TITLE: All possible solutions to $x^y = 8$ and $y^x = 9$
QUESTION [2 upvotes]: I know the obvious solution $(x,y)=(2,3)$. I just want to know if there are any other solutions to above system of equations or is $(x,y)=(2,3)$ the only solution. Also, both $x$ and $y$ should be real numbers.
Please provide a method to solve equations like these if it exists.

REPLY [1 votes]: I wrote it as a comment but i will complete it now ,
well, you have two equation with two variables and because the $\ln(x)$ and $e^x$ is one-to-one functions in the Real domain there is one solution. if you allow $x,y$ to be complex numbers then because $\ln(x),e^x$ are periodic function in the complex domain,i think there could by more than one answer.(actually infinitely many)
for instance : take $x=1+i$ then for all $y=\frac{i \left(2 \pi  c-3 i \log (2)\right)}{\log (1+i)}$ for any integer $c$.
hope its what you are looking for.


TITLE: Common tangent lines of two quadratic functions
QUESTION [8 upvotes]: Find all such lines that are tangent to the following curves:
$$y=x^2$$ and $$y=-x^2+2x-2$$
I have been pounding my head against the wall on this. I used the derivatives and assumed that their derivatives must be equal at those tangent point but could not figure out the equations. An explanation will be appreciated.

REPLY [0 votes]: Let $f(x) = x^2$  and  $g(x) = -x^2 + 2x - 2 \implies \dfrac{d}{dx}(f(x))|_{x = a} = 2a$ and $\dfrac{d}{dx}(g(x))|_{x = b} = -2b + 2$
$2a = -2b + 2 \implies a = 1 - b \implies$
$A: (1 - b, (1 - b)^2), B: (b, -b^2 + 2b - 2)
\implies m_{AB} = \dfrac{2b^2 - 4b + 3}{1 - 2b} = -2b + 2 \implies $
$2b^2 - 2b - 1 = 0 \implies b = \dfrac{1 \pm \sqrt{3}}{2}$
Using $b = \dfrac{1 + \sqrt{3}}{2} \implies $
$A: (\dfrac{1 - \sqrt{3}}{2},\dfrac{2 - \sqrt{3}}{2}), B: (\dfrac{1 + \sqrt{3}}{2},\dfrac{\sqrt{3} - 4}{2}) \implies m_{AB} = 1 - \sqrt{3} \implies $
$ \boxed{y = (1 - \sqrt{3})x + \dfrac{\sqrt{3} - 2}{2}} $
Using $b = \dfrac{1 - \sqrt{3}}{2} \implies A^{'}: (\dfrac{1 + \sqrt{3}}{2}, \dfrac{2 + \sqrt{3}}{2}), B^{'}: (\dfrac{1 - \sqrt{3}}{2}, \dfrac{-\sqrt{3} - 4}{2}) \implies$
$m_{A^{'}B^{'}} = 1 + \sqrt{3} \implies \boxed{y = (1 + \sqrt{3})x - \dfrac{\sqrt{3} + 2 }{2}}. $


TITLE: Is the "space of physical quantities" a field of transcendence degree $6$ or $7$ over the rationals?
QUESTION [4 upvotes]: Excuse my naive question and please let me explain it:
In everyday life we experience 3 spatial "dimensions" + time etc.
Usually the 3 dimensions are represented by a coordinate system and mathematically as the vector space $R^3$.
In constrast, the word "dimension" in dimensional analysis has a more objective realism which is based on physical measurements. For instance let $l$ be a transcendental number.
Then $1,l,l^2,l^3,\cdots$ will be linearly independent over the rational numbers, and hence can serve as a basis in a vector space. For instance, instead of saying the point in spatial dimension has coordinates $(x,y,z)$ we might as well write:
$$x+y \cdot L + z \cdot L^2$$
where $L$ represents a "transcendental number over the rationals" and is associated with meter or kilometer etc.
Take for instance the derived unit $s/t^k$ for $k=0,1,2$ where $s$ denotes length and $t$ denotes time.
Then $s$=length, $s/t =$ velocity, $s/t^2=$ acceleration. If we view these as a basis of a vector space, then this vector space has 3 dimensions. But no one would count subjectively these as dimensions.
On the other hand the seven basic SI-units (or 6 if you do not want to count $mol$), could be seen as a transcendental numbers over the rationals or reals, and hence powers of those transcendental numbers could give a basis for vector spaces. Monomials of these transcendent numbers would correspond, as is done in dimensional analysis, to derived units. Adding and subtracting for instance $LT^{-1}+M$ would give a point in the "space of physical quantities" (Edit: and indicate the measured velocity and mass of an object). As a vector space over the reals this space has infinite dimension but has transcendence degree of $7$. Every derived / basic physical quantity measured by SI-Units would correspond to a point in this space / field of transcendence degree $6$ (or $7$ if you count $mol$ which I will not do):
$$\mathbb{Q}(T,L,M,I,\Theta,J)$$
Excuse my naive question: Is there any reason from physics to discard this point of view?
Edit:
It seems that the main idea of this question can be implemented through the ring (Laurent polynomial ring) $L:=\mathbb{R}[T,T^{-1},L,L^{-1},M,M^{-1},I,I^{-1},\Theta, \Theta^{-1},J,J^{-1}]$
second Edit:
I was asked to give at least one application of this idea, which I would like to do:
Application:
Let $k(x,y) = \frac{xy}{x^2+y^2-xy}$ for $x \neq 0,y \neq 0, x,y \in \mathbb{R}$ be a Jaccard-Similarity / positive definite kernel defined on $\mathbb{R}$.
We can define a similarity and positive definite kernel on the Laurent polynomial ring $L$ as :
$$K(x,y) := \frac{1}{N_x + N_y - N_{xy}} \sum_{X_i^{\alpha_i}=Y_j^{\beta_j}} k(a_i,b_j)$$
for $x = \sum_{i} a_i X^{\alpha_i},y = \sum_{j} b_j X^{\beta_j}$ and $X = (T , L, M, I, \Theta,J)$, $\alpha_i, \beta_j \in \mathbb{Z}^6$ and $X^{\alpha_i},X^{\beta_j}$ are multinomials, and $N_x = $ number of nonzero $a_i$, $N_y =$ number of nonzero $b_j$, $N_{xy} =$ number of $(X^{\alpha_i} = X^{\beta_j})$.
Since $k(ca,cb) = k(a,b)$ for all $a,b,c \neq 0$, we deduce that $K(c \cdot x,c \cdot y) = K(x,y)$ for all $x,y \in L$, $c \neq 0, c \in \mathbb{R}$.
Hence this (or any other similarity and positive definite kernel $k$ with $k(ca,cb) = k(a,b)$. This is to make sure, that rescaling of physical units, does not change the similarity between objects.) gives us a possibility to measure the similarity / inner product
of two physical objects $A,B$ each of which is defined through measurements $x = \sum_{i} a_i X^{\alpha_i}$ and $y = \sum_{j} b_j X^{\beta_j}$.
Since there are different possibilities to measure similarites / define positive definite kernels, there should be different possibilities to define
equality / similarity between two physical objects $A$ and $B$.
Hence aposteriori $L$ is a Hilbert space by the Aaronszajn-Kolmogorov theorem.
Example:
The meaning of this kernel is to compare two physical objects. For instance $A = 10 m/s + 1 kg$, $B = 9 m/s + 2kg$, $C = 1 m/s^2+10kg$. Then $K(A,B) = 226/276 = 0.8278$, $K(A,C) = 10/91= 0.10989$, $K(B,C) = 5/21 = 0.2381$. Hence $A$ is most similar to $B$, $B$ is most similar to $A$, $C$ is most similar to $B$, and $A,C$ are the most dissimilar physical objects in this list.

REPLY [14 votes]: I don't think there are many meaningful situations where physical quantities of different dimension are added together (in fact, this is widely regarded as a taboo, and precisely the sort of mistake that dimensional analysis is supposed to prevent us from doing), so I don't think viewing the space of physical quantities as a field is the most fecund or useful description.
Instead, let me offer the following alternative point of view, which I think better matches what physics does and why it does it, and which is, at any rate, how I think of physical dimensions:
Consider the “group of homogeneities” $G = (\mathbb{R}_+^\times)^k$ which is the product of $k$ copies of the positive real numbers, where $k$ is the number of “base units” (so $k=7$ in the SI system, but this depends on the sort of theory we wish to study: economists might have reasons to consider currencies or commodities as different units, whereas relativists will not consider time and length to be different units): the point is that $G$ acts on physical quantities by multiplying each one of the $k$ base quantities by the corresponding positive real number (e.g., $g=(2,3,1,\ldots)$ might multiply lengths by $2$, times by $3$ and preserve masses).
The crucial physics point is that: insofar as the choice of units is arbitrary, $G$ acts as a group of physics-preserving symmetries (e.g., multiplying all lengths by $2$ and times by $3$ while preserving masses should keep physics the same).  So we seek quantities that are covariant under $G$.  This is the whole point of dimensional analysis.
Now the group $G^*$ of characters of $G$, i.e., the group of continuous morphisms $\lambda\colon G\to \mathbb{R}_+^\times$ is the group $\mathbb{Z}^k$ with $(d_1,\ldots,d_k) \in \mathbb{Z}^k$ sending $g := (g_1,\ldots,g_k) \in G$ to $g_1^{d_1}\cdots g_k^{d_k}$.  Really we should forget about $k$ and care about the group $G$ which might not have a clear “basis of dimensions” (e.g., is it electrical charge or electrical current which is the “base unit”? this is meaningless) and might even incorporate other transformations than positive-real-homogeneities (e.g., changes of sign on certain quantities might be physics-preserving).
So I propose that: a “physical dimension” is an element $\lambda$ of $G^*$ and a quantity of that dimension is an element of the corresponding representation $V_\lambda$ of $G$ (which is $1$-dimensional because $G$ is abelian), i.e., $G$ acts on $V_\lambda$ by $g \mapsto v \mapsto g^\lambda\cdot v$.
This $V_\lambda$ is $1$-dimensional, so it is $\mathbb{R}$ as an $\mathbb{R}$-vector space, but it is so in a non-canonical way, and a unit of measure for the physical dimension $\lambda$ is a basis for $V_\lambda$ (probably a positive basis since $V_\lambda$ still has a natural orientation).
Taking the product of two physical quantities of dimensions $\lambda$ and $\mu$ consists of taking the tensor product $V_\lambda \otimes V_\mu \buildrel\sim\over\to V_{\lambda+\mu}$, while the inverse of a physical quantity $\lambda$ should be seen as living in the dual space $V_\lambda^\vee \buildrel\sim\over\to V_{-\lambda}$.  The sum, on the other hand, is meaningful only insofar as it compatible under the action of $\lambda$, i.e., adding elements of the same $V_\lambda$.  (Of course, you are welcome to consider $V_\lambda \oplus V_{\lambda'}$, in some cases it might be meaningful, e.g., adding different currencies in economics, but the direct sum is then canonically split.)
What you propose to do, by taking arbitrary rational combinations of the base units and considering them as transcendental quantities, would be germane if the physically relevant acting group were the full Cremona group of automorphisms of rational fractions.  I don't think we can make any physical sense of such an action, so this point of view is bound to remain very artificial.
PS: I'd like to add the following remark showing that the group of symmetries (so, in my description, $k$) might very well depend on the kind of theory being considered: Americans like to measure heights in feet and plane distances in miles, which makes some kind of sense in a situation where one might apply different homotheties on both; the “fundamental slope” of $5280\,\mathrm{ft}/\mathrm{mi}$ breaks this symmetry, exactly in the same way the “fundamental speed” known as the speed of light of $299\,792\,458\,\mathrm{m}/\mathrm{s}$ breaks the possibility of applying different homotheties on times and lengths: there is no fundamental difference between $5280\,\mathrm{ft}/\mathrm{mi}$ and $299\,792\,458\,\mathrm{m}/\mathrm{s}$ from the dimensional point of view: they are conversion factors breaking the possibility of choosing two units independently which, from the pre-breaking point of view, appears as a fundamental constant of nature (which can be measured experimentally: if the foot and the mile had been defined separately using a reference height and a reference plane length, the former would be an experimental quantity until redefinition of one of the units, exactly like the speed of light in the SI system).


TITLE: Relating area to a line intersecting with a point.
QUESTION [1 upvotes]: I really could use a hint with this following problem:
If a line L separates a parallelogram into two regions of equal areas, then L contains the point of intersection of the diagonals of the parallelogram.
The figure shows a line L horizontally through the sides of the parallelogram.
This creates two trapezoids and I can intuitively show that if the bases of the trapezoids are not congruent then the areas can not be equal.
I just can not currently see how to relate the intersection with the diagonals.
Any help will be appreciated.

REPLY [1 votes]: WLOG, draw in one of the diagonals, $d$. This will form two triangles using parts of $d$, $L$, and the sides of the parallelogram. Using properties of parallel lines, these triangles must be similar. Moreover, your discovery that the trapezoid bases must be equal implies that these triangles are actually congruent, and therefore $L$ intersects the diagonal at its midpoint (also the point of intersection with the other diagonal).


TITLE: An inequality property of the Fibonacci sequence
QUESTION [2 upvotes]: Given the Fibonacci sequence $F_n$, Wikipedia says (http://en.wikipedia.org/wiki/Fibonacci_number#List_of_Fibonacci_numbers) $$ F_{2n-1} = F_n^2+F^2 _{n-1}$$ so that $$F_{2n-1}>F^2_n$$
What is the smallest such k for which $$F_{n+k}>F^2_n\,\,?$$
I'm not sure where to start or to find smaller values than $2n-1$

REPLY [2 votes]: $k=n-1$ is the minimum possible value.
We have $F_{2n-1}=F_n^2+F_{n-1}^2$.  Rewrite the LHS as $F_{2n-2}+F_{2n-3}$, so $$F_{2n-2}=F_n^2+F_{n-1}^2-F_{2n-3}$$
But now $F_{2n-3}=F_{n-1}^2+F_{n-2}^2$.  Substituting, we get $$F_{2n-2}=F_n^2-F_{n-2}^2<F_n^2$$
Hence even the term immediately preceding $F_{2n-1}$ is less than $F_n^2$.


TITLE: Sequence with limit $\infty$
QUESTION [1 upvotes]: Let $x_n$ be a sequence of real strictly positive numbers. If $\lim_{n\to \infty} x_n=\infty$, is it necessary for the sequence to be strictly increasing? If not, give a counterexample.
Intuitively I think that it is true, but I have no idea how to prove it.

REPLY [1 votes]: Let consider for example
$$x_n=n+2\cos (n\pi)=-1,4,1,6,3,\ldots$$


TITLE: A function $f$ such that the limit of $f(x^2)$ exists but not $f(x)$.
QUESTION [0 upvotes]: I want to show a function $f$ such that $\displaystyle\lim_{x\to x_0}f(x^2)\in\mathbb{R}$ but $\displaystyle\lim_{x\to x_0}f(x)$ doesn't exist.
I only need a suggest of such a function $f$. I can't find an example of this.
Any hint? Thanks.

REPLY [6 votes]: Let $f(x) = \begin{cases} -1,&x < 0,\\1,& x \geq 0, \end{cases}$ with $x_0 = 0$.


