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\title[On endomorphisms of quantum tensor space]{On endomorphisms of quantum tensor space}

\author{G.I. Lehrer and R.B. Zhang}
\address{School of Mathematics and Statistics,
University of Sydney, N.S.W. 2006, Australia}
\email{gusl@maths.usyd.edu.au, rzhang@maths.usyd.edu.au}
\date {23rd June, 2008}

\begin{abstract} We give a presentation of the endomorphism algebra
$\End_{\cU_q(\fsl_2)}(V^{\otimes r})$, where $V$ is the
$3$-dimensional irreducible module for quantum $\fsl_2$ over the
function field $\C(q^{\frac{1}{2}})$. This will be as a quotient of
the Birman-Wenzl-Murakami algebra
$BMW_r(q):=BMW_r(q^{-4},q^2-q^{-2})$ by an ideal generated by a
single idempotent $\Phi_q$. Our presentation is in analogy with the
case where $V$ is replaced by the $2$- dimensional irreducible
$\cU_q(\fsl_2)$-module, the BMW algebra is replaced by the Hecke
algebra $H_r(q)$ of type $A_{r-1}$, $\Phi_q$ is replaced by the
quantum alternator in $H_3(q)$, and the endomorphism algebra is the
classical realisation of the Temperley-Lieb algebra on tensor space.
In particular, we show that all relations among the endomorphisms
defined by the $R$-matrices on $V^{\otimes r}$ are consequences of
relations among the three $R$-matrices acting on $V^{\otimes 4}$.
The proof makes extensive use of the theory of cellular algebras.
Potential applications include the decomposition of tensor powers
when $q$ is a root of unity.
\end{abstract}
\maketitle

\section{Introduction and statement of results}
Let $q^{\frac{1}{2}}$ be an indeterminate over $\C$.
Our objective is to give a presentation of
$\End_{\cU_q(\fsl_2)}(V_q^{\otimes r})$, where $V_q$ is the
irreducible $3$-dimensional representation of quantum $\fsl_2$. This
presentation will be as a quotient of the Birman-Murakami-Wenzl
algebra $BMW_r(q):=BMW_r(q^{-4},q^2-q^{-2})$ by an ideal generated
by a single quasi-idempotent $\Phi_q$. The endomorphism algebra
therefore has the same presentation as $BMW_r(q)$, with the
additional relation $\Phi_q=0$. This paper is a sequel to
\cite{LZ2}, where the main results were conjectured, and by and
large we maintain the notation of that work. Our results may be
stated integrally, i.e. in terms of algebras over the ring
$\C[q^{\pm \frac{1}{2}}]$, and they therefore have the potential to
generalise to the situation where $q$ is a root of unity. We intend
to address that issue in a future work.

\subsection{General notation.}

Denote the function field $\C(q^{\frac{1}{2}})$ by $\CK$. Let
$\cU_q=\cU_q(\fsl_2)$ be the quantised universal enveloping algebra
of $\fsl_2$ over $\CK$, and write $V_q$ for the 3-dimensional
irreducible $\cU_q$ module. More generally, write $V(d)_q$ for the
irreducible $\cU_q$-module with highest weight $d\in\Z_{\geq 0}$, so
that $V_q=V(2)_q$. It is well known that there is a homomorphism

\be\label{braidhom}
\eta:\CK \CB_r\lr \End_{\cU_q}(V(d)_q^{\otimes r}):=E_q(d,r),
\ee
where $\CB_r$ is the $r$-string braid group, and $\eta$ maps the
$i^{\text{th}}$ generator of $\CB_r$ to the $R$-matrix acting on
the $i^{\text{th}}$ and $(i+1)^{\text{st}}$ factors of the tensor
power (cf. \cite[\S\S 3,4]{LZ2}).

It was shown in \cite{LZ1} that $\eta$ is surjective for all $d$ and $r$.
It is our purpose
to use this to give an explicit presentation of $E_q(2,r)$, analogous to the
celebrated presentation of the Temperley-Lieb algebra $E_q(1,r)$ as a quotient
of the Hecke algebra of type $A$ (\cite[Theorem 3.5]{LZ2}). The first step
is to identify a finite dimensional quotient of $\CK\CB_r$ through which
$\eta$ factors.

\subsection{The algebra $BMW_r(\CK)$.} It was shown in \cite[Theorem 4.4]{LZ2}
that $\eta$ factors through the finite dimensional algebra $BMW_r(\CK)$,
which we now proceed to define.

Let $\CA_{y,z}$ be the ring $\C[y^{\pm 1},z]$, where $y,z$ are
indeterminates. The BMW algebra $BMW_r(y,z)$ over $\CA_{y,z}$ is the
associative $\CA_{y,z}$-algebra with generators $g_1^{\pm
1},\dots,g_{r-1}^{\pm 1}$ and $e_1,\dots,e_{r-1}$, subject to the
following relations:

The braid relations for the $g_i$:

\begin{equation}\label{braidgi}
\begin{aligned}
g_ig_j&=g_jg_i\text{ if }|i-j|\geq 2\\
g_ig_{i+1}g_i&=g_{i+1}g_ig_{i+1} \text{ for }1\leq i\leq r-1;\\
\end{aligned}
\end{equation}

The Kauffman skein relations:

\begin{equation}\label{kauffman}
g_i-g_i^{-1}=z(1-e_i)\text { for all }i;
\end{equation}

The de-looping relations:

\begin{equation}\label{delooping}
\begin{aligned}
&g_ie_i=e_ig_i=ye_i;\\\
&e_ig_{i-1}^{\pm 1}e_i=y^{\mp 1}e_i;\\
&e_ig_{i+1}^{\pm 1}e_i=y^{\mp 1}e_i.\\
\end{aligned}
\end{equation}

The next five relations are easy consequences of the previous three.

\begin{eqnarray}
&&e_ie_{i\pm 1}e_i=e_i; \label{bmwtl}\\

&&(g_i-y)(g_i^2-zg_i-1)=0; \label{cubic}\\

&&ze_i^2=(z+y^{-1} -y)e_i,\quad
\label{esquared}\\

&&-yze_i=g_i^2-zg_i-1; \label{equadg}\\

&&yzg_ie_{i+1}e_i=zg_{i+1}^{-1} e_i. \label{combeg}
\end{eqnarray}

It is easy to show that $BMW_r(y,z)$ may be defined using the
relations (\ref{braidgi}), (\ref{delooping}), (\ref{cubic}) and
(\ref{equadg}) instead of (\ref{braidgi}), (\ref{kauffman}) and
(\ref{delooping}), i.e. that (\ref{kauffman}) is a consequence of
(\ref{cubic}) and (\ref{equadg}).

We shall require a particular specialisation of $BMW_r(y,z)$
to a subring $\CA_q$ of $\CK$, which is defined as follows.
Let $\CS$ be the multiplicative subset of
$\C[q, q^{-1}]$
generated by $[2]_q$, $[3]_q$ and $[3]_q-1$. Let $\CA_q:=\C[q,
q^{-1}]_\CS:=\C[q,q\inv, [2]_q\inv,[3]_q\inv, (q^2+q^{-2})\inv]$
be the localisation of $\C[q, q^{-1}]$ at $\CS$.

Now let $\psi:\C[y^{\pm 1},z]\lr\CA_q$ be the homomorphism
defined by $y\mapsto q^{-4}$, $z\mapsto q^2-q^{-2}$. Then
$\psi$ makes $\CA_q$ into an $\CA_{y,z}$-module, and the
specialisation $BMW_r(q):=\CA_q\otimes_{\CA_{y,z}} BMW_r(y,z)$
is the $\CA_q$-algebra with generators which we denote, by abuse of
notation, $g_i^{\pm 1},e_i$ ($i=1,\dots,r-1$) and relations
(\ref{braidgiq}) below, with the relations (\ref{extrarel}) being
consequences of (\ref{braidgiq}).

\begin{equation}\label{braidgiq}
\begin{aligned}
g_ig_j&=g_jg_i\text{ if }|i-j|\geq 2\\
g_ig_{i+1}g_i&=g_{i+1}g_ig_{i+1} \text{ for }1\leq i\leq r-1\\
g_i-g_i^{-1}&=(q^2-q^{-2})(1-e_i)\text { for all }i\\
g_ie_i&=e_ig_i=q^{-4}e_i\\
e_ig_{i-1}^{\pm 1}e_i&=q^{\pm 4}e_i\\
e_ig_{i+1}^{\pm 1}e_i&=q^{\pm 4}e_i.\\
\end{aligned}
\end{equation}

\begin{equation}\label{extrarel}
\begin{aligned}
e_ie_{i\pm 1}e_i&=e_i  \\
(g_i-q^2)(g_i+q^{-2})&=-q^{-4}(q^2-q^{-2})e_i\\
(g_i-q^{-4})(g_i-q^2)(g_i+q^{-2})&=0\\
e_i^2&=(q^2+1+q^{-2})e_i\\
g_{i+1}^{-1} e_i&=q^{-4}g_ie_{i+1}e_i.\\

\end{aligned}
\end{equation}

We shall be concerned with the following two specialisations
of $BMW_r(q)$.

\begin{definition}\label{spec}  Let $\phi_q:\CA_q\lr \CK=\C(q^{\frac{1}{2}})$ be the
inclusion map, and let $\phi_1:\CA_q\lr\C$ be the $\C$-algebra homomorphism
defined by $q\mapsto 1$. Define the specialisations
$BMW_r(\CK):=\CK\otimes_{\phi_q}BMW_r(q)$, and
$BMW_r(1):=\C\otimes_{\phi_1}BMW_r(q)$.
\end{definition}

The following facts are proved in \cite[Lemma 4.2, Theorem
4.4]{LZ2}.
\begin{prop}\label{bmwfact}
\begin{enumerate}
\item The surjective homomorphism $\eta$ of (\ref{braidhom}) factors through
$BMW_r(\CK)$. That is, there is a surjective homomorphism
$$
\eta_q:BMW_r(\CK)\lr E_r(2,q),
$$
in which the generators $g_i$ are mapped to the $R$ matrices on factors $i,i+1$.
\item The specialisation $BMW_r(1)$ is isomorphic to the Brauer algebra $B_r(3)$.
\end{enumerate}
\end{prop}

\subsection{The quasi-idempotent $\Phi_q$.}

Following \cite{LZ2}, we introduce an element $\Phi_q$ of $BMW_r(q)$,
and state its principal properties. Our objective will be to show that
$\Phi_q\in BMW(\CK)$ generates the kernel of $\eta_q$. We begin with the following
elements of $BMW_r(q)$. Note that we regard $BMW_s(q)\subset BMW_{s+1}(q)$ in the
usual way; it is well known that the algebra $BMW_s(q)$ generated by the $e_i$ and $g_i$
with $1\leq i\leq s-1$ and the relations among those generators is a subalgebra
of $BMW_{s+1}(q)$. In terms of diagrams, this subalgebra is spanned by the
diagrams spanning $BMW_s(q)$, with an additional string joining the rightmost nodes.

Let $f_i= -g_i-(1-q^{-2})e_i +q^2$, and set
\begin{eqnarray} \label{defFq}
F_q= f_1 f_3.
\end{eqnarray}
We also define $e_{1 4}=g_3^{-1} g_1 e_2 g_1^{-1} g_3$ and $e_{1 2 3
4} = e_2 g_1 g_3^{-1} g_2 g_1^{-1} g_3$.

\begin{definition}\label{defn:phiq}
Maintaining the above notation, define the following element of
$BMW_4(q)\subseteq BMW_r(q)$:

\begin{eqnarray}\label{defPhiq}
\Phi_q &=&aF_q e_2 F_q - b F_q - c F_qe_2e_{14}F_q +d F_q e_{1 2 3 4} F_q,
\end{eqnarray}
where
\begin{eqnarray}\label{abcd}
\begin{aligned}
a&=1+(1-q^{-2})^2,\\
b&=1+(1-q^{2})^2+(1-q^{-2})^2,\\
c&= \frac{1 +(2+q^{-2})(1-q^{-2})^2 + (1+q^2)
(1-q^{-2})^4}{([3]_q-1)^2},\\
d&=(q-q^{-1})^2=q^2(a-1).\\
\end{aligned}
\end{eqnarray}
\end{definition}

The principal properties of these elements are summarised in the following
statement, which is \cite[Prop. 7.3]{LZ2}.

\begin{prop}\label{propsPhiq}
The elements $F_q, \Phi_q$ have the following properties:
\begin{enumerate}
\item $F_q^2= (q^2+q^{-2})^2 F_q$. \label{PropsPhiq1}
\item $e_i\Phi_q=\Phi_q e_i=0$ for $i=1,2,3$. \label{PropsPhiq2}
\item $\Phi_q^2 = -(q^2+q^{-2})^2 (1+(1-q^{2})^2+(1-q^{-2})^2 )
\Phi_q$. \label{PropsPhiq3}
\item $\Phi_q$ acts as $0$ on $V_q^{\otimes 4}$. \label{PropsPhiq4}
\end{enumerate}
\end{prop}

\subsection{The classical limit $q\lr 1$; the Brauer algebra.}\label{sect:class}
Let $B_r(\delta)$ be the Brauer algebra over a commutative ring $A$,
with $\delta\in A$; this may be defined as follows. It has generators
$\{s_1,\dots,s_{r-1};e_1,\dots,e_{r-1}\}$, with relations
$s_i^2=1,\; e_i^2=\delta e_i,\; s_ie_i=e_is_i=e_i$ for all $i$,
$s_is_j=s_js_i,\;s_ie_j=e_js_i,\;e_ie_j=e_je_i$ if $|i-j|\geq 2$,
and $s_is_{i+1}s_i=s_{i+1}s_is_{i+1},\; e_ie_{i\pm 1}e_i=e_i$ and
$s_ie_{i+1}e_i=s_{i+1}e_i,\; e_{i+1}e_is_{i+1}=e_{i+1}s_{i}$ for all
applicable $i$. We shall assume the reader is familiar with the
diagrammatic representation of a basis of $B_r(\delta)$, and how
basis elements are multiplied by concatenation of diagrams. In
particular, the group ring $A\Sym_r$ is the subalgebra of
$B_r(\delta)$ spanned by the diagrams with $r$ ``through strings'',
and the algebra contains elements $w\in\Sym_r$ which are appropriate
products of the $s_i$.

In this work we shall take $A=\C$ and $\delta=3$; the corresponding
Brauer algebra will be denoted by $B_r(3)$. In the identification
given by Proposition \ref{bmwfact}(2) of the specialisation
$BMW_r(q)\otimes_{\phi_1}\C$ with $B_r(3)$, $g_i\otimes 1$ corresponds
to $s_i$ and $e_i\otimes 1$ corresponds to $e_i\in B_r(3)$.
Accordingly, $F_q$ and $\Phi_q$ specialise respectively to
$F=(1-s_1)(1-s_3)$ and $\Phi=Fe_2F-F-\frac{1}{4}Fe_2e_{1,4}F$,
where $e_{1,4}=s_1s_3e_2s_3s_1$. From the relations in
Proposition \ref{propsPhiq} we see that $F^2=4F$, and $\Phi^2=-4\Phi$.

