# A Poincaré disk model for taxicab hyperbolic geometry

Aaron Fish*, Dylan Helliwell

###### Abstract.

The taxicab plane is a modification of the Euclidean plane that uses an
alternative notion of distance. Similarly, the Poincaré disk is a model of
hyperbolic geometry that consists of a subset of the Euclidean plane with an
alternative notion of distance. In this paper, we merge these two variations
to create a taxicab version of the Poincaré disk in an attempt to create a
type of taxicab hyperbolic space.

*Aaron Fish was supported by the Seattle University Mathematics Early Research REU in 2017.

##  1\. Introduction

The taxicab plane is the set $\mathbb{R}^{2}$ with the taxicab distance $d$
defined as follows: for two points $x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$,

$d_{T}(x,y)=|x_{1}-y_{1}|+|x_{2}-y_{2}|.$

This distance function arises from the taxicab norm

$||v||_{T}=|v_{1}|+|v_{2}|.$

This space was introduced by Hermann Minkowski in the late 19th century as an
alternative to Euclidean geometry and has since enjoyed a fair amount of
interest. See [Kra73, Rey80, BCF+20, FHS23] for details about this space along
with various constructions and objects that are developed within it.

The Poincaré disk is a model of hyperbolic geometry consisting of the set
$D=\\{x\in\mathbb{R}^{2}:x_{1}^{2}+x_{2}^{2}<1\\}$ equipped with the
Riemannian metric

$g_{P}(v,w)=4\frac{v\cdot w}{(1-||x||_{E}^{2})^{2}}$

where $v$ and $w$ are tangent vectors at the point $x\in D$ and
$||\cdot||_{E}$ is the Euclidean norm. This gives rise to a norm on each
tangent space

$||v||_{P}=2\frac{||v||_{E}}{1-||x||_{E}^{2}}$

and from this, the length of a piecewise smooth curve $\gamma:[0,1]\rightarrow
D$ is given by

$\mathcal{L}_{P}(\gamma)=\int_{0}^{1}2\frac{||\gamma^{\prime}(t)||_{E}}{1-||\gamma(t)||_{E}^{2}}\,dt.$

See for example [And05] for more detail about hyperbolic geometry, including
the Poincaré disk as a model.

In this paper, we create a taxicab Poincaré disk by following this development
using taxicab norm wherever possible. Specifically, we define the taxicab
Poincaré disk to be the set

$D_{T}=\\{x\in\mathbb{R}^{2}:||x||_{T}<1\\}=\\{x\in\mathbb{R}^{2}:|x_{1}|+|x_{2}|<1\\}$

equipped with the norm on each tangent space given by

$||v||_{D_{T}}=\frac{||v||_{T}}{1-||x||_{T}^{2}}$

where $v$ is a vector based at $x\in D_{T}$ and the coordinate basis is used
to define the norm on the tangent space. We do not start with a Riemannian
metric because the taxicab norm does not arise from an inner product. We also
do not include a factor of 2. In the case of the standard Poincaré disk, this
factor is included to ensure that the resulting curvature is uniformly equal
to $-1$, and we will not be attempting to compute curvature in $D_{T}$. That
being said, see the comment in Section 5.2 after Theorem D.

This norm allows us to measure the length of curves that admit absolutely
continuous parameterizations. The set of such curves from $p$ to $q$ is
denoted $\Gamma(p,q)$.

We denote by $m(p,q)$ a certain minimal point associated to two given points
$p$ and $q$. This point, defined in Section 2.1 and illustrated in Figure 1,
gives rise to a particular curve $\lambda_{p,q}$, which is just the
concatenation of the segments from $p$ to $m(p,q)$ and from $m(p,q)$ to $q$.

With this, we have:

######  Theorem A.

For $p,q\in D_{T}$ and $\gamma\in\Gamma(p,q)$,
$\mathcal{L}(\gamma)\geq\mathcal{L}(\lambda_{p,q})$ with equality if and only
if $\gamma$ is doubly monotonic and passes through $m(p,q)$.

Using this, a distance function on $D_{T}$ is established:

######  Theorem B.

The distance function on $D_{T}$ arising from the length functional
$\mathcal{L}$ is

$d(p,q)=\tanh^{-1}(|p_{1}|+|p_{2}|)+\tanh^{-1}(|q_{1}|+|q_{2}|)-2\tanh^{-1}(|m_{1}|+|m_{2}|)$

where $m=m(p,q)$.

This allows us to characterize circles and determine the isometry group:

######  Theorem C.

The isometry group for $D_{T}$ is isomorphic to $D_{4}$.

Since the isometry group does not act transitively, $D_{T}$ is not
homogeneous. Despite this, hyperbolicity can still be explored, and from the
perspective of Gromov hyperbolicity, we prove

######  Theorem D.

$D_{T}$ is $\ln(3)$-hyperbolic.

This paper is organized as follows: In Section 2 we introduce some preliminary
constructions and results that support the remainder of the paper. In Section
3 we prove Theorem A, and with this, in Section 4 we prove Theorem B, and use
this to characterize circles and prove Theorem C. In Section 5 we explore the
extent to which $D_{T}$ is actually hyperbolic, proving Theorem D. Finally, we
share some concluding remarks in Section 6.

##  2\. Preliminaries

In this section, we introduce some terminology and discuss the analytical
tools used to prove our results.

###  2.1. Minimal points

We define here a certain point associated to two given points that proves to
be important in identifying length minimizing curves and establishing an
explicit formula for the distance function on $D_{T}$.

Let

$\displaystyle\ell:\mathbb{R}^{2}\longrightarrow\mathbb{R}$
$\displaystyle\ell(x,y)=\begin{cases}\mathrm{sgn}(x)\min\\{|x|,|y|\\}&\mbox{if}\
\mathrm{sgn}(x)=\mathrm{sgn}(y)\\\ 0&\mbox{if}\
\mathrm{sgn}(x)\neq\mathrm{sgn}(y),\end{cases}$

where $\mathrm{sgn}(x)$ is the sign of $x$:

$\mathrm{sgn}(x)=\begin{cases}1&\mbox{if}\ x>0\\\ -1&\mbox{if}\ x<0\\\
0&\mbox{if}\ x=0.\end{cases}$

With this, given $p,q\in D_{T}$ let

$m(p,q)=\bigl{(}\ell(p_{1},q_{1}),\ell(p_{2},q_{2})\bigr{)}.$

We call this point the minimal point associated to $p$ and $q$. For example,
note that if $p$ and $q$ lie in opposite quadrants, $m(p,q)$ is the origin
which we denote with $\theta$. See Figure 1.

$p$$q$$m(p,q)$$r$$s$$m(r,s)$$u$$v$$m(u,v)=\theta$$x=m(x,y)$$y$ Figure 1. Some
points in $D_{T}$, the minimal points associated to various pairs of points,
and the corresponding L-shaped curves.

Given two points $p$ and $q$ in the same quadrant, we say $p$ lies beyond $q$
if $m(p,q)=q$. If, moreover, $p$ and $q$ do not share a coordinate line, we
say $p$ lies strictly beyond $q$.

One way to think about the minimal point for $p$ and $q$ is that it is the
point furthest from the origin with the property that both $p$ and $q$ lie
beyond it.

###  2.2. Isometries

Isometries of $D_{T}$ will be discussed in more detail in Section 4.3. We
mention here only that rotations about the origin by integer multiples of
$\frac{\pi}{2}$ and reflections about the coordinate axes and the lines of
slope $\pm 1$ through the origin are isometries. We leave it to the reader to
confirm this. These isometries will be used regularly and without explicit
mention to simplify various arguments.

###  2.3. Curves

Unless otherwise stated, the domain for curves will be $[0,1]$ and the
codomain will be $D_{T}$. When we say “let $\gamma$ be a curve from $p$ to
$q$,” we mean both components of $\gamma$ are absolutely continuous,
$\gamma(0)=p$, and $\gamma(1)=q$. The set of such curves is denoted
$\Gamma(p,q)$. If both components of a curve $\gamma$ are monotonic, we say
$\gamma$ is doubly monotonic.

Given two curves $\gamma$ and $\phi$, with $\gamma(1)=\phi(0)$, we define the
concatenation $\gamma\cup\phi$ as follows:

$\displaystyle\gamma\cup\phi:[0,1]\longrightarrow D_{T}$
$\displaystyle\gamma\cup\phi(t)=\begin{cases}\gamma(2t)\ &\mbox{if}\ 0\leq
t\leq\frac{1}{2}\\\ \phi(2t-1)\ &\mbox{if}\ \frac{1}{2}<t\leq 1.\end{cases}$

Given two points $p$ and $q$, we define the segment from $p$ to $q$ to be

$\displaystyle\sigma_{p,q}:[0,1]\longrightarrow D_{T}$
$\displaystyle\sigma_{p,q}(t)=\bigl{(}p_{1}+t(q_{1}-p_{1}),p_{2}+t(q_{2}-p_{2})\bigr{)}.$

Given two points $p$ and $q$ and $m=m(p,q)$, we say the L-shaped curve from
$p$ to $q$ is the curve

$\displaystyle\lambda_{p,q}:[0,1]\longrightarrow D_{T}$
$\displaystyle\lambda_{p,q}=\sigma_{p,m}\cup\sigma_{m,q}.$

See Figure 1 for examples of these curves.

We define the length of a curve $\gamma:[a,b]\rightarrow D_{T}$ as follows:

$\mathcal{L}(\gamma)=\int_{a}^{b}\|\gamma^{\prime}(t)\|_{D_{T}}\,dt=\int_{a}^{b}\frac{|\gamma_{1}^{\prime}(t)|+|\gamma_{2}^{\prime}(t)|}{1-(|\gamma_{1}(t)|+|\gamma_{2}(t)|)^{2}}\,dt.$

Note that $\mathcal{L}(\gamma\cup\phi)=\mathcal{L}(\gamma)+\mathcal{L}(\phi)$.

