TITLE: Electricity voltage & current
QUESTION [1 upvotes]: Why does current return to its source?

REPLY [1 votes]: If I'm right, your question is "why does current need to return to the same source?" Electrical current is caused by the flow of electrons, i.e. when an electron moves from the shell of one atom to another, then there is a net current flow. However, this causes the ionization of the other atom, which is naturally in a neutral (non-ionized) state. As a result, this newly positively charged ion attracts an electron. The idea of current returning to the same place isn't exactly right - it's that current in must equal current out (see: Kirchhoff's Laws), most easily represented in a circuit by drawing a loop.


text:
\begin{document}
\title[Reverses of the Continuous Triangle Inequality]{Some Reverses of the
Continuous Triangle Inequality for Bochner Integral of Vector-Valued
Functions in Complex Hilbert Spaces}
\author{Sever S. Dragomir}
\address{School of Computer Science and Mathematics\\
Victoria University of Technology\\
PO Box 14428, MCMC 8001\\
Victoria, Australia.}
\email{sever@csm.vu.edu.au}
\urladdr{http://rgmia.vu.edu.au/SSDragomirWeb.html}
\date{April 16, 2004.}
\subjclass[2000]{Primary 46C05; Secondary 26D15.}
\keywords{Triangle inequality, Diaz-Metcalf inequality, Integral
inequalities, Complex Hilbert spaces.}

\begin{abstract}
Some reverses of the continuous triangle inequality for Bochner integral of
vector-valued functions in complex Hilbert spaces are given. Applications
for complex-valued functions are provided as well.
\end{abstract}

\maketitle

\section{Introduction}

Let $f:\left[ a,b\right] \rightarrow \mathbb{K}$, $\mathbb{K}=\mathbb{C}$ or
$\mathbb{R}$ be a Lebesgue integrable function. The following inequality is
the continuous version of the triangle inequality
\begin{equation}
\left\vert \int_{a}^{b}f\left( x\right) dx\right\vert \leq
\int_{a}^{b}\left\vert f\left( x\right) \right\vert dx,  \label{1.1}
\end{equation}
and plays a fundamental role in Mathematical Analysis and its applications.

It appears, see \cite[p. 492]{MPF}, that the first reverse inequality for (
\ref{1.1}) was obtained by J. Karamata in his book from 1949, \cite{K}:
\begin{equation}
\cos \theta \int_{a}^{b}\left\vert f\left( x\right) \right\vert dx\leq
\left\vert \int_{a}^{b}f\left( x\right) dx\right\vert  \label{1.2}
\end{equation}
provided
\begin{equation*}
-\theta \leq \arg \left[ f\left( x\right) \right] \leq \theta ,\ \ x\in
\left[ a,b\right]
\end{equation*}
for given $\theta \in \left( 0,\frac{\pi }{2}\right) .$

In \cite{SSD1}, the author has extended the above result for Bochner
integrals of vector-valued functions in real or complex Hilbert spaces.

If $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $ is a Hilbert
space over $\mathbb{K}$ $\left( \mathbb{K}=\mathbb{C},\mathbb{R}\right) $
and $f\in L\left( \left[ a,b\right] ;H\right) $, this means that $f:\left[
a,b\right] \rightarrow H$ is Bochner measurable on $\left[ a,b\right] $ and $
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt$ is finite, and there
exists a constant $K\geq 1$ and a vector $e\in H,$ $\left\Vert e\right\Vert
=1$ such that
\begin{equation}
\left\Vert f\left( t\right) \right\Vert \leq K\func{Re}\left\langle f\left(
t\right) ,e\right\rangle \text{ \ for a.e. \ }t\in \left[ a,b\right] ,
\label{1.3}
\end{equation}
then we have the inequality
\begin{equation}
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq K\left\Vert
\int_{a}^{b}f\left( t\right) dt\right\Vert .  \label{1.4}
\end{equation}
This provides a reverse inequality for the well know result for Bochner
integrals and vector-valued functions:
\begin{equation}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert \leq
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt  \label{1.5}
\end{equation}
for any $f\in L\left( \left[ a,b\right] ;H\right) .$

Note that the case of equality holds in (\ref{1.4}) (see \cite{SSD1}) if and
only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\frac{1}{K}\left( \int_{a}^{b}\left\Vert
f\left( t\right) \right\Vert dt\right) e.  \label{1.6}
\end{equation}

For some particular cases of interest, see \cite{SSD1}.

The main aim of the present paper is to point out some newer inequalities
for complex Hilbert spaces under various conditions for both $\func{Re}
\left\langle f\left( t\right) ,e\right\rangle $ and $\func{Im}\left\langle
f\left( t\right) ,e\right\rangle $ $\left( e\in H,\ \left\Vert e\right\Vert
=1\right) $ and in this way improve some earlier results from \cite{SSD1}
that have been stated for real or complex Hilbert spaces. Applications for
complex-valued functions are also provided.

\section{The Case of a Unit Vector}

The following result holds.

\begin{theorem}
\label{t2.1}Let $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $
be a complex Hilbert space. If $f\in L\left( \left[ a,b\right] ;H\right) $
is such that there exists $k_{1},k_{2}\geq 0$ with
\begin{equation}
k_{1}\left\Vert f\left( t\right) \right\Vert \leq \func{Re}\left\langle
f\left( t\right) ,e\right\rangle ,\ \ k_{2}\left\Vert f\left( t\right)
\right\Vert \leq \func{Im}\left\langle f\left( t\right) ,e\right\rangle
\label{2.1}
\end{equation}
for a.e. $t\in \left[ a,b\right] ,$ where $e\in H,$ $\left\Vert e\right\Vert
=1,$ is given, then
\begin{equation}
\sqrt{k_{1}^{2}+k_{2}^{2}}\int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt\leq \left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert .
\label{2.2}
\end{equation}
The case of equality holds in (\ref{2.2}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( k_{1}+ik_{2}\right) \left(
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\right) e.  \label{2.3}
\end{equation}
\end{theorem}

\begin{proof}
Using the Schwarz inequality $\left\Vert u\right\Vert \left\Vert
v\right\Vert \geq \left\vert \left\langle u,v\right\rangle \right\vert ,$ $
u,v\in H;$ in the complex Hilbert space $\left( H;\left\langle \cdot ,\cdot
\right\rangle \right) ,$ we have
\begin{align}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert ^{2}& =\left\Vert
\int_{a}^{b}f\left( t\right) dt\right\Vert ^{2}\left\Vert e\right\Vert ^{2}
\label{2.4} \\
& \geq \left\vert \left\langle \int_{a}^{b}f\left( t\right)
dt,e\right\rangle \right\vert ^{2}=\left\vert \int_{a}^{b}\left\langle
f\left( t\right) ,e\right\rangle dt\right\vert ^{2}  \notag \\
& =\left\vert \int_{a}^{b}\func{Re}\left\langle f\left( t\right)
,e\right\rangle dt+i\left( \int_{a}^{b}\func{Im}\left\langle f\left(
t\right) ,e\right\rangle dt\right) \right\vert ^{2}  \notag \\
& =\left( \int_{a}^{b}\func{Re}\left\langle f\left( t\right) ,e\right\rangle
dt\right) ^{2}+\left( \int_{a}^{b}\func{Im}\left\langle f\left( t\right)
,e\right\rangle dt\right) ^{2}.  \notag
\end{align}
Now, on integrating (\ref{2.1}), we deduce
\begin{equation}
k_{1}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq \int_{a}^{b}
\func{Re}\left\langle f\left( t\right) ,e\right\rangle dt,\quad
k_{2}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq \int_{a}^{b}
\func{Im}\left\langle f\left( t\right) ,e\right\rangle dt  \label{2.5}
\end{equation}
implying
\begin{equation}
\left( \int_{a}^{b}\func{Re}\left\langle f\left( t\right) ,e\right\rangle
dt\right) ^{2}\geq k_{1}^{2}\left( \int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt\right) ^{2}  \label{2.6}
\end{equation}
and
\begin{equation}
\left( \int_{a}^{b}\func{Im}\left\langle f\left( t\right) ,e\right\rangle
dt\right) ^{2}\geq k_{2}^{2}\left( \int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt\right) ^{2}.  \label{2.7}
\end{equation}
If we add (\ref{2.6}) and (\ref{2.7}) and use (\ref{2.4}), we deduce the
desired inequality (\ref{2.2}).

Further, if (\ref{2.3}) holds, then obviously
\begin{align*}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert & =\left\vert
k_{1}+ik_{2}\right\vert \left( \int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt\right) \left\Vert e\right\Vert \\
& =\sqrt{k_{1}^{2}+k_{2}^{2}}\int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt,
\end{align*}
and the equality case holds in (\ref{2.2}).

Before we prove the reverse implication, let us observe that, for $x\in H$
and $e\in H,$ $\left\Vert e\right\Vert =1,$ the following identity is valid
\begin{equation*}
\left\Vert x-\left\langle x,e\right\rangle e\right\Vert ^{2}=\left\Vert
x\right\Vert ^{2}-\left\vert \left\langle x,e\right\rangle \right\vert ^{2},
\end{equation*}
therefore $\left\Vert x\right\Vert =\left\vert \left\langle x,e\right\rangle
\right\vert $ if and only if $x=\left\langle x,e\right\rangle e.$

If we assume that equality holds in (\ref{1.2}), then the case of equality
must hold in all the inequalities required in the argument used to prove the
inequality (\ref{2.2}). Therefore, we must have
\begin{equation}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert =\left\vert
\left\langle \int_{a}^{b}f\left( t\right) dt,e\right\rangle \right\vert
\label{2.7a}
\end{equation}
and
\begin{equation}
k_{1}\left\Vert f\left( t\right) \right\Vert =\func{Re}\left\langle f\left(
t\right) ,e\right\rangle ,\ \ \ k_{2}\left\Vert f\left( t\right) \right\Vert
=\func{Im}\left\langle f\left( t\right) ,e\right\rangle  \label{2.8}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

From (\ref{2.7a}) we deduce
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left\langle \int_{a}^{b}f\left( t\right)
dt,e\right\rangle e,  \label{2.9}
\end{equation}
and from (\ref{2.8}), by multiplying the second equality with $i,$the
imaginary unit, and integrating both equations on $\left[ a,b\right] ,$ we
deduce
\begin{equation}
\left( k_{1}+ik_{2}\right) \int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt=\left\langle \int_{a}^{b}f\left( t\right) dt,e\right\rangle .
\label{2.10}
\end{equation}
Finally, by (\ref{2.9}) and (\ref{2.10}), we deduce the desired equality (
\ref{2.3}).
\end{proof}

The following corollary is of interest.

\begin{corollary}
\label{c2.2}Let $e$ be a unit vector in the complex Hilbert space $\left(
H;\left\langle \cdot ,\cdot \right\rangle \right) $ and $\eta _{1},\eta
_{2}\in \left( 0,1\right) .$ If $f\in L\left( \left[ a,b\right] ;H\right) $
is such that
\begin{equation}
\left\Vert f\left( t\right) -e\right\Vert \leq \eta _{1},\quad \left\Vert
f\left( t\right) -ie\right\Vert \leq \eta _{2}\text{ \ for a.e. \ }t\in
\left[ a,b\right] ,  \label{2.11}
\end{equation}
then we have the inequality
\begin{equation}
\sqrt{2-\eta _{1}^{2}-\eta _{2}^{2}}\int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt\leq \left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert .
\label{2.12}
\end{equation}
The case of equality holds in (\ref{2.12}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \sqrt{1-\eta _{1}^{2}}+i\sqrt{1-\eta
_{2}^{2}}\right) \left( \int_{a}^{b}\left\Vert f\left( t\right) \right\Vert
dt\right) e.  \label{2.13}
\end{equation}
\end{corollary}

\begin{proof}
From the first inequality in (\ref{2.11}) we deduce, by taking the square,
that
\begin{equation*}
\left\Vert f\left( t\right) \right\Vert ^{2}+1-\eta _{1}^{2}\leq 2\func{Re}
\left\langle f\left( t\right) ,e\right\rangle ,
\end{equation*}
implying
\begin{equation}
\frac{\left\Vert f\left( t\right) \right\Vert ^{2}}{\sqrt{1-\eta _{1}^{2}}}+
\sqrt{1-\eta _{1}^{2}}\leq \frac{2\func{Re}\left\langle f\left( t\right)
,e\right\rangle }{\sqrt{1-\eta _{1}^{2}}}  \label{2.14}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