Let $V_1$ be the irreducible $U(\fsl_2)$-module. The following statement
may be found in \cite[\S 6]{LZ2}.

\begin{prop}\label{braueractn}
\begin{enumerate}
\item There is a surjective homomorphism
\be\label{classeta}
\eta:B_r(3)\lr\End_{U(\fsl_2)}(V_1^{\otimes r}):=E(r).
\ee
\item The kernel of $\eta$ contains $\Phi$.
\end{enumerate}
\end{prop}

\subsection{Statements}\label{statements} We are now able to state our main results.

\begin{thm}\label{mainq}
Let $\cU_q$ be the quantised enveloping algebra of $\fsl_2$ over the
field $\CK=\C(q^{\frac{1}{2}})$. Let $V_q$ be the irreducible
three-dimensional $\cU_q$-module (with highest weight $2$), and let
$\eta_q:BMW_r(\CK)\lr E_r(2,q):=\End_{\cU_q}(V_q^{\otimes r})$ be
the surjective homomorphism defined in Proposition \ref{bmwfact}
(1). Then the kernel of $\eta_q$ is the two sided ideal of
$BMW_r(\CK)$ generated by the quasi-idempotent $\Phi_q$, defined in
Definition \ref{defn:phiq}.

In particular, $E_r(2,q)$ has a presentation given as follows. $E_r(2,q)$ has
generators $g_i,e_i$, $i=1,\dots,r-1$, with relations (\ref{braidgiq}), plus
the relation $\Phi_q=0$.
\end{thm}

The classical ($q=1$) analogue of the above statement is as follows.

\begin{thm}\label{mainclass}
Let $U$ be the universal enveloping algebra of $\fsl_2(\C)$, and let $V_1$
be the irreducible $U$-module of dimension three. If $B_r(3)$ is the complex
$r$-string Brauer algebra with parameter $3$ (\S \ref{sect:class}) and
$\eta:B_r(3)\lr \End_{U(\fsl_2)}(V_1^{\otimes r}):=E(r)$ be the surjective homomorphism
of Prop. \ref{braueractn}. Then the kernel of $\eta$ is the two sided ideal of
$B_r(3)$ which is generated by the quasi-idempotent $\Phi$ of
\S \ref{sect:class}.
\end{thm}

It should be noted that both $BMW_r(\CK)$ and $B_r(3)$ are non-semisimple
if $r\geq 5$. This is part of the significance of these results.

We shall prove both theorems together, making use of the following fact.

\begin{prop}\label{redn:class}
Theorem \ref{mainq} is a consequence of Theorem \ref{mainclass}.
\end{prop}
\begin{proof}[Sketch of Proof] This is Corollary 7.12~(1) of \cite{LZ2}.
For the convenience of the reader, we give a brief sketch of the proof here.

The algebra $BMW_r(q)$ has a well known cellular structure
(\cite{X}), with the cells being indexed by the partially ordered
set \be\label{lambda-r} \Lambda_r:=\{{\text {partitions}}\;
\lambda\mid |\lambda|=t,\;0\leq t\leq r,\; t \equiv r (\mod 2)\},
\ee where for any partition
$\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq \lambda_p>0)$, we
write $|\lambda|=\sum_i\lambda_i$. The partial order on $\Lambda_r$
is given by $\mu>\lambda$ if $|\mu|>|\lambda|$ or
$|\mu|=|\lambda|=t$ and $\mu>\lambda$ in the dominance order on
partitions of $t$. This cellular structure is inherited by the
specialisations $BMW_r(\CK)$ and $B_r(3)$ (cf. \cite{GL96}), and by
Prop. 7.1 of \cite{LZ2}. The cell modules, their radicals, and the
irreducible modules for any specialisation of $BMW_r(q)$ are
obtained by specialising the respective $BMW_r(q)$-modules.

Using this, one shows, using the general criteria established in
\cite[\S 5]{LZ2} that if $J_q$ is the ideal of $BMW_r(q)$ generated
by $\Phi_q$, and $J_\CK$ and $J$ are its respective specialisations
to $BMW_r(\CK)$ and $B_r(3)$, then $J_\CK$ contains the radical of
$BMW_r(\CK)$ if and only if $J$ contains the radical of $B_r(3)$.
But the latter are precisely the respective conditions for
$J_\CK$ (resp. $J$) to be the whole kernel of $\eta_q$ (resp. $\eta$),
whence the assertion.
\end{proof}

We therefore turn to the proof of Theorem \ref{mainclass}, which will require
detail concerning the cellular structure of the Brauer algebras. Before discussing
the details in \S\ref{sect:brauer}, we shall recall some basic facts concerning
cellular algebras.

\section{Cellular algebras and their radicals.} In this section we summarise the
main properties of cellular algebras of which we make use below. The
principal references are \cite{GL03,GL04} and \cite[\S 5]{LZ2}.

\subsection{Cellular algebras and cell modules}\label{ca:basic}
Let $\bF$ be any commutative ring,
which later will be taken to be a field. The $\bF$-algebra $B$ is {\it cellular} if
it has a {\it cell datum} $(\Lambda,M,C,{\,}^*)$ with the following properties
\begin{itemize}
\item $\Lambda$ is a finite
partially ordered set
\item $M:\Lambda\to\text{Sets}$ is a function which
associates a finite set $M(\lambda)$ to each $\lambda\in\Lambda$
\item $C$ is a function
$$
C:\amalg_{\lambda\in\Lambda}M(\lambda)\times M(\lambda)\lr B,
$$
whose image is an $\bF$-basis of $B$; for $S,T\in M(\lambda)$,
write $C(S,T):=C^\lambda_{S,T}$.
\item For any element $a\in B$ and $S,S'\in M(\lambda)$ there are scalars
$r_a(S',S)\in\bF$ such that for $S,T\in M(\lambda)$, we have
\be\label{cell-basic}
aC^\lambda_{S,T}\equiv\sum_{S'\in M(\Lambda)}r_a(S',S)C^\lambda_{S',T}\;\;
\mod B(<\lambda),
\ee
where $B(<\lambda)$ is the two-sided ideal of $B$ spanned by
$\{C^\mu_{U,V}\mid \mu<\lambda,\;\;U,V\in M(\mu)\}$.
\item The linear map defined by
${\,}^*:C^\lambda_{S,T}\mapsto {C^\lambda_{S,T}}^*:=C^\lambda_{T,S}$ is
an involutory anti-automorphism of $B$.
\end{itemize}

The {\it cell module} $W(\lambda)$ ($\lambda\in\Lambda$) is the free
$\bF$-module spanned by symbols $C(S)$ ($S\in M(\lambda)$), with the
$B$-action defined by (\ref{cell-basic}), i.e.

\[
a\cdot C(S)=\sum_{S'\in M(\lambda)}r_a(S',S)C(S'), \quad \text{for
$a\in B$}.
\]

Define the bilinear form $\phi_\lambda$ on the basis $\{C(S)\}$ of
$W(\lambda)$ by
$$
{C^\lambda_{S,T}}^2=\phi_\lambda(C(S),C(T))C^\lambda_{S,T}\;\;\mod B(<\lambda).
$$
The principal properties of $\phi_\lambda$ are:
\begin{itemize}
\item $\phi_\lambda$ is well-defined and is symmetric.
\item $\phi_\lambda$ is invariant, in the sense that for $u,v\in W(\lambda)$ and $a\in B$,
we have $\phi_\lambda(a\cdot u,v)=\phi_\lambda(u,a^*\cdot v)$.
\item It follows that $\Rad(\lambda):=\{w\in W(\lambda)\mid \phi_\lambda(w,v)=0\text{ for all }
v\in W(\lambda)\}$ is a submodule of $W(\lambda)$.
\item If $\bF$ is a field, then for any element $w\in W(\lambda)\setminus\Rad(\lambda)$,
we have $W(\lambda)=B\cdot w$. Hence $\Rad(\lambda)$ is the unique maximal submodule
of $W(\lambda)$.
\end{itemize}

The last statement above arises in the following context. In addition
to its structure as a left $B$-module defined above, the $\bF$-module
$W(\lambda)$ also has a right $B$-module structure, where the action
is as above, composed with ${\,}^*$. We define the
$\bF$-module monomorphism $C^\lambda$ by

\be\label{def:clambda}
\begin{aligned}
C^\lambda:&W(\lambda)\otimes_\bF W(\lambda)\lr B\\
&C(S)\otimes C(T)\mapsto C^\lambda_{S,T}.\\
\end{aligned}
\ee

The properties of this map are summarised as follows.
\begin{itemize}\label{props:clambda}
\item There is an $\bF$-module isomorphism
$B\lr\oplus_{\lambda\in\Lambda} \im (C^\lambda)$.
\item If we compose $C^\lambda$ with the
natural map $B\to B/B(<\lambda)$, where $B(<\lambda)$ is the
two-sided ideal defined above, we obtain a monomorphism of
$(B,B)$ bimodules, whose image is, in the obvious notation,
$B(\leq\lambda)/B(<\lambda)$.
\item For $u,v$ and $w$ in $W(\lambda)$, we have
$$
C^\lambda(u\otimes v)\cdot w=\phi_\lambda(v,w)u.
$$
\end{itemize}

If $\bF$ is a field, the cyclic nature of $W(\lambda)$ if $\phi_\lambda
\neq 0$ follows from the last statement.

\subsection{Simple modules and composition factors}

It follows from \S\ref{ca:basic} that for each $\lambda\in\Lambda$,
either $\phi_\lambda=0$ or $L(\lambda):=W(\lambda)/\Rad(\lambda)$
is a simple $B$-module. One of the keys to understanding these modules,
and more generally the composition factors of the cell modules $W(\lambda)$,
is the following fact.

\begin{prop}\label{homs} Let $\lambda,\mu\in\Lambda$, and
suppose we have a non-zero homomorphism $\psi:W(\mu)\to W(\lambda)/M$,
where $M$ is some submodule of $W(\lambda)$. Then
\begin{enumerate}
\item $\mu\geq\lambda$.
\item If $\mu=\lambda$ and $\bF$ is a field then $\psi$ is
realised as multiplication by a scalar.
\end{enumerate}
\end{prop}

Although neither is essential, we shall now make two simplifying
assumptions. First, we assume that {\it  henceforth $\bF$ is a field},
and secondly, that for each $\lambda\in\Lambda$, $\phi_\lambda\neq 0$.
This says that $B$ is quasi-hereditary. The algebras to which we shall apply
the theory satisfy both these assumptions, but it is possible to develop
the theory without them.

Given the assumptions, we have the following summary of the main points
concerning the representation theory of $B$.

\begin{itemize}
\item The modules $L(\lambda)=W(\lambda)/\Rad(\lambda)$ ($\lambda\in\Lambda$)
form a complete set of non-isomorphic simple $B$-modules.

\item If $L(\mu)$ is a composition factor of $W(\lambda)$, then $\mu\geq\lambda$.
\item The following statements are equivalent: (a) $B$ is semisimple;
(b) Each cell module $W(\lambda)$ is simple; (c) For each
$\mu\neq\lambda$, $\Hom_B(W(\mu),W(\lambda))=0$.

\end{itemize}

\subsection{The radical} Suppose $B$ is a cellular algebra over the
field $\bF$ as above, and denote the radical of $B$ by $\CR$. This is a
two sided ideal of $B$ which may be defined by the property that $B/\CR$
is semisimple, and any homomorphism $B\lr A$, where $A$ is a semisimple
$\bF$-algebra, factors through $B/\CR$. In particular, we have an exact
sequence of $\bF$-algebras
$$
0\lr \CR\lr B\overset{\pi}{\lr}
\oplus_{\lambda\in\Lambda}\End_\bF(L(\lambda))\lr 0,
$$
so that $\CR$ may be characterised as the set of elements of $B$ which act
trivially on each of the irreducible $B$-modules. Note that in the exact sequence
above, for each $\lambda\in\Lambda$, the subspace $\im (C^\lambda)$ of $B$ is mapped
by $\pi$ (surjectively) onto $\End_\bF(L(\lambda))$.

Now the key point of the proof of Theorem \ref{mainq} is to prove that a certain
two-sided ideal of $B$ contains $\CR$. With this in mind we state the
next result, which may be found in \cite[Lemma 5.3]{LZ2}.

\begin{prop}\label{prop:rad}
The radical $\CR$ has a filtration
($\CR(\lambda):=\CR\cap B(\leq \lambda)_{\lambda\in\Lambda}$)
with the property that there is a $(B,B)$ bimodule isomorphism
$$
\CR(\lambda)/(\CR\cap B(<\lambda))\lr W(\lambda)\otimes_\bF \Rad(\lambda)
+\Rad(\lambda)\otimes_\bF W(\lambda)\subseteq W(\lambda)\otimes_\bF W(\lambda).
$$
\end{prop}

This leads to the next result, which
is easily deduced from \cite[Theorem 5.4]{LZ2}, and which is
crucial in the proof of our main theorem.

\begin{thm}\label{thm:rad}
Maintaining the notation above, let $J$ be a two-sided ideal of $B$
which satisfies $J^*=J$. Define complementary subsets
$\Lambda^0$ and $\Lambda^1$ of $\Lambda$ by
$\Lambda^0:=\{\lambda\in\Lambda\mid J\cdot L(\lambda)=0\}$ and
$\lambda^1:=\Lambda\setminus\Lambda^0$. Then $J\supseteq \CR$ if
and only if, for each $\lambda\in\Lambda^0$, all composition factors of
the cell module $W(\lambda)$ other than $L(\lambda)$ are isomorphic to some
$L(\mu)$ for $\mu\in\Lambda^1$.
\end{thm}
\begin{proof}
It is a straightforward consequence of Proposition \ref{prop:rad} above and
the general characterisation of $\CR$ as the set of elements of $B$ which
annihilate each $L(\lambda)$, that $J\supseteq\CR$ if and only if,
for each $\lambda\in\Lambda$, $J\cdot W(\lambda)\supseteq \Rad(\lambda)$.
But for $\lambda\in\Lambda^1$, since $J\cdot W(\lambda)\not\subseteq \Rad(\lambda)$,
it follows from the cyclic nature of $W(\lambda)$ (see above) that
$J\cdot W(\lambda)=W(\lambda)\supset \Rad(\lambda)$.

It is therefore evident (cf. \cite[Theorem 5.4(3)]{LZ2}),
that $J\supseteq \CR$ if and only if
for each $\lambda\in\Lambda^0$, $J\cdot W(\lambda)\supseteq \Rad(\lambda)$.
We shall show that the condition in our statement is equivalent to this.

