\\{r\leq\Re(s)\leq
1+\epsilon\\}$ for some $\epsilon>0$. Then

$\langle\mathfrak{d},f\rangle=\frac{1}{2\pi}\int
L(r+s_{1}i,\mathfrak{d})\mathcal{M}f(r+s_{1}i)ds_{1}.$

We shall make some remarks here. First this result can be easily extended to
the situation that $L(s,\mathfrak{d})$ has finite number of poles in the
stripe $\\{r\leq\Re(s)\leq 1+\epsilon\\}$ by including the residue at the
poles in the equation. Care should be taken to manage the distribution of
infinitely many poles. But for the applications to $L$-functions, this theorem
is sufficient. Secondly, this result can also be extended to the left half
plane for $L$-functions. However, in our context, the Mellin transform
$\mathcal{M}f$ is only defined on the right half plane. Hence there is a
nature barrier at the critical line.

For any $\sigma>\frac{1}{2}$, $x^{\sigma-1+is_{1}}$ is a distribution in
$\mathcal{H}^{-\infty}$. We can see this by writing $x^{\sigma-1+is_{1}}$ as
$(x+1)\frac{x^{\sigma+is_{1}-1}}{x+1}$ and notice that $x+1$ is an operator in
$\mu(U(\mathfrak{g}))$.

###### Corollary 4.1

Let $r>\frac{1}{2}$ and $L(s,\mathfrak{d})\in\mathcal{M}\mathcal{R}$. Suppose
that $L(s,\mathfrak{d})$ is holomorphic and bounded by a polynomial on the
strip $\\{r\leq\Re(s)\leq 1+\epsilon\\}$ for some $\epsilon>0$. Then as a
distribution in $\mathcal{H}^{-\infty}$,

$\mathfrak{d}=\frac{1}{2\pi}\int^{*}L(r+s_{1}i,\mathfrak{d})x^{r+s_{1}i-1}ds_{1}.$

The right hand side is a weak-star integral with test functions in
$\mathcal{H}^{\infty}$.

###  4.3 $\langle\mu(a)\mathfrak{d},f\rangle$ in terms of Mellin transform

Observe that

$\mathcal{M}(\mu^{*}(a^{-1})f)(s)=\int_{0}^{\infty}a^{\frac{1}{2}}f(ax)x^{s}\frac{dx}{x}=a^{\frac{1}{2}-s}\mathcal{M}f(s).$

###### Corollary 4.2

Under the same assumption as Theorem 4.2,

$\langle\mu(a)\mathfrak{d},f\rangle=\frac{1}{2\pi}\int
L(r+s_{1}i,\mathfrak{d})\mathcal{M}f(r+s_{1}i)a^{\frac{1}{2}-r-s_{1}i}ds_{1}.$

or equivalently

$\mathcal{M}(\langle\mu(a)\mathfrak{d},f\rangle)(r-\frac{1}{2}+s_{1}i)=L(r+s_{1}i,\mathfrak{d})\mathcal{M}f(r+s_{1}i).$

##  5 Fundamental Inequality

We shall now consider the (generalized) matrix coefficient
$\langle\mu(ab_{t})\mathfrak{d},f\rangle$ with
$f\in(\mathcal{H}^{*})^{\infty}$. We may still regard $f$ as a function in
$\mathcal{H}^{\infty}$ and keep in mind that the action of $G$ will be given
by $\mu^{*}$. We have

$\langle\mu(ab_{t})\mathfrak{d},f\rangle=\langle\mu(b_{t})\mathfrak{d},\mu^{*}(a^{-1})f\rangle=\sum_{n\in\mathbb{Z}^{+}}a^{\frac{1}{2}}d_{n}f(an)\exp(-2\pi
itn).$

Since $f(x)$ decays very fast, $\sum_{n\in\mathbb{N}}$ converges as a Fourier
series. We write this matrix coefficient as $\mu_{\mathfrak{d},f}(ab_{t})$.

###  5.1 Fourier Series Estimates

By Parseval’s theorem, we have

###### Lemma 5.1

Let $\mathfrak{d}$ be of polynomial growth and $f\in\mathcal{H}^{\infty}$, we
have

$\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}dt=a\sum_{n=1}^{\infty}|f(an)d_{n}|^{2}.$

###### Theorem 5.1

Let $L(s,\mathfrak{d})\in\mathcal{R}$, $f\in\mathcal{H}^{\infty}$, and
$\epsilon>0$. Set
$C_{\epsilon}=\sum_{n=1}^{\infty}n^{-1-\epsilon}|d_{n}|^{2}$. Then for any
$\delta\geq 0$, we have

$|d_{1}|^{2}\int_{\delta}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{\delta}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{\delta}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Proof: Since $\mathfrak{d}$ satisfies the Ramanujan condition, $C_{\epsilon}$
is well-defined. The lower bound follows from

$\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})^{2}|dt\geq
a^{\epsilon}|f(a)d_{1}|^{2}.$

The upper bound can be established as follows.