We say two curves $\gamma:[a_{1},b_{1}]\rightarrow D_{T}$ and
$\eta:[a_{2},b_{2}]\rightarrow D_{T}$ are equivalent, and write
$\gamma\sim\eta$, if they trace the same set in $D_{T}$ and
$\mathcal{L}(\gamma)=\mathcal{L}(\eta)$. Note that if there exists an
absolutely continuous monotonic function
$f:[a_{1},b_{1}]\rightarrow[a_{2},b_{2}]$ such that $\gamma=\eta\circ f$, then
$\gamma\sim\eta$. Because of this, the specific domains and exact
parameterizations for various curves are usually not important. Moreover,
through the equivalence, concatenation is associative:
$\gamma\cup(\phi\cup\eta)\sim(\gamma\cup\phi)\cup\eta$. Also note that if $p$
lies beyond $q$ then $\lambda_{p,q}\sim\sigma_{p,q}$.

###  2.4. Absolutely Continuous Functions

As mentioned above, the curves under consideration will be parameterized using
absolutely continuous functions. We work at this level of regularity for a
number of reasons. First, absolutely continuous functions are differentiable
almost everywhere, so the formula for the length of a curve makes sense.
Second, concatenation preserves absolute continuity. Third, absolute
continuity allows for substitution (change of variables) as an integration
technique. Fourth, if a function $f$ is absolutely continuous, then so is
$|f|$. Finally, in some instances a given curve is altered using a
minimization process, and absolute continuity is preserved under this change,
as shown below.

Given $f:[a,b]\rightarrow\mathbb{R}$, we define the cumulative minimum
function $\underline{f}:[a,b]\rightarrow\mathbb{R}$ and residual minimum
function $\overline{f}:[a,b]\rightarrow\mathbb{R}$ as follows:

$\underline{f}(x)=\min_{t\in[a,x]}\\{f(t)\\}$

and

$\overline{f}(x)=\min_{t\in[x,b]}\\{f(t)\\}.$

See Figure 2 for examples of such functions.

######  Proposition 2.1.

Let $f:[a,b]\rightarrow\mathbb{R}$ be absolutely continuous. Then the
cumulative minimum function $\underline{f}$ and residual minimum function
$\overline{f}$ have the following properties:

  * •

Both $\underline{f}$ and $\overline{f}$ are also absolutely continuous;

  * •

$\underline{f}$ is monotonically decreasing, $\overline{f}$ is monotonically
increasing;

  * •

$\underline{f}(a)=f(a)$ and $\overline{f}(b)=f(b)$;

  * •

for all $x\in[a,b]$, $\underline{f}(x)\leq f(x)$ and $\overline{f}(x)\leq
f(x)$;

  * •

for all $x\in[a,b]$, $|\underline{f}^{\prime}(x)|\leq|f^{\prime}(x)|$ whenever
both are defined and $|\overline{f}^{\prime}(x)|\leq|f^{\prime}(x)|$ whenever
both are defined.

###### Proof.

The proofs of all but the first property are left to the reader.

For the first property, focussing first on $\underline{f}$, since $f$ is
absolutely continuous, for any $\varepsilon>0$. there exists a $\delta>0$ such
that for any finite collection of disjoint subsets
$(a_{i},b_{i})\subset[a,b]$, $i=1,\ldots,n$ with

$\sum_{i=1}^{n}(b_{i}-a_{i})<\delta$

it follows that

$\sum_{i=1}^{n}|f(b_{i})-f(a_{i})|<\varepsilon.$

We will show that the same $\delta$ works for $\underline{f}$. Given a
collection of disjoint sets as above, let $c_{i}\in[a_{i},b_{i}]$ be a point
such that $f(x)\geq f(c_{i})$ for all $x\in[a_{i},b_{i}]$. Such a point exists
since $f$ is continuous. Now consider two possibilities. If
$\underline{f}(a_{i})\leq f(c_{i})$, then in fact
$\underline{f}(x)=\underline{f}(a_{i})$ for all $x\in[a_{i},b_{i}]$ and so
$|\underline{f}(b_{i})-\underline{f}(a_{i})|=0$. Indicate these sets with the
new index $j=1,\ldots,n_{0}$. On the other hand, if
$\underline{f}(a_{i})>f(c_{i})$ then $\underline{f}(x)=f(c_{i})$ for all
$x\in[c_{i},b_{i}]$ and
$f(a_{i})\geq\underline{f}(a_{i})>f(c_{i})=\underline{f}(c_{i})$. In this
case, indicate these sets with the new index $k=1,\ldots,n_{1}$ so that
$n_{0}+n_{1}=n$, and subdivide $(a_{i},b_{i})$ into $(a_{i},c_{i})$ and
$(c_{i},b_{i})$. This new family of disjoint subsets will still have total
length less than $\delta$:

$\sum_{j=1}^{n_{0}}(b_{j}-a_{j})+\sum_{k=1}^{n_{1}}\bigl{[}(c_{k}-a_{k})+(b_{k}-c_{k})\bigr{]}=\sum_{i=1}^{n}(b_{i}-a_{i})<\delta$

so by the absolute continuity of $f$,

(1)
$\sum_{j=1}^{n_{0}}|f(b_{j})-f(a_{j})|+\sum_{k=1}^{n_{1}}\bigl{(}|f(c_{k})-f(a_{k})|+|f(b_{k})-f(c_{k})|\bigr{)}<\varepsilon.$

Then,

$\displaystyle\sum_{i=1}^{n}|\underline{f}(b_{i})-\underline{f}(a_{i})|$
$\displaystyle=\sum_{j=1}^{n_{0}}|\underline{f}(b_{j})-\underline{f}(a_{j})|+\sum_{k=1}^{n_{1}}|\underline{f}(b_{k})-\underline{f}(a_{k})|$
$\displaystyle=0+\sum_{k=1}^{n_{1}}|\underline{f}(c_{k})-\underline{f}(a_{k})|$
$\displaystyle=0+\sum_{k=1}^{n_{1}}\bigl{(}|\underline{f}(c_{k})-\underline{f}(a_{k})|+|\underline{f}(b_{k})-\underline{f}(c_{k})|\bigr{)}$
$\displaystyle\leq\sum_{j=1}^{n_{0}}|f(b_{j})-f(a_{j})|+\sum_{k=1}^{n_{1}}\bigl{(}|f(c_{k})-\underline{f}(a_{k})|+|f(b_{k})-f(c_{k})|\bigr{)}$
$\displaystyle\leq\sum_{j=1}^{n_{0}}|f(b_{j})-f(a_{j})|+\sum_{k=1}^{n_{1}}\bigl{(}|f(c_{k})-f(a_{k})|+|f(b_{k})-f(c_{k})|\bigr{)}$
$\displaystyle<\varepsilon$

where the last line follows from Equation (1). Therefore, $\underline{f}$ is
also absolutely continuous.

While a completely analogous proof works for $\overline{f}$, it is perhaps
simpler to note that the residual minimum function is a horizontal reflection
of the cumulative minimum function for a horizontal reflection of $f$. ∎

$a$$b$$f$$\underline{f}$$\overline{f}$ Figure 2. An absolutely continuous
function $f$ in gray, its cumulative minimum function $\underline{f}$ in solid
black, and its residual minimum function $\overline{f}$ in dashed black.

##  3\. Length minimizers

In the Poincaré disk, length minimizers are arcs of circles perpendicular to
the circle at infinity, and these curves provide a great deal of information
about the geometry in the space. We find that length minimizers in $D_{T}$ are
not as consistent as for the Poincaré disk, but they do provide the desired
insight into the geometry of the taxicab Poincaré disk. We have:

Theorem A. _For $p,q\in D_{T}$ and $\gamma\in\Gamma(p,q)$,
$\mathcal{L}(\gamma)\geq\mathcal{L}(\lambda_{p,q})$ with equality if and only
if $\gamma$ is doubly monotonic and passes through $m(p,q)$. _

In general, the length minimizer will not be unique, even up to equivalence,
but $\lambda_{p,q}$ is always a length minimizer, and there are some cases
where the minimizer must be equivalent to $\lambda_{p,q}$. Specifically, if
$p$ and $q$ lie in the same quadrant, and neither point lies strictly beyond
the other, or if $p$ and $q$ are in adjacent quadrants and share a coordinate
line, then the only doubly monotonic curves from $p$ to $q$ are those that are
equivalent to $\lambda_{p,q}$.

To prove Theorem A, we identify the following four cases:

  * •

Case 1: $p$ and $q$ lie in the same quadrant and one point lies beyond the
other;

  * •

Case 2: $p$ and $q$ lie in the same quadrant, and neither point lies beyond
the other;

  * •

Case 3: $p$ and $q$ lie in opposite quadrants;

  * •

Case 4: $p$ and $q$ lie in adjacent quadrants.

We prove these cases sequentially: the proof of Case 2 relies on Case 1 and
the proofs of Cases 3 and 4 each rely on Case 1 and Case 2. Note in particular
that Case 1 includes the situation where $p$ and $q$ lie in the same quadrant
and share a coordinate line. This special case will be important for the proof
of Case 2.

While the two points under consideration may lie on the same side of a
coordinate axis, a curve connecting them need not. In a number of places, our
analysis is simplified if we know the curve connecting two points lies on the
same side of a coordinate axis that the points do. The following lemma shows
that we can alter the curve without changing its length so that the new curve
lies on the same side of a coordinate axis as $p$ and $q$.

######  Lemma 3.1.

Let $p$ and $q$ lie on one side of a coordinate axis. Then for
$\gamma\in\Gamma(p,q)$, there exists a curve $\widehat{\gamma}\in\Gamma(p,q)$
that lies on the same side of the coordinate axis and such that
$\mathcal{L}(\widehat{\gamma})=\mathcal{L}(\gamma)$.

###### Proof.

Without loss of generality, suppose $p_{2}$ and $q_{2}$ are both nonnegative.
Then, define $\widehat{\gamma}=(\gamma_{1},|\gamma_{2}|)$. By Proposition 2.1,
this curve is an element of $\Gamma(p,q)$. The fact that
$\mathcal{L}(\widehat{\gamma})=\mathcal{L}(\gamma)$ is left to the reader. ∎

Note that if $p$ and $q$ lie in the same quadrant, we can apply Lemma 3.1
twice to produce a curve that is also restricted to that quadrant. See Figure
3 for illustrations of these results.

