text:
\begin{document}
\maketitle

\begin{abstract}
We say a graph is $(d, d, \ldots, d, 0, \ldots, 0)$-colorable with $a$ of $d$'s and $b$ of $0$'s if $V(G)$ may be partitioned into $b$ independent sets $O_1,O_2,\ldots,O_b$ and $a$ sets $D_1, D_2,\ldots, D_a$ whose induced graphs have maximum degree at most $d$.   The maximum average degree, $mad(G)$, of  a graph $G$ is the maximum average degree over all subgraphs of $G$.  In this note, for nonnegative integers $a, b$, we show that if $mad(G)< \frac{4}{3}a + b$, then $G$ is $(1_1, 1_2, \ldots, 1_a, 0_1, \ldots, 0_b)$-colorable.
\medskip

\noindent{\bf Keywords:}  Relaxed coloring, maximum average degree, global discharing

\noindent{\bf AMS Subject class:} 05C15
\end{abstract}

\section{Introduction}
We say a graph is $(d, d, \ldots, d, 0, \ldots, 0)$-colorable with $a$ of $d$'s and $b$ of $0$'s if $V(G)$ may be partitioned into $b$ independent sets $O_1, \ldots, O_b$ and $a$ sets $D_1, D_2,\ldots, D_a$ whose induced graphs have maximum degree at most $d$. For convenience, we also call the graph to be $(d_1, d_2, \ldots, d_a, 0_1, \ldots, 0_b)$-colorable.   The maximum average degree, $mad(G)$, of  a graph $G$ is the maximum average degree over all subgraphs of $G$.  This parameter is used to measure how sparse a graph is.

Borodin and Kostochka (2011, \cite{BK11}) showed that $mad(G)\le \frac{12}{5}$, then $G$ is $(1,0)$-colorable, and the upper bound $\frac{12}{5}$ is sharp; and for $d\ge 2$, they (2014,~\cite{BK14}) showed that if $mad(G)\le 3-\frac{1}{d+1}$, then $G$ is $(d,0)$-colorable, and again the upper bound $3-\frac{1}{d+1}$ is sharp.   Borodin, Kostochka, and Yansey~(2013, \cite{BKY13}) also gave the sharp result that if $mad(G)\le \frac{14}{5}$, then $G$ is $(1,1)$-colorable.

Havet and Sereni (2006,~\cite{HS06}) proved that if $mad(G)<a+\frac{ad}{a+d}$, then $G$ is $(d_1, d_2, \ldots, d_a)$-colorable.    For nonnegative integers $a, b, d$, Dorbec-Kaiser-Montassier-Raspaud (2014~\cite{DKMR14}) proved that a graph $G$ is $(d_1, d_2, \ldots, d_a, 0_1, \ldots, 0_b)$-colorable if $mad(G)<a + b + \frac{da(a+1)}{(a+d+1)(a+1) + ab}.$    Montassier and  Ochem (2015,~\cite{MO15}) gave a good survey on the results of this kind.

Clearly $(1_1, 1_2, \ldots, 1_a, 0_1, \ldots, 0_b)$-colorable graphs are also $(d_1, d_2, \ldots, d_a, 0_1, \ldots, 0_b)$-colorable for every $d\ge 1$.  When $d=1$, the above theorem shows that a graph $G$ is $(1_1, 1_2, \ldots, 1_a, 0_1, \ldots, 0_b)$-colorable if $mad(G)<a+b+\frac{a(a+1)}{(a+1)(a+2)+ab}<a+b+1$. In this note, we improve the upper bound by replacing the $1$ by $a/3$, which improves the previous result as long as $a>1$ or $b>0$.

\begin{thm}\label{main}
Let $a, b$ be nonnegative integers  with $a\ge 1$. Then
\begin{center}
if $mad(G)< \frac{4}{3}a + b$, then $G$ is $(1_1, 1_2, \ldots, 1_a, 0_1, \ldots, 0_b)$-colorable.
\end{center}
\end{thm}

Note that a complete graph with $2a+b+1$ vertices has maximum average degree $2a+b$ and is not $(1_1, 1_2, \ldots, 1_a, 0_1, \ldots, 0_b)$-colorable. So the best possible improvement over our result $\frac{4}{3}a+b$ would be $2a+b$. It is an interesting question to find optimal upper bound on maximum average degrees.

Our proof uses a non-traditional discharging method. Instead of distributing charges among local neighborhoods, we define a flow of charges among subsets of vertices.  The global discharging method allows us to prove a stronger result without lengthy discussion.

\section{The proof}
Let $G$ be a counterexample with fewest vertices.  Then for each vertex $x$ in $G$, $G-x$ has a desired coloring.   In a coloring of $G-x$, a vertex is \textit{saturated} if it is colored $0_i$ or colored $1_j$ with $1_j$-colored neighbor, and it is called $0$-saturated and $1$-saturated, respectively.

\begin{lem}\label{lem:min}
$\delta(G) \geq a + b$.
\end{lem}

\begin{proof}
By minimality of $G$, $G-v$ can be colored.  The coloring of $G-v$ cannot be extended to $v$.  Then all the $a+b$ colors must appear in $N(v)$. So $v$ has at least $a+b$ neighbors.
\end{proof}

For a vertex $v\in G$,  let $h(v)= d(v) - (a + b)$.

\begin{lem}\label{lem:satnbr}
Let $v\in V(G)$ be a vertex with $h(v)<a+b$. Then in a coloring of $G-v$,  $v$ has at least $a+b-h(v)$ saturated neighbors whose colors are unique in $N(v)$, and among them, at least $\max\{a - h(v),0\}$ are $1$-saturated.
\end{lem}

\begin{proof}
Let $S$ be the set of neighbors of $v$ whose colors appear only once in $N(v)$.  Note that each vertex in $S$ must be saturated, for otherwise, we may color $v$ with the color of a non-saturated vertex in $S$.

Suppose that $s=|S|<a+b-h(v)$. Since $v$ cannot be colored, all $a+b$ colors should appear in $N(v)$. So $a+b-s$ colors appear at least twice in $N(v)$. Therefore, $d(v)\ge 2(a+b-s)+s=a+b+(a+b-s)>a+b+h(v)=d(v)$, a contradiction.  This shows the first part of the lemma.

Now let $v$ such that $h(v)<a$, and suppose that $v$ has $p<a-h(v)$ $1$-saturated neighbors whose colors appear once in $N(v)$. Note that all $a+b$ colors must appear on $N(v)$, and each of the non-saturated neighbors must share the same color with another neighbor. Therefore, $d(v)\ge 2(a-p)+b+p=a+b+(a-p)>a+b+h(v)=d(v)$, a contradiction.
\end{proof}

\begin{definition}
Let $F_0$ be the set of vertices with degrees at least $2a+2b$.  For $k\ge 0$,  we define $H_k\subseteq V(G)-F_k$ to be a set of vertices with the following properties:

\begin{enumerate}
\item $v\in H_k$ has at least $\max\{a-h(v),0\}$ neighbors in $F_k$, and
\item $v\in H_k$ has at least $a + b - h(v)$ neighbors in the induced subgraph $G[H_k\cup F_k]$.
\end{enumerate}
Let $F_{k+1}=F_k\cup H_k$.
\end{definition}

We should note that $G$ may not have vertices of degree at least $2a+2b$, thus $F_0$ could be an empty set.  However, we shall show in the lemma below that $H_0$ can be chosen to be non-empty (that is, there exists vertices with $h(v)\le a$ and satisfying the second condition).

\begin{lem}\label{lem:flagged}
For some $k\ge 0$, $F_k=V(G)$.
\end{lem}

\begin{proof}
Suppose that $F_k\not=V(G)$ for every $k\ge 0$.   Consider the largest subset $F_k$ in $G$.

In all colorings of $G-v$ with $v\not\in F_k$,  the ones with minimized number of $1$-saturated vertices in $V(G)-F_k$ are called {\em minimum partial coloring} of $G$.  Not every vertex $v$ in $V(G)-F_k$ can make $G-v$ to have a minimum partial coloring of $G$, but since $G$ is finite, some vertices do.   Let $H$ be the set of vertices in $V(G)-F_k$ such that $G-v$ for $v\in H$ has a minimum partial coloring. Then for each $v\in H$,

\begin{enumerate}
\item $v$ has at least $\max\{a-h(v),0\}$ neighbors in $F_k$.

We assume that $a-h(v)>0$, and suppose that $v$ has fewer than $a-h(v)$ neighbors in $F_k$, which includes the case that $F_0=\emptyset$.  Consider a minimum partial coloring $c(G-v)$.  By Lemma~\ref{lem:satnbr}, $v$ has at least $a-h(v)$ $1$-saturated neighbors whose colors appear once in $N(v)$, and so one of them, say $u$, must be in $V(G)-F_k$.  Uncolor $u$ and color $v$ with the color of $u$, we obtain a coloring of $G-u$ with $u\in V(G)-F_k$.  In this coloring, $v$ is not saturated and no other vertices become saturated; so this coloring has fewer $1$-saturated  vertices than the one of $G-v$, a contradiction to the minimality of the coloring. Therefore, $v$ has at least $\max\{a-h(v), 0\}$ neighbors in $F_k$.

\item $v$ has at least $a+b-h(v)$ neighbors in the induced graph $G[H\cup F_k]$.

Since $v\in H$, $v\not\in F_0$, so $d(v)<2a+2b$, thus $a+b-h(v)>0$.    Consider a minimum coloring of $G-v$.  By Lemma~\ref{lem:satnbr}, $v$ has at least $a+b-h(v)$ saturated neighbors whose colors appear once in $N(v)$. We claim that all such neighbors are in $H\cup F_k$. For otherwise, let $w\not\in H\cup F_k$ be such a neighbor.  Uncolor $w$ and color $v$ with the color of $w$, we obtain a coloring of $G-w$ with $w\in V(G)-F_k$. Now, $v$ is not $1$-saturated and no other vertices become $1$-saturated.  So this coloring is also a minimum partial coloring with $w\in V(G)-F_k$.  By the definition of $H$, $w\in H$, a contradiction.
\end{enumerate}

Hence, by definition, $H\subseteq H_k$ (true even if $k=0$ and $F_0$ is empty), a contradiction.
\end{proof}

Now, we are ready to prove Theorem~\ref{main}.

