# Sharp reverse isoperimetric inequalities in nonpositively curved cones

Mikhail G. Katz  M. Katz, Department of Mathematics, Bar Ilan University,
Ramat Gan 5290002 Israel<EMAIL_ADDRESS>and  Stéphane Sabourau  Univ
Paris Est Creteil, CNRS, LAMA, F-94010 Creteil, France Univ Gustave Eiffel,
LAMA, F-77447 Marne-la-Vallée, France<EMAIL_ADDRESS>
###### Abstract.

We prove a pair of sharp reverse isoperimetric inequalities for domains in
nonpositively curved surfaces: (1) metric disks centered at the vertex of a
Euclidean cone of angle at least $2\pi$ have minimal area among all
nonpositively curved disks of the same perimeter and the same total curvature;
(2) geodesic triangles in a Euclidean (resp. hyperbolic) cone of angle at
least $2\pi$ have minimal area among all nonpositively curved geodesic
triangles (resp. all geodesic triangles of curvature at most $-1$) with the
same side lengths and angles.

###### Key words and phrases:

Reverse isoperimetric inequalities, Euclidean cone, nonpositive curvature,
geometric inequalities, area comparison

###### 2020 Mathematics Subject Classification:

Primary 53C21; Secondary 49Q10

Partially supported by the ANR project Min-Max (ANR-19-CE40-0014).

##  1\. Introduction

Isoperimetric inequalities provide upper bounds on the area of domains in a
surface with a fixed metric (typically of constant curvature) in terms of
their boundary length; see [5] for an account on this classical subject.

Often the metric is fixed (flat, hyperbolic, or spherical), but there are a
few instances where isoperimetric inequalities hold for large classes of
metrics satisfying curvature bounds. Thus, André Weil ([12], 1926) developed
such inequalities for nonpositively curved planes, proving the Cartan-Hadamard
isoperimetric conjecture in dimension two.

In this article, we establish a pair of sharp reverse isoperimetric
inequalities providing _lower bounds_ on the area of some domains in terms of
their boundary length.

These geometric inequalities hold for nonpositively curved surfaces. The
optimal metrics for such reverse isoperimetric inequalities, as well as the
extremal domains, can be described in terms of the total curvature of the
domains under consideration. The boundary cases of equality in our optimal
inequalities are attained by Euclidean cones with nonpositive total curvature.
These inequalities provide bounds that are stronger than Euclidean ones.

It seems that such geometric inequalities should have been known already for
some time, but we were unable to find them in the literature. A possible
reason is that previous research focused more on isoperimetric inequalities
for special homogeneous metrics regardless of the total curvature of the
domains considered.

We will now present our two main results. As motivation, note that the area of
a Euclidean cone of angle less than $2\pi$ can be easily decreased among
nonnegatively curved metrics by smoothing off the tip of the cone. Our first
result shows that this is impossible while preserving the nonpositive
curvature condition; see Theorem 1.2 for a more general statement. Namely, one
cannot decrease the area of a metric disk centered at the vertex of a
Euclidean cone of angle at least $2\pi$ among all nonpositively curved disks
of the same perimeter and the same total curvature. We will need the following
definition.

######  Definition 1.1.

Given a surface with a complete Riemannian metric with Gaussian curvature
function $K$, we define the nonnegative and nonpositive parts of $K$ as

$K^{+}=\max\\{K,0\\}\quad\mbox{ and }\quad K^{-}=\max\\{-K,0\\}$

so that $K=K^{+}-K^{-}$.

Our convention is consistent with the corresponding definitions of nonnegative
and nonpositive parts of curvature measures introduced by Yu. Burago in [4];
see Definition 2.1. We can now state our first result.

######  Theorem 1.2.

Let $M$ be a surface with a complete Riemannian metric. Then every disk $D$ of
radius $R$ and boundary length $L$ in $M$ satisfies the bound

${\rm area}(D)\geq{\rm area}(\hat{D})$ (1.1)

where $\hat{D}$ is the disk with the same boundary length $L$ as $D$, centered
at the vertex of the Euclidean cone with total curvature $-\int_{D}K^{-}\,dA$.

