TITLE: square matrix is not invertible if at least one row or column is zero
QUESTION [5 upvotes]: How to show that a square matrix is not invertible if at least one row or column is zero ? I can show if a row is zero, the result C of $AB=C$ can not be the identity matrix because there is a zero row. But for the column case ?
Assume I don't know something about determinants.

REPLY [1 votes]: hint:
theorem. let A be square invertible matrix. then [A,I] can be transformed into [I,A(inverse)] using elementary row operations.
but since A has a zero row or column, you can never transform the ith row or jth column be equal to 1 or that there will always be a zero in the diagonal which is not the identity matrix since [I]= 0 if i is not equal to j and 1 if i=j.


TITLE: Rewrite $-\frac{y}{x}+2\ln|1+\frac{y}{x}|+\ln|x|=\ln|c|$ with separated variables and without logarithms
QUESTION [0 upvotes]: This seems simple maybe to some, but I can't get this right.
I have $-\dfrac{y}{x}+2\ln|1+\frac{y}{x}|+\ln|x|=\ln|c|$
And I  want to rewrite it nicely, with no logarithms and in the form $y=f(x)$
I tried and got
$y=x[\ln c+2\ln(1+y/x)+\ln x]$
alternatively:
$e^{-y/x}+(1+y/x)^2+x-c=0$
but then neither went any further.
Can't say I am able to separate the $x$ and $y$'s.
Any ideas?
Thanks

REPLY [1 votes]: $$-\dfrac{y}{x}+2\ln|\frac{x+y}{x}|+\ln|x|=\ln|c|$$
Analytically solving this equation for $y$ cannot be done with a finite number of elementary functions. A special function is required, namely the LambertW function :
https://mathworld.wolfram.com/LambertW-Function.html
$$-\dfrac{y}{x}+2\ln|\frac{x+y}{x}|-2\ln(2)+\ln|x|=\ln|c|-2\ln(2)$$
$$-\dfrac{y}{x}+2\ln|\frac{x+y}{2x}|=\ln|c|-2\ln(2)-\ln|x|$$
$$-\dfrac{y}{2x}+\ln|\frac{x+y}{2x}|=\frac12\ln|c|-\ln(2)-\frac12\ln|x|$$
$$\dfrac{x+y}{-2x}+\ln|\frac{x+y}{2x}|=\frac12\ln|c|-\ln(2)-\frac12\ln|x|-\frac12$$
$$\dfrac{x+y}{-2x}+\ln|\frac{x+y}{-2x}|=\frac12\ln|c|-\ln(2)-\frac12\ln|x|-\frac12$$
Let $X=\dfrac{x+y}{-2x}\quad;\quad \ln|C|=\frac12\ln|c|-\ln(2)-\frac12\quad$and$\quad \ln|Y|=\ln|C|-\frac12\ln|x| $
$$X+\ln|X|=\ln|Y|\quad\implies\quad Xe^X=Y$$
Which solution is
$$X=W(Y)$$
$$\dfrac{x+y}{-2x}=W\left(\frac{C}{\sqrt{|x|}} \right)$$
$$\boxed{y(x)=-x-2x\;W\left(\frac{C}{\sqrt{|x|}} \right)}$$
This is the closed form solution involving the Lambert W function.
$C$ is a constant related to $c$ with the relationship given above.


text:
\begin{document}

\title{Chromogeometry}
\author{N J Wildberger\\School of Mathematics and Statistics\\UNSW Sydney 2052 Australia}
\maketitle
\date{}

\begin{abstract}
Chromogeometry brings together Euclidean geometry (called \textit{blue}) and
two relativistic geometries (called \textit{red }and \textit{green}), in a
surprising three-fold symmetry. We show how the red and green `Euler lines'
and `nine-point circles' of a triangle interact with the usual blue ones, and
how the three orthocenters form an associated triangle with interesting
collinearities. This is developed in the framework of rational trigonometry
using \textit{quadrance} and \textit{spread} instead of \textit{distance} and
\textit{angle}. The former are more suitable for relativistic geometries.

\end{abstract}

\section*{Introduction}

Three-fold symmetry is at the heart of a lot of interesting mathematics and
physics. This paper shows that it also plays an unexpected role in planar
geometry, in that the familiar Euclidean geometry is only one of a trio of
interlocking metrical geometries. We refer to Euclidean geometry here as
\textit{blue }geometry; the other two geometries, called \textit{red }and
\textit{green,} are relativistic in nature and are associated with the names
of Lorentz, Einstein and Minkowski.

The three geometries support each other and interact in a rich way. This
transcends Klein's Erlangen program, since there are now \textit{three groups
}acting on a space. Remarkable algebraic identities lie at the heart of the explanations.

The results described here are just the tip of an iceberg, leading to many
rich generalizations of results of Euclidean geometry, with much waiting to be
discovered and explored, see for example \cite{Wildberger 4} for applications
to conics and \cite{Wildberger3} for connections with one dimensional metrical geometry.

The basic structure of all three geometries are the same---they are ruled by
the laws of \textit{rational trigonometry} as developed recently in
\cite{Wildberger}, which hold over a general field not of characteristic two.
Although over the rational numbers (or the `real numbers') there are
significant differences between the Euclidean (blue) version and the other two
(red and green), it is the \textit{interaction }of all three which yields the
biggest surprises.

To start the ball rolling, this paper first introduces the phenomenon in the
context of the classical Euler line and nine-point circle of a triangle. Then
we recall the main laws of rational trigonometry, introduce the basic facts
about the three geometries and state some explicit formulas, and then show how
chromogeometry allows us to enlarge our understanding of the geometry of a
triangle. In particular we associate to each triangle $\overline{A_{1}
A_{2}A_{3}}$ in the Cartesian plane a second interesting triangle which we
call the $\Omega$\textit{-triangle }of $\overline{A_{1}A_{2}A_{3}}$.

The results are verified by routine but sometimes lengthy computation, and
they inevitably reduce to algebraic identities, some of which are quite
lovely. The development takes place in the framework of \textit{universal
geometry}, so that we are interested primarily in what happens over
\textit{arbitrary fields}. The paper (\cite{Wildberger2}) shows that universal
geometry also extends to arbitrary quadratic forms, and embraces both
spherical and hyperbolic geometries in a projective version.

\section*{Euler lines and nine-point circles in relativistic settings}

Recall that for a triangle $\overline{A_{1}A_{2}A_{3}}$ the intersection of
the medians is the \textbf{centroid }$G,$ the intersection of the altitudes is
the \textbf{orthocenter }$O$ and the intersection of the perpendicular
bisectors of the sides is the \textbf{circumcenter} $C,$ which is the center
of the circumcircle of the triangle. Remarkably, it was left to Euler to
discover that these three points are collinear, and that $G$ divides
$\overline{OC}$ in the (affine) proportion $2:1.$ Furthermore the center $N$
of the circumcircle of the triangle $\overline{M_{1}M_{2}M_{3}}$ of midpoints
of the sides of $\overline{A_{1}A_{2}A_{3}}$ (called the \textbf{nine-point
circle} of $\overline{A_{1}A_{2}A_{3}}$) also lies on the Euler line, and is
the midpoint of $\overline{OC}.$

\begin{center}
\includegraphics[
height=3.3698in,
width=3.0564in
]
{ChromoEx2Blue.eps}
\end{center}

This is above shown for the triangle $\overline{A_{1}A_{2}A_{3}}$ with points
\[
\begin{tabular}
[c]{lllll}
$A_{1}\equiv\left[  0,0\right]  $ &  & $A_{2}\equiv\left[  6,1\right]  $ &  &
$A_{3}\equiv\left[  2,3\right]  .$
\end{tabular}
\]
The triangle $\overline{A_{1}A_{2}A_{3}}$ is in black, while the circumcircle
and nine-point circle are in blue (the latter more boldly), as are the Euler
line and the points $O,C$ and $N,$ which are given the subscript $b$ for blue,
and henceforth referred to as the \textbf{blue Euler line}, the \textbf{blue
orthocenter }etc.

Planar Euclidean geometry rests on the \textbf{blue quadratic form}
$x^{2}+y^{2}$ (or if you prefer the corresponding symmetric bilinear form, or
dot product). It is also interesting to consider the \textbf{red quadratic
form} $x^{2}-y^{2}$ which figures prominently in two dimensional special
relativity. In this case, two lines are red perpendicular precisely when one
can be obtained from the other by ordinary Euclidean reflection in a
\textbf{red null line, }which is red perpendicular to itself, and has usual
slope $\pm1.$

It turns out that for any triangle $\overline{A_{1}A_{2}A_{3}}$ the three red
altitudes also intersect, now in a point called the \textbf{red orthocenter}
and denoted $O_{r},$ and the three perpendicular bisectors also intersect in a
point called the \textbf{red circumcenter} and denoted $C_{r}$. This latter
point is the center of the unique red circle through the three points of the
triangle, where a \textbf{red circle} is given by an equation of the form
\[
\left(  x-x_{0}\right)  ^{2}-\left(  y-y_{0}\right)  ^{2}=K.
\]
This is what we would usually call a rectangular hyperbola, with axes in the
red null directions.

\begin{center}
\includegraphics[
height=3.2352in,
width=2.9795in
]
{ChromoEx2Red.eps}
\end{center}

This diagram shows the same triangle $\overline{A_{1}A_{2}A_{3}}$, as well as
the red circumcircle, the red nine-point circle and the red orthocenter,
circumcenter, nine-point center and centroid $G$, the latter being independent
of colour.

Note that these points all lie on a line---the \textbf{red Euler line}, and
the affine relationships between these points is exactly the same as for the
blue Euler line, so that for example $N_{r}$ is the midpoint of $\overline
{O_{r}C_{r}}$.

In the classical framework, there are some difficulties in setting up this
relativistic geometry, as `distance' and `angle' are problematic. In universal
geometry one regards the \textit{quadratic form }as primary, not its
\textit{square root}, and by expressing everything in terms of the algebraic
concepts of \textit{quadrance }and \textit{spread}, Euclidean geometry can be
built up so as to allow generalization to the relativistic framework, and
indeed to geometries built from other quadratic forms.

This approach was introduced recently in \cite{Wildberger}, see also
\cite{Wildberger5}, and works over a general field with characteristic two
excluded for technical reasons, as shown in \cite{Wildberger2}. The
possibility of relativistic geometries over other fields seems particularly attractive.

There is a third geometry, that associated to the \textbf{green quadratic
form} $2xy.$ Two lines are green perpendicular when one is the ordinary
Euclidean reflection of the other in a line parallel to the axes, the latter
being a \textbf{green null line}.

Since $x^{2}-y^{2}$ and $2xy$ are conjugate by a simple change of variable, it
should be no surprise that the corresponding relations between the green
orthocenter $O_{g},$ green circumcenter $C_{g}$, green nine-point center
$N_{g}$ and the centroid $G$ hold as well. Here is the relevant diagram for
our triangle $\overline{A_{1}A_{2}A_{3}}.$

\begin{center}
\includegraphics[
height=3.2175in,
width=3.1133in
]
{ChromoEx2Green.eps}
\end{center}

I conjecture that \textit{most theorems of planar Euclidean geometry, when
formulated algebraically in the context of universal geometry, extend to the
red and green situations}. However there are exceptions. For example, over the
`real numbers' there are no equilateral triangles in the red or green
geometries, so Napoleon's theorem and Morley's theorem will not have direct analogs.

Much could be said further to support this conjecture, but this is not what I
wish to pursue here. Instead, let's consider a completely new phenomenon.
Observe what happens when the three diagrams are put together!

\begin{center}
\includegraphics[
height=3.7176in,
width=3.5373in
]
{ChromoEx2.eps}
\end{center}

We get remarkable collinearities, for example between $O_{b},$ $C_{r}$ and
$O_{g},$ and between $C_{b},$ $N_{r}$ and $C_{g}$, with furthermore $C_{r}$
the midpoint of $\overline{O_{b}O_{g}},$ and $N_{r}$ the midpoint of
$\overline{C_{b}C_{g}}$. Also we observe that, for example, $O_{r}$ and $O_{g}
$ lie on the blue circumcircle of $\overline{A_{1}A_{2}A_{3}},$ while $C_{r}$
and $C_{g}$ lie on the blue nine-point circle of $\overline{A_{1}A_{2}A_{3}}$.
The three colours generally interact symmetrically, so the same relations hold
if we permute colours.

However there are some aspects of chromogeometry in which this symmetry is
broken. The blue geometry as we shall see behaves somewhat differently from
the red and the green in certain contexts, and when we come to explicit
formulas we will see that the green geometry is often simpler. The red
geometry seems less inclined to distinguish itself.

\section*{Rational trigonometry}

Let's now proceed more formally, beginning with the main definitions and laws
of rational trigonometry. We work over a fixed field, not of characteristic
two, whose elements will be called \textbf{numbers}. The \textbf{plane }will
consist of the standard vector space of dimension two over this field. A
\textbf{point}, or \textbf{vector}, is an ordered pair $A\equiv\left[
x,y\right]  $ of numbers. The origin is denoted $O\equiv\left[  0,0\right]  $.

A \textbf{line} is a proportion $l\equiv\left\langle a:b:c\right\rangle $
where $a$ and $b$ are not both zero. The point $A\equiv\left[  x,y\right]  $
\textbf{lies on} the line $l\equiv\left\langle a:b:c\right\rangle $, or
equivalently the line $l$ \textbf{passes through} the point $A$, precisely
when
\[
ax+by+c=0.
\]

This is not the only possible convention, and the reader should be aware that
it is prejudiced towards the usual Euclidean (blue) geometry. For any two
points $A_{1}\equiv\left[  x_{1},y_{1}\right]  $ and $A_{2}\equiv\left[
x_{2},y_{2}\right]  $ there is a unique line $l\equiv A_{1}A_{2}$ which passes
through them both. Specifically we have
\[
A_{1}A_{2}=\left\langle y_{1}-y_{2}:x_{2}-x_{1}:x_{1}y_{2}-x_{2}
y_{1}\right\rangle .
\]
Three points $\left[  x_{1},y_{1}\right]  $, $\left[  x_{2},y_{2}\right]  $
and $\left[  x_{3},y_{3}\right]  $ are \textbf{collinear} precisely when they
lie on the same line, which amounts to the condition
\begin{equation}
x_{1}y_{2}-x_{1}y_{3}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{2}y_{1}
=0.\label{AreaFormula}
\end{equation}
Three lines $\left\langle a_{1}:b_{1}:c_{1}\right\rangle $, $\left\langle
a_{2}:b_{2}:c_{2}\right\rangle $ and $\left\langle a_{3}:b_{3}:c_{3}
\right\rangle $ are \textbf{concurrent }precisely when they pass through the
same point, which amounts to the condition
\[
a_{1}b_{2}c_{3}-a_{1}b_{3}c_{2}+a_{2}b_{3}c_{1}-a_{3}b_{2}c_{1}+a_{3}
b_{1}c_{2}-a_{2}b_{1}c_{3}=0.
\]

Fix a symmetric bilinear form, denoted by the dot product $A_{1}\cdot A_{2}$.
In practise we will take this bilinear form to be non-degenerate. The line
$A_{1}A_{2}$ is \textbf{perpendicular} to the line $B_{1}B_{2}$ precisely
when
\[
\left(  A_{2}-A_{1}\right)  \cdot\left(  B_{2}-B_{1}\right)  =0.
\]
A point $A$ is a\textbf{\ null point }or\textbf{\ null vector} precisely when
$A\cdot A=0.$ The \textbf{origin} $O$ is always a null point, but there may be
others. A line $A_{1}A_{2}$ is a \textbf{null line} precisely when the vector
$A_{2}-A_{1}$ is a null vector.

