# A characterization of the polyhedral metrics on triangulated surfaces

Tianqi Wu

###### Abstract.

Given a triangulated surface, a polyhedral metric could be constructed by
gluing Euclidean triangles edge-to-edge. We carefully describe the
construction and prove that such a polyhedral metric is the only intrinsic
metric on the glued surface that preserves the lengths of the curves on the
Euclidean triangles. We also discuss the edge length coordinates of the
Teichmüller space of the polyhedral metrics on a marked surface.

##  1\. Introduction

Polyhedral metrics and (geodesic) triangulations on surfaces have been widely
studied and relied on in mathematical theories and computational geometry.
Given a triangulation of a surface, a polyhedral metric could be constructed
by gluing Euclidean triangles edge-to-edge. Such a construction seems obvious.
In this paper, we carefully describe the construction in details, and
rigorously justify fundamental properties of such a constructed polyhedral
metrics. Specifically, we defined the “glued metric” based on the idea of
quotient metric, and proved that arc lengths are preserved after gluing. We
also discuss the edge length coordinates of the Teichmüller space of
polyhedral metrics, briefly explaning the connection with the Teichmüller
theory of hyperbolic surfaces with cusps and the theory of discrete conformal
geometry.

In Section 2, we give a detailed specific construction of $\Delta$-complexes
by gluing Euclidean triangles. Then in Section 3, we construct polyhedral
metrics on triangulated surfaces and justify that the constructed one is
indeed a metric. In Section 4, we proved the arc lengths preserving property.
In section 5, we define the Teichmüller space $T_{PL}$ of polyhedral metrics,
and introduced a $\mathbb{R}^{n}$-parameterization with the help of the
Teichmüller space of hyperbolic surfaces with cusps. We also discuss the edge
lengths coordianates and the connection with discrete conformal geoemetry.

##  2\. Triangulations of a closed surface

Denote

(a) $D$ as a fixed closed regular triangle in $\mathbb{R}^{2}$ with the edge
length 1,

(b) $V_{0}$ as the set of the three vertices of $D$,

(c) $E_{0}$ as the set of the three closed edges of $D$,

(d) $D^{o}$ as the interior of $D$ in $\mathbb{R}^{2}$, and

(e) $e^{o}=e-V_{0}$ as an open edge for all $e\in E_{0}$.

######  Definition 2.1.

A _triangulation_ $T$ of a connected closed surface $S$ is represented by a
finite index set $I=\\{1,2,...,n\\}$ for some $n\in\mathbb{Z}_{>0}$ and
continuous maps $\sigma_{i}:D\rightarrow S$ for $i\in I$ such that

(a) $\sigma_{i}|_{e^{o}}$ is injective for all $i\in I$ and $e\in E_{0}$,

(b) $\sigma_{i}|_{D^{o}}$ is injective for all $i\in I$,

(c) $S$ is a disjoint union of $\cup_{i}\sigma_{i}(V_{0})$,
$\sigma_{i}(D^{o})$ for $i\in I$, and elements in

$\\{\sigma_{i}(e^{o}):e\in E_{0},~{}i\in I\\},$

and

(d) if $\sigma_{i_{1}}(e_{1}^{o})=\sigma_{i_{2}}(e_{2}^{o})$ for some
$i_{1},i_{2}\in I$ and $e_{1},e_{2}\in E_{0}$, then there exists a linear
homeomorphism (i.e., an isometry) $h:e_{1}\rightarrow e_{2}$ such that

$\sigma_{i_{2}}(h(x))=\sigma_{i_{1}}(x)$

for all $x\in e_{1}$.

We denote such a triangulation as $T=(\sigma_{i})_{i\in I}$.

######  Remark 2.2.

Notice that by part (c) of Definition 2,1, for all
$(e_{1},i_{1}),(e_{2},i_{2})\in E_{0}\times I$,
$\sigma_{i_{1}}(e_{1}^{o})=\sigma_{i_{2}}(e_{2}^{o})$ if
$\sigma_{i_{1}}(e_{1}^{o})\cap\sigma_{i_{2}}(e_{2}^{o})\neq\emptyset$.

######  Remark 2.3.

Such a triangulation $T$ is naturally a $\Delta$-complex, but not necessary a
simplicial complex. Two edges of the same triangle could possibly be glued
together, and two triangles may be glued along more than one pairs of edges.
As a consequence, such a triangulation $T$ is naturally a CW-complex but may
not be regular.

Given a triangulation $T=(\sigma_{i})_{i\in I}$ of a closed surface $S$,

(a) the maps $\sigma_{i}$ for $i\in I$ are called the _characteristic maps_ of
$T$,

(b) $\sigma_{i}(v)$ is called a _vertex_ of $T$ if $v\in V_{0}$ and $i\in I$,

(c) $V(T)=\cup_{i\in I}\sigma_{i}(V_{0})$ denotes the set of vertices of $T$,

(d) $\sigma_{i}(e)$ is called an _edge_ of $T$ if $e\in E_{0}$ and $i\in I$,

(e) $E=E(T)$ denotes the set of edges of $T$,

##  3\. Construction of polyhedral metrics

In this section we will construct polyhedral metrics on $S$, with a given
triangulation and prescribed edge lengths.