Take $\lambda\in\Lambda^0$. Let
$$
W(\lambda)=W_0\supset W_1=\Rad(\lambda)\supset W_2\supset\dots\supset W_s=0
$$
be a composition series of $W(\lambda)$, and write $L_i$ for the simple quotient
$L_i:=W_{i-1}/W_i$ ($i=1,\dots,s$). Then $L_1\cong L(\lambda)$ and
since the $L(\mu)$ ($\mu\in\Lambda$) form a complete set of isomorphism classes
of simple $B$-modules, it follows from Proposition \ref{homs} that for $i>1$,
$L_i\cong L(\mu_i)$ for some $\mu_i\gneq\lambda$. Applying $J$, we obtain a sequence
$$
J\cdot W(\lambda)=J\cdot W_0\supseteq J\cdot W_1=J\cdot \Rad(\lambda)
\supseteq J\cdot W_2\supseteq \dots\supseteq J\cdot W_s=0,
$$
and the quotient $J\cdot W_{i-1}/J\cdot W_i\cong J\cdot L(\mu_i)$.

It is therefore clear that $J\cdot W(\lambda)\supseteq\Rad(\lambda)$ if and only
if, for each $i>1$, $J\cdot L(\mu_i)\neq 0$, i.e. $\mu_i\in\Lambda^1$.
\end{proof}
\section{On the representation theory of the Brauer algebras.}\label{sect:brauer}

We shall require some detailed analysis of the cell modules
of the Brauer algebras and their radicals. Most of the material in this section
is adapted from \cite{GL96,DHW,HW,LZ2}. Since there is no difference
in the arguments, for the purposes of the current section, $\CK$
could be any field of characteristic zero, and we shall consider
the Brauer algebras $B_r(\delta)$,
where $\delta$ is any non-zero element of $\CK$. This restriction is not
essential, but enables us to simplify the exposition; moreover our application
is to this situation ($\delta=3$). We shall write $B_r$ for $B_r(\delta)$
for this section only.

Notation will be similar to that in \cite{LZ2}. For $\lambda\in\Lambda_r$,
we have a cell module $W_r(\lambda)$ which has a canonical symmetric
bilinear invariant form $(-,-)_\lambda$, whose radical $\Rad_r(\lambda)$
is the unique maximal submodule of $W_r(\lambda)$. In our situation
($\delta\neq 0$) we have $(-,-)_\lambda\neq 0$ for all $\lambda\in\Lambda_r$
(the ``quasi-hereditary'' case). The irreducible quotients $L_r(\lambda):=
W_r(\lambda)/\Rad_r(\lambda)$ are pairwise non-isomorphic, and provide
a complete set of irreducible $B_r$-modules.

For the proof of the next Lemma, we shall require the following
explicit description of the cell modules. Recall that for any $t$,
the group algebra $\C\Sym_t$ is a subalgebra of $B_t$, realised as
the span of the diagrams with $t$ through strings (or generated by
the $s_i$). Let $\lambda\in\Lambda_r$ with $|\lambda|=t$, and
$r=t+2k$. Since $B_r$ is a $(B_r,B_r)$ bimodule, the subspace
$\tilde {I_r^t}:=B_re_{t+1}e_{t+3}\dots e_{t+2k-1}$ is a
$(B_r,B_t)$-bimodule, and hence {\it a foriori} a
$(B_r,\C\Sym_t)$-bimodule. Clearly the subspace ${\tilde {I_r^t}}'$
spanned by diagrams with at most $t-2$ through strings is a
sub-bimodule, to be interpreted as $0$ if $t<2$. Define $I_r^t$ as the quotient
$(B_r,\C\Sym_t)$-bimodule ${\tilde {I_r^t}}/{\tilde {I_r^t}}'$.

It is easy to see from the description
in \cite[Def. (4.9)]{GL96} (cf. also \cite{DHW}) that
\be\label{defn:w}
W_r(\lambda)\cong I_r^t\otimes_{\C\Sym_t}S(\lambda),
\ee
where $S(\lambda)$ is the cell (or Specht) module for $\C\Sym_t$ which corresponds
to the partition $\lambda$ of $t$.

\begin{lem}\label{strictineq}
Let $\mu_1,\mu_2\in\Lambda_r$. If $L_r(\mu_1)$ is a composition
factor of $W_r(\mu_2)$, then $|\mu_1|>|\mu_2|$ or $\mu_1=\mu_2$.
\end{lem}
\begin{proof}
Suppose $\phi:W_r(\mu_1)\lr W_r(\mu_2)/M$ is a $B_r$-homomorphism,
where $\mu_1,\mu_2\in\Lambda_r$ and $|\mu_1|=|\mu_2|=t$. We shall
show that if $\mu_1\neq \mu_2$, $\phi=0$.

Consider the subspaces $W_r^0(\mu_i)$ of $W_r(\mu_i)$ ($i=1,2$) defined by
$$W_r^0(\mu_i)=e_{t+1}e_{t+3}\dots e_{r-1}W_r(\mu_i).$$ Since
$e_{t+1}e_{t+3}\dots e_{r-1}\in B_r^t$ and for
$w\in e_{t+1}e_{t+3}\dots e_{r-1}W_r(\mu_i)$ we have
$$e_{t+1}e_{t+3}\dots e_{r-1}w=\delta^kw \;(r=t+2k),$$
it follows
that $W_r^0(\mu_i)$ generates $W_r(\mu_i)$ as $B_r$-module.
Moreover it is evident that $W_r^0(\mu_i)$ is stable under
$\C\Sym_t\subset B_t\subseteq B_r$, and is isomorphic as
$\C\Sym_t$-module to $S(\mu_i)$.

Since $\phi$ respects the $B_r$-action, we have
$\phi(W_r^0(\mu_1))\subseteq e_{t+1}e_{t+3}\dots e_{r-1}(W_r(\mu_2)/M)$.
It follows that the restriction to $W_r^0(\mu_1)$ of $\phi$ defines a
$\C\Sym_t$-homomorphism $\phi^0:S(\mu_1)\to S(\mu_2)$. But since $\CK$ has
characteristic zero, the cell modules $S(\nu)$ of $\C\Sym_t$ are simple,
whence if $\mu_1\neq \mu_2$, $\phi^0=0$. It follows that $\phi=0$ on
a generating subspace of $W_r(\mu_1)$, whence $\phi=0$.
\end{proof}

Note that Lemma \ref{strictineq} is false if the restriction on
the characteristic of $\CK$ is lifted, since then there may be
non-trivial homomorphisms between the cell (Specht) modules $S(\lambda)$
for $\Sym_t$.

This leads to the following alternative characterisation of the
radical $\Rad_r(\lambda)$ of a cell module of $B_r$.

\begin{cor}\label{char-rad}
Let $\lambda\in\Lambda_r$, with $|\lambda|=t$. For any integer
$m$, let $B_r^m$ be the ideal of $B_r$ spanned by diagrams with
at most $m$ through strings. Then
$$
\Rad_r(\lambda)=\{w\in W_r(\lambda)\mid bw=0\text{ for all }b\in B_r^t\}
=\Ann_{W_r(\lambda)}(B_r^t).
$$
\end{cor}
\begin{proof}
By the previous Lemma, any composition factor of $W_r(\lambda)$ other than $L_r(\lambda)$
is of the form $L_r(\mu)$ with $|\mu|>t$. In particular, any composition factor of
$\Rad_r(\lambda)$ is of this form. But if $|\mu|>t$, then $B_r^t$ acts as zero
on $W_r(\mu)$, since the number of through strings of any diagram $bD$ ($b\in B^t_r$)
has at most $t(>r-|\mu|)$ through strings. It follows that $B_r^tL_r(\mu)=0$.
Hence $B_r^t$ acts as zero on each composition factor of $\Rad_r(\lambda)$, and
hence on $\Rad_r(\lambda)$. So $\Rad_r(\lambda)\subseteq\Ann_{W_r(\lambda)}B^t_r$.

But $\Rad_r(\lambda)$ the unique maximal submodule of $W_r(\lambda)$; in fact
any element of $W_r(\lambda)$ which is not in $\Rad_r(\lambda)$ generates $W_r(\lambda)$
(\cite[Prop. 2.5]{GL96}). Hence we have equality.
\end{proof}

Next, we require the following strengthening of \cite[Thm. 5.4]{DHW}. In the
statement, we use the fact that for any positive integer $r$, we have
$\Lambda_r\subset\Lambda_{r+2}$.

\begin{lem}\label{composn-trans}
Let $\mu,\lambda\in\Lambda_r$. Then $L_r(\mu)$ is a composition factor of
$W_r(\lambda)$ if and only if $L_{r+2}(\mu)$ is a composition factor
of $W_{r+2}(\lambda)$
\end{lem}
\begin{proof} The proof will use the functors $F,G$ introduced in \cite{G}
in the context of Schur algebras, and used in the current context in
\cite{DHW,M}. They are defined as follows. Let $B_r-mod$ denote the category
of left $B_r$-modules. Recalling that $B_{r-2}\subset B_r\subset B_{r+2}$
as described above, define $F:B_r-mod\lr B_{r-2}-mod$ and
$G:B_r-mod\lr B_{r+2}-mod$ by
\be\label{defn:F,G}
\begin{aligned}
&F(M):=e_{r-1}M\\
&G(M):=B_{r+2}e_{r+1}\otimes_{B_r}M,\\
\end{aligned}
\ee
for any $B_r$-module $M$.
The main properties of $F,G$ are that $FG=\id$, $F$ is exact, and $G$
is right exact (cf. \cite[Cor 4.4, p. 143]{HWu}).

To prove the lemma, first note that $L_r(\mu)$ is a composition factor of
$W_r(\lambda)$ if and only if there is a non-trivial homomorphism
$\xi:W_r(\mu)\to W_r(\lambda)/N$ for some submodule $N$ of $W_r(\lambda)$.
Applying the functor $G$, we obtain $G(\xi):G(W_r(\mu))\to G(W_r(\lambda)/N)$,
and by \cite[Lemma 5.1]{DHW}, $G(\xi)\neq 0$. Moreover by [{\it loc.~cit.},
Prop. 5.2], $G(W_r(\mu))=W_{r+2}(\mu)$. To prove that $L_{r+2}(\mu)$
is a composition factor of $W_{r+2}(\lambda)$, it will therefore suffice to
show that $G(W_r(\lambda)/N)\simeq W_{r+2}(\lambda)/M$, for some submodule
$M$ of $W_{r+2}(\lambda)$. But the exact sequence
$W_r(\lambda)\to W_r(\lambda)/N\to 0$ is taken by the
(right exact) functor $G$
to an exact sequence of $B_{r+2}$-modules, whence $G(W_r(\lambda)/N)$
is isomorphic to a quotient of $G(W_r(\lambda))=W_{r+2}(\lambda)$. Hence
if $L_r(\mu)$ is a composition factor of
$W_r(\lambda)$, then $L_{r+2}(\mu)$ is a composition factor of
$W_{r+2}(\lambda)$.

Conversely, if $L_{r+2}(\mu)$ is a composition factor of
$W_{r+2}(\lambda)$, there is a non-trivial homomorphism
$\psi:W_{r+2}(\mu)\to W_{r+2}(\lambda)/M$, for some submodule
$M$ of $W_{r+2}(\lambda)$. Applying the functor $F$, we obtain
a homomorphism
$$F(\psi):F(W_{r+2}(\mu))\to F(W_{r+2}(\lambda)/M).$$
But by the exact nature of $F$, we have $F(W_{r+2}(\lambda)/M)\cong
F(W_{r+2}(\lambda))/F(M)$, and since $F(W_{r+2}(\nu))=W_r(\nu)$
for $\nu\in\Lambda_r$ \cite[Prop. 5.3]{DHW}, we obtain
$F(\psi):W_r(\mu)\to W_r(\lambda)/F(M)$. It remains only to prove that
$F(\psi)\neq 0$. For this, consider the exact sequence
$$
0\lr \Ker \psi\lr W_{r+2}(\mu)\overset{\psi}{\lr}\im \psi\lr 0.
$$
Applying $F$, we obtain an exact sequence
$$
0\lr F(\Ker \psi)\lr W_{r}(\mu)\overset{F(\psi)}{\lr} F(\im \psi)\lr 0,
$$
and hence it suffices to show that $F(\Ker \psi)\neq W_r(\mu)$.
But $\Ker \psi\subseteq\Rad_{r+2}(\mu)=\Ann_{W_{r+2}(\mu)}(B_{r+2}^s)$,
where $|\mu|=s$ (by Cor \ref{char-rad}). It follows that
$e_{s+1}e_{s+3}\dots e_{r+1}\Ker \psi=0=e_{s+1}e_{s+3}\dots e_{r-1}F(\Ker \psi)$.
Moreover since $e_{s+1}e_{s+3}\dots e_{r-1}$ generates the two-sided ideal
$B_r^s$ of $B_r$, it follows that $F(\Ker\Psi)\subseteq\Ann_{W_r(\mu)}(B_r^s)
=\Rad_r(\mu)$.
\end{proof}

\begin{cor}\label{cor:shift}(cf. \cite[Thm. 5.4]{DHW})
Let $\lambda,\mu\in\Lambda_r$, with $|\mu|=s$. Then $L_r(\mu)$ is a
composition factor of $W_r(\lambda)$ if and only if $L_s(\mu)$ is a
composition factor of $W_s(\lambda)$.
\end{cor}
This is immediate from repeated application of Lemma \ref{composn-trans}.

We shall require a combinatorial consequence of Corollary
\ref{cor:shift} which is proved in \cite{DHW}. The statement
involves the following details concerning partitions. Associated to
a partition $\lambda:=(\lambda_1\geq\lambda_2\geq \dots\lambda_p>0)$
we have a ``Young diagram'' which is the set of plane lattice points
$Y(\lambda):=\{(i,j)\in\Z_{\geq 0}\times\Z_{\geq 0}\mid 1\leq j\leq
\lambda_i\}$. For any point $P=(i,j)\in Y(\lambda)$, define the {\it
content} $c(P)$ of $P$ by $c(P)=j-i$. Write $c(\lambda)=\sum_{P\in
Y(\lambda)}c(P)$.

\begin{prop}\label{dhwcrit}(\cite[Thms. 3.1 and 3.3]{DHW})
Let $\mu,\lambda\in\Lambda_r$ with $|\mu|=s$. If $L_s(\mu)$
is a composition factor of $W_s(\lambda)$, then
\begin{enumerate}
\item $Y(\lambda)\subseteq Y(\mu)$ (as sets), and
\item $|\mu|-|\lambda|+\sum_{P\in Y(\mu)\setminus Y(\lambda)}c(P)=0$.
\end{enumerate}
\end{prop}

\begin{cor}\label{composcrit}
Let $\mu,\lambda\in\Lambda_r$ with $|\mu|=s$. If $L_r(\mu)$
is a composition factor of $W_r(\lambda)$, then the
conditions (1) and (2) of Proposition \ref{dhwcrit}
hold.
\end{cor}

This is immediate from Corollary \ref{cor:shift}

\section{Proof of the main theorems.}

Recall that we have the surjective homomorphism (\ref{classeta})
$$
\eta:B_r(3)\lr E(r),
$$
where $E(r)$ is the semisimple $\C$-algebra $\End_{U(\fsl_2)}(V_1^{\otimes r})$.
Write $N=\Ker \eta$; then $\Phi\in N$, so that if $J$ is the two sided ideal
of $B_r(3)$ generated by $\Phi$, we have $J\subseteq N$.
Define the subsets $\Lambda_r^0,\Lambda_r^1$ of $\Lambda_r$ by
$\Lambda_r^0=\{(t), (t-1,1),1^3\mid 0\leq t\leq r;\;\;
t\equiv r(\text{mod $2$})\}$, and $\Lambda_r^1:=\Lambda\setminus\Lambda^0$.

\begin{prop}\label{ann-lambdas}
The following are equivalent conditions on an element $\lambda\in\Lambda_r$.
\begin{enumerate}
\item $\lambda\in\Lambda_r^0$.
\item $NL_r(\lambda)=0$.
\item $JL_r(\lambda)=0$.
\end{enumerate}
\end{prop}
\begin{proof}[Sketch of Proof.]
This statement is contained in \cite[Thm. 6.8]{LZ2}; we give a brief sketch
of the argument for the reader's convenience. One first shows that if $\lambda
\notin\Lambda_r^0$ (i.e. $\lambda\in\Lambda_r^1$) then $\Phi L_r(\lambda)\neq 0$;
this involves explicit computation, and a knowledge of the cases $r=4,5$.
It follows that $N$ acts non-trivially on $L_r(\lambda)$ for $\lambda\in\Lambda_r^1$,
since $\Phi\in N$.

