text:
\begin{document}

\allowdisplaybreaks

 \setcounter{tocdepth}{1}
\newtheorem{thm}{Theorem}
\newtheorem{conj}[thm]{Conjecture}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{lemma}[thm]{Lemma}
\newtheorem{corollary}[thm]{Corollary}
\newtheorem*{fact}{Fact}
\theoremstyle{remark}
\newtheorem{rem}[thm]{Remark}
\theoremstyle{definition}
\newtheorem{defn}[thm]{Definition}
\newtheorem*{goal}{Goal}
\newtheorem{assume}{Assumption}
\renewcommand{\theassume}{\Alph{assume}}
\newtheorem{exam}[thm]{Example}
\numberwithin{equation}{section}

\begin{abstract}

The family of Euclidean triangles having some fixed perimeter and area can be identified with a subset of points on a nonsingular cubic plane curve, i.e., an elliptic curve; furthermore, if the perimeter and the square of the area are rational, then the curve has rational coordinates and those triangles with rational side lengths correspond to rational points on the curve. We first recall this connection, and then we develop hyperbolic analogs. There are interesting relationships between the arithmetic on the elliptic curve (rank and torsion) and the family of triangles living on it. In the hyperbolic setting, the analogous plane curve is a quartic with two singularities at infinity, so the genus is still $1$. We can add points geometrically by realizing the quartic as the intersection of two quadric surfaces. This allows us to construct nontrivial examples of rational hyperbolic triangles having the same inradius and perimeter as a given rational right hyperbolic triangle.
\end{abstract}

\maketitle

\section{Introduction}

Connections between families of triangles and elliptic curves have been long studied.
Consider, for example, the famous congruent number problem: What rational numbers $A$
are the area of a right triangle with rational side lengths?
We call such $A$ \emph{congruent numbers}.
Using similar triangles, we may assume $A$ is a squarefree positive integer. For example, 6 is a congruent number since it is the area of
the $(3,4,5)$-right triangle. Euler showed that $7$ is a congruent number, but
$1$, $2$, and $3$, are not. It turns out that a

positive rational number $A$ is a congruent number if and only if the elliptic curve $y^2 = x^3 - A^2x$
has a rational point in the first quadrant; in fact, since the the only torsion
points on this elliptic curve have order dividing two, $A$ being congruent here is
equivalent to the group of rational points on the curve having positive rank.
More generally, one can ask whether $A$
is the area of a \emph{rational triangle} (i.e., a triangle having rational side lengths) one of whose internal angles is some fixed value $\theta \in(0,\pi)$. Such numbers
correspond to the existence of a certain kind of rational point on elliptic curves of the form $y^2 = x(x-A\lambda)(x+A/\lambda)$
where $\lambda = \sin(\theta)/(\cos(\theta) +1)$ as found in Problem 3, Section 2, Chapter I of \cite{Kobl}.

Similarly, one can consider rational numbers $A$ which are the area of a rational triangle having some
fixed perimeter instead of a fixed internal angle.
Again, the existence of such a triangle corresponds to the existence of a rational point on an elliptic curve,
namely, a rational point in the first quadrant
on the curve $s^2xy-A^2=sxy(x+y)$ where $s$ is the \emph{semiperimeter} defined to be half the perimeter.
We will recall this connection in Section \ref{sectone}.
Note that the area $A$ of a triangle is determined via Heron's formula $A = rs$ where $r$ is the inradius, so we are led
to study rational points on the curves $C_{r,s} \colon s(xy-r^2)=xy(x+y)$, which are
similar to those studied in \cite{Camp} and/or \cite{Goin}. From the equation for $C_{r,s}$, it
is clear that similar triangles do, in fact, define isomorphic curves and that the family of curves
can be parameterized by $k = s/r$. We will keep the parameters $r$, $s$ separate, however,
to motivate the hyperbolic analogs which have no chance of exploiting similarity since
similar hyperbolic triangles are actually congruent.

In general, any triangle gives rise to a point on $C_{r,s}$ in the first quadrant where
$r$ is the triangle's inradius and $s$ is its semiperimeter. (Here $s\geq 3\sqrt{3}r >0$, and conversely, for any pair of real numbers $r,s$ satisfying this
inequality, there is a triangle with these parameters.) Such a curve $C_{r,s}$ will be an elliptic curve provided the
triangle we started with was not equilateral or, equivalently, $s>3\sqrt{3}r$. Moreover, if the given triangle is rational, then
the point will have rational coordinates and $C_{r,s}$ will have rational coefficients since here $A^2\in \Q$
even though $A$ itself is not necessarily rational.

In particular, when $s,r^2\in \Q$ and $s> 3\sqrt{3}r >0$, one can ask questions about how the structure of the Mordell-Weil group
$C_{r,s}(\Q)$ is related to the family of rational triangles with parameters $r,s$. For example, ``When do the points coming from rational triangles have infinite order?''
It turns out that many Pythagorean triples give rise to rational points of infinite order (see Theorem \ref{rankthm} below), a fact
which comes from the observation that points corresponding to triangles cannot have odd order. There can be points of even order
coming from triangles, and such a point will have order $2$ or $6$ if and only if the corresponding triangle is isosceles.

We can also play the same game for hyperbolic triangles. The natural curve we get for hyperbolic triangles of a fixed inradius $r$ and fixed semiperimeter $s$, however, is a quartic curve
\begin{align*}
Q_{\rho,\sigma}\colon \sigma(x^2y^2+xy + \rho^2(x^2 + xy + y^2-1)) = (1+\rho^2)(x^2y+xy^2)
\end{align*}
where $\rho = \sinh(r)$ and $\sigma=\tanh(s)$. Again, this will be an elliptic curve in the non-equilateral case (i.e., $\sigma > 3\sqrt{3}\rho\frac{1+\rho^2}{1+9\rho^2}$). Also, certain right triangles generate rational points on curves with rational coefficients. In fact, there is a hyperbolic Pythagorean theorem: for a right hyperbolic triangle
with legs $\ell_1$, $\ell_2$ and hypotenuse $\ell_3$, we have
\begin{align*}
\tilde{\ell}_3^2 = \tilde{\ell}_1^2+\tilde{\ell}_2^2-\tilde{\ell}_1^2\tilde{\ell}_2^2
\end{align*}
where, in general, $\tilde{u} = \tanh(u)$ for $u\in \R$. This motivates our definition of a
\emph{rational hyperbolic triangle} to be one whose side lengths $\ell_1$, $\ell_2$, $\ell_3$
have rational hyperbolic tangents $\tilde{\ell}_1$, $\tilde{\ell}_2$, $\tilde{\ell}_3$. Note, this notion
of a rational hyperbolic triangle is more general than that used by Hartshorne
and van Luijk in \cite{Hart2} where the authors consider the more restrictive condition that the side lengths are natural logs of rational numbers.
In contrast, our definition of a rational hyperbolic triangle is equivalent to requiring its side lengths to be natural logs of square roots of
rational numbers. In particular, the examples of hyperbolic Pythagorean triples derived by Hartshorne
and van Luijk are rational hyperbolic triangles under our definition, and in this case both $\sigma$ and $\rho^2$ are rational.

\section{Euclidean Triangles and Cubic Curves}\label{sectone}

Consider a Euclidean triangle $\Delta$. The angle bisectors are concurrent in a point called the incenter. Furthermore, the distances from the sides of $\Delta$ to the incenter are all equal, so this determines an inscribed circle, called the incircle. We define the inradius $r$ to be the radius of the incircle. The inradii joining the incenter to the three intersection points of the incircle with $\Delta$ determine three central angles $\alpha$, $\beta$, $\gamma$, and three partial side lengths $\a$, $\b$, $\c$, as in Figure \ref{E} below.
\begin{figure}
\begin{center}
\includegraphics[scale=0.23]{Euclidean.png}
\end{center}
\caption{Euclidean Triangle with Angle Bisectors, Incircle, and Inradii}
\label{E}
\end{figure}
The angle bisectors also bisect the central angles, so
\begin{align*}
\a = r\tan(\alpha/2) && \b = r\tan(\beta/2) && \c = r\tan(\gamma/2).
\end{align*}
Hence the Triple Tangent Identity (see \ref{trip} in Appendix \ref{A}) implies
\begin{align*}
\frac{\a\b\c}{\a+\b+\c} = r^2,
\end{align*}
so
\begin{align*}
\c = \frac{r^2(\a+\b)}{\a\b-r^2}.
\end{align*}
Thus the semiperimeter $s$ (half the perimeter) of $\Delta$ satisfies
\begin{align*}
s = \a + \b + \c = \a + \b +  \frac{r^2(\a+\b)}{\a\b-r^2},
\end{align*}
or equivalently
\begin{align}\label{cubic}
s(\a\b-r^2) = \a^2\b+\a\b^2.
\end{align}
One can show that $s\geq 3\sqrt{3}r$ with equality only for an equilateral triangle. Suppose $\Delta$ has side lengths $\ell_1,\ell_2,\ell_3$. Then, without loss of generality,
\begin{align*}
\a + \b=\ell_1  &&  \b+ \c = \ell_2 &&  \c+ \a = \ell_3,
\end{align*}
so
\begin{align}\label{one}
\a = \frac{\ell_1-\ell_2+\ell_3}{2} && \b = \frac{\ell_1+\ell_2-\ell_3}{2} && \c = \frac{-\ell_1+\ell_2+\ell_3}{2}.
\end{align}
Therefore Eq. \ref{cubic} implies that we get a point
\begin{align*}
(\a,\b) = \left(s-\ell_2, s-\ell_3\right)
\end{align*}
on the cubic curve
\begin{align}\label{eq}
C_{r,s}\colon s(xy-r^2) = x^2y+xy^2
\end{align}
where
\begin{align}\label{two}
s=\frac{\ell_1+\ell_2+\ell_3}{2}, && r^2 = \frac{(s-\ell_1)(s-\ell_2)(s-\ell_3)}{s}.
\end{align}
The graph of $C_{r,s}$ in the affine real plane has four components (see Figure \ref{crs}), one in each quadrant.
The point $(\a,\b)$ from $\Delta$ is, of course, in the first quadrant.

\begin{figure}
\begin{center}
\includegraphics[width=12 cm]{crs}
\end{center}
\caption{Graph of $C_{r,s}$ in the affine real plane}
\label{crs}
\end{figure}

In fact, we get up to six points on $C_{r,s}$ coming from $\Delta$ (see Figure \ref{sixpoints}), namely, $(\a,\b)$, $(\b,\c)$, $(\c,\a)$ and the transposes $(\b,\a)$, $(\c,\b)$, $(\a,\c)$.
\begin{figure}
\begin{center}
\includegraphics[width=6.5 cm]{sixpoints}
\end{center}
\caption{Six points on $C_{r,s}$ coming from a triangle}
\label{sixpoints}
\end{figure}
We homogenize Eq. \ref{eq} to view $C_{r,s}$ as a curve in $\P^2(\C)$ with projective coordinates $[x,y,z]$; it is nonsingular provided that $s>3\sqrt{3}r$ (i.e., $\Delta$ is non-equilateral) and has the following three points at infinity: $[1,0,0]$, $[0,1,0]$, and $[1,-1,0]$. There is a natural group law on the points of $C_{r,s}$ which can be interpreted geometrically via the usual tangent/chord construction: for points $P$, $Q$, we take $P+Q$ to be the reflection about the line $y=x$ of the third intersection point $P\ast Q$ of $C_{r,s}$ with the secant/tangent line for $P$, $Q$. In particular, $[1,-1,0]$ is the identity element and $[1,0,0]$, $[0,1,0]$ both have order three with $[1,0,0]+[0,1,0]=[1,-1,0]$. For the affine point $(\a,\b)$ as above, we have $-(\a,\b) = (\b,\a)$, $(\a,\b)+[1,0,0] = (\b,\c)$ and similarly $(\a,\b)+[0,1,0] = (\c, \a)$. If $\Delta$ is isosceles (e.g., when $\a=\b$), then we get three corresponding points on $C_{r,s}$, one of which has order two.

In fact, there are always three real points on $C_{r,s}$ of order two when $s>3\sqrt{3}r$, and exactly two of those points come from distinct isosceles triangles.

There is an easily determined Weierstrass form which gives us a different, but related, structure of an elliptic curve in the non-equilateral case, namely,
\begin{align*}
E_{r,s}\colon Y^2 + sXY+sr^2Y = X^3.
\end{align*}
Explicitly,
\begin{align*}
(X,Y) = \left(-\frac{sr^2}{x}, \frac{sr^2y}{x} \right),
\end{align*}
which we can view as a map in projective coordinates
\begin{align*}
C_{r,s} \longrightarrow E_{r,s} \colon [x,y,z] \mapsto [-sr^2z, sr^2y, x].
\end{align*}
This transformation maps the points of order three $[0,1,0]$ and $[1,0,0]$ for $C_{r,s}$ to the identity $[0,1,0]$ and the point of order three $(0,0)$, respectively, for $E_{r,s}$ and maps the identity $[1,-1,0]$ for $C_{r,s}$ to the point of order three $(0, -sr^2) = -(0,0)$ in $E_{r,s}$. The curves $C_{r,s}$ are very natural and symmetric, but the curves $E_{r,s}$ are better suited for computer computations because of their simple Weierstrass equations.

We say that $\Delta$ is \emph{rational} if $\ell_1,\ell_2,\ell_3\in\Q$. In this case, Eq.s \ref{one} and \ref{two} imply $\a,\b,\c,r^2\in\Q$, so both $C_{r,s}$ and $E_{r,s}$ have coefficients in $\Q$ and $(\a,\b)$ is rational point on $C_{r,s}$ which maps to a rational point $(-sr^2/\a, sr^2\b/\a)$ on $E_{r,s}$; thus we can exploit the group structures on $C_{r,s}(\Q)$ and $E_{r,s}(\Q)$ to generate other triangles with rational side lengths having the same semiperimeter $s$ and inradius $r$.
\begin{thm}
Suppose $s,r^2\in\Q$ with $s\geq 3\sqrt{3}r > 0$. Then the assignment
\begin{align*}
(x_0,y_0) \mapsto (x_0+y_0, s-x_0, s-y_0)
\end{align*}
is a bijection from the rational points $(x_0,y_0)$ on $C_{r,s}$ having $x_0, y_0 > 0$ to the sequences of side lengths of rational Euclidean triangles having semiperimeter $s$ and inradius $r$.

