text:
\begin{document}

\title{The ancient Greeks present: Rational Trigonometry}
\author{N J Wildberger\\School of Mathematics and Statistics \\UNSW Sydney 2052 Australia\\webpages: http://web.maths.unsw.edu/\symbol{126}norman/}
\maketitle

\begin{abstract}
Pythagoras' theorem, the area of a triangle as one half the base times the
height, and Heron's formula are amongst the most important and useful results
of ancient Greek geometry. Here we look at all three in a new and improved
light, using \textit{quadrance} not \textit{distance}. This leads to a simpler
and more elegant trigonometry, in which \textit{angle} is replaced by
\textit{spread}, and which extends to arbitrary fields and more general
quadratic forms.

\end{abstract}

\section*{Three ancient Greek theorems}

There are three classical theorems about triangles that every mathematics
student traditionally meets. To state these, consider a triangle
$\overline{A_{1}A_{2}A_{3}}$ with side lengths $d_{1}\equiv\left\vert
A_{2},A_{3}\right\vert $, $d_{2}\equiv\left\vert A_{1},A_{3}\right\vert $ and
$d_{3}\equiv\left\vert A_{1},A_{2}\right\vert $.

\begin{description}
\item[Pythagoras' theorem] \textit{The triangle }$\overline{A_{1}A_{2}A_{3}}
$\textit{\ has a right angle at }$A_{3}$\textit{\ precisely when}
\[
d_{1}^{2}+d_{2}^{2}=d_{3}^{3}.
\]

\item[Area of triangle] \textit{The area of a triangle is one half the length
of the base times the height.}

\item[Heron's formula] \textit{If }$s\equiv\left(  d_{1}+d_{2}+d_{3}\right)
/2$\textit{\ is the semi-perimeter of a triangle, then its area is}
\[
\mathrm{area}=\sqrt{s\left(  s-d_{1}\right)  \left(  s-d_{2}\right)  \left(
s-d_{3}\right)  }.
\]

\end{description}

In this paper we will recast all three in simpler and more general forms. As a
reward, we find that \textit{rational trigonometry} falls into our laps,
essentially for free. Our reformulation works over a general field (not of
characteristic two), in arbitrary dimensions, and even with an arbitrary
quadratic form---see \cite{Wild1}, \cite{Wild2} and \cite{Wild3}.

\section*{Pythagoras' theorem}

Euclid and other ancient Greeks regarded \textit{area}, not \textit{distance},
as the fundamental quantity in planar geometry. Indeed they worked with a
straightedge and compass in their constructions, not a ruler and protractor. A
line segment was measured by constructing a square on it, and determining the
area of that square. Two line segments were considered equal if they were
\textit{congruent}, but this was independent of a direct notion of distance measurement.

Area is an \textit{affine concept}: more precisely \textit{proportions}
between areas are maintained by linear transformations. Even with a different
metrical geometry, for example a relativistic geometry in which $x^{2}-y^{2}$
plays the role of $x^{2}+y^{2}$, the notion of signed area defined by a
determinant applies. In other words, in planar geometry area is a basic notion
and may be considered prior to any theory of linear measurement.

To Euclid, Pythagoras' theorem is a relation about the areas of squares built
on each of the sides of a right triangle. This insight has largely been lost
in the modern formulation, but with a sheet of graph paper it is still an
attractive way to introduce students to the subject, as the area of many
simple figures can be computed by subdividing, translating and counting cells.

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=6.1734cm,
width=7.1413cm
]
{Pyth4,2BW.eps}
\caption{Pythagoras' theorem using areas}
\label{Pythagoras' theorem}
\end{center}
\end{figure}

The squares on the sides of triangle $\overline{A_{1}A_{2}A_{3}}$ shown in
Figure \ref{Pythagoras' theorem} have areas $5,20$ and $25.$ The largest
square for example can be seen as four triangles which can be rearranged to
get two $3\times4$ rectangles, together with a $1\times1$ square, for a total
area of $25.$ So from this point of view Pythagoras' theorem is a result which
can be established by \textit{counting}, and the use of irrational numbers to
describe lengths is not necessary. This applies to any right triangle with
rational coordinates.

Following the Greek terminology of `quadrature', we define the
\textbf{quadrance} $Q$ of a line segment to be the area of the square
constructed on it. Pythagoras' theorem allows us to assert that if
$A_{1}\equiv\left[  x_{1},y_{1}\right]  $ and $A_{2}\equiv\left[  x_{2}
,y_{2}\right]  ,$ then the quadrance between $A_{1}$ and $A_{2}$ is
\[
Q\left(  A_{1},A_{2}\right)  =\left(  x_{2}-x_{1}\right)  ^{2}+\left(
y_{2}-y_{1}\right)  ^{2}.
\]
So for example the quadrance between the points $\left[  0,0\right]  $ and
$\left[  1,2\right]  $ is $Q=5.$ The usual distance between the points is the
`square root' of the quadrance and requires a prior theory of irrational
numbers. Clearly the irrational number $\sqrt{5}\approx\allowbreak
2.\,\allowbreak236\,067\,977\ldots\allowbreak$ is a far more sophisticated and
complicated object than the natural number $5.$

In statistics, variance is more natural than standard deviation. In quantum
mechanics, wave functions are more basic than probability amplitudes. In
harmonic analysis, $L^{2}$ is more pleasant than $L^{1}.$ In geometry,
\textit{quadrance is more fundamental than distance}.

For a triangle $\overline{A_{1}A_{2}A_{3}}$ we define the quadrances
$Q_{1}=Q\left(  A_{2},A_{3}\right)  $, $Q_{2}=Q\left(  A_{1},A_{3}\right)  $
and $Q_{3}=Q\left(  A_{1},A_{2}\right)  $. Here then is\textit{\ Pythagoras'
theorem as the Greeks viewed it}---and it extends to arbitrary fields, to many
dimensions, and even with general quadratic forms.

\begin{theorem}
[Pythagoras]The lines $A_{1}A_{3}$ and $A_{2}A_{3}$ of the triangle
$\overline{A_{1}A_{2}A_{3}}$ are perpendicular precisely when
\[
Q_{1}+Q_{2}=Q_{3}.
\]

\end{theorem}

\section*{Area of a triangle}

\textit{The area of a triangle is one-half the base times the height}. Let's
see how we might remove some of the irrationalities that occur with this
ancient formula by looking at an example.

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=1.5338in,
width=1.9433in
]
{Euclid2.eps}
\caption{A triangle and an associated parallelogram}
\label{Euclid1}
\end{center}
\end{figure}

The area of the triangle $\overline{A_{1}A_{2}A_{3}}$ in Figure \ref{Euclid1}
is one half of the area of the associated parallelogram $\overline{A_{1}
A_{2}A_{3}A_{4}}.$

The latter area may be calculated by removing from the circumscribed
$12\times8$ rectangle four triangles, which can be combined to form two
rectangles, one $5\times3$ and the other $7\times5.$ The area of
$\overline{A_{1}A_{2}A_{3}}$ is thus $23.$

To apply the one-half base times height rule, the base $\overline{A_{1}A_{2}}
$ by Pythagoras has length
\[
d_{3}=\left\vert A_{1},A_{2}\right\vert =\sqrt{7^{2}+5^{2}}=\sqrt{74}
\approx8.\,\allowbreak602\,325\,267\,\allowbreak04\ldots.
\]
To find the length $h$ of the altitude $\overline{A_{3}F},$ set the origin to
be at $A_{1},$ then the line $A_{1}A_{2}$ has Cartesian equation $5x-7y=0$
while $A_{3}=\left[  2,8\right]  .$ A well-known result from coordinate
geometry then states that the distance $h=\left\vert A_{3},F\right\vert $ from
$A_{3}$ to the line $A_{1}A_{2}$ is
\[
h=\frac{\left\vert 5\times2-7\times8\right\vert }{\sqrt{5^{2}+7^{2}}}
=\frac{46}{\sqrt{74}}\allowbreak\approx5.\,\allowbreak
347\,391\,382\,\allowbreak22\ldots.
\]
If an engineer doing this calculation works with the surd forms of both
expressions, she will notice that the two occurrences of $\sqrt{74}$
conveniently cancel when she takes one half the product of $d_{3}$ and $h$,
giving an area of $23.$ However if she works immediately with the decimal
forms, she may be surprised that her calculator gives
\[
\mathrm{area}\approx\frac{8.\,\allowbreak602\,325\,267\,\allowbreak
04\ldots\times~5.\,\allowbreak347\,391\,382\,\allowbreak22\ldots}{2}
\approx23.000\,000\,000\,\allowbreak01.
\]
The usual formula forces us to descend to the level of irrational numbers and
square roots, even when the eventual answer is a natural number, and this
introduces unnecessary approximations and inaccuracies into the subject. It is
not hard to see how the use of quadrance allows us to reformulate the result.

\begin{theorem}
[Triangle area]The square of the area of a triangle is one-quarter the
quadrance of the base times the quadrance of the corresponding altitude.
\end{theorem}

As a formula, this would be
\[
\mathrm{area}^{2}=\frac{Q\times H}{4}
\]
where $Q$ is the quadrance of the base and $H$ is the quadrance of the
altitude to that base.

\section*{Heron's or Archimedes' Theorem}

The same triangle $\overline{A_{1}A_{2}A_{3}}$ of the previous section has
side lengths
\[
\begin{tabular}
[c]{lllll}
$d_{1}=\sqrt{34}$ &  & $d_{2}=\sqrt{68}$ &  & $d_{3}=\sqrt{74}.$
\end{tabular}
\]
The semi-perimeter $s,$ defined to be one half of the sum of the side lengths,
is then
\[
s=\frac{\sqrt{34}+\sqrt{68}+\sqrt{74}}{2}\allowbreak\approx11.\,\allowbreak
339\,744\,206\,\allowbreak6\ldots.
\]

Using the usual Heron's formula, a computation with the calculator shows that
\[
\mathrm{area}=\sqrt{s\left(  s-\sqrt{34}\right)  \left(  s-\sqrt{68}\right)
\left(  s-\sqrt{74}\right)  }\approx23.000\,000.
\]
Again we have a formula involving square roots in which there appears to be a
surprising integral outcome. Let's now give another form of Heron's formula,
with a new name. Arab sources suggest that Archimedes knew Heron's formula
earlier, and the greatest mathematician of all time deserves credit for more
than he currently gets.

\begin{theorem}
[Archimedes]The area of a triangle $\overline{A_{1}A_{2}A_{3}}$ with
quadrances $Q_{1},Q_{2}$ and $Q_{3}$ is given by
\[
16~\mathrm{area}^{2}=\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}-2\left(  Q_{1}
^{2}+Q_{2}^{2}+Q_{3}^{2}\right)  .
\]

\end{theorem}

In our example the triangle has quadrances $34,68$ and $74,$ each obtained by
Pythagoras' theorem. So Archimedes' theorem states that
\[
16~\mathrm{area}^{2}=\left(  34+68+74\right)  ^{2}-2\left(  34^{2}
+68^{2}+74^{2}\right)  =8464
\]
and this gives an area of $23.$ In rational trigonometry, the quantity
\[
\mathcal{A}=\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}-2\left(  Q_{1}^{2}
+Q_{2}^{2}+Q_{3}^{2}\right)
\]
is the \textbf{quadrea} of the triangle, and turns out to be the single most
important number associated to a triangle. Note that
\begin{align*}
\mathcal{A} &  =4Q_{1}Q_{2}-\left(  Q_{1}+Q_{2}-Q_{3}\right)  ^{2}\\
&  =-
\begin{vmatrix}
0 & Q_{1} & Q_{2} & 1\\
Q_{1} & 0 & Q_{3} & 1\\
Q_{2} & Q_{3} & 0 & 1\\
1 & 1 & 1 & 0
\end{vmatrix}
.
\end{align*}

It is instructive to see how to go from Heron's formula to Archimedes'
theorem. In terms of the side lengths $d_{1},d_{2}$ and $d_{3}$:
\begin{align*}
16~\mathrm{area}^{2}  & =\left(  d_{1}+d_{2}+d_{3}\right)  \left(  d_{1}
+d_{2}-d_{3}\right)  \left(  -d_{1}+d_{2}+d_{3}\right)  \left(  d_{1}
-d_{2}+d_{3}\right) \\
& =\left(  \left(  d_{1}+d_{2}\right)  ^{2}-d_{3}^{2}\right)  \left(
d_{3}^{2}-\left(  d_{1}-d_{2}\right)  ^{2}\right) \\
& =\left(  2d_{1}d_{2}+\left(  d_{1}^{2}+d_{2}^{2}-d_{3}^{2}\right)  \right)
\left(  2d_{1}d_{2}-\left(  d_{1}^{2}+d_{2}^{2}-d_{3}^{2}\right)  \right) \\
& =4d_{1}^{2}d_{2}^{2}-\left(  d_{1}^{2}+d_{2}^{2}-d_{3}^{2}\right)  ^{2}\\
& =4Q_{1}Q_{2}-\left(  Q_{1}+Q_{2}-Q_{3}\right)  ^{2}\\
& =\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}-2\left(  Q_{1}^{2}+Q_{2}^{2}
+Q_{3}^{2}\right)  .
\end{align*}

Archimedes' theorem implies another formula of considerable importance.

\begin{theorem}
[Triple quad formula]The three points $A_{1},A_{2}$ and $A_{3}$ are collinear
precisely when
\[
\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}=2\left(  Q_{1}^{2}+Q_{2}^{2}+Q_{3}
^{2}\right)  .
\]

\end{theorem}

The proof is of course immediate, as collinearity is equivalent to the area of
the triangle being zero.

