TITLE: Why are even numbers closed under multiplication?
QUESTION [0 upvotes]: I'm a little bit confused about why even numbers are closed under multiplication.
Closure under multiplication is $$\lambda u \ \in V   $$ where $V$ is a vector space. If $\lambda = 1.5$, then $\lambda u$ is an odd number right? Therefore, it should not be closed under multiplication?

REPLY [2 votes]: A closed binary operation merely means that the elements remain in the same set, which is to say the operation is a function of the form $X \times X \rightarrow X$. So for example the natural numbers are closed under addition because when you add two naturals numbers together the answer is still a natural number.
In this case the even numbers are closed under multiplication because $2m \times 2n = 2(2mn)$ which is even.


TITLE: Property of the dot product?
QUESTION [0 upvotes]: If I have vectors $v_1, v_2$ and scalar, real $a$
How is :

$$\overrightarrow{v_1} \cdot a\overrightarrow{v_{2}} = a\overrightarrow{v_1} \cdot \overrightarrow{v_{2}}$$

I don't see how this property holds?

REPLY [0 votes]: Suppose
$\overrightarrow{v_{1}} = x_{1} \hat{\textbf{i}} + y_{1} \hat{\textbf{j}} + z_{1} \hat{\textbf{k}} $
and $\overrightarrow{v_{2}} = x_{2} \hat{\textbf{i}} + y_{2} \hat{\textbf{j}} + z_{2} \hat{\textbf{k}} $ , then consider the L.H.S
\
$\overrightarrow{v_{1}}.a\overrightarrow{v_{2}} = (x_{1} \hat{\textbf{i}} + y_{1} \hat{\textbf{j}} + z_{1} \hat{\textbf{k}}).(ax_{2} \hat{\textbf{i}} + ay_{2} \hat{\textbf{j}} + az_{2} \hat{\textbf{k}}) = x_{1}ax_{2} + y_{1}ay_{2}+z_{1}az_{2}$ which is the same when we consider the right hand side , because here in the result the multiplication is commutative as it is in real system.


text:
\begin{document}

\title{Modified Jacobi forms of index zero}

\author{Ja Kyung Koo}

\address{Department of Mathematical Sciences, KAIST}

\curraddr{Daejeon 373-1, Korea} \email{jkkoo@math.kaist.ac.kr}

\thanks{}

\author{Dong Hwa Shin}
\address{Department of Mathematical Sciences, KAIST}
\curraddr{Daejeon 373-1, Korea} \email{shakur01@kaist.ac.kr}
\thanks{}

\subjclass[2010]{11F11, 11F50}

\keywords{Jacobi forms, Klein forms, modular forms.
\newline The first author was partially supported by Basic Science Research Program through
the NRF of Korea funded by MEST (2010-0001654).}

\begin{abstract}
By modifying a slash operator of index zero we define
\textit{modified Jacobi forms} of \textit{index zero}. Such forms
play a role of generating nearly holomorphic modular forms of
integral weight. Furthermore, by observing a relation between the
coefficients of Fourier development of a modified Jacobi form we
construct a family of finite-dimensional subspaces.
\end{abstract}

\maketitle

\section {Introduction}

Let $\mathfrak{H}$ denote the complex upper half-plane
$\{\tau\in\mathbb{C}:\mathrm{Im}(\tau)>0\}$. The letters $\tau$ and
$z$ will always stand for variables in $\mathfrak{H}$ and
$\mathbb{C}$, respectively. For fixed integers $k$ and $m$ ($\geq0$)
we define two slash operators on a function
$\phi:\mathfrak{H}\times\mathbb{C}\rightarrow\mathbb{C}$ as
\begin{eqnarray}
\bigg(\phi|_{k,m}\left[\begin{matrix}a&b\\c&d\end{matrix}\right]\bigg)(\tau,z)&:=&(c\tau+d)^{-k}
e^m\bigg(\frac{-cz^2}{c\tau+d}\bigg)\phi\bigg(\frac{a\tau+b}{c\tau+d},\frac{z}{c\tau+d}\bigg)
\quad\bigg(\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb{R})\bigg)\nonumber\\
(\phi|_m\left[\begin{matrix}\lambda &
\mu\end{matrix}\right])(\tau,z)&:=&e^m(\lambda^2\tau+2\lambda
z+\lambda\mu)\phi(\tau,z+\lambda\tau+\mu)\quad((\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{R}^2)\label{secondslash}
\end{eqnarray}
where $e^m(x):=\exp(2\pi imx)$ (and $e(x):=e^1(x)$). Then we have
the relations
\begin{eqnarray}
(\phi|_{k,m}M)|_{k,m}M'&=&\phi|_{k,m}(MM')\quad(M,~M'\in\mathrm{SL}_2(\mathbb{R}))\nonumber\\
(\phi|_m X)|_m
X'&=&e^m\bigg(\det\begin{pmatrix}X\\X'\end{pmatrix}\bigg)\phi|_m(X+X')\quad(X,~X'\in\mathbb{R}^2)\nonumber\\
(\phi|_{k,m}M)|_m(XM)&=&(\phi|_mX)|_{k,m}M\quad(M\in\mathrm{SL}_2(\mathbb{R}),~X\in\mathbb{R}^2).\label{slash3}
\end{eqnarray}
(\cite{E-Z} $\S$1). From now on, we let
$\Gamma_1:=\mathrm{SL}_2(\mathbb{Z})$ throughout this paper. A
\textit{Jacobi form} of \textit{weight} $k$ and \textit{index} $m$
on a subgroup $\Gamma\subset\Gamma_1$ of finite index is a
holomorphic function
$\phi:\mathfrak{H}\times\mathbb{C}\rightarrow\mathbb{C}$ satisfying
\begin{itemize}
\item[(i)] $\phi|_{k,m}M=\phi$ for all $M\in\Gamma$,
\item[(ii)] $\phi|_m X=\phi$ for all $X\in\mathbb{Z}^2$,
\item[(iii)] for each $M\in\Gamma_1$, $\phi|_{k,m}M$ has a Fourier
development of the form
\begin{equation*}
\sum_{\begin{smallmatrix}n,r\in\mathbb{Z}\\
r^2\leq 4nm\end{smallmatrix}
}c(n,r)q^n\zeta^r\quad(q:=e(\tau),~\zeta:=e(z)).
\end{equation*}
\end{itemize}
The $\mathbb{C}$-vector space of all such functions $\phi$ is
denoted by $J_{k,m}(\Gamma)$. Then, as is well-known,
$\sum_{k,m}J_{k,m}(\Gamma)$ forms a bigraded ring (\cite{E-Z}
Theorem 1.5). And, Eichler-Zagier further developed the following
theorems.

\begin{theorem}[\cite{E-Z} Theorem 1.3]\label{EZtheorem}
Let $\phi\in J_{k,m}(\Gamma)$ and
$(\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{Q}^2$. Then the
function
\begin{equation*}
(\phi|_m\left[\begin{matrix}\lambda&\mu\end{matrix}\right])(\tau,0)=e^m(\lambda^2\tau)\phi(\tau,\lambda\tau+\mu)
\end{equation*}
is a modular form of weight $k$ on some subgroup of $\Gamma$ of
finite index depending only on $\Gamma$ and
$(\begin{matrix}\lambda&\mu\end{matrix})$ (in the sense of
\cite{Koblitz}).
\end{theorem}

\begin{theorem}[\cite{E-Z} Theorem 1.1]\label{EZtheorem2}
$J_{k,m}(\Gamma)$ is a finite-dimensional space.
\end{theorem}

On the other hand, a Jacobi form of weight $k$ and index $0$ is
independent of $z$ by the third condition for a Jacobi form, and
hence it is simply an ordinary modular form of weight $k$ in $\tau$.
In this paper we shall first modify the slash operator described in
(\ref{secondslash}) when $m=0$, and define so called
\textit{modified Jacobi forms} of \textit{weight $k$} and
\textit{index $0$}. And we shall obtain an analogue of Theorem
\ref{EZtheorem} (Theorem \ref{main}). This construction of modified
Jacobi forms is in fact motivated by the Weierstrass
$\sigma$-function, which will give us Klein forms as nearly
holomorphic modular forms of integral weight by virtue of Theorem
\ref{main} (Remark \ref{Klein}).
\par
We shall further construct certain subspaces of modified Jacobi
forms of index $0$ and show that they are indeed finite-dimensional
(Theorem \ref{main2}), which could be an analogue of Theorem
\ref{EZtheorem2}.

\section {Modification of a slash operator}

For a lattice
$L=[\omega_1,\omega_2]=\mathbb{Z}\omega_1+\mathbb{Z}\omega_2$ in
$\mathbb{C}$ the \textit{Weierstrass $\wp$-function} is defined by
\begin{equation*}
\wp(z,L):=\frac{1}{z^2}+\sum_{\omega\in
L-\{0\}}\bigg(\frac{1}{(z-\omega)^2}-\frac{1}{\omega^2}\bigg),
\end{equation*}
and the \textit{Weierstrass $\sigma$-function} is defined as
\begin{equation*}
\sigma(z,L):=z\prod_{\omega\in
L-\{0\}}\bigg(1-\frac{z}{\omega}\bigg)e^{(z/\omega)+(z/\omega)^2/2}.
\end{equation*}
Taking the logarithmic derivative we come up with the
\textit{Weierstrass $\zeta$-function}
\begin{equation*}
\zeta(z,L):=\frac{\sigma'(z,L)}{\sigma(z,L)}=\frac{1}{z}+\sum_{\omega\in
L-\{0\}}\bigg(\frac{1}{z-\omega}+\frac{1}{\omega}+\frac{z}{\omega^2}\bigg).
\end{equation*}
Differentiating the function $\zeta(z+\omega,L)-\zeta(z,L)$ for
$\omega\in L$ results in $0$ because
$\frac{d}{dz}\zeta(z,L)=-\wp(z,L)$ and the $\wp$-function is
periodic with respect to $L$. Hence there is a constant
$\eta(\omega,L)$ satisfying
$\zeta(z+\omega,L)=\zeta(z,L)+\eta(\omega,L)$. So we define the
\textit{Weierstrass $\eta$-function} by $\mathbb{R}$-linearity,
namely, if $z=r_1\omega_1+r_2\omega_2$ with $r_1$,
$r_2\in\mathbb{R}$, then
\begin{equation*}
\eta(z,L):=r_1\eta(\omega_1,L)+ r_2\eta(\omega_2,L).
\end{equation*}
We further define a function $\psi(z,L)$ on $\mathbb{C}$ by
\begin{equation*}
\psi(z,L):=\left\{\begin{array}{ll}-1 & \textrm{if}~z\in L-2L\\
1 & \textrm{otherwise}.\end{array}\right.
\end{equation*}
If $X=(\begin{matrix}\lambda &\mu\end{matrix})\in\mathbb{R}^2$ is
fixed, then the value $\psi(\lambda\tau+\mu,[\tau,1])$  does not
depend on $\tau\in\mathfrak{H}$. Therefore we will simply write
$\psi(X)$ for $\psi(\lambda\tau+\mu,[\tau,1])$.

\begin{lemma}\label{transform}
Let $L$ be a lattice in $\mathbb{C}$.
\begin{itemize}
\item[(i)]
$\sigma(z,L)$, $\eta(z,L)$ and $\psi(z,L)$ are homogeneous of degree
$1$, $-1$ and $0$, respectively. Namely, for any
$\lambda\in\mathbb{C}-\{0\}$ we have
\begin{equation*}
\sigma(\lambda z,\lambda L)=\lambda\sigma(z,L),~\eta(\lambda
z,\lambda L)=\frac{1}{\lambda}\eta(z,L)~\textrm{and}~ \psi(\lambda
z,\lambda L)=\psi(z,L).
\end{equation*}
\item[(ii)]
If $L=[\omega_1,\omega_2]$ and $(\begin{matrix}\lambda &
\mu\end{matrix})\in\mathbb{Z}^2$, then
\begin{equation*}
\psi(\lambda\omega_1+\mu\omega_2,L)=(-1)^{\lambda\mu+\lambda+\mu}.
\end{equation*}
\item[(iii)] Let $X\in\mathbb{R}^2$ and
$M\in\mathrm{SL}_2(\mathbb{Z})$. Then $\psi(XM)=\psi(X)$.
\end{itemize}
\end{lemma}
\begin{proof}
One can obtain (i) and (ii) directly from the definitions of
$\sigma(z,L)$, $\eta(z,L)$ and $\psi(z,L)$.
\par
Now, let $X=(\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{R}^2$
and
$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\mathrm{SL}_2(\mathbb{Z})$.
We derive that
\begin{eqnarray*}
\psi(XM)&=&\psi((\lambda a+\mu c)\tau+(\lambda b+\mu
d),[\tau,1])~\textrm{for any
$\tau\in\mathfrak{H}$}\\
&=&\psi((\lambda a+\mu c)\tau+(\lambda b+\mu d),[a\tau+b,c\tau+d])
~\textrm{by the fact $[a\tau+b,c\tau+d]=[\tau,1]$}\\
&=&\psi(\lambda(a\tau+b)+\mu(c\tau+d),[a\tau+b,c\tau+d])\\
&=&\psi\bigg(\lambda\frac{a\tau+b}{c\tau+d}+\mu,\bigg[\frac{a\tau+b}{c\tau+d},
1\bigg]\bigg)~\textrm{by (i)}\\
&=&\psi(\lambda\tau+\mu,[\tau,1])=\psi(X).
\end{eqnarray*}
This proves (iii).
\end{proof}

\begin{lemma}[Legendre Relation]\label{Legendre}
Let $L=[\omega_1,\omega_2]$ be a lattice in $\mathbb{C}$ with
$\omega_1/\omega_2\in\mathfrak{H}$. Then we have
\begin{equation*}
\eta(\omega_2,L)\omega_1-\eta(\omega_1,L)\omega_2=2\pi i.
\end{equation*}
\end{lemma}
\begin{proof}
See \cite{Lang} p. 241 or \cite{Silverman} p. 41.
\end{proof}

