Approaching the isoperimetric problem in HCm$H_{{\mathbb {C}}}^m$ via the hyperbolic log-convex density conjecture

Abstract

We prove that geodesic balls centered at some base point are uniquely isoperimetric sets in the real hyperbolic space HRnHRnH_{{\mathbb {R}}}^n endowed with a smooth, radial, strictly log-convex density on the volume and perimeter. This is an analogue of the result by G. R. Chambers for log-convex densities on RnRn{\mathbb {R}}^n. As an application we prove that in any rank one symmetric space of non-compact type, geodesic balls are uniquely isoperimetric in a class of sets enjoying a suitable notion of radial symmetry.

Full text

Calc. Var. (2024) 63:11
https://doi.org/10.1007/s00526-023-02617-0
Calculus of Variations
Approaching the isoperimetric problem in Hm
Cvia the
hyperbolic log-convex density conjecture
Lauro Silini1
Received: 10 October 2022 / Accepted: 2 November 2023 / Published online: 30 November 2023
© The Author(s) 2023
Abstract
We prove that geodesic balls centered at some base point are uniquely isoperimetric sets in
the real hyperbolic space Hn
Rendowed with a smooth, radial, strictly log-convex density on
the volume and perimeter. This is an analogue of the result by G. R. Chambers for log-convex
densities on Rn. As an application we prove that in any rank one symmetric space of non-
compact type, geodesic balls are uniquely isoperimetric in a class of sets enjoying a suitable
notion of radial symmetry.
Mathematics Subject Classification 49Q20 ·49Q05 ·53A10
1 Introduction
We denote by (Hn
R,gH)the real hyperbolic space of dimension n∈Nwith constant sectional
curvature equal to −1. Call dHthe induced Riemannian distance. Choose an arbitrary base
point o∈Hn
R. We say that a function f:Hn
R→R>0is (strictly) radially log-convex if
ln(f(x)) =h(dH(o,x)),
for a smooth, (strictly) convex and even function h:R→R. We define the weighted
perimeter and volume of a set with finite perimeter E⊂Hn
Ras
Vf(E):= E
fdHn,and Pf(E)=∂∗E
fdHn−1.
Here, following the notation in [14], ∂∗Edenotes the reduced boundary of E. A set of finite
perimeter Ewith volume Vf(E)=v>0 is called isoperimetric if it solves the minimization
problem
J(v) := inf Pf(F):Vf(F)=v, F⊂Hn
Rof finite perimeter.(1.1)
Communicated by Andrea Mondino.
BLauro Silini
lauro.silini@math.ethz.ch
1Department of Mathematics, ETH Zürich, Zurich, Switzerland
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11 Page 2 of 27 L. Silini
The first goal of this paper is to show the following characterization of the isoperimetric sets,
which will be developed in Sect.3.
Theorem 1.1 For any strictly radially log-convex density f , geodesic balls centered at o ∈
Hn
Runiquely minimize the weighted perimeter for any given weighted volume with respect
to Pfand V f.
Our main motivation in proving such result is the tight relation of this problem with
the (unweighted) isoperimetric problem in the complex hyperbolic spaces Hm
C, the quater-
nionic spaces Hm
Hand the Cayley plane H2
Orestricted to a family of sets sharing a particular
symmetry that we define as follows.
Definition 1.2 (Hopf-symmetric sets) Let K∈{C,H,O},d=dim(K)∈{2,4,8}and
(M,g)=(Hm
K,g)be the associated rank one symmetric space of non-compact type of real
dimension n=dm,m=2ifK=O. Fix an arbitrary point o∈Mand let Nbe the unit length
radial vector field emanating from o. Then, up to renormalization of the metric, the Jacobi
operator R(·,N)Narising from the Riemannian curvature tensor Ris a self adjoint operator
of TM, and has exactly three eigenvalues: {0,−1,−4}.The(−4)-eigenspace defines at every
point x= oa distribution Hxof real dimension d−1. A C1-set E⊂Mwith normal vector
field νis said to be Hopf-symmetric if ν(x)is orthogonal to Hxat each point x∈∂E,o/∈∂E.
Remark 1.3 Let h:Sn−1→KPm−1be the celebrated Hopf fibration. Then, for any C1-
profile ρ:Sn−1→(0,+∞)so that ρis constant along the fibres of h, the set with
boundary
∂E:= {expo(ρ (x)x):x∈Sn−1⊂ToM},
is Hopf-symmetric, where expois the exponential map of Mat an arbitrary point o∈M.
Remark 1.4 Being Hopf-symmetric has not to be confused with the standard notion of being
Hopf in Hm
C, that is a set with principal curvature along the characteristic directions Jν,
where Jdenotes the associated complex structure. It is worth saying that spheres are the only
Hopf, compact, embedded constant mean curvature surfaces in Hm
C,asitisprovenbyA.A.
Borisenkoinin[3]. The natural generalization of this concept when K∈{H,O}is being a
curvature-adapted hypersurface, that is, the normal Jacobi operator R(·,ν)ν commutes with
the shape operator.
We adopt the notation of Definition 1.2 for the rest of the paper. Let Pand Vbe the perime-
ter and volume functionals induced by gin Hm
K. Consider the (unweighted) isoperimetric
problem
infP(F):V(F)=v, F⊂Hm
KHopf-symmetric.(1.2)
We dedicate Sect. 4to the proof of the following theorem.
Theorem 1.5 If geodesic balls centered at o ∈Hn
Rare isoperimetric with respect to Problem
(1.1) for the strictly radial log-convex density
f(x)=cosh(dH(o,x)))d−1,d=dim(K),
then geodesic balls in H m
Kare optimal with respect to the isoperimetic Problem (1.2).
The explicit expression of the perimeter for Hopf-symmetric sets that we will develop in the
proof of Theorem 1.5,andTheorem1.1 will lead to the following consequence.
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Approaching the isoperimetric problem in Hm
C... Page 3 of 27 11
Corollary 1.6 In the class of Hopf-symmetric sets, geodesic balls are the unique isoperimetric
sets in H m
K.
In the past two decades, numerous researchers have shown great interest in studying the
isoperimetric problem within manifolds with positive densities on the perimeter and volume.
In the context of radial weights on Rn, C. Rosales, A. Cañete, V. Bayle and F. Morgan
established the existence of isoperimetric sets by imposing certain growth conditions on
the weight. Furthermore, they proved that spheres centered at the origin are stable if and
only if the weight is log-convex, see [20, Theorem 3.10]. Under this last assumption, K.
Brakke conjectured that balls centered at the origin are not only stable, but in fact global
minimizers of the weighted perimeter. In the aforementioned paper, the authors completely
proved this conjecture in the one dimensional case, and in all dimensions for the particular
weight exp(|x|2)via a symmetrization argument, see [20, Corollary 4.11, Theorem 5.2]. In
[19] A. Pratelli and F. Morgan provided several important new results and examples related
to this topic. In particular, existence, boundedness and mean convexity of the isoperimetric
sets are ensured in all dimensions if the density is log-convex. The conjecture was proved
in the large volume regime by A. Kolesnikov and R. Zhdanov in [12] through an ingenious
application of the divergence theorem, proving that large isoperimetric sets have to be level
sets of the given radial weight. The small volume regime was then proved by A. Figalli and F.
Maggi in [8] via a rescaling argument taking advantage of deep quantitative stability estimates
on the spheres. Moreover, the authors established the general result when the weight is close
enough in norm to the before mentioned exponential function exp(|x|2)via an interpolation
argument. The conjecture was finally proven true by the remarkable article by G. R. Chambers
[5], who obtained a complete characterization of the isoperimetric sets. The analysis relies
on a meticulous examination of the generating profile of spherical symmetrized sets. In fact,
the first and main part of this paper is an adaptation of the method to our negatively curved
case. It is worth saying that this strategy was moreover successfully employed by W. Boyer,
B. Brown, G. R. Chambers, A. Loving and S. Tammen in [4] to show the surprising fact
that for all volumes balls whose boundary passes through the origin are isoperimetric with
respect to the radial polynomial weight |x|p,forallp>0.
For what concerns curved ambient spaces, various results have been obtained. In [18]F.
Morgan, M. Hutchings, and H. Howards focused their attention on the plane enowed with
a smooth, rotationally symmetric metric with radially increasing Gauss curvature, proving
that an isoperimetric set in this case is either a circle, a complement of a circle or an annulus.