\section*{Spread between lines}

An \textit{angle }is the ratio of a \textit{circular distance} to a
\textit{linear distance,} and this is a\textit{\ }complicated concept. To
define an angle properly \textit{you require calculus}, an important point
essentially understood by Archimedes. Vagueness about angles, and the
accompanying ambiguities in the definition of the circular functions
$\cos\theta,$ $\sin\theta$ and $\tan\theta$ weaken most calculus texts.

There is a reason that classical trigonometry is painful to students---it is
\textit{based on the wrong notions}. As a result, mathematics teachers are
forced to continually rely on $90-45-45$ and $90-60-30$ triangles for examples
and test questions, which makes the subject very narrow and repetitive.

Rational trigonometry, developed in \cite{Wild1}, see also \cite{Wild3}, shows
how to simplify and enrich the subject, leading to greater accuracy and
quicker computations. We want to show that the basic results of this new
theory \textit{follow naturally from the above presentation of Pythagoras'
theorem, the Triangle area theorem, and Archimedes' theorem}.

The key innovation is to replace angle with a completely algebraic concept.
The separation between lines $l_{1}$ and $l_{2}$ is captured rather by the
notion of \textit{spread}, which may be defined as the ratio of two quadrances
as follows.

Suppose $l_{1}$ and $l_{2}$ intersect at the point $A.$ Choose a point $B\neq
A$ on one of the lines, say $l_{1},$ and let $C$ be the foot of the
perpendicular from $B$ to $l_{2}$.

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=1.3379in,
width=1.5244in
]
{SpreadDef.eps}
\caption{Spread $s$ between two lines $l_{1}$ and $l_{2}$}
\label{SpreadRatio}
\end{center}
\end{figure}

Then the \textbf{spread} $s$ between $l_{1}$ and $l_{2}$ is
\[
s=s\left(  l_{1},l_{2}\right)  =\frac{Q\left(  B,C\right)  }{Q\left(
A,C\right)  }=\frac{Q}{R}.
\]
This ratio is independent of the choice of $B,$ according to Thales, and is
defined \textit{between lines, not rays}. Parallel lines are defined to have
spread $s=0,$ while perpendicular lines have spread $s=1.$

The spread between the lines $l_{1}$ and $l_{2}$ with equations $a_{1}
x+b_{1}y+c_{1}=0$ and $a_{2}x+b_{2}y+c_{2}=0$ turns out to be
\begin{equation}
s\left(  l_{1},l_{2}\right)  =\frac{\left(  a_{1}b_{2}-a_{2}b_{1}\right)
^{2}}{\left(  a_{1}^{2}+b_{1}^{2}\right)  \left(  a_{2}^{2}+b_{2}^{2}\right)
}.\label{SpreadFormula}
\end{equation}
Since this is a rational expression, spread becomes a useful concept also over
general fields, although in this paper we stick to the usual situation over
the decimal numbers.

You may check that the spread corresponding to $30^{\circ}$ or $150^{\circ} $
or $210^{\circ}$ or $330^{\circ}$ is $s=1/4,$ the spread corresponding to
$45^{\circ}$ or $135^{\circ}$ etc. is $s=1/2$, and the spread corresponding to
$60^{\circ}$ or $120^{\circ}$ etc. is $3/4.$

The following \textit{spread protractor} was created by M. Ossmann
\cite{Ossmann}.

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=1.6071in,
width=2.8755in
]
{SpreadProtractor.eps}
\caption{A spread protractor}
\label{SpreadProt}
\end{center}
\end{figure}

We use the notation that a triangle $\overline{A_{1}A_{2}A_{3}}$ has
quadrances $Q_{1},Q_{2}$ and $Q_{3},$ as well as spreads $s_{1},s_{2}$ and
$s_{3},$ labelled as in Figure \ref{SpreadsQuads}. Note the diagrammatic
conventions that help us distinguish these quantities from distance and angle.

\begin{figure}
[h]
\begin{center}
\includegraphics[
height=1.1836in,
width=1.6202in
]
{Fig1.4.eps}
\caption{Quadrances and spreads of a triangle}
\label{SpreadsQuads}
\end{center}
\end{figure}

\section*{Rational trigonometry}

Let's see how to combine the three ancient Greek theorems as restated above to
derive the main laws of rational trigonometry, independent of classical
trigonometry, and without any need for transcendental functions. If $H_{3}$ is
the quadrance of the altitude from $A_{3}$ to the line $A_{1}A_{2},$ then the
Triangle area theorem and the definition of spread give
\[
\mathrm{area}^{2}=\frac{Q_{3}\times H_{3}}{4}=\frac{Q_{3}Q_{2}s_{1}}{4}
=\frac{Q_{3}Q_{1}s_{2}}{4}.
\]
By symmetry, we get the following analog of the Sine law.

\begin{theorem}
[Spread law]For a triangle with quadrances $Q_{1},Q_{2}$ and $Q_{3},$ and
spreads $s_{1},s_{2}$ and $s_{3},$
\[
\frac{s_{1}}{Q_{1}}=\frac{s_{2}}{Q_{2}}=\frac{s_{3}}{Q_{3}}=\frac
{4~\mathrm{area}^{2}}{Q_{1}Q_{2}Q_{3}}.
\]

\end{theorem}

By equating the formulas for $16~\mathrm{area}^{2}$ obtained from the Triangle
area theorem and Archimedes' theorem, we get
\begin{align*}
4Q_{2}Q_{3}s_{1}  & =\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}-2\left(  Q_{1}
^{2}+Q_{2}^{2}+Q_{3}^{2}\right) \\
& =2Q_{1}Q_{2}+2Q_{1}Q_{3}+2Q_{2}Q_{3}-Q_{1}^{2}-Q_{2}^{2}-Q_{3}^{2}.
\end{align*}
Rearranging gives the following analog of the Cosine law.

\begin{theorem}
[Cross law]For a triangle with quadrances $Q_{1},Q_{2}$ and $Q_{3},$ and
spreads $s_{1},s_{2}$ and $s_{3},$
\[
\left(  Q_{1}-Q_{2}-Q_{3}\right)  ^{2}=4Q_{2}Q_{3}\left(  1-s_{1}\right)  .
\]

\end{theorem}

Now substitute $Q_{1}=s_{1}D,$ $Q_{2}=s_{2}D$ and $Q_{3}=s_{3}D,$ where
$D=Q_{1}Q_{2}Q_{3}/4~\mathrm{area}^{2}$ from the Spread law into the Cross
law, and cancel the common factor of $D^{2}$. The result is the relation
\[
\left(  s_{1}-s_{2}-s_{3}\right)  ^{2}=4s_{2}s_{3}\left(  1-s_{1}\right)
\]
between the three spreads of a triangle, which can be rewritten more
symmetrically as follows.

\begin{theorem}
[Triple spread formula]
\[
\left(  s_{1}+s_{2}+s_{3}\right)  ^{2}=2\left(  s_{1}^{2}+s_{2}^{2}+s_{3}
^{2}\right)  +4s_{1}s_{2}s_{3}.
\]

\end{theorem}

This formula is a deformation of the Triple quad formula by a single cubic
term, and is the analog in rational trigonometry to the classical fact that
the three angles of a triangle sum to $\allowbreak3.\,\allowbreak
141\,592\,653\,\allowbreak59\ldots.$

The \textit{Triple quad formula}, \textit{Pythagoras' theorem}, the
\textit{Spread law}, the \textit{Cross law} and the \textit{Triple spread
formula} are the five main laws of rational trigonometry. We now see these are
closely linked to the geometrical work of the ancient Greeks.

As demonstrated at some length in \cite{Wild1}, these formulas and a few
additional ones suffice to solve the majority of trigonometric problems,
usually more simply, more accurately and more elegantly than the classical
theory involving transcendental circular functions and their inverses. As
shown in \cite{Wild2} and \cite{Wild4}, the same formulas extend to geometry
over arbitrary fields (not of characteristic two) and with general quadratic forms.

In retrospect, the blind spot first occurred with the Pythagoreans, who
initially believed that all of nature should be expressible in terms of
natural numbers and their proportions. When they discovered that the ratio of
the length of a diagonal to the length of a side of a square was the
incommensurable proportion $\sqrt{2}:1,$ legend has it that they tossed the
exposer of the secret overboard while at sea.

Had they maintained their beliefs in the workings of the Divine Mind, and
stuck with the \textit{squares of the lengths as the crucial quantities in
geometry,} then mathematics would have had a significantly different history,
Einstein's special theory of relativity would possibly have been discovered
earlier, algebraic geometry would have quite another aspect, and students
would today be studying a simpler and more elegant trigonometry---much more happily!


TITLE: functions composition and surjective functions
QUESTION [0 upvotes]: Given  $$(f◦g)(x)=x$$ (from $\Bbb R$ to $\Bbb R$ for any $x$ in $\Bbb R$)
And $g(x)$ is onto!
does it mean that also $$(g◦f)(x)=x$$
It seems like it does but how can I prove it?

REPLY [3 votes]: It seems the following.
Since the map $g$ is onto, $x=g(y)$ for some $y\in R$. Then $$(g\circ f)(x)= (g\circ f)(g(y))=g\circ f\circ g(y)=g(y)=x.$$


text:
\begin{document}

\title*{On Hodge polynomials of Singular Character Varieties}

\author{Carlos Florentino\inst{1}\and
Azizeh Nozad\inst{2}\and Jaime Silva\inst{3}\and Alfonso Zamora\inst{4}}

\institute{Departamento de Matem\'{a}tica, Faculdade de Ci\^encias, Univ. de Lisboa, Campo Grande, Edf. C6, Lisbon, Portugal
\texttt{caflorentino@ciencias.ulisboa.pt}
\and School of Mathematics, Institute for Research in Fundamental Sciences (IPM), P.O.Box: 19395-5746, Tehran, Iran
\texttt{anozad@ipm.ir}
\and Centro de Matem\'atica da Universidade do Porto, Faculdade de Ci\^encias da Universidade do Porto, Rua do Campo Alegre, 4169-007 Porto, Portugal\texttt{jaime.silva@fc.up.pt}
\and Departamento de Matem\'atica Aplicada a las TIC, ETSI Inform\'aticos, Universidad Polit\'ecnica de Madrid, Campus de Montegancedo,
28660, Madrid, Spain
\texttt{alfonso.zamora@upm.es}}

\maketitle

\subsection*{Abstract}
Let $\mathcal{X}_{\Gamma}G:=\hom(\Gamma,G)\quot G$ be the $G$-character
variety of $\Gamma$, where $G$ is a complex reductive group and
$\Gamma$ a finitely presented group. We introduce new techniques
for computing Hodge-Deligne and Serre polynomials of $\mathcal{X}_{\Gamma}G$,
and present some applications, focusing on the cases when $\Gamma$
is a free or free abelian group. Detailed constructions and proofs
of the main results will appear elsewhere.