Let $k$ be an integer and
$\phi:\mathfrak{H}\times\mathbb{C}\rightarrow\mathbb{C}$ be a
function. We define two operators $|'_k$ and $|''_k$ on $\phi$ as
follows.
\begin{eqnarray*}
(\phi|'_k M)(\tau,z)&:=&(\phi|_{k,0}M)(\tau,z)\quad(M\in\mathrm{SL}_2(\mathbb{R}))\\
(\phi|''_k
\left[\begin{matrix}\lambda&\mu\end{matrix}\right])(\tau,z)&:=&
\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k\\
&&\times(\phi|_0
\left[\begin{matrix}\lambda&\mu\end{matrix}\right])(\tau,z)
\quad((\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{R}^2).
\end{eqnarray*}

\begin{proposition}\label{split}
\begin{itemize}
\item[(i)]
For any $X$, $X'\in\mathbb{R}^2$ we get
\begin{equation*}
(\phi|''_k X)|''_k X'=\bigg(\psi(X)\psi(X') \psi(X+X')
e\bigg(\frac{1}{2}\det\begin{pmatrix}X'\\X\end{pmatrix}\bigg)\bigg)^k(\phi|''_k
(X+X')).
\end{equation*}
In particular, if $X$, $X'\in\mathbb{Z}^2$, then
\begin{equation*}
(\phi|''_k X)|''_k X'=\phi|''_k (X+X').
\end{equation*}
\item[(ii)] For any $M\in\mathrm{SL}_2(\mathbb{Z})$ and $X\in\mathbb{R}^2$ we have
\begin{equation*}
(\phi|'_k M)|''_k (XM)= (\phi|''_k X)|'_k M.
\end{equation*}
\end{itemize}
\end{proposition}
\begin{proof}
(i) Let $X=(\begin{matrix}\lambda & \mu\end{matrix})$ and
$X'=(\begin{matrix}\lambda' & \mu'\end{matrix})$. If we set
$\omega=\lambda\tau+\mu$ and $\omega'=\lambda'\tau+\mu'$, then
\begin{eqnarray*}
&&(\phi|''_kX)|''_k X'\\
&=&\bigg(\bigg(\psi(X)\exp(\eta(\omega,[\tau,1])(z+\tfrac{1}{2}\omega))\bigg)^k\phi(\tau,z+\omega)\bigg)|''_k
X'\\
&=&\bigg(\psi(X)\psi(X')
\exp(\eta(\omega,[\tau,1])(z+\omega'+\tfrac{1}{2}\omega)
+\eta(\omega',[\tau,1])(z+\tfrac{1}{2}\omega'))\bigg)^k
\phi(\tau,z+\omega+\omega')\\
&=&\bigg(\psi(X)\psi(X') \psi(X+X')
\exp(\tfrac{1}{2}(-\eta(\omega',[\tau,1])\omega+\eta(\omega,[\tau,1])\omega'))\bigg)^k
(\phi|''_k(X+X'))\\
&=&\bigg(\psi(X)\psi(X)\psi(X+X')
\exp(\tfrac{1}{2}(\lambda'\mu-\lambda\mu')(\tau\eta(1,[\tau,1])-\eta(\tau,[\tau,1]))\bigg)^k
(\phi|''_k (X+X'))\\
&=&\bigg(\psi(X)\psi(X')\psi(X+X')\exp(\pi
i(\lambda'\mu-\lambda\mu'))\bigg)^k(\phi|''_k (X+X'))~\textrm{by
Lemma \ref{Legendre}}.
\end{eqnarray*}
This yields the first part of (i). Moreover, if $X$,
$X'\in\mathbb{Z}^2$, then
\begin{eqnarray*}
&&\psi(X)\psi(X')\psi(X+X')
\exp(\pi i(\lambda'\mu-\lambda\mu'))\\
&=&
(-1)^{\lambda\mu+\lambda+\mu}(-1)^{\lambda'\mu'+\lambda'+\mu'}(-1)^{(\lambda+\lambda')(\mu+\mu')+(\lambda+\lambda')+(\mu+\mu')}
(-1)^{\lambda'\mu-\lambda\mu'}=1~\textrm{by Lemma
\ref{transform}(ii)},
\end{eqnarray*}
which proves the second part of (i).\\
(ii) Let $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}$,
$X=(\begin{matrix}\lambda &\mu\end{matrix})$ and
$(\begin{matrix}\lambda' &\mu'\end{matrix})=XM$. Then
\begin{eqnarray*}
&&(\phi|'_k M)|''_k(XM)\\
&=&\bigg(\psi(XM)
\exp(\eta(\lambda'\tau+\mu',[\tau,1])(z+\tfrac{1}{2}(\lambda'\tau+\mu'))\bigg)^k
((\phi|_{k,0}M)|_0(XM))\\
&=&\bigg(\psi(X)\exp(\eta(\lambda'\tau+\mu',[\tau,1])
(z+\tfrac{1}{2}(\lambda'\tau+\mu'))\bigg)^k((\phi|_0 X)|_{k,0}
M)~\textrm{by Lemma \ref{transform}(iii) and (\ref{slash3})}.
\end{eqnarray*}
On the other hand, we achieve that
\begin{eqnarray*}
&&(\phi|''_k X)|'_k M\\
&=&
\bigg(\bigg(\psi(X)\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k
\phi|_0 X\bigg)|_{k,0}M\\
&=&\bigg(\psi(X)
\exp\bigg(\eta\bigg(\lambda\frac{a\tau+b}{c\tau+d}+\mu,\bigg[\frac{a\tau+b}{c\tau+d},1\bigg]\bigg)
\bigg(\frac{z}{c\tau+d}+\frac{1}{2}\bigg(\lambda\frac{a\tau+b}{c\tau+d}+\mu\bigg)\bigg)\bigg)\bigg)^k
((\phi|_0X)|_{k,0}M)
\\
&=&\bigg(\psi(X)
\exp(\eta(\lambda(a\tau+b)+\mu(c\tau+d),[a\tau+b,c\tau+d])
(z+\tfrac{1}{2}(\lambda(a\tau+b)+\mu(c\tau+d)))\bigg)^k\\
&&\times(\phi|_0X)|_{k,0}M~\textrm{by Lemma \ref{transform}(i)}\\
&=&\bigg(\psi(X)\exp(\eta(\lambda'\tau+\mu',[\tau,1])
(z+\tfrac{1}{2}(\lambda'\tau+\mu'))\bigg)^k
((\phi|_0X)|_{k,0}M)~\textrm{by the
fact}~[a\tau+b,c\tau+d]=[\tau,1].
\end{eqnarray*}
Hence we obtain the assertion (ii).
\end{proof}

Define a function $\rho(\tau,z)$ by
\begin{equation*}
\rho(\tau,z):=\exp(\tfrac{1}{2}\eta(1,[\tau,1])z^2-\pi iz).
\end{equation*}
Since $\eta(1,[\tau,1])$ has the expansion
\begin{equation*}
\eta(1,[\tau,1])=\frac{(2\pi
i)^2}{12}\bigg(-1+24\sum_{n=1}^\infty\frac{nq^n}{1-q^n}\bigg)
\end{equation*}
(\cite{Lang} p. 249), $\rho(\tau,z)$ is a holomorphic function on
$\mathfrak{H}\times\mathbb{C}$.

\begin{definition}
A \textit{modified Jacobi form} of \textit{weight} $k$ and
\textit{index $0$} on a subgroup $\Gamma\subset\Gamma_1$ of finite
index is a holomorphic function
$\phi:\mathfrak{H}\times\mathbb{C}\rightarrow\mathbb{C}$ satisfying
\begin{itemize}
\item[(i)] $\phi|'_k M=\phi$ for all $M\in\Gamma$,
\item[(ii)] $\phi|''_k X=\phi$ for all $X\in\mathbb{Z}^2$,
\item[(iii)] for each $M\in\Gamma_1$, the function  $\rho^k(\phi|'_k M)$ has a Fourier
development of the form
\begin{equation*}
\sum_{n\geq0}\sum_{|r|\leq r_0(n)}c(n,r)q^n\zeta^r
\end{equation*}
for some increasing sequence $\{r_0(n)\}_{n\geq0}$ of nonnegative
integers such that
\begin{equation*}
\frac{r_0(n)}{n}\rightarrow0~\textrm{as}~n\rightarrow\infty.
\end{equation*}
Here we allow that $c(n,r)$ could be zero even if $|r|\leq r_0(n)$.
\end{itemize}
In what follows, we denote the $\mathbb{C}$-vector space of all such
functions $\phi$ by $J_k(\Gamma)$. And we let
$J_*(\Gamma):=\sum_{k\in\mathbb{Z}}J_k(\Gamma)$.
\end{definition}

\begin{proposition}\label{graded}
If $\Gamma$ is a subgroup of $\Gamma_1$ of finite index, then
$J_*(\Gamma)$ is a graded ring.
\end{proposition}
\begin{proof}
Let $\phi\in J_{k}(\Gamma)$ and $\phi'\in J_{k'}(\Gamma)$ for some
$k$, $k'\in\mathbb{Z}$. If $M\in\Gamma$ and $X\in\mathbb{Z}^2$, then
one can readily show by the definitions of slash operators that
$(\phi\phi')|'_{k+k'}M=\phi\phi'$ and
$(\phi\phi')|''_{k+k'}X=\phi\phi'$.
\par
Now let $M\in\Gamma_1$ and suppose that $\rho^k(\phi|'_k M)$ and
$\rho^{k'}(\phi'|'_{k'}M)$ have Fourier developments
\begin{equation*}
\sum_{n\geq0}\sum_{|r|\leq r_0(n)}c(n,r)q^n\zeta^r
~\textrm{and}~\sum_{m\geq0}\sum_{|s|\leq s_0(m)}d(m,s)q^m\zeta^s,
\end{equation*}
respectively. Here $\{r_0(n)\}_{n\geq0}$ and $\{s_0(m)\}_{m\geq0}$
are increasing sequences of nonnegative integers such that both
$r_0(n)/n$ and $s_0(m)/m$ tend to $0$ as $n$, $m\rightarrow\infty$.
We then have
\begin{eqnarray*}
\rho^{k+k'}((\phi\phi')|'_{k+k'}M)&=&\rho^k(\phi|'_k
M)\rho^{k'}(\phi'|'_{k'}M)~\textrm{by the
definition of $|'$}\\
&=&\sum_{\begin{smallmatrix}\ell=n+m\\
n\geq0,~m\geq0\end{smallmatrix}}\sum_{\begin{smallmatrix}t=r+s
\\|r|\leq r_0(n),~|s|\leq s_0(m)\end{smallmatrix}}
c(n,r)d(m,s)q^{\ell}\zeta^t.
\end{eqnarray*}
Note that for a given $\ell\geq0$ there are only finitely many
nonnegative integers $n$ and $m$ such that $\ell=n+m$. So there are
only finitely many integers $t$ which contribute terms of the form
$q^\ell\zeta^t$. Furthermore, we have
\begin{equation*}
\frac{|t|}{\ell}\leq\frac{r_0(n)+s_0(m)}{n+m}
\leq2\max\bigg(\frac{r_0(n+m)}{n+m},\frac{s_0(n+m)}{n+m}\bigg)\rightarrow0~\textrm{as}~
n+m\rightarrow\infty.
\end{equation*}
Hence $\phi\phi'$ satisfies the third condition for a modified
Jacobi form of weight $k+k'$. This proves the proposition.
\end{proof}

\begin{proposition}\label{zero}
Let $\phi$ be a modified Jacobi form of weight $k$. Assume that  the
function $z\mapsto\phi(\tau_0,z)$ is not identically zero for a
fixed point $\tau_0\in\mathfrak{H}$. Let $F$ be a fundamental domain
for the torus $\mathbb{C}/[\tau_0,1]$, whose boundary does not have
any zeros of $\phi(\tau_0,z)$(, such $F$ always exists). Then
$\phi(\tau_0,z)$ has exactly $-k$ zeros (counting multiplicity) in
$F$.
\end{proposition}
\begin{proof}
It follows from the second condition for a modified Jacobi form
$\phi$ that
\begin{equation}\label{2ndcondition}
\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k
\phi(\tau,z+\lambda\tau+\mu)=\phi(\tau,z)
\quad((\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{Z}^2).
\end{equation}
Differentiating the above equation with respect to $z$ we have
\begin{equation}\label{differentiation}
\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k
\bigg(k\eta(\lambda\tau+\mu,[\tau,1])\phi(\tau,z+\lambda\tau+\mu)
+\phi_z(\tau,z+\lambda\tau+\mu)\bigg)=\phi_z(\tau,z)
\end{equation}
where $\phi_z=\frac{d}{dz}\phi$. Dividing the equation
(\ref{differentiation}) by (\ref{2ndcondition}) we obtain
\begin{equation*}
k\eta(\lambda\tau+\mu,[\tau,1])+\frac{\phi_z}{\phi}(\tau,z+\lambda\tau+\mu)=\frac{\phi_z}{\phi}(\tau,z).
\end{equation*}
If we put $(\begin{matrix}\lambda &\mu\end{matrix})=
(\begin{matrix}1&0\end{matrix})$ and
$(\begin{matrix}0&1\end{matrix})$ in the previous equation, then we
get that
\begin{equation}\label{putting}
k\eta(\tau,[\tau,1])+\frac{\phi_z}{\phi}(\tau,z+\tau)=\frac{\phi_z}{\phi}(\tau,z)~\textrm{and}~
k\eta(1,[\tau,1])+\frac{\phi_z}{\phi}(\tau,z+1)=\frac{\phi_z}{\phi}(\tau,z),
\end{equation}
respectively. Now, we set $\partial F=\partial F_1+\partial
F_2+\partial F_3+\partial F_4$ as follows:
\begin{equation*}
\begindc{\undigraph}[7]
\obj(1,1){$z_0$}[\west]

\obj(11,1){$z_0+1$}[\east]

\obj(15,8){$z_0+1+\tau_0$}[\east]

\obj(5,8){$z_0+\tau_0$}[\west]

\mor{$z_0$}{$z_0+1$}{$\partial F_1$}[\atright,\solidarrow]

\mor{$z_0+1$}{$z_0+1+\tau_0$}{$\partial F_2$}[\atright,\solidarrow]

\mor{$z_0+1+\tau_0$}{$z_0+\tau_0$}{$\partial
F_3$}[\atright,\solidarrow]

\mor{$z_0+\tau_0$}{$z_0$}{$\partial F_4$}[\atright,\solidarrow]
\enddc
\end{equation*}
By the Residue Theorem we derive that the number of zeros of
$\phi(\tau_0,z)$ in $F$ is equal to
\begin{eqnarray*}
\frac{1}{2\pi i} \oint_{\partial
F}\frac{\phi_z}{\phi}(\tau_0,z)dz&=&\bigg( \frac{1}{2\pi i}
\oint_{\partial F_1}\frac{\phi_z}{\phi}(\tau_0,z)dz +\frac{1}{2\pi
i} \oint_{\partial F_3}\frac{\phi_z}{\phi}(\tau_0,z)dz \bigg)\\
&&+ \bigg( \frac{1}{2\pi i} \oint_{\partial
F_2}\frac{\phi_z}{\phi}(\tau_0,z)dz
+\frac{1}{2\pi i} \oint_{\partial F_4}\frac{\phi_z}{\phi}(\tau_0,z)dz \bigg)\\
&=&\bigg( \frac{1}{2\pi i} \oint_{\partial
F_1}\frac{\phi_z}{\phi}(\tau_0,z)dz -\frac{1}{2\pi i}
\oint_{\partial F_1}\frac{\phi_z}{\phi}(\tau_0,z+\tau_0)dz \bigg)
\\
&&+ \bigg(-\frac{1}{2\pi i} \oint_{\partial
F_4}\frac{\phi_z}{\phi}(\tau_0,z+1)dz
+\frac{1}{2\pi i} \oint_{\partial F_4}\frac{\phi_z}{\phi}(\tau_0,z)dz \bigg)\\
&=&\frac{1}{2\pi i}\oint_{\partial F_1}k\eta(\tau_0,[\tau_0,1])dz +
\frac{1}{2\pi i}\oint_{\partial F_4}k\eta(1,[\tau_0,1])dz~\textrm{by
(\ref{putting})}\\
&=&\frac{k}{2\pi
i}(\eta(\tau_0,[\tau_0,1])-\tau_0\eta(1,[\tau_0,1]))=-k~\textrm{by
Lemma \ref{Legendre}}.
\end{eqnarray*}
This completes the proof.
\end{proof}

\begin{corollary}
Let $\Gamma$ be a subgroup of $\Gamma_1$ of finite index. For all
positive integers $k$ we have $J_k(\Gamma)=\{0\}$.
\end{corollary}
\begin{proof}
Assume that there exists a nonzero element $\phi$ of $J_k(\Gamma)$
for some $k>0$. Take a point $\tau_0\in\mathfrak{H}$ such that
$\phi(\tau_0,z)$ is not identically zero. And, consider a
fundamental domain $F$ for $\mathbb{C}/[\tau_0,1]$ whose boundary
has no zeros of $\phi(\tau_0,z)$. Then $\phi(\tau_0,z)$ has $-k<0$
zeros in $F$ by Proposition \ref{zero}, which is impossible. Hence
$J_k(\Gamma)=\{0\}$ for all $k>0$, as desired.
\end{proof}

\section{Construction of nearly holomorphic modular forms}

In this section we shall show that through a modified Jacobi form
one can generate nearly holomorphic modular forms of integral
weight.