Given $l\in\mathbb{R}^{E_{0}}_{>0}$ satisfying the triangle inequality, there
exists a linear map $f:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2}$ such that
$f(D)$ is a Euclidean triangle where the edge length of $f(e)$ is $l_{e}$ for
all $e\in E_{0}$. Let $d[l]$ be the induced metric on $D$, defined by

$d[l](x,y)=|f(x)-f(y)|_{2}.$

Notice that the metric $d[l]$ is independent on the choice of $f$. Suppose
$T=(\sigma_{i})_{i\in I}$ is a triangulation of $S$, denote
$\mathbb{R}^{E(T)}_{\Delta}$ as a real-valued function on $E(T)$ satisfying
the triangle inequalities. Specifically,

$\mathbb{R}^{E(T)}_{\Delta}=\\{l\in\mathbb{R}_{>0}^{E(T)}:l_{\sigma_{i}(e_{1})}+l_{\sigma_{i}(e_{2})}>l_{\sigma_{i}(e_{3})}\text{
if $\\{e_{1},e_{2},e_{3}\\}=E_{0}$ and $i\in I$}\\}.$

Suppose $l\in\mathbb{R}^{E(T)}_{\Delta}$ and
$l^{(i)}\in\mathbb{R}^{E_{0}}_{>0}$ denotes the restricted edge length on
$\sigma_{i}(D)$, i.e.,

$l^{(i)}_{e}=l_{\sigma_{i}(e)}$

for all $e\in E_{0}$.

A potential metric on $\sigma_{i}(D)$ could be

$d_{i}(x,y)=\min\\{d[l^{(i)}](x^{\prime},y^{\prime}):x^{\prime}\in\sigma_{i}^{-1}(x),~{}y^{\prime}\in\sigma_{i}^{-1}(y)\\}$

for all $x,y\in\sigma_{i}(D)$. Notice that in general $\sigma_{i}$ is not
injective on $D$ and $\sigma_{i}^{-1}(x)$ may contain more than one point.
Such $d_{i}$ actually may not be a metric on $\sigma_{i}(D)$. This is fine and
we can still construct a metric $d[l]$ on $S$ as

$d[l](x,y)=\inf\bigg{\\{}\sum_{j=1}^{k}d_{i_{j}}(x_{j-1},x_{j}):x_{j-1},x_{j}\in\sigma_{i_{j}}(D)\text{
for some $i_{j}\in I$},~{}x=x_{0},~{}y=x_{k}\bigg{\\}}$

for all $x,y\in S$. To justify that $d[l]$ is indeed a metric on $S$, we need
to prove that for all $x,y,z\in S$,

(a) $d[l](x,y)<\infty$ ,

(b) $d[l](x,y)\geq 0$,

(c) $d[l](x,y)=0$ only if $x=y$,

(d)

$d[l](x,y)=d[l](y,x),$

and

(e)

$d[l](x,z)\leq d[l](x,z)+d[l](y,z).$

Here part (a) is indeed Proposition 3.1 below. Part (b) is obvious from the
definition. Part (c) is an immediate consequence of Proposition 3.4 below. It
is straightforward to verify part (d) and (e) from the definition of $d[l]$.

Furthermore, $d[l]$ is compatible with the quotient topology on $S$ (see
Proposition 3.5).

######  Proposition 3.1.

For all $x,y\in S$, $d[l](x,y)<\infty$, i.e., there exists

$x_{0},x_{1},...,x_{k}\in S$

such that $x_{0}=x$, $x_{k}=y$, and for all $j=1,...,k$,
$x_{j-1},x_{j}\in\sigma_{i_{j}}(D)$ for some $i_{j}\in I$.

###### Proof.

Given $x\in S$, denote

$C(x)=\\{y\in S:d[l](x,y)<\infty\\}.$

It suffices to show that $C(x)$ is the whole space $S$. Since $S$ is connected
and $x\in C(x)$, we only need to show that $C(x)$ is both open and closed. It
is easy to see that $\sigma_{i}(D)\subset C(x)$ if $\sigma_{i}(D)\cap
C(x)\neq\emptyset$. Let

$J=\bigg{\\{}i\in I:~{}\sigma_{i}(D)\cap C(x)\neq\emptyset\bigg{\\}}.$

and then

$C(x)=\bigcup_{i\in J}\sigma_{i}(D)$

is compact and closed. For all $i\in I-J$,

$C(x)\cap\sigma_{i}(D)=\emptyset.$

So

$\bigcup_{i\in I-J}\sigma_{i}(D)$

is the complement of $C(x)$ in $S$ and is compact. So $C(x)$ is open.

∎

For all $x\in D$, denote

$K(x)=\cup\\{e\in E_{0}:x\notin e\\}$

as the union of the edges not containing $x$. In particular,

(a) if $x\in D^{o}$, then $K$ is the union of all the three edges of $D$,

(b) if $x\in e^{o}$ for some $e\in E_{0}$, then $K$ is the union of the other
two edges of $D$, and

(c) if $x\in V_{0}$, then $K$ is the edge opposite to $x$ in $D$.
For all $x\in D$ and $i\in I$, denote $r_{i}(x)>0$ as the distance between
$K(x)$ and $x$ in $(D,d[l^{(i)}])$. For all $x\in S$, let

$r(x)=\min\bigg{\\{}r_{i}(x^{\prime}):i\in
I,~{}x^{\prime}\in\sigma_{i}^{-1}(x)\bigg{\\}}.$