\section{Introduction}

Let $G$ be a connected reductive complex algebraic group, and $\Gamma$
be a finitely presented group. The $G$-character variety of $\Gamma$
is defined to be the (affine) geometric invariant theory (GIT) quotient
\[
\mathcal{X}_{\Gamma}G=\hom(\Gamma,G)\quot G.
\]
The most well studied families of character varieties include the
cases when the group $\Gamma$ is the fundamental group of a Riemann
surface $\Sigma$, and its ``twisted'' variants. In these cases,
the non-abelian Hodge correspondence (see, for example \cite{Si})
shows that (components of) $\mathcal{X}_{\Gamma}G$ are homeomorphic
to certain moduli spaces of $G$-Higgs bundles which appear in connection
to important problems in Mathematical-Physics: for example, these
spaces play an important role in the quantum field theory interpretation
of the geometric Langlands correspondence, in the context of mirror
symmetry (\cite{KW}).

The study of geometric and topological properties of character varieties
is an active topic and there are many recent advances in the computation
of their Poincar\'e polynomials and other invariants. For the surface
group case ($\Gamma=\pi_{1}(\Sigma)$ and related groups) the calculations
of Poincar\'e polynomials started with Hitchin and Gothen, and have
been pursued more recently by Hausel, Lettelier, Mellit, Rodriguez-Villegas,
Schiffmann and others, who also considered the parabolic version of
these character varieties (see \cite{HRV,Me,Sc}). Those recent results
use arithmetic methods: it is shown that the number of points of the
corresponding moduli space over finite fields is given by a polynomial,
which turns out to coincide with the $E$-polynomial of $\mathcal{X}_{\Gamma}G$
(\cite[Appendix]{HRV}). Then, in the smooth case, the pure nature
of the cohomology of Higgs bundles moduli spaces allows the derivation
of the Poincar\'e polyomial from the $E$-polynomial.

On the other hand, for many important classes of \emph{singular} character
varieties, explicitly computable formulas for the $E$-polynomials
(also called Serre, or Hodge-Euler polynomials) are very hard to obtain. In
the articles of Logares, Mu\~noz, Newstead and Lawton \cite{LMN}, \cite{LM}
(using geometric methods) and of Baraglia and Hekmati \cite{BH} (using
arithmetic methods), the $E$-polynomials are computed for several
character varieties, with $G=GL(n,\mathbb{C})$, $SL(n,\mathbb{C})$
and $PGL(n,\mathbb{C})$ for small values of $n$, but the computations
quickly become intractable for $n$ higher than $3$.

In this short article, we describe some of the techniques and constructions
that we have recently developed for computations of $E$-polynomials
of singular character varieties, and present some of their main applications.

The outline of the article is as follows. Section 2 covers notations
and preliminaries on mixed Hodge and $E$-polynomials and on character
varieties in the context of GIT. In section 3, we explain how to use
equivariant mixed Hodge structures to study (the identity component
of) $\mathcal{X}_{\Gamma}G$ when $\Gamma$ is a free abelian group
and $G$ a classical group. These character vareities have orbifold
singularities and we can obtain their full mixed Hodge polynomials.
In section 4, for arbitrary $\Gamma$, we define a stratification
of $GL(n,\mathbb{C})$-character varieties (which also exists for
$G=SL(n,\mathbb{C})$ or $PGL(n,\mathbb{C})$) which allows writing
down an explicit plethystic exponential relation between generating
functions of the $E$-polynomials of $\mathcal{X}_{\Gamma}GL(n,\mathbb{C})$
and of its locus of irreducible representations $\mathcal{X}_{\Gamma}^{irr}GL(n,\mathbb{C})$.
Finally, in Section 5, we consider the free group $\Gamma=F_{r}$
of rank $r$, and announce the solution of a conjecture of Lawton
and Mu\~noz: the $E$-polynomials of $\mathcal{X}_{F_{r}}SL(n,\mathbb{C})$
and of $\mathcal{X}_{F_{r}}PGL(n,\mathbb{C})$ coincide, for every
$n\in\mathbb{N}$. For lack of space, the proofs are omitted and will
be published elsewhere.

\subsection*{Acknowledgements }

We would like to thank S. Lawton, M. Logares, S. Mozgovoy, V. Mu\~noz,
A. Oliveira and F. Rodriguez-Villegas for several interesting and
very useful conversations on topics around mixed Hodge structures
and $E$-polynomials. We also thank Alexander Schmitt for the invitation
to deliver a talk at the Special Session on Complex Geometry of the
conference ISAAC 2019, where these results were presented.

This work was partially supported by CAMGSD and CMAFcIO of the University of Lisbon, the projects PTDC/MAT-GEO/2823/2014 and PTDC/MAT-PUR/30234/2017, FCT Portugal, a grant from IPM (Iran) and ICTP (Italy), and the project MTM2016-79400-P by the Spanish government.

\section{Preliminaries on Hodge-Deligne polynomials, Affine GIT and Character
Varieties}

In this article, all algebraic varieties are defined over $\mathbb{C}$,
$G$ is a connected reductive algebraic group, and $\Gamma$ is a finitely presented group.

Let $X$ be a quasi-projective variety (not necessarily irreducible),
of complex dimension $\leq d$.
Deligne showed that the compactly supported
cohomology $H_{c}^{*}(X):=H_{c}^{*}(X,\mathbb{C})$ can be endowed
with a mixed Hodge structure whose mixed Hodge numbers are given by
\[
h^{k,p,q}(X):=\dim_{\mathbb{C}}H_{c}^{k,p,q}(X)\in\mathbb{N}_{0},
\]
for $k,p,q\in\{0,\cdots,2d\}$, and we call $(p,q)$ the $k$-weights
of $X$, if $h^{k,p,q}\neq0$ (c.f. \cite{De}, \cite{PS}).

Mixed Hodge numbers are symmetric in the weights, $h^{k,p,q}=h^{k,q,p}$,
and $\dim_{\mathbb{C}}H_{c}^{k}(X)=\sum_{p,q}h^{k,p,q}$. Therefore,
they provide the (compactly supported) Betti numbers, yielding the
usual Betti numbers, by Poincar\'e duality, in the non-singular case. They
are also the coefficients of the mixed Hodge polynomial of $X$ on
three variables,
\begin{equation}
\mu(X;\,t,u,v):=\sum_{k,p,q}h^{k,p,q}(X)\ t^{k}u^{p}v^{q}\in\mathbb{\mathbb{N}}_{0}[t,u,v],\label{eq:mu}
\end{equation}
which specializes to the (compactly supported) Poincar\'e polynomial
by setting $u=v=1$, $P_{t}^{c}(X):=\mu(X;\,t,1,1)$ (and provides
the usual Poincar\'e polynomial in the smooth situation). Plugging $t=-1$,
mixed Hodge polynomials convert into the $E$-polynomial of $X$,
or the \emph{Serre polynomial} of $X$, given by
\[
E(X;\,u,v)=\sum_{k,p,q}(-1)^{k}h^{k,p,q}(X)\ u^{p}v^{q}\in\mathbb{Z}[u,v].
\]
From the $E$-polynomial we can compute the (compactly supported)
Euler characteristic of $X$ as $\chi^{c}(X)=E(X;\,1,1)=\mu(X;\,-1,1,1)$.

Serre polynomials satisfy an \emph{additive property with respect
to stratifications} by locally closed (in the Zariski topology) strata:
if $X$ has a closed subvariety $Z\subset X$ we have (see, eg. \cite{PS}),
\[
E(X)=E(Z)+E(X\setminus Z).
\]
The $E$-polynomial also satisfies (c.f. \cite{DL,LMN}) a \emph{multiplicative
property for fibrations. }Namely, for a given algebraic fibration
$F\hookrightarrow X\to B$, we have
\[
E(X)=E(F)\cdot E(B)
\]
in any of the following three situations:
\begin{itemize}
\item[(i)]  the fibration is locally trivial in the Zariski topology of $B$,
\item[(ii)]  $\,$ $F$, $X$ and $B$ are smooth, the fibration is locally trivial
in the complex analytic topology, and $\pi_{1}(B)$ acts trivially
on $H_{c}^{*}(F)$, or
\item[(iii)] $\, \,$ $X$, $B$ are smooth and $F$ is a complex connected Lie group.
\end{itemize}
We say $X$ is of \emph{Hodge-Tate type} (also called \emph{balanced}
type) if all the $k$-weights are of the form $(p,p)$ with $p\in\{0,\cdots,k\}$,
in which case the sum in $\mu(X)$ reduces to a one-variable sum.
In particular, the $E$-polynomials of Hodge-Tate type varieties depend
only the product $uv$, so we write $x=uv$ and use the notation $E(X;\,x):=E(X;\,\sqrt{x},\sqrt{x})\in\mathbb{Z}[x].$

Now let $X$ be an affine algebraic variety, and
let the reductive group $G$ act algebraically on $X$. The induced action of $G$ on
the ring $\mathbb{C}[X]$ of regular functions on $X$ defines the
(affine) GIT quotient
\[
X\quot G:=Spec\left(\mathbb{C}[X]^{G}\right),
\]
where $\mathbb{C}[X]^{G}$ is the subring of $G$-invariants in $\mathbb{C}[X]$.
This quotient identifies $G$-orbits whose closures intersect, such
that each point in the quotient classifies an equivalence class of
orbits, leading to a stability condition. Let $G_{x}\subset G$ be
the stabilizer of $x\in X$ and consider the orbit map through $x$,
$\psi_{x}:G\to X;\;g\mapsto g\cdot x$. We define $x\in X$ to be
\emph{stable} if $\psi_{x}$ is a proper map and \emph{polystable}
if the orbit $G\cdot x$ is closed in $X$. Stability implies polystability,
but not conversely.

GIT shows that the stable locus $X^{s}\subset X$ is a Zariski open
set (hence dense, when non-empty) and that the restriction of the
affine quotient map $\Phi:X\to X\quot G$ to the stable locus, $X^{s}\to X^{s}/G$,
is a geometric quotient (or an orbit space), where $\Phi(X^{s})$
is Zariski open in $X\quot G$.

Now, consider a finitely presented group $\Gamma$. The (generally
singular) algebraic variety of representations of $\Gamma$ in $G$
is
\[
\mathcal{R}_{\Gamma}G=\hom(\Gamma,G).
\]
Each $\rho\in\mathcal{R}_{\Gamma}G$ is determined by $\rho(\gamma)$,
for each generator $\gamma\in\Gamma$, and satisfying the relations
of the group $\Gamma$. There is an algebraic action of $G$ on the
variety $\mathcal{R}_{\Gamma}G$ by conjugation of representations,
$g^{-1}\rho g$, yielding the $G$-character variety of $\Gamma$,
\[
\mathcal{X}_{\Gamma}G:=\hom(\Gamma,G)\quot G,
\]
as the GIT quotient.