Since, obviously
\begin{equation}
2\left\Vert f\left( t\right) \right\Vert \leq \frac{\left\Vert f\left(
t\right) \right\Vert ^{2}}{\sqrt{1-\eta _{1}^{2}}}+\sqrt{1-\eta _{1}^{2}},
\label{2.15}
\end{equation}
hence, by (\ref{2.14}) and (\ref{2.15}) we get
\begin{equation}
0\leq \sqrt{1-\eta _{1}^{2}}\left\Vert f\left( t\right) \right\Vert \leq
\func{Re}\left\langle f\left( t\right) ,e\right\rangle  \label{2.16}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

From the second inequality in (\ref{2.11}) we deduce
\begin{equation*}
0\leq \sqrt{1-\eta _{2}^{2}}\left\Vert f\left( t\right) \right\Vert \leq
\func{Re}\left\langle f\left( t\right) ,ie\right\rangle
\end{equation*}
for a.e. $t\in \left[ a,b\right] .$ Since
\begin{equation*}
\func{Re}\left\langle f\left( t\right) ,ie\right\rangle =\func{Im}
\left\langle f\left( t\right) ,e\right\rangle
\end{equation*}
hence
\begin{equation}
0\leq \sqrt{1-\eta _{2}^{2}}\left\Vert f\left( t\right) \right\Vert \leq
\func{Im}\left\langle f\left( t\right) ,e\right\rangle  \label{2.17}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

Now, observe from (\ref{2.16}) and (\ref{2.17}), that the condition (\ref
{2.1}) of Theorem \ref{t2.1} is satisfied for $k_{1}=\sqrt{1-\eta _{1}^{2}},$
$k_{2}=\sqrt{1-\eta _{2}^{2}}\in \left( 0,1\right) ,$ and thus the corollary
is proved.
\end{proof}

The following corollary may be stated as well.

\begin{corollary}
\label{c2.3}Let $e$ be a unit vector in the complex Hilbert space $\left(
H;\left\langle \cdot ,\cdot \right\rangle \right) $ and $M_{1}\geq m_{1}>0,$
$M_{2}\geq m_{2}>0.$ If $f\in L\left( \left[ a,b\right] ;H\right) $ is such
that either
\begin{equation}
\func{Re}\left\langle M_{1}e-f\left( t\right) ,f\left( t\right)
-m_{1}e\right\rangle \geq 0,\ \ \ \text{ \ }\func{Re}\left\langle
M_{2}ie-f\left( t\right) ,f\left( t\right) -m_{2}ie\right\rangle \geq 0
\label{2.18}
\end{equation}
or, equivalently,
\begin{align}
\left\Vert f\left( t\right) -\frac{M_{1}+m_{1}}{2}e\right\Vert & \leq \frac{1
}{2}\left( M_{1}-m_{1}\right) ,  \label{2.19} \\
\left\Vert f\left( t\right) -\frac{M_{2}+m_{2}}{2}ie\right\Vert & \leq \frac{
1}{2}\left( M_{2}-m_{2}\right) ,  \notag
\end{align}
for each a.e. $t\in \left[ a,b\right] ,$ then we have the inequality
\begin{equation}
2\left[ \frac{m_{1}M_{1}}{\left( M_{1}+m_{1}\right) ^{2}}+\frac{m_{2}M_{2}}{
\left( M_{2}+m_{2}\right) ^{2}}\right] ^{1/2}\int_{a}^{b}\left\Vert f\left(
t\right) \right\Vert dt\leq \left\Vert \int_{a}^{b}f\left( t\right)
dt\right\Vert .  \label{2.20}
\end{equation}
The equality holds in (\ref{2.20}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=2\left( \frac{\sqrt{m_{1}M_{1}}}{M_{1}+m_{1}}
+i\frac{\sqrt{m_{2}M_{2}}}{M_{2}+m_{2}}\right) \left( \int_{a}^{b}\left\Vert
f\left( t\right) \right\Vert dt\right) e.  \label{2.21}
\end{equation}
\end{corollary}

\begin{proof}
Firstly, remark that, for $x,z,Z\in H,$ the following statements are
equivalent.

\begin{enumerate}
\item[(i)] $\func{Re}\left\langle Z-x,x-z\right\rangle \geq 0$

and

\item[(ii)] $\left\Vert x-\frac{Z+z}{2}\right\Vert \leq \frac{1}{2}
\left\Vert Z-z\right\Vert .$
\end{enumerate}

Using this fact, we may simply realise that (\ref{2.18}) and (\ref{2.19})
are equivalent.

Now, from the first inequality in (\ref{2.18}), we get
\begin{equation*}
\left\Vert f\left( t\right) \right\Vert ^{2}+m_{1}M_{1}\leq \left(
M_{1}+m_{1}\right) \func{Re}\left\langle f\left( t\right) ,e\right\rangle
\end{equation*}
implying
\begin{equation}
\frac{\left\Vert f\left( t\right) \right\Vert ^{2}}{\sqrt{m_{1}M_{1}}}+\sqrt{
m_{1}M_{1}}\leq \frac{M_{1}+m_{1}}{\sqrt{m_{1}M_{1}}}\func{Re}\left\langle
f\left( t\right) ,e\right\rangle  \label{2.22}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

Since, obviously,
\begin{equation}
2\left\Vert f\left( t\right) \right\Vert \leq \frac{\left\Vert f\left(
t\right) \right\Vert ^{2}}{\sqrt{m_{1}M_{1}}}+\sqrt{m_{1}M_{1}},
\label{2.23}
\end{equation}
hence, by (\ref{2.22}) and (\ref{2.23})
\begin{equation}
0\leq \frac{2\sqrt{m_{1}M_{1}}}{M_{1}+m_{1}}\left\Vert f\left( t\right)
\right\Vert \leq \func{Re}\left\langle f\left( t\right) ,e\right\rangle
\label{2.24}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$

Using the same argument as in the proof of Corollary \ref{c2.2}, we deduce
the desired inequality. We omit the details.
\end{proof}

\section{The Case of Orthonormal Vectors}

In the early paper \cite{SSD1}, we pointed out the following reverse of the
continuous triangle inequality for real or complex Hilbert spaces $\left(
H;\left\langle \cdot ,\cdot \right\rangle \right) .$

\begin{theorem}
\label{t3.1}Let $\left\{ e_{1},\dots ,e_{n}\right\} $ be a family of
orthonormal vectors in $H$, $k_{i}\geq 0,$ $i\in \left\{ 1,\dots ,n\right\} $
and $f\in L\left( \left[ a,b\right] ;H\right) $ such that
\begin{equation}
k_{i}\left\Vert f\left( t\right) \right\Vert \leq \func{Re}\left\langle
f\left( t\right) ,e_{i}\right\rangle   \label{3.1}
\end{equation}
for each $i\in \left\{ 1,\dots ,n\right\} $ and for a.e. $t\in \left[ a,b
\right] .$ Then
\begin{equation}
\left( \sum_{i=1}^{n}k_{i}^{2}\right) ^{\frac{1}{2}}\int_{a}^{b}\left\Vert
f\left( t\right) \right\Vert dt\leq \left\Vert \int_{a}^{b}f\left( t\right)
dt\right\Vert ,  \label{3.2}
\end{equation}
where the case of equality holds if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \int_{a}^{b}\left\Vert f\left(
t\right) \right\Vert dt\right) \sum_{i=1}^{n}k_{i}e_{i}.  \label{3.3}
\end{equation}
\end{theorem}

In what follows, we improve this result for the case of complex Hilbert
spaces. The following result holds.

\begin{theorem}
\label{t3.2}Let $\left\{ e_{1},\dots ,e_{n}\right\} $ be a family of
orthonormal vectors in the complex Hilbert space $\left( H;\left\langle
\cdot ,\cdot \right\rangle \right) $. If $k_{j},h_{j}\geq 0,$ $j\in \left\{
1,\dots ,n\right\} $ and $f\in L\left( \left[ a,b\right] ;H\right) $ are
such that
\begin{equation}
k_{j}\left\Vert f\left( t\right) \right\Vert \leq \func{Re}\left\langle
f\left( t\right) ,e_{j}\right\rangle ,\quad h_{j}\left\Vert f\left( t\right)
\right\Vert \leq \func{Im}\left\langle f\left( t\right) ,e_{j}\right\rangle
\label{3.4}
\end{equation}
for each $j\in \left\{ 1,\dots ,n\right\} $ and a.e. $t\in \left[ a,b\right]
,$ then
\begin{equation}
\left[ \sum_{j=1}^{n}\left( k_{j}^{2}+h_{j}^{2}\right) \right] ^{\frac{1}{2}
}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq \left\Vert
\int_{a}^{b}f\left( t\right) dt\right\Vert .  \label{3.5}
\end{equation}
The case of equality holds in (\ref{3.5}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \int_{a}^{b}\left\Vert f\left(
t\right) \right\Vert dt\right) \sum_{j=1}^{n}\left( k_{j}+ih_{j}\right)
e_{j}.  \label{3.6}
\end{equation}
\end{theorem}

\begin{proof}
Before we prove the theorem, let us recall that, if $x\in H$ and $
e_{1},\dots ,e_{n}$ are orthonormal vectors, then the following identity
holds true:
\begin{equation}
\left\Vert x-\sum_{j=1}^{n}\left\langle x,e_{j}\right\rangle
e_{j}\right\Vert ^{2}=\left\Vert x\right\Vert ^{2}-\sum_{j=1}^{n}\left\vert
\left\langle x,e_{j}\right\rangle \right\vert ^{2}.  \label{3.7}
\end{equation}
As a consequence of this identity, we have the \textit{Bessel inequality}
\begin{equation}
\sum_{j=1}^{n}\left\vert \left\langle x,e_{j}\right\rangle \right\vert
^{2}\leq \left\Vert x\right\Vert ^{2},x\in H.  \label{3.8}
\end{equation}
in which the case of equality holds if and only if
\begin{equation}
x=\sum_{j=1}^{n}\left\langle x,e_{j}\right\rangle e_{j}.  \label{3.9}
\end{equation}
Now, applying Bessel's inequality for $x=\int_{a}^{b}f\left( t\right) dt,$
we have successively
\begin{align}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert ^{2}& \geq
\sum_{j=1}^{n}\left\vert \left\langle \int_{a}^{b}f\left( t\right)
dt,e_{j}\right\rangle \right\vert ^{2}=\sum_{j=1}^{n}\left\vert
\int_{a}^{b}\left\langle f\left( t\right) ,e_{j}\right\rangle dt\right\vert
^{2}  \label{3.10} \\
& =\sum_{j=1}^{n}\left\vert \int_{a}^{b}\func{Re}\left\langle f\left(
t\right) ,e_{j}\right\rangle dt+i\left( \int_{a}^{b}\func{Im}\left\langle
f\left( t\right) ,e_{j}\right\rangle dt\right) \right\vert ^{2}  \notag \\
& =\sum_{j=1}^{n}\left[ \left( \int_{a}^{b}\func{Re}\left\langle f\left(
t\right) ,e_{j}\right\rangle dt\right) ^{2}+\left( \int_{a}^{b}\func{Im}
\left\langle f\left( t\right) ,e_{j}\right\rangle dt\right) ^{2}\right] .
\notag
\end{align}
Integrating (\ref{3.4}) on $\left[ a,b\right] ,$ we get
\begin{equation}
\int_{a}^{b}\func{Re}\left\langle f\left( t\right) ,e_{j}\right\rangle
dt\geq k_{j}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt
\label{3.11}
\end{equation}
and
\begin{equation}
\int_{a}^{b}\func{Im}\left\langle f\left( t\right) ,e_{j}\right\rangle
dt\geq h_{j}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt
\label{3.12}
\end{equation}
for each $j\in \left\{ 1,\dots ,n\right\} .$

Squaring and adding the above two inequalities (\ref{3.11}) and (\ref{3.12}
), we deduce
\begin{multline*}
\sum_{j=1}^{n}\left[ \left( \int_{a}^{b}\func{Re}\left\langle f\left(
t\right) ,e_{j}\right\rangle dt\right) ^{2}+\left( \int_{a}^{b}\func{Im}
\left\langle f\left( t\right) ,e_{j}\right\rangle dt\right) ^{2}\right] \\
\geq \sum_{j=1}^{n}\left( k_{j}^{2}+h_{j}^{2}\right) \left(
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\right) ^{2},
\end{multline*}
which combined with (\ref{3.10}) will produce the desired inequality (\ref
{3.5}).

