text:
\section{Introduction}
\subsection{Quantum mechanics on a complex domain}\label{context}
The motion of a quantum particle on a one-dimensional path is described by the Hamiltonian operator
\[ \left(\frac{\partial}{\partial z}\right)^2 - \frac{q}{2}, \]
where $z$ is the position variable and $q/2$ is the function that gives the potential energy of the particle at each point on the path. This kind of operator is called a Hill's operator~\cite{geom-kdv}.

Sometimes, a particle on a path has a chance of escaping the path altogether. For example, a photon in an optical fiber might escape through the side of the fiber, or a neutron passing through an atom's nucleus might be absorbed into the nucleus. One way of modeling this, widely used in nuclear physics~\cite{nuclear-optical}, is to make the potential energy function complex-valued, with non-positive imaginary part. The imaginary part of the potential gives the particle's chance of escape at each point on the path.

In nuclear physics, the complex potential often extends to a holomorphic function on a neighborhood $\Omega \subset \C$ of the real interval parameterizing the path~\cite[\S 3.3]{semiclass-scat}. When it does, we can treat $z$ as a complex variable and $\tfrac{\partial}{\partial z}$ as a complex derivative, reinterpreting the Hamiltonian as an operator on holomorphic functions over $\Omega$. Each holomorphic eigenfunction, if it's tame enough,\footnote{The details depend on which rigged Hilbert space we use to define unbound states~\cite[\S 8.4]{capri-nr-qm}\cite[\S 1.4, end of \S 10.1]{qm-modern}. In good cases, the tame holomorphic eigenfunctions correspond one-to-one with the generalized eigenfunctions we find in the rigged Hilbert space. See Appendix~\ref{gen-eigfns} for some examples.} describes a stationary or steadily decaying state of the particle---a bound state if it restricts to an $L^2$ function on the path, and an unbound state\footnote{Or, equivalently, a scattering state: for stationary and steadily decaying states, typical notions of unbound and scattering states coincide. You can find this point of view in the table at the end of Chapter~IV of \cite{dirac-gamow-gelfand}, or the discussion after equation~(1.2) of \cite{rhs-gamow}. Some typical notions of scattering states are described in \cite[\S 2.1]{phys-math-gamow}, \cite[\S 2]{leaky-modes}, and \cite[\S\S 16.3, 16.5]{qm-modern}.} otherwise. We can therefore study the motion of the particle using techniques from complex analysis and geometry.

One fruitful approach, called the {\em exact WKB method}, is rooted in the asymptotic analysis of solutions of complex differential equations~\cite{semiclass-scat}\cite{quartic-return}\cite{alg-anal-sing}\cite{exact-wkb-cluster}. The aim of this article is to illustrate another approach, based on the geometry of complex projective surfaces. Our main results are loosely modeled on the exact WKB ``transformation theory'' results of Kamimoto, Koike, Aoki, Kawai, and Takei \cite{borel-tfm-ser}. They're meant to serve as a foundation for efforts to gain a more geometric understanding of the exact WKB method.

Results similar to ours could be achieved in a more analytic way using Olver's error bounds for the Liouville-Green approximation~\cite[\S 6]{asymp-special}. Both approaches have their own advantages, and I think they could be usefully combined. I'll say more after stating the results.
\subsection{Conformal maps and the WKB method}
The first step of the WKB method is to transform the Hill's operator under consideration into something resembling the Hamiltonian of a free particle. The transformation that accomplishes this is an example of a {\em Liouville transformation} \cite[\S 6.1]{asymp-special}. On a complex domain, Liouville transformations pick up an especially geometric flavor, motivating the geometric point of view this article takes.

Say we have a Hill's operator $\mathcal{H}$ defined on an open region $\Omega \subset \C$, and another Hill's operator
\[ \tilde{\mathcal{H}} = \left(\frac{\partial}{\partial z}\right)^2 - \frac{\tilde{q}}{2} \]
defined on another open region $\tilde{\Omega} \subset \C$. A {\em Liouville transformation} that relates $\mathcal{H}$ to $\tilde{\mathcal{H}}$ is a conformal map $y \maps \Omega \to \tilde{\Omega}$ with the property that
\[ \mathcal{H} = y_z^{3/2} \circ \left[ \left(\frac{\partial}{\partial y}\right)^2 - \frac{y^* \tilde{q}}{2} \right] \circ y_z^{1/2}. \]

Those odd-looking powers of $y_z$ give Liouville transformations many nice features, discussed in Sections \ref{geom-cp} and \ref{deform-cp}. Their usefulness is perplexing if you treat $\mathcal{H}$ and $\tilde{\mathcal{H}}$ as operators on holomorphic functions over $\C$, as we'll do throughout this article. You can make geometric sense of them, however, by letting $\mathcal{H}$ and $\tilde{\mathcal{H}}$ act on holomorphic sections of the anti-tautological bundle $\mathcal{O}(1)$ over $\CP{1}$, sending them to holomorphic sections of $\mathcal{O}(-3)$.
\subsection{A free enough particle on a complex strip}
A basic system in quantum mechanics is the free particle: a particle that moves along the real line with constant potential energy, and no chance of escape. Its Hamiltonian is
\[ \left(\frac{d}{dz}\right)^2 - \frac{1}{4} \]
when we set the potential to $\tfrac{1}{4}$.

Let's consider a particle that's merely ``free enough'': its complex potential can vary with position, but not by too much. We'll express its Hamiltonian as
\[ \left(\frac{d}{dz}\right)^2 - \left(\frac{1}{4} + \frac{p}{2}\right), \]
where $p$ is a ``small enough'' function of position. Let's say $p$ is holomorphic on a horizontal strip $B \subset \C$ of height at least $\pi$.

In order to state our results cleanly, it will be useful to introduce a ``doubly extended'' complex plane $\hat{\C}$ with two extra points, $-\infty$ and $+\infty$. The left half-plane is a neighborhood of $-\infty$, and the right half-plane is a neighborhood of $+\infty$. Let $\hat{B}$ be the closure of $B$ in $\hat{\C}$. It comprises the strip $B$, the edges of the strip, and the points $-\infty$ and $+\infty$. If $B$ has finite height, $\hat{B}$ is topologically a closed disk.

Now we can state our two main results. We'll present them here in analytic language, with the Hamiltonian operators appearing explicitly. Before we prove these results, we'll recast them in geometric language as Theorems \ref{bicorn-qual-perturb} and \ref{bicorn-quant-perturb}, which can be found in Section~\ref{main-results}. The geometric statements are sleeker, for two reasons. First, by viewing the Hamiltonian operators as geometric structures, we avoid the need to write them out explicitly, as discussed in Section~\ref{oper-as-geom}. Second, the strange-looking function bounding the size of $p$ will turn out to be a coordinate description of a natural Riemannian metric on $B$, which we'll meet in Section~\ref{thurston-metric}.
\begin{thm}\label{elem-qual-perturb}
Consider a Hill's operator
\[ \mathcal{H} = \left(\frac{d}{dz}\right)^2 - \left(\frac{1}{4} + \frac{p}{2}\right) \]
on a horizontal strip $B \subset \C$ of height at least $\pi$, with $p$ holomorphic. Suppose $p$ is small in the sense that
\[ \left|\frac{p}{2}\right| < \begin{cases}
\left(\frac{1}{2 \sin \ell}\right)^2 & \ell \le \frac{\pi}{2} \\
\frac{1}{4} & \ell \ge \frac{\pi}{2},
\end{cases} \]
where $\ell$ is the function that gives the distance to the edge of $B$. Then we can find a new coordinate $y \maps B \to \C$ with the following properties.
\begin{itemize}
\item It extends to extends to a topological embedding $\hat{B} \to \hat{\C}$ which fixes $\pm\infty$.
\item It's a Liouville transformation that relates $\mathcal{H}$ to the Hamiltonian of a free particle with potential $\tfrac{1}{4}$. In other words,
\[ \mathcal{H} = y_z^{3/2} \circ \left[ \left(\frac{d}{dy}\right)^2 - \frac{1}{4} \right] \circ y_z^{1/2}. \]
\end{itemize}
\end{thm}
The strength of this theorem is that the new coordinate $y$ extends continuously to $\hat{B}$, even though $p$ can blow up at the boundary of $B$. If we improve the bound on $p$, we can get a more quantitative version of this result.
\begin{thm}\label{elem-quant-perturb}
Assume the setting of Theorem~\ref{elem-qual-perturb}. Suppose $p$ satisfies the stronger bound
\[ \left|\frac{p}{2}\right| < \begin{cases}
M\left(\frac{1}{2 \sin \ell}\right)^2 & \ell \le \frac{\pi}{2} \\
M\frac{1}{4} & \ell \ge \frac{\pi}{2}
\end{cases} \]
for some $M \in (0, \tfrac{1}{3})$. Then any coordinate $y$ with the properties offered by Theorem~\ref{elem-qual-perturb} is close to a translation. Specifically, for any two points $\mathfrak{m}, \mathfrak{n} \in \hat{B} \smallsetminus \{\pm\infty\}$, the displacements $r = z(\mathfrak{n}) - z(\mathfrak{m})$ and $r' = y(\mathfrak{n}) - y(\mathfrak{m})$ are kept close to each other by the inequality
\[ |r' - r| \le \tfrac{2M}{1-M}(|\Re r| + \tfrac{3}{2}j\pi), \]
where $j \in \{1, 2, 3, \ldots\}$ and $|\Im r| \in [0, j\pi]$.
\end{thm}
For the applications to the exact WKB method that I have in mind, $p$ will have nothing worse than isolated power-law singularities on the boundary of $B$. In this case, we could get similar results from Olver's error bounds for the Liouville-Green approximation~\cite[\S 6.11]{asymp-special}, using the tricks in \cite[\S 6.4]{asymp-special} to show that the error control function has bounded variation along a progressive path to each singularity. The bound on $|r' - r|$ might be more complicated near the singularities, especially regular singularities. However, it might also be tighter, especially where $p$ is small. I expect that Theorem~\ref{elem-quant-perturb} could be improved by using our geometric approach near the boundary of $B$, in the region $\ell \le \tfrac{\pi}{2}$, and switching to Olver's approach in the region $\ell \ge \tfrac{\pi}{2}$.