By definition, polystable representations are representations $\rho:\Gamma\to G$
whose orbits $G\cdot\rho:=\{g\rho g^{-1}:\ g\in G\}$ are Zariski
closed in $\mathcal{R}_{\Gamma}G$. Alternatively, a representation
$\rho$ is polystable if and only if it is completely reducible (i.e,
if $\rho(\Gamma)\subset P\subset G$ for some proper parabolic $P$
of $G$, then $\rho(\Gamma)$ is contained in a Levi subgroup of $P$).
Denote the subset of polystable representations in
$\mathcal{R}_{\Gamma}G$ by $\mathcal{R}_{\Gamma}^{ps}G\subset\mathcal{R}_{\Gamma}G$,
which is a Zariski locally-closed subvariety containing the stable
locus $\mathcal{R}_{\Gamma}^{s}G\subset\mathcal{R}_{\Gamma}G$.
\begin{proposition}

\cite{FL1}\label{prop:character-polystable representation} There
is a bijective correspondence:
\[
\mathcal{X}_{\Gamma}G=\mathcal{R}_{\Gamma}G\quot G\cong\mathcal{R}_{\Gamma}^{ps}G/G,
\]
where the right hand side is called the polystable quotient.
\end{proposition}

We say that $\rho$ is \emph{irreducible} if $\rho(\Gamma)$
is not contained in a proper parabolic subgroup of $G$.
Alternatively, $\rho$ is irreducible  if it is polystable and
$Z_{\rho}$, the centralizer of $\rho(\Gamma)$ inside $G$, is a finite
extension of the center $ZG\subset G$.
Denote by $\mathcal{R}_{\Gamma}^{irr}G\subset\mathcal{R}_{\Gamma}^{ps}G$
the subset of irreducible representations (being a Zariski open subset
of $\mathcal{R}_{\Gamma}G$, $\mathcal{R}_{\Gamma}^{irr}G$ is a quasi-projective
variety), and since irreducibility is well defined on $G$-orbits,
denote by
\begin{equation}
\mathcal{X}_{\Gamma}^{irr}G:=\mathcal{R}_{\Gamma}^{irr}G/G\label{eq:X-irr}
\end{equation}
the\emph{ $G$-irreducible character variety of $\Gamma$}, which
is a geometric quotient, as it happens with the stable locus. In fact,
it can be proved that irreducibility is equivalent to GIT stability
for character varieties (see \cite[Thm. 1.3(1)]{CF}).

\section{The Free Abelian Case}

In this section, we are concerned with the determination of the mixed
Hodge polynomials of character varieties $\mathcal{X}_{\Gamma}G$
of the free abelian group of rank $r$, $\Gamma\cong\mathbb{Z}^{r}$.
As we always work over $\mathbb{C}$, we abbreviate the notation of
the classical groups such as the linear group, special linear, special
orthogonal and symplectic to $GL_{n}$, $SL_{n}$, $SO_{n}$ and $Sp_{n}$,
respectively (instead of $GL(n,\mathbb{C})$, etc).

The topology and geometry of the character varieties $\mathcal{X}_{\mathbb{Z}^{r}}G$
was studied in \cite{FL2,Sk}, among others. Most important for
us are the following facts:
\begin{itemize}
\item[(i)]  there is only one irreducible component containing the trivial representation,
that we denote by $\mathcal{X}_{\mathbb{Z}^{r}}^{0}G$ \cite[Theorem 2.1]{Sk},
\item[(ii)] $\,$ if the semisimple part of $G$ is a classical group (ie, one of
$SL_{n}$, $SO_{n}$ and $Sp_{n}$), there exists an algebraic isomorphism
\begin{eqnarray}
\mathcal{X}_{\mathbb{Z}^{r}}^{0}G & \cong & \left(T_{G}\right)^{r}/W_{G}\label{eq:algis}
\end{eqnarray}
where $T_{G}$ is a maximal torus of $G$, and $W_{G}$ its Weyl group
\cite[Theorem 2.1]{Sk},
\item[(iii)] $\, \,$ the irreducibility of the free abelian character varieties $\mathcal{X}_{\mathbb{Z}^{r}}G$
can be characterized, in terms of $G$: for example, if the semisimple
part of $G$ is a product of $SL_{n}$'s and $Sp_{n}$'s then $\mathcal{X}_{\mathbb{Z}^{r}}G$
is irreducible, so that $\mathcal{X}_{\mathbb{Z}^{r}}G\cong\mathcal{X}_{\mathbb{Z}^{r}}^{0}G$
\cite[Theorem 1.2]{FL2}.
\end{itemize}
We now focus on the determination of the mixed Hodge numbers of $\mathcal{X}_{\mathbb{Z}^{r}}G$
when it is irreducible, or of $\mathcal{X}_{\mathbb{Z}^{r}}^{0}G$
when the algebraic isomorphism \eqref{eq:algis} applies. We start
by explaining how mixed Hodge numbers transform under finite quotients.

Let $X$ be a complex quasi-projective variety and $F$ a finite group
acting algebraically on it. The action of $F$ on $X$ induces an
action on its cohomology. Since $F$ acts by algebraic isomorphisms,
it also induces an action on the mixed Hodge components. Then we can
regard $H^{k,p,q}\left(X\right)$ as $F$-modules, that we denote
by $\left[H^{k,p,q}\left(X\right)\right]_{F}$. As in equation \eqref{eq:mu}
for the mixed Hodge polynomial, we codify these in the \emph{equivariant
mixed Hodge polynomial}, defined by
\begin{eqnarray*}
\mu_{F}(X;\,t,u,v) & \coloneqq & \sum_{k,p,q}\left[H^{k,p,q}\left(X\right)\right]
t^{k}u^{p}v^{q}\in R\left(F\right)\left[t,u,v\right]
\end{eqnarray*}
whose coefficients belong to $R(F)$, the representation ring of $F$.
The polynomial $\mu_{F}(X;\,t,u,v)$ may also be seen as a polynomial-weighted
representation. For instance, one can consider equivariant
cohomology to obtain an isomorphism
\begin{eqnarray}
H^{*}\left(X/F\right) & \cong & H^{*}\left(X\right)^{F}\label{eq:quoinv}
\end{eqnarray}
that respects mixed Hodge structures. In particular,
this isomorphism allows us to identify the mixed Hodge polynomial
of the quotient $X/F$ as the coefficient of the trivial representation
of $\mu_{F}(X;\,t,u,v)$ when written on a basis of irreducible representations
of $F$. Another important consequence for us is the inequality
$h^{k,p,q}\left(X\right)\geq h^{k,p,q}\left(X/F\right)$,
which holds since $H^{k,p,q}\left(X/F\right)$ is given by the $F$-invariant
part of $H^{k,p,q}\left(X\right)$. We conclude that if $X$ is, for
instance, a balanced variety, or if its mixed Hodge structure is actually
\emph{pure} (that is, if $h^{k,p,q}\neq0$ then $k=p+q$), then the same
holds for $X/F$.

We now summarize our strategy to obtain the mixed Hodge polynomials
of $\mathcal{X}_{\mathbb{Z}^{r}}^{0}G$, in the cases when the isomorphism
\eqref{eq:algis} holds (so, these character varieties are
isomorphic to finite quotients of algebraic tori).
The only non-zero Hodge numbers of the maximal torus $T_{G}\cong\left(\mathbb{C}^{*}\right)^{n}$
are $h^{k,k,k}\left(T_{G}\right)$. Moreover, its natural mixed Hodge structure satisfies:
\begin{eqnarray*}
H^{k,k,k}\left(T_{G}\right) & \cong & \bigwedge^{k} \, \, H^{1,1,1}\left(T_{G}\right).
\end{eqnarray*}
So, the action of $W_G$ on the cohomology ring can be understood from the
one on the mixed Hodge component $H^{1,1,1}\left(T_{G}\right)$.
The next three theorems are proved
in \cite{FS}.
\begin{theorem}
\label{thm:det-formula} For a reductive group $G$ satisfying \eqref{eq:algis},
we have
\begin{eqnarray*}
\mu(\mathcal{X}_{\mathbb{Z}^{r}}^{0}G;\,t,u,v) & = & \frac{1}{|W_{G}|}\sum_{g\in W_{G}}\left[\det\left(I+tuv\,A_{g}\right)\right]^{r}
\end{eqnarray*}
where $A_{g}$ is the automorphism of $H^{1,1,1}(T_{G})$ induced
by the action of $g\in W_{G}$.
\end{theorem}
The proof starts by establishing the $r=1$ case, and using the diagonal
action for higher $r$ as well as the isomorphism \eqref{eq:algis},
together with the multiplicative relation for the equivariant polynomials
$\mu_{W_{G}}\left(T_{G}^{r}\right)=\mu_{W_{G}}\left(T_{G}\right)^{\otimes r}.$
We remark that Theorem \ref{thm:det-formula} generalizes a formula for the Poincar\'e
polynomial of $\mathcal{X}_{\mathbb{Z}^{r}}^{0}G$, recently obtained
in \cite{St}.

To further work with Theorem \ref{thm:det-formula}, we examine the
induced action of $W_{G}$ on $H^{1,1,1}\left(T_{G}\right)$ for some
classical groups. In the case $G=GL_{n}$, the Weyl group is the symmetric
group $S_{n}$ on $n$ letters, which acts on $H^{1,1,1}\left(T_{G}\right)\cong\mathbb{C}^{n}$
by permutation of coordinates, and we obtain a general formula in
terms of partitions of $n$.

A partition of $n\in\mathbb{N}$ is denoted by $[k]=[1^{k_{1}}\cdots j^{k_{j}}\cdots n^{k_{n}}]$
where the exponent $k_{j}\geq0$ is the number of parts of size $j\in\{1,\cdots,n\}$,
so that $n=\sum_{j=1}^{n}j\cdot k_{j}$. Let $\mathcal{P}_{n}$ denote
the finite set of partitions of $n$.
\begin{theorem}
The mixed Hodge polynomials of $\mathcal{X}_{\mathbb{Z}^{r}}GL_{n}$
and of $\mathcal{X}_{\mathbb{Z}^{r}}SL_{n}$ satisfy
\[
\mu(\mathcal{X}_{\mathbb{Z}^{r}}GL_{n};\,t,u,v)\ =\ \mu(\mathcal{X}_{\mathbb{Z}^{r}}SL_{n};\,t,x)(1+tuv)^{r}=\sum_{[k]\in\mathcal{P}_{n}}\prod_{j=1}^{n}\frac{(1-(-tuv)^{j})^{k_{j}r}}{k_{j}!\,j^{k_{j}}},
\]

\end{theorem}
By using similar considerations as for the $GL_{n}$ case, we can
also deduce a concrete formula for $Sp_{n}$ in terms of \emph{bipartitions}.
A bipartition of $n$, denoted $[a,b]\in\mathcal{B}_{n}$ consists
of two partitions $[a]\in\mathcal{P}_{k}$ and $[b]\in\mathcal{P}_{l}$,
such that $0\leq k,l\leq n$ with $k+l=n$.
One can show that bipartitions of $n$ are in one-to-one correspondence
with conjugacy classes in $W_{Sp_{n}}$, the Weyl group of $Sp_{n}$.
\begin{theorem}
The mixed Hodge polynomial of $\mathcal{X}_{\mathbb{Z}^{r}}Sp_{n}\mathbb{C}$
is given by
\begin{eqnarray*}
\mu(\mathcal{X}_{\mathbb{Z}^{r}}Sp_{n};\,t,u,v) & = & \frac{1}{2^{n}n!}\sum_{[a,b]\in\mathcal{B}_{n}}c_{[a,b]}\prod_{i=1}^{k}(1-(-tuv)^{i})^{a_{i}r}\prod_{j=1}^{l}(1+(-tuv)^{j})^{b_{j}r}
\end{eqnarray*}
where $c_{[a,b]}$ is the size of the conjugacy class in $W_{Sp_{n}}$,
corresponding to $[a,b]\in\mathcal{B}_{n}$.
\end{theorem}
The same method allows to obtain explicit expressions for
$\mu(\mathcal{X}_{\mathbb{Z}^{r}}^{0}G)$
in the case of other reductive $G$; the special orthogonal groups $SO_n$ will be
addressed in a future work.

\section{Generating functions for $E$-polynomials}

In this section we consider character varieties with arbitrarily bad
singularities. In this case, there are formidable difficulties in
computing the corresponding Poincar\'e polynomials in general, and previous
explicit methods have dealt with the $E$-polynomials for low dimensional
groups such as $SL_{2}$ and $SL_{3}$ (\cite{LMN,LM,BH}).

By using the additive and multiplicative properties of $E$-polynomial,
for $G=GL_{n}$ we now address our new approach on $E$-polynomial
computations based on a stratification of $\mathcal{X}_{\Gamma}G$
that we term \emph{by partition type}, and which works for arbitrary
$\Gamma$.

