# Weak crossed product orders arising from partitions of a finite group

Ch.Lamprakis<EMAIL_ADDRESS>Th.Theohari-Apostolidi
<EMAIL_ADDRESS>School of Mathematics, Aristotle University of
Thessaloniki, Thessaloniki 54124, Greece

###### Abstract

Let $R$ be a complete discrete valuation ring with quotient field $K$, $L$ a
finite Galois extension of $K$ with Galois group $G$ and $S$ the integral
closure of $R$ in $L$. In this article, using elements of the monoid $Sl(G)$
introduced in [8], we construct certain crossed product orders in the case
that the extension $S/R$ is unramified. We associate an element of $Sl(G)$ to
every crossed product order and give a criterion for inflating idempotent
$2$-cocycles.

###### keywords:

Galois extension , Galois cohomology , Crossed product orders , Weak
$2$-cocycle , Weak crossed product algebra , Semigroups , Inflation of
idempotent $2$-cocycles.

###### MSC:

[2010] Primary 16S35 , 16H05 , 11S20 , 11S45 , Secondary 16W50

††journal: Journal of Algebra

##  1 Introduction

Let $R$ be a complete discrete valuation ring with quotient field $K$, $L$ a
finite Galois extension of $K$ with Galois group $G$, $S$ the integral closure
of $R$ in $L$ and $\pi S$ (resp.$\pi_{0}R$) the unique maximal ideal of $S$
(resp.$R$), $L^{*}:=L\setminus\\{0\\}$, $S^{*}:=S\setminus\\{0\\}$ and $U(S)$
the group of units of $S$.

A function $f:G\times G\longrightarrow L^{*}$ is called normalized $2$-cocycle
when the following conditions are satisfied:

$\displaystyle
f(\sigma,\tau)f(\sigma\tau,\rho)=f^{\sigma}(\tau,\rho)f(\sigma,\tau\rho),\quad\sigma,\tau\rho\in
G,$ (1.1) $\displaystyle f(1,\sigma)=f(\sigma,1)=1,\quad\text{for}\;\sigma\in
G,$ (1.2)

where $l^{\sigma}$ means $\sigma(l)$, for $\sigma\in G$ and $l\in L$, and $1$
is the unit element of $G$. Associated to a $2$-cocycle $f$ there is a
$K$-algebra $A_{f}$ called the crossed product algebra associated to $f$. The
$K$-algebra $A_{f}$ is defined as an $L$-vector space

$A_{f}=\bigoplus_{\sigma\in G}Lx_{\sigma}$

having as an $L$-basis the symbols $x_{\sigma}$, $\sigma\in G$, and
multiplication defined by the rules

$x_{\sigma}l=l^{\sigma}x_{\sigma}\quad\text{and}\quad
x_{\sigma}x_{\tau}=f(\sigma,\tau)x_{\sigma\tau},$

for $\sigma,\tau\in G$ and $l\in L$. The cocycle condition (1.1) guarantees
that $A_{f}$ is an associative $K$-algebra with unit element $x_{1}$, that we
denote also by $1$, and moreover $A_{f}$ is a central simple $K$-algebra
([11],Theorem 29.6).

The set $Z^{2}(G,L^{*})$ of all $2$-cocycles of $G$ on $L^{*}$ becomes a group
under the pointwise multiplication called the group of boundaries of $G$ on
$L^{*}$ ([3]). The subgroup $B^{2}(G,L^{*})$ of $Z^{2}(G,L^{*})$ consisting of
$2$-cocycles $b:G\times G\longrightarrow L^{*}$ such that

$b(\sigma,\tau)=a(\sigma)a^{\sigma}(\tau)a(\sigma\tau)^{-1},$

for $\sigma,\tau\in G$ and for a function $a:G\longrightarrow L^{*}$, is
called the group of coboundaries of $G$ on $L^{*}$ and the quotient group
$Z^{2}(G,L^{*})/B^{2}(G,L^{*})$ is the second cohomology group $H^{2}(G,L)$
([11],Section 29).

If we allow $f$ to take values on $L$ instead of $L^{*}$, then we are
reffering to a weak $2$-cocycle of $G$ on $L$. D.E. Haile, R.G. Larson and
M.E. Sweedler in [4] studied the weak $2$-cocycles and they introduced a new
cohomology theory based on them. The set $Z^{2}(G,L)$ of $2$-cocycles of $G$
on $L$ becomes a monoid under the pointwise multiplication and

$M^{2}(G,L):=Z^{2}(G,L)/B^{2}(G,L^{*})$

is also a monoid, the invertible elements of which are exactly the elements of
$H^{2}(G,L)$. Moreover every class $[f]\in M^{2}(G,L)$ contains a unique
idempotent $2$-cocycle, i.e. that takes values on the set $\\{0,1\\}$ of the
idempotent elements of $L$.

A function $f:G\times G\longrightarrow S^{*}$ satisfying relations (1.1) and
(1.2) is called a normalized weak $2$-cocycle of $G$ on $S^{*}$. The ring

$\Lambda_{f}:=\bigoplus_{\sigma\in G}Sx_{\sigma}:=(S/R,f)$

is an $R$-order in the crossed product $K$-algebra $A_{f}$ and is called the
weak crossed product order associated to $f$. D.E. Haile introduced and
studied the weak crossed product order $\Lambda_{f}$ in case $R$ is a discrete
valuation ring and the extension $L/K$ is unramified ([6]). Let
$Z^{2}(G,S^{*})$ be the set of boundaries of $G$ on $S^{*}$, that is, the set
of all weak $2$-cocycles of $G$ on $S^{*}$. Then $Z^{2}(G,S^{*})$ becomes a
monoid under the pointwise multiplication having as a subgroup the set
$B^{2}(G,U(S))$ consisting of all functions $b:G\times G\longrightarrow U(S)$
such that

$b(\sigma,\tau)=a(\sigma)a^{\sigma}(\tau)a(\sigma\tau)^{-1},$

for $\sigma,\tau\in G$ and a function $a:G\longrightarrow U(S)$, called the
group of coboundaries of $G$ on $U(S)$. The set $Z^{2}(G,S^{*})/B^{2}(G,U(S))$
of equivalent classes, denoted $N^{2}(G,S)$, is a monoid under the pointwise
multiplication. Since every element of $Z^{2}(G,L)$ or of $Z^{2}(G,S^{*})$ is
cohomologous to a normalized $2$-cocycle, in the whole of the article we
always take normalized $2$-cocycles.

In this article we assume that the extension $L/K$ is unramified with finite
residue field. Then the extension $S/R$ is also a Galois extension and $G\cong
Gal(S/R)\cong Gal(\overline{S}/\overline{R})$, where $\overline{S}:=S/\pi S$
and $\overline{R}:=R/\pi_{0}R$.

In [8], the authors introduced the monoid $Sl(G)$, for a finite group $G$,
consisting of all functions $r:G\longrightarrow\mathbb{N}$ satisfying the
relations $r(1)=0$ and $r(\sigma\tau)\leq r(\sigma)+r(\tau)$, for all
$\sigma,\tau\in G$. These functions arose naturally from some partitions of
$G$ and were used to construct idempotent $2$-cocycles by the rule

${}e_{r}(\sigma,\tau)=\begin{cases}1,&\text{if}\;r(\sigma\tau)=r(\sigma)+r(\tau),\\\
0,&\text{if}\;r(\sigma\tau)<r(\sigma)+r(\tau).\end{cases}$ (1.3)

([8],Theorem 5.2).

In this article, using the elements of $Sl(G)$ we establish a relation between
weak crossed product orders of $G$ on $S^{*}$ and weak crossed product
algebras of $G$ on $L$. It consists of five sections including Introduction.
In Section 2 we construct new elements of $Sl(G)$ from a given one. In Section
3 we define the monoid $B^{2}(G,S^{*})\subseteq Z^{2}(G,S^{*})$ and we prove
that the weak crossed product orders associated to elements of this set are
not hereditary. Moreover we prove that the monoid $Sl(G)$ is all we need to
describe up to equivalence the monoid $B^{2}(G,S^{*})$. In Section 4 we
associate an element of $Sl(G)$ to a weak $2$-cocycle $f:G\times
G\longrightarrow S^{*}$ and we give two applications. Finally in Section 5,
for a normal subgroup $H$ of $G$, we give a necessary and sufficient condition
for inflating idempotent weak $2$-cocycles from $G/H$ to $G$ and specialize
this procedure using elements of $Sl(G)$.

We refer to [4] and [11] for Galois theory and Galois cohomology theory of
groups, to [11] for crossed product algebras, to [2] for orders and crossed
product orders, to [6] and [7] for weak crossed product orders, to [12] for
local fields and ramification theory and to [2], [9], [10] for ring theory and
algebras.

##  2 Preliminaries

Throughout the rest of the article we consider the notation and the
assumptions of Section 1. Moreover we denote by $E^{2}(G,L)$ the set of
idempotent $2$-cocycles of $G$ on $L$.

One of the main results of ([4],Section 7) is that

$M^{2}(G,L)=\bigcupdot M_{e}^{2}(G,L),$

where the union is over all idempotent $2$-cocycles $e$ and

$\displaystyle M_{e}^{2}(G,L):$ $\displaystyle=$ $\displaystyle\\{[f]\in
M^{2}(G,L):[f][e]=[f]\;\text{and}\exists[g]\in M^{2}(G,L)$
$\displaystyle\text{such that}\;[f][g]=[e]\\},$

which is a group with unit the class $[e]$. For an element $f\in Z^{2}(G,L)$,
let $H(f)=\\{\sigma\in G:f(\sigma,\sigma^{-1})\neq 0\\}$. It was shown in
([4],Section 10) that $H(f)$ is a subgroup of $G$, called the inertial group
of $f$. Moreover $H(f)=H(e)$, where $e$ is the unique idempotent $2$-cocycle
such that $[f]\in M_{e}^{2}(G,L)$. Similarly if $f\in Z^{2}(G,S^{*})$, then
the set $\\{\sigma\in G:f(\sigma,\sigma^{-1})\in U(S)\\}$ is also a subgroup
of $G$ ([5]). We call this subgroup also the inertial group of the $2$-cocycle
$f$ and we denote it also as $H(f)$. Occasionally we use the notion $H_{G}(f)$
for $H(f)$ to emphasise the group $G$.

