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% An Isoperimetric Inequality for the Heisenberg Groups
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% Daniel Allcock
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% Daniel Allcock (allcock@math.utah.edu)
% 17 December 1996
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%-----------------beginning-of-document---------------------------
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\title{An Isoperimetric Inequality for the Heisenberg Groups}
\author{Daniel Allcock\footnote*{Supported by a National
Science Foundation Graduate Fellowship.}}
\date{July 21, 1995; revised 5 February 1997}
% revised 21 September 1995
\address{Deptartment of Mathematics\par University of Utah\par
Salt Lake City, UT 84112}
\email {allcock@math.utah.edu}
\homepage{http://www.math.utah.edu/$\sim$allcock}
\subject{20F32 (53C20, 22E25)}
%\note {}
\plaintitlepage
%
% Definitions for cross-references
%
% Do not tamper with the following line.
% ---begin deftags---
\deftag{1}{sec-intro}
\deftag{2}{sec-isotropic-disks}
\deftag{2.1}{thm-symplectic-orthogonal-homotopy}
\deftag{2.2}{thm-swooping-lemma}
\deftag{2.1}{eq-ang-momentum}
\deftag{2.3}{thm-isotropic-disks}
\deftag{3}{sec-Heisenberg-gps}
\deftag{4}{sec-Heisenberg-isoper-ineq}
\deftag{4.1}{thm-Heisenberg-isoper-ineq}
\deftag{4.2}{thm-homotopic-to-horiz-then-vert}
\deftag{4.3}{thm-vert-to-horiz}
\deftag{5}{sec-Heisenberg-apps}
\deftag{5.1}{thm-chn-and-qhn-isoper-ineq}
\deftag{5.2}{thm-disk-extension}
% ---end deftags---
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% ---begin defcites---
\defcite{1}{alon:isoper-ineq}
\defcite{2}{bur:geom-and-comb-dehn-funcs}
\defcite{3}{dbae:wd-proc-gps}
\defcite{4}{farb:thesis}
\defcite{5}{sger:dehn-func-l1-norm}
\defcite{6}{ger:isoper-and-isodiam}
\defcite{7}{wmg:cx-hyp-geom}
\defcite{8}{gromov:hyperbolic-groups}
\defcite{9}{gromov:asymp-invts}
\defcite{10}{gromov:carnot-caratheodory}
\defcite{11}{gromov:geom-top}
\defcite{12}{lee:metric-properties-lagrangian-surfaces}
\defcite{13}{olshanskii:subquadratic-isoperimetric-inequalities}
\defcite{14}{scott:3-mflds}
% ---end defcites---
% Do not tamper with the preceeding line.
% ABBREVIATIONS
\def\del{\partial}
\def\dGds {\del\Gamma/\del s}
\def\dGdt {\del\Gamma/\del t}
\def\dGdth {\del\Gamma/\del\theta}
\def\heis {\calh ^{2n+1}}
\def\Heisenberg {\calh ^{2n+1}}
\def\heisa {{\frak h}^{2n+1}}
\def\frakz {\frak z}
\def\CH{\C H}
\def\H{{\bbb H}}
\def\HH{\H H}
\def\O{{\bbb O}}
\def\OH{\O H}
\abstract
We show that the Heisenberg groups $\calh ^{2n+1}$ of dimension
five and higher, considered as Riemannian manifolds, satisfy a
quadratic isoperimetric inequality. (This means that each loop
of length $L$ bounds a disk of area $\sim L^2$). This implies several
important results about isoperimetric inequalities for discrete
groups that act either on $\calh ^{2n+1}$ or on complex
hyperbolic space, and provides interesting examples in geometric
group theory. The proof consists of explicit construction of a
disk spanning each loop in $\Heisenberg$.
\section{\Tag{sec-intro} Introduction}
The Heisenberg groups $\calh ^3, \calh ^5, \calh ^7,\ldots$ are
a sequence of nilpotent Lie groups that arise in geometry in
several ways. For example, $\calh ^3$ is known to
three-dimensional geometers as {\it Nilgeometry}, and arises in the study of
Seifert-fibered three-manifolds \cite{scott:3-mflds}. The
$\Heisenberg$ also
appear in hyperbolic geometry: a horosphere in complex
hyperbolic space $\CH^n$ is a copy of $\calh^{2n-1}$ (see
\cite{wmg:cx-hyp-geom}, \cite{gromov:geom-top}).
Thurston \cite{dbae:wd-proc-gps}, without proof, and Gromov
\cite{gromov:asymp-invts}\cite{gromov:carnot-caratheodory},
outlining a proof, have stated
that $\calh ^5, \calh ^7, \ldots$ satisfy quadratic
isoperimetric inequalities (defined below). This paper provides
a new and complete proof of this theorem, by explicitly exhibiting
spanning disks and estimating their areas.
Our main purpose in
proving this
theorem is to obtain isoperimetric
inequalities for certain finitely presented groups, in
particular, for the discrete Heisenberg groups and for nonuniform
lattices in ${\rm Isom}\,(\CH^n)$. In a sense, we use
elementary differential geometry to obtain results in the
geometric theory of discrete groups. Our inequalities for
$\heis\>(n>1)$ contrast with the case of $\calh ^3$, which
satisfies a cubic (but no quadratic) isoperimetric inequality;
see \cite{dbae:wd-proc-gps}. For more information about
isoperimetric inequalities for finitely presented groups, see
\cite{sger:dehn-func-l1-norm}, \cite{ger:isoper-and-isodiam}.
A Riemannian manifold $M$ is said to satisfy the isoperimetric
inequality $f$, where $f$ is a function from the positive real
numbers to themselves, if any smooth closed curve in $M$ of
length at most $L$ bounds a disk in $M$ of area at most
$f(L)$. One says that $M$ satisfies a quadratic isoperimetric
inequality if $ f$ may be taken to be a quadratic polynomial,
and one makes similar statements for cubic bounds of $f$, etc.
It turns out that our problem may be reduced to a problem in
symplectic geometry which is interesting in its own right.