A formula for the area of $\hat{D}$ is given in Section 2.

Comparison geodesic triangles play an important role in nonpositive curvature
geometry. For our second result, we consider a geodesic triangle of Gaussian
curvature at most $\lambda_{0}\leq 0$ along with its comparison triangle (in
the strong sense) in a cone $\mathcal{C}_{\lambda_{0}}^{\theta}$ of constant
nonpositive curvature $\lambda_{0}$ with angle $\theta$. Here, a comparison
triangle is taken in the following strong sense: both the angles and the side
lengths of the two triangles are the same. The following theorem asserts that
the area of the initial triangle is bounded from below by the area of its
comparison triangle in the cone. (For simplicity, one can assume that
$\lambda_{0}=0$.)

######  Theorem 1.3.

Let $\Delta$ be a geodesic (two-dimensional) triangle in a surface with a
complete Riemannian metric of Gaussian curvature $K\leq\lambda_{0}$ for some
constant $\lambda_{0}\leq 0$. Suppose $\bar{\Delta}$ is a geodesic (two-
dimensional) triangle with the same side lengths and the same angles $\alpha$,
$\beta$, $\gamma$ as $\Delta$ in the cone $\mathcal{C}_{\lambda_{0}}^{\theta}$
of constant curvature $\lambda_{0}$ with angle
$\theta=3\pi-(\alpha+\beta+\gamma)$. Then

${\rm area}(\Delta)\geq{\rm area}(\bar{\Delta})$

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$.

Our proof of Theorem 1.3 also shows that the area of a geodesic triangle of
Gaussian curvature at most $\lambda_{0}$ is bounded from below by the area of
a comparison triangle having the same base length and the same adjacent angles
in the plane ${\mathbb{H}}_{\lambda_{0}}$ of constant curvature
$\lambda_{0}\leq 0$; see Proposition 3.1.

We would like to provide some context for our study of reverse isoperimetric
inequalities. We found the geometric inequalities of this paper while working
on a proof that systolically extremal nonpositively curved surfaces are flat
with finitely many conical singularities; see [7]. In that context, it was
clear that it should be imposssible to round off a conical singularity of
angle greater than $2\pi$ in a nonsystolic region in order to decrease the
area (while keeping the nonpositive curvature condition) by any cut-and-paste
argument with metric disks of the same perimeter. This observation is
formalized by our Theorem 1.2. Theorem 1.3 is a variation on this theme, while
trying to cut-and-paste triangles instead of disks. Though we did not use
these reverse isoperimetric inequalities in our argument, they confirmed our
intuition that piecewise flat metrics with conical singularities should play a
role in extremal systolic geometry through their local extremal features.

_Acknowledgment._ The second author would like to thank the Fields Institute
and the Department of Mathematics at the University of Toronto for their
hospitality while this work was completed.

##  2\. Reverse isoperimetric inequality for metric disks

The reverse isoperimetric inequality for metric disks established in this
section shows the optimality of the nonpositively curved Euclidean cones for
the area with respect to compact deformations keeping the same curvature sign.

Before proving this result, we need to extend the notion of curvature to
singular spaces.

######  Definition 2.1.

Associated to a surface $M$ with a complete Riemannian metric $g$ of Gaussian
curvature $K$ is the _curvature measure_ $K\,dA$, where $dA$ is the area
measure of $M$. The notion of curvature measure extends to piecewise flat
surfaces with conical singularities (and more generally to Alexandrov
surfaces), where it is denoted by $\omega$. The curvature measure $\omega$ is
a signed measure which can be decomposed as $\omega=\omega^{+}-\omega^{-}$,
where $\omega^{+}$ and $\omega^{-}$ are the nonnegative and nonpositive parts
of $\omega$, and are both nonnegative measures. For smooth metrics, we have
$\omega^{\pm}=K^{\pm}\,dA$.

We refer to [1] and [9] for a precise definition of the curvature measure; see
also [11] for a modern exposition on Alexandrov surfaces.