In warped product manifolds a significant result is due to S. Howe, who in addition of
generalizing the aforementioned result by A. Kolesnikov and R. Zhdanov, established several
situations in which the fibres minimize the vertical volume, see [11]. For what concerns the
hyperbolic setting, Brakke’s conjecture in the two dimensional case was proved according
to I. McGuillivray in [15]. L. Di Giosia, J. Habib, L. Kenigsberg, D. Pittman, and W. Zhu
proved in [6] via a direct application of Chamber’s theorem, that balls centered at the origin
are isoperimetric for smooth log-convex radially symmetric densities in Hn
Rintroducing an
additional weight on the perimeter to counteract the impact arising from the negatively curved
metric. The work of E. Bongiovanni, A. Diaz, A. Kakkar, and N. Sothanaphan in [2] provide
an affermative answer to Brakke’s conjecture for large volume sets containing the origin,
in the general setting of two dimensional surfaces of revolution, in which the product of
the metric factor with the given volume density is eventually log-convex. This applies for
instance to the hyperbolic plane with density equal to exp(dH(x,o)2)for some fixed base
point o∈H2
R,see[2, Corollary 5.10]. Finally, very recently in [13]H.LiandB.Xushowed
sharp isoperimetric inequalities in Hn
Rendowed with radial density of the form
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11 Page 4 of 27 L. Silini
φ(sinh(dH(x,o)) cosh(dH(x,0)),
for φeven, log-convex, and o∈Hn
Rany base point. The proof, that generalizes the result by
J. Scheuer and C. Xia in [21], cleverly applies the result of G. R. Chambers by projecting the
hyperbolic space onto Rnand employing a comparison argument. This result proves Theorem
1.5 in the case of the complex hyperbolic space by simply taking φ≡1. Our contribution
consists in a further generalization: observe that the density
f(x):= φ(sinh(dH(x,o)) cosh(dH(x,0))
is always strictly log-convex, but the converse is not true: for instance when f(x)=
cosh(dH(x,o))d−1for d>2 (like in Theorem 1.5), the associated function
φ(R):= cosh(arsinh(R))d−1
cosh(arsinh(R)) =cosh(arsinh(R))d−2
is not log-convex.
Remark 1.7 In extending the proof of Brakke’s conjecture from the Euclidean space to the
hyperbolic space, we decided for simplicity to assume the weight to be strictly log-convex
rather than simply log-convex. This choice was motivated by the technical difficulties arising
from the presence of regions with constant weight. It is worth noting that this restriction has
no bearing on the application being studied.
2 Preliminaries
In what follows, we will always assume E⊂Hn
Rto be an isoperimetric set with respect to
the weighted problem (1.1).
2.1 Qualitative properties of the isoperimetric sets
The main argument of this paper is grounded in the principles of existence and regularity
of isoperimetric sets in manifolds with densities. We refer to the work of E. Milman [16,
Section 2.2 and 2.3] as a very general reference. Existence, boundedness and mean-convexity
of isoperimetric sets in Rnendowed with a various family of densities was the focus of
the article by F. Morgan and A. Pratelli [19]. For completeness, the detailed application
of their arguments to our hyperbolic setting can be found in the Appendix 1, Theorems
A.2,A.3,andA.4. Regularity of area minimizing surfaces has been the object of study of
geometric measure theorists for many decades. The result ensuring smoothness away from
a singular set of Hausdorff dimension at most n−8 is by now a well-established and widely
acknowledged fact. For a presentation of the historical background, we recommend referring
to [17, Chapter 8], and [16, Section 2.2] for numerous references on the subject. In analogy
with the unweighted case, the last crucial property of the isoperimetric sets Eis to have
constant weighted mean curvature
Hf:= H+∂νln(f), (2.1)
at each regular point of ∂E. Here, Hdenotes the unaveraged inward Riemannian mean
curvature, and νthe outward pointing unit normal. The peculiar form of Hfis obtained via a
direct computation of the volume preserving first variation of the perimeter, see [20, Section
3]. The next theorem summarizes all the before mentioned properties of isoperimetric sets.
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Approaching the isoperimetric problem in Hm
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Theorem 2.1 (Existence and regularity) For any volume v>0there exists a set E ⊂Hn
Rof
finite perimeter and weighted volume V f(E)=vsolving the isoperimetric problem (1.1).
Moreover, E enjoys the following properties:
–∂E is a bounded C ∞embedded hypersurface outside a singular set of Hausdorff
dimension at most n −8.
– There exists λ∈Rsuch that at any regular point x ∈∂E,Hf(x)=λ. As a consequence,
∂E is mean-convex at each regular point y ∈∂E , that is H(y)≥(n−1).
– If the tangent cone at x ∈∂E lies in a halfspace, then it is a hyperplane, and therefore
∂E is regular at x. In particular, ∂E is regular at points x ∈∂E satisfying dH(x,o)=
supx∈∂EdH(x,o).
2.2 The Poincaré model of Hn
R
Adopting the Poincaré model, Hn
Ris conformal to the open Euclidean unit ball. At a point
x∈Hn
Rthe metric is
gH=4
(1−r2)2gflat,
where r=|x|will always denote the Euclidean distance of xfrom the origin, and gflat the
usual Euclidean metric of Rn. The hyperbolic distance from the origin is then given by
dH(x,0)=2artanh(r).
We define the boundary at infinity ∂∞Hn
Rof Hn
Rto be the Euclidean unit sphere ∂B(0,1)=
Sn−1. We will identify the base point o∈Hn
Rof the radial density fwith the origin 0 in
B(0,1).
2.3 Isometries and special frames in H2
R
Denote by e1and e2the horizontal and vertical Cartesian axes in the two dimensional Poincaré
disk model. Also, let (H2
R)+be the intersection of H2
Rwith the closed upper half-plane
having e1as boundary. The isometry group of (H2
R,gH)is completely determined (up to
orientation) by the Möbius transformations preserving the boundary ∂∞H2
R. Hence, geodesic
circles coincide with Euclidean circles completely contained in H2
R. Their curvature lies in
(1,+∞). Circles touching ∂∞H2
Rin a point are called horocycles, and have curvature equal
to 1. Geodesics are arcs of (possibly degenerate) circles hitting ∂∞H2
Rperpendicularly in two
points. Arcs of (possibly degenerate) circles that are not geodesics are called hypercycles,
and have constant curvature in (−1,1)\{0}. It will be convenient to work with a particular
frame: define
S:(H2
R)+→R,
to be the hyperbolic distance of a point in (H2
R)+from the horizontal axis e1.SetX=∇S,
where we naturally extend by continuity Xat e1. Then, denoting with X⊥the counterclock-
wise rotation of Xby π
2radians, since the level sets of Sare equidistant to each other,
{X,X⊥}forms an orthonormal frame of (H2
R)+,seeFig.1. The integral curves of Xare all
geodesic rays hitting e1perpendicularly. For each l∈[0,1),letδlbe the integral curve of
X⊥so that δl(0)=(0,l). Then, (δl)l∈[0,1)is a family of equidistant hypercycles foliating
(H2
R)+, crossing e2perpendicularly and with constant curvature which coincides with the
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11 Page 6 of 27 L. Silini
Fig. 1 The special frame {X,X⊥}
Eucliden one: K1=2l
1+l2=1
R(l),where R(l)∈(0,+∞]is the radius of the Euclidean circle
representing the curve. Similarly, set {N,N⊥}to be the orthonormal frame on H2
R\{0}where
Nis the radial unit length vector field emanating from the origin. Then, integral curves of N
are rays of geodesics, and integral curves of N⊥are concentric geodesic cycles. Notice that
the frame {X,X⊥}is invariant under the one-parameter subgroup of hyperbolic isometries
fixing e1(X⊥is the infinitesimal generator of the action by translations) and, up to reverse
the orientation, under the reflections with respect to any geodesic integral curve of X. Finally,
notice that on e1and e2,{X,X⊥}is a positive rescaling of {(0,1), (−1,0)}.
For a regular curve parametrized by arc length ηwe denote with κη(t)the inward signed
curvature of ηat η(t). We recall the identity
κη˙η⊥=∇˙η˙η,
where here ∇denotes the standard Levi-Civita connection associated to gH.
2.4 Reduction to H2
R
From now on, let Ebe an isoperimetric set with arbitrary weighted volume. Since both the
density fand the conformal term of gHare radial, the coarea formula implies that spherical
symmetrization pointed at the origin preserves the weighted volume and does not increase
the weighted perimeter of E(see [19, Theorem 6.2]). For this reason, we will assume E
spherically symmetric with respect to the e1axis. Now, intersecting Ewith the the Euclidean
plane spanned by {e1,e2}, we obtain a spherically symmetric profile ⊂H2
R.Letxbe
the furthest point of lying in the positive part of the e1axis (this is always possible by
reflecting with respect to the e2geodesic). Let γ:[−a,a]→H2
Rbe a counter-clockwise,
arclength parametrization of the boundary of the connected component of containing x,
so that γ(0)=x,seeFig.2.Thecurveγenjoys the following properties:
–γis smooth on (−a,a).Indeed,ifthereexistsa∗∈(−a,a)such that γ(a∗)is not regular,
then ∂Econtains a singular set of Hausdorff dimension n−2, but this is impossible
because of Theorem 2.1.
–γis symmetric with respect to the axis e1.
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Fig. 2 The spherical symmetrization
–Thecurveγforms a simple, closed curve.