Here are some instructions for translating the analytic statements above into the geometric statements of Section~\ref{main-results}. They won't be immediately usable, because they rely on the geometric language we'll introduce on the way from here to Section~\ref{free-enuf-geom}. I've placed them here as a guide to how that geometric language is relevant to the analytic problem we originally set out to solve.
\begin{proof}[Reduction of Theorem~\ref{elem-qual-perturb} to Theorem~\ref{bicorn-qual-perturb}]
An open region in $\C$ equipped with a Hill's operator can be seen as a chart for a complex projective surface (Section~\ref{oper-as-geom}). The complex plane, equipped with our free particle Hamiltonian, is the image of the global chart $u \maps U \to \C$ for the infinite bicorn $U$ (Section~\ref{free-geom}). The strip $B$, equipped with the same Hamiltonian, is a global chart for a bicorn (Section~\ref{bicorn-defn}).

Let $\varepsilon$ be the quadratic differential $-p\,dz^2$ on $B$. Using a coordinate expression for the Thurston metric of $B$ (Section~\ref{thurston-metric}), you can see that the bound on $p$ in Theorem~\ref{elem-qual-perturb} is equivalent to the bound on $\varepsilon$ in Theorem~\ref{bicorn-qual-perturb}. The conformal embedding $\gamma \maps B \to U$ that we get from Theorem~\ref{bicorn-qual-perturb} appears in coordinates, by composition with $u$, as a conformal map $y \maps B \to \C$.

The space $\hat{\C}$ is a coordinate description of $\hat{U}$, so $\gamma$ extends to a topological embedding $\hat{B} \to \hat{U}$ if and only if $y$ extends to a topological embedding $\hat{B} \to \hat{\C}$. Using the properties of Schwarzian derivatives outlined at the end of Section~\ref{deform-cp}, you can work out that $\gamma$ has Schwarzian derivative $\varepsilon$ if and only if $y$ is a Liouville transformation that relates $\mathcal{H}$ to our free particle Hamiltonian.
\end{proof}
\begin{proof}[Reduction of Theorem~\ref{elem-quant-perturb} to Theorem~\ref{bicorn-quant-perturb}]
We give the infinite bicorn a translation structure by declaring $u$ to be a translation chart (Section~\ref{tras-struct}). Expressed in terms of this chart, the conclusion of Theorem~\ref{bicorn-quant-perturb} becomes the conclusion of Theorem~\ref{elem-quant-perturb}.
\end{proof}
\subsection{Acknowledgements}
This line of work was suggested by Marco Gualtieri, informed by Shinji Sasaki, supported by the University of Toronto, and pursued in part at the Iowa City Public Library. I'm grateful to Francis Bischoff, Eduard Duryev, Marco Gualtieri, Omar Kidwai, Tatsuya Koike, Leonid Monin, Nikita Nikolaev, Shinji Sasaki, and Israel Michael Sigal for their feedback on the results and presentation of this paper.
\section{Hill's operators as geometric structures}\label{oper-as-geom}
\subsection{Overview}
The analysis of holomorphic functions on open regions in $\C$ leads naturally to the geometric study of Riemann surfaces and conformal maps. The analysis of Hill's operators on open regions in $\C$ leads to its own kind of geometry---the study of {\em complex projective surfaces} and {\em M\"{o}bius maps}~\cite{geom-kdv}.

When you take a geometric point of view, the problem of finding a Liouville transformation that relates one Hill's operator to another becomes the problem of finding a M\"{o}bius map from one complex projective surface to another. This is useful because complex projective surfaces, like Riemann surfaces, all look the same on small scales; the only obstacles to finding M\"{o}bius maps between them are large-scale mismatches. Complex projective surfaces are also very rigid: M\"{o}bius maps are determined by their local behavior, and their local behavior has few degrees of freedom. These properties aren't obvious from an analytic point of view. We'll see how they arise in Section~\ref{geom-cp}.

The perturbation theory of Hill's operators appears geometrically as a rich deformation theory of complex projective structures. Its basic ideas are outlined in Section~\ref{deform-cp}. The WKB method is a perturbative analysis, and the main results of this paper are, correspondingly, deformation theory results.
\subsection{Analytic definition}
A conformal structure on a surface is an atlas of charts to $\C$ whose transition maps are conformal. A {\em complex projective structure} is a conformal structure that comes with a Hill's operator on the image of each chart. Its transition maps are required to be Liouville transformations.

A {\em M\"{o}bius map} is a structure-preserving map between complex projective surfaces: a conformal map which becomes a Liouville transformation when composed with charts on both sides.
\subsection{Geometric simplification}\label{geom-cp}
You might expect small pieces of complex projective surfaces to come in many different shapes, arising from Hill's operators with different sorts of potentials. Surprisingly, they don't! All complex projective surfaces are locally isomorphic. In other words, all Hill's operators are locally Liouville-equivalent. The construction behind this observation dates back to Kummer, although the observation may not~\cite[\S 2]{old-and-new}. It's reviewed in Appendix~\ref{kummer-tfm}.

Using Kummer's trick, you can describe any complex projective structure in terms of {\em projective charts}, which come with the operator $(\frac{\partial}{\partial z})^2$ on their images. The Liouville transformations relating this operator to itself are the M\"{o}bius transformations. If you stick to projective charts, you can forget about Hill's operators entirely, and say that a complex projective structure is an atlas of charts to $\CP{1}$ whose transition maps are M\"{o}bius transformations. A M\"{o}bius map is a conformal map which becomes a M\"{o}bius transformation when composed with projective charts on both sides.
\subsection{Deformations}\label{deform-cp}
If you have a conformal map $\zeta$ from one complex projective surface to another, you can see how it deforms the complex projective structure by measuring its {\em Schwarzian derivative} $\mathcal{S}\zeta$. This holomorphic quadratic differential tracks how $\zeta$ deviates from its best M\"{o}bius approximation at each point. Its description here follows the more detailed treatment in \cite[\S 4.2]{warping-geom}, which in turn is based on an exposition by Thurston~\cite{zippers}.

Pick a point $\mathfrak{m}$ in the domain of $\zeta$. To reduce clutter---hopefully without adding too much confusion---we'll also use $\mathfrak{m}$ to denote the ideal of holomorphic functions vanishing at $\mathfrak{m}$. For any projective chart $v$ vanishing at $\mathfrak{m}$, you can find a projective chart $\tilde{v}$ around $\zeta(\mathfrak{m})$ with
\[ \zeta^* \tilde{v} \in v + \mathfrak{m}^3. \]
Intuitively, this shows that $\zeta$ is a M\"{o}bius map through second order near $\mathfrak{m}$.

To see how far $\zeta$ is from actually being a M\"{o}bius map, let's look at its third-order expansion~\cite[\S 4.2.2]{warping-geom}. Rewriting the second-order expansion above as
\[ \zeta^* \tilde{v} \in v(1 + \mathfrak{m}^2), \]
and expanding to the next order inside the parentheses, we get
\[ \zeta^* \tilde{v} \in v(1 + \tfrac{1}{6} p_\mathfrak{m}\,v^2 + \mathfrak{m}^3) \]
for some $p_\mathfrak{m} \in \C$. More abstractly,\footnote{The factor of $\tfrac{1}{6}$ gives the standard normalization for the Schwarzian derivative. The normalization used in \cite[\S 4.2.2]{warping-geom} is not standard.}
\[ \zeta^* \tilde{v} \in v(1 + \tfrac{1}{6}\,\mathcal{S}\zeta_\mathfrak{m}) \]
for some coset $\mathcal{S}\zeta_\mathfrak{m} \in \mathfrak{m}^2/\mathfrak{m}^3$, which can be written in coordinates as $p_\mathfrak{m}\,v^2 + \mathfrak{m}^3$. This coset, which turns out to be the same for any choice of $v$~\cite[\S 4.2.3]{warping-geom}, is the Schwarzian derivative of $\zeta$ at $\mathfrak{m}$.

You can identify the coset $v^2 + \mathfrak{m}^3$ with the square of the cotangent vector $dv_\mathfrak{m}$~\cite[\S 4.2.2]{warping-geom}. This leads to the expression $\mathcal{S}\zeta_\mathfrak{m} = p_\mathfrak{m}\,dv_\mathfrak{m}^2$, and the interpretation of $\mathcal{S}\zeta$ as a quadratic differential. If you fix the chart $v$, the approximating chart $\tilde{v}$ depends holomorphically on $\mathfrak{m}$, so $\mathcal{S}\zeta$ is holomorphic. In coordinates, $\mathcal{S}\zeta = p\,dv^2$ for some holomorphic function $p$.

The Schwarzian derivative provides a useful perspective on Kummer's construction of local projective charts. To see it, you need to know two things. One is that when you compose two conformal maps between complex projective surfaces, their Schwarzian derivatives add: $\mathcal{S}(\omega \circ \zeta) = \mathcal{S}\zeta + \zeta^* \mathcal{S}\omega$~\cite[p. 188]{zippers}\cite[\S 3.1]{cp-structs}. The other is that you can find the Schwarzian derivative of a chart for a complex projective structure by looking at the Hill's operator on its image. A chart $f$ that comes with the operator
\[ \mathcal{H} = \left(\frac{\partial}{\partial z}\right)^2 - \frac{q}{2} \]
on its image has Schwarzian derivative $f^*q\,df^2$.

