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\newtheorem{prop}{Proposition}
\newtheorem{cor}[prop]{Corollary}
\newtheorem{lem}[prop]{Lemma}
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\begin{document}

\section{Appendix : Isoperimetric profiles of wedges}

\subsection{Itai Benjamini's problem}

\textbf{Definition.} Let $M$ be a Riemannian manifold with boundary. Define the \emph{isoperimetric profile} of $M$ as the largest function $I:(0,vol(M))\to\R_+$ such that for every compact set $S\subset M$,
\begin{eqnarray*}
area((\partial S)\setminus(\partial M))\geq I(vol(S)).
\end{eqnarray*}

\bigskip

\textbf{Classical example.} When $M$ is a Euclidean ball of radius $R$, 
\begin{eqnarray*}
I(v)\geq\textrm{const.}R^{-1}\,\min\{v,vol(M)-v\}.
\end{eqnarray*}

\bigskip

\textbf{Question.} Let $f:\R_+ \to \R_+$ be nondecreasing. Assume $f$ grows less than linearly. Consider the wedge in $R^3$,
\begin{eqnarray*}
W_f = \{(x,y,z)\,|\, x>0,\, 0<z< f(x)\}.
\end{eqnarray*}View $W_f$ as a manifold with boundary.
What is the isoperimetric profile of $W_f$ ?

\bigskip

\textbf{Guess.} Since, at large scales, $W_{f}$ looks like a thickened half-plane, minimizers should look like thickened disks.

\subsection{The result}
Let $C_{R}=W_{f}\cap\{x^{2}+y^{2}<R^{2}\}$ be the thickened disk of radius $R$ in $W_{f}$. Then
\begin{eqnarray*}
\frac{1}{3} R^{2}f(R) < vol(C_{R}) < \frac{\pi}{2} R^{2}f(R)
\end{eqnarray*}
and
\begin{eqnarray*}
2Rf(R) < area(\partial C_{R}) < \pi Rf(R).
\end{eqnarray*}
Furthermore,
\begin{eqnarray*}
\lim_{R\to +\infty}\frac{vol(C_{R})}{\frac{\pi}{2}R^{2}f(R)}=1,\quad \lim_{R\to +\infty}\frac{area(\partial C_{R})}{\pi Rf(R)}=1.
\end{eqnarray*}

\bigskip

\textbf{Notation.} Denote by $R(v)$ the solution of $\frac{\pi}{2}R^{2}f(R)=v$. Set $J(v)=\pi R(v)f(R(v))$. By definition of the isoperimetric profile $I$ of the wedge,
\begin{eqnarray*}
area(\partial C_R )\geq I(vol(C_R )).
\end{eqnarray*}
This implies
\begin{eqnarray*}
\limsup_{v\to +\infty} \frac{I(v)}{J(v)}\leq 1.
\end{eqnarray*}

\bigskip

\textbf{Conjecture.} \emph{The isoperimetric profile $I$ of the wedge satisfies }
\begin{eqnarray*}
\lim_{v\to +\infty} \frac{I(v)}{J(v)}=1.
\end{eqnarray*}

In this appendix, only a weaker statement is proven. Fortunately, it is sufficient for Itai's needs.

\begin{thm}
\label{profile}
Let $f:\R_+ \to \R_+$ be nondecreasing. Assume $x\mapsto f(2x)/f(x)$ and $x\mapsto f(x)/x$ are bounded near $+\infty$. Then for $v$ large enough, the isoperimetric profile $I$ of the wedge $W_f = \{(x,y,z)\,|\, x>0,\, 0<z< f(x)\}$ satisfies 
\begin{eqnarray*}
I(v)\geq \textrm{const.}\,J(v).
\end{eqnarray*}
\end{thm}

\bigskip

\textbf{Remark.} A concave function satisfies the assumptions of Theorem \ref{profile}.