######  Lemma 3.2.

If $x,z\in\sigma_{i}(D)$, $d_{i}(x,z)<r(x)$, and
$z^{\prime}\in\sigma_{j}^{-1}(z)$, then there exists
$x^{\prime}\in\sigma_{j}^{-1}(x)$ such that

$d[l^{(j)}](x^{\prime},z^{\prime})=d_{i}(x,z).$

###### Proof.

Assume

$d_{i}(x,z)=d[l^{(i)}](x^{\prime\prime},z^{\prime\prime})<r(x)$

for some $x^{\prime\prime}\in\sigma_{i}^{-1}(x)$,
$z^{\prime\prime}\in\sigma_{i}^{-1}(z)$. The conclusion is trivial if $x=z$ or
$(z^{\prime\prime},i)=(z^{\prime},j)$. So we may assume that $x\neq z$ and
$(z^{\prime\prime},i)\neq(z^{\prime},j)$. By the definition of $r(x)$,

$z^{\prime\prime}\notin K(x^{\prime\prime})\cup\\{x^{\prime\prime}\\}\supset
V_{0}.$

Then by $(z^{\prime},i)\neq(z^{\prime\prime},j)$ and part (c) of the
definition of a triangulation, there exists $e,e^{\prime}\in E_{0}$ such that

$z^{\prime\prime}\in e^{o},$ $z^{\prime}\in e^{\prime},$

and

$\sigma_{i}(e)=\sigma_{j}(e^{\prime}).$

Furthermore, $e\not\subset K(x^{\prime\prime})$ and thus $x^{\prime\prime}\in
e$.

By part (d) of the definition of a triangulation, there exists a linear
homeomorphism $h:e\rightarrow e^{\prime}$ such that
$\sigma_{j}(h(w))=\sigma_{i}(w)$ for all $w\in e$. Such $h$ is an isometry
from $(e,d[l^{(i)}])$ to $(e^{\prime},d[l^{(j)}])$ since

(a) $l_{\sigma_{i}(e)}=l_{\sigma_{j}(e^{\prime})}$,

(b) the metrics $d[l^{(i)}],d[l^{(j)}]$ are induced from linear maps to
Euclidean triangles in $\mathbb{R}^{2}$, and

(c) the map $h$ is linear.
Then $h(z^{\prime\prime})=z^{\prime}$,
$h(x^{\prime\prime})\in\sigma_{j}^{-1}(x)$, and

$d[l^{(j)}](h(x^{\prime\prime}),z^{\prime})=d[l^{(j)}](h(x^{\prime\prime}),h(z^{\prime\prime}))=d[l^{(i)}](x^{\prime\prime},z^{\prime\prime})=d_{i}(x,z).$

∎

######  Corollary 3.3.

If $x,z\in\sigma_{i}(D)$, $d_{i}(x,z)<r(x)$, and $z\in\sigma_{j}(D)$, then

$x\in\sigma_{j}(D)$

and

$d_{i}(x,z)=d_{j}(x,z).$

###### Proof.

Let $z^{\prime}\in\sigma_{j}^{-1}(z)$. By Lemma 3.2, there exists
$x^{\prime}\in\sigma_{j}^{-1}(x)$ such that

$d_{j}(x,z)\leq d[l^{(j)}](x^{\prime},z^{\prime})=d_{i}(x,z)<r.$

By the same reason

$d_{i}(x,z)\leq d_{j}(x,z)$

and thus

$d_{i}(x,z)=d_{j}(x,z).$

∎

######  Proposition 3.4.

Given $x,y\in S$,

$d[l](x,y)\geq r(x)$

or

$d[l](x,y)=d_{i}(x,y)$

for some $i\in I$ with $x,y\in\sigma_{i}(D)$.

###### Proof.

Let us prove by contradiction and assume that this is not true. By the
definition of $d[l]$ there exists $y\in S$ and $x_{0},...,x_{k}$ such that

(a) for all $j\in\\{1,...,k\\}$, $x_{j-1},x_{j}\in D_{i_{j}}$ for some
$i_{j}\in I$,

(b) $x_{0}=x$ and $x_{k}=y$,

(c)

$\sum_{j=1}^{k}d_{i_{j}}(x_{j-1},x_{j})<r,$

and

(d)

$\sum_{j=1}^{k}d_{i_{j}}(x_{j-1},x_{j})<d_{i}(x,y)$

if $i\in I$ and $x,y\in\sigma_{i}(D)$.

Now let $m\geq 1$ be the maximum integer in $\\{1,...,k\\}$ such that for all
$j=1,...,m$,

(a) $x\in\sigma_{i_{j}}(D)$, and

(b) $d_{i_{j}}(x,x_{j-1})<r(x).$
If $m=k$, then $x,y\in\sigma_{i_{m}}(D)$ and

$d_{i_{m}}(x,x_{m})=d_{i_{m}}(x,y).$

If $k^{\prime}<k$, then

$d_{i_{m}}(x,x_{m})\geq r(x),$

since otherwise by Corollary 3.3 we have that

$x\in\sigma_{i_{m+1}}(D)$

and

$d_{i_{m+1}}(x,x_{m})<r(x)$

contradicting with the maximality of $m$.