(a)$p$$q$$\gamma$$\widehat{\gamma}$(b)$p$$q$$\gamma$$\widehat{\gamma}$$\widehat{\widehat{\gamma}}$
Figure 3. (a) When $p$ and $q$ lie on the same side of a coordinate axis, we
can restrict a curve connecting them to the same side without changing its
length. (b) When $p$ and $q$ lie in the same quadrant, we can restrict a curve
connecting them to the same quadrant without changing its length.

###  3.1. Length minimizers when $p$ lies beyond $q$

We start with an important computation. Then, after a special case is
resolved, Theorem A, Case 1 is proved.

######  Lemma 3.2.

Let $p$ lie beyond $q$ and let $\gamma\in\Gamma(p,q)$ be doubly monotonic.
Then

$\mathcal{L}(\gamma)=\tanh^{-1}(|p_{1}|+|p_{2}|)-\tanh^{-1}(|q_{1}|+|q_{2}|).$

See Figure 4.

###### Proof.

Let
$\widetilde{\gamma}=(\widetilde{\gamma}_{1},\widetilde{\gamma}_{2})=\bigl{(}\bigl{|}\gamma_{1}\bigr{|},\bigl{|}\gamma_{2}\bigr{|}\bigr{)}$.
Note that this is a curve from $\bigl{(}|p_{1}|,|p_{2}|\bigr{)}$ to
$\bigl{(}|q_{1}|,|q_{2}|\bigr{)}$ and that
$\mathcal{L}(\widetilde{\gamma})=\mathcal{L}(\gamma)$. Then

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\widetilde{\gamma})$
$\displaystyle=\int_{0}^{1}\frac{|\widetilde{\gamma}_{1}^{\prime}(t)|+|\widetilde{\gamma}_{2}^{\prime}(t)|}{1-(|\widetilde{\gamma}_{1}(t)|+|\widetilde{\gamma}_{2}(t)|)^{2}}\,dt$
$\displaystyle=-\int_{0}^{1}\frac{\widetilde{\gamma}_{1}^{\prime}(t)+\widetilde{\gamma}_{2}^{\prime}(t)}{1-(\widetilde{\gamma}_{1}(t)+\widetilde{\gamma}_{2}(t))^{2}}\,dt$
$\displaystyle=-\int_{\widetilde{\gamma}_{1}(0)+\widetilde{\gamma}_{2}(0)}^{\widetilde{\gamma}_{1}(1)+\widetilde{\gamma}_{2}(1)}\frac{1}{1-u^{2}}\,du$
$\displaystyle=-\tanh^{-1}(u)\,\Bigr{|}^{\widetilde{\gamma}_{1}(1)+\widetilde{\gamma}_{2}(1)}_{\widetilde{\gamma}_{1}(0)+\widetilde{\gamma}_{2}(0)}$
$\displaystyle=-\tanh^{-1}(u)\,\Bigr{|}^{|q_{1}|+|q_{2}|}_{|p_{1}|+|p_{2}|}$
$\displaystyle=\tanh^{-1}(|p_{1}|+|p_{2}|)-\tanh^{-1}(|q_{1}|+|q_{2}|).$

∎

$p$$q$$\sigma_{p,q}$ Figure 4. Given $p$ and $q$ in the same quadrant, with
$p$ lying beyond $q$, doubly monotonic curves from $p$ to $q$ all have the
same length.

######  Lemma 3.3.

Let $p$ lie beyond $q$ and let $\gamma\in\Gamma(p,q)$ have the property that
$\gamma(t)$ lies beyond $q$ for all $t\in[0,1]$. Let $\psi\in\Gamma(p,q)$ be
doubly monotonic. Then $\mathcal{L}(\gamma)\geq\mathcal{L}(\psi)$ with
equality if and only if $\gamma$ is doubly monotonic.

###### Proof.

Since, by Lemma 3.2, all doubly monotonic curves have the same length $L$, it
is sufficient to show that $\mathcal{L}(\gamma)>L$ if $\gamma$ is not doubly
monotonic.

Without loss of generality, suppose $p$ and $q$ lie in the first quadrant.
Define the doubly monotonic shadow of $\gamma$ to be

$\displaystyle\underline{\gamma}:[0,1]\longrightarrow D_{T}$
$\displaystyle\underline{\gamma}(t)=\left(\underline{\gamma_{1}}(t),\underline{\gamma_{2}}(t)\right).$

Note that by Proposition 2.1, $\underline{\gamma}$ is absolutely continuous,
doubly monotonic, and starts at $p$. The fact that it ends at $q$ follows from
the fact that $\gamma(t)$ lies beyond $q$ for all $t\in[0,1]$. Hence, by Lemma
3.2, $\mathcal{L}(\underline{\gamma})=L$. By Proposition 2.1,

$\underline{\gamma_{i}}(t)\leq\gamma_{i}(t)$

for all $t\in[0,1]$, so that

$\frac{1}{1-(|\underline{\gamma_{1}}(t)|+|\underline{\gamma_{2}}(t)|)^{2}}\leq\frac{1}{1-(|\gamma_{1}(t)|+|\gamma_{2}(t)|)^{2}}.$

Also by Proposition 2.1,

$|\underline{\gamma_{i}}^{\prime}(t)|\leq|\gamma_{i}^{\prime}(t)|$

whenever both are defined. Moreover, if $\gamma$ is not doubly monotonic,
there will exist an interval where at least one of these inequalities is
strict.

These two estimates imply

$\displaystyle\mathcal{L}(\underline{\gamma})$
$\displaystyle=\int_{0}^{1}\frac{|\underline{\gamma_{1}}^{\prime}(t)|+|\underline{\gamma_{2}}^{\prime}(t)|}{1-(|\underline{\gamma_{1}}(t)|+|\underline{\gamma_{2}}(t)|)^{2}}\,dt$
$\displaystyle\leq\int_{0}^{1}\frac{|\gamma_{1}^{\prime}(t)|+|\gamma_{2}^{\prime}(t)|}{1-(|\gamma_{1}(t)|+|\gamma_{2}(t)|)^{2}}\,dt$
$\displaystyle=\mathcal{L}(\gamma),$

and if $\gamma$ is not doubly monotonic, the inequality on the second line
will be strict. ∎

See Figure 5 for an illustration of the doubly monotonic shadow of a curve
$\gamma$.

$p$$q$$\gamma$$\underline{\gamma}$ Figure 5. A curve $\gamma$ from $p$ to $q$
with the property that $\gamma(t)$ lies beyond $q$ for all $t\in[0,1]$, in
gray, and its doubly monotonic shadow $\underline{\gamma}$, in black.
Piecewise, $\underline{\gamma}$ is either equal to $\gamma$, a segment with
$\gamma$ lying beyond it, or stationary, waiting for $\gamma$ to “catch up.”
In the first case, the lengths coincide. In the second and third cases,
$\underline{\gamma}$ is strictly shorter.

We now prove Theorem A in the case when one point lies beyond the other.

###### Proof of Theorem A, Case 1.

Suppose, without loss of generality, $p$ and $q$ lie in the first quadrant
and, in light of Lemma 3.1, suppose $\gamma$ lies completely in the first
quadrant as well. Let

$r=\left(\min_{t\in[0,1]}\gamma_{1}(t),\min_{t\in[0,1]}\gamma_{2}(t)\right)$

and consider the curve $\widetilde{\gamma}=\gamma\cup\sigma_{q,r}$. By Lemma
3.3 the doubly monotonic shadow $\underline{\widetilde{\gamma}}$ is shorter
than $\widetilde{\gamma}$ and the same length as the doubly monotonic curve
$\sigma_{p,q}\cup\sigma_{q,r}$. Putting these facts together, we have

$\displaystyle\mathcal{L}(\gamma)+\mathcal{L}(\sigma_{q,r})$
$\displaystyle=\mathcal{L}(\widetilde{\gamma})$
$\displaystyle\geq\mathcal{L}(\underline{\widetilde{\gamma}})$
$\displaystyle=\mathcal{L}(\sigma_{p,q}\cup\sigma_{q,r})$
$\displaystyle=\mathcal{L}(\sigma_{p,q})+\mathcal{L}(\sigma_{q,r}).$

Cancelling the $\mathcal{L}(\sigma_{q,r})$ on both sides yields the result.
See Figure 6 for an illustration of this construction. ∎

$p$$p$$q$$q$$r$$\gamma$$\gamma$$\sigma_{q,r}$

Figure 6. When $p$ lies beyond $q$, the curve $\gamma$ connecting $p$ to $q$
on the left, and the augmented curve $\gamma\cup\sigma_{q,r}$ on the right
used in the proof of Theorem A, Case 1.

.

###  3.2. Length minimizers when $p$ and $q$ lie in the same quadrant, and
neither point lies beyond the other

Next we explore the situation where two points lie in the same quadrant, but
neither lies beyond the other. As in the previous case, our first lemma
identifies the length minimizing curves, and then later results establish that
they are in fact the minimizers.

######  Lemma 3.4.

Suppose $p$ and $q$ lie in the same quadrant and neither point lies beyond the
other, and let $m=m(p,q)$. Let $\gamma\in\Gamma(p,q)$ have the property that
for all $t\in[0,1]$, $\gamma(t)$ lies beyond $m$. Then
$\mathcal{L}(\gamma)\geq\mathcal{L}(\lambda_{p,q})$ with equality if and only
if $\gamma\sim\lambda_{p,q}$.

See Figure 7.