\begin{lemma}\label{Legendre2}
If $(\begin{matrix}\lambda&\mu\end{matrix})\in\mathbb{R}^2$, then
\begin{equation*}
\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau
+\mu)))\rho(\tau,z)\rho(\tau,z+\lambda\tau+\mu)^{-1}=
e(\tfrac{1}{2}\mu(1-\lambda))q^{\lambda(1-\lambda)/2}\zeta^{-\lambda}.
\end{equation*}
\end{lemma}
\begin{proof}
We achieve that
\begin{eqnarray*}
&&\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau
+\mu)))\rho(\tau,z)\rho(\tau,z+\lambda\tau+\mu)^{-1}\\
&=&\exp\bigg((\lambda\eta(\tau,[\tau,1])+\mu\eta(1,[\tau,1]))
(z+\tfrac{1}{2}(\lambda\tau+\mu))
+\tfrac{1}{2}\eta(1,[\tau,1])z^2-\pi
iz\\
&&-\tfrac{1}{2}\eta(1,[\tau,1])
(z+\lambda\tau+\mu)^2+\pi i(z+\lambda\tau+\mu)\bigg)\\
&=&\exp\bigg((\tau\eta(1,[\tau,1])-\eta(\tau,[\tau,1]))(-\lambda
-\tfrac{1}{2}\lambda^2\tau-\tfrac{1}{2}\lambda\mu)+\pi i\lambda\tau
+\pi i\mu\bigg)\\
&=&\exp\bigg(2\pi i(-\lambda
-\tfrac{1}{2}\lambda^2\tau-\tfrac{1}{2}\lambda\mu)+\pi i\lambda\tau
+\pi i\mu\bigg)~\textrm{by Lemma \ref{Legendre}}\\
&=&e(\tfrac{1}{2}\mu(1-\lambda))q^{\lambda(1-\lambda)/2}\zeta^{-\lambda}
\end{eqnarray*}
as desired.
\end{proof}

\begin{theorem}\label{main}
Let $\Gamma$ be a subgroup of $\Gamma_1$ of finite index. Let
$\phi\in J_k(\Gamma)$ for some integer $k$. Then for
$X\in\mathbb{Q}^2$ the function
\begin{equation*}
\phi_X(\tau):=(\phi|''_k X)(\tau,0)
\end{equation*}
is a nearly holomorphic modular form (that is, holomorphic on
$\mathfrak{H}$ and meromorphic at every cusp) of weight $k$ on some
subgroup of $\Gamma$ of finite index depending only on $\Gamma$ and
$X$.
\end{theorem}
\begin{proof}
Let $X=(\begin{matrix}\lambda &\mu\end{matrix})$. For any $X'\in
2\mathbb{Z}^2$ we have
\begin{eqnarray*}
\phi_{X+X'}(\tau)&=&(\phi|''_k(X+X'))(\tau,0)
=(\phi|''_k(X'+X)(\tau,0)\\
&=&\bigg(\psi(X')\psi(X) \psi(X'+X)
e\bigg(\frac{1}{2}\det\begin{pmatrix}X\\X'\end{pmatrix}\bigg)\bigg)^{-k}
((\phi|''_k X')|''_k X)(\tau,0)~\textrm{by Proposition \ref{split}(i)}\\
&=&\bigg(\psi(X) \psi(X)
e\bigg(\frac{1}{2}\det\begin{pmatrix}X\\X'\end{pmatrix}\bigg)\bigg)^{-k}
((\phi|''_k X')|''_k X)(\tau,0)~\textrm{because
$X'\in 2\mathbb{Z}^2$}\\
&=&e\bigg(-\frac{k}{2}\det\begin{pmatrix}X\\X'\end{pmatrix}\bigg)
\phi_X(\tau)~\textrm{from the second condition for $\phi(\tau,z)$}.
\end{eqnarray*}
On the other hand, for any
$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma$ we deduce that
\begin{eqnarray*}
(c\tau+d)^{-k}\phi_X\bigg(\frac{a\tau+b}{c\tau+d}\bigg)&=&
((\phi|''_k X)|'_k M)(\tau,0)~\textrm{by the definitions of $\phi_X$ and slash operators}\\
&=&((\phi|'_k M)|''_k (XM))(\tau,0)~\textrm{by Proposition \ref{split}(ii)}\\
&=&(\phi|''_k (XM))(\tau,0)~\textrm{by the first condition for
$\phi(\tau,z)$}\\
&=&\phi_{XM}(\tau).
\end{eqnarray*}
Thus the above observations indicate that $\phi_X(\tau)$ behaves
like a modular form with respect to the congruence subgroup
\begin{eqnarray*}
&&\bigg\{M\in\Gamma~:~XM\equiv X\pmod{2\mathbb{Z}^2},~\frac{k}{2}
\cdot\det\begin{pmatrix}X\\XM-X\end{pmatrix}\in\mathbb{Z}
\bigg\}\\
&=&\bigg\{\begin{pmatrix} a&b\\c&d
\end{pmatrix}\in\Gamma~:~
(a-1)\lambda+c\mu,~b\lambda+(d-1)\mu,~k(b\lambda^2+(d-a)\lambda\mu-c\mu^2)\in
2\mathbb{Z} \bigg\}
\end{eqnarray*}
which contains $\Gamma\cap\Gamma(\frac{2N^2}{\gcd(N,k)})$ if $X\in
N^{-1}\mathbb{Z}^2$ for some integer $N\geq1$ and
$\Gamma(\frac{2N^2}{\gcd(N,k)})$ is the principal congruence
subgroup of level $\frac{2N^2}{\gcd(N,k)}$.
\par
Next, let $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1$ and
suppose that $\rho^k(\phi|'_k M)$ has a Fourier development
\begin{equation*}
\sum_{n\geq0}\sum_{|r|\leq r_0(n)}c(n,r)q^n\zeta^r
\end{equation*}
such that $r_0(n)/n\rightarrow0$ as $n\rightarrow\infty$. Then we
get that
\begin{eqnarray}
&&(c\tau+d)^{-k}\phi_X\bigg(\frac{a\tau+b}{c\tau+d}\bigg)\nonumber\\
&=&((\phi|''_k X)|'_k M)(\tau,0)~\textrm{by the definitions of $\phi_X$ and slash operators}\nonumber\\
&=&((\phi|'_k M)|''_k (XM))(\tau,0)~\textrm{by Proposition \ref{split}(ii)}\nonumber\\
&=&
\bigg(\psi(XM)\exp(\tfrac{1}{2}\eta(\lambda'\tau+\mu',[\tau,1])(\lambda'\tau+\mu'))\bigg)^k
(\phi|'_k M)(\tau,\lambda'\tau+\mu')~\textrm{where}~
(\begin{matrix}\lambda'&\mu'\end{matrix})=XM\nonumber\\
&=&\bigg(\psi(XM)\exp(\tfrac{1}{2}\eta(\lambda'\tau+\mu',[\tau,1])(\lambda'\tau+\mu'))\bigg)^k
\bigg(\rho^{-k}\sum_{n\geq0}\sum_{|r|\leq r_0(n)}c(n,r)q^n\zeta^r\bigg)(\tau,\lambda'\tau+\mu')\nonumber\\
&=&\bigg(\psi(XM)e(\tfrac{1}{2}\mu'(1-\lambda'))q^{\lambda'(1-\lambda')/2}\bigg)^k\sum_{n\geq0}\sum_{|r|\leq
r_0(n)} c(n,r)e(\mu'r)q^{n+\lambda'r}~\textrm{by Lemma
\ref{Legendre2}}.\label{sum}
\end{eqnarray}
Now, we observe that
\begin{equation*}
n+\lambda' r\geq
n-|\lambda'|r_0(n)=n\bigg(1-|\lambda'|\frac{r_0(n)}{n}\bigg)\rightarrow\infty
~\textrm{as}~n\rightarrow\infty,
\end{equation*}
from which it follows that for a given rational number $\ell$ there
are only finitely many $n$ and $r$ which contributes the term
$q^{\ell}$ in (\ref{sum}). Hence $\phi_X(\tau)$ is meromorphic at
each cusp. This completes the proof.
\end{proof}

We are ready to introduce well-known Klein forms in view of
Weierstrass $\sigma$-function which will be a concrete example of
modified Jacobi form.

\begin{lemma}\label{sigmafunction}
\begin{itemize}
\item[(i)] Let $L$ be a lattice in $\mathbb{C}$.
If $\omega\in L$, then
\begin{equation*}
\frac{\sigma(z+\omega,L)}{\sigma(z,L)}=\psi(\omega,L)\exp(\eta(\omega,L)(z+\tfrac{1}{2}\omega)).
\end{equation*}
\item[(ii)]
The function $\sigma(\tau,z):=\sigma(z,[\tau,1])$ has the infinite
product expansion
\begin{equation*}
\sigma(\tau,z)=-\frac{1}{2\pi
i}\rho(\tau,z)(1-\zeta)\prod_{n=1}^\infty
\frac{(1-q^n\zeta)(1-q^{n}\zeta^{-1})}{(1-q^n)^2}.
\end{equation*}
\end{itemize}
\end{lemma}
\begin{proof}
See \cite{Lang} p. 241, p. 247 or \cite{Silverman} p. 44, p. 53.
\end{proof}

\begin{theorem}
The function $\sigma(\tau,z)$ belongs to $J_{-1}(\Gamma_1)$.
\end{theorem}
\begin{proof}
First, we note that $\sigma(\tau,z)$ is a holomorphic function by
Lemma \ref{sigmafunction}(ii).
\par
Let $M=\begin{pmatrix}a&b\\c&d\end{pmatrix}\in\Gamma_1$. Then we
derive
\begin{eqnarray*}
(\sigma|'_{-1}M)(\tau,z)&=&(c\tau+d)\sigma\bigg(\frac{z}{c\tau+d},\bigg[\frac{a\tau+b}{c\tau+d},1\bigg]\bigg)\\
&=&\sigma(z,[a\tau+b,c\tau+d])~\textrm{by Lemma
\ref{transform}(i)}\\
&=&\sigma(\tau,z)~\textrm{by the fact $[a\tau+b,c\tau+d]=[\tau,1]$}.
\end{eqnarray*}
\par
And, for $X=(\begin{matrix}\lambda &\mu\end{matrix})\in\mathbb{Z}^2$
we achieve that
\begin{eqnarray*}
(\sigma|''_{-1}X)(\tau,z)&=&\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})
\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^{-1}
\sigma(\tau,z+\lambda\tau+\mu)\\
&=&\sigma(\tau,z)~\textrm{by Lemma \ref{sigmafunction}(i)}.
\end{eqnarray*}
\par
Finally, let $M\in\Gamma_1$. Then we have
\begin{eqnarray*}
\rho(\tau,z)^{-1}(\sigma|'_{-1}M)(\tau,z)&=&\rho(\tau,z)^{-1}\sigma(\tau,z)~\textrm{by
the first part of the proof}\\
&=&-\frac{1}{2\pi i}(1-\zeta)\prod_{n=1}^\infty
\frac{(1-q^n\zeta)(1-q^{n}\zeta^{-1})}{(1-q^n)^2}~\textrm{by Lemma
\ref{sigmafunction}(ii)}.
\end{eqnarray*}
Hence it is easy to check that $\sigma(\tau,z)$ satisfies the third
condition for a modified Jacobi form by adopting the argument in the
proof of Proposition \ref{graded}. Therefore $\sigma(\tau,z)$
belongs to $J_{-1}(\Gamma_1)$.
\end{proof}

\begin{remark}\label{Klein}
It follows that if $(\begin{matrix}\lambda &\mu\end{matrix})\in
N^{-1}\mathbb{Z}^2$ for some integer $N\geq1$, then the function
\begin{equation*}
\mathfrak{k}_{(\begin{matrix}\lambda &\mu\end{matrix})}(\tau):=
(\sigma|''_{-1}\left[\begin{matrix}\lambda
&\mu\end{matrix}\right])(\tau,0)=
\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(-\tfrac{1}{2}\eta(\lambda\tau+\mu,[\tau,1])(\lambda\tau+\mu))
\sigma(\tau,\lambda\tau+\mu)
\end{equation*}
is a nearly holomorphic modular form of weight $-1$ on
$\Gamma(2N^2)$ by Theorem \ref{main}. This function is called a
\textit{Klein form} (indexed by $(\begin{matrix}\lambda
&\mu\end{matrix})$) whose infinite product expansion is
\begin{equation*}
\mathfrak{k}_{(\begin{matrix}\lambda
&\mu\end{matrix})}(\tau)=e(\mu(\lambda-1)/2)q^{\lambda(\lambda-1)/2}
(1-e(\mu)q^\lambda) \prod_{n=1}^\infty\frac{(1-e(\mu)q^{n+\lambda})
(1-e(-\mu)q^{n-\lambda})}{(1-q^n)^2}
\end{equation*}
if $(\begin{matrix}\lambda&\mu\end{matrix})\not\in\mathbb{Z}^2$ (and
identically zero otherwise). The modularity of a product of finitely
many Klein forms on $\Gamma(N)$ are intensively studied by Kubert
and Lang (\cite{K-L}). On the other hand, in a recent paper
\cite{E-K-S} the authors present a sufficient condition for a
product of Klein forms to be a nearly holomorphic modular form on
$\Gamma_1(N)$.
\end{remark}

\section {Finite-dimensional subspaces}

In this section we shall consider a family of finite-dimensional
subspaces of $J_k(\Gamma_1)$.

\begin{definition}
Let $k$ be an integer. For each integer $m>0$ we let $J_k^m$ be the
subspace of $J_k$ ($:=J_k(\Gamma_1)$) consisting of $\phi$ for which
$\rho^k\phi=\sum_{n\geq0}\sum_{|r|\leq r_0(n)}c(n,r)q^n\zeta^r$ and
\begin{equation}\label{condition}
\min\{n-r_0(n)~:~n\geq0\}+\frac{k}{8}\geq-m.
\end{equation}
\end{definition}

\begin{remark}
\begin{itemize}
\item[(i)]
Note that since
\begin{equation*}
n-r_0(n)=n\bigg(1-\frac{r_0(n)}{n}\bigg)\rightarrow\infty~\textrm{as}~n\rightarrow\infty,
\end{equation*}
the minimum of $n-r_0(n)$ exists. Thus we have a filtration
$J_k^1\subseteq J_k^2\subseteq\cdots$ and $J_k=\cup_{m=1}^\infty
J_k^m$.
\item[(ii)]
Since $\rho^{-1}\sigma$ has the following Fourier development
\begin{eqnarray*}
-\frac{1}{2\pi
i}(1-\zeta)\prod_{n=1}^\infty\frac{(1-q^n\zeta)(1-q^n\zeta^{-1})}{(1-q^n)^2}
&=&-\frac{1}{2\pi i}\bigg\{
(1-\zeta)+(-\zeta^{-1}+3-3\zeta+\zeta^2)q\\
&&\hspace{0.8cm}+(-3\zeta^{-1}+9-9\zeta+3\zeta^2)q^2\\
&&\hspace{0.8cm}+(\zeta^{-2}-9\zeta^{-1}+22-22\zeta+9\zeta^2-\zeta^3)q^3\\
&&\hspace{0.8cm}+(51-51\zeta-22\zeta^{-1}+22\zeta^2+3\zeta^2-3\zeta^3)q^4\\
&&\hspace{0.8cm}+(9\zeta^{-2}-51\zeta^{-1}+108-108\zeta+51\zeta^2-9\zeta^3)q^5+\cdots
\bigg\},
\end{eqnarray*}
one can readily check that the Weierstrass $\sigma$-function in fact
lies in $J_{-1}^2$.
\end{itemize}
\end{remark}

\begin{proposition}
Let $\phi\in J_k$ for some integer $k$ with
$\rho^k\phi=\sum_{n,r}c(n,r)q^n\zeta^r$. Then we have
\begin{equation}\label{coeffs}
c(n,r)=(-1)^{\lambda
k}c(n-r\lambda-\tfrac{1}{2}k(\lambda^2+\lambda), r+\lambda k)
\end{equation}
for all integers $n$, $r$ and $\lambda$.
\end{proposition}
\begin{proof}
If $(\begin{matrix}\lambda &\mu\end{matrix})\in\mathbb{Z}^2$, then
we derive that
\begin{eqnarray*}
&&\sum_{n,r}c(n,r)q^n\zeta^r=\rho^k\phi(\tau,z)
=\rho^k(\phi|''_k\left[\begin{matrix}\lambda&\mu\end{matrix}\right])\\
&=&\rho(\tau,z)^k
\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k
\phi(\tau,z+\lambda\tau+\mu)\\
&=&\rho(\tau,z)^k
\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})\exp(\eta(\lambda\tau+\mu,[\tau,1])(z+\tfrac{1}{2}(\lambda\tau+\mu)))\bigg)^k
\rho(\tau,z+\lambda\tau+\mu)^{-k}(\rho^k\phi)(\tau,z+\lambda\tau+\mu)\\
&=&\bigg(\psi(\begin{matrix}\lambda&\mu\end{matrix})e(\tfrac{1}{2}\mu(1-\lambda))
q^{\lambda(1-\lambda)/2}\zeta^{-\lambda}\bigg)^k\sum_{n,r}c(n,r)q^{n+r\lambda}\zeta^r
~\textrm{by Lemma \ref{Legendre2}}\\
&=&(-1)^{\lambda
k}\sum_{n,r}c(n,r)q^{n+r\lambda+k\lambda(1-\lambda)/2}\zeta^{r-\lambda
k}~\textrm{by Lemma \ref{transform}(ii)}\\
&=&(-1)^{\lambda
k}\sum_{n',r'}c(n'-r'\lambda-\tfrac{1}{2}k(\lambda^2+\lambda),r'+\lambda
k)q^{n'}\zeta^{r'}\\
&&\textrm{by letting $r':=r-\lambda k$ and
$n':=n+r'\lambda+\tfrac{1}{2}k(\lambda^2+\lambda)$}.
\end{eqnarray*}
We get the assertion by comparing the coefficients of Fourier
developments.
\end{proof}

\begin{remark}\label{relationrk}
If we put $\lambda=1$ in (\ref{coeffs}), then we have
\begin{equation}\label{relation}
c(n,r)=(-1)^kc(n-r-k,r+k).
\end{equation}
\end{remark}