The set $Sl(G)$ becomes an additive monoid with operation
$(r_{1}+r_{2})(\sigma)=r_{1}(\sigma)+r_{2}(\sigma)$, for $r_{1},r_{2}\in
Sl(G)$ and $\sigma\in G$, with unit element the function $r(\sigma)=0$, for
all $\sigma\in G$ ([8],Proposition 8.1). Moreover the set $M_{r}:=\\{\sigma\in
G:r(\sigma)=0\\}$ is a subgroup of $G$ ([8],Proposition 4.2).

The next proposition is a criterion for checking the equality of elements of
$Sl(G)$.

###### Proposition 2.1

Let $r_{1},r_{2}\in Sl(G)$. Then $r_{1}=r_{2}$ if and only if

$r_{1}(\sigma)+r_{1}(\tau)-r_{1}(\sigma\tau)=r_{2}(\sigma)+r_{2}(\tau)-r_{2}(\sigma\tau),$

for $\sigma,\tau\in G$.

Proof: If $r_{1}=r_{2}$, then the equality is obvious. Now let
$r_{1}(\sigma)+r_{1}(\tau)-r_{1}(\sigma\tau)=r_{2}(\sigma)+r_{2}(\tau)-r_{2}(\sigma\tau)$,
for $\sigma,\tau\in G$. First we show that $M_{r_{1}}=M_{r_{2}}$. Both sets
are not empty since they contain the unit of $G$. Let $\sigma\in M_{r_{1}}$.
Then, for $\tau=\sigma^{-1}$, we have
$r_{1}(\sigma)+r_{1}(\sigma^{-1})-r_{1}(1)=r_{2}(\sigma)+r_{2}(\sigma^{-1})-r_{2}(1)$.
Since $r_{1},r_{2}\in Sl(G)$ it follows that $r_{1}(1)=r_{2}(1)=0$. Also
$\sigma^{-1}\in M_{r_{1}}$ so $r_{1}(\sigma^{-1})=0$. We deduce that
$r_{2}(\sigma)+r_{2}(\sigma^{-1})=0$ and since both numbers in the left side
of this equation are positive it follows that
$r_{2}(\sigma)=r_{2}(\sigma^{-1})=0$, from where $\sigma\in M_{r_{2}}$. So
$M_{r_{1}}\subseteq M_{r_{2}}$. With the same argument we prove that
$M_{r_{2}}\subseteq M_{r_{1}}$ and so $M_{r_{1}}=M_{r_{2}}$.

Suppose now that $r_{1}\neq r_{2}$. We set $M=M_{r_{1}}=M_{r_{2}}$. Then there
exists $\rho\in G\setminus M$ for which $r_{1}(\rho)\neq r_{2}(\rho)$ (of
course $\rho\notin M)$. Suppose that $r_{1}(\rho)<r_{2}(\rho)$ (otherwise we
interchange the roles of $r_{1}$ and $r_{2}$). Firstly we show by induction
that $r_{1}(\rho^{k})<r_{2}(\rho^{k})$, for every
$k\in\mathbb{N}^{*}:=\mathbb{N}\setminus\\{0\\}$. For $k=1$, we have nothing
to prove. We suppose the above relation holds for some $k\in\mathbb{N}^{*}$.
By the assumption, setting $\sigma=\rho$ and $\tau=\rho^{k}$, we have
$r_{1}(\rho)+r_{1}(\rho^{k})-r_{1}(\rho^{k+1})=r_{2}(\rho)+r_{2}(\rho^{k})-r_{2}(\rho^{k+1})$.
From the induction hypothesis, $r_{1}(\rho^{k})<r_{2}(\rho^{k})$ and also
$r_{1}(\rho)<r_{2}(\rho)$, so

$r_{2}(\rho^{k+1})-r_{1}(\rho^{k+1})=(r_{2}(\rho^{k})-r_{1}(\rho^{k}))+(r_{2}(\rho)-r_{1}(\rho))>0,$

therefore $r_{1}(\rho^{k+1})<r_{2}(\rho^{k+1})$ and the induction is
completed.

The group $G$ is finite so, for $\rho\in G\setminus\\{1\\}$, $ord(\rho)-1$
belongs to $\mathbb{N}^{*}$ and then
$r_{1}(\rho^{ord(\rho)-1})<r_{2}(\rho^{ord(\rho)-1})$. On the other hand by
the assumption, setting $\sigma=\rho$ and $\tau=\rho^{-1}$ we have

$\displaystyle
r_{1}(\rho)+r_{1}(\rho^{-1})-r_{1}(\rho\rho^{-1})=r_{2}(\rho)+r_{2}(\rho^{-1})-r_{2}(\rho\rho^{-1})\Rightarrow$
$\displaystyle
r_{1}(\rho^{-1})-r_{2}(\rho^{-1})=r_{2}(\rho)-r_{1}(\rho)>0\Rightarrow
r_{1}(\rho^{ord(\rho)-1})>r_{2}(\rho^{ord(\rho)-1}),$

which is a contradiction. Hence $r_{1}=r_{2}$. $\square$

Let $e$ be an idempotent $2$-cocycle and $H=H(e)$, then associated to $e$ is a
partial ordering on the $G$-set $G/H$ given by

$\sigma H\leq\tau H\Leftrightarrow e(\sigma,\sigma^{-1}\tau)=1,$

for $\sigma,\tau\in G$ ([4],Section 7). It is shown in ([4],Theorem 7.13) that
there is a one-to-one correspondence between the idempotent $2$-cocycles with
inertial group $H$ and the partial orderings on $G/H$, with least element $H$
which are “lower subtractive”, that is, satisfy

$\sigma H\leq\tau H\Rightarrow(\sigma H\leq\rho H\leq\tau
H\Leftrightarrow\sigma^{-1}\rho H\leq\sigma^{-1}\tau H),$

for $\sigma,\tau,\rho\in G$. The partial ordering associated to the idempotent
$e$ gives rise to a left graph with vertices the left cosets of $H$ in $G$ in
the obvious way ([4],Section 7). Let $r\in Sl(G)$, $e_{r}$ as in relation
(1.3) and $H=H(e_{r})$, then $M_{r}=H$ ([8],Theorem 5.2) and

$\sigma H\leq\tau H\Leftrightarrow r(\tau)=r(\sigma)r(\sigma^{-1}\tau),$

for $\sigma,\tau\in G$ ([8],Remark 5.3). This relation allows us to construct
the corresponding graphs on the left cosets.

By the following proposition we can construct new elements of $Sl(G)$ from a
given one.

###### Proposition 2.2

Let $r\in Sl(G)$, $e_{r}$ be as in relation (1.3), $H=H(e_{r})$ and

$N_{1}(e_{r})=\\{\sigma\in G:u_{\sigma}\in
rad(A_{e_{r}})\;\text{and}\;u_{\sigma}\notin rad(A_{e_{r}})^{2}\\}.$

Then, for an element $a\in N_{1}(e_{r})$, the function
$r^{\prime}_{a}:G\rightarrow\mathbb{N}$ defined by the rule

$r^{\prime}_{a}(\sigma)=\begin{cases}r(\sigma),&\text{if}\;\sigma\notin
HaH,\\\ r(\sigma)+1,&\text{if}\;\sigma\in HaH,\end{cases}$

is an element of $Sl(G)$ with $M_{r^{\prime}_{a}}=H$.

Proof: Let $r^{\prime}_{a}$ be as in the statement of the proposition and let
us denote $r^{\prime}_{a}=r^{\prime}$. We remark that from the definition of
$r^{\prime}$ it holds that $r(\sigma)\leq r^{\prime}(\sigma)$, for $\sigma\in
G$. From ([8],Lemma 3.1) we see that $a\notin H$ and consequently $1\notin
HaH$. Hence $r^{\prime}(1)=r(1)=0$. Now, for $\sigma,\tau\in G$, if
$\sigma\tau\notin HaH$, then $r^{\prime}(\sigma\tau)=r(\sigma\tau)\leq
r(\sigma)+r(\tau)\leq r^{\prime}(\sigma)+r^{\prime}(\tau)$. So it remains to
examine the case $\sigma\tau\in HaH$, for $\sigma,\tau\in G$. In this case
$r^{\prime}(\sigma\tau)=r(\sigma\tau)+1$. We have the following three
subcases: i) $\sigma\in H$, ii) $\tau\in H$ and iii) $\sigma,\tau\notin H$. If
$\sigma\in H$, then $\tau\in HaH$ and so $r^{\prime}(\sigma)=r(\sigma)$ and
$r^{\prime}(\tau)=r(\tau)+1$. Hence $r^{\prime}(\sigma\tau)\leq
r^{\prime}(\sigma)+r^{\prime}(\tau)$. If $\tau\in H$, then similarly we get
$r^{\prime}(\sigma\tau)\leq r^{\prime}(\sigma)+r^{\prime}(\tau)$. Finally, if
$\sigma,\tau\notin H$, then, since $HaH\subseteq N_{1}(e_{r})$ ([8],Remark
3.2) and from ([8],Lemma 3.1) we get $e_{r}(\sigma,\tau)=0$ and hence
$r(\sigma\tau)<r(\sigma)+r(\tau)$. Therefore
$r^{\prime}(\sigma\tau)=r(\sigma\tau)+1<r(\sigma)+r(\tau)+1\leq
r^{\prime}(\sigma)+r^{\prime}(\tau)+1$ and since all terms are natural numbers
we get $r^{\prime}(\sigma\tau)\leq r^{\prime}(\sigma)+r^{\prime}(\tau)$. So we
conclude that $r^{\prime}\in Sl(G)$.

