text:
\begin{document}
\title{Weak Markov Operators}
\author{H\={u}lya Duru \and Serkan Ilter}

\institute{H. Duru \at
              Istanbul University, Faculty of Science, Mathematics Department,
Vezneciler-Istanbul, 34134, Turkey \\
              Tel.: +90-212-4555700 (15315)\\
              \email{hduru@istanbul.edu.tr}

           \and
           S. Ilter \at
              Istanbul University, Faculty of Science, Mathematics Department,
Vezneciler-Istanbul, 34134, Turkey \\
              \email{ilters@istanbul.edu.tr}
}

\date{Received: date / Accepted: date}

\maketitle

\begin{abstract}
Let $A$ and $B$ be $f$-algebras with unit elements $e_{A}$ and $e_{B}$
respectively. A positive operator $T$ from $A$ to $B$ satisfying $T\left(
e_{A}\right) =e_{B}$ is called a Markov operator. In this definition we
replace unit elements with weak order units and, in this case, call $T$ to
be a weak Markov operator. In this paper, we characterize extreme points of
the weak Markov operators.
\end{abstract}

\subclass{47B38 \and 46B42}
\keywords{f-algebra \and lattice homomorphism \and weak order unit \and
weak Markov operator}

\section{INTRODUCTION}

Let $A$ and $B$ be $f$-algebras with point separating order duals and unit
elements $e_{A}$ and $e_{B}$ respectively. A positive linear operator $T$
from $A$ to $B$ satisfying $T\left( e_{A}\right) =e_{B}$ is called a Markov
operator. In this definition we replace unit elements with weak order units
and, in this case, call $T$ to be a weak Markov operator.The set of all weak
Markov operators is convex. In this paper, we characterize extreme points of
the weak Markov operators. In this regard, first we give an alternate and
quick proof of \cite{Ref6}\textrm{\ }for $f$-algebras instead of
order complete real vector spaces (Proposition \ref{p3}). Then we give an alternate
proof of Theorem $5.7$ in \cite{Ref5}\textrm{\ }(Theorem \ref{p3}). In
addition we present another necessary and sufficient condition to this
Theorem. Then we show that a weak Markov operator is a lattice homomorphism
if and only if it is an extreme point in the collection of all weak Markov
operators from $A$ into $B$ provided $B$ is order complete.

\section{PRELIMINARIES}

For unexplained terminology and the basic results on vector lattices and
semiprime $f$-algebras we refer to \cite{Ref1,Ref10,Ref13}. Let us recall some definitions.

\begin{definition}\label{d1}
The real algebra $A$ is called a Riesz algebra or lattice-ordered algebra if
$A$ is a Riesz space such that $ab\in A$ whenever $a,b$ are positive
elements in $A.$ The Riesz algebra is called an $f$-algebra if $A$ satisfies
the condition that
\begin{equation*}
a\wedge b=0\text{ implies }ac\wedge b=ca\wedge b=0\text{ for all }0\leq c\in
A\text{.}
\end{equation*}
\end{definition}

In an Archimedean $f$-algebra $A$, all nilpotent elements have index $2$.
Throughout this paper \thinspace $A$ will show an Archimedean semiprime $f$
-algebra with point separating order dual. By definition, if zero is the
unique nilpotent element of $A$, that is, $a^{2}=0$ implies $a=0$, $A$ is
called semiprime $f$-algebra. It is well known that every $f$-algebra with
unit element is semiprime.

\begin{definition}\label{d2}
Let $A$ be an Archimedean Riesz space and let $\widehat{A}$ be the Dedekind
completion of $A$. The closure of $A$, $\overline{A}$, in $\widehat{A}$ with
respect to the relatively uniform topology \cite{Ref8}\textrm{\ }is
so called that relatively uniformly completion of $A$ \cite{Ref9}
\textrm{. }
\end{definition}