– Writing γ=(γ1,γ
2)in cartesian coordinates, one has that sgn(γ2(t)) =sgn(t).In
particular, γ:[0,a)→(H2
R)+.
–˙γ(0)=X(γ (0)).
To translate Eq. (2.1) as a property of the profile γ, we need the following definition.
Definition 2.2 For any t∈[0,a), denote by
–Ct=Ct(s)the (possibly degenerated) oriented circle tangent to γ(t), with center on
e1, parametrized by arclength and such that Ct(0)=γ(t).Denotebyκ(Ct)its constant
curvature.
–ct=ct(s)the (possibly degenerated) oriented circle tangent to γ(t), parametrized by
arclength, such that ct(0)=γ(t)and κ(ct)=κγ(t).
–x(Ct)and x(ct)the hyperbolic center of Ctand ctrespectively. Similarly, let x1(Ct)and
x1(ct)be the first Euclidean coordinate of x(Ct)and x(ct)respectively.
Remark 2.3 Let F⊂B(0,1)⊂Rn. Then, at every regular point x∈∂F, the mean curvature
His related with the Euclidean mean curvature Hflat by
H(x)=1−r2
2Hflat(x)+(n−1)gflat (x,˜ν),
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11 Page 8 of 27 L. Silini
where ˜νis the outward normal vector to ∂Fwith Euclidean norm equal to one. In particular,
when n=2, denoting with κflat the usual Euclidean curvature, one has that
κη=1−|γ(t)|2
2κflat
η+gflat(η, ˜ν).
Therefore, κflat(ct)=κflat
γ, that is comparison circles ctand Ctin the hyperbolic setting
coincide with comparison circles with respect to the Euclidean metric. From this formula,
we also deduce that for any Euclidean circle C
κC=1
21−|x0|2
τ0+τ0=coth(τ ),
where x0and τ0are the Euclidean center and radius, and τis the hyperbolic radius.
Lemma 2.4 On t ∈[0,a)it holds
H(t)=κγ(t)+(n−2)κ(Ct).
In particular,
Hf(t)=κγ(t)+(n−2)κ(Ct)+h(dH(o,γ(t)))gH(ν(t), N(γ (t))) =λ,
where ν=−˙γ⊥.
We call H1(t):= ∂ν(ln(f))(γ (t)) =h(dH(o,γ(t)))gH(ν(t), N(γ (t))) the term coming
from the log-convex density.
Proof In [5, Proposition 3.1] it is shown that the Euclidean mean curvature of the spherically
symmetric set Ecan be computed as
Hflat =κflat
γ+(n−2)κflat(Ct).
Thanks to Remark 2.3 we have that
H(γ (t)) =1−|γ(t)|2
2Hflat(γ (t)) +(n−1)gflat (γ (t), ν)
=1−|γ(t)|2
2κflat
γ+(n−2)κflat(Ct)+(n−1)gflat(γ (t), ν )
=κγ+(n−1)κ(Ct).

3 The proof
We have seen that existence, boundedness, and regularity of isoperimetric sets (Theorems
2.1,A.2,andA.3) together with the radial nature of the density fallows us to assume
the optimal set Eto be bounded and spherically symmetric with generating curve smooth
away from the axis of symmetrization. Consequently, the problem is reduced to a planar
situation, in which the profile curve γsolves the ordinary differential equation induced
by the constant weighted mean curvature Hfof the original isoperimetric set, as stated in
Lemma 2.4. Adapting Chamber’s analysis to our specific situation presents difficulties as the
nonexistence of a natural choice of a frame on the tangent space as in the Euclidean plane.
Consequently, to carry out a rigorous curvature-comparison analysis (for instance Lemma
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Fig. 3 The curl described in Lemma 3.2
3.15), it is crucial to carefully select a frame that appropriately accommodates the geometry
of our particular case, as we did in Sect. 2.3. The proof of Theorem 1.1 relies on showing
that γrepresents a circumference centered at the origin. The argument presented shows that
refuting this possibility leads to a surprising consequence: the curve γmust make a curl, as
represented in Fig. 3, contradicting the fact that γis the parameterization of a spherically
symmetric set. More rigorously, the contradiction arises as the combination of the next two
lemmas.
Lemma 3.1 For every t ∈(0,a)
gH(N,˙γ) ≤0.
Proof The fact that the set is spherically symmetric implies that t→ gflat(γ (t), γ (t)) is
non increasing. Differentiating in tgives the desired sign of the angle between Nand ˙γ.
Sect. 3.1 is devoted to the proof of the next lemma.
Lemma 3.2 (The tangent lemma) If γis not a circle centered in the origin, there exist 0<
a0<a1<a2<a such that ˙γ(a0)=X⊥(γ (a0)),˙γ(a1)=−X(γ (a1)) and ˙γ(a2)=
X(γ (a2)).
Assuming now that Lemma 3.2 holds true, the proof of the main result goes as follows.
Proof of Theorem 1.1 If γis a circle centered at the origin we are done. Otherwise, Lemma
3.2 ensures the existence of 0 <a2<asuch that ˙γ(a2)=X(γ (a2)). This violates the
inequality of Lemma 3.1, because at a2
gH(N,˙γ) =gH(N,X)>0.
Therefore, the profile curve γhas to be a circumference centered in the origin. Uniqueness
is established by observing that up to measure zero the only set which, when spherically
symmetrized, results in a centered ball, is a centered ball itself. 
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11 Page 10 of 27 L. Silini
3.1 Proof of the tangent lemma
The proof is made by following the behaviour of γstep by step: first we show that γhas
to arch upwards with curvature strictly greater than one. The endpoint of this arc will be
γ(a0),where ˙γ(a0)=X⊥(γ (a0)). Then, it goes down curving strictly faster than before,
and this result about curvature is the tricky point to generalize in the hyperbolic setting. It
turns out that the special frame given by the hypercyclical foliation (δl)l∈[0,1)is the good
one. Then, arguing by contradiction, we will show that this behaviour must end at a point
0<a0<a1<a,where ˙γ(a1)=−X(γ (a1)). Finally, we prove the existence of a2so
that ˙γ(a2)=X(γ (a2)) taking advantage of the mean-curvature convexity of .Westartby
looking at what happens at the starting point.
Lemma 3.3 One has that ˙γ(0)=X(γ (0)),˙κγ(0)=0and κγ(0)≥κ(C0)>1.
Proof This is a consequence of the symmetry of γwith respect to the e1axis, and of the fact
that γ(0)represent the furthest point from the origin of .
Lemma 3.4 If there exists t∗∈[0,a)such that x1(Ct∗)=0and κγ(t∗)=κ(Ct∗),thenγ
has to be a centered circle.
Proof In this case γ(t)and Ct∗(s)solve the same ODE, with same initial data. Therefore,
they have to coincide locally, and hence globally. 
Definition 3.5 Call α:[0,a)→[π, −π) theorientedanglemadeby ˙γwith X⊥.Wesay
that ˙γ(t)is in the I, II, III and IV quadrant if α(t)belongs to [π/2,π],[0,π/2],[0,−π/2]
and [−π/2,−π]respectively. We add strictly if ˙γis not collinear to Xand X⊥.
Lemma 3.6 If for some t ∈[0,a),˙γ(t)belongs to the II quadrant, then x1(Ct)≥0.
Proof We first treat the case t∈(0,a). Expressing N(γ (t)) in the {X,X⊥}frame, we have
thanks to Lemma 3.1 that
0≥gH(N,˙γ) =gH(X,N)gH(X,˙γ) +gH(X⊥,N)gH(X⊥,˙γ)
=gH(X,N)sin(α) +gH(X⊥,N)cos(α). (3.1)
If α=π/2, then
0≥gH(X,N),
which is possible only when γ2(t)=0, that is t∈{0,a}.Ifα∈[0,π/2)then cos(α) > 0,
implying that gH(X⊥,N)≤−g(X,N)tan(α) ≤0. Notice that this is possible only if
γ1(t)≥0. Calling −ϑ<0 the angle that Nmakes with X,wegetbyEq.(3.1)that
tan(α) ≤tan(ϑ ). (3.2)
Now, observe that the two geodesic rays σγ,σNstating at γ(t)with initial velocities ˙σγ(0)=
˙γ⊥(t)and ˙σN(0)=N⊥(γ (t)), together with the axis e1and the geodesic orthogonal to e1
passing from γ(t)bound two geodesic triangles γand N.Calldthe distance between
γ(t)and e1. Then, the length of the sides γand Nof γand Nrespectively, lying on e1
are given via hyperbolic trigonometric laws by
tanh(γ)=tan(α) sinh(d)≤tan(ϑ) sinh(d)=tanh(N).
But this implies that x(Ct), which is the intersection of σγwith e1, has first coordinate
positive, as claimed. If t=0, then C0=c0and approximates γ(0)up to the fourth order.