\bigskip

The proof of Theorem \ref{profile} occupies the rest of the appendix. The main trick consists in cutting the wedge into portions $Q_j$ which are biLipschitz equivalent to \emph{beams}, i.e. products of rectangles with lines. By a reflection principle,  isoperimetry of beams reduces to isoperimetry of flat manifolds of the form torus$\times$line, which is known. The edge of the wedge requires a specific treatment.

\bigskip

\textbf{Remark.} If $f$ increases faster than linearly, then one expects the isoperimetric profile of $W_f$ to be $I(v)\sim v^{2/3}$. It is so, for instance, if $f$ has its derivative bounded below, since then $W_f$ is biLipschitz to a half-space.

\subsection{Profiles for flat manifolds}

{\bf Notation.} $M_{L,h}=\R/L\Z\times\R\times\R/h\Z$. We use the following result due to L. Hauswirth, J. P\'erez, P. Romon, and A. Ros, \cite{HPRR} (see also \cite{RR}).
 

\begin{prop}
\label{beam}
{\em (\cite{HPRR})}. If $L\geq \frac{9\pi}{16}h$, the isoperimetric profile of $M_{L,h}$ is achieved by spheres, cylinders or strips bounded by parallel planes. In other words, for every $S\subset M_{L,h}$,
\begin{eqnarray*}
vol(S)\leq \frac{4\pi}{81}h^{3}&\Rightarrow& area(\partial S)\geq (36\pi)^{1/3}\,vol(S)^{2/3},\\
\frac{4\pi}{81}h^{3}\leq vol(S)\leq \frac{1}{\pi}L^{2}h&\Rightarrow& area(\partial S)\geq (4\pi)^{1/2}h^{1/2}\,vol(S)^{1/2},\\
vol(S)\geq \frac{1}{\pi}L^{2}h&\Rightarrow& area(\partial S)\geq 2Lh.
\end{eqnarray*}
\end{prop}


\begin{cor}
\label{flat}
Assume $L\geq h$. For $S\subset M_{L,h}$,
\begin{eqnarray*}
vol(S)\leq \frac{1}{32\pi}\,h^3 &\Rightarrow & area(\partial S)\geq \frac{1}{4}\, vol(S)^{2/3}.\\
vol(S)\leq \frac{1}{2\pi}\,L^{2}h &\Rightarrow & area(\partial S)\geq \frac{1}{2}\,L^{-1}vol(S).
\end{eqnarray*}
\end{cor}

\proof
First, assume that $L\geq\frac{9\pi}{16}h$. If $vol(S)\geq \frac{4\pi}{81}h^{3}$, then 
\begin{eqnarray*}
area(\partial S)&\geq& (4\pi)^{1/2}h^{1/2}\,vol(S)^{1/2}\\
&=&(4\pi)^{1/2}(\frac{L^{2}h}{vol(S)})^{1/2}L^{-1}vol(S)\\ 
&\geq&2\,L^{-1}vol(S).
\end{eqnarray*}
Otherwise, if $\frac{4\pi}{81}h^{3}\leq vol(S)\leq \frac{1}{\pi}L^{2}h$,
\begin{eqnarray*}
area(\partial S)&\geq& (36\pi)^{1/3}\,vol(S)^{2/3}\\
&=&(36\pi)^{1/3}(\frac{L^{3}}{vol(S)})^{1/3}\,L^{-1}vol(S)\\
&\geq&(288\pi)^{1/3}(\frac{h^{3}}{vol(S)})^{1/3}\,L^{-1}vol(S)\\
&\geq&18\,L^{-1}vol(S).
\end{eqnarray*}
In both cases, $area(\partial S)\geq 2\,L^{-1}vol(S)$. 

If $h\leq L\leq\frac{9\pi}{16}h$, use the affine 2-biLipschitz diffeomorphism of $M_{L,h}$ to $M_{L',h}$ with $L'=2L\geq \frac{9\pi}{16}h$. This sends $S$ to $S'$ with $vol(S')\leq 8\,vol(S)$. 