Namely, we consider loops in $\R^{2n}$, which we equip with both
its usual Euclidean metric and its usual symplectic form. The
problem of finding spanning disks of appropriate area in $\heis$
reduces to the following problem in $\R^{2n}$: given a loop
$\gamma$ in $\R^{2n}$, enclosing zero symplectic area and having
length $L$, does $\gamma$ bound an isotropic disk with Euclidean
area of
order $L^2$? The answer is yes if $n>1$, as we prove in
section~\tag{sec-isotropic-disks}.
Section~\tag{sec-Heisenberg-gps} introduces the
Heisenberg groups and explains the connections between them and
the symplectic geometry, and
section~\tag{sec-Heisenberg-isoper-ineq} obtains the advertised
quadratic isoperimetric inequality for $\heis$. Finally,
section~\tag{sec-Heisenberg-apps} offers applications of our
results to complex and quaternionic hyperbolic geometry and to
geometric group theory; it also compares our techniques with
those of Gromov \cite{gromov:carnot-caratheodory} and
Lee \cite{lee:metric-properties-lagrangian-surfaces}.
The
proofs are arranged so that a reader interested only in the
existence of isotropic spanning disks (with or without the area
estimate) need read only a minimal amount.
The author is pleased to S. Gersten and J. Burillo for working
through the paper in their seminar and for suggesting
improvements. This paper is derived from part of the author's
Ph.D. thesis at U.C. Berkeley and the author would like to
thank his advisor, A. Casson, for his suggestions on the
exposition.
\section{\Tag{sec-isotropic-disks} Isotropic Disks in $\R^{2n}$}
We consider $\R^{2n}$ as both a symplectic
and Euclidean space, with the orthonormal basis
$x_1$, $y_1$, \dots, $x_n$,$y_n$, and the standard symplectic form
$\omega$, defined by
$$
\omega(x_i,y_i)=-\omega(y_i,x_i)=1,
$$
with all other pairings of basis vectors vanishing.
(Recall that a symplectic form is a nondegenerate
antisymmetric bilinear form on a vector space.) As usual, we
also consider $\R^{2n}$ as a manifold and $\omega$ as a
differential 2-form thereon. If $\a$ is a smooth closed path
$\a:[0,1]\rightarrow \R^{2n}$, we may choose a smooth disk $D$
spanning $\a$ and evaulate $\int_D \omega$. We call this
quantity the symplectic area enclosed by $\a$; since $\omega$ is
exact, this area does not depend on the disk chosen.
We say that a smooth map from a manifold, $f:M\rightarrow
\R^{2n}$, is isotropic if $f^*\omega=0$; in the following
constructions $M$ will be the unit disk or a rectangle. Stokes'
theorem implies that two closed loops joined by an isotropic
homotopy enclose the same symplectic area.
The goal of this section is to prove
theorem~\tag{thm-isotropic-disks} below, and the strategy is to
first prove two lemmas that provide us with `legal moves' on
loops in $\R^{2n}$. That is, they provide a variety of
isotropic homotopies with small Euclidean areas. We will
prove theorem~\tag{thm-isotropic-disks} by weaving these moves
together.
\beginproclaim Lemma \Tag{thm-symplectic-orthogonal-homotopy}.
Suppose $\R^{2n}=V_1\oplus V_2$, where $V_1$ and $V_2$ are
symplectically orthogonal subspaces, and that
$\a:[0,1]\rightarrow \R^{2n}$ is a smooth loop based at the
origin, with $\a(s)=\a_1(s)+\a_2(s)$ for all $s$, where the
image of $\a_i$ lies in $V_i$. Then there is a smooth isotropic
homotopy between $\a$ and the loop obtained by first traversing
$\a_1$ and then $\a_2$.
If $\,V_1$ and $V_2$ are orthogonal (with respect to the Euclidean
metric) and
$\a$ has length $L$,
then the isotropic homotopy may be taken to have
(Euclidean) area at most $L^2$.
\endproclaim
\beginproof{Proof: }
By reparameterizing and extending its domain, we may (and do)
suppose that $\a$ is a smooth map from $\R$ to $\R^{2n}$ which
vanishes except on $[0,1]$.
%(This reparameterization may be realized by a smooth
%homotopy of the path $\a$, which can be smoothly joined to the homotopy
%constructed below.)
Consider the homotopy
$$
\Gamma(s,t)=\a_1(s+t) + \a_2(s),
$$
where $t\in[0,1]$ and $s\in [-1,1]$. $\Gamma(s,0)$ is the
restriction of $\a$ to $[-1,1]$, and $\Gamma(s,1)$ is the
loop obtained by traversing first $\a_1$ and then $\a_2$. We compute
$$\eqalign{
\dGdt & = \a_1'(s+t) \cr
\dGds & = \a_1'(s+t) + \a_2'(s);
\cr}$$
$\Gamma$ is isotropic: $\omega(\dGds, \dGdt) = 0 $ because $\omega$
is antisymmetric and because the $\a_i'$ lie in symplectically
orthogonal subspaces.
We estimate the area of the homotopy as follows. Writing
$X$ for $[0,1]\times[-1,1]$, we have
$$\eqalign{
{\rm Area}(\Gamma(X)) & = \int_X ds\,dt \sqrt{
\left\| \dGdt \right\|^2
\left\| \dGds \right\|^2
-((\dGds)\cdot (\dGdt))^2} \cr
& = \int_X ds\, dt\sqrt{\|\a_1'(s+t)\|^2 \[\strut\|\a_1'(s+t)\|^2 +
\|\a_2'(s)\|^2\] -\|\a_1'(s+t)\|^4} \cr
& = \int_X \|\a_1'(s+t)\|\, \|\a_2'(s)\| ds\,dt \cr
& = \int_0^1 \|\a_2'(s)\| dt \int_{-1}^1 \|\a_1'(s+t)\| ds \leq L^2.
\cr}$$
In the last step the integrals are bounded by the lengths of
$\a_1$ and $\a_2$, respectively, which are in turn bounded by $L$.
\endproof
\beginproclaim Lemma \Tag{thm-swooping-lemma}.