Putting these facts together, we see that a composition $y \circ f$ is a M\"{o}bius map if and only if $\mathcal{S}y = -q\,dz^2$. Our discussion of Kummer's trick in Appendix~\ref{kummer-tfm} can therefore be reframed as a proof that if $\psi$ and $\tilde{\psi}$ satisfy the Hill's equations $\mathcal{H}\psi = 0$ and $\mathcal{H}\tilde{\psi} = 0$, their ratio $\tilde{\psi}/\psi$ has Schwarzian derivative $-q\,dz^2$. This is what Kummer originally observed~\cite[\S 2]{old-and-new}.
\section{The geometry of the free particle}\label{free-geom}
\subsection{Overview}
Let's look at a free particle from the geometric viewpoint of Section~\ref{oper-as-geom}. Equipped with the Hamiltonian
\[ \left(\frac{d}{dz}\right)^2 - \frac{1}{4}, \]
the complex plane becomes a global chart for some complex projective surface $U$. We'll see in this section that $U$ can be described very simply in geometric terms. We'll also parlay the translation symmetry of the free particle into extra geometric structure on $U$, which complements the complex projective structure.
\subsection{Geometric definition}
The twice-punctured Riemann sphere $\CP{1} \smallsetminus \{0, \infty\}$, also known as $\C^\times$, gets a natural complex projective structure from its inclusion into $\CP{1}$. Its universal cover $U \to \C^\times$ inherits that complex projective structure in the usual way. I'll call $U$ the {\em infinite bicorn}, because the covering map rolls it up like the brim of a bicorn hat.
\begin{center}
\begin{tikzpicture}[nodes={align=center}]
\matrix[column sep=3cm, nodes={outer sep=2mm}]{
\inftbicorn{1}{bicorn}
&
\punkdsphere{1}{sphere}
\\
};
\path[
  commutative diagrams/.cd, every arrow, every label,
  nodes={fill=white, inner sep=1mm, outer sep=0.5mm}
]
  (bicorn) edge (sphere);
\end{tikzpicture}
\end{center}
For convenience, mark a point $\star \in U$ which lies above $1 \in \CP{1}$.
\subsection{Riemann surface description}
As a covering map between Riemann surfaces, $U \to \C^\times$ must be equivalent to the exponential map $\C \to \C^\times$. That means there's a unique conformal equivalence $u \maps U \to \C$ that sends $\star$ to $0$ and makes the diagram
\begin{center}
\begin{tikzpicture}[nodes={align=center}]
\matrix[row sep=2cm, column sep=1cm, nodes={outer sep=2mm}]{
\inftbicorn{0.5}{bicorn} & &
\node[rectangle, fill=smoked!40, minimum size=0.5*4cm] (0, 0) (plane) {};
\draw[mauve, line width=0.2mm] (0, -1) -- (0, 1);
\\
&
\punkdsphere{0.5}{sphere}
\\
};
\path[
  commutative diagrams/.cd, every arrow, every label,
  nodes={fill=white, inner sep=1mm, outer sep=0.5mm}
]
  (bicorn) edge node {$u$} (plane)
  (bicorn) edge (sphere)
  (plane) edge node {$\exp$} (sphere);
\end{tikzpicture}
\end{center}
commute. We'll think of $u$ as a global chart for $U$, using the end of Section~\ref{deform-cp} to find out which Hill's operator it comes with. Working backwards from the fact that $\exp u$ is a M\"{o}bius map, you can infer that $u$ carries the free particle Hamiltonian
\[ \left(\frac{d}{dz}\right)^2 - \frac{1}{4} \]
on its image, and thus has Schwarzian derivative $\tfrac{1}{2} du^2$.
\subsection{Boundary points}
We can extend the projective cover $U \to \C^\times$ to a topological branched cover $\hat{U} \to \CP{1}$ by adding a pair of points $-\infty$ and $+\infty$ that map to $0$ and $\infty$, respectively. I'll call these new points the {\em tips} of the infinite bicorn. Give $\hat{U}$ the coarsest topology that makes the map $\hat{U} \to \CP{1}$ continuous.
\subsection{Translation structure}\label{tras-struct}
We saw earlier that $U$ is conformally equivalent to the flat plane, and we know that all conformal automorphisms of the flat plane are composed of translations, rotations, and scalings. Among these, translations and half-turn rotations are the only ones that preserve the Schwarzian derivative $\tfrac{1}{2} du^2$. The projective automorphisms of $U$ are therefore composed of translations and half-turn rotations---the local symmetries of a {\em half-translation surface}~\cite[\S 2]{billiards-problems}\cite[\S 1.8]{rat-flat}.

It's now clear that all the projective automorphisms of $U$ extend to topological automorphisms of $\hat{U}$, which either fix or exchange the tips. The ones that fix the tips are the pure translations.

In light of all this, let's define a {\em translation structure} on $U$ by declaring $u$ to be a translation chart~\cite[\S 2]{billiards-problems}. The global translations of $U$ are the projective automorphisms that fix the tips. Translation of $\hat{U}$ by $r \in \C$ commutes with multiplication of $\CP{1}$ by $e^r \in \C^\times$. In particular, vertical translations of $\hat{U}$ correspond to rotations of $\CP{1}$, so it's natural to think of vertical distances on $U$ as angular distances.

On a small region $\Omega \subset U$, the complex projective structure is more flexible than the translation structure: there are M\"{o}bius maps $\Omega \to U$ which are not translation maps. On a region that connects the tips, we can stiffen the complex projective structure by keeping the tips fixed.
\begin{prop}\label{bicorn-region-rigid}
Consider a connected open subset $\Omega \subset U$ and a M\"{o}bius map $\omega \maps \Omega \to U$. If the tips of $U$ are in the closure of $\Omega$, and $\omega$ extends to a continuous map $\Omega \cup \{\pm\infty\} \to \hat{U}$ that fixes the tips, then $\omega$ is a translation map.
\end{prop}
\begin{proof}
Any two M\"{o}bius maps from a connected complex projective surface to $\CP{1}$ differ by a M\"{o}bius transformation, so there's a M\"{o}bius transformation $\theta$ that makes the diagram
\begin{center}
\begin{tikzcd}
\Omega \arrow[d] \arrow[r, "\omega"] & U \arrow[d] \\
\CP{1} \arrow[r, "\theta"'] & \CP{1}
\end{tikzcd}
\end{center}
commute. Since $\omega$ extends continuously to the tips of $U$, and fixes each tip, $\theta$ must fix $0$ and $\infty$. That means there's a translation $\omega'$ of $U$ that makes the diagram
\begin{center}
\begin{tikzcd}
U \arrow[d] \arrow[r, "\omega'"] & U \arrow[d] \\
\CP{1} \arrow[r, "\theta"'] & \CP{1}
\end{tikzcd}
\end{center}
commute. We can choose $\omega'$ to agree with $\omega$ at some point in $\Omega$. It follows, by the uniqueness of lifts along covering maps, that $\omega'$ agrees with $\omega$ everywhere in $\Omega$. Hence, $\omega$ is a translation map.
\end{proof}
\subsection{Horizontal strips}\label{bicorn-defn}
A horizontal strip in the infinite bicorn $U$ will be called a {\em bicorn}---a {\em finite} one if it's vertically bounded, and a {\em half-infinite} one if it extends infinitely in one vertical direction. A bicorn of height $2\pi$ or less maps injectively to $\CP{1}$ along the covering map.
\begin{center}
\begin{tikzpicture}[nodes={align=center}]
\matrix[column sep=2.5cm, nodes={outer sep=2mm}]{
\node (bicorn) at (4, 0) {};
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The image of a taller bicorn overlaps itself.
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\begin{tikzpicture}[nodes={align=center}]
\matrix[column sep=2.5cm, nodes={outer sep=2mm}]{
\node (bicorn) at (4, 0) {};
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  (bicorn) edge (sphere);
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A bicorn of height $\pi$ will be called a {\em bicorn disk}, because it maps to a disk in $\CP{1}$.
\begin{center}
\begin{tikzpicture}[nodes={align=center}]
\matrix[column sep=2.5cm, nodes={outer sep=2mm}]{
\node (bicorn) at (4, 0) {};
\bicornplane
\fill[mauve] (0, -1) rectangle (4, 0);
\fill[mauve!80] (0, -1/2) rectangle (4, 0);
&
\node (sphere) at (-2, 0) {};
\twopunkplane{mauve}{mauve!80}{smoked!40}{smoked!20}
\twopunks
\\
};
\path[
  commutative diagrams/.cd, every arrow, every label,
  nodes={fill=white, inner sep=1mm, outer sep=0.5mm}
]
  (bicorn) edge (sphere);
\end{tikzpicture}
\end{center}
Bicorn disks have many special features, which will play an important role in this paper.

A bicorn comes with a complex projective structure, a translation structure, and a boundary---its topological boundary in $\hat{U}$. I'll call this boundary the {\em circular boundary} to avoid ambiguity. The closure in $\hat{U}$ of a bicorn $B$ will be called its {\em circular closure}, and denoted $\hat{B}$.

In addition to the tips $-\infty$ and $+\infty$, the circular boundary of a finite bicorn contains two horizontal lines in $U$, which I'll call the {\em edges}. The circular boundary of a half-infinite bicorn has one edge, and the circular boundary of the infinite bicorn has no edges.

You can describe a bicorn $B$ intrinsically by giving its complex projective structure, the topological details of its circular closure $B \hookrightarrow \hat{B}$, and the locations of its tips $-\infty$ and $+\infty$ in $\hat{B}$. This description, by Proposition~\ref{bicorn-region-rigid}, determines the inclusion $B \hookrightarrow U$ up to translation, which is to say it determines the translation structure of $B$. The intrinsic point of view lets us handle a bicorn as an independent object, not equipped with any particular inclusion into $U$.

We just observed that, up to translation, there's only one M\"{o}bius map $B \to U$ that sends each tip of $B$ to the corresponding tip of $U$. We get the same kind of rigidity for any conformal map $B \to U$ with a fixed Schwarzian derivative.

\begin{prop}\label{bicorn-deform-rigid}
Consider two conformal maps from a bicorn $B$ to the infinite bicorn $U$. Suppose they extend to continuous maps $B \cup \{\pm\infty\} \to U \cup \{\pm\infty\}$ that send each tip of $B$ to the corresponding tip of $U$. If the two maps have the same Schwarzian derivative, they differ only by a translation of $U$.
\end{prop}
\begin{proof}
Call the two maps $\zeta, \zeta' \maps B \to U$. Let $\omega \maps \zeta(B) \to U$ be the conformal map for which $\omega \circ \zeta = \zeta'$. Observe that $\omega$ extends continuously to the tips of $U$, fixing each tip.