TITLE: Show this expression transform like a covariant tensor
QUESTION [0 upvotes]: Show that $g_{\mu v} V^{v}$ transform like a covariant vector.
$$g_{\mu' v'} V^{v'} = g_{\mu v} V^{v} \Lambda_{\mu'}^{\;\; \mu} \Lambda_{\;\; v'}^{v} \Lambda_{\;\; v}^{v'} = g_{\mu v} V^{v} \Lambda_{\mu'}^{\;\; \mu} \times(\Lambda \Lambda)^{v}_{\;v}$$
Now, the problem is this last term. Tecnically it should be the identity, but i am almost sure this is not the case, so i think i have committed an error, but where?

REPLY [2 votes]: Note that$$g_{\mu^\prime\nu^\prime}V^{\nu^\prime}=\frac{\partial x^\mu}{\partial x^{\mu^\prime}}\frac{\partial x^\nu}{\partial x^{\nu^\prime}}g_{\mu\nu}V^{\nu^\prime}=\frac{\partial x^\mu}{\partial x^{\mu^\prime}}g_{\mu\nu}V^{\nu}.$$Feel free to rewrite the partial derivatives as $\Lambda$ entries.


text:
\begin{document}
\title[Nichols algebras which are free algebras]
 {A characterization of Nichols algebras of diagonal type which are free algebras}
\author{I. Heckenberger}
\address{Philipps-Universit\"at Marburg,
FB Mathematik und Informatik,
Hans-Meerwein-Stra\ss e,
35032 Marburg, Germany.}
\email{heckenberger@mathematik.uni-marburg.de}

\author{Y. Zheng}
\address{Department of mathematics East China Normal University, Shanghai 200241, China.}
\email{52150601007@ecnu.cn}
\thanks{The second named author was supported by China Scholarship Council}

\date{}

\begin{abstract}
This paper is devoted to explore the freeness of Nichols algebras of diagonal type and to determine
the dimension of the kernel of the shuffle map considered as an operator acting on the free algebra.
Our proof is based on an inequality for the number of Lyndon words and on an identity for the shuffle map.
For a particular family of examples, the freeness of the Nichols algebra is characterized in terms of
solutions of a quadratic diophantine equation.

\textit{Keywords}: {Nichols algebras, free algebras, shuffle map, Lyndon words}
\end{abstract}
\maketitle

\section{Introduction}

Since their introduction in the late 70ies by W.~Nichols \cite{MR0506406},
the theory of Nichols algebras enjoyed increasing interest because of its
deep interrelation to different research areas. For an overview we refer to \cite{MR3728608}.
The strongest results have been
obtained for finite-dimensional Nichols algebras of diagonal type, mainly due to
the existence of the root system which was introduced in \cite{H2006}, based on
deep results of V.~Kharchenko \cite{MR1763385} on the structure of certain Hopf algebras
generated by group-like and skew-primitive elements.

A general, very difficult question is, what are the roots and their multiplicities
of a given Nichols algebra of diagonal type. In the case of finite-dimensional
Nichols algebras the answer is known: The roots are the real roots with respect to
the action of the Weyl groupoid, and their multiplicity is one.
The other extreme case is the one of the free algebra, where the root vectors are
parametrized by Lyndon words and appropriate powers of them.
Roots of the form $m\alpha_1+\alpha_2$ with $m\ge 0$ are determined using Rosso's
lemma \cite{MR1632802}.
Roots of the form $m\alpha_1+2\alpha_2$ and their multiplicities have been
determined by the authors in \cite{H2018root}. In this paper we address the question
when the multiplicity of a root is smaller than in the tensor algebra.
In particular, we provide a criterion to decide whether
a given Nichols algebra of diagonal type is a free algebra in terms of
polynomial equations for the entries of the braiding matrix.

The defining ideal of a Nichols algebra is spanned by the kernels of
the braided symmetrizer \cite{schauenburg1996}, which decomposes into
a product of shuffle maps. In \cite{duchamp1997},
the authors study identities involving shuffle maps. We use these identities
to study the freeness of Nichols algebras of diagonal type and to determine
the dimension of the kernel of the shuffle map.
With our results we relate the freeness of Nichols algebras of diagonal
type with braiding matrix $(q^{m_{ij}})_{1\le i,j\le n}$, $m_{ij}\in \ndZ$
for all $i,j$, to solutions of a diophantine equation.

In Section~\ref{basic}
we define a family $(P_{\dlm})_{\dlm\in \ndN_0^n,|\dlm|\ge 2}$ of elements
in the polynomial ring $\ndZ[p_{ij}\mid 1\le i,j\le n]$, where
$|(m_1,\dots,m_n)|=\sum_{i=1}^nm_i$.
Let now $\NA(V)$ be a Nichols algebra of diagonal type of rank $n$ with braiding matrix
$\bq =(q_{ij})_{1\le i,j\le n}\in (\fie^\times)^{n\times n}$, where $\fie $ is a field.

\begin{theo} (see Theorem \ref{theo})
We have $\NA(V)=T(V)$ if and only if $P_{\dlm}(\bq )\ne 0$ for all $\dlm\in \ndN_0^n$
with $|\dlm|\ge 2$.
\end{theo}

Assume that $\fie $ has characteristic $0$. If $P_{\dlm}(\bq)=0$ for some
$\dlm \in \ndN_0^n$, then in Section~\ref{upperbound}
two numbers $n_1(\bq),n_2(\bq)\in \ndN_0$ are defined.

\begin{theo} (see Theorem \ref{theo dim})
Assume that $\fie $ has characteristic $0$.
Let $\dlm =(m_1,\dots,m_n)\in \ndN_0^n$ and $m=\sum_{i=1}^nm_i$ such that
$m\ge 2$, $P_{\dlm}(\bq)=0$, and $P_{\dll }(\bq)\ne0$ for all $\dll<\dlm$.
Then
\begin{align*}
\dim(\ker(\rho_m(S_{1,m-1})|V_{\dlm}))=n_1(\bq)-n_2(\bq).
\end{align*}
\end{theo}

The organization of the paper is as follows.
In Section \ref{basic}, we recall some basic notions about Lyndon words and introduce notations.
We also recall the inequalities on the number of Lyndon words, which the paper is based on.
In Section \ref{shuffles}, we discuss the notion of a free prebraided module over a commutative ring and
compute the determinant of the shuffle map.
In Section \ref{freeness}, we formulate and prove our first main theorem.
In Section \ref{upperbound}, we determine an upper bound for the dimension of the kernel of shuffle map.
In Section \ref{dim}, we prove that this upper bound is a lower bound.

The paper was written during the visit of the second named author to Marburg University
supported by China Scholarship Council. The second named author thanks the department of
FB Mathematik and Informatik of Marburg University for hospitality.

\section{Basic Definitions and properties}
\label{basic}

Throughout this paper we write $\ndN$ and $\ndZ$ for the set of positive integers
and the set of integers, respectively. Let $\ndN_0=\ndN\cup\{0\}$.

We start with recalling necklaces and Lyndon words, and collect some notations.

Let $n\in \ndN$. For any $\dlm=(m_1,\dots,m_n)\in \ndN_0^n$ we write
$|\dlm|=\sum_{i=1}^nm_i$. If additionally $\dlm\ne 0$, then let
$\gcd(\dlm)$ be the greatest common divisor of $m_1,\dots,m_n$,
and if $|\dlm|\ge 2$, then let
$$N(\dlm)=\gcd\{m_i(m_i-1),m_jm_k \mid 1\le i,j,k\le n,j<k\}. $$
For any $1\le i\le n$ let
$\underline{e}_i=(\delta_{ij})_{1\le j\le n}\in \ndN_0^n$,
and for any $k\in \ndN_0$ and any
$\dlm =(m_1,\dots,m_n)\in \ndN_0^n\setminus \{0\}$
let $\dlm/k=(m_1/k,m_2/k,\ldots,m_n/k)$.

There is a partial ordering on $\ndN_0^n$
denoted by $\le$: $\dlm\le\dll$ if and only if $m_i\le l_i$ for all $1\le i\le n$.

Let $B$ be a set (called the alphabet) of $n$ elements denoted by
$b_1,b_2,\ldots,b_n$,
and let $\mathbb{B}$ and $\mathbb{B}^{\times}$ be the set of words and
non-empty words, respectively, with letters in $B$.
For $w=b_{i_1}b_{i_2}\cdots b_{i_s} \in \mathbb{B}$,
in which $b_j$ occurs  $m_j$ times, $1\le j\le n$, we write
 $\deg w=(m_1,m_2,\ldots,m_n)$ and call $\deg w$ the \textbf{degree} of $w$.

We fix a total order $\le $ on $B$. There is a total order $\lelex $
on $\mathbb{B}$ induced by $\le $, called the lexicographic order: For $u,v\in \mathbb{B}$,
one lets $u\lelex v$ if and only if either $v=uw$ for some $w\in \mathbb{B}$,
or there exist $w,u',v'\in \mathbb{B}$ and $x,y\in B$ such that $u=wxu'$, $v=wyv'$,
$x\le y$, and $x\ne y$.

A word $w\in\mathbb{B}^{\times}$ is called a \textbf{necklace} if for any decomposition
$w=uv$ with $u,v\in\mathbb{B}^{\times}$, $w \lelex vu$.
A word $w\in\mathbb{B}^{\times}$ is \textbf{Lyndon} if for any decomposition
$w=uv$, $u,v\in \mathbb{B}^{\times}$, $w \lelex v$.
For any $\dlm \in \ndN_0^n$ let $N_{\dlm }$ and $\ell_{\dlm}$ denote the number
of necklaces and Lyndon words, respectively, of degree $\dlm $.

\begin{rema}\label{reLy}
Any Lyndon word is a necklace, and for any necklace $w$ there is a unique pair
$(v,k)\in \mathbb{B}\times \ndN $ such that $v$ is Lyndon and $w=v^k$. Thus,
for any $\dlm \in \ndN_0^n\setminus\{0\}$,
\begin{align}\label{NL}
N_{\dlm}=\sum_{d\mid\gcd(\dlm)}\ell_{\dlm/d}.
\end{align}
\end{rema}

\begin{rema} \label{re:smallell}
  In\cite{HJ2018} and \cite{Joe1999} one can find explicit formulas
  for $N_{\dlm}$ and $\ell_{\dlm}$ for any $\dlm \in \ndN_0^n$.
  In particular,
  \begin{align}
  \label{eq:numbL1}
  &\ell_{\dle_i+k\dle_j}=1, ~\text{for all}~ k\in \ndN_0;\\
  \label{eq:numbL2}
  &\ell_{k\dle_j}=\delta_{k,1},~\text{for all}~ k\in \ndN_0.
  \end{align}
\end{rema}

In the remaining part of this section we will introduce and study
some polynomials, which are crucial for the paper.
For any ring $R$ and any $q\in R$ let $(0)_q=0$ and
$(m)_q=1+q+\cdots+q^{m-1}$ for any $m\in \ndN$.

\begin{defi}\label{def:Pm}
 For any $\dlm=(m_1,m_2,\ldots,m_n) \in \ndN_0^n$ with $|\dlm|\ge 2$
 let $P_{\dlm}\in \ndZ[p_{ij}\mid 1\le i,j\le n]$ be as follows:
 \begin{enumerate}
  \item If $\dlm=m_i\dle_i$, where $1\le i\le n$, and $m_i\in \ndN$, let
    $$P_{\dlm}=(m_i)_{p_{ii}};$$
  \item If $\dlm=\dle_i+m_j\dle_j$, where $1\le i,j\le n$, $i\ne j$,
  $m_j\in \ndN$, let
    $$P_{\dlm}=1-p_{jj}^{m_j-1}p_{ij}p_{ji};$$
  \item If $\dlm=2\dle_i+m_j\dle_j$, where $1\le i,j\le n$, $i\ne j$,
    $m_j\in \ndN$, let
    $$P_{\dlm}=1+p_{jj}^{m_j(m_j-1)/2}(-p_{ij}p_{ji})^{m_j}p_{ii};$$
  \item If $\dlm=3\dle_i+3\dle_j$, where $1\le i,j\le n$, $i\ne j$, let
       $$P_{\dlm}=(3)_{p_{ii}^2(p_{ij}p_{ji})^3p_{jj}^2};$$
  \item If $\dlm=3\dle_i+4\dle_j$, where $1\le i,j\le n$, $i\ne j$, let
  $$P_{\dlm}=(1-p_{ii}^2(p_{ij}p_{ij})^4p_{jj}^4)(3)_{p_{ii}(p_{ij}p_{ji})^2p_{jj}^2};$$
  \item If $\dlm=3\dle_i+6\dle_j$, where $1\le i,j\le n$, $i\ne j$, let
  $$P_{\dlm}=(1-p_{ii}(p_{ij}p_{ji})^3p_{jj}^{5})(3)_{p_{ii}^{2}(p_{ij}p_{ji})^6p_{jj}^{10}};$$
  \item If $\dlm=4\dle_i+4\dle_j$, where $1\le i,j\le n$, $i\ne j$, let
  $$P_{\dlm}=(1+p_{ii}^{3}(p_{ij}p_{ji})^4p_{jj}^{3})(1+p_{ii}^{6}(p_{ij}p_{ji})^8p_{jj}^{6});$$
  \item Otherwise, let
   $$P_{\dlm}=1-\prod_{1\le i\le n}p_{ii}^{m_i(m_i-1)}
   \prod_{1\le i<j\le n}(p_{ij}p_{ji})^{m_im_j}.$$
 \end{enumerate}
  Moreover, let
  $$ Q_{\dlm}=\prod_{1\le i\le n}p_{ii}^{m_i(m_i-1)/N(\dlm)}
  \prod_{1\le i<j\le n}(p_{ij}p_{ji})^{m_im_j/N(\dlm)}.$$
\end{defi}

\begin{rema}\label{Qfactor}
Let $\dlm \in \ndN_0^n$ with $|\dlm|\ge 2$.
By definition of $N(\dlm)$,
$Q_{\dlm}$ is a well-defined non-constant
monomial in $\ndZ[p_{ij}\mid 1\le i,j\le n]$,
and $Q_{\dlm}$ is not a non-trivial power of any other monomial.
Moreover,
$P_{\dlm}$ divides $1-Q_{\dlm}^{N(\dlm)}$. In particular,
$P_{\dlm}=1-Q_{\dlm}^{N(\dlm)}$ in the last case of Definition~\ref{def:Pm}.
\end{rema}

For any $i\in \ndN$ let $\Phi_i\in \ndZ[x]$ denote the $i$-th cyclotomic
polynomial, that is, the minimal polynomial of any primitive $i$-th root of $1$
in the complex numbers. Clearly, $x^k-1=\prod_{i\mid k}\Phi_i$ for any $k\in
\ndN$.