Let $\phi\in J_k$ for some integer $k$ with
$\rho^k\phi=\sum_{n,r}c(n,r)q^n\zeta^r$. Then, for
$X=(\begin{matrix}u&v\end{matrix})\in\mathbb{Q}^2$ we defined in
Theorem \ref{main}
\begin{eqnarray}
&&\phi_X(\tau):=(\phi|''_k X)(\tau,0)\nonumber\\
&=&\bigg\{\bigg(\psi(\begin{matrix}u&v\end{matrix})\exp
(\eta(u\tau+v,[\tau,1])(z+\tfrac{1}{2}(u\tau+v)))\bigg)^k
\rho(\tau,z+u\tau+v)^{-k}(\rho^{k}\phi)(\tau,
z+u\tau+v)\bigg\}(\tau,0)\nonumber\\
&=&\bigg\{
\bigg(\psi(\begin{matrix}u&v\end{matrix})e(\tfrac{1}{2}v(1-u))q^{u(1-u)/2}\zeta^{-u}\bigg)^k
\sum_{n,r}c(n,r)e(vr)q^{n+ur}\zeta^r\bigg\}
(\tau,0)~\textrm{by Lemma \ref{Legendre2}}\nonumber\\
&=&\bigg(\psi(\begin{matrix}u
&v\end{matrix})e(\tfrac{1}{2}v(1-u))\bigg)^k\sum_{n,r} c(n,r)e(v
r)q^{n+ur+ku(1-u)/2}.\label{expression}
\end{eqnarray}

\begin{proposition}\label{01}
Let $X,~Y\in\mathbb{Q}^2$ with $X\equiv Y\pmod{\mathbb{Z}^2}$. Then
$\phi_X(\tau)=\xi\phi_Y(\tau)$ for some root of unity $\xi$.
\end{proposition}
\begin{proof}
If $X=(\begin{matrix}u&v\end{matrix})$ and $Y=(\begin{matrix}
u&v+1\end{matrix})$, then one can readily get from the expression
(\ref{expression}) that $\phi_X(\tau)=\xi\phi_Y(\tau)$ for some root
of unity $\xi$.
\par
Now, let $X=(\begin{matrix}u&v\end{matrix})$ and
$Y=(\begin{matrix}u+1&v\end{matrix})$. We obtain from the expression
(\ref{expression}) that
\begin{eqnarray*}
\phi_Y(\tau)&=& \bigg(\psi(\begin{matrix}u+1
&v\end{matrix})e(-\tfrac{1}{2}vu)\bigg)^k\sum_{n,r} c(n,r)e(v
r)q^{n+(u+1)r-k(u+1)u/2}\\
&=&\bigg(\psi(\begin{matrix}u+1&v\end{matrix})e(-\tfrac{1}{2}vu)\bigg)^k
\sum_{n',r'}c(n'-r'-k,r'+k)e(vr')e(vk)q^{n'+ur'+ku(1-u)/2}\\
&&\textrm{by letting $r':=r-k$ and $n':=n+r'+k$}\\
&=&\bigg(\psi(\begin{matrix}u+1&v\end{matrix})e(-\tfrac{1}{2}vu)\bigg)^k
e(vk)(-1)^k\sum_{n',r'}c(n',r')e(vr')q^{n'+ur'+ku(1-u)/2}
~\textrm{by Remark \ref{relationrk}}\\
&=&\xi\phi_X(\tau)~\textrm{for some root of unity $\xi$}.
\end{eqnarray*}
This proves the proposition.
\end{proof}

Now we are ready to prove our main theorem about dimension.

\begin{theorem}\label{main2}
Let $k<0$ and $m>0$ be integers. Then $J_k^m$ is finite-dimensional.
\end{theorem}
\begin{proof}
Pick any $-k+1$ distinct pairs of
$X_j=(\begin{matrix}u_j&v_j\end{matrix})\in\mathbb{Q}^2$ with
$0<u_j,~v_j<1$. Let $\phi\in J_k^m$ with
$\rho^k\phi=\sum_{n,r}c(n,r)q^n\zeta^r$. Then the functions
\begin{equation*}
\phi_{X_j}(\tau)=(\phi|''_k X_j)(\tau,0)\quad(j=1,\cdots,-k+1)
\end{equation*}
are nearly holomorphic modular forms on some congruence subgroups
$\Gamma_j$ depending on $X_j$ by Theorem \ref{main}. If
$M=\begin{pmatrix} a&b\\c&d\end{pmatrix}\in\Gamma_1$, then we have
\begin{eqnarray}
(c\tau+d)^{-k}\phi_{X_j}\bigg(\frac{a\tau+b}{c\tau+d}\bigg)&=&
(\phi|''_k X_j|'_k M)(\tau,0)~\textrm{by the definitions of slash
operators}\nonumber\\
&=&(\phi|'_k M)|''_k(X_jM)(\tau,0)~\textrm{by Proposition \ref{split}(ii)}\nonumber\\
&=&(\phi|''_k(X_jM))(\tau,0)~\textrm{by the first condition for a modified Jacobi form}\nonumber\\
&=&\phi_{X_jM}(\tau).\label{cusp}
\end{eqnarray}
Set $(\begin{matrix}u_j'&v_j'\end{matrix}):=X_jM$. We may assume
that $0\leq u_j',~v_j'<1$ to estimate
$\mathrm{ord}_q\phi_{X_jM}(\tau)$ by Proposition \ref{01}. Note that
from the expression (\ref{expression}) one can deduce
\begin{equation*}
\phi_{X_jM}(\tau)=\xi \sum_{n\geq0} \sum_{|r|\leq r_0(n)}
c(n,r)e(v_j'r)q^{n+u_j'r+ku_j'(1-u_j')/2}
\end{equation*}
for some root of unity $\xi$. It follows that
\begin{eqnarray}
\mathrm{ord}_q\phi_{X_jM}(\tau)&\geq&
\min\{n+u_j'r+ku_j'(1-u_j')/2~:~n\geq0,~|r|\leq r_0(n)\}\nonumber\\
&\geq&\min\{n-r_0(n)+k/8~:~n\geq0\}~\textrm{because $0\leq
u_j',~v_j'<1$}\nonumber\\
&\geq&-m~\textrm{by the condition (\ref{condition}) for $\phi\in
J_k^m$}.\label{inequality}
\end{eqnarray}
\par
On the other hand, we consider a map
\begin{eqnarray*}
g~:~J_k^m&\rightarrow&\bigoplus_{j=1}^{-k+1}M_{k+12m}(\Gamma_j)\nonumber\\
\phi&\mapsto&(\phi_{X_j}(\tau)\Delta(\tau)^m)_{j=1}^{-k+1}
\end{eqnarray*}
where $M_{k+12m}(\Gamma_j)$ is the space of modular forms of weight
$k+12m$ on $\Gamma_j$, and $\Delta(\tau):=(2\pi
i)^{12}q\prod_{n=1}^\infty(1-q^n)^{24}$ is the modular discriminant
function. As is well-known, each $M_{k+12m}(\Gamma_j)$ is of finite
dimension (\cite{Shimura} Theorem 2.23) and $\Delta(\tau)$ is a
modular form of weight $12$ on $\Gamma_1$ which does not vanish on
$\mathfrak{H}$ and has $\mathrm{ord}_q\Delta(\tau)=1$
(\cite{Silverman} Chapter I $\S$3). Hence the identity (\ref{cusp})
and the inequality (\ref{inequality}) imply that $g$ is
well-defined.
\par
If $\phi$ and $\phi'$ are two distinct elements of $J_k^m$, then
there exists a point $\tau_0\in\mathfrak{H}$ such that the function
$(\phi-\phi')(\tau_0,z)$ is not identically zero and has no zeros on
the boundary of the fundamental domain generated by $\tau_0$ and $1$
(for the torus $\mathbb{C}/[\tau_0,1]$). Suppose that $\phi$ and
$\phi'$ have the same image via $g$. Then for every $j$
\begin{equation*}
0=(\phi_{X_j}\Delta^m-\phi'_{X_j}\Delta^m)(\tau_0)
=(\phi_{X_j}-\phi'_{X_j})(\tau_0)\Delta(\tau_0)^m.
\end{equation*}
Since $\Delta(\tau_0)\neq0$, we get
\begin{eqnarray*}
0&=&(\phi_{X_j}-\phi'_{X_j})(\tau_0)=(\phi|''_k X_j-\phi'|''_k
X_j)(\tau_0,0)\\
&=&\bigg(\psi(\begin{matrix}u_j&v_j\end{matrix})
\exp(\eta(u_j\tau_0+v_j,[\tau_0,1])\tfrac{1}{2}(u_j\tau_0+v_j))\bigg)^k
(\phi-\phi')(\tau_0,u_j\tau_0+v_j).
\end{eqnarray*}
This implies $(\phi-\phi')(\tau_0,u_j\tau_0+v_j)=0$. Thus the
function $(\phi-\phi')(\tau_0,z)$ has at least $-k+1$ distinct zeros
in the fundamental domain $F$ generated by $\tau_0$ and $1$. (When
$(\phi-\phi')(\tau_0,z)$ has zeros on the boundary $\partial F$, we
slightly move $F$ into a new domain $F'$ so that the points
$u_j\tau_0+v_j$ still lie inside $F'$ and the boundary $\partial F'$
has no zeros of $(\phi-\phi')(\tau_0,z)$.) But, this contradicts
Proposition \ref{zero}. Therefore, $g$ is injective; hence we obtain
\begin{equation*}
\mathrm{dim}_\mathbb{C}J_k^m\leq\sum_{j=1}^{-k+1}\mathrm{dim}_\mathbb{C}
M_{k+12m}(\Gamma_j)<\infty
\end{equation*}
as desired.
\end{proof}

\bibliographystyle{amsplain}

config: arxiv


TITLE: Quantum field theory, interpretation of commutation relation
QUESTION [1 upvotes]: Let $\phi$ be the quantum field
$$
\phi(x) = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_\mathbf{p}}} \Big[ b_\mathbf{p}e^{-ip\cdot x} + c_\mathbf{p}^\dagger e^{ip\cdot x} \Big]
$$
with commutation relations
$$
[b_\mathbf{p}, b_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
$$
[c_\mathbf{p}, c_\mathbf{q}^\dagger] = (2\pi)^3\delta^{(3)}(\mathbf{p}-\mathbf{q}),
$$
all other commutators zero. Let $Q$ be the charge operator
$$
Q = \int \frac{d^3\mathbf{p}}{(2\pi)^3} \Big[c_{\mathbf{p}}^\dagger c_{\mathbf{p}} - b_{\mathbf{p}}^\dagger b_{\mathbf{p}} \Big].
$$
We calculate the commutator $[Q,\phi] = \phi$. The question is what is an interpretation of this commutation relation? We know that $Q$ is the number of antiparticles minus the number of particles.

REPLY [1 votes]: Q is a charge for the boson field  $\phi$, which, e.g., annihilates a $\pi^-$ or creates a $\pi^+$. Its eigenvalue q for a state is that state's charge: the number of $\pi^+$s minus the number of $\pi^-$s.
Thus, by virtue of the commutation relation $[Q,\phi] = \phi$, the quantum field $\phi$ raises the charge of a state by 1, since
$$ Q|\alpha\rangle= q|\alpha\rangle \implies Q(\phi|\alpha\rangle)= \phi (Q+1) |\alpha\rangle =(q+1) (  \phi|\alpha\rangle). $$
Conjugately, $[Q,\phi^\dagger] = -\phi^\dagger$,  so $\phi^\dagger$ has the opposite charge: it kills a $\pi^+$ or creates a $\pi^-$.


TITLE: Mass and flavour eigenstates in SM
QUESTION [2 upvotes]: From what I understand, if we develop the terms in the Lagrangian of SM containing the $SU(2)$ quark doublets
$$Q_{i,L}=\begin{pmatrix}
u_{i,L}\\
d_{i,L}
\end{pmatrix}$$
with $u_i=u,c,t$ and $d_i=d,s,b$, the terms that are responsible for the mass of the quarks mixes different quarks together. We then say this doesn't diagonalize the masses matrix, it's in the flavour eigenstates.
On the other hand, it's possible to do a linear combination of the quarks such that the mass terms don't mixes the quarks together anymore. We then say that it diagonalize the masses matrix and that it's in the mass eigenstates.
Here are my questions:

Is this explanation correct ?
Why do we call those two ways of writing the Lagrangian eigenstates? We don't talk about states (kets) here.
Apperently, the CKM matrix allow us to go from one point of view to the other, how exactly?

I've followed QFT and SM courses so you can use notions of those.