So it suffices to show that

$\sum_{j=1}^{m}d_{i_{j}}(x_{j-1},x_{j})\geq d_{i_{m}}(x,x_{m}).$

Since

$d_{i_{1}}(x_{0},x_{1})=d_{i_{1}}(x,x_{1}),$

it suffices to show that for all $j=2,...,m$,

$d_{i_{j}}(x_{j-1},x_{j})\geq d_{i_{j}}(x,x_{j})-d_{i_{j-1}}(x,x_{j-1}),$

which by Corollary 3.3 is equivalent to

$d_{i_{j}}(x,x_{j-1})+d_{i_{j}}(x_{j-1},x_{j})\geq d_{i_{j}}(x,x_{j}).$

Here we prove a general fact that

$d_{i}(x,z)+d_{i}(z,w)\geq d_{i}(x,w)$

if $x,z,w\in\sigma_{i}(D)$ and $d_{i}(x,z)<r(x)$.

Suppose $x^{\prime}\in\sigma_{i}^{-1}(x)$,
$z^{\prime},z^{\prime\prime}\in\sigma_{i}^{-1}(x)$,
$w^{\prime\prime}\in\sigma_{i}^{-1}(w)$ are such that

$d_{i}(x,z)=d[l^{(i)}](x^{\prime},z^{\prime})<r(x)$

and

$d_{i}(z,w)=d[l^{(i)}](z^{\prime\prime},w^{\prime\prime}).$

Then by Lemma 3.2, there exists $x^{\prime\prime}\in\sigma^{-1}_{i}(x)$ such
that

$d[l^{(i)}](x^{\prime\prime},z^{\prime\prime})=d[l^{(i)}](x^{\prime},z^{\prime})=d_{i}(x,z).$

So

$d_{i}(x,z)+d_{i}(z,w)=d[l^{(i)}](x^{\prime\prime},z^{\prime\prime})+d[l^{(i)}](z^{\prime\prime},w^{\prime\prime})\geq
d[l^{(i)}](x^{\prime\prime},w^{\prime\prime})\geq d_{i}(x,w).$

∎

######  Proposition 3.5.

Given a triangulation $T$ of $S$ and $l\in\mathbb{R}^{E(T)}_{\Delta}$, the
metric $d[l]$ is compatible with the original topology on $S$.

###### Proof.

Let $\Omega$ be the original topology on $S$. Let $\Omega_{d}$ be the topology
on $S$ induced by $d[l]$. Specifically

$\Omega_{d}=\\{U\subset S:\text{ for all $x\in U$ there exists $\epsilon>0$
such that $y\in U$ if $d[l](x,y)<\epsilon$}\\}.$

We will first prove that $(S,d[l])$ induces a weaker topology, i.e.,
$\Omega_{d}\subset\Omega$. It suffices to prove that for all $x\in S$ and
$r>0$,

$B(x,r):=\\{y\in S:d[l](x,y)<r\\}$

contains some open set $U\ni x$ in $(S,\Omega)$. Fix $x\in S$ and let

$K_{i}=\bigg{\\{}y^{\prime}\in D:d[l^{(i)}](x^{\prime},y^{\prime})\geq r\text{
for all $x^{\prime}\in\sigma_{i}^{-1}(x)$}\bigg{\\}}$

for all $i\in I$. Notice that $K_{i}=D$ if $x\notin\sigma_{i}(D)$. Then
$K_{i}$ is compact and

$\sigma_{i}(D)-\sigma_{i}(K_{i})\subset\sigma_{i}\bigg{(}\big{\\{}y^{\prime}\in
D:d[l^{(i)}](x^{\prime},y^{\prime})<r\text{ for some
$x^{\prime}\in\sigma_{i}^{-1}(x)$}\big{\\}}\bigg{)}\subset B(x,r)$

for all $i\in I$. It is easy to see that $x$ is contained in the open set

$S-\bigcup_{i\in I}\sigma_{i}(K_{i})\subset\bigcup_{i\in
I}(\sigma_{i}(D)-\sigma_{i}(K_{i}))\subset B(x,r).$

So we proved that $\Omega_{d}$ is a weaker topology than $\Omega$.

Then $\Omega=\Omega_{d}$ since the identity map

$Id_{S}:(S,\Omega)\rightarrow(S,\Omega_{d})$

is a homeomorphism, as a continuous bijection from a compact space to a
Hausdorff space.

Here we give a detailed direct argument. We only need to show that any closed
subset in $(S,\Omega)$ is compact in $(S,d[l])$ and thus is also closed in
$(S,d[l])$. Suppose $Y$ is a closed subset in the compact space $(S,\Omega)$,
and

$Y=\cup_{\lambda\in\Lambda}U_{\lambda}$

is an open cover of $Y$ in $(S,\Omega_{d})$. Then $Y$ is compact in
$(S,\Omega)$ and $Y=\cup_{\lambda\in\Lambda}U_{\lambda}$ is also an open cover
in the stronger topology $\Omega$. Then there exists a finite subcover of
$Y=\cup_{\lambda\in\Lambda}U_{\lambda}$. So $Y$ is compact in $\Omega_{d}$. ∎

##  4\. Lengths-preserving property

It is straightforward to verify that $d[l]$ constructed in the previous
section is an _intrinsic metric_ , meaning that the distance between two
points $x,y$ is the infimum of the lengths of the curves connecting $x,y$.
Specifically,

$d[l](x,y)=\inf\bigg{\\{}L(\gamma):\gamma\in
C([0,1],S),~{}\gamma(0)=x,~{}\gamma(1)=y\bigg{\\}}$

where $C([0,1],S)$ is the set of the continuous maps from $[0,1]$ to $S$ and
$L(\gamma)=L(\gamma,d)\in[0,\infty]$ measures the length of $\gamma$ as

$L(\gamma)=\sup\bigg{\\{}\sum_{i=1}^{k}d[l](\gamma(t_{i-1}),\gamma(t_{i})):0=t_{0}<t_{1}<...<t_{k}=1\bigg{\\}}.$

Since we view $d[l]$ as the metric on a surface glued from Euclidean triangles
$(D,d[l^{(i)}])$, $d[l]$ should preserve the lengths of the curves on
$(D,d[l^{(i)}])$. Theorem 4.1 says that $d[l]$ is indeed the only intrinsic
metric on $S$ preserving these lengths.