Now, if (\ref{3.6}) holds true, then
\begin{align*}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert & =\left(
\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\right) \left\Vert
\sum_{j=1}^{n}\left( k_{j}+ih_{j}\right) e_{j}\right\Vert \\
& =\left( \int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\right)
\left( \left\Vert \sum_{j=1}^{n}\left( k_{j}+ih_{j}\right) e_{j}\right\Vert
^{2}\right) ^{\frac{1}{2}} \\
& =\left( \int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\right)
\left[ \sum_{j=1}^{n}\left( k_{j}^{2}+h_{j}^{2}\right) \right] ^{\frac{1}{2}
},
\end{align*}
and the case of equality holds in (\ref{3.5}).

Conversely, if the equality holds in (\ref{3.5}), then it must hold in all
the inequalities used to prove (\ref{3.5}) and therefore we must have
\begin{equation}
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert
^{2}=\sum_{j=1}^{n}\left\vert \left\langle \int_{a}^{b}f\left( t\right)
dt,e_{j}\right\rangle \right\vert ^{2}  \label{3.13}
\end{equation}
and
\begin{equation}
k_{j}\left\Vert f\left( t\right) \right\Vert =\func{Re}\left\langle f\left(
t\right) ,e_{j}\right\rangle \text{ \ and \ }h_{j}\left\Vert f\left(
t\right) \right\Vert =\func{Re}\left\langle f\left( t\right)
,e_{j}\right\rangle  \label{3.14}
\end{equation}
for each $j\in \left\{ 1,\dots ,n\right\} $ and a.e. $t\in \left[ a,b\right]
.$

From (\ref{3.13}), on using the identity (\ref{3.9}), we deduce that
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\sum_{j=1}^{n}\left\langle
\int_{a}^{b}f\left( t\right) dt,e_{j}\right\rangle e_{j}.  \label{3.15}
\end{equation}
Now, multiplying the second equality in (\ref{3.14}) with the imaginary unit
$i,$ integrating both inequalities on $\left[ a,b\right] $ and summing them
up, we get
\begin{equation}
\left( k_{j}+ih_{j}\right) \int_{a}^{b}\left\Vert f\left( t\right)
\right\Vert dt=\left\langle \int_{a}^{b}f\left( t\right)
dt,e_{j}\right\rangle  \label{3.16}
\end{equation}
for each $j\in \left\{ 1,\dots ,n\right\} .$

Finally, utilising (\ref{3.15}) and (\ref{3.16}), we deduce (\ref{3.6}) and
the theorem is proved.
\end{proof}

The following corollaries are of interest.

\begin{corollary}
\label{c3.2}Let $e_{1},\dots ,e_{m}$ be orthonormal vectors in the complex
Hilbert space $\left( H;\left\langle \cdot ,\cdot \right\rangle \right) $
and $\rho _{k},\eta _{k}\in \left( 0,1\right) ,$ $k\in \left\{ 1,\dots
,n\right\} .$ If $f\in L\left( \left[ a,b\right] ;H\right) $ is such that
\begin{equation*}
\left\Vert f\left( t\right) -e_{k}\right\Vert \leq \rho _{k},\qquad
\left\Vert f\left( t\right) -ie_{k}\right\Vert \leq \eta _{k}
\end{equation*}
for each $k\in \left\{ 1,\dots ,n\right\} $ and for a.e. $t\in \left[ a,b
\right] ,$ then we have the inequality
\begin{equation}
\left[ \sum_{k=1}^{n}\left( 2-\rho _{k}^{2}-\eta _{k}^{2}\right) \right] ^{
\frac{1}{2}}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert .  \label{3.17}
\end{equation}
The case of equality holds in (\ref{3.17}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \int_{a}^{b}\left\Vert f\left(
t\right) \right\Vert dt\right) \sum_{k=1}^{n}\left( \sqrt{1-\rho _{k}^{2}}+i
\sqrt{1-\eta _{k}^{2}}\right) e_{k}.  \label{3.18}
\end{equation}
\end{corollary}

The proof follows by Theorem \ref{t3.2} and is similar to the one from
Corollary \ref{c2.2}. We omit the details.

\begin{corollary}
\label{c3.3}Let $e_{1},\dots ,e_{m}$ be as in Corollary \ref{c3.2} and $
M_{k}\geq m_{k}>0,$ $N_{k}\geq n_{k}>0,$ $k\in \left\{ 1,\dots ,n\right\} .$
If $f\in L\left( \left[ a,b\right] ;H\right) $ is such that either
\begin{equation*}
\func{Re}\left\langle M_{k}e_{k}-f\left( t\right) ,f\left( t\right)
-m_{k}e_{k}\right\rangle \geq 0,\ \ \func{Re}\left\langle
N_{k}ie_{k}-f\left( t\right) ,f\left( t\right) -n_{k}ie_{k}\right\rangle
\geq 0
\end{equation*}
or, equivalently,
\begin{align*}
\left\Vert f\left( t\right) -\frac{M_{k}+m_{k}}{2}e_{k}\right\Vert & \leq
\frac{1}{2}\left( M_{k}-m_{k}\right) ,\  \\
\left\Vert f\left( t\right) -\frac{N_{k}+n_{k}}{2}ie_{k}\right\Vert & \leq
\frac{1}{2}\left( N_{k}-n_{k}\right)
\end{align*}
for each $k\in \left\{ 1,\dots ,n\right\} $ and a.e. $t\in \left[ a,b\right]
,$ then we have the inequality
\begin{equation}
2\left\{ \sum_{k=1}^{m}\left[ \frac{m_{k}M_{k}}{\left( M_{k}+m_{k}\right)
^{2}}+i\frac{n_{k}N_{k}}{\left( N_{k}+n_{k}\right) ^{2}}\right] \right\} ^{
\frac{1}{2}}\int_{a}^{b}\left\Vert f\left( t\right) \right\Vert dt\leq
\left\Vert \int_{a}^{b}f\left( t\right) dt\right\Vert .  \label{3.19}
\end{equation}
The case of equality holds in (\ref{3.19}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=2\left( \int_{a}^{b}\left\Vert f\left(
t\right) \right\Vert dt\right) \sum_{k=1}^{n}\left( \frac{\sqrt{m_{k}M_{k}}}{
M_{k}+m_{k}}+i\frac{\sqrt{n_{k}N_{k}}}{N_{k}+n_{k}}\right) e_{k}.
\label{3.20}
\end{equation}
\end{corollary}

The proof employs Theorem \ref{t3.2} and is similar to the one in Corollary
\ref{c2.3}. We omit the details.

\section{Applications for Complex-Valued Functions}

The following reverse of the generalised triangle inequality for
complex-valued functions that improves Karamata's result (\ref{1.2}) holds.

\begin{proposition}
\label{p4.1}Let $f\in L\left( \left[ a,b\right] ;\mathbb{C}\right) $ with
the property that
\begin{equation}
0\leq \varphi _{1}\leq \arg f\left( t\right) \leq \varphi _{2}<\frac{\pi }{2}
\label{4.1}
\end{equation}
for a.e. $t\in \left[ a,b\right] .$ Then we have the inequality
\begin{equation}
\sqrt{\sin ^{2}\varphi _{1}+\cos ^{2}\varphi _{2}}\int_{a}^{b}\left\vert
f\left( t\right) \right\vert dt\leq \left\vert \int_{a}^{b}f\left( t\right)
dt\right\vert .  \label{4.2}
\end{equation}
The equality holds in (\ref{4.2}) if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \cos \varphi _{2}+i\sin \varphi
_{1}\right) \int_{a}^{b}\left\vert f\left( t\right) \right\vert dt.
\label{4.3}
\end{equation}
\end{proposition}

\begin{proof}
Let $f\left( t\right) =\func{Re}f\left( t\right) +i\func{Im}f\left( t\right)
.$ We may assume that $\func{Re}f\left( t\right) \geq 0,$ $\func{Im}f\left(
t\right) >0,$ for a.e. $t\in \left[ a,b\right] ,$ since, by (\ref{4.1}), $
\frac{\func{Im}f\left( t\right) }{\func{Re}f\left( t\right) }=\tan \left[
\arg f\left( t\right) \right] \in \left[ 0,\frac{\pi }{2}\right) $, for a.e.
$t\in \left[ a,b\right] .$ By (\ref{4.1}), we obviously have
\begin{equation*}
0\leq \tan ^{2}\varphi _{1}\leq \left[ \frac{\func{Im}f\left( t\right) }{
\func{Re}f\left( t\right) }\right] ^{2}\leq \tan ^{2}\varphi _{2},\qquad
\text{for a.e. }t\in \left[ a,b\right] ,
\end{equation*}
from where we get
\begin{equation*}
\frac{\left[ \func{Im}f\left( t\right) \right] ^{2}+\left[ \func{Re}f\left(
t\right) \right] ^{2}}{\left[ \func{Re}f\left( t\right) \right] ^{2}}\leq
\frac{1}{\cos {}^{2}\varphi _{2}},
\end{equation*}
for a.e. $t\in \left[ a,b\right] ,$ and
\begin{equation*}
\frac{\left[ \func{Im}f\left( t\right) \right] ^{2}+\left[ \func{Re}f\left(
t\right) \right] ^{2}}{\left[ \func{Im}f\left( t\right) \right] ^{2}}\leq
\frac{1+\tan ^{2}\varphi _{1}}{\tan ^{2}\varphi _{1}}=\frac{1}{\sin \varphi
_{1}},
\end{equation*}
for a.e. $t\in \left[ a,b\right] ,$ giving the simpler inequalities
\begin{equation*}
\left\vert f\left( t\right) \right\vert \cos \varphi _{2}\leq \func{Re}
\left( f\left( t\right) \right) ,\quad \left\vert f\left( t\right)
\right\vert \sin \varphi _{1}\leq \func{Im}\left( f\left( t\right) \right)
\end{equation*}
for a.e. $t\in \left[ a,b\right] .$

Now, applying Theorem \ref{t2.1} for the complex Hilbert space $\mathbb{C}$
endowed with the inner product $\left\langle z,w\right\rangle =z\cdot \bar{w}
$ for $k_{1}=\cos \varphi _{2},$ $k_{2}=\sin \varphi _{1}$ and $e=1,$ we
deduce the desired inequality (\ref{4.2}). The case of equality is also
obvious and we omit the details.
\end{proof}

Another result that has an obvious geometrical interpretation is the
following one.

\begin{proposition}
\label{p4.2}Let $e\in \mathbb{C}$ with $\left\vert e\right\vert =1$ and $
\rho _{1},\rho _{2}\in \left( 0,1\right) .$ If $f\left( t\right) \in L\left(
\left[ a,b\right] ;\mathbb{C}\right) $ such that
\begin{equation}
\left\vert f\left( t\right) -e\right\vert \leq \rho _{1},\quad \left\vert
f\left( t\right) -ie\right\vert \leq \rho _{2}\qquad \text{for a.e. }t\in
\left[ a,b\right] ,  \label{4.5}
\end{equation}
then we have the inequality
\begin{equation}
\sqrt{2-\rho _{1}^{2}-\rho _{2}^{2}}\int_{a}^{b}\left\vert f\left( t\right)
\right\vert dt\leq \left\vert \int_{a}^{b}f\left( t\right) dt\right\vert ,
\label{4.6}
\end{equation}
with equality if and only if
\begin{equation}
\int_{a}^{b}f\left( t\right) dt=\left( \sqrt{1-\rho _{1}^{2}}+i\sqrt{1-\rho
_{2}^{2}}\right) \int_{a}^{b}\left\vert f\left( t\right) \right\vert dt\cdot
e.  \label{4.7}
\end{equation}
\end{proposition}