If $A$ is an semiprime $f$-algebra then the multiplication in $A$ can be
extended in a unique way into a lattice ordered algebra multiplication on $
\overline{A}$ such that $A$ becomes a sub-algebra of $\overline{A}$ and $
\overline{A}$ is an relatively uniformly complete semiprime $f$-algebra. We
also recall that $\overline{A}$ satisfies the Stone condition (that is, $
x\wedge nI^{\ast }\in \overline{A}$, for all $x\in \overline{A}$, where $I$
denotes the identity on $A$ of $OrthA$) due to Theorem $2.5$ in \cite{Ref4}.
For $a\in A,$ the mapping $\pi _{a}:A\rightarrow OrthA,$ defined
by $\pi _{a}\left( b\right) =a.b$ is an orthomorphism on $A.$ Since $A$ is a
Archimedean semiprime $f$-algebra,$,$ the mapping $\pi :A\rightarrow OrthA,$
defined by $\pi \left( a\right) =\pi _{a}$ is an injective $f$-algebra
homomorphism. Hence we shall identify $A$ with $\pi \left( A\right) .$The
ideal center $Z\left( A\right) $of $A$ is defined as the order ideal in $
OrthA$ generated by the identity mapping $I$ which is a unital f-algebra.

\begin{proposition}\label{p1}
\textbf{\ }Let $A$ and $B$ be semiprime $f$-algebras and $T:A\rightarrow B$
an order bounded linear operator. If $T\left( A\cap Z\left( A\right) \right)
=\{0\}$ then $T\left( A\right) =\{0\}$.
\end{proposition}

\begin{proof}
$T$ can be regarded as an element of the collection of all order bounded
linear operators from $A$ to $\widehat{B}$. Therefore there exist two order
bounded positive operators $T_{1}$ and $T_{2}$ such that $T=T_{1}-T_{2}$.
Thus we can assume that $T$ is positive. Since $\widehat{B}$ is relatively
uniformly complete, by Theorem $3.3$ in \cite{Ref11} there exists a
unique positive relatively uniformly continuos extension $\overline{T}:
\overline{A}\rightarrow \widehat{B}$ of $T$ to the relatively uniformly
completion $\overline{A}$ of $A$, defined by,
\begin{equation*}
\overline{T}\left( x\right) =\sup \left\{ Ta:0\leq a\leq x\right\}
\end{equation*}
for $x\in \overline{A}$. Let $0\leq x\in \overline{A}\cap Z\left( \overline{A
}\right) $ and $a\in A\cap \left[ 0,x\right] $. Then $a\leq \lambda I_{A}$
holds for some positive real number $\lambda $. By the hypothesis, $Ta=0$.
From here we conclude that $\overline{T}\left( x\right) =0$ for all $x\in
\overline{A}\cap Z\left( \overline{A}\right) $. Let $0\leq a\in A$. Then $
\left( a\wedge n\overline{I}\right) _{n}$ is a sequencein $\overline{A}\cap
Z\left( \overline{A}\right) $ and it is converging to $a$ by Proposition $
2.1\left( i\right) $ in \cite{Ref5}. It follows from the relatively
uniformly continuity of $\overline{T}$ that $0=\overline{T}\left( a\wedge
nI_{A}\right) \rightarrow \overline{T}\left( a\right) =T\left( a\right) $
(relatively uniformly). Thus $T\left( a\right) =0$ for all $a\in A$.
\end{proof}

\begin{proposition}\label{p2}
Let $A$ and $B$ be $f$-algebras with unit elements $e_{A}$ and $e_{B}$
respectively and $T:A\rightarrow B$ an order bounded linear operator
satisfying $T\left( e_{A}\right) =e_{B}$. If $\mid Ta\mid \leq e_{B}$
whenever $\mid a\mid \leq e_{A}$, then $T$ is positive.
\end{proposition}

\begin{proof}
$T$ can be regarded as an element of the collection of all order bounded
linear operators from $A$ to $\widehat{B}$. As $\widehat{B}$ is Dedekind
complete, $\mid Ta\mid \leq \mid T\mid \left( a\right) =\sup \left\{ \mid
Tb\mid :\mid b\mid \leq a,b\in A\right\} $ holds for all $0\leq a\in A$. Let
$a\in A$ be an element such that $\mid a\mid \leq e_{A}$. Then by the
hypothesis, $\mid Ta\mid \leq e_{B}$. This shows that $\mid Te_{A}\mid
=e_{B}\leq \mid T\mid \left( e_{A}\right) \leq e_{B}$. Thus $\mid T\mid
\left( e_{A}\right) =e_{B}$. Let $a$ $\in A\cap Z\left( A\right) $. Then
there exists a positive real number $\lambda $ such that $\mid a\mid \leq
\lambda e_{A}$. From here we derive that
\begin{equation*}
\left( \mid T\mid -T\right) \left( \mid a\mid \right) \leq \lambda \left(
\mid T\mid -T\right) \left( e_{A}\right) =0
\end{equation*}
Thus the order bounded operator $\mid T\mid -T$ vanishes on $A\cap Z\left(
A\right) $. By proposition $1$, we conclude that $\mid T\mid =T$ on $A$, so $
T$ is positive.
\end{proof}