But the semisimple algebra $E(r)$ has $r+1$ simple components, whence $N$ acts as
zero on precisely $r+1$ of the  modules $L_r(\lambda)$. Since $|\Lambda_r^0|=r+1$,
it follows that $N$ acts trivially on $L_r(\lambda)$ for $\lambda\in \Lambda_r^0$.
The statement is now clear.
\end{proof}

\begin{cor}\label{redn:rad}
Let $\CR$ be the radical of $B_r$. Then $N=J$ if and only if $J$ contains $\CR$.
\end{cor}
\begin{proof} We have $E(r)\cong B_r/N$, and
by Prop. \ref{ann-lambdas}, $B_r/N$ and $B_r/J$ have the same maximal semisimple
quotient (viz. $E(r)$). Hence $N=J+\CR$.
\end{proof}

\begin{proof}[Proof of Theorem \ref{mainq}.]

It remains only to show that $J\supseteq \CR$. Since the element $\Phi$ satisfies
$\Phi^*=\Phi$, we may apply Theorem \ref{thm:rad} above, which implies that it
suffices to show that for $\lambda\in\Lambda_r^0$, $JW_r(\lambda)=\Rad_r(\lambda)$.
Note that by Prop. \ref{ann-lambdas}, we have for $\lambda\in\Lambda_r^0$,
$JL_r(\lambda)=0$, whence $JW_r(\lambda)\subseteq\Rad_r(\lambda)$. We shall
prove that we have equality by showing, for $\lambda\in\Lambda_r^0$,
\be\label{crux}
{\text{If $L_r(\mu)$ is a composition factor of $W_r(\lambda)$ and
$\mu\neq\lambda$, then }}\mu\in\Lambda_r^1.
\ee

Given (\ref{crux}), it follows that for $\lambda\in\Lambda_r^0$, all composition
factors $L_r(\mu)$ of $\Rad_r(\lambda)$ satisfy $\mu\in\Lambda_r^1$, whence
$JL_r(\mu)=L_r(\mu)$. It follows that $J\Rad_r(\lambda)=\Rad_r(\lambda)$, whence by
Theorem \ref{thm:rad}, $J\supseteq \CR$.

It therefore remains only to prove (\ref{crux}). We do this by invoking the
criteria in Prop. \ref{dhwcrit} for $L_r(\mu)$ to be a composition factor
of $W_r(\lambda)$. There are three cases.

First, if $\lambda=(t)$ and $\mu\in\Lambda_r^0$ with $Y(\mu)\supset
Y(\lambda)$, then there are three possibilities for $\mu$: (a)
$\mu=(s)$, $s>t$; (b) $\mu=(s-1,1)$, $s>t$; (c) $t=1$ and
$\mu=(1^3)$. In cases (a) and (b), $|\mu|-|\lambda|+\sum_{P\in
Y(\mu)\setminus Y(\lambda)}c(P)>0$, while in case (c),
$|\mu|-|\lambda|+\sum_{P\in Y(\mu)\setminus Y(\lambda)}c(P)=-1\neq
0$.

Secondly, if $\lambda=(t-1,1)$, the only possibility for
$\mu\in\Lambda_r^1$ with $Y(\mu)\supset Y(\lambda)$ is $\mu=(s-1,1)$
with $s>t$. In this case, again $|\mu|-|\lambda|+\sum_{P\in
Y(\mu)\setminus Y(\lambda)}c(P)>0$.

Finally, if $\lambda=(1^3)$, there is no $\mu\in\Lambda_r^0$ with $Y(\mu)\supset
Y(\lambda)$.

This proves Theorem \ref{mainclass}, and hence by Prop. \ref{redn:class}
completes the proof of Theorem \ref{mainq}.
\end{proof}

\section{An integral form for the endomorphism algebra.}
Let $\CA$ be the ring $\C[q^{\pm 1}]$,

Lusztig \cite{L1,L2}
has defined an $\CA$-form $\cU_\CA$ of the quantised
universal enveloping algebra $\cU_q(\fg)$ of a Lie algebra $\fg$, which involves the
divided powers of the generators $e_i,f_i$ of $\cU_q$. He also showed
how to construct $\CA$-forms of higest weight modules for $\cU_\CA$
by applying these divided powers to highest weight vectors of the
corresponding $\cU_q$-modules. Define $\phi_1:\CA\to\C$
by $\phi_1(q)=1$ and let $\phi_q:\CA\to\CK$ be the inclusion map.

We shall define an $\CA$-form of the
$\CK$-algebra $E_r(2,q)\cong BMW_r(\CK)/\langle \Phi_q\rangle$
which specialises to the endomorphism algebras $E_r(2)$ and $E_r(2,q)$
respectively when we map $\CA$ to $\C$ and to $\CK$ via $\phi_1$ and $\phi_q$
respectively.

The definition is as follows. Let $BMW_r(\CA)$ be the $\CA$-algebra generated
by the set $\{g_1^{\pm 1},\dots g_{r-1}^{\pm 1};e_1,\dots,e_{r-1}\}$
subject to the relations
(\ref{braidgiq}). It is explained in \cite[p.285]{X} that $BMW_r(\CA)$ is
free as an $\CA$-module, with basis a set of ``tangle diagrams''.

Let $a,b,c$ and $d$ be the elements of $\CK$ defined in (\ref{abcd}),
and write $\tilde a=(q^2+q^{-2})a$, $\tilde b=(q^2+q^{-2})b$,
$\tilde c=(q^2+q^{-2})c$ and $\tilde d=(q^2+q^{-2})d$.
Then $\tilde a,\tilde b,\tilde c,\tilde d\in\CA$,
and we define
$\tilde \Phi_q$ as
$\tilde \Phi_q=(q^2+q^{-2})\Phi_q$, i.e. (cf. (\ref{defPhiq})
$$
\tilde\Phi_q =\tilde aF_q e_2 F_q - \tilde b F_q -
\tilde c F_qe_2e_{14}F_q +\tilde d F_q e_{1 2 3 4} F_q.
$$

Then $\tilde\Phi_q\in BMW_r(\CA)$.

\begin{definition}\label{def:er} Define the $\CA$-algebra $\CE_r(\CA)$ by
$\CE_r(\CA):=BMW_r(\CA)/\langle \tilde\Phi_q \rangle$.
\end{definition}

\begin{prop}\label{specend}
Let $\CE_r(\CA)$ be the $\CA$-algebra of (\ref{def:er}), and let $\phi_1:\CA\to\C$
and $\phi_q:\CA\to\CK$ be the homomorphisms defined above, which make $\C$ and $\CK$
into $\CA$-modules. Then we have isomorphisms
$$
\CE_r(\CA)\otimes_{\CA}\C\overset{\simeq}{\lr}\End_{\cU(\fsl_2)}V_1^{\otimes r},
$$
and
$$
\CE_r(\CA)\otimes_{\CA}\CK\overset{\simeq}{\lr}\End_{\cU_q(\fsl_2)}V_q^{\otimes r},
$$
where $V_1$ is the irreducible $\fsl_2(\C)$ module of highest weight $2$ (dimension $3$),
and $V_q$ is its $\cU_q$-analogue.
\end{prop}
\begin{proof}
We prove the second statement. the proof of the first is similar.
We have an exact sequence of $\CA$-algebras
\be
0\lr \langle\tilde\Phi_q\rangle\lr BMW_r(\CA)\lr\CE_r(\CA)\lr 0.
\ee
Moreover we know that $BMW_r(\CA)$ is free as $\CA$-module, whence
so is the kernel $\langle\tilde\Phi_q\rangle$.
Applying the functor $-\otimes_{\CA}\CK$, which by (cf. \cite[Cor 4.4, p. 143]{HWu})
is right exact, we obtain an exact sequence
\be
BMW_r(\CK)\lr\CE_r(\CA)\otimes_{\CA}\CK\lr 0,
\ee
and the kernel of the first map contains $\Phi_q$.

Now since $\tilde\Phi_q$ acts trivially on Lusztig's $\CA$-form
of $V^{\otimes r}$, it follows that there is a homomorphism
$\CE_r(\CA)\lr\End_{\cU_\CA(\fsl_2)}(V_\CA^{\otimes r})$, whose image
is the algebra of endomorphisms which are generated by the $R$-matrices,
and whose kernel is, by Theorem \ref{mainq}, a torsion submodule.
The statement follows.
\end{proof}

\section{Interpretation in terms of orthogonal Lie algebras.}

It is well known that the Lie algebras $\fsl_2(\C)$ and
$\fso_3(\C)$ are abstractly isomorphic. In this section we shall
make explicit the interpretation of our main theorem in terms of $\fso_3$.
We shall discuss the classical ($q=1$) situation; the quantum
case may be discussed entirely similarly.

The three-dimensional representation of $\fsl_2(\C)$ may be realised
as follows. Let $V=\C^3$ be the subspace of the polynomial ring
$\C[x,y]$ with basis $x^2,xy,y^2$. Then $\fsl_2(\C)$ acts via
$e=x\frac{\partial}{\partial y}$, $f=y\frac{\partial}{\partial x}$
and $h=[e,f]=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}$.
With respect to the given basis, $e,f$ and $h$ are represented by the
matrices
$\left[\begin{smallmatrix}0&1&0\\0&0&2\\0 &0 & 0\\ \end{smallmatrix}
\right]$,
$\left[\begin{smallmatrix}0 & 0 & 0\\2 & 0 & 0\\0 & 1 & 0\\ \end{smallmatrix}
\right]$ and
$\left[\begin{smallmatrix}2 &0 & 0\\0&0 &0\\0&0&-2\\ \end{smallmatrix}
\right]$.

Now $\fso_3(\C)$ may be identified with the space of skew symmetric
matrices. If we write $(a,b,c)$ for the matrix
$\left[\begin{smallmatrix}0&-a &-b\\a& 0 &-c\\b&c&0\\ \end{smallmatrix}
\right]$, the Lie product $[(a_1,a_2,a_3), (b_1,b_2,b_3)]=(c_1,c_2,c_3)$,
where $c_i=(-1)^{i-1}\det M_i$, where $M_i$ is the $2\times 2$ matrix
obtained from
$\left[\begin{smallmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\ \end{smallmatrix}
\right]$ by deleting the $i^{\text{th}}$ column.
Writing $E=(i,0,1),F=(i,0,-1)$ and $H=(0,2i,0)$, where $i=\sqrt -1$,
it is easily verified that $E,F$ and $H$ satisfy the $\fsl_2$ relations,
and that
$T=\left[\begin{smallmatrix}1&0&1\\0&i&0\\i&0&-i\\
\end{smallmatrix}\right]$
conjugates $e,f,h$ into $E,F,H$ respectively, i.e. that $TeT\inv=E$ etc.

Inverting $T$, it follows that with respect to the basis
$w_1=x^2+y^2$, $w_2=-2ixy$, $w_3=-ix^2+iy^2$, $e,f$ and $h$ have
skew symmetric matrices. This identifies the 3-dimensional
representation of $\fsl_2(\C)$ explicitly with the natural
representation of $\fso_3(\C)$. The quantum case is similar.

It follows that our main theorems (\ref{mainclass} and \ref{mainq})
may be stated in terms of the natural representation of
$\fso_3$. Now the natural representation $V_n$ of
quantum $\fso_{2n+1}$ shares
with the case $n=1$ the properties
that it is strongly multiplicity free,
and that the surjection $\C(q)\cB_r
\to\End_{\cU_q(\fso_{2n+1})}(V_n^{\otimes r})$, where $\cB_r$ is the $r$-string
braid group, factors through a specialisation of the
algebra $BMW_r(y,z)$. It may be reasonable to speculate that
a result similar to ours holds in this generality.


TITLE: How did Frobenius prove the Cayley-Hamilton theorem for a general matrix?
QUESTION [3 upvotes]: I heard Cayley did the case $n=2$ and then Hamilton generalized the result. How is this done? Assume it works for an $n\times n$ matrix and show $n+1?$
Edit: Sorry, Cayley showed it for $3\times 3$, I mean Frobenius.

REPLY [4 votes]: The concept of the minimal polynomial was introduced by Frobenius in his landmark 1878 paper. He used the language of bilinear forms and the theory of elementary divisors introduced by Weierstrass. He applied the minimal polynomial (called by him 'the degree of equation of the smallest degree $\varphi(A) = 0$') to give one of the first complete proofs of the Cayley-Hamilton theorem.

Hamilton had shown this result for quarternions and Cayley had proved it for matrices of orders $N=2$ and $N=3$. Frobenius however proved the theorem by showing that the minimal polynomial $\varphi$ of a matrix(or bilinear form) $A$ is a divisor of the characteristic polynomial $\psi$. Hence $\varphi(A) =0 \Rightarrow \psi(A) = 0$.
You can also refer to "Frobenius and the symbolical algebra of matrices" in the Archive for History of Exact Sciences here. Hope it helps.


TITLE: If $|xH|$ has order $n$, then there is an element $y$ with $|y|=n$ and $xH=yH$
QUESTION [2 upvotes]: Let $G$ be a group, and let $H$ be a normal subgroup with $|H|=m$. Suppose $n$ and $m$ are relatively prime. If $|xH|$ has order $n$, we wish to find an element $y$ with $|y|=n$ and $xH=yH$.
It is easy to see that $|xH|=n$ divides $|x|$, say $|x|=nt$. (We note also that $x^{n} \in H$.) Of course, we now have that $|x^{t}|=n$, one of our desired conditions. But, I don't see that $xH=x^{t}H$. So maybe this isn't our $y$.
Well, we know also that there are integers $a$ and $b$ such that $1=ma+nb$. Now, the element $x^{ma}$ has the property that $xH=x^{ma}H$, but I don't see how to say much more about its order.