\end{thm}
\begin{proof}
We have already seen from the above construction that sequences of side lengths of rational triangles give rise to rational points on $C_{r,s}$ in the first quadrant; in particular, the map is onto. Conversely, suppose $(x_0,y_0)$ is a rational point on $C_{r,s}$ in the first quadrant. Then we claim $(x_0+y_0, s-x_0, s-y_0)$ is a sequence of side lengths of a triangle. It suffices to check the inequalities
\begin{align*}
(x_0+y_0) + (s-x_0) = s+y_0 > s-y_0 \\
(x_0+y_0) + (s-y_0) = s+x_0 > s-x_0 \\
(s-x_0)+(s-y_0) = 2s - (x_0+y_0) > x_0 + y_0
\end{align*}
where the last inequality follows from
\begin{align*}
\frac{s}{x_0+y_0} = \frac{x_0y_0}{x_0y_0-r^2} > 1.
\end{align*}
It remains only to note that $(x_0+y_0, s-x_0, s-y_0) = (x_0'+y_0', s-x_0', s-y_0') \Rightarrow (x_0, y_0) =(x_0', y_0')$.
\end{proof}
In light of the above result, we say that a point on $C_{r,s}$ is a \emph{triangle point} if it lies in the first quadrant of the affine real plane.
\begin{thm}
The sum of two triangle points on $C_{r.s}$ is not a triangle point, and the sum of a triangle point with a non-triangle point is a triangle point. It follows that the sum
of an odd (resp. even) number triangle points is (resp. is not) a triangle point.
\end{thm}
\begin{proof}
This follows from the observation that the component of $C_{r,s}$ in the first quadrant of the affine real plane is the boundary of a convex region.
In particular, any line intersecting this component will intersect in exactly two points counting multiplicity.
\end{proof}
Since the identity $[1,-1,0]$ on $C_{r,s}$ is not a triangle point, we immediately obtain the following.
\begin{corollary}\label{odd}
A triangle point on $C_{r,s}$ cannot have odd order.
\end{corollary}
By a \emph{Pythagorean triple} we mean a set of 3 integers which together form the side
lengths of a right triangle. A Pythagorean triple $\ell_1$, $\ell_2$, $\ell_3$ is said to be \emph{primitive} if
there is no prime which divides every $\ell_i$. Every primitive Pythagorean triple is of the form $m^2-n^2$, $2mn$, $m^2+n^2$
where $m>n$ are relatively prime positive integers, exactly one of which is even; the inradius and semiperimeter of such a Pythagorean triple are given by $r = n(m-n)$ and $s = m(m+n)$, respectively.
\begin{thm}\label{rankthm}
Let $m>n$ be relatively prime positive integers. Suppose that $m$ is neither an odd perfect cube nor twice a perfect cube.
Then the curve $\mathcal{C}_{m,n} \mathrel{\mathop :}= C_{n(m-n), m(m+n)}$ corresponding to
the Pythagorean triple $m^2-n^2$, $2mn$, $m^2+n^2$ has the property
\begin{align*}
\rank_{\Z} (\mathcal{C}_{m,n}(\Q)) \geq 1.
\end{align*}
\end{thm}
\begin{proof}
By Corollary \ref{odd}, it suffices to show that the condition on $m$ implies $\mathcal{C}_{m,n}$ cannot have a rational point of order
two. We do this by contradiction. Suppose $P$ is a rational of order two on $\mathcal{C}_{m,n}$. Then $P = (x,x)$
for some $x\in \Q$, so $s(x^2-r^2) = 2x^3$. By the rational root test, $x = t/2$ for some $t\in\Z$. Hence $t^3 = s(t+2r)(t-2r)$. Let $p$ be an odd prime dividing $m$.
Then $p$ also divides $t^3$, so the exact power of $p$ dividing $t^3$ is $3k$ for some positive integer $k$. Note that
both $r = n(m-n)$ and $s/m= m+n$ are relatively prime to $m$, so the exact power of $p$ dividing $m$ is also $3k$. Since $m$
is not an odd perfect cube, this forces $m=2^ij^3$ where $j$ is an odd integer and $i$ is a positive integer. On the other hand, $r$ is odd and $t$ is even since $m$ is even, so $3\cdot \ord_2(t) = i + \ord_2(t+2r) + \ord_2(t-2r) = i+2$. Therefore $i\equiv -2 \equiv 1 \pmod{3}$,
which contradicts the assumption that $m$ is not twice a perfect cube.
\end{proof}
\begin{exam}
Consider the familiar right triangle with side lengths $3$, $4$, $5$. The semiperimeter is $s = (3+4+5)/2 = 6$ and the inradius is $r=(6-3)(6-4)(6-5)/6=1$. We can take $\a = (3+4-5)/2 = 1$ and $\b = (-3+4+5)/2= 3$, which yields a point $(1,3)$ on $C_{1,6}$. This point maps to the point $P=(-6/1, 6\cdot 3/1) = (-6, 18)$ on the elliptic curve $E_{1,6}$. We can compute $3P = (-35/9, 343/27)$ in $E_{1,6}(\Q)$ either by hand or with SAGE, which comes from the point $(x_0,y_0)=(54/35, 49/15)$ on $C_{1,6}$. Thus we can construct a rational triangle with lengths $x_0+ y_0 = 101/21$,  $s-x_0 = 156/35$, and $s-y_0 = 41/15$. This produces a non-right triangle which also has inradius $1$ and semiperimeter $6$. In particular, we have two non-congruent triangles with rational side lengths having the same perimeter $12$ and area $6$. We should note that it is not \emph{a priori} obvious why rational Euclidean triangles having the same area and perimeter are not necessarily congruent (see \cite{Rose} for further discussion of this point).
\begin{figure}
\includegraphics[height=10cm]{345}
\hspace{.5 in}
\includegraphics[height=10cm]{345alt}
\end{figure}
\noindent In fact, $E_{r,s}(\Q)$ has rank $1$ with $E_{r,s}(\Q) = \Z\cdot P$ as checked by SAGE, and we can generate arbitrarily many such examples in the same way by taking odd multiples of $P$.
\end{exam}

\section{Hyperbolic Triangles and Quartic Curves}

Consider now a hyperbolic triangle $\Delta$ in the Poincar\'{e} upper half-plane, so each side is either a vertical line segment or the arc of a circle whose center lies on the boundary axis. Again, the angle bisectors are concurrent in a point called the incenter, and the hyperbolic distances from the sides of the triangle to the incenter are all equal, so this determines an inscribed circle, called the incircle. Note that a hyperbolic circle is a Euclidean circle, but the hyperbolic center is not the Euclidean center. We define the inradius $r$ to be the hyperbolic radius of the incircle. The inradii joining the incenter to the three intersection points of the incircle with $\Delta$ determine three central angles $\alpha$, $\beta$, $\gamma$, and three partial side lengths $\a$, $\b$, $\c$, as in Figure \ref{H} below.

\begin{figure}
\begin{center}
\includegraphics[scale=0.25]{Hyperbolic.png}
\end{center}
\caption{Hyperbolic Triangle with Angle Bisectors, Incircle, and Inradii}
\label{H}
\end{figure}
The angle bisectors also bisect the central angles, so
\begin{align*}
\tanh(\a) = \rho\tan(\alpha/2) && \tanh(\b) = \rho\tan(\beta/2) && \tanh(\c) = \rho\tan(\gamma/2)
\end{align*}
where $\rho = \sinh(r)$. For convenience we define the notation
\begin{align*}
\tilde{u} \mathrel{\mathop:}= \tanh(u) \, \, \, \, \mbox{for all } u\in \R,
\end{align*}
so, similar to the Euclidean case, the Triple Tangent Identity implies
\begin{align*}
\tilde{\c} = \frac{\rho^2(\tilde{\a}+\tilde{\b})}{\tilde{\a}\tilde{\b}-\rho^2}.
\end{align*}
Thus using \ref{tanh} in Appendix \ref{A} shows the semiperimeter $s$ of $\Delta$ satisfies
\begin{align*}
\sigma \mathrel{\mathop:}= \tilde{s} = \tanh(\a+\b+\c) = \frac{ \tilde{\a}+\tilde{\b}+\tilde{\c}+\tilde{\a}\tilde{\b}\tilde{\c}}{1 + \tilde{\a}\tilde{\b}+\tilde{\b}\tilde{\c}+\tilde{\c}\tilde{\a}} = \cfrac{ \tilde{\a}+\tilde{\b}+(1+\tilde{\a}\tilde{\b})\cfrac{\rho^2(\tilde{\a}+\tilde{\b})}{\tilde{\a}\tilde{\b}-\rho^2}}{1 + \tilde{\a}\tilde{\b}+(\tilde{\b}+\tilde{\a})\cfrac{\rho^2(\tilde{\a}+\tilde{\b})}{\tilde{\a}\tilde{\b}-\rho^2}}
\end{align*}
or, equivalently,
\begin{align}\label{quartic}
 \sigma(\tilde{\a}^2\tilde{\b}^2+\tilde{\a}\tilde{\b} + \rho^2(\tilde{\a}^2 + \tilde{\a}\tilde{\b} + \tilde{\b}^2-1)) = (1+\rho^2)(\tilde{\a}^2\tilde{\b}+\tilde{\a}\tilde{\b}^2).
\end{align}
One can show that $\sigma \geq 3\sqrt{3}\rho\frac{1+\rho^2}{1+9\rho^2}$ with equality only for an equilateral triangle. Suppose $\Delta$ has side lengths $\ell_1,\ell_2, \ell_3$. Then, without loss of generality,
\begin{align*}
\a + \b= \ell_1  &&  \b+ \c = \ell_2 &&  \c+ \a = \ell_3,
\end{align*}
so
\begin{align*}
\tilde{\ell_1} = \frac{\tilde{\a}+\tilde{\b}}{1+\tilde{\a}\tilde{\b}} && \tilde{\ell_2} = \frac{\tilde{\b}+\tilde{\c}}{1+\tilde{\b}\tilde{\c}} &&  \tilde{\ell_3} =   \frac{\tilde{\c}+\tilde{\a}}{1+\tilde{\c}\tilde{\a}}.
\end{align*}
Therefore Eq. \ref{quartic} implies that we get a point
\begin{align}\label{point}
(\tilde{\a},\tilde{\b}) = \left(\frac{\sigma - \tilde{\ell}_2}{1-\sigma\tilde{\ell}_2}, \frac{\sigma - \tilde{\ell}_3}{1-\sigma\tilde{\ell}_3}\right)
\end{align}
on the quartic curve
\begin{align}\label{heq}
Q_{\rho,\sigma}\colon \sigma(x^2y^2+xy + \rho^2(x^2 + xy + y^2-1)) = (1+\rho^2)(x^2y+xy^2)
\end{align}
where
\begin{align}\label{sigma}
\sigma = \frac{1 + \tilde{\ell}_1\tilde{\ell}_2 + \tilde{\ell}_2\tilde{\ell}_3+\tilde{\ell}_3\tilde{\ell}_1 + \sqrt{(1-\tilde{\ell}_1^2)(1-\tilde{\ell}_2^2)(1-\tilde{\ell}_3^2)}}{\tilde{\ell}_1 + \tilde{\ell}_2 + \tilde{\ell}_3+\tilde{\ell}_1\tilde{\ell}_2\tilde{\ell}_3},
\end{align}
\begin{align}\label{rho}
 \rho^2 = \left( \frac{(1-\sigma\tilde{\ell}_2)(1-\sigma\tilde{\ell}_3)}{(\sigma - \tilde{\ell_2})(\sigma - \tilde{\ell_3})} + \frac{(1-\sigma\tilde{\ell}_1)(1-\sigma\tilde{\ell}_3)}{(\sigma - \tilde{\ell_1})(\sigma - \tilde{\ell_3})} + \frac{(1-\sigma\tilde{\ell}_1)(1-\sigma\tilde{\ell}_2)}{(\sigma - \tilde{\ell_1})(\sigma - \tilde{\ell_2}))}\right)^{-1}.
\end{align}
Here the formula for $\sigma$ comes from the identity
\begin{align*}
 \frac{ \tilde{\ell}_1+\tilde{\ell}_2+\tilde{\ell}_3+\tilde{\ell}_1\tilde{\ell}_2\tilde{\ell}_3}{1 + \tilde{\ell}_1\tilde{\ell}_2+\tilde{\ell}_2\tilde{\ell}_3+\tilde{\ell}_3\tilde{\ell}_1} = \tanh(\ell_1+\ell_2 + \ell_3) = \tanh(2s) = \frac{2\sigma}{1+\sigma^2},
\end{align*}
and the formula for $\rho^2$ comes from
\begin{align*}
\frac{1}{\rho^2} = \frac{\tilde{\a}+\tilde{\b}+\tilde{\c}}{\tilde{\a}\tilde{\b}\tilde{\c}} = \frac{1}{\tilde{\b}\tilde{\c}} + \frac{1}{\tilde{\a}\tilde{\c}} + \frac{1}{\tilde{\a}\tilde{\b}}.
\end{align*}
The graph of $Q_{\rho,\sigma}$ in the affine real plane has five components (see Figure \ref{qrs}).
The point $(\tilde{\a},\tilde{\b})$ from $\Delta$ is, of course, on the component contained in $(0,1)\times(0,1)$.

\begin{figure}
\begin{center}
\includegraphics[width=13 cm]{qrs}
\end{center}
\caption{Graph of $Q_{\rho,\sigma}$ in the affine real plane}
\label{qrs}
\end{figure}
As in the Euclidean case, we get up to six points on $Q_{\rho,\sigma}$, namely, $(\tilde{\a},\tilde{\b})$, $(\tilde{\b},\tilde{\c})$, $(\tilde{\c},\tilde{\a})$ and the transposes $(\tilde{\b},\tilde{\a})$, $(\tilde{\c},\tilde{\b})$, $(\tilde{\a},\tilde{\c})$. We homogenize Eq. \ref{heq} to view $Q_{\rho,\sigma}$ as a curve in $\P^2(\C)$ with projective coordinates $[x,y,z]$; the points at infinity $[1,0,0]$ and $[0,1,0]$ are the only singular points. These singular points both have multiplicity $2$, so the genus of $Q_{\rho, \sigma}$ is
\begin{align*}
\frac{(4-1)(4-2)}{2} - \frac{2(2-1)}{2} - \frac{2(2-1)}{2} = 3 - 1 - 1 = 1.
\end{align*}
There is a natural change of coordinates $(X,Y) = (xy, x+y)$ so that $\alpha^2-Y\alpha + X = (\alpha-x)(\alpha - y)$. Thus if $X$, $Y$ are in a subfield $F$ of $\C$, then $x,y$ will be in $F$ precisely when $Y^2-4X = Z^2$ for some $Z\in F$. This allows us to view our curve as the intersection of two quadric surfaces
\begin{align}\label{intersection}
H_{\rho,\sigma}\colon
\left\{\begin{array}{l}
\sigma(X^2+X+\rho^2(Y^2-X-1))-(\rho^2+1)XY=0\\
4X-Y^2+Z^2=0
\end{array}\right.
\end{align}
\begin{figure}
\begin{center}
\includegraphics[width=16 cm]{surfaces1}
\end{center}
\caption{Graph of $H_{\rho,\sigma}$ with base point $(-1,0,2)$}
\label{six}
\end{figure}
Thus the curve can be visualized, as in Figure \ref{six}, as the intersection of a hyperbolic cylinder (red) and a hyperbolic paraboloid (blue).
We homogenize Eq. \ref{intersection} and view $H_{\rho,\sigma}$ as a curve in $\P^3(\C)$ with projective coordinates $[X,Y,Z,T]$. There is a geometric way of combining points on $H_{\rho,\sigma}$ with respect to the point $\OO = [-1,0,2,1]$: for points $P$, $Q$, we take $P + Q$ to be the fourth intersection point of $H_{\rho,\sigma}$ with the plane spanned by $\OO$ and the secant/tangent line for $P$, $Q$. See \cite{Wang} for a more detailed exposition of curves arising from the intersections of two quadric surfaces.