The following proposition was proved in \cite{Ref6}\textrm{\ }for an
order complete real vector space. In the following proposition we will give
an alternate and quick proof for Archimedean semiprime $f$-algebras.

\begin{proposition}\label{p3}
Let $A$ be a semiprime $f$-algebra. A positive element $a_{0}$ in $A$ is a
weak order unit if and only if \ $Inf\left\{ \mid a-\lambda a_{0}\mid
:\lambda \in

\mathbb{R}

\right\} =0$ for all $a\in A$.
\end{proposition}

\begin{proof}
Let $0<a_{0}$ $\in $ $A$ be a weak order unit. Assume that, on the contrary,
there exist $a\in A$ and $b$ $\in $ $\widehat{A}$ satisfying $0<b\leq \mid
a-\lambda a_{0}\mid $for all $\lambda \in

\mathbb{R}

$. Then
\begin{equation*}
\lambda ^{2}a_{0}^{2}-2\lambda aa_{0}+a^{2}-b^{2}\geq 0\text{ for all }
\lambda \in

\mathbb{R}

\end{equation*}
Using Proposition $3.3$ in \cite{Ref3}, we get $\left( a_{0}b\right)
^{2}=0$. Since $\widehat{A}$ is semiprime, the last equality implies that
\begin{equation*}
0=a_{0}b=\sup \left\{ aa_{0}\in A:a\leq b,\text{ }a\in A\right\} \text{.}
\end{equation*}
As $a_{0}$ is weak order unit, $aa_{0}=0$ implies that $a=0$ for all $a\in A$
satisfying $a\leq b$. Thus $b=0$. Conversely, suppose that there exists $
0\neq b\in A$ such that $ba_{0}=$ $0$. Then since $0<\frac{\mid b\mid }{2}
\leq \mid b-\lambda a_{0}\mid $ for all $\lambda \in

\mathbb{R}

$, we get $Inf\left\{ \mid b-\lambda a_{0}\mid :\lambda \in

\mathbb{R}

\right\} \neq 0$.
\end{proof}

A positive linear operator $T$ between two unital $f$-algebras $A$ and $B$
is \ said to be a Markov operator for which $T\left( e_{A}\right) =e_{B}$
where $e_{A}$, $e_{B}$ are the unit elements of $A$ and $B$ respectively. In
both \cite[Theorem~5.7]{Ref5}\textrm{, }it is proved that an
operator is an extreme point of Markov operators if and only if it is a
lattice homomorphism. In the following theorem we shall give an alternate
proof of this theorem and another sufficient and necessary condition for the
extreme point of a Markov operator.

\begin{theorem}\label{t1}
Let $A$ and $B$ be $f$-algebras with unit elements $e_{A}$ and $e_{B}$
respectively and $T:A\rightarrow B$ an order bounded linear operator
satisfying $T\left( e_{A}\right) =e_{B}$ and $\mid Ta\mid \leq e_{B}$
whenever $\mid a\mid \leq e_{A}$. Then the following are equivalent;\newline
$\left( i\right) $ $T$ is an extreme point of Markov operators.\newline
$\left( ii\right) $ $Inf\left\{ T\mid a-\lambda e_{A}\mid :\lambda \in

\mathbb{R}

\right\} =0$ for all $a\in A$.\newline
$\left( iii\right) $ $T$ is a lattice homomorphism.
\end{theorem}

\begin{proof}
$\left( i\right) \Rightarrow \left( ii\right) :$ By Proposition  \ref{p2}, $T$ is
positive. Suppose, on the contrary, that $Inf\left\{ T\mid a_{0}-\lambda
e_{A}\mid :\lambda \in

\mathbb{R}

\right\} \neq 0$ for some $0\leq a_{0}\in A$. Then there exists $b\in
\widehat{B}$ such that $0<b\leq T\mid a_{0}-\lambda e_{A}\mid $ for all $
\lambda \in

\mathbb{R}

$. Taking square from both side and applying the Schwarz inequality \cite[Corollary~3.5] {Ref3}, one can get easily
\begin{equation*}
\lambda ^{2}e_{B}-2\lambda Ta_{0}+T\left( a_{0}^{2}\right) -b^{2}\geq 0\text{
for all }\lambda \in