###### Proof.

Suppose, without loss of generality, that $p$ and $q$ lie in the first
quadrant, and that $p_{1}<q_{1}$ and $p_{2}>q_{2}$. Define the curves

$\displaystyle\phi^{1}:[0,1]\longrightarrow D_{T}$
$\displaystyle\phi^{1}(t)=\bigl{(}m_{1},\underline{\gamma_{2}}(t)\bigr{)}$
$\displaystyle\phi^{2}:[0,1]\longrightarrow D_{T}$
$\displaystyle\phi^{2}(t)=\bigl{(}\overline{\gamma_{1}}(t),m_{2}\bigr{)},$

Note that $\phi^{1}\sim\sigma_{p,m}$ and $\phi^{2}\sim\sigma_{m,q}$, so
$\phi^{1}\cup\phi^{2}\sim\lambda_{p,q}$. Then using the properties established
by Proposition 2.1,

$\displaystyle\mathcal{L}(\lambda_{p,q})$
$\displaystyle=\mathcal{L}(\phi^{1})+\mathcal{L}(\phi^{2})$
$\displaystyle=\int_{0}^{1}\frac{0+|\underline{\gamma_{2}}^{\prime}(t)|}{1-\bigl{(}|p_{1}|+|\underline{\gamma_{2}}(t)|\bigr{)}^{2}}\,dt+\int_{0}^{1}\frac{|\overline{\gamma_{1}}^{\prime}(t)|+0}{1-\bigl{(}|\overline{\gamma_{1}}(t)|+|q_{1}|\bigr{)}^{2}}\,dt$
$\displaystyle\leq\int_{0}^{1}\frac{0+|\gamma_{2}^{\prime}(t)|}{1-\bigl{(}|p_{1}|+|\gamma_{2}(t)|\bigr{)}^{2}}\,dt+\int_{0}^{1}\frac{|\gamma_{1}^{\prime}(t)|+0}{1-\bigl{(}|\gamma_{1}(t)|+|q_{1}|\bigr{)}^{2}}\,dt$
$\displaystyle\leq\int_{0}^{1}\frac{0+|\gamma_{2}^{\prime}(t)|}{1-\bigl{(}|\gamma_{1}(t)|+|\gamma_{2}(t)|\bigr{)}^{2}}+\frac{|\gamma_{1}^{\prime}(t)|+0}{1-\bigl{(}|\gamma_{1}(t)|+|\gamma_{2}(t)|\bigr{)}^{2}}\,dt$
$\displaystyle=\mathcal{L}(\gamma)$

and at least one of the inequalities will be strict unless
$\gamma\sim\lambda_{p,q}$. ∎

$p$$q$$m(p,q)$$\gamma$$\phi^{1}$$\phi^{2}$ Figure 7. Any curve $\gamma$
connecting $p$ and $q$ and lying in the gray region is at least as long as
$\lambda_{p,q}\sim\phi^{1}\cup\phi^{2}$.

With this, we are ready to handle Case 2.

###### Proof of Theorem A, Case 2.

Suppose, without loss of generality, that $p$ and $q$ lie in the first
quadrant, and that $p_{1}<q_{1}$ and $p_{2}>q_{2}$. In light of Lemma 3.1,
suppose also that $\gamma$ lies in the first quadrant. Let $t_{1}$ be the time
such that $\gamma_{1}(t_{1})=p_{1}$ and $\gamma_{1}(t)>p_{1}$ for all
$t\in(t_{1},1]$. Let $\gamma^{1}$ be the restriction of $\gamma$ to
$[0,t_{1}]$, let $\gamma^{2}$ be the restriction to $[t_{1},1]$, and let
$r=\gamma(t_{1})$.

Now consider two scenarios:

  * •

Scenario 1: $r_{2}\leq m_{2}$. Then by Theorem A, Case 1,
$\mathcal{L}(\gamma^{1})\geq\mathcal{L}(\sigma_{p,r})\geq\mathcal{L}(\sigma_{p,m})$
and the inequality is strict unless $\gamma^{1}\sim\sigma_{p,m}$. Similarly,
$\mathcal{L}(\gamma^{2})\geq\mathcal{L}(\sigma_{r,m}\cup\sigma_{m,q})\geq\mathcal{L}(\sigma_{m,q})$
and the inequality is strict unless $\gamma^{2}\sim\sigma_{m,q}$. Combining
these, we have

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\gamma^{1}\cup\gamma^{2})$
$\displaystyle\geq\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,q})$
$\displaystyle=\mathcal{L}(\lambda_{p,q}),$

and the inequality is strict unless $\gamma\sim\lambda_{p,q}$. See Figure 8
(a).

  * •

Scenario 2: $r_{2}>m_{2}$. Let $t_{2}$ be the time such that
$\gamma_{2}(t_{2})=q_{2}$ and $\gamma_{2}(t)\neq q_{2}$ for all
$t\in[t_{1},t_{2})$. Let $\gamma^{2a}$ be the restriction of $\gamma^{2}$ to
$[t_{1},t_{2}]$, let $\gamma^{2b}$ be the restriction to $[t_{2},1]$, and let
$s=\gamma(t_{2})$.

By the Intermediate Value Theorem, $\gamma^{2a}(t)$ lies beyond $m$ for all
$t\in[t_{1},t_{2}]$. Hence by Lemma 3.4,
$\mathcal{L}(\gamma^{2a})>\mathcal{L}(\lambda_{r,s})$. Moreover, by Theorem A,
Case 1, $\mathcal{L}(\gamma^{1})\geq\mathcal{L}(\sigma_{p,r})$ and
$\mathcal{L}(\gamma^{2b})\geq\mathcal{L}(\sigma_{s,q})$, and so

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\gamma^{1}\cup\gamma^{2a}\cup\gamma^{2b})$
$\displaystyle>\mathcal{L}(\sigma_{p,r}\cup\lambda_{r,s}\cup\sigma_{s,q})$
$\displaystyle=\mathcal{L}(\lambda_{p,q}).$

See Figure 8 (b).

∎

$p$$p$$q$$q$$m$$m$$\gamma^{1}$$\gamma^{2}$$r$$r$$s$$\gamma^{1}$$\gamma^{2a}$$\gamma^{2b}$(a)(b)
Figure 8. When neither point lies beyond the other, the curve $\gamma$ from
$p$ to $q$ may cross $\\{x_{1}=p_{1}\\}$ for the last time at or below $m$ as
on the left so that $\gamma=\gamma^{1}\cup\gamma^{2}$, or it may cross above
$m$ as on the right so that $\gamma=\gamma^{1}\cup\gamma^{2a}\cup\gamma^{2b}$.
These scenarios require slightly different analysis, as seen in the proof of
Theorem A, Case 2.

###  3.3. Length minimizers when $p$ and $q$ lie in opposite quadrants

Together, Cases 1 and 2 of Theorem A completely determine the length
minimizing curves connecting two points that are both in a single quadrant.
Handling points in different quadrants does not require any new technical
results.

###### Proof of Theorem A, Case 3.

First note that every point lies beyond $\theta=(0,0)$ so if $\gamma$ passes
through the origin, then by Theorem A, Case 1 it can be made shorter by
replacing with a concatenation of doubly monotonic curves from $p$ to $\theta$
and $\theta$ to $q$.

If $\gamma$ does not pass through the origin, it must cross the $x_{1}$-axis
at a point $a=(a_{1},0)$, and it must cross the $x_{2}$-axis at a point
$b=(0,b_{2})$, where both $a_{1}$ and $b_{2}$ are non-zero. Then by Theorem A,
Case 2, the part of $\gamma$ connecting $a$ to $b$ can be shortened by
replacing it with $\lambda_{a,b}$. This new shorter curve passes through the
origin and then, as above, can be shortened by replacing with a doubly
monotonic curve from $p$ through $\theta$ to $q$. See Figure 9 for an
illustration of these steps. ∎

Figure 9. The steps in the proof of Theorem A, Case 3 showing that a length
minimizing curve connecting points in opposite quadrants must be doubly
monotonic and pass through the origin.

###### Proof of Theorem A, Case 4.

Suppose, without loss of generality, that $p$ lies in the first quadrant, $q$
lies in the second quadrant, $p_{2}\leq q_{2}$, and, noting that $p_{1}=0$ is
covered by Case 1, suppose $p_{1}>0$. In light of Lemma 3.1, suppose also that
$\gamma_{2}(t)\geq 0$ for all $t\in[0,1]$. By the Intermediate Value Theorem,
there exists a time $t_{1}$ such that $\gamma_{1}(t_{1})=0$. Let
$r=\gamma(t_{1})=(0,\gamma_{2}(t_{1}))$, let $\gamma^{1}$ be the part of
$\gamma$ from $p$ to $r$, and let $\gamma^{2}$ be the part of $\gamma$ from
$r$ to $q$. Finally, let $m=m(p,q)=(0,p_{2})$, let $n=(0,q_{2})$, and consider
three scenarios:

  * •

Scenario 1: $r_{2}\leq m_{2}$. Then $p$ and $q$ both lie beyond $r$, so by
Theorem A, Case 1,

$\mathcal{L}(\gamma^{1})\geq\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,r})$

and

$\mathcal{L}(\gamma^{2})\geq\mathcal{L}(\sigma_{r,m}\cup\sigma_{m,q})$

with equality if and only if $\gamma^{1}$ and $\gamma^{2}$ are each doubly
monotonic. Combining these inequalities, we have

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\gamma^{1})+\mathcal{L}(\gamma^{2})$
$\displaystyle\geq\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,r})+\mathcal{L}(\sigma_{r,m}\cup\sigma_{m,q})$
$\displaystyle=\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,q})+2\mathcal{L}(\sigma_{m,r})$
$\displaystyle\geq\mathcal{L}(\lambda_{p,q})$

and equality is achieved if and only if, as above $\gamma^{1}$ and
$\gamma^{2}$ are both doubly monotonic, and in addition, $r=m$.

  * •

Scenario 2: $m_{2}<r_{2}\leq n_{2}$. Then neither $p$ nor $r$ lies beyond the
other so by Theorem A, Case 2,

$\mathcal{L}(\gamma^{1})\geq\mathcal{L}(\lambda_{p,r})$

with equality if and only if $\gamma^{1}$ is doubly monotonic and passes
through $m$. Also, since $q$ lies beyond $r$, by Theorem A, Case 1,

$\mathcal{L}(\gamma^{2})\geq\mathcal{L}(\sigma_{r,q})$

and equality is achieved if and only if $\gamma^{2}$ is doubly monotonic.