Now, suppose $\zeta$ and $\zeta'$ have the same Schwarzian derivative. Then $\omega$ is a M\"{o}bius map. Proposition~\ref{bicorn-region-rigid} guarantees that $\omega$ is a translation map, which extends to a translation of $U$.
\end{proof}
\subsection{The Thurston metric}\label{thurston-metric}
Every complex projective surface of hyperbolic type comes with a Riemannian metric called the Thurston metric. It's defined pointwise as the infimum of the hyperbolic metrics of all projectively immersed disks~\cite[\S 4.3]{cp-structs}.\footnote{Following the convention of our sources, we'll use the hyperbolic metric of curvature $-4$.} When one disk sits inside another, the hyperbolic metric of the larger disk is smaller at every point. That means the Thurston metric is determined by the maximal disks---the disks with no larger disks containing them.\footnote{When I say one projectively immersed disk sits inside another, I mean there's a M\"{o}bius map from one to the other that commutes with the immersions.}

The Thurston metric of a disk is the same as the hyperbolic metric. On the bicorn disk $\Im u \in (0, \pi)$ in $U$, it can be written as
\[ \left[\frac{|du|}{2\sin(\Im u)}\right]^2. \]
The maximal disks of the infinite bicorn are precisely the bicorn disks. From this fact, and the expression above, you can deduce that the Thurston metric of the infinite bicorn is $\tfrac{1}{4}|du|^2$---a constant multiple of the flat metric that comes from the translation structure. The Thurston metric of a general bicorn is
\[ \begin{cases}
\left(\frac{|du|}{2 \sin \ell}\right)^2 & \ell \le \frac{\pi}{2} \\
\frac{1}{4}|du|^2 & \ell \ge \frac{\pi}{2},
\end{cases} \]
where $\ell$ is the function that gives the distance to the circular boundary.
\newcommand{3}{3}
\pgfmathsetmacro{\lout}{0.25}
\pgfmathsetmacro{\lin}{pi/2}
\pgfmathsetmacro{\rin}{3+pi/2}
\pgfmathsetmacro{\rout}{3+pi-0.25}
\pgfmathsetmacro{\graphheight}{1/pow(2*sin(\lout r),2)}
\newcommand{0.25}{0.25}
\begin{center}
\begin{tikzpicture}
\fill[smoked!40] (0, 0) rectangle ({3+pi}, \graphheight);
\fill[smoked!20] (\lin, 0) rectangle (\rin, \graphheight);

\node[below] at (\lin/2, 0) {$\ell \le \pi/2$};
\node[below] at (\lin+3/2, 0) {$\ell \ge \pi/2$};
\node[below] at (\rin+\lin/2, 0) {$\ell \le \pi/2$};

\draw[very thick] plot[domain=\lout:\rout, samples=80, smooth] function{
  x < \lin ? 1/(2*sin(x))**2 : (x < \rin ? 1/4. : 1/(2*sin(x-3-pi))**2)
};
\end{tikzpicture}
\end{center}
The graph above shows how the size of the Thurston metric depends on your vertical position in a bicorn. Size is measured relative to $|du|^2$.
\section{The geometry of a free enough particle}\label{free-enuf-geom}
\subsection{Foundations}
Our main results are based on two classic extension theorems for local conformal maps from a complex projective surface to the Riemann sphere. The first, by Gehring and Pommerenke, extends a conformal map on a disk to a topological embedding on a larger domain. The second, by Ahlfors and Weill, bounds the maximal dilatation of the extension, making it quasiconformal.

We'll use these theorems through the following corollaries, chosen for their resemblance to the results we aim to achieve. The second theorem has been combined with Teichm\"{u}ller's observation that quasiconformal maps are close to M\"{o}bius maps, in the sense that they don't distort cross ratios too much.
\begin{cor}[Gehring--Pommerenke]\label{gp}
Consider a holomorphic quadratic differential $\varepsilon$ on a disk $D$. Write the hyperbolic metric on $D$ as $\eta^2$. If $|\varepsilon| < 2\eta^2$, there's a conformal embedding $D \to \CP{1}$ with the following properties.
\begin{itemize}
\item It extends to a topological embedding $\hat{D} \to \CP{1}$.
\item Its Schwarzian derivative is $\varepsilon$.
\end{itemize}
\end{cor}
\begin{cor}[Ahlfors--Weill, Teichm\"{u}ller]\label{awt}
If the quadratic differential $\varepsilon$ in Corollary~\ref{gp} satisfies the stronger bound $|\varepsilon| < 2M\eta^2$ for some $M \in (0, 1)$, the embedding $\hat{D} \to \CP{1}$ that we get is close to a M\"{o}bius map. Specifically, the cross ratio $\rho$ of four distinct points in $\hat{D}$ and the cross ratio $\rho'$ of their images in $\CP{1}$ are kept close to each other by the inequality
\[ d_{\therefore}(\rho, \rho') \le \tfrac{1}{2} \log \tfrac{1 + M}{1 - M}, \]
where $d_\therefore$ is the hyperbolic distance on $\CP{1} \smallsetminus \{0, 1, \infty\}$.
\end{cor}
\begin{proof}[Proof of Corollary~\ref{gp}]
Identify $\hat{D}$ with the closed unit disk in $\CP{1}$. Since $D$ is simply connected, there's a conformal map $D \to \CP{1}$ with Schwarzian derivative $\varepsilon$~\cite[Theorem~II.1.1]{unival-teich}. If $|\varepsilon| < 2\eta^2$ this map extends to a homeomorphism $\CP{1} \to \CP{1}$ by Gehring and Pommerenke's theorem~\cite[Theorem II.5.4]{unival-teich}. Restricting to $\hat{D}$ gives the desired topological embedding.
\end{proof}
\begin{proof}[Proof of Corollary~\ref{awt}]
As we did in the proof of Corollary~\ref{gp}, identify $\hat{D}$ with the closed unit disk in $\CP{1}$, and find a conformal map $D \to \CP{1}$ with Schwarzian derivative $\varepsilon$. If $|\varepsilon| < 2\eta^2$, this map extends to a quasiconformal homeomorphism $\mathring{\delta} \maps \CP{1} \to \CP{1}$ with a known Beltrami differential, found by Ahlfors and Weill~\cite[Theorem II.5.1]{unival-teich}. If $|\varepsilon| < 2M\eta^2$, the magnitude of the Beltrami differential is less than $M$ almost everywhere. It follows that $\mathring{\delta}$ is $\tfrac{1 + M}{1 - M}$-quasiconformal~\cite[Theorem I.4.1]{unival-teich}. Teichm\"{u}ller's theorem on the quasi-invariance of cross ratios then bounds $d_{\therefore}(\rho, \rho')$ as stated~\cite{teich-thm}.
\end{proof}
\subsection{Main results}\label{main-results}
\begin{thm}\label{bicorn-qual-perturb}
Consider a holomorphic quadratic differential $\varepsilon$ on a bicorn $B$ wide enough to contain a bicorn disk. Write the Thurston metric on $B$ as $\lambda^2$. If $|\varepsilon| < 2\lambda^2$, there's a conformal embedding $B \to U$ with the following properties.
\begin{itemize}
\item It extends to a topological embedding $\hat{B} \to \hat{U}$ that sends each tip of $B$ to the corresponding tip of $U$.
\item Its Schwarzian derivative is $\varepsilon$.
\end{itemize}
\end{thm}
\begin{thm}\label{bicorn-quant-perturb}
If the quadratic differential $\varepsilon$ in Theorem~\ref{bicorn-qual-perturb} satisfies the stronger bound $|\varepsilon| < 2M\lambda^2$ for some $M \in (0, \tfrac{1}{3})$, the embedding $\hat{B} \to \hat{U}$ that we get is close to a translation map. Specifically, the displacement $r$ between two points in $\hat{B} \smallsetminus \{\pm\infty\}$ and the displacement $r'$ between their images in $U$ are kept close to each other by the inequality
\[ |r' - r| \le \tfrac{2M}{1-M}(|\Re r| + \tfrac{3}{2}j\pi), \]
where $j \in \{1, 2, 3, \ldots\}$ and $|\Im r| \in [0, j\pi]$. Displacement is measured with respect to the standard translation structures on $B$ and $U$, as described in Section~\ref{tras-struct}.
\end{thm}
\subsection{Proofs}
\begin{proof}[Proof of Theorem~\ref{bicorn-qual-perturb}]
Since $B$ is wide enough to contain a bicorn disk, we can cover it with bicorn disks $D_1, \ldots, D_n$, ordered counterclockwise around $-\infty$. For convenience, let's arrange for the only overlaps to be between $D_\nu$ and $D_{\nu+1}$ for each $\nu$. Write the hyperbolic metric on $D_\nu$ as $\eta_\nu^2$.

Suppose $|\varepsilon| \le 2\lambda^2$. Since $D_\nu \subset B$ is a maximal disk, the definition of the Thurston metric ensures that $\lambda^2 \le \eta_\nu^2$. Corollary~\ref{gp} then gives us a conformal embedding $\delta \maps D_\nu \to \CP{1}$ with the following properties.
\begin{itemize}
\item Its Schwarzian derivative is $\varepsilon$.
\item It extends to a topological embedding $\hat{\delta} \maps \hat{D}_\nu \to \CP{1}$.
\end{itemize}
By composing with a suitable M\"{o}bius transformation, we can arrange for $\hat{\delta}_\nu$ to send $-\infty$ and $+\infty$ to $0$ and $\infty$, respectively. Then we can lift it along the branched covering $\hat{U} \to \CP{1}$ to a topological emedding $\hat{\gamma}_\nu \maps \hat{D}_\nu \to \hat{U}$.

By Proposition~\ref{bicorn-deform-rigid}, we can translate the embeddings $\hat{\gamma}_1, \ldots, \hat{\gamma}_n$ so that each one agrees with the previous one on the overlap between their domains. Then they fit together into one big topological embedding $\hat{B} \to \hat{U}$, which restricts to the desired conformal embedding $B \to U$.
\end{proof}
To prove Theorem~\ref{bicorn-quant-perturb}, we'll use a version of Corollary~\ref{awt} that not only tells us the embedding that we get is close to a M\"{o}bius map, but also shows us an efficient way to reach one.
\begin{lem}\label{hty-awt}
If the quadratic differential $\varepsilon$ in Corollary~\ref{gp} satisfies the stronger bound $|\varepsilon| < 2M\eta^2$ for some $M \in (0, \tfrac{1}{3})$, there's a homotopy $[0, 1] \times D \to \CP{1}$ with the following properties.
\begin{itemize}
\item It's conformal at each time.
\item It's a M\"{o}bius map at time zero.
\item It has Schwarzian derivative $\varepsilon$ at time one.
\item It extends to a homotopy $[0, 1] \times \hat{D} \to \CP{1}$ which is a topological embedding at each time.
\item Its extension deforms cross ratios Lipschitz-continuously, with Lipschitz constant $\arctanh M$.
\end{itemize}
To state the last property precisely, pick four distinct points in $\hat{D}$, and let $\rho_a$ be the cross ratio of their images in $\CP{1}$ at time $a \in [0, 1]$. The property is that
\[ d_{\therefore}(\rho_a, \rho_b) \le |b - a| \arctanh M \]
for all $a, b \in [0, 1]$, no matter which four points we picked.
\end{lem}
\begin{proof}
Identify $\hat{D}$ with the closed upper half-plane in $\CP{1}$. Find a conformal map $\delta \maps D \to \CP{1}$ with the properties listed in Corollary~\ref{gp}, using the method from the proof of the corollary. We know, from the details of the proof, that $\delta$ extends to a $\frac{1+M}{1-M}$-quasiconformal homeomorphism $\mathring{\delta} \maps \CP{1} \to \CP{1}$. By composing with a suitable M\"{o}bius transformation, we can arrange for $\mathring{\delta}$ to fix $0$, $1$, and $\infty$.