\mathbb{R}

\text{.}
\end{equation*}
Taking into account Proposition $3.3$ in \cite{Ref3}, we derive that

\begin{equation}
\left( Ta_{0}\right) ^{2}\leq T\left( a_{0}^{2}\right) -b^{2} \label{eq1}.
\end{equation}

Let, $a_{n}=\frac{a_{0}}{n}\wedge e_{A}$ $\left( n=1,2,...\right) $. Then $
0<a_{n}$, as $Inf\left\{ T\mid a_{0}-\lambda e_{A}\mid :\lambda \in

\mathbb{R}

\right\} \neq 0$, for each $n$. Define,
\begin{equation}
T_{a_{n}}\left( b\right) =T\left( a_{n}b\right) -T\left( a_{n}\right)
T\left( b\right) \label{eq2}
\end{equation}
for $b\in A$. Clearly the operators $T\mp T_{a_{n}}$ are positive and $
T_{a_{n}}\left( e_{A}\right) =0$. Hence $T\mp T_{a_{n}}$ are Markov
operators. From here we conclude that $T_{a_{n}}=0$, as $T$ is the extreme
point of Markov operators. Hence the equality (\ref{eq2}) implies
that
\begin{equation*}
nT\left( a_{n}b\right) =nT\left( a_{n}\right) T\left( b\right)
\end{equation*}
and by setting $a_{n}$, we have $T(\left( a_{0}\wedge ne_{A}\right)
a_{0})=T\left( a_{0}\wedge ne_{A}\right) T\left( a_{0}\right) $. Applying
Proposition $2.1\left( i\right) $ in \cite{Ref5}\textrm{\ }and using
the positivity of $T$, one may get,
\begin{equation*}
T\left( a_{0}^{2}\right) =\left( T\left( a_{0}\right) \right) ^{2}\text{.}
\end{equation*}
Thus the ineaquality (\ref{eq1}) implies that $b=0$, as $\overset{
\wedge }{B}$ is semiprime whenever $B$ is semiprime.\newline
$\left( ii\right) \Rightarrow \left( iii\right) :$ Let $a\in A$ and $\lambda
\in

\mathbb{R}

$. Then
\begin{equation*}
T\mid a\mid \leq T\mid a-\lambda e_{A}\mid +T\mid \lambda e_{A}\mid =T\mid
a-\lambda e_{A}\mid +\mid T(\lambda e_{A})\mid \leq 2T\mid a-\lambda
e_{A}\mid +\mid Ta\mid
\end{equation*}
it follows from $\left( ii\right) $ that $T\mid a\mid \leq \mid Ta\mid $ and
since $T$ is positive, $T\mid a\mid =\mid Ta\mid $ which implies that $T$ is
a lattice homomorphism.\newline
$\left( iii\right) \Rightarrow \left( i\right) :$ By Proposition \ref{p3}, we get
that $Inf\left\{ \mid Ta-\lambda e_{B}\mid :\lambda \in

\mathbb{R}

\right\} =0$ for all $a\in A$, as $e_{B}$ is a weak order unit in $B$. To
prove that $T$ is an extreme point of Markov operators it is enough to show
that for any Markov operator $S$ and $0<\alpha \in

\mathbb{R}

$ satisfying $\alpha T-S\geq 0$ implies that $T=S$. Let $a\in A$ and $
\lambda \in

\mathbb{R}

$. It follows from
\begin{equation*}
\mid Sa-S\left( \lambda e_{A}\right) \mid \leq S\mid a-\lambda e_{A}\mid
\leq \alpha T\mid a-\lambda e_{A}\mid =\alpha \mid Ta-\lambda e_{B}\mid
\end{equation*}
that
\begin{equation*}
\mid Ta-Sa\mid \leq \mid Ta-\lambda e_{B}\mid +\mid Sa-\lambda e_{B}\mid
\leq \left( 1+\alpha \right) \mid Ta-\lambda e_{B}\mid
\end{equation*}
Hence $Ta=Sa$ for $a\in A$.
\end{proof}

\begin{definition}\label{d3}
Let $A$ and $B$ be $f$-algebras with point separating order duals and weak
order units $e_{A}$ and $e_{B}$ respectively. In this case, we call a
positive linear operator $T$ from $A$ to $B$ satisfying $T\left(
e_{A}\right) =e_{B}$ to be a weak Markov operator $\left( \text{briefly WMO}
\right) $.
\end{definition}