For each vertex $v\in G$, let $\mu(v)=d(v)-(\frac{4}{3}a+b)=h(v)-\frac{1}{3}a$.  Since $mad(G)<\frac{4}{3}a+b$, $$\sum_{v\in V(G)} \mu(v)=\sum_{v\in V(G)} (d(v)-(\frac{4}{3}a+b))=2|E(G)|-(\frac{4}{3}a+b)|V(G)|<0.$$

We distribute the charges among vertices by the following rule:
\begin{enumerate}
\item[(R)] if $v\in F_k$ with $k=0$ or $v\in F_{k} - F_{k-1}$ for some $k > 0$, then $v$ gives $\frac{1}{3}$ to each $u \in N(v) - F_k$.
\end{enumerate}

\medskip

For vertex $v\in F_0$, $v$ gives at most $\frac{1}{3}$ to each of its neighbors. Note that $d(v)\ge 2a+2b$,  so the final charge of $v$ is $$\mu^*(v)\ge d(v)-(\frac{4}{3}a+b)-\frac{1}{3} d(v)\ge \frac{1}{3}(2(2a+2b)-4a-3b)>0.$$

Let $v\in F_k-F_{k-1}$ with $k\ge 1$. Then $v\in H_k$.  By the rule and definition of $H_k$, $v$ receives at least $\frac{1}{3}\max\{a-h(v),0\}$ from the neighbors in $F_{k-1}$ and gives out $\frac{1}{3}$ to each of its neighbors not in $F_k$.  By definition, $v$ has at most $d(v)-(a+b-h(v))=2h(v)$ neighbors not in $F_k$.    So the final charge of $v$ is
\begin{align*}
\mu^*(v)&\ge (h(v)-\frac{1}{3}a)+\frac{1}{3}\max\{a-h(v),0\}-\frac{1}{3}\cdot 2h(v)\\
&\ge \begin{cases}
	(h(v)-\frac{1}{3}a)+\frac{1}{3}(a-h(v))-\frac{2}{3}h(v)=0, \text{ if $a\ge h(v)$,}\\
	(h(v)-\frac{1}{3}a)+0-\frac{2}{3}h(v)=\frac{1}{3}(h(v)-a)>0, \text{ if $a<h(v)$.}
	\end{cases}
\end{align*}

Therefore, every vertex in $G$ has a nonnegative final charge.  But $$\sum_{v\in V(G)}\mu^*(v)=\sum_{v
\in V(G)} \mu(v)<0,$$ we reach a contradiction.  This contradiction shows the truth of Theorem~\ref{main}.

\bigskip

{\bf Acknowledgement:}
 We thank the referees for their helpful comments.


TITLE: Understanding Stirling no of first kind
QUESTION [0 upvotes]: I was reading about Stirling no of the first kind, so $ \left[ \frac{n}{k} \right] $ represent no of k cycles of n items, so in  $ \left[ \frac{4}{2} \right] $  there would be 11 such combinations, of this $ \left[ 1, 4 \right]  \left[ 2, 3 \right] $, I think is same as $ \left[ 2, 3 \right]  \left[ 1, 4 \right] $, however is it same as $ \left[ 1, 4 \right]  \left[ 3, 2 \right] $ i.e does the internal permutation matter?
Also for $ \left[ \frac{3}{1} \right] $, the similar looking cycles, $ \left[ 1,2, 3 \right] $ and $ \left[2, 3,1 \right]$ are considered different right?

REPLY [0 votes]: I have read about this now and try to explain the answer to the first part.
In Stirling no of first kind we first distribute the no into indistinguishable boxes and then the inner circular permutations are counted, so if we have elements $\left\lbrace1,2,3,4\right\rbrace$ and $2$ boxes then we can distribute them first as
$$ \left\lbrace 1,2 \right\rbrace \left\lbrace 3,4 \right\rbrace \text{,} \left\lbrace 1,3 \right\rbrace \left\lbrace 2,4 \right\rbrace \text{,} \left\lbrace 1,4 \right\rbrace \left\lbrace 2,3 \right\rbrace \text{,}$$ $$\left\lbrace 1,2,3 \right\rbrace \left\lbrace 4 \right\rbrace \text{,} \left\lbrace 1,2,4 \right\rbrace \left\lbrace 3 \right\rbrace \text{,} \left\lbrace 1,3,4 \right\rbrace \left\lbrace 2 \right\rbrace \text{,} \left\lbrace 2,3,4 \right\rbrace \left\lbrace 1 \right\rbrace $$
Now since they count circular permutation $\left\lbrace 1,4 \right\rbrace $ will be same as $\left\lbrace 4,1 \right\rbrace$ but $\left\lbrace 1,2,3 \right\rbrace$ will be different from $\left\lbrace 1,3,2 \right\rbrace$, thus all the boxes containing 3 elements can be permuted in 2 ways, so total,  $S[4,2] =4 . 2 +3 = 11 $ ways.


text:
\section{Introduction}
\label{intro}

Various emerging systems and applications work by analyzing
the data submitted by their users in order to serve them; we call such systems  \emph{user-data driven} (UDD) \amir{we need a name. how's this?} services. Examples of UDD services include  smart cities, connected vehicles, smart homes, and connected healthcare devices, which are
 revolutionizing our lives in different ways.
Unfortunately, 	the sheer volume of user data collected by these systems
compromises users' privacy ostensibly~\cite{FTC2015}.
Even the employment of  standard
privacy-protection mechanisms (e.g.,  anonymization of user identities and obfuscation of submitted data) does not mitigate the privacy leakage as the adversaries are able to use powerful statistical inference techniques to infer sensitive private information~\cite{FTC2015,0Quest2016, 2nia2016comprehensive, 3ukil2014iot, 4Hosseinzadeh2014,iotCastle,matching}.

To illustrate   their privacy leakage, consider the following three popular UDD services.
(1) {\em Health care:}  Wearable monitors that constantly track user health variables can be invaluable in assessing individual health trends and responding to emergencies.  However, such monitors produce long time-series of user data uniquely matched to the health characteristics of each  user; (2) {\em Smart homes:}  Emerging smart-home technologies such as fine-grained power measurement systems can help users and utility providers to address one of the key challenges of the twenty-first century:  energy conservation.  But the measurements of power by such devices can be  mapped to users and reveal their lifestyle habits; and, (3)
	{\em Connected vehicles:}  The location data provided by connected vehicles promises to greatly improve everyday lives by reducing congestion and traffic accidents.  However, the matching of such location traces to prior behavior not only allows for user tracking, but also reveals a user's habits.
In summary, despite their critical importance to the society and their emerging popularity, these services have one thing in common:
they collect  long time-series of user data, for them to operate,
that puts users' privacy at significant risk.

There are two main approaches to augment privacy in UDD services:
 \emph{identity perturbation (anonymization)}, and \emph{location perturbation (obfuscation)}.  In anonymization techniques, privacy is obtained by concealing the mapping between users and data, and the mapping is changed periodically to  thwart statistical inference attacks that try to deanonymize the anonymized data traces by matching user data to
 known  user profiles.
 On the other hands,  obfuscation mechanisms aim at protecting privacy by perturbing user data, e.g., by adding noise to user locations.

Anonymization and obfuscation improve user privacy at the cost of user utility.  In anonymization, we need to change these pseudonym mappings frequently to achieve high privacy by reducing the length of time series exploited by statistical analysis.  However, this frequent change could decrease usability and functionality by concealing the temporal relation between a user's locations, which may be critical in the utility of some systems, e.g., a dining recommendation system that makes suggestions based on the dining places visited by a user in the past.  For obfuscation-based mechanisms, the added noise to the reported values of user locations will degrade the application utility; for example, user utility would be degraded by location obfuscation in ride-sharing systems.  Thus, choosing the right level of privacy-protection mechanism is an important question, and understanding what levels of anonymization and obfuscation can provide theoretical guarantees of privacy is of interest.

In this paper, we derive the fundamental limits of user privacy in UDD services in the presence of both anonymization and obfuscation protection mechanisms. This work is built on our previous work on formalizing privacy in location-based services~\cite{montazeri2016defining, Mont1610Achieving,tifs2016, ciss2017}, by extensively expanding not just the application area but also user models and settings.
Particularly, our previous work introduced the notion of
\emph{perfect privacy} for location-based services, and derives the rate at which an anonymization mechanism should change the pseudonyms in order to
achieve the defined perfect privacy.
In this work, we expand the notion of perfect privacy to UDD services in general, and derive the conditions for it to hold when \emph{both}
anonymization and obfuscation-based protection mechanisms are employed.
 We study both achievability and converse results.

\subsection{Summary of the Results}

Given $n$, the total number of the users in a network, their degree of privacy depends
on two parameters: (1) The number of observations $m=m(n)$ by the adversary per user for a fixed anonymization mapping (i.e., the number of observations before the pseudonyms are changed);  and (2) the value of the noise added by the obfuscation technique (as defined in Section~\ref{sec:framework}, we quantify the obfuscation noise with a parameter $a_n$).
Intuitively, smaller $m(n)$ and larger $a_n$ result in stronger privacy, at the expense of lower utility for the users.

Our goal is to identify values of $a_n$ and $m(n)$  that satisfy perfect privacy as the number of users grows ($n \rightarrow \infty$).