We'll now travel from the identity to $\mathring{\delta}$ along a clever path found by Dao-shing, and studied further by Gehring and Reich~\cite{parametric-rep}\cite[Lemma~3.3]{area-distortion}\cite[\S 2]{planar-qc}. Write $\mu$ for the Beltrami differential of $\mathring{\delta}$. The family of $(-1, 1)$-forms
\[ \mu_a = \frac{\mu}{|\mu|} \tanh (a \arctanh |\mu|), \]
parameterized by $a \in [0, 1]$, goes continuously from $0$ to $\mu$. (To resolve the ambiguity in the formula, set $\mu_a$ to zero wherever $\mu$ is zero.) Let $\mathring{\Delta}_a \maps \CP{1} \to \CP{1}$ be the map which has Beltrami differential $\mu_a$ and fixes $0$, $1$, and $\infty$. Using an argument by Lehto,\footnote{Write out the Beltrami differential of $\mathring{\Delta}_b \circ \mathring{\Delta}_a^{-1}$ using formula (I.4.4) of \cite{unival-teich}, and simplify using the angle addition identity for the hyperbolic tangent. You'll see that the magnitude of the Beltrami differential is bounded by $\tanh [(b - a) \arctanh M]$ almost everywhere. Taking the hyperbolic arctangent of both sides, and then writing the hyperbolic arctangent in terms of the natural logarithm, you'll get the desired bound on the dilatation of $\mathring{\Delta}_b \circ \mathring{\Delta}_a^{-1}$. Lehto uses the same kind of argument to prove Theorem~I.4.7 of \cite{unival-teich}.} you can show that $\mathring{\Delta}_b \circ \mathring{\Delta}_a^{-1}$ is $\big(\tfrac{1+M}{1-M}\big)^{|b - a|}$-quasiconformal\footnote{This formula conflicts with the second displayed equation in \cite[\S 2]{planar-qc}, which would give $e$ as the base of the exponent.} for all $a, b \in [0, 1]$. Teichm\"{u}ller's theorem on the quasi-invariance of cross ratios then bounds $d_{\therefore}(\rho_a, \rho_b)$ as stated~\cite{teich-thm}, just as in the proof of Corollary~\ref{awt}.

Each $\mu_a$ vanishes on $D$, because $\mu$ vanishes there. Each quasiconformal map $\mathring{\Delta}_a \maps \CP{1} \to \CP{1}$ thus restricts to a conformal map $\Delta_a \maps D \to \CP{1}$, whose Schwarzian derivative we'll call $\varepsilon_a$. It's apparent from the formula for $\mu_a$ that $|\mu_a| \le |\mu|$, and we know from the proof of Corollary~\ref{awt} that $|\mu| < M$ almost everywhere. This implies that $|\varepsilon_a| < 6M\eta^2$, as proven by K\"{u}hnau and by Lehto~\cite[\S II.3.3]{unival-teich}. Hence, by Corollary~\ref{gp}, $\mathring{\Delta}_a$ restricts to a topological embedding $\hat{D} \to \CP{1}$.

Our bound on the deformation of cross ratios ensures that $\mathring{\Delta}_a$ varies continuously with respect to $a$. Hence, $\mathring{\Delta}$ is a homotopy from the identity to $\mathring{\delta}$. Its restriction $\Delta$ is the homotopy we want.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{bicorn-quant-perturb} for $|\Im r| \le \pi$]
Consider two points $\mathfrak{m}_\text{start}, \mathfrak{m}_\text{end} \in \hat{B} \smallsetminus \{\pm\infty\}$ separated by a vertical distance of $\pi$ or less. Pick an embedding $\hat{\gamma} \maps \hat{B} \to \hat{U}$ of the kind produced by Theorem~\ref{bicorn-qual-perturb}. Let $r$ be the displacement from $\mathfrak{m}_\text{start}$ to $\mathfrak{m}_\text{end}$, and let $r'$ be the displacement from $\hat{\gamma}(\mathfrak{m}_\text{start})$ to $\hat{\gamma}(\mathfrak{m}_\text{end})$. By Proposition~\ref{bicorn-deform-rigid}, $r'$ doesn't depend on our choice of $\hat{\gamma}$.

Suppose $|\varepsilon| < 2M\lambda^2$ for some $M \in (0, \tfrac{1}{3})$. For convenience, we'll declare $\tfrac{1+M}{1-M}$ as a new parameter $K \in (1, \infty)$, which varies monotonically with $M$. We want to show that
\[ |r' - r| \le (K-1)(|\Re r| + \tfrac{3}{2}\pi). \]
It's easy to see that both $r$ and $r'$ are zero when $\mathfrak{m}_\text{start}$ and $\mathfrak{m}_\text{end}$ are the same, so let's restrict ourselves to the case where $\mathfrak{m}_\text{start}$ and $\mathfrak{m}_\text{end}$ are distinct.

Because $\mathfrak{m}_\text{start}$ and $\mathfrak{m}_\text{end}$ are separated by a vertical distance of $\pi$ or less, we can find a bicorn disk $D \subset B$ whose circular closure $\hat{D}$ contains both of them. Pick a homotopy $\Delta \maps [0, 1] \times D \to \CP{1}$ with the properties listed in Lemma~\ref{hty-awt}. Let $\hat{\Delta}$ be its extension to $[0, 1] \times \hat{D}$. By composing with a suitable time-dependent M\"{o}bius transformation, we can arrange for $\hat{\Delta}$ to always send $-\infty$ and $+\infty$ to $0$ and $\infty$, respectively. Then we can lift it along the branched covering $\hat{U} \to \CP{1}$ to a homotopy $\hat{\Gamma} \maps [0, 1] \times \hat{D} \to \hat{U}$ which is a topological emedding at each time.

For each time $a \in [0, 1]$, let $\rho_a$ be the cross ratio
\[ (0, \hat{\Delta}_a(\mathfrak{m}_\text{start}), \hat{\Delta}_a(\mathfrak{m}_\text{end}), \infty) = \left. \frac{\hat{\Delta}_a(\mathfrak{m}_\text{end}) - 0}{\hat{\Delta}_a(\mathfrak{m}_\text{start}) - 0} \middle/ \frac{\hat{\Delta}_a(\mathfrak{m}_\text{end}) - \infty}{\hat{\Delta}_a(\mathfrak{m}_\text{start}) - \infty} \right., \]
which simplifies to $\hat{\Delta}_a(\mathfrak{m}_\text{end})/\hat{\Delta}_a(\mathfrak{m}_\text{start})$. Let $r_a$ be the displacement from $\hat{\Gamma}_a(\mathfrak{m}_\text{start})$ to $\hat{\Gamma}_a(\mathfrak{m}_\text{end})$, and observe that $\exp r_a = \rho_a$. In other words, $r_a$ is a lift of the path $\rho_a$ along the covering map
\begin{center}
\begin{tikzcd}
\C \smallsetminus 2\pi i\Z \arrow[r, "\exp"] & \CP{1} \smallsetminus \{0, 1, \infty\}.
\end{tikzcd}
\end{center}

Write the hyperbolic metric on $\CP{1} \smallsetminus \{0, 1, \infty\}$ and the pullback of that metric to $\C \smallsetminus 2\pi i\Z$ as $\eta_\therefore^2$ and $\eta_{\ldots}^2$, respectively. Because the homotopy $\hat{\Delta}$ was produced by Lemma~\ref{hty-awt}, the path $\rho_a$ is Lipschitz with respect to $\eta_\therefore^2$, with Lipschitz constant $\arctanh M$. The lift $r_a$ is therefore Lipschitz with respect to $\eta_{\ldots}^2$, with the same Lipschitz constant. In particular,
\[ d_{\ldots}(r_0, r_1) \le \arctanh M, \]
where $d_{\ldots}$ is the distance induced by the metric $\eta_{\ldots}^2$.

The homotopy $\hat{\Gamma}$ restricts to a homotopy to $\Gamma \maps [0, 1] \times D \to U$ which is conformal at each time. The initial map $\Gamma_0$ is a M\"{o}bius map, so $\hat{\Gamma}_0$ is a translation map, by Proposition~\ref{bicorn-region-rigid}. Hence, $r_0 = r$. The final map $\Gamma_1$ has Schwarzian derivative $\varepsilon$, so the embedding $\hat{\gamma} \maps \hat{B} \to \hat{U}$ that we chose at the beginning matches $\hat{\Gamma}_1$ up to a translation, thanks to Proposition~\ref{bicorn-deform-rigid}. Hence, $r_1 = r'$. The bound above now says that
\[ d_{\ldots}(r, r') \le \arctanh M. \]

To get the weaker but more explicit bound stated in the theorem, we use the inequality~\cite[Theorem 14.3.1]{local-viewpoint}
\[ \frac{|dz|}{2|z|\big(\big|\log |z|\big| + \tfrac{3}{2}\pi\big)} \le \eta_\therefore, \]
which pulls back along $\exp$ to the inequality
\[ \frac{|dz|}{2(|\Re z| + \tfrac{3}{2}\pi)} \le \eta_{\ldots}. \]
The left-hand side, which I'll call $\eta_\text{tract}$, has a nice hyperbolic interpretation. Under the metric $\eta_\text{tract}^2$, the complex plane becomes the universal cover of a tractricoid---a pair of hyperbolic cusps glued together along a horocycle. You can see this concretely by identifying the closed left and right half-planes in $(\C, \eta_\text{tract}^2)$ with the horodisks $\Re z \le -\tfrac{3}{2}\pi$ and $\tfrac{3}{2}\pi \le \Re z$ in the left and right half-plane models of $\mathbb{H}$.

We've shown that $r'$ is in the closed $\eta_{\ldots}^2$ ball around $r$ of radius $\arctanh M$, which can also be expressed as $\tfrac{1}{2} \log K$. We know from the inequality above that this $\eta_{\ldots}^2$ ball is inside the closed $\eta_\text{tract}^2$ ball around $r$ of the same radius. Using the hyperbolic interpretation, you can show that this $\eta_\text{tract}^2$ ball is inside the closed $|dz|^2$ ball around $r$ of radius $(K-1)(|\Re r| + \tfrac{3}{2}\pi)$. This bounds $|r' - r|$ as desired.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{bicorn-quant-perturb} in general]
Consider any two points $\mathfrak{m}_\text{start}, \mathfrak{m}_\text{end} \in \hat{B} \smallsetminus \{\pm\infty\}$. Pick an embedding $\hat{\gamma} \maps \hat{B} \to \hat{U}$ of the kind produced by Theorem~\ref{bicorn-qual-perturb}, and define $r$ and $r'$ like we did in the proof of the base case. For convenience, let's restrict our attention to the case where $\Im r \ge 0$. The argument we'll use adapts straightforwardly to the other case.