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Approaching the isoperimetric problem in Hm
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Therefore, if x1(Ct)<0, then there exists ε>0 such that γ|(ε,2ε) lies outside the ball
of centered in the origin and with radius dH(γ (0), o). This is a contradiction because by
construction γ(0)is the furthest point of from o.
Our next goal is to show four important properties of the curve γ. The proof is made by
comparison with the circles ctand Ct, and the preservation of the weighted mean curvature
Hf. For this reason, we need the following preliminary lemma.
Lemma 3.7 Let η=η(s)be an arc-length, counter-clockwise parametrization of a circle
centered at (0,y)such that η(0)=(τ, y)and η(L)=(0,y+τ).Let O =(−˜o,0)be an
arbitrary point lying on e1with ˜o∈[0,1), and ν(s)the outward pointing normal to η(s).
Then, setting
˜
H1(s):= ∂ν(h(dH(O,x)))|x=η(s),
if y =0,then
˜
H
1(s)≤0,in (0,L],(3.3)
and
˜
H
1(0)≤0.(3.4)
Both inequalities are strict if ˜o= 0.Ify ∈(0,1)and ˜o= 0,then
˜
H
1(L)<0.(3.5)
Proof Let T:H2
R→H2
Rbe the unique isometry fixing e1and sending the origin to O.
Then,
˜
H1(s)=∂ν(h(dH(O,x)))|x=η(s)=h(dH(O,η(s)))gH(ν(s), T∗N(η(s))).
For the sake of exposition, we will omit the arguments in the following computations.
Differentiating one time in swe have that
˜
H
1(s)=hgH(T∗N,˙η)gH(ν, T∗N)+hd
ds gH(ν, T∗N)
=hgH(T∗N,˙η)gH(ν, T∗N)−hgH(∇˙η˙η⊥,T∗N)+gH(˙η⊥,∇˙ηT∗N)
=hgH(T∗N,˙η)gH(ν, T∗N)−h−gH(T∗N,˙η)κη+gH(T∗N,˙η)gH(T∗N⊥,˙η)κ1,
where κ1is the curvature of the integral curve of T∗N⊥passing through η(s),whichisa
geodesic sphere centered at O. Suppose first that y=0and ˜o= 0. Then, ˙η=N⊥,and
˜
H
1(s)=hgH(T∗N,N⊥)gH(N,T∗N)−hgH(T∗N,N⊥)(−κη+gH(T∗N,N)κ1)<0,
because h >0, h>0, gH(T∗N,N⊥)<0, gH(T∗N,N)>0andκη>κ
1since ˜o= 0.
This proves Equation (3.3)when ˜o= 0. The same holds in the context of Eq. (3.5)since
˙η(L)=N⊥. Up to relaxing the inequalities, the proof when ˜o=0 is exactly the same. To
prove Eq. (3.4), we differentiate ˜
H1one more time, obtaining
˜
H
1(s)=hgH(T∗N,˙η)2gH(ν, T∗N)+h d
ds gH(T∗N,˙η)gH(ν, T∗N)
+hgH(T∗N,˙η) d
ds gH(ν, T∗N)+hgH(T∗N,˙η) d
ds gH(ν, T∗N)
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11 Page 12 of 27 L. Silini
+hd2
ds2gH(ν, T∗N).
Observe that in zero gH(T∗N,˙η) =0, hence only the second and last term survive
˜
H
1(0)=h d
dss=0gH(T∗N,N⊥)gH(T∗N,N)+hd2
ds2s=0gH(T∗N,N).
Taking advantage of the explicit expression for d
ds g(T∗N,N)we obtained before, we get
d2
ds2s=0gH(T∗N,N)=−d
dss=0gH(T∗N,N⊥)−κη+gH(T∗N,N)κ1
=κη−gH(T∗N,N)κ1d
dss=0gH(T∗N,N⊥),
which implies that
˜
H
1(0)=hgH(T∗N,N)+hκη−hgH(T∗N,N)κ1
 
>0
d
dss=0gH(T∗N,N⊥).
Hence, we are left to show that d
ds s=0gH(T∗N,N⊥)<0. Developing again we get
d
dss=0gH(T∗N,N⊥)=gH(∇N⊥T∗N,N⊥)|s=0+gH(T∗N,∇N⊥N⊥)|s=0
=gH(T∗N,N)2κ1|s=0−gH(TN,N)κη|s=0
=κ1−κη<0.

We are now ready to prove the next result.
Lemma 3.8 The following four points hold.
iIffort∈(0,a)one has that κγ(t)≥κ(Ct)>1,thent → x1(Ct)is smooth and
d
dt x1(Ct)≥0.
ii If γis not a centered circle, then ¨κγ(0)>0.
iii If for t ∈(0,a),˙γ(t)is in the II quadrant and κγ(t)=κ(Ct)>1,then ˙κγ(t)≥0.
Moreover, if ˙γ(t)= X⊥(γ (t)) and Ctis not centered in the origin, then ˙κγ(t)>0.
iv If for t ∈(0,a)one has that ˙γ(t)=X⊥(γ (t)),γ1(t)>0and κγ(t)≥κ(Ct)>1,then
˙κγ(t)>0.
Proof We start with point i. Observe that since ctapproximates γup to the third order around
γ(t), it suffices to prove d
dt x1(Ct)≥0 replacing γwith ct. Also, we can suppose x(ct)on
e2by composing with the unique hyperbolic isometry translating x(ct)on e2and fixing e1.
The curvature condition κγ(t)≥κ(Ct)>1 ensures that x(ct)∈(H2
R)+. By monotonicity
of the function tanh(·/2), it suffices to prove the claim for the Euclidean center of Ct. Thus,
we have reduced the problem to an explicit computation in the Euclidean plane, that can be
found in [5, Lemma 5.3]. Thanks to Lemma 3.7 the proofs of the other points go exactly as
in [5, Lemma 3.4, Lemma 3.5 and Lemma 3.7]. We show point ii. Differentiating Hftwice,
we get that
¨κγ(0)=−(n−2)¨κ(C0)−H
1.
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By symmetry, c0=C0. Moreover, since ˙κγ(0)=0, we have that both c0and C0approximate
γup to the fourth order near zero. Hence, ¨κ(C0)=0. Therefore, it suffices to determine
the sign of H
1replacing γwith C0.LetT:H2
R→H2
Rbe the unique isometry fixing e1
that moves x(C0)to the origin. The result follows by Equation (3.4) of Lemma 3.7 setting
O=T(0), and noticing that T(0)= 0 by Lemma 3.4. The proofs of points iii.andiv.are
similar: in the first case the condition κγ(t)=κ(Ct)implies that ct=Ctapproximates
γnear tup to the third order, the same holds if ˙γ(t)=X⊥(γ (t)) by symmetry. Hence,
substituting γwith ctand differentiating one time Hf,wehavetodeterminethesignofH
1
in the case of a circle, via Lemma 3.7.
We are now ready to analyse the first behaviour of γ.
Definition 3.9 (Upper curve) The upper curve is the (possibly empty) maximal connected
interval IU⊂[0,a)such that 0 ∈IUand for all t∈IU
a˙γ(t)is in the II quadrant,
bκγ(t)≥κ(Ct)>1,
ct→ x1(Ct)is smooth and d
dt x1(Ct)≥0.
We set
a0:= sup IU.
In the discussion, we will sometimes identify the upper curve with its image through γ.
Definition 3.10 We sa y t ha t a c ur ve ηis graphical with respect to the hypercyclic foliation
(δl)l∈[0,1)if ηmeets each δlat most once.
Notice that the upper curve (if non empty) is graphical with respect to the hypercyclical
foliation because ˙γis in the II quadrant
Proposition 3.11 The upper curve is non empty and enjoys the following properties
i. 0<a0<a,
ii. a0∈IU,
iii. γ1(a0)>0,
iv. ˙γ(a0)=X⊥(γ (a0)).
Proof Thanks to Lemma 3.8, the proof goes exactly as [5, Lemma 3.11 and Proposition 3.12].
We sketch for completeness the idea behind each point. We start by showing that the upper
curve is non empty.
By Lemma 3.3 we know that ˙γ(0)=X(γ (0)),˙κγ(0)=0, and κγ(0)≥κ(C0)>1.
Moreover, by Lemma 3.8 point ii. since by assumption γis not a centered circle, we get
that ¨κγ(0)>0. By continuity, since c0=C0approximates γup to the fourth order near
zero, we have that there exists ε>0 such that for all t∈[0,ε) points a.andb. in Definition
3.9 are satisfied. Finally, point c. of Definition 3.9 follows from Lemma 3.8 point i.which
asserts that κγ(t)≥κ(Ct)>1 implies that x1(Ct)is smooth and d
dt x1(Ct)≥0. Hence,
[0,ε) ⊂IU, proving that the upper curve cannot be empty.