REPLY [1 votes]: The preimage of $\langle xH\rangle$ under the natural projection $G\rightarrow \frac{G}{H}$ is a group with $mn$ elements. It contains $\langle x \rangle$ which implies $x$ has order $nd$ with $d|m$ and since $(n,m)=1$ we have $(d,n)=1$. Consider the powers of $x$:
$x,x^2\dots x^n,x^{n+1}\dots x^{nd}=e$. The elements that are in $xH$ are those which have exponent congruent to $1\bmod n$. On the other hand the elements that have order $n$ are those in which the exponent is $dj$ with $(j,n)=1$.
So the element of $xH$ of order $n$ will be $x^{dj}$ so that $dj\equiv 1 \bmod n$, this is clearly possible, just take $j\equiv d^{-1}\bmod n$. The inverse exists since $(d,n)=1$.


TITLE: Geometrical interpretation of pictures transforms and other "high dimensional everyday objects"
QUESTION [4 upvotes]: During the preparation of a general audience talk on why mathematicians use dimensions higher than three (or four) even for concrete applications, I came up with the following enjoyable observation : Taking the negative of a picture is a central reflexion of the hypercube.  I'm pretty sure many people already observed this, so my question is :
Are you aware of other similar transformations for "everyday life high dimensional objects" which have a nice geometric interpretation?
To provide more details, consider a picture of $240\times 240$ pixels, where each pixel has a grey level ranging from $0$ to $255$. Each grey level can be written as a $8$ digit number in base $2$ (and now your computer understands what grey level means). With this identification, this picture is a vertex  of the $$
240\times240\times8=460800\mbox{-dimensional hypercube: }\quad [0,1]^{
460800}.$$
Now, the negative of a picture is the picture you obtain by reversing the grey scale, meaning mapping a grey level $\alpha$ to $255-\alpha$ for every pixels. In binary this means you switch every zero to one and vice versa. Thus the negative picture, seen as a vertex of the hypercube, is the central reflexion of your initial picture with respect to the center of the hypercube.
I would like to see if one can find other examples of nice geometric interpretations for (colored or not) image transformations, sound effects (i.e. Fourier series of reasonable functions), etc.

REPLY [1 votes]: Changing the tune/pitch of a sound-piece is a scaling/dilation of the Fourier transform.
Similarly, zooming of an image is also a scaling/dilation of the (now 2D) Fourier transform.
JPEG compression is quantization after a "rotation" of images patches (more precisely: a DCT of 8x8 pixels block) - well plus a number of other tricks that are not that geometrical…
Some image denoising techniques rely on the idea of (non-linear) diffusion, e.g. interpreting the gray value as heat and letting the heat diffuse according to some (non-linear, heat- or heat-gradient-dependent) way.


TITLE: Series $\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$
QUESTION [8 upvotes]: Find the sum of the series to infinity$$\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots$$
Attempt-
I wrote the general term as $$\frac{\binom{2n}{n}}{2^{2n}\cdot (n+1)}$$
I don't know what to do next

REPLY [3 votes]: $$S = 2\left[\frac{1}{2\cdot 4}+\frac{1\cdot 3}{2\cdot 4\cdot 6}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6\cdot 8}+\cdots\right]$$
Here $$T_{n}\bf{(n^{th}) term } = 2\cdot \left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)(2n+2)}\right]$$
$$ = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6...(2n)(2n+2)}((2n+2)-(2n+1))\right]$$
So $$T_{n} = 2\left[\frac{1\cdot 3 \cdot 5......(2n-1)}{2\cdot 4 \cdot 6.........(2n)}-\frac{1\cdot 3 \cdot 5......(2n+1)}{2\cdot 4 \cdot 6.........(2n)}\right]$$
So $$S_{\infty} = 2\sum^{\infty}_{r=1}\left[\frac{1\cdot 3 \cdot 5......(2r-1)}{2\cdot 4 \cdot 6.........(2r)}-\frac{1\cdot 3 \cdot 5......(2r+1)}{2\cdot 4 \cdot 6.........(2r)}\right] = 2\cdot \frac{1}{2} = 1$$


TITLE: I don't understand this system of inequalities word problem
QUESTION [1 upvotes]: I don't get how to come up with the 2nd inequality in the answer explanation (not the $b+f$ inequality). Help is greatly appreciated.

REPLY [1 votes]: To make one cake it is necessary to have $4$ cups of batter and $\frac{7}{2}$ cups of frosting. For two cakes: $8$ and $7$ cups, for three cakes: $12$ and $\frac{21}{2}$ cups, for $n$ cakes: $4n$ and $\frac{7}{2}n$ cups, respectively. So if there are $b=4n$ batters, there must be $f=\frac{7}{2}n$ frosts. Having more frosts than required is also possible, which implies leftover frosts.
Now if there are $b$ batters and $f$ frosts, then  it can be made $n=\frac{b}{4}$ cakes provided there are enough frosts, that is $\frac{f}{\frac72}\ge n$. Now substituting first into second you get:
$$\frac{f}{\frac72}\ge \frac{b}{4}.$$


TITLE: Deduce number of components from chromatic polynomial
QUESTION [0 upvotes]: How can one deduce the number of connectivity components from the chromatic polynomial of a graph?

REPLY [1 votes]: Let $\chi_G(x) = \sum_{i = k}^{n + 1} a_i x^i$ be a chromatic polynomial of graph $G$. If $a_k \ne 0$ then graph $G$ has exactly $k$ components.
It is easy to see that chromatic polynomial of a graph is product of chromatics polynomials of its connected components. For each connected graph chromatic polynomial is divisible by $x$ and is not divisible by $x^2$.
The first is true because we can color all vertices in breads first search order (or the same for depth first search order) and the first vertex colored always has all $x$ colors available.
I wouldn't give strict proof that chromatic polynomial of connected graph is not divisible by $x^2$, but some reason for it is the following. Coloring vertices in BFS order we will have all $x$ colors available for the first colored vertex only. All other vertices will have less than $x$ colors, then sum over all choices for all vertices in sense of "one of previously used or a new color" can't be divisible by $x$ for all $x$.
Full proof of even more strong claim can be found here.


TITLE: Explanation on the Proof of $W(t)=W(t_0)\exp\left(\int_{t_0}^{t} \text{tr}(\underline{A}(s)) \ ds\right)$
QUESTION [2 upvotes]: In my study of Floquet Theory, I have been given a sketch proof on the following definition of the Wronskian, $$W(t)=W(t_0)\exp\left(\int_{t_0}^{t} \text{tr}(\underline{A}(s)) \ ds\right).$$

Proof:\begin{align}
\underline{x}(t)&=\underline{x}(t_0)+(t-t_0)\underline{x'}(t_0)+O(t-t_0)^2 \\
&=\underline{x}(t_0)+(t-t_0)\underline{A}(t_0)\underline{x}(t_0)+O(t-t_0)^2,
\end{align}
  as $\underline{x'}=\underline{A}\underline{x}$ and thus $\underline{X'}=\underline{A}\underline{X}$ where $\underline{X}$ denotes the fundamental matrix.
  Now, \begin{align}
W(t)&=\det(\underline{X}(t)) \\
&=\det((\underline{I}+(t-t_0)\underline{A}(t_0))\underline{X}(t_0)+O(t-t_0)^2) \tag{1}\\
&=W(t_0)(1+(t-t_0)\text{tr}(\underline{A}(t_0))+O(t-t_0)^2 \tag{2}.
\end{align}
  Using Taylor expansion:
  $$W(t)=W(t_0)+(t-t_0)W'(t_0)+O(t-t_0)^2. \tag{3}$$ Letting $t\rightarrow t_0$, $$W'(t)=W(t)\text{tr}(\underline{A}(t))\implies W(t)=W(t_0)\exp\left(\int_{t_0}^{t} \text{tr}(\underline{A}(s)) \ ds\right)$$

There are many parts that I do not understand. Are there any resources that could help explain this? For instance, how is the last line derived (where did $W'(t)$ appear from)?

REPLY [2 votes]: So, firstly the Taylor series expansion of $X(t)$ near $t=t_0$ is given by
$$X(t)=X(t_0)+X'(t_0)(t-t_0)+\mathcal{O}\big((t-t_0)^2\big).$$
Next, we can use $X'(t_0)=A(t_0)X(t_0)$ to get
$$X(t)=X(t_0)\big(I+(t-t_0)A(t_0)\big)+\mathcal{O}\big((t-t_0)^2\big). $$
To take the determinant of both sides, we can use $\det(X(t_0)B)=\det(X(t_0))\det B$:
$$ \begin{array}{ll} W(t) & =W(t_0)\det\big(I+(t-t_0)A(t_0)+\mathcal{O}((t-t_0)^2)\big). \\ & =W(t_0)\det\big(I+(\color{Red}{t-t_0})\big[\color{Blue}{A(t_0)+\mathcal{O}(t-t_0)}\big]\big) \end{array}$$
Note when you factor $X(t_0)$ out of $\mathcal{O}((t-t_0)^2)$, you just get $\mathcal{O}((t-t_0)^2)$ since $X(t_0)$ is constant.
Next we can use the fact $\det(I+\color{Red}{\varepsilon} \color{Blue}{X})=1+\color{Red}{\varepsilon}\mathrm{tr}(\color{Blue}{X})+\mathcal{O}(\color{Red}{\varepsilon}^2)$ (which follows from the Leibniz formula for the determinant, itself the final result of expansion by minors), where $\varepsilon=t-t_0$. Note factoring $(t-t_0)$ out of $\mathcal{O}((t-t_0)^2)$ will be $\mathcal{O}(t-t_0)$, but everything gets absorbed back in the end:
$$ \begin{array}{ll} W(t) & =W(t_0)\big[1+(\color{Red}{t-t_0})\mathrm{tr}\big(\color{Blue}{A(t_0)+\mathcal{O}(t-t_0)}\big)+\mathcal{O}((\color{Red}{t-t_0})^2)\big] \\ & =W(t_0)\big[1+(t-t_0)\mathrm{tr}\,A(t_0)+\mathcal{O}((t-t_0)^2)\big]. \end{array}$$
On the other hand, the Taylor expansion of $W(t)$ is
$$ W(t)=W(t_0)+W'(t_0)(t-t_0)+\mathcal{O}((t-t_0)^2). $$
Equating coefficients of $(t-t_0)$ yields $W'(t_0)=W(t_0)\mathrm{tr}\,A(t_0)$. Equivalently, the other expansion of $W(t)$ we can subtract $W(t_0)$ from both sides, divide by $t-t_0$ to obtain
$$ \frac{W(t)-W(t_0)}{t-t_0}=W(t_0)\mathrm{tr}\,A(t_0)+\mathcal{O}(t-t_0). $$
Letting $t\to t_0$ yields $W'(t_0)=W(t_0)\mathrm{tr}\,A(t_0)$. Now replace $t_0$ with $t$ (since it was arbitrary):
$$ W'(t)=A(t)W(t). $$
This is a separable ODE which can be solved with the integrating factor method. First divide by $W(t)$ and then notice $W'(t)/W(t)$ is the derivative of $\ln W(t)$, so integrate from $t_0$ to $t$:
$$ W'(t)/W(t)=\mathrm{tr}\,A(t) $$
$$ \ln W(t)-\ln W(t_0)=\int_{t_0}^t \mathrm{tr}\,A(s)\,\mathrm{d}s $$
Notice so the LHS is $\ln(W(t)/W(t_0))$, so exponentiate and multiply by $W(t_0)$ to get
$$ W(t)=W(t_0)\exp\left(\int_{t_0}^t \mathrm{tr}\,A(s)\,\mathrm{d}s\right). $$


TITLE: Prove that if a sequence $\{a_{n}\}$ converges then $\{\sqrt a_{n}\}$ converges to the square root of the limit.
QUESTION [10 upvotes]: My attempt and the question:Can you tell if I am right :)? thank you

REPLY [0 votes]: Here is a quick answer when $a\neq 0$. Write
\begin{equation}
\lim (a_n^{1/2} - a^{1/2})\lim(a_n^{1/2} + a^{1/2}) = \lim(a_n-a)=0.
\end{equation}
Thus, $\lim(a_n^{1/2} + a^{1/2}) \neq 0$ and
\begin{equation}
\lim (a_n^{1/2} - a^{1/2}) = 0 \rightarrow\lim a_n^{1/2}=a^{1/2}.
\end{equation}


TITLE: how can I find the character table of $(Z/9Z)^\times$?
QUESTION [1 upvotes]: I tried using sagemath. But I don't think sagemath is supporting character table of multiplicative groups of $(Z/nZ)^\times$. Also it would be great if you can suggest a way to calculate character table of $(Z/9Z)^\times$ which I can generalise into character table of $(Z/p^2Z)^\times$, where p is prime.
Thanks in advance.

REPLY [1 votes]: Since you asked about doing it in Sagemath, note that a similar answer to this is available at this question.
sage: H = Integers(9)
sage: G = H.unit_group()
sage: G
Multiplicative Abelian group isomorphic to C6
sage: GG = gap(G)
sage: C = gap.CharacterTable(GG)
sage: gap.Display(C)
CT1

     2  1   1  1  1  1   1
     3  1   1  1  1  1   1

       1a  6a 3a 2a 3b  6b

X.1     1   1  1  1  1   1
X.2     1  -1  1 -1  1  -1
X.3     1   A /A  1  A  /A
X.4     1  -A /A -1  A -/A
X.5     1  /A  A  1 /A   A
X.6     1 -/A  A -1 /A  -A

A = E(3)^2
  = (-1-Sqrt(-3))/2 = -1-b3


TITLE: If $H$ is a subgroup of a finite goup $G$, is there a subgroup $K$ s.t. $G=HK$ and $|G|=|H||K|$?
QUESTION [1 upvotes]: $G$ is a finite group $G$, $H$ is a subgroup of $G$.
I know that there must be a $A \subseteq G$, $G=AH$.
However, must there be a subgroup $K$ of $G$ s.t. $G=HK$ and $|G|=|H||K|$?

REPLY [2 votes]: Several simple counterexamples have been given in the comments. One of these is: let $G=A_4$, the alternating group on four letters. Then $G$ has order 12, and $G$ has several elements of order 2, each of which generates a subgroup of order 2, but $G$ has no subgroup of order 6.


TITLE: Does a curve's genus depend on its base field?
QUESTION [4 upvotes]: Let $C$ be a projective plane curve, defined by a polynomial in $Z[x,y]$, over a field $K$.  Does the geometric genus of $C$ depend on the choice of $K$?
I think the answer to this question is obviously 'yes', since the genus can change when you reduce modulo a prime (i.e, change the base field from $Q$ to $F_p$), i.e. "bad reduction".
I guess my real question is: can the genus change when I change the base field from from $Q$ to $\bar{Q}$ (algebraic closure of $Q$)?
My first answer is still 'yes', because the genus-degree formula tells us that we can write the genus as $\frac{(d-1)(d-2)}{2} - \sum \delta_P$, where $\delta_P$ is the delta invariant at a singular point $P$, so I'm thinking that changing my base field from $Q$ to $\bar{Q}$ might introduce some new singularities.
On the other hand, it might be possible that all of the singularities over $\bar{Q}$ are already present over $Q$.
Anybody know for sure?