A set $\left\{  A_{1},A_{2},A_{3}\right\}  $ of three distinct non-collinear
points is a \textbf{triangle} and is denoted $\overline{A_{1}A_{2}A_{3}}$. The
\textbf{lines} of the triangle are $l_{3}\equiv A_{1}A_{2},$ $l_{2}\equiv
A_{1}A_{3}$ and $l_{1}\equiv A_{2}A_{3}.$ A triangle is \textbf{non-null}
precisely when each of its lines is non-null. A \textbf{side} of the triangle
is a subset of $\left\{  A_{1},A_{2},A_{3}\right\}  $ with two elements, and
is denoted $\overline{A_{1}A_{2}}$ etc. A \textbf{vertex} of the triangle is a
subset of $\left\{  l_{1},l_{2},l_{3}\right\}  $ with two elements, and is
denoted $\overline{l_{1}l_{2}}$ etc.

The \textbf{quadrance} between the points $A_{1}$ and $A_{2}$ is the number
\[
Q\left(  A_{1},A_{2}\right)  \equiv\left(  A_{2}-A_{1}\right)  \cdot\left(
A_{2}-A_{1}\right)  .
\]
The line $A_{1}A_{2}$ is a null line precisely when $Q\left(  A_{1}
,A_{2}\right)  =0$.

The \textbf{spread} between the non-null lines $A_{1}A_{2}$ and $B_{1}B_{2}$
is the number
\[
s\left(  A_{1}A_{2},B_{1}B_{2}\right)  \equiv1-\frac{\left(  \left(
A_{2}-A_{1}\right)  \cdot\left(  B_{2}-B_{1}\right)  \right)  ^{2}}{Q\left(
A_{1},A_{2}\right)  Q\left(  B_{1},B_{2}\right)  }.
\]
This is independent of the choice of points lying on the two lines. Two
non-null lines are perpendicular precisely when the spread between them is
$1.$

Here are the five main laws of planar rational trigonometry in this general
setting, replacing the usual Sine law, Cosine law etc. Proofs can be found in
\cite{Wildberger2}. Aside from giving new directions to geometry, these laws
have the potential to change the teaching of high school mathematics, because
they are simpler, and allow faster and more accurate calculations in practical
problems. But the advantage for us here is that they \textit{hold for general
quadratic forms}, and in particular for each of the blue, red and green geometries.

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

\end{theorem}

\begin{theorem}
[Pythagoras' theorem]For $A_{1},$ $A_{2}$ and $A_{3}$ three distinct points,
$A_{1}A_{3}$ is perpendicular to $A_{2}A_{3}$ precisely when the quadrances
$Q_{1}\equiv Q\left(  A_{2},A_{3}\right)  $, $Q_{2}\equiv Q\left(  A_{1}
,A_{3}\right)  $ and $Q_{3}\equiv Q\left(  A_{1},A_{2}\right)  $ satisfy
\[
Q_{1}+Q_{2}=Q_{3}.
\]

\end{theorem}

\begin{theorem}
[Spread law]Suppose the non-null triangle $\overline{A_{1}A_{2}A_{3}}$ has
quadrances $Q_{1}\equiv Q\left(  A_{2},A_{3}\right)  $, $Q_{2}\equiv Q\left(
A_{1},A_{3}\right)  $ and $Q_{3}\equiv Q\left(  A_{1},A_{2}\right)  $, and
spreads $s_{1}\equiv s\left(  A_{1}A_{2},A_{1}A_{3}\right)  $, $s_{2}\equiv
s\left(  A_{2}A_{1},A_{2}A_{3}\right)  $ and $s_{3}\equiv s\left(  A_{3}
A_{1},A_{3}A_{2}\right)  $. Then
\[
\frac{s_{1}}{Q_{1}}=\frac{s_{2}}{Q_{2}}=\frac{s_{3}}{Q_{3}}.
\]

\end{theorem}

\begin{theorem}
[Cross law]Suppose the non-null triangle $\overline{A_{1}A_{2}A_{3}}$ has
quadrances $Q_{1}\equiv Q\left(  A_{2},A_{3}\right)  $, $Q_{2}\equiv Q\left(
A_{1},A_{3}\right)  $ and $Q_{3}\equiv Q\left(  A_{1},A_{2}\right)  $, and
spreads $s_{1}\equiv s\left(  A_{1}A_{2},A_{1}A_{3}\right)  $, $s_{2}\equiv
s\left(  A_{2}A_{1},A_{2}A_{3}\right)  $ and $s_{3}\equiv s\left(  A_{3}
A_{1},A_{3}A_{2}\right)  $. Then
\[
\left(  Q_{1}+Q_{2}-Q_{3}\right)  ^{2}=4Q_{1}Q_{2}\left(  1-s_{3}\right)  .
\]

\end{theorem}

Note that the Cross law includes as special cases both the Triple quad formula
and Pythagoras' theorem. The next result is the algebraic analog to the sum of
the angles in a triangle formula.

\begin{theorem}
[Triple spread formula]Suppose the non-null triangle $\overline{A_{1}
A_{2}A_{3}}$ has spreads $s_{1}\equiv s\left(  A_{1}A_{2},A_{1}A_{3}\right)
$, $s_{2}\equiv s\left(  A_{2}A_{1},A_{2}A_{3}\right)  $ and $s_{3}\equiv
s\left(  A_{3}A_{1},A_{3}A_{2}\right)  $. Then
\[
\left(  s_{1}+s_{2}+s_{3}\right)  ^{2}=2\left(  s_{1}^{2}+s_{2}^{2}+s_{3}
^{2}\right)  +4s_{1}s_{2}s_{3}.
\]

\end{theorem}

A useful observation is that the Triple spread formula shows that $s_{3}=1$
implies that
\[
s_{1}+s_{2}=1.
\]

\section*{Three fold symmetry}

The vectors $A_{1}\equiv\left[  x_{1},y_{1}\right]  $ and $A_{2}\equiv\left[
x_{2},y_{2}\right]  $ are \textbf{parallel} precisely when
\[
x_{1}y_{2}-x_{2}y_{1}=0.
\]
We will be interested in three main examples of symmetric bilinear forms.
Define the \textbf{blue} dot product
\[
\left[  x_{1},y_{1}\right]  \cdot_{b}\left[  x_{2},y_{2}\right]  \equiv
x_{1}x_{2}+y_{1}y_{2},
\]
the \textbf{red} dot product
\[
\left[  x_{1},y_{1}\right]  \cdot_{r}\left[  x_{2},y_{2}\right]  \equiv
x_{1}x_{2}-y_{1}y_{2}
\]
and the \textbf{green} dot product
\[
\left[  x_{1},y_{1}\right]  \cdot_{g}\left[  x_{2},y_{2}\right]  =x_{1}
y_{2}+x_{2}y_{1}.
\]

Note that between them these four expressions give all possible bilinear
expressions in the two vectors that involve only coefficients $\pm1,$ up to
sign. Two lines $l_{1}$ and $l_{2}$ are \textbf{blue},\textbf{\ red} and
\textbf{green perpendicular} respectively precisely when they are
perpendicular with respect to the blue, red and green forms. For lines
$l_{1}\equiv\left\langle a_{1}:b_{1}:c_{1}\right\rangle $ and $l_{2}
\equiv\left\langle a_{2}:b_{2}:c_{2}\right\rangle $ these conditions amount to
the respective conditions
\[
a_{1}a_{2}+b_{1}b_{2}=0~~\mathrm{[blue]}
\]
\textbf{\ }
\[
a_{1}a_{2}-b_{1}b_{2}=0~~\mathrm{[red]}
\]
and
\[
a_{1}b_{2}+b_{1}a_{2}=0~~\mathrm{[green].}
\]

In terms of coordinates, the formulas for the\textbf{\ blue}, \textbf{red} and
\textbf{green quadrances} between points $A_{1}\equiv\left[  x_{1}
,y_{1}\right]  $ and $A_{2}\equiv\left[  x_{2},y_{2}\right]  $ are
\begin{align*}
Q_{b}\left(  A_{1},A_{2}\right)   & =\left(  x_{2}-x_{1}\right)  ^{2}+\left(
y_{2}-y_{1}\right)  ^{2}\\
Q_{r}\left(  A_{1},A_{2}\right)   & =\left(  x_{2}-x_{1}\right)  ^{2}-\left(
y_{2}-y_{1}\right)  ^{2}\\
Q_{g}\left(  A_{1},A_{2}\right)   & =2\left(  x_{2}-x_{1}\right)  \left(
y_{2}-y_{1}\right)  .
\end{align*}

\begin{theorem}
[Coloured quadrances]For any points $A_{1}$ and $A_{2}$ let $Q_{b}$, $Q_{r}$
and $Q_{g}$ be the blue, red and green quadrances between $A_{1}$ and $A_{2}$
respectively. Then
\[
Q_{b}^{2}=Q_{r}^{2}+Q_{g}^{2}.
\]

\end{theorem}

\begin{proof}
This is a consequence of the identity
\[
\left(  r^{2}+s^{2}\right)  ^{2}=\left(  r^{2}-s^{2}\right)  ^{2}+\left(
2rs\right)  ^{2}.
\]

\end{proof}

The formulas for the \textbf{blue}, \textbf{red} and \textbf{green spreads}
between lines $l_{1}\equiv\left\langle a_{1}:b_{1}:c_{1}\right\rangle $ and
$l_{2}\equiv\left\langle a_{2}:b_{2}:c_{2}\right\rangle $ are
\begin{align*}
s_{b}\left(  l_{1},l_{2}\right)   & =1-\frac{\left(  a_{1}a_{2}+b_{1}
b_{2}\right)  ^{2}}{\left(  a_{1}^{2}+b_{1}^{2}\right)  \left(  a_{2}
^{2}+b_{2}^{2}\right)  }=\allowbreak\frac{\left(  a_{1}b_{2}-a_{2}
b_{1}\right)  ^{2}\allowbreak}{\left(  a_{2}^{2}+b_{2}^{2}\right)  \left(
a_{1}^{2}+b_{1}^{2}\right)  }\\
s_{r}\left(  l_{1},l_{2}\right)   & =1-\frac{\left(  b_{1}b_{2}-a_{1}
a_{2}\right)  ^{2}}{\left(  b_{1}^{2}-a_{1}^{2}\right)  \left(  b_{2}
^{2}-a_{2}^{2}\right)  }=-\frac{\left(  a_{1}b_{2}-a_{2}b_{1}\right)
^{2}\allowbreak}{\left(  a_{2}^{2}-b_{2}^{2}\right)  \left(  a_{1}^{2}
-b_{1}^{2}\right)  }\\
s_{g}\left(  l_{1},l_{2}\right)   & =1-\frac{\left(  -a_{1}b_{2}-a_{2}
b_{1}\right)  ^{2}}{4a_{1}a_{2}b_{1}b_{2}}=\allowbreak-\frac{\left(
a_{1}b_{2}-a_{2}b_{1}\right)  ^{2}\allowbreak}{4a_{1}a_{2}b_{1}b_{2}}.
\end{align*}
Note carefully the \textit{minus signs} that precede the final expressions in
the red and green cases.

\begin{theorem}
[Coloured spreads]For any lines $l_{1}$ and $l_{2}$ let $s_{b}$, $s_{r}$ and
$s_{g}$ be the blue, red and green spreads between $l_{1}$ and $l_{2}$
respectively. Then
\[
\frac{1}{s_{b}}+\frac{1}{s_{r}}+\frac{1}{s_{g}}=2.
\]

\end{theorem}

\textbf{Proof.} This is a consequence of the identity
\[
\left(  a_{1}^{2}+b_{1}^{2}\right)  \left(  a_{2}^{2}+b_{2}^{2}\right)
-\left(  a_{1}^{2}-b_{1}^{2}\right)  \left(  a_{2}^{2}-b_{2}^{2}\right)
-4a_{1}a_{2}b_{1}b_{2}=2\left(  a_{1}b_{2}-a_{2}b_{1}\right)  ^{2}.

{\hspace{.1in} \rule{0.5em}{0.5em}}

\]

\section*{Quadrances}

The most important single quantity associated to a triangle $\overline
{A_{1}A_{2}A_{3}}$ with quadrances $Q_{1},Q_{2}$ and $Q_{3}$ is the
\textbf{quadrea} $\mathcal{A}$ defined by
\[
\mathcal{A}\equiv\left(  Q_{1}+Q_{2}+Q_{3}\right)  ^{2}-2\left(  Q_{1}
^{2}+Q_{2}^{2}+Q_{3}^{2}\right)  .
\]
By the Triple quad formula this is a measure of the non-collinearity of the
points $A_{1},A_{2}$ and $A_{3}$. We denote by $\mathcal{A}_{b},\mathcal{A}
_{r}$ and $\mathcal{A}_{g}$ the respective \textbf{blue},\textbf{\ red
}and\textbf{\ green quadreas} of a triangle $\overline{A_{1}A_{2}A_{3}}.$

\begin{theorem}
[Quadrea ]For three points $A_{1}\equiv\left[  x_{1},y_{1}\right]  $,
$A_{2}\equiv\left[  x_{2},y_{2}\right]  $ and $A_{3}\equiv\left[  x_{3}
,y_{3}\right]  $, the three quadreas $\mathcal{A}_{b},\mathcal{A}_{r}$ and
$\mathcal{A}_{g}$ satisfy
\[
\mathcal{A}_{b}=-\mathcal{A}_{r}=-\mathcal{A}_{g}=4\left(  x_{1}y_{2}
-x_{1}y_{3}+x_{2}y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{2}y_{1}\right)  ^{2}.
\]

\end{theorem}

\begin{proof}
A calculation.
\end{proof}

So each quadrea of a triangle is $\pm16$ times the square of its signed area,
the latter being defined purely in an affine setting, without any need for
metrical choices.

We now adopt the convention that if no proof is given, `a calculation' is to
be assumed.

\section*{Altitudes}

\begin{theorem}
[Altitudes to a line]For any point $A$ and any line $l,$ there exist unique
lines $n_{b},n_{r}$ and $n_{g}$ through $A$ which are respectively blue, red
and green perpendicular to $l$. If $A\equiv\left[  x_{0},y_{0}\right]  $ and
$l\equiv\left\langle a:b:c\right\rangle $ then
\begin{align*}
n_{b}  & =\left\langle b:-a:-bx_{0}+ay_{0}\right\rangle \\
n_{r}  & =\left\langle b:a:-bx_{0}-ay_{0}\right\rangle \\
n_{g}  & =\left\langle a:-b:-ax_{0}+by_{0}\right\rangle .