If $vol(S)\leq \frac{1}{32\pi}\,h^3$ then $vol(S')\leq\frac{4\pi}{81}\,h^3$, thus $area(\partial S')\geq (36\pi)^{1/3}vol(S')^{2/3}$ and finally $area(\partial S)\geq \frac{1}{4}\,vol(S)^{2/3}$.

If $vol(S)\leq \frac{1}{2\pi}\,L^{2}h$ then $vol(S')\leq\frac{1}{\pi}\,L'^{2}h$, thus $area(\partial S')\geq 2\,L'^{-1}vol(S')$ and finally $area(\partial S)\geq \frac{1}{2}\,L^{-1}vol(S)$.\qed

\subsection{Selecting a scale $R$}

Let $f$ be a nondecreasing function on $\R_{+}$. Assume that $f(2x)/f(x)\leq\lambda$ for $x$ large enough. Assume too that $f(x)/x$ is bounded near $+\infty$. Up to multiplying $f$ by a constant (which only changes the constant in the final result), one can assume that $f(x)/x \leq 1$ for $x$ large enough.

Given $v>0$, let $R$ be defined by $R^{2}f(R)=16\pi\lambda^3 v$. Fix $v$ large enough so that $f(R)/R\leq 1$ and $f(2R)/f(R)\leq \lambda$. 

Note that $R\leq 16 \lambda^{3/2}R(v)$. Indeed, if $n$ is the smallest integer larger than $\log(64 \lambda^3 )/\log 4$, then $4^{n-1}\leq 64 \lambda^3 \leq 4^n$,
\begin{eqnarray*}
\frac{\pi}{2}(2^{-n}R)^2 f(2^{-n}R)\leq \frac{\pi}{2}\,4^{-n}R^2 f(R)
\leq 16\pi \lambda^3 R^2 f(R)=v=\frac{\pi}{2}R(v)^2 f(R(v)),
\end{eqnarray*}
which implies $2^{-n}R\leq R(v)$. By construction, $2^n \leq (256 \lambda^3 )^{1/2}$, thus $R\leq 16 \lambda^{3/2}R(v)$.

\subsection{Cutting sets into 3 pieces}

Let $S\subset W_{f}$ have volume $v$. The goal is to estimate $area(\partial S)$ by $v/R$. Let $k$ be the largest integer such that $f(kR)\leq R$.

Split $S$ into 3 parts,
\begin{eqnarray*}
A=S\cap\{0<x<R\},\quad B=S\cap\{R<x<(k+1)R\},
\end{eqnarray*}
\begin{eqnarray*}
C=S\cap\{x>(k+1)R\}.
\end{eqnarray*}
\subsection{Estimate for $vol(A)$}

Consider the projection along the first factor $proj(x,y,z)=(y,z)$. It decreases areas. Note that the restriction of $proj$ to $\partial S$ has global degree zero. At points of $\partial S \cap\partial W_f$, $proj$ has local degree 1. Therefore every point of $T=proj(\partial S)$ is the projection of at least one point of $\partial S$ which does not belong to $\partial W_f$, $T=proj(\partial S\cap W_f )$. It follows that $area(T)\leq area(\partial S\cap W_f )$. Now $A\subset [0,R]\times T$, thus
\begin{eqnarray*}
vol(A)\leq R\,area(T)\leq R\,area((\partial S)\cap W_f ).
\end{eqnarray*}