$\begin{split}&\int_{\delta}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}ta^{\epsilon}\frac{da}{a}\\\
=&\int_{\delta}^{\infty}\sum_{n=1}^{\infty}|f(an)d_{n}|^{2}a^{\epsilon}da\\\
=&\sum_{n=1}^{\infty}(|d_{n}|^{2}\int_{\delta}^{\infty}a^{\epsilon}|f(an)|^{2}da)\\\
=&\sum_{n=1}^{\infty}(|d_{n}|^{2}\int_{n\delta}^{\infty}a^{\epsilon}n^{-1-\epsilon}|f(a)|^{2}da)\\\
\leq&(\sum_{n=1}^{\infty}n^{-1-\epsilon}|d_{n}|^{2})\int_{{\delta}}^{\infty}a^{\epsilon}|f(a)|^{2}da.\end{split}$
(2)

$\Box$.

Notice that the domain we worked with $\\{\delta<a,0\leq t\leq 1\\}$ is
classically known as the Siegel domain. However, the measure we use is
different. In particular, if $\delta=0$, we have

$|d_{1}|^{2}\int_{\mathbb{R}^{+}}a^{\epsilon}|f(a)|^{2}da\leq\int_{\mathbb{R}^{+}}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{\mathbb{R}^{+}}a^{\epsilon}|f(a)|^{2}da.$

###  5.2 Generalized Matrix Coefficients as functions

Given any $f\in\mathcal{H}^{-\infty}$, we have shown in [4] that generalized
matrix coefficient $\langle\mu(g)\mathfrak{d},f\rangle$ is a distribution on
$G$. The question then arises, for which $f$,
$\langle\mu(g)\mathfrak{d},f\rangle$ will be a locally integrable function.
One simple guess is $f\in\mathcal{H}$? This turns out not to be the case for
general $\mathfrak{d}$ satisfying the Ramanujan condition ([12]). However, if
we make assumption that for some $\epsilon>0$ and $K>0$,

$\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da<\infty$

then $\mu_{\mathfrak{d},f}(ab_{t})$ will be a locally integrable function. We
first prove a weaker version of this.

###### Theorem 5.2

Let $f$ be a locally square integrable function in $\mathcal{H}^{-\infty}$
such that for some $\epsilon>0$ ,

$\int_{0}^{\infty}a^{\epsilon}|f(a)|^{2}da<\infty.$

Then $\mu_{\mathfrak{d},f}(ab_{t})$ is a well-defined locally square
integrable function. In addition,

$|d_{1}|^{2}\int_{0}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{0}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{0}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Proof: Let $\\{f_{n}\\}$ be a sequence of function in
$C_{c}^{\infty}(\mathbb{R}^{+})\subseteq\mathcal{H}^{\infty}$ such that

$\int|f_{n}(a)-f(a)|^{2}a^{\epsilon}da\rightarrow
0\qquad(as\,\,n\rightarrow\infty).$

In particular, $\\{f_{n}\\}$ is a Cauchy sequence in
$L^{2}(\mathbb{R}^{+},x^{\epsilon}dx)$. By Theorem 5.1,
$\\{\mu_{\mathfrak{d},f_{n}}(ab_{t})\\}$ is a Cauchy sequence in
$L^{2}(G,a^{\epsilon}dt\frac{da}{a})$. It has a $L^{2}$-limit $\psi$ under the
measure $a^{\epsilon}dt\frac{da}{a}$. Hence, as distributions in
$\mathbf{D}(G)^{\prime}$, $\mu_{\mathfrak{d},f_{n}}(g)\rightarrow\psi(g)$
under the weak star topology. By [4],
$\mu_{\mathfrak{d},f}(g)=\psi(g)\,\,a.e.$. For every $n$,

$|d_{1}|^{2}\int_{0}^{\infty}a^{\epsilon}|f_{n}(a)|^{2}da\leq\int_{0}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f_{n}}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{0}^{\infty}a^{\epsilon}|f_{n}(a)|^{2}da.$

By taking $n\rightarrow\infty$, our inequalities follow. $\Box$

Knowing that $\mu_{\mathfrak{d},f}(ab_{t})$ is a square integrable function,
by [4], we can approximate $\mu_{\mathfrak{d},f}(ab_{t})$ by
$\mu_{\mathfrak{d},f_{n}}(ab_{t})$ as long as $f_{n}\rightarrow f$ under the
weak star topology.