Next we describe the irreducible factors of the polynomials
$P_{\dlm}$.

\begin{lemm} \label{le:mmatrix}
	Let $D$ be a Euclidean domain, let
	$\dlm =(m_1,\dots,m_n)$ be a non-zero vector in $D^n$, and let
	$d=\gcd(m_1,\dots,m_n)$. Then there is a matrix $M\in D^{n\times n}$
	with $\dlm$ as its first row and determinant $d$.
\end{lemm}

\begin{proof}
	View $\dlm$ as a $1\times n$-matrix. Choose a composition $f$
	of elementary column transformations which maps $\dlm$
	to the vector $(d,0,\dots,0)$. Let $M'$ be the diagonal matrix with
	diagonal entries $(d,1,\dots,1)$. Then $M=f^{-1}(M')$ satisfies the desired
	properties.
\end{proof}

\begin{rema}\footnote{Thanks to Ben Anthes for this remark.}
  Lemma\ref{le:mmatrix} also holds for principal ideal domains $D$.
  On the other hand, let $D=\fie [x,y]$ for some field $\fie $
  and let $\dlm=(x,y)$. Then $\gcd(\dlm)=1$, but there are no $a,b\in D$
  with $xb-ya=1$. Hence Lemma~\ref{le:mmatrix} does not hold for this $D$.
\end{rema}

\begin{lemm} \label{le:Laurentauto}
	Let $\dlm=(m_1,\dots,m_n)$ be a non-zero vector in $\ndZ^n$ with
	$\gcd(m_1,\dots,m_n)=1$.
	Then there is a ring automorphism $\varphi $ of the Laurent polynomial ring
	$\ndZ[x_1^{\pm 1},\dots,x_n^{\pm 1}]$
	with $\varphi (x_1)=x_1^{m_1}\cdots x_n^{m_n}$.
\end{lemm}

\begin{proof}
	By Lemma~\ref{le:mmatrix} there is a matrix $M\in \ndZ^{n\times n}$ with
	$\dlm$ as its first row and with determinant $1$. Then
	$\varphi (x_i)=x_1^{m_{i1}}x_2^{m_{i2}}\cdots x_n^{m_{in}}$ for $1\le i\le n$
	defines a ring automorphism of
	$\ndZ[x_1^{\pm 1},\dots,x_n^{\pm 1}]$ as desired.
\end{proof}

\begin{lemm} \label{le:PhikQirred}
	For any $k\in \ndN$ and any $\dlm \in \ndN_0^n$ with $|\dlm|\ge 2$,
	the polynomial $\Phi_k(Q_{\dlm})\in \ndZ[p_{ij}\mid 1\le i,j\le n]$
	is irreducible. In particular, $Q_{\dlm}^k-1=\prod_{i\mid k}\Phi_i(Q_{\dlm})$
	is the unique factorization of $Q_{\dlm}^k-1$ into irreducibles, and each
	irreducible factor of $P_{\dlm}$ is of the form $\Phi_l(Q_{\dlm})$ for some
	$l\mid N(\dlm)$.
\end{lemm}

\begin{proof}
	Let $k\in \ndN$ and $\dlm \in \ndN_0^n$ with $|\dlm|\ge 2$.
	For any $1\le i,j,l\le n$ let
	$$m_{ii}=m_i(m_i-1)/N(\dlm),\quad m_{jl}=m_jm_l/N(\dlm).$$
	Then $\gcd(m_{ij}\mid 1\le i,j\le n)=1$ and $Q_{\dlm}=\prod_{1\le i,j\le
	n}p_{ij}^{m_{ij}}$
	by construction. By Lemma~\ref{le:Laurentauto}
	there is a ring automorphism $\varphi $ of the Laurent polynomial ring
	$\ndZ[p_{ij}^{\pm 1}\mid 1\le i,j\le n]$ with $\varphi (p_{11})=
	Q_{\dlm}$.
	Thus $\Phi_k(Q_{\dlm})=\varphi (\Phi_k(p_{11}))$ is irreducible in
	$\ndZ[p_{ij}^{\pm 1}\mid 1\le i,j\le n]$.
	Since $\Phi_k(Q_{\dlm})$ is not divisible in
	$\ndZ[p_{ij}\mid 1\le i,j\le n]$ by any $p_{ij}$ with $1\le i,j\le n$,
	the polynomial $\Phi_k(Q_{\dlm})$ is irreducible.
\end{proof}

\begin{lemm}\label{le:prime}
  Let $\dlm,\dll\in \ndN_0^n$ with $|\dll|\ge 2$.
Suppose that there exist $1\le i<j\le n$ such that $m_i,m_j\ne 0$.
Then $P_{\dlm}$ and $P_{\dll}$
are relatively prime if and only if $\dlm\ne\dll$.
In particular,  $P_{\dlm}$ and $P_{\dll}$
are relatively prime whenever $\dll<\dlm$.
\end{lemm}

\begin{proof}
	Recall that $P_{\dlm}$ is not constant. Thus, if $P_{\dlm}$
  and $P_{\dll}$ are relatively prime, then $\dlm\ne\dll$.

  Conversely, suppose that $P_{\dlm}$ and $P_{\dll}$ are not relatively prime.
  Then, by Remark~\ref{Qfactor}, $Q_{\dlm}^{N(\dlm)}-1$ and
	$Q_{\dll}^{N(\dll)}-1$
	are not relatively prime. Let $f$ be a non-constant
	common factor of $Q_{\dlm}^{N(\dlm)}-1$ and $Q_{\dll}^{N(\dll)}-2$.
  Lemma~\ref{le:PhikQirred} implies that there exist non-constant
	monic polynomials $p_1,p_2\in \ndZ[x]$ with $f=p_1(Q_{\dlm})=p_2(Q_{\dll})$.
	In particular, $Q_{\dlm}=Q_{\dll}$.
  Let $1\le i<j\le n$ with $m_i,m_j\ne 0$. Then $l_i,l_j\ne 0$, and
	the following equations hold:
\begin{align}
\label{Eq1}
&\frac{m_i(m_i-1)}{N(\dlm)}=\frac{l_i(l_i-1)}{N(\dll)},\\
\label{Eq2}
&\frac{m_j(m_j-1)}{N(\dlm)}=\frac{l_j(l_j-1)}{N(\dll)},\\
\label{Eq3}
&\frac{m_im_j}{N(\dlm)}=\frac{l_il_j}{N(\dll)}.
\end{align}
From Equation \eqref{Eq1} and \eqref{Eq3}, one gets
\begin{align*}
l_jm_i-l_im_j=l_j-m_j.
\end{align*}
Similarly, using Equation \eqref{Eq2} and \eqref{Eq3}, one gets
\begin{align}
\label{Eq4}
l_jm_i-l_im_j=m_i-l_i.
\end{align}
Thus $m_i+m_j=l_i+l_j$. Let $t=m_i+m_j=l_i+l_j$. Replacing $m_j$ with $t-m_i$ and
$l_j$ with $t-l_i$ in Equation \eqref{Eq4}, we get
$t(m_i-l_i)=m_i-l_i$, and hence $m_i=l_i$ because of $t>1$. Thus
$m_j=l_j$. It follows that $\dlm=\dll$.
\end{proof}

Now we pass to another family of polynomials, which are the main reason for our
interest in the family $(P_{\dlm})_{\dlm \in \ndN_0^n,|\dlm|\ge 2}$.

\begin{defi} \label{defAm}
For any $\dlm=(m_1,m_2,\dots,m_n)\in \ndN_0^n$, $|\dlm|\ge 2$, let
$$A_{\dlm}=\frac{\prod_{i:m_i>0}\prod_{k\mid \gcd(\dlm-\dle_i)}
(1-Q_{\dlm}^{N(\dlm)/k})^{\ell_{(\dlm-\dle_i)/k}}}
{\prod_{k\mid \gcd(\dlm)}(1-Q_{\dlm}^{N(\dlm)/k})^{\ell_{\dlm/k}}}.
$$
\end{defi}

Note that any $k$ with $k\mid \gcd(\dlm-\dle_i)$ divides $m_i-1$ and any $m_j$,
$j\ne i$, and hence $k\mid N(\dlm)$. Similarly, $k\mid \gcd(\dlm)$ implies that
$k\mid N(\dlm)$. Therefore the numerator and the denominator of $A_{\dlm}$ are
polynomials. If $\dlm=m_i\dle_i$ for some $1\le i\le n$ and $m_i\ge 2$, then
$Q_{\dlm}=p_{ii}$ and
\begin{align} \label{eq:Amiei}
	A_{m_i\dle _i}=\frac{1-Q_{\dlm}^{N(\dlm)/(m_i-1)}}{1-Q_{\dlm}^{N(\dlm)/m_i}}
	=\frac{1-Q_{\dlm}^{m_i}}{1-Q_{\dlm}^{m_i-1}}=\frac{(m_i)_{p_{ii}}}{(m_i-1)_{p_{ii}}}.
\end{align}
In order to show that every other $A_{\dlm}$ is a polynomial, we use
some results in \cite{HJ2018} about the number of Lyndon words.
Using Equation~\eqref{NL}, these can be restated as follows.

\begin{theo}(\cite[Lemma~4.1,~4.2,~Theorem~1.2]{HJ2018})\label{basic theo}
Let $\dlm\in \ndN_0^n$. Assume that $m_s,m_t\ne0$ for some $1\le s<t\le n$. Then,
\begin{enumerate}
\item
\begin{align*}
  \sum_{k\mid\gcd(\dlm)}\ell_{\dlm/k}\le\sum_{1\le i\le n,m_i>0}\ell_{\dlm-\dle_i}.
\end{align*}
\item
\begin{align}
&\sum_{k\mid \gcd(\dlm)}\ell_{\dlm/k}=\sum_{1\le i\le
n,m_i>0}\ell_{\dlm-\dle_i}
\end{align}
if and only if $\dlm$ is one of the cases $(2)$,$(3)$,$(4)$,$(5)$,$(6)$,$(7)$ in
Definition~\ref{def:Pm} different from $\dlm=\dle_s+\dle_t$.
\end{enumerate}
\end{theo}

\begin{lemm}\label{le:Am}
 Let $\dlm=(m_1,m_2,\dots,m_n)\in \ndN_0^n$.
 If there exist $i,j\in\{1,2,\ldots,n\}$, $i\ne j$, such that $m_i\ne0$, $m_j\ne0$, then
  $A_{\dlm}$ is a polynomial in $\ndZ[p_{ij},1\le i,j\le n]$.
\end{lemm}

\begin{proof}
	The numerator of $A_{\dlm}$ is a multiple of
	$\prod_{i:m_i>0}(1-Q_{\dlm}^{N(\dlm)})^{\ell_{\dlm-\dle_i}}$
	and the denumerator of $A_{\dlm}$ is a divisor of
	$\prod_{k\mid \gcd(\dlm)}(1-Q_{\dlm}^{N(\dlm)})^{\ell_{\dlm/k}}$.
	Thus the claim follows from Theorem~\ref{basic theo}(1).
\end{proof}

\begin{prop} \label{pr:Am=0}
Let $\dlm=(m_1,m_2,\ldots,m_n) \in \ndN_0^n$ with $m_s,m_t>0$ for some $1\le
s<t\le n$. Then $P_{\dlm}$ is the product of the irreducible factors of
$A_{\dlm}$.
\end{prop}

\begin{proof}
	We follow Definition~\ref{def:Pm} case by case to compare $A_{\dlm}$ and
	$P_{\dlm}$. Then the claim follows directly from Lemma~\ref{le:PhikQirred}.

	(1) If $\dlm=m_i\dle_i$ for some $1\le i\le n$, $m_i\ge 2$, then the
	assumptions of the lemma are not fulfilled.

	(2)
	Assume that $\dlm=\dle_i+m_j\dle_j$, $1\le i,j\le n$, $i\ne j$, $m_j>0$.
	Then $N(\dlm)=m_j$, $Q_{\dlm}=p_{ij}p_{ji}p_{jj}^{m_j-1}$, and
\begin{align*}
	A_{\dlm}&=\frac{(1-Q_{\dlm}^{m_j})^{\ell_{\dle_i+(m_j-1)\dle_j}}
	\prod_{k\mid m_j}(1-Q_{\dlm }^{m_j/k})^{\ell_{m_j\dle_j/k}}}
	{(1-Q_{\dlm }^{m_j})^{\ell_{\dlm}}}\\
	&=\frac{(1-Q_{\dlm}^{m_j})(1-Q_{\dlm})}{1-Q_{\dlm}^{m_j}}
	=1-Q_{\dlm}=1-p_{ij}p_{ji}p_{jj}^{m_j-1}
  =P_{\dlm},
\end{align*}
where we used Equations \eqref{eq:numbL1} and \eqref{eq:numbL2}.

(3) Assume that $\dlm=2\dle_i+m_j\dle_j$, $1\le i,j\le n$, $i\ne j$, $m_j>0$.
Then $N(\dlm)=2$ and $Q_{\dlm}=p_{ii}(p_{ij}p_{ji})^{m_j}p_{jj}^{m_j(m_j-1)/2}$.
Moreover,
\begin{align*}
	&A_{\dlm}=\frac{(1-Q_{\dlm}^{N(\dlm)})^{\ell_{\dlm-\dle_i}}
	\prod_{k\mid\gcd(2,m_j-1)}(1-Q_{\dlm}^{N(\dlm)/k})^{\ell_{(\dlm-\dle_j)/k}}}
	{\prod_{k\mid \gcd(2,m_j)}(1-Q_{\dlm}^{N(\dlm)/k})^{\ell_{\dlm/k}}}.
\end{align*}
If $m_j$ is even, then
$\ell_{\dlm/2}+\ell_{\dlm}=\ell_{\dlm-\dle_i}+\ell_{\dlm-\dle_j}$ by
Theorem~\ref{basic theo}(2). Thus
\begin{align*}
A_{\dlm}&=\frac{(1-Q_{\dlm}^2)^{\ell_{\dlm-\dle_i}}
                (1-Q_{\dlm}^2)^{\ell_{\dlm-\dle_j}}}
               {(1-Q_{\dlm}^2)^{\ell_{\dlm}}(1-Q_{\dlm})^{\ell_{\dlm/2}}}
							 =1+Q_{\dlm}=P_{\dlm}
\end{align*}
by Equation \eqref{eq:numbL1}.

If $m_j$ is odd, then
$\ell_{(\dlm-\dle_i)}+\ell_{(\dlm-\dle_j)}=\ell_{\dlm}$ by
Theorem~\ref{basic theo}(2). Therefore
\begin{align*}
A_{\dlm}&=\frac{(1-Q_{\dlm}^2)^{\ell_{\dlm-\dle_i}}
(1-Q_{\dlm}^2)^{\ell_{\dlm-\dle_j}}(1-Q_{\dlm})^{\ell_{(\dlm-\dle_j)/2}}}
               {(1-Q_{\dlm}^2)^{\ell_{\dlm}}}
							 =1-Q_{\dlm}=P_{\dlm}
\end{align*}
by Equation \eqref{eq:numbL1}.