We will make use of the following result on bi-Lipschitz metric approximation,
announced by Reshetnyak [8] and proved by Yu. Burago [4, Lemma 6], in the more
general setting of Alexandrov surfaces; see also [5, Theorem 3.1.1], [11] and
[7, §3].

######  Proposition 2.2 (See [8] and [4, Lemma 6]).

Let $M$ be a compact surface (possibly with boundary) with a Riemannian
metric. Then there is a sequence $M_{i}$ of piecewise flat surfaces with
conical singularities converging to $M$ in the Lipschitz topology such that
the nonnegative and nonpositive parts $\omega^{\pm}_{i}$ of the curvature
measure $\omega_{i}$ of $M_{i}$ weakly converge to their counterparts
$\omega^{\pm}$ for the curvature measure $\omega$ on $M$.

Consider a sequence $(g_{i})$ of piecewise flat metrics with conical
singularities approximating a given complete Riemannian metric $g$ on a
surface for every compact domain as in Proposition 2.2. Denote by $M$ the
surface with the complete Riemannian metric $g$, and by $M_{i}$ the same
surface with the piecewise flat metric $g_{i}$.

######  Proposition 2.3.

Fix $p\in M$. Let $D\subseteq M$ and $D_{i}\subseteq M_{i}$ be the disks of
radius $R$ centered at the same point $p$. Then

  1. (1)

the area of the symmetric difference $D\Delta D_{i}$ tends to zero (for every
area measure). That is, $|D\Delta D_{i}|\to 0$;

  2. (2)

$\lim{\rm area}(D_{i},g_{i})={\rm area}(D,g)$;

  3. (3)

$\lim\omega^{\pm}_{i}(D_{i})=\omega^{\pm}(D)$;

  4. (4)

$\liminf{\rm length}(\partial D_{i},g_{i})\geq{\rm length}(\partial D,g)$.

###### Proof.

(1) By bilipschitz convergence of the metrics, the symmetric difference
$D\Delta D_{i}$ is contained in the $\varepsilon$-tubular neighborhood
$U_{\varepsilon}(\partial D)$ of $\partial D$ for $i$ large enough. Since
$\partial D$ is $1$-rectifiable, the area of this tubular neighborhood tends
to zero (see [6, Theorem 3.2.39]), and the result follows.

(2) By bilipschitz convergence of the metrics, we have the inclusions

$D-\varepsilon\subseteq D_{i}\subseteq D+\varepsilon$

for $i$ large enough, where $D\pm\varepsilon\subseteq M$ are the balls of
radius $R\pm\varepsilon$ centered at $p$. We also have weak convergence of the
area measures. Taking the area with respect to $g_{i}$ in the previous double
inclusion between $D_{i}$ and $D\pm\varepsilon$, and using the weak
convergence of the area measure, we obtain

${\rm area}(D-\varepsilon,g)-\varepsilon\leq{\rm area}(D_{i},g_{i})\leq{\rm
area}(D+\varepsilon,g)+\varepsilon$

for $i$ large enough. Since the area of the $\varepsilon$-tubular neigborhood
$U_{\varepsilon}(\partial D)$ of $\partial D$ tends to zero, the result is
immediate.

(3) As in the proof of item (2), using the weak convergence of the curvature
measure instead of the area measures, we obtain

$\omega^{\pm}(D-\varepsilon)-\varepsilon\leq\omega^{\pm}_{i}(D_{i})\leq\omega^{\pm}(D+\varepsilon)+\varepsilon$

for $i$ large enough. Since the curvature measure $\omega^{\pm}=K^{\pm}\,dg$
is absolutely continuous with respect to the area measure, the curvature
measure of the $\varepsilon$-tubular neighborhood $U_{\varepsilon}(\partial
D)$ of $\partial D$ tends to zero. Hence the result.