The proof is obvious by Corollary \ref{c2.2} applied for $H=\mathbb{C}$ and
we omit the details.

\begin{remark}
If we choose $e=1,$ and for $\rho _{1},\rho _{2}\in \left( 0,1\right) $ we
define
\begin{equation*}
\bar{D}\left( 1,\rho _{1}\right) :=\left\{ z\in \mathbb{C}|\left\vert
z-1\right\vert \leq \rho _{1}\right\} ,\ \ \bar{D}\left( i,\rho _{2}\right)
:=\left\{ z\in \mathbb{C}|\left\vert z-i\right\vert \leq \rho _{2}\right\} ,
\end{equation*}
then obviously the intersection domain
\begin{equation*}
S_{\rho _{1},\rho _{2}}:=\bar{D}\left( 1,\rho _{1}\right) \cap \bar{D}\left(
i,\rho _{2}\right)
\end{equation*}
is nonempty if and only if $\rho _{1}+\rho _{2}>\sqrt{2}.$

If $f\left( t\right) \in S_{\rho _{1},\rho _{2}}$ for a.e. $t\in \left[ a,b
\right] ,$ then (\ref{4.6}) holds true. The equality holds in (\ref{4.6}) if
and only if
\begin{equation*}
\int_{a}^{b}f\left( t\right) dt=\left( \sqrt{1-\rho _{1}^{2}}+i\sqrt{1-\rho
_{2}^{2}}\right) \int_{a}^{b}\left\vert f\left( t\right) \right\vert dt.
\end{equation*}
\end{remark}

config: arxiv


TITLE: Sum of Power of Two Fibonacci reciprocals
QUESTION [3 upvotes]: Evaluate $$\sum_{k=0}^{\infty} \frac{1}{F_{2^n}} \;.$$
I'm thinking of using a relation from a term to another.

REPLY [1 votes]: This series is known as the Millin series. One can show that, $$\sum_{n=0}^\infty\frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2}$$ by first proving, using induction, that: $$\sum_{n=0}^\ell \frac{1}{F_{2^n}}=3-\frac{F_{2^\ell-1}}{F_{2^\ell}},$$ and then taking the limit as $\ell\to\infty$, knowing that $\lim\limits_{\ell\to\infty}\dfrac{F_{\ell+1}}{F_\ell}=\frac{1+\sqrt{5}}{2}$,
$$\sum_{n=0}^\infty\frac{1}{F_{2^n}}=\lim_{\ell\to\infty}\left(3-\frac{F_{2^\ell-1}}{F_{2^\ell}}\right)=3-\dfrac{1}{(1+\sqrt{5})/2}=\dfrac{7-\sqrt{5}}{2}.$$


TITLE: Software for geometrical optics
QUESTION [16 upvotes]: Is there any good software for construction optical path's in geometrical optics. More specifically I want features like:

draw $k \in \mathbb{N}$ objects $K_1,\dots,K_n$ with indices of refraction $n_1,\dots,n_k$ and light sources $l \in \mathbb{N}$ light sources $L_1,\dots,L_l$

define some light sources (or just light rays) and the program constructs the optical path of those rays through $K_1,\dots,K_n$.
define a point and the program constructs the optical path between the light sources and the given point

draw some specific optical elements like lenses, (concave, convex) mirrors and do things like above
construct automatically virtual images
3D drawings would be fine

I would prefer free software for linux.

REPLY [1 votes]: http://www.onshape.com with Ray Tracer custom feature.
I had some success simulating geometrical light propagation via a set of lenses and prisms using www.onshape.com, which is free, online 3D modelling environment (see attached screenshot). I used a custom feature, named Ray Tracer, created by one of OnShape users. You can define refractive indices of optical components, trace refracted and reflected rays by specifying their origin and initial direction and account for total internal reflection. Generally speaking a CAD program is a natural choice for geometrical ray tracing.


TITLE: Best Fake Proofs? (A M.SE April Fools Day collection)
QUESTION [240 upvotes]: In honor of April Fools Day $2013$, I'd like this question to collect the best, most convincing fake proofs of impossibilities you have seen.
I've posted one as an answer below. I'm also thinking of a geometric one where the "trick" is that it's very easy to draw the diagram wrong and have two lines intersect in the wrong place (or intersect when they shouldn't). If someone could find and link this, I would appreciate it very much.

REPLY [6 votes]: We shall prove that $x=e^{\pi/2} \approx 4.8$ is a solution to $x^x=x$.
First, we compute $i^i = e^{i \log i} = e^{i(\pi/2i)} = e^{- \pi/2}$. Next we compute $i^{i^i} = e^{-pi/2 i} = -i$. Then $x=i^{i^{i^i}}=(-i)^i=e^{i\log (-i)}=e^{i(-\pi/2 i)}=e^{\pi/2}$.
Finally, $x^i=i^{i^{i^{i^i}}}=e^{i \pi/2}=i$.
Raise each side of the previous equation to the $i^{i^i}$th power. We get
$$ x^{i^{i^{i^i}}}={i^{i^{i^i}}}$$
or
$$x^x=x$$
where $x=e^{\pi/2}$.


text:
\begin{document}
\title[Finite Quotients of $\B_3$]
{Finite Linear Quotients of $\B_3$ of Low Dimension}

\author{Eric C. Rowell}
\email{rowell@math.tamu.edu}
\address{Department of Mathematics\\
    Texas A\&M University \\
    College Station, TX 77843-3368\\
    U.S.A.}
\author{Imre Tuba}
\email{ituba@mail.sdsu.edu}
\address{San Diego State University\\
Imperial Valley Campus\\
720 Heber Ave.\\
Calexico, CA 92231\\
U.S.A.}
\subjclass[2000]{Primary 20F36; Secondary 20C15}
\thanks{Eric Rowell was partially supported by NSA grant H98230-08-1-0020.}

\begin{abstract}
We study the problem of deciding whether or not the image of an irreducible representation of the braid group $\B_3$ of degree
$\leq 5$ has finite image if we are only given the eigenvalues of a generator.   We provide a partial algorithm that determines when the images are finite or
infinite in all but finitely many cases, and use these results to study examples coming from quantum groups.  Our technique uses two classification theorems and the computational group theory package GAP.
\end{abstract}
\maketitle
\section{Introduction}
Let $\B_3$ denote Artin's braid group on $3$ strands with generators $\sigma_1$ $\sigma_2$, satisfying
 $$\sigma_1\sigma_{2}\sigma_1=\sigma_{2}\sigma_1\sigma_{2}.$$

We consider the following problem.
Suppose $\phi:\B_3\rightarrow GL(V)$ is a $d$-dimensional complex representation  for which we are given:
\begin{enumerate}
\item $\rho$ is irreducible and
\item $\Spec(\rho(\sigma_1))=\{\la_1,\ldots,\la_d\}.$
\end{enumerate}
Can we determine if $G:=\rho(\B_3)$ is a finite or infinite group from this
information?  In this paper we
determine some conditions under which $G$ is finite or infinite, assuming that
$d\le 5$.

The general question of determining the image of complex braid group representations
seems first to have been studied by Jones \cite{Jones1} for the representations
associated with the Jones polynomial.  Indeed, the Burau representation was essentially the only representation of $\B_n$ that was known until the 1980s.  More recently this question has been studied extensively for unitary representations
obtained from solutions to the Yang-Baxter equation \cite{FRW}, Hecke algebras
\cite{FLW1,FLW2}, and BMW-algebras \cite{LRW,LR}.  In all of these cases one has a tower  of (generally reducible) representations $\rho_n$ such that $$\rho_3(\B_3)\subset\cdots\subset\rho_n(\B_n)\subset\rho_{n+1}(\B_{n+1})\subset\cdots.$$  So clearly it is enough to show that $\rho_3(B_3)$ is infinite to conclude the same for the tower.  One motivating application of this is to answer the question of universality in the setting of topological quantum computation (see \cite{FLW2}).

A generalization of the above towers of representations comes from \emph{ribbon categories}, see \cite{TurBook} for the definition.  For any object $X$ in a ribbon category $\CC$ one obtains
a tower of representations of $\B_n$ acting on $\End(X^{\ot n})$.  If the object is a self-dual object (i.e. $X \cong X^*$) then $\B_3$ acts on $V=\Hom(X,X^{\ot  3})$, often irreducibly.  As an application, we will apply our results to ribbon categories coming from quantum groups in Section \ref{aps}.
\section{Main Result}
Throughout this paper let $\rho:\B_3\rightarrow\GL(V)$ be a $d$-dimensional irreducible representation with $2\leq d\leq 5$ and set $A=\rho(\sigma_1)$ and $B=\rho(\sigma_2)$.  Let $G$ denote the image  $\rho(\B_3)=\lan A,B\ra$ i.e. the group generated by $A$ and $B$ and
define $\mS:=\Spec(A)=\Spec(B)$.  The analysis naturally breaks into imprimitive and primitive cases (defined in Section \ref{genres}).
Our results are summarized in the following:
\begin{theorem}\label{maintheorem}  Let $\rho$, $G$, $A$ and $\mS$ be defined as
in the previous paragraph.
Let $\mS=\{\la_1,\ldots,\la_d\}$, and define the projective order of $A$ by
$$po(A):=\min\{t: (\la_1)^t=(\la_2)^t=\cdots=(\la_d)^t\}.$$   We use the convention
that each successive statement excludes the hypotheses of all
of the preceding cases.
\begin{enumerate}
\item[(a)] Suppose some $\la_i$ is not a root of unity, or $\la_i=\la_j$ for some $i\neq j$.  Then $G$ is \textbf{infinite}.
\item[(b)] Suppose $po(A)\leq 5$.  Then $G$ is \textbf{finite}.
\item[(c)] Suppose $G$ is imprimitive.  Then $\mS$ is of the form:
\begin{enumerate}
 \item[(i)] $\{\pm\chi,\alpha\}$ or $\chi\{1,\omega,\omega^2\}\cup\{\alpha\}$
with $\omega$ a primitive 3rd root of unity and $G$ is finite or
\item[(ii)] $\{\pm r,\pm s\}$.  In this case if $u=r/s$
is a root of unity of order $o(u)\in \{7,8,9\}\cup[11,\infty)$ then $G$ is \textbf{infinite}, if $o(u)=6$, $G$ is \textbf{finite} and if $o(u)=5$ or $10$
one cannot decide $|G|$ without further information.
\end{enumerate}
\item[(d)] Suppose $G$ is primitive. Then:
\begin{enumerate}
\item[(i)] If $d=2$ then $G$ is \textbf{infinite}.
\item[(ii)] If $d=3$ and $po(A)\geq 8$ then $G$ is \textbf{infinite}.  If $po(A)=7$, and $\frac{1}{\la_1}\mS$ is Galois conjugate to $\{1,e^{2\pi\im/7},e^{2k\pi\im/7}\}$ with $k$ even, $G$ is \textbf{infinite}, whereas if
$k$ is odd, $G$ is \textbf{finite}.
\item[(iii)] If $d=4$ and $po(A)\not\in\{6,\ldots,10,12,15,20,24\}$ then $G$ is \textbf{infinite}.
\item[(iv)] If $d=5$ and $po(A)\in\{7,8\}\cup[13,\infty)$ then $G$ is \textbf{infinite}.
\end{enumerate}
\end{enumerate}
\end{theorem}

\begin{rmk}
Notice that this theorem covers all but the following cases: $G$ primitive with
\begin{enumerate}
\item $d=4$ with $po(A)\in\{6,\ldots,10,12,15,20,24\}$.
\item $d=5$ with $po(A)\in\{6,9,10,11,12\}$.
\end{enumerate}
So Theorem \ref{maintheorem} can be used as an algorithm to decide the if $G$ is finite or not, at least for $d\le 3$ or $G$ imprimitive.
\end{rmk}

There are two main ingredients to our approach.  Necessary
conditions can be derived from the classification of finite primitive linear groups of low degree (see \cite{Ft}), while
sufficient conditions can often be gleaned from the following classification of irreducible representations of $\B_3$ for $2\leq d\leq 5$ found in \cite{TbWz}:
\begin{theorem}[Tuba-Wenzl]\label{tw}

 \begin{enumerate}
  \item Suppose $\rho$ is an irreducible representation of $\B_3$ with eigenvalues
$\mS=\{\la_1,\ldots,\la_d\}$
as above.  Then,
\begin{enumerate}
\item $\rho$ is uniquely determined up to equivalence by $\mS$ up to a choice of $\gamma^2$ for $d=4$ and a choice of $\gamma$ for $d=5$ where $\gamma:=\det(A)^{1/d}$.
\item there exists a basis for $\C^d$ so that the matrices $A$ and $B$ are in a triangular form, given in
\cite{TbWz}.
\end{enumerate}
\item There exists an irreducible representation of $\B_3$ with eigenvalues $\mS$ if only if the $\la_i$ and $\gamma$ do not satisfy certain polynomials.  In particular for $d=3$ and $d=4$ these polynomials are:
\begin{enumerate}
\item $d=3$ $\la_r^2+\la_s\la_t$ for $r,s,t$ distinct and
\item $d=4$ $(\la_r^2+\gamma^2)(\gamma^2+\la_r\la_s+\la_t\la_u)$ for $r,s,t,u$ distinct.
\end{enumerate}
\end{enumerate}
\end{theorem}
We will refer to this theorem as the \emph{TW classification}.