(4)
Assume that $\dlm=3\dle_i+3\dle_j$ with $1\le i<j\le n$. Then $N(\dlm)=3$,
$Q_{\dlm}=p_{ii}^2(p_{ij}p_{ji})^3p_{jj}^2$, and $\ell_{(1,1)}+\ell_{(3,3)}
=\ell_{(2,3)}+\ell_{(3,2)}$ by Theorem~\ref{basic theo}(2). Thus
\begin{align*}
	A_{\dlm}&=\frac{\prod_{k\mid\gcd(2,3)}(1-Q_{\dlm }^{3/k})^{\ell_{(2,3)/k}}
	\prod_{k\mid\gcd(3,2)}(1-Q_{\dlm}^{3/k})^{\ell_{(3,2)/k}}}
	{\prod_{k\mid\gcd(3,3)}(1-Q_{\dlm}^{3/k})^{\ell_{(3,3)/k}}}\\
	&=\frac{(1-Q_{\dlm }^3)^{\ell_{(2,3)}+\ell_{(3,2)}}}
	{(1-Q_{\dlm }^3)^{\ell_{(3,3)}}(1-Q_{\dlm})^{\ell_{(1,1)}}}
	=\frac{(1-Q_{\dlm}^3)^{\ell_{(1,1)}}}{(1-Q_{\dlm})^{\ell_{(1,1)}}}
	=(3)_{Q_{\dlm}}=P_{\dlm}.
\end{align*}

(5)
If $\dlm=3\dle_i+4\dle_j$ with $1\le i,j\le n$, $i\ne j$, then
$N(\dlm)=6$, $Q_{\dlm}=p_{ii}(p_{ij}p_{ji})^2p_{jj}^2$, and
$\ell_{(3,4)}=\ell_{(2,4)}+\ell_{(3,3)}$. Thus
\begin{align*}
	A_{\dlm}&=\frac{\prod_{k\mid\gcd(2,4)}(1-Q_{\dlm}^{6/k})^{\ell_{(2,4)/k}}
	\prod_{k\mid \gcd(3,3)}(1-Q_{\dlm}^{6/k})^{\ell_{(3,3)/k}}}
	{\prod_{k\mid\gcd(3,4)}(1-Q_{\dlm}^{6/k})^{\ell_{(3,4)/k}}}\\
	&=\frac{(1-Q_{\dlm}^6)^{\ell_{(2,4)}}(1-Q_{\dlm}^3)^{\ell_{(1,2)}}
	(1-Q_{\dlm}^6)^{\ell_{(3,3)}}(1-Q_{\dlm}^2)^{\ell_{(1,1)}}}
	{(1-Q_{\dlm}^6)^{\ell_{(3,4)}}}\\
	&=(1-Q_{\dlm}^2)(1-Q_{\dlm}^3)=(1-Q_{\dlm})^2(1+Q_{\dlm})(3)_{Q_{\dlm}}
	=(1-Q_{\dlm})P_{\dlm}.
\end{align*}

(6)
If $\dlm=3\dle_i+6\dle_j$ with $1\le i,j\le n$, $i\ne j$, then
$N(\dlm)=6$, $Q_{\dlm}=p_{ii}(p_{ij}p_{ji})^3p_{jj}^5$, and
$\ell_{(3,6)}+\ell_{(1,2)}=\ell_{(2,6)}+\ell_{(3,5)}$. Thus
\begin{align*}
	A_{\dlm}&=\frac{\prod_{k\mid\gcd(2,6)}(1-Q_{\dlm}^{6/k})^{\ell_{(2,6)/k}}
	\prod_{k\mid \gcd(3,5)}(1-Q_{\dlm}^{6/k})^{\ell_{(3,5)/k}}}
	{\prod_{k\mid\gcd(3,6)}(1-Q_{\dlm}^{6/k})^{\ell_{(3,6)/k}}}\\
	&=\frac{(1-Q_{\dlm}^6)^{\ell_{(2,6)}}(1-Q_{\dlm}^3)^{\ell_{(1,3)}}
	(1-Q_{\dlm}^6)^{\ell_{(3,5)}}}
	{(1-Q_{\dlm}^6)^{\ell_{(3,6)}}(1-Q_{\dlm}^2)^{\ell_{(1,2)}}}
	=\frac{(1-Q_{\dlm}^6)(1-Q_{\dlm}^3)}{1-Q_{\dlm}^2}\\
	&=(1-Q_{\dlm})(3)_{Q_{\dlm}}^2(3)_{-Q_{\dlm}}
	=(3)_{Q_{\dlm}}P_{\dlm}.
\end{align*}

(7)
If $\dlm=4\dle_i+4\dle_j$ with $1\le i,j\le n$, $i\ne j$, then
$N(\dlm)=4$, $Q_{\dlm}=p_{ii}^3(p_{ij}p_{ji})^4p_{jj}^3$, and
$\ell_{(1,1)}+\ell_{(2,2)}+\ell_{(4,4)}=\ell_{(3,4)}+\ell_{(4,3)}$. Thus
\begin{align*}
	A_{\dlm}&=\frac{\prod_{k\mid\gcd(3,4)}(1-Q_{\dlm}^{4/k})^{\ell_{(3,4)/k}}
	\prod_{k\mid \gcd(4,3)}(1-Q_{\dlm}^{4/k})^{\ell_{(4,3)/k}}}
	{\prod_{k\mid\gcd(4,4)}(1-Q_{\dlm}^{4/k})^{\ell_{(4,4)/k}}}\\
	&=\frac{(1-Q_{\dlm}^4)^{\ell_{(3,4)}}(1-Q_{\dlm}^4)^{\ell_{(4,3)}}}
	{(1-Q_{\dlm}^4)^{\ell_{(4,4)}}(1-Q_{\dlm}^2)^{\ell_{(2,2)}}
	(1-Q_{\dlm})^{\ell_{(1,1)}}
  }
	=\frac{(1-Q_{\dlm}^4)^2}{(1-Q_{\dlm}^2)(1-Q_{\dlm})}\\
	&=(1+Q_{\dlm}^2)^2(1+Q_{\dlm})=(1+Q_{\dlm}^2)P_{\dlm}
\end{align*}
since $\ell_{(2,2)}=1$.

(8)
Now we suppose $\dlm$ is not equal to any of the above cases.
By following the proof of Lemma~\ref{le:Am} and using Theorem~\ref{basic theo}
we conclude that $Q_{\dlm}^{N(\dlm)}-1$ divides $A_{\dlm}$. Moreover, every
irreducible factor of $A_{\dlm}$ is a factor of $P_{\dlm}=1-Q_{\dlm }^{N(\dlm)}$.
Since $P_{\dlm}$ is a product of pairwise non-associated irreducible factors,
we conclude that $P_{\dlm}$ contains every irreducible factor of $A_{\dlm}$
precisely once.

Thus the proof is completed.
\end{proof}

\begin{coro}\label{cor:prime}
Let $\dlm,\dll\in \ndN_0^n$ be different from $m\dle_i$ for all $m\ge 0$ and
$1\le i\le n$.
Then $A_{\dlm}$ and $A_{\dll}$
are relatively prime if and only if $\dlm\ne\dll$. In particular,
$A_{\dlm}$ and $A_{\dll}$
are relatively prime whenever $\dll<\dlm$.
\end{coro}

\begin{proof}
The claim follows from Lemma~\ref{le:prime} and Proposition~\ref{pr:Am=0}.
\end{proof}

\section {The shuffle map over commutative rings}
\label{shuffles}

In this section let $R$ be a unital commutative ring. We calculate
the determinant of the shuffle map over $R$.

\begin{defi}
Let $\overline{V}$ be a finitely generated free module over $R$
and let $\bar{c}:\overline{V}\ot _R\overline{V}\rightarrow \overline{V}\ot _R\overline{V}$
be an $R$-module endomorphism of $\overline{V}\ot _R\overline{V}$.
We say that $(\overline{V},\bar{c})$ is a \emph{free prebraided module of diagonal type over} $R$,
if there exist a basis $x_1,x_2,\ldots,x_n$ of $\overline{V}$ and
$(\overline{q}_{ij})_{1\le i,j\le n}\in R^{n\times n}$ with
  $$\bar{c}(x_i\otimes x_j)=\overline{q}_{ij} x_j\otimes x_i \quad \text{for all $i,j$.} $$
\end{defi}

Let $n\in \ndN$ and let $(\overline{V},\overline{c})$ be a free prebraided module
of diagonal type with basis $x_1,\dots,x_n$. Let $I=\{1,2,\ldots,n\}$ and
$(\overline{q}_{ij})_{i,j\in I}\in R^{n\times n}$. Assume that
$$\overline{c}(x_i\otimes x_j)=\overline{q}_{ij} x_j\otimes x_i$$
for all $i,j\in I$. Let $\overline{V}^{\otimes k}$ denote the $k$-fold
tensor product of $\overline{V}$ over $R$ and let
$T(\overline{V})=\bigoplus_{k=0}^{\infty} \overline{V}^{\otimes k}$.
Note that $\overline{V}^{\otimes k}$ is a free module over $R$
for all $k\in \ndN$.

For any $\dlm\in \ndN^n$ let $\mathbb{X}_{\dlm}$ denote the set of words over $I$
of degree $\dlm$.
Let $\alpha_1,\alpha_2,\ldots,\alpha_n$ be the standard basis of $\ndZ^n$. Then $T(\overline{V})$ admits a $\ndZ^n$-grading given by $\deg x_i=\alpha_i$, for all $i\in I$. Thus
for any $i_1i_2\cdots i_l\in \mathbb{X}_{\dlm}$, the degree of
$x_{i_1}x_{i_2}\cdots x_{i_l}$ is $\sum_{j=1}^l \alpha_{i_j}$,
and we write $\deg x$ for the degree
of any homogeneous element $x$ of $T(\overline{V})$.
For any $\dlm \in \ndN_0^n$
let $\olV_{\dlm}$ denote the $\ndZ^n$-homogeneous
component of $T(\overline{V})$ of degree $\dlm$.

For any $k\ge 2$, let $\mathbb{B}_k$ denote the monoid which is generated by generators
$\sigma_1,\sigma_2,\ldots \sigma_{k-1}$ and relations
\begin{enumerate}
  \item $\sigma_i\sigma_j=\sigma_j\sigma_i$ for
  $i,j\in\{1,2,\ldots, k-1\}$ with $|i-j|\ge 2$, and
  \item $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_{i}\sigma_{i+1}$
  for $1\le i\le k-2$.
\end{enumerate}
Let $S_{1,0}=1 \in \mathbb{B}_k$, and for any $1\le m<k$ let
 $$ S_{1,m}=1+\sigma_1+\sigma_2\sigma_1+\cdots
     +\sigma_m\sigma_{m-1}\cdots \sigma_1. $$
 A variant of the following equation appeared already in \cite[Lemma~6.12]{duchamp1997}.
\begin{lemm}\label{basic Lemm}
For any $k>m\ge 1$ the following equation holds in $\BG k$.
\begin{align*}
		(1-\sigma_m\cdots \sigma_2\sigma_1)S_{1,m}
		=S_{1,m-1}(1-\sigma_m\cdots \sigma_2\sigma_1^2).
\end{align*}
\end{lemm}

\begin{proof}
It is easy to check that
$$(\sigma_m\sigma_{m-1}\cdots\sigma_2\sigma_1)\sigma_{i}=
\sigma_{i-1}(\sigma_m\sigma_{m-1}\cdots\sigma_2\sigma_1)
$$
for all $2\le i\le m<k$. Thus
\begin{align*}
&\sigma_m\cdots \sigma_2\sigma_1S_{1,m}\\
&=\sum_{t=1}^m \sigma_m\cdots \sigma_2\sigma_1 \sigma_{t}\sigma_{t-1}\cdots \sigma_2\sigma_1
+\sigma_m\cdots \sigma_2\sigma_1\\
&=\sum_{t=1}^{m-1}\sigma_{t}\sigma_{t-1}\cdots \sigma_2\sigma_1
(\sigma_m\cdots \sigma_2\sigma_1)\sigma_1+\sigma_m\cdots \sigma_2\sigma_1^2+
\sigma_m\cdots \sigma_2\sigma_1\\
&=S_{1,m-1}(\sigma_m\cdots \sigma_2\sigma_1^2)+\sigma_m\cdots \sigma_2\sigma_1.
\end{align*}
Hence,
\begin{align*}
&(1-\sigma_m\cdots \sigma_2\sigma_1)S_{1,m}\\
&=S_{1,m}-S_{1,m-1}(\sigma_m\cdots \sigma_2\sigma_1^2)
-\sigma_m\cdots \sigma_2\sigma_1\\
&=S_{1,m-1}-S_{1,m-1}(\sigma_m\cdots \sigma_2\sigma_1^2)\\
&=S_{1,m-1}(1-\sigma_m\cdots \sigma_2\sigma_1^2).
\end{align*}
This proves the lemma.
\end{proof}

For any $m\ge 2$ let $R\BG m$ denote the monoid ring of $\BG m$ over $R$ and let
$\overline{\rho}_m:R\BG m\rightarrow\End_R(\overline{V}^{\otimes m})$ be the ring
homomorphism such that $\overline{\rho}_m(\sigma_i)$ is the prebraiding
$\overline{c}$ applied to the $i$-th and $i+1$-th tensor factors of
$\overline{V}^{\otimes m}$.

Let $f:\ndZ[p_{ij}\mid i,j\in I]\rightarrow R$ be the ring homomorphism such that
 $$f(p_{ij})=\overline{q}_{ij} \quad \text{for all $i,j\in I$.} $$

\begin{lemm}\label{det1}
For any $\dlm\in \ndN_0^n$ with $\dlm\ne0$ and $m=|\dlm|$ we have
\begin{align*}
\det(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)\big|\overline{V}_{\dlm})
  =\prod_{k\mid \gcd(\dlm)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{\dlm/k}}.
\end{align*}
Moreover, if $R$ is a field, $d=\mathrm{ord}(f(Q_{\dlm}))$, and $d\mid
N(\dlm)$, then
\begin{align*}
	\dim (\ker
	(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)\big|\olV_{\dlm}))
	=\sum_{k\mid \gcd(\dlm),k\mid N(\dlm)/d}\ell _{\dlm/k}.
\end{align*}
\end{lemm}

\begin{proof}
Consider the action of $\ndZ$ on $\mathbb{X}_{\underline{m}}$ given by
   $$1\cdot i_1 \cdots i_m=i_2\cdots i_mi_1.$$
In any $\ndZ$-orbit of $\mathbb{X}_{\underline{m}}$ there is a unique element $v^k$,
where $v$ is a Lyndon word and $k \in \ndN$ with $k|m_i$ for each $1\le i\le n$.

To any $\ndZ$-orbit $\mathcal{O}$ of $\mathbb{X}_{\dlm}$
we attach the submodule $\overline{V}_{\mathcal{O}}$ of $\overline{V}_{\underline{m}}$ generated by
$x_{i_1}x_{i_2}\cdots x_{i_m}$ with $i_1i_2 \cdots i_m \in \mathcal{O}$. Then
\begin{align}\label{decom1}
 \overline{V}_{\underline{m}}=\bigoplus_{\mathcal{O}}\overline{V}_{\mathcal{O}}.
\end{align}
Let $\mathcal{O}$ be a $\ndZ$-orbit and let $k\ge 1$ and $v=i_1i_2\cdots i_l$ be the
Lyndon word such that $v^k\in \mathcal{O}$. Then $l=m/k$. Let
$\widetilde{v}=x_{i_1}x_{i_2}\cdots x_{i_l}\in T(\overline{V})$.
Then
\begin{align} \label{eq:O1basis}
x_{i_t}x_{i_{t+1}}\cdots x_{i_l}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}}, \quad
1\le t\le l,
\end{align}
is a basis of $\overline{V}_{\mathcal{O}}$, where $x_{i_1}\cdots x_{i_0}=1$. Moreover,
for all $1\le t\le l$,
\begin{align*}
&\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)
(x_{i_t}\cdots x_{i_{l}}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}})\\
&=x_{i_t}\cdots x_{i_l}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}}
    -\lambda_tx_{i_{t+1}}\cdots x_{i_{l}}\widetilde{v}^{k-1}x_{i_1}
   \cdots x_{i_{t}},
\end{align*}
where $\lambda_t=\overline{q}_{i_ti_1}^{k}\overline{q}_{i_ti_2}^{k}\cdots
   \overline{q}_{i_ti_{t-1}}^{k}\overline{q}_{i_ti_{t}}^{k-1}
   \overline{q}_{i_ti_{t+1}}^{k}\cdots \overline{q}_{i_ti_{l}}^{k}$
   and $x_{i_{t+1}}\cdots x_{i_t}=1$.