{\hspace{.1in} \rule{0.5em}{0.5em}}

\end{align*}

\end{theorem}

The lines $n_{b},n_{r},n_{g}$ are respectively the \textbf{blue},\textbf{\ red
}and\textbf{\ green altitudes} from $A$ to $l$, and they intersect $l$ at the
\textbf{feet, }provided that $l$ is non-null.

\begin{theorem}
[Perpendicularity of altitudes]For any point $A$ and any line $l,$ let
$n_{b},n_{r},n_{g}$ be the blue, red and green altitudes from $A$ to $l$
respectively. Then $n_{b}$ and $n_{r}$ are green perpendicular, $n_{r}$ and
$n_{g}$ are blue perpendicular, and $n_{g}$ and $n_{b}$ are red perpendicular.
$

{\hspace{.1in} \rule{0.5em}{0.5em}}

$

\begin{center}
\includegraphics[
height=1.6212in,
width=2.3248in
]
{Altitudes.eps}
\end{center}

\end{theorem}

The figure shows an example of the three colour altitudes from a point $A$ to
a line $l,$ and their feet $F_{b},$ $F_{r}$ and $F_{g}.$

\begin{theorem}
[Pythagorean means]Let $l\equiv\left\langle a:b:c\right\rangle $ be a line
which is non-null in each of the three geometries. If $A$ is a point and
$F_{b},F_{r}$ and $F_{g}$ are the respective feet of the altitudes
$n_{b},n_{r}$ and $n_{g}$ from $A$ to $l$, then we have the affine relation
\[
F_{b}=\frac{\left(  a^{2}-b^{2}\right)  ^{2}}{\left(  a^{2}+b^{2}\right)
^{2}}F_{r}+\frac{4b^{2}a^{2}}{\left(  a^{2}+b^{2}\right)  ^{2}}F_{g}.
\]

\end{theorem}

\begin{proof}
Suppose that $A\equiv\left[  x_{0},y_{0}\right]  $ and $l\equiv\left\langle
a:b:c\right\rangle $. Elimination yields
\begin{align*}
F_{b}  & =\left[  \frac{b^{2}x_{0}-aby_{0}-ac}{a^{2}+b^{2}},\frac
{-abx_{0}+a^{2}y_{0}-bc}{a^{2}+b^{2}}\right] \\
F_{r}  & =\left[  \frac{-b^{2}x_{0}-aby_{0}-ca}{a^{2}-b^{2}},\frac
{abx_{0}+a^{2}y_{0}+bc}{a^{2}-b^{2}}\right] \\
F_{g}  & =\left[  \frac{ax_{0}-by_{0}-c}{2a},\frac{-ax_{0}+by_{0}-c}
{2b}\right]
\end{align*}
from which we deduce the result.
\end{proof}

Note again the connection with Pythagorean triples.

\section*{Anti-symmetric polynomials}

We use notation for anti-symmetric polynomials introduced in \cite[page
28]{Wildberger}. If $m$ is a monomial in the variables $x_{1},x_{2}
,x_{3},y_{1},y_{2},y_{3},z_{1},z_{2},z_{3},\cdots$ with all indices in the
range $1,2$ and $3$, then we define $\left[  m\right]  ^{-}$ to be the
\textit{anti-symmetric }polynomial consisting of the sum of all monomials
obtained from $m$ by performing all six permutations of the indices and
multiplying each term by the sign of the corresponding permutation.

We often write such polynomials in the order described by the successive
transpositions
\[
\left(  23\right)  ,\left(  12\right)  ,\left(  23\right)  ,\left(  12\right)
,\left(  23\right)  .
\]

For example
\begin{align*}
\left[  x_{1}y_{2}\right]  ^{-}  & \equiv x_{1}y_{2}-x_{1}y_{3}+x_{2}
y_{3}-x_{3}y_{2}+x_{3}y_{1}-x_{2}y_{1}\\
\left[  x_{1}^{2}x_{2}y_{2}\right]  ^{-}  & \equiv x_{1}^{2}x_{2}y_{2}
-x_{1}^{2}x_{3}y_{3}+x_{2}^{2}x_{3}y_{3}-x_{3}^{2}x_{2}y_{2}+x_{3}^{2}
x_{1}y_{1}-x_{2}^{2}x_{1}y_{1}\\
\left[  x_{1}^{3}y_{1}\right]  ^{-}  & \equiv x_{1}^{3}y_{1}-x_{1}^{3}
y_{1}+x_{2}^{3}y_{2}-x_{3}^{3}y_{3}+x_{3}^{3}y_{3}-x_{2}^{3}y_{2}=0.
\end{align*}
The polynomial $\left[  x_{1}y_{2}\right]  ^{-}$ is of particular importance,
since it occurs in (\ref{AreaFormula}), is twice the signed area of the
triangle $\overline{A_{1}A_{2}A_{3}}$, appears in the Quadrea theorem, and is
often a denominator in formulas in the subject.

\section*{Orthocenters}

Given a triangle $\overline{A_{1}A_{2}A_{3}},$ for each point $A_{m}$, $m=1,2
$ and $3$ we may construct the blue, red and green altitudes $a_{m}^{b}
,a_{m}^{r}$ and $a_{m}^{g}$ respectively to the opposite side.

\begin{theorem}
[Orthocenter formulas]The three blue altitudes $a_{1}^{b},a_{2}^{b}$ and
$a_{3}^{b}$ meet in a point $O_{b}$ called the \textbf{blue orthocenter}. The
three red altitudes $a_{1}^{r},a_{2}^{r}$ and $a_{3}^{r}$ meet in a point
$O_{r}$ called the \textbf{red orthocenter}. The three green altitudes
$a_{1}^{g},a_{2}^{g}$ and $a_{3}^{g}$ meet in a point $O_{g}$ called the
\textbf{green orthocenter}. For $A_{1}\equiv\left[  x_{1},y_{1}\right]  $,
$A_{2}\equiv\left[  x_{2},y_{2}\right]  $ and $A_{3}\equiv\left[  x_{3}
,y_{3}\right]  $ these points are given by
\begin{align*}
O_{b}  & =\left[  \frac{\left[  x_{1}x_{2}y_{2}\right]  ^{-}+\left[
y_{1}y_{2}^{2}\right]  ^{-}}{\left[  x_{1}y_{2}\right]  ^{-}},\frac{\left[
x_{1}y_{1}y_{2}\right]  ^{-}+\left[  x_{1}^{2}x_{2}\right]  ^{-}}{\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
O_{r}  & =\left[  \frac{\left[  x_{1}x_{2}y_{2}\right]  ^{-}-\left[
y_{1}y_{2}^{2}\right]  ^{-}}{\left[  x_{1}y_{2}\right]  ^{-}},\frac{\left[
x_{1}y_{1}y_{2}\right]  ^{-}-\left[  x_{1}^{2}x_{2}\right]  ^{-}}{\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
O_{g}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}+\left[  x_{1}
x_{2}y_{1}\right]  ^{-}}{\left[  x_{1}y_{2}\right]  ^{-}},\frac{\left[
x_{1}y_{2}^{2}\right]  ^{-}-\left[  x_{1}y_{1}y_{2}\right]  ^{-}}{\left[
x_{1}y_{2}\right]  ^{-}}\right]  .

{\hspace{.1in} \rule{0.5em}{0.5em}}

\end{align*}

\end{theorem}

\section*{Circumcenters}

When $A_{1}$ and $A_{2}$ are distinct points with $l=A_{1}A_{2},$ and $A$ is
the midpoint of $A_{1}$ and $A_{2}$, then the blue, red and green altitudes
from $A$ to $l$ are respectively called the \textbf{blue},\textbf{\ red
}and\textbf{\ green perpendicular bisectors }of the side $\overline{A_{1}
A_{2}}$.

\begin{theorem}
[Perpendicular bisectors]If $A_{1}\equiv\left[  x_{1},y_{1}\right]  $ and
$A_{2}\equiv\left[  x_{2},y_{2}\right]  $ are distinct points then the blue,
red and green perpendicular bisectors of $\overline{A_{1}A_{2}}$ have
respective equations
\begin{align*}
\left(  x_{1}-x_{2}\right)  x+\left(  y_{1}-y_{2}\right)  y  & =\frac
{x_{1}^{2}-x_{2}^{2}+y_{1}^{2}-y_{2}^{2}}{2}\\
\left(  x_{1}-x_{2}\right)  x-\left(  y_{1}-y_{2}\right)  y  & =\frac
{x_{1}^{2}-x_{2}^{2}-y_{1}^{2}+y_{2}^{2}}{2}\\
\left(  y_{2}-y_{1}\right)  x+\left(  x_{2}-x_{1}\right)  y  & =y_{2}
x_{2}-x_{1}y_{1}.

{\hspace{.1in} \rule{0.5em}{0.5em}}

\end{align*}

\end{theorem}

Given a triangle $\overline{A_{1}A_{2}A_{3}}$ we may construct the blue, red
and green perpendicular bisectors of the three sides, denoted by $b_{m}
^{b},b_{m}^{r}$ and $b_{m}^{g}$ respectively for $m=1,2$ and $3,$ where
$b_{1}^{b}$ for example is the blue perpendicular bisector of the side
$\overline{A_{2}A_{3}}$ and so on.$\allowbreak$

\begin{theorem}
[Circumcenter formulas]The three blue perpendicular bisectors $b_{1}^{b}
,b_{2}^{b}$ and $b_{3}^{b}$ meet in a point $C_{b}$ called the \textbf{blue
circumcenter}. The three red perpendicular bisectors $b_{1}^{r},b_{2}^{r}$ and
$b_{3}^{r}$ meet in a point $C_{r}$ called the \textbf{red circumcenter}. The
three green perpendicular bisectors $b_{1}^{g},b_{2}^{g}$ and $b_{3}^{g}$ meet
in a point $C_{g}$ called the \textbf{green circumcenter}. For $A_{1}
\equiv\left[  x_{1},y_{1}\right]  $, $A_{2}\equiv\left[  x_{2},y_{2}\right]  $
and $A_{3}\equiv\left[  x_{3},y_{3}\right]  $, these points are given by
\begin{align*}
C_{b}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}+\left[  y_{1}
^{2}y_{2}\right]  ^{-}}{2\left[  x_{1}y_{2}\right]  ^{-}},\frac{\left[
x_{1}y_{2}^{2}\right]  ^{-}+\left[  x_{1}x_{2}^{2}\right]  ^{-}}{2\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
C_{r}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}-\left[  y_{1}
^{2}y_{2}\right]  ^{-}}{2\left[  x_{1}y_{2}\right]  ^{-}},\frac{\left[
x_{1}y_{2}^{2}\right]  ^{-}-\left[  x_{1}x_{2}^{2}\right]  ^{-}}{2\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
C_{g}  & =\left[  \frac{\left[  x_{1}x_{2}y_{2}\right]  ^{-}}{\left[
x_{1}y_{2}\right]  ^{-}},\frac{\left[  x_{1}y_{1}y_{2}\right]  ^{-}}{\left[
x_{1}y_{2}\right]  ^{-}}\right]  .

{\hspace{.1in} \rule{0.5em}{0.5em}}

\end{align*}

\end{theorem}

\begin{theorem}
[Circumcenters as midpoints]For any triangle a coloured circumcenter is the
midpoint of the two orthocenters of the other two colours.
\end{theorem}

\begin{proof}
This follows from the Orthocenter formulas and Circumcenter formulas.
\end{proof}

\section*{Nine-point centres}

Suppose that the respective midpoints of a triangle $\overline{A_{1}A_{2}
A_{3}}$ are $M_{m}$ for $m=1,2$ and $3$, where $M_{1}$ is the midpoint of the
side $\overline{A_{2}A_{3}}$ and so on. We let $N_{b},N_{r}$ and $N_{g}$ be
the blue, red and green circumcenters respectively of the triangle
$\overline{M_{1}M_{2}M_{3}}$, and call these the \textbf{blue},\textbf{\ red
}and\textbf{\ green nine-point centers} of the original triangle
$\overline{A_{1}A_{2}A_{3}}$.

\begin{theorem}
[Nine-point center formulas]For $A_{1}\equiv\left[  x_{1},y_{1}\right]  $,
$A_{2}\equiv\left[  x_{2},y_{2}\right]  $ and $A_{3}\equiv\left[  x_{3}
,y_{3}\right]  $, the blue, red and green nine-point centers of $\overline
{A_{1}A_{2}A_{3}}$ are
\begin{align*}
N_{b}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}-\left[  y_{1}
^{2}y_{2}\right]  ^{-}+2\left[  x_{1}x_{2}y_{2}\right]  ^{-}}{4\left[
x_{1}y_{2}\right]  ^{-}},\frac{\left[  x_{1}y_{2}^{2}\right]  ^{-}-\left[
x_{1}x_{2}^{2}\right]  ^{-}+2\left[  x_{1}y_{1}y_{2}\right]  ^{-}}{4\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
N_{r}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}+\left[  y_{1}
^{2}y_{2}\right]  ^{-}+2\left[  x_{1}x_{2}y_{2}\right]  ^{-}}{4\left[
x_{1}y_{2}\right]  ^{-}},\frac{\left[  x_{1}y_{2}^{2}\right]  ^{-}+\left[
x_{1}x_{2}^{2}\right]  ^{-}+2\left[  x_{1}y_{1}y_{2}\right]  ^{-}}{4\left[
x_{1}y_{2}\right]  ^{-}}\right] \\
N_{g}  & =\left[  \frac{\left[  x_{1}^{2}y_{2}\right]  ^{-}}{2\left[
x_{1}y_{2}\right]  ^{-}},\frac{\left[  x_{1}y_{2}^{2}\right]  ^{-}}{2\left[
x_{1}y_{2}\right]  ^{-}}\right]  .

{\hspace{.1in} \rule{0.5em}{0.5em}}

\end{align*}

\end{theorem}

\begin{theorem}
[Nine-point centers as midpoints]In any triangle a coloured nine-point center
is the midpoint of the two circumcenters of the other two colours.
\end{theorem}

\begin{proof}
This follows from the Circumcenter formulas and Nine-point center formulas.
\end{proof}

\section*{The $\Omega$-triangle and the Euler lines}

The $\Omega$\textbf{-triangle }of\textbf{\ }a triangle $\overline{A_{1}
A_{2}A_{3}}$ is the triangle $\Omega\equiv\Omega\left(  \overline{A_{1}
A_{2}A_{3}}\right)  \equiv\overline{O_{b}O_{r}O_{g}}$ of orthocenters of
$\overline{A_{1}A_{2}A_{3}}$. From the theorems of the last two sections, the
corresponding midpoints of the sides of $\Omega$ are $C_{b},C_{r}$ and
$C_{g},$ with $C_{b}$ the midpoint of $O_{r}$ and $O_{g}$ etc., and the
midpoints of the triangle $\overline{C_{b}C_{r}C_{g}}$ are $N_{b},N_{r}$ and
$N_{g},$ with $N_{b}$ the midpoint of $C_{r}$ and $C_{g}$ etc. We also know
that the centroid of $\Omega$ is the same as the centroid $G$ of the original
triangle $\overline{A_{1}A_{2}A_{3}}$.

\begin{theorem}
[Blue Euler line]The points $O_{b},N_{b},G$ and $C_{b}$ lie on a line called
the \textbf{blue Euler line}. Furthermore $N_{b}$ is the midpoint of $O_{b}$
and $C_{b},$ and we have the affine relations
\[
G=\frac{1}{3}O_{b}+\frac{2}{3}C_{b}=\frac{1}{3}C_{b}+\frac{2}{3}N_{b}.
\]

\end{theorem}

\begin{proof}
This follows from the Orthocenter, Circumcenter and Nine-point center formulas.
\end{proof}

\begin{theorem}
[Red Euler line]The points $O_{r},N_{r},G$ and $C_{r}$ lie on a line called
the \textbf{red Euler line}. Furthermore $N_{r}$ is the midpoint of $O_{r}$
and $C_{r},$ and
\[
G=\frac{1}{3}O_{r}+\frac{2}{3}C_{r}=\frac{1}{3}C_{r}+\frac{2}{3}N_{r}.
\]

\end{theorem}

\begin{proof}
Likewise.
\end{proof}

\begin{theorem}
[Green Euler line]The points $O_{g},N_{g},G$ and $C_{g}$ lie on a line called
the \textbf{green Euler line}. Furthermore $N_{g}$ is the midpoint of $O_{g}$
and $C_{g}, $ and
\[
G=\frac{1}{3}O_{g}+\frac{2}{3}C_{g}=\frac{1}{3}C_{g}+\frac{2}{3}N_{g}.
\]

\end{theorem}

\begin{proof}
Likewise.
\end{proof}

The geometry of the $\Omega$-triangle clarifies the various ratios occurring
along points on the Euler lines, since these are just the medians of $\Omega$.
The lines joining the circumcenters are the lines of the medial triangle of
$\Omega,$ and so are parallel to the lines of $\Omega$.