######  Theorem 4.1.

(a) For all $i\in I$ and all continuous maps $\gamma:[0,1]\rightarrow D$,

$L(\gamma,d[l^{(i)}])=L(\sigma_{i}\circ\gamma,d[l]).$

(b) Suppose $d$ is an intrinsic metric on $S$. If for all $i\in I$ and all
continuous maps $\gamma:[0,1]\rightarrow D$,

$L(\gamma,d[l^{(i)}])=L(\sigma_{i}\circ\gamma,d),$

then $d=d[l]$.

###### Proof.

(a) It is straightforward to verify that

$L(\gamma,d[l^{(i)}])\geq L(\sigma_{i}\circ\gamma,d[l]),$

since

$d[l^{(i)}](x,y)\geq d_{i}(\sigma_{i}(x),\sigma_{i}(y))\geq
d[l](\sigma_{i}(x),\sigma_{i}(y))$

for all $x,y\in D$.

Now we show

$L(\gamma,d[l^{(i)}])\leq L(\sigma_{i}\circ\gamma,d[l]).$

It suffices to show that for all $0=t_{0}<t_{1}<...<t_{k}=1$,

$\sum_{j=1}^{k}d[l^{(i)}](\gamma(t_{j-1}),\gamma(t_{j}))\leq
L(\sigma_{i}\circ\gamma,d[l])=\sum_{j=1}^{k}L(\sigma_{i}\circ\gamma|_{[t_{j-1},t_{j}]},d[l]).$

Comparing the terms on both sides and by a scaling it suffices to show that
for all $i\in I$ and $\alpha\in C([0,1],D)$,

$d[l^{(i)}](\alpha(0),\alpha(1))\leq L(\sigma_{i}\circ\alpha,d[l]).$

For all $x\in S$, let

$r^{\prime}(x)=\min\bigg{\\{}\frac{1}{2}d[l^{(i)}](x^{\prime},x^{\prime\prime}):i\in
I,~{}x^{\prime},x^{\prime\prime}\in\sigma^{-1}_{i}(x),~{}x^{\prime}\neq
x^{\prime\prime}\bigg{\\}},$

and

$r^{\prime\prime}(x)=\min\\{r(x),r^{\prime}(x)\\}>0.$

Then if $x^{\prime},z^{\prime}\in D$ and

$d[l^{(i)}](x^{\prime},z^{\prime})<r^{\prime\prime}(x),$

by Lemma 3.2 there exists $x^{\prime\prime}\in D$ with
$\sigma_{i}(x^{\prime\prime})=\sigma_{i}(x)$ such that

$d[l^{(i)}](x^{\prime\prime},z^{\prime})=d_{i}(\sigma_{i}(x^{\prime}),\sigma_{i}(z^{\prime}))<r^{\prime\prime}(x)$

and then by the definition of $r^{\prime}(x)$ we have
$x^{\prime}=x^{\prime\prime}$ and

$d[l^{(i)}](x^{\prime},z^{\prime})=d_{i}(\sigma_{i}(x^{\prime}),\sigma_{i}(z^{\prime})).$

We will find $0=t_{0}<...<t_{k}=1$ such that

$d[l^{(i)}](\alpha(t_{j-1}),\alpha(t_{j}))<\max\\{r^{\prime\prime}(\sigma_{i}\circ\alpha(t_{j-1})),r^{\prime\prime}(\sigma_{i}\circ\alpha(t_{j}))\\}$

for all $j=1,...,k$, and then by the above discussion, Proposition 4.4, and
Corollary 4.3,

$d[l^{(i)}](\alpha(t_{j-1}),\alpha(t_{j}))=d_{i}(\sigma_{i}(\alpha(t_{j-1})),\sigma_{i}(\alpha(t_{j})))=d[l](\sigma_{i}\circ\alpha(t_{j-1}),\sigma_{i}\circ\alpha(t_{j}))$

and we are done by the triangle inequality on $(D,d[l^{(i)}])$ and the
definition of $L(\sigma_{i}\circ\alpha,d[l])$.

Let $A$ be the set of $t_{*}\in[0,1]$ such that there exist
$0=t_{0}<...<t_{k}=t_{*}$ with

$d[l^{(i)}](\alpha(t_{j-1}),\alpha(t_{j}))<\max\\{r^{\prime\prime}(\sigma_{i}\circ\alpha(t_{j-1})),r^{\prime\prime}(\sigma_{i}\circ\alpha(t_{j}))\\}.$

If $1\in A$ we are done. Suppose $1\notin A$ and $\bar{t}=\sup A$. Then it is
elementary to verify that

$\max\bigg{\\{}1,t_{*}+\frac{1}{2}r^{\prime\prime}(\sigma_{i}\circ\alpha(t_{*}))\bigg{\\}}\in
A,$

contradicting with $1\notin A$ or $t_{*}=\sup A$.