We say that $\Delta$ is \emph{rational} if $\tilde{\ell}_1, \tilde{\ell}_2, \tilde{\ell}_3\in\Q$. Note that $\Delta$ being
rational does not imply $\sigma\in\Q$ since, e.g., taking $\tilde{\ell}_1 = 1/2 = \tilde{\ell}_2$ and $\tilde{\ell}_3=1/3$ gives $\sigma = \frac{19+ 6\sqrt{2}}{17}\notin\Q$ by Equation \ref{sigma}.
However, if $\Delta$ is a rational right hyperbolic triangle with hypotenuse $\ell_3$, then recall that we have the following hyperbolic Pythagorean identity
\begin{align*}
\tilde{\ell}_3^2 = \tilde{\ell}_1^2+\tilde{\ell}_2^2-\tilde{\ell}_1^2\tilde{\ell}_2^2,
\end{align*}
so $1-\tilde{\ell}_3^2 = (1-\tilde{\ell}_1^2)(1-\tilde{\ell}_2^2)$ which combined with Equation \ref{sigma} implies
\begin{align*}
\sigma = \frac{1 + \tilde{\ell}_1\tilde{\ell}_2 + \tilde{\ell}_2\tilde{\ell}_3+\tilde{\ell}_3\tilde{\ell}_1 + 1-\tilde{\ell}_3^2}{\tilde{\ell}_1 + \tilde{\ell}_2 + \tilde{\ell}_3+\tilde{\ell}_1\tilde{\ell}_2\tilde{\ell}_3}\in\Q.
\end{align*}
In any case, it is clear from Eq.s \ref{point} and \ref{rho} that if $\Delta$ is rational and $\sigma\in\Q$, then we must also have $\tilde{a}, \tilde{b}, \tilde{c}, \rho^2\in\Q.$
\begin{thm}
Suppose $\sigma,\rho^2\in\Q$ with $\sigma \geq 3\sqrt{3}\rho\frac{1+\rho^2}{1+9\rho^2} > 0$. Then the assignment
\begin{align*}
(x_0,y_0) \mapsto \left(\tanh^{-1}\left(\frac{x_0+y_0}{1+x_0y_0}\right), \tanh^{-1}\left(\frac{\sigma-x_0}{1-\sigma x_0}\right), \tanh^{-1}\left(\frac{\sigma-y_0}{1-\sigma y_0}\right)\right)
\end{align*}
is a bijection from the rational points $(x_0,y_0)$ on $Q_{\rho,\sigma}$ having $1 > x_0, y_0 > 0$ to the sequences of side lengths of rational hyperbolic triangles having semiperimeter $s = \tanh^{-1}(\sigma)$ and inradius $r = \sinh^{-1}(\rho)$.
\end{thm}
\begin{proof}
We have already seen from the above construction that sequences of side lengths of rational hyperbolic triangles give rise to rational points on $Q_{\rho,\sigma}$ in $(0,1)\times (0,1)$; in particular, the map is onto. Conversely, suppose $(x_0,y_0)$ is a rational point on $Q_{\rho,\sigma}$ in $(0,1)\times (0,1)$. Then we claim
\begin{align*}
\left(\tanh^{-1}\left(\frac{x_0+y_0}{1+x_0y_0}\right), \tanh^{-1}\left(\frac{\sigma-x_0}{1-\sigma x_0}\right), \tanh^{-1}\left(\frac{\sigma-y_0}{1-\sigma y_0}\right)\right) \\
= (\tanh^{-1}(x_0) + \tanh^{-1}(y_0), s - \tanh^{-1}(x_0), s - \tanh^{-1}(y_0) )
\end{align*}
is a sequence of side lengths of a rational hyperbolic triangle. We have the inequalities
\begin{align*}
(\tanh^{-1}(x_0)+\tanh^{-1}(y_0) ) + (s-\tanh^{-1}(x_0)) = s + \tanh^{-1}(y_0) > s - \tanh^{-1}(y_0) \\
(\tanh^{-1}(x_0)+\tanh^{-1}(y_0) ) + (s-\tanh^{-1}(y_0)) = s + \tanh^{-1}(x_0) > s-\tanh^{-1}(x_0).
\end{align*}
Also,
\begin{align*}
0>(x_0-1)(1-y_0)  = x_0+y_0-(x_0y_0+1),
\end{align*}
so $(x_0y_0+1) > (x_0+y_0)^2/(1+x_0y_0)$ and
\begin{align*}
1<\frac{(1+\rho^2)x_0y_0}{\left(x_0y_0 +\rho^2 \left( \frac{(x_0+y_0)^2}{1+x_0y_0} -1\right)\right)} = \frac{\sigma(1+x_0y_0)}{x_0+y_0}.
\end{align*}
Thus
\begin{align*}
&\hspace{0.2 in} (s-\tanh^{-1}(x_0)) + (s-\tanh^{-1}(y_0)) \\
&= 2\tanh^{-1}(\sigma)-(\tanh^{-1}(x_0) +\tanh^{-1}(y_0)) \\
&>2\tanh^{-1}\left(\frac{x_0+y_0}{1+x_0y_0}\right) -(\tanh^{-1}(x_0) +\tanh^{-1}(y_0))\\
&=2(\tanh^{-1}(x_0) +\tanh^{-1}(y_0)) - (\tanh^{-1}(x_0) +\tanh^{-1}(y_0)) \\
&=\tanh^{-1}(x_0) +\tanh^{-1}(y_0).
\end{align*}
It remains only to note that $(\tanh^{-1}(x_0)+\tanh^{-1}(y_0), s-\tanh^{-1}(x_0), s-\tanh^{-1}(y_0)) =(\tanh^{-1}(x'_0)+\tanh^{-1}(y'_0), s-\tanh^{-1}(x'_0), s-\tanh^{-1}(y'_0)) \Rightarrow (x_0, y_0) =(x_0', y_0')$.
\end{proof}
\begin{exam}
There is a construction, discussed in \cite{Hart2} for instance, which allows one to write down rational\footnote{As mentioned in the introduction, the notion of a rational hyperbolic triangle in Hartshorne and van Luijk's article \cite{Hart2} is a stronger notion than ours; namely, they require that the side lengths are logs of rational numbers, whereas we only require that the hyperbolic tangents of the side length are rational numbers or, equivalently, that the side lengths are the logs of square roots of rational numbers.} hyperbolic Pythagorean triples. For example, there is a hyperbolic right triangle with side lengths $\tanh^{-1}(672/697)$, $\tanh^{-1}(104/185)$, $\tanh^{-1}(40/41)$. We have $\sigma = 312/317$ and $\rho^2 = 112^2/242201$ in this case by Equations \ref{sigma} and \ref{rho}, so by Equation \ref{point} we get a rational point
\begin{align*}
(\tilde{\a},\tilde{\b}) = \left(\frac{1456}{1541}, \frac{112}{517} \right)
\end{align*}
on the rational quartic $Q_{\rho, \sigma}$. This gives us a rational point
\begin{align*}
P=\left(\frac{163072}{796697}, \frac{925344}{796697}, \frac{580160}{796697}\right)
\end{align*}
in the affine patch $(X,Y,Z) = [X,Y,Z,1]$ of $H_{\rho,\sigma}$. In order to compute $P + P$ on $H_{\rho,\sigma}$, we first compute the tangent line at $P$ by intersecting the tangent planes at $P$ on the two surfaces defining $H_{\rho,\sigma}$. This tangent line at $P$ is parallel to a vector $v$ obtained by taking a cross product of normal vectors of the tangent planes. We find
\begin{align*}
v \times (P-\OO) =  \left(315005821528688640, -\frac{1615149902619671040}{11}, \frac{1807175612011392000}{11}\right),
\end{align*}
which is the normal vector of the plane determined by the point $\OO = (-1,0,2)$ and the tangent line for $P$. We find the fourth intersection of this plane with $H_{\rho,\sigma}$ to be
\begin{align*}
P + P = \left(\frac{157101469162847924}{6671549185609843471}, -\frac{3620500406298490680}{6671549185609843471}, \frac{2985897265044714172}{6671549185609843471}\right).
\end{align*}
This gives us a rational point on $Q_{\rho,\sigma}$, but does not correspond to a triangle, of course, since the point has negative $y$-coordinate. However, we can now compute $P + (P + P)$. We compute $((P + P)-\OO) \times (P-\OO) =$
\begin{align*}
\left(\frac{13252388009067818908542000}{5315203221527805463815287},
-\frac{274681775539499477051700}{483200292866164133074117},
\frac{208376488637078243567400}{113089430245272456676921}\right),
\end{align*}
which is the normal vector for the plane determined by the point $\OO$ and the secant line for $P$, $P + P$. The fourth intersection point of this plane with $H_{\rho,\sigma}$ has coordinates
\begin{align*}
& \left(\frac{5263691075333761370098794452493746497879860068608}{16384416048645387501313685205281020081739448108365},\right. \\
& \hspace{0.17 in} \frac{21085154660473767875635316117377007639096311317792}{16384416048645387501313685205281020081739448108365}, \\
& \hspace{0.15 in}  \left. \frac{9980667760067897618538531202408325397579645579328}{16384416048645387501313685205281020081739448108365}\right).
\end{align*}
This point does indeed give a point on $Q_{\rho,\sigma}$ in the first quadrant with coordinates smaller than $1$:
\begin{align*}
(x_0,y_0) = \left(\frac{2072869433189638375660592}{2186502887201310556520693}, \frac{2539325917520154646338224}{7493434444816664924429305} \right)
\end{align*}
This point corresponds to a non-right rational hyperbolic triangle with semiperimeter $s = \tanh^{-1}(312/317)$ and inradius $r = \sinh^{-1}(112/\sqrt{242201})$, namely, the triangle with side lengths
\begin{align*}
&\tanh^{-1}\left(\frac{4938503954557916283489312}{5070357052721862942058853} \right), \\
&\tanh^{-1}\left(\frac{25089290485693528550048552}{46386152087648273210954977} \right),\\
&\tanh^{-1}\left(\frac{1532985230928910433532726152}{1583149032740594531386563797}\right).
\end{align*}
This example would have been very difficult to find by hand, i.e., without exploiting the elliptic curve structure.
\end{exam}

\appendix

\section{Two Trigonometric Identities}\label{A}

The key identity we used in deriving our plane curves came from relations
among the tangents of half the central angles.
\begin{thm}[Triple Tangent Identity]\label{trip}
Suppose $\theta_1 + \theta_2 + \theta_3 = \pi$. Then

\begin{align*}
\tan(\theta_1) + \tan(\theta_2) + \tan(\theta_3) = \tan(\theta_1)\tan(\theta_2)\tan(\theta_3).
\end{align*}
\end{thm}

\begin{proof}
Using oddness, $\pi$-periodicity, and the addition law, we get
\begin{align*}
-\tan(\theta_3) = \tan(\pi - \theta_3) = \tan(\theta_1+\theta_2) = \frac{\tan(\theta_1)+\tan(\theta_2)}{1-\tan(\theta_1)\tan(\theta_2)}.
\end{align*}
\end{proof}

In the hyperbolic case, the trigonometric formulas for right triangles involved hyperbolic tangents
of side lengths, so we needed to make use of an iterated addition law.

\begin{thm}\label{tanh}
For $A,B,C \in\R$, we have
\begin{align*}
\tanh(\tanh^{-1}(A) + \tanh^{-1}(B) + \tanh^{-1}(C)) = \frac{A+B +C + ABC}{1+AB+BC+CA}.
\end{align*}
\end{thm}
\begin{proof}
Use the definition of the hyperbolic tangent function $\tanh(z) = \displaystyle{\frac{e^{z}-e^{-z}}{e^{z}+e^{-z}} =  \frac{e^{2z}-1}{e^{2z}+1} }$ to get
\begin{align*}
\frac{\tanh(A)+\tanh(B)}{1+\tanh(A)\tanh(B)}
&= \frac{(e^{2A}-1)(e^{2B}+1) +  (e^{2B}-1)(e^{2A}+1)}{(e^{2A}+1)(e^{2B}+1) + (e^{2A}-1)(e^{2B}-1)} \\
&=  \frac{2e^{2(A+B)}  -2}{2e^{2(A+B)} +2} = \tanh(A + B),
\end{align*}
so
\begin{align*}
\tanh(\tanh^{-1}(A) + \tanh^{-1}(B)) = \frac{A+B}{1+AB}.
\end{align*}
Thus
\begin{align*}
\tanh(\tanh^{-1}(A) + \tanh^{-1}(B) + \tanh^{-1}(C)) &= \frac{\tanh(\tanh^{-1}(A) + \tanh^{-1}(B)) +C }{1+ \tanh(\tanh^{-1}(A) + \tanh^{-1}(B))C} \\
&=  \frac{A+B +C(1+AB) }{1+AB+(A+B)C}.
\end{align*}
\end{proof}

\bibliographystyle{amsalpha}
\bibliography{References}

\end{document}


TITLE: Which knots appear as the singular locus of a polyhedral metric on the 3-sphere?
QUESTION [18 upvotes]: What can be said about a knot $K\subseteq S^3$ for which there exists a (Euclidean) polyhedral metric (aka Euclidean cone-manifold structure) on $S^3$ whose singular locus is precisely $K$? I'm particularly interested in the case where the conical angle $\alpha$ around $K$ is less than $2\pi$.
The classic example of this is the figure-8 knot. If you quotient $\mathbb E^3$ by the crystallographic group $\mathrm P2_13$, you get a Euclidean orbifold structure on $S^3$ whose singular locus is the figure-8 knot with angle $2\pi/3$. Dunbar (pp.82–86) gives a classification of all Euclidean orbifold structures on $S^3$, and this is the only one whose singular locus is a knot. Therefore, we know that if $K$ is not the figure-8 knot, then $\alpha$ cannot equal $2\pi/n$. That's a starting point, at least!
Dunbar, William D., Geometric orbifolds, Rev. Mat. Univ. Complutense Madr. 1, No. 1-3, 67-99 (1988). ZBL0655.57008.