\section*{Circles}

A \textbf{blue, red or green circle} is an equation $c$ in $x$ and $y$ of the
form
\begin{align*}
\left(  x-x_{0}\right)  ^{2}+\left(  y-y_{0}\right)  ^{2}  & =K\\
\left(  x-x_{0}\right)  ^{2}-\left(  y-y_{0}\right)  ^{2}  & =K\\
2\left(  x-x_{0}\right)  \left(  y-y_{0}\right)   & =K
\end{align*}
respectively, where the point $\left[  x_{0},y_{0}\right]  $ is then unique
and called the \textbf{centre} of $c$, and $K$ is the \textbf{quadrance} of
$c$. A blue circle is an ordinary Euclidean circle. Red and green circles are
more usually described as rectangular hyperbolas. A red circle has asymptotes
parallel to the lines with equations $y=\pm x,$ and a green circle has
asymptotes parallel to the coordinate axes.

\begin{theorem}
[Circumcircles]If $A_{1},A_{2}$ and $A_{3}$ are three distinct non-collinear
points, then there are unique blue, red and green circles passing through
$A_{1},A_{2}$ and $A_{3}$. $

{\hspace{.1in} \rule{0.5em}{0.5em}}

$
\end{theorem}

The circles above will be called respectively the \textbf{blue},\textbf{\ red
}and\textbf{\ green circumcircles} of the triangle $\overline{A_{1}A_{2}A_{3}
},$ while the circumcircles of the triangle of midpoints $M_{1},M_{2},M_{3}$
of the triangle $\overline{A_{1}A_{2}A_{3}}$ will be called respectively the
\textbf{blue},\textbf{\ red }and \textbf{green nine-point circles }of the
triangle $\overline{A_{1}A_{2}A_{3}}$.

\begin{theorem}
[Orthocenters on circumcircles]Any coloured orthocenter of a triangle
$\overline{A_{1}A_{2}A_{3}}$ lies on the circumcircles of the other two
colours. $

{\hspace{.1in} \rule{0.5em}{0.5em}}

$
\end{theorem}

\begin{theorem}
[Nine-point circles]Any coloured nine-point circle of a triangle
$\overline{A_{1}A_{2}A_{3}}$ passes through the feet of the altitudes of that
colour, as well as the midpoints of the segments from the same coloured
orthocenter to the points $A_{1},A_{2}$ and $A_{3}.$ In addition it passes
through the circumcenters of $\overline{A_{1}A_{2}A_{3}}$ of the other two
colours. $

{\hspace{.1in} \rule{0.5em}{0.5em}}

$
\end{theorem}

The following figure shows some of the other points on the nine-point circles
of different colours. Others are off the page.

\begin{center}
\includegraphics[
height=3.716in,
width=5.0959in
]
{ChromoNinePoint.eps}
\end{center}

Hopefully this taste of chromogeometry will encourage others to explore this
rich new realm. See \cite{Wildberger3} and \cite{Wildberger 4} for more in
this direction.

config: arxiv


TITLE: Are such functions differentiable?
QUESTION [23 upvotes]: In my recent researches, I encountered functions $f$ satisfying the following functional inequality:
$$
(*)\; f(x)\geq f(y)(1+x-y) \; ; \; x,y\in \mathbb{R}.
$$
Since $f$ is convex (because $\displaystyle f(x)=\sup_y [f(y)+f(y)(x-y)]$), it is left and right differentiable. Also, it is obvious that all functions of the form $f(t)=ce^t$ with $c\geq 0$ satisfy
$(*)$. Now, my questions:
(1) Is $f$ everywhere differentiable?
(2) Are there any other solutions for $(*)$?
(3) Is this functional inequality well-known (any references
(paper, book, website, etc.) for such functional inequalities)?
Thanks in advance

REPLY [43 votes]: Replace $x$ with $x+y$ to get
$f(x+y)\ge f(y)(1+x)$ or $f(x+y)-f(y)\ge xf(y)$.
Replace $y$ with $x+y$ and then interchange $x$ and $y$ to get $f(x+y)-f(y)\le xf(x+y)$.
Together,
$$
xf(y)\le f(x+y)-f(y)\le xf(x+y).
$$
Dividing by $x$ and taking the limit as $x\to0$ implies that $f$ is differentiable with $f'=f$.

REPLY [28 votes]: For any $x$ and for sufficiently large $n$ such that $1+x/n>0$, it holds that
\begin{align}
f(x) &\ge f\left (\frac{(n-1)x}n \right) (1+x/n)\\
&\ge f\left (\frac{(n-2)x}n \right)(1+x/n)^2 \\
&\ge \cdots \ge f(0) \left(1+ \frac x n\right)^n.
\end{align}
by substituting $(x,y)=(x,(n-1)x/n), ((n-1)x/n, (n-2)x/n),...$ in the given equation.
In other words,
$$
f(x) \ge \lim_{n\rightarrow \infty} f(0) \left(1+ \frac x n\right)^n = f(0)\cdot e^x.
$$
On the other hand, for any $y$ and for sufficiently large $n$ such that $1-y/n>0$, we can similarly get the following inequality.
\begin{align}
f(y) &\le f\left( \frac{(n-1)y} n\right) / (1-y/n)\\
&\le \cdots \le f(0)/(1-y/n)^n.
\end{align}
It implies $f(y)\le f(0) \cdot e^y$. Combining these inequalities, we get that $f(x)=f(0) \cdot e^x$ is the only solution as you wanted.


TITLE: Double integral of a product in calculus of variations
QUESTION [5 upvotes]: Let's say I have an integral of the form
$$
  V(u) = \iint\limits_{[0,T]^2}f(x,y)u(x)u(y)\mathrm dx\mathrm dy
$$
which I would like to optimize over smooth functions $u$. For the variation I get
$$
  \delta V(u) \approx \iint\limits_{[0,T]^2}f(x,y)[u(x)\delta u(y) + u(y)\delta u(x)]\mathrm dx\mathrm dy
$$
since I hope the term $\delta u(x)\delta u(y)$ can be disregarded. How can I find the optimal $u$ now? In the 1d case I'd have
$$
  F(u) = \int_0^T G(x,u(x))\mathrm dx \quad \implies \quad \delta F(u) \approx \int_0^TG_u(x, u(x))\delta u(x)\mathrm dx
$$
and by fundamental lemma of calculus of variations I would get $G_u(x, u(x)) = 0$. However, I am not sure whether this passage is extendable to my case. Fwiw, my original case is a bit more complicated: I deal with an integral
$$
  W(u) = V(u) + F(u),
$$
and since I know how to variate the latter part, I hope it is enough to know how to variate the former. Would be happy to hear any suggestions. I tag it as LA, since the $V$ functional is a bilinear form, so maybe there are some results in LA known for that case.

REPLY [1 votes]: I wanted to provide a little more detail on the answer given by Siminore.  I don't have a reference, as I worked this out on my own, but I think it is correct.
From the original problem statement, we may take the variation of $W(u)$ (I am using the more general problem), yielding
\begin{align*}
\delta W(u) &= \delta V(u) + \delta F(u) \\
&= \int_0^T\int_0^Tf(x,y)\left[ u(x)\delta u(y) + u(y)\delta u(x) \right]\ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx.
\end{align*}
Distribute the $f(x,y)$ to the terms in the brackets and break the integrals into two:
\begin{equation*}
\delta W(u) = \int_0^T\int_0^Tf(x,y)u(x)\delta u(y) \ dx\ dy + \int_0^T\int_0^Tf(x,y)u(y)\delta u(x) \ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx.
\end{equation*}
In the first set of integrals, we note that the integration variables are just dummy indices, and may be swapped [e.g., $(x,y)\rightarrow (y,x)$] without changing the result.  Furthermore, the limits of integration are the same for both integrals and are constants, and so we may arbitrarily change the order of integration.  Making use of these observations, we may write
\begin{align*}
\delta W(u) &= \int_0^T\int_0^Tf(y,x)u(y)\delta u(x) \ dx\ dy + \int_0^T\int_0^Tf(x,y)u(y)\delta u(x) \ dx\ dy \\
&\hspace{10mm}+ \int_0^T G_u(x,u)\delta u(x)\ dx \\
&= \int_0^T\int_0^T\left[f(y,x) + f(x,y)\right]u(y)\delta u(x) \ dx\ dy + \int_0^T G_u(x,u)\delta u(x)\ dx \\
&= \int_0^T\left( \int_0^T\left[f(x,y) + f(y,x)\right]u(y)\ dy + G_u(x,u) \right)\delta u(x)\ dx. \tag{1}
\end{align*}
Then, by the fundamental lemma of variational calculus you would conclude
\begin{equation*}
\int_0^T\left[f(x,y) + f(y,x)\right]u(y)\ dy + G_u(x,u) = 0,
\end{equation*}
which is an integral equation that may be solved for $u$.  If we make the additional assumption made by Siminore, i.e. that $f(x,y)=f(y,x)$, and let $G=0$ then the equation simplifies to
\begin{equation*}
\int_0^Tf(x,y)u(y)\ dy = 0,
\end{equation*}
which is what Siminore obtained.
As a special case that I was interested in, consider $f(x,y)=h(x)h(y)$ and $G=0$.  In this case we may write $W(u)=V(u)=H^2(u)$ where
\begin{equation*}
H(u) = \int_0^Th(x)u(x)\ dx.
\end{equation*}
Note that the variation of $H$ may be written as
\begin{equation*}
\delta H(u) = \int_0^Th(x)\delta u(x)\ dx.
\end{equation*}
From Eq. (1) above, we may write
\begin{align*}
\delta W(u) = \delta H^2(u) &= \int_0^T\int_0^T h(x)h(y)u(y)\delta u(x)\ dx\ dy \\
&= \int_0^Th(y)u(y)\ dy \int_0^T h(x)\delta u(x)\ dx \\
&= 2H(u)\delta H(u).
\end{align*}
Thus, the simple differentiation rule applies to these variations.


TITLE: How to calculate a group probability
QUESTION [0 upvotes]: Problem:

If I have 10 monkeys and 30 bananas, and the monkeys are all
  equally likely to eat each banana.  What is the probability that every
  monkey eats at least 2 bananas?

Could someone walk me through this?

REPLY [0 votes]: This is the same to ask the ways to sum 30 with 10 numbers each one equal or bigger than 2. And this is the same amount that distribute 20 bananas on 10 monkeys, i.e., normalize the numbers to 1 as minimum subtracting 1 to each monkey.
The ways to sum a quantity S with N positive numbers (with repetition) is $\binom{S-1}{N-1}$ (see here for more details).
So the probability will be valid cases divided by the total possible cases
$$p=\frac{\binom{20-1}{10-1}}{\binom{30-1}{10-1}}=\frac{(19)_9}{(29)_9}=\frac{646}{70035}\approx 1\%$$
P.S.: I supposed at least one banana for each monkey. If you can let some monkey with zero bananas then the calculus is different.


TITLE: Leray's theorem up to some degree
QUESTION [6 upvotes]: I am interested in the proof of Leray's theorem that relates Čech cohomology and sheaf cohomology.
The theorem states that if we have a space $X$, a sheaf $\mathcal{F}$ and a covering of $X$ such that our sheaf is acyclic on the covering, i.e. Čech cohomology  is $0$ on every finite intersection of elements in the covering, then these two cohomologies agree.
Now, my question. All the proves that I've found are by induction and assume that cohomology vanishes in every degree, so $\check{\mathrm{H}}^p(\mathcal{U}, \mathcal{F})=0$ for every $p>0$.
This looks like a rather strong assumption to check, especially if one is interested in computing cohomology just in low degrees (in particular I am interested in degree 0 and 1).
Looking at the proves that I found, indeed, it seems to me that we just need that Čech cohomology vanishes up to degree $p$ to prove the isomorphism in this degree (for example Climbing Mount Bourbaki).
I may be very wrong on this point, so the question is

Can someone provide a counter example in which we really need that cohomology vanishes in all the degrees and not just up to what we are interested in?

or

Can someone point out a reference which clearly states that we can just check up to degree $p$ (or $p+1$, depending on your indexing) to prove the isomorphism?

Thank you in advance,
Davide
EDIT: Despite the 5 "ups" I received no answer, so I decided to ask the same question also on Math Overflow.

REPLY [0 votes]: Answering in order to remove the question from the list of unanswered questions. See the answer on MathOverflow.


TITLE: Calculate $\sum\limits_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$
QUESTION [1 upvotes]: Calculate the sum of the series $$\sum_{n=1}^\infty \frac{1}{(n+2)(n+4)^2}$$

I have tried partial fraction decomposition.
$$\sum_{n=1}^\infty\frac{1}{4(n+2)}- \sum_{n=1}^\infty\frac{1}{4(n+4)}-\sum_{n=1}^\infty\frac{1}{2(n+4)^2}$$
Is this correct?  What is the sum?

REPLY [4 votes]: Your partial fraction decomposition looks OK, but it should be written as
$$\sum_{n=1}^\infty\left({1\over4(n+2)}-{1\over4(n+4)}\right)-\sum_{n=1}^\infty{1\over2(n+4)^2}$$
instead of being split into three infinite series.  That's because $\sum_{n=1}^\infty{1\over4(n+2)}$ and $\sum_{n=1}^\infty{1\over4(n+4)}$ are each divergent.  But combined they give the convergent, telescoping series
$$\left({1\over4\cdot3}-{1\over4\cdot5}\right)+\left({1\over4\cdot4}-{1\over4\cdot6}\right)+\left({1\over4\cdot5}-{1\over4\cdot7}\right)+\left({1\over4\cdot6}-{1\over4\cdot8}\right)+\cdots\\={1\over4\cdot3}+{1\over4\cdot4}={7\over48}$$
The other series you need to recognize as
$${1\over2}\left({1\over5^2}+{1\over6^2}+\cdots\right)={1\over2}\sum_{n=1}^\infty{1\over n^2}-{1\over2}\left(1+{1\over2^2}+{1\over3^2}+{1\over4^2}\right)={1\over2}\left(\pi^2\over6\right)-{1\over2}\left(205\over144\right)$$
The trick is understanding the hat that the $\pi^2/6$ rabbit came from, but I'm assuming you've seen it somewhere.
I'll leave it to you to put the pieces together.