###### Theorem 5.3

Let $f$ be a locally integrable function in $\mathcal{H}^{-\infty}$ such that
for some $K>0$ and $\epsilon>0$

$\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da<\infty.$

Then $\mu_{\mathfrak{d},f}(ab_{t})$ is a well-defined locally integrable
function. In addition,

$|d_{1}|^{2}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{K}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Proof: We may write $f$ as a sum of two locally integrable functions
$f_{1}+f_{2}$ with $f_{1}(x)=0\,\,(\forall x\geq K)$ and
$f_{2}(x)=0\,\,(\forall x<K)$. Then
$\mu_{\mathfrak{d},f}(ab_{t})=\mu_{\mathfrak{d},f_{1}}(ab_{t})+\mu_{\mathfrak{d},f_{2}}(ab_{t})$.
By Theorem 5.2, $\mu_{\mathfrak{d},f_{2}}(ab_{t})$ is well-defined, locally
square integrable. Since $f_{1}(x)=0$ for any $x\geq K$,
$\mu_{\mathfrak{d},f_{1}}(ab_{t})=\sum_{n\in\mathbb{Z}^{+}}a^{\frac{1}{2}}d_{n}f(an)\exp(-2\pi
itn)$ is locally a finite sum of locally integrable function, hence locally
integrable. In addition, $\mu_{\mathfrak{d},f_{1}}(ab_{t})=0$ for any $a\geq
K$. It follows that
$\mu_{\mathfrak{d},f}(ab_{t})=\mu_{\mathfrak{d},f_{2}}(ab_{t})$ for any $a\geq
K$. By essentially the same proof as Theorem 5.2,

$|d_{1}|^{2}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{K}^{\infty}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{\epsilon}\frac{da}{a}\leq
C_{\epsilon}\int_{K}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

$\Box$

###  5.3 Fundamental Inequality

Define the subset of $G$, $\Omega_{T_{1}}=\\{b_{T}a\mid
T\in[0,T_{1}],a\in[1,\infty)\\}$. In terms of $ab_{t}$ coordinates, we have

$\Omega_{T_{1}}=\\{ab_{t}\mid t\in[0,aT_{1}],a\in[1,\infty)\\}.$

###### Theorem 5.4

Let $L(s,\mathfrak{d})\in\mathcal{R}$, $T_{1}\geq 1$ and $\epsilon>0$. Let
$f\in L^{2}(\mathbb{R}^{+},(1+x^{\epsilon})dx)$. Set
$C_{\epsilon}=\sum_{n=1}^{\infty}n^{-1-\epsilon}|d_{n}|^{2}$. We have

$\frac{1}{2}T_{1}C_{\epsilon}\int_{1}^{\infty}a^{\epsilon}|f(a)|^{2}da\leq\int_{1}^{\infty}\int_{0}^{T_{1}}|\mu_{\mathfrak{d},f}(b_{T}a)|^{2}dTa^{\epsilon}\frac{da}{a}\leq
2T_{1}C_{\epsilon}\int_{1}^{\infty}a^{\epsilon}|f(a)|^{2}da.$

Proof: If $f(x)\in L^{2}(\mathbb{R}^{+},(1+x^{\epsilon})dx)$, then Theorem 5.2
holds. To prove the upper bound, we observe

$\begin{split}&\int_{1}^{\infty}\int_{0}^{T_{1}}|\mu_{\mathfrak{d},f}(b_{T}a)|^{2}dTa^{\epsilon}\frac{da}{a}\\\
=&\int_{1}^{\infty}\int_{0}^{aT_{1}}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}a^{-1}dta^{\epsilon}\frac{da}{a}\\\
\leq&\int_{1}^{\infty}\lceil aT_{1}\rceil
a^{\epsilon-1}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}dt\frac{da}{a}\\\
\leq&\int_{1}^{\infty}2aT_{1}a^{\epsilon-1}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}dt\frac{da}{a}\\\
=&2T_{1}\int_{1}^{\infty}a^{\epsilon}\int_{0}^{1}|\mu_{\mathfrak{d},f}(ab_{t})|^{2}dt\frac{da}{a}\\\
\leq&2T_{1}C_{\epsilon}\int_{1}^{\infty}a^{\epsilon}|f(a)|^{2}da\end{split}$
(3)

Here $\lceil aT_{1}\rceil$ is the minimal integer greater or equal to
$aT_{1}$. We use the fact $\lceil aT_{1}\rceil\leq 2aT_{1}$. The lower bound
follows similarly. $\Box$

If $T_{1}$ is any positive real number, the inequalities still hold if we
replace $T_{1}C_{\epsilon}$ be a constant $C_{T_{1}}C_{\epsilon}$. In
addition, we can allow $a\in[\delta,\infty)$, then the inequalities will
depend on $\delta$. However, we will not have any similar bounds if
$a\in(0,\infty)$ since the behavior of $f(ab_{t})$ near $a=0$ is hard to
control without any additional assumption. Nevertheless, If the group is
$SL(2)$ and $\mathfrak{d}$ comes from the Fourier coefficients of an
automorphic form, we can bound the $L^{2}$-norm of
$\mu_{\mathfrak{d},f}(B_{T}a)$ over the whole space $a\in(0,\infty)$ ([5]).