\section{General Results}\label{genres}
In this section we state some general results that will be used later.

\begin{lemma}Suppose $|G|<\infty$.  Then, $\mS$ consists of $d$ \emph{distinct} roots of unity.
\end{lemma}
\begin{proof}
 It is clear that if $A$ is of finite order, it is diagonalizable and its eigenvalues must be roots of unity.  Corollary 2.2 of  \cite{TbWz} states that the minimal and characteristic polynomials of $A$ coincide (for $d\leq 5$) so that the eigenvalues of $A$ must also be distinct.
\end{proof}

The following result due to Coxeter \cite{Cox} dates back to the 1950s:
\begin{prop}\label{coxprop}
The quotient of $\B_n$ by the normal subgroup generated by
$\sigma_1^p$ is finite if and only if $1/p+1/n>1/2$.
In particular, the quotient of $\B_3$ by the normal subgroup generated by $\sigma_1^p$
is finite if and only if
$p=2,3,4$ or $5$, where the quotient groups are
\begin{eqnarray}\label{qgps}
\B_3/\lan \sigma_1^p\ra\cong\begin{cases}S_3 & p=2\\
\SL(2,3) & p=3\\
H & p=4\\
\SL(2,5)\times\Z_5 & p=5
                             \end{cases}\end{eqnarray}
where $H$ is a non-split central extension of $S_4$ by $\Z_4$.
\end{prop}

We have the following immediate consequence:
\begin{corollary}\label{coxcor}
 If $\mS\subset\chi\{\zeta_p^i:i=1...p\}$ where $\chi$ is any root of unity and $\zeta_p$ is a primitive
$p$th root of unity with $2\leq p\leq 5$ then $|G|$ is finite.
\end{corollary}
\begin{proof}
 The given hypotheses imply that $G$ is contained in a central extension of a quotient of one of the finite groups in (\ref{qgps}) by the finite cyclic group generated by $\chi$.
\end{proof}

Irreducible finite linear groups naturally break up into two distinct classes: primitive and imprimitive groups.

\begin{defn}
A linear group $\Gamma\subset\GL(V)$ is \textbf{imprimitive} if $V$ is irreducible
and can be expressed as a direct sum of
subspaces $V_i$ which $\Gamma$ permutes nontrivially.  Otherwise, we say that $\Gamma$ is \textbf{primitive}.
\end{defn}

The following observation can be found in \cite{LRW}:
\begin{lemma}\label{nocyclem}
 Suppose $G$ is an irreducible imprimitive finite linear group of dimension $d$.  Then $\mS$ contains a $\C$-coset of the group $C_r=\{\zeta_r^i: 0\leq i\leq r-1\}$ of roots of unity of degree
$r\leq d$.
\end{lemma}

The imprimitive linear groups that appear as images of irreducible representations of $\B_3$ will be analyzed in Section \ref{imp}, and the primitive cases will be covered in Section \ref{prim}.

\section{Imprimitive Groups}\label{imp}

For $2\leq d\leq 5$, there are only two ways that a group $\Gamma\subset \GL(V)$ can be imprimitive:
\begin{enumerate}
\item[Case (1)] $d=4$ and $V=V_1\oplus V_2$ with $\dim(V_i)=2$ and $\Gamma$ permutes $V_1$ and $V_2$ non-trivially and this block structure has no
refinement to $1\times 1$ blocks or
\item[Case (2)] $\Gamma$ is a isomorphic to a monomial group, i.e. a subgroup of $M(d):=S_d\ltimes D(d)$ where $D(d)$ is the group of $d\times d$ diagonal matrices and $S_d$ is identified with the $d\times d$ permutation matrices, and acts of $D(d)$ by permuting the entries.
\end{enumerate}

Case (2) is covered by the following:
\begin{theorem}\label{monom}  Let $2\leq d\leq 5$, and assume $\mS$ consists of roots of unity.  If $G$ is an irreducible imprimitive monomial group then $G$ is finite and $\mS$ is of the form:
\begin{enumerate}
 \item[(a)] $\mS=\chi\{\zeta_d^i: 1\leq i\leq d\}$,
\item[(b)] $\mS=\{\alpha,\pm\chi\}$ or
\item[(c)] $\mS=\{\alpha\}\cup\chi\{1,\zeta_3,\zeta_3^2\}$.
\end{enumerate}
\end{theorem}
\begin{proof}
In each of the cases (a)-(c) one checks that the image of $G$ is indeed finite,
in case (a) it follows from Proposition \ref{coxcor}, while in cases (b) and (c) one employs the TW classification (in case (c) one has two cases to check depending on the sign of $\det(A)^{1/2}$).  By Lemma \ref{nocyclem} we need only
show that if $\mS$ contains a coset of roots of unity but is not in case (a)-(c) the corresponding representation is not irreducible.  Since $G$ is monomial, we may assume that its generators $A$ and $B$ are generalized permutation matrices, that is, of the form $A=D_1P_1$, $B=D_2P_2$ where $D_i$ is a diagonal matrix and $P_i$ is a permutation matrix.  It is clear that $G$ is reducible if the action of $g:=\lan P_1, P_2\ra$ on the standard basis for $\C^d$ is intransitive, since the span of any orbit will be an invariant subspace for $G$.  For the excluded
cases we may assume either $d=4$ and $P_1$ is $(1,2)$ or $(1,2)(3,4)$,
or $d=5$ and $P_1$ is $(1,2)$, $(1,2,3)$, $(1,2)(3,4)$, $(1,2,3)(4,5)$ or $(1,2,3,4)$.  One then uses the braid relation to determine
the possible cycle forms of $P_2$.  We record them in Table \ref{mapletable} in disjoint cycle notation.
In each case it is clear that $P_1$ and $P_2$ generate an intransitive subgroup
of $S_d$.
\end{proof}

\begin{table}\caption{Intransitive Cases}\label{mapletable}
\begin{tabular}{*{3}{|l}|}
\hline
$d$ & $P_1$ & $P_2$  \\
\hline\hline
4& $(1,2)$ & $(1,i)$ or $(2,j)$\\
\hline
4 & $(1,2)(3,4)$ & $P_1$\\
\hline
5 & $(1,2)$ & $(1,i)$ or $(2,j)$\\
\hline
5 & $(1,2)(3,4)$ & $(1,2)(j,k)$ or $(3,4)(r,s)$\\
\hline
5 & $(1,2,3)$ & $P_1$, $(1,j,2)$, $(2,j,3)$ or $(1,3,j)$ $j=4$ or $5$\\
\hline
5 & $(1,2,3)(4,5)$ & $P_1$ \\
\hline
5& $(1,2,3,4)$ & $(i,j,k,\ell)\in S_4$\\
\hline
\end{tabular}
\end{table}

Now consider Case (1), so that $A$ and $B$ permute two dimension $2$ vector spaces $V_1$ and $V_2$.  Since $G$ is assumed to be irreducible, by choosing an ordered basis consisting of the union of bases of $V_1$ and $V_2$, we may assume either $A$ or $B$ is block skew-diagonal with blocks of size $2$.  Otherwise both $A$ and $B$ would be block diagonal with respect to this basis, violating irreducibility.  Now the characteristic polynomial of such a block skew-diagonal matrix is a polynomial in $x^2$, so that eigenvalues occur in pairs $\pm r$ and $\pm s$.  If $V_1\oplus V_2$ has a refinement to $1\times 1$ blocks then $\mS=\chi\{\pm 1,\pm i\}$ and is covered by Theorem \ref{monom}.
 Thus we may assume $\mS=\{\pm r,\pm s\}$ with $r/s$ not a $4$th root of unity.    For each pair $(r,s)$ there are
two inequivalent irreducible $4$-dimensional representations of $\B_3$ with $\mS=\{\pm r,\pm s\}$ depending on a choice of $D=\pm\sqrt{\la_2\la_3/(\la_1\la_4)}=\pm 1$ (see \cite[Prop. 2.6]{TbWz}).  Setting $$p_1(u,t):=u^2t^4+(u+u^2+u^3)t^3+(1+2u+2u^2+2u^3+u^4)t^2+(u+u^2+u^3)t+u^2$$ we have the following:
\begin{theorem}
Suppose $G$ is imprimitive and irreducible with unrefinable blocks of size 2 so that $\mS=\{\pm r,\pm s\}$ with $u=r/s$ not a $4$th root of unity.  Denote the order of $u$ by $o(u)$.  Then:
\begin{enumerate}
\item[(a)] The characteristic polynomial of $AB^{-1}$ is either $p_1(u,t)/u^2$
 or
$p_1(-u,t)/u^2$ where $D=-1$ corresponds to $p_1(u,t)/u^2$ and $D=1$ corresponds
to $p_1(-u,t)/u^2$.
\item[(b)] If $o(u)\not\in\{3,5,6,10\}$, then $G$ is \textbf{infinite}.
\item[(c)] If $(D,o(u))\in\{(1,5),(-1,10)\}$, $G$ is \textbf{infinite}.
\item[(d)] If $o(u)\in\{3,6\}$ $G$ is \textbf{finite}.
\item[(e)] If $(D,o(u))\in\{(-1,5),(1,10)\}$ $G$ is \textbf{finite}.

\end{enumerate}
\end{theorem}

\begin{proof} We may replace $A$ by $\frac{1}{s}A$ and $B$ by $\frac{1}{s}B$
without changing the finiteness of $G$ or the characteristic polynomial of $AB^{-1}$, and doing so we obtain the following matrices from \cite{TbWz}:
$$A=\begin{pmatrix}1&-({D}^{-2}+D^{-1}+1)&
\left( {D}^{-2}+D^{-1}+1 \right) u&-u\\0&-1&
\left( D^{-1}+1 \right) u&-u\\0&0&u&-u
\\0&0&0&-u\end {pmatrix}$$
and
$$B=\begin{pmatrix} -u&0&0&0\\-u&u&0&0
\\-D&D+1&-1&0\\-{D}^{3}&{D}^{3}+{D
}^{2}+D&-{D}^{2}-D-1&1\end {pmatrix}$$
and (a) follows by computation.
If the eigenvalues of $AB^{-1}$ are not roots of unity then $G$ is infinite.
Let us assume that $D=-1$ and the roots $t$ of $p_1$ (i.e. the eigenvalues of $AB^{-1}$) are roots of unity.  Set $x=t+1/t$ and $y=u+1/u$ and consider
$p_1(u,t)/(ut)^2=x^2+(y+1)x+(y^2-2+2y)$.  By applying a Galois automorphism of the field $\Q[u,t]$ we may assume that $u=e^{2\pi \im/\ell}$ for $\ell=o(u)$. Notice that $y=2\Re(u)=2\cos(2\pi/\ell)$.  If we assume $\ell>6$
then $y>1$, and the discriminant of $x^2+(y+1)x+(y^2-2+2y)$ is:
$-3y^2-6y+9<0$ so that $x\not\in\R$, contradicting $x=2\Re(t)$.
Next assume $D=1$.  The argument is essentially the same, except we get
the polynomial $x^2+(1-y)x+(y^2-2-2y)$ with discriminant $9-3y^2+6y$ when we replace $u$ by $-u$ in $p_1$.
Using a Galois automorphism we may assume that $u=e^{2\pi \im k/\ell}$ where $k$ is chosen so that $u$ is nearest $-1$.  Now $y=2\Re(u)<-1$ provided $\ell>4$ and $\ell=\neq 6$ or $10$, the latter exclusion coming from the fact that the primitive $10$th root of unity nearest $-1$ is $e^{6\pi \im/10}$ which has real part $-\cos(2\pi/5)>-1/2$.  Again, one obtains a contradiction since $9-3y^2+6y<0$
contradicting $x\in\R$.  This proves (b) and (c).