Notice that 0 <a0cannot be equal to asince otherwise the curve γdoes not close itself
on e1, simply because ˙γbelongs to the II quadrant by definition of IU. By the regularity of
γand that IUis defined by three close conditions, we have that a0∈IU. By composing with
the unique hyperbolic isometry sending γ(a0)on e2fixing e1,wecanseethatx1(Ca0)≤0
because ˙γ(a0)belongs to the II quadrant. Lemma 3.4 and Lemma 3.6 imply that x1(C0)>0
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11 Page 14 of 27 L. Silini
Fig. 4 The curvature comparison
and since d
dt x1(Ct)≥0inIU, one must have that γ1(a0)>0. The last point is proved by
contradiction: if ˙γ(a0)= X⊥(γ (a0)),thena0∈IUimplies that ˙γ(a0)is strictly in the II
quadrant. If κγ(a0)=κ(Ca0)>1, then ca0=Ca0approximates γto the third order and
Lemma 3.8 point iii. implies that there exists some δ>0 such that κγ(t)≥κ(Ct)>1
for t∈[a0,a0+δ). The same holds if κγ(a0)>κ(Ca0)>1 by continuity. This means
that [a0,a0+δ) ⊂IU, which is not possible by the very definition of a0. Hence, ˙γ(a0)=
X⊥(γ (a0)).
Definition 3.12 (Lower curve) The lower curve is the maximal connected interval IL⊂
[a0,a)such that for all t∈IL
a. a0∈IL,
b. ˙γ(t)is in the III quadrant,
c. calling ¯
t∈IUthe unique time such that S(γ (t)) =S(γ ( ¯
t)) we have that κγ(t)≥κγ(¯
t).
We set
a1:= sup IL.
Notice that a0truly belongs to IL,soIL=∅. Also, the lower curve is graphical with respect
to the hypercyclical foliation because ˙γis in the III quadrant. Our next goal is to prove that
a1<a. Again, we proceed by contradiction, and the intuition is the following: suppose that
a1=a.Ifκγ(t)=κγ(¯
t)for all t∈IL, then the lower curve is nothing else than the upper
curve reflected with respect to the geodesic orthogonal to e1and passing through γ(a0).
Hence, limt→a+α(t)=−
π
2. Otherwise, if the γ|ILcurves strictly faster than the upper
curve at some point, then the angle of incidence limt→a+α(t)<−π
2(see Fig. 4). But this
cannot be true, because it contradicts the regularity of ∂Epointed out in Theorem 2.1.To
prove that the lower curve curves strictly faster than the upper curve we need first to express
the curvature with respect to the {X,X⊥}frame, and next prove three comparison lemmas.
Lemma 3.13 Let ηany regular curve parametrized by arclength such that ˙η(t)is not collinear
to X(η(t)). Then,
−κη(t)=˙
β(t)−K1(η(t)) cos(β(t)),
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where β(t)denotes the angle between ˙ηand X⊥, and K 1is the curvature of the leaf δlpassing
through η(t).
Proof Decompose ˙η=AX +BX⊥. Then, since κη˙η⊥=∇˙η˙η,wegetthat
−cos(β)κη=gH∇˙η˙η, X=∂tsin(β )−gH˙η, ∇˙ηX.
Now, keeping in mind that ∇XX=0andgH∇X⊥X⊥,X=−K1(η(t)),weget
gHAX +BX⊥,∇AX+BX⊥X=B2gHX⊥,∇X⊥X=cos(β)2K1(η(t)).
Dividing both sides by cos(β) we get the desired identity. 
We can prove our first curvature comparison lemma.
Lemma 3.14 (κcomparison lemma) Let η1:(0,A1]→H2
Rand η2:(0,A2]→H2
Rbe two
hypercyclical graphical curves parametrized by arclength and with velocity vectors in the II
quadrant. Suppose that there exists l0∈[0,1)such that
lim
t→0+η1(t)and lim
t→0+η2(t),
exist and belong to the same leaf δl0. Also, suppose that η1(A1)=η2(A2),˙η1(A1)=˙η2(A2),
and that if S(η1(t)) =S(η2(τ )) then κη1(t)≥κη2(τ ). Then, calling α1and α2the angle
made by ˙η1and ˙η2with X⊥we have that
lim
t→0+α1(t)≥lim
t→0+α2(t).
Moreovoer, if for some t and τsuch that S (η1(t)) =S(η2(τ )), one has that κ1(t)>κ
2(τ ),
then
lim
t→0+α1(t)> lim
t→0+α2(t).
Proof Since the curves are graphical with respect to the hypercyclical foliation we can operate
a change of variable: we observe that the two height functions s1(t):= S(η1(t)) and s2(τ) =
S(η2(τ )) are bijections with same image of the form (l0,L]. By hypothesis κη1(s−1
1(l)) ≥
κη2(s−1
2(l)) for every l∈(l0,L]. Comparing the two curves in the l∈(l0,L]variable, since
s
i=gH(∇S,˙ηi)=gH(X,˙ηi)=sin(αi),i=1,2, we get by Lemma 3.13 that
0≤κη1(l)−κη2(l)=˙α2(l)sin(α2(l)) −˙α1(l)sin(α1(l)) −2l
1+l2cos(α2(l)) −cos(α1(l)).
Multiplying by (1+l2)and integrating we finally get that
0≤L
l0
(1+l2)(cos(α1)cos(α2))+2l(cos(α1)−cos(α2)) dl
=L
l0
d
dl (1+l2)(cos(α1)−cos(α2))dl =lim
l→l+
0
(1+l2)(cos(α2(l)) −cos(α1(l))).
If the two curvatures are different somewhere, then the inequality between the two angles is
strict. 
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11 Page 16 of 27 L. Silini
Fig. 5 The curvature comparison for Ct
Lemma 3.15 (κ(Ct)comparison lemma) Let η1,η2be as in Lemma 3.14. Then, for any two
points η1(t1)and η2(t2)on the same leaf δl, calling C1and C 2the comparison circles at
η1(t1)and η2(t2)as in Definition 2.2, we have that
κ(C1)≤κ(C2).
Proof For i=1,2, the hyperbolic radius of Citogether with e1and the geodesic starting from
ηi(ti)and hitting e1perpendicularly bound a geodesic triangle i.Letd1
ibe the hyperbolic
radius, d2
ibe the side touching e1and d3
ithe the remaining side of i. Similarly, for i=1,2
and j=1,2,3, call βj
ithe angle opposite to dj
i,andj
ithe length of dj
i. We refer to Fig. 5.By
construction β1
1=β1
2=π
2,β2
i=αi, and since η1(t1)and η2(t2)are in the same hypercycle
by hypothesis, we get 3
1=3
2. Then, by the hyperbolic law of cosines and by Lemma 3.14
we get
κ(C1)=coth(1
1)=cos(α1)
tanh(3
1)≤cos(α2)
tanh(3
2)=coth(1
2)=κ(C2). (3.6)

Lemma 3.16 (H1comparison lemma) Let η1,η2be as in Lemma 3.14 and let ˜η1be the
reflection of η1with respect to the geodesic passing through η1(A1)and crossing e1perpen-
dicularly. Reverse its parametrization, so that the angle that ˙
˜η1makes with X⊥is equal to
˜α1=−α1. Moreover, suppose that
gH(N,˙η2)≤0.
Denote the unitary outward pointing normals to η1and η2by ˜ν1and ν2. Then, for any two
points ˜η1(t1)and η2(t2)on the same leaf δlwe have that
gH(N(˜η1(t1)), ˜ν1(t1)) ≤gH(N(η2(t2)), ν2(t2)),
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Approaching the isoperimetric problem in Hm
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with equality if and only if ˙η2(t2)and ˙
˜η1(t1)are tangent to the same circle centered in the
origin and ˙
˜η1(t1)=−˙η2(t2).
Proof Parametrize δl:R→H2
Rby arclength in the X⊥direction, so that δlintersects e2
at δl(0).Letϑ(s)be the angle that Nmakes with X⊥at δl(s)and s2<s1be such that
δl(s1)=˜η1(t1)and δl(s2)=η2(t2).Let1,
2∈[0,π
2]be the angles that Nmakes with
ν1(t1)and ν2(t2)respectively. Then, 1:= ϑ(s1)−˜α1−π
2,2:= ϑ(s2)−α2−π
2,
gH(N(˜η1(t1)), ˜ν1(t1)) =cos(1),
and
gH(N(η2(t2)), ν2(t2)) =cos(2).
We need to investigate if 2≤1,andwhen2<
1.Lets∗
2∈Rbe such that the unit
vector at δl(s∗
2)that forms an angle of α2with X⊥is tangent to a circle centered in the origin.
The value s∗
2exists in the interval [s2,0)because by Lemma 3.6,Eq.(3.2), we have that
ϑ(s2)−π
2≥α2and, in the intersection of δlwith e2we have that ϑ(0)−π
2=0≤α2.By
continuity there must be a point s2≤s∗
2<0 such that ϑ(s∗
2)−π
2=α2.Then
2=−ϑ(s∗
2)+ϑ(s2)=−s∗
2
s2˙
ϑ(s)ds.