To prove that $H\subseteq M_{r^{\prime}}$ we remark that if $\sigma\in H$,
then $\sigma\notin HaH$ and hence $r^{\prime}(\sigma)=r(\sigma)=0$. For the
other direction if $\sigma\in M_{r^{\prime}}$, then $r^{\prime}(\sigma)=0$ and
since $r(\sigma)\leq r^{\prime}(\sigma)$, for all $\sigma\in G$, we get that
$r(\sigma)=0$ and so $M_{r^{\prime}}\subseteq H$. $\square$

##  3 Constructing weak crossed product orders from elements of $Sl(G)$

We consider the $\pi S$-adic valuation
$v_{\pi}:S\longrightarrow\mathbb{Z}\cup\\{\infty\\}$, for
$v_{\pi}(s)=max\\{z\in\mathbb{Z}:s\in\pi^{z}S\\}$. Let
$\sigma(\pi)=u(\sigma)\pi$, where $u(\sigma)\in U(S)$, for $\sigma\in G$, and
we define the function $u:G\longrightarrow U(S)$, $\sigma\mapsto u(\sigma)$.
We refer to ([12]) for more details.

###### Definition 3.1

We define $B^{2}(G,S^{*})$ to be the set of all functions $b:G\times
G\longrightarrow S^{*}$ such that there exists a function $a:G\longrightarrow
S^{*}$ with $a(1)=1$, for which

$b(\sigma,\tau)=a(\sigma)a^{\sigma}(\tau)a(\sigma\tau)^{-1},\quad\text{for}\;\sigma,\tau\in
G.$

¿From the definition of $B^{2}(G,S^{*})$ it is clear that
$a(\sigma\tau)^{-1}\in L^{*}$ and so $b(\sigma,\tau)$ is not in general an
element of $S^{*}$. However, we are interested only in functions $a$ forcing
$b$ to take values in $S^{*}$. For example, if the function $a$ takes values
in $U(S)\subseteq S^{*}$, then the function $b$ takes values in $U(S)$ and
then $b$ is an element of $B^{2}(G,U(S))$. This demonstrates that

$B^{2}(G,U(S))\subseteq B^{2}(G,S^{*})\subseteq B^{2}(G,L^{*}).$

It is easy to show that the set $B^{2}(G,S^{*})$ is a submonoid of
$Z^{2}(G,S^{*})$.

###### Proposition 3.2

Let $r\in Sl(G)$, then the function $b_{r}:G\times G\longrightarrow S^{*}$
defined by then rule

${}b_{r}(\sigma,\tau)=u(\sigma)^{r(\tau)}\pi^{r(\sigma)+r(\tau)-r(\sigma\tau)},\quad\text{for}\;\sigma,\tau\in
G,$ (3.1)

is an element of $B^{2}(G,S^{*})$ with inertial group $M_{r}$.

Proof: For any $r\in Sl(G)$ consider the function $a_{r}:G\longrightarrow
S^{*}$ defined by the rule $a_{r}(\sigma)=\pi^{r(\sigma)}$. Then
$a_{r}(\sigma)a_{r}^{\sigma}(\tau)a_{r}(\sigma\tau)^{-1}=b_{r}(\sigma,\tau)$.
Since $r\in Sl(G)$ and so $r(\sigma)+r(\tau)\geq r(\sigma\tau)$, it follows
that $b_{r}(\sigma,\tau)$ is a $2$-cocycle taking values in $S^{*}$, for
$\sigma,\tau\in G$, so $b_{r}\in B^{2}(G,S^{*})$. Finally for the inertial
group $H(b_{r})$ of $b_{r}$ we have

$H(b_{r})=\\{\sigma\in G:b_{r}(\sigma,\sigma^{-1})\in U(S)\\}=\\{\sigma\in
G:r(\sigma)+r(\sigma^{-1})=0\\}=M_{r}.\square$

In what follows, for $r\in Sl(G)$, we use the notation

${}\epsilon_{r}(\sigma,\tau)=\pi^{r(\sigma)+r(\tau)-r(\sigma\tau)},$ (3.2)

so that relation (3.1) takes the form

${}b_{r}(\sigma,\tau)=u(\sigma)^{r(\tau)}\epsilon_{r}(\sigma,\tau).$ (3.3)

The weak $2$-cocycle $b_{r}$ defines a weak crossed product order
$\Lambda_{b_{r}}=(S/R,b_{r})$ and the idempotent $2$-cocycle $e_{r}$ of
relation (1.3) defines a weak crossed product algebra $A_{e_{r}}=(L/K,e_{r})$.
By reducing $mod\pi S$ the values of $b_{r}$ of relation (3.1) we get a new
weak $2$-cocycle $\overline{b}_{r}$ of $G$ in $\overline{S}$. More explicitly,

$\overline{b}_{r}(\sigma,\tau)=\begin{cases}u(\sigma)^{r(\tau)}+\pi
S,&\text{if}\;r(\sigma\tau)=r(\sigma)+r(\tau),\\\
0,&\text{if}\;r(\sigma\tau)<r(\sigma)+r(\tau).\end{cases}$

###### Example 3.3

Let $G=\mathbb{Z}/10\mathbb{Z}$ be the cyclic group of order $10$. Consider
the standard partition of $G\setminus\\{\overline{0}\\}$ derived from the
generating set $\\{\overline{1},\overline{6}\\}$ ([1],Section 3 and
[8],Section 10):

$\delta=\\{\\{\overline{1},\overline{6}\\},\\{\overline{2},\overline{7}\\},\\{\overline{3},\overline{8}\\},\\{\overline{4},\overline{9}\\},\\{\overline{5}\\}\\},$

from where we can construct $r=\\{0,1,2,3,4,5,1,2,3,4\\}\in Sl(G)$ with
$N_{1}(e_{r})=\\{\overline{1},\overline{6}\\}$ ([8],Remark 10.9 ). With the
procedure described in Proposition 2.2, we can construct the function

$r_{1}=r^{\prime}_{\overline{1}}=\\{0,2,2,3,4,5,1,2,3,4\\},$

which is again an element of $Sl(G)$. From relation (3.2), the values of the
non-invertible term $\epsilon_{r_{1}}:G\times G\longrightarrow S^{*}$ are
given by the following table:

$\begin{array}[]{|c|cccccccccc|}\hline\cr(\sigma,\tau)&\overline{0}&\overline{1}&\overline{2}&\overline{3}&\overline{4}&\overline{5}&\overline{6}&\overline{7}&\overline{8}&\overline{9}\\\
\hline\cr\overline{0}&1&1&1&1&1&1&1&1&1&1\\\
\overline{1}&1&\pi^{2}&\pi&\pi&\pi&\pi^{6}&\pi&\pi&\pi&\pi^{6}\\\
\overline{2}&1&\pi&1&1&\pi^{5}&\pi^{5}&1&1&\pi^{5}&\pi^{4}\\\
\overline{3}&1&\pi&1&\pi^{5}&\pi^{5}&\pi^{5}&1&\pi^{5}&\pi^{4}&\pi^{5}\\\
\overline{4}&1&\pi&\pi^{5}&\pi^{5}&\pi^{5}&\pi^{5}&\pi^{5}&\pi^{4}&\pi^{5}&\pi^{5}\\\
\overline{5}&1&\pi^{6}&\pi^{5}&\pi^{5}&\pi^{5}&\pi^{10}&\pi^{4}&\pi^{5}&\pi^{5}&\pi^{5}\\\
\overline{6}&1&\pi&1&1&\pi^{5}&\pi^{4}&1&1&1&1\\\
\overline{7}&1&\pi&1&\pi^{5}&\pi^{4}&\pi^{5}&1&1&1&\pi^{5}\\\
\overline{8}&1&\pi&\pi^{5}&\pi^{4}&\pi^{5}&\pi^{5}&1&1&\pi^{5}&\pi^{5}\\\
\overline{9}&1&\pi^{6}&\pi^{4}&\pi^{5}&\pi^{5}&\pi^{5}&1&\pi^{5}&\pi^{5}&\pi^{5}\\\
\hline\cr\end{array}$

As in the relation (3.3), multiplying the values of
$\epsilon_{r_{1}}(\sigma,\tau)$ with the invertible term
$u(\sigma)^{r_{1}(\tau)}$, we get the $2$-cocycle $b_{r_{1}}$ which has
trivial inertial group.

Notice that $r_{1}$ defines also an idempotent $2$-cocycle $e_{r_{1}}$ as in
relation (1.3) with the following graph on (left) cosets:

$\textstyle{\overline{4}}$$\textstyle{\overline{5}}$$\textstyle{\overline{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{9}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{1}}$$\textstyle{\overline{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

###### Proposition 3.4

Every element of $B^{2}(G,S^{*})$ with inertial group $H$ can be written
uniquely as a product of an element of $B^{2}(G,U(S))$ and a $2$-cocycle
$b_{r}$, for some $r\in Sl(G)$ with $M_{r}=H$.

Proof: Let $b\in B^{2}(G,S^{*})$. By the definition of $B^{2}(G,S^{*})$ there
exists $a:G\longrightarrow S^{*}$ defined by
$a(\sigma)=w(\sigma)\pi^{v_{\pi}(a(\sigma))}$, with $w(\sigma)\in U(S)$ for
$\sigma\in G$ and $a(1)=1$, such that
$b(\sigma,\tau)=a(\sigma)\sigma(a(\tau))a(\sigma\tau)^{-1}$, for
$\sigma,\tau\in G$. Substituting $a$ we get:

$b(\sigma,\tau)=w(\sigma)\sigma(w(\tau))w(\sigma\tau)^{-1}u(\sigma)^{v_{\pi}(a(\tau))}\pi^{v_{\pi}(a(\sigma))+v_{\pi}(a(\tau))-v_{\pi}(a(\sigma\tau))}.$

We set $c(\sigma,\tau)=w(\sigma)\sigma(w(\tau))w(\sigma\tau)^{-1}$ and
$r=v_{\pi}\circ a$. It is obvious that $c\in B^{2}(G,U(S))$. Also since
$b(\sigma,\tau)\in S^{*}$, for $\sigma,\tau\in G$, it must be
$r(\sigma)+r(\tau)-r(\sigma\tau)\geq 0$. Since $r(1)=v_{\pi}(a(1))=0$ we have
$r\in Sl(G)$. So $b(\sigma,\tau)=c(\sigma,\tau)b_{r}(\sigma,\tau)$, with $c\in
B^{2}(G,U(S))$ and $r\in Sl(G)$.