\label{sec:stratification-partition} Using standard arguments in
GIT, any character variety admits a stratification by the dimension
of the stabilizer of a given representation. When $G$ is the general
linear group $GL_{n}$ (as well as the related groups $SL_{n}$ and
$PGL_{n}$), there is a more convenient refined stratification that
gives a lot of information on the corresponding character varieties
$\mathcal{X}_{\Gamma}G$ which we call \emph{stratification by partition
type}.

\begin{definition}
Let $G=GL_{n}$ and $[k]\in\mathcal{P}_{n}$. We say that $\rho\in\mathcal{R}_{\Gamma}G=\hom(\Gamma,G)$
is $[k]$-polystable if $\rho$ is conjugated to $\bigoplus_{j=1}^{n}\rho_{j}$
where each $\rho_{j}$ is, in turn, a direct sum of $k_{j}>0$ \emph{irreducible}
representations of $\mathcal{R}_{\Gamma}(GL_{j})$, for $j=1,\cdots,n$
(by convention, if some $k_{j}=0$, then $\rho_{j}$ is not present
in the direct sum).
\end{definition}

We denote $[k]$-polystable representations by $\mathcal{R}_{\Gamma}^{[k]}G$
and use similar terminology/notation for equivalence classes under
conjugation $\mathcal{X}_{\Gamma}^{[k]}G\subset\mathcal{X}_{\Gamma}G$.
It is to be noted that the trivial partition $[n]=[n^{1}]\in\mathcal{P}_{n}$ corresponds exactly to the irreducible (or stable) locus:
$\mathcal{X}_{\Gamma}^{[n]}G=\mathcal{X}_{\Gamma}^{irr}G$.

\begin{proposition}
\label{prop:locally-closed-GLn}Fix $n\in\mathbb{N}$, and let $G=GL_{n}$.
Then $\mathcal{X}_{\Gamma}G=\bigsqcup_{[k]\in\mathcal{P}_{n}}\mathcal{X}_{\Gamma}^{[k]}G,$
as a disjoint union of locally closed quasi-projective varieties.
\end{proposition}

The next result relates, by the plethystic exponential, the generating
functions of the $E$-polynomials $E(\mathcal{X}_{\Gamma}GL_{n})$
to the corresponding generating functions of the $E$-polynomials
of the irreducible character varieties $E(\mathcal{X}_{\Gamma}^{irr}GL_{n})$.

The \emph{plethystic exponential} of a formal power series $f(x,y,z)=\sum_{n\geq0}f_{n}(x,y)\,z^{n}\in\mathbb{Q}[x,y][[z]]$
is denoted by $\pexp(f)$, and defined formally (in terms of the usual exponential)
as $\pexp(f):=e^{\Psi(f)}\in\mathbb{Q}[x,y][[z]],$ where $\Psi$
acts on monomials as: $\Psi(x^{i}y^{j}z^{k})=\sum_{l\geq1}\frac{x^{li}y^{lj}z^{lk}}{l},$
where $(i,j,k)\in\mathbb{N}_{0}^{3}\setminus\{(0,0,0)\}$, and is $\mathbb{Q}$-linear
on $\mathbb{Q}[x,y][[z]]$. This exponential plays a prominent role
in the combinatorics of symmetric functions, and has applications
in counting of gauge invariant operators in supersymmetric quantum
field theories (see eg. \cite{FHH}).

\begin{theorem}
\label{thm:A-PExpB}Let $\Gamma$ be any finitely presented group.
Then:
\[
\sum_{n\geq0}E(\mathcal{X}_{\Gamma}GL_{n};u,v)\,t^{n}=\pexp\left(\sum_{n\geq1}E(\mathcal{X}_{\Gamma}^{irr}GL_{n};u,v)\,t^{n}\right).
\]
\end{theorem}
The proofs of Theorem \ref{thm:A-PExpB} and Proposition \ref{prop:locally-closed-GLn}
are detailed in \cite{FNZ}; they allow to write explicit expressions for $E(\mathcal{X}_{\Gamma}GL_{n})$, for any group $\Gamma$, for which we have a formula for $E(\mathcal{X}_{\Gamma}^{irr}GL_{m})$, for all $m\leq n$, by a simple finite algorithm (and vice-versa). The formula
of Theorem \ref{thm:A-PExpB} generalizes a formula of \cite{MR}
to an arbitrary group $\Gamma$,
even if the corresponding $GL_{n}$-character variety is not of polynomial type.

\section{The Free Group Case}

In this last section, we describe applications of the above methods
to the case of the free group of rank $r$, $\Gamma=F_{r}$; for simplicity
we adopt the notations $\mathcal{X}_{r}GL_{n}$, $\mathcal{X}_{r}SL_{n}$,
etc, for the corresponding character varieties. In \cite{MR}, it
was shown that $\mathcal{X}_{r}^{irr}GL_{n}$ and $\mathcal{X}_{r}GL_{n}$
are of polynomial type. Moreover, by counting points over finite fields
and using a theorem of Katz (\cite[Appendix]{HRV}), Mozgovoy and
Reineke found a formula for the $E$-polynomial of $\mathcal{X}_{r}^{irr}GL_{n}$
that can be written as follows (dropping the $x$ variable in $E(X;\,x)$,
and using $|[k]|:=k_{1}+\cdots+k_{d}$ for the length of a partition $[k]\in\mathcal{P}_{d}$).
\begin{proposition}
\cite{MR,FNZ}\label{prop:E-irr} For $r,n\geq2$, we have:
\[
E(\mathcal{X}_{r}^{irr}GL_{n})=(x-1)\sum_{d|n}\frac{\mu(n/d)}{n/d}\,\sum_{[k]\in\mathcal{P}_{d}}\frac{(-1)^{|[k]|}}{|[k]|}\binom{|[k]|}{k_{1},\cdots,k_{d}}\prod_{j=1}^{d}b_{j}(x^{n/d})^{k_{j}}x^{\frac{n(r-1)k_{j}}{d}\binom{j}{2}},
\]
where $\mu$ is the M\"obius function, and the $b_{j}(x)$ are polynomials defined
by:\\[-3mm]
\begin{equation}
(1+\sum_{n\geq1}b_{n}(x)\,t^{n})\left(1+\sum_{n\geq1}\big((x-1)(x^{2}-1)\ldots(x^{n}-1)\big)^{r-1}\,t^{n}\right)=1.\label{eq:F(t)}
\end{equation}

\end{proposition}
Using Propositions \ref{prop:locally-closed-GLn} and \ref{prop:E-irr}
and Theorem \ref{thm:A-PExpB}, we are able to write down very explicit
expressions for $E(\mathcal{X}_{r}^{[k]}GL_{n})$, the $E$-polynomials
of all polystable strata of $\mathcal{X}_{r}GL_{n}$ (see \cite[Secs. 5 and 6]{FNZ},
where we also compute $E(\mathcal{X}_{\Gamma}^{irr}GL_{n})$ for other
$\Gamma$ and low $n$).

We now provide a few lines on a forthcoming proof of the equality
between the $E$-polynomials of $\mathcal{X}_{r}SL_{n}$ and of $\mathcal{X}_{r}PGL_{n}$
for all $n\in\mathbb{N}$. This has been conjectured in Lawton-Mu\~noz
in \cite{LM}, who proved by explicit computation the cases $n=2$
and 3.

In a analogous way as for $GL_{n}$ (see Section~\ref{sec:stratification-partition}),
we can define the $[k]$-polystable loci $\mathcal{X}_{r}^{[k]}SL_{n}$
and $\mathcal{X}_{r}^{[k]}PGL_{n}$ as follows. For a partition $[k]\in\mathcal{P}_{n}$,
the $[k]$-stratum of $\mathcal{X}_{r}SL_{n}$ is defined by restriction
of the corresponding one for $GL_{n}$:
\[
\mathcal{X}_{r}^{[k]}SL_{n}:=\{\rho\in\mathcal{X}_{r}^{[k]}GL_{n}\,|\,\det\rho=1\},
\]
where the determinant of a representation is an element of $\mathcal{R}_{r}\mathbb{C}^{*}$.
By considering the action $\mathcal{R}_{r}\mathbb{C}^{*}\times\mathcal{X}_{r}GL_{n}\to\mathcal{X}_{r}GL_{n}$
given by multiplication of (conjugacy classes of) representations,
which is well defined on the GIT quotients and preserves the stratification
of $GL_{n}$, we can define
\begin{equation}
\mathcal{X}_{r}^{[k]}PGL_{n}:=\mathcal{X}_{r}^{[k]}GL_{n}/\mathcal{R}_{r}\mathbb{C}^{*}=\mathcal{X}_{r}^{[k]}GL_{n}/(\mathbb{C}^{*})^{r}.\label{eq:GLn-PGLn}
\end{equation}

\begin{theorem}
\label{thm:SLn-PGLn}\cite{FNZ2} For the free group $F_{r}$, we
have the equalities:
\begin{eqnarray*}
E(\mathcal{X}_{r}SL_{n}) & = & E(\mathcal{X}_{r}PGL_{n})=E(\mathcal{X}_{r}GL_{n})(x-1)^{-r}\\
E(\mathcal{X}_{r}^{[k]}SL_{n}) & = & E(\mathcal{X}_{r}^{[k]}PGL_{n})=E(\mathcal{X}_{r}^{[k]}GL_{n})(x-1)^{-r},
\end{eqnarray*}
for every $r,\;n$ and partition $[k]\in\mathcal{P}_{n}$.
\end{theorem}
The proof of Theorem \ref{thm:SLn-PGLn} uses geometric methods and
has two parts. The easy part is the relation between the $E$-polynomials
of $\mathcal{X}_{r}^{[k]}PGL_{n}$ and of $\mathcal{X}_{r}^{[k]}GL_{n}$,
which follows from the locally trivial (in the Zariski topology) fibration
corresponding to the quotient \eqref{eq:GLn-PGLn}. The difficult part
is the relation between the strata $\mathcal{X}_{r}^{[k]}PGL_{n}$
and $\mathcal{X}_{r}^{[k]}SL_{n}$ which involves finite quotients: it
requires the proof of the triviality of the action of the center
$\mathbb{Z}_{n}\subset SL_{n}$
on the cohomology (with compact support) of all the
strata $\mathcal{X}_{r}^{[k]}SL_{n}$; for this
we use equivariant cohomology and a deformation retraction
between $\mathcal{X}_{r}^{irr}SL_{n}$ and the smooth part of the
semialgebraic set $\hom(F_r,SU(n))/SU(n)$ (see \cite{FL}).

config: arxiv


TITLE: Gravity's effect on time
QUESTION [0 upvotes]: How can we say the time gets faster or slower when we are measuring time using tangible objects (like a caesium clock) .
Can't we say that gravity is affecting the oscillation of the caesium clock and hence it's showing error in time?
Also does this mean that time is nothing if we don't measure it or time exist only if we measure it?
And our body cells have their own biological clock so why would they even respond to the change in time and they'll complete their cell cycle normally {24 hours in human (but that's what we have calculated cell doesn't have information for the time  it works on it's own pace)} so will they age less than others (I'm supposing that their time gets slowed down)?
And if this should happen of course due to gravity then does it mean that even DNA replication all the way down to sub atomic level is affected by the gravity and the enzyme gets slower and slower?

REPLY [1 votes]: For simplicity, let's take two cesium atoms, one on the suface of Earth, one in space.
Now gravitational time dilation will cause the two atoms to go through relatively different amounts of hyperfine transitions during a certain period of time.
When viewed from Earth, the atom in space will undergo more hyperfine transitions, then the atom on the surface of Earth.
The atom inside Earth's gravitational field can only undergo relatively less hyperfine transitions. You are asking why? You are saying that the gravitational field of Earth does have some effect on the atom inside it (inside the gravitational field) and causes some error.
You can call it whatever you want, but we see in the experiments that there is this effect of gravity on the atom.
According to the equivalence principle, this effect of gravity is indistinguishable from acceleration.
Thus, we can say that the atom inside the gravitational field undergoes acceleration, and this causes this atom to undergo less hyperfine transitions relatively then the atom in space.
You are asking why? There is a example with photon clocks as they accelerate, analogously. The reasoning is that the movement of the photon is affected by the acceleration, thus the photon in the gravitational field has to travel allover longer paths relatively to the photon in space. Thus the photon clock in the gravitational field will relatively tick slower then the photon clock in space.

In atomic physics, hyperfine structure is defined by small shifts and splittings[clarification needed] in the energy levels of atoms, molecules, and ions, due to interaction between the state of the nucleus and the state of the electron clouds.
  Hyperfine structure, with energy shifts typically orders of magnitudes smaller than those of a fine-structure shift, results from the interactions of the nucleus (or nuclei, in molecules) with internally generated electric and magnetic fields.