(4) The flat distance between the one-cycles $\partial D$ and $\partial D_{i}$
is bounded by the mass of the $2$-current defined as the difference $D-D_{i}$,
see [6] for precise definitions. This mass is equal to the area of $D\Delta
D_{i}$. Thus, by (1), the sequence $\partial D_{i}$ converges to $\partial D$
in the flat topology. The desired result follows from the lower semicontinuity
of the mass (here, the length); see [6]. ∎

We can now proceed to the proof of the following theorem. Recall that a disk
of radius $R$ centered at the vertex of a Euclidean cone of angle $\theta$ has
perimeter $L=\theta R$ and area
$A=\frac{\theta}{2}R^{2}=\frac{L^{2}}{2\theta}$. Thus, we have

$A=\frac{L^{2}}{2(2\pi-\mathcal{K})}$ (2.1)

where $\mathcal{K}=2\pi-\theta<0$ is the total curvature of the Euclidean
cone.

With this formula, Theorem 1.2 can be restated as follows.

######  Theorem 2.4.

Let $M$ be a surface with a complete Riemannian metric. Then every disk $D$ of
radius $R$ of boundary length $L$ in $M$ satisfies

${\rm area}(D)\geq\frac{L^{2}}{4\pi+2\mathcal{K}_{D}^{-}}$

where $\mathcal{K}_{D}^{-}=\int_{D}K^{-}\,dA\geq 0$ is the total mass of the
nonpositive part of the curvature measure of $D$.

###### Proof.

Proposition 2.3 shows that it is sufficient to prove Theorem 2.4 for piecewise
flat metrics with conical singularities. The desired result for Riemannian
metrics will follow by piecewise flat metric approximation; see Proposition
2.2. This enables us to avoid regularity issues.

Let $p$ be the center of the disk $D\subseteq M$ of radius $R$. Consider the
(closed) disk $D_{t}$ and the circle $\mathcal{C}_{t}$ of radius $R-t$
centered at $p$. Note that $D_{0}=D$ and $\mathcal{C}_{0}=\partial D$. Since
the metric on $M$ is piecewise flat, the circle $\mathcal{C}_{t}$ is a
piecewise smooth curve (possibly with several connected components) bounding
the metric disk $D_{t}$ in $M$. Moreover, the length function $t\mapsto
L(\mathcal{C}_{t})$ is differentiable except for a finite number of values of
$t$, and its derivative is given by the first variation formula; see [5, Lemma
3.2.3]. Namely, as long as $D_{t}$ is nonempty, we have

$L^{\prime}(\mathcal{C}_{t})=-\int_{\mathcal{C}_{t}}\kappa(s)\,ds-S_{t}$

for almost every $t$, where $\kappa$ is the geodesic curvature of the curve
$\mathcal{C}_{t}$ and $S_{t}$ is the sum of the angular difference of the
tangent vectors at the corner points of $\mathcal{C}_{t}$.

By the Gauss–Bonnet formula for polyhedral metrics (see [9, Theorem 5.3.2]),
we derive

$L^{\prime}(\mathcal{C}_{t})=-2\pi\chi(D_{t})+\omega(D_{t}).$

Since $D_{t}$ is a connected region with boundary, its Euler characteristic
$\chi(D_{t})$ is at most $1$, that is, $\chi(D_{t})\leq 1$. Since
$D_{t}\subseteq D$ and $\omega=\omega^{+}-\omega^{-}$, we have

$\omega(D_{t})\geq-\omega^{-}(D).$

Combining these two bounds, we deduce that

$L^{\prime}(\mathcal{C}_{t})\geq-2\pi-\omega^{-}(D).$

Integrating this relation leads to

$L(\mathcal{C}_{t})\geq L(\partial D)-\left(2\pi+\omega^{-}(D)\right)\,t.$
(2.2)

In particular, the domain $D_{t}$ is nonempty for every $t<t_{0}$, where

$t_{0}=\frac{L(\partial D)}{2\pi+\omega^{-}(D)}.$

By the coarea formula, integrating the inequality (2.2) between $0$ and
$t_{0}$, we obtain

${\rm area}(D)\geq L(\partial
D)\,t_{0}-\frac{1}{2}\,\left(2\pi+\omega^{-}(D)\right)t_{0}^{2}.$

In other words, we have

${\rm area}(D)\geq\frac{L(\partial D)^{2}}{2\left(2\pi+\omega^{-}(D)\right)}$

where the right-hand side represents the area of the disk centered at the
vertex of the Euclidean cone with total curvature $\mathcal{K}=-\omega^{-}(D)$
and with the same boundary length $L(\partial D)$ as $D$; see formula (2.1). ∎