Let $\a$ and $\b$ be two smooth paths in $\R^{2n}$. If their
images lie in symplectically orthogonal subspaces and they are
parameterized such that
$$
\omega(\a(s),\a'(s))=\omega(\b(s),\b'(s)),
\eqno\eqTag{eq-ang-momentum}
$$
then there is a smooth isotropic homotopy between them. If $\a$
and $\b$ are closed paths, then this may be taken to be a
homotopy through closed paths.
If $\a$ has length $L$ and there are constants $A$ and $B$ such
that for all $s$ we have
$\|\b'(s)\| \leq \|\a'(s)\|$, $\|\a(s)\| \leq A$ and $\|\b(s)\|
\leq B$, then the homotopy may be taken to have (Euclidean) area at most
$(A+B)\pi L$.
\endproclaim
\remark{Remark:}
The quantity on the left side of \eqtag{eq-ang-momentum} may be
interepreted geometrically as (twice) the rate at which the family of
segments joining $0$ and $\a(s)$ sweeps out symplectic area. If
$\a$ lies in a symplectic plane then it may also be interpreted
as the angular momentum (about the origin) of a particle moving
along $\a$.
Similar interpretations in terms of $\b$ describe the right hand
side.
\beginproof{Proof: }
Define the homotopy
$$
\Gamma(s,t)=\a(s)\cos t + \b(s)\sin t,
$$
where $t\in [0,\pi/2]$. It is obvious that if $\a$ and $\b$ are
loops then $\Gamma$ is a homotopy through loops. We have
$$\eqalign{
\dGdt & = -\a(s)\sin t + \b(s)\cos t \cr
\dGds & = \a'(s)\cos t + \b'(s)\sin t.
\cr}$$
To evaluate $\Gamma^*\omega$, one computes $\omega(\dGds,
\dGdt)$, which vanishes, establishing that $\Gamma$ is
isotropic. To estimate the area of the homotopy under the
conditions of the last claim of the theorem, we observe
$$\eqalign{
\|\dGdt\| \leq & \|\a(s)\| + \|\b(s)\| \leq A+B \cr
\|\dGds\| \leq & \|\a'(s)\| + \|\b'(s)\| \leq 2 \|\a'(s)\|.
\cr}$$
Writing $X$ for $[0,1]\times [0,\pi/2]$, we have
$$\eqalign{
{\rm Area}(\Gamma(X)) & \leq \int_X \|\dGdt\| \> \|\dGds\| ds \, dt \cr
& \leq \int_X 2(A+B)\|\a'(s)\| ds \, dt \cr
& = (A+B)\pi L.
\cr}$$
\endproof
\beginproclaim Theorem \Tag{thm-isotropic-disks}.
Let $\gamma:S^1\rightarrow \R^{2n}$ be a smooth loop enclosing
zero symplectic area, and suppose $n>1$. Then there is a smooth
isotropic map of the unit disk $f:D\rightarrow \R^{2n}$ such that
$f|\partial D = \gamma$.
If $\gamma$ has length $L$, then the spanning disk
$f(D)$ may be chosen to have Euclidean area at most $kL^2$, where $k$ is a
constant.
\endproclaim
\remark{Remark: }
The proof below shows that $k$ may be taken to be
$1+\pi (2+\sqrt{5})$.
\beginproof{Proof: }
The proof proceeds by applying a sequence of isotropic
homotopies to $\c$; their composition shrinks $\c$ to a
point.
STEP 1: Consider $\c$ to be a closed path $\c=\c_0:I\rightarrow \R^{2n}$,
and suppose that $\c_0(0)=0$. By
lemma~\tag{thm-symplectic-orthogonal-homotopy}, $\c_0$ is
homotopic by a smooth isotropic homotopy to a smooth curve
$\c_1$ which is a composition of two loops, the first being the
projection of $\c_0$ to the $x_1,y_1$ plane and the second the
projection of $\c_0$ to the span of $x_2,y_2,\ldots,x_n,y_n$.
STEP 2: If $\a(s)=(x(s),y(s),0,0)$ is a smooth curve in $\R^4$
then define $\b(s)=(0,0,x(s),y(s))$, and apply
lemma~\tag{thm-swooping-lemma} to deduce that there is an
isotropic homotopy carrying $\a$ to $\b$. Applying this to the
first of the two loops comprising $\c_1$ produces an isotropic
homotopy between $\gamma_1$ and a loop $\c_2$ lying entirely in
the span of $x_2,y_2,\ldots,x_n,y_n$.
STEP 3: Define $\gamma_3(s)=(L/2,h(s),0,\ldots,0)$, where $h(s)$
is defined by the conditions $h(0)=0$ and
$$
\omega(\c_3(s),\c_3'(s))=\omega(\c_2(s),\c_2'(s)).
$$
The latter condition is equivalent to
$h'(s)=2\omega(\c_2(s),\c_2'(s))/L$.
Since the symplectic area enclosed by $\gamma_2$ is zero, we
have $h(1)=0$, so $\gamma_3$ is a closed path. Applying
lemma~\tag{thm-swooping-lemma} to $\c_2$ and $\c_3$ yields an
isotropic homotopy between them.
STEP 4: Steps 1 through 3 have provided an isotropic homotopy
between our given loop $\c_0=\gamma$ and the loop $\c_3$, which
lies in a line of $\R^{2n}$. As our
last step, we contract the loop $\c_3$ to a point in this
one-dimensional subspace by a linear (and obviously isotropic)
homotopy. By changing the parametrization of the homotopy
parameter $s$, we may paste the four homotopies together to
obtain a smooth isotropic disk $f:D\rightarrow \R^{2n}$ bounding $\gamma$.
This completes the proof of the first part of the theorem.
To prove the second part, we must estimate the areas of the
homotopies used above. By
lemma~\tag{thm-symplectic-orthogonal-homotopy}, the homotopy in
step one has area at most $L^2$, and we observe that $\c_1$
consists of two loops, each of length at most $L$.
This implies that $\|\c_1(s)\|\leq L$ for all $s$.