Edit. Mednykh and Rasskazov construct a very simple family of fundamental polyhedra $\mathcal P(\theta)$ for the polyhedral manifolds $\mathcal C(\theta)$ ($S^3$ with conical angle $\theta$ along the figure-8 knot), for $\theta\in[0,4\pi/3)$. They show that, as $\theta$ increases, $\mathcal P(\theta)$ goes from hyperbolic to spherical, passing through Euclidean at $\theta=2\pi/3$. The reason I mention this is that they note (p.446):

...that [their] approach is general and can be applied to other two-bridge links and knots.

And again, they (p.448):

...describe an algorithm for the construction of the fundamental set $\mathcal P$ for the figure-eight orbifold which effectively works in every space of constant sectional curvature. With slight modifications this algorithm can also be used to construct the fundamental set of any 2-bridge link orbifold.

Mednykh, Alexander; Rasskazov, Alexey, Volumes and degeneration of cone-structures on the figure-eight knot, Tokyo J. Math. 29, No. 2, 445-464 (2006). ZBL1124.57008.

Final Edit. According to a recent paper of Mednykh, the two-bridge knots $4_1,5_2,6_1,6_2,6_3,7_2,7_3,7_4,7_5,7_6$ and $7_7$ all admit polyhedral metrics with a given conical angle $\alpha_0<\pi$. With the exception of $7_5,7_6$ and $7_7$, the value of $\alpha_0$ is given implicitly, but exactly.
Do all two-bridge knots have this property?

REPLY [9 votes]: The ones that have a Euclidean cone metric with cone angle $\leq \pi$ should only be the spherical Montesinos knots (including 2-bridge knots, excluding the (2,n) torus knots). This should follow from the proof of the orbifold theorem, but one would need to search through the proof to sort out the logic. As indicated in the comments, Porti hinted at this in his paper on 2-bridge links. If there’s a spherical metric, then the $\pi$ orbifold with knot as the singular locus is spherical, and the representation into $SO(4)$ can be deformed to decrease the cone angle until it collapses to a Euclidean cone metric. Actually, the proof proceeds in the other direction: a hyperbolic metric is deformed to increase the cone angle until it collapses to a (rescaled) Euclidean cone metric. Conversely, if there’s a metric with cone angle $\leq\pi$, then the double branched cover admits an metric of non-negative Alexandrov  curvature, and therefore must be modeled on $S^3$, $S^2\times R$, or be Euclidean. One can rule out the last two cases in the case of knots (Dunbar classified the knots whose $\pi$ orbifold admits exceptional geometries).
For the general case with cone angle $> \pi$, I think this is wide open.
There are some restrictions though.
The case of cone angle $\geq 2\pi$ is impossible, since this corresponds to a CAT(0) metric on the 3-sphere which doesn't exist.
Similarly, if the cone angle is $> \pi$, the double branched cover will admit a CAT(0) metric. This is only possible if it is hyperbolic. If there’s an essential torus that meets the branch locus, then it can be isotoped to be a flat plane invariant under the involution, and there’s an essential Conway sphere for the knot, and the cone angle would be $= \pi$, a contradiction. An essential sphere cannot exist, since there is a CAT(0) metric so $\pi_2$ vanishes. If the torus does not meet the singular locus, then the knot was a satellite. But then the torus may be isotoped to be totally geodesic (again by the flat plane theorem). The pattern knot would then admit a flat metric with totally geodesic torus boundary. But the only such manifold is the twisted I-bundle over a Klein bottle, which is not a knot complement. So one should restrict to knots whose double cover is either spherical (and the cone angle is $\leq \pi$) or hyperbolic (and the cone angle is betweeen $\pi$ and $2\pi$). The knot will be prime and not satellite and have no essential Conway sphere. Most arborescent knots, such as the (-2,3,7) pretzel considered in Sam Nead’s answer, will be ruled out.
A possible obstruction would come from the representation of $\pi_1(S^3-K)$ to $Isom^+(\mathbb{E}^3)\cong \mathbb{R}^3 \rtimes SO(3)$. If the action of this group has no global fixed point (so is not conjugate into $SO(3)$), then  the projection of this representation into $SO(3)$ gives a representation with non-trivial cohomology with coefficients in $\mathbb{R}^3$ twisted by the representation. Any knot $K$ will have many $SO(3)$ representations of  $\pi_1(S^3-K)$, but one could try to look for ones for which this cohomology always vanishes and for which the holonomy of the meridian is elliptic (this means the translational part of the cocycle must be perpendicular to the axis of rotation of the $SO(3)$ holonomy of the meridian). For a Euclidean cone structure, I would suspect that this representation does not have a global fixed point, since otherwise the developing map would have precompact image, which shouldn't happen for a Euclidean cone structure (a geodesic in a generic direction should miss the cone locus by general position and hence have non-compact developing map).


TITLE: Let $n=apq+1$. Prove that if $pq \ | \ \phi(n)$ then $n$ is prime.
QUESTION [5 upvotes]: Let $p$ and $q$ be distinct odd primes and $a$ be a positive integer with $a<p<q$. I need to prove that if $pq \ | \ \phi(n) $ then  $n$ is prime.
The proof for the trivial case when $a=2$ is given below.
Proof.  Let $n=2pq+1$. Assume $pq \  | \ \phi(n)$. We may write $\phi(n) = cpq$ for some positive integer $c \le 2$. We know that $\phi(n)$ is even for all $n>2$ therefore we must have $c=2$ otherwise $\phi(n) $ is odd. $\phi(n) =2pq=n-1$ which shows that $ n $ is prime. This completes the proof for the trivial case $a=2$.
How do I prove that the proposition holds for an arbitrary positive integer $a$?

REPLY [3 votes]: The equation for $n$ is given as
$$n = apq + 1 \tag{1}\label{eq1A}$$
As you've already indicated, if $n$ is prime, then $\varphi(n) = n - 1 = apq$, so $pq \mid \varphi(n)$.
Consider the opposite direction, i.e., $pq \mid \varphi(n)$. With the definition of Euler's totient function, since $\gcd(pq, n) = 1$, this means $pq$ must divide $\prod_{p_i \mid n}(p_i - 1)$, so either $p$ and $q$ divide $2$ different factors, or $pq$ divides just $1$ factor, among the $p_i - 1$ factors, where the $p_i$ are the prime factors of $n$. Thus, there's two cases to consider.

Case #$1$:
Here, $n$ is not a prime, with there being two odd primes $p_{1}$ and $p_{2}$ where
$$p_{1}p_{2} \mid n \implies n = bp_{1}p_{2}, \; b \ge 1 \tag{2}\label{eq2A}$$
$$p \mid p_{1} - 1 \implies p_{1} = cp + 1, \; c \ge 2 \tag{3}\label{eq3A}$$
$$q \mid p_{2} - 1 \implies p_{2} = dq + 1, \; d \ge 2 \tag{4}\label{eq4A}$$
Substituting \eqref{eq3A} and \eqref{eq4A} into \eqref{eq2A}, and equating the result to \eqref{eq1A}, gives
$$\begin{equation}\begin{aligned}
b(cp + 1)(dq + 1) & = apq + 1 \\
(bcd)pq + bcp + bdq + b & = apq + 1 \\
bcp + bdq + b - 1 & = (a - bcd)pq
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
The left side is positive, so the right side must be as well. This means
$$a \gt bcd \tag{6}\label{eq6A}$$
From \eqref{eq6A}, plus that $c \ge 2$ from \eqref{eq3A} and $d \ge 2$ from \eqref{eq4A}, we also get $bc \lt \frac{a}{d} \le \frac{a}{2}$, $bd \lt \frac{a}{c} \le \frac{a}{2}$ and $b \lt a$. Using this, along with $p \le q - 2$, in the left side of \eqref{eq5A} gives
$$\begin{equation}\begin{aligned}
bcp + bdq + b - 1 & \lt \frac{ap}{2} + \frac{aq}{2} + a \\
& = a\left(\frac{p + q}{2} + 1\right) \\
& \le a\left(\frac{q - 2 + q}{2} + 1\right) \\
& = a\left(q - 1 + 1\right) \\
& = aq
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
However, since the left side of \eqref{eq5A} must be equal to a positive multiple of $pq$, this gives
$$aq \gt pq \implies a \gt p \tag{8}\label{eq8A}$$
which contradicts the requirement of $a \lt p$. Thus, this case is not valid.

Case #$2$:
Here, there is an odd prime $p_{3}$ where
$$p_{3} \mid n \implies n = ep_{3}, \; e \ge 1 \tag{9}\label{eq9A}$$
$$pq \mid p_{3} - 1 \implies p_{3} = fpq + 1, \; f \ge 2 \tag{10}\label{eq10A}$$
Substituting \eqref{eq10A} into \eqref{eq9A}, and equating the result to \eqref{eq1A}, gives
$$\begin{equation}\begin{aligned}
e(fpq + 1) & = apq + 1 \\
(ef)pq + e & = apq + 1 \\
e - 1 & = (a - ef)pq
\end{aligned}\end{equation}\tag{11}\label{eq11A}$$
Since $pq \mid e - 1$, but $pq \gt a \ge ef$ so $e \lt pq$, then $e = 1$ is the only possibility. This then gives $n = p_{3}$ in \eqref{eq9A}, which means $n$ is a prime.

Only case #$2$ can apply, with it giving that $n$ must be a prime, so this concludes the proof in the opposite direction.


text:
\section{Repeated Poles}
\label{sec:ext}
The results in the previous sections were stated for LTI systems with simple poles.
The restriction simplified the exposition, because the dominant dynamics of $G_{[j]}$ take a more complicated form in the general case. For the Hankel operator, however, the additional effort is minimal. To see this, let $G(z)$ have $l$ distinct poles with highest multiplicity $m_j$, $1 \leq j \leq l$, this is,
\begin{align*}
G(z) = \sum_{a=1}^{l} \sum_{b=1}^{m_a} \frac{r_{ba}}{(z-p_a)^b}.
\end{align*}
Then, external positivity requires that $r_{1m_1} > 0$ and $p_1 \geq 0$. Assuming that $m_1 > 1$ and $p_1 > 0$, it follows then that the dominant dynamics of $g_{[2]}$ are provided by $G_1(z) := \frac{r_{1m_1}}{(z-p_1)^{m_1}}$. Since $g_1$ not log-convex on any interval, $\Hank{g}$ cannot be $\tp{2}$. Further, it is easy to see that the only FIR system that is Hankel $\tp{2}$ is a first order system.
Similarly, assuming that $m_1 = \dots = m_{k-2} = 1$ and $m_{k-1} > 1$, it follows by (\ref{eq:ss_compound}) that the dominant dynamics of $G_{[k-1]}$ have a multiple pole in $\prod_{i=1}^{k-1}p_i$. However, since $g_{[k-1]}$ also needs to be log-convex on $\mathds{Z}_{\geq k-1}$, $\Hank{g}$ cannot be $k$-positive then.
\begin{prop}
	Let $G(z) = \sum_{a=1}^{l} \sum_{b=1}^{m_a} \frac{r_{ba}}{(z-p_a)^b}$ be such that $\Hank{g}$ is $k$-positive. Then, $m_1 = \dots = m_{k-1} = 1$.
\end{prop}

config: arxiv


TITLE: "Class" of functions whose inverse, where defined, is the same "class"
QUESTION [9 upvotes]: Please excuse the amateurish use of the term "class", I don't know what the exact term is for what I'm looking for.
Anyway, details.
I'm asking specifically about real-valued functions on the real domain ($\mathbb{R}\to\mathbb{R}$).  To keep things simple, let's assume that the function is defined on some interval of interest, and is continuous and strictly monotonic in that interval, so that there is an inverse function that's also continuous and monotonic.
I am looking for a "class" of functions where the inverse is of the same "class".  By "class", I mean a set of functions with a finite number parameters that, if you changed them, the function would still be in the same "class".  (An obvious example of what I mean by a "class" is the polynomials: you can change the coefficients but the function is still a polynomial.)  Again I apologize if this is omitting a detail or if there's a nice little word for this that I don't know.
I know of a few examples of "classes" that meet these criteria, including:

Linear functions
Piecewise linear functions

Polynomials, of course, do not fit this criteria in general: the inverse of a polynomial is generally not a polynomial.  I don't think rational functions do either, but I'm not sure.
For the record, I am asking partly out of curiosity, and partly because I have a nice application in mind.  I have a application where I need a function that can approximate a curve with perfect round-tripping ($f(f'(x))=x$, exactly).  We're using piecewise linear approximation for now, but it's desirable to be smooth as well.
Thanks.