##  3\. Area comparison for triangles with the same base

In order to prove Theorem 1.3, we will need the following result, which may be
of independent interest. This result provides a lower bound on the area of a
geodesic triangle of Gaussian curvature at most $\lambda_{0}$, in terms of the
area of a comparison triangle having the same base length with the same
adjacent angles in the plane ${\mathbb{H}}_{\lambda_{0}}$ of constant
curvature $\lambda_{0}\leq 0$.

######  Proposition 3.1.

Let $\Delta$ be a geodesic (two-dimensional) triangle with vertices $A$, $B$,
$C$ in a surface $M$ with a complete Riemannian metric of Gaussian curvature
$K\leq\lambda_{0}$ for some constant $\lambda_{0}\leq 0$. Let $\bar{\Delta}$
be a geodesic (two-dimensional) triangle with distinct vertices $\bar{A}$,
$\bar{B}$, $\bar{C}$ in the plane ${\mathbb{H}}_{\lambda_{0}}$ of constant
curvature $\lambda_{0}$ such that

  * •

the sides $\bar{A}\bar{B}$ and $AB$ have the same length;

  * •

the angles at $\bar{A}$ and $A$ are the same;

  * •

the angles at $\bar{B}$ and $B$ are the same.

Then

${\rm area}(\Delta)\geq{\rm area}(\bar{\Delta})$ (3.1)

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$.

We first establish the proposition when one of the angles at $A$ or $B$, say
$B$, is greater or equal to $\frac{\pi}{2}$; see Lemma 3.2. We will then
derive the general result from this particular case.

######  Lemma 3.2.

Let $\Delta$ and $\bar{\Delta}$ be as in Proposition 3.1. Let $\alpha$,
$\beta$, $\gamma$ be the angles of $\Delta$ at $A$, $B$, $C$. Suppose that
$\beta\geq\frac{\pi}{2}$. Then

${\rm area}(\Delta)\geq{\rm area}(\bar{\Delta})$

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$.

###### Proof.

Let $\hat{\Delta}$ be a geodesic triangle of ${\mathbb{H}}_{\lambda_{0}}$ with
the same side lengths as $\Delta$. Denote by $\hat{A}$, $\hat{B}$, $\hat{C}$
the vertices of $\hat{\Delta}$ and by $\hat{\alpha}$, $\hat{\beta}$,
$\hat{\gamma}$ their angles. We can (and will) assume that $\hat{A}=\bar{A}$
and $\hat{B}=\bar{B}$. By [2, Proposition II.1.7.(4)], we have

$\begin{cases}\hat{\alpha}\geq\alpha\\\ \hat{\beta}\geq\beta.\end{cases}$
(3.2)

In particular, $\hat{\beta}\geq\frac{\pi}{2}$. Since the sum of the angles of
a geodesic triangle in a nonpositively curved surface is at most $\pi$; see
[2, Proposition II.1.7.(4)], we derive that $\hat{\gamma}\leq\frac{\pi}{2}$.

The geodesic triangle of ${\mathbb{H}}_{\lambda_{0}}$ with side
$\bar{A}\bar{B}$, and angles $\alpha$ and $\beta$ at $\bar{A}$ and $\bar{B}$
can be isometrically identified to $\bar{\Delta}$. It follows from the
relations (3.2) that the triangle $\bar{\Delta}$ lies in $\hat{\Delta}$; see
Figure 1. Since $\hat{\gamma}\leq\frac{\pi}{2}$, we deduce that

$|\bar{A}\bar{C}|\leq|\hat{A}\hat{C}|=|AC|.$

This relation holds for any point $D$ lying in the segment $BC$ by replacing
$\hat{C}$ with $\hat{D}$ and $\bar{C}$ with $\bar{D}$ (note that $\bar{D}$
lies in the segment $\bar{B}\bar{C}$).