Lemma~\tag{thm-swooping-lemma} then proves that the second homotopy
has area at most $\pi L^2$. We will also apply
lemma~\tag{thm-swooping-lemma} to the homotopy of step three.
Since $\|\c_2(s)\| \leq L/2$ for all $s$, we have $|h'(s)| \leq
\|\c_2'(s)\|$, which implies that the length of $\c_3$ is at
most $2L$, and hence that $|h(s)|\leq L$ and $\|\c_3(t)\| \leq
L\sqrt{5}/2$. Since the length of $\c_2$ is at most $2L$,
lemma~\tag{thm-swooping-lemma} shows that the area of this
homotopy is at most $\pi L^2(1+\sqrt{5})$. Observing that the
fourth homotopy has zero area, we add all these areas together
to obtain the advertised bound.
\endproof
\section{\Tag{sec-Heisenberg-gps} The Heisenberg Groups}
For background on the Heisenberg groups, see
\cite{wmg:cx-hyp-geom}. The Heisenberg group $\heis$ is the
connected and simply connected Lie group with Lie algebra
$\heisa$, which has a basis consisting of the $2n+1$ vectors
$x_1,y_1,\ldots,x_n,y_n,z,$ and Lie bracket defined by the
relations that $[x_i,y_i]=z$ and that all other pairings of
basis elements vanish. The center $\frakz$ of $\heisa$ is the
span of $z$, and we denote by $\calz$ the center of the group
$\heis$. Let $\pi:\heis \rightarrow \heis /\calz \cong \R^{2n}$
be the canonical projection map.
We may equip $\heis$ with a left-invariant Riemannian metric $g$
by declaring the $2n+1$ basis vectors given above to be an
orthonormal basis for $\heisa$, and translating this inner
product to other points of the group by (left)
multiplication. This metric is not
canonical, but in this investigation, the choice of metric does
not matter. When $n>1$, theorem~\tag{thm-Heisenberg-isoper-ineq} below
provides a quadratic isoperimetric inequality
for $\heis$ equipped with the metric $g$. Since any
other (left invariant) metric on $\heis$ disagrees with $g$
about lengths and areas by some uniformly bounded amount, we
will be able to conclude that $\heis$ satisfies a quadratic
isoperimetric inequality when equipped with {\it any}
(left-invariant) Riemannian metric.
Two important notions are the {\it vertical} and {\it
horizontal} subspaces at a point $x \in \heis$. The vertical
subspace $V_x$ is the set of vectors in $T_x\heis$ that are
tangent to the coset of $\calz$ passing through $x$. (Note that
the left and right cosets of $\calz$ coincide.) $V_x$ is one-%
dimensional. We define the horizontal subspace $H_x$ to be the
orthogonal complement of $V_x$ in $T_x\heis$; the
distribution of $2n$-dimensional horizontal subspaces is
invariant under the action of (left) group multiplication. We
may use the horizontal subspaces to equip the central quotient
$\R^{2n}$ of $\heis$ with a Riemannian metric: if $x \in \R^{2n}$,
then choose any representative $\tilde{x}$ for $x$ in $\heis$,
and observe that the projection $\pi$ establishes an isomorphism
between $H_{\tilde{x}}$ and $T_x\R^{2n}$. Define the inner
product on $T_x\R^{2n}$ via this identification; this is
independent of the choice of $\tilde{x}$, and yields a
`quotient' metric on $\R^{2n}$. It is not hard to see that this
is the standard Euclidean metric.
We say that a vector in $T_x\heis$ is
horizontal (resp. vertical) if it lies in $H_x$ (resp. $V_x$).
If $M$ is a manifold and
$f:M\rightarrow \heis$ is smooth, then we say that $f$ and
$f(M)$ are horizontal if every element of $f_*TM$ is
horizontal. (In our applications, $M$ will be an interval or a
surface.)
We may define a
symplectic form $\omega$ on the vector space
$H_1\subseteq{\heisa}$ by identifying $\frakz$ with $\R$ (by
identifying $z$ with $1\in \R$), and setting
$\omega(v,w)=[v,w]\in \frakz \cong \R$, for $v,w\in H_1$. We may
translate this structure to other points of $\heis$ by (left)
group multiplication, and then follow the procedure above to
equip the quotient $\R^{2n}$ with a symplectic structure, which
we also call $\omega$. It is easy to see that $\omega$ is the
standard symplectic form relative to the basis
$x_1,y_1,\ldots,x_n,y_n$.
When equipped with both its Euclidean and symplectic structures,
$\R^{2n}$ encodes much information about $\heis$, and the family of
horizontal subspaces allows us to `lift' objects in $\R^{2n}$ to
objects in $\heis$. Namely, if $\a:[0,1]\rightarrow \R^{2n}$ is a
smooth curve and we select a point $x \in \pi^{-1}(\a(0))$,
then there is a unique horizontal lift
$\tilde{\a}:[0,1]\rightarrow\heis$ having the properties that
$\tilde{\a}(0)=x$ and that $\pi\circ\tilde{\a}(t)=\a(t)$ for all
$t \in [0,1]$. If $\a$ is a closed loop in $\R^{2n}$, then
$\tilde{\a}$ typically fails to be a closed loop in $\heis$; the
symplectic structure on $\R^{2n}$ tells us the amount by which it
fails to close. The vertical distance between $\tilde{\a}(0)$
and $\tilde{\a}(1)$ is equal to the absolute value of the
symplectic area enclosed by the loop $\a$ in $\R^{2n}$, and the
sign of the symplectic area tells us which of $\tilde{\a}(0)$
and $\tilde{\a}(1)$ is `above' the other.
More generally, it is natural to ask whether can we find a
horizontal lift of a map $f:M \rightarrow \R^{2n}$, where $M$ is
some manifold. If $M$ is simply
connected, $f:M\rightarrow \R^{2n}$ is isotropic, $x\in M$, and
$\tilde{x}\in \pi^{-1}(f(x))$, then there is a unique horizontal
map $\tilde{f}:M \rightarrow \heis$ with the properties that
$\tilde{f}(x)=\tilde{x}$ and that for all $y\in M$,
$\pi\circ\tilde{f}(y)=f(y)$.