Combining these inequalities, and noting that
$\mathcal{L}(\sigma_{m,r}\cup\sigma_{r,q})=\mathcal{L}(\sigma_{m,q})$ we have

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\gamma^{1}\cup\gamma^{2})$
$\displaystyle\geq\mathcal{L}(\lambda_{p,r}\cup\sigma_{r,q})$
$\displaystyle=\mathcal{L}(\lambda_{p,q})$

and equality is achieved if and only if $\gamma$ both doubly monotonic and
passes through $m$.

  * •

Secenario 3: $r_{2}>n_{2}$. Then neither $p$ nor $r$ lies beyond the other and
neither $r$ nor $q$ lies beyond the other, so by Theorem A, Case 2,

$\mathcal{L}(\gamma^{1})\geq\mathcal{L}(\lambda_{p,r})$

and

$\mathcal{L}(\gamma^{2})\geq\mathcal{L}(\lambda_{r,q}).$

Also note that, since $r_{2}>n_{2}$,

$\displaystyle\mathcal{L}(\lambda_{p,r})$
$\displaystyle=\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,r})$
$\displaystyle>\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,n})$

and

$\displaystyle\mathcal{L}(\lambda_{r,q})$
$\displaystyle=\mathcal{L}(\sigma_{r,n}\cup\sigma_{n,q})$
$\displaystyle>\mathcal{L}(\sigma_{n,q}).$

Combining these inequalities, we have

$\displaystyle\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\gamma^{1}\cup\gamma^{2})$
$\displaystyle\geq\mathcal{L}(\lambda_{p,r}\cup\lambda_{r,q})$
$\displaystyle>\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,n}\cup\sigma_{n,q})$
$\displaystyle=\mathcal{L}(\sigma_{p,m}\cup\sigma_{m,q})$
$\displaystyle=\mathcal{L}(\lambda_{p,q}).$

See Figure 10 for an illustration of these three scenarios.

∎

$p$$q$$m$$r$$p$$q$$m$$r$$p$$q$$m$$r$$n$ Figure 10. The three scenarios in the
proof of Theorem A, Case 4 showing that a length minimizing curve connecting
points in adjacent quadrants must be the concatenation of a coordinate segment
and a doubly monotonic curve.

##  4\. Distance, Circles, Isometries

In this section, we use the length minimizers to define a distance function on
the taxicab Poincaré disk, use this distance to characterize circles, and
determine the isometries of $D_{T}$.

###  4.1. Distance function for $D_{T}$

In the usual Poincaré disk $D$, the curves that minimize the length functional
$\mathcal{L}_{P}$ are arcs of circles that are perpendicular to $\partial D$.
These curves are used to define the distance between points $p,q\in D$,
resulting in

$d_{P}(p,q)=\cosh^{-1}\left(1+2\frac{||p-q||_{E}^{2}}{(1-||p||_{E}^{2})(1-||q||_{E}^{2})}\right).$

See [And05] for the development of length and distance in the Poincaré disk.

Similarly, in $D_{T}$, the curves that minimize the length functional
$\mathcal{L}$ are given by Theorem A and these curves are used to define a
distance function on $D_{T}$:

Theorem B. _The distance function on $D_{T}$ arising from the length
functional $\mathcal{L}$ is_

(2)
$d(p,q)=\tanh^{-1}(|p_{1}|+|p_{2}|)+\tanh^{-1}(|q_{1}|+|q_{2}|)-2\tanh^{-1}(|m_{1}|+|m_{2}|)$

_where $m=m(p,q)$. _

###### Proof.

Since $\lambda_{p,q}=\sigma_{p,m}\cup\sigma_{m,q}\in\Gamma(p,q)$ is always a
minimizing curve from $p$ to $q$, $\sigma_{p,m}$ and $\sigma_{m,q}$ each lie
in a single quadrant, and $p$ and $q$ both lie beyond $m$, by Lemma 3.2,

$\displaystyle d(p,q)$
$\displaystyle=\min_{\gamma\in\Gamma(p,q)}\mathcal{L}(\gamma)$
$\displaystyle=\mathcal{L}(\lambda_{p,q})$
$\displaystyle=\mathcal{L}(\sigma_{p,m})+\mathcal{L}(\sigma_{m,q})$
$\displaystyle=\tanh^{-1}(|p_{1}|+|p_{2}|)+\tanh^{-1}(|q_{1}|+|q_{2}|)$
$\displaystyle\qquad\qquad\qquad-2\tanh^{-1}(|m_{1}|+|m_{2}|).$

∎

Note that $d(p,\theta)=\tanh^{-1}(|p_{1}|+|p_{2}|)$, so Equation (2) can be
rewritten as follows:

(3) $d(p,q)=d(p,\theta)+d(q,\theta)-2d(m,\theta).$

###  4.2. Circles

With a distance function established for $D_{T}$, our next goal is to
characterize circles in this space. Circles in the usual Poincaré disk are
Euclidean circles, although the hyperbolic center does not coincide with the
Euclidean center. We find here that, unfortunately, most circles in $D_{T}$
are not taxicab circles, although there are a number of interesting facts
about them that help us better understand the space.

Let $C_{r}(p)$ be the circle of radius $r$ centered at $p$. Also, let $r_{p}$
be the radius of the circle centered at $(0,0)$ that contains $p$. Note that,
using this notation, Equation (3) becomes

(4) $d(p,q)=r_{p}+r_{q}-2r_{m}.$

######  Lemma 4.1.

The circles centered at the origin are taxicab circles.

###### Proof.

The formula for the circle $C_{r}(\theta)$ is
$d(p,\theta)=\tanh^{-1}(|p_{1}|+|p_{2}|)=r$ which can be rewritten

$|p_{1}|+|p_{2}|=\tanh(r),$

which in turn is the formula of the taxicab circle of radius $\tanh(r)$
centered at $\theta$. ∎

Given the point $p$, let $c^{1}(p)=(p_{1},0)$ and $c^{2}(p)=(0,p_{2})$ The
coordinate lines through $p$ and the coordinate lines through the origin
intersect at $c^{i}=c^{i}(p)$ and subdivide $D_{T}$ into nine open regions,
some of which will be empty if $p$ lies on a coordinate axis. These regions
comprise four quadrants $Q_{\theta}$, $Q_{p}$, $Q_{c^{1}}$, $Q_{c^{2}}$, a
central rectangle $R$, and four strips $S_{\theta,c^{1}}$, $S_{\theta,c^{2}}$,
$S_{p,c^{1}}$, $S_{p,c^{2}}$. See Figure 11.

Note that each of the quadrants intersects exactly one edge of the circle at
infinity. We say a segment in one of these quadrants is admissible if its
endpoints lie on the coordinate lines defining the quadrant and it is parallel
to the corresponding edge at infinity. The edge at infinity corresponding to
$Q_{p}$ can also be associated to the rectangle $R$ and we say a segment in
$R$ is admissible if its endpoints lie on the boundary of $R$ and it is
parallel to the corresponding edge at infinity.

$p$$c^{1}$$c^{2}$$\theta$$Q_{p}$$Q_{\theta}$$Q_{c^{1}}$$Q_{c^{2}}$$R$$S_{p,c^{1}}$$S_{p,c^{2}}$$S_{\theta,c^{1}}$$S_{\theta,c^{2}}$
Figure 11. The points $c^{i}=c^{i}(p)$, $i=1,2$, and various regions
associated to $p$.

######  Theorem 4.2.

If a circle $C_{r}(p)$ intersects a quadrant $Q$ or the rectangle $R$, it does
so along an admissible segment. If it intersects a strip $S$, it does so along
the following curves:

  * •

in $S_{p,c^{1}}$,

$|x_{1}|=\frac{\kappa[1+(|p_{1}|+|x_{2}|)^{2}]+2(|p_{1}|+|x_{2}|)}{1+(|p_{1}|+|x_{2}|)^{2}+2\kappa(|p_{1}|+|x_{2}|)}-|x_{2}|;$

  * •

in $S_{p,c^{2}}$,

$|x_{2}|=\frac{\kappa[1+(|x_{1}|+|p_{2}|)^{2}]+2(|x_{1}|+|p_{2}|)}{1+(|x_{1}|+|p_{2}|)^{2}+2\kappa(|x_{1}|+|p_{2}|)}-|x_{1}|;$

  * •

in $S_{0,c^{1}}$,

$|x_{2}|=\frac{\kappa(1+|x_{1}|^{2})+2|x_{1}|}{1+|x_{1}|^{2}+2\kappa|x_{1}|}-|x_{1}|;$

  * •

in $S_{0,c^{2}}$,

$|x_{1}|=\frac{\kappa(1+|x_{2}|^{2})+2|x_{2}|}{1+|x_{2}|^{2}+2\kappa|x_{2}|}-|x_{1}|;$

where $\kappa=\tanh(r-r_{p})$.

The circle $C_{r}(p)$ always intersects $Q_{p}$, $S_{p,c^{1}}$, and
$S_{p,c^{2}}$. It intersects $R$ if and only if $r<r_{p}$ and it intersects
$Q_{\theta}$ if and only if $r>r_{p}$. It intersects $Q_{c^{1}}$ and
$S_{\theta,c^{1}}$ if and only if $r>d(p,c^{1})$ and it intersects $Q_{c^{2}}$
and $S_{\theta,c^{2}}$ if and only if $r>d(p,c^{2})$.

###### Proof.

Let $x\in C_{r}(p)\cap Q$. Note that the minimal point $m=m(p,x)$ is the
nearest vertex of $R$, which is independent of $x$. Hence the equation for the
circle $d(p,x)=r$ can be rewritten as:

$r_{x}=r+2r_{m}-r_{p}$

and since $m$ is independent of $x$, this implies that all such points lie on
some fixed circle centered at 0. Similarly, if $x$ lies in $R$, then $m=x$ and
the equation for the circle can be rewritten as

$r_{x}=r_{p}-r$

and this implies that all such points lie on some fixed circle centered at
$\theta$. In either case, by Lemma 4.1 this intersection will be an admissible
segment.