REPLY [1 votes]: The arithmetic genus of a curve $X/k$ is defined by $p_a(X) = 1 - \chi(\mathcal{O}_X)$. By base change and cohomology this is invariant under extension of scalars, so the arithmetic genus does not change.


TITLE: What is image of $f(z)=\tan(z)$ where $\Im(z)=cst$?
QUESTION [0 upvotes]: Can you help me figure out what is the image of line segments ${z =x+iy: -π/2<x<π/2, y=const}$ under $f(z)=\tan(z)$.
I've got
$tan(x+iy) = sin(2x)/(ch(2y)+cos(2x)) + i sh(2y)/(ch(2y)+cos(2x)) = u + iv$
and
$tan(z)=w => z=-i1/2\log((wi+1)/(1-wi))$
What should I do next? Is that a good way to start?

REPLY [1 votes]: You can develop $\tan(x+iy)=\dfrac{\tan(x)+i\tanh(y)}{1-i\tan(x)\tanh(y)}$
And let's call $a=\tanh(y)$ constant and $t=\tan(x)\in]-\infty,+\infty[$
We have the curve $X+iY=\dfrac{t+ia}{1-iat}\iff\begin{cases}X=\dfrac{(a^2-1)t}{1+a^2t^2}\\Y=\dfrac{a(t^2+1)}{1+a^2t^2}\end{cases}$
When $t=0$ then the curve passes by $(0,a)$ and at infinity by $(0,\frac 1a)$.
We can verify by calculation, that it is a circle tangent to the lines $Y=a$ and $Y=\frac 1a$ of origin $Y_0=\dfrac{a+\frac 1a}2=\dfrac{a^2+1}{2a}$
$X^2+(Y-Y_0)^2=R^2$ with $R=\left|\dfrac{a^2-1}{2a}\right|$
https://www.desmos.com/calculator/7l3l66kuyq

Now if you consider the set $\Omega=\{x+iy\mid x\in(-\frac\pi 2,\frac\pi 2),y\in\mathbb R\}\sim\{(a,t)\mid a\in(-1,1),t\in\mathbb R\}$
Then $\tan(\Omega)$ is a family of circles covering the plane.

in fact the points $(0,\pm 1)$ cannot be reached since $|a|<1$
also since $|t|<\infty$ the points $\{(0,y)\mid |y|>1\}$ cannot be reached either.

I enhanced the effect on the visual below by limiting $|t|<1000$ and $|a|<0.9$.
You can see the "areas" not reached.
This is of course exaggerated, in reality $\tan(\Omega)=\mathbb C\setminus\{iy\mid |y|\ge 1\}$.


TITLE: Inverting a fraction
QUESTION [0 upvotes]: I'm doing a complex analysis question and it is as follows

$$\int\limits_{c}\frac{1}{z(e^z-1)}dz$$

Using Maclaurin Series I simplified this expression upto this point:

$$f(z) = \frac{1}{z(e^z-1)} = \frac{1}{z^2+\frac{z^3}{2!}+\frac{z^4}{3!}+ \cdots}$$

To use the residue theorem I need to get $z^{-1}$ terms. Some guy at my class inverted (flipped) this. Can I have a little guidance how to get the $z^{-1}$ from this? :)
This is the flipped version of this fraction I found:

$$\frac{1}{z^2}\left(1-\left(\frac{z}{2!}+\frac{z^2}{3!}+\frac{z^3}{4!}\right)+\left(\frac{z}{2!}+\frac{z^2}{3!}+\frac{z^3}{4!}+\cdots\right)+\cdots\right)$$

Is this correct? Then how to get this?

REPLY [3 votes]: $$f(z)=\left(\frac1{z^2}\right)\left(1+\frac z2+\frac{z^2}6+\cdots\right)^{-1}\\
=\left(\frac1{z^2}\right)\left(1-(\frac z2+\frac{z^2}6+\cdots)+(\frac z2+\frac{z^2}6+\cdots)^2-\cdots\right)\\
=z^{-2}(1-\frac z2+\frac {z^2}{12}+...)$$


TITLE: Example of a Group Action of group $D_6$ on set $A = \{1,2,3\}$
QUESTION [3 upvotes]: I am an undergraduate currently taking Abstract Algebra. We have learned about groups and examples of groups (dihedral, symmetric, general linear, etc.). We are just starting to learn about subgroups in more detail. I am reading through Dummit and Foote's Abstract Algebra 3ed, and I will refer to their section 1.7 on Group Actions.
I am confused because an arbitrary group action is a map $\eta: G \times A \rightarrow A$ (written $\eta(g, a) = g \cdot a$) that satisfies:
$$g_2 \cdot (g_1 \cdot a) = (g_1g_2) \cdot a,$$
$$1 \cdot a = a.$$
I understand that $\cdot$ is not the group operation of $G$. However, what can it be? How can we guarantee that an operation defined by us can consistently combine elements $g \in G$ and $a \in A$ to produce another element of $A$? Or, is this just a natural challenge of coming up with group actions?
In particular, let's say that I want to define a group action that describes how elements of $D_6$ permute the vertices of a regular triangle. I will represent the vertices of the regular triangle by a set $A = \{1,2,3\}$. Is it possible to write the group action I am describing with just the elements of $D_6$?
Or, must we observe that $D_6$ is isomorphic to $S_3$ and define our action using $S_3$ (because $S_3$ specifically contains all bijections from the set A to A and so elements of $S_3$ can naturally "affect" elements of $A$). Additionally, what is this process of representing an initial group by another group in order to more naturally define a group action?

REPLY [1 votes]: Another thing is that the two conditions for a group action,  "compatibility" and "identity", as they are often referred to,  guarantee a homomorphism from the group into the symmetric group on as many points or vertices as elements in the $G$-set (the set being acted upon).
So in this case,  we have $h:D_6\to S_3$.  Thus the number of actions is precisely the number of such homomorphisms.
Since $D_6$ is $S_3$, we are considering all endomorphisms of $S_3$.  Now, $S_3=\langle (12),(123)\rangle $.  There's $4$ choices where to send $(12)$ and $3$ choices for $(123)$.  That gives $4\cdot3=12$ total  (don't forget the ones where one or both generators go to the identity).
Also,  you can recover the action from the endomorphism:
$$\sigma \cdot x=h(\sigma) (x)$$.

REPLY [1 votes]: These are great questions! First of all your suspicion that it’s a “challenge of coming up with group actions” is partially correct. An arbitrary map $\eta: G \times A \to A$ is not likely to be group action. Those properties must be satisfied for all $g \in G$ and all $a \in A$, so usually the action “comes from” somewhere else, often a geometric or combinatorial structure.
To understand this better, you need to know that there’s an alternative way to define a group action: a representation. Here’s how.
Given a set $A$, we can look at the set of all symmetries (i.e. bijections) $A \to A$. But this set naturally has the structure of a group! If you compose two such symmetries, the resulting map is again a symmetry, and composition is associative. Moreover, there is always the identity symmetry and any bijection naturally has an inverse. Great! In fact, historically, this is the definition of a group. The modern definition abstracts these properties from those that are always satisfied by symmetries of a set.
So how do we turn a group action on a set into a representation in the symmetries of that set? And vice versa? Let’s set notation $\operatorname{Sym}(A) = \{ f: A \to A \mid f \text{ is a bijection} \}$, for the group of symmetries of $A$ described above. A representation of the group $G$ is a map $\rho: G \to \operatorname{Sym}(A)$. So for any $g \in G$, we need to understand $\rho(g): A \to A$. It’s typical to write this symmetry as $\rho_g = \rho(g)$. Let’s define
$$
\rho_g(a) = g \cdot a.
$$
What do the properties of the action tell us about this representation?
First, the identity property gives us
$$
\rho_1(a) = 1 \cdot a = a
$$
for all $a \in A$. This means that $\rho_1$ is the identity in $\operatorname{Sym}(A)$. In other words, $\rho$ satisfies
$$
\rho(1_G) = 1_{\operatorname{Sym}(A)}.
$$
What about the “associative” property?
$$
\rho_g \bigl( \rho_h(a) \bigr)
= \rho_g ( h \cdot a )
= g \cdot ( h \cdot a )
= (gh) \cdot a
= \rho_{gh}(a)
$$
for all $a \in A$. In other words, $\rho_g \circ \rho_h = \rho_{gh}$ as symmetries.
These two properties, taken together, define $\rho$ as a group homomorphism! In other words, via the group action, the representation exhibits a shadow of the structure of the group in the symmetries of the set $A$.
But we can reverse the correspondence. Given a representation, that is a group homomorphism $\rho:G \to \operatorname{Sym}(A)$, we can define an action by the same equation, but with information flowing in the opposite direction:
$$
g \cdot a = \rho_g(a)
$$
for all $g \in G$ and all $a \in A$. The homomorphism properties of the map $g \mapsto \rho_g$ immediately give us the properties that define a group action. These two notions are tautologically the same thing!

To your example, where you define the symmetric group $S_3$ as group of bijections of the set $A = \{1, 2, 3\}$ under composition, the homomorphism $\rho: D_3 \to S_3$ (which happens to be an isomorphism) is the representation of $D_3$ on $A$ that corresponds to the action you desire!
$^*$I prefer to denote by $D_n$ the group of symmetries of the regular $n$-gon of order $2n$. Some authors prefer to call this group $D_{2n}$ to reflect its order instead.


TITLE: Showing a Lipschitz function is differentiable
QUESTION [2 upvotes]: Let $f : \mathbb{R} \to \mathbb{R}$ be a Lipschitz function.  Suppose
$$ \lim_{n\to \infty} n[f(x + \frac{1}{n}) - f(x)] = 0.$$
Prove that $f$ is differentiable.
I am tempted to just let $n = \frac{1}{h},$ then $$\lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = 0,$$
meaning that the limit exists and therefore $f$ is differentiable.  However, I feel like the issue may be more subtle than this.  Could someone give a more rigourus proof that might not involve using $n = \frac{1}{h}?$

REPLY [6 votes]: Knowing the limit $\frac{f(x+h_n)-f(x)}{h_n} \to 0$ along a single sequence $h_n = \frac{1}{n}\to 0$ is not enough to be able to conclude that the limit along all possible sequences $h_n\to 0$ exists and is equal zero (which is needed to prove differentiabillity).

To solve the problem let $[h]$ denote the closest number to $h$ on the from $\frac{1}{n}$ with $n\in\mathbb{N}$ and write: $$\frac{f(x+h) - f(x)}{h} = \frac{f(x+h) - f(x+[h])}{h - [h]} \cdot \frac{h-[h]}{h} + \frac{f(x+[h]) - f(x)}{[h]}\cdot \frac{[h]}{h}$$ Now take $h\to 0$. The first term can be bounded using the Lipschitz condition, $\frac{|f(x)-f(y)|}{|x-y|}\leq C$, and the second term can be bounded using the given limit. You also need to show that $\frac{h-[h]}{h}\to 0$ and that $\frac{[h]}{h}$ is bounded as $h\to 0$ to reach the desired conclusion but this is fairly straight forward from the definition of $[h]$.


TITLE: How many distinct ways are there to traverse the edges of an N-dimensional hypercube?
QUESTION [3 upvotes]: I have an N-dimensional hypercube, and would like to traverse the edges so that I reach each vertex exactly once. One path that satisfies this property is known as the Gray code, but there are many others. How many such paths are there? Paths that can be mapped to each other by the symmetries of the hypercube may or may not be counted as distinct, depending on your preference.

REPLY [1 votes]: This really is a sketch of a lower bound--but is too long for the comments
The lower bound for the number $X_n$ of Hamiltonian cycles in $H_n$ is at least something like $2^{2^n}$. In fact, $X_{n+1} \geq 2^{2^n}X_n$.
Indeed, let $Q = v^0v^1\ldots v^{2^n}$ be a Hamiltonian cycle in $H_{n}$. We will construct $2^{2^{n}}$ Hamilonian cycles in $H_{n+1}$ from $Q$.

Let $w^0w^1 \ldots w^{2^n}$ be vertices such that (a) the first $n$ bits of $w^i$ are the same as the first $n$ bits of $v^i$, but the $n+1$-th bit of $w^i$ is free (for $i=0,1,2, \ldots, 2^n-1$). So $2^{2^n}$ ways to chose the $w^i$s.
Let $H^i_2$ be the 4-vertex Hamiltonian path in the 4-vertex hypercube $\{v^i,v^{i+1},v^i+ {\bf{e}}_{n+1}, v^{i+1}+{\bf{e}}_{n+1} \}$ starting from $w^i$ and ending at $w^{i+1}$, where ${\bf{e}}_l$ is the element where the $l$-th coordinate is 1 and the rest are all 0. [Note that $w^i \in \{v^i,v^i+ {\bf{e}}_{n+1}\}$, and $w^{i+1} \in \{v^{i+1},v^{i+1}+ {\bf{e}}_{n+1}\}$]
Then let $Q$ be the Hamiltonian cycle consisting of the $H^i_2$s concatonated. This specifies $2^{2^n}$ distinct  Hamiltoninan cycles in $H_{n+1}$ from one Hamiltonian cycle $Q$ in $H_n$. In fact, if $X_n$ is the number of Hamiltoninan cycles in $H_n$ then  $X_{n+1} \geq 2^{2^n} X_n$.

The (trivial) upper bound $U_n$ for the number of Hamiltonian cycles in $H_n$ is $n^{2^n} = 2^{2^n \log n}$ (given the $i$-th vertex of the Hamiltonian cycle there are only at most $n$ choices for the $(i+1)$-th vertex), so letting $L_n$ be the lower bound derived above, we note that $\log \log U_n \le \log \log L_n + \log \log n$.


TITLE: Does ZF prove that topological groups are completely regular?
QUESTION [17 upvotes]: Let $\mathbf{G} = \langle G,\cdot,\mathcal{T}\;\rangle$ be a topological group. Let $\mathbf{e}$ be the identity element of $\langle G,\cdot \rangle$.
Assume $\{\mathbf{e}\}$ is closed in $\langle G,\cal{T}\;\rangle$. Then, I have managed to convince myself that:

ZF proves that $\langle G,\cal{T}\;\rangle$ is regular Haudorff.
ZF + (Dependent Choice) proves that $\langle G,\cal{T}\;\rangle$ is completely regular.

My questions are:

Are those right?
Does ZF prove that $\langle G,\cal{T}\;\rangle$ is completely regular?
If no to question 2, does assuming one or more of following suffice for ZF to conclude that $\langle G,\cal{T}\,\rangle$ is completely regular?