Now we remark that in the last step $\left( iii\right) \Rightarrow \left(
i\right) $ of the above proof we proved the following Corollary as well;

\begin{corollary}\label{c1}
Let $A$ and $B$ be a semiprime $f$-algebras with weak order units $e_{A}$
and $e_{B}$ respectively and let $T:A\rightarrow B$ be a weak Markov
operator. If $T$ is a lattice homomorphism, then it is an extreme point in
the set of all weak Markov operators.
\end{corollary}

In \cite[Lemma~4.2] {Ref2}, it was proved that a positive
contractive Stone operator between two Stone $f$-algebras is a lattice
homomorphism. At this point we remark that this result is true for all
positive linear operator between any semiprime $f$-algebras. For the
completeness, we repeat this proof in the following Proposition.

\begin{proposition}\label{p4}
Let $A$ and $B$ be semiprime $f$- algebras and $T:A\rightarrow B$ a positive
linear operator. If there exists $0\leq a_{0}\in A$ and $0\leq b_{0}\in B$
such that $T\left( a\wedge a_{0}\right) =T\left( a\right) \wedge b_{0}$, for
all $a\in A$, then $T$ is a lattice homomorphism.
\end{proposition}

\begin{proof}
We remark that, for all elements $a$ in $A$ or $B$, the following equations
hold;
\begin{eqnarray*}
\left( a\wedge a_{0}\right) ^{+} &=&a^{+}\wedge a_{0} \\
\left( a\wedge a_{0}\right) ^{-} &=&a^{-}
\end{eqnarray*}
Since $T$ is positive, in order to prove that $T$ is a lattice homomorphism
it is enough to show that $T(a^{+})\leq (Ta)^{+}$ for all $a\in A$. By above
remark,
\begin{eqnarray*}
Ta &=&T\left( a^{+}-a^{-}-(a\wedge \frac{1}{n}a_{0})^{+}+(a\wedge \frac{1}{n}
a_{0})^{-}\right) +T(a\wedge \frac{1}{n}a_{0}) \\
&=&T\left( a^{+}-(a\wedge \frac{1}{n}a_{0})^{+}\right) +(T(a\wedge \frac{1}{n
}a_{0}))
\end{eqnarray*}
and
\begin{eqnarray*}
Ta &=&\left( Ta\right) ^{+}-\left( Ta\right) ^{-}-(T(a\wedge \frac{1}{n}
a_{0}))^{+}+(T(a\wedge \frac{1}{n}a_{0}))^{-}+T(a\wedge \frac{1}{n}a_{0}) \\
&=&\left( Ta\right) ^{+}-(T(a\wedge \frac{1}{n}a_{0}))^{+}+(T(a\wedge \frac{1
}{n}a_{0}))
\end{eqnarray*}
combining these results, we get
\begin{equation*}
T\left( a^{+}\right) -\left( Ta\right) ^{+}=^{-}T\left( a^{+}\wedge \frac{1}{
n}a_{0}\right) -\left( Ta\right) ^{+}\wedge \frac{1}{n}b_{0}\leq \frac{1}{n}
T\left( a_{0}\right)
\end{equation*}
Passing limit as $n\rightarrow \infty $, we have the desired result.
\end{proof}

\begin{corollary}\label{c2}
\label{7}Let $A$ and $B$ be semiprime $f$- algebras with weak order units $
e_{A}$ and $e_{B}$ respectively and $T:A\rightarrow B$ a weak Markov
operator. If there exists $0\leq a_{0}\in A$ and $0\leq b_{0}\in B$ such
that $T\left( a\wedge a_{0}\right) =T\left( a\right) \wedge b_{0}$, for all $
a\in A$, then $T$ is an extreme point of weak Markov operators.
\end{corollary}

\begin{proof}
By Proposition \ref{p4}, $T$ is a lattice homomorphism and by Corollary \ref{c1}, $T$
is an extreme point of weak Markov operators.
\end{proof}

\begin{theorem}\label{t2}
Let $A$ and $B$ be a semiprime $f$-algebras with weak order units $e_{A}$
and $e_{B}$ respectively and $T:A\rightarrow B$ a weak Markov operator. If $
B $ is order complete, then $T$ is an extreme point of the set of all weak
Markov operators if and only if $T$ is a lattice homomorphism.
\end{theorem}

\begin{proof}
First we remark that any weak Markov operator is an extension of the
operator $f:M\rightarrow B$, defined by $f\left( \lambda e_{A}\right)
=\lambda e_{B}$, where $M=\left\{ \lambda e_{A}:\lambda \in