To prove (d) we observe that in case $o(u)=3$ or $6$, the projective order of $A$ is $6$, and the element $AB^{-1}$ also has projective order $6$.  These together with the braid relation $ABA=BAB$ are enough to conclude that, modulo the center, $G$ is a quotient of
a group of order $648$.

For (e)  we consider the normal subgroup $H$ generated by $A^5$ and $B^5$.  Observe that $H$ is of finite index by the result of Coxeter above.  A set of generators for $H$ is $\{A^5,B^5,AB^5A^{-1},BA^5B^{-1}\}$.  Set $M:=\begin{pmatrix} 1&-1\\0&-1\end{pmatrix}$, $N:=\begin{pmatrix} -1&0\\-1&1\end{pmatrix}$ and
$T:=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$.  Then if $o(u)=5$ and $D=-1$ we compute $A^5=M\oplus M$, $B^5=N\oplus N$ and $AB^5A^{-1}=BA^5B^{-1}=T\oplus T$.  One can easily see that $M,N$ and $T$ generate
$S_3$, from which it follows that $G$ is finite.  For the case $D=1$ and $o(u)=10$ the matrices involved are slightly more complicated, such as $\begin{pmatrix} 1 & -3\\0&-1\end{pmatrix}$, $\begin{pmatrix} -2 & 3\\-1&2\end{pmatrix}$ and $\begin{pmatrix} 1 & 0\\1&-1\end{pmatrix}$, but give an equivalent representation of $S_3$ and we conclude that $G$ is finite in this case as well.

\end{proof}

This completes the proof of Theorem \ref{maintheorem}(c).

 \section{Primitive Groups}\label{prim}
 In this section we
assume that $G$ is a finite primitive irreducible group.  Then by rescaling $A$ and $B$ by a choice of a root of unity $(\det(A))^{-1/d}$, we may assume that
$G$ is unimodular without changing $po(A)$.   Thus we can determine the possible values of $po(A)$ by computing the projective orders of elements in the groups on Feit's list of finite unimodular primitive irreducible linear groups of degree $5$ or less \cite{Ft}.

\begin{lemma}\label{orders}
Suppose $g$ is an element in a primitive unimodular irreducible finite group $H$ of dimension $2\leq d\leq 5$ and let $t$ be the order of $g$ modulo $Z(H)$.  Then
\begin{enumerate}
\item[(a)] if $d=2$, $1\leq t\leq 5$,
\item[(b)] if $d=3$, $1\leq t\leq 7$,
\item[(c)] if $d=4$, $t\in\{1,\ldots,10,12,15,20,24\}$ and
\item[(d)] if $d=5$, $t\in\{1,\ldots,6,9,\ldots,12\}$.
\end{enumerate}
\end{lemma}
\begin{proof}
The list of such  primitive unimodular irreducible finite groups is given in \cite[Section 8.5]{Ft}, and the only work to do is to construct the groups using \cite{GAP} and compute the orders of elements in the quotient $H/Z(H)$.  Modulo their centers, the dimension $2$ groups are  $S_4, A_4$, and $A_5$, and for $d=3$ one has $A_5, A_6, \PSL(2,7)$ and subgroups of the Hessian group of order $216$.  For $d=4$ there are 11 types of groups.  Four of these come from direct products of dimension $2$ linear groups and their extensions by an order $2$ outer automorphism.  One also has the explicit groups $A_5, S_5, A_6, S_6, A_7, \SL(2,5), \SL(2,7)$ and $\Sp(4,3)$.  The last class of dimension $4$ groups are certain subgroups of extraspecial $2$-groups of order $2^5$ by their automorphism groups.
For $d=5$ Feit's list yields the following groups (with trivial centers):
\begin{enumerate}
\item Subgroups of $\SL(2,5)\ltimes(\Z_5\times\Z_5)$: orders $(1,2,3,4,5,6,10)$
\item $A_5$: orders $(1,2,3,5)$
\item $S_5$: orders $(1,2,3,4,5,6)$
\item $A_6$: orders $(1,2,3,4,5)$
\item $S_6$: orders $(1,2,3,4,5,6)$
\item $\PSp(4,3)$: orders $(1,2,3,4,5,6,9,12)$
\item $\PSL(2,11)$: orders $(1,2,3,5,6,11)$
\end{enumerate}

To illustrate how one uses GAP to get the information listed in the theorem, we give the following sample code for determining the possible orders in
the class of $4$ dimensional groups mentioned above.:\\

\noindent\texttt{gap> H:=ExtraspecialGroup(32,"-");\\
<pc group of size 32 with 5 generators>\\
gap> T:=AutomorphismGroup(H);\\
 <group of size 1920 with 2 generators>\\
 gap> M:=SemidirectProduct(T,H);\\
 <permutation group with 7 generators>\\
gap> N:=M/Center(M);\\
<permutation group of size 30720 with 11 generators>\\
gap> s:=Elements(N);;\\
gap> ords:=List(s,Order);;\\
gap> Set(ords);\\
$[$ 1, 2, 3, 4, 5, 6, 8, 12 $]$ }

\end{proof}

Parts (i),(iii) and (iv) of Theorem \ref{maintheorem}(d) are immediate from Lemma \ref{orders}.  We proceed by cases to prove part (ii).
\begin{proof}(of Theorem 2.1(d)(ii)).
For $d=3$, Lemma \ref{orders} implies that $G$ is infinite if $po(A)\geq 8$, so we need only consider $\ell:=po(A)=6$ and $7$.
Let $\mS=\{\la_1,\la_2,\la_3\}$.  Clearly $G$ is finite and irreducible if and only if the group generated by $A/\la_1$ and $B/\la_1$ is finite and irreducible, so we may
assume $\mS=\{1,\theta,\phi\}$.  By applying a Galois automorphism we may further assume that $\theta=e^{2\pi\im/\ell}$ and $\phi=\theta^k$ with $2\leq k\leq \ell-1$.
From \cite[Proposition 2.5]{TbWz} we obtain:
$$A=\begin{pmatrix} 1&\phi/\theta+\theta & \theta\\
            0&\theta & \theta\\
            0&0& \phi\end{pmatrix},
B=\begin{pmatrix} \phi & 0& 0\\
        -\theta & \theta & 0\\
        \theta & -\phi/\theta-\theta & 1
  \end{pmatrix}.$$

If $po(A)=6$, the cases $k=2,5$ correspond to reducible representations (with infinite image), since $\theta\cdot 1+\phi^2=e^{\pi\im/3}+e^{2k\pi\im/3}=0$ for $k=2,5$.  For $k=3,4$ the eigenvalues case are $\pm 1,e^{\pi\im/3}$ and $1,\pm e^{\pi\im/3}$ respectively.  Theorem \ref{monom} suggests these give us imprimitive groups.  Writing down the $A$ and $B$ matrices in monomial form one checks easily that they have no common eigenvector and hence the representation is irreducible and imprimitive. Thus $po(A)=6$ implies $G$ is either reducible or imprimitive.

If $po(A)=7$, the cases $k=2,4$ or $6$ each give infinite groups as can be seen either by checking that the eigenvalues of $AB^{-1}$ are not roots of unity, or by observing that that the only dimension $3$ irreducible imprimitive group with elements of order 7 is $\PSL(2,7)$, which also has elements of orders $1,2,3$ and $4$.  So we need only check that $AB^{-1}$ does not have one of these orders, and for $k=2,4$ or $6$ this computation yields the desired result. For $k=3,5$ one finds that $(AB^{-1})^4=I$ which, together with the braid relation and $A^7=B^7=I$, is enough to show that $G$ is a quotient of a group of order $1176$ and hence finite.
\end{proof}

This completes the proof of Theorem \ref{maintheorem}(d).

\begin{rmk}
We would like to point out the limitations of our approach, in particular why we do not get complete results for dimensions $4$ and $5$ with $G$ primitive.  Firstly, the representations are not uniquely determined by the eigenvalues: for $d=4$ there are two choices, and for $d=5$ there are $5$.  Secondly, the sets of eigenvalues $\mS$ with $po(A)=t$ for some $t$ found in Lemma \ref{orders} can be quite large particularly if $t$ is composite.  Finally, the only technique we have for showing that $G$ is infinite in these cases is to show that some element (such as $AB^{-1}$) has infinite order.  Moreover, if the order happens to be finite, this relation might not be sufficient to conclude that
$G$ is finite, in which case we must resort to further \emph{ad hoc} means.  For $d\geq 6$, there are further issues.  A representation of degree $6$ or higher is not determined by $\mS$ and $\gamma$.  Moreover, the imprimitive cases are significantly more delicate.  Given the eigenvalues of an irreducible representation $\rho$ of $\B_3$ of degree $d=6$ or $7$ one can sometimes use our approach as follows: first verify that $\rho(\B_3)$ is primitive by using Lemma \ref{nocyclem}.  Then use Feit's list \cite{Ft} to determine which $po(A)$ can appear in finite linear groups of degree $d$.  For example, if $d=7$, $\mS$ does not contain a full coset of $C_r\subset\C$, $r\leq 7$ and $p\mid po(A)$ for $13\neq p\geq 11$ a prime, then $\rho(\B_3)$ is infinite.
\end{rmk}

\section{Applications}\label{aps}
We apply our results to ribbon categories obtained as subquotients of $\Rep(U_q\g)$ where $q=e^{\pi\im/\ell}$, see \cite{TurBook} for details.  These categories
will be denoted $\CC(\g,q,\ell)$ as in \cite{Rsurvey}.  In particular the representation of $\B_3$ acting on $\Hom(V,V^{\ot 3})$ is irreducible provided the eigenvalues of $A$ are distinct by \cite[Lemma 3.2]{TbWz}.  We can compute the
eigenvalues by applying results of Reshetikhin found in \cite[Corollary 2.22(3)]{LeducRam}.  Since the cases $d=2$ and $d=3$ were already considered in \cite{FLW2} and \cite{LRW} respectively, we focus on the cases $d=4$ and $5$.
Since the order of $G=\rho(\B_3)$ is invariant under Galois automorphisms, we can safely assume $q=e^{\pi\im/\ell}$ without loss of generality.  The computation involves two steps: first we use standard Lie theory techniques to decompose $V^{\ot 2}$ as a direct sum of simple objects $V_{\la_i}$.  The eigenvalues of the
action of $\rho(\sigma_1)$ are then computed (up to an overall scale factor)
as $\pm q^{\lan \la_i+2\delta,\la_i\ra/2}$ where the sign is positive if $V_{\la_i}$ appears in the symmetrization of $V^{\ot 2}$ and is negative otherwise, and $\delta$ is $1/2$ the sum of the positive weights.