$\bar{A}=\hat{A}$
---
$\alpha$
---
$\beta$
---
$\bar{B}=\hat{B}$
---
$\hat{C}$
---
$\bar{C}$
---
Figure 1. The comparison triangles $\bar{\Delta}$ and $\hat{\Delta}$

Consider the two exponential maps $\exp_{A}:T_{A}M\to M$ and
$\exp_{\bar{A}}:T_{\bar{A}}{\mathbb{H}}_{\lambda_{0}}\to{\mathbb{H}}_{\lambda_{0}}$.
Define $\sigma:{\mathbb{H}}_{\lambda_{0}}\to M$ as

$\sigma=\exp_{A}\circ I\circ\exp_{\bar{A}}^{-1}$

where $I:T_{\bar{A}}{\mathbb{H}}_{\lambda_{0}}\to T_{A}M$ is a linear
isometry. We can choose $I$ so that the map $\sigma$ takes the geodesic rays
$[\bar{A}\bar{B})$ and $[\bar{A}\bar{C})$ to the geodesic rays $[AB)$ and
$[AC)$. By construction, the map $\sigma$ sends every segment $\bar{A}\bar{D}$
joining $\bar{A}$ to a point $\bar{D}$ in the opposite side $\bar{B}\bar{C}$
of $\bar{\Delta}$ to a subarc of the geodesic arc $AD$. In particular, the map
$\sigma$ sends $\bar{\Delta}$ into $\Delta$.

Since $K\leq\lambda_{0}$, the Rauch theorem implies that the map $\sigma$ is
distance-nondecreasing; see [3, Theorem 6.5.4]. Therefore,

${\rm area}(\bar{\Delta})\leq{\rm area}(\Delta)$

with equality if and only if the map $\sigma$ is an isometry between
$\bar{\Delta}$ and $\Delta$. ∎

We can now derive Proposition 3.1 from the previous lemma.

###### Proof of Proposition 3.1.

Let $\alpha$, $\beta$, $\gamma$ be the angles of $\Delta$ at $A$, $B$, $C$.
The cases where $\alpha\geq\frac{\pi}{2}$ or $\beta\geq\frac{\pi}{2}$ are
covered by Lemma 3.2. Thus, we can assume that $\alpha<\frac{\pi}{2}$ and
$\beta<\frac{\pi}{2}$. This implies that the projection $H$ of $C$ to the
segment $AB$ strictly lies between $A$ and $B$. Furthermore, both angles
$\measuredangle{AHC}$ and $\measuredangle{BHC}$ are equal to $\frac{\pi}{2}$.
Thus, the height $CH$ decomposes $\Delta$ into two right triangles
$\Delta^{\prime}$ and $\Delta^{\prime\prime}$. Denote by $\gamma^{\prime}$ and
$\gamma^{\prime\prime}$ the angles of $\Delta^{\prime}$ and
$\Delta^{\prime\prime}$ at $C$. Observe that
$\gamma=\gamma^{\prime}+\gamma^{\prime\prime}$. Since the angles of
$\Delta^{\prime}$ and $\Delta^{\prime\prime}$ at $H$ are right, we can apply
Lemma 3.2 to the triangles $\Delta^{\prime}=AH^{\prime}C^{\prime}$ and
$\Delta^{\prime\prime}=BH^{\prime\prime}C^{\prime\prime}$, where
$H=H^{\prime}=H^{\prime\prime}$ and $C=C^{\prime}=C^{\prime\prime}$. Thus, the
areas of the triangles $\Delta^{\prime}$ and $\Delta^{\prime\prime}$ are
bounded from below by the areas of the right triangles
$\bar{\Delta}^{\prime}=\bar{A}\bar{H}^{\prime}\bar{C}^{\prime}$ and
$\bar{\Delta}^{\prime\prime}=\bar{B}\bar{H}^{\prime\prime}\bar{C}^{\prime\prime}$
in ${\mathbb{H}}_{\lambda_{0}}$. We can glue together these two right
triangles of ${\mathbb{H}}_{\lambda_{0}}$ along their sides
$\bar{H}^{\prime}\bar{C}^{\prime}$ and
$\bar{H}^{\prime\prime}\bar{C}^{\prime\prime}$ so that $\bar{H}^{\prime}$ and
$\bar{H}^{\prime\prime}$ coincide. Since the angles at $\bar{H}^{\prime}$ and
$\bar{H}^{\prime\prime}$ are right, the two segments $\bar{A}\bar{H}^{\prime}$
and $\bar{H}^{\prime\prime}\bar{B}$ form a long segment $\bar{A}\bar{B}$ of
the same length as $AB$; see Figure 2.