TITLE: Counting pair in a poker hand - counting confusion
QUESTION [0 upvotes]: I want to be good enough to do math competitions, and one of the main topics in them is always combinatorics. So I'm going through a combinatorics book, "A Walk through Combinatorics."
The confusion I have is with the following problem:
Find the prob that a poker hand has a pair( and nothing better than a pair).
My attempt:
We know that a poker hand has five cards from a deck of 52 cards. Each card is equally likely so the total combinations is:  $${52} \choose 5$$
The number of hands that contain only a pair and nothing better than a pair are:
$${{13} \choose {4}} * {{4} \choose {2}}* {{4} \choose {1}}^3$$
There are 13 cards of each card. We choose 4 the 13 values. Then from one of the groups we want a pair, and from the remaining group we want a single.
So, the probability I got was:
$$\frac {{{13} \choose {4}} * {{4} \choose {2}}* {{4} \choose {1}}^3}{{52} \choose 5}$$.
However, according to  the answers:
The correct probability should be :
$\frac {{{13} \choose {1}} *{{12} \choose {3}} * {{4} \choose {2}}* {{4} \choose {1}}^3} {{52} \choose 5}$
Where is my logic flawed? Where am I going wrong? Am I double counting somewhere? I think my answer produces a smaller probability so am I undercounting?

REPLY [1 votes]: The difference is your ${13 \choose 4}$ versus their ${13 \choose 1} {12 \choose 3}$ (your ${4 \choose 2}$ and ${4 \choose 1}$ factors are correct).  You're choosing the four of 13 ranks in the hand, but not specifying which rank has the pair.  They choose the rank for the pair with ${13 \choose 1}$ and then the three other ranks with ${12 \choose 3}$.


TITLE: Fisher Information: While calculating the expectation of score function, why do we integrate with dx?
QUESTION [0 upvotes]: Let $$S=\frac{d}{dy} \left[\log f(x|y)\right]$$ where $y$ is the parameter and $S$ is score function
Now in text books,
              Ey[S] = integrate (S*f(x|y)*dx)

While I think it should be
              Ey[S] = integrate (S*y*dy)

my reasoning goes like this, the expectation is defined to be the weighted mean of variable and its function at that point. Here the function is S and it depends on y so the expectation is the weighted mean run over all y's.
To get a link of my problem visit
http://people.missouristate.edu/songfengzheng/Teaching/MTH541/Lecture%20notes/Fisher_info.pdf

REPLY [2 votes]: No, when you write $E_y[S]$, it means that you fix the parameter of the distribution of X being equal to $y$ and then you calculate the expected value with this parameter. So you integrate with respect to the density of $X$ when its parameter is $y$, which is $f(x|y)$.


TITLE: What is the mass of photon?
QUESTION [2 upvotes]: I'm sorry if this question is asked before, but I searched through the site and none satisfied me.
In most of the books I've come across, they just write "rest mass of photon is zero." But never talk about the relativistic Mass. Even in other answers on this site they have written exactly the same.
And once in my class there was some discussion on which I said that Mass of photon is zero, but my teacher corrected me, saying "Rest Mass of photon is 0".
So, what is the real Mass of photon? Or does there even exist something as relativistic Mass of photon?
I know the equation $m\gamma$ gives indeterminate form thus can't be used for photons. And I've no confusion on energy momentum relation which uses the rest mass.

REPLY [3 votes]: The mass of the photon is zero. The end.
Relativistic mass is a hazardous concept, and many authors refuse to use it. It makes the increase in kinetic energy of an object with velocity appear to be connected with some change in the internal structure of the object. See also this question on SE.

REPLY [0 votes]: The concept of relativistic mass has been abandoned as it is just $E/c^2 $. We can all agree that $E\neq 0$ for a photon. The modern concept of mass is $E/c^2 $ in the rest frame, or for the mathematically inclined, the limit to the rest frame. The experimental upper limit on photon mass is extremely small, see https://en.m.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass. Btw I consider reference 36 a fundamental mistake.


TITLE: parallel translation within geometrical form with two different heights
QUESTION [0 upvotes]: Please see attached images.
I need a formula for c' and to be honest I have no clue how to proceed with this problem.
The graphic construction of this problem is easy, but I need a formula.
So, when having the "inner form" I add 1cm to each side and connect the lines.
What I need is the mathematical construction for this.
Thanks for any help!
Best,
Jules

Rephrased to use MathJax:
$a' = a + 2 $
$b' = b + 1 $
$c' = ?$

REPLY [1 votes]: $c'=c+x$ and $x=1+y$, but:
$$
y:(c-b)=2:a,
\quad\hbox{hence:}\quad
y=2{c-b\over a}.
$$


text:
\begin{document}

\title[Compactness and asymptotic behavior]
 {Compactness and asymptotic behavior in nonautonomous linear parabolic equations with unbounded coefficients in $\R^d$}

\author[A. Lunardi]{Alessandra Lunardi}

\address{
Dipartimento di Matematica\\
Universit\`a di Parma\\
Viale G.P. Usberti, 53/A\\
43100 Parma, Italy}

\email{alessandra.lunardi@unipr.it}

\thanks{This work was partially supported by the GNAMPA reseach project 2009 "Equazioni differenziali ellittiche e paraboliche di ordine 2}

\subjclass{Primary 35D40; Secondary 28C10, 47D07}

\keywords{Evolution operator, compactness, asymptotic behavior}

 \dedicatory{Dedicated to Herbert Amann}

\begin{abstract}
We consider a class of second order linear nonautonomous parabolic equations in $\R^d$ with time periodic unbounded coefficients. We give sufficient conditions for the evolution operator $G(t,s)$ be compact in $C_b(\R^d)$ for $t>s$, and describe the asymptotic behavior of $G(t,s)f$ as $t-s \to \infty$ in terms of a family of measures $\mu_s$, $s\in \R$, solution of the associated Fokker-Planck equation.
\end{abstract}

\maketitle

\section{Introduction}

Linear nonautonomous parabolic equations in $\R^d$ are a classical subject in the mathematical literature. Most papers and books  about regular solutions are devoted to the case of bounded coefficients (e.g., \cite{LSU,friedman}, but the list is very long), and recently the interest towards unbounded coefficients grew up. The standard motivations to the study of unbounded coefficients are  on one side the well known connections with stochastic ODEs with unbounded nonlinearities, and on the other side  the changes of variables that transform bounded into unbounded coefficients, occurring in different mathematical models.
However, only for a few equations with unbounded coefficients
 it is possible to recover the familiar results about the bounded coefficients case. Many of them exhibit very different, and at first glance surprising, aspects. Therefore, a third motivation    is the
interest in new phenomena in PDEs.

This paper deals with one of these new phenomena, giving sufficient conditions in order that  the  evolution operator $G(t,s)$ associated to a class of second order parabolic equations  is a  compact  contraction in $C_b(\R^d)$ for $t>s$. Precisely,   Cauchy problems   such as

\begin{equation}
u_t(t,x) = \A(t)u(t, \cdot)(x), \quad t>s, \;x\in \R^d,
\label{equazione}
\end{equation}

\begin{equation}
u (s,x) =\varphi(x), \quad x\in \R^d
\label{Cauchy}
\end{equation}

will be considered, where the elliptic operators $\A(t)$ are defined by

\begin{align}
\label{A(t)}
(\A(t)\varphi) (x) &: =\sum_{i,j=1}^d
q_{ij}(t,x)D_{ij}\varphi (x) + \sum_{i=1}^d b_i(t,x) D_i\varphi
(x)\notag\\
&: = \tr \left( Q(t,x) D^2 \varphi (x)\right) + \langle b(t,x),
\nabla \varphi (x) \rangle ,
\end{align}

and the (smooth enough) coefficients $q_{ij}$, $b_i$ are allowed to be unbounded.
If $\varphi$ is smooth and it has compact support, a  classical bounded solution to  \eqref{equazione}--\eqref{Cauchy} is readily constructed, as the limit as $R\to \infty$ of the solutions $u_R$ of Cauchy-Dirichlet problems in the balls $B(0,R)$. However, classical bounded solutions  need not be unique. Under assumptions that guarantee positivity preserving in \eqref{equazione}--\eqref{Cauchy} (and hence, uniqueness of its bounded classical solution), a basic study of the evolution operator $G(t,s)$ for \eqref{equazione} in $C_b(\R^d)$ is in the paper \cite{KLL}. The evolution operator turns out to be markovian, since it has the representation

$$G(t,s)\varphi(x) = \int_{\R^d} \varphi(y) p_{t,s,x}(dy), \quad t>s, \;x\in \R^d, \;\varphi\in C_b(\R^d),$$

where the probability measures $p_{t,s,x}$ are given by $p_{t,s,x}(dy) = g(t,s,x,y)dy$ for a positive function $g$.

It is easy to see that if a markovian $G(t,s)$ is compact  in $C_b(\R^d)$, then it does not preserve $C_0(\R^d)$, the space of the continuos functions vanishing as $|x|\to \infty$, and it cannot be extended to a bounded operator in  $L^p(\R^d, dx)$ for $1\leq p<\infty$. Therefore, much of the theory developed for bounded coefficients fails.

When a parabolic problem is not well posed in $L^p$ spaces with respect to the Lebesgue measure, it is natural to look for other measures $\mu$, and in particular to weighted Lebesgue measures, such that $G(t,s)$ acts in $L^p(\R^d, \mu)$. This is well understood in the autonomous case $\A(t)\equiv \A$, where the dynamics is held  by a semigroup $T(t)$ and $G(t,s) = T(t-s)$. Then, an important role is played by {\em invariant measures}, that are Borel probability measures $\mu$ such that
$$\int_{\R^d} T(t)\varphi\,d\mu = \int_{\R^d}\varphi\,d\mu, \quad \varphi \in C_b(\R^d).$$
If a Markov semigroup has an invariant measure $\mu$, it can be extended in a standard way to a contraction semigroup in all the spaces $L^p(\R^d, \mu)$, $1\leq p<\infty$.
Under broad assumptions the invariant measure is unique, and it is strongly related with the asymptotic behavior of $T(t)$, since $\lim_{t\to \infty} T(t)\varphi = \int_{\R^d}\varphi \,d\mu$, locally uniformly  if $\varphi\in C_b(\R^d)$ and in $L^p(\R^d, \mu)$ if $\varphi \in  L^p(\R^d, \mu)$, $1<p<\infty$.
In the nonautonomous case the role of the invariant measure is played by families of measures $\{ \mu_s:\; s\in \R\}$,  called {\em evolution systems of measures}, that satisfy

\begin{equation}
\int_{\R^d} G(t,s)\varphi\,d\mu_t = \int_{\R^d}\varphi\,d\mu_s, \quad t>s, \; \varphi \in C_b(\R^d).
\label{invariance}
\end{equation}

If \eqref{invariance} is satisfied, the function $s\mapsto \mu_s$ satisfies (at least, formally) the Fokker-Planck equation

$$D_s \mu_s + \A(s)^*\mu_s = 0, \quad s\in \R, $$

which is  a parabolic equation for measures without any initial, or final, condition. Therefore it is natural to have infinitely many solutions, and to look for uniqueness of special solutions. For instance, in the autonomous case the unique stationary solution is the invariant measure, in the periodic case $\A(t) = \A(t+T)$ under reasonable assumptions there is a unique $T$-periodic solution, etc.
Arguing as in the autonomous case, it is easy to see that if \eqref{invariance} holds then $G(t,s)$ may be extended to a contraction from $L^p(\R^d, \mu_s)$ to $L^p(\R^d, \mu_t)$ for $t>s$. Therefore, it is natural to investigate asymptotic behavior of $G(t,s)$ not only in $C_b(\R^d)$ but also in these $L^p$ spaces.

A basic study of the evolution operator for parabolic equations with (smooth enough) unbounded coefficients is in \cite{KLL}. In its sequel \cite{LLZ} we studied asymptotic behavior of $G(t,s)$ in the case of time-periodic coefficients.

In this paper   sufficient conditions will be given for the evolution operator $G(t,s)$ be compact in $C_b(\R^d)$. Then, compactness will be used to obtain asymptotic behavior results in the case of time-periodic coefficients. Indeed, compactness implies that there exists a unique $T$-periodic evolution system of measures $\{ \mu_s:\; s\in \R\}$, and that
denoting by $m_s\varphi $ the mean value

\begin{equation}
\label{m_s}
m_s\varphi  := \int_{\R^d} \varphi(x) \mu_s(dx), \quad s\in \R ,
\end{equation}

 there is $\omega <0$ such that for each $\eps >0$ we have

\begin{equation}
\label{CompAsInfty} \| G(t,s)\varphi - m_s\varphi\|_{\infty}
\leq M_{\eps} e^{(\omega +\eps) (t-s)} \|\varphi\|_{\infty} , \quad t>s,
\; \varphi \in C_b(\R^d),
\end{equation}

for some $ M_{\eps} >0$. As a consequence,  for
every $p\in (1, \infty)$ and $\eps >0$  we get

\begin{equation}
\label{CompAsExp} \| G(t,s)\varphi - m_s\varphi\|_{L^p(\R^d, \mu_t)}
\leq Me^{(\omega +\eps) (t-s)} \|\varphi\|_{L^p(\R^d, \mu_s)} , \quad t>s,
\; \varphi \in L^p(\R^d, \mu_s),
\end{equation}

for some $M= M(p, \eps)>0$. Note that while the constant $M$ may depend on $p$, the exponential rate of decay is independent of $p$.
These results complement the asymptotic behavior results of \cite{LLZ}, where \eqref{CompAsExp} was obtained under different assumptions.

\section{Preliminaries: the evolution operator  $G(t,s)$.}

We use standard notations. $C_b(\R^d)$ is the space of the bounded continuous functions from $\R^d$ to $\R$, endowed with the sup norm.
$C_0(\R^d)$ is the space of the continuous functions that vanish at infinity. For $k\in \N\cup \{0\}$, $C^k_c(\R^d)$ is the space of $k$ times differentiable functions with compact support. For $0<\alpha <1$, $a<b$ and $R>0$, $C^{\alpha/2, \alpha}([a,b]\times B(0,R))$ and $C^{1+\alpha/2, 2+ \alpha}([a,b]\times B(0,R))$ are the usual parabolic H\"older spaces in the set $[a,b]\times B(0,R)$;
$C^{\alpha/2, \alpha}_{loc}(\R^{1+d})$ and $C^{1+\alpha/2, 2+ \alpha}_{loc}(\R^{1+d})$ are the subspaces of  $C_b(\R^{1+d})$ consisting of functions whose restrictions to $[a,b]\times B(0,R)$ belong to $C^{\alpha/2, \alpha}([a,b]\times B(0,R))$ and to $C^{1+\alpha/2, 2+ \alpha}([a,b]\times B(0,R))$, respectively, for every $a<b$ and $R>0$.

\vspace{2mm}

In this section we recall some results from \cite{KLL}  about the evolution operator for parabolic equations with unbounded coefficients.
They were proved under standard regularity and ellipticity assumptions, and nonstandard qualitative assumptions.

\begin{hyp}
\label{hyp1}
\begin{itemize}
\item[(i)]
The coefficients $q_{ij}$,  $b_i$ $(i,j=1,\ldots,d)$ belong to $C^{\alpha/2, \alpha}_{loc}(\R^{1+d})$ for some $\alpha\in (0, 1)$.