\mathbb{R}

\right\} $. It is not difficult to see that the extreme point of the
collection of all weak Markov operators is the same of the extreme point of
all positive extensions of $f$ to the whole $A$. Let $T~$\ be an extreme
point in the set of all weak Markov operators. Then $T$ is an extreme point
of the set of all positive extensions of $f$ to $A$. Since $f$ is a lattice
homomorphism, by Theorem $2\left( a\right) $ in \cite{Ref7},\textrm{\
}we derive that $T$ is a lattice homomorphism. Conversely assume that $T$ is
a lattice homomorphism. Taking into account Proposition \ref{p3} and Theorem $
2\left( b\right) $ in \cite{Ref7}\textrm{,} we conclude that $T$ is
an extreme point of the weak Markov operators.
\end{proof}


TITLE: Surface: intersection of 2 polar curves
QUESTION [1 upvotes]: I have these two polar curves:
$$
C_1: r = 2 - \cos(\theta)\\
C_2: r = 3 \cos(\theta)
$$
Plots: C1 and C2.
I need to find the surface of $D = C_1 \cap C_2$.
I started by finding the solution to $C_1 = C_2$. I found $\pm \pi\over{3}$.
I tried this integral:
$$
\int_{-\pi\over{3}}^{\pi\over{3}} \int_{3\cos(\theta)}^{2-\cos(\theta)} r \;\operatorname d\!r \operatorname d\!\theta
$$
However, I'm not sure about that $r$, and I get a negative answer.
I'd appreciate if someone could point me in the right direction!

REPLY [1 votes]: The problem arises because the angular interval over which you want to integrate the area between the two curves does not correspond for each curve.  This is because the curve $r = 3 \cos \theta$ is traversed twice when $r = 2 - \cos \theta$ is traversed only once for $\theta \in [0,2\pi)$:
To rectify this, we would use the method shown in JimmK4542's answer:  over $\theta \in [0,\pi/3]$, we integrate the sector traced by the blue curve, and then from $\theta \in [\pi/3, \pi/2]$, we integrate the slice traced by the red curve.  However, you can "rescue" your calculation by computing the negative area as you did, and then simply add the area of the red circle:  $$A = \pi \bigl(\tfrac{3}{2}\bigr)^2 + \int_{\theta = -\pi/3}^{\pi/3} \int_{r = 3 \cos \theta}^{2 - \cos \theta} r \, dr \, d\theta.$$


TITLE: Give an example of an open cell in a cell complex for which the image of the characteristic map is not a closed cell.
QUESTION [2 upvotes]: Let $(X, \Gamma)$ be a cell complex, and let $e \in \Gamma$ be a open cell of dimension $n \geq 1$.
I want to try and construct an example of a characteristic map $\Phi : D \to X$ (where $D$ is some closed $n$-cell) for which $\Phi[D] = \operatorname{Cl}_X(e)$ is not a closed cell. To do that I need to find a $\Phi$ that is not injective on the boundary of $D$.
I am not sure how to construct such an example (and in this example, I'd need to choose, $X$, $\Gamma$, $e$ and $\Phi$ appropriately so that $\Phi$ is not injective on the boundary)
The biggest problem is having that $\Phi$ must be continuous, for example we could take $X = [0, 1]$ and $\Gamma = \{(0, 1) \{0\}, \{1\}\}$ and define $\Phi : [0, 1] \to [0, 1]$ by $\Phi(x) = x$ for $x \in (0, 1)$ and $\Phi(x) = 0$ on $x \in \{0, 1\}$, and $\Phi$ satisfies all the criteria for being a characteristic map except for being continuous.

Note that this is not a homework problem, but rather a problem that I came up with while reading a claim in a book and wanting to find an explicit example for myself

REPLY [1 votes]: Here's the simplest example.  Let $X$ be the circle $\{z\in\mathbb{C}:|z|=1\}$ and put a cell complex structure on it as follows.  We have one $0$-cell, which is just $\{1\}$.  We have one $1$-cell, whose characteristic map $\Phi:[0,1]\to X$ is defined by $\Phi(t)=e^{2\pi i t}$.  This is not injective, since $\Phi(0)=\Phi(1)$.