If the heights $\bar{H}^{\prime}\bar{C}^{\prime}$ and
$\bar{H}^{\prime\prime}\bar{C}^{\prime\prime}$ have the same length, the union
of these two triangles form a large triangle which satisfies the same
geometric features as $\bar{\Delta}$ and so can be identified with
$\bar{\Delta}$.

If one of these heights, say $\bar{H}^{\prime}\bar{C}^{\prime}$, is shorter
than the other, we extend the hypothenuse $\bar{A}\bar{C}^{\prime}$ until it
intersects the other hypothenuse $\bar{B}\bar{C}^{\prime\prime}$ at some point
$\bar{O}$; see Figure 2. As previously, the triangle $\bar{A}\bar{O}\bar{B}$
can be identified with $\bar{\Delta}$. Moreover, the area of this triangle is
bounded by the sum of the two triangles $\bar{\Delta}^{\prime}$ and
$\bar{\Delta}^{\prime\prime}$.

$\bar{A}$$\bar{H}^{\prime}=\bar{H}^{\prime\prime}$$\bar{C}^{\prime}$$\bar{C}^{\prime\prime}$$\bar{O}$$\bar{B}$
Figure 2. The two triangles $\bar{\Delta}^{\prime}$ and
$\bar{\Delta}^{\prime\prime}$

In either case, we have

${\rm area}(\bar{\Delta})\leq{\rm area}(\bar{\Delta}^{\prime})+{\rm
area}(\bar{\Delta}^{\prime\prime})\leq{\rm area}(\Delta)$

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$. ∎

##  4\. Area comparison for triangles with the same side lengths and angles

We can now prove our second main result.

######  Theorem 4.1.

Let $\Delta$ be a geodesic (two-dimensional) triangle in a surface with a
complete Riemannian metric of Gaussian curvature $K\leq\lambda_{0}$ for some
constant $\lambda_{0}\leq 0$. Suppose $\bar{\Delta}$ is a geodesic (two-
dimensional) triangle with the same side lengths and the same angles $\alpha$,
$\beta$, $\gamma$ as $\Delta$ in the cone $\mathcal{C}_{\lambda_{0}}^{\theta}$
of constant curvature $\lambda_{0}$ with angle
$\theta=3\pi-(\alpha+\beta+\gamma)$. Then

${\rm area}(\Delta)\geq{\rm area}(\bar{\Delta})$ (4.1)

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$.

######  Remark 4.2.

Unlike the situation with Proposition 3.1, the comparison triangle
$\bar{\Delta}$ in $\mathcal{C}_{\lambda_{0}}^{\theta}$ in Theorem 4.1 do not
necessarily exist in general.

###### Proof of Theorem 4.1.

Let $A$, $B$ and $C$ be the vertices of $\Delta$, with angle $\alpha$ at the
vertex $A$, angle $\beta$ at $B$, and angle $\gamma$ at $C$. Denote by
$\bar{A}$, $\bar{B}$, $\bar{C}$ the corresponding vertices of $\bar{\Delta}$.

By [2, Proposition II.1.7.(4)], the angle $\theta$ of the conical singularity
of the cone is at least $2\pi$. Moreover, we can assume that $\theta>2\pi$ and
that the conical singularity lies in $\bar{\Delta}$. Otherwise, the sum
$\alpha+\beta+\gamma$ of the angles of $\Delta$ and $\bar{\Delta}$ would be
equal to $\pi$ and so the triangle $\Delta$ would be flat isometric to
$\bar{\Delta}$ by [2, Proposition II.2.9].