REPLY [0 votes]: Answering my own question, much later. The application I needed this for is still an issue, and I finally thought of something that I think will work.  (Wait, what, you mean I want to use math for something?  Weird.)
I realized that there is a pretty generic way to obtain a function that meets my requirements, though I can't say I have a proof that it can work in general for any desired interval of interest even if the function is monotonic in that interval.
Suppose we have a real-valued function $F(x,y;A,B,C,D,...)$, defined on some region of the $x,y$-plane in terms of a finite number of real-valued parameters $A,B,C,D,...$, such that there exists another set of real-valued parameters $Z,Y,X,W,...$ for which $F(x,y;A,B,C,D,...)=F(y,x;Z,Y,X,W,...)$.  In other words, a function of $x$ and $y$ where you can swap $x$ and $y$ and the function will be of the same except for the value of the parameters.  I'm sure there's a spiffy name for this like incomplete symmetry or something like that.
Now, if the equation $F(x,y;A,B,C,D,...)=0$ has a general closed-form solution for $y$ over some area where the curve is single-valued and monotonic in terms of $x$, then it will have the same general closed-form solution for $y$, except for the value of the parameters.
Thus, one can obtain a function that meets the requirements I set out.
Example.  The following equation is the general form of an equation representing conic sections in the real plane:
$$
Ax^2+Bxy+Cy^2+Dx+Ey+F=0
$$
The general solution to this equation for $x$, over a region where the curve is single-valued and monotonic in terms of $y$, is:
$$
y=Zx^2+Yx+X+W\sqrt{x^2+Vx+U}
$$
Likewise, the general solution to this equation for $y$ is:
$$
x=Zy^2+Yy+X+W\sqrt{y^2+Vy+U}
$$
Thus the function $f(x)=Zx^2+Yx+X+W\sqrt{x^2+Vx+U}$, in a region where it's monotonic, has an inverse is of the same form.
The only thing is, I don't think this necessarily holds for arbitrary intervals, even if $f$ is monotonic in that interval and thus invertible.  However, as a practical matter, for my use case, I think it will suffice for a simple reason: I can patch several of these together if I need to.  If, for example, there is a point where the parameters have to change, I can put a break at that point.
So my general answer for this is piecewise patched conics, the inverse of which is also piecewise patched conics.


TITLE: Cyclic vectors of a companion matrix
QUESTION [0 upvotes]: Let $A=\begin{bmatrix}
0 & 0 & \dots & 0 & p \\
1 & 0 & \dots & 0 & 0 \\
0 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & \dots & 1 & 0
\end{bmatrix}$ be the $n\times n$ companion matrix whose characteristic function is $x^n-p$ (For convenience let's assume $p$ is prime so that the polynomial is irreducible, this might be true in more general cases). I wonder if any nonzero vector $v\in \mathbb Q^n$ is a cyclic vector. i.e. the set $\{v, Av, A^2v, ..., A^{n−1}v\}$ is linearly independent.
Let $x=(x_1,\dots,x_n)$ be a nonzero vector in $\mathbb Q^n$. Then basically $A$ permutes the entries of $x$ around and multiply the first coordinate by $p$. It seems to me this will create a linear independent set $\{v, Av, A^2v, ..., A^{n−1}v\}$ but I don't know how to prove it.
I think $p$ being a prime is vey crucial. In the case when $n=2$, $(x,y)$ and $(y,px)$ being linearly independent implies that $px^2=y^2$ never holds. In $n=3$ case, the nonzero determinant means $x^3-p^2z^3-py^3-3pxyz=0$ has no nontrivial rational solutions.

REPLY [1 votes]: The answer is yes.
Let $v\in \mathbb{Q}^n\setminus\{0\}$; the vector-space $U=span((A^iv)_i)$ is $A$-invariant.
When $gcd(m,n)=1$, the polynomial $x^n-p^m$ is irreducible over $\mathbb{Q}$. Thus, here, $\chi_A$ is irreducible over $\mathbb{Q}$ and $A$ has no proper invariant vector-space over $\mathbb{Q}$; finally $U=\mathbb{Q}^n$ and we are done.


TITLE: Quantum measurement limit of decoherence and scattering: why are position pointer said classical?
QUESTION [2 upvotes]: In decoherence theory, we explain the decoherence by hamiltonian evolutions between a system and its environment. Calling $H$ the total hamiltonian, I have:
$$H=H_S + H_E + H_{SE} $$
A pointer state $|s\rangle$ is a state of the system $S$ for which the associated observable $|s\rangle \langle s |$ will commute with the total Hamiltonian. Then, coherences between pointer states will be killed under the Hamiltonian evolution.
One particular limit is the quantum-measurement limit for which $H \approx H_{SE}$. As typical interactions between a system and its environment are of the form:
$$H_{SE} = X \otimes E$$
Where $X$ is the position operator of the system, we figure out that the pointer states are the states $|x\rangle$, and thus the coherence between position eigenstates will be killed.
My question:
Now, a particular example of this situation if I understood is the quantum scattering. Basically our system $S$ (a molecule for example) will interact with environmental molecules under $H_{SE}$. And this interaction will kill superposition in the position basis as I explained. And from this we say that it matches the classical behavior in which the molecule doesn't have position superposition.
However I am puzzled by this. Indeed for me the classical limit should be a packet that is narrow both in position and momentum, like a coherent state.
Thus I am not sure to really understand why quantum scattering, as it is a particular case of quantum measurement limit, would fit as a good quantum to classical correspondance. If the particle is localized in space it is spread in momentum and thus highly non classical.
Does that mean that experimentally, we really see that molecules in a gaz for example are well defined in position, but very poorly in momentum (and then this explanation would match the decoherence theory) ?
A paper in which is explicitly said that collisional decoherence is a particular case of the quantum measurement limit (i.e hamiltonian dominated by interacting part) is The quantum-to-classical transition and decoherence on page 5, beginning of second column.

REPLY [1 votes]: the classical limit should be a packet that is narrow both in position and momentum, like a coherent state.

That's correct. The approximation $H\approx H_{SE}$ can be used to show the tendency for pointer states to be localized in position, but that's just an approximation. The kinetic term $H_S$ tends to make a particle's wavefunction spread out, which competes with the tendency of $H_{SE}$ to make it localized. The balance of these two trends results in the particle being somewhat localized in position, but still spread out enough that it can have a somewhat well-defined momentum, too, so its location can move like a classical particle's location does.
The "measurement limit" is the limit in which the decoherence in one observable's eigenbasis is so rapid and so complete that competing terms don't have a chance. This is the defining feature of a perfect measurement. Natural measurements, like the mutual "measurements" incurred by the molecules in a macroscopic sample of gas, are far enough away from being perfect that both position and momentum can be approximately well-defined (without violating the uncertainty principle $\Delta x\Delta p\gtrsim\hbar$, of course).


TITLE: Is $C(\mathbb{R}_+ , E)$, space of all continuous mappings $f$ of $[0,\infty)$ into a Polish space $E$, locally compact?
QUESTION [1 upvotes]: Let $C(\mathbb{R}_+ , E)$ be the space of all continuous mappings $f$ of $[0,\infty)$ into a Polish space $E$, endowed with the topology of uniform convergence on compact subsets of $\mathbb{R}_+$. I know this space is Polish, but is it also locally compact Hausdorff?

REPLY [2 votes]: I don't know what polish space is, but in case if $\Bbb{R} $ is a Polish space then $C(\Bbb R_+ , \Bbb R) $ is not locally compact, since $B(0, \delta)$, the closed ball,  is not compact
Since $B(0, \delta)$ is not compact in $C[0,1].$ Therefore there is a sequence of continuous functions, say $\{ f_n \}_{n \in N}$ which does not have convergent subsequence in $B(0, \delta)  \subseteq C[0,1] $ Now define
$$ g_n (x) =\left\{\begin{matrix}
 f_n (x)&  x \in [0,1]  \\
 f_n(1)&   x \in (1,+\infty)
\end{matrix}\right. $$
Then we have $ \{ g \}_{n \in N}  \subseteq B(0, \delta)  \subseteq C(\Bbb R_{+} , \Bbb R ) $ does not  have convergent subsequence in $B(0,  \delta) \subseteq C(\Bbb R_{+} , \Bbb R ) $
.


TITLE: Different characterizations of Liouville numbers
QUESTION [5 upvotes]: Usually, Liouville numbers are defined as follows:
$x$ is Liouville if for ever $i\in\mathbb N$ there exist $n,m\in\mathbb Z$ such that
\begin{equation}
\left|x-\frac nm\right|<\frac1{m^i}.
\end{equation}
In their 1982 paper on almost-periodic Schrödinger operators, however, Avron and Simon use the following definition:
$x$ is Liouville if for ever $i\in\mathbb N$ there exist $n,m\in\mathbb Z$ such that
\begin{equation}
\left|x-\frac nm\right|<\frac1{i^m}.
\end{equation}
Do these sets of numbers agree? If yes, how can one show that?

REPLY [1 votes]: This seems to be a long-time misuse of terminology. The second definition is not of a Liouville number, it is much stronger. Liouville number is a number which can be approximated by rationals at any power rate. The second definition assumes that a number can be approximated at any exponential rate.


TITLE: A way for cosine of a decimal to equal to a cosine of a whole number?
QUESTION [1 upvotes]: I wasn't really sure where to put this since this relates to programming as well, and I'm sorry if this question is off-topic or plain ignorant. I was hoping to find a method of having a cosine of a whole number equal to a cosine of a decimal. Here's an example
$\cos(0.8) = \cos(x), x \in \mathbb{Z}$
So that an cosine of an integer is equal to cosine of 0.8. The reason being is that I'm scripting in Batch, and it does not handle any floating point arithmetic, but I'm trying to calculate the cosine of an angle. If it helps at all, I have a cosine function that calculates the value of cosine *10000.
SET "_SIN=a-a*a/1920*a/312500+a*a/1920*a/15625*a/15625*a/2560000-a*a/1875*a/15360*a/15625*a/15625*a/16000*a/44800000"
SET "cos=(a=(15708 - (x) * 31416 / 180)%%62832, c=(a>>31|1)*a, a-=(((c-47125)>>31)+1)*((a>>31|1)*62832)  +  (-((c-47125)>>31))*( (((c-15709)>>31)+1)*(-(a>>31|1)*31416+2*a)  ), %_SIN%)"

Many thanks.

REPLY [3 votes]: $\cos(\alpha)=\cos(\beta)$ if and only if there is $n\in\mathbb{Z}$ such that $\alpha + \beta = 2\pi n$ or $\alpha - \beta = 2\pi n.$ Now assume $\alpha = 0.8$ and $\beta\in\mathbb{Z}.$ Then we would have $\pi = \frac{\alpha \pm\beta}{2n}\in\mathbb{Q}.$ This is a contradiction, because $\pi$ is irrational. Therefore, there is no $x\in\mathbb{Z}$ such that $\cos(0.8)=\cos(x).$


TITLE: Determine the largest area of an ellipse enclosed by the hyperbolas ($xy=1$ and $xy=-1$)
QUESTION [0 upvotes]: Question: An elipse with equation $$ {x^2\over a^2} + {y^2\over b^2} = 1 $$
    is enclosed by the hyperbolas given by $xy=1$ and $xy=-1$. , Determine the largest area of an ellipse enclosed by the hyperbolas. Note that the area of the ellipse is $$ A = πab$$

I know that the answer is $A=2\pi$ by solving it this way:
Considering the hyperbola $y=\frac{1}{x} \Leftrightarrow y^2 = \frac{1}{x^2} $
Now
Ellipse is
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2}= 1 $$
$$ x^2b^2 + y^2a^2-a^2b^2=0 $$
$$ x^2b^2 + \frac{a^2}{x^2}-a^2b^2=0$$
$$x^4b^2 - a^2b^2x^2+a^2=0$$
As this is a biquadratic and with the ellipse it appears to be a 4 point symmetric (i.e. tangent) hence discriminant must be equal to zero.
so
$$ b^2-4ac=0$$
$$ a^4b^4-4b^2a^2=0$$
$$ a^2b^2(a^2b^2-4)=0$$
$$ ab = 2 $$
Hence max area is $A=2\pi$
But I was wondering if there is an alternative way of finding out this area through integration or parametric equations, surely there is more than one way?

REPLY [2 votes]: $u = \frac xa\\
v = \frac yb\\
\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1 \to
u^2 + v^2 = 1\\
xy = 1 \to uv = \frac {1}{ab}$
$uv = k$ touches $u^2 + v^2 = 1$ when $k =\frac 12$
For any ellipse that touches the hyperbola, $ab = 2,$ and its area $= 2 \pi$


TITLE: Real Analysis - Limits and Continuity of Piecewise Function
QUESTION [1 upvotes]: Let $f(x) = \begin{cases}0, \text{if $x \in \mathbb{Q}$} \\ 1, \text{if $x \in \mathbb{R}-\mathbb{Q}$} \end{cases}$
Show that, for any $x_0 \in \mathbb{R}$, $\lim \limits_{x \to x_0}$ $f(x)$ does not exist.
Show that $f$ is nowhere continuous, i.e. for any $x_0 \in \mathbb{R}$, $f$ is not continuous at $x_0$.

I understand that the graph of $f$ is constantly jumping between $0$ and $1$ as we progress down the x-axis; this means that the LHS and RHS of the limit are not equal (because of the constant jumping), and therefore the limit cannot exist. Now this is obviously not a proof, and I know I need to use the epsilon-delta definition of a limit to prove this. How do I formulate the LHS limit $\ne$ RHS limit to show this? Is that even the right approach?
For the second part, it follows from the first, right? Since for any $x_0 \in \mathbb{R}$, the limit does not exist, which means the limit does not exist anywhere for $f$, which means $f$ cannot be continuous?

REPLY [2 votes]: How do I formulate the LHS limit $\neq$ RHS limit to show this? Is that even the right approach?

Neither LHS limit nor RHS limit does not exist. So your idea does not make sense.
Instead try to show that the limit does not exist using definition. It is enough to show that for any $L$, we can find $\epsilon>0$ such that for all $\delta>0$, the condition "$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon$" does not hold.
Let us try. Fix a real number $L$ and set $\epsilon=\tfrac{1}{3}$. We need to show that for all $\delta>0$, the condition "$0<|x-x_0|<\delta \Rightarrow |f(x)-L|<\epsilon$" does not hold. To this end, it suffices to find an $x$ such that $0<|x-x_0|<\delta$ but $|f(x)-L|\geq\epsilon=\tfrac{1}{3}$.
We already know that there are infinitely many rational numbers and infinitely many irrational numbers between $x_0$ and $x_0+\delta$. If $L$ is closer to $0$ than $1$, choose $x$ an irrational number between $x_0$ and $x_0+\delta$ so that $f(x)=1$. Otherwise choose $x$ a rational number between $x_0$ and $x_0+\delta$ so that $f(x)=0$.
In both case we have $|f(x)-L|\geq\tfrac{1}{3}$. Finally we found $x$ such that $0<|x-x_0|<\delta$ but $|f(x)-L|\geq\epsilon=\tfrac{1}{3}$. Therefore any real number $L$ cannot be a limit of $f(x)$ at $x_0$.

Your argument is right.