TITLE: Analytical solution to non-linear 2nd order ODE
QUESTION [2 upvotes]: Is there an analytical solution to the following non-linear ODE?
$$y''(x) = A \frac{y}{B+\alpha y} ... (1)$$ where $A,B,\alpha$ are constants.
The boundary conditions are:
$$y'(x = 0) = y'(x = L) = 0$$

REPLY [1 votes]: $y(x)=0$ is a solution to the problem.


TITLE: Finding the number of distinct elements in a group generated by a permutation and a function over tuple.
QUESTION [1 upvotes]: Okay, I'm not sure if the title is correct but is there any general way of finding the number of distinct elements that are generated by some function (for example $*$) and a permutation (in this case the permutation group that shifts all number to the right, $p$) and an $n$-tuple (in this case a 4-tuple for example $(1,0,-1,0)$ ).
So we have all combinations of $p(1,0,-1,0)*p(1,0,-1,0)$ so that:
$$ p_0(1,0,-1,0)*p_1(1,0,-1,0) = (1,0,-1,0)*(0,1,0,-1) = (0,0,0,0) $$
where p_n denotes n shifts of the tuple to the right.
Is there anyway of enumerating the combination such that the result of the multiplications, here $(0,0,0,0)$, are all distinct.

REPLY [3 votes]: This answers names a few things and shows you a program that makes it easy to work with things like this.
Names
If you have a single permutation p and a single vector v and a single function f, then the values you are looking for are called the image under f of the orbit of v under the action of the permutation group generated by p.
In your particular case, the standard algorithm is going to be very natural but not very impressive. Find all the possible rotated n-tuples, and then multiply them. :-)
In general, you can have more complicated permutations and then the standard algorithm is much more impressive, and will find you the distinct n-tuples quickly and tell you how it got them using Schreier vectors.
I'll explain them in reverse order: the permutation group G generated by a permutation p includes not only $p=p_1$, but also $p\cdot p = p_2$, and $p \cdot p \cdot p = p_3$. This might seem like a very large thing, but in your example it doesn't take long to finish: $p_4$ of a 4-tuple actually just leaves everything alone. $p_5 = p_1$ effectively shifts things to the right by 1, rather than worrying about shifting 5 times.  If you have more than one permutation p, then you need to allow applying any of the permutations over and over.
The orbit of v under the action of G is just everything that can happen to v. You apply every element of G to v once, and you get all the possible v. Alternatively you apply your (short) list of important permutations to v and keep track of the different answers in a simple way.
The image of the orbit under f is pretty simple: you take the tuples and you apply f to them. In your case, you just take pairs of n-tuples and multiply them.
If f has any relationship to G, then this last step can be improved. However, in your case there is no real relationship, and so one might as well do the naive thing.
Easy program
Should you want to do this in GAP you would use:
gap> p := (1,2,3,4);;
gap> G := Group( p );;
gap> v := [1,0,-1,0];;
gap> orbit := Orbit( G, v, Permuted );;
gap> f := function( x, y ) return List( [1..Size(x)], i -> x[i] * y[i] ); end;;
gap> values := Set( Tuples( orbit, 2 ), xy -> f( xy[1], xy[2] ) );
[ [ -1, 0, -1, 0 ], [ 0, -1, 0, -1 ], [ 0, 0, 0, 0 ], [ 0, 1, 0, 1 ], [ 1, 0, 1, 0 ] ]

If you allowed swapping q the first two elements in the vector too, then you'd do:
gap> p := (1,2,3,4);;
gap> q := (1,2);;
gap> G := Group( p, q );;
gap> v := [1,0,-1,0];;
gap> orbit := Orbit( G, v, Permuted );;
gap> f := function( x, y ) return List( [1..Size(x)], i -> x[i] * y[i] ); end;;
gap> values := Set( Tuples( orbit, 2 ), xy -> f( xy[1], xy[2] ) );;
gap> PrintArray(values);
[ [  -1,  -1,   0,   0 ],
  [  -1,   0,  -1,   0 ],
  [  -1,   0,   0,  -1 ],
  [  -1,   0,   0,   0 ],
  [   0,  -1,  -1,   0 ],
  [   0,  -1,   0,  -1 ],
  [   0,  -1,   0,   0 ],
  [   0,   0,  -1,  -1 ],
  [   0,   0,  -1,   0 ],
  [   0,   0,   0,  -1 ],
  [   0,   0,   0,   0 ],
  [   0,   0,   0,   1 ],
  [   0,   0,   1,   0 ],
  [   0,   0,   1,   1 ],
  [   0,   1,   0,   0 ],
  [   0,   1,   0,   1 ],
  [   0,   1,   1,   0 ],
  [   1,   0,   0,   0 ],
  [   1,   0,   0,   1 ],
  [   1,   0,   1,   0 ],
  [   1,   1,   0,   0 ] ]