$\mathbf{G}$ is two-sided complete
$\langle G,\cdot \rangle$ is abelian
Countable Choice

REPLY [3 votes]: I'm not sure about the proof I gave, But as I checked, it didn't use full AC, but as Asaf mentioned in a comment it uses DC. The following theorem is due to Pontrjagin. See Book by Montgomery and Zippin(Page 29).
I will give a sketch of proof.
Note: Maybe it's needed to add some separation axiom to the following theorem. please edit it, if needed.
Theorem: Every $T_{0}$ topological group is completely regular.
Proof.
It's enough to prove that a given topological group $(G,*)$ is completely regular at $e$. Let $F$ be a closed set not containing $e$. Put $O=F^{c}$. Choose symmetric open neighborhoods  of $e$, $U_{n}$ By continuity of $*$, such that $U_{0}=O$(w.l.o.g assume $O$ is symmetric) $U_{n+1}^2 \subseteq U_{n} \cap O ~~~~n=0,1,2...$.
Now for rational numbers of the form $r=\frac{k}{2^n}~~k \in \{1,2,3,...2^n \},~~ n \in \{0,1,2,...\}$ inductively define open neighborhood $V_{r}$ of $e$ such that:
1) $V_{\frac{1}{2^n}}=U_{n}~~~~~\forall n$
2) $V_{\frac{2k}{2^{n+1}}}=V_{\frac{k}{2^n}}$
3) $V_{\frac{2k+1}{2^{n+1}}}=V_{\frac{1}{2^{n+1}}}V_{\frac{k}{2^n}}$
The definition does not depend to the representation of $r$ and the family $V_{r}$ has the following properties:
4) $V_{\frac{1}{2^n}}V_{\frac{m}{2^{n}}} \subseteq V_{\frac{m+1}{2^n}}~~~~~~~~~m+1 \leq 2^n$
5) $V_{r} \subseteq V_{s}~~~~~$if $~r <s<1$
Now define the function $f:G \longrightarrow [0, 1]$ as follows:
$f(x)=0~$ if $x \in \bigcap \limits_{r} V_{r}$
$f(x)=1$ if $~~x \notin V_{1}$, and in other cases define:
$f(x)= \sup \{r \leq 1 : x \notin V_{r}\}$
Claim: $f$ is as required.


TITLE: How do I calculate the mass of a moving proton?
QUESTION [0 upvotes]: When a proton moves it's no longer its rest mass, it now has more mass because of the Kinetic Energy. So if we're to find the new protons mass if it was moving at some speed, how would we find it? Would we convert the kinetic energy to mass then plus that on the rest mass? Or is it just Kinetic Energy + Rest mass?

REPLY [6 votes]: Suppose we accelerate a proton to a speed $v$ then crash it into some large mass $M$ so that the proton comes to rest inside the large mass. Then we measure the speed $u$ that the large mass + proton are moving. We'll assume the final speed $u \ll c$ so that regular Newtonian mechanics applies.
We know conservation of momentum always applies, and the momentum of the large mass + proton is $p_M = (M+m_p)u$, and this has to be equal to the initial momentum of the proton $p_p$. So now we know the momentum that the proton had when it was moving at the initial speed $v$.
If we do this experiment we will find that the momentum of the proton moving at speed $v$ was:
$$ p_p = \gamma m_p v \tag{1} $$
where $m_p$ is the rest mass of the proton and $\gamma$ is the Lorentz factor:
$$ \gamma = \frac{1}{\sqrt{1 - v^2/c^2}} $$
Now there are two ways of looking at equation (1). One option is to define a relativistic mass as:
$$ m_r = \gamma m_p \tag{2} $$
Then the equation for the momentum becomes $p = m_r v$ and this looks just like the Newtonian equation for the momentum that we all learned at school. If you take this option then equation (2) defines the mass of the moving proton as you asked.
But I strongly recommend you do not do this as the relativistic mass is not a concept used in modern physics and you will not be taught it if you do a physics degree. For more on this see Why is there a controversy on whether mass increases with speed?
Instead we take the term mass of the proton to only ever mean its rest mass, and we accept that the Newtonian expression for the momentum is an approximation that needs to be modified to equation (1) for speeds comparable to the speed of light.
Since you mentioned the energy, we can show that the total energy of a moving object with rest mass is given by:
$$ E = \gamma mc^2 $$
Thus includes the rest energy given by Einstein's famous equation $E = mc^2$, so to get the kinetic energy we subtract off $mc^2$ to get the kinetic energy:
$$ KE = (\gamma - 1)mc^2 $$
At speeds much less than $c$ this is approximately equal to the Newtonian expression:
$$ KE = \tfrac12 mv^2 $$

REPLY [0 votes]: Relativistic mass $= m_{rest}+ \frac{E_{kin}}{c^2}$
Relativistic mass$ =  \frac{1}{\sqrt{ 1-v^2/c^2 }} * m_{rest}$
Relativistic mass of proton is good for:

Calculating force needed to steer the proton (to change its direction while not changing its speed)

Calculating the rest mass of a box with a moving proton inside.


TITLE: Pigeonhole principle, choosing 1-8 numbers out of 27
QUESTION [2 upvotes]: prove that for every 8 choosen numbers from 10 to 36 you can always make equalities.
number can be used once.
examples.
let say that the choosen numbers are
10, 11, 12, 15, 18, 25, 32, 36
you can write
11+25=36 or 10+12+18=15+25.
i tried to prove for summation for every 2 numbers
$$\binom{8}{2}=28 $$
but there are 51 different summation of 2 differend numbers
Lets say that the nombers are
a b c d e f g h
2 out of 8 is valid because the subsets (ab) and (ac) cannot be equal so they cannot be compared
But 3 of 8 can be (abc) and (ade) and they can be equal so it is not valid
So i can use only use 2 out of 8
Baybe i am in the wrong direction.
that's the closest i got

REPLY [1 votes]: Let $S\subset\{10,11,12,\ldots,36\}$ be an arbitray subset of eight integers. The set $S$ itself has $2^8=256$ subsets. Of these $255$ are non-empty. For any subset $U\subset S$, $U\neq\emptyset$, let's denote by $\Sigma(U)$ the sum of elements of $U$. The smallest possible value of $\Sigma(U)$ is clearly
$$
\Sigma(\{10\})=10,
$$
and the largest possible value of $\Sigma(U)$ is
$$
\Sigma(\{36,35,34,\ldots,29\})=4\cdot65=260.
$$
These cannot occur for subsets of the same set $S$, but we don't need to worry about that. No matter what the set $S$ is, the $255$ numbers $\Sigma(U),U\subseteq S, U\neq\emptyset$, are all integers in the range $[10,260]$. That range has $251$ integers, so by the pigeonhole principle there exists two distinct subsets $U_1,U_2\subseteq S$ with the property that
$$
\Sigma(U_1)=\Sigma(U_2).
$$
Getting warmer, but let's not forget that a single element of $S$ can only be used once. In other words, the two sets $U_1,U_2$ should be disjoint. But this does not really form an obstacle for us. For if $V=U_1\cap U_2$ is non-empty, then, denoting $U_1'=U_1\setminus V$ and $U_2'=U_2\setminus V$, we see that
$$\Sigma(U_i)=\Sigma(U_i')+\Sigma(V)$$ for $i=1,2$.
Therefore
$$
\Sigma(U_1')=\Sigma(U_1)-\Sigma(V)=\Sigma(U_2)-\Sigma(V)=\Sigma(U_2').
$$
Thus the disjoint subsets $U_1'$ and $U_2'$ have equal set sums. Because obviously neither of $U_1,U_2$ can be a subset of the other, both $U_1'$ and $U_2'$ are non-empty.


TITLE: Does $f_n(x) = \sqrt[n]{1+x^n}$ converge uniformly on $[0,\infty)$?
QUESTION [1 upvotes]: Does $f_n(x) = \sqrt[n]{1+x^n}$ converge uniformly on $[0,\infty)$?

First, let's investigate point-wise convergence:

For $x>1$: $\lim_{n\to\infty} \sqrt[n]{1+x^n} = x$
For $x\le 1$: $\lim_{n\to\infty} \sqrt[n]{1+x^n} = 1$

So there is a point-wise convergence only for $x\ge 1$.
Basically, the question asks if there's a uniform convergence at $[0,\infty)$ so the answer is no because there isn't even a point-wise convergence at $[0,1)$. Am I right?
But (assuming I'm right so far) let's assume the question asked about uniform convergence at $[1,\infty)$:
We should look for $$\sup_{x\in [1,\infty)} \left( \sqrt[n]{1+x^n} -x \right)$$
Let's find extremum points of $f_n(x)$:
$$f_n'(x) = (1+x^n)^{\frac{1}{n}-1}x^{n-1}$$
Here I got a little stuck, tough I think there's only minimum point at $x=0$ so $f_n(x)$ grows with $x$.
Could you help me finish up this proof? (And validate the first part.)

REPLY [2 votes]: The sequence of functions
$$ f_n(x) = (1+x^n)^{\frac{1}{n}} $$
uniformly converges to the function $g(x)=\max(1,x)$ over $\mathbb{R}^+$. If $x\in[0,1]$ we have:
$$0\leq f_n(x)-1 \leq 2^{\frac{1}{n}}-1\leq \frac{1}{n} \tag{1}$$
through the inequality:
$$ u>v>0 \Longrightarrow u^{\frac{1}{n}}-v^{\frac{1}{n}}\leq \frac{u-v}{n v^{\frac{n-1}{n}}}\tag{2}$$
that also gives, over $[1,+\infty)$,
$$ 0\leq f_n(x)-x \leq \frac{1}{n x^{n-1}}. \tag{3}$$


text:
\begin{document}

\ \hfill{{{\small To appear in } {\em Topology Appl.}}}

\

\begin{center}

\section*{Equivalence of a Bridged Link Calculus\\ and Kirby's Calculus of
Links\\
on Non-Simply Connected 3-Manifolds}

\vspace*{.5cm}
\bigskip

{\large Thomas Kerler}\\
\medskip

 November 1994 (Revised Version April 1997)

\end{center}
 \vspace*{.5cm}

{\small \noindent{\bf Abstract} : We recall an extension of Kirby's
Calculus on non-simply connected 3-manifolds given in [FR], and the
 surgery calculus of bridged links from [Ke], which involves only local
moves. We give a short combinatorial proof that the two calculi
are equivalent,  and thus describe the same classes of 3-manifolds.
This makes the proofs for the validity of surgery calculi in [FR] and [Ke]
interchangeable.  }

\paragraph*{1.) \underline{The Extended Kirby Calculus} :}

A popular  way of presenting a closed, compact, connected, and oriented
3-manifold, $\, N^{(3)}\,$,
is by use of framed links in $\,S^3\,$. To
any such link, $\,L\,$, we can associate a three-fold, $\,S^3_L\,$,
by doing {\em surgery} along its components. It is long been known
[LW] that  any  closed $\, N^{(3)}\,$ is homeomorphic to
 $\,S^3_L\,$ for some link, $\,L\,$.

 In [Ki] Kirby shows  that $\,S^3_{L_1}\,$ is homeomorphic to
$\,S^3_{L_2}\,$, if and only if $\,L_1\,$ can be changed into  $\,L_2\,$
by a sequence of certain {\em moves}, i.e., if   $\,L_1\,$  and  $\,L_2\,$
are {\em $\delta$-equivalent } in the language of [Ki]. The moves are either
ambient isotopies, the {\em signature  move} ${\cal O}_1$, or the {\em two-handle
slide} $\,{\cal O}_2\,$. Thickening the links into ribbons, in order to
indicate their framings, the $\,{\cal O}_2\,$-move is depicted in Figure 1,
where $\,R_2\,$ is slid over $\,R_1\,$.

\begin{center}\

\epsfbox{F-1.eps}
\newline{{\sc Figure 1:} $\,{\cal O}_2\,$-{\em move}}
\end{center}

Hence Kirby's calculus establishes a one-to-one correspondence between the
set of homeomorphism classes of closed 3-manifolds and the combinatorial
set of  $\delta$-equivalence classes of links in $\,S^3\,$.
\smallskip

In a more general situation we can consider framed links $\,L\,$ embedded into a connected, compact, and oriented 3-manifold, $\,M=M^{(3)}\,$, which may have a boundary or be non-simply connected. For  two
framed links, $L_1$
and $\,L_2\,$, in $\,M\,$, we can again consider the correspondingly surgered manifolds
$\,M_{L_1}\,$ and $\,M_{L_2}\,$. It was suggested by Kirby
and proven by Fenn and Rourke [FR] that $\,M_{L_1}\,\cong\,M_{L_2}\,$, if and only
if $\,L_1\,$ is {\em $\alpha$-equivalent}
 to $\,L_2\,$, where the  $\alpha$-equivalences
are generated by isotopies, the  ${\cal O}_1\,$-,  ${\cal O}_2\,$-, and
${\cal O}_3\,$-move. The new ${\cal O}_3$-move (also called $\eta$-move in [Ke])
is given by including or
deleting a two-component configuration from a link, as shown in Figure 2.
It consist of an arbitrary component, $\,R\,$, together with an annulus,
$\,A\,$,
that encircles $\,R\,$.

\begin{center}\

\epsfbox{F-2.eps}
\newline{{\sc Figure 2:} $\,{\cal O}_3\,$-{\em move}}
\end{center}

For the purpose of this letter let us consider the slightly smaller
equivalent classes, where we   omit the ${\cal O}_1$-move as a generator of
equivalences. The space of classes of framed links in a given
compact manifold, $\,M\,$, modulo  ${\cal O}_2$- and ${\cal O}_3$-moves
shall be  denoted $\,{\cal L}_M\,$.

It can be taken from  arguments in [Ki], [FR], and [Ke] that, in fact, it
suffices to choose a much smaller set of
  ${\cal O}_3$-moves by selecting  two
long ribbons $\,R\,$ for each class in $\,[S^1, M]\,$, with different framings
${\rm mod}\, 2\,$. All other ${\cal O}_3$-moves can be reduced to these special ones
by combining them with isotopies and ${\cal O}_2$-moves.
The ${\cal O}_3$-move for the trivial element in $\,[S^1, M]\,$ consists then of adding or deleting isolated Hopf links, $^0\bigcirc\mkern-13mu\bigcirc^{0,1}\,$, with framings
as indicated.  Applying ${\cal O}_2$-moves they can both be substituted for the
pair of  ${\cal O}_1$-moves, in which  $\bigcirc^{1}\sqcup\bigcirc^{-1}$ is
added or deleted.

In order to see the topological interpretation of $\,{\cal L}_M\,$, recall
that a framed link $\,L\,\subset\,M\,$ defines also a four dimensional
cobordism $\,W_L:M\to M_L\,$, with $\partial W_L\cong M \cup_{\partial M}M_L\,$.
It is now easily extracted from [Ki] or [FR] that $\,L_1\,$ and  $\,L_2\,$
are equivalent via  ${\cal O}_2\,$-, and
${\cal O}_3\,$-moves, if and only if $\,M_{L_1}\cong M_{L_2}\,$ {\em and}
$\,{\rm sign}(W_{L_1})={\rm sign}(W_{L_2})\,$ for the signatures. The
${\cal O}_1$-move connects a ${\bf CP}^2$ to $\,W_L\,$ and thus increases
${\rm sign}(W_L)\,$ by one. (Note, however, that
 connecting ${\bf CP}^2\#\overline{{\bf CP}}^2$
or $S^2\times S^2\,$, as in
the $^0\bigcirc\mkern-13mu\bigcirc^{0,1}$-moves, does not change the signature).