For points in the strips, $m$ depends on both $p$ and $x$, so the formulas for
the circle in these regions do not reduce to that of a circle centered at the
origin and the resulting curve is not a straight line.

The equation for $x\in C_{r}(p)$ is

$r_{x}+r_{p}-2r_{m}=r$

which can be rewritten as

$|x_{1}|+|x_{2}|=\tanh(r-r_{p}+2r_{m})=\frac{\tanh(r-r_{p})[1+(|m_{1}|+|m_{2}|)^{2}]+2(|m_{1}|+|m_{2}|)}{1+(|m_{1}|+|m_{2}|)^{2}+2\tanh(r-r_{p})(|m_{1}|+|m_{2}|)}.$

Note that $\tanh(r-r_{p})$ is constant for a given circle and within each
strip, $m$ can be determined explicitly in terms of the coordinates of $x$ and
$p$. Substituting and rearranging yields the desired formulas.

Finally, the regions $Q_{p}$, $S_{p,c^{1}}$, and $S_{p,c^{2}}$ each contain
points of all distances from $p$. On the other hand, the points in $R$ are all
less than $r_{p}$ away from $p$ while the points in $Q_{\theta}$ are all
greater than $r_{p}$ away from $p$. Similarly, for all points $q\in
Q_{c^{1}}\cup S_{\theta,c^{1}}$, $d(p,q)>d(p,c^{1})$ and for all points $q\in
Q_{c^{2}}\cup S_{\theta,c^{2}}$, $d(p,q)>d(p,c^{2})$. ∎

While Theorem 4.2 provides explicit formulas for circles in $D_{T}$, it only
provides partial information about what such circles actually look like.
Figures 12, 13, and 14 illustrate various circles and here, we present some
observations that help to fill out our understanding. These observations are
about their “extrinsic” geometry as viewed from the perspective of the
Euclidean plane. These results are somewhat removed from our main goal and the
full proofs are a bit technical, so, while we indicate the main elements of
the proofs, the details are left to the interested reader.

First, Theorem 4.2 shows that in each strip, one coordinate variable can be
determined as a function of the other, and a computation shows that the second
derivative has constant sign resulting in the fact that these circles are all
boundaries of extrinsically convex sets. On the other hand, many circles are
not boundaries of intrinsically geodesically convex sets (for any reasonable
definition). For example, if a circle $C$ has nonempty intersection with $R$,
then for $p$ and $q$ inside $C$ and near $C\cap R$, $\lambda_{p,q}$ will often
lie partially outside $C$.

Second, circles lying in a single quadrant posses extrinsic reflective
symmetry relative to the line through the circle’s center and perpendicular to
the corresponding edge at infinity. Somewhat more generally, even if a circle
does not lie in a single quadrant, those points that do lie in a single
quadrant enjoy the indicated symmetry. See Figure 12.

$\gamma$ Figure 12. Three circles centered at the same point. The inner two
circles lie completely in the first quadrant and have reflective symmetry
about the line $\gamma$. The outer circle does not lie in a single quadrant
and does not have reflective symmetry, but if a point and its reflection both
lie in the first quadrant, then that point lies in the circle if and only if
its reflection lies in the circle.

Third, suppose $C_{r}(p)$ and $C_{r}(q)$ each lie in a single quadrant and
that $r_{p}=r_{q}$. Then these two circles have the same extrinsic shape. See
Figure 13. This can be checked algebraically, but thinking in terms of the
radii involved leads to more insight. Rewrite the equation for the circle
$C_{r}(p)$ as

$(r_{x}-r_{m})+(r_{p}-r_{m})=r$

where $m=m(x,p)$. Since this equation is written completely in terms of
distances from the origin, the locations of the points in $C_{r}(p)$ are not
important as long as they posses the correct radii.

Figure 13. Two circles with the same radii and with centers the same distance
from the origin. $p$$q$$s$ Figure 14. The circle $C_{r}(p)$ in gray, with
circles $C_{r^{\prime}}(q)$ and $C_{r^{\prime\prime}}(s)$ in black, where
$r^{\prime}=r-d(p,q)$ and $r^{\prime\prime}=r-d(p,s)$.

Finally, let $p$ and $q$ lie in the same quadrant, and share a coordinate line
$\ell$, with $p$ lying beyond $q$. Let $\tilde{\ell}$ be the coordinate line
through $q$ and perpendicular to $\ell$ and let $H$ be the half-plane with
boundary $\tilde{\ell}$ and not containing $p$. Let $r>d(p,q)$ and let
$r^{\prime}=r-d(p,q)$. Then $C_{r^{\prime}}(q)\cap H=C_{r}(p)\cap H$. See
Figure 14. To justify this, suppose, without loss of generality, that $p$ lies
in the first quadrant and that $0\leq q_{1}\leq p_{1}$ and $q_{2}=p_{2}$.
Then, for any point $s$ such that $s_{1}\leq q_{1}$,
$m^{\prime}=m(s,q)=m(s,p)$. Note that $r^{\prime}=r-r_{p}+r_{q}$, so

$r_{s}+r_{q}-2r_{m^{\prime}}=r^{\prime}$

becomes

$\displaystyle r_{s}-2r_{m}$ $\displaystyle=r-r_{p}+r_{q}-r_{q}$
$\displaystyle=r-r_{p}$

which implies that $s$ solves the equation for $C_{r}(q)$ if and only if it
solves the equation for $C_{r}(p)$.

###  4.3. Isometries

The isometry group for $(\mathbb{R}^{2},d)$ is isomorphic to
$\mathbb{R}^{2}\rtimes D_{4}$ where the transformations associated to
$\mathbb{R}^{2}$ are translations and the transformations associated to
$D_{4}$ are reflections about coordinate lines and lines with slope $\pm 1$,
and rotations by integer multiples of $\frac{\pi}{2}$ [Sch84, KO98].
Meanwhile, the isometry group for the Poincaré Disk is $SU(1,1)$ which
comprises the set of Möbius transformations in the complex plane that map the
unit disk to itself, the reflections across lines through the origin, and
their compositions [Nee23].

Theorem C. _The isometry group for $D_{T}$ is isomorphic to $D_{4}$. _

To prove that this is all there is, we proceed in four steps, showing
successively that an isometry must be progressively more constrained. In all
four steps, we use the fact that any isometry must map circles of a given
radius bijectively to other circles of the same radius.

###### Proof.

The fact that rotations by multiples of $\frac{\pi}{2}$ about the origin and
reflections across the coordinate axes and the lines $x_{2}=\pm x_{1}$ are
isometries is left to the reader.

In the other direction, let $\Psi:D_{T}\rightarrow D_{T}$ be an isometry.
Suppose first that $\Psi$ does not fix the origin. Let $\Psi(p)=\theta$ and
suppose, without loss of generality, that $p_{1}>0$ and $p_{2}\geq 0$. Choose
$r$ small enough that $C_{r}(p)$ lies to the right of the $x_{2}$-axis. Then a
portion $\widetilde{\gamma}$ of the circle lies in $S_{p,c^{2}}$, with
endpoints $\widetilde{q}$ to the left of $p$ and $\widetilde{s}$ above $p$.
Note that $\widetilde{s}$ lies beyond $\widetilde{q}$ so by the mean value
theorem, there exists a point, and hence an arc on $\widetilde{\gamma}$ where
its slope is positive. Mapping this part of $\widetilde{\gamma}$ by $\Psi$ may
not result in an edge in a single quadrant, so if necessary, restrict the arc
further to produce a curve $\gamma$ from $q$ to $s$ such that $\Psi(\gamma)$
lies in a single quadrant. See Figure 15.

$\longrightarrow$$\Psi$$p$$C_{r}(p)$$\gamma$$q$$s$$\theta=\Psi(p)$$\Psi(\gamma)$$\Psi(q)$$\Psi(s)$
Figure 15. Arcs of circles in bijective correspondence under a proposed
isometry $\Psi$ in the first part of the proof of Theorem C.

Since $\gamma$ is doubly monotonic, by Theorem A, Case 1,
$d(q,s)=\mathcal{L}(\gamma)$, and since $\Psi$ is an isometry,
$\mathcal{L}(\gamma)=\mathcal{L}(\Psi(\gamma))$. But neither $\Psi(q)$ nor
$\Psi(s)$ can lie beyond the other, so by Theorem A, Case 2,
$\mathcal{L}\bigl{(}\Psi(\gamma)\bigr{)}>d\bigl{(}\Psi(q),\Psi(s)\bigr{)}$,
arriving at a contradiction.

Next, suppose $\Psi$ does not map a ray of a coordinate axis to another ray of
a coordinate axis. Let $p=(p_{1},0)$ with $p_{1}>0$ and, without loss of
generality, suppose $\Psi(p)=q$ where $q_{1}\geq q_{2}>0$. Let $r>0$ be small
enough that $C_{r}(q)$ lies only in the first quadrant. Let $s$ be the unique
point on the $x_{1}$-axis that lies in $C_{r}(p)$ and lies between $\theta$
and $p$. Since $\Psi$ fixes the origin, there is a bijective correspondence
between $\\{x\in C_{r}(p):r_{x}=r_{s}\\}$ and $\\{x\in
C_{r}(q):r_{x}=r_{s}\\}$, but the first set is the singleton $\\{s\\}$ and the
second set is a segment. See Figure 16.

$\longrightarrow$$\Psi$$C_{r}(p)$$C_{r}(q)$$\gamma$$s$ Figure 16. The point
$s$ and the segment $\gamma$ would need to be in bijective correspondence
under a proposed isometry $\Psi$ in the second part of the proof of Theorem C.

The next step is to show that a coordinate axis must map into a single
coordinate axis; that is, a coordinate axis cannot be bent at the origin.
Since the origin is fixed, by continuity, each coordinate ray based at the
origin must be preserved. Suppose, without loss of generality, that $\Psi$
maps the positive $x_{1}$-axis to itself and maps the positive $x_{2}$-axis to
the negative $x_{1}$-axis.