We obtain that the matrix of $\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)
\big|\overline{V}_{\mathcal{O}}$ with respect to the basis \eqref{eq:O1basis}
is $A=(a_{st})_{1\le s,t\le l}$, where
$$a_{st}=
\begin{cases}
     1& \text{if $s=t$,}  \\
     -\lambda_t & \text{if $s=t+1$, $1\le t\le l-1$,} \\
     -\lambda_l & \text{if $s=1$, $t=l$,} \\
     0& \text{otherwise.}
\end{cases}
$$
Therefore,
\begin{align*}
&\det(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)\big|\overline{V}_{\mathcal{O}})\\
&=1+(-1)^{l+1}(-1)^l\lambda_1\lambda_2\cdots\lambda_l\\
&=1-\prod_{1\le t \le l}\overline{q}_{i_ti_t}^{k-1}\prod_{1\le t<s\le l}(\overline{q}_{i_ti_s}
\overline{q}_{i_si_t})^k\\
&=1-\prod_{1\le t \le n}\overline{q}_{tt}^{m_t(m_t-1)/k}\prod_{1\le t<s\le n}
(\overline{q}_{st}\overline{q}_{ts})^{m_tm_s/k}\\
&=1-f(Q_{\dlm})^{N(\dlm)/k}.
\end{align*}
Hence
\begin{align*}
\det(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1)\big|\olV_{\dlm})
  =\prod_{k\mid \gcd(\dlm)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{\dlm/k}}
\end{align*}
because of the decomposition of $\olV_{\dlm}$ in \eqref{decom1}.

If $R$ is a field, then the matrix $A$ above has corank $0$ or $1$.
Moreover, $A$ has corank $1$ if and only if
$f(Q_{\dlm})^{N(\dlm)/k}=1$, that is, if and only if $d\mid N(\dlm)/k$,
where $d=\mathrm{ord}(f(Q_{\dlm}))$. This implies the last claim.
\end{proof}

\begin{lemm}\label{det2}
For any $\dlm\in \ndN_0^n$ with $m=|\dlm|\ge 2$ we have
\begin{align*}
&\det(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1^2)\big|\olV_{\dlm})\\
&\quad =\prod_{ i:m_i>0}\prod_{k\mid \gcd(\dlm-\dle_i)}
(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{(\dlm-\dle_i)/k}}.
\end{align*}
Moreover, if $R$ is a field, $d=\mathrm{ord}(f(Q_{\dlm}))$, and $d\mid
N(\dlm)$, then
\begin{align*}
	\dim (\ker
	(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1^2)\big|\olV_{\dlm}))
	=\sum_{i:m_i>0}\sum_{k\mid \gcd(\dlm-\dle _i),k\mid N(\dlm)/d}\ell
	_{(\dlm-\dle_i)/k}.
\end{align*}
\end{lemm}

\begin{proof}
 Let us consider the $\ndZ$-action on $\mathbb{X}_{\dlm}$ given by
   $$1\cdot i_1i_2\cdots i_m=i_1i_3i_4\cdots i_mi_2.$$
Then $(m-1)\cdot i_1i_2\cdots i_m=i_1i_2\cdots i_m$.
In any $\ndZ$-orbit of $\bX_{\dlm}$
there is a unique element $jv^k$, where $j\in I$, $v$ is a Lyndon word,
and $k\ge 1$. Moreover, then $k\mid m_j-1$ and $k\mid m_t$ for each $1\le t\le n$
with $t\ne j$.

Again, to any $\ndZ$-orbit $\cO $ we attach the submodule $\overline{V}_{\cO}$
of $\overline{V}_{\dlm}$
generated by the monomials $x_{i_1}\cdots x_{i_{m}}$, where $i_1\cdots
i_{m}\in \mathcal{O}$.
Then
\begin{align}
\overline{V}_{\dlm} =\bigoplus_{\cO} \overline{V}_{\cO}.
\end{align}
Let $v=i_1i_2\cdots i_l$ be a Lyndon word, $j\in \{1,\dots,n\}$, and $k\ge 1$.
Assume that $\deg jv^k=\dlm$. Then $l=(m-1)/k$.
Let $\widetilde{v}=x_{i_1}\cdots x_{i_l}\in T(\overline{V})$.
Then the monomials
\begin{align} \label{eq:O2basis}
  x_jx_{i_t}x_{i_{t+1}}\cdots x_{i_l}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}},
\quad 1\le t\le l,
\end{align}
form a basis of $\overline{V}_{\cO}$ for the $\ndZ$-orbit $\cO$ of $jv^k$,
where $x_{i_1}\cdots x_{i_0}=1$.

For any $1\le t\le l$ one obtains that
\begin{align*}
&\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1^2)
(x_jx_{i_t}\cdots x_{i_{l}}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}})\\
&=x_jx_{i_t}\cdots x_{i_{l}}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t-1}}
    -\overline{q}_{ji_t}\overline{q}_{i_tj}\lambda_tx_jx_{i_{t+1}}\cdots
    x_{i_{l}}\widetilde{v}^{k-1}x_{i_1}\cdots x_{i_{t}},
\end{align*}
where $x_{i_{l+1}}\cdots x_{i_l}=1$ and
$\lambda_t=\overline{q}_{i_ti_1}^{k}\overline{q}_{i_ti_2}^{k}\cdots
  \overline{q}_{i_ti_{t-1}}^{k}\overline{q}_{i_ti_{t}}^{k-1}
  \overline{q}_{i_ti_{t+1}}^{k}\cdots \overline{q}_{i_ti_{l}}^{k}$
   for all $1\le t\le l$.
Thus the matrix of $\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1^2)|\overline{V}_{\cO}$
with respect to the basis \eqref{eq:O2basis} is $B=(b_{st})_{1\le s,t\le l}$, where
$$b_{st}=
\begin{cases}
     1 & \text{if $s=t$,}  \\
     -\overline{q}_{ji_t}\overline{q}_{i_tj}\lambda_t & \text{if $s=t+1$, $1\le t\le l-1$,} \\
     -\overline{q}_{ji_l}\overline{q}_{i_lj}\lambda_l & \text{if $s=1$, $t=l$,} \\
     0 & \text{otherwise.}
\end{cases}
$$
Hence,
\begin{align*}
&\det(\overline{\rho}_m(1-\sigma_{m-1}\cdots\sigma_2\sigma_1^2)\big|\overline{V}_{\cO})\\
&=1+(-1)^{l+1}(-1)^l\overline{q}_{ji_1}\overline{q}_{i_1j}\cdots
\overline{q}_{ji_l}\overline{q}_{i_lj}\lambda_1\lambda_2\cdots\lambda_l\\
&=1-\prod_{1\le t \le l}(\overline{q}_{ji_t}\overline{q}_{i_tj})
\prod_{1\le t\le l}\overline{q}_{i_ti_t}^{k-1}
\prod_{1\le t<s\le l}(\overline{q}_{i_ti_s}\overline{q}_{i_si_t})^k\\
&=1-\prod_{1\le t \le n}\overline{q}_{tt}^{m_t(m_t-1)/k}\prod_{1\le t<s\le n}(\overline{q}_{st}\overline{q}_{ts})^{m_tm_s/k}\\
&=1-f(Q_{\dlm})^{N(\dlm)/k}.
\end{align*}
This implies the first claim.

If $R$ is a field, then the matrix $B$ above has corank $0$ or $1$.
Moreover, $B$ has corank $1$ if and only if
$f(Q_{\dlm})^{N(\dlm)/k}=1$, that is, if and only if $d\mid N(\dlm)/k$,
where $d=\mathrm{ord}(f(Q_{\dlm}))$. This implies the last claim.
\end{proof}

\begin{lemm}\label{le:detshuffle}
Let $\dlm\in \ndN_0^n$ with $m=|\dlm|\ge 2$. Then
\begin{align*}
	&\det (\overline{\rho}_m(S_{1,m-1})|\overline{V}_{\dlm})
	\prod_{k\mid \gcd(\dlm)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{\dlm/k}}\\
	&=\prod_{i:m_i>0}\Big(\det
	(\overline{\rho}_{m-1}(S_{1,m-2})|\overline{V}_{\dlm-\dle_i})\\
  &\qquad \quad
   \prod_{k\mid \gcd(\dlm-\dle_i)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{(\dlm-\dle_i)/k}}
   \Big).
\end{align*}
\end{lemm}

\begin{proof}
From Lemma \ref{basic Lemm} we conclude that
\begin{align*}
&\det (\overline{\rho}_m(S_{1,m-1})|\overline{V}_{\underline{m}})
  \det(\overline{\rho}_m(1-\sigma_{m-1}\cdots \sigma_{2}\sigma_{1})|\overline{V}_{\dlm})\\
  &=\det (\overline{\rho}_m(S_{1,m-2})|\overline{V}_{\underline{m}})
  \det(\overline{\rho}_m(1-\sigma_{m-1}\cdots \sigma_{2}\sigma_{1}^2)|\overline{V}_{\dlm}).
\end{align*}
Because of Lemma \ref{det1} and \ref{det2} the above equality is equivalent to
\begin{align*}
&\det (\overline{\rho}_m(S_{1,m-1})|\overline{V}_{\dlm})
  \prod_{k\mid \gcd(\dlm)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{\dlm/k}}\\
  &=\det (\overline{\rho}_m(S_{1,m-2})|\overline{V}_{\dlm})
 \prod_{i:m_i>0}\prod_{k\mid \gcd(\dlm-\dle_i)}(1-f(Q_{\dlm})^{N(\dlm)/k})^{\ell_{(\dlm-\dle_i)/k}}.
\end{align*}
This implies the lemma.
\end{proof}

\begin{prop}\label{pr:detshuffle2}
 Let $\dlm =(m_1,\dots,m_n)\in \ndN_0^n$ with $m=|\dlm|\ge 2$.
\begin{enumerate}
  \item If $\dlm=m_i\dle_i$ with $1\le i\le n$
  then $\overline{\rho}_m(S_{1,m-1})|\overline{V}_{\dlm}=(m_i)_{\overline{q}_{ii}}\id$.
  \item If there exist $1\le i<j\le n$ with $m_i,m_j\ne0$, then
\begin{align}
\label{detshuf}
\det (\overline{\rho}_m(S_{1,m-1})|\overline{V}_{\underline{m}})
	=f(A_{\dlm})\prod_{i:m_i>0}\det (\overline{\rho}_{m-1}(S_{1,m-2})|\overline{V}_{\dlm-\dle_i}).
\end{align}
\end{enumerate}
\end{prop}

\begin{proof}
Claim (1) follows directly from the definition of $S_{1,m-1}$.

In order to prove part (2) of the Proposition it suffices to
consider $R=\ndZ [p_{ij}\mid 1\le i,j\le n]$ and $f=\id $.
In this case the claim follows from
Lemma~\ref{le:detshuffle} and Lemma~\ref{le:Am}.
\end{proof}

\section{Nichols algebras which are free algebras}
\label{freeness}

In the remaining part of this paper let $\fie$ be a field,
let $\fie^{\times}=\fie\backslash \{0\}$,
and let $(V,c)$ be an $n$-dimensional braided vector space of diagonal type
with basis $x_1,x_2, \ldots, x_n$
and braiding matrix $\bq \in (\fie^{\times})^{n\times n}$.
Let $T(V)$ and $\NA(V)$ denote the tensor algebra and the Nichols algebra of $V$, respectively.

For the basic theory of Nichols algebras we refer to \cite{AS2002}.

In this section we determine when $\NA (V)$ is a free algebra, that is, $\NA (V)=T(V)$.

For all $k\ge 2$ there is a unique group homomorphism $\tau:\BG k\to \BG{k+1}$
with $\tau(\sigma_i)=\sigma_{i+1}$ for all $1\le i<k$. We also write $\tau $
for the induced algebra maps $\fie \BG k\to \fie \BG{k+1}$.

For all $m\ge 2$ let
$\rho_m:\fie\BG m\longrightarrow \End(V^{\otimes m})$
be the representation of $\fie\BG m$ introduced in Section \ref{shuffles} as $\overline{\rho}_m$,
and let
$$ S_m=S_{1,m-1}\tau(S_{1,m-2})\tau^2(S_{1,m-3})\cdots \tau^{m-2}(S_{1,1})\in \fie\BG m.$$
Then, by \cite{schauenburg1996},
\begin{align} \label{schauenburg}
\NA(V)=\fie \oplus V \oplus \bigoplus_{m=2}^{\infty} V^{\ot m}/\ker(\rho _m(S_m)).
\end{align}

\begin{lemm}\label{det=0}
Let $\dlm\in \ndN_0^n $ with $m=|\dlm|\ge 2$.

(1) If $P_{\dlm}(\bq)=0$, then $\det (\rho_m(S_{1,m-1})|V_{\dlm})=0$.

(2) If $\det (\rho_m(S_{1,m-2})|V_{\dlm})\ne0$ and
$\det (\rho_m(S_{1,m-1})|V_{\dlm})=0$
then $P_{\dlm}(\bq)=0$.
\end{lemm}

\begin{proof}
 If $\dlm=m_i\dle_i$ for some $1\le i\le n$, $m_i\in\ndN$,
 then the claim holds because of Proposition~\ref{pr:detshuffle2}(1).

 Assume now that there exist $1\le i<j\le n$ with $m_i,m_j\ne0$.
 Then Proposition~\ref{pr:Am=0} implies that
 $A_m(\bq)=0$ if and only if $P_{\dlm}(\bq)=0$.
 Hence the lemma follows from Proposition~\ref{pr:detshuffle2}(2).
\end{proof}

\begin{prop}\label{prop:main}
Let $\dlm\in \ndN_0^n $ with $m=|\dlm|\ge 2$.
\begin{enumerate}
   \item If $P_{\dlm}(\bq)=0$, then there is a non-trivial relation in
   $\NA(V)$ of degree $\dlm$.
   \item If $P_{\dll}(\bq)\ne0$ for all $\dll\le \dlm$ with $|\dll|\ge 2$,
   then there is no non-trivial relation
   in $\NA(V)$ of degree $\dlm$.
\end{enumerate}
\end{prop}

\begin{proof}
(1) Assume that $P_{\dlm}(\bq)=0$.
Then $\det (\rho_m(S_{1,m-1})|V_{\dlm})=0$ by Lemma \ref{det=0}(1). Thus
$\det (\rho_m(S_{m})|V_{\dlm})=0$ by the definition of $S_{m}$,
and the claim follows from Equation~\eqref{schauenburg}.