\begin{example} Let $\g=\g_2$ and let $V$ be the simple object labelled by $\la_1=(\ve_1-\ve_3)$, i.e. the highest weight of the $7$ dimensional fundamental representation of $\g_2$.  First suppose that $3\mid\ell$ so that the corresponding category is a unitary modular category (see \cite{rowellJPAA}).  Then if $18\leq\ell$, $\dim\Hom(V,V^{\ot 3})=4$ and the image of $\sigma_1$ has eigenvalues $\{q^{-12},q^2,-q^{-6},-1\}$.  Thus the projective order of the image of $\sigma_1$ is $\ell$ if $\ell$ is even and
$2\ell$ if $\ell$ is odd.  Thus by Theorem \ref{maintheorem}(d)(iii) the image of $\B_3$ is infinite unless or $24$.  Notice however, that for $\ell=24$ we have repeated eigenvalues, which implies the image is infinite by Theorem \ref{maintheorem}(a).  Now suppose that $3\nmid\ell$.  In this case
$\dim\Hom(V,V^{\ot 3})=4$ for $10\leq\ell$.  The eigenvalues of the image of $\sigma_1$ are as above, so that by Theorem \ref{maintheorem}(d)(iii) the image of $\B_3$ is
infinite unless $\ell=10$ or $20$.
\end{example}

\begin{example} Now let $\g=\mathfrak{f}_4$, and $V$ be the simple object in $\CC(\mathfrak{f}_4,q,\ell)$ analogous to the vector representation of $\mathfrak{f}_4$.  If $22\leq\ell$ and even, $\dim\Hom(V,V^{\ot 3})=5$ and the eigenvalues of the image of $\sigma_1$ are $\{q^{-24},q^{-12},q^2,-1,-q^{-6}\}$.  Thus Theorem \ref{maintheorem}(d) implies that the image of $\B_3$ is infinite for $22\leq\ell$ since the projective order of the image of $\sigma_1$ is $\ell$ in these cases.  Notice that in the case $\ell=24$ we have repeated eigenvalues.  Next assume that $\ell$ is odd.  We have $\dim\Hom(V,V^{\ot 3})=5$ when $15\leq\ell$, and the projective order of the image of $\sigma_1$ is $2\ell$ so the image of $\B_3$ is again always infinite by Theorem \ref{maintheorem}(d).
\end{example}

\begin{example}
Consider $\g=\mathfrak{so}_7$ and let $V$ be the simple object in $\CC(\mathfrak{so}_7,q,\ell)$  corresponding to the fundamental spin representation of $\mathfrak{so}_7$.
When $\ell$ is even and $14\leq\ell$ we have $\dim\Hom(V,V^{\ot 3})=4$ and the eigenvalues of the image of $\sigma_1$ are $\chi\{1,q^{12},-q^6,-q^{10}\}$.  Since $q=e^{\pi\im/\ell}$ with $\ell$ even, $po(A)=\ell/2\geq 7$.  So provided $\ell/2\not\in\{7,8,9,10,12,15,15,20,24\}$, $G$ is infinite.  There are two representations with these eigenvalues corresponding to the two choices of
$D=\pm \sqrt{\frac{\la_2\la_3}{\la_1\la_4}}=\pm q^4$.  Choosing $D=q^4$, we obtain the following matrices from \cite{TbWz}:
$$A=\begin {pmatrix} 1& \left( {q}^{8}+{q}^{4}+1 \right) {q}^{4}&-{\frac {{q}^{8}+{q}^{4}+1}{{q}^{2}}}&-{q}^{10}\\0
&{q}^{12}&- \left( {q}^{4}+1 \right) {q}^{2}&-{q}^{10}
\\0&0&-{q}^{6}&-{q}^{10}\\0&0&0&-{
q}^{10}\end {pmatrix}$$
$$B=\begin {pmatrix} -{q}^{10}&0&0&0\\{q}^{6}&-{q}^{6}&0&0\\{q}^{16}&- \left( {q}^{4}+1 \right)
{q}^{12}&{q}^{12}&0\\-{q}^{12}&{q}^{12}+{q}^{8}+{q}^
{4}&-{q}^{8}-{q}^{4}-1&1\end {pmatrix}.$$

We first consider the case $\ell=14$.  In this case we have $A^7=B^7=I$.  We compute that (projectively) $(AB^{-1})^4=I$.  This implies that $G$ is indeed finite!  We compute that $|G/Z(G)|=168$ so that, projectively, $G$ is $\PSL(2,7)$.  We further note that there is an object $U$ so that $\dim\Hom(U,V^{\ot 3})=3$ and the corresponding braid group representation is
irreducible, given in \cite{West}.  The eigenvalues are (up to scaling and Galois conjugation) $\{1,e^{2\pi \im/7},e^{10\pi \im/7}\}$ so that by Theorem
\ref{maintheorem}(d)(ii), this reprensentation also has finite image.

Next consider the case $\ell=18$.  Under this substitution
the eigenvalues are of the form $\{1,\omega,\omega^2,\alpha\}$ where $\omega$ is a primitive $3$rd root of unity.  So for either choice of $D$ we find that $G$ is a finite imprimitive group by Theorem \ref{maintheorem}(c)(i) and the TW classification.
\end{example}

\begin{example}
Consider $\g=\mathfrak{so}_9$ and let $V$ be the simple object in $\CC(\mathfrak{so}_9,q,\ell)$  corresponding to the fundamental spin representation of $\mathfrak{so}_9$. When $\ell$ is even and $18\leq\ell$ we have $\dim\Hom(V,V^{\ot 3})=5$ and the eigenvalues of the image of $\sigma_1$ are
$\chi\{1,q^8,-q^{14},-q^{18},q^{20}\}$, and $\gamma=\det(A)^{1/5}=\zeta_5q^{12}$ where $\zeta_5$ is a primitive $5$th root of unity.  For simplicity we will assume $\gamma=q^{12}$, although this assumption certainly can affect $|G|$.

For $\ell=18$ the matrix $A$ has repeated eigenvalues, (namely $1$ and $-q^{18}=1$) so Lemma 3.2 of \cite{TbWz}
fails and we cannot conclude that the representation is irreducible in this case.
In fact, the image for $\ell=18$ satisfy relations $A^9=B^9=(A^4(ABA)A^5(ABA))^2$
which, together with the braid relation, imples that the projective image $G$ is a quotient of a group of order $324$.  This implies that the representation is reducible, since any such group cannot have an irreducible representation of dimension $5$.

For $\ell\geq 20$ one does have distinct eigenvalues, so that the corresponding representation is irreducible.  We see that $po(A)=\ell/2$ in these cases, so
provided $\ell/2\not\in\{10,11,12\}$ $G$ is infinite.  The matrices, $A$ and $B$ we obtain are:
$$\begin {pmatrix} 1&{q}^{8}-{q}^{6}+{q}^{4}-{q}^{2}&-{q}^{14}+{q}^{12}-2\,{q}^{10}+{q}^{8}-{q}^{6}&-{q}^{16}+{q}^{14}-{q}^{12}+{
q}^{10}&{q}^{16}\\0&{q}^{8}&-{q}^{14}+{q}^{12}-{q}^{
10}&-{q}^{16}+{q}^{14}-{q}^{12}&{q}^{16}\\0&0&-{q}^{
14}&-{q}^{16}+{q}^{14}&{q}^{16}\\0&0&0&-{q}^{18}&{q}
^{18}\\0&0&0&0&{q}^{20}\end {pmatrix}$$
$$\begin {pmatrix} {q}^{20}&0&0&0&0\\{q}^{18}&-{q}^{18}&0&0&0\\{q}^{16}&-{q}^{16}+{q}^{14}&-{
q}^{14}&0&0\\{q}^{16}&-{q}^{16}+{q}^{14}-{q}^{12}&-{
q}^{14}+{q}^{12}-{q}^{10}&{q}^{8}&0\\{q}^{16}&-{q}^{
16}+{q}^{14}-{q}^{12}+{q}^{10}&-{q}^{14}+{q}^{12}-2\,{q}^{10}+{q}^{8}-
{q}^{6}&{q}^{8}-{q}^{6}+{q}^{4}-{q}^{2}&1\end {pmatrix}.$$
The case $\ell=22$ is interesting: If we set $S=A$ and $T=ABA$, we find that the $\PSL(2,11)$ relations $S^{11}=T^2=(S^4TS^6T)^2=I$ hold (projectively), so that the projective image $G$ is $\PSL(2,11)$.  This is not too surprising since $\PSL(2,\Z)$ is a quotient of $\B_3$.  The cases $\ell=20$ and $\ell=24$ do not yield finite groups.
\end{example}


text:
\section{Problem Formulation and Related work}
\label{sec:related_work}

We focus on supervised learning tasks where the goal is to solve
a non-convex stochastic optimization problem of the form:
\begin{equation}\label{eq:basic_problem}
    \min_{\theta}\loss(\theta) = \frac1N\sum_{i=1}^N l_i(x_i, y_i; \theta),
\end{equation}
where $\theta\in\sR^d$ denotes the model parameters, $l_i(x_i, y_i; \theta)$ is the loss function, $(x_i,y_i)$ is the paired input data and its corresponding ground truth label, and
$N$ is the total number of data points in the training dataset.
Furthermore, we denote the gradient of the loss w.r.t. model parameters as $\g=\frac{1}{N_B}\sum_{i=1}^{N_B}\frac{\partial l_i}{\partial \theta}$, and the corresponding
second derivative (i.e., Hessian) as $\H=\frac{1}{N_B}\sum_{i=1}^{N_B}\frac{\partial^2 l_i}{\partial \theta^2}$, where $N_B$ is the size of one mini-batch.

Solving~\eref{eq:basic_problem} for a real learning problem (and not a simple model) is a very challenging task. Despite years
of research, we have not yet been able to resolve several seemingly ad-hoc tricks
that are required to converge (hopefully) to a \emph{good} solution.
Next, we briefly discuss the different popular optimization methods proposed in recent years
to address the challenges associated with solving~\eref{eq:basic_problem}.
This is by no means a comprehensive review, and we refer the interested reader
to~\cite{bottou2018optimization} for a thorough review.

\subsection{Adaptive First Order Methods}

Due to their simplicity and effectiveness,
first order optimization methods~\cite{robbins1951stochastic,nesterov1983method,duchi2011adaptive,zeiler2012adadelta,kingma2014adam,loshchilov2017decoupled} have become the de-facto algorithms used in deep learning.
There are multiple variations, but these methods can be represented using the following general update formula:
\begin{equation}
    \theta_{t+1} = \theta_t - \eta_t m_t / v_t,
    \label{e:sgd}
\end{equation}
where $\eta_t$ is the learning rate, and $m_t,\text{ and } v_t$ denote the so called first and second moment terms, respectively.
A simple and popular update method is SGD, originally proposed in 1951 as a root-solving algorithm~\cite{robbins1951stochastic}:
\begin{equation}
    m_t = \beta m_{t-1} + (1-\beta)\g_t~~~\text{and}~~~v_t \equiv 1.
\end{equation}
Here, $\g_t$ is the gradient of a mini-batch at $t$-th iteration and $\beta$ is the momentum hyperparameter.