We show that when the users' data sets are governed by an i.i.d.\ process,
the $m(n)- a_n$ plane can be divided into two areas.  In the first area, all users have perfect privacy~\cite{tifs2016} (as defined in Section~\ref{sec:framework}), and, in the second area, users have no privacy.
Figure~\ref{fig:region} shows the limits of privacy in the entire $m(n)- a_n$.
As the figure shows, in regions $1$, $2$, and $3$, users have perfect privacy, while in region $4$ users have no privacy.
\begin{figure}[h]
	\centering
	\includegraphics[width=0.8\linewidth, height=0.8 \linewidth]{fig/region}
	\caption{Limits of privacy in the entire $m(n)-a_n$ plane: in regions $1$, $2$, and $3$, users have perfect privacy, and in region $4$ users have no privacy.}
	\label{fig:region}
\end{figure}

For the case where the users' data sets are governed by irreducible and aperiodic Markov chains with $r$ states and $|E|$ edges, we show that users will have no privacy if $m =cn^{\frac{2}{|E|-r} +  \alpha}$ and $a_n =c'n^{-\left(\frac{1}{|E|-r}+\beta \right)}$, for some constants $c>0$, $c'>0$, $\alpha>0$, and $\beta>\frac{\alpha}{4}$. We also provide some insights for the opposite direction (under which conditions users have perfect privacy) for the case of Markov chains.

config: arxiv


TITLE: Symmetry of Bloch Hamiltonian
QUESTION [11 upvotes]: If a crystal system preserve a symmetry C, why its Bloch Hamiltonian satisfy
$H(C\vec k)=CH(\vec k)C^{-1} $

REPLY [10 votes]: That a Hamiltonian preserves a symmetry means
$$ [H, C] = 0  \Rightarrow CHC^\dagger = CHC^{-1} = H$$
For a unitary symmetry operator $C$ (or anti-unitary if it is time reversal).
The Hamiltonian of a crystalline condensed matter system written in terms of the Bloch matrix is:
$$ H = \sum_{\vec k} \psi^\dagger(\vec k) H(\vec k) \psi(\vec k) $$
Where $$\psi(\vec k) = \begin{pmatrix} \vdots \\ c_i(\vec k) \\ \vdots \end{pmatrix}.$$
And $c_i(\vec k)$ are the annihilators for electrons at $\vec k$ in state $i$ (which is a combined index of spin and band). As $C$ is applied to $\vec k$, it must be a spatial symmetry (that is a lattice symmetry).
The transformation of $\psi(\vec k)$ under $C$ is easily obtained by argument.
$C$ is a spatial transformation, thus $C\psi(\vec k)C^{-1}$ does the following:

the coordinates are transformed.
a particle with momentum $\vec k$ is destroyed in the new coordinates.
the space rotation is undone leaving a particle with momentum $C \vec k$ destroyed in the original coordinates.

I effect this means that $C\psi(\vec k)C^{-1} = \psi(C\vec k)$.
Using the above results:
$$ H = CHC^{-1} = \sum_{\vec k} C\psi^\dagger(\vec k)C^{-1}C H(\vec k) C^{-1}C \psi(\vec k)C^{-1}. $$
As $C\psi(\vec k)C^{-1} = \psi(C\vec k)$, $C H(\vec k) C^{-1} = H(C\vec k)$ must hold, in order to assure the equality to $H$ (which is then shown by relabelling the summation variable $\vec k' = C\vec k$).


TITLE: Smaller groups generating a larger group
QUESTION [1 upvotes]: How could I  show that the 3 groups
$$x\to x+a$$ $$x\to \exp(b) x$$ $$x\to {x\over 1-cx}$$ where $a,b,c\in \mathbb R$ and also $x\in \mathbb R$,
generate a group $$x\to {\alpha x+\beta\over \gamma x+\delta}$$ where $\alpha\delta-\beta\gamma=1$?
Thank you.

I can see how the 3 smaller groups are subgroups of the bigger group, but I don't know how to show that they generate the larger group.

REPLY [2 votes]: Given,
$$ f : x \to x+a, \quad g : x \to e^bx, \quad h : x \to \frac{x}{1-cx}, $$
then we compose as
$$ h(g(f(x)))=h(g(x+a))=h(e^b(x+a))=\frac{e^b(x+a)}{1-c(e^b(x+a))}\,. $$
Now, simplify the above and you will get $\alpha,\beta,\gamma,$ and $\delta$.


TITLE: Expectation of sum of $n$ random variables.
QUESTION [0 upvotes]: Let $X_1, X_2, X_3 \ldots, X_n$ be n random variables which take values from $\{+1,-1\}$ uniformly. Let $S_n$ be defined as $$S_n =|X_1+ X_2+ X_3+ \dots +X_n| $$
Find the expectation $E(S_n)$ of the random variable.
Now I can find the closed form solution to be equal to $n 2^{1-n}{{n-1}\choose{(n-1)/2}}$
But I can not see how the book says that $E(S_n) = ((2/\pi)^\frac{1}{2} + o(1))\sqrt{2}$.

REPLY [1 votes]: $X_1 + \ldots + X_n = 2 Y - n$ where $Y$ is a binomial random variable with parameters $n$ and $1/2$, and
$E[S_n] = 2 \sum_{j=1}^n j P(Y = (n+j)/2)$.   I'll do the case where $n=2m$ is even, so we sum over even $j = 2i$.
$$E[S_n] = 4 \sum_{i=1}^{m} i {2m \choose m+i} 2^{-n} = (m+1) {2m \choose m+1} 2^{1-2m}$$
Using Stirling's approximation, as $m \to \infty$ we have
$$E[S_n] =  2 \sqrt{\dfrac{m}{\pi}}- \dfrac{1}{4 \sqrt{\pi m}}+ \ldots
 = \sqrt{\dfrac{2n}{\pi}} + o(1)$$
This must be what the book meant (or you copied it wrong).


TITLE: Calculating the $n$th generation with $p_{n}=\frac{p_{n-1}(1-p_{n-1})}{1-{p_{n-1}}^2}$?
QUESTION [0 upvotes]: The formula for the next generation $p$ is as following:
$$p_{n}=\frac{p_{n-1}(1-p_{n-1})}{1-{p_{n-1}}^2}$$

$p_n$ is the percentage of the new allele in the next generation
Also $0 <p_{n-1} < 1$

I was wondering if there is just a standard formula for $p_n$, because I was asked to calculate the percentage of the $50$th generation. My only conclusion is that I have to enter the value 50 times into the formula, but I don't know if there is a more efficient way to calculate this.

REPLY [1 votes]: Assuming $p_{n-1}<1$, a factor of $1-p_{n-1}$ can be pulled out of both the numerator and denominator to get
$$p_n=\frac{p_{n-1}}{1+p_{n-1}} \tag{1}$$
Defining a reciprocal series $q_n=\frac{1}{p_n}$ we get by inverting (1)
$$q_n=q_{n-1}+1 \tag{2}$$
Hence for $k\ge1$ we have by repeated application of (2):
$$q_n=q_{n-k}+k \tag{3}$$
so, for example, $q_{50}=q_0+50$ whence $p_{50}=\left(\frac{1}{p_0}+50\right)^{-1}$.

REPLY [1 votes]: Let $p_n = \frac{ p_{n-1}}{1+ mp_{n-1}}$.
Then for $m=1$ we have the correct formula. However, this gives us
\begin{eqnarray}
p_{n+1} &=& \frac{ p_{n}}{1+ m p_n} \\
&=& \frac{ \frac{p_{n-1}}{1+m p_{n-1}}}{1+ \frac{p_{n-1}}{1+ m p_{n-1}}}\\
&=&\frac{ \frac{p_{n-1}}{1+m p_{n-1}}}{\frac{1 + mp_{n-1}+p_{n-1}}{1+ m p_{n-1}}}\\
&=& \frac{ p_{n-1}}{ 1 + (m+1)p_{n-1}}
\end{eqnarray}
So if we start with $p_2 = \frac{p_1}{1+ p_1}$ we find $p_{50} = \frac{ p_1}{1+49p_1}$.


text:
\begin{document}
\maketitle

\begin{abstract}
For each natural number $n$, we construct an example of a graph manifold supporting at least $n$ different Anosov flows that are not orbit equivalent.
Our construction is reminiscent of the Thurston-Handel construction \cite{HandelThurston}: we cut a geodesic flow on a surface of constant negative curvature into two pieces, modify the flow in each piece by pulling back to finite covers, and glue together compatible pairs of pullback flows along their boundary tori to get many distinct flows on the resulting graph manifold.
\end{abstract}

\section{Introduction}
An Anosov flow, also called a hyperbolic flow, is a flow for which some directions are expanded and others are contracted. Such flows are fundamental examples for (idealized) chaotic dynamical systems.
The study of Anosov flows in dimension three is well-connected to the study of the topology of the three-manifold carrying such flows.
For instance,  a three-manifold carrying an Anosov flow is always irreducible \cite{Palmeira}, has a fundamental group of exponential growth \cite{PlanteThurston}, and carries a tight contact structure \cite{Mitsumatsu}.

In some cases, one can classify the Anosov flows supported by a three-manifold if the manifold has a simple geometry. For instance \cite{ghys2007knots} and \cite{Barbot_Waldhausen} proved that a flow on a Seifert fibered three-manifold is equivalent to a geodesic flow up to finite covers, while \cite{PlanteSolvable} proved that an Anosov flow on a Solv manifold is a suspension of an Anosov diffeomorphism of the torus.

Although any three-manifold carrying an Anosov flow is irreducible, it may be toroidal, i.e. it can contain essential embedded tori.
When this is the case, a fundamental tool of three-dimensional topology allows one to cut the manifold along a collection of the essential tori, obtaining pieces with boundary that have better understood topological and geometric properties.

Therefore, it makes sense to cut the manifold along essential tori and analyse the resulting flows on the geometric pieces.
This is the subject of a number of seminal papers by Barbot and Fenley. In particular, they prove in \cite{barbotfenleytoroidal} that every essential torus is homotopic to one which is either transverse to the flow, or is  \emph{quasi-transverse}: It is transverse except along a finite number of periodic orbits.
The manifold can then be cut along these transverse or quasi-transverse essential tori, from which they have developed a variety of classification results for the Anosov flows on the resulting pieces.

 Essential tori have also been used in the converse direction, namely to glue pieces of Anosov flows together to produce
new Anosov flows with surprising qualities.
See for instance the examples of Fried-Williams, who construct a non-transitive Anosov flow, and Handel-Thurston who construct an Anosov flows which is transitive, but is neither a geodesic flow nor a suspension.

A central problem in the field has been to determine the number of Anosov flows that can be supported by a single manifold.  For example, this appears as Problem 3.53 of Kirby's problem list, where he asks: Given an integer $N$, does there exist a hyperbolic $3$-manifold with at least $N$ Anosov flows which are topologically inequivalent?