We will use this to build
horizontal lifts of the isotropic disks $f:D\rightarrow \R^{2n}$
obtained in theorem~\tag{thm-isotropic-disks}.
The lift $\tilde{f}$ may be defined
as follows: if $y\in M$, choose a smooth path $\a:[0,1]\to M$
from
$x$ to $y$. Define $\tilde{f}(y)$ to be the endpoint
$\widetilde{f\circ\a}(1)$ of the horizontal lift
$\widetilde{f\circ\a}$ satisfying
$\widetilde{f\circ\a}(0)=\tilde{x}$. This definition does not
depend on the choice of path $\a$ because if $\b$ is another
path in $M$ from $x$ to $y$, then $\a$ and $\b$ together bound a
disk in $M$, and so $f\circ \a$ and $f\circ \b$ together bound a
disk in $\R^{2n}$ with zero symplectic area. Hence, any
horizontal lift of the boundary of this disk is a closed loop,
which is to say that
$\widetilde{f\circ\a}(1)=\widetilde{f\circ\b}(1)$.
We note the simple yet important fact that if $\a$ is a
horizontal path in $\heis$, then its length measured with
respect to $g$ is the same as the Euclidean length of its
projection in $\R^{2n}$. Similarly, the area of a horizontal
surface in $\heis$ is the same as the Euclidean area of its
image in $\R^{2n}$. These properties hold because $\pi$ carries each
horizontal space $H_x$ isometrically to $T_{\pi(x)}\R^{2n}$.
\section{\Tag{sec-Heisenberg-isoper-ineq} An Isoperimetric
Inequality for $\heis$}
\beginproclaim Theorem \Tag{thm-Heisenberg-isoper-ineq}.
The Heisenberg groups $\heis$ with $n>1$ satisfy
a quadratic isoperimetric inequality.
\endproclaim
\remark{Remarks: } The proof below show that if a loop has
length $L$, then it spans a disk of area less than $(1+k)L^2 +
4\pi^{1/2}(k+1/3)L^{3/2} + (4\pi k + 2)L$, where $k$ is the
constant of theorem~\tag{thm-isotropic-disks}. The proof uses the following
lemmas, whose proofs appear after the proof of the theorem.
\beginproclaim Lemma \Tag{thm-homotopic-to-horiz-then-vert}.
A path in $\heis$ $(n\geq 1)$ of length $L$ is homotopic (rel
endpoints) to a path which is the composition of two paths, each
of length at most $L$, the first being horizontal and the second
being vertical; such a homotopy may be chosen with area at most
$L^2$.
\endproclaim
\beginproclaim Lemma \Tag{thm-vert-to-horiz}.
A vertical path in $\heis$ $(n\geq 1)$ of length $L$ is
homotopic (rel endpoints) to a horizontal path of length $2(\pi
L)^{1/2}$, by a homotopy of area $\leq 2L +
4\pi^{1/2}L^{3/2}/3$.
\endproclaim
\beginproof{Proof of theorem~\tag{thm-Heisenberg-isoper-ineq}: }
First suppose that $\c$ is a smooth horizontal loop of length
$L$.
Then $\pi\circ\c$ in $\R^{2n}$ also has
length $L$, and since one of its horizontal lifts is the closed
loop $\c$, it encloses zero symplectic area. By
theorem~\tag{thm-isotropic-disks}, $\pi\circ\c$ bounds a smooth
isotropic disk $D$ with area at most $kL^2$. A horizontal lift
of this isotropic disk has the same area as $D$, and one of the
horizontal lifts of $D$ bounds $\c$. This completes the proof in
the case that $\c$ is horizontal.
Now suppose that $\c$ is an arbitrary smooth loop of length $L$
in $\heis$. Using lemma~\tag{thm-homotopic-to-horiz-then-vert}
and then applying lemma~\tag{thm-vert-to-horiz} in the obvious way, we
apply a homotopy carrying $\c$ to a horizontal loop $\c'$ of
length at most $L+2\sqrt{\pi L}$; the area of this homotopy may
be taken to be $\leq L^2 +2L+4\pi^{1/2}L^{3/2}/3$. Applying the
horizontal case to $\c'$, and adding together the areas of the
homotopies used, we find that $\c$ spans a disk
of area at most the bound given in the remark.
\endproof
The geometric image to keep in mind for the proofs of the two
lemmas is that if $\d$ is a simple smooth path in $\R^{2n}$ then
$\pi^{-1}(\d)$ is locally isometric to $\R^2$, having two
orthogonal foliations. These are given by the horizontal lifts
of $\d$ and the preimages under $\pi$ of the points of $\d$.
\beginproof{Proof of lemma~\tag{thm-homotopic-to-horiz-then-vert}:}
Fix an isometry $\phi:\R\rightarrow\calz$ that carries $0\in \R$
to the identity element of $\calz$;
$\phi$ is also a group homomorphism. Let
$\c:[0,1]\rightarrow\heis$ be a path and define $\c_1$ to be the
horizontal lift of $\pi\circ\c$ with $\c_1(0)=\c(0)$. Consider
the map $f:[0,1]\times \R\rightarrow\heis$ given by
$f(s,t)=\c_1(s)\phi(t)$, juxtaposition denoting multiplication
in $\heis$. All computations will take place in $[0,1]\times
\R$, so we establish some facts regarding it. We define a
horizontal line to be a set of the form $[0,1]\times\{x\}\>(x\in
\R)$, and a vertical line to be a set of the form $\{x\}\times
\R\>(x\in[0,1])$. With respect to the pullback ``metric'' $f^*g$, vertical
and horizontal lines are orthogonal, and vertical line segments
have their natural lengths.
(Note that $f^*g$ may fail to be a metric by being degenerate at
some points.) Length along horizontal lines
depends on $\|\c_1'\|$, but all we will need about the behavior
of $f^*g$ on horizontal lines is that the total length of any
horizontal line is the length of $\c_1$, which is bounded by
$L$.