Starting at $\mathfrak{m}_\text{start}$, walk upward in steps of length $\pi$ until your vertical distance to $\mathfrak{m}_\text{end}$ is $\pi$ or less. This produces a sequence of points $\mathfrak{m}_1, \ldots, \mathfrak{m}_j$, with $\mathfrak{m}_1 = \mathfrak{m}_\text{start}$. Notice that $|\Im r| \in [0, j\pi]$.

By construction, the displacement from each point $\mathfrak{m}_\nu$ to $\mathfrak{m}_{\nu+1}$ is $i\pi$. Let $r'_\nu$ be the displacement from $\hat{\gamma}(\mathfrak{m}_\nu)$ to $\hat{\gamma}(\mathfrak{m}_{\nu+1})$. Let $r_\text{fin}$ be the displacement from $\mathfrak{m}_j$ to $\mathfrak{m}_\text{end}$, and let $r'_\text{fin}$ be the displacement from $\hat{\gamma}(\mathfrak{m}_j)$ to $\hat{\gamma}(\mathfrak{m}_\text{end})$.

Suppose $|\varepsilon| < 2M\lambda^2$ for some $M \in (0, \tfrac{1}{3})$. As before, we'll declare $\tfrac{1+M}{1-M}$ as a new parameter $K \in (1, \infty)$. Since the vertical distance from each point $\mathfrak{m}_\nu$ to $\mathfrak{m}_{\nu+1}$ is $\pi$, we can apply the base case of the theorem and conclude that
\begin{align*}
|r'_\nu - i\pi| & \le (K-1)(|\Re i\pi| + \tfrac{3}{2}\pi) \\
& = (K-1)\tfrac{3}{2}\pi.
\end{align*}
Applying the base case to the step from $\mathfrak{m}_j$ to $\mathfrak{m}_\text{end}$, we learn that
\[ |r'_\text{fin} - r_\text{fin}| \le (K-1)(|\Re r_\text{fin}| + \tfrac{3}{2}\pi). \]

Taking the difference between the expressions
\begin{align*}
r & = i\pi + \ldots + i\pi + r_\text{fin} \\
r' & = r'_1 + \ldots + r'_{j-1} + r'_\text{fin}
\end{align*}
and applying the triangle inequality, we get the bound
\[ |r' - r| \le |r'_1 - i\pi| + \ldots + |r'_{j-1} - i\pi| + |r'_\text{fin} - r_\text{fin}|. \]
Combining this with the step-by-step bounds from above, we learn that
\begin{align*}
|r' - r| & \le (K-1)\tfrac{3}{2}(j-1)\pi + (K-1)(|\Re r_\text{fin}| + \tfrac{3}{2}\pi) \\
& = (K-1)(|\Re r_\text{fin}| + \tfrac{3}{2}j\pi).
\end{align*}
Noting that $\Re r_\text{fin} = \Re r$, we end up with the desired bound.
\end{proof}

redpajama_set_name: RedPajamaArXiv


Chris and Arnie Galin explore abstract and representational photography - they are interested in capturing "Nature in Motion," through the use of panning, zooming, tilting and other camera techniques - to really "see" through the eyes of the camera. Implementing a process called dye sublimation they then transfer original photographs to many different substrates including tile, glass, aluminum, wood, and porcelain. The Downtown Artist Series students for this exhibition will be poet Sally Familia and artist Aaron Scott.


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The eccentric engineer: William Playfair and his many, varied and unappreciated careers
By Justin Pollard
Published Sunday, March 22, 2020
The story of an adventurous and revolutionary engineer, draughtsman, inventor, silversmith, merchant, banker, statistician, pamphleteer, editor, speculator... and convict.
Not many children who grapple with pie charts, bar charts and time-series plots know they can thank an engineer for their misery. Nor would William Playfair have probably wanted to be remembered as such, but few can choose where fate takes them - and Playfair was taken further than many.
William was born in 1759 at the dawn of one of the great eras of engineering. The son of a Scottish clergyman, he became apprenticed to one of the major figures of the Agricultural Revolution, Andrew Meikle, the inventor of the threshing machine, before becoming personal assistant to James Watt at the Boulton and Watt engine works in Birmingham in 1777.
Just five years later, he left to form the first of his many companies in London. Sadly, business didn't seem to come easily to Playfair and his silversmithing company failed, though not before he had invented the first industrial process for silver-plating spoons.
Playfair then returned to one of his first loves, draughtsmanship, which he learned at the drawing tables of Meikle and Watt. Keen to write about another passion, economics, he found it hard to visually get his ideas across so he meticulously 'diagrammised' his arguments. In the process, he laid the foundations of statistical graphics. His 'Commercial and Political Atlas', published in 1786, contained not a single map, but instead was illustrated with 43 'time-series' plots and a bar chart – the first bar chart ever printed. This was revolutionary, but received little publicity in England.
Feeling somewhat underappreciated, he decided to move to Paris in 1787 where he found his work much more widely read, even by the King, who was fond of geography. Unfortunately, France was becoming rather revolutionary and political events began to overtake him. Not one to be left out, and not showing much loyalty to his royal patron, Playfair decided to join in as part of the hastily formed St Antoine Quarter militia, and took part in the storming of the Bastille.
However, politics proved no easier a career than any of his others, and he was widely seen as an English aristocratic Royalist – despite being the Scottish son of a vicar. Things weren't helped by his involvement in a new business – a scheme to send French settlers to America which had, typically for Playfair, gone bankrupt. With the heat rising in Paris, particularly after he rescued a judge who had invested in his scheme from the mob, he decided to head for Germany.
Here, he tried his hand again at engineering, deftly copying a description of the recently invented telegraph and sending a model of the apparatus to the Duke of York along with a new alphabet for telegraphy. Sadly, the Duke of York seemed unimpressed and no commission followed, so Playfair turned to yet another career. Now back in London, he turned to banking, setting up the 'Security Bank' using banking techniques he had seen in Paris. However, following a spat with the Bank of England, this business also folded.
Yet Playfair was not quite done with banking and managed finally to attract the attention of the government in the form of Home Secretary Henry Dundas. Disillusioned with the Revolution and the subsequent conflict with Britain, Playfair suggested that rather than sending men and weapons to fight the French, he could defeat them with money. His scheme involved forging one hundred million 'assignats' – the French financial instruments used during the Revolution – and flooding the country with them. His plan was that, as bad money drives out good, this would cripple the French economy and prevent them waging war. The forged notes were introduced with huge success, and within two years the assignat was worthless.
Sadly, he received no credit for this work, and instead turned to writing and pamphleteering, warning the government (among other things) of the likely escape of Napoleon from Elba. But Playfair proved something of a Cassandra in official circles and was ignored. Forced into debt, he ended up in the Fleet Prison for three years.
With the restoration of the Bourbons, he decided to try his hand again in France as a magazine editor but was prosecuted for libel and had to escape back to England to avoid another prison sentence. Here he returned to his graph paper and publishing, inventing his pièce de résistance, the pie chart. By this time, he had been an engineer, draughtsman, inventor, silversmith, merchant, banker, statistician, pamphleteer, editor, speculator and convict, but it would be his graphs that would live on in every school maths book, every economics paper and news programme.
eccentric engineer
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redpajama_set_name: RedPajamaCommonCrawl


text:
Kościół ewangelicki w Pierśćcu – kościół ewangelicko-augsburski w Pierśćcu, w województwie śląskim w Polsce. Należy do parafii ewangelicko-augsburskiej w Skoczowie.

Historia
Pierwsza kaplica ewangelicka w Pierśćcu istniała do 28 października 1616 r., kiedy to uległa spaleniu. Nową świątynię wybudowano w 1618 r. na terenie obecnego cmentarza katolickiego. Funkcjonowała ona do czasu jej odebrania w 1654 r. Kaplica wróciła jednak tymczasowo do ewangelików, odebrano im ją ponownie w 1718 r. Po tej dacie luteranie spotykali się potajemnie w domach prywatnych oraz w leśnych kościołach. Po wybudowaniu kościoła ewangelickiego w Drogomyślu mieszkańcy wsi uczęszczali tam na nabożeństwa.

W 1865 r. Pierściec przyłączono do parafii ewangelicko-augsburskiej w Skoczowie. Nowa kaplica na terenie miejscowości powstała w 1914 r. na działającym od 1858 r. cmentarzu ewangelickim. 11 października 1914 r. odbyło się jej poświęcenie. Kaplica została rozebrana po 1994 r.

Kościół ewangelicki wybudowano w latach 1984-1989. Poświęcenia dokonał 15 października 1989 r. ksiądz biskup Janusz Narzyński razem z proboszczem skoczowskim Janem Nogą i księdzem Andrzejem Czyżem. Towarzyszyli im również księża Ryszard Janik z Jaworza, Karol Macura z Drogomyśla oraz Jan Gross z Mikołowa.

Przypisy

Świątynie luterańskie w województwie śląskim
Kościoły w diecezji cieszyńskiej
Obiekty sakralne w gminie Skoczów
Kościoły w powiecie cieszyńskim
Kościół
Budynki w Polsce oddane do użytku w 1989