The first explicit examples of manifolds supporting multiple Anosov flows were constructed in \cite{Barbot_Waldhausen}, where Barbot constructs a family of graph manifolds that each support two distinct Anosov flows.
The surgery techniques of Goodman \cite{Goodman_Surgery} may also produce two distinct Anosov flows on a manifold if the periodic orbit used for the Dehn surgery admits two distinct purely cosmetic surgery slopes.
More recently, Bonatti, Beguin and Yu proved a general theorem allowing  them to glue many pieces with transverse toral boundaries \cite{beguin2014building}. They use this new technique to construct, for each $n >0$, a non-geometric three-manifold $M_n$ consisting of two hyperbolic pieces and one Seifert fibered piece which supports at least $n$ distinct Anosov flows. A similar result was recently obtained by Bowden and Mann using an analysis of the rigidity properties of certain fundamental group actions \cite{mannbowden}, but their manifolds $M_n$ are hyperbolic (thus answering the question from Kirby's problem list).

A recurrent theme in these examples is that a three manifold must have increasingly ``complicated topology" in order to support an increasing number of distinct Anosov flows.  For example, in \cite{beguin2014building}, this manifests itself as a choice of increasingly complex pseudo-Anosov diffeomorphisms of a surface in order to construct certain hyperbolic pieces in the JSJ decomposition of their manifold $M_n$.

The main result of this work is to produce new examples of three-manifolds admitting arbitrarily many non-equivalent Anosov flows, with each manifold having particularly simple topology.  For each $n>0$ we construct, via modifications to a geodesic flow, a graph manifold which supports arbitrarily many Anosov flows, no two of which are orbit equivalent.

The manifolds we construct are therefore topologically simpler than those in the literature, in the sense that the examples of \cite{mannbowden, beguin2014building} both use hyperbolic pieces to achieve the necessary complexity to support arbitrarily many Anosov flows.
Moreover, we use the simplest possible graph manifolds, each having only two pieces, each piece being the exterior of a trefoil. Indeed, in order to increase the number of Anosov flows supported by the manifolds arising from our construction, we need only increase the shear factor of a certain shear mapping used to identify the boundary tori of the two pieces.

Our technique is inspired by the examples of Handel-Thurston.  They begin by fixing a closed hyperbolic surface $S$, and equipping the unit tangent bundle $T^1S$ with the geodesic flow $\psi$.  Next, they choose a simple closed geodesic $g(t)$ in $S$, and cut $S$ along $g(t)$ to produce two pieces $P_1$ and $P_2$, and equip each of $N_1 = T^1P$ and $N_2 = T^1P_2$ with $\psi|N_i$.  Last, they build a gluing map $F: \partial T^1N_1 \rightarrow \partial T^1N_2$ which is distinct from the identity, but which behaves very will with respect to the flow on each piece, so that the pieces can be assembled into a new manifold $N_1 \cup_F N_2$ equipped with an Anosov flow built from the flows $\psi|N_i$.
For any of these gluings, the resulting flow is not automatically Anosov. The difficulty is that an Anosov flow carries two invariant foliations, one uniformly (exponentially) attracting and one uniformly repelling.
The gluing usually cannot be made to preserve both these foliations, the smoothness of the flow, and the uniformity of the attraction.
Therefore, one must typically appeal to cone-field arguments to prove that the resulting flow is indeed Anosov.

Our method of proof is as follows: We consider the geodesic flow on the modular surface, which is a flow on the trefoil complement.
 We then observe that the trefoil complement admits an $k$-sheeted self-covering $p_k: M \rightarrow M$ for certain values of $k$.  Fixing a carefully chosen gluing map $F_n$ for $n>0$, and closely following the Handel-Thurston analysis of the invariant directions for the geodesic flow, we are able to find infinitely many pairs of integers $(k, m)$ with corresponding coverings such that the pullback of the geodesic flow from $M$ along $p_k$ and $p_m$ are ``compatible" with the map $F_n$.  With such a choice, the gluing arguments of Handel--Thurston show that the lifted geodesic flows may be glued to produce an Anosov flow on $M \cup_{F_n} M$.
For distinct pairs of integers $(k,m)$ corresponding to compatible pullbacks, the resulting flows on $M \cup_{F_n} M$ are never orbit equivalent. It follows that:

\begin{theorem}\label{thm:Main}
For every natural number $n$, there exists a graph manifold $M_n$ supporting at least $n$ Anosov flows, no two of which are orbit equivalent.
\end{theorem}

Note that it is impossible to obtain a similar result by fixing a geodesic flow on a \emph{closed} surface and varying the covering maps, as these coverings will have different domains. In fact, in a recent result of Barbot and Fenley \cite{barbotfenley_contact}, they prove that any cover, along the fiber direction, of a unit tangent bundle can carry at most two distinct Anosov flows up to topological equivalence. Thus, it is cardinal here to consider Anosov flows on manifolds with boundary.

\subsection*{Acknowledgments}
We are grateful to Fran\c{c}ois B\'eguin, Christian Bonatti, Sergio Fenley, Kathryn Mann and Bin Yu for making comments on earlier versions of this paper.

\section{background}
\subsection{The geodesic flow}\label{sec:geodesic}
Let $S$ be a hyperbolic surface constructed as $\HH^2/\Gamma$ for some discrete subgroup $\Gamma$ of $\text{Isom}^+(\HH^2)$.
The surface $S$ thus inherits a Riemannian metric from $\HH^2$, which allows one to define the geodesic flow on $S$,
defined on the unit tangent bundle $T^1(S)\cong \PSL(\mathbb{R})/\Gamma$.

A flow $\phi^t$ on a three manifold $M$ is called \emph{Anosov} if there is a continuous $D\phi$-invariant decomposition of the tangent bundle $TM=E^s\oplus E^u\oplus E^t$ and constants $A>0$, $\lambda>1$, so that for all $t \in \mathbb{R}$:
\begin{itemize}
    \item $E^t$ is tangent to $\phi^t$ at any point $x\in M$,
    \item for any $x\in M$ and any $v\in E^u_x$,
    $||D\phi^t(v)_{\phi^t(x)}||\geq A\lambda^t||v_x||,$
    \item for any $x\in M$ and any $v\in E^s_x$,
    $||D\phi^t(v)_{\phi^t(x)}||\leq A\lambda^{-t}||v_x||.$
\end{itemize}
In this decomposition, $E^u$ is called the \emph{strong unstable} direction, and $E^s$ the \emph{strong stable} direction.

Anosov \cite{anosov_geodesic_flows_67} proved that the geodesic flow on $\HH^2$ is Anosov, and that this property descends to any hyperbolic surface $S$.  We can describe the geodesic flow, together with its stable and unstable directions, as follows:

Each point in $T^1\HH^2$ consists of a point $x\in\HH^2$ and a unit vector $v\in T_x\HH^2$. There exists a unique geodesic $g$ in $\HH^2$ passing through $x$ and tangent to $v$. Denote the emanating point of $g$ on $S^1=\partial\HH^2$ by $v^-$, and its terminating point by $v^+$. We denote the point $x$ with direction $v$ by $(x,v)$. The strong stable direction at $(x,v)$ is the direction of the horocycle passing through $x$ based at $v^+$ with a direction perpendicular to the horocycle itself at any point. The strong unstable direction is likewise given by the horocycle based at $v^-$ with a similar perpendicular direction, see Figure~\ref{fig:Horo}.

The geodesic flow moves $(x,v)$ forward along $g$ by a time $t$, while mapping the stable horocycle to another stable horocycle based at $v^+$ (exponentially contracting in $t$ the distance along the horocycle), and the unstable horocycle to an unstable horocycle based at $v^-$.

\begin{figure}[ht!]
\centering
\begin{overpic}[width=5cm]{Horocycles.pdf}
\put(-10,48){$v^-$}
\put(100,48){$v^+$}
\put(32,63){$(x,v)$}
\put(46,26){${E}^u$}
\put(78,67){${E}^s$}
\end{overpic}
\caption{A point $(x,v)\in T^1\HH^2$ together with the horocycles determining the stable and unstable manifolds at $(x,v)$.}\label{fig:Horo}
\end{figure}

In order to perform the gluings in Section~\ref{sec:Gluing}, we need to identify the directions of the Anosov decomposition $TM=E^s\oplus E^u\oplus E^t$ for $M=T^1\HH^2$, and later determine these directions upon passing to a quotient $T^1S$ for a particular choice of surface $S$.
To this end, we analyze the action of the flow on $T(T^1\HH^2)$ by first introducing a coordinate system used in \cite{HandelThurston}.