TITLE: Can a point outside a set in the co-countable topology be the limit point of a sequence in that set?
QUESTION [0 upvotes]: Consider the co-countable topology on $\mathbb{R}$, and $A \subset \mathbb{R}$ such that $A = \mathbb{R} \setminus \{p\}$. I was able to show that for any sequence $(a_n) \in \mathbb{R}$ to converge in the co-countable topology, it should eventually be constant. We can easily find a point in $A$ that is the limit point of a sequence in $A$.
But, my question is, can a point outside $A$ be the limit point of a sequence in $A$?
Here is what I've thought of but I don't know if I'm on the right track: The only point that lies outside $A$ is $p$. So, can $p$ be a limit point of a sequence $(a_n) \in A$? For $p$ to be the limit point, if we take an open set $U$, $p \in U$, $U \setminus \{p\} \cap A \neq \emptyset$.  But we can never take an open set around $p$ since $\{p\}$ is a closed singleton under this topology. Hence the limit point can never be outside $A$?
I'm new to topology and would appreciate any help!

REPLY [2 votes]: The answer to the central question is: "no".
No matter how you interpret "limit point" it should imply "in the closure of".
If $C=\{a_n:n\in\mathbb{N}\}$ is a sequence, a countable set, in $A$ then $C$ is closed in the co-countable topology and if $p\notin A$ then $\mathbb{R}\setminus C$ is a neighbourhood of $p$ that is disjoint from $C$.
So even if $p$ outside $A$ is in the closure of $A$ it is not in the closure of any countable subset of $A$.


TITLE: How to compute expected value
QUESTION [0 upvotes]: How do I solve the expected value of this problem, if I have already calculated the pmf?

Let $X$ be a random variable with cumulative distribution function given
  below:
  $$F_X(x) = \begin{cases} 0 & : x < 0
\\ 1/2 & : 0\leqslant x < 1
\\ 3/5 & : 1 \leqslant x < 2
\\ 4/5 & : 2 \leqslant x < 3
\\ 9/10 & : 3 \leqslant x < 3.5
\\ 1 & : x \geqslant 3.5
\end{cases}$$
  Compute $\mathsf E[X]$

REPLY [2 votes]: You probability mass function should look like: $f_X(x) = \begin{cases}
 \Box & : x=0
\\ \Box & : x=1
\\ \Box & : x=2
\\ \Box & : x=3
\\ \Box & : x= 3.5
\\ 0 & : \text{elsewhere}
\end{cases}$
Since you have filled in the boxes for this, then you can just evaluate:
$$\begin{align}\mathsf E[X] &= \sum_{x\in\{0,\,1,\,2,\,3,\,3.5\}} x\, f_X(x)
\\[1ex] & = 0\, f_X(0) + 1\, f_X(1) + 2\, f_X(2) + 3\,f_X(3)+3.5\,f_X(3.5)
\end{align}$$


TITLE: An application of Weierstrass M-Test to determine sets of uniform convergence.
QUESTION [0 upvotes]: let $\sum_{n=1}^{\infty}fn(x) $ = $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}(\frac{x}{1+x})^n$
I am attempting to apply Weierstrass M-Test on [a,$\infty$)
I get that
$|\frac{1}{\sqrt{n}}(\frac{x}{1+x})^n|$<$|\frac{1}{\sqrt{n}}|$
but the RHS doesnt converge, so we cant use M-Test.
What other methods can I use?

REPLY [1 votes]: Let $g_m(x)=\sum_{n=1}^m \frac {1}{\sqrt n}\left(\frac {1}{1+x}\right)^n.$
If $S$ is a subset of $\Bbb R$ with no upper bound in $\Bbb R$ then $g_n$ does not converge uniformly on $S.$
Proof: We show that for any $m\in \Bbb N$ there exist $m'>m$ and $x\in S$ with $g_{m'}(x)-g_m(x)>1/(2\sqrt 2),$ as follows:
Let $m'=2m.$ Now take $x\in S$ where $x>0$ and $x$ is large enough that $(x/(1+x))^{2m}>1/2.$ Then $(x/(1+x))^n>1/2$ when $m<n\le 2m,$ so $$g_{m'}(x)-g_m(x)=g_{2m}(x)-g_n(x)=\sum_{n=m+1}^{2m}(1/\sqrt n)(x/(1+x))^n>$$ $$>\sum_{n=m+1}^{2m}(1/\sqrt n)(1/2)\ge$$ $$\ge \sum_{n=m+1}^{2m}(1/\sqrt {2m})(1/2)=$$ $$=m(1/\sqrt {2m})(1/2)=(\sqrt m)/(2\sqrt 2)\ge 1/(2\sqrt 2).$$


TITLE: Isomophism between two irrational rotations
QUESTION [3 upvotes]: Define two dynamical systems $([0,1), \mathbb{B}_1, \mathbb{L}, T_1(x)=x+\alpha_1\mod 1)$ and  $([0,1), \mathbb{B}_2, \mathbb{L}, T_2(x)=x+\alpha_2\mod 1)$, where $\alpha_1,\alpha_2 $ are two irrational numbers, $\mathbb{L}$ is the Lebesgue measure and $\mathbb{B}_i$ are the sigma algebras. Now if these two systems are isomorphic, then what the relationship between $\alpha_1$ and $\alpha_2 $? I do  know that the irrational rotation is isomorphic to the Sturmian orbit dynamical systems.

REPLY [4 votes]: The first major classification result in ergodic theory was found by Halmos and von Neumann in 1942 (Operator methods in classical mechanics II, Ann. of
Math. (2) 43 (1942), 332–350.). They showed that for ergodic rotations of compact abelian groups, spacial isomorphism (i.e. measure-theoretic isomorphism) is the same as spectral isomorphism (i.e. unitary equivalence of the associated operators). For irrational rotations $\alpha_1$ and $\alpha_2$ of the circle, this amounts  saying that the subgroups generated by $\alpha_1$ and $\alpha_2$ are equal. From this it is easy to deduce that $\alpha_1 = \pm \alpha_2$. This is simply an expanded version of Anthony's initial answer.


TITLE: Boundary of Set defined with $\sin(\frac{1}{x})$
QUESTION [3 upvotes]: Let $S=\{(x,y) \in (0,\infty) \times (-1,\infty) \mid y \geq \sin(\frac{1}{x})\}$ in $(\mathbb{R}^2,\lVert \cdot \rVert_\infty)$. I think that the boundary $\partial S= \{ (0,y) \mid y \in (-1, \infty) \} \cup \{ (x,y) \mid x \in (0, \infty), y =\sin(\frac{1}{x}) \}$, but I'm not sure about this as $\lim_{x \to 0} \sin(1/x)$ is undefined. And how can I prove, that $\forall x \in \partial S \; \forall \varepsilon > 0 \; \exists y_1,y_2 \in B(x,\varepsilon) : y_1 \in S, y_2 \in S^\complement$?

REPLY [0 votes]: For proving, that $\partial S$ is a subset of the boundary I have to show: Let $x \in (\partial S)^C$, it is $x \in S^o$ or $x \in (S^C)^o$.

$x_1 < 0, x_2 \in \mathbb{R}$. $\Rightarrow x \in S^C$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\frac{|x_1|}{2}$. $y_1 < x_1+\varepsilon < 0$. $\Rightarrow y \in S^C$.
$x_1 = 0, x_2<-1$. $\Rightarrow x \in S^C$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\frac{|x_2+1|}{2}$. $y_2 < x_2+\varepsilon < 1$. $\Rightarrow y \in S^C$.
$x_1 > 0, x_2 > \sin(\frac{1}{x_1})$. $\Rightarrow x \in S$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\min\{ \frac{|x_1|}{2}, \frac{1}{2}\min_{x>0} \{ \sqrt{(x-x_1)^2+(\sin(\frac{1}{x})-x_2)^2} \} \}$.$y_1 > x_1-\varepsilon > 0$. $y_2 > x_2-\varepsilon \geq \sin(\frac{1}{y_1})$. $\Rightarrow y \in S$.
$x_1 > 0, x_2 < \sin(\frac{1}{x_1})$. $\Rightarrow x \in S^C$. Let $y \in B(x, \varepsilon)$ with $\varepsilon=\min_{x>0} \{ \sqrt{(x-x_1)^2+(\sin(\frac{1}{x})-x_2)^2} \}$. $y_2 < x_2+\varepsilon \leq \sin(\frac{1}{y_1})$. $\Rightarrow y \in S^C$.

Can I simplify the cases in any way?


TITLE: Can anyone explain me how to find the value of $\left(\frac{1}{3}\right)!$?
QUESTION [0 upvotes]: I know how to calculate the value of $\left(\dfrac{1}{2}\right)!$ using the gamma function, but I don't know how to find the value of $\left(\dfrac{1}{3}\right)!$ or $\left(\dfrac{1}{6}\right)!$ using the gamma function.

REPLY [1 votes]: $$ \Gamma\left(\frac{1}{3}\right)=\frac{2^{7/9}\pi^{2/3}}{3^{1/12}{\rm agm}(2,\sqrt{2+\sqrt{3}})^{1/3}} $$
where $\rm agm$ is the arithmetic-geometric mean. One can then deduce that
$$ \Gamma\left(\frac{1}{6}\right)=\frac{2^{14/9}3^{1/3}\pi^{5/6}}{{\rm agm}(1+\sqrt{3},\sqrt{8})^{2/3}} $$
See : Upper bound on integral: $\int_1^\infty \frac{dx}{\sqrt{x^3-1}} < 4$.
To get the values of $\Gamma\left(\frac{4}{3}\right)$ and $\Gamma\left(\frac{7}{6}\right)$, you can use the formula $$\Gamma\left(n+\frac{1}{p}\right)=\Gamma\left(\frac{1}{p}\right)\frac{(pn-p+1)!^{(p)}}{p^n}$$


TITLE: finite tor dimension
QUESTION [2 upvotes]: Hi.  Can, every one, give me  an example of finite surjective morphism of finite tor dimension (but not flat!) between reduced schemes or complex analytic spaces... Thank you.

REPLY [3 votes]: Consider a smooth surface $Y$ with a point $p\in Y$. Let $X$ be obtained by gluing two copies of $Y$ at $p$, with the obvious morphism $X \to Y$. This is surjective and finite, and has finite Tor-dimension (because $Y$ is regular, hence every morphism to $Y$ has finite Tor dimension). However, it is not flat (for example, because $X$ is not Cohen-Macaulay).


text:
\begin{document}

\maketitle
\begin{abstract}
We consider the periodic Manhattan lattice with alternating orientations going north-south and
east-west. Place obstructions on vertices independently with probability $0<p<1$. A particle is moving on the edges with unit speed following the orientation of the lattice and it will turn only when encountering an obstruction. The problem is that for which value of $p$ is the trajectory of the particle closed almost surely. We prove this for $p>\frac{1}{2}-\varepsilon$ with some $\varepsilon>0$.
\end{abstract}

\section{Introduction}
We consider the Manhattan pinball problem which was previously proposed and analysed in \cite{beamond2003quantum}. See also \cite{spencer2012duality} and \cite{cardy2010quantum}. There are equivalent ways to state the problem formally and we state it by using bond percolation. Consider the $\Z^{2}$ lattice embedded into the plane $\R^{2}$. Denote the tilted lattice by $$\widetilde{\Z^{2}}=\left\{\left(x+\frac{1}{2},y+\frac{1}{2}\right):\text{$x,y$ are integers and $x-y$ is even}\right\},$$
and for any $a,b\in \widetilde{\Z^{2}}$, $a$ and $b$ are connected by an edge if and only if $|a-b|=\sqrt{2}$. Here, $|\cdot|$ is the Euclidean distance on $\R^{2}$.
See Figure \ref{fig:2} for an illustration.
\begin{figure}
    \centering
    \includegraphics{2.eps}
    \caption{The black dots in the picture illustrate the lattice $\widetilde{\Z^{2}}$. The dashed lines are edges of $\widetilde{\Z^{2}}$. The arrows indicate the orientation of the Manhattan lattice.}
    \label{fig:2}
\end{figure}
Given $0<p<1$, we consider the Bernoulli bond percolation on $\widetilde{\Z^{2}}$. We declare each edge of $\widetilde{\Z^{2}}$ to be closed with probability $p$ and open with probability $1-p$, independent of all other edges. We use $\omega$ to denote a configuration of open and closed edges. We place a two-sided plane mirror on each closed edge. Suppose a ray of light starts from the origin and initially moves towards east. When the light reaches an open edge, it passes through without deflection. When the light reaches a closed edge, it is deflected through a right angle by the mirror on the edge. Let $L(\omega)$ denote the trajectory of the light.

Our main result is the following
\begin{theorem}\label{thm:main}
Let $\mathcal{E}_{n}$ denote the event that the diameter of $L(\omega)$ is at most $n$. Then there exists $\varepsilon_{0}>0$ such that following holds. For each $p>\frac{1}{2}-\varepsilon_{0}$, there are $\alpha,c>0$ such that for $n>\alpha$, we have
\begin{equation}
    \Prob_{p} [\mathcal{E}_{n}]\geq 1-\exp(-c n).
\end{equation}
\end{theorem}
Theorem \ref{thm:main} implies that almost surely, the trajectory of the light is finite when $p>\frac{1}{2}-\varepsilon_{0}$. This result
is previously known to hold for $p>\frac{1}{2}$ by using bond percolation, see e.g. \cite{beamond2003quantum}, \cite{spencer2012duality}. It is suggested by physicists that the same happens for any $p\in (0,1)$, see e.g. \cite{beamond2003quantum}, \cite{cardy2010quantum}.

\section{Enhancement}
\begin{figure}
    \centering
    \includegraphics{1.eps}
    \caption{The left figure illustrates the configuration $G_{0}$ where solid black edges are closed and dashed edges are open. In the right hand side, after placing mirrors on closed edges of $G_{0}$, the light comes from the west of the red edge will travel in the blue path before leaving.}
    \label{fig:1}
\end{figure}

The proof of Theorem \ref{thm:main} uses an \emph{enhancement} procedure which we define now.
Denote by $G_{0}$ the finite configuration illustrated in the left hand side of Figure \ref{fig:1} and its caption.
Given any configuration $\omega$, we enhance it by the following procedure. For any translated  copy of $G_{0}$ which appears in $\omega$, we close the open edge which coincides with the red edge (illustrated in Figure \ref{fig:5}). We denote the configuration after this procedure by $\tilde{\omega}$.