Let $p$ lie on the positive $x_{1}$-axis and let $q$ lie on the positive
$x_{2}$-axis with $r_{p}=r_{q}$. Let $s=\Psi(q)$ so that $s$ lies on the
negative $x_{1}$-axis and $r_{s}=r_{p}$. Also, $\Psi(\sigma_{p,q})$ is a curve
that crosses the $x_{2}$-axis, and in fact it must be an arc of a circle
centered at the origin since $\sigma_{p,q}$ is an arc on a circle centered at
the origin. This implies there is a time $t\in(0,1)$ such that
$\Psi(\sigma_{p,q})_{1}(t)=0$, but this is a contradiction by the previous
step since there is no such time when $(\sigma_{p,q})_{1}(t)=0$. See Figure
17.

$\longrightarrow$$\Psi$$p$$q$$\gamma$$\Psi(p)=p$$\Psi(q)=s$ Figure 17. Arcs of
circles in bijective correspondence under a proposed isometry $\Psi$ in the
third part of the proof of Theorem C.

Finally since $\Psi$ sends each coordinate axis to a coordinate axis, suppose,
without loss of generality, that $\Psi$ is the identity map on the axes. Let
$p$ be any point not on the axes and, without loss of generality, suppose $p$
is in the first quadrant. Let $q$ and $s$ lie on the positive $x_{1}$-axis and
positive $x_{2}$-axis respectively such that $r_{s}=r_{q}=r_{p}$. Note that
$\Psi(q)=q$ and $\Psi(s)=s$. Let $r=d(q,p)$ and let $\tilde{r}=d(s,p)$. The
fact that $p$ is the unique point that lies on the three circles $C_{r}(q)$,
$C_{\tilde{r}}(s)$, and $C_{r_{p}}(\theta)$ is left as an exercise for the
reader. Since $\Psi$ maps these three circles bijectively to themselves, $p$
must also be preserved. Therefore, $\Psi$ must be the identity. See Figure 18.

Since, throughout this process, the only transformations used to constrain
$\Psi$ are elements of $D_{4}$, $\Psi$ must be an element of $D_{4}$.

$p$$q$$s$ Figure 18. Since $\Psi$ preserves $q$, $s$, and the origin, it must
also preserve the circles centered at these points and containing $p$, which
in turn implies that $\Psi$ must preserve $p$.

∎

##  5\. Hyperbolicity

It is somewhat disappointing that the isometry group for $D_{T}$ is so small.
The fact that this space is not homogeneous indicates that it is perhaps not
the best candidate as a representative for taxicab hyperbolic geometry. Here,
we explore the extent to which $D_{T}$ is nonetheless hyperbolic. Since we are
not working with curvature, we consider alternatives that are more accessible.

###  5.1. Playfair’s Axiom

In Euclidean geometry, Playfair’s Axiom states that given a line $\ell$ and a
point $p$ not on $\ell$, there exists at most one line passing through $p$ and
not intersecting $\ell$. In $D_{T}$, we find that, using a suitable definition
of “line,” given a line $\ell$ and a point $p$, there exist infinitely many
lines passing through $p$ and not intersecting $\ell$.

In the taxicab plane $(\mathbb{R}^{2},d)$, since length minimizing curves are
not unique, Playfair’s axiom fails unless an additional constraint is imposed.
Since Euclidean segments are minimizers, we can define a line in
$(\mathbb{R}^{2},d)$ to be a Euclidean line, and then Playfair’s axiom holds.

Similarly, since $\lambda_{p,q}$ is always a viable minimizer in $D_{T}$, we
could define lines to be extensions of $\lambda$’s such that no new corners or
bends are introduced. Then two points $p$ and $q$ define a unique line in the
sense that

$\Lambda:(D_{T}\times D_{T})\backslash\Delta\rightarrow\\{\mbox{lines}\\}$

is a well defined function. Here, $\Delta=\\{(p,p)\\}\subset D_{T}\times
D_{T}$. However $s\in\Lambda(p,q)$ does not imply $p\in\Lambda(s,q)$ so we do
not have uniqueness in the Euclidean sense. See Figure 19 for an illustration
of the set of lines through a point $p$ using this definition.

$p$ Figure 19. The set of lines through a point $p$. The thick segments and
rays are used by multiple lines.

With this definition, we have the following:

######  Theorem 5.1.

Given a line $\ell$ and a point $p$ not on $\ell$, there are infinitely many
lines through $p$ that do not intersect $\ell$.

Since $D_{T}$ lacks homogeneity, a complete proof would require considering a
number of cases. See Figure 20 for some examples. The reader is encouraged to
consider other scenarios as well.

$\ell$$p$$\ell$$p$$\ell$$p$$\ell$$p$ Figure 20. Various scenarios of a line
$\ell$, a point $p$ not on $\ell$, and two lines passing through $p$ and not
intersecting $\ell$. The reader is encouraged to find other lines to justify
the fact that there are infinitely many such lines through $p$.

###  5.2. Gromov hyperbolicity

For a somewhat more sophisticated perspective, we consider Gromov
hyperbolicity. For a given pair of points $x$ and $y$, and base point $z$, let
$G(x,y;z)$ be the Gromov product

$G(x,y;z)=\frac{1}{2}\left[d(x,z)+d(y,z)-d(x,y)\right].$

Introduced by Gromov (see for example [Gro87]), a metric space is said to be
$\delta$-hyperbolic if there exists $\delta\geq 0$ such that for all
$x,y,z,w$, the following inequality holds:

(5) $G(x,y;w)\geq\min\left\\{G(x,z;w),G(y,z;w)\right\\}-\delta.$

The usual hyperbolic plane, with sectional curvature -1, is 2-hyperbolic.
Meanwhile, trees are 0-hyperbolic. The Gromov hyperbolicity of $D_{T}$ is
determined here, giving us a new concrete example in this area.

Theorem D. _ $D_{T}$ is $\ln(3)$-hyperbolic. _

As mentioned in the introduction, the norm being used for $D_{T}$ differs from
that of the Poincar’e disk by a factor of 2. If we were to rescale the norm
for $D_{T}$ to coincide, this would also double the $\delta$ in Theorem D, so
in this sense, $D_{T}$ is slightly less hyperbolic than usual hyperbolic
space.

###### Proof.

In [CDP90], it is shown that if Inequality (5) holds for a single base point
$w_{0}$ and $\delta$, then it holds for all $w$ and at most $2\delta$. With
this in mind, let $w_{0}=\theta$ and note that by Equation (4)

$\displaystyle G(p,q;\theta)$
$\displaystyle=\frac{1}{2}\left[d(p,\theta)+d(q,\theta)-d(p,q)\right]$
$\displaystyle=\frac{1}{2}\left[r_{p}+r_{q}-\left(r_{p}+r_{q}-2r_{m(p,q)}\right)\right]$
$\displaystyle=r_{m(p,q)}.$

Our goal is to show that with $\delta=\frac{1}{2}\ln(3)$, for any three points
$p$, $q$, and $s$

$G(p,q;\theta)\geq\min\left\\{G(p,s;\theta),G(q,s;\theta)\right\\}-\delta$

which, using the formula above, can be written

(6) $r_{m(p,q)}\geq\min\left\\{r_{m(p,s)},r_{m(q,s)}\right\\}-\delta.$

In general, if $m(p,q)$ lies beyond $m(p,s)$ or $m(q,s)$, then $\delta=0$ will
suffice. This configuration occurs often and as the following four cases are
considered, depending on the relative location of the points $p$, $q$, and
$s$, it is left to the reader to verify some of the simpler such scenarios.

  1. Case 1:

All three points are in the same quadrant. In this case, the only scenario
where $\delta=0$ is not sufficient is when neither $p$ nor $q$ lies beyond the
other, and $s$ lies beyond $m(p,q)$. Suppose, without loss of generality, that
all three points are in the first quadrant with $p_{1}<q_{1}$ and
$p_{2}>q_{2}$. Then $s_{1}>p_{1}$ so $m(p,s)=(p_{1},\min\\{p_{2},s_{2}\\})$
and $(p_{1},s_{2})$ always lies beyond this point. Similarly, $s_{2}>q_{2}$ so
$m(q,s)=(\min\\{q_{1},s_{1}\\},q_{2})$ and $(s_{1},q_{2})$ always lies beyond
this point. Hence, since $s_{1}+s_{2}<1$

$\displaystyle\min\\{r_{m(p,s)},r_{m(q,s)}\\}$
$\displaystyle\leq\min\\{r_{(p_{1},s_{2})},r_{(s_{1},q_{2})}\\}$
$\displaystyle=\min\\{\tanh^{-1}(p_{1}+s_{2}),\tanh^{-1}(s_{1}+q_{2})\\}$
$\displaystyle\leq\min\\{\tanh^{-1}(p_{1}+1-s_{1}),\tanh^{-1}(s_{1}+q_{2})\\}.$

The right hand side is maximized when $p_{1}+1-s_{1}=s_{1}+q_{2}$ so that
$s_{1}=\frac{1+p_{1}-q_{1}}{2}$. Therefore, to satisfy Inequality (6), since
$m(p,q)=(p_{1},q_{2})$, we need

$\tanh^{-1}(p_{1}+q_{2})\geq\tanh^{-1}\left(\frac{1+p_{1}+q_{2}}{2}\right)-\delta.$

Solving for $\delta$, we get

$\displaystyle\delta$
$\displaystyle\geq\tanh^{-1}\left(\frac{1+p_{1}+q_{2}}{2}\right)-\tanh^{-1}(p_{1}+q_{2})$
$\displaystyle=\tanh^{-1}\left(\frac{1-(p_{1}+q_{2})}{2-(1+p_{1}+q_{2})(p_{1}+q_{2})}\right).$

Since $0\leq p_{1}+q_{2}<1$, this is maximized when $p_{1}+q_{2}=0$ and
$\delta=\tanh^{-1}\left(\frac{1}{2}\right)$. See Figure 21.