Define $\b_1,\b:[0,1]\rightarrow[0,1]\times \R$ to be the unique
maps such that $f\circ\b_1=\c_1$ and $f\circ\b=\c$. We have
$\b_1(s)=(s,0)$ and $\b(s)=(s,u(s))$ for some function
$u:[0,1]\rightarrow \R$; this function measures the difference
between $\c(s)$ and the horizontal path
$\c_1(s)$. Let $B$ be the homotopy which pushes
$\b$ along fibers to $\b_1$. Explicitly, $B(s,t)=(s,(1-t)u(s))$,
for $t\in[0,1]$. The length of the track~$s$ of $B$ is just
$u(s)$, which is obviously bounded by the arc length of
$\b([0,s])$, which is in turn bounded by $L$.
Since $|u(s)|\leq L$ for all $s$, we deduce that the
area of $B$ is at most $L\cdot{\rm length}(\b_1) \leq L^2$.
Observe that $B$ leaves $\b(0)$ fixed, while moving $\b(1)$
toward $\b_1(1)$; by reparameterizing $B$, we may regard it as a
homotopy rel endpoints between $\b$ and the path obtained by
first traversing $\b_1$ and then the vertical arc (of length
$\leq L$) from $\b_1(1)$ to $\b(1)$. The homotopy promised in
the lemma is $f\circ B$. Lengths and areas of objects in
$[0,1]\times \R$, measured with respect to $f^*g$, coincide with
the lengths and areas of their images under $f$, measured with
respect to $g$. This delivers the promised bounds for length and area.
\endproof
\beginproof{Proof of lemma~\tag{thm-vert-to-horiz}}
The homotopy we will construct lies entirely in
$\calh^3$. For each $t\in[0,\infty)$ let $S_t(\theta)$ (for
$\theta\in[0,2\pi]$) be the path in $\R^2$ that begins at $0$
and travels counterclockwise with constant speed around the
circle with center $(-t,0)$ and radius $t$. As $t$ varies between
$0$ and any given $T>0$, the $S_t$ sweep out a disk in $\R^2$
and the $\tilde{S_t}$ sweep out a disk $\tilde{D}(T)$ in
$\calh^3$, where $\tilde{S_t}$ denotes the horizontal lift of
$S_t$ beginning at the identity of $\calh^3$. For all $t$,
$\tilde{S_t}(2\pi)$ lies over $0\in\R^2$, so the path
$\tilde{S_t}(2\pi)$ (with $t\in[0,T]$) is a vertical path.
Setting $T=(L/\pi)^{1/2}$ we see that $\tilde{D}(T)$ provides a
homotopy (rel endpoints) between a vertical path of length $L$
and the horizontal path $\widetilde{S_T}$ of length $2(\pi
L)^{1/2}$.
It remains to bound the area of $\tilde{D}(T)$. We may
parameterize the homotopy by
$$\displaylines{
\Gamma:[0,T]\times[0,2\pi]\to\calh^3\cr
\Gamma(t,\theta)=\tilde{S_t}(\theta).
\cr}$$
We observe
$\|\dGdth\|=\|\del\tilde{S_t}/\del\theta\|=
\|\del S_t/\del\theta\|=t$.
We write $(\dGdt)(t,\theta)=h(t,\theta)+v(t,\theta)$ where
$h(t,\theta)$ and $v(t,\theta)$ are horizontal and
vertical vectors at $\Gamma(t,\theta)$, respectively. We have
$\|h(t,\theta)\|=\|(\del S_t/\del t)(t,\theta)\|$.
Since $S_t(\theta)=(-t,0)+(t,0)\cos\theta+(0,t)\sin\theta$ we
see that $\|h(t,\theta)\|\leq2$ for all $t$ and $\theta$.
We bound $\|v(t,\theta)\|$ by observing that $v(t,\theta)$ is given by the
infinitesimal area of the region bounded by the arcs
$S_t([0,\theta])$ and $S_{t+dt}([0,\theta])$ and the
infinitesimal segment joining $S_t(\theta)$ and
$S_{t+dt}(\theta)$. Formally, letting $A(t,\theta)$ be the area
of $\pi\circ\Gamma([0,t]\times[0,\theta])$, we observe that
$$
\|v(t,\theta)\|=\|(\del A/\del t)(t,\theta)\|.
$$
The right hand side is obviously bounded by
$\|(\del A/\del t)(t,2\pi)\|=2\pi t$. Our bounds on the norms
of $h$ and $v$ show that $\|\dGdt\|\leq 2+2\pi t$.
Finally, writing $X$ for $[0,T]\times[0,2\pi]$, we have
$$\eqalign{
{\rm Area}(\Gamma(X))&\leq
\int_X \|\dGdth\|\,\|\dGdt\|\,dt\,d\theta\cr
&\leq \int_X t(2+2\pi t)dt\,d\theta\cr
&=2\pi(T^2+2\pi T^3/3)\cr
&=2L+4\pi^{1/2}L^{3/2}/3.
\cr}$$
\endproof
\section{\Tag{sec-Heisenberg-apps} Applications and Remarks}
Theorem~\tag{thm-Heisenberg-isoper-ineq} has several
applications in the field of geometric group theory. Since
$\calh^5,\>\calh^7,\ldots$ satisfy a quadratic isoperimetric
inequality, so does any group which acts cocompactly,
discretely, and isometrically on one of them (see
\cite{bur:geom-and-comb-dehn-funcs}). The main example, the discrete
Heisenberg group $\Heisenberg_\Z$, has
generators $x_1,\ldots,x_n,y_1,\ldots,y_n,z$ and relations
asserting that $[x_i,y_i]=z$ and that all other pairs of
generators commute. $\Heisenberg_\Z$
is a cocompact discrete subgroup of $\heis$, and we may conclude
that this group satisfies a quadratic isoperimetric inequality
when $n>1$. For more information about isoperimetric
inequalities for finitely presented groups, see
\cite{sger:dehn-func-l1-norm}, \cite{ger:isoper-and-isodiam}. It
is easy to see that with respect to the presentation above,
$\calh ^3_\Z$ is isometrically embedded in $\heis_\Z$; it is also
true that $\calh ^3$ is isometrically embedded in $\heis$. It is
surprising that efficient spanning of loops in the isometrically
embedded $\calh ^3$ {\it requires} disks that do not lie in
$\calh ^3$. Finally, by proving
theorem~\tag{thm-Heisenberg-isoper-ineq}, we have justified the
claim in \cite{dbae:wd-proc-gps} that the groups $\heis_\Z$ for
$n>1$ provide examples of finitely presented groups that satisfy
quadratic isoperimetric inequalities while not admitting
automatic structures.