Using \sgd to solve~\eref{eq:basic_problem} is often very challenging, as the convergence of the iterative formulae in~\eref{e:sgd}
is very sensitive to the right choice of the learning rate, its decay schedule, and the momentum parameter.
To address this, several methods have been proposed to take into account the knowledge of the geometry of the
data by scaling gradient coordinates, using the past gradient information.
This can be viewed in one of two equivalent ways: either as automatically adjusting the
learning rate in~\eref{e:sgd}; or as an adaptive \emph{preconditioning} of the gradient.
One notable method is \adagrad~\cite{duchi2011adaptive,mcmahan2010adaptive}, which accumulates
all the gradients from the first iteration and applies the square root of the result to
precondition the current gradient. The update formulae in this case become\footnote{Throughout the paper, without further notification, for two vectors, e.g., $a$ and $b$, we use both $ab$ and $a\odot b$ to denote the element-wise product, and $\langle a,b \rangle$ denotes the inner product.}:
\begin{equation}
\label{e:adagrad}
    m_t = \g_t~~~~~~\text{and}~~~~~v_t = \sqrt{\sum_{i=1}^t\g_i\g_i}.
\end{equation}

While \adagrad works well for sparse settings, its performance significantly degrades for dense settings, which
is the case for many machine learning tasks.
In particular, this stems from the accumulation of all previous gradients for the preconditioner~\eref{e:adagrad}.
This results in a monotonic increase in the magnitude of the second moment, $v_t$, which effectively translates
into a rapid decay of the learning rate.
To address this, several methods have been proposed where the intuition is to limit the accumulation
to a small window of past iterations, and in particular exponentially reduce the weight of earlier iterations.
Notable works incorporating this method are RMSProp, ADADelta, and \adam~\cite{tieleman2012lecture,zeiler2012adadelta,kingma2014adam}.
In particular, for \adam~\cite{kingma2014adam}, the two moments for the update rule are the following:
\begin{equation}
\begin{split}
    m_t = \frac{(1-\beta_1)\sum_{i=1}^t\beta_1^{t-i}\g_i }{1-\beta_1^t}, \\
    v_t = \sqrt{\frac{(1-\beta_2)\sum_{i=1}^t \beta_2^{t-i}\g_i\g_i}{1-\beta_2^t}},
\end{split}
\end{equation}
where $0<\beta_1,~\beta_2<1$ are two hyperparameters sometimes referred to as first and second moment coefficients.
In particular, note that the sum over past gradients is scaled by $\beta_2$ which exponentially reduces
the contribution of early gradients.
A summary of the different $m_t$ and $v_t$ used by common first-order optimizers is given in~\tref{tab:optimizer_summary}.
A notable variant here is \adamw~\cite{loshchilov2017decoupled}, which shows that decoupling weight decay from the update equation of \adam can lead to a noticeable performance improvement.
Recently, \adamw has become the preferred optimizer for NLP tasks, and in particular for training
transformers~\cite{vaswani2017attention}.
There are also many other variants of adaptive first order methods~\cite{chaudhari2019entropy,zhang2015deep,loshchilov2016sgdr,shazeer2018adafactor,singh2015layer}.

Despite all these attempts, it is still not clear which optimizer should work for a \emph{new} learning task/model.
This is in fact one of the main baffling practical issues in machine learning, and one for which theory has little to say.
For example, \sgd is currently the best performing optimizer for some CV tasks.
That is, using other variants such as \adamw leads to significantly worse generalization performance.
However, for NLP tasks, \adamw has the best performance by a large margin as compared to \sgd.
The point here is that even the choice of the optimizer has effectively become a hyperparameter.

\begin{table}[t]
\caption{\footnotesize
Summary of the first and second moments used in different optimization
algorithms for updating model parameters ($\theta_{t+1} = \theta_t - \eta m_t / v_t$).
Here $\beta_1$ and $\beta_2$ are first and second moment hyperparameters.
}
\label{tab:optimizer_summary}
\centering

\setlength\tabcolsep{2.pt}
\begin{adjustbox}{width=1\linewidth}
\begin{tabular}{lcccccc}
\toprule
\ha Optimizer & $m_t$  & $v_t$   \\
\midrule
\ha \sgd~\cite{robbins1951stochastic}   & $\beta_1 m_{t-1} + (1-\beta_1)\g_t $  & 1 \\[8pt]
\ha \adagrad~\cite{duchi2011adaptive}   & $\g_t$                            & $\sqrt{\sum_{i=1}^t \g_i\g_i}$ \\[8pt]
\ha \adam~\cite{kingma2014adam}         & $\frac{(1-\beta_1)\sum_{i=1}^t\beta_1^{t-i}\g_i }{1-\beta_1^t}$   & $\sqrt{\frac{(1-\beta_2)\sum_{i=1}^t \beta_2^{t-i}\g_i\g_i}{1-\beta_2^t}}$ \\[8pt]
\ha RMSProp~\cite{tieleman2012lecture}  & $\g_t$                                                            & $\sqrt{\beta_2 v_{t-1}^2 + (1-\beta_2)\g_t\g_t}$ \\[8pt]
\midrule

\hc \OURS                               & $\frac{(1-\beta_1)\sum_{i=1}^t\beta_1^{t-i}\g_i }{1-\beta_1^t}$   & $\sqrt{\frac{(1-\beta_2)\sum_{i=1}^t \beta_2^{t-i}\mD_i^{(s)}\mD_i^{(s)}}{1-\beta_2^t}}$ \\
\bottomrule
\end{tabular}
\end{adjustbox}
\end{table}

\subsection{Second Order Methods}

Second order methods are among the most powerful optimization methods that have been designed, and
there have been several attempts to use their many advantages for training NNs.
Second order methods are designed to address \emph{ill-conditioned} optimization problems
by automatically rotating and rescaling the gradient. This allows one to choose a better descent direction, and to adjust automatically the learning rate for each parameter.
There have also been multiple theoretical studies showing better convergence
rate of second order based methods~\cite{bollapragada2019exact,yao2018inexact,roosta2016sub,xu2016sub,xu2017newton,xu2017second,conn2000trust,agarwal2016second,wang2018giant,agarwal2016finding,carmon2018accelerated,chen2019fast}.
In particular, second order methods can guarantee convergence to
second order critical points, while the vast majority
of first order methods lack such guarantees.
For example, theoretically it has
been shown that
some first order methods can only converge to an approximate second order critical point~\cite{ge2015escaping,jin2017escape,levy2016power,reddi2017generic}.

\begin{figure*}[t]
\begin{center}

  \includegraphics[width=.90\linewidth]{figures/diagonal_illustration2.pdf}
\end{center}
 \caption{\footnotesize
A simple model with N layers (first column); with the convolutional blocks of the N-1 layer shown (second column); and the loss landscape of each block (third column), which can be calculated by perturbing the convolutions's parameters
in two different eigendirections.
(See~\cite{yao2019pyhessian} for details of how to construct loss landscape.)
Note the different loss landscape topologies.
First order methods do not explicitly capture this difference.
The entries (3D tensors) colored in orange show the components used for calculating the spatial average of Hessian.
The part of the gradient (fourth panel) highlighted in the orange box is the corresponding gradient of the orange convolution kernel; and the part of the Hessian diagonal (fifth panel) highlighted in the orange box is used to compute the spatial average.
}
\label{fig:block_hessian_vis}
\end{figure*}

Newton's method is a classical second order method where one solves a linear system, essentially meaning that the inverse of the local Hessian is used at every
iteration to precondition the gradient vector.
\footnote{To be clear, when we refer to computing an inverse, we mean that we use a numerical method that performs a linear equation solve that effectively amounts to working with the inverse implicitly.  Of course, one would never actually compute the Hessian inverse explicitly.}
One major challenge with this
approach is that it can be expensive to solve the linear system, na\"{\i}vely requiring cubic computational complexity, not including the cost of forming the Hessian itself and
the corresponding quadratic memory complexity.
However, the overhead of such a na\"{\i}ve implementation can be improved by
using so-called matrix free methods, where the Hessian matrix is never explicitly formed (addressing quadratic
memory cost), and its inverse is approximately and only implicitly applied (addressing the cubic computational complexity).

One seminal work here is the limited memory BFGS (LBFGS)~\cite{byrd1995limited} method which has
a desirable linear computational and memory complexity. This approach approximates the
Hessian as a series sum of first order information from prior iterations. As such, these
approaches that do not directly use the Hessian operator are referred to as \emph{Quasi-Newton} methods.
While this approach works well for many optimization problems, it does not work
well for many machine learning problems. One reason for this is that LBFGS method
requires full batch gradients, as stochastic gradients can lead to drastic
errors in the approximation~\cite{bollapragada2018progressive}. This is one
of the main challenges with Quasi-Newton methods applied to machine learning problems~\cite{aghazadeh2020bear}.
Other approaches such as approximating the Hessian as the Kronecker product of vectors
have also been explored~\cite{martens2015optimizing}.

There has also been work on enhancing first order methods by incorporating the Fisher
information matrix~\cite{gupta2018shampoo}. The main idea is to use the Fisher
information instead of the squared norm of the gradient. A na\"{\i}ve use of Fisher information
has computational and memory overhead, but it is possible to also approximate the Fisher
information matrix using low rank decomposition~\cite{gupta2018shampoo}.

Another line of work has been to incorporate automatically the Hessian operator itself, instead
of approximating it using first order information. A

work here
is~\cite{schaul2013no} which uses Gauss-Newton Hessian diagonal to
adjust adaptively the learning rate.
The work of~\cite{xu2017second} also directly incorporates
the Hessian using a trust region~method.

While the above approaches are very interesting
and result in a good performance for simple models, they do not achieve
comparable results for more complex NN architectures.
One of the reasons that second order methods have not been successful yet for machine learning,
as opposed to other domains such as scientific computing, is due to the stochastic
nature of the problem. Such stochastic noise leads to an erroneous approximation of the Hessian, leading
to suboptimal descent directions.
SGD is more robust to such noise since we can efficiently incorporate moving averages and momentum.
Ideally, if there was a way to apply the same moving average method to the Hessian, then that would help smooth out local curvature noise to get a better approximation to the non-noisy
curvature of the loss landscape. However, such an approximation is challenging
since the Hessian is a matrix that cannot be explicitly formed to be averaged, whereas
it is easy to form the gradient vector.

As we show below, \OURS addresses this problem by incorporating the Hutchinson's method
along with spatial averaging to reduce the impact of the stochastic noise. The result exceeds
the performance of all the above methods for machine learning tasks.
Next, we formally introduce the \OURS algorithm.


TITLE: Energy change as ball rolls down a curved ramp
QUESTION [0 upvotes]: At the top of the ramp, the ball has gravitational potential energy. As the ball rolls down the ramp, the potential energy converts to kinetic energy. Due to friction, some energy is also converted to thermal energy. The ramp curves upwards at the bottom. So the ball then moves upwards. So the kinetic energy turns back to potential energy. The final height is less that initial height. To describe this change, which option is most accurate:

potential → kinetic → potential + thermal
potential → kinetic + thermal → potential + thermal

My guess is 1, l know that friction is caused due to motion, so kinetic energy should be converting to thermal, and thus kinetic is converted to potential and thermal at the end of the experiment.
But I think 2 could also be correct as friction is caused simultaneously as potential is converted to kinetic, and after maximum potential energy is converted, the kinetic starts to convert back to potential, also simultaneously producing heat.
I understand the topic, but I am not sure how to express on paper. Please give an answer and explain why the other is wrong.

REPLY [2 votes]: I assume > means “goes to” not “greater than”.
Not sure what the sequence is intended to describe but on the way down PE is simultaneously converted to  KE and heat dissipated by friction. At the bottom of the path the kinetic energy will be
$$KE_{bottom} = PE_{start}-Q_1$$
Where $Q_1$ = energy dissipated in the environment in the form of friction heating on the way down.
At the end on the way up
$$PE_{end} = KE_{bottom}-Q_2$$
Where $Q_{2}$ = energy dissipated in the environment in the form of friction heating on the way up.
For overall conservation of energy we have
$$PE_{end}=PE_{start}-(Q_{1}+Q_{2})$$
So I think option 2 is most accurate qualitatively realizing PE at the end is less than PE at the start. Also, friction heat on the way up is less than on the way down assuming the ramp is symmetrical about the bottom since friction work depends on the length of the path.
Hope this helps


TITLE: Solving the system of three equations $x+y=1, y+z=7, z-x=6$
QUESTION [0 upvotes]: $$x+y=1\\
y+z=7\\
z-x=6$$
I found an answer, but I don't know how to show my work to get it. I must've gotten it on a fluke, because I don't know how I got there.
Can anyone explain to me how to get the answer to this problem?
And what type of problem is this called? I couldn't find any examples like it online.
Thanks.

REPLY [1 votes]: This problem is a system of linear equations. You can use substitution to solve it. From the first equation, $x = 1 - y$, so substituting into the third equation we get $z - (1-y)=6$, that is, $y+z=7$. Notice that this is precisely the second equation, so that equation does not provided additional constraints on $(x,y,z)$.  So we have 2 equations and 3 unknowns, thus infinite possible solution triplets. Further substitution shows the pattern of the triplets:
$y +z = 7 \implies y = 7 - z$, and
$x+y = 1 \implies x = 1 -y = 1 - (7 - z) = z - 6$.
So, for any given $z$, the solution will look like $(z-6,7-z,z)$.