Now, let $b(\sigma,\tau)=c^{\prime}(\sigma,\tau)b_{r^{\prime}}(\sigma,\tau)$
be another expression of $b$ with $c^{\prime}\in B^{2}(G,U(S))$ and
$r^{\prime}\in Sl(G)$. Since the expression of $b(\sigma,\tau)$ as a product
of a unit and a power of $\pi$ is unique, we must have that
$r(\sigma)+r(\tau)-r(\sigma\tau)=r^{\prime}(\sigma)+r^{\prime}(\tau)-r^{\prime}(\sigma\tau)$,
for every $\sigma,\tau\in G$, and
$c(\sigma,\tau)u(\sigma)^{r(\tau)}=c^{\prime}(\sigma,\tau)u(\sigma)^{r^{\prime}(\tau)}$.
From Proposition 2.1 we deduce that $r=r^{\prime}$ and as a consequence we get
$c=c^{\prime}$.

Finally, let $\sigma\in M_{r}$. Then $r(\sigma)=r(\sigma^{-1})=0$ and so
$b(\sigma,\sigma^{-1})=c(\sigma,\sigma^{-1})u(\sigma)^{r(\sigma^{-1})}\pi^{0}\in
U(S)$ from where we conclude that $\sigma\in H(b)=H$. On the other hand, if
$\sigma\in H$, then $b(\sigma,\sigma^{-1})\in U(S)$ from where we deduce that
$r(\sigma)+r(\sigma^{-1})=0$ and so $r(\sigma)=0$. Hence $H=M_{r}$. $\square$

###### Corollary 3.5

Let $b\in B^{2}(G,S^{*})$ and $H(b)\subsetneqq G$ be the inertial group of
$b$. Then the weak crossed product order $\Lambda_{b}$ is not hereditary.

Proof: Let $b\in B^{2}(G,S^{*})$ and let $b=cb_{r}$ be the unique expression
of $b$ in accordance to Proposition 3.4, for some $c\in B^{2}(G,U(S))$ and a
weak $2$-cocycle $b_{r}:G\times G\longrightarrow S^{*}$. Then we have
$v_{\pi}(b(\sigma,\sigma^{-1})=r(\sigma)+r(\sigma^{-1})$, for $\sigma\in G$.
Since $H(b)\subsetneqq G$, let $\tau\in G\setminus H$. Then $r(\tau)\geq 1$
and $r(\tau^{-1})\geq 1$, so $r(\tau)+r(\tau^{-1})\geq 2$. Now the result
follows from ([14],Theorem 2.15(1)) by which $\Lambda_{b}$ is hereditary if
and only if $v_{\pi}(b(\sigma,\sigma^{-1}))\leq 1$ for every $\sigma\in G$.
$\square$

Inside the monoid $B^{2}(G,S^{*})$ we consider the equivalence relation

$b_{1}\sim b_{2}\Leftrightarrow b_{1}B^{2}(G,U(S))=b_{2}B^{2}(G,U(S)).$

Let $\overline{b}$ denote the class of $b$. We write
$B^{2}(G,S^{*})/B^{2}(G,U(S))$ for the set of classes of elements of
$B^{2}(G,S^{*})$ which is a monoid under the operation
$\overline{b_{1}}\cdot\overline{b_{2}}=\overline{b_{1}b_{2}}$.

###### Theorem 3.6

There is a monoid isomorphism

$B^{2}(G,S^{*})/B^{2}(G,U(S))\simeq Sl(G)$

Proof: Define the mapping $\phi$ by the rule $\phi(\overline{b})=r$, for all
$b\in B^{2}(G,S^{*})$ where $b=c\cdot b_{r}$ with $c\in B^{2}(G,U(S))$ and
$r\in Sl(G)$ as in Proposition 3.4. The uniqueness of Proposition 3.4 implies
that $\phi$ is well defined. Also, if
$\phi(\overline{b_{1}})=\phi(\overline{b_{2}})$, then $r_{1}=r_{2}$ and so
$b_{1}=c_{1}b_{r}$, $b_{2}=c_{2}b_{r}$ from where it follows that
$b_{1}=c_{1}c_{2}^{-1}c_{2}b_{r}=c_{1}c_{2}^{-1}b_{2}$ and $\phi$ is
injective. To see that $\phi$ is a homomorphism it is enough to notice that
$b_{r_{1}+r_{2}}=b_{r_{1}}b_{r_{2}}$. If we set $c_{1}(\sigma,\tau)=1$, for
every $\sigma,\tau\in G$, and $r^{*}(\sigma)=0$, for every $\sigma\in G$, then
we get the identity $2$-cocycle. Finally, for $r\in Sl(G)$, there exists
$b=c_{1}b_{r}$ such that $\phi(\overline{b})=r$ and $\phi$ is onto. $\square$

##  4 Elements of $Sl(G)$ defined by a valuation

In this section we construct new elements of $Sl(G)$ from elements of
$Z^{2}(G,S^{*})$.

###### Lemma 4.1

Let $f\in Z^{2}(G,L^{*})$. Then

$v_{\pi}(f(\sigma,\tau))+v_{\pi}(f(\tau^{-1},\sigma^{-1}))=v_{\pi}(f(\sigma,\sigma^{-1}))+v_{\pi}(f(\tau,\tau^{-1}))-v_{\pi}(f(\sigma\tau,\tau^{-1}\sigma^{-1})).$

Proof: For an element $f\in Z^{2}(G,L^{*})$ we consider successively the
cocycle condition (1.1) for the elements $\sigma,\tau,(\sigma\tau)^{-1}$ and
$\tau,\tau^{-1},\sigma^{-1}$ of $G$. We get two equations and then we
substitute the value $f(\tau,\tau^{-1}\sigma^{-1})$ of the second equation in
the first equation. In both sides of the resulting equation we apply the
valuation $v_{\pi}$ and we get the result, using the fact that
$v_{\pi}(l^{\sigma})=v_{\pi}(l)$, for $\sigma\in G$ and $l\in L$. $\square$

###### Proposition 4.2

Let $f\in Z^{2}(G,S^{*})$. Then the function

$r_{f}:G\longrightarrow\mathbb{N},\quad
r_{f}(\sigma)=v_{\pi}(f(\sigma,\sigma^{-1})),$

is an element of $Sl(G)$ with $M_{r_{f}}=H(f)$.

Proof: First we see that $r_{f}(1)=0$. Now we remark that since
$f(\sigma,\tau)\in S^{*}$, for $\sigma,\tau\in G$, then $r_{f}(\sigma)\geq 0$,
$\sigma\in G$. So from Lemma 4.1 we get that

$r_{f}(\sigma)+r_{f}(\tau)-r_{f}(\sigma\tau)\geq 0,$

hence $r_{f}\in Sl(G)$. Finally

$M_{r_{f}}=\\{\sigma\in G:r_{f}(\sigma)=0\\}=\\{\sigma\in
G:f(\sigma,\sigma^{-1})\in U(S)\\}=H(f).\quad\square$

Let $f\in Z^{2}(G,S^{*})$, $r\in Sl(G)$ and

${}f(\sigma,\tau)=c(\sigma,\tau)\pi^{v_{\pi}(f(\sigma,\tau))},$ (4.1)

for some $c(\sigma,\tau)\in U(S)$ and $\sigma,\tau\in G$. We define the
function $h:G\times G\rightarrow S^{*}$ by the rule

${}h(\sigma,\tau)=[c(\sigma,\tau)]^{-1}u(\sigma)^{r(\tau)}\pi^{r(\sigma)+r(\tau)-r(\sigma\tau)-v_{\pi}(f(\sigma,\tau))},$
(4.2)

for $\sigma,\tau\in G$ and $c(\sigma,\tau)$ is as in the relation (4.1). The
function $h\in Z^{2}(G,L^{*})$. Indeed, it is easy to see that $fh=b_{r}$,
where $b_{r}$ is as in Proposition 3.2. Hence $h=f^{-1}b_{r}$, where
$f^{-1}(\sigma,\tau)=f(\sigma,\tau)^{-1}$. In the sequel we prove that, for
$f\in Z^{2}(G,S^{*})$, we can find an element $h\in Z^{2}(G,S^{*})$ such that
$fh\in B^{2}(G,S^{*})$.

###### Proposition 4.3

Let $f\in Z^{2}(G,S^{*})$ and $r_{f}$ as in Proposition 4.2. Then the function
$h:G\times G\longrightarrow S^{*}$ defined by the rule

$h(\sigma,\tau)=[c(\sigma,\tau)c(\tau^{-1},\sigma^{-1})]^{-1}u(\sigma)^{r_{f}(\tau)}f(\tau^{-1},\sigma^{-1}),$

for $c(\sigma,\tau)$ as in (4.1) for $\sigma,\tau\in G$, is an element of
$Z^{2}(G,S^{*})$ such that $fh=b_{r_{f}}$.

Proof: We put in the relation (4.2) the function $r_{f}$ instead of $r$. Using
Lemma 4.1 and the relation
$f(\tau^{-1},\sigma^{-1})=c(\tau^{-1},\sigma^{-1})\pi^{v_{\pi}(f(\tau^{-1},\sigma^{-1}))}$
we get

$h(\sigma,\tau)=[c(\sigma,\tau)c(\tau^{-1},\sigma^{-1})]^{-1}u(\sigma)^{r_{f}(\tau)}f(\tau^{-1},\sigma^{-1}).$

The result follows from the fact that $v_{\pi}(f(\tau^{-1},\sigma^{-1})\geq
0$. $\square$

For the next application of the function $r_{f}$ we need a new transformation
of elements of $Sl(G)$.