(b) We first prove that

$d(x,y)\leq
d[l](x,y)=\inf\bigg{\\{}\sum_{j=1}^{k}d_{i_{j}}(x_{j-1},x_{j}):x_{j-1},x_{j}\in\sigma_{i_{j}}(D),~{}x=x_{0},~{}y=x_{k}\bigg{\\}}.$

Suppose $x_{0},...,x_{k}$ are such that $x_{0}=x$, $x_{k}=y$, and for all
$j=1,...,k$, $x_{j-1},x_{j}\in\sigma_{i_{j}}$ for some $i_{j}\in I$. Then
construct a curve $\gamma$ by connecting $x_{j-1},x_{j}$ by a “shortest arc”
for all $j=1,...,k$. Specifically, for all $t\in[0,1/k]$ and $j=1,...,k$, let

$\gamma\bigg{(}\frac{j-1}{k}+t\bigg{)}=\sigma_{j}\bigg{(}ktx_{j}^{\prime\prime}+(1-kt)x_{j-1}^{\prime}\bigg{)}$

where
$x_{j-1}^{\prime\prime}\in\sigma_{i_{j}}^{-1}(x_{j-1}),x_{j}^{\prime}\in\sigma_{i_{j}}^{-1}(x_{j})$
are such that

$d_{i_{j}}(x_{j-1},x_{j})=d[l^{(i_{j})}](x_{j-1}^{\prime\prime},x_{j}^{\prime}).$

Then

$d(x,y)\leq L(\gamma,d)$ $=\sum_{j=1}^{k}L(\gamma|_{[(j-1)/k,j/k]},d)$
$=\sum_{j=1}^{k}L\bigg{(}\big{(}t\mapsto(ktx_{j}^{\prime\prime}+(1-kt)x_{j-1}^{\prime})\big{)}|_{[0,1/k]},~{}d[l^{(i_{j})}]\bigg{)}$
$=\sum_{j=1}^{k}d[l^{(i_{j})}](x_{j-1}^{\prime\prime},x_{j}^{\prime})$
$=\sum_{j=1}^{k}d_{i_{j}}(x_{j-1},x_{j}).$

Notice that the above inequality only uses the fact that $\gamma$ is map from
$[0,1]$ to $(S,d)$ and does not require that $\gamma$ is continuous from
$[0,1]$ to $(S,d)$.

Now we prove that

$d(x,y)\geq d[l](x,y)$

for all $x,y\in D$. Since $d(x,y)\leq d[l](x,y)$ for all $x,y\in D$,

$Id:(S,d[l])\rightarrow(S,d)$

is a continuous bijection from a compact space to a Hausdorff space, so it is
a homeomorphism and $(S,d)$ is also compact. Let us prove by contradiction and
assume

$d(x,y)<d[l](x,y)$

for some $x,y\in D$. Since $(S,d)$ is compact, by the Arzelà–Ascoli Theorem it
is easy to see that there exists a shortest arc in $(S,d)$ connecting $x,y$.
Speicifically, there exists a continuous map $\gamma:[0,1]\rightarrow S$ such
that $\gamma(0)=x$, $\gamma(1)=y$, and $L(\gamma,d)=d(x,y)$. As a shortest
arc, $\gamma$ is injective. Let

$0=t_{0}\leq t_{1}<t_{2}<....<t_{k-1}\leq t_{k}=1$

be such that

$\\{t_{1},...,t_{k-1}\\}=\\{t\in[0,1]:\gamma(t)\in V(T)\\}.$

Then by the triangle inequality and the additivity of the length of a curve,

$d[(\gamma(t_{j-1}),\gamma(t_{j}))]=L(\gamma|_{[t_{j-1},t_{j}]})<d[l](\gamma(t_{j-1}),\gamma(t_{j}))$

for some $j\in\\{1,...,k\\}$. So without loss of generality, by replacing
$\gamma$ with $\gamma|_{[t_{j-1},t_{j}]}$ we may assume that

$\gamma(0,1)\cap V(T)=\emptyset.$

Then by the continuity, we can replace $\gamma$ with
$\gamma|_{[\epsilon,1-\epsilon]}$ for sufficiently small $\epsilon>0$ we may
assume that

$\gamma[0,1]\cap V(T)=\emptyset.$

If $\gamma(t)\in\sigma_{i}(D^{o})$ for some $i\in I$ and $t\in[0,1]$, then

(a) there exists a maximal open interval $A=A_{t}\subset[0,1]$ such that $t\in
A$ and

$\gamma(A)\subset\sigma_{i}(D^{o}),$

(b) $\sigma_{i}^{-1}\circ\gamma(A)$ is an open straight line since $\gamma$ is
locally shortest,

(c)

$L(\gamma|_{A},d)=L(\sigma_{i}^{-1}\circ\gamma|_{A},d[l^{(i)}])\geq\epsilon$

for some constant $\epsilon=\epsilon(S,T,l,\gamma)>0$ by a standard
compactness argument,

(d) the length of the interval $A$ is at least $\delta$ for some
$\delta=\delta(S,T,l,\gamma)>0$ by the uniform continuity of $\gamma$,

(e)

$\mathcal{A}=\\{A_{t}:t\in[0,1],~{}\gamma(t)\in\sigma_{i}(D^{o})\text{ for
some $i\in I$}\\}$

is finite,

(f) $[0,1]-\sup_{A\in\mathcal{A}}A$ is a finite disjoint union of points
$\\{t_{1},...,t_{k-1}\\}$ and closed nondetenerate intervals
$\\{B_{1},...,B_{m}\\}$.