Set s∗
1:= −s∗
2, and notice that ϑ(−s)=π−ϑ(s)for every s≥0. Then,
1=ϑ(s1)−˜α1−π
2=ϑ(s1)+α2−(˜α1+α2)−π
2
=ϑ(s1)+ϑ(s∗
2)−π−(˜α1+α2)
=ϑ(s1)−ϑ(s∗
1)−(˜α1+α2)
=−s∗
1
s1˙
ϑ(s)ds −(˜α1+α2).
Hence,
2−1=(˜α1+α2)−s∗
2
s2˙
ϑ(s)ds +s∗
1
s1˙
ϑ(s)ds ≤−s∗
2
s2˙
ϑ(s)ds +s∗
1
s1˙
ϑ(s)ds,
since ˜α1+α2≤0 by Lemma 3.14.Let=2artanh(l)be the hyperbolic distance of δlfrom
e1. We claim that
˙
ϑ(s)=−
sinh() cosh() cosh2(sech()s)+coth() sinh2(sech()s)−1<0.(3.7)
Assuming that this identity holds, letting δ:= s∗
2−s2,wehavethat
2−1≤−s∗
2
s∗
2−δ˙
ϑ(s)ds +s∗
1
s∗
1−δ˙
ϑ(s)ds +s∗
1−δ
s1˙
ϑ(s)ds
=δ
0˙
ϑ(s∗
2−τ)−˙
ϑ(s∗
2+τ)dτ+s∗
1−δ
s1˙
ϑ(s)ds
≤s∗
1−δ
s1˙
ϑ(s)ds ≤0,
whereweusedthats∗
1=−s∗
2and ˙
ϑis a strictly negative, even function increasing in
[0,+∞). This implies that 2≤1, with equality if and only if δ=0andα1=α2,that
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11 Page 18 of 27 L. Silini
is when ˙η2and ˙
˜η1are tangent to the same circle centered in the origin. We are left to prove
Equation (3.7). Let β(s)be the angle that Xmakes with Nin δl(s).Sinceβ(s)+π
2=ϑ(s),
it suffices to compute ˙
β(s). The hypercycle δl(s)has curvature 2l
1+l2=tanh().Thecircle
centered in the origin passing through δl(s)has curvature coth(dH(0,δ
l(s))) =cos(β(s))
tanh() ,by
the hyperbolic trigonometric laws (as in Equation (3.6)). Now, we obtain an ODE for β(s)
arguing as in Lemma 3.13:atanytimes∈Rwe have that
−tanh() cos(β) =−tanh()gH(˙
δl,N⊥)=gH(∇˙
δl˙
δl,N)
=d
ds(gH(˙
δl,N)) −gH(˙
δl,∇˙
δlN)
=−cos(β ) ˙
β+gH(˙
δl,N⊥)2gH(∇N⊥N⊥,N)
=−cos(β ) ˙
β−cos(β)3
tanh() .
Dividing both sides by cos(β) = 0 it follows that
˙
β(s)=tanh() −cos(β(s))2
tanh() ,s∈R,
β(0)=0.
By integration, one can compute the explicit solution
β(s)=−arctancsch() tanh(sech()s),
that by differentiation gives
˙
ϑ(s)=˙
β(s)=−
sinh() cosh() cosh2(sech()s)+coth() sinh2(sech()s)−1,
proving Eq. (3.7). 
We can prove the main result about the lower curve.
Proposition 3.17 It γis not a circle centered in the origin, the lower curve is contained in
[a0,a), that is 0<a1<a. Furthermore, a1∈ILand ˙γ(a1)=−X(γ (a1)).
Proof By property iv. of Lemma 3.8,a1>a>0. Suppose by contradiction that a1=a.Set
˜η1to be the (reparametrized) lower curve and η2the upper curve. Choose any point t∈IL
with corresponding ¯
t∈IU. Applying Lemma 3.15 and Lemma 3.16 to ˜η1(t)and η2(¯
t),and
taking advantage of the expression for HfgiveninLemma2.4,wegetthat
Hf(γ (¯
t)) =Hf(γ (t)) =κγ(t)+(n−2)κ(Ct)+h(dH(0,γ(t)))gH(ν(t), N)
<κ
γ(t)+(n−2)κ(C¯
t)+h(dH(0,γ(t)))gH(ν(t), N).
We have th at
dH(0,γ(t)) ≤dH(0,γ(¯
t)).
This can be verified again via the trigonometric rules for hyperbolic triangles: fix t∈IU,
and call βand ¯
βthe angle that Nmakes with X(γ (t)) and X(γ ( ¯
t)) respectively. Notice that
0≤β≤¯
β. Then, calling dthe distance of γ(t)and γ(¯
t)from e1,wegetthat
tanh(dH(0,γ(t))) =tanh(d)
cos(β) ≤tanh(d)
cos(¯
β) =tanh(dH(0,γ(¯
t))).
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Approaching the isoperimetric problem in Hm
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Hence
Hf(γ (¯
t)) < κγ(t)+(n−2)κ(C¯
t)+h(dH(0,γ(¯
t)))gH(ν(¯
t), N),
implying
κγ(¯
t)<κ
γ(t),
since his strictly convex. This is a contradiction because Lemma 3.14 tells us that the lower
curve hits the e1axis with an angle strictly smaller than −π
2. Therefore, a1<a.Sinceγis
smooth in a1<a, and the conditions on ILare closed, we deduce that a1∈IL. Suppose now
that α(a1)is strictly in the III quadrant. Since a1∈IL, we can apply again the comparison
lemmas to γ(a1)and γ(¯a1)to infer
κγ(¯a1)<κ
γ(a1).
By continuity of κγand ˙γaround a1, we get that there exists a neighbourhood of a1in which
˙γis in the III quadrant and the above inequality holds in the not strict sense. But this implies
that a1is not the supremum of IL. Therefore, the velocity vector of γat a1has to be equal
to −X.
We prove the last part of the tangent lemma.
Proposition 3.18 If γis not a centered circle, then there exists 0<a1<a2such that
˙γ(a2)=X(γ (a2)).
Proof If a2exists we are done. Otherwise, we show that the non existence contradicts the
mean-curvature convexity of .Let
Ic:= {t∈[a1,a):˙γis in the I or IV quadrant}.
Here the index stands for curl curve.Set˜a2:= sup Ic.Sinceκ(a1)>1wehavethata1<˜a2.
If ˜a2<a, then the mean convexity of implies that
κγ(˜a2)≥(n−1)−(n−2)κ (C˜a2)>0.
To see this, move γ(˜a2)on e2as in Lemma 3.8. Then, C˜a2is oriented clockwise, and hence
has negative curvature. But this implies that we can extend Icafter ˜a2, contradicting the
definition of ˜a2. So, we need to rule out the situation in which ˜a2=a. If it is the case, then
again for mean-convexity one has that in Ic
κγ(t)>1.
Moreover, for t∈Ic\{a1}we have that ˙γlies in the IV quadrant, because otherwise this
implies together with κγ(t)>0thatγcannot close at e1. Therefore α(t)lies in the IV
quadrant and it is strictly increasing, implying that
lim
t→a+α(t)<−π
2.
This cannot happen because of the before mentioned regularity properties of isoperimetric
sets. 
The proof of the tangent lemma is then the collection of the results we showed in this section.
Proof of Lemma 3.2 The existence of the chain 0 <a0<a1<a2<ais ensured by
Proposition 3.11, Proposition 3.17 and Proposition 3.18.
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11 Page 20 of 27 L. Silini
4 Symmetric sets in Hm
K
Consider any rank one symmetric space of non-compact type (Mn,g)=(Hm
K,g),K∈
{C,H,O}.Setd=dim(K)∈{2,4,8}so that the real dimension of Mis n=md. Recall
that if K=O, we only have the Cayley plane H2
O. As classical references on symmetric
spaces we cite the books of Eberlein [7] and Helgason [10]. Fix an arbitrary base point o∈M,
and let Nbe the unit-length, radial vector field emanating from it. As in Definition 1.2,let
Hbe the distribution on M\{o}induced by the (−4)-eigenspace of the Jacobi operator
R(·,N)N. Denote with Vthe orthogonal complement of Hwith respect to g.Forevery
x∈M\{o}, we have the orthogonal splitting
TxM=Hx⊕Vx,
with orthogonal projections (·)Hand (·)V.Letnow(¯
Mn,gH)=(Hn
R,gH), and choose an
arbitrary base point ¯oin it. The isometric identification of ToMwith T¯o¯
Maccording to the
flat metrics (expM
o)∗g|0and (exp ¯
M
¯o)∗gH|0, induces a well defined diffeomorphism
=exp ¯
M
¯o◦(exp M
o)−1:M→¯
M.