\subsection{Chopping and reflecting}

Chop $B$ and $C$ into pieces
\begin{eqnarray*}
P_{j}=S\cap\{jR\leq x\leq (j+1)R\},
\end{eqnarray*}
so that 
\begin{eqnarray*}
B=\bigcup_{j\leq k}P_j ,\quad C=\bigcup_{j\geq k+1}P_j .
\end{eqnarray*}
$P_{j}$ is contained in a portion $Q_{j}=W_{f}\cap\{jR\leq x\leq (j+1)R\}$ which is biLipschitz to a beam $B_{R,f(jR)}=\{jR\leq x\leq (j+1)R,\,0\leq z\leq f(jR)\}$, where the Lipschitz constant is at most $f((j+1)R)/f(jR)\leq \lambda$. Reflect $P_j$ in the hyperplanes $\{x=jR\}$, $\{z=0\}$, and map it to the quotient manifold $M_{L_{j},h_{j}}$ with $L_{j}=2R$, $h_{j}=2f(jR)$. The obtained set $S_{j}\subset M_{L,h}$ has
\begin{eqnarray*}
4\lambda^{-3}\leq\frac{vol(S_{j})}{vol(P_j )}\leq 4\lambda^{3},\quad
4\lambda^{-2}\leq\frac{area(\partial S_{j})}{area(\partial P_j \setminus \partial Q_j )}\leq 4\lambda^{2}.
\end{eqnarray*}
In particular, for all $j$, $vol(S_j )\leq 4\lambda^3 vol(S)$.

\subsection{Estimate for $vol(B)$}

For $j\leq k$, $vol(S_{j})\leq \frac{1}{4\pi}R^{2}f(R)\leq \frac{1}{2\pi}\,L_{j}^{2}h_{j}$ and $h_{j}\leq L_{j}$. Corollary \ref{flat} gives
\begin{eqnarray*}
area(\partial S_{j})\geq \frac{1}{2}\,(2R)^{-1}vol(S_{j}).
\end{eqnarray*}
It follows that
\begin{eqnarray*}
vol(P_j )&\leq& \frac{1}{4}\lambda^3 vol(S_j )\\
&\leq&2\lambda^3 R\,area(\partial S_{j})\\
&\leq&8\lambda^5 R\,area(\partial P_j \setminus \partial Q_j ),
\end{eqnarray*}
and finally
\begin{eqnarray*}
vol(B)&\leq&8\lambda^5 R\,area((\partial S) \cap W_f ).
\end{eqnarray*}

Combining estimates for $A$ and $B$ leads to
\begin{eqnarray*}
vol(A)+vol(B)&\leq&(8\lambda^5 +1)R\,area((\partial S) \cap W_f )\\
&\leq&9\lambda^5 \,R\,area((\partial S) \cap W_f ).
\end{eqnarray*}

\subsection{Estimate for $vol(C)$}

For $j\geq k+1$, $vol(S_{j})\leq \frac{1}{4\pi}R^{2}f(R)\leq \frac{1}{32\pi}\,(2R)^{3}\leq \frac{1}{32\pi}\,\min\{h_{j},L_{j}\}^{3}$. Corollary \ref{flat} gives
\begin{eqnarray*}
area(\partial S_{j})\geq \frac{1}{4}\,vol(S_{j})^{2/3}.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
vol(P_j )^{2/3}&\leq& (\frac{1}{4})^{2/3}\lambda^2 vol(S_j )^{2/3}\\
&\leq&4^{1/3}\lambda^2  \,area(\partial S_{j})\\
&\leq&4^{-2/3}\lambda^4 \,area(\partial P_j \setminus \partial Q_j ),
\end{eqnarray*}
and finally
\begin{eqnarray*}
vol(C)^{2/3}&=&(\sum_{j\geq k+1}vol(P_j ))^{2/3}\\
&\leq&\sum_{j\geq k+1}vol(P_j )^{2/3}\\
&\leq&4^{-2/3}\lambda^4 \,\sum_{j\geq k+1}area(\partial P_j \setminus \partial Q_j )\\
&\leq&4^{-2/3}\lambda^4 \,area((\partial S) \cap W_f ).
\end{eqnarray*}