\item[(ii)]
For every $(s,x)\in  \R^{1+d}$,
the matrix $Q(s,x)$ is symmetric and there exists a function
$\eta:  \R^{1+d}\to \R$ such that
$0<\eta_0:=\inf_{\R^{1+d}}\eta$ and
\begin{eqnarray*}
\langle Q(s,x)\xi , \xi \rangle \geq \eta(s,x) |\xi |^2, \qquad\;\,
\xi \in \R^d,\;\,(s,x) \in  \R^{1+d}.
\end{eqnarray*}

\item[(iii)]
There exist a positive function $W  \in C^2(\R^d)$ and a number $\lambda \in \R$ such that

\begin{eqnarray*}
\qquad\;\, \lim_{|x|\to \infty} W (x) = \infty \quad \mbox{
and } \quad \sup_{s\in \R, \,x\in \R^d}(\A(s) W)( x) -\lambda W(x)< 0.
\end{eqnarray*}
\end{itemize}
\end{hyp}

Assumptions (i) and (ii) imply that for every $s\in \R$ and $\varphi\in C_b(\R^d)$, the Cauchy problem

\begin{equation}
\label{CP}
\left\{ \begin{array}{rcll}
\displaystyle  D_tu(t,x) &  =  & \A (t)u(t,x), & t>s,\; x \in   \R^{d},\\[2mm]
u(s,x) &  =  & \varphi(x), & x\in \R^d,
\end{array}\right.
\end{equation}

has a bounded classical solution.
Assumption (iii) implies that the bounded classical solution to \eqref{CP} is unique  (in fact, a maximum principle that yields uniqueness is proved in \cite{KLL} under a slightly weaker assumption). The evolution operator $G(t,s)$ is defined by

\begin{equation}
\label{(t,s)}
G(t,s)\varphi   = u(t, \cdot), \quad t\geq s \in \R ,
\end{equation}

where $u $ is the unique bounded solution to \eqref{CP}. Some of the properties of $G(t,s)$  are in next theorem.

\begin{thm}
\label{Th:G(t,s)}
Let Hypothesis  \ref{hyp1} hold. Define $ \Lambda
:= \{(t,s,x)\in \R^{2+d}: t > s, \;x\in \R^{d}\} $. Then:
\begin{itemize}
\item[(i)]
for every $\varphi \in C_b(\R^d)$, the function $(t,s,x)\mapsto
G(t,s)\varphi(x)$ is continuous in $\overline{\Lambda }$. For each
$s\in \R$,   $(t,x)\mapsto G(t,s)\varphi(x)$ belongs to
$C^{1+\alpha/2, 2+\alpha}_{loc}((s, \infty) \times \R^d )$;
\item[(ii)]
for every $\varphi\in C^{2}_{c}( \R^d )$, the function
$(t,s,x)\mapsto G(t,s)\varphi(x)$ is continuously differentiable
with respect to $s$ in $\overline{\Lambda }$ and
$D_sG(t,s)\varphi(x) =   -G(t,s)\A(s)\varphi(x)$ for any
$(t,s,x)\in\overline\Lambda$;
\item[(iii)]
for each $(t,s,x)\in \Lambda$ there exists a Borel probability measure $p_{t,s,x}$ in $\R^d$ such that

\begin{equation}
\label{representation}
G(t,s)\varphi(x) = \int_{\R^d}\varphi (y)
p_{t,s,x}(dy), \quad \varphi \in C_b(\R^d).
\end{equation}

Moreover, $ p_{t,s,x}(dy) = g(t,s,x,y)dy$ for a positive function
$g$.
\item[(iv)]
$G(t,s)$ is strong Feller;  extending it to  $L^{\infty}(\R^d, dx)$ through  formula \eqref{representation}, it maps
  $L^{\infty}(\R^d, dx)$ (and, in particular, $B_b(\R^d)$) into $C_b(\R^d)$ for $t>s$, and
\begin{eqnarray*}
\|G(t,s)\varphi \|_{\infty} \leq \|\varphi \|_{\infty}, \quad
\varphi \in L^{\infty}(\R^d, dx), \;\,t>s.
\end{eqnarray*}

\item[(v)]  If $(\varphi_n)$ is a bounded sequence in  $C_b(\R^d)$ that converges uniformly to $\varphi $ in each compact set $K\subset\R^d$, then for each $s\in \R$ and $T>0$, $G(\cdot,s)\varphi _n$
converges to $G(\cdot , s )\varphi $ uniformly in
$[s,s+T] \times K$, for each   compact set $K \subset \R^d$.

\item[(vi)]
For every $s\in \R$ and $R>0$, $0<\eps <T$ there is $C= C(s,\eps, T, R)>0$ such that
$$\sup_{ s+\eps \leq t\leq s+T}\|G(t,s)\varphi \|_{C^2(B(0,R))}  \leq C\|\varphi\|_{\infty} , \quad \varphi \in C_b(\R^d).$$
\end{itemize}
\end{thm}

Statements (i) to (v) are explicitly mentioned in \cite{KLL} (Thm. 2.1, Prop. 2.4, Cor. 2.5, Prop. 3.1, Lemma 3.2). Statement (vi) is hidden in the proof of Theorem 2.1, where $G(t,s)\varphi$ is obtained by an approximation procedure, in three steps: first, for $\varphi\in C^{2+\alpha}_{c}(\R^d)$, then for
$\varphi\in C_{0}(\R^d)$, and then for $\varphi \in C_b(\R^d)$. At each step, we have interior Schauder estimates for a sequence $u_n$ that approaches $G(t,s)\varphi$, namely for  $s\in \R$ and $R>0$, $0<\eps <T$ there is $C= C(s,\eps, T, R)>0$ such that
$$\|u_n\|_{C^{1+\alpha/2 , 2+\alpha}([s+\eps, s+T]\times B(0,R))} \leq C\|\varphi\|_{\infty}, \quad n\in \N , $$
and $u_n$ converges to $G(t,s)\varphi$ locally uniformly. This yields (vi).

\vspace{3mm}

 To get evolution system of measures we have to  strenghten  assumption \ref{hyp1}(iii).   The following theorem is proved in \cite{KLL}.

\begin{thm}
\label{evsyst}
Under  Hypotheses \ref{hyp1}, assume in addition that  there exist a positive function $W  \in C^2(\R^d)$ and numbers
$a$,  $c>0$    such that

\begin{equation}
\label{V}
\qquad\;\, \lim_{|x|\to \infty} W (x) = +\infty \quad \mbox{
and } \quad (\A(s) W)( x) \leq a - c W(x), \quad (s,x)\in   \R^{1+d}.
\end{equation}

Then there exists a tight$^($\footnote{i.e.,   $\forall \eps
>0$   $\exists R=R(\eps)>0$ such that
$
\mu_s(B(0,R)) \geq 1-\eps$, for all $s\in\R$.}$^)$
 evolution system of measures $\{ \mu_s: s\in
\R\}$ for $G(t,s)$.  Moreover,

\begin{equation}
\label{G(t,s)V}
G(t,s)W(x) := \int_{\R^d}W(y) p_{t,s,x}(dy) \leq
W(x) + \frac{a}{c}, \quad t>s, \;x\in \R^d,
\end{equation}
and
\begin{equation}
\label{m_tV}
  \int_{\R^d}W(y)\mu_t(dy) \leq \min W +
\frac{a}{c}, \quad t\in \R .
\end{equation}

\end{thm}

\section{Compactness in $C_b(\R^d)$}

A necessary and sufficient condition for $G(t,s)$ be compact in
$C_b(\R^d)$ for $t>s$ is very similar to the corresponding condition
in the autonomous case (\cite{MPW}).

\begin{prop}
\label{tight-compattezza} Under Hypothesis \ref{hyp1}
the following statements are equivalent:
\begin{itemize}
\item[(a)]
for any $t>s$, $G(t,s):  C_b(\R^d)\to C_b(\R^d)$ is compact.
\item[(b)]
for any $t>s$ the family of measures $\{p_{t,s,x}(dy):\; x\in
\R^d\}$ is tight, i.e., for every  $\eps >0$ there exists
$R=R(t,s,\eps)>0$ such that
\begin{eqnarray*}
p_{t,s,x}(B(0,R)) \geq 1-\eps, \quad x\in \R^d.
\end{eqnarray*}
\end{itemize}
\end{prop}

\begin{proof}
We follow the proof given in \cite[Prop.~ 3.6]{MPW} for the autonomous case.

Let statement (a) hold. For every $R >0$ let $\varphi_R:\R^d\to\R$
be a continuous function such that $\one _{B(0,R)} \leq
\varphi_R\leq \one _{B(0,R+1)}$. Since $\|\varphi_R\|_{\infty} \leq
1$ and $G(t,s)$ is compact, there is a  sequence $G(t,s)
\varphi_{R_n}$ that converges uniformly in the whole $\R^d$ to a
limit function $g$. Since $\varphi_R$ goes to $\one$ as $R\to
+\infty$, uniformly on each compact set and $\|\varphi_R\|_{\infty}
\leq 1$ for every $R$, by Theorem \ref{Th:G(t,s)}(v)  $\lim_{R\to \infty} G(t,s)
\varphi_R  = G(t,s)\one = \one$ uniformly on each compact set.
Then, $g\equiv 1$ and $\lim_{R\to +\infty} \|G(t,s) \varphi_R
-\one\|_{\infty} =0$. Therefore, fixed any $\eps >0$, we have
\begin{eqnarray*}
p_{t,s,x}(B(0,R) ) = (G(t,s)\one _{B(0,R)})(x)  \geq
(G(t,s)\varphi_{R})(x) \geq 1-\eps , \quad x\in\R^d,
\end{eqnarray*}
for $R$ large enough.

Let now statement (b) hold. For $t>s$ fix $r\in (s,t)$ and recall
that (see formula \eqref{representation})
\begin{eqnarray*}
(G(t,s)\varphi)(x) = (G(t,r)G(r,s)\varphi)(x)=\int_{\R^d}
(G(r,s)\varphi) (y) p_{t,r,x}(dy),
 \end{eqnarray*}

for any $\varphi\in C_b(\R^d)$ and any  $x\in \R^d$. For every $R>0$
set

\begin{eqnarray*}
(G_{R}\varphi)(x) =  \int_{B(0,R)} (G(r,s)\varphi) (y)
p_{t,r,x}(dy), \quad x\in \R^d.
\end{eqnarray*}

Each $G_R: C_b(\R^d)\to C_b(\R^d)$ is a compact operator, since it
may be written as $G_{R} = {\mathcal S}\circ {\mathcal R} \circ
G(r,s)$ where $G(r,s): C_b(\R^d)\to C_b(\R^d)$ is continuous,
${\mathcal R}: C_b(\R^d)\to C( B(0,R))$ is the
restriction operator, and $ {\mathcal S}:
C(B(0,R)) \to C_b(\R^d)$ is defined by

\begin{eqnarray*}
{\mathcal S} \psi(x) =  \int_{B(0,R)} \psi(y)  p_{t,r,x}(dy) =
(G(t,r)\widetilde{\psi})(x),\qquad\;\,x\in B(0,R),
\end{eqnarray*}

where $\widetilde{\psi}(x)$ is the null extension of $\psi$ to the
whole $\R^d$. Now, ${\mathcal R}\circ G(s,r): C_b(\R^d)\to
C(B(0,R))$ is compact by Theorem
\ref{Th:G(t,s)}(v),
and $ {\mathcal S}$ is continuous from $C(B(0,R))$ to
$C_b(\R^d)$ because $G(t,r)$ is strong Feller by Theorem
\ref{Th:G(t,s)}(iv).

Moreover, $G_R\to G(t,s)$ in ${\mathcal L}(C_b(\R^d))$, as
$R\to+\infty$. Indeed, for $\eps >0$ there is  $R_0>0$   such that $
p_{t,r,x}(B(0,R))\geq 1-\eps $ for each $x\in \R^d$ and $R\geq R_0$,
and consequently
\begin{eqnarray*}
|(G(t,s)\varphi)(x) - (G_R\varphi)(x) |  \leq
\|G(r,s)\varphi\|_{\infty}\int_{\R^d\setminus B(0,R)}p_{t,r,x}(dy)
\leq \eps \|\varphi\|_{\infty},
\end{eqnarray*}

for $R\geq R_0$ and for each $x\in \R^d$.

Being limit of compact operators, $G(t,s)$ is compact. \end{proof}

\begin{rem}
\label{rem:comp}
{\rm Some remarks are in order.
\begin{itemize}
\item[(i)]
An insight in the proof shows that if  $G(t,s)$ is compact for some
$t>s$, then the family   $\{p_{t,s,x}(dy):\; x\in \R^d\}$ is tight;
conversely if for some $r>s$ the family $\{p_{r,s,x}(dy):\; x\in
\R^d\}$ is tight then $G(t,s)$ is compact for each $t>r$.
\item[(ii)]
If for some $r>s$ the family $\{p_{r,s,x}(dy):\; x\in \R^d\}$ is
tight, then the family $\{p_{t,s,x}(dy):\; t\geq r, \;x\in \R^d\}$
is tight. Indeed, for every $R>0$ we have

$$\begin{array}{lll}
p_{t,s,x}(\R^d\setminus B(0,R)) &= &(G(t,s)\one _{\R^d\setminus B(0,R)})(x)
\\
\\
&=&(G(t,r)G(r,s)\one _{\R^d\setminus B(0,R)})(x)
\\
\\
&\leq &\|G(r,s)\one _{\R^d\setminus B(0,R)}\|_{\infty},
\end{array}$$

so that, if $p_{r,s,x}(\R^d\setminus B(0,R)) = G(r,s)\one
_{\R^d\setminus B(0,R)}(x)\leq \eps$ for every $x$, also
$p_{t,s,x}(\R^d\setminus B(0,R))\leq \eps$ for every $x$.
\item[(iii)]
As in the autonomous case (\cite{MPW}), if $G(t,s)$ is compact
in $C_b(\R^d)$, it does not preserve $L^p(\R^d,dx)$ for any $p\in
[1, +\infty)$ and it does not preserve $C_0(\R^d)$. Indeed, let
$R>0$ be so large that $p_{t,s,x}(B(0,R)) \geq 1/2$ for every $x\in
\R^d$, and let $\varphi \in C_c(\R^d)$ be such that $\varphi \geq
\one_{B(0,R)}$. Then,
\begin{eqnarray*}
(G(t,s)\varphi)(x) \geq (G(t,s)\one_{B(0,R)})(x) =
p_{t,s,x}(B(0,R))\geq \frac{1}{2},
\end{eqnarray*}
for every $x$, so
that $G(t,s)\varphi $ does not belong to any space $L^p(\R^d,dx)$
and to $C_0(\R^d)$.
\item[(iv)]
A similar argument shows that $\inf G(t,s)\varphi >0$ for each $t>s$ and $\varphi \in C_b(\R^d)\setminus \{0\}$, $\varphi \geq 0$. Indeed, if $\varphi(x) >0$ for each $x$,
and $R>0$ is as before, then $(G(t,s)\varphi)(x) \geq \delta (G(t,s)\one_{B(0,R)})(x) \geq \delta/2$, with $\delta = \min_{|x|\leq R}\varphi(x) >0$.
If $\varphi(x) \geq 0$ for each $x$, it is sufficient to recall that $G(t,s)\varphi = G(t,(s+t)/2) G((s+t)/2, s)\varphi$ and that $G((s+t)/2, s)\varphi (x)>0$ for each $x$ by Theorem \ref{Th:G(t,s)} (iii).
\end{itemize}
}
\end{rem}

However,  to check the tightness condition of Proposition \ref{tight-compattezza} is not obvious, since the measures $p_{t,s,x}$ are not explicit, in general. In the case of time depending Ornstein-Uhlenbeck operators (e.g. \cite{DPL}),

$$(\A(t)\varphi)(x) = \frac12 \tr  \left(Q(t)D_x^2\varphi   (x)\right)+ \langle A(t)x+f(t),\nabla \varphi(x)\rangle ,\quad x\in\R^d,$$

the measures $p_{t,s,x}$ are explicit Gaussian measures and it is possible to see that the tightness condition does not hold. Alternatively, one can check that $G(t,s)$ maps $L^p(\R^d, dx)$ into itself for every $p\in (1, \infty)$ and therefore it cannot be compact in $C_b(\R^d)$.