Now you are asking whether this acceleration is causing some error in the atom, making it relatively do less hyperfine transitions. You could imagine that the hyperfine transition requires the nucleus and the electrons in the atom to interact. This requires time. Analogously to the photon clock, this interaction between the nucleus and electrons (the hyperfine transition) seems to require relatively more time.
In reality we do not know what the underlying mechanism (it is all QM) is that causes the interaction to take more time, but it might be that it is similar to the photon clock, and the interaction between the nucleus and electrons requires more time simply because of the acceleration (caused by the gravitational field) requiring longer paths and that is (information propagation) ultimately limited by the speed of light.
If you accept that the underlying world is quantum mechanical and information propagation is limited by the speed of light, then the answer is that the gravitational field causes (has the same effect as) acceleration and that causes the atom to undergo relatively less hyperfine transitions.


TITLE: how to prove complex rational function has this property
QUESTION [1 upvotes]: Let's say we have a rational function $f$ (i.e polynom divided by polynom) , and assume that $f$ has no poles in the upper plane $\{z;Imz \geq 0\}$.
we have to prove that: $$sup\{|f(z)|; Imz \geq 0\} = sup\{|f(z)|; Imz = 0\}$$

I assume that we are dealing in cases where the suprimum exists.
every approach I tried, I can't really understand what is the significant of function being rational. I know, for example, that there is finite number of zero's, and finite number of poles, but I can't see how it is helpful.

REPLY [1 votes]: We know that the function is holomorphic on the upper half plane and thus does not have any local maxima on the interior of the upper half plane (maximum principle). Since $f$ is rational, it's behavior at infinity is easy to deal with. Let's consider the cases:
First case: $f(z) \to \infty$ as $z \to \infty$. Then the supremeum on the real line and on the upper half plane $H$ are both $\infty$.
Next case: $f(z) \to c \in \mathbb{C}$ as $z \to \infty$. If $|c|= \sup_\mathbb{R} |f|$ then it is also $\sup_{H} |f|$. Otherwise, $|c|< \sup_\mathbb{R} |f|$ and there is a a big ball $B$. where $|f| < \sup_\mathbb{R} |f|$ outside $B$. But $\overline{B} \cap H$ is compact and so $f$ achieves is max on the boundary. The max doesn't occur on the upper semicircle by construction of $B$, so it must occur on $\mathbb{R} \cap \overline{B}$. This max is then the max on $H$ of $|f|$ because $|f|$ is smaller than $|c|$ outside $B$. Hence $$\max_H |f| = \max_{H \cap \overline{B}} |f| = \max_{\mathbb{R} \cap \overline{B}} |f|= \max_\mathbb{R} |f|$$


TITLE: Calculating $[H,\vec r]$
QUESTION [0 upvotes]: I'm doing Griffiths' Introduction to Quantum Mechanics. In a question, it introduces the modified Schrodinger equation in which the Hamiltonian, $$ H~=~-\frac{\hbar^2}{2m}\nabla^2 + V $$is replaced by the classical Hamiltonian, $$ H~=~\left[\frac{1}{2m}(\frac{\hbar}{i}\nabla-q\vec {\mathbf A})^2 +q\phi\right]$$ which I expanded out to $$ H= \left[ \frac{1}{2m}\left\{-\hbar^2\nabla^2+q^2A^2+i\hbar q(\vec\nabla\cdot\vec {\mathbf A}+\vec {\mathbf A}\cdot\vec \nabla)\right \}+q\phi \right]$$
Here $\vec{\mathbf A}$ is the vector potential for the magnetic field $(\vec{\mathbf B}=\vec\nabla \times\vec{\mathbf A}) $ and $\phi$ is the scalar potential for the electric field $(\vec{\mathbf E}=-\nabla\phi-\frac{\partial\vec{\mathbf A}}{\partial t})$.
Now the first part of the question requires one to calculate $[H,\vec {\mathbf r}]$. But that requires me to take the Laplacian of $ \vec {\mathbf r}\psi$ which is a vector function (due to the first term in the expression). Is this legal?
I know that one way around this is to work component wise. But how to make this work?

REPLY [2 votes]: Your problem (as I understand it) boils down to calculating the vector Laplacian for the vector field function $\mathbf{r} \psi$.  This calculation can be performed in any coordinate system you choose;  if you know how to apply divergences, gradients, and curls to an arbitrary vector field in that coordinate system, then the identity
$$
\nabla^2 \mathbf{A} \equiv \nabla \left( \nabla \cdot \mathbf{A} \right) - \nabla \times \left( \nabla \times \mathbf{A} \right)
$$
allows for the calculation of the vector Laplacian.
In Cartesian components, this reduces to the particularly nice
$$
\nabla^2 \mathbf{A} = \left( \nabla^2_s A_x \right) \hat{x} +  \left( \nabla^2_s A_y \right) \hat{y} + \left( \nabla^2_s A_z \right) \hat{z},
$$
where $\nabla^2_s$ is the familiar "scalar Laplacian".  It should therefore be possible to calculate $\nabla^2 (\mathbf{r} \psi)$ by setting $A_x = x \psi$ in the above, and similarly for $A_y$ and $A_z$.  The result will then be expressible in terms of vectors and gradients, which you can then put back into a coordinate-invariant form and re-express in terms of spherical coordinates if you like.
Alternately, if you want to stay in spherical coordinates the whole time, you can use the above identity to express $\nabla^2 \mathbf{A}$ in terms of the spherical coordinates of $\mathbf{A}$.  In your case, you will have $\mathbf{A} = \mathbf{r} \psi$, and so $A_r = r \psi$ while $A_\theta = A_\phi = 0$.  However, the vector Laplacian is not nearly as nice in spherical coordinates as it is in Cartesian coordinates;  if you want to see the full expression, it can be found in this table on Wikipedia.  When the dust settles, though, you should find an expression in terms of $\psi$, the spherical coordinates $\{r, \theta, \phi\}$, and the derivatives of the former with respect to the latter.
Both of the above methods are equivalent, and should yield the same result;  try them both out if you need to while away a rainy afternoon.  And, of course, both are equivalent to the method (suggested by gented in the comments) of expressing $H = \frac{1}{2m}(\mathbf{p} - q \mathbf{A})^2 + q \phi$ and using the canonical commutation relations $[x_i, x_j] = [p_i, p_j] = 0$, $[x_i, p_j] = \delta_{ij}$.  I suspect that after going through the Laplacian calculations above, you will appreciate the elegance of the operator method.


TITLE: Electric field and potential between two charges
QUESTION [0 upvotes]: There are two point charges on the $x$ axis and $x'$ is a place where the potential, relative to infinity, is the biggest. Why is the electric field zero at this point?

REPLY [0 votes]: The statement that $V$ is at an extremum can be written $\partial_i V=0$. Now think about the mathematical relation between $V$ and $E$ in electrostatics.


TITLE: Show that the following is only continuous at x=0
QUESTION [3 upvotes]: What is this question asking?
How does rational and irrational affect it? What are the absolute values used for in this solution?
Why is the series irrational when $c\ne0$ for any nonzero rational number but then the series is made to be all irrational values?
What is this question?

REPLY [2 votes]: hint
$ f $ is continuous at $ 0 $ because
$$(\forall x\in \Bbb R)\;\;|f(x)|=|x|$$
and
$$\lim_{x\to 0}f(x)=0=f(0)$$
Let $ a\ne 0$ rational.
Take, for $ n $ great enough,
$$a_n=a+\frac{\pi}{n}\notin \Bbb Q$$
then
$$\lim_{n\to+\infty}f(a_n)=\lim_{n\to+\infty}(-a_n)=-a\ne f(a)$$
Let $ b $ be an irrational number.
Take
$$b_n=\frac{\lfloor 10^nb\rfloor}{10^n}\in \Bbb Q$$
then
$$\lim_{n\to+\infty}f(b_n)=\lim_{n\to+\infty}b_n=b\ne f(b)$$


TITLE: Find the ratio of the shaded areas in a square and circle.
QUESTION [0 upvotes]: In both pictures, the circle has radius .
Find the ratio of the shaded area in (i) to the shaded area in (ii).
Answer in the form 1 : , where  is a decimal to 3 significant figures.
(i) Area of square = $4r^2$, area of circle = $r^2$, shaded area = $4r^2 - r^2$
(ii) Area of circle = $r^2$, area of square = $2r^2$, shaded area = $r^2 - 2r^2$
Going round in "circles" here to determine the ratio of these seemingly disparate fractions; any ideas?
Compare shaded area areas directly:
$\frac{4r^2 - r^2}{r^2 - 2r^2} = \frac{4 - }{ - 2} = 0.752$
Ratio 1 : 0.752

REPLY [2 votes]: To avoid possible confusion plot the areas together on either side of common circle periphery

Considering area outside and inside of the circle radius $r$ their ratio
$$ \frac {A_{outgreen}}{A_{inyellow}}= \frac {4r^2- \pi r^2}{\pi r^2- (\sqrt 2 r )^2}=  \frac {4-\pi}{\pi-2}=0.751938  $$


TITLE: Resonance of a pipe closed back on itself
QUESTION [1 upvotes]: Would the math formula for resonant air columns open at both ends apply to a hollow tube shaped like a donut?
No ends are closed because there are no ends.  Would it be considered an infinitely long open air column?
What would the math formula be to determine the fundamental?

REPLY [2 votes]: A donut shaped pipe can have a standing wave whose wavelength is equal to an integer fraction of the circumference of the pipe - you would have one wave traveling clockwise, and another traveling anticlockwise, and where they result in node and antinodes depends on their relative phase.
So for a pipe of length $L$ bent into a torus, the radius $R = \frac{L}{2\pi}$ and the wavelengths would be
$$\lambda_n=\frac{L}{n}$$
The frequency is of course the speed of sound $c$ divided by the wavelength $\lambda$, so
$$f_n = \frac{n c}{L}$$
Or in terms of the major radius of the torus,
$$f_n = \frac{n c}{2\pi R}$$
Where the fundamental frequency is the value obtained when n=1


TITLE: Lagrange multiplier question: finding a counterexample.
QUESTION [11 upvotes]: I'm helping a student through a course in mathematics. In the course text, we came across the following problem concerning the Lagrange multiplier technique.
Given a differentiable function with continuous partial derivatives $f:\mathbb{R}\to\mathbb{R}:(x,y)\mapsto f(x,y)$ that has to be extremized and a constraint given by an implicit relation $g(x,y)=0$ with $g$ likewise having continuous partial derivatives. The Lagrange multiplier technique looks for points $(x^*,y^*,\lambda^*)$ such that
$$\nabla f(x^*,y^*)=\lambda^* \nabla g(x^*,y^*)$$
$\lambda$ is the so-called Lagrange multiplier.
The technique rests upon the fact that the gradient of the constraint is different from the zero vector: $\nabla g(x^*,y^*) \neq 0$. Then, the text says that if that condition is not fulfilled, it is possible to not have solutions of the system of equations resulting from the technique while there can actually be a constrained extremum.
I have tried to construct such an example, but until now unsuccessfully as every attempt I make with a $\nabla g(x^*,y^*) = 0$ turns out to have a solution. Does anyone know of a nice counterexample?
And does one know of a counterexample when there are $n$ variables and $m$ constraints with $n>m>1$ ? For the latter, the condition becomes that the gradients of the constraints have to be linearly independent. So the question is: can we find an example where there is no solution for the Lagrange multiplier method, but there is an extremum and the gradients of the constraints are linearly dependent?