REPLY [3 votes]: I should think your QFT or SM text, or WP
describe the linear-algebraic unitary rotation of mixing adequately. States and quantum fields (e.g. acting on the vacuum) are equivalent descriptions for the same thing.
None of these does (/should) call what you call "flavor eigenstates" this way, and if they did, you should switch text.
Mass eigenstates are the ones that correspond to the eigenvalues of the mass operator, so they correspond to the u,c,t;d,s,b quarks. These states also define what we call flavor. An s quark is the state that propagates with the mass $m_s$, and has well-defined strangeness. These are what appear in generation charts, where  the generation index   increases with the mass of the respective quarks. (Such charts are conceptually arbitrary, and group particles in generations by mass; there is no extant deep reason you could not reassign the s to the first generation and the d to the second, perversely, provided you properly accounted for this change of basis--below.)
Now these states do not couple to the charged weak current simply: Every "up-like" quark (charge 2/3) couples to all three  quarks d,s,b and a $W^+$. By convention, and nothing deeper, we denote the linear combination of d,s,b coupling to u as d'; to c, as s'; and to t as b'.
The respective coefficients comprise the rows of the CKM matrix,
$$
\begin{bmatrix}  d^\prime  \\  s^\prime  \\  b^\prime  \end{bmatrix} = \begin{bmatrix} V_{ud} & V_{us} & V_{ub} \\ V_{cd} & V_{cs} & V_{cb} \\ V_{td} & V_{ts} & V_{tb} \end{bmatrix} \begin{bmatrix}  d  \\  s  \\  b  \end{bmatrix}.
$$
One sometimes calls these primed linear combination states "weak eigenstates" since they characterize the charged weak coupling. But they do not have a well-defined flavor, and thus generation. (This underlies the comparably oracular statement that "charged weak interactions violate flavor"...) However, above,
it appears your are using $d_i$ for the down-like primed states, and not the mass eigenstates!
There is nothing special about the down-like quark mixing; it is a feature of convention: we could have kept the down-like mass eigenstates, and defined up-like primed combinations (weak eigenstates)  by the hermitian conjugate of the above CKM matrix, without conceptual stress. (But don't ever do it...)
The majority component of these primed states is the homonymous unprimed one, so the biggest component of d' is d, etc... (but this cannot be done in the formally identical case of lepton mixing, which has often been the source of confusion).  (I hope it is clear to you now why calling them, instead, "flavor eigenstates" is an invitation for confusion. A dangerous term to use. Flavor is a feature of masses. People have used the term "flavor basis" on occasion, but it is sadistically meant to send your head spinning, so reach for your holster/cudgel when you hear it used.)
For the neutral weak couplings, there are no such complications, and you simply use mass eigenstates,  unprimed states.

Fine print/geeky. You might well wonder where this flavor-versus-mass nonsense started. It did so in the lepton sector, where elusive neutrinos are exclusively studied through their production and decay to charged leptons, so experimentally the primed states are characterized/identified by the flavor of the charged lepton they interact with, $e,\mu,\tau$. This is radically different usage than what you see for quarks above, and "neutrino flavor" means not mass eigenstate: it means a primed
weak-coupling state. So "flavor oscillations" induced by neutrinos means mutation of the charged leptons involved in production versus decay of the neutrinos! From then on, one  drifted to neutrino flavor states, the flavor basis, and, lunatically, to "flavor eigenstates", designed to confuse. The lunacy is receding with the PDG and SM chart makers supplanting mass eigenstates $\nu_L,\nu_M,\nu_H$, where they can.


TITLE: Value of $x^2 \sin (\frac1x)$ at $x=0$
QUESTION [2 upvotes]: What is the value of $y=x^2 \sin(1/x)$ at $x=0$?

I see that $x^2 =0$ but $\sin (1/x)$ is undefined.
More generally: if a function made up of a product of functions,  like $y= a(x)b(x)c(x)d(x)\dots$, then at a specific value of $x$ if one of the functions is $0$ and all the other functions are undefined, does it still mean that $y=0$?

REPLY [7 votes]: The function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x = 0$ because "$1/0$" does not exist. Note also that this function is continuous on $\mathbb{R}^{\ast}$.
However, this is not all we can say. Since the function $x \, \longmapsto \, \sin \Big( \frac{1}{x} \Big)$ is bounded, one can prove that :
$$ \lim \limits_{x \to 0} x^{2} \sin \Big( \frac{1}{x} \Big) = 0.$$
As a consequence, even though the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ is not defined at $x=0$, it has a finite limit as $x$ goes to $0$. So, this function can be extended to a new function, say $\overline{f}$, defined on $\mathbb{R}$ the following way :
$$ \overline{f}(x) =
\begin{cases} x^{2} \sin \Big( \frac{1}{x} \Big) & \text{if } x \neq 0 \\[2mm]
0 & \text{if } x = 0
\end{cases}
$$
This new function $\overline{f}$ is defined at $x=0$ and continuous on $\mathbb{R}$. In a very unformal, non-mathematical way, we could say that the value of the function $x \, \longmapsto \, x^{2} \sin \Big( \frac{1}{x} \Big)$ at $x=0$ is $0$. But this is not really the truth !


text:
\begin{document}

\baselineskip=17pt

\title{Automaticity of the sequence of the last nonzero digits of $n!$ in a fixed base}

\author{Eryk Lipka}
\address{Graduate of Institute of Mathematics\\ Jagiellonian University\\
Kraków, Poland}
\email{eryklipka0@gmail.com}

\begin{abstract}
In 2011 Deshouillers and Ruzsa (\cite{deshruzsa}) tried to argument that the sequence of the last nonzero digit of $n!$ in base 12 is not automatic. This statement was proved few years later by Deshoulliers in \cite{desh2}. In this paper we provide alternate proof that lets us generalize the problem and give an exact characterization in which bases the sequence of the last nonzero digits of $n!$ is automatic.
\end{abstract}

\keywords{automatic sequence, factorial, the last nonzero digit}
\subjclass[2010]{Primary 11B85; Secondary 11A63, 68Q45, 68R15}

\maketitle
\section{Introduction}
Let $\lns{b}$ be the sequence of last nonzero digits of $n!$ in base $b$, in this paper we will answer the question for which values of $b$ is this sequence automatic. It was known that $\lns{b}$ is automatic in many cases including bases being primes or powers of primes, one can also prove that $\lns{b}$ is automatic for some small bases that have more prime factors, like 6 or 10. In general, for base of the form $b=p_1^{a_1}p_2^{a_2}$ where $p_1\neq p_2,\; p_1, p_2 \in \mathbb{P},\; a_1,a_2\in\mathbb{N}_+$, it can be shown that $\lns{b}$ is automatic when $a_1\left(p_1-1\right)\neq a_2\left(p_2-1\right)$. The smallest base for which the answer is unclear is 12. This was the case analysed by Deshouillers and Ruzsa in \cite{deshruzsa}. They conjectured that $\lns{12}$ can not be automatic, despite the fact that it is equal to some automatic sequence nearly everywhere. An attempt to prove that conjecture was done by Deshouliers in his paper \cite{desh1}, and few years later he answered the question by proving the following, stronger result
\begin{thm}{(Deshouillers \cite{desh2})}
For $a \in \{3, 6, 9\}$, the characteristic sequence of $\{n\, ; \ell_{12}\left(n!\right) = a\}$ is not automatic.
\end{thm}
Another way of proving similar fact (but for $a \in \{4,8\}$) was provided recently by Byszewski and Konieczny in \cite{kuba}. It seems that both proofs can be generalized to all cases when $a_1\left(p_1-1\right)=a_2\left(p_2-1\right)$, however it is not obvious if, or how, can it be extended to bases with more than two prime factors. This was our main motivation to write this paper, and we provide complete characterization in which bases is this sequence automatic, including those with many prime factors.

In this paper we will use the following notation: the string of digits of $n$ in base $k$ will be denoted $\left[n\right]_k$, by $v_b\left(n\right)$ we mean the largest integer $t$ such that $b^t|n$, and $s_b\left(n\right)$ is the sum of digits of $n$ in base $b$.
This paper is composed of two main parts, first we recall some basic facts about automatic sequences for readers not familiar with the topic, in the latter part we present our results about automaticity of $\lns{b}$.

We would like to thank Piotr Miska and Maciej Ulas for proof reading and helpful suggestions while preparing this paper.

\section{Basics of automatic sequences}
In this section we will give short summary of topics from automatic sequence theory that we will be using later. If the reader is interested in getting more insight into this topic, we strongly recommend book of Allouche and Shallit \cite{alloucheshallit} that covers all important topics in this area.
\begin{defin}
{\bf Deterministic finite automaton with output} is a 6-tuple $\left(Q, \Sigma, \rho, q_0, \Delta, \tau \right)$ such that
\begin{itemize}
\item$Q$ is a {\bf finite} set of states;
\item$\Sigma$ is an input alphabet;
\item$\rho: Q\times\Sigma \rightarrow Q$ is a transition function;
\item$q_0 \in Q$ is an initial state;
\item$\Delta$ is an output alphabet (finite set);
\item$\tau:Q\rightarrow\Delta$ is an output function.
\end{itemize}
Transition function can be generalized to take strings of characters instead of single ones. For string $s_1s_2s_3\ldots$ we define $\rho\left(q, s_1s_2s_3\ldots\right)=\rho\left(\ldots\rho\left(\rho\left(q, s_1\right), s_2\right)\ldots\right)$.
\end{defin}

\begin{defin}
For any finite alphabet $\Sigma$, function  $f: \Sigma^* \rightarrow \Delta$ is called a {\bf finite-state function} if there exists a deterministic finite automaton with output $\left(Q, \Sigma, \rho, q_0, \Delta, \tau \right)$ such that $f(\omega) = \tau\left(\rho\left(q_0,\omega\right)\right)$.
\end{defin}

\begin{lem} \label{reverse}
If $f: \Sigma^* \rightarrow \Delta$ is a finite-state function then function $g: \Sigma^* \rightarrow \Delta$ defined as $g\left(\omega\right) = f\left(\omega^R\right)$ is also finite-state. ($ ^R$ denotes taking reverse of a word).
\end{lem}
\begin{proof}
(Sketch) Let $\left(Q, \Sigma, \rho, q_0, \Delta, \tau \right)$ be automaton that is related to $f$, we will define another automaton $\left(Q', \Sigma, \rho', q_0', \Delta, \tau' \right)$. Let $Q' = \Delta^Q$ be all functions from $Q$ to $\Delta$ and $q_0'\equiv \tau$. For any $g\in Q'$ we define $\tau'\left(g\right)=g\left(q_0\right)$, and for any $\sigma \in \Sigma, q\in Q$ we put $\rho'\left(g,\sigma\right)\left(q\right)=g\left(\rho\left(q,\sigma\right)\right)$. By induction on length of word $\omega\in\Sigma^*$ one can prove that equation
$$\rho'\left(g,\omega\right)\left(q\right)=g\left(\rho\left(q,\omega^R\right)\right)$$
holds for any $g \in Q', q\in Q$. And finally $$g\left(\omega\right)=\tau'\left(\rho'\left(q_0',\omega\right)\right)=\rho'\left(q_0',\omega\right)\left(q_0\right)=g\left(\rho\left(q_0,\omega^R\right)\right)=f\left(\omega^R\right).$$
\end{proof}

\begin{defin}
$\left(a\left(n\right)\right)_{n\in\mathbb{N}}$ is an {\bf $k$-automatic sequence} if function $\left[n\right]_k \rightarrow a_n$ is finite-state. By Lemma \ref{reverse} it is not important whether we read representation of $n$ from the right or from the left side.
\end{defin}

Now we present some simple examples of sequences that are automatic.
\begin{exa} \label{simpleexa}
The sequence $a_n = n \pmod{m}$ is $k$-automatic for any $k\geq 2, m\in \mathbb{Z}_+$. In order to see this it is enough to take $Q=\left\{0,1,\ldots,m-1\right\}, \rho\left(q,\sigma\right) = kq+\sigma \pmod{m}$ and read input "from left to right".\\
The sequence $a_n = s_k\left(n\right) \pmod{m}$ is $k$-automatic for any $k\geq 2, m\in \mathbb{Z}_+$. Take $Q=\left\{0,1,\ldots,m-1\right\}$ and $\rho\left(q,\sigma\right) = q+\sigma \pmod{m}$.\\
For any $k\geq 2$ and $x\in \mathbb{N}$, the characteristic sequence $a_n = \delta_x\left(n\right)$ is $k$-automatic. Automaton that computes it can be constructed by taking $\lceil \log_k\left(x\right) \rceil$ states that count how many digits were correct plus one "sinkhole" state that accepts all numbers other than $x$.
\end{exa}
We can also obtain automatic sequences by modifying existing ones.
\begin{exa}
If $\left(a\left(n\right)\right)_{n\in\mathbb{N}}$ is $k$-automatic sequence then so is $b_n = f(a_n)$ for any function $f$ taking values from the image of $a_n$. The difference will be only in the output function of related automaton.\\
If $\left(a\left(n\right)\right)_{n\in\mathbb{N}}, \left(b\left(n\right)\right)_{n\in\mathbb{N}}$ are $k$-automatic sequences, then so is $c_n = f(a_n, b_n)$ for any function $f$ as long as it is well defined on all possible pairs $(a_n, b_n)$. To obtain such automaton $\left(Q_c, \Sigma, \rho_c, q_c, \Delta_c, \tau_c \right)$ we can take the "product" of automatons $\left(Q_a, \Sigma, \rho_a, q_a, \Delta_a, \tau_a \right)$, $\left(Q_b, \Sigma, \rho_b, q_b, \Delta_b, \tau_b \right)$ defined by
\begin{itemize}
\item $Q_c=Q_a \times Q_b$;
\item $\rho_c\left(a,b\right) = \left(\rho_a\left(a\right),\rho_b\left(b\right)\right)$;
\item $q_c = \left(q_a,q_b\right)$;
\item $\Delta_c=f(\Delta_a\times \Delta_b)$;
\item $\tau_c\left(a,b\right) = f\left(\tau_a\left(a\right),\tau_b\left(b\right)\right)$.
\end{itemize}
This can be easily generalized to the case with $f$ taking any finite number of sequences as an input.
\end{exa}

By combining above examples together we can get some additional facts.
\begin{lem} Let $k\in \mathbb{N}_{\geq2}$ be fixed, then:
\begin{itemize}
\item characteristic sequence of a finite set is $k$-automatic;
\item if sequence $\left(a\left(n\right)\right)_{n\in\mathbb{N}}$ differs from $\left(b\left(n\right)\right)_{n\in\mathbb{N}}$ only on finitely many terms and one of them is $k$-automatic so does the other one;
\item periodic sequence is $k$-automatic;
\item ultimately periodic sequence is $k$-automatic;
\end{itemize}
\end{lem}

Of course this does not exhaust all possible automatic sequences, but is enough to give some insight and be useful in our work. We should also notice what is the relation between automaticity in different bases.

\begin{lem} \label{automatonpower}
Sequence $\left(a\left(n\right)\right)_{n\in\mathbb{N}}$ is $k$-automatic if and only if it is $k^m$-automatic for all $m\in\mathbb{N}_{\geq 2}$.
\end{lem}
\begin{proof} (Sketch)
If we have $k$-automaton generating a sequence, then we can easily manipulate it to create $k^m$-automaton generating the same sequence, main idea is to take transition function to be $m$-th composition of the original transition function with itself (digit in base $k^m$ can be seen as $m$ digits in base $k$). \\
On the other hand, let $Q$ be set of states of the $k^m$-automaton generating a sequence, and $\rho$ be its transition function. We take $Q' = Q\times\{0,1,\ldots,k^{m-1}-1\}\times\{0,1,\ldots,m-1\}$ and
$$\rho'\left(\left(q,r,s\right),\sigma\right)=\begin{cases}
(q,kr+\sigma,s+1) & \text{if } s<m-1\\
(\rho(q,kr+\sigma),0,0) & \text{if } s=m-1\end{cases},$$
this way we accumulate base $k$ digits until we collect $m$ of them and then use the original transition function.
\end{proof}