(2) Assume that the Nichols algebra
$\NA (V)$ has a non-trivial relation in degree $\dlm $.
Let $\dll \le \dlm $ be such that $\NA (V)$ has a
non-trivial relation in degree $\dll $ and no non-trivial relation in
any degree $< \dll $. Let $l=|\dll |$. Then $l\ge 2$,
$\ker (\rho_l(S_{1,l-2})|V_{\dll})=0$,
and $\ker (\rho_l(S_{1,l-1})|V_{\dll})\ne 0$
by \eqref{schauenburg} and by the definition of $S_l$.
Hence $P_{\dll}(\bq)=0$ by Lemma~\ref{det=0}(2). This proves (2).
\end{proof}

Based on the above proposition we obtain our first main Theorem as follows.
\begin{theo}\label{theo}
We have $\NA(V)=T(V)$ if and only if $P_{\dlm}(\bq)\ne0$
for all $\dlm\in \ndN^n$ with $|\dlm|\ge 2$.
\end{theo}

\begin{proof}
 The claim follows immediately from Proposition~\ref{prop:main}.
\end{proof}

\begin{exam}(Diophantine equation)\label{Diophequation}
Assume that the characteristic of $\fie $ is neither $2$ nor $3$
and that $\bq=(q^{a_{ij}})_{i,j\in I}$ with $q\in \fie ^\times $ not a root of $1$
and $(a_{ij})_{i,j\in I}\in \ndZ^{n\times n}$.
For any $\dlm =(m_1,\dots,m_n)\in \ndN_0^n$ let
$$K(\dlm)=\sum_{i,j=1}^{n}a_{ij}m_im_j,\quad  \lambda(\dlm)=\sum_{i=1}^{n}a_{ii}m_i.$$
Then $P_{\dlm}(\bq)\ne 0$ in the following cases:
\begin{enumerate}
\item $\dlm=m\dle_i$, $m\ge 2$, $1\le i\le n$,
\item $\dlm=2\dle_i+2m\dle_j$, $m\ge 1$, $i,j\in I$, $i\ne j$,
\item $\dlm=3\dle_i+3\dle_j$ or $\dlm=4\dle_i+4\dle_j$, $i,j\in I$, $i\ne j$.
\end{enumerate}
Moreover, for any other $\dlm\in \ndN_0^n$ with $|\dlm|\ge 2$,
$$ P_{\dlm}(\bq)=0 \text{ if and only if }K(\dlm)=\lambda (\dlm). $$
Hence, by Theorem~\ref{theo},
$\NA(V)=T(V)$ if and only if there is no solution of the diophantine equation
$K(\dlm)=\lambda(\dlm)$
with $\dlm\in \ndN_0^n$, $|\dlm|\ge 2$,
$\dlm \notin \{m\dle_i\mid m\ge 2\}\cup \{2\dle_i+2m\dle_j,3\dle_i+3\dle_j,4\dle_i+4\dle_j
\mid m\ge 1,i,j\in I, i\ne j\}$.

We now provide concrete examples of Nichols algebras of diagonal type which are identified
by this example as a free algebra.
Let $n=2$ and let $a,b$ be positive integers with $a>b$.
Let $q\in \fie^\times $ be not a root of $1$ and let
$\bq=(q_{ij})_{1\le i,j\le 2}$ with $q_{11}=q_{22}=q^a$ and $q_{12},q_{21}\in q^{\ndZ}$
with $q_{12}q_{21}=q^{-b}$. For any $\dlm=(m_1,m_2)\in \ndN_0^2$ with $|\dlm|\ge2$
we have
$$K(\dlm)=am_1^2-bm_1m_2+am_2^2,\quad \lambda(\dlm)=am_1+am_2$$
and hence
\begin{align*}
 &K(\dlm)-\lambda(\dlm)=am_1^2-bm_1m_2+am_2^2-(am_1+am_2)\\
 &\quad =a(m_1-m_2)(m_1-m_2-1)+(2a-b)m_2\Big(m_1-2+\frac{2(a-b)}{2a-b}\Big).
\end{align*}
Assume that $K(\dlm)=\lambda (\dlm)$ and $m_1,m_2>0$. By symmetry of $m_1,m_2$,
without loss of generality we may assume that $m_1\ge m_2$.
Then
$$2(m_1-m_2)(m_1-m_2-1)\ge 0, \quad m_2>0, $$
and
$m_1-2+\frac{2(a-b)}{2a-b}>m_1-2\ge 0$ for $m_1\ge 2$ since $a>b$.
Hence $K(\dlm)=\lambda(\dlm)$ implies that $m_1=1$ and hence $m_2=1$. However,
$K(1,1)-\lambda(1,1)=-b\ne 0$. Therefore $K(\dlm)\ne \lambda (\dlm)$
for all pairs $\dlm=(m_1,m_2)$ with $m_1,m_2>0$. Thus $\NA (V)=T(V)$.
\end{exam}

config: arxiv


text:
\section{Reverse Triangle Inequality/Normed Vector Space}
Tags: Triangle Inequality

\begin{theorem}
Let $\struct {X, \norm {\, \cdot \,} }$ be a [[Definition:Normed Vector Space|normed vector space]].
Then:
:$\forall x, y \in X: \norm {x - y} \ge \size {\norm x - \norm y}$
\end{theorem}

\begin{proof}
Let $d$ denote the [[Definition:Metric Induced by Norm|metric induced by $\norm {\, \cdot \,}$]], that is:
:$\map d {x, y} = \norm {x - y}$
From [[Metric Induced by Norm is Metric]] we have that $d$ is indeed a  [[Definition:Metric|metric]].
Then, from the [[Reverse Triangle Inequality]] as applied to [[Definition:Metric Space|metric spaces]]:
:$\forall x, y, z \in X: \size {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$
Then:
:$\forall x, y \in X: \size {\norm x - \norm y} = \size {\norm {x - 0} - \norm {y - 0} } \le \norm {x - y}$.
{{qed}}
\end{proof}

config: wiki


TITLE: Find the radius of convergence of the alternating power series
QUESTION [2 upvotes]: Consider the alternating series $ \sum_{n=1}^{\infty} (-1)^n \frac{5}{n^3} x^n $.
Find the radius of convergence of it.
Answer:
$a_n=(-1)^n \frac{5}{n^3} $
Root test:
$ \lim_{n \to \infty} \sqrt[n]{|a_n|}=\lim_{n \to \infty} \sqrt[n]{\frac{5}{n^3}}=5$
Ratio test:
$ \lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|=\lim_{n \to \infty} |\frac{(n+1)^3}{n^3}|=1 $
So both limits are not equal.
But we know that if limits exists , then they will be equal.
So I am doing something wrong.
Help me

REPLY [3 votes]: By roots test we have
$$ \lim_{n \to \infty} \sqrt[n]{|a_n|}=\lim_{n \to \infty} \sqrt[n]{\frac{5}{n^3}}=1$$
then the radius of convergence is $1$.
To check for the convergence we need to consider separately the cases $x=1$ and $x=-1$.


TITLE: Calculating a limit with trigonometric and quadratic function
QUESTION [9 upvotes]: $$\lim_{n->\infty}\left(1+{1\over{n^2+\cos n}}\right)^{n^2+n}$$

I vaguely get the idea that since $\cos n$ and $n$ dont really matter compared to $n^2$, this must evaluate to $e$. But not sure how to prove this. Further do all the limits of form $$\lim_{n->\infty}\left(1+{1\over{n^2+f(n)}}\right)^{n^2+g(n)}$$
evaluate to e given that $f'(x),g'(x)<2x;\forall x>0$?

REPLY [0 votes]: Taking $\log$ and using a well-known equivalence:
$$
\log\lim_{n\to\infty}\left(1 + {1\over{n^2 + \cos n}}\right)^{n^2+n} =
\lim_{n\to\infty}\log\left(1 + {1\over{n^2 + \cos n}}\right)^{n^2+n} =
$$
$$
\lim_{n\to\infty}(n^2 + n)\log\left(1 + {1\over{n^2 + \cos n}}\right) =
\lim_{n\to\infty}\frac{n^2 + n}{n^2 + \cos n} =
\lim_{n\to\infty}\frac{1 + 1/n}{1 + (\cos n)/n^2} = 1,
$$
so
$$\lim_{n\to\infty}\left(1 + {1\over{n^2 + \cos n}}\right)^{n^2+n} = e.
$$


TITLE: Does $X\cdot\langle2\rangle\cdot\langle3\rangle$ cover $\Bbb Z\left[\frac16\right]\cap(0,\infty)$? Is there a similar method using a Cantor set?
QUESTION [2 upvotes]: Does the set given by $X\cdot\langle2\rangle\cdot\langle3\rangle$ exactly cover $\Bbb Z\left[\frac16\right]\cap(0,\infty)$?  Is there a similar method using a Cantor set?

Where $X=\{x\in\Bbb N:x\equiv\pm1\pmod 6\}$ and $\langle p\rangle=\{p^i:i\in\Bbb Z\}$
I think it does because if we exclude from the positive integers the multiples of $2$ and $3$, we get the numbers equivalent to $\pm1\pmod 6$ then reintroducing the factors $2^i\cdot3^i$ restores $\Bbb N>0$ and also all the dyadic and ternary denominators.
Is there a similar cover using the middle two-thirds Cantor set $C_{2/3}$ in place of $X$?  The middle two-thirds Cantor set is the set of strings in base $6$ constructed out of 5s and 0s, so ends in $\pm1$. So I imagine there is some similar way of constructing $\Bbb Z[\frac16]$ using $C_{2/3}$, $\langle2\rangle$ and $\langle3\rangle$.
Obviously a direct replacement doesn't work but is there a relatively simple way of swapping the Cantor set in, such as e.g. a sum of two cantor sets?

REPLY [3 votes]: Yes $X.\langle2\rangle.\langle3\rangle$ covers $\mathbb{Z}[\frac{1}{6}] \cap (0, \infty)$. You can show this bit pretty easily by seeing that any element $x \in \mathbb{Z}[\frac{1}{6}] \cap (0, \infty)$ will be of the form $\frac{z}{6^n} = \frac{2^a3^bz'}{6^n} = 2^{a'}3^{b'}z'$ (where $z'$ and 6 are coprime), which cleary are contained in $X.\langle2\rangle.\langle3\rangle$.


TITLE: Thomae's function, doubt in continuous proof in the irrationals.
QUESTION [2 upvotes]: I was studying about this proof and i almost understand all of it, i just have one doubt there, the proof i found is the following;
Let f be defined by;
$$
\begin{align}
f(x) = \begin{cases}
0 & \text{if $x$ is irrational}\\
\frac{1}{q} & \text{if $x = \frac{p}{q}$ where $(p,q) = 1$ and q > 0}
\end{cases}
\end{align}
$$
Let r be irrational. Then f(r)=0. Let m be a positive integer. Then r is in a unique interval of the form ( k/m,(k+1)/m). Let dm = min{|r −k/m|, |r −(k+1)/m|} and let δm = min{d1, d2, . . . , dm}. Notice δm < 1/m.
Let ε > 0 be given. Choose m so that 1/m < ε. Let δ=δm. If x is a rational number with |x − r| < δ then x =p/q with gcd(p,q)=1 and q > m. Hence $0<f(x)=1/q < 1/m < ε$ . If x is irrational, f(x) = 0. So for any x, with |x − r| < δ, |f(x) − f(r)| < ε . This proves that f is continuous at any irrational number.
i think i understand the most part of it, but i just have a big question quen it says that q < m, i don't know how he can ensure that, because we are taking m large enough to be 1/m < ε, so, how you ensure that m > q is not going to happen?

REPLY [1 votes]: If we had $q\le m$, then $\delta_m \le d_q \le |r-x|$ (since $x=\frac pq$); but $|r-x| < \delta = \delta_m$, a contradiction.


TITLE: A proof of a basic conclusion in operator algebra
QUESTION [0 upvotes]: There is a quotation below: (in a book named "C*-algebras Finite-Dimensional Approximations")
Lemma 2.3.4. Let $A$ be a Banach space, $\mathbb{B}(A)$ be the space of all bounded linear maps from $A$ to $A$ and $C\subset B(A)$ be any convex set. Then the point-weak and point-norm closures of $C$ coincide
Proof.
Step 1. Let $T\in \mathbb{B}(A)$ be a map such that there exists a net $\{T_{i}\}_{i\in I}\subset C$ with the property that for every $a\in A$ and functional $f\in A^{*}$,
$$f(T_{i}(a))\rightarrow f(T(a)).$$
We must show that for each finite set $F=\{a_{1},...,a_{k}\}\subset A$ and $ \epsilon>0$ there exists $S\in C$ such that $||S(a_{j})-T(a_{j})||<\epsilon$ for $1\leq j\leq k$.
Step 2. This is a standard application of the Hahn-Banach Theorem since the net of maps (k-fold direct sum)
$$T_{i}\oplus...\oplus T_{i}$$
converges point-weakly to
$$T\oplus...\oplus T$$
in the Banach space $\mathbb{B}(A\oplus...\oplus A)$ (take $l^{p}$-norm on $A\oplus...\oplus A$).
Thus the element
$$T(a_{1})\oplus...\oplus T(a_{k})$$
belongs to the weak closure of
$$\{T_{i}(a_{1})\oplus...\oplus T_{i}(a_{k}\}_{i\in I}$$
and hence, by the Hahn-Banach Theorem, also to the norm closure of the convex hull.
Step 3. This implies that we can find a single convex combination of the $T_{i}$ which is simultaneously close to $T$ on all of $F$.
I have several incomprehensions on the proof above:

Why does the author consider the k-fold direct sum of $T_{i}$ in the step 2?
How to use Hahn-Banach Theorem, in step 2, to get the norm closure of the convex hull?

REPLY [1 votes]: It is a basic consequence of the Hahn-Banach theorem that the weak and norm closures of a convex set agree. This is the key result used in the argument.
Here they want to use that fact, but they have the problem that they need to be able to choose a net of operators in $C$. That's the role of the finite sets $F$: they will be the indices of the net: for each $F$, their argument produces a $T_F$ (denoted $S$ in the argument).
To find such operator, they want to use the result in the first paragraph, but now they have to make it work on $a_1,\ldots,a_k$. A typical trick in such situation is to use the direct sum to work with a single element containing all the information.


TITLE: How can $n^4 + 2n^3 + n^2$ be written in sum notation?
QUESTION [0 upvotes]: This problem was given in a math contest, and the following answers were provided

$\sum_{k=1}^n4k^3$
$\sum_{k=1}^n2k^3$
$\sum_{k=1}^n6k^2$
$\sum_{k=1}^n4k^2$
$\sum_{k=1}^n2k^4$
none of the above

Thanks!
EDIT: I'm sorry, the first term should be n^4.
OK so what I've considered so far is factoring down the expression to $(n^2)(n+1)^2$
From there, I'm not sure if there's a way to recognize that that's a summation or something (Similar to $n*(n+1)/2$)

REPLY [2 votes]: Plugging in $n=1$ and $n=2$ shows that only $\sum_{k=1}^n4k^3$ is a candidate. So we need to show that $\sum_{k=1}^n4k^3 = n^4+2n^3+n^2$
Proof by induction:
$n=1$: already done
$n\to n+1$:
$$\sum_{k=1}^{n+1}4k^3 - ((n+1)^4+2(n+1)^3+(n+1)^2)\\= \sum_{k=1}^{n}4k^3+4(n+1)^3 - (n^4+4n^3+6n^2+4n+1+2(n^3+3n^2+3n+1)+n^2+2n+1)\\=
n^4+2n^3+n^2+4(n^3+3n^2+3n+1) - (n^4+6n^3+13n^2+12n+4)\\=0$$
Note that with this kind of series it is easier to show $l-r=0$ instead of $l=…=r$.