The procedure above is a special case of the general enhancement (see e.g. \cite[Section 3.3]{grimmett1999percolation}). Intuitively, we have added closed edges to $\omega$ in a systematic way and we expect that the resulting configuration $\tilde{\omega}$ is more ``percolative'' than $\omega$.
This is the content of \cite[Theorem (3.16)]{grimmett1999percolation}. For \cite[Theorem (3.16)]{grimmett1999percolation} to hold, the enhancement is required to be \emph{essential} (a concept defined in \cite[Page 64]{grimmett1999percolation}) which is true for our case (see Figure \ref{fig:5} and its caption).

\begin{figure}
    \centering
    \includegraphics{5.eps}
    \caption{The picture contains a doubly-infinite closed path if and only if the enhancement is activated so that the red edge is turned to be closed. This means the enhancement is essential.}
    \label{fig:5}
\end{figure}
\begin{rem}
There are other choices of the enhancement $G_{0}$ for our proof of Theorem \ref{thm:main}. We will see in Section \ref{sec:proof} that, our choice of $G_{0}$ makes the proof easier because of an important feature contained in Lemma \ref{lem:translation} below.
\end{rem}
The following lemma can be directly checked from Figure \ref{fig:5}.
\begin{lemma}\label{lem:translation}
Suppose $\omega$ is a configuration and $\overline{G_{0}}$, $\overline{\overline{G_{0}}}$ are two translated copies of $G_{0}$ that appear in $\omega$. Then any open edge in $\overline{G_{0}}$ does not coincide with the red edge in $\overline{\overline{G_{0}}}$.
\end{lemma}
Denote
$$Q_{n}=\left\{(x,y)\in \widetilde{\Z^{2}}: \text{$|x+y-1|\leq n$ and $|x-y|\leq n$}  \right\}$$
to be the tilted box centered at $(\frac{1}{2},\frac{1}{2})$. For any edge $e$ of $\widetilde{\Z^{2}}$, we say $e$ is \emph{inside} $Q_{n}$ if $e$ connects two vertices in $Q_{n}$, otherwise we say it is \emph{outside} $Q_{n}$.
The following proposition follows from the proof of \cite[Theorem (3.16)]{grimmett1999percolation} (more precisely, the argument near \cite[Equation (3.8)]{grimmett1999percolation}) and the fact that the enhancement which we constructed is essential.
\begin{prop}[Theorem (3.16) in \cite{grimmett1999percolation}]\label{prop:event-A_n}
There exists $\varepsilon_{1}>0$ such that, for each $p>\frac{1}{2}-\varepsilon_{1}$,
\begin{equation}
    \Prob_{p}[\{\omega : \tilde{\omega}\in A_{n}\}] \geq \Prob_{\frac{1}{2}+\varepsilon_{1}}[\{\omega\in A_{n}\}]
\end{equation}
for large enough $n$. Here, $A_{n}$ denotes the event that there exists a path of closed edges joining $(\frac{1}{2},\frac{1}{2})$ to some vertex in $\widetilde{\Z^{2}}\setminus Q_{n}$.
\end{prop}

In fact, by directly adapting the proof of \cite[Theorem (3.16)]{grimmett1999percolation}, we can substitute $A_{n}$ in Proposition \ref{prop:event-A_n} by $A'_{n}$ which denotes the crossing event in a $4n\times n$ tilted rectangle. Thus we have
\begin{prop}\label{prop:event-A'_n}
There exists $\varepsilon_{1}>0$ such that, for each $p>\frac{1}{2}-\varepsilon_{1}$,
\begin{equation}
    \Prob_{p}[\{\omega : \tilde{\omega}\in A'_{n}\}] \geq \Prob_{\frac{1}{2}+\varepsilon_{1}}[\{\omega\in A'_{n}\}]
\end{equation}
for large enough $n$. Here, $A'_{n}$ denotes the event that there is a path of closed edges in the tilted rectangle
$$T_{n}=\{(x,y)\in \widetilde{\Z^{2}}:1\leq x+y-1\leq n, |x-y| \leq 2n \}$$ joining some vertex on northwest side $\left\{(x,y)\in T_{n}: x-y= -2n\right\}$ to some vertex on southeast side $\left\{(x,y)\in T_{n}: x-y= 2n \right\}$. See Figure \ref{fig:3} for an illustration.
\end{prop}
The following well-known lemma states that in the supercritical phase (i.e. $p>\frac{1}{2}$), the crossing event $A'_{n}$ happens with high probability.
\begin{lemma}\label{lem:prob-crossing}
For any $p>\frac{1}{2}$, there is constant $c>0$ such that $$\Prob_{p}[\{\omega\in A'_{n}\}]\geq 1-\exp(-c n)$$ for large enough $n$.
\end{lemma}
\begin{proof}
Follow the argument prior to \cite[Lemma (11.22)]{grimmett1999percolation}.
\end{proof}
Together with Proposition \ref{prop:event-A'_n} and Lemma \ref{lem:prob-crossing}, we have
\begin{figure}
    \centering
    \includegraphics{3.eps}
    \caption{The picture illustrates a closed path in $T_{n}$ which joins some vertex on northwest side to southeast side.}
    \label{fig:3}
\end{figure}
\begin{prop}\label{prop:circuit}
There exists $\varepsilon_{1}>0$ such that, for each $p>\frac{1}{2}-\varepsilon_{1}$,
\begin{equation}
    \Prob_{p}[\{\omega : \tilde{\omega}\in A''_{n}\}] \geq 1-\exp(-c_{p} n).
\end{equation}
for a constant $c_{p}>0$ and large enough $n$. Here, $A''_{n}$ denotes the event that $Q_{n}$ lies in the interior of a circuit which consists of closed edges in $Q_{2n}$.
\end{prop}
\begin{proof}
Consider the four tilted rectangles
\begin{enumerate}
    \item $T^{(1)}=\{(x,y)\in \widetilde{\Z^{2}}: n+1\leq x+y-1\leq 2n, |x-y|\leq 2n\}$,
    \item $T^{(2)}=\{(x,y)\in \widetilde{\Z^{2}}: -2n\leq x+y-1\leq -n-1, |x-y|\leq 2n\}$,
    \item  $T^{(3)}=\{(x,y)\in \widetilde{\Z^{2}}:n+1 \leq x-y\leq 2n,  |x+y-1|\leq 2n\}$,
    \item  $T^{(4)}=\{(x,y)\in \widetilde{\Z^{2}}:-2n\leq x-y\leq -n-1, |x+y-1|\leq 2n\}$.
\end{enumerate}
See Figure \ref{fig:4} for an illustration. If these four tilted rectangles are crossed in the `long direction' by closed paths as indicated in Figure \ref{fig:4}, then $Q_{n}$ lies in the interior of a closed circuit. By symmetry, each of these four crossing events has the same probability as $A'_{n}$.
Thus our proposition follows from Proposition \ref{prop:event-A'_n} and Lemma \ref{lem:prob-crossing}.
\end{proof}
\begin{figure}
    \centering
    \includegraphics{4.eps}
    \caption{The four green lines are four closed paths that cross tilted rectangles $T^{(i)}(i=1,2,3,4)$. If such paths exist, then $Q_{n}$ (the yellow region) lies in the interior of a closed circuit.}
    \label{fig:4}
\end{figure}
\section{Proof of Theorem \ref{thm:main}}\label{sec:proof}
\begin{proof}[Proof of Theorem \ref{thm:main}]

By Proposition \ref{prop:circuit}, it suffices to prove $A''_{n}\subset \mathcal{E}_{2n+10}$ for $n>100$. Thus we suppose $\omega\in A''_{n}$. Let configuration $\omega_{0}$ be obtained by $\omega_{0}=\omega$ inside $Q_{100}$ and $\omega_{0}=\tilde{\omega}$ outside $Q_{100}$. Recall that $L(\omega_{0})$ is the trajectory of the light under the configuration $\omega_{0}$. By definition of $A''_{n}$ and $n>100$, the closed circuit gives rise to an enclosure of mirrors surrounding the origin and thus traps the light. Hence, $L(\omega_{0})$ is contained inside $Q_{2n}$.

Note that, for each open edge $e$ in $\omega$,  $e$ will be closed in the configuration $\omega_{0}$ if and only if
\begin{enumerate}
    \item $e$ is outside $Q_{100}$,
    \item $e$ is at the position of the red edge in a translated  copy of $G_{0}$.
\end{enumerate}
By Lemma \ref{lem:translation}, if $\overline{G_{0}}$ is a translated  copy of $G_{0}$ which appears in $\omega$, then only the red edge will be closed under the enhancement and other edges in $\overline{G_{0}}$ will remain unchanged.
We denote the set of edges (outside $Q_{100}$) which are closed under the enhancement by $E$. We now start from $\omega_{0}$ and open edges in $E$ to reach the initial configuration $\omega$ and we keep track of the trajectory. Only those edges in $E$ which intersects with $L(\omega_{0})$ will affect the trajectory by adding translated copies of the blue path (or its inverse, see Figure \ref{fig:1}) to $L(\omega_{0})$. Here, we used the fact that the edges in $E$ are outside $Q_{100}$ and thus if the light comes to some $e\in E$, it must come from west or south (otherwise, $L(\omega_{0})$ is a translated  copy of the blue loop and is outside $Q_{90}$ which contradicts with the fact that $L(\omega_{0})$ starts from the origin). Hence, $L(\omega)$ is ``trapped'' inside $Q_{2n+10}$ (see Figure \ref{fig:6}) and our theorem follows.
\end{proof}

\begin{figure}
    \centering
    \includegraphics{7.eps}
    \caption{$L(\omega)$ is obtained from $L(\omega_{0})$ (the pink path) by attaching translated copies of the blue path (or its inverse). If $L(\omega_{0})$ is inside $Q_{2n}$, then $L(\omega)$ is inside $Q_{2n+10}$.}
    \label{fig:6}
\end{figure}
\section*{Acknowledgement}
The author thanks Lingfu Zhang for telling the problem to him and having several discussions. The author also thanks Professor Jian Ding for useful comments on writing. The author is partially supported by NSF grant  DMS-1757479.
\bibliographystyle{style}
\bibliography{bib}

\end{document}


TITLE: Analyzing linear operator convergence through transformations
QUESTION [2 upvotes]: Assume $A_n: C[0,1] \rightarrow C[0,1]$, $n \in \mathbb{N}$, is a linear operator, having the following form:
$$A_nx(t) = \int_{0}^{1} \sqrt{(t-s)^2 + \frac{1}{n}} \cdot x(s)ds$$
I'm trying to analyze its (strong) convergence through checking whether
$$ ||A_n - A_m|| = \sup_{||x|| \leq 1} ||A_n x - A_m x|| \rightarrow 0  $$
holds, when $n,m \rightarrow \infty$.
While it seems that there are a few possible ways one could continue from here, my question revolves around dealing with the squared root. The simplest way that comes to mind is analysing the operator $A_n^{2}$, from where it should follow that
$$ ||A_n^2 - A_m^2|| = \sup_{||x|| \leq 1} ||A_n^2 x - A_m^2 x||  \leq \sup_{||x|| \leq 1} ||x||^2  |\frac{1}{n} - \frac{1}{m} | \rightarrow 0 $$
My question: Does such convergence of squared operators $(A_n^{2})$ imply convergence of non-transformed operators $(A_n)$? I'm sure that it doesn't hold in every case, although here the square transformation seems as a safe transformation with being a bijection on $C[0,1]$.
Is there anything else I'm missing? The main goal I'm looking for is to understand whether we can (and how, if so) transform the expressions of various linear operators to allow for an easier/cleaner analysis.

REPLY [1 votes]: It is unclear how to get the above control of $A_n^2$, which should be given by an even "uglier" formula. Even so, such a formula "for the squares" (correctly, rather for the compositions) does not say too much about a formula "for the initial sequence". Consider for instance the sequence $1,-1,1,-1,\dots$. (Here, $1$ is the identity, if we really need operators acting on a functions space.)
A way to proceed is as follows.
Let $A_n,A_m$ be two operators in the sequence, fix an $x$ in $C[0,1]$.
Let $I=[0,1]$.
Then:
$$
\begin{aligned}
((A_n-A_m)x)(t)
&=
\int_I
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
\; x(s)\; ds
\\
&=
\int_I
\frac{
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
\left(
\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
\right)
}
{\phantom{
\left(
\sqrt{(t-s)^2+\frac 1n}
-
\sqrt{(t-s)^2+\frac 1m}
\right)
}
\left(
\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
\right)
}
\; x(s)\; ds
\\
&=
\int_I
\frac{
(t-s)^2+\frac 1n
-
(t-s)^2-\frac 1m
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\
&=
\left(\frac 1n-\frac 1m\right)
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\[3mm]
|\ ((A_n-A_m)x)(t)\ |
&=
\left|\frac 1n-\frac 1m\right|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1n}
+
\sqrt{(t-s)^2+\frac 1m}
}
\; x(s)\; ds
\\
&\le
\left|\frac 1n-\frac 1m\right|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1M}
}
\; x(s)\; ds
\qquad
\text{ with }
M = \min(m,n)
\\
&\le
\left|\frac 1n-\frac 1m\right|\;\|x\|
\int_I
\frac{
1
}
{\sqrt{(t-s)^2+\frac 1M}
}
\; ds
\\
&=
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot2\cdot
\frac 12
\Big(
\text{arcsinh} (t\sqrt M)
+
\text{arcsinh} ((1-t)\sqrt M)
\Big)
\\
&\le
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot
2\cdot
\text{arcsinh}
\left(
\frac 12
\left(
t\sqrt M
+
(1-t)\sqrt M
\right)
\right)
\\
&
\qquad\qquad \text{ concavity of $\text{arcsinh}$ on $(0,\infty)$}
\\
&=
\left|\frac 1n-\frac 1m\right|\;\|x\|
\cdot
2\text{arcsinh} \left(\frac 12\sqrt M\right)
\\
&\in \mathcal O\left(\frac 1M\ln M\right)\ .
\end{aligned}
$$
We have bounded the expression by a sequence converging to $0$ for $M\to\infty$, $M\le m,n$.