\section{New results}
Lets start with some facts that we will be using in our proof
\begin{prop}(Legendre's formula \cite{legendre}) \label{legendre}
for any prime $p$ and positive integers $a,n$, we have
$$
v_{p^a}(n!) = \left\lfloor\frac{n-s_p\left(n\right)}{a\left(p-1\right)}\right\rfloor.
$$
\end{prop}
\begin{prop} \label{sum} (Result from \cite{stewart})
For any positive integers $b, c$ such that $\frac{\ln(b)}{\ln(c)}\not\in\mathbb{Q}$ there exists a constant $d$ such that for each integer $n>25$ there holds
$$
s_b\left(n\right) + s_{c}\left(n\right) > \frac{\log\log n}{\log \log \log n + d}-1.
$$
\end{prop}
Next proposition is known fact, but We haven't found it clearly stated anywhere, it can be easily proven using Dirichlet's approximation theorem or Equidistribution theorem.
\begin{prop} \label{word} For any positive integers $a, b, c$ such that $\frac{\ln(b)}{\ln(c)}\not\in\mathbb{Q}$ there exist infinitely many triples of non-negative integers $d,e,f$ with $1\leq f<b^e$ such that
$$
c^d = a\cdot b^e +f.
$$
In other words, there are infinitely many powers of $c$ with base $b$ notation starting with given string of digits.
\end{prop}
After such introduction we can finally state our results. The following lemma and theorem are the main steps in proving when $\lns{b}$ is not automatic.

\begin{lem} \label{alwaysword}
Let $P$ be a non-empty finite set of prime numbers and $p$ be its biggest element. Let $a>0, k>1$ be integers. Then there exist an integer $a'$ such that $\max_{i\in P}\left\{s_i\left(a'\right)\right\}=s_p\left( a'\right)$ and $\left[a\right]_k$ is prefix of $\left[a'\right]_k$.
\end{lem}
\begin{proof}
If $k$ is not a power of $p$, then by Proposition \ref{word} there exist infinitely many triples $(d,e,f)$ of non-negative integers with $1\leq f+1<k^e$ such that
$$p^d = a\cdot k^e + (f+1).$$
Furthermore we have
$$s_p\left(p^d-1\right) = d\left(p-1\right)>\left(p-1\right)\frac{\ln\left(p^d-1\right)}{\ln(p)},$$
and from the definition of $s_q$, for any prime $q$ the following holds
$$s_q\left(p^d-1\right) < \left(q-1\right)\left(\frac{\ln\left(p^d-1\right)}{\ln(q)}+1\right).$$
Because $p$ is the biggest number in $P$, then for any $q\in P, q\neq p$, we have
$$s_q\left(p^d-1\right) - s_p\left(p^d-1\right)< \ln\left(p^d-1\right)\left(\frac{q-1}{\ln\left(q\right)}-\frac{p-1}{\ln\left(p\right)}\right) + q -1.$$
Right side of this inequality is negative for $d$ big enough, so because $0\leq f<k^e$ we can take $a' = p^d-1$. When $k=p^t$ we can notice that for any integer $d$
$$s_p\left( a\cdot p^{td}+p^{td}-1\right) =s_p\left(a\right)+ td\left(p-1\right)>\left(p-1\right)\left(\frac{\ln\left(a\cdot p^{td}+p^{td}-1\right)}{\ln(p)}-\frac{\ln\left(a\right)}{\ln(p)}-1\right),$$
and by similar argument it is enough to take $a'=a\cdot p^{td}+p^{td}-1$ for $d$ sufficiently large.
\end{proof}

\begin{thm} \label{sets}
Let $P$ be a finite set of prime numbers with at least two elements and $p$ be its biggest element, also let $c>0$ be a real number. Let us define sets
\begin{align*}
A_- &= \left\{n\in\mathbb{Z}_+:\;\max_{i\in P}\left\{s_i\left(n\right)\right\}=s_p\left(n\right)\right\},\\
A_+ &= \left\{n\in\mathbb{Z}_+:\;\max_{i\in P}\left\{s_i\left(n\right)\right\}-s_p\left(n\right)\geq c\right\}.
\end{align*}
Then there does not exist deterministic finite automaton with output that assigns one value to integers in $A_-$ and other value to those in $A_+$.
\end{thm}
\begin{proof}
Lets suppose that we have such an automaton $\left(Q, \Sigma_k, \rho, q_0, \Delta, \tau \right)$ for some $k$. Because $Q$ is finite, there exists some internal state $\mathcal{S}\in Q$ such that for infinitely many positive integers $c_1 < c_2 < \ldots$ we have $\rho\left(q_0,\left[p^{c_i}\right]_k\right) = \mathcal{S}$. Now, by Lemma \ref{alwaysword}, there exists an integer $a'\in A_-$ which can be obtained from $p^{c_1}$ by appending some suffix. Hence we can fix positive integers $e,f < k^e$ such that $a'=p^{c_1}\cdot k^e+f$. Let the sequence of digits $\left(f_1, f_2, \ldots, f_e\right)$ be a representation of $f$ in base $k$, possibly with added leading zeros. By $\mathcal{T}\in Q$ we denote an internal state such that $$\mathcal{T} = \rho\left(q_0, \left[a'\right]_k\right) = \rho\left(\mathcal{S},f_1f_2\ldots f_e\right).$$
This means that for every $i\in \mathbb N _+$ we have
$$\rho\left(q_0, \left[p^{c_i}\cdot k^e +f\right]_k\right) =\rho\left(q_0, \left[p^{c_i}\right]_kf_1f_2\ldots f_e\right) =\rho\left(\mathcal{S},f_1f_2\ldots f_e\right) = \mathcal{T},$$
and this implies that $\tau\left(\rho\left(q_0, \left[p^{c_i}\cdot k^e +f\right]_k\right)\right)=\tau\left(\mathcal{T}\right)$ does not depend on value of $i$.\\
On the other hand, when $c_i > \left\lceil \log_p(f)\right \rceil$ we have $s_p\left(p^{c_i}\cdot k^e +f\right)=s_p\left(k^e\right)+s_p\left(f\right)$ which is a constant. However, due to Proposition \ref{sum} we know that for any $q\in P, q \neq p$, the value of $s_q\left(p^{c_i}\cdot k^e+f\right)$ is increasing with $c_i$. Hence for $c_i$ big enough there holds $p^{c_i}\cdot k^e+f \in A_+$.\\
All but finitely many integers of the form $p^{c_i}\cdot k^e +f$ are elements of $A_+$ but at least one (namely $p^{c_1}\cdot k^e +f$) is an element of $A_-$. This proves that such automaton cannot assign different values to members of those two sets.
\end{proof}

Now we will show that $l_b\left(n!\right)$ can be automatic for some $b$.

\begin{lem} \label{bpa}
If $b=p^a, p\in \mathbb{P}, a\in \mathbb{N}$ then the sequence $\left(\ell_b\left(n!\right)\right)_{n\in\mathbb{N}}$ is $b$-automatic.
\end{lem}
\begin{proof}
First, we notice that $\ell_b\left(xy\right)=\ell_b\left(\ell_b\left(x\right)\ell_b\left(y\right)\right)$, so
$$\ell_b\left((bn)!\right) =\ell_b\left(\ell_b\left(n!\right)\prod_{i=n+1}^{bn} \ell_b\left(i\right) \right).$$
Because $\ell_b\left(bx\right) = \ell_b\left(x\right)$ we can rewrite the product in the following way
$$\ell_b\left((bn)!\right) =\ell_b\left(\ell_b\left(n!\right)\prod_{\substack{i=1 \\ b \nmid i}}^{bn} \ell_b\left(i\right) \right)=\ell_b\left(\ell_b\left(n!\right)\prod_{i=1}^{n} \ell_b\left(\prod_{j=1}^{j=b-1}j\right) \right).$$
We denote $m_i=\ell_b\left(i!\right)$ and obtain $\ell_b\left((bn)!\right) =\ell_b\left(\ell_b\left(n!\right) m_{b-1}^n \right)$. Now we take the string of digits $n_1n_2\ldots n_l = \left[n\right]_b$ and obtain the following formula
$$\ell_b\left(n!\right)=\ell_b\left(\left(n_1n_2\ldots n_l\right)!\right)=\ell_b\left(m_{n_l}\ell_b\left(\left(n_1n_2\ldots n_{l-1}\right)!\right)m_{b-1}^{\left(n_1n_2\ldots n_{l-1}\right)}\right),$$
which by iteration leads to
\begin{equation} \label{eq:factorial1}
\ell_b\left(\left(n_1n_2\ldots n_l\right)!\right)=\ell_b\left(m_{n_1}m_{n_2}\ldots m_{n_l} \ell_b\left(m_{b-1}^r\right)\right),
\end{equation}
Where $r=\left(n_1n_2\ldots n_{l-1}\right) +\ldots +\left(n_1n_2\right) + \left(n_1\right)$. Now, by Euler's Theorem $m_{b-1}^{\varphi(b)}\equiv m_{b-1}^{p^a-p^{a-1}}\equiv1 \pmod b$ so we only need to know the value of $r \pmod{p^a-p^{a-1}}$.
\begin{equation} \label{eq:factorial2}
r =\sum_{i=1}^{l-1}\left(b^{i-1}\sum_{j=1}^{l-i}n_j\right)
\equiv\sum_{i=1}^{l-1}n_i + p^{a-1}\sum_{i=1}^{l-2}\left(l-1-i\right)n_i \pmod{\left(p^a-p^{a-1}\right)}.
\end{equation}
Finally, we can define an automaton $\left(Q, \Sigma_b, \rho, q_0, \Delta, \tau \right)$ generating the sequence $\left(\ell_b\left(n!\right)\right)_{n\in\mathbb{N}}$ in the following way:
\begin{itemize}
\item the input alphabet $\Sigma_b=\{0,1,2,\ldots,b-1\}$;
\item the output alphabet $\Delta = \{1,2,\ldots,b-1\}$;
\item the set of states $Q=\Delta\times\Sigma_{p^a-p^{a-1}}\times\Sigma_{p-1}$;
\item the initial state $q_0=\left(1,0,0\right)$;
\item the output function $\tau\left(u,v,w\right)=\ell_b\left(u \cdot m_{b-1}^{v+p^{a-1}w}\right)$;
\item the transition function $$\rho\left(\left(u,v,w\right),s\right)=\left(\ell_b\left(u\cdot m_s\right),\left(v+s\right) \pmod{\left(p^a-p^{a-1}\right)},\left(w+v\right) \pmod{\left(p-1\right)}\right).$$
\end{itemize}
With such definition we have $\rho\left(q_0, \left[n\right]_b\right) = \left(u,v,w\right)$ where
\begin{itemize}
\item $u = \ell_b\left(m_{n_1}m_{n_2}\ldots m_{n_l}\right)$;\\
\item $v = \sum_{i=1}^{l-1}n_i \pmod {\left(p^a-p^{a-1}\right)}$; \\
\item $w = \sum_{i=1}^{l-2}\left(l-1-i\right)n_i \pmod{\left(p-1\right)}$.
\end{itemize}
Hence using equations \eqref{eq:factorial1} and \eqref{eq:factorial2} we see, that $\ell_b\left(n!\right) =\tau\left(u,v,w\right)$.
\end{proof}

Now we are ready to prove the following
\begin{thm}Let $b=p_1^{a_1}p_2^{a_2}\ldots$ with $a_1\left(p_1-1\right)\ge a_2\left(p_2-1\right)\ge \ldots$. The sequence $\left(\ell_b\left(n!\right)\right)_{n\in\mathbb{N}}$ is $p_1$-automatic if $a_1\left(p_1-1\right)> a_2\left(p_2-1\right)$ or $b=p_1^{a_1}$ and not automatic otherwise.
\end{thm}
\begin{proof}
Let $n \gg 0$. For $b={p_1}^{a_1}$ the sequence is $b$-automatic from Lemma \ref{bpa}, by Lemma \ref{automatonpower} it is also $p_1$-automatic. If $b$ has more than one prime factor and $a_1\left(p_1-1\right)> a_2\left(p_2-1\right)$ we take $b' = \frac{b}{p_1^{a_1}}$ so $p_1 \nmid b'$. From Proposition \ref{legendre} and the definition of $v_{b'}$ we have
$$v_{b'}\left(n!\right)=\min_{i>1}v_{p_i^{a_i}}\left(n!\right)=\min_{i>1}\left\lfloor\frac{n-s_{p_i}\left(n\right)}{a_i\left(p_i-1\right)}\right\rfloor > \left\lfloor\frac{n-s_{p_1}\left(n\right)}{a_1\left(p_1-1\right)}\right\rfloor=v_{p_1^{a_1}}\left(n!\right),$$
which leads to $b'|\ell_b\left(n!\right)$. Thus $\ell_b\left(n!\right)\in\left\{b',2b',3b',\ldots\left(p_1^{a_1}-1\right)b'\right\}$, so value of $\ell_b\left(n!\right)$ can be computed from value of $\ell_b\left(n!\right) \pmod{p_1^{a_1}}$. We also know, that there exist integers $c_1, c_2$ satisfying the equation
$$n! = b^{\left(v_{p_1^{a_1}}\left(n!\right)\right)} \ell_b\left(n!\right) +  b^{\left(v_{p_1^{a_1}}\left(n!\right)+1\right)} c_1 = p_1^{a_1\left(v_{p_1^{a_1}}\left(n!\right)\right)} \ell_{p_1^{a_1}}\left(n!\right) +  p_1^{a_1\left(v_{p_1^{a_1}}\left(n!\right)+1\right)} c_2.$$
After division of the above equality by $p_1^{a_1\left(v_{p_1^{a_1}}\left(n!\right)\right)}$ we obtain the following
$$\left(b'\right)^{\left(v_{p_1^{a_1}}\left(n!\right)\right)} \ell_b\left(n!\right) + p_1^{a_1} \left(b'\right)^{\left(v_{p_1^{a_1}}\left(n!\right)+1\right)} c_1 =  \ell_{p_1^{a_1}}\left(n!\right) +  p_1^{a_1} c_2.$$
Now, we can notice that $\ell_b\left(n!\right){\left(b'\right)}^{v_{p_1^{a_1}}\left(n!\right)} \equiv \ell_{p_1^{a_1}}\left(n!\right) \pmod{p_1^{a_1}}$, hence to finish this part of proof we just need to construct an $p_1$-automaton that returns the value of $v_{p_1^{a_1}}\left(n!\right) \pmod{\varphi(p_1^{a_1})}$. By Proposition \ref{legendre} this value can be computed from $\left(n-s_{p_1}(n)\right) \pmod{\varphi(p_1^{a_1})\cdot a_1(p_1-1)}$, and such expression is $p_1$-automatic as we already mentioned in Example \ref{simpleexa}.

Now, in the last case, when $a_1\left(p_1-1\right)=a_2\left(p_2-1\right)$, let $I = \{i : a_i\left(p_i-1\right) =a_1\left(p_1-1\right)\}$. Without loss of generality we can assume $p_1=\max_{i\in I} p_i$. By Legendre formula (Proposition \ref{legendre}) we have
\begin{align*}
\max_{i\in I}s_{p_i}\left(n\right)=s_{p_1}\left(n\right) &\Rightarrow v_{p_1^{a_1}}\left(n!\right) = \min_{i\in I} v_{p_i^{a_i}}\left(n!\right) \Rightarrow p_1^{a_1} \nmid \ell_b\left(n!\right),\\
\max_{i\in I}s_{p_i}\left(n\right)> a_1(p_1-1) + s_{p_1}\left(n\right) &\Rightarrow v_{p_1^{a_1}}\left(n!\right) > \min_{i\in I} v_{p_i^{a_i}}\left(n!\right) \Rightarrow p_1^{a_1} | \ell_b\left(n!\right).
\end{align*}
Hence, by Theorem \ref{sets}, there is no finite automaton that can, for given $n$, tell whether $p_1^{a_1}$ divides $\ell_b\left(n!\right)$ or not. This completes the proof, as finite automaton generating the sequence $\left(\ell_b\left(n!\right)\right)_{n\in\mathbb{N}}$ should distinguish those two sets.
\end{proof}


TITLE: generalized eigenvectors - questions
QUESTION [0 upvotes]: We introduced generalized eigenvectors and then we were given this example:
Given an endomorphism $\varphi: V \to V$ with $\chi_\varphi(t) = (-1)^n (t-\lambda)^n$, then $U(\lambda) = V$ since $(\varphi-\lambda · \mathrm{id})^n = 0$.
I don't really get this. Why is it that $(\varphi-\lambda · \mathrm{id})^n = 0$? If this is the case then clearly $U(\lambda) = V$ but I can't justify this.