Another important reason to study the Heisenberg groups is that they
occur as horospheres in complex hyperbolic space $\CH^n$.
Another family of nilpotent Lie groups, the quaternionic
Heisenberg groups $\H\calh ^{4n-1}$, appear as the horospheres in
quaternionic hyperbolic space $\HH^n$ (see
\cite{gromov:geom-top}), and if $n>4$ then the techniques of
\S\S~\tag{sec-isotropic-disks}--\tag{sec-Heisenberg-isoper-ineq}
can be applied directly to yield quadratic isoperimetric
inequalities for them. In the case $n=4$, the techniques can
also be applied, but more care is required. If a group $G$ acts
(meaning that it acts discretely and isometrically) on $\CH^n$ or
$\HH^n$ with noncompact finite volume quotient, then each of its
cusp groups acts cocompactly on a copy of $\calh ^{2n-1}$ or
$\H\calh ^{4n-1}$. In his work on relatively hyperbolic groups,
Farb \cite{farb:thesis} has shown that under these
conditions, $G$ and its cusp groups satisfy isoperimetric
inequalities of the same degrees. This proves
\beginproclaim Theorem \Tag{thm-chn-and-qhn-isoper-ineq}.
A group $G$ which acts on $\CH^n\>(n>2)$ or $\HH^n\>(n>3)$ with
noncompact finite-volume quotient satisfies a quadratic
(but no subquadratic) isoperimetric inequality.
\endproclaim
\remark{Remark: } The optimality follows from work of
Gromov \ecite{gromov:hyperbolic-groups}{p.~104} and Olshanskii
\cite{olshanskii:subquadratic-isoperimetric-inequalities}, that
any group satisfying
a subquadratic isoperimetric inequality is word hyperbolic.
Since a word hyperbolic group cannot contain a
subgroup isomorphic to $\Z^2$, and $G$ contains
Heisenberg groups and hence copies of $\Z^2$, $G$ cannot satisfy
a subquadratic inequality. If a group
acts as in the theorem on real hyperbolic space $H^n$ then it
satisfies a quadratic inequality (and a linear one if $n=1$),
and if on $\CH^2$ it satisfies a cubic inequality. Again these
are optimal. The only hyperbolic spaces for which such precise
results are not yet known are $\HH^2$, $\HH^3$, and the hyperbolic
plane $\OH^2$ defined over the alternative field of octaves. (See
\cite{gromov:geom-top} for a description of these spaces.)
We hope the reader will find merit in the following comments
upon the proofs. The heart of the proof of
theorem~\tag{thm-Heisenberg-isoper-ineq} is contained in its
first paragraph. It is known that $\heis$ is quasi-isometric to
the metric space
$\heis_{\rm Carnot}$, in which the distance between two points
is the infimum of the lengths of {\it horizontal} paths joining
them. Since quasi-isometric spaces tend to satisfy
isoperimetric inequalities of the same degrees (see, e.g.,
\cite{alon:isoper-ineq}), one might hope to deduce the
isoperimetric inequality for $\heis$ from this, but it's not
clear what `area' means in $\heis_{\rm Carnot}$.
We began this work by trying to fill loops in a Cayley graph for
$\heis_\Z$, working combinatorially. After a while, it became
clear that two words in the generators $x_i,y_i$ commute exactly
when the parallelogram they span in the central quotient
$\Z^{2n}$ of $\heis_\Z$ encloses zero symplectic
area. Then it seemed more natural to work with polygonal paths
in $\R^{2n}$, and it was in this setting that the proof was
completed. It is something of a bonus that the technique applies
to smooth loops, and in fact is phrased most naturally in terms
of them. Most of our constructions have
analogues in $\heis_\Z$. The process used in
the proof of lemma~\tag{thm-homotopic-to-horiz-then-vert}
corresponds to the operation on words in $\heis_\Z$ of commuting
each $z$ all the way to the far end.
Lemma~\tag{thm-vert-to-horiz} performs in our smooth setting the
same sort of service as the combinatorial operation of replacing
each $z$ by $x_1y_1x_1^{-1}y_1^{-1}$. The proof of
lemma~\tag{thm-symplectic-orthogonal-homotopy} uses a smooth
version of the process ``commute all the $x_1$'s and $y_1$'s to
the beginning of a word.'' Only lemma~\tag{thm-swooping-lemma}
seems to have no neat combinatorial analogue. In fact, even if
the given loops are polygonal, it produces a homotopy between
them that isn't polygonal: the tracks of the homotopy are
elliptical arcs. Anyone wishing to devise a combinatorial
algorithm for contracting loops in $H^{2n+1}_\Z$ might first
describe how to span a polygonal loop in $\R^{2n}$ (that encloses
zero symplectic area) with a polygonal isotropic disk of
appropriately small area.
\smallskip
Gromov \cite{gromov:carnot-caratheodory} has developed his
arguments of \cite{gromov:asymp-invts} to
give another proof of theorem~\tag{thm-Heisenberg-isoper-ineq}. He derives the result from
his theorem~3.5D, the ``disk extension theorem'', a version of
which we reproduce here.
\beginproclaim Theorem
\Tag{thm-disk-extension}.
If $V$ is a simply connected compact Riemannian contact manifold of
dimension $2n+1\geq5$ then there exists $C>0$ such that the
following holds. For every Lipschitz function
$f_0:S^1\to V$ there exists a Lipschitz extension
$f:D^2\to V$ of $f_0$ such that the Lipschitz constant of $f$ is
bounded by $C$ times that of $f_0$.