For all $B_{j}$, suppose $A_{j}$ is an open interval adjacent to $B_{j}$, then
it is easy to find a map $\tilde{\gamma}:A_{j}\cup B_{j}\rightarrow D$ such
that

$\sigma_{i}\circ\tilde{\gamma}=\gamma$

on $A_{j}\cup B_{j}$. Then $\tilde{\gamma}$ is not locally shortest, and then
neither is $\gamma=\sigma_{i}\circ\tilde{\gamma}$. So $\\{B_{1},...,B_{m}\\}$
is actually empty. Let $t_{0}=0,t_{k}=1$ and we have that
$(t_{j-1},t_{j})=\emptyset$ or $(t_{j-1},t_{j})\in\mathcal{A}$. Then

$d(x,y)=L(\gamma,d)=\sum_{j=1}^{k}L(\gamma|_{[t_{j-1},t_{j}]},d)=\sum_{j=1}^{k}L(\sigma^{-1}_{i_{j}}\circ\gamma|_{(t_{j-1},t_{j})},d[l^{(i)}])$
$\geq\sum_{j=1}^{k}d_{i_{j}}(\gamma(t_{j-1}),\gamma(t_{j}))\geq\sum_{j=1}^{k}d[l](\gamma(t_{j-1}),\gamma(t_{j}))\geq
d[l](x,y).$

∎

##  5\. Teichmüller space of polyhedral metrics

$(S,V)$ is called a _marked surface_ if $S$ is a connected closed surface and
$V$ is a finite subset of $S$. $T$ is a _triangulation of a marked surface_
$(S,V)$ if $T$ is a triangulation of $S$ with $V(T)=V$. By an elementary
calculation,

$|E(T)|=3|V|-3\chi(S)$

where $|X|$ denotes the number of elements in the set $X$. From now on, we
always assume that $|V|-\chi(S)>0$.

A metric $d$ on $S$ is called a _polyhedral metric of a marked surface_
$(S,V)$ if there exists a triangulation $T$ of $(S,V)$ and
$l\in\mathbb{R}^{E(T)}_{\Delta}$ such that $d=d_{T}[l]$. There is an
equivalent definition of the polyhedral metrics.

A _standard cone of angle $\theta\in(0,\infty)$_ is defined to be the quotient
topological space

$C_{\theta}=\\{(r,t):r\geq
0,t\in\mathbb{R}/\theta\mathbb{Z}\\}/_{(0,t)\sim(0,t^{\prime})}$

with a Riemannian metric $ds^{2}=dr^{2}+r^{2}dt^{2}$ on
$C^{o}_{\theta}=C_{\theta}-\\{[(0,0)]\\}$. This Riemannian metric $ds^{2}$
naturally induces an intrinsic metric $d_{\theta}$ on $C_{\theta}$.

######  Definition 5.1.

An intrinsic metric $d$ on a closed surface $(S,V)$ is called _polyhedral_ if

(a) for all $x\in S-V$ there exists an open neighborhood of $x$ isometric to
some open domain in the Euclidean plane, and

(b) for all $x\in V$ there exists an open neighborhood of $x$ isometric to a
neighborhood of $[(0,0)]$ in $(C_{\theta},d_{\theta})$ with $x$ mapped to
$[(0,0)]$ for some $\theta>0$.

It is intuitively correct and routine to verify that the metric $d[l]$
constructed in Section 3 is a polyhedral metric defined in Definition 5.1. It
is also well-known that a polyhedral metric defined in Definition 5.1 could
always be represented by $d[l]$ for some triangulation $T$ on $(S,V)$ and
$l\in\mathbb{R}^{E(T)}_{\Delta}$. For example, [1] describes a construction of
such representation. In a summary,

######  Theorem 5.2.

Given $(S,V)$,

$\\{\text{polyhedral metrics on $(S,V)$}\\}$
$=\\{d[l]:\text{$l\in\mathbb{R}^{E(T)}_{\Delta}$ where $T$ is a triangulation
of $(S,V)$}\\}$

In many cases, we do not care about the differences between two isometric
polyhedral metrics on a marked surface $(S,V)$. So it would be convenient and
useful to reduce the space of polyhedral metrics to a finitely dimensional
space.

######  Definition 5.3.

Given a polyhedral metric $d$ on $(S,V)$, denote

$[d]=\\{f^{*}d:f\in\mathcal{H}_{0}(S,V)\\}$

where

(a) $f^{*}d$ is the pullback metric on $S$ defined as

$f^{*}d(x,y)=d(f(x),f(y)),$

and

(b)

$\mathcal{H}_{0}(S,V)=\\{f:S\rightarrow S\big{|}\text{ $f$ is a homeomorphism
and $f|_{S-V}$ is isotopic to $Id|_{S-V}$}\\}.$

The resulted quotient set

$\\{[d]:\text{$d$ is a polyhedral metric on $(S,V)$}\\}$

is denoted as $T_{PL}(S,V)$ and called the _Teichmüller space_ of the
polyhedral metrics on $(S,V)$.