REPLY [1 votes]: This is by the Cayley–Hamilton theorem: $0=\chi_\varphi(\varphi)=(-1)^n(\varphi-\lambda I)^n$


TITLE: Methods for calculating the mean and variance of a distribution created from the addition of two normally distributed quantities
QUESTION [1 upvotes]: I'm trying to understand how to interpret the following which refers to determination of the mean and variance of a distribution that's the result of adding two normally distributed random variables.
$\sum_{i=1}^{n}X_i \sim N(\sum_{i=1}^{n}\mu_i, \sum_{i=1}^{n}\sigma^{2}_i)$
Let's imagine I've a bag of 100 lotto balls each with some number printed on the side, and the values of these numbers are distributed normally about the mean 1 with a variance of 1.
(OK, so normal lotto balls are integers, but let's assume a continuous notion of 'value' here)
Now, if I add another bag of 100 lotto balls, which are also normally distributed about the mean, this time of 100, with a variance of 1.
What is my expected mean and variance of the new population?
If I start from scratch and calculate the new mean, it's clear that the mean is now $101/2 = 50.5$, but from the addition of distribution formula above, I'd expect the new mean to be $\sum_{i=1}^n\mu_{i}$ which would, in this case be 101.
Similarly, the variance of such a distribution would not be 2 (i.e. $1^2 + 1^2$ ) but something much larger.
So my question here is, how am I misinterpreting this formula?
Is there a more appropriate function that will provide me the new mean and variation of a random variable that's the result of adding or mixing two known populations, calculable from totals, means, variances or other descriptive data, without resorting to recalculation from the elemental data?
And part two of my question is, under what circumstances would you use the distribution addition formula above? I can see how by adding heights of (say) paired elements I might find the results described, but how does this work if the distributions describe populations of different sizes?

REPLY [1 votes]: Your question (and to an extent also the Answer by @ZoranLoncarevic) fails to distinguish explicitly between the ideas of 'adding' random variables and 'mixing' their distributions. If I add $X \sim Norm(\mu=0, \sigma^2 = 4)$ and $Y \sim Norm(\mu=8, \sigma^2 = 4)$, then I get the random variable $T = X + Y \sim Norm(\mu = 8, \sigma^2 = 8)$. If I make a 50-50 mixture of these two distributions, I get a bimodal distribution with mean 4. In the
latter case the PDF of the mixture is an average of the PDFs of
$X$ and $Y$. (The mixture is bimodal if the means are far enough
apart; here more than twice the common SD $\sigma = 2.$)
The two histograms below illustrate the distinction. At the
end of the program, approximate means, variances, and SDs of
the two simulated distributions are shown.
 x = rnorm(10^5, 0, 2);  y = rnorm(10^5, 8, 2)  # SDs are 2
 t = x + y;  # sum
 h = rbinom(10^5, 1, 1/2); v = h*x + (1-h)*y    # 50-50 mixture
 par(mfrow=c(1,2))
 hist(t, br=20, prob=T, col="wheat", main="Sum of Two Normals")
   curve(dnorm(x, 8, sqrt(8)), lwd=2, col="blue", add=T)
 hist(v, prob=T, col="wheat", main="Mixture of Two Normals")
   curve(.5*dnorm(x, 0, 2) + .5*dnorm(x, 8, 2), lwd=2, col="darkgreen", add=T)
 par(mfrow=c(1,1))

 mean(t);  var(t);  sd(t)
 ## 7.991537  # Approx E(T) = 8
 ## 8.026132  # Approx Var(T) = 8
 ## 2.833043  # Approx SD(T)
 mean(v);  var(v);  sd(v)
 ## 3.988405  # Approx E(V) = 4
 ## 19.99138  # Approx Var(V)
 ## 4.471172  # Approx SD(V)


TITLE: If $A$ is independent of $B$ and $C$ then why is it not necessarally independent of $B\cap C$?
QUESTION [3 upvotes]: I'm attempting to acquire an intuitive understanding of why the content in the question of the title is correct, however I am unable to do so. Is there way of thinking about the result that makes sense?

REPLY [1 votes]: Consider the following Venn diagram.  $P(A)=P(A|B)=P(A|C)=\frac 58$ but $P(A|BC)=1$


TITLE: Are coherent modules hopfian?
QUESTION [2 upvotes]: As it is well-known, a noetherian module over an arbitrary ring is hopfian.

Are coherent modules also hopfian?

REPLY [1 votes]: You can even find a ring $R$ such that $_RR$ is coherent but not Hopfian.  (Keep in mind for rings being Hopfian amounts to being Dedekind finite.)
One such example is the endomorphism ring of an infinite dimensional vector space.  It's left coherent because it's left semihereditary.  It's well-known to be non-Dedekind finite.


text:
\section{Introduction}

\subsection{What is {\tt LattE}?} \label{intro}

The name ``{\tt LattE}'' is an abbreviation for ``{\bf Latt}ice point
{\bf E}numeration.'' So what exactly does {\tt LattE} do? The
software's main function is to count the lattice points contained in
convex polyhedra defined by linear equations and inequalities with
integer coefficients. The polyhedra can be of any (reasonably small)
dimension, and {\tt LattE} uses an algorithm that runs in polynomial
time for fixed dimension: Barvinok's algorithm \cite{BarviPom}. To
learn more about the exact details of our implementation and
algorithmic techniques involved, the interested reader can consult
\cite{latte,newlatte,latte3} and the references listed therein. Here we
give a rather short description of the mathematical objects used by
{\tt LattE}, {\bf Barvinok's Rational Functions}:

\noindent
Given a convex polyhedron $P = \{u\in\R^d:Au\leq b\}$, where $A$ and
$b$ are integral, the fundamental object that we compute is a short
representation of the infinite power series:
\[
f(P;x) \quad = \sum_{\alpha\in P\cap\Z^d} x_1^{\alpha_1}
x_2^{\alpha_2} \ldots x_d^{\alpha_d}.
\]
Here each lattice point is given by one monomial. Note that this can be
a rather long sum, in fact for a polyhedral cone it can be infinite, but
the good news is that it admits short representations.

\noindent {\bf Example:} Let $P$ be the quadrangle with vertices
$V_1=(0,0)$, $V_2=(5,0)$, $V_3=(4,2)$, and $V_4=(0,2)$.

\begin{figure}[thb]
\begin{center}
     \includegraphics[scale=.79]{examplebrion.eps}
\end{center}
\end{figure}

{\small
\noindent
$f(P;x,y)={x}^{5}+{x}^{4}y+{x}^{4}+{x}^{4}{y}^{2}+y{x}^{3}+{x}^{3}+
{x}^{3}{y}^{2}+y{x}^{2}+{x}^{2}+{x}^{2}{y}^{2}+xy+x+x{y}^{2}+y+1+
{y}^{2}$
}

The fundamental theorem of Barvinok (circa 1993, see \cite{BarviPom})
says that you can write $f(P;x)$ as a sum of short rational functions,
in polynomial time when the dimension of the polyhedron is fixed.
In our running example we easily see that the 16 monomial polynomial
can be written as shorter rational function sum:

\noindent $f(P;x,y)=f(K_{V_1};x,y)+f(K_{V_2};x,y)+f(K_{V_3};x,y)+f(K_{V_4};x,y)$

\noindent where

$f(K_{V_1};x,y)={\frac {1}{\left (1-x\right )\left (1-y\right )}} \quad f(K_{V_2};x,y)=\frac{({x}^{5}+{x}^{4}y)}{ (1-{x}^{-1}) (1-y^2x^{-1})}$

$f(K_{V_3};x,y)=\frac{({x}^{4}{y}^{2}+{x}^{4})}{ (1-{x}^{-1})
(1-xy^{-2})}
 \quad  f(K_{V_4};x,y)=\frac{y^{2}}{(1-{y}^{-1} )(1-x )}$
\vskip .5cm

$ f(P; 1,1)=16$

Counting the lattice points in convex polyhedra is a powerful tool which
allows many applications in areas such as Combinatorics, Statistics,
Optimization, and Number Theory.

\subsection{What can {\tt LattE} compute?}

In the following we list the operations that {\tt LattE} v1.1 can
perform on bounded convex polyhedra (more commonly referred to as
\textit{polytopes}). For the reader's convenience, we already include
the basic commands to actually do the tasks. Let us assume that a
description of a polytope $P$ is given in the file ``fileName'' (see
Section \ref{Input Files} for format) and that a cost vector is
specified in the file ``fileName.cost'' (needed for the optimization
part, see Section \ref{Input Files} for format).

Tasks performed by {\tt LattE} v1.1:
\begin{enumerate}
\item Count the number of lattice points in $P$.
\begin{verbatim}
     ./count fileName
\end{verbatim}
\item Count the number of lattice points in $nP$, the dilation of $P$
  by the integer factor $n$.
\begin{verbatim}
     ./count dil n fileName
\end{verbatim}
\item Calculate a rational function that encodes the \textit{Ehrhart
  series} associated with the polytope. By definition, the $n$-th
  coefficient in the Ehrhart series equals the number of lattice
  points in $nP$. For more details on Ehrhart counting functions see,
  for example, Chapter 4 of \cite{stanley}.
\begin{verbatim}
     ./ehrhart fileName
\end{verbatim}
\item Calculate the first $n+1$ terms of the Ehrhart series associated
  with the polytope.
\begin{verbatim}
     ./ehrhart n fileName
\end{verbatim}
\item Maximize or minimize a given linear function of the lattice
  points in $P$.
\begin{verbatim}
     ./maximize fileName
     ./minimize fileName
\end{verbatim}
\end{enumerate}

In addition to these basic functions, there are more specific calls to
{\tt LattE}. For example to use the homogenized Barvinok algorithm
instead of the original one in order to count the lattice points. These
details will be explained in Section \ref{Running LattE}.

\newpage

\section{Downloading and Installing {\tt LattE}}

{\tt LattE} is downloadable from the following website:

\makebox[12 cm]{http://www.math.ucdavis.edu/$\sim$latte/downloads/}

\textbf{Step 1: Create directory for {\tt LattE}}

\begin{verbatim}
     mkdir latte
\end{verbatim}

\textbf{Step 2: Download ``latte\_v1.1.tar.gz'' to directory ``latte''}

\begin{verbatim}
     Download ``latte_v1.1.tar.gz'' from
\end{verbatim}

\makebox[12 cm]{http://www.math.ucdavis.edu/$\sim$latte/downloads/}

(If you have never downloaded a file from the internet: A click with
your right mouse button onto the file name on the webpage should do
the trick. In any case, if you do not succeed, ask your system
administrator, a friend, or send us an email.)

\textbf{Step 3: Change to directory for ``latte''}

\begin{verbatim}
     cd latte
\end{verbatim}

\textbf{Step 4: Unzip and untar the archive}

\begin{verbatim}
     gunzip latte_v1.1.tar.gz
     tar xvf latte_v1.1.tar
\end{verbatim}

\textbf{Step 5: Make ``install'' executable}

\begin{verbatim}
     chmod 700 install
\end{verbatim}

\textbf{Step 6: Install {\tt LattE}}

\begin{verbatim}
     ./install
\end{verbatim}

\newpage

\section{Input Files}\label{Input Files}

\subsection{{\tt LattE} Input Files}

\subsubsection{Inequality Description}
For computations involving a polytope $P$ described by a
system of inequalities $Ax\leq b$, where $A\in\Z^{m\times d}$,
$A=(a_{ij})$, and $b\in\Z^m$, the {\tt LattE} readable input file
would be as follows:
\begin{verbatim}
m d+1
b  -A
\end{verbatim}

\textbf{EXAMPLE.}
Let $P=\{(x,y): x\leq 1, y\leq 1, x+y\leq 1, x\geq 0, y\geq 0\}$.
Thus
\[
\begin{array}{ccc}
A=\left(
\begin{array}{rr}
 1 &  0 \\
 0 &  1 \\
 1 &  1 \\
-1 &  0 \\
 0 & -1 \\
\end{array}
\right)
& , &
b = \left(
\begin{array}{r}
1 \\
1 \\
1 \\
0 \\
0 \\
\end{array}
\right)
\end{array}
\]
and the {\tt LattE} input file would be as such:
\begin{verbatim}
5 3
1 -1  0
1  0 -1
1 -1 -1
0  1  0
0  0  1
\end{verbatim}

\subsubsection{Equations}
In {\tt LattE}, polytopes are represented by {\bf linear constraints},
i.e. equalities or inequalities. By default a constraint is an
inequality of type $ax\leq b$ unless we specify, by using a single
additional line, the line numbers of constraints that are linear
equalities.

\textbf{EXAMPLE.}
Let $P$ be as in the previous example, but require $x+y=1$ instead of
$x+y\leq 1$, thus,
$P=\{(x,y): x\leq 1, y\leq 1, x+y=1, x\geq 0, y\geq 0\}$.
Then the {\tt LattE} input file that describes $P$ would be as such:
\begin{verbatim}
5  3
1 -1  0
1  0 -1
1 -1 -1
0  1  0
0  0  1
linearity 1 3
\end{verbatim}
The last line states that among the $5$ inequalities one is to be
considered an equality, the third one.

\subsubsection{Nonnegativity Constraints}
For bigger examples it quickly becomes cumbersome to state all
nonnegativity constraints for the variables one by one. Instead, you
may use another short-hand.

\textbf{EXAMPLE.}
Let $P$ be as in the previous example, then the {\tt LattE} input file
that describes $P$ could also be described as such:
\begin{verbatim}
3  3
1 -1  0
1  0 -1
1 -1 -1
linearity 1 3
nonnegative 2 1 2
\end{verbatim}
The last line states that there are two nonnegativity constraints and
that the first and second variables are required to be nonnegative.
{\bf NOTE} that the first line reads ``3 3'' and not ``5 3'' as above!

\subsubsection{Cost Vector}
The functions maximize and minimize solve the integer linear programs
\[
\max\{c^\intercal x: x\in P\cap\Z^d\}
\]
and
\[
\min\{c^\intercal x: x\in P\cap\Z^d\}.
\]
Besides a description of the polyhedron $P$, these functions need a
linear objective function given by a certain cost vector $c$. If the
polyhedron is given in the file ``fileName''
\begin{verbatim}
4  4
1 -1  0  0
1  0 -1  0
1  0  0 -1
1 -1 -1 -1
linearity 1 4
nonnegative 3 1 2 3
\end{verbatim}
the cost vector must be given in the file ``fileName.cost'', as for
example in the following three-dimensional problem:
\begin{verbatim}
1 3
2 4 7
\end{verbatim}
The first two entries state the size of a $1\times n$ matrix (encoding
the cost vector), followed by the $1\times n$ matrix itself. Assuming
that we call maximize, this whole data encodes the integer program
\[
\max\{2x_1+4x_2+7x_3: x_1+x_2+x_3=1, x_1,x_2,x_3\in\{0,1\}\}.
\]

\subsection{{\tt cdd} Input Files}
In addition to the formats described above, {\tt LattE} can also
accept input files in standard {\tt cdd} format. (See Subsection
\ref{Command Syntax} for
details on how to run {\tt LattE} on a {\tt cdd} input file.) Below is
an example of {\tt cdd} input that is readable into {\tt LattE}.
\begin{verbatim}
H-representation
begin
4  4  integer
2 -2  4 -1
3 -2 -2  3
6  2 -4 -3
1  2  2  1
end
\end{verbatim}
It is important to note that {\tt LattE} can only read
{\em integer} input.  Clearly, {\tt cdd}'s rational data files can be
converted into integer files by multiplying by the right constants. In
the packaged release of {\tt LattE} we include a binary version of
{\tt cdd}.