TITLE: Proving that the discrete topology satisfies the first axiom of countability but not the second one
QUESTION [0 upvotes]: Could anyone please check my proof of the second assertion in the title and prove the first one? (It will be enough to work with the set of all real numbers). Let the topological space be denoted as $T = (R^1, \tau)$ where $\tau$ consists of the "set of all subsets of $R^1$."
My proof of the second part:
i) Since every single-element set is considered open in this topological space and since a base is a collection of open sets where every open set in $T$ is a union of some sets of this collection, any base must contain all the single-element sets (all real numbers). But this is naturally uncountable.
ii) For the first part, we would need to prove that given any $x \in R^1$ there is a countable family of open sets such that for every open set $G$ that contains $x$ there is a set $O$ in this family that is contained in $G$. We know that any set $G$ is already an open set in $T$. But if the statement is true wouldn't it imply that the set of all sets in $R^1$ that contain an arbitrary pont $x$ is countable? Is this corect?
Happy Good Friday :)

REPLY [1 votes]: In fact $\{\{x\}\}$ is a (finite, hence countable) local base at $x$.
If $\mathcal{B}$ is any base for $\tau$, then for all $x$ there must be a $B \in \mathcal{B}     $ such that $x \in B \subseteq \{x\}$, which implies that $B =\{x\} \in \mathcal{B}$, so that the base has size at least $|\Bbb R|$. So  an uncountable set in the discrete topology doesn’t have a countable base.


TITLE: A single die is rolled 7 times. What is the probability that a six is rolled exactly once, if it is known that at least one six is rolled?
QUESTION [0 upvotes]: A single die is rolled 7 times. What is the probability that a six is rolled exactly once, if it is known that at least one six is rolled?
Could you walk me through the concepts?

REPLY [1 votes]: To calculate $P(\text{exactly one } 6) $, the denominator should be the total number of outcomes, which is $6^7$. Looking at the outcomes with one $6$, there are seven slots to put the 6. From the remaining six slots, there are $5^6$ possible outcomes. Hence,
$$
P(\text{exactly one } 6)=\frac{7\times5^6}{6^7}.
$$
For $P(\text{at least one } 6)$, use the complement law. The answer is then
\begin{align}
P(\text{exactly one }6|\text{at least one }6)&=\frac{P(\text{exactly one }6\text{ and at least one }6)}{P(\text{at least one }6)} \\
&=\frac{P(\text{exactly one }6)}{P(\text{at least one }6)}.
\end{align}


text:
\begin{document}

\title[Homological Stability of Non-orientable Mapping Class Groups]{Homological Stability of Non-orientable Mapping Class Groups with Marked Points}
\date{}
\author{Elizabeth Hanbury}

\address{Mathematical Institute, University of Oxford, 24-29 St Giles', Oxford, OX1 3LB}
\curraddr{Department of Mathematics, National University of Singapore, 2 Science Drive 2, Singapore 117543}
\email{liz.hanbury@gmail.com}

\subjclass[2000]{Primary 57N05. Secondary 20F38}

\begin{abstract}
Wahl recently proved that the homology of the non-orientable mapping class group stabilizes as the genus increases. In this short note we analyse the situation where the underlying non-orientable surfaces have marked points.
\end{abstract}

\maketitle

\section{Introduction}
Every non-orientable surface is homeomorphic to $S_{g,n}=\#_{g} \mathbb{R}P^{2} \setminus \coprod_{n}D^{2}$ for some unique $g,n$. The object of study in this note is the associated mapping class group $\mathcal{N}_{g,n}:=\pi_{0}\mathrm{Diff}(S_{g,n}, \partial)$ - the connected components of the group of diffeomorphisms of $S_{g,n}$ that fix the boundary pointwise. In \cite{Nathalie} Wahl proved that for $g \geq 4r +3$, $H_{r}(\mathcal{N}_{g,n})$ is independent of $g$ and $n$. This is the non-orientable version of Harer's well-known stability theorem for the orientable mapping class group \cite{Harer}.

Here we will look at non-orientable surfaces with marked points - the surface $S_{g,n}$ with $k$ marked points will be denoted $S_{g,n}^{k}$. We can consider the group of diffeomorphisms of $S_{g,n}^{k}$ that fix the marked points or the group of diffeomorphisms that may permute them. The corresponding mapping class groups will be denoted by $\mathcal{N}_{g,n}^{k}$ and $\mathcal{N}_{g,n}^{(k)}$ respectively. These groups fit into an exact sequence
\[
0 \to \mathcal{N}^{k}_{g,n} \to \mathcal{N}^{(k)}_{g,n} \to \Sigma_{k} \to 0.
\]

The first result of this note is the following corollary of Wahl's theorem.

\begin{theorem}
\label{oops}
For any $k \geq 0$, $H_{r}(\mathcal{N}_{g,n}^{k})$ and $H_{r}(\mathcal{N}_{g,n}^{(k)})$ are independent of $g$ and $n$ when $g$ is sufficiently large compared to $r$.
\end{theorem}

We show how to deduce this from Wahl's theorem. As she notes in her introduction, the original proof of stability could also be adapted to deal with marked points.

There are also stable versions of the non-orientable mapping class groups: for $n \geq 1$, $\mathcal{N}_{\infty,n}^{k}$ is defined to be the colimit of the system
\[
\mathcal{N}_{g,n}^{k} \stackrel{\alpha}{\to} \mathcal{N}_{g+1,n}^{k} \stackrel{\alpha}{\to} \mathcal{N}_{g+2,n}^{k} \stackrel{\alpha}{\to} \ldots
\]
where $\alpha$ is given by attaching $\mathbb{R}P^{2}\setminus (D^{2} \sqcup D^{2})$ to the underlying surface and extending diffeomorphisms by the identity. Similarly we have $\mathcal{N}_{\infty,n}^{(k)}$. Our second result is the following splitting theorem for the classifying spaces of these groups. This is analogous to Theorem 3.1 in \cite{Bod-Till} where the orientable case was considered.

\begin{theorem}
There are homology isomorphisms
\begin{eqnarray*}
B\mathcal{N}_{\infty,n}^{k} \to B\mathcal{N}_{\infty,1} \times B(O(2))^{k} & and &
B\mathcal{N}_{\infty,n}^{(k)} \to B\mathcal{N}_{\infty,1} \times B(\Sigma_{k} \wr O(2))
\end{eqnarray*}
\end{theorem}

In section \ref{suspect} we move on to look at how the homology of $\mathcal{N}_{g,n}^{(k)}$ depends on $k$. We prove
\begin{theorem}
\label{equal}
$H_{r}(\mathcal{N}_{g,n}^{(k)})$ is independent of $k$ (as well as $g$ and $n$) when $g$ and $k$ are sufficiently large relative to $r$.
\end{theorem}
Our techniques could also be applied to the orientable case to prove
\begin{theorem}
\label{more-equal}
$H_{r}(\Gamma_{g,n}^{(k)})$ is independent of $k$ when $g$ and $k$ are sufficiently large relative to $r$.
\end{theorem}
Here $\Gamma_{g,n}^{(k)}$ is the orientable mapping class group in which diffeomorphisms are allowed to permute marked points.

Hatcher and Wahl have recently proved versions of Theorems \ref{equal} and \ref{more-equal} in which they don't require the genus to be large \cite{HW}. They obtain these results as special cases of the more general theory of homological stability of mapping class groups of $3$-manifolds.

The proof of Theorem \ref{equal} is based on the following theorem on the Borel construction. This theorem is well-known but we are not aware of any published proof.
\begin{theorem}
\label{buttmunch}
If $X$ is a path-connected space then for $k \geq 2r+6$ there is an isomorphism
\[
H_{r}(E\Sigma_{k} \times_{\Sigma_{k}} X^{k}) \to H_{r}(E\Sigma_{k+1} \times_{\Sigma_{k+1}} X^{k+1}).
\]
\end{theorem}
This theorem is based on the homological stability of symmetric groups with twisted coefficients \cite{Betley}. The proof is contained in the appendix.

\vspace{\headheight}
\section{Homological Stability with Marked Points}

For any $n\geq 1$ there is a stabilization map $\alpha:\mathcal{N}_{g,n} \to \mathcal{N}_{g+1,n}$ as described in the introduction. Similarly there are maps $\beta:\mathcal{N}_{g,n} \to \mathcal{N}_{g,n+1}$, given by attaching a pair of pants, and $\gamma:\mathcal{N}_{g,n} \to \mathcal{N}_{g,n-1}$ given by attaching a disc to one of the boundary components. The map $\gamma$ forms a left-inverse to $\beta$ for $n \geq 2$. The main result of \cite{Nathalie} is the following.

\begin{theorem}[Wahl]
\label{julie}
For any $n\geq 1$
\begin{itemize}
\item[(i)] $\alpha_{\ast}: H_{r}(\mathcal{N}_{g,n}) \to H_{r}(\mathcal{N}_{g+1,n})$ is an isomorphism provided $g \geq 4r+3$
\item[(ii)] $\beta_{\ast}: H_{r}(\mathcal{N}_{g,n}) \to H_{r}(\mathcal{N}_{g,n+1})$ is an isomorphism provided $g \geq 4r+3$
\item[(iii)] $\gamma_{\ast}: H_{r}(\mathcal{N}_{g,1}) \to H_{r}(\mathcal{N}_{g,0})$ is an isomorphism provided $g \geq 4r+5$.
\end{itemize}
\end{theorem}

The stabilization maps $\alpha, \beta$ and $\gamma$ also define maps on $\mathcal{N}_{g,n}^{k}$ and $\mathcal{N}_{g,n}^{(k)}$. Theorem \ref{oops} follows from the following extension of Wahl's theorem.

\begin{theorem}
\label{eat}
Theorem \ref{julie} still holds if we replace $\mathcal{N}_{g,n}$ by $\mathcal{N}_{g,n}^{k}$ or $\mathcal{N}_{g,n}^{(k)}$.
\end{theorem}

\begin{proof}
We will give the proof of (i) for $\mathcal{N}_{g,n}^{(k)}$ - the other cases are entirely analogous.

Let $\mathrm{Diff} (S_{g,n}^{(k)}, \partial)$ denote the group of diffeomorphisms of $S_{g,n}^{k}$ which fix the boundary and may permute the marked points. Fix a framing at each marked point i.e. a pair of orthonormal tangent vectors. By considering the effect of a diffeomorphism on the marked points and their framings we obtain a map
\begin{equation}
\label{squidge}
\mathrm{Diff} (S_{g,n}^{(k)},\partial) \stackrel{\rho}{\to} \Sigma_{k} \wr Gl_{2} (\mathbb{R}) \simeq \Sigma_{k} \wr O(2)
\end{equation}
which is a fibration onto its image. The fiber is homotopic to the group of diffeomorphisms which fix a neighbourhood of each marked point i.e. $\mathrm{Diff}(S_{g,n+k},\partial)$.

We claim that $\rho$ is in fact surjective. By considering diffeomorphisms of $D^{2}$ with $k$ points removed we can see that every element of $\Sigma_{k} \wr Gl_{2}^{+}(\mathbb{R})$ lies in the image - we complete the proof of the claim by showing that $\iota \times - \mathrm{id}$ also lies in the image where $\iota$ denotes the identity permutation. This amounts to displaying a diffeomorphism of $S_{g,n}^{k}$ which reverses the orientation at a chosen marked point and fixes everything else. Such a diffeomorphism is given by a `dragging function': fix a curve $C$ which passes through the chosen marked point and such that a neighbourhood $U$ of $C$ is homeomorphic to a M\"obius band. Choose $U$ to avoid the other marked points. The diffeomorphism we require is obtained by dragging the marked point all the way around $C$ and is the identity outside $U$. Explicitly we can define this as a composition of $m$ diffeomorphisms $\{ \epsilon_{j}\}_{j=1}^{m}$ which each move the marked point $\frac{1}{m}$th of the way around $C$ and are the identity outside $U$. See Figure 1 for a representation of $\epsilon_{j}$.

\begin{figure}
\includegraphics{mobius}
\caption{The diffeomorphism $\epsilon_{j}$}
\label{}
\end{figure}

Now we take classifying spaces in (\ref{squidge}) and use the fact that the components of the diffeomorphism groups are contractible \cite{EE, ES} to obtain a homotopy fibration
\begin{equation}
\label{excess}
B\mathcal{N}_{g,n+k} \to B\mathcal{N}_{g,n}^{(k)} \to B(\Sigma_{k} \wr O(2)).
\end{equation}
The map $\alpha$ induces a map from this fibration to
\[
B\mathcal{N}_{g+1,n+k} \to B\mathcal{N}_{g+1,n}^{(k)} \to B(\Sigma_{k} \wr O(2))
\]
which is the identity on the base spaces.

The induced map on the $E^{2}$-page of the associated Serre spectral sequences is the obvious map
\begin{eqnarray*}
E_{p,q}^{2} = H_{p}(B(\Sigma_{k} \wr O(2)); H_{q}(B\mathcal{N}_{g,n+k})) \\
\to \tilde{E}_{p,q}^{2} = H_{p}(B(\Sigma_{k} \wr O(2)); H_{q}(B\mathcal{N}_{g+1,n+k})).
\end{eqnarray*}

By Theorem \ref{julie} the map on the coefficients is an isomorphism for $g \geq 4q+3$. Hence the map on the $E_{2}$-term is an isomorphism for all $p$ and for $q \leq \frac{g-3}{4}$ so by the Zeeman comparison theorem (see section 1 in \cite{Ivanov}) the map on the abutments
\[
H_{r}(\mathcal{N}_{g,n}^{(k)}) \to H_{r}(\mathcal{N}_{g+1,n}^{(k)})
\]
is an isomorphism for $r \leq \frac{g-3}{4}$.
\end{proof}

\vspace{\headheight}
\section{Splitting of the Classifying Spaces}

In this section we will establish a splitting result for the classifying spaces of $\mathcal{N}_{\infty,n}^{k}$ and $\mathcal{N}_{\infty,n}^{(k)}$ cf. \cite{Bod-Till}.
\begin{theorem}
\label{patricia}
There are homology isomorphisms
\begin{itemize}
\item[(i)] $B\mathcal{N}_{\infty,n}^{k} \to B\mathcal{N}_{\infty,1} \times B(O(2))^{k}$
\item[(ii)] $B\mathcal{N}_{\infty,n}^{(k)} \to B\mathcal{N}_{\infty,1} \times B(\Sigma_{k} \wr O(2))$.
\end{itemize}
\end{theorem}

\begin{proof}
We give the proof of (ii), the proof of (i) is completely analogous. Consider the fibration (\ref{excess}) from the previous section
\[
B\mathcal{N}_{g,n+k} \to B\mathcal{N}_{g,n}^{(k)} \stackrel{B\rho}{\longrightarrow} B(\Sigma_{k} \wr O(2)).
\]
There is a map from this fibration to the trivial fibration
\[
B\mathcal{N}_{g,1} \to B\mathcal{N}_{g,1} \times B(\Sigma_{k} \wr O(2)) \to B(\Sigma_{k} \wr O(2)).
\]
On the total space this is given in the first factor by forgetting marked points and applying $\gamma$ and in the second factor by $B\rho$. On the fiber it's induced by $\gamma$.