###### Proposition 4.4

Let $r\in Sl(G)$. Then the function $r_{/}:G\longrightarrow\mathbb{N}$ defined
by the rule

$r_{/}(\sigma)=\begin{cases}k,&r(\sigma)=2k,\\\
k+1,&r(\sigma)=2k+1\end{cases}$

is an element of $Sl(G)$ with $M_{r_{/}}=M_{r}$.

Proof: It is easy to see that $r_{/}(1)=0$. For $\sigma,\tau\in G$, we
distinguish six cases in accordance to $r(\sigma)$, $r(\tau)$ or
$r(\sigma\tau)$ is even or odd natural number. Let $r(\sigma)=2k_{1}$,
$r(\tau)=2k_{2}$, $r(\sigma\tau)=2k_{3}$. Then $r_{/}(\sigma)=k_{1}$,
$r_{/}(\tau)=k_{2}$, $r_{/}(\sigma\tau)=k_{3}$. Since $r\in Sl(G)$, we get
$2k_{3}\leq 2k_{1}+2k_{2}$ from where $k_{3}\leq k_{1}+k_{2}$ and so
$r_{/}(\sigma\tau)\leq r_{/}(\sigma)+r_{/}(\tau)$. Similarly we examine the
other cases. Finally, $\sigma\in M_{r_{/}}\Leftrightarrow
r_{/}(\sigma)=0\Leftrightarrow r(\sigma)=0$ so $M_{r_{/}}=M_{r}$. $\square$

###### Corollary 4.5

Let $r\in Sl(G)$. Then the function $r^{\prime}:G\longrightarrow\mathbb{N}$
defined by the rule

$r^{\prime}(\sigma)=\begin{cases}r(\sigma),&r(\sigma)=2k,\\\
r(\sigma)+1,&r(\sigma)=2k+1\end{cases}$

is an element of $Sl(G)$ with $M_{r^{\prime}}=M_{r}$.

Proof: Let $r\in Sl(G)$ and $r_{/}\in Sl(G)$ be as in Proposition 4.4. We
remark that if $r(\sigma)=2k$, then $r_{/}(\sigma)=k$ and so
$r^{\prime}(\sigma)=2k=2r_{/}(\sigma)$, for $\sigma\in G$. Similarly for
$r(\sigma)=2k+1$, $\sigma\in G$. The result follows from ([8],Proposition
8.3). $\square$

###### Example 4.6

We consider Example 3.3 and let $r=\\{0,2,2,3,4,5,1,2,3,4\\}\in Sl(G)$. Then
$r_{/}=\\{0,1,1,2,2,3,1,1,2,2\\}$ and $r^{\prime}=\\{0,2,2,4,4,6,2,2,4,4\\}$.
The graph on left cosets of the corresponding idempotent $2$-cocycles
$e_{r_{/}}$ and $e_{r^{\prime}}$ is common and it is the following:

$\textstyle{\overline{5}}$$\textstyle{\overline{9}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{4}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{1}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{7}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{\overline{0}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

###### Theorem 4.7

For $f\in Z^{2}(G,S^{*})$, there exist a $2$-cocycle $h\in Z^{2}(G,L^{*})$ and
$r\in Sl(G)$ such that $fh=b_{r}$ with $0\leq
v_{\pi}(h(\sigma,\sigma^{-1}))\leq 1$, for $\sigma\in G$.

Proof: Let $f\in Z^{2}(G,S^{*})$ and $r=r_{f}$ as in Proposition 4.2. For the
$2$-cocycle $h$ of relation (4.2) and $r_{/}\in Sl(G)$ as in Proposition 4.4,
we have that $fh=b_{r_{/}}$ and so
$v_{\pi}(h(\sigma,\sigma^{-1}))=r_{/}(\sigma)+r_{/}(\sigma^{-1})-r(\sigma)$.
From ([14],Lemma 2.6) we have that
$v_{\pi}(f(\sigma,\sigma^{-1}))=v_{\pi}(f(\sigma^{-1},\sigma))$, so
$r(\sigma)=r(\sigma^{-1})$ and consequently
$r_{/}(\sigma)=r_{/}(\sigma^{-1})$, for $\sigma\in G$. But then

$\displaystyle r_{/}(\sigma)+r_{/}(\sigma^{-1})-r(\sigma)$ $\displaystyle=$
$\displaystyle\begin{cases}k+k-2k,&if\;r(\sigma)=2k,\\\
k+1+k+1-2k-1,&if\;r(\sigma)=2k+1,\end{cases}$ $\displaystyle=$
$\displaystyle\begin{cases}0,&if\;r(\sigma)=2k,\\\
1,&if\;r(\sigma)=2k+1,\end{cases}$

from where $0\leq v_{\pi}(h(\sigma,\sigma^{-1}))\leq 1$, for all $\sigma\in
G$. $\square$

##  5 Inflating idempotent $2$-cocycles

Let $N$ be a normal subgroup of G, $L^{N}$ be the fixed field of $N$ and $g\in
Z^{2}(G/N,L^{N})$. Then we can form a new normalized weak $2$-cocycle
$\widehat{g}\in Z^{2}(G,L)$ defined by the relation

$\widehat{g}(\sigma,\tau)=g(\sigma N,\tau N),\quad\text{for}\;\sigma,\tau\in
G,$

called the inflation of $g$. The main result in this section is Theorem 5.3 by
which a relation is established between the inertial groups of $\varepsilon$
and $\widehat{\varepsilon}$. First we need some lemmata. Before these let us
remark that as a consequence of the definition of the inertial group $H(f)$,
for $f\in Z^{2}(G,L)$, we get that $f(\sigma,h)\neq 0$ and $f(h,\sigma)\neq
0$, for $\sigma\in G$ and $h\in H(f)$.

###### Lemma 5.1

Let $f\in Z^{2}(G,L)$ and $N$ be a normal subgroup of $G$ such that
$N\subseteq H(f)$. Then, for $\sigma,\tau\in G$, and $n_{1},n_{2}\in N$, we
have:

  * (i)

If $f(\sigma,\tau)=0$, then $f(\sigma n_{1},\tau n_{2})=0$,

  * (ii)

If $f(\sigma,\tau)\neq 0$, then $f(\sigma n_{1},\tau n_{2})\neq 0$.

Proof: (i) Let $n_{1},n_{2}\in N$. First we prove that $f(\sigma
n_{1},\tau)=0$. Since $N$ is a normal subgroup of $G$, there exists $n\in N$
such that $\sigma n_{1}=n\sigma$. From the relation (1.1) and the elements
$n,\sigma,\tau$, we have the relation
$f(n,\sigma)f(n\sigma,\tau)=f^{n}(\sigma,\tau)f(n,\sigma\tau)$. Since $n\in
H(f)$ we have that $f(n,\sigma)\neq 0$. From the assumption $f(\sigma,\tau)=0$
so $f(n\sigma,\tau)=f(\sigma n_{1},\tau)=0$. Again from the relation (1.1) and
for the elements $\sigma n_{1},\tau,n_{2}$ we have $f(\sigma
n_{1},\tau)f(\sigma n_{1}\tau,n_{2})=f^{\sigma n_{1}}(\tau,n_{2})f(\sigma
n_{1},\tau n_{2})$. Since $f(\sigma n_{1},\tau)=0$ and $f(\tau,n_{2})\neq 0$,
it follows that $f(\sigma n_{1},\tau n_{2})=0$.
(ii) Let $n_{1},n_{2}\in N$. For $n\in N$ and the elements $\sigma,\tau,n$,
the relation (1.1) becomes
$f(\sigma,\tau)f(\sigma\tau,n)=f^{\sigma}(\tau,n)f(\sigma,\tau n)$. Since
$f(\sigma,\tau)\neq 0$ and $f(\sigma\tau,n)\neq 0$ we get $f(\sigma,\tau
n)\neq 0$. Since $N$ is a normal subgroup of $G$, there exists $n^{\prime}\in
N$ such that $n_{1}\tau n_{2}=\tau n^{\prime}$. Then from the relation (1.1)
we get

$f(\sigma,n_{1})f(\sigma n_{1},\tau n_{2})=f^{\sigma}(n_{1},\tau
n_{2})f(\sigma,n_{1}\tau n_{2})=f^{\sigma}(n_{1},\tau n_{2})f(\sigma,\tau
n^{\prime}),$

since $f(n_{1},\tau n_{2})\neq 0$. Then $f(\sigma,\tau n^{\prime})\neq 0$, so
we get $f(\sigma n_{1},\tau n_{2})\neq 0$. $\square$

In what follows, for $f\in Z^{2}(G,L)$ and $M$ a subgroup of $G$ we write
$res_{M}(f):M\times M\longrightarrow L$ for the restriction of $f$ to $M$,
which is again a normalized weak $2$-cocycle. Also let $E^{2}(G,L;H)$ be the
set of idempotent $2$-cocycles of $G$ in $L$ with inertial group $H$.

###### Lemma 5.2

Let $e\in E^{2}(G,L;H)$ and $M$ be a subgroup of $G$. Then $res_{M}(e)=1$ if
and only if $M\leq H$.