\subsection{Conclusion}

Let $\tau=vol(C)$, $v=vol(S)$. We have shown that
\begin{eqnarray*}
area((\partial S) \cap W_f )\geq \frac{1}{2}(4^{2/3}\lambda^{-4}\tau^{2/3}+\frac{1}{9}\lambda^{-5} R^{-1}(v-\tau)).
\end{eqnarray*}
The right hand side is a concave function of $\tau\in[0,v]$, which achieves its minimum either at 0 or at $v$. One concludes that
\begin{eqnarray*}
area((\partial S) \cap W_f )\geq \min\{2^{1/3}\lambda^{-4}v^{2/3},\frac{1}{18}\lambda^{-5}R^{-1}v\}.
\end{eqnarray*}
It turns out that the minimum is $\frac{1}{18}\lambda^{-5}R^{-1}v$. Therefore
\begin{eqnarray*}
area((\partial S) \cap W_f )&\geq& \frac{1}{24}\lambda^{-5}R^{-1}v\\
&\geq&\frac{1}{24}\lambda^{-5}\frac{1}{16}\lambda^{-3/2}R(v)^{-1}v\\
&=&\frac{1}{776}\lambda^{-13/2}J(vol(S)).
\end{eqnarray*}

\subsection{Remark}

Only a coarser version, with unsharp constants, of the result of \cite{HPRR} is needed. A coarser proof can be given, along the following lines.

For every Riemannian manifold with a cocompact isometry group, extremal domains exist, so that the profile never vanishes. Extremal domains with small enough volume look like small balls. In a flat manifold, they are exactly balls. This occurs at a scale of the order of the injectivity radius. This gives the first statement of Corollary \ref{flat}, with an unsharp constant $c$,
\begin{eqnarray*}
vol(S)\leq c\,h^3 &\Rightarrow & area(\partial S)\geq (36\pi)^{1/3}\, vol(S)^{2/3}.
\end{eqnarray*}

Up to rescaling, one can assume that $L=1$. Up to biLipschitz equivalence, one can assume that $1/h$ is an integer. Let $M_{1,1}\to M_{L,h}$ be the $1/h$-fold cover of $M_{L,h}$. If $S\subset M_{L,h}$ has volume $c\,L^{2}h$, its inverse image $\tilde{S}$ in $M_{1,1}$ has volume $\leq c$, thus
\begin{eqnarray*}
\frac{L}{h}\,area(\partial S)=area(\partial\tilde{S})\geq (36\pi)^{1/3}\, \,vol(\tilde{S})^{2/3}=(36\pi)^{1/3}\, \,vol(S)^{2/3}(\frac{L}{h})^{2/3}.
\end{eqnarray*}
Since $vol(S)\leq c\,L^{2}h$, $\frac{h}{L}\geq vol(S)/cL^{3}$, $(\frac{h}{L})^{1/3}vol(S)^{2/3}\geq vol(S)/c^{1/3}L$, which leads to
\begin{eqnarray*}
area(\partial S)\geq \textrm{const.}\,L^{-1}vol(S),
\end{eqnarray*}
as announced.\qed

\begin{thebibliography}{}

\bibitem[HPRR]{HPRR} L. HAUSWIRTH, J. P\'EREZ, P. ROMON, A. ROS,
{\sl The periodic isoperimetric problem.}
Trans. Amer. Math. Soc. ${\bf 356}$, $2025-2047$ $(2004)$.

\bibitem[RR]{RR} M. RITOR\'E, A. ROS,
{\sl Stable constant mean curvature tori and the isoperimetric problem in three-space forms.}
Comment. Math. Helv. ${\bf 67}$, $293-305$ $(1992)$.

\end{thebibliography}

\vskip1cm
\noindent
Pierre Pansu\\
Laboratoire de Math\'ematique d'Orsay\\
UMR 8628 du C.N.R.S.\\
B\^atiment 425\\
Universit\'e Paris-Sud 11 - 91405 Orsay (France)\\
\smallskip\noindent
{\tt\small Pierre.Pansu@math.u-psud.fr}\\
http://www.math.u-psud.fr/$\sim$pansu

\end{document}