Fix a point $x \in\HH^2$ and $v\in T^1_x\HH^2$.
Following \cite{HandelThurston}, we identify
$T_{(x,v)}(T^1\HH^2)$ with $T_x\HH^2\times\R$ as follows:
At any point $y\in\HH^2$ we define the angle $\sphericalangle(w,u)$ between any two vectors $w,u\in T_y^1\HH^2$ to be the angle measured counterclockwise from $w$ to $u$.
Given a vector $w\in T_x\HH^2$ and a real number $\rho\in\R$, define $c_{w,\rho}(t)=(\exp(tw),u(\alpha+t\rho)\,)$ where $\alpha=\sphericalangle(w, v)$ and $u(\alpha+t\rho)\in T^1_{\exp(tw)}\HH^2$ is the vector satisfying $\sphericalangle(T\exp(tw), u(\alpha+t\rho))=\alpha+t\rho$ (see Figure~\ref{fig:Path}). The path $c(t): = c_{w, \rho}(t)$ in $T^1\HH^2$ satisfies $c(0)=(x,v)$, and we identify the point $(w, \rho) \in T_x\HH^2 \times \R$ with $\left.\frac{dc}{dt}\right|_{t=0}$ in $T_{(x,v)}(T^1\HH^2)$.

\begin{figure}[ht]
\centering
\begin{overpic}[width=8cm, trim=0 0 0 -60, clip]{path.pdf}
\put(0,-2){$x$}
\put(90,12){$\exp(wt)$}
\put(27,38){$w$}
\put(0,36){$v$}
\put(38,63){$u(\alpha+t\rho)$}
\put(100,32){$T\exp(tw)$}
\put(5,16){$\alpha$}
\put(70,40){$\alpha+t\rho$}
\end{overpic}
\caption{The path $c(t)$ in $T^1\HH^2$.}\label{fig:Path}
\end{figure}

For any point $(y,w)\in T^1\HH^2$, define the vector $w^\perp$ to be the vector in $T_y^1\HH^2$ satisfying $\sphericalangle(w^\perp,w)=\pi/2$.
Consider again Figure~\ref{fig:Horo} showing the Anosov directions. The strong stable and unstable manifolds are both tangent to $v^\perp$ (that is, both are pointing downwards from $x$ in the figure). For the unstable manifold (that is, the horocycle on the left), the direction vector at a point rotates clockwise as one proceeds along the horocycle, while for the stable manifold the rotation is counterclockwise.  Thus in our coordinate system:

\begin{itemize}
    \item $E_{(x,v)}^t$ is generated by $(v,0)$,
    \item $E_{(x,v)}^u$ is generated by $(v^\perp,-1)$,
    \item $E_{(x,v)}^s$ is generated by $(v^\perp,1)$.
\end{itemize}

In particular, the directions in the $(v^\perp,\rho)$ plane corresponding to the unstable manifold lie in the second and fourth quadrants.  We will need this information for the gluing arguments in Section \ref{sec:Gluing}.

\begin{remark}\label{rem:Trivializing}  Fixing a geodesic $g$ in $\HH^2$, we can consider the unit tangent bundle to $g$. At any point $(x,v)\in T^1g$,
parallel transport along $g$ preserves the angle $\alpha=\sphericalangle(Tg,v)$. One may identify $T^1g$ with $\R\times\R/2\pi$ by $(t,\theta)\mapsto(g(t),v)$ where $v$ is the vector in $T^1_{g(t)}\HH^2$ satisfying $\sphericalangle(Tg,v)=\theta$. In these coordinates the angles of the Anosov directions above do not depend on the point t, but only on the angle $\alpha$.
\end{remark}

\subsection{The modular surface}\label{sec:modular}
Let $\Gamma$ denote the group $\PSL(\mathbb{Z})$, with generating set
$$\left\{ \begin{pmatrix}
0 &1\\
-1& 0
\end{pmatrix},

\begin{pmatrix}
0 &-1\\
1& 1
\end{pmatrix} \right\}.$$
In the  standard representation of  $\text{Isom}^+(\HH^2)$ as $\PSL(\R)$ acting by M\"obius transformations, these generators of $\Gamma \subset \PSL(\R)$  act respectively as rotation by $\pi$ about the point $p=i$, and rotation by $2\pi/3$ about the point $q=\frac{1}{2}+\frac{\sqrt{3}}{2}i$.
The modular surface $\Mod$ is the hyperbolic orbifold $\Mod=\HH^2/\Gamma$, a surface with one cusp at infinity and two cone points, one of order 2 and one of order 3.

Increasing the distance between the centers of rotations $p$ and $q$ in $\HH^2$ as in \cite{ghys2007knots} turns the cusp into a funnel. There is then a unique closed geodesic around the funnel, that we denote by $g$. The greater the distance between $p$ and $q$, the longer the geodesic $g$. We cut the surface along $g$ to obtain a compactification of $\Mod$ which is a surface with a boundary. The resulting surface $\M$ is depicted in Figure~\ref{fig:Modular}.

\begin{figure}[ht]
\centering
\begin{overpic}[width=4cm]{modular.pdf}
\put(-2,0){$2$}
\put(86,0){$3$}
\put(10,80){$g$}
\end{overpic}
\caption{The compactified modular surface, $\M$.}\label{fig:Modular}
\end{figure}

The unit tangent bundle $T^1\M$ is the bundle of all $(x,v)$ where $x\in \M$ and $v\in T_x\M$ satisfies $||v||=1$.
Consider the boundary torus $\partial T^1\M=T^1g$.
We may trivialize the fiber direction along this torus using the tangent to the geodesic $g$ at each point as a section, as in Remark~\ref{rem:Trivializing}. If $g$ has length $L$, this results in the coordinates $[0,L]\times[0,2\pi]$ on the boundary torus, $(s,\theta)\mapsto(g(s),v)$, where $g$ is parametrized by arc-length, and $v\in T^1_{g(s)} \M$ satisfies $\sphericalangle(Tg,v)=\theta$.
We call these the orbit-fiber coordinates. As the geodesic flow is structurally stable, it follows that it is independent of the length $L$ of the boundary geodesic. Hence, we may choose the length to be equal to $2\pi$ so that $0\leq s\leq 2\pi$.

It is uncommon to define Anosov flows on a manifolds with boundary, so for ease of exposition we avoid introducing such a definition and instead introduce our objects of study as pieces of an Anosov flow on a closed $3$-manifold cut along essential tori.

Consider the surface in Figure \ref{fig:Double}, which is a sphere with four cone points, two of order $2$ and two of order $3$, denoted by $S_{2,2,3,3}$. It is the double of the modular surface $\M$ and is also a hyperbolic surface. As in Section~\ref{sec:geodesic}, there is a geodesic flow over $S_{2,2,3,3}$, which is an Anosov flow defined on $T^1 S_{2,2,3,3}$. Cutting the three manifold $T^1 S_{2,2,3,3}$ along the two dimensional torus $T^1g$ as in Figure \ref{fig:Double} one obtains two copies of $T^1\M$, and the geodesic flow on $\M$ is thus a piece of an Anosov flow on a closed manifold as in \cite{barbot_fenley_free}.

\begin{figure}[ht]
\centering
\begin{overpic}[width=7cm]{double.pdf}
\put(0,5){$3$}
\put(0,43){$2$}
\put(93,5){$3$}
\put(93,45){$2$}
\end{overpic}
\caption{The double $S_{2,2,3,3}$ of $\M$ with the curve $g$.}\label{fig:Double}
\end{figure}

We may use the coordinate system defined in Section~\ref{sec:geodesic} for $T(T^1\M)$ to identify the Anosov directions over the boundary torus $\partial T^1\M$.
In our coordinate system $(s,\theta)$ corresponds to the point $(g(s),v)$ where $v$ is the rotation $R_\theta$ by $\theta$ of $Tg$ along the fiber at the point $g(s)$. In particular,
The flow direction at $(g(s),v)$ is generated by $(v,0)=(R_\theta(Tg),0)$.
The strong unstable direction $E_{(g(s),v)}^u$ is generated by $(v^\perp,-1)$, where $v^\perp$ is the direction in the plane perpendicular to $v$, and the unstable direction
 $E_{(x,v)}^s$ is generated by the same vector and slope $+1$,  $(v^\perp,1)$.
Thus, the Anosov directions are all rotations of the directions along $g$, all by the same angle $\theta$ in a horizontal (tangent to the surface) direction.
In particular, the Anosov directions depend solely on $\theta$ and not on $s$.
These directions are the same for any hyperbolic surface $\HH^2/\Gamma$, and $g$ a closed geodesic in $S$.

\subsection{The trefoil complement}
In this section we will show how to identify $T^1\M$ with the trefoil complement, and will identify the closed orbit on the boundary of $T^1\M$ with a curve on the boundary of the trefoil complement expressed in meridan/longitude coordinates.

 The trefoil complement and its Seifert fibration can be described as follows.  There is an action of $S^1$ on $S^3$ given by $\lambda \cdot (z_1, z_2)$ = $(z_1 \lambda^2, z_2 \lambda^3)$ for all $\lambda \in S^1$.  Each orbit of this action is a trefoil knot lying on one of the tori $\{(z_1, z_2) \mid |z_1|^2=r\}_{0<r<1}$, except for the two orbits $$S^1 \times \{0\}=\{(z_1,z_2)\mid |z_1|=1, z_2=0\}
\mbox{ and } \{0\} \times S^1=\{(z_1,z_2)\mid z_1=0, |z_2|=1\}.$$
\begin{figure}[ht]
\centering
\begin{overpic}[width=8cm]{trefoil2.pdf}
\put(78,68){$\mu$}
\put(79,40){$h$}
\end{overpic}
\caption{The complement of the trefoil knot.}\label{fig:Trefoil}
\end{figure}
Fixing a choice of orbit $T_{2,3} \subset T=\{ (z_1, z_2) \mid |z_1|^2 = |z_2|^2\}$, we may remove an open neighbourhood of $T_{2,3}$ from $S^3$ consisting of a union of fibers (orbits) to get  $M = S^3\setminus \NN (T_{2,3})$, a compact manifold with boundary, as in Figure \ref{fig:Trefoil}.  The decomposition of $M$ into orbits under the $S^1$ action gives $M$ the structure of a Seifert fiber space, where the orbit $S^1 \times \{0\}$ admits a fibered torus neighbourhood with invariants $(2,1)$ and $\{0\} \times S^1$ admits a fibered torus neighbourhood with invariants $(3,1)$.  The orbit surface for this fibration, i.e. the space obtained by collapsing each fiber to a point, is a copy of $S^2$ minus a disk, with two cone points of orders $2$ and $3$, i.e. it is the surface $\M$.

The trefoil complement also has a fibration over $S^1$: Consider the punctured torus $\text{T}_0$ composed of two disks in the solid torus $\{(z_1, z_2) \mid |z_1|^2 \leq |z_2|^2 \}$ connected by three half twisted bands in the solid torus $\{(z_1, z_2) \mid |z_1|^2 \geq |z_2|^2 \}$ (see \cite{Rolfsenknots} for more details, including an explicit parameterization of this surface).  On the boundary torus $\partial M$, we can choose a basis composed of the meridian $\mu$ of the torus $\NN (T_{2,3})$, and a regular fiber $h$ of the Seifert fibration (see Figure~\ref{fig:Trefoil}). The longitude $\lambda$ is the boundary of the Seifert surface  $\partial \text{T}_0\subset \partial M$.