\endproclaim
\noindent
(A contact $(2n+1)$-manifold is a manifold equipped with a
hyperplane field locally equivalent to that of $\Heisenberg$,
and the term Lipschitz refers to the Carnot metric on $V$
induced by the Riemannian metric and the hyperplane
distribution. A smooth map is called horizontal if its derivative takes
values in the distribution, and smooth horizontal maps from
$S^1$ are automatically Lipschitz.)
This immediately implies that a horizontal loop $\c$ in $V$ of
length $L$ spans a horizontal disk of area $\leq C^2L^2/4\pi$,
which yields a quadratic isoperimetric inequality for horizontal
loops in $V$. Despite the hypothesis that $V$ be compact, one
may apply the theorem to prove our theorem~\tag{thm-Heisenberg-isoper-ineq}. One takes $V$
to be a simply connected compact manifold-with-boundary neighborhood of
$1\in\Heisenberg$. If $\c$ is any horizontal loop in
$\Heisenberg$ of
length $L$ then let $\c'$ be the image of $\c$ under the
``dilation'' $x_i\mapsto tx_i$, $y_i\mapsto ty_i$, $z\mapsto
t^2z$ for $t$ small enough so that $\c'\sset V$. Because the
dilation preserves the field of horizontal hyperplanes, $\c'$
is horizontal and has length $tL$. By the theorem, it spans a
horizontal disk of area $\leq C^2(tL)^2/4\pi$. Then by taking
the image of the disk under the inverse of the dilation we see
that $\c$ spans a disk of area $\leq C^2L^2/4\pi$. This argument
does not address the issue of non-horizontal loops in
$\Heisenberg$, but one is generally not interested in such
things when thinking about Carnot geometry. In any case, to
prove that any loop bounds a disk of Riemannian area quadratic
in the loop's length, one can reduce to the horizontal case and
apply the argument above. The
reduction to the horizontal case is easy (say, our
lemmas~\tag{thm-homotopic-to-horiz-then-vert} and
\tag{thm-vert-to-horiz}.)
Gromov's arguments for theorem~\tag{thm-disk-extension} are part of a systematic
study of Lipschitz maps to Carnot-Carath\'eodory manifolds (a
generalization of a contact manifold). Among other things, he
treats the approximation of continuous functions by horizontal Lipschitz
functions, which can be used to provide the reduction to the
case of horizontal loops. He also treats the problem of
extending a piecewise smooth and horizontal map satisfying
bounds on its higher derivatives by functions satisfying similar
conditions.
Our methods differ from Gromov's primarily in that the
explicit homotopies we employ use the global structure of $\heis$
rather than just the local contact structure. (Our homotopies
may wander far from the original curves.)
The proofs of
lemmas~\tag{thm-homotopic-to-horiz-then-vert} and
\tag{thm-vert-to-horiz} could be
rewritten to show that for any curve $\c$ in $\Heisenberg$ there
exists a horizontal curve of approximately the same length
within (say) distance 1 of $\c$. However, the key to our
construction, lemma~\tag{thm-swooping-lemma}, seems to require a
``global homotopy''. As a consequence, it seems unlikely that
the argument could be modified to work for general
Carnot-Carath\'eodory manifolds. On the other hand, our explicit approach
has the advantage of directness and provides a uniform
isoperimetric inequality for all the $\Heisenberg$ with $n>1$.
\smallskip
\def\H{{\bf(H)}}
Yng-Ing Lee \cite{lee:metric-properties-lagrangian-surfaces} has
also proven a version of
theorem~\tag{thm-isotropic-disks}. Namely, that there is a
constant $c>0$ such that
if $\c$ is a smoothly embedded loop of length $L$ in $\R^4$ such
that
\rom1 the center of mass of $\c$ is the origin,
\rom2 $\c$ encloses zero symplectic area, and
\rom3 $\c$ satisfies condition \H\ below,
then $\c$ bounds a smoothly embedded disk
in $\R^4$ of area $\leq cL^2$. We say that $\c$ satisfies
condition \H\ if there is a smooth homotopy through smoothly
embedded loops $\c_t$, each enclosing zero symplectic area, that
carries $\c=\c_1$ to $\c_0=C$, a standard circle of
circumference $L$ in the $(x_1,x_2)$-plane. Below, we compare
this with her original definition. Her construction is
completely explicit (given the homotopy) and has the advantage
of yielding an embedded spanning disk. However, her method only
works for smoothly embedded loops satisfying \H. This is not a
serious drawback: she sketches an argument that any loop
enclosing zero symplectic area may be approximated by a loop
satisfying \H.
Her argument is more in the style of pure symplectic geometry
than ours: it proceeds by realizing the given homotopy as the
track of $\c$ under the flow of a suitable time-dependent
Hamiltonian vector field $V_t$. Reversing the flow carries the
obvious disk $D$ spanning $C$ to a disk bounded by $\c$. Since
the flow of a Hamiltonian vector field is a symplectomorphism,
this yields an isotropic disk spanned by $\c$. The key to her
argmument is that one can choose the vector fields $V_t$ in such a way that
they and their first derivatives with respect to the $x_i$ and
$y_i$ are all bounded by some universal constant. This lets one
estimate how much the disk $D$ is distorted as it flows along
the $V_t$, and thus yields a bound on the area of the disk
spanning $\c$. She chooses the $V_t$ by explicitly writing down
a time-dependent Hamiltonian function, and then checking that is
satisfies the various required properties.
We close by indicating why our version of condition\H\ implies
hers. The additional criteria she imposes are that all the
$\c_t$ have the same length and that there be $\d>0$ such that
there is an embedded $\d$-tubular neighborhood of $\c_t$ for all
$t$. If $\c$ satisfies our version of \H, with the length of
$\c_t$ being $L_t$, then replacing each $\c_t$ with the scaled
loop ${L\over L_t}\c_t$, we recover her first extra
condition. The existence of the tubular neighborhoods for some
$\d$ follows
from the facts that
each $\c_t$ is smoothly embedded and that
there is a uniform bound on
the extrinsic curvature of the paths $\c_t$. (The latter claim
follows from the compactness of the homotopy.)
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% ---end bibentries---
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\bye