###  5.1. Manifold structure of $T_{PL}$ and the edge lengths coordinates

By a standard argument, one can show that

######  Proposition 5.4.

Given a triangulation $T$ of $(S,V)$, $l\mapsto\big{[}d_{T}[l]\big{]}$ is
injective from $\mathbb{R}^{E(T)}_{\Delta}$ to $T_{PL}(S,V)$.

By relating to the Teichmüller space of hyperbolic metrics, we will briefly
show that

(a) $T_{PL}(S,V)$ is naturally homeomorphic to $\mathbb{R}^{3|V|-3\chi(S)}$
when $|V|-\chi(S)>0$, and

(b) $l\mapsto\big{[}d_{T}[l]\big{]}$ induces an atlas on $T_{PL}(S,V)$.

Given a marked surface $(S,V)$ and a complete hyperbolic metrics $d$ of finite
area on $S-V$, denote

$[d]=\\{f^{*}d:f\in\mathcal{H}_{0}(S-V)\\}$

where $\mathcal{H}_{0}(S-V)$ is the collection of homeomorphisms from $S-V$ to
$S-V$ that are isotopic to $id|_{S-V}$. We denote

$T(S-V)=\\{[d]:\text{$d$ is a complete hyperbolic metrics on $(S-V)$ of finite
area}\\}$

and recall that $T(S-V)$ is called the _Teichmüller space of the complete
hyperbolic metrics of finite area_ on $S-V$. It is well-known that $T(S-V)$ is
homeomorphic to $\mathbb{R}^{2|V|-3\chi(S)}$.

Given a polyhedral metric $d$ on $(S,V)$, denote $d_{h}$ as the unique
complete hyperbolic metric on $S-V$ that is conformal to $d$ on $S-V$. By
estimating the extremal lengths, it is routine to verify that all the
topological ends of $(S-V,d_{h})$ are cusp ends. So $d_{h}$ is a complete
hyperbolic metric on $S-V$ of finite area and $[d_{h}]\in T(S-V)$.

Given a polyhedral metric $d$ on $(S,V)$, denote $A(d)$ as the area of $(S,d)$
and $\theta(d)\in\mathbb{R}^{V}_{>0}$ as the cone angles on $V$. By the
Gauss–Bonnet theorem

$\sum_{v\in V}(2\pi-\theta_{v}(d))=2\pi\chi(S).$

Troyanov (1986) proved that every compact Riemann surface carries a unique, up
to homothety, Euclidean (flat) conformal metric with prescribed conical
singularities of given angles, provided the Gauss-Bonnet relation is
satisfied. In other words, he proved that

######  Theorem 5.5 (Troyanov).

$d\mapsto(d_{h},\theta(d),A(d))$ is a bijection from $\tilde{T}_{PL}(S,V)$ to

$\tilde{T}(S-V)\times GB\times\mathbb{R}_{>0}$

where

(a) $\tilde{T}_{PL}(S,V)$ is the set of the polyhedral metric on $(S,V)$,

(b) $\tilde{T}(S-V)$ is the set of the complete hyperbolic metric of finite
area on $S-V$, and

(c)

$GB=\\{\theta\in\mathbb{R}^{V}_{>0}:\sum_{v\in
V}(2\pi-\theta_{i}(d))=2\pi\chi(S)\\}.$

As a consequence, we have that

######  Corollary 5.6.

$[d]\mapsto([d_{h}],\theta(d),A(d))$ is a bijection from $T_{PL}(S,V)$ to

$T(S-V)\times GB\times\mathbb{R}_{>0}.$

This bijection naturally assign $T_{PL}(S,V)$ a topology that is homeomorphic
to $\mathbb{R}^{3|V|-3\chi(S)}$. We will see that for any triangulation $T$ of
$(S,V)$, $l\mapsto d_{T}[l]$ is an embedding from $\mathbb{R}^{E(T)}_{\Delta}$
to $T_{PL}(S,V)$. So $l\mapsto d_{T}[l]$ naturally induces an atlas on
$T_{PL}(S,V)$. To see that $l\mapsto d_{T}[l]$ is an embedding, by Proposition
5.4 and the open mapping theorem it suffices to show that
$l\mapsto(d_{T}[l])_{h}$ is continuous. Fix
$\bar{l}\in\mathbb{R}^{E(T)}_{\Delta}$ and then

$Id|_{S}:(S,d_{T}[\bar{l}])\rightarrow(S,d_{T}[l])$

is $K_{l}$-quasiconformal where $K_{l}\rightarrow 1$ as $l\rightarrow\bar{l}$.
So the Teichmüller distance between $[(d_{T}[l])_{h}]$ and
$[(d_{T}[\bar{l}])_{h}]$ goes to $0$ as $l\rightarrow\bar{l}$.

###  5.2. Another approach using discrete conformal geometry

Gu et al. [5] introduced the discrete uniformization theorem in the
Teichmüller space of polyhedral metrics on marked surfaces. This discrete
uniformization theorem gives another $\mathbb{R}^{n}$-parameterization of
$T_{PL}$, by a similar argument. Related work on discrete conformality and
geodesic triangulations could be found in [18, 11, 15, 12, 2, 16, 6, 10, 8,
17, 9, 7, 14, 13, 5, 4, 3].

## References

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