TITLE: How can a rigid body's weight do work on it to make it rotate?
QUESTION [4 upvotes]: Consider a cylinder that rolls without sliding on an inclined plane. If it's placed at the top of the plane, with its center of mass at a height $h$ from the bottom, it will have a potential energy $mgh$ (considering the bottom of the plane as the zero point). Then, the cylinder will start rolling down the plane, as its weight does work on it, causing its potential energy to be transformed into rotational and translational kinetic energy.
I'm confused about the origin of the rotational kinetic energy of the cylinder. Since the static friction and the normal force are applied at the point of contact with the plane, neither of them can do work on the cylinder (which is why its mechanical energy is conserved), since the cylinder is, by constraint, not sliding. This seems to imply that the cylinder's rotational kinetic energy comes from the work done by its weight.
However, I don't understand how this happens, since a rigid body's weight can be seen as being applied on its center of mass, which means it can exert no torque on it. Instead, the only torque exerted with respect to its center of mass comes from static friction.
Moreover, if the inclined plane was frictionless, there would be no torque exerted on the cylinder, so it would not rotate. Yet, I don't see how this situation could be distinguished from the previous one by only looking at the cylinder's mechanical energy (since the normal force would still be doing no work).
So, my question is: how is gravitational potential energy converted into rotational kinetic energy, and what is friction's role on this?

REPLY [6 votes]: It is important to keep in mind that there are three different conserved quantities of interest here: linear momentum, angular momentum, and energy. Rotational kinetic energy is not itself a conserved quantity.
Each of the conserved quantities has an associated rate of change, or “flow”. The rate of change of linear momentum is the force, the rate of change of the angular momentum is the torque, and the rate of change of the energy is the power.
Each interaction can produce all three: force, torque, and power. It is not necessary that the force which delivers torque also deliver power.
For a disk rolling without slipping there are two interactions, the gravitational interaction and the friction interaction. Assuming no dissipation, it is straightforward to show that the change in the linear momentum is equal to the sum of the frictional and gravitational forces, that the change in the angular momentum (about the center of mass) is equal to the torque from the friction only, and that the change in the energy is equal to the power from gravity only.
It is incorrect to assume that the torque must provide any energy. It does not. It only provides angular momentum. Only the power provides energy, and that comes entirely from the gravitational interaction. The frictional interaction provides torque. It does not provide power, although it does provide a constraint which splits the power into rotational and translational KE. Providing such a constraint does not itself require energy.


TITLE: Can this string of inequalities take on arbitrary values?
QUESTION [0 upvotes]: Let $(a,b,c,d)$ be a 4-tuple of real numbers such that $a \leq b \leq c \leq d$, but is otherwise arbitrary. Are there real sequences $x_n$ and $y_n$, such that lim inf $x_n$ $+$ lim inf $y_n$ $= a$, lim inf $(x_n + y_n) = b$, lim sup $(x_n + y_n) = c$, and lim sup $x_n$ $+$ lim sup $y_n$ $= d$?

REPLY [3 votes]: No. If $a=0$, then we may add a constant to $x_i$ and subtract it from $y_i$ to achieve $\liminf x_n=\liminf y_n=0$; this does not affect $b$, $c$, and $d$. Without loss of generality, assume that $\limsup x_n\geq d/2$; then $c\geq d/2$ as well.
Similarly, since $\limsup y_n\leq d/2$, we get $b\leq d/2$.
Conversely, it seems that any 4-tuple with $b\leq (a+d)/2\leq c$ is reachable. Set $\liminf x_n=\liminf y_n=a/2$,  $\limsup x_n=\limsup y_n=d/2$, achieved on sparse pairwise  disjoint sequences of indices. Now it is easy to reach $b$ and $c$.