REPLY [12 votes]: Here is an example, albeit with two constraints: We want to find the minimum of the function
$$f(x,y,z):=y\ ,$$
given the constraints
$$F(x,y,z):=x^6-z=0\ ,\qquad G(x,y,z):=y^3-z=0\ .$$
The constraints define the curve
$$\gamma:\quad x\mapsto (x,x^2, x^6)\qquad(-\infty<x<\infty)\ ,$$
and it is easy to see that the minimum of $f\restriction \gamma$  is taken at the origin. On the other hand
$$\nabla f(0,0,0)=(0,1,0)\ ;\quad\nabla F(0,0,0)=\nabla G(0,0,0)=(0,0,-1)\ ;$$
whence $\nabla f(0,0,0)$ is not a linear combination of $\nabla F(0,0,0)$ and $\nabla G(0,0,0)$. This means that using Lagrange's method the origin would not have shown up as a conditionally stationary point.


text:
\section{Mutual information-characterization of correlated recovery}
\label{app:MIcorr}

\newcommand{\Unif}{\mathrm{Unif}}

We consider a general setup:  Let the number of communities $k$ be a constant. Denote the membership vector by $\sigma=(\sigma_1,\ldots,\sigma_n) \in [k]^n$ and the observation is $A=(A_{ij}: 1 \leq i < j \leq n)$. Assume the following conditions:
\begin{enumerate}[label=A\arabic*]

	\item \label{A1}
	For any permutation $\pi\in S_k$, $(\sigma,A)$ and $(\pi(\sigma),A)$ are equal in law, where $\pi(\sigma)\triangleq (\pi(\sigma_1),\ldots,\pi(\sigma_n))$;

	\item \label{A2}
	For any $i \neq j \in [n]$, $I(\sigma_i,\sigma_j;A)= I(\sigma_1,\sigma_2;A)$;

	\item \label{A3}
	For any $z_1,z_2 \in [k]$, $\prob{\sigma_1=z_1,\sigma_2=z_2} = \frac{1}{k^2} + o(1)$ as $n\to \infty$.
\end{enumerate}
These assumptions are satisfied for example for $k$-community SBM (where each pair of vertices $i$ and $j$ are connected independently with probability $p$ if $\sigma_i=\sigma_j$ and $q$ otherwise),\index{Stochastic block model (SBM)! $k$ communities}
 and the membership vector
$\sigma$ can be either uniformly distributed on $[k]^n$ or the set of equal-sized $k$-partition of $[n]$.

Recall that correlated recovery entails the following:
For any $\sigma , \hat{\sigma} \in [k]^n$, define the overlap:
\begin{align}
o\left( \sigma, \hat{\sigma} \right) = \frac{1}{n} \max_{\pi \in S_k}
\sum_{i \in [n] } \left( \indc{ \pi\left(\sigma_i \right) = \hat{\sigma}_i} - \frac{1}{k} \right).
\end{align}
We say an estimator $\hat{\sigma} = \hat{\sigma}(A)$ achieves correlated recovery if\footnote{For the special case of $k=2$, \prettyref{eq:corro} is equivalent to
$\frac{1}{n}\Expect[|\Iprod{\sigma}{\hat \sigma}|] = \Omega(1)$, where $\sigma , \hat{\sigma}$ are assumed to be $\{\pm\}^n$-valued.}
\begin{equation}
\expect{o\left( \sigma, \hat{\sigma} \right)}=\Omega(1),
\label{eq:corro}
\end{equation}
 that is, the misclassification rate, up to a global permutation, outperforms random guessing.
Under the above three assumptions, we have the following characterization of correlated recovery:
\begin{lemma}
\label{lmm:MIcorr}
Correlated recovery is possible if and only if
$I(\sigma_1, \sigma_2 ; A) = \Omega(1)$.
\end{lemma}

\begin{proof}
We start by recalling the relation between mutual information and total variation.
For any pair of random variables $(X,Y)$, define the so-called $T$-information \cite{Csiszar96}:
$T(X;Y) \triangleq \TV(P_{XY}, P_XP_Y) = \Expect[\TV(P_{Y|X}, P_Y)]$.
For $X \sim \Bern(p)$, this simply reduces to
\begin{equation}
T(X;Y) = 2p(1-p) \TV(P_{Y|X=0}, P_{Y|X=1}).
\label{eq:Tbern}
\end{equation}
Furthermore, the mutual information can be bounded by the $T$-information,
by Pinsker's and Fano's inequality, as follows \cite[Eq.~(84) and Prop.~12]{PW14a}
\begin{equation}
 2 T(X;Y)^2 \leq I(X;Y)  \leq \log (M-1) T(X;Y) + h(T(X;Y))
	\label{eq:TI}
\end{equation}
where in the upper bound $M$ is the number of possible values of $X$, and $h$ is the binary entropy function in \prettyref{eq:binaryentropy}.

We prove the ``if'' part.
Suppose

$I(\sigma_1, \sigma_2; A) = \Omega(1)$.
We first claim that assumption \ref{A1} implies that
\begin{equation}
I (\indc{\sigma_1 =\sigma_2}; A)=I(\sigma_1, \sigma_2; A)
\label{eq:ss}
\end{equation}
that is,
 $A$ is independent of $\sigma_1,\sigma_2$ conditional on $\indc{\sigma_1 =\sigma_2}$.

Indeed,
for any $z \neq z'\in [k]$, let $\pi$ be any permutation such that

$\pi(z')=z.$

Since $P_{\sigma, A} =  P_{\pi(\sigma), A}$,
we have $P_{A|\sigma_1=z,  \sigma_2=z} =  P_{A| \pi(\sigma_1)=z,  \pi(\sigma_2)=z }$, i.e.,
$P_{A |\sigma_1=z, \sigma_2=z} =  P_{A| \sigma_1 =z',  \sigma_2 =z' }$.
Similarly, one can show that
$P_{A|\sigma_1=z_1, \sigma_2=z_2} =  P_{A| \sigma_1 =z_1',  \sigma_2 =z_2'}$,
for any $z_1 \neq z_2$ and $z_1'\neq z_2'$, and this proves the claim.

Let $x_j=\indc{\sigma_1 =\sigma_j}$.
By the symmetry assumption \ref{A2},
$I(x_j; A) = I(x_2; A) = \Omega(1)$ for all $j \neq 1$.
Since $\prob{x_j = 1} = \frac{1}{k} + o(1)$ by assumption \ref{A3}, applying \prettyref{eq:TI} with $M=2$ and in view of \prettyref{eq:Tbern}, we have
$\TV(P_{A|x_j=0},P_{A|x_j=1})=\Omega(1)$.
Thus, there exists an estimator $\widehat{x}_j \in \{0,1\}$ as a function of $A$, such that
\begin{align}
\prob{\widehat{x}_j = 1 \mid x_j =1 } +
\prob{\widehat{x}_j = 0 \mid x_j =0 } \ge 1+\TV(P_{A|x_j=0},P_{A|x_j=1})
= 1+ \Omega(1).  \label{eq:estimator_x_hat}
\end{align}

Define $\hat\sigma$ as follows: set $\hat\sigma_1= 1 $; for $j \neq 1$, set $\hat \sigma_j = 1 $ if $\widehat{x}_j = 1$
and draw $\hat \sigma_j $ from $\{2,\ldots,k\}$ uniformly at random
if $\widehat{x}_j = 0$.
Next, we show that $\hat\sigma$ achieves correlated recovery.
Indeed, fix a permutation $\pi \in S_k$ such that $\pi(\sigma_1)=1$. It follows from
the definition of overlap that
\begin{equation}
\Expect[o\left(\sigma, \hat{\sigma} \right)]
\ge \frac{1}{n} \sum_{j \neq 2} \prob{\pi(\sigma_j) =\hat\sigma_j} - \frac{1}{k}.
\label{eq:olb}
\end{equation}
Furthermore, since $\pi(\sigma_1)=1$, we have, for any $j\neq 1$,
\[
\prob{\pi(\sigma_j) = \hat\sigma_j,x_j=1}=
\prob{\hat x_j=1,x_j=1}
\]
and
\[
\prob{\pi(\sigma_j) = \hat\sigma_j,x_j=0}=
\prob{\pi(\sigma_j) = \hat\sigma_j, \hat x_j=0,x_j=0}=
\frac{1}{k-1} \prob{\hat x_j=0,x_j=0},
\]
where the last step is because conditional on $\hat{x}_j=0$,
$\hat{\sigma}_j$ is chosen  from $\{2,\ldots,k\}$ uniformly and
independently of everything else.
Since $\prob{x_j = 1} = \frac{1}{k} + o(1)$, we have
\[
\prob{\pi(\sigma_j) =\hat\sigma_j} = \frac{1}{k}(\prob{\widehat{x}_j = 1 \mid x_j =1 } +
\prob{\widehat{x}_j = 0 \mid x_j =0 }) +o(1) \overset{\prettyref{eq:estimator_x_hat}}{\ge} \frac{1}{k}+ \Omega(1).
\]
By \prettyref{eq:olb}, we conclude that $\hat{\sigma}$ achieves correlated recovery
of $\sigma$.

Next we prove the ``only if'' part.
Suppose $I(\sigma_1, \sigma_2; A) = o(1)$ and we aim to show

$\expect{o\left(\sigma, \hat{\sigma} \right)}=o(1)$ for any estimator $\hat\sigma$.
By the definition of overlap, we have
\begin{align*}
o\left(\sigma, \hat{\sigma} \right)
\le
\frac{1}{n} \sum_{\pi \in S_k}
\left| \sum_{i \in [n] }
\left( \indc{ \pi\left(\sigma_i \right) = \hat{\sigma}_i} - \frac{1}{k}  \right) \right|.

\end{align*}
Since there are $k!=\Omega(1)$ permutations in $S_k$, it suffices to show
for any fixed permutation $\pi$,
$$
\expect{ \left| \sum_{i \in [n] } \left( \indc{ \pi\left(\sigma_i \right) = \hat{\sigma}_i} - \frac{1}{k}  \right)
\right| } = o(n).
$$
Since $I(\pi(\sigma_i), \pi(\sigma_j); A)=I(\sigma_i, \sigma_j; A)$, without loss of generality,
we assume $\pi=\text{id}$ in the following. By the Cauchy-Schwarz inequality, it further suffices to show
\begin{equation}
\expect{ \left( \sum_{i \in [n] } \left( \indc{ \sigma_i =\hat{\sigma}_i } - \frac{1}{k} \right)\right)^2 } =o(n^2).
\label{eq:overlap2}
\end{equation}
Note that
\begin{align*}
& \expect{  \left( \sum_{i \in [n] } \left(
   \indc{ \sigma_i  = \hat{\sigma}_i } - \frac{1}{k} \right)   \right)^2 } \\
 & = \sum_{i, j \in [n] }
 \expect{ \left( \indc{ \sigma_i = \hat{\sigma}_i }- \frac{1}{k}  \right)
 \left( \indc{ \sigma_j  = \hat{\sigma}_j }- \frac{1}{k}  \right) }  \\
 & =  \sum_{i, j \in [n] }  \prob{  \sigma_i = \hat{\sigma}_i , \sigma_j  = \hat{\sigma}_j }
 - \frac{2n}{k} \sum_{i \in [n] }\prob{ \sigma_i = \hat{\sigma}_i } + \frac{n^2}{k^2}.
\end{align*}
For the first term in the last displayed equation,
let $\sigma'$ be identically distributed as $\hat\sigma$ but independent of $\sigma$.
Since $I(\sigma_i,\sigma_j;\hat\sigma_i,\hat\sigma_j) \le I(\sigma_i,\sigma_j;A)=o(1)$ by the data processing inequality, it follows from the lower bound in
\prettyref{eq:TI} that $\TV(P_{\sigma_i,\sigma_j,\hat\sigma_i,\hat\sigma_j}, P_{\sigma_i,\sigma_j,\sigma_i',\sigma_j'})=o(1)$.
Since
$\pprob{  \sigma_i = {\sigma}_i', \sigma_j  = {\sigma}_j' } \leq
\max_{a,b \in [k]} \prob{  \sigma_i = a, \sigma_j  = b} \leq \frac{1}{k^2}+o(1)$ by assumption \ref{A3},
we have
$$
\prob{  \sigma_i = \hat{\sigma}_i , \sigma_j  = \hat{\sigma}_j }
\leq \frac{1}{k^2} + o(1),
$$
Similarly, for the second term, we have
$$
\prob{ \sigma_i = \hat{\sigma}_i } = \frac{1}{k} +o(1),
$$
where the last equality holds due to $I(\sigma_i; A) =o(1).$
Combining the last three displayed equations gives \prettyref{eq:overlap2} and completes the proof.
\end{proof}


TITLE: Incorrect Solution for Problem 7 of Pinter's Book of Abstract Algebra, Chapter 2?
QUESTION [5 upvotes]: I'm just getting started with Pinter's A Book of Abstract Algebra, please be kind. The book solution for Chapter 2, Problem 7 claims that the following operator is non-associative:
$$ x * y = \frac{xy}{x+y+1}$$
The book solution:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{xyz(x+y+1)}{x+y+z+xy+yz+xz+1}$$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right) = \frac{xyz(y+z+1)}{x+y+z+xy+yz+xz+1}$$
However the $(x+y+1)$ and $(y+z+1)$ terms appear to actually cancel out:
$$ (x * y) * z = \left(\frac{xy}{x+y+1}\right) * z = \frac{\left(\frac{xy}{x+y+1}\right)z}{\frac{xy}{x+y+1} + z + 1} $$
$$ x * (y * z) = x * \left(\frac{yz}{y+z+1}\right)  =\frac{x\left(\frac{yz}{y+z+1}\right)}{x+\frac{yz}{y+z+1} + 1} $$
So the denominators cancel out, the two groupings are equivalent, the operator is associative.  This result is duplicated in Mathematica, but I'd like to make sure I'm not missing something.  Sorry for bothering people with such a basic question.