The geodesic rays joining the vertices of $\bar{\Delta}$ to its conical
singularity decompose each angle around the vertices of $\bar{\Delta}$ into
two angles. In particular, the angle $\alpha$ splits into two angles
$\alpha^{\prime}$ and $\alpha^{\prime\prime}$, where
$\alpha=\alpha^{\prime}+\alpha^{\prime\prime}$. The same holds with $\beta$
and $\gamma$. Observe that these geodesic rays decompose $\bar{\Delta}$ into
three small triangles.

We would like to carry out a similar construction for $\Delta$, except that
there is no conical singularity to rely on. Instead, we consider the geodesic
rays of $\Delta$ emanating from the vertices of $\Delta$ and splitting each
angle as in $\bar{\Delta}$. These three geodesic rays do not necessarily
intersect at a single point as in $\bar{\Delta}$. Nevertheless, they decompose
$\Delta$ into three triangles $\Delta_{a}$, $\Delta_{b}$, $\Delta_{c}$ as in
Figure 3. Recall that two geodesic rays in a nonpositively curved surface
intersect at most once. There may be a small triangular region

$X=\Delta\setminus(\Delta_{a}\cup\Delta_{b}\cup\Delta_{c})$

lying in $\Delta$ which is not covered by the triangles $\Delta_{a}$,
$\Delta_{b}$, $\Delta_{c}$.

$A$$\alpha^{\prime}$$\alpha^{\prime\prime}$$A^{\prime}$$B$$C$$\Delta_{a}$$\beta^{\prime}$
Figure 3. Decomposition of $\Delta$ into triangles

Let $A^{\prime}$ be the vertex of $\Delta_{a}$ different from $B$ and $C$.
Denote by $\bar{\Delta}_{a}$ the triangle of
$\mathcal{C}_{\lambda_{0}}^{\theta}$ with vertices $\bar{A}^{\prime}$,
$\bar{B}$, $\bar{C}$ such that

  * •

the sides $\bar{B}\bar{C}$ and $BC$ have the same length;

  * •

the angles at $\bar{B}$ and $B$ are the same (equal to
$\beta^{\prime\prime}$);

  * •

the angles at $\bar{C}$ and $C$ are the same (equal to $\gamma^{\prime}$).

Similarly, we define $\bar{\Delta}_{b}$ and $\bar{\Delta}_{c}$. By
construction, the three triangles $\bar{\Delta}_{a}$, $\bar{\Delta}_{b}$ and
$\bar{\Delta}_{c}$ are isometric to the three smaller triangles forming
$\bar{\Delta}$ and delimited by the geodesic segments joining the conical
singularity of $\bar{\Delta}$ to its vertices. By Proposition 3.1, we have

${\rm area}(\bar{\Delta}_{a})\leq{\rm area}(\Delta_{a})$

with equality if and only if $\Delta_{a}$ is isometric to $\bar{\Delta}_{a}$.
The same holds with $\bar{\Delta}_{b}$ and $\bar{\Delta}_{c}$. Since the
triangle $\bar{\Delta}$ is partitioned into $\bar{\Delta}_{a}$,
$\bar{\Delta}_{b}$ and $\bar{\Delta}_{c}$, we derive that

$\displaystyle{\rm area}(\bar{\Delta})$ $\displaystyle={\rm
area}(\bar{\Delta}_{a})+{\rm area}(\bar{\Delta}_{b})+{\rm
area}(\bar{\Delta}_{c})$ $\displaystyle\leq{\rm area}(\Delta_{a})+{\rm
area}(\Delta_{b})+{\rm area}(\Delta_{c})$ $\displaystyle={\rm
area}(\Delta)-{\rm area}(X)$ $\displaystyle\leq{\rm area}(\Delta)$

with equality if and only if $\Delta$ is isometric to $\bar{\Delta}$. ∎

######  Remark 4.3.

It follows from the proof of Theorem 4.1 that the difference between ${\rm
area}(\Delta)$ and ${\rm area}(\bar{\Delta})$ is bounded from below by the
area of the small triangle
$X=\Delta\setminus(\Delta_{a}\cup\Delta_{b}\cup\Delta_{c})$ lying in $\Delta$.

## References

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