text:
\subsection{Verification of convergence rate}\label{convergence_study}
We use the method of the manufactured solution to verify the fourth order convergence rate of the proposed scheme. We choose the mapping of the coarse domain $\Omega^c$ as
\[ {\bf x} = {\bf X}^c({\bf r}) = \left(\begin{array}{c}
2\pi r^{(1)} \\
2\pi r^{(2)} \\
r^{(3)}\theta_i\big(r^{(1)},r^{(2)}\big) + (1-r^{(3)})\theta_b\big(r^{(1)},r^{(2)}\big) \end{array}\right), \]
where $0\leq r^{(1)}, r^{(2)}, r^{(3)}\leq 1$, $\theta_i$ represents the interface surface geometry,
\begin{equation}\label{iterface_geometry}
\theta_i\big(r^{(1)},r^{(2)}\big) = \pi+0.2\sin(4\pi r^{(1)})+0.2\cos(4\pi r^{(2)}),
\end{equation}
and $\theta_b$ is the bottom surface geometry,
\begin{equation*}\label{bottom_geometry}
\theta_b\big(r^{(1)},r^{(2)}\big) = 0.2\exp\left(-\frac{(r^{(1)}-0.6)^2}{0.04}\right)+0.2\exp\left(-\frac{(r^{(2)}-0.6)^2}{0.04}\right).
\end{equation*}
As for the fine domain $\Omega^f$, the mapping is chosen to be
\[ {\bf x} = {\bf X}^f({\bf r}) = \left(\begin{array}{c}
2\pi r^{(1)}\\
2\pi r^{(2)}\\
r^{(3)}\theta_t\big(r^{(1)},r^{(2)}\big) + (1-r^{(3)})\theta_i\big(r^{(1)},r^{(2)}\big) \end{array}\right), \]
where $0\leq r^{(1)}, r^{(2)}, r^{(3)}\leq 1$ and $\theta_t$ is the top surface geometry,
\begin{equation*}\label{top_geometry}
\theta_t\big(r^{(1)},r^{(2)}\big) = 2\pi+0.2\exp\left(-\frac{(r^{(1)}-0.5)^2}{0.04}\right)+0.2\exp\left(-\frac{(r^{(2)}-0.5)^2}{0.04}\right).
\end{equation*}

In the entire domain, we choose the density
\begin{equation*}\label{density_function}
\rho(x^{(1)},x^{(2)},x^{(3)}) = 2 + \sin(x^{(1)}+0.3)\sin(x^{(2)}+0.3)\sin(x^{(3)}-0.2),
\end{equation*}
and material parameters $\mu, \lambda$
\begin{equation*}\label{mu_function}
\mu(x^{(1)},x^{(2)},x^{(3)}) = 3 + \sin(3x^{(1)}+0.1)\sin(3x^{(2)}+0.1)\sin(x^{(3)}),
\end{equation*}
and
\begin{equation*}\label{lambda_function}
\lambda(x^{(1)},x^{(2)},x^{(3)})  = 21+ \cos(x^{(1)}+0.1)\cos(x^{(2)}+0.1)\sin^2(3x^{(3)}).
\end{equation*}
 In addition, we impose a boundary forcing on the top surface and Dirichlet boundary conditions for the other boundaries. The external forcing, top boundary forcing ${\bf g}$ and initial conditions are chosen such that the solutions for both fine domain ($\bf F$) and coarse domain ($\bf C$) are ${\bf F}(\cdot,t) = {\bf C}(\cdot,t) = {\bf u}(\cdot,t) = (u_1(\cdot,t),u_2(\cdot,t),u_3(\cdot,t))^T$ with
\begin{align*}
u_1(\cdot,t) &= \cos(x^{(1)}+0.3)\sin(x^{(2)}+0.3)\sin(x^{(3)}+0.2)\cos(t^2),\\
u_2(\cdot,t) &= \sin(x^{(1)}+0.3)\cos(x^{(2)}+0.3)\sin(x^{(3)}+0.2)\cos(t^2),\\
u_3(\cdot,t) &= \sin(x^{(1)}+0.2)\sin(x^{(2)}+0.2)\cos(x^{(3)}+0.2)\sin(t).
\end{align*}
For example, for the boundary forcing at the top surface, we have
\begin{equation*}\label{traction_force}
{\bf g} = (g_1,g_2,g_3)^T = \sum_{i=1}^3\left(\sum_{j = 1}^3 M_{ij}^f\frac{\partial{\bf u}}{\partial x^{(j)}}\right) n^{f,+,i}_3,
\end{equation*}
where, $M_{ij}^f$ and $n^{f,+,i}_3$ are defined in (\ref{M_definition}) and (\ref{outward_normal}), respectively.

The problem is evolved until final time $T = 0.5$. In Table \ref{convergence_rate}, we use $L_2$ to represent the $L^2$ error in the entire domain $\Omega = \Omega^c\cup\Omega^f$. The notations $L_2^f$ and $L_2^c$ represent the $L^2$ error in the fine domain $\Omega^f$ and coarse domain $\Omega^c$, respectively. The convergence rates are shown in the parentheses in Table \ref{convergence_rate}. We observe that the convergence rate is fourth order for all cases. Even though the boundary accuracy of the SBP operator is only second order, the optimal convergence rate is fourth order. For a more detailed analysis of the convergence rate, we refer to \cite{Wang2017, Wang2018b}.  To solve the linear system for ghost point values, we use a block Jacobi iterative method. In the following section, we study two more iterative methods and compare them in terms of the condition number and the number of iterations.

\begin{table}[htb]
	\begin{center}
		\begin{tabular}{|c|c c c|}
			\hline
			$2h_1 = 2h_2 = 2h_3 = 2h$   & $L_2$ & $L_2^f$ & $L_2^c$  \\
			\hline
			$2\pi/24$ &2.2227e-03 ~~~~~~~~ & 8.0442e-04 ~~~~~~~~ & 2.0720e-03 ~~~~~~~~\\
			\hline
			$2\pi/48$ &1.4142e-04 (3.97) & 5.1478e-05 (3.97) & 1.3171e-04 (3.98)\\
			\hline
			$2\pi/96$ &8.6166e-06 (4.04) & 3.0380e-06 (4.08) & 8.0632e-06 (4.03)\\
			\hline
		\end{tabular}
	\end{center}
	\caption{The $L^2$ error and corresponding convergence rates of the fourth order SBP method.}\label{convergence_rate}
\end{table}


TITLE: Modifying Feynman-Wheeler absorber theory to work with arbitrary potentials?
QUESTION [6 upvotes]: I'm trying to consider relativistic multi-body dynamics in special relativity. In classical mechanics, it's easy to write a simple $n$-body system with arbitrary potential $V$:
\begin{equation}
m \ddot{x}_ i=\sum_ j - \nabla_ {x_ i} V(|x_ i-x_ j|). \tag{1}
\label{1}
\end{equation}
In special relativity, it's tempting to just replace this with the retarded potential, where $x_ j$ is evaluated at the time where $c |\Delta t|=|x_ i-x_ j|$. However this ends up in solutions blowing up over time. I want to find an action for a 2-body system which reduces to equation \ref{1} in the limit $v\ll c$, but which also has correct and physically meaningful conservation laws.
Since this is all within the realm of radiation reaction, I figure a surefire starting point is to consider things from a Lagrangian Feynman-Wheeler type system (Classical Electrodynamics in Terms of Direct Interparticle Action), since its symmetries will pretty directly give conservation laws (albeit with some speed of light delays). I label the two particles $a$ and $b$, and I'm working with $c=1$, unit charges and masses, signature $(- + + +)$, and $t$ an arbitrary parameter labeling the world lines. Then the action is:
$$A=-\sum_{i=a,b}\int dt \sqrt{-\dot x_i^\mu \dot x_{i\mu}} - \iint \delta((x_a-x_b)^2) \dot x_a^\mu \dot x_{b\mu}dt_1 dt_2 \label{2}\tag{2}$$
Note that $dt \sqrt{-\dot x_i^\mu \dot x_{i\mu}}$ should really be considered as $\sqrt{-dx_i^\mu dx_{i\mu}}$, and that the double integral should really be considered as $dx_a^\mu dx_{b\mu}$. So we really are reparameterization invariant, and we really are integrating with respect to the world lines. (Also note: "$x^2$" in the delta function means $x^\mu x_\mu$.)
It's easy to see that this gives the Coulomb force: Fix particle $b$ to the origin so that $x_b^\mu(t)=(t,\vec{0})$. Then for $x_a^\mu(t)=(t,\vec{x}_a(t))$, we find $\dot x_a^\mu \dot x_{b\mu}=1$.  Apply the delta function identity $\delta(g(x))=\sum_{g(x_0)=0} \delta(x-x_0)/|g'(x_0)|$ and integrate with respect to $t_2$ to get
$$\iint \delta((x_a-x_b)^2) \dot x_a^\mu \dot x_{b\mu}dt_1 dt_2 =\int dt_1 \sum_{t_2=t_a,t_r}\frac{1}{|2(x_a^\mu-x_b^\mu) \dot x_{b\mu}|}=\int dt_1 \sum_{t_2=t_a,t_r}\frac{1}{|2\Delta t|}.\label{3}\tag{3}$$
$t_a$ and $t_r$ are the advanced and retarded times with $|\Delta t|=|\Delta x|$, so summing over the two we get the action of a single particle in a Coulomb potential $$\int dt_1 \frac{1}{|\Delta x|}$$
So the term $|(x_a^\mu-x_b^\mu) \dot x_{b\mu}|$ turned into a vector difference $|\Delta \vec{x}|$. This leads to the idea: just multiply the interaction term by terms like that. The corrected action term might look something like this:
$$\iint F(|(x_a^\mu-x_b^\mu) \dot x_{b\mu} /\sqrt{- \dot x_b^\nu\dot x_{b\nu}}|) \delta((x_a-x_b)^2) \dot x_a^\mu \dot x_{b\mu}dt_1 dt_2. \label{4}\tag{4}$$
If $F(x)=xV(x)$ and particle $b$ is fixed at the origin, this gives the correct limit, and is Lorentz covariant and reparameterization invariant (that's what the $\sqrt{-\ldots}$ term is for), but it also favors $x_a$ over $x_b$! Symmetrizing with respect to $a$ and $b$ also seems OK, because for $|\frac{d}{dt} \vec{x}_a| \ll 1$ we should have $\dot x_{a\mu} /\sqrt{- \dot x_a^\nu\dot x_{a\nu}}\approx (1,\vec{0})$, but it feels like there should be a more simple route to go down.
Does anyone know of a way to do this, or have any better ideas on how to modify the interaction term?
Lorentz covariance and reparameterization invariance put some heavy restrictions on the action, so maybe it's not possible to get a very elegant action with the desired properties.

REPLY [2 votes]: Have a look at the following paper:

Louis-Martinez, D. J. (2006). Relativistic non-instantaneous action-at-a-distance interactions. Physics Letters B, 632(5-6), 733-739, doi:10.1016/j.physletb.2005.11.029, arXiv:hep-th/0510193.

From the action generalizing the Feynman–Wheeler's direct interparticle interaction by imposing conditions of Poincaré invariance and additional requirements that the parameters along the worldlines were the proper times of the particles and that mass must be scalar quantity, the author was able to show that the only form of potential allowed by that conditions is the combination of Coulomb's potential and a linearly rising potential: $V(r)=\alpha r + \beta/r$.


text:
/-
Copyright (c) 2020 Aaron Anderson, Jalex Stark. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Aaron Anderson, Jalex Stark
-/
import linear_algebra.matrix.charpoly.coeff
import field_theory.finite.basic
import data.matrix.char_p

/-!
# Results on characteristic polynomials and traces over finite fields.
-/

noncomputable theory

open polynomial matrix
open_locale polynomial

variables {n : Type*} [decidable_eq n] [fintype n]

@[simp] lemma finite_field.matrix.charpoly_pow_card {K : Type*} [field K] [fintype K]
  (M : matrix n n K) : (M ^ (fintype.card K)).charpoly = M.charpoly :=
begin
  casesI (is_empty_or_nonempty n).symm,
  { cases char_p.exists K with p hp, letI := hp,
    rcases finite_field.card K p with ⟨⟨k, kpos⟩, ⟨hp, hk⟩⟩,
    haveI : fact p.prime := ⟨hp⟩,
    dsimp at hk, rw hk at *,
    apply (frobenius_inj K[X] p).iterate k,
    repeat { rw iterate_frobenius, rw ← hk },
    rw ← finite_field.expand_card,
    unfold charpoly, rw [alg_hom.map_det, ← coe_det_monoid_hom,
      ← (det_monoid_hom : matrix n n K[X] →* K[X]).map_pow],
    apply congr_arg det,
    refine mat_poly_equiv.injective _,
    rw [alg_equiv.map_pow, mat_poly_equiv_charmatrix, hk, sub_pow_char_pow_of_commute, ← C_pow],
    { exact (id (mat_poly_equiv_eq_X_pow_sub_C (p ^ k) M) : _) },
    { exact (C M).commute_X } },
  { -- TODO[gh-6025]: remove this `haveI` once `subsingleton_of_empty_right` is a global instance
    haveI : subsingleton (matrix n n K) := subsingleton_of_empty_right,
    exact congr_arg _ (subsingleton.elim _ _), },
end

@[simp] lemma zmod.charpoly_pow_card {p : ℕ} [fact p.prime] (M : matrix n n (zmod p)) :
  (M ^ p).charpoly = M.charpoly :=
by { have h := finite_field.matrix.charpoly_pow_card M, rwa zmod.card at h, }

lemma finite_field.trace_pow_card {K : Type*} [field K] [fintype K]
  (M : matrix n n K) : trace (M ^ (fintype.card K)) = trace M ^ (fintype.card K) :=
begin
  casesI is_empty_or_nonempty n,
  { simp [zero_pow fintype.card_pos, matrix.trace], },
  rw [matrix.trace_eq_neg_charpoly_coeff, matrix.trace_eq_neg_charpoly_coeff,
       finite_field.matrix.charpoly_pow_card, finite_field.pow_card]
end

lemma zmod.trace_pow_card {p : ℕ} [fact p.prime] (M : matrix n n (zmod p)) :
  trace (M ^ p) = (trace M)^p :=
by { have h := finite_field.trace_pow_card M, rwa zmod.card at h, }

subset_name: curated


TITLE: How can we define temperature in integrable systems?
QUESTION [5 upvotes]: For a thermalised system, at equilibrium, the usual quantum statistical ensembles are valid and we can define temperature at an equilibrium between system and its bath. But for integrable systems (non-thermalisable) we don't have those ensembles, so how do we define a temperature?
PS- I had this doubt specific to many body localized state. And my intution is that the ensemble for a many body localised state would be similar to any other integrable system. Please correct me if I am wrong.

REPLY [2 votes]: To the best of my knowledge, to describe thermalization in integrable models one should to pass to the so-called "generalized Gibbs ensemble". For a prototypical integrable model, the XXX spin chain, the former was constructed by Ilievski et al arXiv:1507.02993; see also arXiv:1610.06911. (For the dynamics, away from thermal equilibrium, the key concept is "generalized hydrodynamics", see e.g. J. Dubail's Viewpoint in Physics)