REPLY [4 votes]: Let $f(x)=1+\frac 1x$. Then $f^{-1}(z)=\frac{1}{z-1}$ and show that:
$$x*y = f^{-1}(f(x) f(y))$$
This means that:
$$\begin{align}(x*y)*z &= f^{-1}(f(x)f(y))* z\\
&= f^{-1}\left(f\left(f^{-1}(f(x)f(y))\right)f(z)\right)\\
&= f^{-1}(f(x)f(y)f(z))
\end{align}$$
Since $*$ is commutative, we get:
$$\begin{align}x*(y*z)&= (y*z)*x\\
&= f^{-1}(f(y)f(z)f(x))
\end{align}$$
Clearly, these two are equal.
This is true in general for any invertible function $f$ if we define $*$ in this way.
Most easily defined associative operators on the real numbers are of this form. Even addition can be defined this way in terms of multiplication, with $f(x)=e^x$ and $f^{-1}(x)=\log x$, then:
$$x+y = \log(e^xe^y)$$
Alternatively, of course, you can define positive multiplication in terms of addition:
$$xy = e^{\log x + \log y}$$


TITLE: Conservation of energy and angular momentum
QUESTION [1 upvotes]: I'm writing a java program to simulate the solar system. All planets are modelled as point masses. How do I check if my solar system is conserving energy? I'm not sure how to calculate the energy of the system at the start of the simulation, let alone at the end!
And given that my model is with point masses, presumably I can't calculate angular momentum at all?

REPLY [1 votes]: The conserved quantities in your problem are: total mechanic energy $T+U$, total linear momentum $\vec P$, and total angular momentum $\vec L$ of the $N$ point masses.
They are defined as:
$$E=T+U \quad\text{ total mechanical energy}$$
$$U=-G\sum_{i}^N \sum_{j}^N  \frac{m_i m_j}{|\vec r_i -\vec r_j|}\quad\text{ total potential energy}$$
$$T=\frac12\sum_i^N m_i v_i^2\quad\text{ total kinetic energy}$$
$$\vec P=\sum_i^N m_i \vec v_i \quad\text{ total linear momentum}$$
$$\vec L=\sum_i^N m_i (\vec r_i\times \vec v_i) \quad\text{ total angular momentum}$$
where $G$ is the gravitational constant (the gravitational force between 2 masses is $F=G \frac{m_i m_j}{(\vec r_i -\vec r_j)^2}$).
Note that:
$m_i$, $\vec r_i$, and $\vec v_i$ are the mass, the position, and the velocity of the point mass $i$.
only the total mechanical energy $E=T+U$ is conserved, while the total kinetic and total potential energies are not conserved separately;
the total linear and angular momenta are vectors, while total energies are scalars;
you must include also the sun in the sums over point masses $i=1,...,N$ and $j=1,...,N$ in the definition of the energies and momenta.
A point mass has a well defined angular momentum $m (\vec r \times \vec v)$ which is non-zero if the mass is rotating around another mass (or any point in the space).


TITLE: Proofcheck to show that $\bigcap\limits_{n \in \mathbb N} A_{n} = \{x\}$
QUESTION [3 upvotes]: Let $X$ be a complete metric space and $\varnothing \neq A_{n}\subseteq X$ closed for all $n \in \mathbb N$ so that $A_{n+1} \subseteq A_{n}$ and $\operatorname{diam{(A_{n})}}\xrightarrow{n \to \infty}0$. Show there exists unique $x \in X$ so that $\bigcap\limits_{n \in \mathbb N} A_{n} = \{x\}$.
My idea: We can construct a cauchy sequence $(x_{n})_{n}$ where $x_{n} \in A_{n}$ and $x_{m} \in A_{m}$ so that, for any $\epsilon > 0$ there is a $N \in\mathbb N$ so that for any $n>m \geq N$: we get $d(x_{m}, x_{n})\leq \operatorname{diam{(A_{N})}}<\epsilon$ thus $(x_{n})_{n}$ is a cauchy sequence and for completeness of $X$, there exists limit point $x \in X$. In order to show that $x$ is indeed unique. We assume $\exists y\neq x \in X$ that also lies in $\bigcap\limits_{n \in \mathbb N} A_{n}$, since we are in a metric space, $X$ is Hausdorff and thus there exists balls $B_{\delta_{y}}(y), B_{\delta_{x}}(x)$ so that $B_{\delta_{y}}(y) \cap B_{\delta_{x}}(x)=\varnothing$ but this immediately leads to a contradiction as we choose $N\in \mathbb N$ so that $\operatorname{diam{(A_{N})}}<\min{\{ \delta_{y}, \delta_{x}\}}$ but we already know that $x\in A_{N}$ and thus $y\notin A_{N}\Rightarrow y \notin \bigcap\limits_{n \in \mathbb N}A_{n}$. Is my proof, particularly the second part ok?

REPLY [1 votes]: Example style. Not meant as a criticism of the OP.
(1). Take some $a_n\in A_n$ for each $n.$
(2). For $\epsilon >0$ take $m$ such that diam($A_m)<\epsilon$. Then for all $n,n'$ we have $$n,n'\ge m \implies \{a_n,a_{n'}
\}\subset A_m \implies d(a_n,a_{n'})\le  \operatorname{diam}(A_m)<\epsilon.$$
So the sequence $ (a_n)_n$ is Cauchy. And $d$ is complete  so $(a_n)_n$ converges  to some $x\in X$.
(3). For each $n$ we have $x\in \overline {\{a_m:m\ge n\}}\subset \overline {A_n}=A_n .$ So $x\in \cap_nA_n.$
(4). For all $y$  we have $$y\in \cap_n A_n \implies \forall n\,(\,\{x,y\}\subset A_n\,) \implies$$ $$\implies \forall n\,(\,d(x,y)\le \operatorname{diam}(A_n)\,)\implies$$ $$\implies d(x,y)\le \inf_n \operatorname{diam}(A_n)=0\implies$$ $$\implies d(x,y)=0\implies y=x.$$
(5). By (3) and (4) we have $\{x\}\subset \cap_n A_n\subset \{x\}.$ Therefore $\cap_n A_n=\{x\}.$


TITLE: On Hurwitz Square (r, s, t)-Identities examples
QUESTION [3 upvotes]: By (r, s, t)-identity I mean any sort of such identity:
$$
(x_1^2+\ldots + x_r^2)(y_1^2+\ldots +y_s^2)=(z_1(x,y)^2+\ldots + z_t(x,y)^2),
$$
where $z_i(x,y)$ is a polynomial for every $i$.
See this for some further reading: https://en.wikipedia.org/wiki/Hurwitz_problem
I can't find in the web any non-standard examples of such identities like Adem's example of (3,5,7)-identity or some others for small r, s and t. Can anyone provide several such expressions?

REPLY [3 votes]: From pages 137-138 of Rajwade, "Notes on Chapter 10"
the Hurwitz-Radon theorem gives
$$  (2,2,2), \; (4,4,4), \; (8,8,8), \; (9,16,16), \; (10,32,32), \ldots  $$
K. Y. Lam found $(10,10,16)$ in 1966.
In 1975, Adem found
$$  (3,5,7), \; (10,10,16), \; (12,12,28), \; (13,13,28), \; (17,18,32), \ldots  $$
In 1982, Yuzvinski found
$$   (10,10,16), \; (12,20,32), \; (14,40,64), \ldots  $$
He recommends a 1989 survey article by Shapiro called Products of Sums of Squares
https://people.math.osu.edu/shapiro.6/lec1.pdf
https://people.math.osu.edu/shapiro.6/lec2.pdf
https://people.math.osu.edu/shapiro.6/lec3.pdf


TITLE: Rational $F_n$ Limit $\to$ Find Denominator Polynomial
QUESTION [0 upvotes]: Given: $P(x)$ is a polynomial such that the following apply:

I)  $\lim_\limits{x \to 1} \dfrac{P(x)}{(x-1)^3} = 6$
II). $\lim_\limits{x \to -2} \dfrac{P(x)}{x^2+x-2} = 9$
III) $\lim_\limits{x \to -\infty}\dfrac{P(x)}{(x+7)^6} = 0$

Find explicit formula for $p(x)$?
A) Calculation:
1) By the rule that if the degree of the numerator is less than the degree of the denominator $\lim_\limits{x \to -\infty}=0$ holds for III.  Therefore, $P(x)$ is less than a 6th degree polynomial.
2) Factoring the denominator of II leads to $(x+2)(x-1)$.  Therefore, there is a discontinuity at $x=-2$ and $x=1$ for $P(x)$.
B)
Further more:
3) I tried to make a polynomial in the numerator which would divide out the discontinuities in the denominator so made a polynomial of $(x+2)(x-1)^3$ which seems to work for II and III on graphing calculator.
4) In addition, the limits of I and II are factors of $3$.  This leads me to believe there is a factor in the numerator which will involve $3$.

REPLY [2 votes]: It should be $$P(x)=(x-1)^3(x+2)(ax+b),$$
which gives the system:
$$-2a+b=-\frac{1}{3}$$ and $$a+b=2.$$
I got $$P(x)=\frac{1}{9}(x-1)^3(x+2)(7x+11).$$


TITLE: Hilbert's basis theorem
QUESTION [0 upvotes]: In first proof of wikipedia:

Why the $deg(f_i)$ is a non-decreasing sequence of naturals?

REPLY [1 votes]: $f_n$ and $f_{n+k}$ are both polynomials in $\mathfrak a\setminus\mathfrak b_n$ and, since $f_n$ has minimal (technically, least) degree, $\deg f_n\le \deg f_{n+k}$.


TITLE: maximizing multivariate polynomial
QUESTION [1 upvotes]: Consider $J = \sum_{i=0}^{N}y_{i-1}x_{i}y_{i+1}$ where $+$ and $-$ in the indices are mod $N+1$. Let $x_{i} = 1 - y_{i} \in \{0,1\}$. What are some of the tools useful and relaxation techniques available to maximize $J$ or any other symmetric multivariate polynomial?

REPLY [1 votes]: The optimization is a summation of triples over a cycle. You can enumerate the value of a few binary variables to break the cycle down to a (second-order) Markov chain. Then dynamic programming can be used to solve this problem efficiently.
To elaborate, consider enumerating $x_0$ and $x_N$, which are both binary variables. Then the cycle becomes a chain in which the first term contains only $x_1$ (since $x_0$ and $x_N$ are known), the second term contains only $x_1$ and $x_2$ ($x_0$ is known), and similarly for the last two terms. The rest terms still contain three x's.
Then if we consider state $s_i = [x_i, x_{i+1}]$ (which has four possible choices), then the term $x_{i-1}y_ix_{i+1}$ can be written in $\phi_i(s_{i-1}, s_i)$ and the entire summation can be written as
$J = \sum_{i=2}^{N-2} \phi_i(s_{i-1}, s_{i})$
which can be solved efficiently by dynamic programming.