With a slight abuse of notation, we still denote with gHthe metric ∗gH, that makes M
isometric to ¯
M. The following lemma allows us to compare gwith gH.
Lemma 4.1 For every x ∈M\{o}, the splitting
TxM=Hx⊕Vx,
is orthogonal with respect to gH. In particular, letting dHbe the Riemannian distance induced
by gHon M , one has that
g(v, w) =cosh2(dH(o,x))gH(vH,w
H)+gH(vV,w
V), (4.1)
for all v, w ∈TxM.
Proof Fix an arbitrary unit direction No∈ToM,andletVo∈ToMbe any vector orthogonal
to it with respect to g|o=gH|o. Since the radial geodesics emanating from oare the same
for gand gH, the Jacobi field Y(t)along the geodesic σ:t→ exp M
o(tN
o),determinedby
the initial conditions Y(0)=0, ˙
Y(0)=Vois the same for both metrics. Let V(t)and VH(t)
be the parallel transport of Voalong σwith respect to gand gH, respectively. By the very
definition of symmetric spaces, the curvature tensor Ris itself parallel along geodesics. This
implies that
sinh(t)VH(t)=Y(t)=sinh(√−κt)
√−κV(t),
provided Vobelongs to the κ-eigenspace of the Jacobi operator R(·,No)No. Therefore, par-
allel vector fields in the eigenspaces are collinear for the two metrics. Hence, for t>0the
linear subspaces Hσ(t)and Vσ(t)are nothing else than the parallel transport of the corre-
sponding eigenspaces of R(·,No)Noalong σ. It follows that the splitting TxM=Hx⊕Vx
is orthogonal not only with respect to g, but also with respect to the hyperbolic metric gH.
Equation (4.1) is a direct consequence of this fact and the definition of the distribution H.
We can now prove Theorem 1.5.
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Approaching the isoperimetric problem in Hm
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Proof of Theorem 1.5 Let E⊂Mbe a Hopf-symmetric set with outward pointing normal
vector field νwith respect to g. By the very definition of Hopf-symmetry, νH≡0. Therefore,
thanks to Lemma 4.1,νis orthonormal to ∂Ealso with respect to gH. Let vol and vol Hthe
volume forms associated to gand gH.Wehavethat
P(E)=∂E
ινvol =∂E
coshd−1(dH(o,x))ινvol H(x),
where ι:(M)p→(M)p−1denotes the interior product ιXα(·)=α(X,·).Thevolume
of Eis given by the formula
V(E)=E
vol =E
coshd−1(dH(o,x)) vol H(x).
Hence, the volume and perimeter of Hopf-symmetric sets in Mcorrespond to the volume
and perimeter of (E)in Hn
Rwith density equal to f(x)=coshd−1(dH(o,x)), concluding
the proof. 
Acknowledgements The author would like to thank Urs Lang and Alessio Figalli for their precious guidance
and constant support. A special thanks to Raphael Appenzeller for the very enjoyable and instructive exchanges
about the hyperbolic plane. The author would like to thank Miguel Domínguez Vázquez for suggesting an
efficient way to include the octonionic case in the definition of Hopf-symmetric, and Frank Morgan, for
the useful insight about the earlier contributions to the problem. Finally, the author would like to thank the
anonymous referees for their time, diligent review of the manuscript, and insightful comments. The author has
received funding from the European Research Council under the Grant Agreement No. 721675 “Regularity
and Stability in Partial Differential Equations (RSPDE)”.
Funding Open access funding provided by Swiss Federal Institute of Technology Zurich.
Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which
permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give
appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence,
and indicate if changes were made. The images or other third party material in this article are included in the
article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is
not included in the article’s Creative Commons licence and your intended use is not permitted by statutory
regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.
To view a copy of this licence, visit http://creativecommons.org/licenses/by/4.0/ .
Appendix A: Existence, boundedness and mean convexity
The objective of this section is to establish the existence, boundedness, and mean convexity
of the isoperimetric sets in the hyperbolic space Hn
Rwhen equipped with a radial density
function f:Hn
R→R>0. Expressing
ln(f(x)) =h(dH(o,x))
for some h:R→R, it will be sufficient to assume hlower-semicontinuous and divergent
to infinity to ensure existence, and non-decreasing to ensure boundedness. We will take
advantage of the log-convexity to establish the mean-convexity of the isoperimetric sets. The
proof is a direct application of the arguments employed by Morgan and Pratelli in the flat
case [19, Theorem 3.3, Theorem 4.3, Theorem 5.9, Theorem 6.5]. We recall that we work in
the Poincaré model, that makes Hn
Rconformal to the unit ball in Rn. The metric at a point
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11 Page 22 of 27 L. Silini
x∈Hn
Ris given by
gH=4
(1−r2)2gflat,
where r=|x|will always denote the Euclidean distance of xfrom the origin. In this
coordinate system, the hyperbolic distance from the origin is given by
dH(x,0)=2artanh(r).
We will denote with ˜
f(r):= exp(h(2artanh(r))) the profile of the radial weight in Poincaré
coordinates. Then, one can check that for k∈{n−1,n}the k-dimensional weighted Hausdorff
measures associated to gHand fcan be expressed as
dHk
f:= fdHk=˜
f(r)2
1−r2k
dHk
flat,(A.1)
where we denote with the flat index the Hausdorff measures associated to gflat in the Poincaré
model. To simplify the exposition, let us define the function
ω(r):= 2
1−r2.
Finally we will denote with B(r)the ball centered at the origin in Hn
Rwith Euclidean radius
r∈(0,1), and with Sn−1(r)its boundary.
We start by proving that the isoperimetric profile is monotone. This step will be important
to show boundedness later on.
Theorem A.1 (Monotonicity of the isoperimetric profile) Let h :R≥0→Rbe a lower-
semicontinuous non-decreasing function and let f :Hn
R→R>0be defined through f (x)=
exp(h(dH(x,o))) for some base point o ∈Hn
R. Then, the isoperimetric profile Jdefined in
(1.1)as
J(v) := inf Pf(F):Vf(F)=v, F⊂Hn
Rof finite perimeter5
is non-decreasing in v∈[0,+∞). Moreover, Jis strictly increasing if there exist
isoperimetric sets for all volumes.
Proof Let Ebe any set of finite perimeter with finite volume Vf(E)=v. We claim that for
all r>0 such that E(r):= E∩B(r)Eone has that
Pf(E(r)) < Pf(E). (A.2)
If Eq. (A.2) holds, then it suffices to notice that for every 0 <v
<vthere exists r∈(0,1)
such that
Vf(E(r)) =v,
which implies that
J(v)≤Pf(E(r)) < Pf(E).
If Eis isoperimetric, then we have immediately that J(v)<J(v) for all 0 <v<v
.
Otherwise, for every ε>0letEεbe a set of finite perimeter such that Vf(E)=vand
Pf(Eε)≤J(v) +ε. From the inequality
J(v)< Pf(E)≤J(v) +ε
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Approaching the isoperimetric problem in Hm
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we infer that J(v) ≤J(v)for all 0 <v <v
. We are left to prove Eq. (A.2). Let π:
∂E\B(r)→Sn−1(r)be the normal projection on the sphere of radius r. Notice that πis
strictly 1-Lipschitz with respect to the Euclidean distance. Then,
π(∂E\B(r)) ⊇∂E(r)\∂E.(A.3)
In fact, the set Econtains the (possibly empty) cone
C={λx:λ∈[1,r−1), x∈H},
where H=(∂ E(r)\∂E)\π(∂E\B(r)), and the dilation λxis to be understood with respect
to the Euclidean structure in the Poincaré model. Since the density is non-decreasing, it
follows that Vf(C)=+∞unless Hn−1
flat (H)=0. By assumption, the volume of Eis finite,
and therefore Equation (A.3) must hold up to a set of measure zero. By the coarea formula
(see for instance [14, Chapter 13]) we finally get that
Pf(E(r)) =∂E∩B(r)˜
f(r)ω(r)n−1dHn−1
flat +∂E(r)\∂E˜
f(r)ω(r)n−1dHn−1
flat
≤∂E∩B(r)˜
f(r)ω(r)n−1dHn−1
flat +π(∂E\∂B(r)) ˜
f(r)ω(r)n−1dHn−1
flat
<∂E∩B(r)˜
f(r)ω(r)n−1dHn−1
flat +∂E\∂B(r)˜
f(|π(x)|)ω(|π(x)|)n−1dHn−1
flat
≤Pf(E),
where ˜
f(r)=exp(h(2artanh(r))).
We are now ready to establish existence.
Theorem A.2 (Existence of isoperimetric sets) Let h :R→Rbe a lower-semicontinuous
function that diverges to infinity and let f :Hn
R→R>0be defined through f (x)=
exp(h(dH(x,o))) for some base point o ∈Hn
R. Then, for all volumes there exists a set
attaining the isoperimetric infimum in Eq. (1.1).