The Seifert fibers intersect the surface $\text{T}_0$ transversely, with each regular fiber intersecting it six times.  We assume that these intersections happen at regular intervals along the fiber (that is, if $\lambda, \lambda' \in S^1$ are such that $\lambda \cdot (z_1, z_2) = \lambda'\cdot (z_1', z_2')$ for $(z_1, z_2), (z_1', z_2') \in \text{T}_0$, then $\lambda^{-1} \lambda' = e^{2k \pi i / 6}$
for some $k \in \{0, \ldots, 5\} $).  As such, every point in $M$ can be written uniquely as $\lambda \cdot (z_1, z_2)$ for $(z_2, z_2) \in \text{T}_0$ and $\lambda \in S^1$ with $0 \leq \arg(\lambda) < \pi/3$. The singular fibers intersect the surface $2$ and $3$ times respectively.
Sliding $\text{T}_0$ along the fibers rotates $\text{T}_0$ through $S^3$ until it returns to its initial position as a set---in other words, $e^{2\pi i/6} \cdot \text{T}_0 = \text{T}_0$. Note this will shift each intersection point of a regular Seifert fiber with $\text{T}_0$ to the next intersection point along the fiber, permuting cyclically the two disks on the outside, and permuting the three bands on the inside. This yields a map $\varphi : \text{T}_0 \rightarrow \text{T}_0$ given by $\varphi(x) = e^{2\pi i/6} \cdot x$ for all $x \in \text{T}_0$.

As each Seifert fiber has $2$, $3$ or $6$ intersection points with the punctured torus, $\varphi^6=\text{id}$.  Moreover, $\varphi$ serves as the generator of the group of deck transformations for the natural branched covering map $p : \text{T}_0 \rightarrow \M$ as in Figure~\ref{fig:Cover}.

\begin{figure}[ht]
\centering
\begin{overpic}[width=9cm]{cover.pdf}
\put(10,0){$\pi$}
\put(43,0){$2\pi/3$}
\end{overpic}
\caption{The cover $p:\text{T}_0 \rightarrow \Mod$, here $\text{T}_0$ is identified with the surface $T^2 \setminus D^2$ for ease of illustration.}\label{fig:Cover}
\end{figure}

The unit tangent bundle to $T^1\M$ is composed of two unit tangent bundles, each one a unit tangent bundle to a closed neighbourhood of a cone point, glued along the unit tangent bundle to a segment. It is thus homeomorphic to the (compactified) trefoil complement $M$ \cite{milnor1968singularities}.

\begin{proposition}[Ghys]
\label{orbit_slope}
The identification of $T^1 \M$ with the trefoil complement $M$ carries the closed orbit of the geodesic flow on $\partial T^1 \M$ to the curve $\mu \subset \partial M$.
\end{proposition}

\begin{proof}
The proof is given in \cite{ghys2007knots}, and we include it here for completeness. The covering map $p$ induces a cover $P :T^1 \text{T}_0  \rightarrow M\cong T^1(\M)$ and $\varphi$ induces $\Phi: T^1 \text{T}_0 \rightarrow T^1F$.
As the unit tangent bundle $T^1 \text{T}_0$ is trivial, it has a section, say $s$.  Set $\bar{\lambda} = s( \partial \text{T}_0)$, then we can assume $\bar{\lambda}$ appears
as in Figure \ref{fig:Longitude}, where the black arrows indicate the direction of the nonsingular vector field defined by $s$. The images $\Phi^k \circ s$ are disjoint for $k \in \{0, \ldots, 5\} $. Thus, the image under $P$ of any such section is a copy of $\text{T}_0$ embedded in the trefoil complement  $M$ with $\partial \text{T}_0 \subset \partial M$. Hence, we may identify the image of $P \circ s$ with the Seifert surface for the trefoil, and in particular, choosing the orientation for $\lambda$ accordingly, $P(\bar{\lambda})=\lambda$.

We can also identify a curve $\bar{h}$ in $\partial(T^1 \text{T}_0)$ that covers a regular fiber $h$ of the Seifert fibration of $M$.  Fixing $x \in \partial \text{T}_0$, set $\bar{h} = \{(x, v) \mid ||v||=1 \}$, oriented so that counterclockwise rotation of the vector $v$ is the positive direction along $\bar{h}$.  Choosing an appropriate orientation for $h$, we get $P(\bar{h}) = h$.

Next, we define a curve $\beta$ representing the class $[\bar{h}] + [\bar{\lambda}] \in H^1(\partial T^1 \text{T}_0)$ by first supposing that $\partial \text{T}_0$ is identified with a curve $\alpha:[0,L] \rightarrow \text{T}_0$, and choosing a point $t_0 \in [0,L]$ for which the section is a direction tangent to the curve, i.e.  $s(\alpha(t_0)) = (\alpha(t_0), \frac{T\alpha(t_0)}{||T\alpha(t_0)||})$.

Supposing the longitude is traveled counterclockwise as well, as in Figure \ref{fig:Longitude},

such a point corresponds to the rightmost point of the curve $\bar\lambda$ in Figure \ref{fig:Longitude}, and reparameterizing if necessary, we assume $t_0= 0$.  Our curve $\beta: [0,L] \rightarrow T^1 \text{T}_0$ is $\alpha$ with its tangent direction, $\beta(t) = (\alpha(t), \frac{T\alpha(t)}{||T\alpha(t)||})$.  To see that $[\beta] =[\bar{h}]+ [\bar{\lambda}]$, note that if $\bar{\lambda}(t) = s \circ \alpha(t)= (\alpha(t), v_t)$ for some unit vector $v_t \in T_{\alpha(t)} \text{T}_0$, then
$\sphericalangle(v_t, T \alpha(t)) = \frac{2 \pi t}{L}$.  Thus the vector $v_t$ makes one complete rotation relative to $T \alpha(t)$ as $t$ ranges over $[0,L]$, so that $\beta(t) \sim \bar{\lambda} \circ \bar{h}$.

\begin{figure}[ht]
\centering
\begin{overpic}[width=6cm]{longitude1.pdf}
\put(43,16){$\bar\lambda$}
\end{overpic}
\caption{The curve $\bar\lambda$ on $\partial T^1 \text{T}_0$ in red, as the curve $\partial \text{T}_0$ embedded into $T^1 \text{T}_0$ by assigning a direction to each point.}\label{fig:Longitude}
\end{figure}

Considering Figure \ref{fig:Cover}, we see that the cover $p:\text{T}_0 \rightarrow \Mod$ wraps $\partial \text{T}_0$ six times around the boundary curve $g$ of $\M$, and thus $P(\beta) = (g, \frac{Tg}{||Tg||})$, where $\beta$ wraps six times around its image under $P$.  On the other hand, from Figure~\ref{fig:Trefoil} it follows that the fiber $h$ has linking number $6$ with the trefoil, hence $[h]=\pm[\lambda]\pm6[\mu]$.
However our choices of orientations for $\lambda$ and $h$ yield $[\beta] = [\bar{\lambda}] + [\bar{h}]$, so $[P(\beta)] = [\lambda] + [h] = \pm 6 [\mu]$.

Therefore, the closed orbit $(g, \frac{Tg}{||Tg||})\subset \partial T^1\M$ is exactly the curve you Dehn fill to get $S^3$ (c.f. \cite{ghys2007knots}).
\end{proof}

\vskip .5cm
\section{Lifting}
\label{sec:Covers}

With $M$ as in the previous section, recall that the $S^1$ action which determined the Seifert fibration allowed every point in $M$ to be written uniquely as $\lambda\cdot (z_1, z_2)$ for $(z_2, z_2) \in \text{T}_0$ and $\lambda \in S^1$ with $0 \leq \arg(\lambda) < \pi/3$.  Then for each $d = \pm 1 + 6k$ with $k \in \mathbb{N}_{>0}$ we can construct a $d$-fold covering map $p_d : M \rightarrow M$ as $p_d(\lambda \cdot (z_1, z_2)) = (z_1 \lambda^{2d}, z_2 \lambda^{3d})$ if $d = 1+6k$ and $p_d(\lambda \cdot (z_1, z_2)) = (z_1 \lambda^{-2d}, z_2 \lambda^{-3d})$ if $d = -1+6k$.  From this description, it is clear that under the cover $p_d$ each regular fiber of the Seifert fibration upstairs wraps $d$ times around each regular fiber downstairs (either preserving or reversing orientation depending on the degree of the cover).

This same covering map can also be described in terms of mapping cylinders.  Recall that $M \cong (\text{T}_0 \times [0,1])/ \sim$ , where $(x, 0) \sim (\varphi(x), 1)$ for all $x \in \text{T}_0$ and $\varphi$ is of order $6$.  When $d =1 + 6k$, consider the $d$-fold cyclic cover constructed from $d$ ``puzzle pieces", each homeomorphic to $M$ cut open along a Seifert surface as in \cite{Rolfsenknots}. The resulting manifold is homeomorphic to the mapping cylinder with respect to
$\varphi^d$. As $\varphi^6=\text{id}$, $\varphi^d = \varphi^{1}$ and the cover  manifold is again homeomorphic to $M$. It wraps, via the covering map $p_d$, $d$ times around itself.  It follows from this description that the preimage $p_d^{-1}(\lambda) = p_d^{-1}(\partial \text{T}_0)$  is $d$ disjoint copies of $\lambda$.

\vskip.5cm

These descriptions and the observations about the action of $p_d$ on the fiber $h$ and the longitude $\lambda$ are sufficient to completely describe the behaviour of the maps $p_d$ upon restriction to the boundary torus $\partial M$.

\begin{lemma}
\label{peripheral map}
Let $M$ be the complement of the trefoil knot in $S^3$, and $p_d: M \rightarrow M$ denote a covering map of order
 $d = 1 + 6k$,
 where $k \in \mathbb{N}$.
 Then the preimage of the curve $\mu$ under the map $p_d$ is a single curve, of homology  $[\mu] + k [\lambda]$.
 \end{lemma}

\begin{proof}
From the above descriptions of the covering map $p_d$, the induced map $p_d^{*}:\pi_1(\partial M) \rightarrow \pi_1(\partial M)$
 satisfies
$(p_d^{*})^{-1}([h]) = [h]$ and $(p_d^{*})^{-1}([\lambda]) = d [\lambda]$.
Thus, as $[h] = 6[\mu]  - [\lambda]$:
\[
(p_d^{*})^{-1}(6[\mu]) = (p_d^{*})^{-1}([\lambda] ) + (p_d^{*})^{-1}([h])   = d [\lambda] +[h] = 6[\mu]+(d-1)[\lambda]
\]
so that  $(p_d^{*})^{-1}(6[\mu])=6[\mu]+6k[\lambda]$ and
\[
(p_d^{*})^{-1}([\mu])=[\mu]+k[\lambda]. \]

What remains to be shown is that $p_d^{-1}(\mu)$ consists of  single curve.