$p=m(p,s)$$m(p,q)$$q$$s$$(s_{1},q_{2})=m(q,s)$$(p_{1},s_{2})$ Figure 21. The
points of interest in the proof of $\delta$-hyperbolicity when all points are
in one quadrant. In this example, $p$ and $s$ are positioned such that
$m(p,s)$ is just $p$, and $(p_{1},s_{2})$ lies beyond this point. Meanwhile,
neither $q$ nor $s$ lie beyond the other and $m(q,s)=(s_{1},q_{2})$.

  2. Case 2:

Two points are in one quadrant and the third point is in an adjacent quadrant.
If $p$ and $q$ are in the same quadrant, then $m(p,q)$ will always lie beyond
either $m(p,s)$ or $m(q,s)$, so $\delta=0$ will suffice. If $p$ and $q$ are in
adjacent quadrants, then suppose, without loss of generality, that $s$ and $q$
lie in the first quadrant and $p$ lies in the second quadrant. Then
$m(p,q)=(0,\min\\{p_{2},q_{2}\\})$ and $m(p,s)=(0,\min\\{p_{2},s_{2}\\})$.
From this, if $p_{2}\leq q_{2}$ or $s_{2}\leq q_{2}$, Inequality (6) is
satisfied with $\delta=0$. Otherwise, $m(q,s)=(\min\\{q_{1},s_{1}\\},q_{2})$
and the point $(s_{1},q_{2})$ lies beyond $m(q,s)$, and similarly the point
$(0,s_{2})$ lies beyond $m(p,s)=(0,\min\\{p_{2},s_{2}\\})$ which in turn lies
beyond $m(p,q)=(0,q_{2})$. Hence, again using the fact that $s_{1}+s_{2}<1$,

$\displaystyle\min\\{r_{m(p,s)},r_{m(q,s)}\\}$
$\displaystyle\leq\min\\{\tanh^{-1}(s_{2}),\tanh^{-1}(s_{1}+q_{2})\\}$
$\displaystyle\leq\min\\{\tanh^{-1}(1-s_{1}),\tanh^{-1}(s_{1}+q_{2})\\}$

which is maximized when $1-s_{1}=s_{1}+q_{2}$ so that
$s_{1}=\frac{1-q_{1}}{2}$. Therefore, Inequality (6) is satisfied if

$\tanh^{-1}(q_{2})\geq\tanh^{-1}\left(\frac{1+q_{2}}{2}\right)-\delta.$

Solving for $\delta$, we find the largest $\delta$ needed is

$\delta=\tanh^{-1}\left(\frac{1+q_{2}}{2}\right)-\tanh^{-1}(q_{2})=\tanh^{-1}\left(\frac{1}{2+q_{2}}\right)$

which is its largest when $q_{2}=0$ and
$\delta=\tanh^{-1}\left(\frac{1}{2}\right)$. See Figure 22.

$p$$m(p,s)$$m(p,q)$$q$$s$$(s_{1},q_{2})=m(q,s)$$(0,s_{2})$ Figure 22. The
points of interest in the proof of $\delta$-hyperbolicity when two points are
in one quadrant and one point is in an adjacent quadrant. In this example, $p$
and $s$ are positioned such that $m(p,s)=(0,p_{2})$, and $(0,s_{2})$ lies
beyond this point. Meanwhile, neither $q$ nor $s$ lie beyond the other and
$m(q,s)=(s_{1},q_{2})$.

  3. Case 3:

Two points are in one quadrant and the third point is in the opposite
quadrant. Then two of the three minimal points $m(p,q)$, $m(p,s)$, and
$m(q,s)$ are at the origin, so Inequality (6) holds with $\delta=0$.

  4. Case 4:

Each point is in its own quadrant. If $p$ and $q$ are in adjacent quadrants,
then $m(p,s)$ or $m(q,s)$ lies at the origin and Inequality (6) holds with
$\delta=0$. If $p$ and $q$ are in opposite quadrants, then suppose, without
loss of generality that $s$ lies in the first quadrant, $p$ lies in the
second, and $q$ lies in the fourth. Then $m(p,q)=\theta$,
$m(p,s)=(0,\min\\{p_{2},s_{2}\\})$, and $m(q,s)=(\min\\{q_{1},s_{1}\\},0)$.
Hence $r_{m(p,s)}\leq\tanh^{-1}(s_{2})$ and $r_{m(q,s)}\leq\tanh^{-1}(s_{1})$,
so

$\min\left\\{r_{m(p,s)},r_{m(q,s)}\right\\}\leq\min\left\\{\tanh^{-1}(s_{1}),\tanh^{-1}(s_{2})\right\\}$

and since $s_{1}+s_{2}<1$, the right hand side is bounded by
$\tanh^{-1}\left(\frac{1}{2}\right)$. See Figure 23.

$p$$m(p,s)$$m(p,q)$$q$$s$$(s_{1},q_{2})$$=m(q,s)$$(0,s_{2})$ Figure 23. The
points of interest in the proof of $\delta$-hyperbolicity when each point is
in its own quadrant. In this example, $p$ and $s$ are positioned such that
$m(p,s)=(0,p_{2})$, and $(0,s_{2})$ lies beyond this point. Meanwhile,
$m(q,s)=(s_{1},0)$.

The work up to now shows that for all $p$, $q$, and $s$ in $D_{T}$,

$G(p,q;\theta)\geq\min\\{G(p,s;\theta),G(q,s;\theta)\\}-\tanh^{-1}\left(\frac{1}{2}\right).$

Hence, from [CDP90], for all $p$, $q$, $w$ and $s$ in $D_{T}$,

$G(p,q;w)\geq\min\\{G(p,s;w),G(q,s;w)\\}-2\tanh^{-1}\left(\frac{1}{2}\right).$

We show here that $\delta=2\tanh^{-1}\left(\frac{1}{2}\right)=\ln(3)$ is
necessary. Let $p$, $q$, $w$, and $s$ each lie in their own quadrant with $p$
and $q$ in opposite quadrants and $w$ and $s$ in opposite quadrants. Then
$m(p,q)=m(w,s)=\theta$ so

$\displaystyle G(p,q;w)$
$\displaystyle=\frac{1}{2}\left[d(p,w)+d(q,w)-d(p,q)\right]$
$\displaystyle=\frac{1}{2}\left[r_{p}+r_{w}-2r_{m(p,w)}+r_{q}+r_{w}-2r_{m(q,w)}-(r_{p}+r_{q}-2r_{m(p,q)})\right]$
$\displaystyle=r_{w}-r_{m(p,w)}-r_{m(q,w)},$

and similarly

$\displaystyle G(p,s;w)$
$\displaystyle=\frac{1}{2}\left[d(p,w)+d(s,w)-d(p,s)\right]$
$\displaystyle=\frac{1}{2}\left[r_{p}+r_{w}-2r_{m(p,w)}+r_{s}+r_{w}-2r_{m(s,w)}-(r_{p}+r_{s}-2r_{m(p,s)})\right]$
$\displaystyle=r_{w}-r_{m(p,w)}+r_{m(p,s)},$

and

$G(q,s;w)=r_{w}-r_{m(q,w)}+r_{m(q,s)}.$

Hence the desired inequality reduces to

$r_{w}-r_{m(p,w)}-r_{m(q,w)}\geq\min\\{r_{w}-r_{m(p,w)}+r_{m(p,s)},r_{w}-r_{m(q,w)}+r_{m(q,s)}\\}-\delta$

which, canceling the $r_{w}$ from each side, reduces to

$-r_{m(p,w)}-r_{m(q,w)}\geq\min\\{-r_{m(p,w)}+r_{m(p,s)},-r_{m(q,w)}+r_{m(q,s)}\\}-\delta.$

Now let $s=(t,t)$, $p=(-t,t)$, $q=(t,-t)$, and $w=(-t,-t)$. Then the various
minimal points here are $(\pm t,0)$ and $(0,\pm t)$ and their corresponding
radii are all $\tanh^{-1}(t)$. Hence the inequality reduces to

$-2\tanh^{-1}(t)\geq 0-\delta.$

Since $t$ can approach $\frac{1}{2}$, we need
$\delta=2\tanh^{-1}\left(\frac{1}{2}\right)$. See Figure 24.

$p$$m(p,s)$$m(p,w)$$q$$s$$w$$m(q,s)$$m(p,s)$ Figure 24. The points of interest
in the proof of $\delta$-hyperbolicity showing that a factor of two is
necessary when $w\neq 0$.

∎

##  6\. Final thoughts and next steps

We finish with some ideas for future exploration related to our taxicab
Poincaré disk and taxicab hyperbolic geometry more generally.

The Poincaré disk is one of many models for hyperbolic space. It would be
interesting to look at other models from the taxicab perspective presented
here. The upper half-plane model would be the next natural choice. We expect
that identifying length minimizing curves in this setting would require
analysis comparable to what was done for $D_{T}$. Interestingly, the isometry
group for an upper half-plane model would be infinite, allowing at least for
horizontal translations.

In $D_{T}$, the result shown in Figure 13 is reflective of the fact that,
while the global isometry group is small, there are local isometries involving
translations along lines of slope $\pm 1$. These local isometries may be
related to the translational isometries expected in an upper half plane model,
thus providing a connection between the two models somewhat analogous to that
of the upper half plane model and Poincaré disk model for usual hyperbolic
space.

Alternatively, developing a taxicab hyperbolic space could be attempted from
the perspective of the isometry group. The isometry group for hyperbolic space
is isomorphic to the group of Möbius transformations in the complex plane that
map the unit circle to itself, $SU(1,1)$. It would be interesting to search
for a group similar to $SU(1,1)$ that captures the desired properties for a
taxicab space. It should act transitively on some set, the elements that fix a
point should be isomorphic to $D_{4}$, and the resulting space should be
hyperbolic in some sense.

Finally, the structure of the Poincaré disk is closely related to inversion.
Some initial attempts by the authors to develop a comparable transformation in
the taxicab setting were unsuccessful, but more work is warranted in this
direction.

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