Proof Fix v>0andlet(Ej)j≥1⊂Hn
Rbe a sequence of smooth sets of weighted volume v
whose perimeter converges to the infimum of Eq. (1.1). Without loss of generality, we can
suppose Pf(Ej)<J(v) +1. Intersecting this sequence with balls of growing radii rj→1,
the sequence splits into
Ej=(Ej∩B(rj)) ∪(Ej\B(rj)) =EC
j∪ED
j.
Up to taking a subsequence, a standard argument of compactness (see [9, Theorem 1.19] and
[17, Theorem 13.4]) shows that EC
jconverges to an isoperimetric set, whose volume is equal
to vif and only if there is no volume escaping to infinity, that is
lim
R→1lim sup
j→+∞
Vf(Ej\B(R)) =0.
To establish our argument, we will proceed by contradiction. Let us assume that, after select-
ing a subsequence if necessary, there exists a positive value ε>0 such that for every R>0,
there exists an index j=j(R)satisfying the inequality
Vf(Ej\B(R)) ≥ε. (A.4)
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11 Page 24 of 27 L. Silini
Fix 0 <R<1 a number very close to 1 yet to define, and j=j(R). Thanks to (A.1)we
can rewrite Equation (A.4)as
1
R
ω(r)n˜
f(r)Sj(r)dr ≥ε, (A.5)
where Sj(r):= Hn−1
flat (∂ E∩Sn−1(r)) and ˜
f(r)=exp(h(2artanh(r))). Then, denoting
Mj(R):= supr∈[R,1)ω(r)n−1Sj(r)and m(R)=infr∈[R,1)˜
f(r)we gave that
Pf(Ej)≥Mj(R)m(R). (A.6)
In particular, since Pf(Ej)is uniformly bounded, up to taking Rclose enough to 1, we can
suppose m(R)large enough, so that
Sj(r)≤
Hn−1
flat (Sn−1(r))
2
for all r∈[R,1). By the classical isoperimetric inequality on the sphere, there exists a
dimensional constant cn>0 such that
Hn−2
flat (∂(Ej∩Sn−1(r))) ≥cnSj(r)n−2
n−1,
for all r∈[R,1). By Vol’pert theorem (see [1, Theorem 3.108]), for almost every r∈(0,1)
one has that
∂(Ej∩Sn−1(r))) =∂Ej∩Sn−1(r).
Therefore, the coarea formula (see for instance [14, Chapter 13]) allows us to obtain the
following estimate on the weighted perimeter:
Pf(Ej)≥1
R
ω(r)n−1˜
f(r)Hn−2
flat (∂ Ej∩Sn−1(r)) dr
=1
R
ω(r)n−1˜
f(r)Hn−2
flat (∂(Ej∩Sn−1(r))) dr
≥cn1
R
ω(r)n−1˜
f(r)Sj(r)n−2
n−1dr
≥cn1
R
ω(r)n˜
f(r)Sj(r)(Sj(R)ω(r)n−1)−1
n−1dr
≥cnM−1
n−1
jε,
where in the last line we used assumption (A.5). On the other hand, thanks to (A.6)weget
that
(J(v) +1)n
n−1≥Pf(Ej)Pf(Ej)1
n−1≥cnεm(R)1
n−1.
But this is impossible because m(R)diverges to infinity as R→1. 
After existence, we prove boundedness of the isoperimetric set.
Theorem A.3 (Boundedness of the isoperimetric sets) Let h :R≥0→Rbe a lower-
semicontinuous non-decreasing function and let f :Hn
R→R>0be defined through
f(x)=exp(h(dH(x,o))) for some base point o ∈Hn
R. Then, every set attaining the
isoperimetric infimum in Equation (1.1) is bounded.
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Proof We proceed by contradiction. Let Ebe an unbounded isoperimetric set. Let r∈(0,1)
be close enough to 1 so that
E(r):= E∩B(r)E.
Define Er:= E∩Sn−1(r), and the two functions
Vf(r):= Vf(E\B(r)), Pf(r):= Hn−1
f(∂ E\B(r)).
Notice that Vf(r)and Pf(r)tend to zero as rtends to 1. Thanks to Theorems A.1 and A.2,
we have that
Pf(E)>Pf(E(r)) =Pf(E)−Pf(r)+Hn−1
f(Er),
implying that
Pf(r)>Hn−1
f(Er). (A.7)
Up to taking rcloser to 1, we can assume that
Hn−1
flat (Er)≤1
2
Hn−2
flat (∂ Er),
where the boundary of Eris taken inside the sphere Sn−1(r). Therefore, by classic
isoperimetric inequality on the sphere there exists a dimensional constant cn>0such
that
Hn−2
flat (∂ Er)≥cnHn−1
flat (Er)n−2
n−1,(A.8)
which by Eq. (A.7) leads to
˜
f(r)ω(r)n−2Hn−2
flat (∂ Er)≥cn˜
f(r)1
n−1ω(r)n−1˜
f(r)Hn−1
flat (Er)n−2
n−1
≥cn˜
f(0)1
n−1Hn−1
f(Er)n−2
n−1
>cn˜
f(0)1
n−1Pf(r)−1
n−1Hn−1
f(Er),
whereweusedthat ˜
f(r)=exp(h(2artanh(r))) is non-decreasing. Since
−P
f(r)=˜
f(r)ω(r)n−1Hn−2
flat (∂ Er), −V
f(r)=ω(r)Hn−1
f(Er),
we obtain that
−(Pf(r)n
n−1)>−cn˜
f(0)1
n−1V
f(r),
which integrated from rto1leadsto
Pf(r)n
n−1>2cVf(r), (A.9)
with 2c=cn˜
f(0)1
n−1. The last thing we need to do is to take advantage of the optimality of
the set Eby operating a small perturbation of its boundary. Let K⊂Hn
Rbe a compact subset
of Hn
Rand :(−ε0,ε
0)×Hn
R→Hn
Rbe any variation inside K,thatis f(0,·)=id and for
all ε∈(−ε0,ε
0)the map (ε,·)is a smooth diffeomorphism equal to the identity outside
K. Then the first order expansion of the volume and perimeter operators can be computed as
Vf(Eε)=Vf(E)+ε∂E
gH(ν, X)dHf+o(ε),
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11 Page 26 of 27 L. Silini
Pf(Eε)=Pf(E)+ε∂E
HfgH(ν, X)dHf+o(ε),
where X(x):= ∂
∂ε (0,x),Eε:= (ε, E),andHf=H+∂νln(f)stands for the weighted
mean curvature of E, which is constant at every regular point if Eis isoperimetric. We refer
to [14, Chapter 17.3] and [20, Section 3] for the careful proof of this fact. This allows us
to perturb the set Einside K=B(r0)for some r0close enough to 1 according to a small
parameter ε∈(0,ε
0), such that the resulting perturbed sets (Eε)ε∈(0,ε0)satisfy
Vf(Eε)=Vf(E)+ε,
and
Pf(Eε)≤Pf(E)+ε(Hf+1),
Choose now ε1
n<c
Hf+1and R0>r0so that ε=Vf(r)<ε
0. Then, F=Eε∩B(R0)has
weighted volume equal to E, and from
Pf(F)=Pf(Eε)−Pf(r)+Hn−1
f(Er)≤Pf(E)+ε(Hf+1)−2cεn−1
n+Hn−1
f(Er)
<Pf(E)−cεn−1
n+Hn−1
f(Er),
and the optimality of E, we infer that
Hn−1
f(Er)>cεn−1
n.
Hence
−V
f(r)=ω(r)Hn−1
f(Er)>cω(r)ε n−1
n=cω(r)Vf(r)n−1
n,
implies that
−(Vf(r)1
n)>cω(r)=c(2artanh(r)).
Integrating both sides, since artanh(r)tends to +∞ as rtends to 1, we obtain a contradiction
with the assumption that Vf(r)>0foreveryr∈(0,1).
We complete the section by proving the Riemannian mean-convexity of the isoperimetric
sets.
Theorem A.4 (Mean-convexity) Let h :R→Rbe a convex and even function, and let
f:Hn
R→R>0be defined through f (x)=exp(h(dH(x,o))) for some base point o ∈Hn
R.
Then, every set attaining the isoperimetric infimum in Eq. (1.1) is mean-convex.
Proof Let Ebe an isoperimetric set. Thanks to Theorem A.3,Eis bounded, and therefore
there exists z∈∂Emaximizing the distance from the base point o. By the regularity properties
summarized in Theorem 2.1,zis a regular point, and
H(z)≥(n−1),
where H(z)denotes the unweighted mean curvature of ∂Eat z.Letnowx∈∂Ebe another
regular point. Since the weighted mean curvature Hf=H+∂νln(f)is constant, we have
in particular that
H(x)=H(z)+∂νln(f)(z)−∂νln(f)(x)
≥(n−1)+h(dH(o,z)) −h(dH(o,x)) ≥(n−1),
where in the last inequality we used the convexity of the exponent h.
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Approaching the isoperimetric problem in Hm
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