We know that $p_d^{-1}(\lambda)$ consists of $d$ disjoint (parallel) curves, and we can assume that $\mu \cap \lambda$ is a single point.

It follows that $|p_d^{-1}(\lambda\cap\mu)|=d$ and $\mu$ must be covered by a single curve intersecting each of the $d$ components of $p_d^{-1}(\lambda)$ exactly once.
\end{proof}

\begin{proposition}\label{pro:TwoOrbits}
Suppose $d = 1  + 6k$ and let $\psi_d$ denote the pullback of the geodesic flow along the covering map $p_{d}: M \rightarrow M$.  Then $\psi_d$ has exactly two periodic orbits of slope $[\mu] + k [\lambda]$.  Moreover, if $m$ and $n$ are positive integers with $m, n \equiv 1 \mod 6$, then $\psi_n$ and $\psi_m$ are orbit equivalent if and only if $n=m$.
\end{proposition}

\begin{proof}
That $\psi_d$ has exactly two periodic orbits of the given slope follows from Lemma \ref{peripheral map}, and Proposition \ref{orbit_slope}.

Now suppose that $H :M \rightarrow M$ is a diffeomorphism carrying the orbits of $\psi_m$ to the orbits of $\psi_n$. As the symmetry group of the pair $(S^3, T_{p,q})$ is $\mathbb{Z}_2$, if $H$ is not isotopic to the identity then it is isotopic to a map that generates the symmetry group.  Such a map induces, up to conjugation, the homomorphism $\phi : \langle a, b \mid a^3 =b^2\rangle \rightarrow \langle a, b \mid a^3 =b^2\rangle$ given by $\phi(a) = a^{-1}, \phi(b) = b^{-1}$.

The homomorphism $\phi$ is conjugate to a homomorphism whose restriction to the peripheral subgroup $\pi_1(\partial M)$ has action $[\mu] \mapsto -[\mu]$ and $[\lambda] \mapsto -[\lambda]$.  As such, a closed orbit $\mu + k \lambda$ will be carried by $H$ to a closed orbit isotopic to $-\mu - k \lambda$, meaning that the closed orbits of $\psi_n$ and $\psi_m$ are never identified by $H$ unless $m = n$.
 \end{proof}

\begin{remark}\label{rem:fiber}
It follows that there is a choice of longitude $\lambda$ that intersects the orbit at a single point for any cover of degree $d=1+6k$ where $k\in\mathbb{N}_{>0}$.  Thus we can always choose the longitude $[\lambda]$ and a periodic orbit $g_d$ such that $[g_d]$ is a basis for $H_1(\partial M)$. In this basis we have, by the equation in the proof of Lemma \ref{peripheral map}, $
[h]=-d[\lambda]+6[g_d].
$
 \end{remark}

\section{Gluing pieces}\label{sec:Gluing}
In this section we complete the proof of Theorem \ref{thm:Main}.  For each $n\in\mathbb{N}$ we begin by defining the map $F_n:\partial M \rightarrow \partial M$ as a diffeomorphism inducing the following map on the first homology of $\partial M$:
\[ F_n^*([\mu]) = [\mu]+2n[\lambda], \mbox{ and } F_n([\lambda]) = -[\lambda].
\]

\begin{theorem}\label{thm:Glue}
The manifold $M_n=M\cup_{F_{n}} M$
supports $n+1$ Anosov flows, no two of which are orbit equivalent.

\end{theorem}

\begin{proof}
For the geodesic flow $\psi$, we may switch from the orbit-fiber coordinates $(t,\theta)$ to orbit-longitude coordinates $(t,\tau)$ where $0\leq\tau\leq2\pi$ is an arc-length parametrisation of the longitude $\lambda$, normalized to have length $2\pi$.
Recall that the Anosov directions are independent of $t$, and they are a horizontal rotation by $\theta$ of the directions at $\theta=0$.
Thus, they are also independent of $t$ in the $(t,\tau)$ coordinates, and consist of a rotation by $\tau$ of the directions at points with $\tau=0$.

Next consider the flow $\psi_d$ corresponding to a self-cover of the trefoil complement of degree $d=1+ 6k$. By the previous section, a tangent orbit $g_d$ is a curve of homology $[\mu] + k[\lambda]$ so that $\{[g_d], [\lambda]\}$ is a basis for the homology of the boundary torus.
The curve $g_d$ is $d$ times the length of the boundary geodesic $g_1$ of $\M$.
Since the geodesic flow on $M$ is independent of the length of $g_1$ (see Section~\ref{sec:modular}), we may choose its length to be $L=2\pi/d$, so that an arc-length parameter for $g_d$ ranges from 0 to $2\pi$ for any $d$.

 Fix $n > 0$.  For each $k$ with $0 \leq k \leq 2n$, the map $F_n$ is chosen so that
 \[ F_n^*([g_d]) = F_n^*([\mu] + k [\lambda]) = [\mu] + (2n-k) [\lambda] = [g_{d'}]
 \]
 where $d' = 1+6(2n-k)$. In other words, $F_n$ carries any periodic orbit $g_d$ of the flow $\psi_{1+ 6k}$ to a curve isotopic to a periodic orbit $g_{d'}$ of the flow $\psi_{1+ 6(2n-k)}$, with the same orientation.

At the same time, the periodic orbits of $\psi_d$  for any $d$ divide $\partial M$ into two Birkhoff annuli, $A^d_{out}$ and $A^d_{in}$, where the flow $\psi_d$ points outward from $M$ and inward respectively. The longitude $\lambda$ intersects $g_d$ transversely precisely once for any $d$, pointing in the direction opposite the fiber, i.e. into $A^{d'}_{in}$ and $A^{d}_{in}$. Since $F_n(\lambda) = -\lambda$, we see that $F_n(A^d_{out}) = A^{d'}_{in}$ and $F_n(A^d_{in}) = A^{d'}_{out}$.

Moreover, the normal direction to $\partial M$ is unchanged by the gluing, i.e. the $(v,v^\perp)$ plane in one copy of $M$ is matched with the $(v,v^\perp)$ plane in the other.

We define the flow $\Psi_k$ to be the smooth flow resulting from gluing $\psi_{1+ 6k}$ to $\psi_{1+ 6(2n-k)}$ via $F_n$.

Although the orbit-fiber coordinates are now not a basis for the homology of the boundary torus, as the regular fiber intersects an orbit $g_d$ multiple times when $d>1$, the coordinates $(v,v^\perp,\rho)$ are still a local basis for the unit tangent bundle at each point.

By Remark \ref{rem:fiber}, the fiber $h$ has the form $[h]=-d[\lambda_2]+6[g_{d}]$. It is thus mapped to $[F_n(h)]=d[\lambda_1]-6[g_d']$ for the  corresponding degree $d'$.
In particular, a fiber is not glued to a fiber and the resulting manifold is not globally Seifert fibered but is a graph manifold.

As $d'[\lambda]=6[g_d']-[h]$, the slope of $\lambda$ is $(\frac{6}{d'}, -\frac{1}{d'})$ in the local orbit-fiber directions on the boundary torus for $d'=1+6(2n-k)$.
Thus, $[F_n(h)]=-d[\lambda]-6[g_d']$ is of slope $(\frac{-6d}{d'}-6,\frac{d}{d'})$.
Therefore, the image of the fiber direction $DF(\rho)$ is always tangent to the boundary torus, and is in the
second and fourth quadrant in the (local) orbit-fiber basis.

This direction falls in the second and fourth quadrant in the $(v^\perp,\rho)$ plane along the core curve of the
Birkhoff annulus $A_{out}^d$.

In the
second Birkhoff annulus $A_{in}^d$, one flows inward to get back to the original copy, and thus the gluing is performed via $F_n^{-1}$.

\vskip .5cm
\begin{claim} The flow $\Psi_k$ is Anosov for any $k$.
\end{claim}

This follows immediately from \cite[Propositions 3 and 4]{HandelThurston}, and we give the idea of the proof here for sake of completeness.
We first show there exist two continuous plane fields, $F^u$ and $F^s$, that intersect along the direction $F^t$ tangent to the flow $\Psi_k$, and are invariant under $D\Psi_k$.

Consider at each point $x$ of $M_n$ the union of the second and fourth quadrants in the $(v^{\perp},h)$ plane, product with the $v$ direction. That is, the portion of the tangent space $T_xM_n$ consisting of the two infinite wedges
\[
\mathcal{W}^u=\big\{(av+bv^\perp,c)\,|\, bc\leq0\big\}.
\]
Within each piece, the action of $D\Psi_k$ on the tangent directions is given by the action of $\psi_d$ and $\psi_{d'}$. Under these actions, both $v^\perp$ and $h$ are moved towards the unstable direction, and thus into the interiors of the second and fourth quadrants.
When passing through the gluing, one uses the action of $DF$ when passing from the back copy to the front one, and the action of $DF^{-1}$ when passing from the front copy to the back one.
In both of these cases, by the computation above the claim, $v^\perp$ is invariant, while $h$ is mapped into the interior of the second quadrant.
It follows that
\[
D\Psi_k^t(\mathcal{W}^u_x)\subset \mathcal{W}^u_{\Psi_k^t(x)}.
\]

By fixing any point $x\in N$ and considering
$D\Psi_k^n(\mathcal{W}^u_{\Psi_k^{-n}(x)})$, these is a sequence of closed sets (each corresponding to a closed set of possible slopes in the $(v^\perp,h)$ plane), that are each contained in all its prequels. Thus, there is an invariant plane field $W^{u}_x=(av+bv^\perp,-\eta(x)b\,)_x\subset T_xN$ at any point $x$ with some finite slope function $\eta(x)>0$. This plane field is the weak stable direction.