\newpage

\section{Running {\tt LattE}}\label{Running LattE}

\subsection{Command Syntax}\label{Command Syntax}
The basic syntax to invoke the various functions of {\tt LattE} is:
\begin{verbatim}
./count fileName
./ehrhart fileName
./maximize fileName
./minimize fileName
\end{verbatim}
Note that the last two functions require a cost vector specified in
the file ``fileName.cost''!

Additionally, a variety of options can be used. All options should be
space-delimited in the command.

One option that can be set in addition to the options given below is
``cdd'' which tells {\tt LattE} to read its input from a {\tt cdd}
input file. Thus, the above invocations for {\tt cdd} input files would be
\begin{verbatim}
./count cdd fileName
./ehrhart cdd fileName
./maximize cdd fileName
./minimize cdd fileName
\end{verbatim}

\subsection{Counting}
\begin{itemize}
\item Count the number of lattice points in polytope $P$, where $P$
  is given in ``fileName''.
\begin{verbatim}
     ./count fileName
\end{verbatim}
\item Count the number of lattice points in $nP$, the dilation of $P$
  by the integer factor $n$.
\begin{verbatim}
     ./count dil n fileName
\end{verbatim}
\item Count the number of lattice points in the interior of the
  polytope $P$, where $P$ is given in ``fileName''.
\begin{verbatim}
     ./count int fileName
\end{verbatim}
\item Use the homogenized Barvinok algorithm \cite{latte3} to count
  the number of lattice points in the polytope $P$, where $P$ is given
  in ``fileName''. Use if number of vertices of $P$ is big compared to
  the number of constraints.
\begin{verbatim}
     ./count homog fileName
\end{verbatim}
\end{itemize}

\subsection{Ehrhart Series}
\begin{itemize}
\item Compute the Ehrhart series encoded as a rational function for the
  polytope given in ``fileName''. Writes the unsimplified rational function
  to file ``fileName.rat''.
\begin{verbatim}
     ./ehrhart fileName
\end{verbatim}
\item Compute the Ehrhart series encoded as a rational function for
  the polytope given in ``fileName''. {\bf NEEDS} Maple for
  simplification of terms. Writes the simplified rational function to
  file ``fileName.rat''.
\begin{verbatim}
     ./ehrhart simplify fileName
\end{verbatim}
\item Compute the Taylor series expansion of Ehrhart generating
  function up to degree $n$ for the polytope given in ``fileName''.
\begin{verbatim}
     ./ehrhart n fileName
\end{verbatim}
\end{itemize}

\subsection{Optimizing}
This functions {\bf NEEDS} a cost vector specified in ``fileName.cost''!!!
\begin{itemize}
\item Maximizes/Minimizes given linear cost function over the lattice
  points in the polytope given in ``fileName''. Digging algorithm
  \cite{latte3} is used. Optimal point and optimal value is returned.
\begin{verbatim}
     ./maximize fileName
     ./minimize fileName
\end{verbatim}
\item Maximizes/Minimizes given linear cost function over the lattice
  points in the polytope given in ``fileName''. Binary search
  algorithm is used. Only optimal value is returned.
\begin{verbatim}
     ./maximize bbs fileName
     ./minimize bbs fileName
\end{verbatim}
\end{itemize}

\newpage

\section{A Brief Tutorial}
In this section we invite the reader to follow along a few examples
that show how to use {\tt LattE} and also how to counter-check
results.

\subsection{Counting Magic Squares}
Our first example deals with counting magic $4\times 4$ squares. We
call a $4\times 4$ array of nonnegative numbers a magic square if the
sums of the $4$ entries along each row, along each column and along
the two main diagonals equals the same number $s$, the magic
constant. Let us start with counting magic $4\times 4$ squares that
have the magic constant $1$. Associating variables $x_1,\ldots,x_{16}$ with
the $16$ entries, the conditions of a magic $4\times 4$ square of
magic sum $1$ can be encoded into the following input file
``EXAMPLES/magic4x4'' for {\tt LattE}.
\begin{verbatim}
10 17
1 -1 -1 -1 -1  0  0  0  0  0  0  0  0  0  0  0  0
1  0  0  0  0 -1 -1 -1 -1  0  0  0  0  0  0  0  0
1  0  0  0  0  0  0  0  0 -1 -1 -1 -1  0  0  0  0
1  0  0  0  0  0  0  0  0  0  0  0  0 -1 -1 -1 -1
1 -1  0  0  0 -1  0  0  0 -1  0  0  0 -1  0  0  0
1  0 -1  0  0  0 -1  0  0  0 -1  0  0  0 -1  0  0
1  0  0 -1  0  0  0 -1  0  0  0 -1  0  0  0 -1  0
1  0  0  0 -1  0  0  0 -1  0  0  0 -1  0  0  0 -1
1 -1  0  0  0  0 -1  0  0  0  0 -1  0  0  0  0 -1
1  0  0  0 -1  0  0 -1  0  0 -1  0  0 -1  0  0  0
linearity 10 1 2 3 4 5 6 7 8 9 10
nonnegative 16 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
\end{verbatim}
Now we simply invoke the counting function of {\tt LattE} by typing:
\begin{verbatim}
    ./count EXAMPLES/magic4x4
\end{verbatim}

The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Total Unimodular Cones: 418
Maximum number of simplicial cones in memory at once: 27

*****  Total number of lattice points: 8  ****

Computation done.
Time: 1.24219 sec
\end{verbatim}
Therefore, there are exactly $8$ magic $4\times 4$ squares that
have the magic constant $1$. This is not yet impressive, as we could
have done that by hand. Therefore, let us try and find the
corresponding number for the magic constant $12$. Since this problem
is a dilation (by factor $12$) of the original problem, we do not have
to create a new file. Instead, we use the option ``dil'' to indicate
that we want to count the number of lattice points of a dilation of
the given polytope:
\begin{verbatim}
    ./count dil 12 EXAMPLES/magic4x4
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Total Unimodular Cones: 418
Maximum number of simplicial cones in memory at once: 27

*****  Total number of lattice points: 225351  ****

Computation done.
Time: 1.22656 sec
\end{verbatim}
Therefore, there are exactly $225351$ magic $4\times 4$ squares that
have the magic constant $12$. (We would NOT want to do THAT one by
hand, would we?!)

Here is some amazing observation: the running time of {\tt LattE}
is roughly the same for counting magic squares of sum $1$ and of sum
$12$. This phenomenon is due to the fact that the main part of the
computation, the creation of the generating function that encodes all
lattice points in the polytope, is nearly identical in both cases.

Although we may be already happy with these simple counting results,
let us be a bit more ambitious and and let us find a counting formula
that, for given magic sum $s$, returns the number of magic
$4\times 4$ squares that have the magic constant $s$.

For this, simply type (note that {\tt LattE} invokes {\tt Maple} to
simplify intermediate expressions):
\begin{verbatim}
    ./ehrhart simplify EXAMPLES/magic4x4
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen looks
as follows:
\begin{verbatim}
Rational function written to EXAMPLES/magic4x4.rat

Computation done.
Time: 0.724609 sec
\end{verbatim}
We are informed that this call created a file ``EXAMPLES/magic4x4.rat''
containing the Ehrhart series as a rational function:
{\small
\begin{verbatim}
(t^8+4*t^7+18*t^6+36*t^5+50*t^4+36*t^3+18*t^2+4*t+1)/(-1+t)^4/(-1+t^2)^4
\end{verbatim}
}
Now we could use {\tt Maple} (or your favorite computer algebra
software) to find a series expansion of this expression.
\begin{eqnarray*}
& &
\frac{t^8+4*t^7+18*t^6+36*t^5+50*t^4+36*t^3+18*t^2+4*t+1}{(-1+t)^4(-1+t^2)^4}\\
& = & 1+8t^1+48t^2+200t^3+675t^4+1904t^5+4736t^6+10608t^7+21925t^8+\\
& & 42328t^9+77328t^{10}+134680t^{11}+225351t^{12}+364000t^{13}+570368t^{14}+\\
& & 869856t^{15}+{O}(t^{16})\\
\end{eqnarray*}
The summands $8t$ and $225351t^{12}$ reconfirm our previous
counts.

Although this rational function encodes the full Ehrhart series, it is
not always as easy to compute as for magic $4\times 4$ squares. As it
turns out, adding and simplifying rational functions, although in just
one variable $t$, can be extremely costly due to the high powers in
$t$ and due to long integer coefficients that appear.

However, even if we cannot compute the full Ehrhart series, we can at
least try and find the first couple of terms of it.
\begin{verbatim}
    ./ehrhart 15 EXAMPLES/magic4x4
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Memory Save Mode: Taylor Expansion:
1
8t^1
48t^2
200t^3
675t^4
1904t^5
4736t^6
10608t^7
21925t^8
42328t^9
77328t^10
134680t^11
225351t^12
364000t^13
570368t^14
869856t^15
Computation done.
Time: 1.83789 sec
\end{verbatim}
Again, our previous counts are reconfirmed.

Nice, but the more terms we want to compute the more time-consuming
this task becomes. Clearly, if we could find sufficiently many
terms, we could compute the full Ehrhart series expansion in terms of
a rational function by interpolation.

\subsection{Counting Lattice Points in the $24$-Cell}
Our next example deals with a well-known combinatorial object, the
$24$-cell. Its description is given in the file ``EXAMPLES/24\_cell'':
\begin{verbatim}
24 5
2 -1  1 -1 -1
1  0  0 -1  0
2 -1  1 -1  1
2 -1  1  1  1
1  0  0  0  1
1  0  1  0  0
2  1 -1  1 -1
2  1  1 -1  1
2  1  1  1  1
1  1  0  0  0
2  1  1  1 -1
2  1  1 -1 -1
2  1 -1  1  1
2  1 -1 -1  1
2  1 -1 -1 -1
1  0  0  1  0
2 -1  1  1 -1
1  0  0  0 -1
2 -1 -1  1 -1
1  0 -1  0  0
2 -1 -1  1  1
2 -1 -1 -1  1
2 -1 -1 -1 -1
1 -1  0  0  0
\end{verbatim}
Now we invoke the counting function of {\tt LattE} by typing:
\begin{verbatim}
    ./count EXAMPLES/24_cell
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Total Unimodular Cones: 240
Maximum number of simplicial cones in memory at once: 30

*****  Total number of lattice points: 33  ****

Computation done.
Time: 0.429686 sec
\end{verbatim}
Therefore, there are exactly $33$ lattice points in the $24$-cell. We
get the same result by using the homogenized Barvinok algorithm:
\begin{verbatim}
    ./count homog EXAMPLES/24_cell
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Memory Save Mode: Taylor Expansion:

****  Total number of lattice points is: 33  ****

Computation done.
Time: 0.957031 sec
\end{verbatim}
But how many of these $33$ points lie in the interior of the
$24$-cell?
\begin{verbatim}
    ./count int EXAMPLES/24_cell
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Reading .ext file...

*****  Total number of lattice points: 1 ****
\end{verbatim}
Therefore, there only one of the $33$ lattice points in the $24$-cell
lies in the interior.

\subsection{Maximizing Over a Knapsack Polytope}
Finally, let us solve the problem ``cuww1'' \cite{cuww,latte3}. Its
description is given in the file ``EXAMPLES/cuww1'':
\begin{verbatim}
1 6
89643482 -12223 -12224 -36674 -61119 -85569
linearity 1 1
nonnegative 5 1 2 3 4 5
\end{verbatim}
The cost function can be found in the file ``EXAMPLES/cuww1.cost'':
\begin{verbatim}
1 5
213 -1928 -11111 -2345 9123
\end{verbatim}
Now let us maximize this cost function over the given knapsack
polytope. Note that by default, the digging algorithm as described in
\cite{latte3} is used.
\begin{verbatim}
    ./maximize EXAMPLES/cuww1
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
{\small
\begin{verbatim}
Finished computing a rational function.
Time: 0.158203 sec.

There is one optimal solution.

No digging.
An optimal solution for [213 -1928 -11111 -2345 9123] is: [7334 0 0 0 0].
The projected down opt value is: 191928257104
The optimal value is: 1562142.
The gap is: 7995261.806
Computation done.
Time: 0.203124 sec.
\end{verbatim}
}
The solution $(7334,0,0,0,0)$ is quickly found. Now let us try to
find the optimal value again by a different algorithm, the binary
search algorithm.
\begin{verbatim}
    ./maximize bbs EXAMPLES/cuww1
\end{verbatim}
The last couple of lines that {\tt LattE} prints to the screen
look as follows:
\begin{verbatim}
Total of Iterations: 26
The total number of unimodular cones: 125562
The optimal value: 1562142

The number of optimal solutions: 1
Time: 0.042968
\end{verbatim}
Note that we get the same optimal value, but no optimal solution is
provided.


TITLE: Automorphisms of genus 6 surfaces
QUESTION [14 upvotes]: I have found in Wikipedia that the number of automorphisms of a Riemann surface of genus 6 does not exceed 150. The page offers neither a proof nor a reference. Can someone help me?

REPLY [3 votes]: The question is purely combinatorial - the Hurwitz bound comes from the observation that the quotient of a surface by its automorphism group is a hyperbolic orbifold, and the hyperbolic orbifold of smallest area is the $(2, 3, 7)$ triangle orbifold, Dividing the area of your surface by the area of the orbifold gives you the Hurwitz bound. However, for some genera, there is no triangulation where the vertices have the right degrees, then you have to look for bigger triangle groups. For the surface of genus $6,$ you get the $(2, 3, 10)$ triangle group, but in any case, this is just a combinatorial exercise (for any fixed genus).


TITLE: Prove that if a set $A \subset \mathbb R^n$ is connected, then it has the Intermediate Value Property.
QUESTION [1 upvotes]: Prove that if a set $A \subset \mathbb R^n$ is connected, then it has the Intermediate Value Property.
Intermediate Value Property: let $f$ be a real-valued continuous function on a domain $A$ if $a,b \in A$  and $f(a)<c<f(b)$ , then  $c \in f(A)$
I know that $A \subset \mathbb R^n$ is connected meaning there exist no disjoint, non empty, open sets $U$ and $V$ such that $A=U \cup V$. This means the function $f:A \to \mathbb R$ should be continuous on $A$. therefore, $A$ should have the Intermediate Value Property. but I don't know hwo to prove it formally.

REPLY [1 votes]: Note $f^{-1} (U)$ is open if $U$ is open in $\mathbb{R}$ because $f$ is a continuous function. If your set $A$ doesn't have a point $M$ for which $f(M)=0$. Let $N$ be the set of negative reals and $P$ the set of positive reals. Then the union of the opens $f^{-1} (N)\cap A \neq \emptyset$ and $f^{-1} (P)\cap A\neq \emptyset$ is $A$. So that means $A$ is disconnected. Which is a contradiction.


TITLE: An integral inequality involving increasing function
QUESTION [1 upvotes]: Let $0\leq a< b \leq \pi/2$
Let $f:[a,b]\to\mathbb R$ be a positive, increasing function.
Prove that $\left|\int_a^b f(t)\cos(t)dt\right|\leq f(b)(b+\sin(b))-f(a)(b+\sin(a))$

I haven't made any progress. I tried integration by parts, to no avail. I noted the inequality isn't symetric w.r.t $a$ and $b$.
Thanks for any hint.

REPLY [1 votes]: If $f$ is increasing is almost everywhere differentiable. Since:
$$\frac{d}{db}\left[f(b)(b+\sin(b))-f(a)(b+\sin a)\right]=(f(b)-f(a))+f(b)\cos b+(b+\sin b)\,f'(b)$$
and $f(b)-f(a)>0,b+\sin b>0,f'(b)\geq 0$ since $f$ is increasing, by integrating both sides of:
$$ \frac{d}{db}\left[f(b)(b+\sin(b))-f(a)(b+\sin a)\right] > f(b)\cos b $$
we get our claim.