Proof: Let $e\in E^{2}(G,L;H)$, $e_{1}=res_{M}(e)=1$ and $\sigma\in M$. Then
$\sigma^{-1}\in M$ and $e(\sigma,\sigma^{-1})=e_{1}(\sigma,\sigma^{-1})=1$, so
$\sigma\in H$ and $M\subseteq H$. On the other hand, if $M\subseteq H$ and
$\sigma\in M$, then $e_{1}(\sigma,\sigma^{-1})=e(\sigma,\sigma^{-1})\neq 0$
and since $e$ takes only the values 0 or 1, we get
$e_{1}(\sigma,\sigma^{-1})=1$. Then, for $\sigma,\tau,\tau^{-1}\in M$, the
$2$-cocycle condition (1.1) yields
$e_{1}(\sigma,\tau)e_{1}(\sigma\tau,\tau^{-1})=e_{1}(\tau,\tau^{-1})e_{1}(\sigma,\tau\tau^{-1})$,
so $e_{1}(\tau,\tau^{-1})=1$. By definition $e_{1}(\sigma,1)=1$. Therefore
$e_{1}(\sigma,\tau)=1$, for every $\sigma,\tau\in M$. Hence $e_{1}=1$.
$\square$

The following is a more general statement of ([13],Theorem 3.5):

###### Theorem 5.3

Let $N$ be a normal subgroup of $G$ and $e\in E^{2}(G,L;H)$. Then $N\leq H$ if
and only if there exists $\varepsilon\in E^{2}(G/N,L^{N})$ such that
$e=\widehat{\varepsilon}$, that is $e(\sigma,\tau)=\varepsilon(\sigma N,\tau
N)$, for $\sigma,\tau\in G$. In this case

$H_{G/N}(\varepsilon)=H_{G}(\widehat{\varepsilon})/N=H/N.$

Proof: If $\sigma\in G$, then we write $\overline{\sigma}$ for $\sigma N\in
G/N$. Firstly, let $N\leq H$ and $\varepsilon:G/N\times G/N\longrightarrow
L^{N}$ defined by the rule
$\varepsilon(\overline{\sigma},\overline{\tau})=e(\sigma,\tau)$, for
$\overline{\sigma},\overline{\tau}\in G/N$. We prove that the function
$\varepsilon$ is well defined. For this, let $\sigma,\tau\in G$,
$\sigma_{1}=\sigma n_{1}$ and $\tau_{1}=\tau n_{2}$ representatives of
$\overline{\sigma}$ and $\overline{\tau}$ respectively, for $n_{1},n_{2}\in
N$. Then we get the following:

i) If $e(\sigma,\tau)=1$, from Lemma 5.1(ii) it follows that $e(\sigma
n_{1},\tau n_{2})\neq 0$, so $e(\sigma_{1},\tau_{1})=e(\sigma n_{1},\tau
n_{2})=1=e(\sigma,\tau)$.

ii) If $e(\sigma,\tau)=0$, from Lemma 5.1(i) it follows that $e(\sigma
n_{1},\tau n_{2})=0$, so $e(\sigma_{1},\tau_{1})=e(\sigma n_{1},\tau
n_{2})=0=e(\sigma,\tau)$.

In any case
$\varepsilon(\overline{\sigma}_{1},\overline{\tau}_{1})=e(\sigma_{1},\tau_{1})=e(\sigma,\tau)=\varepsilon(\overline{\sigma},\overline{\tau})$,
so the value of $\varepsilon$ is independent of the representatives of
$\overline{\sigma}$ and $\overline{\tau}$.

For $\overline{\sigma},\overline{\tau},\overline{\rho}\in G/N$, we get

$\varepsilon(\overline{\sigma},\overline{\tau})\varepsilon(\overline{\sigma}\cdot\overline{\tau},\overline{\rho})=e(\sigma,\tau)e(\sigma\tau,\rho)=e(\tau,\rho)e(\sigma,\tau\rho)=\varepsilon(\overline{\tau},\overline{\rho})\varepsilon(\overline{\sigma},\overline{\tau}\cdot\overline{\rho}),$

so $\varepsilon$ is an idempotent $2$-cocycle. The $2$-cocycle $e$ is
normalized, so for $\overline{\sigma},\overline{\tau}\in G/N$ it holds that
$\varepsilon(\overline{1},\overline{\tau})=e(1,\tau)=1$ and
$\varepsilon(\overline{\sigma},\overline{1})=e(\sigma,1)=1$. So $\varepsilon$
is also normalized.

For the opposite direction suppose that there exists $\varepsilon\in
E^{2}(G/N,L^{N})$ such that $e=\widehat{\varepsilon}$. Since
$H_{G/N}(\varepsilon)\leq G/N$, there exists a subgroup $G_{1}$ of $G$
containing $N$ such that $H_{G/N}(\varepsilon)=G_{1}/N$. From Lemma 5.2 it
follows that $res_{G_{1}/N}(\varepsilon)=1$. Hence, for $\sigma,\tau\in
G_{1}$, we have:

$res_{G_{1}}(e)(\sigma,\tau)=e(\sigma,\tau)=\varepsilon(\sigma N,\tau
N)=res_{G_{1}/N}(\varepsilon)(\sigma N,\tau N)=1,$

since $\sigma N,\tau N\in G_{1}/N$. Then $res_{G_{1}}(e)=1$ and again from
Lemma 5.2 we get $G_{1}\leq H$ from where $N\leq H$.

Finally we will show that $G_{1}=H$ and then the equality of the statement of
the theorem is obvious. If $\sigma\in H$, then $e(\sigma,\sigma^{-1})\neq 0$.
Since $N\trianglelefteq G$ it holds that $(\sigma N)^{-1}=\sigma^{-1}N$ and so

$\varepsilon(\sigma N,(\sigma N)^{-1})=\varepsilon(\sigma
N,\sigma^{-1}N)=e(\sigma,\sigma^{-1})\neq 0.$

But then, $\sigma N\in H_{G/N}(\varepsilon)=G_{1}/N$ and so $\sigma\in G_{1}$.
Hence $G_{1}=H$. $\square$

A special case of Theorem 5.3 is when $N$ equals the inertial group of $e$.
Then $e$ comes always from a lifting of an idempotent $2$-cocycle with trivial
inertial group, as the next corollary shows.

###### Corollary 5.4

Let $e\in E^{2}(G,L;H)$. If $H$ is a normal subgroup of $G$, then there exists
$\varepsilon\in E^{2}(G/H,L^{H};1)$ such that $e=\widehat{\varepsilon}$.

Proof: From Theorem 5.3, setting $N=H$, it follows that there exists
$\varepsilon\in E^{2}(G/H,L^{H})$ such that $e=\widehat{\varepsilon}$.
Furthermore
$H_{G/H}(\varepsilon)=H_{G}(\widehat{\varepsilon})/H=H_{G}(e)/H=1$. $\square$

The opposite is also true.

###### Corollary 5.5

Let $H\unlhd G$ and $\varepsilon\in E^{2}(G/H,L^{H};1)$. Then there exists
$e\in E^{2}(G,L;H)$ such that $e=\widehat{\varepsilon}$.

Proof: By inflating $\varepsilon$ from $G/H$ to $G$, we construct $e\in
E^{2}(G,L)$ such that $e=\widehat{\epsilon}$. We only have to show that the
inertial group of $e$ is $H$. Indeed

$\displaystyle H_{G}(e)$ $\displaystyle=$ $\displaystyle\\{\sigma\in
G:e(\sigma,\sigma^{-1})=1\\}=\\{\sigma\in G:\varepsilon(\sigma
H,\sigma^{-1}H)=1\\}$ $\displaystyle=$ $\displaystyle\\{\sigma\in G:\sigma
H\in H_{G/H}(\varepsilon)\\}=H.\quad\square$

Let $H\unlhd G$ and $r\in Sl(G/H)$. In this case, lifting the idempotent
$2$-cocycle $\varepsilon_{r}$ from $G/H$ to $G$ can also be accomplished using
([8], Proposition 8.6 and Corollary 8.7). Specifically, if we set
$\widehat{r}(\sigma)=r(\sigma H)$, for $\sigma\in G$, then $\widehat{r}\in
Sl(G)$ with $M_{\widehat{r}}=H$. For the corresponding idempotent $2$-cocycles
we have the relation $e_{\widehat{r}}=\widehat{\varepsilon}_{r}$.

###### Example 5.6

Let

$G=D_{8}=<\sigma,a:a^{4}=\sigma^{2}=1,\sigma a\sigma^{-1}=a^{-1}>$

be the dihedral group of order $8$ and identity element $1$. Let
$H=\\{1,a^{2}\\}$. It is a normal subgroup of $G$ with $G/H=\\{H,aH,\sigma
H,a\sigma H\\}$. In order to construct an element $r$ of $Sl(G/H)$ with
trivial inertial group, we start with a generating set of
$(G/H)^{*}=G/H\setminus\\{H\\}$ as in ([8],Section 10), say
$A_{1}=\\{aH,\sigma H\\}$. Then $A_{2}=\\{a\sigma H\\}$ and $r$ takes the
following values:

$r(H)=0,\;r(aH)=1,\;r(\sigma H)=1,\;r(a\sigma H)=2.$

The corresponding idempotent $2$-cocycle $\varepsilon_{r}:G/H\times
G/H\longrightarrow L^{H}$ takes the following values:

$\begin{array}[]{|r|cccc|}\hline\cr(\sigma,\tau)&H&aH&\sigma H&a\sigma H\\\
\hline\cr H&1&1&1&1\\\ aH&1&0&1&0\\\ \sigma H&1&1&0&0\\\ a\sigma H&1&0&0&0\\\
\hline\cr\end{array}$

As in ([8],Proposition 8.6) we construct $\widehat{r}\in Sl(G)$ with the
following values:

$\begin{array}[]{ll}\widehat{r}(1)=\widehat{r}(a^{2})=r(H)=0,&\widehat{r}(a)=\widehat{r}(a^{3})=r(aH)=1,\\\
\widehat{r}(\sigma)=\widehat{r}(a^{2}\sigma)=r(\sigma
H)=1,&\widehat{r}(a\sigma)=\widehat{r}(a^{3}\sigma)=r(a\sigma
H)=2.\end{array}$

¿From the relation (1.3) we construct the $2$-cocyle $e_{\widehat{r}}:G\times
G\longrightarrow L$ with the following values:

$\begin{array}[]{c|c|cc|cc|cc|cc|}&&H&&aH&&\sigma H&&a\sigma H&\\\
\hline\cr&(\sigma,\tau)&1&a^{2}&a&a^{3}&\sigma&a^{2}\sigma&a\sigma&a^{3}\sigma\\\
\hline\cr H&1&1&1&1&1&1&1&1&1\\\ &a^{2}&1&1&1&1&1&1&1&1\\\ \hline\cr
aH&a&1&1&0&0&1&1&0&0\\\ &a^{3}&1&1&0&0&1&1&0&0\\\ \hline\cr\sigma
H&\sigma&1&1&1&1&0&0&0&0\\\ &a^{2}\sigma&1&1&1&1&0&0&0&0\\\ \hline\cr a\sigma
H&a\sigma&1&1&0&0&0&0&0&0\\\ &a^{3}\sigma&1&1&0&0&0&0&0&0\\\
\hline\cr\end{array}$