The continuity of $\eta$ and $W^u$ follows from the continuity of the $D\Psi_k^t$ action.
By the same argument applied to the action of $\Psi_k^{-t}$ on the first and third quadrants, a continuous invariant plane field $W^{s}$ exists as well.

Next we show there exist continuous $D\Psi_k$ invariant line fields $E^{u}\subset W^u$ and $E^{s}\subset W^s$ which together with $E^t$ yield the Anosov directions for $\Psi_k^t$. We again use a cone field argument.

Any vector in the plane field $W^u$ can be expressed as $(av+bv^\perp,-\eta(x)b\,)_x$. A vector $(bv^\perp,-\eta(x)b\,)_x$ in the intersection of the plane field with the $(v^\perp,\rho)$ plane is shifted along the $v$ direction some bounded distance, as $\rho$ is shifted. As one increases the $v$ component, so $|a|$ becomes larger, the amount of the shift decreases, as both $v$ and $v^\perp$ are invariant under the gluing.

On the other hand, within each piece, the flow takes a vector $(av+bv^\perp,-\eta(x)b\,)_x$ exponentially fast towards the unstable direction $(v^\perp,-1)$ in the intersection of the $(v^\perp,\rho)$ plane and the plane field (as we already know the plane field is invariant).
It follows that for such a vector with $|a|$ large enough, its image under  $D\Psi_k$ is a vector with $|a|$ exponentially decreasing.
Therefore, we can define the cone field within $W^u$:

\[
\mathcal{C}^u=\big\{\big(a(x)v+bv^\perp,-\eta(x)b\,\big)_x\,|\, b>0\big\}.
\]
Here, $a(x)$ is a function that is a large enough constant $a_0$ on most of $M_n$, decreasing on each piece when approaching $A_{out}^d$, and we can fix this decrease to be slower than the decrease induced by  $D\Psi_0$, but still small enough at the annulus, so that the image under the gluing satisfies $|a|<a_0$.

It follows that the cone field is invariant under the flow
\[
D\Psi_k^t(\mathcal{C}^u_x)\subset \mathcal{C}^u_{\Psi_k^t(x)},
\]

and there exist invariant line fields $E^u$ and $E^s$.
Thus, the resulting flow is indeed Anosov, as the stable and unstable directions are globally defined.

\vskip .5cm
\begin{claim}
The flows $\Psi_k$ and $\Psi_m$, where $0\leq k,m \leq 2n$ are conjugate only if $m=k$ or $m = 2n-k$.
\end{claim}

Considering the manifold $M_n=M\cup_{F_n} M$, since $F$ does not map a regular fiber in one copy of $M$ to the regular fiber in the second copy, $M_n$ is a graph manifold that is not a Seifert fibered manifold. Thus, its JSJ torus is unique up to isotopy \cite{JacoShalen1979, johannson}.

Suppose that $H$ were a self-homeomorphism of $M_n$, carrying $\Psi_k$ to $\Psi_m$, demonstrating that they are topologically equivalent flows.  For ease of exposition set $c = 1+ 6m$ and $c' = 1+6(2n-m)$ so that $\Psi_m$ is the result of gluing $\psi_c$ to $\psi_{c'}$.

Let $T\subset N$ be the embedding of $T^1\partial \M$ into $N$. By construction, $T$ is a Birkhoff torus for $\Psi_k$.
The homeomorphism $H$ takes $T$ to a Birkhoff torus $H(T)$ for $\Psi_m$. At the same time, $T$ is also a Birkhoff torus for $\Psi_m$ and they are isotopic by the uniqueness of the JSJ torus. As the only tangent orbits in the geodesic flow that can be isotoped into the boundary $\partial M$ are $g$ and $g^{-1}$, this is also true for all of its covers. Thus, $T$ and $H(T)$ share the same tangent orbits, and $H(T)$ can be isotoped to $T$ along the flowlines transverse to $T$.

This implies that the two pieces of $M_n \setminus H(T)$, equipped with $\Psi_m$, are orbit equivalent to $(M,\psi_c)\cup(M,\psi_{c'})$.
Thus, $H$ carries $\{(M,\psi_d),(M, \psi_{d'})\}$ to $\{(M, \psi_c),(M, \psi_{c'})\}$ and it follows from Proposition~\ref{pro:TwoOrbits} that
$\{d,d'\}=\{c,c'\}$ and therefore either $k=m$ or $k = 2n-m$.
\end{proof}

\begin{remark}
 Handel--Thurston point out in \cite{HandelThurston} that one can take covers of the geodesic flows in the pieces, which for them are denoted by $N_i$:   ``The family we have described can be enlarged by taking the $n$-fold cover of $N_i$ ($n$ independent of $i$) corresponding to the $S^1$ fiber, before gluing the $N_i$ together." One might ask if this could lead to a generalization of our constructions presented here.  However it is not clear which basis to use in the general setting, and how to control the manifold obtained by gluing (the orbit-fiber coordinates do not work for this setting, as the orbit and the fiber necessarily intersect multiple times in the cover).
 \end{remark}
 \begin{remark}
 It is possible to greatly generalize our techniques. Any orbifold or surface with boundary will have a unit tangent bundle that is a suspension with a periodic (or trivial) monodromy, and thus will cover itself in infinitely many different ways, with some covers being orientation reversing. For any hyperbolic surface this yields infinitely many Anosov flows on a manifold with boundary.
To glue such pieces to each other, one must compute the slopes of the periodic orbits on the boundary, the number of orbits for each lift of the geodesic flow, as well as the orientations of the flows through the Birkhoff annuli. In general, this may depend on the choice of a section.
\end{remark}

\bibliographystyle{plain}
\bibliography{Bibliography.bib}

\end{document}


TITLE: Likelihood function is joint measurable
QUESTION [0 upvotes]: Here is the problem in Theory of Statistics Schervish:

Let $\Theta$ be a parameter with parameter space $(\Omega, \tau)$ ($\tau$ is the $\sigma$ algebra). Let $f_{X|\Theta}(x|\theta) = dP_\theta/d\nu$ for every $\theta$. Let $\mu_\Theta$ be a prior on $(\Omega, \tau)$. Let $Q$ be the joint distribution of $(X,\Theta)$. Show that $f_{X|\Theta}= dQ/(d\nu\times \mu_\Theta)$, a.s. $\nu\times \mu_\Theta$. This means for every prior distribution, we can find a version of $f_{X|\Theta}(x|\theta)$ which is joint measurable $(x,\theta)$.

Let $A\in \mathcal{F}$, $B\in \tau$.
\begin{align}
Q(A\times B) = \int_{B}\int_{A}f_{X|\Theta}(x|\theta)\ d\nu d\mu_\Theta
\end{align}
The RHS can be extended uniquely to a measure on $(X,\Omega)$, so it is equal to $Q$. However, I am having trouble proving $Q<< \nu\times \mu_\Theta$. If $f_{X|\Theta}(x|\theta)\in L^1$ then we can conclude this immediately. However, in this case $f_{X|\Theta}(x|\theta)$ is not even joint measurable, we cannot use Tonelli's Theorem to rewrite the iterated integral. Does anyone know how to proceed?
Thanks in advance!

REPLY [0 votes]: You are saying, that you are having trouble proving, that $Q << \nu \otimes \mu_\Theta$, but it would seem that you already have. Remember by definition of absolute continuity, that  $Q << \nu \otimes \mu_\Theta$ if and only if
$$\forall C \in \mathcal{F}\otimes \tau \: : \: \Big(\nu\otimes\mu_\Theta(C)=0 \Rightarrow Q(C) = 0\Big),$$
but you have already proven this for the special case $C=A\times B$ and it can easily be extended to all $C$ (since $\mathcal{F}\otimes \tau = \sigma(\{A\times B \: | \: A \in \mathcal{F},B\in \tau\})$). Assuming that $\nu$ is $\sigma$-finite, then the Radon-nikodym theorem ensures the existence of $\frac{dQ}{d\nu \otimes \mu_\Theta}$. It clearly follows that
$$\int_A \int_B \frac{dQ}{d\nu\otimes \mu_\Theta} \: d\nu d\mu_\Theta = \int_A\int_B f_{X|\Theta} \: d\nu d\mu_\Theta,$$
for all $A \in \mathcal{F}$ and $B\in \tau$. Again we can use that the product sets $A\times B$ generate the $\sigma$-algebra $\mathcal{F}\otimes \mu_\Theta$, and the two functions must thus be equal $\nu \otimes \mu_\Theta$ almost surely.


TITLE: Let $f$ be a non-negative differentiable function such that $f'$ is continuous and $\int_{0}^{\infty}f(x)\,dx$ and $\int_{0}^{\infty}f'(x)\,dx$ exist.
QUESTION [2 upvotes]: Let $f$ be a non-negative differentiable function such that $f'$ is continuous and
  $\displaystyle\int_{0}^{\infty}f(x)\,dx$ and $\displaystyle\int_{0}^{\infty}f'(x)\,dx$ exist.
Prove or give a counter example: $f'(x)\overset{x\rightarrow
\infty}{\rightarrow} 0$

Note: I think it is not true but I couldn't find a counter example.

REPLY [3 votes]: Let $g_n(x) = \sin (n^2 x) 1_{[0,{2 \pi \over n^2}]}$ and note that $g_n$ is continuous, $\int |g_n| = {2 \over n^2}$, $\int g_n = 0$ and if
$f_n(x)=\int_0^x g_n$, then $f_n(x) \ge 0$ for all $x$. Furthermore,
$\int |f_n| \le {4 \pi \over n^3}$.
Define $f(x) = \sum_{n \ge 1} f_n(x-n)$, note that $f$ is integrable and non negative. Furthermore, $f'(x) = \sum_{n \ge 1} g_n(x-n)$ and it is straightforward to check that $f'$ is integrable as well.