We know from Theorem \ref{julie} that the maps on the fiber and on the base are $H_{r}$-isomorphisms for $g$ sufficiently large. Considering the associated Serre spectral sequences and applying the Zeeman comparison theorem we are able to conclude that the map on total spaces is an $H_{r}$-isomorphism for sufficiently large $g$. Letting $g$ tend to infinity we obtain the result.

\end{proof}

\section{Changing the Number of Marked Points}
\label{suspect}

We now move on to investigating how the homology of $\mathcal{N}_{g,n}^{k}$ and $\mathcal{N}_{g,n}^{(k)}$ depend on $k$. We can see from Theorem \ref{patricia} that the homology of $\mathcal{N}_{g,n}^{k}$ is dependent on $k$ no matter how large $g$ or $k$ become. However we have maps $\mu:\mathcal{N}_{g,n}^{k} \to \mathcal{N}_{g,n}^{k+1}$ and $\nu:\mathcal{N}_{g,n}^{k+1} \to \mathcal{N}_{g,n}^{k}$ where $\mu$ is given by gluing on a cylinder with a marked point and extending maps by the identity and $\nu$ is given by forgetting the $(k+1)$st marked point. The map $\nu$ is a left-inverse to $\mu$ so we have

\begin{lemma}
$\mu_{\ast}:H_{r}(\mathcal{N}_{g,n}^{k}) \to H_{r}(\mathcal{N}_{g,n}^{k+1})$ is injective.
\end{lemma}

When we are dealing with $\mathcal{N}_{g,n}^{(k)}$ we can say much more. In fact in the next theorem we will prove that $H_{r}(\mathcal{N}_{g,n}^{(k)})$ is independent of $k$ provided $g$ and $k$ are sufficiently large. The proof relies on Betley's result on the homological stability of the symmetric groups with twisted coefficients \cite{Betley}.

\begin{theorem}
\label{cherie}
When $g \geq 4r+3$ and $k \geq 2r+8$, gluing on a cylinder with a marked point induces an isomorphism
\[
H_{r}(\mathcal{N}_{g,n}^{(k)}) \stackrel{\cong}{\longrightarrow} H_{r}(\mathcal{N}_{g,n}^{(k+1)}).
\]
\end{theorem}

\begin{proof}
Recall the fibration (\ref{excess}) from above
\[
B\mathcal{N}_{g,n+k} \to B\mathcal{N}_{g,n}^{(k)} \longrightarrow B(\Sigma_{k} \wr O(2)).
\]
Gluing on a cylinder with a marked point induces a map from this fibration to
\[
B\mathcal{N}_{g,n+k+1} \to B\mathcal{N}_{g,n}^{(k+1)} \longrightarrow B(\Sigma_{k+1} \wr O(2)).
\]
There is an induced map of the associated Serre spectral sequences and on the $E^{2}$-term  it is
\begin{eqnarray*}
\label{criticise}
E^{2}_{p,q} = H_{p}(B(\Sigma_{k} \wr O(2)); H_{q}(\mathcal{N}_{g,n+k})) \\
\to \tilde{E}^{2}_{p,q} = H_{p}(B(\Sigma_{k+1} \wr O(2)); H_{q}(\mathcal{N}_{g,n+k+1})).
\end{eqnarray*}
The map on the coefficients is an isomorphism for $g \geq 4q+3$ by Theorem \ref{julie}. We also have that the $\pi_{1}(B(\Sigma_{k} \wr O(2)))$-action on $H_{q}(\mathcal{N}_{g,n+k})$ is trivial for $g$ in this range - this is proved in the same way as Lemma 3.3 in \cite{Bod-Till} - so the coefficients are not twisted. Since $B(\Sigma_{k} \wr O(2)) \simeq E\Sigma_{k} \times_{\Sigma_{k}} BO(2)^{k}$ then, for $g \geq 4q+3$, we can apply Theorem \ref{buttmunch} and deduce that the map on the $E^{2}$-term is an isomorphism for $p \leq \frac{k-6}{2}$, $q \leq \frac{g-3}{4}$. Applying the Zeeman comparison theorem yields the result.
\end{proof}

\vspace{1mm}
\section{Appendix: Applying Betley's Stability Result}

Let $\Phi$ be the category whose objects are the finite pointed sets $[n]=\{ 0,1,\ldots, n \}$ and whose morphisms are pointed maps with the property that the pre-image of each non-zero element contains at most one element. Let $Ab$ denote the category of abelian groups. A coefficient system is a functor $T: \Phi \to Ab$ with $T([0])=0$. Thus $\Sigma_{n}$ acts on $T([n])$ for each $n$ and we have the homology groups $H_{i}(\Sigma_{n}; T([n]))$.

In order to define the notion of degree of coefficient systems, we
consider the maps $e_{s}:[n] \to [n-1]$ for $1 \leq s \leq n$ given
by $e_{s}(j)=j$ for $j < s$, $e_{s}(s)=0$ and $e_{s}(j)=j-1$ for
$j>s$.  The coefficient system $T$ is said to be of degree $d$ if
\[
T_{d+1}:= ker \left( \prod_{s=1}^{d+1}(e_{s})_{\ast}: T([d+1]) \to \prod_{s=1}^{d+1}T([d]) \right)
\]
is zero but $T_{d}$ is non-zero.

Betley's main result was the following
\begin{theorem}[Betley]
\label{spam}
For any coefficient system $T$ of degree $\leq d$ the natural map
\[
H_{i}(\Sigma_{n}; T([n])) \to H_{i}(\Sigma_{n+1}; T([n+1]))
\]
is an isomorphism for $2i + d \leq n$.
\end{theorem}

Betley actually defined a coefficient system to be a functor on the category $\Gamma$ of finite pointed sets with pointed maps, which contains $\Phi$. Thus Theorem \ref{spam} is a slightly stronger result than Betley originally proved. We'll summarize his proof to check that it's still valid in this more general situation.

\begin{proof}[Proof of Theorem \ref{spam}.]
There is an isomorphism of abelian groups;
\begin{equation}
\label{angel}
T([n]) \cong \bigoplus_{k} T^{k}_{n}
\end{equation}
where for given $k$, $T^{k}_{n}=\bigoplus_{\gamma}T_{k}$, the latter sum being taken over all order-preserving, pointed injections $\gamma:[k] \to [n]$. This is formula (1.1.1) in Betley's paper. A proof can be found in \cite{Eilenberg-MacLane}, Theorem 9.1, and only requires the morphisms from $\Phi$.

The next step is to identify the $\Sigma_{n}$-action on $\bigoplus_{k}T^{k}_{n}$ under (\ref{angel}). The action respects the decomposition. Each $T_{k} \leq T([k])$ has a natural $\Sigma_{k}$-action. Careful consideration of the bijection (\ref{angel}) gives an isomorphism of $\Sigma_{n}$-modules
\begin{equation}
\label{arithmetic}
T_{n}^{k} \cong T_{k} \otimes_{\mathbb{Z}[\Sigma_{k} \times \Sigma_{n-k}]} \mathbb{Z}[\Sigma_{n}]
\end{equation}
where $\Sigma_{n-k}$ acts trivially on $T_{k}$. This is in the proof of Betley's Lemma 3.1.

Now if $T$ has degree $d$ and $n>d$ we have
\[
H_{i}(\Sigma_{n};T([n])) \cong H_{i}(\Sigma_{n}; \bigoplus_{k=1}^{d} T_{n}^{k}) \cong \bigoplus_{k=1}^{d} H_{i}(\Sigma_{n}; T_{n}^{k}).
\]
By (\ref{arithmetic}), Shapiro's lemma and the K\"unneth theorem, this is isomorphic to
\[
\bigoplus_{k=1}^{d} H_{i}(\Sigma_{k} \times \Sigma_{n-k}; T_{k}) \cong \bigoplus_{k=1}^{d} \{ \bigoplus_{j+l=i} H_{j} (\Sigma_{n-k}; H_{l}(\Sigma_{k};T_{k})) \}.
\]

The same formula holds for $H_{i}(\Sigma_{n+1}; T([n+1]))$. The stabilization map is induced by the canonical inclusion $\Sigma_{n-k} \to \Sigma_{n-k+1}$ and is the identity on the coefficients. The result is now immediate by the homological stability of the symmetric groups.
\end{proof}

The proof of Theorem \ref{buttmunch} requires the following proposition
\begin{proposition}
\label{balula}
If $X$ is a path-connected space with basepoint and
$F$ is any field then $[n] \mapsto H_{q}(X^{n},F)$ defines a
coefficient system of degree $\leq q$.
\end{proposition}

\begin{proof}
For ease of notation, we'll suppress the coefficient field $F$. First we'll check that this assignment has the necessary functoriality. Let $\lambda:[k] \to [n]$ be a morphism in $\Phi$, there is an induced map $\lambda_{\ast}:X^{k} \to X^{n}$ given by $(x_{1}, \ldots, x_{k}) \to (x_{\lambda^{-1}(1)}, \ldots, x_{\lambda^{-1}(n)})$ where $x_{\lambda^{-1}(j)}:=\ast$ if $\lambda^{-1}(j)=\emptyset$. Clearly $(\lambda \circ \mu)_{\ast}=\lambda_{\ast} \circ \mu_{\ast}$ so this gives us functoriality.

Now we move on to checking that the coefficient system has finite degree. By the definition of degree, our task is to prove that the kernel of the map
\begin{equation}
\label{joseph}
\prod_{s=1}^{n}(e_{s})_{\ast}: H_{q}(X^{n}) \to \prod_{s=1}^{n} H_{q}(X^{n-1})
\end{equation}
is trivial when $n > q$.

Since we're working with field coefficients, the K\"unneth theorem tells us that,
\[
H_{q}(X^{n}) \cong \bigoplus_{i_{1}+ \ldots + i_{n}=q} H_{i_{1}}(X) \otimes_{F} \ldots \otimes_{F} H_{i_{n}}(X).
\]

The induced map $(e_{s})_{\ast}:H_{q}(X^{n}) \to  H_{q}(X^{n-1})$
takes the summand corresponding to $(i_{1}, \ldots, i_{n})$ to zero
if $i_{s} \neq 0$ and isomorphically to the summand corresponding to
$(i_{1}, \ldots, \hat{i_{s}}, \ldots, i_{n})$ if $i_{s}=0$. To see
this, think of $(e_{s})_{\ast}: X^{n} \to X^{s-1} \times \{\ast\}
\times X^{n-s} \cong X^{n-1}$ as the map sending the $s$th factor to
the basepoint. Then the induced map on homology takes $H_{i_{r}}(X)$
by the identity to $H_{i_{r}}(X)$ when $r \neq s$ and takes
$H_{i_{s}}(X)$ to $H_{i_{s}}(\ast)$ by the obvious induced map. The
latter is an isomorphism if $i_{s}=0$ (because $X$ is
path-connected) and is the zero map otherwise. Thus the kernel of
$(e_{s})_{\ast}$ is
\[
\bigoplus_{\stackrel{i_{1}+ \ldots + i_{n}=q}{i_{s} \neq 0}} H_{i_{1}}(X) \otimes_{F} \ldots \otimes_{F} H_{i_{n}}(X)
\]
The kernel of $\prod_{s=1}^{n}(e_{s})_{\ast}$ is the intersection of these kernels and hence it is the direct sum of those summands corresponding to $n$-tuples $(i_{1}, \ldots, i_{n})$ with $i_{1}+ \ldots + i_{n}=q$ and every $i_{s} \neq 0$. When $n > q$ there are no such $n$-tuples so the kernel is trivial.
\end{proof}

\vspace{2mm}
\begin{proof}[Proof of Theorem \ref{buttmunch}.]
Consider the map of fibrations
\begin{eqnarray*}
\begin{diagram}
\node{X^{k}} \arrow{e} \arrow{s} \node{E\Sigma_{k} \times_{\Sigma_{k}} X^{k}} \arrow{e} \arrow{s} \node{B\Sigma_{k}} \arrow{s}\\
\node{X^{k+1}} \arrow{e} \node{E\Sigma_{k+1} \times_{\Sigma_{k+1}} X^{k+1}} \arrow{e} \node{B\Sigma_{k+1}}
\end{diagram}
\end{eqnarray*}
and the associated Serre spectral sequences with coefficients in some field $F$. There is an induced map of the spectral sequences and on the $E^{2}$-term it is given by
\[
E_{p,q}^{2} = H_{p}(\Sigma_{k}; H_{q}(X^{k},F)) \to \tilde{E}_{p,q}^{2} = H_{p}(\Sigma_{k+1}; H_{q}(X^{k+1},F)).
\]
But by Proposition \ref{balula}, $\{H_{q}(X^{k},F)\}_{k}$ is a coefficient system of degree $\leq q$. Hence Betley's stability result implies that this map is an isomorphism for $2p+q \leq k$. Now by the Zeeman comparison theorem, the induced map on the abutments
\[
H_{r}(E\Sigma_{k} \times_{\Sigma_{k}} X^{k},F) \to H_{r}(E\Sigma_{k+1} \times_{\Sigma_{k+1}} X^{k+1},F)
\]
is an isomorphism for $r \leq \frac{k}{2}-1$. Since $E\Sigma_{k} \times_{\Sigma_{k}} X^{k} \to E\Sigma_{k+1} \times_{\Sigma_{k+1}} X^{k+1}$ induces an isomorphism on $H_{r}(-,F)$ for any field $F$ when $r \leq \frac{k}{2}-1$ it also induces an isomorphism on $H_{r}(-,\mathbb{Z})$ for $r \leq \frac{k}{2}-3$. This is a consequence of the universal coefficient theorem.
\end{proof}

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\end{document}


TITLE: Mathematical induction on a finite subset of $\mathbb{N}$
QUESTION [2 upvotes]: I'm trying to prove the following using induction (or well-ordering principle). (Induction says that whenever $ 0 \in S $ and $ n \in S \implies (n+1) \in S$, then $S = \mathbb{N} $. )
Let $ \{ P_{k} \}$  be a sequence of propositions such that
$P_{1}$ is true
The truth of $P_{i-1} $ implies the truth of $P_{i}$ for all $(i-1) \in \{ 1,2,...(n-1) \} $
Then show that ${P_{k}}$ is true for all $ k \in \{ 1,2,...n \} $
Clearly, this is a modified form of induction (perhaps a weaker form) restricted to the first $n$ naturals, so the knowledge of induction should easily allow me to prove this. Yet I am stuck, what's the trick?

REPLY [1 votes]: We use the Well-Ordering Principle.
Let $T = \{k \in \{1, 2, 3, \ldots, n\} \mid P_k~\text{is false}\}$.  We will show that $T = \emptyset$.
Suppose otherwise.  Then since $T \subseteq \mathbb{N}$ and $T \neq \emptyset$, there exists a least element $m \in T$ such that $P_m$ is false.  Since $P_1$ holds, $m \neq 1$.  Since $m > 1$ and $m - 1 < m \leq n$,  $P_{m - 1}$ holds.  However, $P_{m - 1} \implies P_m$ holds.  Thus, $m \notin T$, contrary to our assumption that $m$ is the least element of $T$.  Hence, $T$ does not have a least element, so $T = \emptyset$.  Therefore, if $P_1$ holds and $P_{k - 1} \implies P_k$ for all $k \in \{1, 2, 3, \ldots, n - 1\}$, $P_k$ holds for each $k \in \{1, 2, 3, \ldots, n\}$.