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Posted on 30th April 2011 by Penny
Route Constraints
A couple of interesting (in a HS2-interesting sort of way) videos have caught my attention recently.
The first is from a BBC report about the HS2 summit organised by Bucks CC in April. In the second part of the report, they have an interview with Hugh Jaeger from Thames Valley Rail Future. He is opposed to the HS2 proposal on the grounds that is the wrong route at the wrong speed.
In particular, he says that the then Labour Government insisted that the route should go through Old Oak Common to provide a link to Heathrow: Hugh says that this doubles the amount of tunnel and adds £1 billion to the cost, even though only 1.4% of HS2 passengers would be going to Heathrow.
The second video is from a talk given by Christian Wolmar to the Chiltern Society and posted on his website.
He also talks about the route of HS2: in this case he points out that the HS2 analysis shows a huge business case for stopping at Milton Keynes. However, he says, this was the wrong answer for HS2 Ltd: they didn't want it to stop there, so came up with some sleight-of-hand reasoning to change the route.
But there is another major criticism of the way the route is planned. Many people in favour of high speed rail talk about how HS2 could be part of the continental high speed rail network. To do this it has to join up with HS1. But when HS2 Ltd were coming up with the design for the route of HS2, they did not include this link in their original plans, almost as if they had forgotten it.
So when looking at HS2 Consultation Question 4, maybe it's worth asking why HS2 Ltd constrained the route like this? – to exclude a station which many thousands of people would use every day and add in a station used by a very small percentage of passengers? And why ignore the very obvious need for a connection to HS1.
Tagged Chiltern Society, Christian Wolmar, Eurostar, Heathrow Airport, Hugh Jaeger, Old Oak Common interchange, Question four
8 comments to "Route Constraints"
ggrrllaa
30th April 2011 at 7:02 pm
As for Christian Wolmar, I have to say he goes down further in my estimation every time he opens his mouth. For a self-appointed railway expert, he seems to have very sparse knowledge of how railways actually operate.
High speed lines work best like a pipeline, when all the trains on them are running non-stop at the same speed.
The HS2 plan sees up to 18 trains per hour operating on the route between London and the West Midlands once the "Y" network is built. Let us imagine there is a station at Milton Keynes as he suggests (ignoring for the moment that the HS2 route doesn't go through MK). The times of heaviest demand from MK are in the morning and evening peaks. That is also the time of heaviest demand on the long distance trains. MK commuters would stand to benefit by about 10 minutes if a long-distance HS train were to be stopped there, whilst everybody already on the train would be penalised by about 8 minutes, because of the time lost in braking + station dwell + acceleration. So the MK commuters would need to outnumber the existing passengers by 5:4 just to break even in journey times (which means either the long distance train would need to be running half empty from its point of origin – unlikely in the peak period – or there would be a lot of uncomfortable overcrowding on the train).
Then let us consider the effect on route capacity of making an intermediate station stop in the middle of a high speed route. You don't do this instantaneously, it takes time to slow down and time to accelerate again. So instead of consuming 1 non-stop train path, your stopping train will consume at least 2 and maybe 3. On a route that is full (as 18 trains per hour would be) that would mean CANCELLING 2 or 3 long-distance trains in order to insert a single stop at Milton Keynes. Does it make sense to cancel a train from (say) Leeds in the morning peak so that MK commuter can save 10 minutes? Emphatically no.
Wolmar's failure to demonstrate any understanding of the basics of railway timetabling, capacity and operations undermine his entire credibility on the subject of HS2.
Sorry, make that "cancelling 1 or 2" as I double-counted the one path consumed by the train stopping at MK itself.
And of course, lets not forget the near 100% increase in trains that Milton Keynes does get when HS2 is built…….the vast majority of passengers from MK head to London as commuters anyway.
are you saying that at 300 kph you still need a straight track?
On a route designed for 300 kph the curvature can be sharper than one designed for 400 kph. However that is not the point being made in the interview. He was saying it would cost significantly more to build it for 400 kph than for 300 kph, and I do not believe this to be the case. Incremental cost from 200 kph to 400 kph is less than 10% according to HS2 (i.e. the difference between building a conventional railway and an HS railway).
An interesting interview with Mr Jaeger, but he seems to have got so much of it badly wrong.
The only available surface rail route out of London towards Birmingham and the North West is the former Great Western / Great Central Joint Line from Old Oak Common via Northolt and Ruislip. Alternatives such as following the M1 out of London would be massively more destructive to people and the environment than the HS2 proposals. Therefore they would have to go in tunnel instead. So it is Mr Jaeger's alternative that would massively increase tunnelling, not the HS2 proposals as he claims.
I believe Heathrow is a complete distraction to the HS2 project. It is a waste of time and money sending the line there. People want to go to / from London, not Heathrow, so why delay trains, reduce capacity, and increase costs by serving Heathrow? On that I agree with Mr Jaeger.
That said, I think Old Oak Common station has many benefits as a West London interchange station in its own right regardless of anything to do with Heathrow. Connecting with Crossrail and Great Western Main Line is far more important.
As for costs of designing and building the route for 400 kph versus 300 kph, the only difference would be in tunnels if you wanted to maintain 400 kph you would need a bigger tunnel diameter. HS2 is only 400 kph in north of the Chiltern Tunnel in any case, so I think he(Mr Jaeger) has got his facts wrong there as well.
Railfuture lol!!!!!!!!!!!!! Dear me!!
At first glance, the website appears to be advocating rail travel all over……..but once you delve into it, you realise that a lot of contributors are actually either current or ex rail employees, who are probably one of the most argumentative group of people on the web.!!
Whilst I was doing some research a few weeks ago, I came across one of their " blogs " …..a bunch of them were arguing about the definition of a " train conductor " ( I kid you not !! ). One of the posters spat his dummy out , posted a load of expletives, and promptly got himself banned from the site !!
1st May 2011 at 4:18 pm
Declaration: Employee of railway industry, living in the West Midlands.
(Apologies – this should be attached to each of my posts on this and other topics).


This is great and simple.
It may be enough to kill this blog.
I love Adrian Underhill who is of course the author of Sound Foundations one of the most helpful books for teaching teachers and students about how their mouths work and how pronunciation classes work.
This won't put him or us out of business. This will just make teaching English a hell of a lot easier when you have the internet in the classroom. Having a great resource doesn't make better teachers but it does learning to be a better English student and learning to be a better English teacher much easier.
And that's the whole point, isn't it?


BA Tours light is a booking theme developed for travel agencies and tour operators of any size. It's a first theme based on a true fast and flexible free booking engine – BA Book Everything plugin. You no longer need any shopping plugins, so this theme doesn't support Woocommerce. The booking engine allows you to create different booking rules and categories for different tours (with or without ages in prices), set seasonal prices and discounts, use promo codes (discount coupons), sell additional services with booking (extra tickets etc.), and add as many taxonomies (attributes) for filtering tours in search requests as you like. Clear landing design looks great on both desktop and mobile screens. Modern responsive design, based on Bootstrap 4 framework, touch-ready and responsive sliders, Fontawesome 5 beautiful icons, customizable theme typography, and more.


With the addition of three new members, NCT is back as an 18 member boy band in 2018. SM Entertainment clearly has a thing for numbers and the underlying meanings behind them.
Many people have been guessing for a long time that the three new members from SMROOKIES would eventually debut as part of the NCT lineup. It has finally happened and 2018 seems like it will be the year where NCT 2018 will take their stardom to the next level.
Interesting Facts: Attended Seoul School of Performing Arts. Dislikes science. Likes playing the piano.
Interesting Facts: Can play the piano. Is arachnophobic. Is a fan of EXO member Lay.
Interesting Facts: Was born in Chicago. Was an idol trainee for EXO. Can play the piano. Is friends with Kai, ChanYeol, SuHo, and SeHun of EXO.
Interesting Facts: Likes playing soccer. Is a fan of EXO's Kai.
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Interesting Facts: White is his favorite color. Is a fan of Kai of EXO.
Interesting Facts: Is a fan of IU. Is a fan of Leonardo DiCaprio.
Interesting Facts: Blue is his favorite color. Is a fan of Beyonce and Chris Brown.
Interesting Facts: Is half Thai, half Chinese. Appeared in TEN's MV "Dream In A Dream'.
Interesting Facts: Attended Beijing Contemporary Music Institute. Can play the piano.
Interesting Facts: 1 is his favorite number. Is a fan of SHINee.
Interesting Facts: Favorite food is hot pot. Is a fan of the Marvel series 'Avengers'.
Interesting Facts: Yellow is his favorite color. Can speak Japapnese and Korean.
Interesting Facts: Black and white are his favorite colors. Is a fan of actor Kim SooHyun.
Interesting Facts: 10 is his favorite number. His favorite color is black. Can speak Thai, English, and Korean. Likes to surf and snowboard.
Interesting Facts: Likes photography. Black is his favorite color. Is a fan of rapper Drake. Wanted to become a firefighter.


text:
Michael Sollbauer, né le  à Sankt Veit an der Glan, est un footballeur autrichien. Il évolue au poste de défenseur central au Rapid Vienne.

Biographie

En club

Avec l'équipe du Wolfsberger AC, il joue quatre matchs en Ligue Europa lors de la saison 2015-2016, inscrivant un but.

Le , il est prêté à Barnsley.

Le , il rejoint Dynamo Dresde.

En équipe nationale

Palmarès
 Champion d'Autriche de D2 en 2012 avec le Wolfsberger AC

Notes et références

Liens externes

Footballeur autrichien
Naissance en mai 1990
Naissance à Sankt Veit an der Glan
Joueur de l'Austria Kärnten
Joueur du Wolfsberger AC
Joueur du Barnsley FC
Joueur du Dynamo Dresde
Joueur du Rapid Vienne


Lovely, remodeled home in the desirable neighborhood of Finn Hill in Kirkland. This spacious floorplan boasts 1,940 s.f. and features 4 bedrooms, 1.75 baths on a 7,345 s.f. lot. The list of recent updates include: Outside – Freshly painted exterior, resealed driveway, newer composite deck and double pane windows. Inside – remodeled kitchen and bathrooms, new paint & doors, gleaming oak hardwoods and new carpet.
The living room features wood burning fireplace and sunny windows. Dining room with sliding glass door to back deck. Remodeled kitchen includes cherry wood cabinetry, quartz counters, extra deep soaking sink, custom tile backsplash, oak hardwood floor, designer lighting and updated fixtures – newer stainless steel appliances included! Master bedroom with oak hardwood floors and remodeled continental bath. Remodeled bath features new vanity, bath tub wrapped in beautiful tile, tile floor and updated fixtures. Two additional spacious bedrooms with oak hardwood floors. Additional storage in the hall – coat closet at the entry and linen closet adjacent to the master.
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Lush, fully fenced rear yard with large composite deck, flat & sunny lawn, beautiful garden space and large storage shed. Covered parking – 1 space extra deep carport. The home is part of Kirkland's NE 138th Street Rain Garden project. Garden absorbs rainwater runoff, filters toxins, replenishes ground and provides wildlife habitat.
Outstanding Lake Washington schools – Thoreau Elementary, Finn Hill Middle School & Juanita High.
Address: 8836 NE 138th St.
Happy Client Returns for 4th Successful Transaction and are now Headed to Retirement!
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text:
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The counting of votes for Jalgaon Municipal Corporation Election 2018 Result began at 10 AM on Friday, August 3. Jalgaon Maha Nagar Palika Elections 2018 Results are expected to be declared somewhere around noon. The voting for 75-member Jalgaon Municipal Corporation (JMC) began at 7.30 AM on Wednesday, August 1, and continued till 5.30 PM. Northern Mumbai's Jalgaon Municipal Corporation (JMC) recorded 55 per cent voter turnout. Jalgaon Municipal Corporation has a total of 3,65,072 eligible voters. The poll for the seat in Guhagar Municipal Council was won by a local alliance, which is being called the 'shahar vikas aghadi'. Neither of the Congress, the Nationalist Congress Party or Shiv Sena managed to win a single municipal out of the six, whereas the BJP registered a clean sweep in the Jamner Council, where it has won 24 out of the 24 available municipal seats.