The corresponding graphs on left cosets are:

$\textstyle{a\sigma H}$$\textstyle{\sigma
H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{aH\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$
$\textstyle{a\sigma}$$\textstyle{a^{3}\sigma}$$\textstyle{\sigma\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a^{2}\sigma\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a^{3}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{1\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\textstyle{a^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$

Finally, the values of the function $\epsilon_{\widehat{r}}:G\times
G\longrightarrow S^{*}$ from relation (3.2) are the following:

$\begin{array}[]{c|c|cc|cc|cc|cc|}&&H&&aH&&\sigma H&&a\sigma H&\\\
\hline\cr&(\sigma,\tau)&1&a^{2}&a&a^{3}&\sigma&a^{2}\sigma&a\sigma&a^{3}\sigma\\\
\hline\cr H&1&1&1&1&1&1&1&1&1\\\ &a^{2}&1&1&1&1&1&1&1&1\\\ \hline\cr
aH&a&1&1&\pi^{2}&\pi^{2}&1&1&\pi^{2}&\pi^{2}\\\
&a^{3}&1&1&\pi^{2}&\pi^{2}&1&1&\pi^{2}&\pi^{2}\\\ \hline\cr\sigma
H&\sigma&1&1&1&1&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{2}\\\
&a^{2}\sigma&1&1&1&1&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{2}\\\ \hline\cr a\sigma
H&a\sigma&1&1&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{4}&\pi^{4}\\\
&a^{3}\sigma&1&1&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{2}&\pi^{4}&\pi^{4}\\\
\hline\cr\end{array}$

We remind that $\epsilon_{\widehat{r}}$ is not a $2$-cocycle. In order
$\epsilon_{\widehat{r}}$ to become an element of $B^{2}(G,S^{*})$, every
element $\epsilon_{\widehat{r}}(\sigma,\tau)$, with $\sigma,\tau\in G$, must
be multiplied with the invertible term $u(\sigma)^{r(\tau)}$.

References

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Spectral Radius of Biased Random Walks
on Regular Trees

Song He School of Mathematics and Statistics, Huaiyin Normal University
Huaian City, 223300, P. R. China
<EMAIL_ADDRESS>

###### Abstract

We consider biased random walk on regular tree and we obtain the spectral
radius, first return probability and $n$-step transition probability.

Keywords: Biased random walk, spectral radius, regular tree.

##  1 Introduction

Let $G=(V(G),\,E(G))$ be a locally finite, connected infinite graph, where
$V(G)$ is the set of its vertices and $E(G)$ is the set of its edges. Fix a
vertex $o$ of $G$ as the root, we assume that $o$ has at least one neighbor.
For any vertex $x$ of $G$ let $|x|$ denote the graph distance between $x$ and
$o$. Let $\mathbb{N}:=\\{1,\,2,\ldots\\}$ and
$\mathbb{Z}_{+}=\mathbb{N}\cup\\{0\\}$. For any $n\in\mathbb{Z}_{+}$:

$B_{G}(n)=\\{x\in V(G):\ |x|\leq n\\},\qquad\partial B_{G}(n)=\\{x\in V(G):\
|x|=n\\}.$

Let $d\in\mathbb{N}$, $d\geq 2$, $\mathbb{T}_{d}$ denotes $d$-regular trees.
Fix a vertex $o$ of $\mathbb{T}_{d}$ as the root. For $\lambda>0$, if an edge
$e=\\{x,y\\}$ is at distance $n$ from $o$, its conductance is defined as
$\lambda^{-n}$. Denote by ${\mathrm{RW}}_{\lambda}$ the nearest-neighbour
random walk $(X_{n})_{n=0}^{\infty}$ among such conductances and call it the
$\lambda$-biased random walk. In other words, ${\mathrm{RW}}_{\lambda}$ on
$\mathbb{T}_{d}$ has the following transition probabilities: for $v\sim u$
(i.e., if $u$ and $v$ are adjacent on $\mathbb{T}_{d}$),

$\displaystyle
p(v,u):=p_{\lambda}^{\mathbb{T}_{d}}=\begin{cases}\frac{1}{d}&\mathrm{if}\
v=o,\\\ \frac{\lambda}{d+\lambda-1}&\mathrm{if}\ u\in\partial
B_{\mathbb{T}_{d}}(|v|-1)\ \mathrm{and}\ v\neq o,\\\
\frac{1}{d+\lambda-1}&\mathrm{otherwise}.\end{cases}$ (1.1)

Notice ${\mathrm{RW}}_{1}$ is just the simple random walk on $\mathbb{T}_{d}.$
By Rayleigh’s monotonicity principle (see [1], p. 35), there is a critical
parameter $\lambda_{c}(G)\in(0,\,\infty]$ such that ${\mathrm{RW}}_{\lambda}$
is transient for $\lambda<\lambda_{c}(G)$ and is recurrent for
$\lambda>\lambda_{c}(G)$. In the following we will introduce some basic
notations. Write

$p^{(n)}(x,\,y):=p^{(n)}_{\lambda}(x,\,y)=\mathbb{P}_{x}(X_{n}=y),$

where $\mathbb{P}_{x}:=\mathbb{P}_{x}^{G}$ is the law of
${\mathrm{RW}}_{\lambda}$ starting at $x$. The Green function is given by

$\mathbb{G}(x,\,y\,|\,z):=\mathbb{G}_{\lambda}(x,\,y\,|\,z)=\sum_{n=0}^{\infty}p^{(n)}(x,\,y)z^{n},~{}x,~{}y\in
V(G),~{}z\in\mathbb{C},\ |z|<R_{\mathbb{G}}\,,$

where $R_{\mathbb{G}}=R_{\mathbb{G}}(\lambda)=R_{\mathbb{G}}(\lambda,x,y)$ is
its convergence radius. Recall [1] Exercise 1.2,

$R_{\mathbb{G}}=R_{\mathbb{G}}(\lambda)=\frac{1}{\limsup_{n\to\infty}\sqrt[n]{p^{(n)}(x,y)}}$

is independent of $x$, $y$ when ${\mathrm{RW}}_{\lambda}$ is irreducible,
i.e., $\lambda>0$. Call

$\rho_{\lambda}=\rho(\lambda)=\frac{1}{R_{\mathbb{G}}}=\limsup_{n\to\infty}p^{(n)}(x,x)^{1/n}=\limsup_{n\to\infty}p^{(n)}(o,o)^{1/n}$

the spectral radius of ${\mathrm{RW}}_{\lambda}$. Let $M_{n}=|\partial
B_{G}(n)|$ be the cardinality of $\partial B_{G}(n)$ for any
$n\in\mathbb{Z}_{+}.$ Define the growth rate of $G$ as

$\mathrm{gr}(G)=\liminf_{n\rightarrow\infty}\sqrt[n]{M_{n}}.$

Since the sequence $\\{M_{n}\\}_{n=0}^{\infty}$ is submultiplicative, the
limit $\mathrm{gr}(G)=\lim\limits_{n\rightarrow\infty}\sqrt[n]{M_{n}}$ exists
indeed.

The motivation for introducing ${\mathrm{RW}}_{\lambda}$ was to design a
Monte-Carlo algorithm for self-avoiding walks by Berretti and Sokal [2]. See
[3, 4, 5] for refinements of this idea. Due to interesting phenomenology and
similarities to concrete physical systems ([6, 7, 8, 9, 10]), biased random
walks and biased diffusions in disordered media have attracted much attention
in mathematical and physics communities since the 1980s.

In the 1990s, Lyons ([11, 12, 13]), and Lyons, Pemantle and Peres ([14, 15])
made series of achievements in the study of ${\mathrm{RW}}_{\lambda}$’s.
${\mathrm{RW}}_{\lambda}$ has also received attention recently, see [16, 17,
18, 19] and the references therein. Ben Arous and Fribergh publish a survey on
biased random walks on random graphs see[20]. For spectral radius, R. Lyons
[12] showed that the critical parameter for $RW_{\lambda}$ on a general tree
is exactly the exponential of the Hausdorff dimension of the tree boundary.
And R. Lyons [13] proved that for Cayley graphs and degree bounded transitive
graphs, the growth rate is exactly the critical parameter of the
$RW_{\lambda}.$ This paper focuses on a specific properties of spectral radius
of ${\mathrm{RW}}_{\lambda}$’s on non-random infinite graphs.

We are ready to state our main results. The proofs will be presented in
Section 2.

###### Theorem 1.1

For the $d$-regular tree $\mathbb{T}_{d}$, the following holds:

$\rho_{\mathbb{T}_{d}}(\lambda)=\frac{2\sqrt{(d-1)\lambda}}{d-1+\lambda},\qquad\lambda\in(0,\,\lambda_{c}(\mathbb{T}_{d})]=(0,\,d-1],$

and for $\lambda\in(0,\,\infty)$ and $n\to\infty$,

$f^{(2n)}_{\lambda}(o,\,o)\sim\frac{1}{\sqrt{\pi}}\left(\frac{2\sqrt{(d-1)\lambda}}{d-1+\lambda}\right)^{\\!\\!2n}n^{-3/2}.$
(1.2)

Moreover,

$p^{(2n)}_{\lambda}(o,o)\sim\begin{cases}\frac{(d-1-\lambda)^{2}}{16(\pi\lambda)^{1/2}(d-1)^{3/2}}\rho_{\mathbb{T}_{d}}(\lambda)^{2n}n^{-3/2}&\hbox{if}\
\lambda\in(0,\,d-1),\\\ \frac{1}{\sqrt{\pi n}}&\hbox{if}\
\lambda=d-1.\end{cases}$ (1.3)

###### Remark 1.2

Since for the case $\lambda>\lambda_{c}(\mathbb{T}_{d})=d-1$,
${\mathrm{RW}}_{\lambda}$ is recurrence, it means that $\rho_{\lambda}=1$.
Hence, the spectral radius $\rho_{\lambda}$ is continuous in
$\lambda\in(0,\infty)$, and $\rho(\lambda_{c}(\mathbb{T}_{d}))=1$.