 Moreover, if  $\,N\,$ is any other compact, oriented, and connected
3-manifold with $\,\partial M\cong \partial N\,$, then there is a framed link
$\,L\,$, such that $\,M_L\cong N\,$, extending the given homeomorphism on the
boundaries. Hence the classes in $\,{\cal L}_M\,$ are
in one-to-one correspondence
with pairs $\,(N,n)\,$, where $\,N\,$ is a homeomorphism class of 3-manifolds
with boundary $\,\cong\partial M\,$, and $n\in{\bf Z}\,$ is given by $\,n={\rm sign}(W_L)\,$.
Let us denote the set of all such pairs by $\,{\cal M}_{\partial M}\,$.
\smallskip

The proof in [FR], although referring to results from [Ki], is quite involved since
$\alpha$-equivalence of two links is first expressed as  $\delta$-equivalence
together with an isomorphism between the corresponding $\pi_1(W_L)\,$ and
the vanishing of an additional obstruction class. A shorter proof that builds
directly on the one in  [Ki] was given very recently by Roberts [R]. Several
central arguments used there to generalize the steps in the proof of [Ki] had
already  been given  in [Ke] and an earlier version of this letter.

\paragraph*{2.) \underline{Calculus of Bridged Links} :}
In [Ke] we derive an alternative calculus of links, starting in a
similar way as in [Ki] but following a different strategy to single out
moves that generate the necessary equivalences. In order to include
the possibility of index-1-surgery we consider
{\em bridged links} $\,L\subset M\,$ in a connected, compact, and oriented
3-manifold,  $M\,$. Thus $\,L\,$ not only consists of ribbons but also pairs
of balls, and a ribbon can enter one ball of  a pair and reemerge at
a corresponding spot at the other ball.

In contrast to the non-local moves ${\cal O}_2$ and ${\cal O}_3$ of the
extended Kirby calculus, the inclusion of 1-handle data allows us to
use instead two {\em local} moves. The motivation in [Ke] is that
locality is essential, e.g.,  for efficient presentations of cobordism categories,
Hopf-algebraic interpretations of elementary cobordisms, and  thus
constructions of TQFT's
by means of straight forward structural comparisons
 rather than involved calculations, see [KL].

  The easier of the two moves is the {\em isolated cancellation move}, which we
shall call here ${\cal A}_1\,$. As depicted in Figure 3 the
${\cal A}_1\,$-move adds or deletes an isolated configuration from
the rest of the link, which consists of one ribbon, and one pair of
balls. The ribbon passes through the surgery spheres exactly once.

\begin{center}\

\epsfbox{F-3.eps}
\newline{{\sc Figure 3:} $\,{\cal A}_1\,$-{\em move}}
\end{center}

The ${\cal A}_2\,$-move, more commonly called {\em modification} or {\em handle trading}, is
described as follows. Any 0-framed component, $\,A\,$, bounding a disc,
can be replaced by a pair of surgery balls, $B_1$ and $B_2$, as indicated
in Figure 4. Other strands of the links that run through the disc bounded
by $\,A\,$ are redirected through the surgery spheres.

\begin{center}\
\epsfysize=2.8in
\epsfbox{F-4.eps}
\newline{{\sc Figure 4:} $\,{\cal A}_2\,$-{\em move}}
\end{center}

Let us denote by $\,{\cal BL}_M\,$ the space of bridged links modulo equivalences
generated by isotopies,
$\,{\cal A}_1\,$-, and $\,{\cal A}_2\,$-moves.
\medskip

 In [Ke] it was shown that $\,{\cal BL}_M\,$ for a connected, compact, and
oriented 3-manifold is, like $\,{\cal L}_M\,$, in
one-to-one correspondence with the set $\,{\cal M}_{\partial M}\,$ of pairs $\,(N,n)\,$,
where $\,N\,$ is a homeomorphism class of 3-manifolds with
$\partial N\cong\partial M\,$ and
$\,n\in{\bf Z}\,$.

The correspondence is again given by the assignment
$\,L\,\mapsto\,(M_L,{\rm sign}(W_L))\,$, only now we obtain $\,M_L\,$
by first performing index-1-surgeries along the pairs of balls
(yielding $\breve M\cong \,M\# S^1\times S^2\#\ldots\# S^1\times S^2\,$) before we do
the usual index-2-surgeries along the remaining framed link in $\,\breve M\,$.
Similarly, $\,W_L\,$ is obtained, by also adding 1-handles instead of
just 2-handles to $\,M\times [0,1]\,$.

Since both link spaces have the same topological interpretation in
terms  of $\,{\cal M}_{\partial M}\,$ it is obvious that there has to be a one-to-one
correspondence also between $\,{\cal L}_M\,$ and $\,{\cal BL}_M\,$ for
a given connected $\,M\,$ as above.

The main purpose of this letter is to present a direct and concise of
proof of the equivalence of  $\,{\cal L}_M\,$ and $\,{\cal BL}_M\,$ as
{\em formal} link calculi in $\,M\,$, and thereby show, how the two calculi
can be translated into each other  purely combinatorially. More precisely,
we prove the following:

\begin{theorem}\

The identification of a framed link in $\,M\,$ as a special bridged link
 in $\,M\,$ factors into a map on the equivalence classes of links
$$
I\,:\;{\cal L}_M\,\longrightarrow\,{\cal BL}_M\;.
$$
Moreover, the map $\,I\,$ is a bijection.
\end{theorem}

 In the proof of Theorem 1 we shall only use the definition of
$\,{\cal L}_M\,$ and $\,{\cal BL}_M\,$ as sets of classes of
links as given in Paragraphs 1 and 2. In particular,
none of the arguments in the proof will involve the topological
interpretation of
the links as  surgery data.

In the next theorem let us relate this combinatorial result to the
results
on presentations of 3-manifolds outlined previously. The first part
 is obvious from the fact that the moves
${\cal O}_2\,$,  ${\cal O}_3\,$, ${\cal A}_1\,$, and ${\cal A}_2\,$
can be interpreted as manipulations of $\,W_L\,$, which do not change
the homeomorphisms type of the boundary piece  $\,M_L\,$.

\begin{theorem}\
Let $\,M\,$ be a compact, connected, and oriented 3-manifold.
\begin{enumerate}
\item The interpretations of (bridged) links as surgery prescriptions
factor into  maps:
$$
S_{\rm KFR}\,:\;{\cal L}_M\,\longrightarrow\,{\cal M}_{\partial M}\qquad\;
{\rm and }\qquad\;
S_{\rm BL}\,:\;{\cal BL}_M\,\longrightarrow\,{\cal M}_{\partial M}
$$
with\vspace*{-.9cm}

$$\; S_{\rm KFR}\;=\;S_{\rm BL}\circ I\quad.$$

\item Both  $\,S_{\rm KFR}\,$ and $\,S_{\rm BL}\,$ are bijections.
\end{enumerate}
\end{theorem}

The second part reveals an interesting application of Theorem 1 in that
bijectivity of $\,I\,$ allows us to exchange the  proofs for the
two surgery calculi. I.e., we can use the proofs in [FR] or [Ki]
in order to establish the validity
of the bridged link calculus, or , alternatively, infer the extended
Kirby calculus starting from the results in [Ke].

\paragraph*{4.) \underline{Proof of Theorem 1} :\ }\

{\em 4.1) Existence of $\,I\,$: } In order to see that $\,I\,$ is well defined,
 we need to
show that if two links lie in the same class in  $\,{\cal L}_M\,$, then they are
also equivalent as links in $\,{\cal BL}_M\,$. For example, if two links differ
by an $\,{\cal O}_3\,$-move, we see from Figure 5 that they can also be
related by  a combination of  $\,{\cal A}_1\,$- and $\,{\cal A}_2\,$-moves,
and thus belong to the same class in $\,{\cal BL}_M\,$.

\bigskip

\begin{center}\
\epsfxsize=5.8in
\epsfbox{F-5.eps}
\newline{{\sc Figure 5:}  $\,{\cal O}_3\,$ {\em from} $\,{\cal A}_1\,$ {\em and}
$\,{\cal A}_2\,$ }
\end{center}

A more general version, $\,{\cal A}^{gen}_1\,$, of the cancellation move depicted in
Figure 3 is defined by letting additional strands pass through the surgery balls.
It is shown in Figure 6 that, with the $\,{\cal O}_3\,$-move now found to be an
equivalence in $\,{\cal BL}_M\,$, also $\,{\cal A}^{gen}_1\,$ is an equivalence in
$\,{\cal BL}_M\,$.

\bigskip

\begin{center}\
\epsfxsize=5.8in
\epsfbox{F-6.eps}
\newline{{\sc Figure 6:} $\,{\cal A}_1^{gen}\,$ {\em from} $\,{\cal O}_3\,$ {\em and}
$\,{\cal A}_2\,$ }
\end{center}

Moreover, as depicted in Figure 7,
the $\,{\cal O}_2\,$-move can be expressed as a combination of
a general cancellation and uncancellation $\,{\cal A}^{gen}_1\,$, and
an isotopy, where the surgery ball $\,B\,$ is pushed all the way along
the  second ribbon until it meets $\,B'\,$ again at the other end.

\bigskip

{\em 4.2) Surjectivity of $\,I\,$: } It is  easy to see that every
class in $\,{\cal BL}_M\,$ has an ordinary, framed link as a representative.
Given an arbitrary bridged link, such a representative in the same class
can be found by applying an $\,{\cal A}_2\,$-move to each pair of balls after
they have been moved together in the connected manifold $\,M\,$.
The choice of isotopy, by which the balls are moved
together, is conveniently expressed by a {\em recombination band} between
the two balls in the original bridged link. See for example Figure 8, where
we indicated the isotopy and $\,{\cal A}_2\,$-move by the recombination band
$\,r^B\,$ between $\,B\,$ and  $\,B'\,$.

\begin{center}\
\epsfxsize=6.1in
\epsfbox{F-7.eps}
\newline{{\sc Figure 7:} $\,{\cal O}_2\,$ {\em from} $\,{\cal A}_1^{gen}\,$}
\end{center}

{\em 4.3) Injectivity of $\,I\,$: } We verify injectivity by constructing
a left inverse, $\,P\,$,  for $\,I\,$.  The image of $\,P\,$ of a class in
$\,{\cal BL}_M\,$ shall be defined by taking a representing bridged link,
recombining all pairs of surgery balls via $\,{\cal A}_2\,$-moves,
as described in {\em 4.2)}, and
then taking the class of that link in $\,{\cal L}_M\,$. To see that  $\,P\,$
is well defined by this prescription, we first need to show that the image in
$\,{\cal L}_M\,$ does not depend on the way we apply the $\,{\cal A}_2\,$-moves,
i.e., the choice of recombination ribbons, and is also invariant under the
equivalences in $\,{\cal BL}_M\,$.

\begin{center}\
\epsfxsize=5.9in
\epsfbox{F-8.eps}
\newline{{\sc Figure 8:} Recombination along $\,r^B\,$}
\end{center}

Consider for a pair of balls in a bridged link two choices, $r_1^B\,$ and $r_2^B\,$,
of recombination ribbons. As depicted in Figure 9 we obtain either the
link on the left with annulus $A_2\,$ or the link on the right  with annulus $A_1\,$.
To both diagrams we can apply an $\,{\cal O}_3\,$-move as indicated, where
the long ribbon $\,C\,$ runs along $r_1^B\,$ and $r_2^B\,$, and the short one
is the
respective opposite annulus. The resulting links differ only by
$\,{\cal O}_2\,$-slides over $\,C\,$ as depicted in the bottom row of
Figure 9. Hence the links obtained from
either recombining along $r_1^B\,$ or along $r_2^B\,$ are in the same class in
$\,{\cal L}_M\,$.
\bigskip

\begin{center}\
\epsfxsize=6in
\epsfbox{F-9.eps}

\newline{{\sc Figure 9:} Recombinations related by  $\,{\cal O}_2\,$ {\em and}
$\,{\cal O}_3\,$}
\end{center}

Furthermore, if two bridged links differ by an $\,{\cal A}_2\,$-move, as in
Figure~4, and we choose the recombination ribbon to be a straight strip between
the balls, the images of the two links in Figure~4 under $\,P\,$ are
tautologically the same.

Finally, suppose two bridged links differ by an  $\,{\cal A}_1\,$-move, i.e.,
by an isolated configuration as in Figure 3. This configuration is mapped by
$\,P\,$ to the Hopf link $^0\bigcirc\mkern-13mu\bigcirc^0\,$ for an appropriately
chosen recombination. As described in Paragraph 1 removing or adding this link
is a special case of the $\,{\cal O}_3\,$-move.
\smallskip

 We can thus conclude that $\,P\,$ is well defined on $\,{\cal BL}_M\,$. It is
also obvious that $\,P\circ I\,$ is the identity in  $\,{\cal L}_M\,$ so that
$\,I\,$ must be injective. Thus $\,I\,$ is a bijection as asserted in Theorem 1.

\ \hfill{\rule{3mm}{3mm}}
\bigskip

\paragraph*{Acknowledgments:} I thank
Robion Kirby and Justin Roberts for interest, and encouragement, and
the referee for comments that improved  this letter in content and style.
The author was partially supported by NSF grant DMS-9305715.

\newpage

\paragraph*{References :}\

\begin{enumerate}

\item[[FR]] Fenn, R., Rourke, C.: {On Kirby's Calculus of Links},
{\em Topology}, Vol.  {\bf 18} (1979), 1-15.

\item[[Ke]] Kerler, T.: Bridged Links and Tangle Presentations
of Cobordism Categories. \newline
 {\em Adv. Math.} to appear.(Preprint 1994).\newline
Available at {\tt http://www.math.ohio-state.edu/$\sim$kerler/papers/BL/}.

\item[[Ki]] Kirby, R.: A Calculus for framed Links in $S^3\,$,
{\em Invent. Math.} {\bf 45} (1978), 35-56.

\item[[KL]] Kerler, T., Lyubashenko, V.: Non-semisimple, Extended TQFT's.
   Preprint.

\item[[LW]] Lickorish, W.B.R.: A Representation of Orientable 3-Manifolds.
 {\em Ann. Math.} {\bf 76} (1962) 531-540.

Wallace, A.H.: Modifications and Cobounding Manifolds, {\em Cam. J. Math.}
{\bf 12 } (1960), 503-528.

\item[[R]] Roberts, J.: Kirby Calculus in Manifolds with Boundary. Preprint 1996.

\end{enumerate}

\bigskip

\bigskip

{\small {\sc Harvard University, Cambridge, MA, USA}

present address:

{\small {\sc The Ohio State University, Columbus,  OH, USA}

{\em E-mail-address:} kerler@math.ohio-state.edu }

\end{document}

config: arxiv