###  5.4 Inequality in terms of Mellin Transform

For any locally integrable function $u(x)$ on $\mathbb{R}^{+}$, write

$u(x)=u_{+}(x)+u_{-}(x)$

such that $u_{+}(x)$ is supported on $[1,\infty)$ and $u_{-}(x)$ is supported
on $(0,1]$. Let $f(x)$ be a function in
$L^{2}(\mathbb{R}^{+},(1+x^{\epsilon})dx)$. Then Theorem 5.4 asserts that

$\int_{0}^{T_{1}}|\mu_{\mathfrak{d},\exp(-2\pi
ixT)f}(a)_{+}|^{2}_{L^{2}(\mathbb{R}^{+},a^{\epsilon}\frac{da}{a})}dT\leq
2T_{1}C_{\epsilon}\|f_{+}\|^{2}_{L^{2}(\mathbb{R}^{+},a^{\epsilon}\frac{da}{a})}.$

By Fubini’s theorem, for almost all $T\in[0,T_{1}]$,
$\mu_{\mathfrak{d},\exp(-2\pi ixT)f}(a)_{+}\in
L^{2}(\mathbb{R}^{+},a^{\epsilon}\frac{da}{a})$.

Recall that for $0<\sigma\leq\frac{\epsilon}{2}$, The Fourier-Mellin transform
$\mathcal{M}^{\sigma}f$ is well-defined and

$\|\mathcal{M}^{\sigma}f\|^{2}=\frac{1}{2\pi}\int_{\mathbb{R}^{+}}|f(x)|^{2}x^{2\sigma-1}dx$

We can now restate Theorem 5.4.

###### Theorem 5.5

Let $L(s,\mathfrak{d})\in\mathcal{R}$, $T_{1}\geq 1$ and $\epsilon>0$. Let
$f\in L^{2}(\mathbb{R}^{+},(1+x^{\epsilon})dx)$. There exists a constant
$C_{\epsilon,T_{1},\mathfrak{d}}$ such that

$\int_{0}^{T_{1}}\|\mathcal{M}^{\frac{\epsilon}{2}}(\mu_{\mathfrak{d},\exp(-2\pi
iTx)f}(a)_{+})(s_{1})\|_{L^{2}(\mathbb{R})}^{2}dT\leq
C_{\epsilon,T_{1},\mathfrak{d}}\|\mathcal{M}^{\frac{1+\epsilon}{2}}(f_{+})(s_{1})\|_{L^{2}(\mathbb{R})}^{2}.$

Recall that

$\mathcal{M}^{\frac{\epsilon}{2}}(\mu_{\mathfrak{d},\exp(-2\pi
iTx)f}(a))(s_{1})=L(\frac{1+\epsilon}{2}+s_{1}i,\mathfrak{d})\mathcal{M}^{\frac{1+\epsilon}{2}}\\{f(x)\exp-2\pi
iTx\\}(s_{1})$

The Mellin transform $\mathcal{M}\\{\mu_{\mathfrak{d},\exp(-2\pi
iTx)f}(a)_{+}\\}(\frac{\epsilon}{2}+s_{1}i)$, which lives in the Hardy space
of a right half plane, may be obtained explicitly from The Mellin transform

$\mathcal{M}\\{\mu_{\mathfrak{d},\exp(-2\pi
iTx)f}(a)\\}(\frac{\epsilon}{2}+s_{1}i)$

by solving a Riemann-Hilbert problem. We indeed get an inequlity in terms of
the Dirichlet series $L(\frac{1+\epsilon}{2}+s_{1}i,\mathfrak{d})$.
Nevertheless, in order to obtain a bound on $L(s,\mathfrak{d})$ we will have
to control

$\int_{0}^{T_{1}}\|\mathcal{M}^{\frac{\epsilon}{2}}(\mu_{\mathfrak{d},\exp(-2\pi
iTx)f}(a)_{-})(s_{1})\|_{L^{2}(\mathbb{R})}^{2}dT$

with $a\in(0,1]$. This can be achieved when $\mathfrak{d}$ comes from an
automorphic distribution for an arithmetic subgroup of $SL(2,\mathbb{Z})$ ([7]
[1] [5]).

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