text:
Great Britain Guidebook
Rick's picks for sights, eating, sleeping
In-depth coverage of our favorite destinations in England, Scotland, and Wales
Great self-guided neighborhood walks
Includes handy full-color foldout map
Buy your Great Britain Guidebook:
In Rick Steves Great Britain you'll find in-depth Rick-tested information on:
London and nearby
Greater London, including Greenwich and the Docklands
Glastonbury and Wells
Stonehenge and Avebury
Salisbury and nearby
Chipping Campden and nearby
Stow-on-the-Wold and nearby
Moreton-in-Marsh and nearby
Blenheim Palace
Ironbridge Gorge
Keswick and the North Lake District
Ullswater Lake Area
South Lake District
Durham and Northeast England
Conwy and Beaumaris
Snowdonia National Park (including Betws-y-Coed, Beddgelert, and Llanberis)
Ruthin and Llangollen
Stirling and nearby
Oban and nearby
Kilmartin Glen
Inner Hebrides
Isle of Mull
Isle of Iona
Isle of Staffa
Glencoe and Fort William
Inverness and nearby
Culloden Battlefield
Clava Cairns
Cawdor Castle
Self-guided walks and museum tours
Easy-to-follow maps (including color maps of Great Britain, West London, East London, Bath, and Edinburgh)
Trip planning: When to go, pre-trip checklist, festivals and holidays, recommended books and movies
Recommended three-week itinerary and city-specific advice on planning your time and avoiding lines
Transportation: Trains, buses, city transit, and driving (including route tips)
British history, culture, and cuisine, plus Scottish and Welsh language
Handy British–Yankee vocabulary list
Detailed advice on managing money, communicating, reserving rooms, handling emergencies, and other helpful hints
Also included: a 14½" x 19½" full-color foldout Planning Map that shows all of Britain and Ireland, plus key driving distances/times, and handy close-ups (showing major sights) of Edinburgh, York, the Cotswolds, Bath, and Dublin — and a large city map of London on the back. (Note: This is a slightly smaller version of Rick's stand-alone Britain & Ireland Planning Map.)
Dimensions: 8" x 4 ½" x 1 ¼"
Publication date: May 2018 (22nd edition)
Next edition arrives: April 2020
What's the difference between this Great Britain book and either Rick's England guidebook or Rick's Best of England guidebook?
Rick Steves Great Britain covers Scotland, Wales, and our top picks of English destinations. Both Rick Steves Best of England and the complete Rick Steves England exclude Scotland and Wales, and just focus on England (with one exception: Best of England does cover Edinburgh). While the Best Of book covers fewer destinations in England than the Great Britain book, the complete Rick Steves England covers some English destinations not included in the Great Britain guidebook (Canterbury, Dover and southeast England, Brighton, Portsmouth, Dartmoor, Cornwall, Oxford, Blackpool, and some of North Yorkshire).
If you're going to Scotland as well as some of the destinations that are covered in our England books but not covered in the bigger Great Britain book, you could opt to forgo the Great Britain book and instead supplement your England book with Rick Steves Scotland. (If Edinburgh is your only stop in Scotland, you could instead grab Rick Steves Snapshot: Edinburgh.)
Should I get the London guidebook (or London Pocket guide) in addition to this Britain guidebook?
If you're spending only a few days in London, Rick Steves Great Britain is all you're likely to need.
If, however, you're spending four days or more in London, the extra information in Rick Steves London can be worthwhile. It offers more in-depth sightseeing information, several additional self-guided city walks and museum/sight tours, and more specifics on shopping and nightlife, plus activities for kids. Other information is generally the same — the London chapter of Rick Steves Great Britain has nearly all the hotels, restaurants, and nitty-gritty practical advice for London that you'll find in the Rick Steves London guidebook.
It would make less sense to supplement Rick Steves Great Britain with the smaller Pocket London guide, which includes just one walk and one museum tour more than what's included in the London chapter of the Rick Steves Great Britain guidebook, but otherwise offers far more limited coverage of the city's sights, hotels, and restaurants.
Updates and Feedback
Latest updates: When we learn of critical changes to the information in this guidebook, we post them. (Of course, it's still smart to reconfirm critical transportation and sightseeing details locally.) Armed with a Rick Steves guidebook and these late-breaking updates, you're set for a great trip!
Submit feedback: Once you've used this book in Europe, we'd love to hear your feedback — good or bad — about our advice on sights, hotels, restaurants, and travel tips. We're also interested in any tips or discoveries you made while in Europe. Your comments help us improve our guidebooks for future travelers.
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25 years after the original production won Best New Play at Dublin Theatre Festival, Brothers of the Brush by Jimmy Murphy returns in an entertaining and thought-provoking new production from Verdant.
Starring Luke Griffin (The Clinic, Pure Mule), Stephen Cromwell (Red Rock), Stephen Jones (Love/Hate, Red Rock) and Gerard Byrne (Fair City), Murphy's play tells the story of three house painters, making a meagre living, overseen by an arrogant and shifty boss who is running a small business. Underneath the humour and playfulness of the relationships, a tough world is emerging, where workers are losing faith in old ways.
Set just at the dawn of the "Celtic Tiger" as the onset of apartment building kicks off in the city, Heno and Lar have returned from London now that work is readily available in Ireland. Yet, the need to survive has become far greater than anything else in their lives, and manipulation is now the new order of the day as fragile loyalties are tested and taken to limit.


text:
Trump's complaints about process are just whining
In case you had any doubt about the shallowness of Donald Trump's complaints about how he got beat in Colorado because of the Republican party, let's isolate two moments from Tuesday night's town hall on CNN, where he appeared with his children.
"You talk about the RNC," host Anderson Cooper said to Trump after the businessman had complained about the process being "stacked against him." "Reince Priebus, head of the RNC, tweeted yesterday, said, quote, the rules were set last year. Nothing mysterious. Nothing new. The rules have not changed. The rules are the same, nothing different."
"Anderson, they changed the rules a number of months ago," Trump replied. "About eight months ago," Cooper replied.
"Well, that's not very long ago!" Trump countered.
Just to draw a line through this, Trump is saying that eight months is not enough time for his team to have gotten up to speed on rule changes delineating how the nomination process works.
Let's now jump forward a bit in the conversation, to the point at which Ivanka, Trump's daughter, is responding to having failed to register as a Republican in time to vote for her father in next week's New York primary.
"I'm an independent and I've always voted based on the candidate as opposed to based on the party," Ivanka said. "We're not a family of politicians. We haven't been in politics very long. New York has one of the most onerous rules in terms of registration, and it required us to register a long time ago, almost -- close to a year ago. And we . . . we didn't do that."
The "year ago" line was used by Trump when the topic was first broached earlier this week. But the rules don't mandate registering a year in advance. The deadline was actually in October, about six months ago.
Which, per Donald Trump, is not very long ago at all -- when he wants it not to be.
Trump is right when he says that the way the process is set up to reward candidates embraced by the Republican Party. There's not much of a secret to this; there was a book called "The Party Decides," which details how the party establishment has guided eventual nominee picks.
That's different than saying the party is cheating. Priebus is right that the rules were set well in advance, and critics of Trump are right in saying he simply got outworked by Cruz's campaign in Colorado. As was the case after he did surprisingly poorly in Iowa, Trump is blaming others for his campaign's failure to do much more than hold rallies and cross their fingers.
The nomination process is less NASCAR's Sprint Cup than it is a decathlon. Trump did well at the 100 meters and the 400 meters -- the straight-up votes-lead-to-delegates contests. But in the long jump and the high jump contests, different sorts of competition, he's getting beaten. He got beat in Colorado and he's losing fights over unbound delegates in other places. Trump's argument, in essence, is that he also ran very quickly down the runway toward the long jump and that's what should count. But that's not how a decathlon works.
His excuses to Anderson Cooper make that obvious. The rules were changed practically yesterday when he needs to demonstrate that he couldn't possibly have been prepared for them. The deadline was so long ago when he needs to explain away his family's failure to get geared up for the campaign.
A better and likely more accurate explanation derives from Ivanka Trump's note that they are not a family of politicians. He's new to this, to the process of figuring out how to cobble together a delegate total and how to organize a campaign in a number of states at once. The process is clunky and complex, which is not entirely unintentional. It's a series of different events testing a different range of skills.
Trump's a good sprinter. That may not be enough to win the gold at the decathlon.
• 04/11/16: Trump's terrible night in Colorado exemplifies his campaign's Achilles' heel
• 04/04/16: In the year of Trump, a made-up news website run by an ex-convict finds success
• 03/30/16: Is Donald Trump a politician? An investigation
• 03/29/16: California could hand Donald Trump the GOP nomination --- and then doom him on election night
• 03/28/16: Let's uproot the pernicious, unproven claim that Ted Cruz attacked Donald Trump's wife


Hi y'all! I was told about this site by Italian LX. What a great guy! And since I've gotten such a warm welcome from him and 95fiveoh, I thought I'd post up and introduce myself.
My name is Shannon, I live in Denton, Texas and I love cars....mostly Mustangs, but I have respect for all in their own way. My daily is a 2000 Mustang V6 and my 10-year work in progress is an 84 Fox Body, which has the usual mods of cam, intake, exhaust, etc. It'll probably be another 10 years and still not fixed! LOL! Time and money are just two things I don't have right now, the latter being the bigger issue.
I mostly hang out on Girlscantoo.com and that's where I met Italian LX. I think this will be a great site to refer our girls that may have any technical questions that they might not be able to get answered over there.
:welcome: I was registered on the GC2 site but I mainly post here since my hubby ownz it :D Nice to see more car girls over here.
Haha! I was hesitant in saying that's where I met him! Actually we have quite a few active male members over there. Very supportive guys.
You know that gay people are the ones who surround them self with alot of girls... ahahahahahaaaaaaaa!!!!!! Damn I'm terrible.
Thanks for the welcome Edgar.
Don't give Brian such a hard time.
OH! Poor Baby! Damnit, stop giving Brian a hard time, ye all here!!
By the way Brian, today is Cheo's Birthday, big old 16yrs old, I'm going to get him a CD for his car and taking him to get his drivers Lic today, that's part of his present.
:rofl: at Brian. Can't say I'm surprised.
Ahahahahaaaaaa!! I can see the Cop giving the test and climbing thru the cage and see him strap in the 5 way harness, then a 3 point trun with the spool, that will be Priceless!!!
That would be awesome!!!! :metal: Hope he has a good birthday! He's a good kid. You gonna let him race it at the track???
You gonna let him race it at the track???