###### Remark 1.3

For $\lambda\in(0,\,d-1)$, the derivative of $\rho_{\mathbb{T}_{d}}(\lambda)$:

$\rho^{{}^{\prime}}_{\mathbb{T}_{d}}(\lambda)=\frac{\sqrt{(d-1)/\lambda}(d-1+\lambda)}{(d-1-\lambda)^{2}}>0.$

It means that $\rho_{\mathbb{T}_{d}}(\lambda)$ is strictly increasing for
$\lambda\in(0,\,d-1)$.

##  2 Proof of Therom 1.1

Proof of Theorem 1.1 Assume $\lambda>0.$ Notice that ${\mathrm{RW}}_{\lambda}$
$(X_{n})_{n=0}^{\infty}$ must return to $o$ in even steps, and that
$\\{|X_{n}|\\}_{n=0}^{\infty}$ with $|X_{0}|=0$ is a Markov chain on
$\mathbb{Z}_{+}$ with transition probabilities given by

$p(x,\,y)=\left\\{\begin{array}[]{cl}1&{\rm if}\ x=0,\ y=1\\\
\frac{\lambda}{d-1+\lambda}&{\rm if}\ y=x-1\ \mbox{and}\ x\neq 0,\\\
\frac{d-1}{d-1+\lambda}&{\rm otherwise}.\end{array}\right.$

Recall for any $n\in\mathbb{N}$ and $k\in\mathbb{Z}_{+}$,

$f^{(2n)}_{\lambda}(o,\,o)=\mathbb{P}_{o}\left(\tau_{o}^{+}=2n\right),\qquad
f^{(2n-1)}_{\lambda}(o,\,o)=0,\ \lambda\in(0,\,\infty),$

and the $k$th Catalan number given by $c_{k}=\frac{1}{k+1}{{2k}\choose{k}}$,
with the associated related generating function

$\mathcal{C}(x):=\sum_{k=0}^{\infty}c_{k}x^{k}=\frac{1-\sqrt{1-4x}}{2x},\qquad
x\in\left[-\frac{1}{4},\,\frac{1}{4}\right].$ (2.1)

Note the number of all $2n$-length nearest-neighbour paths
$\gamma=w_{0}w_{1}\cdots w_{2n}$ on $\mathbb{Z}_{+}$ such that

$w_{0}=w_{2n}=0,\qquad w_{j}\geq 1,\ 1\leq j\leq 2n-1$

is precisely $c_{n-1}$. Hence for any $\lambda>0,$

$f_{\lambda}^{(2n)}(o,\,o)=c_{n-1}\left(\frac{d-1}{d-1+\lambda}\right)^{n-1}\left(\frac{\lambda}{d-1+\lambda}\right)^{n},\
n\in\mathbb{N},$

which readily yields (1.2) by means of Stirling’s formula.

By definition, for $\lambda>0$,

$\displaystyle U_{\lambda}(o,\,o\,|\,z)$
$\displaystyle=\sum_{n=1}^{\infty}f^{(2n)}_{\lambda}(o,\,o)z^{2n}=\sum_{n=1}^{\infty}c_{n-1}\left(\frac{d-1}{d-1+\lambda}\right)^{n-1}\left(\frac{\lambda}{d-1+\lambda}\right)^{n}z^{2n}$
$\displaystyle=\frac{\lambda}{d-1+\lambda}z^{2}\,\mathcal{C}\left(\frac{\lambda(d-1)z^{2}}{(d-1+\lambda)^{2}}\right),$

which, in view of (2.1), implies that for
$|z|\leq\frac{d-1+\lambda}{2\sqrt{\lambda(d-1)}}$,

$U_{\lambda}(o,\,o\,|\,z)=\frac{(d-1+\lambda)-\sqrt{(d-1+\lambda)^{2}-4\lambda(d-1)z^{2}}}{2(d-1)}.$
(2.2)

Taking $z=1$ gives that

$\mathbb{P}_{o}\left(\tau_{o}^{+}<\infty\right)=U_{\lambda}(o,\,o\,|\,1)=\frac{\lambda\wedge(d-1)}{d-1}.$

Notice from (2.2) that when $0<\lambda\leq d-1$,

$U_{\lambda}\left(o,\,o\,\Big{|}\,\frac{d-1+\lambda}{2\sqrt{\lambda(d-1)}}\right)=\frac{d-1+\lambda}{2(d-1)}\leq
1.$

Hence, for $|z|<\frac{d-1+\lambda}{2\sqrt{\lambda(d-1)}}$ and $0<\lambda\leq
d-1$,

$\displaystyle\mathbb{G}_{\lambda}(o,\,o\,|\,z)$
$\displaystyle=\frac{1}{1-U_{\lambda}(o,\,o\,|\,z)}$
$\displaystyle=\frac{2(d-1)}{2(d-1)-(d-1+\lambda)+\sqrt{(d-1+\lambda)^{2}-4\lambda(d-1)z^{2}}}.$
(2.3)

This implies that the convergence radius for
$\mathbb{G}_{\lambda}(o,\,o\,|\,z)$ is
$\frac{d-1+\lambda}{2\sqrt{\lambda(d-1)}}$. In other words,

$\rho(\lambda):=\rho_{\mathbb{T}_{d}}(\lambda)=\frac{2\sqrt{\lambda(d-1)}}{d-1+\lambda},\qquad
0<\lambda\leq d-1.$

It remains to show (1.3) for $\lambda\in(0,\,d-1)$. Write
$a(\lambda)=\frac{2(d-1)}{d-1+\lambda}$ and
$b(\lambda)=\frac{d-1-\lambda}{d-1+\lambda}.$ Then for any $|z|\leq
R_{\mathbb{G}}(\lambda)=\frac{1}{\rho(\lambda)},$

$\mathbb{G}_{\lambda}(o,\,o\,|\,z)=\frac{2(d-1)}{d-1+\lambda}\frac{1}{\frac{d-1-\lambda}{d-1+\lambda}+\sqrt{1-\rho(\lambda)^{2}z^{2}}}=\frac{a(\lambda)}{b(\lambda)+\sqrt{1-\rho(\lambda)^{2}z^{2}}}.$

noindent Let

$\Phi(t):=\Phi_{\lambda}(t)=\frac{-a(\lambda)b(\lambda)+\sqrt{a(\lambda)^{2}+\rho(\lambda)^{2}(1-b(\lambda)^{2})t^{2}}}{1-b(\lambda)^{2}},\qquad
t\in\mathbb{R}.$

Then for any $|z|\leq R_{\mathbb{G}}(\lambda)$,

$\mathbb{G}_{\lambda}(o,\,o\,|\,z)=\Phi\left(z\,\mathbb{G}_{\lambda}(o,\,o\,|\,z)\right).$

Define

$\Psi(u,\,v):=\Phi(uv)-v,\qquad u,\,v\in\mathbb{R}\,.$

Then

$\displaystyle\frac{\partial\Psi(u,\,v)}{\partial
v}\Big{|}_{(u,\,v)=\left(\frac{1}{\rho(\lambda)},\,\mathbb{G}_{\lambda}(o,\,o\,|\,\frac{1}{\rho(\lambda)})\right)}=0,$
$\displaystyle c_{1}(\lambda):=\frac{\partial^{2}\Psi(u,\,v)}{\partial
v^{2}}\Big{|}_{(u,\,v)=\left(\frac{1}{\rho(\lambda)},\,\mathbb{G}_{\lambda}(o,\,o\,|\,\frac{1}{\rho(\lambda)})\right)}=\frac{(d-1-\lambda)^{3}}{2(d-1)(d-1+\lambda)^{2}}\not=0,$
$\displaystyle c_{2}(\lambda):=\frac{\partial\Psi(u,\,v)}{\partial
u}\Big{|}_{(u,\,v)=\left(\frac{1}{\rho(\lambda)},\,\mathbb{G}_{\lambda}(o,\,o\,|\,\frac{1}{\rho(\lambda)})\right)}=\frac{2\rho(\lambda)(d-1)}{d-1-\lambda}\not=0.$

Applying the method of Darboux (see [21] Theorem 5), we obtain that

$p^{(2n)}_{\lambda}(o,\,o)\sim\left(\frac{c_{1}(\lambda)}{2\pi\rho(\lambda)c_{2}(\lambda)}\right)^{1/2}\rho(\lambda)^{2n}{(2n)}^{-3/2}=\frac{(d-1-\lambda)^{2}}{16(\pi\lambda)^{1/2}(d-1)^{3/2}}\rho(\lambda)^{2n}n^{-3/2}.$

The idea of using the method of Darboux to establish the asymptotics for
$p^{(2n)}_{\lambda}(o,\,o)$ is not new. For example, in Woess [22] Chapter III
Section 17 pp. 181–189, examples of random walk on groups are given such that
$p^{(n)}(o,\,o)\sim c\rho^{n}n^{-3/2}$ for some constant $c>0$. The exact
value of $c$ is not known in general.

For $z\in(-1,\,1),$
$\mathbb{G}_{d-1}(o,\,o\,|\,z)=\frac{1}{\sqrt{1-z^{2}}}=\sum_{n=0}^{\infty}\frac{(2n)!}{2^{2n}(n!)^{2}}z^{2n}$.
Thus

$p_{d-1}^{(2n)}(o,\,o)=\frac{(2n)!}{2^{2n}(n!)^{2}}\sim\frac{1}{\sqrt{\pi
n}}.$

Acknowledgements. The authors would like to thank an anonymous referee and
Zhan Shi, Kainan Xiang, Longmin Wang for valuable comments and suggestions to
improve the quality of the paper.HS’s research is supported by Jiangsu
University Natural Science Research Project 21KJB110003 and by Xiang Yu Ying
Cai 31SH002.

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