If the assumptions of Theorem \ref{evsyst} hold,  estimate \eqref{G(t,s)V} implies that the family $\{
p_{t,s,x}: t>s, \;x\in B(0,r)\}$ is tight for every $r>0$. However, this is not enough for compactness.
To obtain compactness we have to strenghten condition \eqref{V} on the auxiliary function $W$.

\begin{thm}
\label{Giorgio} Let Hypotheses  \ref{hyp1}   hold.
Assume in addition that there exist a $C^2$ function $W:\R^d\mapsto \R$, such that $\lim_{|x|\to \infty} W(x) = +\infty$,
a number
$R>0$ and a convex increasing
function $g: [0, +\infty)\to \R$ such that   $1/g$ is in $L^1(a,
+\infty)$ for  large $a $, and
\begin{equation}
(\A (s) W)(x) \leq -g(W(x)), \qquad\;\,s\in \R ,\;\, |x| \geq R.
\label{stima-conv}
\end{equation}

Then, for every $\delta >0$ there is $C=C(\delta) >0$ such that
$(G(t,s)W)(x) \leq C$ for  every $x\in \R^d$ and  $s\leq t-\delta $.
Consequently,  the family of probabilities $\{ p_{t,s,x}(dy): s\leq
t-\delta, \,x\in \R^d\}$ is   tight, and $G(t,s)$ is compact in
$C_b(\R^d)$  for $t>s$.
\end{thm}

\begin{proof}
As a first step we show that

 \begin{equation}
 \label{maggV}
(G(t,s)W)(x)  - (G(t,r)W)(x)  \geq - \int_{r}^{s} (G(t,
\sigma)\A(\sigma) W)(x)d\sigma, \quad r<s<t, \;x\in \R^d.
\end{equation}

Let $\varphi \in C^{\infty}(\R)$ be a nonincreasing function such that $\varphi\equiv 1$ in $(-\infty , 0]$, $\varphi \equiv 0$ in $[1, + \infty)$, and define $\psi_n (t) = \int_{0}^{t} \varphi (s-n)ds$ for each $n\in \N$. The functions  $\psi_n$ are smooth and enjoy the following properties:

\begin{itemize}
\item
$\psi_n(t) = t$ for $t \in [0, n]$,
\item
$\psi_n(t) \equiv \mathrm{const.}$ for $ t \geq n+1$,
\item
$0 \leq \psi_n' \leq 1$ and $\psi_n'' \leq 0$,
\item
for every $t\geq 0$, the sequence $(\psi_n'(t))$ is increasing.
\end{itemize}

Then, the function $W_n := \psi_n\circ W$ belongs to
$C^2_b(\R^d)$ and it is constant outside a compact set. By Theorem
\ref{Th:G(t,s)}(ii),  applied to $W_n - c$, we have

\begin{align*}
(G(t,s)W_n)(x) -&(G(t,r)W_n)(x) =   - \int_r^s (G(t,\sigma)\A(\sigma )W_n)(x)\, d\sigma\notag\\
&= - \int_r^s G(t,\sigma) \left \{\psi_n'(W) (\A(\sigma)
W)  + \psi_n''(W) \langle Q  \nabla W , \nabla W \rangle\right\} (x) d\sigma\notag\\
&\ge - \int_r^s G(t,\sigma)( \psi_n'(W) \A(\sigma )W)(x)d\sigma \notag\\
& = - \int_r^sd\sigma\int_{E_{\sigma}}\psi_n'(W(y))(\A(\sigma
)W)(y)p_{t,\sigma,x}(dy)
\\
&- \int_r^sd\sigma\int_{\R^d\setminus
E_{\sigma}}\psi_n'(W(y))(\A(\sigma )W)(y)p_{t,\sigma,x}(dy),
\end{align*}

where $E_{\sigma}=\{x\in\R^d: (\A(\sigma )W)(x)>0\}$.
Letting $n\to +\infty$, the left-hand side goes to  $(G(t,s)W)(x)
-(G(t,r)W)(x)$. Concerning the right-hand side,  both integrals converge by monotone convergence. We have to prove that their limits are finite.  The first term  converges to $-\int_r^sd\sigma\int_{E_{\sigma}}(\A(\sigma
)W)(y)p_{t,\sigma,x}(dy)$, which is finite since the sets
$E_{\sigma}$ are equibounded in $\R^d$ (recall that  the function ${\mathcal A}(\sigma)W$ tends to $-\infty$
as $|x|\to +\infty$, uniformly with respect to $\sigma\in [r,s]$). The second term may be estimated
by

\begin{align*}
&-\int_r^sd\sigma\int_{\R^d\setminus
E_{\sigma}}\psi_n'(W(y))(\A(\sigma )W)(y)p_{t,\sigma,x}(dy)\\
\le &\int_r^sd\sigma\int_{E_{\sigma}}\psi_n'(W(y))(\A(\sigma
)W)(y)p_{t,\sigma,x}(dy)+ (G(t,s)W_n)(x) -(G(t,r)W_n)(x).
\end{align*}
Letting $n\to +\infty$, we  obtain that
$\int_r^sd\sigma\int_{\R^d\setminus E_{\sigma}}(\A(\sigma
)W)(y)p_{t,\sigma,x}(dy)$ is finite.

Summing up,  the function
$\sigma\mapsto (G(t,\sigma)(\A(\sigma)W))(x)$ is in $L^1(r,s)$ and
\eqref{maggV} follows.

Possibly replacing $g$ by $\widetilde{g} = g-C$ for a suitable
constant $C$, we may assume that  $(\A(s)W)(x) \leq -g(W(x))$ for
every $s\in \R$ and $x\in \R^d$.

Fix $x\in \R^d$, $t\in \R$, and set
\begin{eqnarray*}
\beta(s) := (G(t, t-s)W)(x), \quad s\geq 0.
\end{eqnarray*}

Then $\beta$ is measurable, since it is the  limit  of the sequence of  continuous
functions $s\mapsto G(t,t-s)W_n(x)$. Inequality \eqref{maggV} implies

$$
\beta(b)-\beta(a) \leq -\int_{t-b}^{t-a}
(G(t,\sigma)g(W))(x)d\sigma, \qquad\;\, a<b,
$$

and, since $g$ is convex,

$$\begin{array}{l}
(G(t, \sigma)g(W))(x) = \displaystyle{ \int_{\R^d} g(W(y))p_{t,\sigma, x}(dy) }
\\
\\
\geq
g\bigg(  \displaystyle{ \int_{\R^d}  W(y)p_{t,\sigma, x}(dy) } \bigg) = g((G(t,
\sigma)W)(x))
\end{array}$$

so that

\begin{equation}
\label{eq:beta}
\begin{array}{lll}
 \beta(b)-\beta(a) & \le &
 \displaystyle{-\int_{t-b}^{t-a}g((G(t,\sigma)W)(x))d\sigma }
 \\
 \\
&   = &
 \displaystyle{-\int_{t-b}^{t-a}g(\beta(t-\sigma))d\sigma }
 \\
 \\
 & = &  \displaystyle{ -\int_{a}^{b}
g(\beta(\sigma))d\sigma, }
\end{array}
\end{equation}

for any $a<b$. Then, for every  $s\geq 0$, $\beta(s)\leq \zeta(s)$,
where $\zeta$ is the solution of the Cauchy problem

\begin{eqnarray*}
\left\{ \begin{array}{l}
\zeta'(s) = -g(\zeta (s)), \quad s\geq 0,
\\[3mm]
\zeta(0) = W(x).
\end{array}\right.
\end{eqnarray*}

Indeed, assume by contradiction that there exists $s_0>0$ such that
$\beta(s_0)>\zeta(s_0)$, and denote by $I$ the largest interval
containing $s_0$ such that $\beta(s )>\zeta(s )$ for each $s\in
I$.

\noindent Inequality \eqref{eq:beta} implies that $ \beta(b)-\beta(a) \leq -m (b-a)$ for $b>a$, with $m :=\min g$. In other words, the function $s\mapsto \beta(s) + ms $ is decreasing.
This implies that  $I$ contains some left neighborhood of $s_0$. Indeed, since $s\mapsto \beta(s)+ms $ is   decreasing, then

$$\lim_{s\to  s_0^-} \beta(s)+ms \geq \beta(s_0) + ms_0 > \zeta(s_0) + ms_0 = \lim_{s\to  s_0^-}  \zeta(s)+ms$$

so that $\lim_{s\to  s_0^-} ( \beta(s)- \zeta(s)) >0$, which yields $\beta>\zeta$ in a left neighborhood of $s_0$.

Let    $a= \inf I $. Then $a< s_0$, and  there is a sequence $(s_n)\uparrow a$ such that $\beta(s_n) \leq \zeta(s_n) $, so that $\beta(a) + ma  \leq \lim_{n\to \infty}\beta(s_n) + ms_n  \leq \zeta(a) + ma$, that is $\beta(a) \leq \zeta(a)$. On the other hand, for each $s\in I$ we have

\begin{eqnarray*}
\beta(s) - \beta(a) \leq \int_{a}^{s} -g(\beta(\sigma))d\sigma,
\quad \zeta(s) -\zeta(a) = \int_{a}^{s} -g(\zeta(\sigma))d\sigma ,
\end{eqnarray*}

so that

\begin{eqnarray*}
\beta(s) - \zeta(s) \leq   \int_{a}^{s} [-g(\beta(\sigma))+ g(\zeta (\sigma))] \,d\sigma
, \quad s\in I.
\end{eqnarray*}

Since $\beta(\sigma) > \zeta(\sigma)$ for every $\sigma \in I$ and
$g$ is increasing, the integral in the right-hand side is
nonpositive, a contradiction. Therefore, $\beta(s)\leq \zeta(s)$ for
every $s\geq 0$.

By standard arguments about ODE's, for every $\delta >0$ there
is
 $C=C(\delta )$ independent on the initial datum $W(x)$ such that
$\zeta(s)\leq C$ for every $s\geq \delta$. Therefore,

$$\beta (s)=
(G(t,t-s)W)(x)\leq C, \quad s\geq \delta ,$$

with $C$ independent of $t$. This implies that for every $\delta >0$ the family of probabilities $p_{t,s,x}(dy)$
with $s\leq t-\delta$ and $x\in \R^d$ is   tight, because for every $R>0$ we have

\begin{align*}
 p_{t,s,x}(\R^d \setminus B(0,R)) &= \int_{\R^d \setminus B(0,R)}p_{t,s,x}(dy)\\
&\le \frac{1}{\inf \{W(y):\;|y|\geq R\}}  \int_{\R^d \setminus B(0,R)}W(y)p_{t,s,x}(dy)\\
&\le \frac{1}{\inf \{W(y):\;|y|\geq R\}} (G(t,s)W)(x) \\
&\le   \frac{C}{\inf \{W(y):\;|y|\geq R\}}
\end{align*}

and $\inf \{W(y):\;|y|\geq R\} $ goes to $+\infty $ as $r\to +\infty$. So,
condition (b) of Proposition \ref{tight-compattezza} is satisfied.
\end{proof}

\begin{ex}[As in the autonomous case]
If there is $R>0$ such that

\begin{eqnarray*}
{\rm Tr}\; Q(s,x) + \langle b(s,x),x\rangle - \frac{2}{|x|^2}
\langle Q(s,x)x, x\rangle \leq -c |x|^2(\log |x|)^{\gamma}, \quad s\in
\R ,\;\, |x|\geq R,
\end{eqnarray*}
with $c>0$ and $\gamma >1$, then the condition \eqref{stima-conv} is
satisfied by any $W$ such that $W(x) = \log |x|$ for $|x|\ge R$,
with $g(s) = cs^{\gamma}$. If the regularity and ellipticity assumptions  \ref{hyp1}(i)(ii) hold, Theorem \ref{Giorgio} implies that the evolution operator $G(t,s)$  is compact in $C_b(\R^d)$ for $t>s$.
\end{ex}

 \section{Compactness and asymptotic behavior}

In this section we derive asymptotic behavior results from compactness of $G(t,s)$ in $C_b(\R^d)$.

Throughout the section we assume that  Hypothesis \ref{hyp1}  holds, and that
the coefficients $q_{ij}$ and $b_i$, $i,j=1, \ldots, d$ are periodic in time, with period $T>0$. Then the asymptotic behavior of $G(t,s)$ is driven by the spectral properties of the operators

$$V(s) : = G(s+T, s), \quad s\in \R.$$

This is well known in the case of evolution operators associated to families $A(t)$ of generators of analytic semigroups, see e.g.  \cite[\S7.2]{Henry}, \cite[Ch.~6]{L}, \cite{F}. Most of the arguments are independent of analyticity and will be adapted to our situation.

To begin with, since each $V(s)$ is a contraction in $C_b(\R^d)$, its spectrum is contained in the unit circle. Its spectral radius is $1$, since $1$ is an eigenvalue. The nonzero eigenvalues of $V(s)$ are independent of $s$, since the equality $G(t,s)V(s) = V(t)G(t,s)$ implies that for each eigenfunction $\varphi $ of $V(s)$, $G(t,s)\varphi \neq 0$ is an eigenfunction of $V(t)$ with the same eigenvalue, for $t>s$.

 If $G(t,s)$ is compact in $C_b(\R^d)$ for $t>s$, then  $\sigma (V(s))\setminus \{0\}$ consists of isolated eigenvalues, hence it is independent of $s$. Therefore,

\begin{equation}
\label{r}
\sup\{|\lambda| : \;\lambda\in \sigma(V(s)), \; |\lambda|<1, \;s\in \R\} := r <1.
\end{equation}

Denoting  by $Q(s)$ the spectral projection

$$Q(s) = \frac{1}{2\pi i} \int_{\partial B(0,a)} (\lambda I- V(s))^{-1}d\lambda , \quad s\in \R , $$

with any $a\in (r, 1)$,  it is not difficult to see that for every $\eps >0$ there is $M_{\eps}>0$ such that

\begin{equation}
\label{decexp}
\|G(t,s)Q(s)\varphi  \|_{\infty} \leq M_{\eps}e^{(t-s)(\log r(s) +\eps)/T}\|\varphi\|_{\infty}, \quad t>s, \;\varphi \in C_b(\R^d).
\end{equation}

(The proof may be obtained from the proof of  \eqref{decexp2}  in Proposition \ref{comp_as_Lp}, replacing the $L^p$ spaces considered there by $C_b(\R^d)$).

In the proof of the next proposition we use an important corollary of the Krein-Rutman Theorem, whose proof may be found in e.g. \cite[Ch.~1]{Du}.

 \begin{thm}
\label{KR}
Let $K$ be a cone with nonempty interior part $\oo{K}$ in a Banach space $X$, and let $L:X\mapsto X$ be a linear compact operator such that $L\varphi \in \oo{K}$ for each $\varphi \in K\setminus\{0\}$. Then the spectral radius $r$ of $L$ is a simple eigenvalue of $L$, and all the other eigenvalues
have modulus $<r$.
\end{thm}

\begin{prop}
\label{pr:compact}
If $G(t,s)$ is compact in $C_b(\R^d)$ for $t>s$, then $1$ is a simple eigenvalue of $V(s)$ for each $s$, and it is the unique  eigenvalue on the unit circle. The spectral projection $P(s) = I - Q(s)$ is given by
$$P(s)\varphi(x) = \int_{\R^d} \varphi(y)\mu_s(dy), \quad \varphi \in C_b(\R^d), \;s\in \R , x\in \R^d, $$

where $\{ \mu_s :\;s\in \R\}$ is a $T$-periodic  evolution system of measures.
\end{prop}
\begin{proof}
Let $K= \{\varphi \in C_b(\R^d):\; \forall x\in \R^d, \;\varphi(x) \geq 0\}$ be the cone of the nonnegative functions in $C_b(\R^d)$. By Remark \ref{rem:comp}(iv),  if $\varphi \in C_b(\R^d)\setminus \{0\}$ is such that $\varphi(x)\geq 0$ for each $x$, then
$\inf V(s)\varphi   = \inf G(s+T,s)\varphi(x) >0$. In other words, $V(s)$ maps   $K\setminus \{0\}$ into the interior part of $K$.
Theorem \ref{KR}  implies that the spectral radius $1$ of $V(s)$ is a simple eigenvalue, and it is  the unique eigenvalue of $V(s)$ on the unit circle. The associated spectral projection  $P(s) = I- Q(s)$, with $Q(s)$ defined above, may be expressed as

 $$P(s)\varphi = m_s\varphi \, \one , \quad \varphi\in C_b(\R^d) , $$

for some $m_s$ in the dual space of $C_b(\R^d)$. To prove that $m_s\varphi = \int_{\R^d} \varphi(y)\mu_s(dy)$ for some   measure $\mu_s$ we use the Stone--Daniell Theorem (e.g., \cite[Thm.~4.5.2]{Dudley}): it is enough to check that $m_s\varphi \geq 0$ if $\varphi \geq 0$, and that
for each sequence  $(\varphi_n) \subset C_b(\R^d)$   such that   $(\varphi_n(x))$ is decreasing and converges to $0$ for each $x\in \R^d$, we have $\lim_{n \to \infty} m_s\varphi_n  = 0$. In this case,  $\mu_s$ is a probability measure for every $s$, because $P(s)\one = \one$.

By the general spectral theory, $P(s) = \lim_{\lambda \to  1^-}V_{\lambda}$, where
$V_{\lambda}:=  (\lambda -1)(\lambda I - V(s))^{-1}$. In its turn,  $(\lambda I - V(s))^{-1} = \sum_{k=0}^{\infty}V(s)^k/\lambda^{k+1}$ maps nonnegative functions into nonnegative functions because  $V(s)$ does. Therefore, $m_s\varphi \, \one = P(s)\varphi \geq 0$ for each $\varphi \geq 0$.

Let now  $\varphi_n \downarrow 0$. We claim that $V(s) \varphi_n$ converges to $0$ uniformly. Indeed, since the measures $\{ p_{s+T,s,x}:\;x\in \R^d\}$ are tight, for each $\varepsilon >0$ there is $R>0$ such that $\int_{\R^d\setminus B(0,R)}  p_{s+T,s,x}(dy) \leq \eps$, for each $x\in \R^d$. On the other hand, $\varphi_n$ converges to $0$ uniformly on $B(0,R)$ by the Dini Monotone Convergence Theorem, so that for $n$ large, say $n\geq n_0$, we have $  \varphi_n(y) \leq \eps$, for $|y|\leq R$.  Therefore, for $n\geq n_0$ we have

$$\begin{array}{lll}
0\leq   V(s)  \varphi_n(x) & = & \displaystyle{\int_{B(0,R)}  \varphi_n(y)p_{s+T,s,x}(dy) +  \int_{\R^d\setminus B(0,R)}  \varphi_n(y)p_{s+T,s,x}(dy) }
\\
\\
& \leq &\eps + \|  \varphi_n\|_{\infty} \eps
\end{array}$$

for each $x\in \R^d$.
Since $V(s) \varphi_n$ converges to $0$ uniformly, then $P(s)V(s)  \varphi_n = V(s) P(s) \varphi_n$ converges to $0$ uniformly. But $V(s)$ is the identity on the range of $P(s)$. Then, $P(s) \varphi_n$ converges uniformly to  $0$, which implies that  $\lim_{n \to \infty} m_s\varphi_n  = 0$.

Let us prove that $\{ \mu_s :\;s\in \R\}$ is a $T$-periodic  evolution system of measures. Since $s\mapsto P(s)$ is $T$-periodic, then $\mu_s = \mu_{s+T}$ for each $s\in \R$. Moreover, since $V(t)G(t,s) = G(t,s)V(s)$, then $P(t)G(t,s)\varphi =G(t,s)P(s)\varphi$, for each
$\varphi\in C_b(\R^d)$. This means

$$\int_{\R^d} G(t,s)\varphi \,d\mu_t \one =  G(t,s) \bigg( \int_{\R^d} \varphi \, d\mu_s \,\one \bigg) , \quad \varphi\in  C_b(\R^d),$$

and since $G(t,s)\one = \one$, then

$$\int_{\R^d} G(t,s)\varphi \,d\mu_t =   \int_{\R^d} \varphi \, d\mu_s,  \quad \varphi\in  C_b(\R^d),$$

so that $\{ \mu_s :\;s\in \R\}$ is an  evolution system of measures. \end{proof}

\begin{cor}
\label{decay_infty}
Assume that  $G(t,s)$ is compact in $C_b(\R^d)$ for $t>s$. Then:
\begin{itemize}
\item[(i)] There exists a unique $T$-periodic evolution system of measures $\{ \mu_s :\;s\in \R\}$;
\item[(ii)] Setting $\omega_0 = \log r /T $, where $r $ is defined in \eqref{r}, for each $\omega >\omega_0$ there exists $M = M(\omega)>0$ such that

\begin{equation}
\| G(t,s)\varphi -   \int_{\R^d} \varphi \, d\mu_s\|_{\infty} \leq M  e^{\omega  (t-s)}\|\varphi\|_{\infty}, \quad t>s, \;\varphi\in C_b(\R^d),
\label{decexp1}
\end{equation}

while for $\omega<\omega_0$ there is no $M$ such that  \eqref{decexp1} holds.
\end{itemize}

\end{cor}
\begin{proof}
Let $\{ \mu_s :\;s\in \R\}$ be the $T$-periodic evolution system of measures given by Proposition \ref{pr:compact}. Since $P(s) \varphi   =\int_{\R^d} \varphi \, d\mu_s \one$, estimate \eqref{decexp} implies \eqref{decexp1}.  Since there exist eigenvalues of $V(s) $ with modulus $r$,
\eqref{decexp1} cannot hold for  $\omega<\omega_0$.  Indeed, if $V(s)\varphi = r\varphi $ then $P(s)\varphi  =0$ and $G(s+nT,s)\varphi = r^n\varphi = e^{\omega_0nT}\varphi$ for each $n\in \N$, so that $\| G(s+nT, s)\varphi -  m_s \varphi  \|_{\infty} = \| G(s+nT,s)\varphi    \|_{\infty} =
e^{\omega_0nT}\|\varphi\|_{\infty}$.

If $\{ \nu_s :\;s\in \R\}$ is another $T$-periodic evolution system of measures, fix $t\in \R$ and $\varphi \in C_b(\R^d)$. Since $G(t,s)\varphi -   \int_{\R^d} \varphi  \, d\mu_s$ goes to zero uniformly as $s\to -\infty$, then
$$ 0 = \lim_{s\to -\infty} \int_{\R^d} \bigg( G(t,s)\varphi -   \int_{\R^d} \varphi \, d\mu_s\bigg) d\nu_t =   \lim_{s\to -\infty}\bigg(  \int_{\R^d}  \varphi \, d\nu_s -   \int_{\R^d} \varphi \, d\mu_s\bigg).$$

Since $s\mapsto  \int_{\R^d}  \varphi \, d\nu_s -   \int_{\R^d} \varphi \, d\mu_s$ is $T$-periodic and goes to $0$ as $s\to -\infty$, it vanishes in $\R$. By the arbitrariness of  $\varphi$,  $\nu_s = \mu_s$ for every $s\in \R$.
\end{proof}

Once we have an evolution system of measures $\{ \mu_s :\;s\in \R\}$, $G(t,s)$ is extendable to a contraction (still denoted by $G(t,s)$) from $L^p(\R^d, \mu_s)$ to $L^p(\R^d, \mu_t)$ for $t>s$.
Compactness and asymptotic behavior results  in $C_b(\R^d)$ imply compactness and asymptotic behavior results in such $L^p$ spaces, as the next proposition shows.

\begin{prop}
\label{comp_as_Lp}
Let $G(t,s)$ be compact in $C_b(\R^d)$ for $t>s$. Then  for every $p\in (1, \infty)$,  $G(t,s): L^p(\R^d, \mu_s)\mapsto L^p(\R^d, \mu_t)$ is compact  for $t>s$. Moreover, for every $\omega \in (\omega_0, 0)$ and $p\in (1, \infty)$ there exist $M = M(\omega, p)>0$ such that

\begin{equation}
\| G(t,s)\varphi -   \int_{\R^d} \varphi \, d\mu_s\|_{L^p(\R^d, \mu_t)} \leq M e^{\omega (t-s)}\|\varphi\|_{L^p(\R^d, \mu_s)}, \quad t>s, \;\varphi\in L^p(\R^d, \mu_s),
\label{decexp2}
\end{equation}

and for every $\omega<\omega_0$ there is no $M$ such that \eqref{decexp2} holds. Here $\omega_0 $ is given by Corollary \ref{decay_infty}(ii).
\end{prop}
\begin{proof}
Let us prove that $G(t,s): L^p(\R^d, \mu_s)\mapsto L^p(\R^d, \mu_t)$ is compact  for $t>s$.
We have
$G(t,s) = G(t, (t+s)/2) G((t+s)/2, s)$ where $G((t+s)/2, s)$ is bounded from $L^{\infty}(\R^d)$ to $C_b(\R^d)$ by Theorem \ref{Th:G(t,s)}(iv), and $G(t, (t+s)/2)$ is compact in $C_b(\R^d)$. Therefore, $G(t,s)$ is compact in $L^{\infty}(\R^d)$.

Now, if $\mu_1$ and $\mu_2$ are probability measures and a linear operator is bounded from $L^1(\R^d, \mu_1)$ to $L^1(\R^d, \mu_2)$ and compact from $L^{\infty}(\R^d, \mu_1)$ to $L^{\infty}(\R^d, \mu_2)$, then it is compact from $L^p(\R^d, \mu_1)$ to $L^p(\R^d, \mu_2)$, for every $p\in (1, \infty)$ (the proof is the same as in \cite[Prop.~4.6]{MPW}, where only one probability measure was considered).

Let us prove \eqref{decexp2}. Since   $L^{\infty}(\R^d, \mu_t)  = L^{\infty}(\R^d, \mu_s)   = L^{\infty}(\R^d, dx)$ by \cite[Prop.~5.2]{KLL}, interpolating   \eqref{decexp1}
and $\| G(t,s)\varphi -   m_s\varphi\|_{L^1(\R^d, \mu_t)} \leq 2\|\varphi\|_{L^1(\R^d , \mu_s)} $ we obtain that
$\| G(t,s)  -  m_s  \|_{{\mathcal L}(L^p(\R^d, \mu_s), L^p(\R^d , \mu_t))} $ decays exponentially as $(t-s)\to \infty$. However, the decay rate that we obtain by interpolation depends on $p$. To prove \eqref{decexp2} it is enough to show  that the spectrum of the operators $V(s)$ in $L^p(\R^d, \mu_s)$ does not depend on $s$, and coincides with the spectrum in $C_b(\R^d)$. Since $V(s) = G(s+T,s)$, then $V(s)$ is compact  in $L^p(\R^d, \mu_s)$. Therefore, its $L^p$ spectrum (except zero) consists of eigenvalues, that are independent of $s$.
They are independent of $p$ too, as well as the associated spectral projections, by \cite[Cor.~1.6.2]{davies}.

The statement follows now as in the case of  evolution operators in a fixed  Banach space as in the mentioned references \cite{Henry,L,F}. Note however that our Banach spaces $L^p(\R^d, \mu_s)$ vary with $s$, so that the classical theory  cannot be used verbatim. For the reader's convenience we give the proof below.

Let  $t-s = \sigma + kT$, with $k\in \N$ and $\sigma \in [0, T)$. We have
$$\begin{array}{l}
\|G(t,s)\varphi -   m_s \varphi \|_{L^p(\R^d, \mu_t)} = \|G(t, t-\sigma) V(s)^k (I-P(s))\varphi \|_{L^p(\R^d, \mu_t)}
\\
\\
\leq \| V(s)^k (I-P(s))\varphi \|_{L^p(\R^d, \mu_s)} = \| [V(s)(I-P(s))]^k \varphi \|_{L^p(\R^d, \mu_s)}.
\end{array}$$

For $\omega>\omega_0$ let $\eps>0$ be such that $\log (r+\eps)\leq \omega $, and let
 $k(s)\in \N$ be such that $ \| [V(s)  (I-P(s))]^k\varphi \|_{L^p(\R^d, \mu_s)}\leq (r+\eps)^k$ for each $k > k(s)$. Therefore, if the integer part $[(t-s)/T]$ is larger than $k(s)$ we have

$$\|G(t,s)\varphi -  m_s\varphi \|_{L^p(\R^d, \mu_t)} \leq (r+\eps)^k\|\varphi \|_{L^p(\R^d, \mu_s)}
\leq e^{(t-s) \omega} \|\varphi \|_{L^p(\R^d, \mu_s)}, $$

for each $  \varphi  \in L^p(\R^d, \mu_s)$. Using the obvious inequality  $\|G(t,s)\varphi -   m_s\varphi \|_{L^p(\R^d, \mu_t)} \leq 2 \|\varphi \|_{L^p(\R^d, \mu_s)}$ for
$[(t-s)/T]\leq k(s)$ we arrive at

$$\|G(t,s)\varphi -  m_s\varphi \|_{L^p(\R^d, \mu_t)} \leq M_s e^{(t-s) \omega} \|\varphi \|_{L^p(\R^d, \mu_s)}, \quad \varphi  \in L^p(\R^d, \mu_s)$$

for some $M_s>0$.  It remains to show that $M_s$ can be taken independent of $s$. Since $V$ is $T$-periodic, we may take $k(s)= k(s+T)$  and hence $M_s = M_{s+T}$ for every $s\in \R$. Therefore it is enough to show that $M_s$ can be taken independent of $s$ for $s\in [0,T)$.
For $0\leq s<T$ and $t\geq T$ we have $m_T  G(T,s)\varphi  = \int_{\R^d} G(T,s)\varphi\,d\mu_T = m_s\varphi$,  hence

$$\begin{array}{l}
\|G(t,s)\varphi -  m_s\varphi  \|_{L^p(\R^d, \mu_t)} = \| (G(t, T)-m_T) G(T,s)\varphi  \|_{L^p(\R^d, \mu_t)}
\\
\\
\leq M_Te^{\omega(t-T)} \| G(T,s)\varphi \|_{L^p(\R^d, \mu_T)} \leq  M_Te^{|\omega| T}e^{\omega(t-s)} \|  \varphi \|_{L^p(\R^d, \mu_s)}
\end{array}$$

So, we can take $M_s = M_Te^{|\omega| T}$ for $0\leq s<T$. \eqref{decexp2} follows.
\end{proof}