A soap film wants to be as small as possible. If you dip a wire loop into
soap water, the film that forms is (to very good approximation) the surface
of least area among all surfaces that have the wire as boundary. The oldest
question in the subject is: how small can that surface be?
Start in the flat plane. A closed curve of length L bounds a disk of area
L^2/(4 pi), and no surface with that boundary has less area, because a disk
has the least area among all sets with given perimeter (perimeter is the
two dimensional word for length of the boundary; area is the volume of the
region). For a curve that is not a circle the disk of the same length is
strictly bigger, so L^2/(4 pi) is an upper bound for the least filling area
of any closed curve of length L, and it is attained only by the round one.
The same question in every dimension. A closed n-dimensional surface of
area A bounds an (n+1)-dimensional region of volume V. Euler's constant
pi is the first of a family: write
omega_j = volume of the unit ball of R^j
s_n = area of the unit sphere S^n inside R^(n+1)
= (n+1) * omega_(n+1)
so omega_1 = 2, omega_2 = pi, omega_3 = 4 pi / 3, and s_0 = 2, s_1 = 2 pi,
s_2 = 4 pi, s_3 = 2 pi^2. The Euclidean isoperimetric inequality says
V <= A^((n+1)/n) / ( (n+1)^((n+1)/n) * omega_(n+1)^(1/n) ) . (0.1)
The ball is the equality case: for the sphere of radius R we have
A = s_n R^n and V = omega_(n+1) R^(n+1), and (0.1) holds with equality. So
(0.1) cannot be improved, and the "sharp constant"
c_n = 1 / ( (n+1) * s_n^(1/n) ) (0.2)
is the best possible coefficient in the inequality V <= c_n A^((n+1)/n).
In numbers: c_2 = 1/(6 sqrt(pi)) = 0.09403..., and
c_3 = (128 pi^2)^(-1/3) = 0.09248...
Now let the room be curved. Our question becomes: if the ambient space is
not flat R^(n+1) but some curved space X, and a closed n-dimensional surface
inside X has area A, how much volume must a filling of it have? Equivalently
and more usefully: if the filling has n+1 dimensional volume V, how much
area must its boundary have?
The main theorem of this file. The precise statement is [T1] in section 3;
here it is.
Let X be a proper CAT(0) space and let n >= 2. Let T be an "integral
n-cycle" of compact support in X, with mass A. Then T bounds an
integral current S of compact support with
boundary of S = T, mass of S <= c_n A^((n+1)/n), (0.3)
where c_n is the Euclidean constant (0.2). There is no restriction on
the dimension of X, on the topology or singularities of T, or on the
integer multiplicities that T carries.
Two words need a promise behind them.
"CAT(0) space" means a metric space in which distance is measured along
geodesics and triangles are no fatter than their Euclidean comparison
triangles. The plane, all hyperbolic spaces, all trees, Hilbert space and
every convex subset of any of these are CAT(0). Nonpositive curvature in
the classical Riemannian sense implies CAT(0). Section 1.2 gives the exact
definition and section 1.3 gives the examples.
"Integral current" means an oriented n-dimensional surface with an integer
weight at each point, allowing pieces that are only Lipschitz (so, corners
and kinks are allowed) and allowing weights bigger than one (so a surface
can be covered twice). Its mass is the weighted area. Its boundary is the
oriented (n-1)-dimensional surface that is "the edge", and a cycle is a
current whose boundary is zero. Section 2 builds this up from nothing.
So the theorem says: in any CAT(0) space, however singular, and for cycles
of any dimension n >= 2, the Euclidean price for filling is still the true
price, even though there is no linear structure, no differentiable charts,
no normal bundle, and no smoothness anywhere.
Three separate reasons, and each one is a section of the proof.
(a) Curvature can make filling worse, and there is a cheap example showing
the exponent can change completely. Take X to be an infinite tree with
one branch point where three unit segments meet (a tripod). Take n = 1:
a 1-cycle is a formal difference of two points, and its only fillings
are the paths joining them, so filling two points at distance 2 costs 2.
But the Euclidean bound for n = 1 would cost 2^2/(4 pi) = 0.32. False by
a factor of six. So the theorem is genuinely about n >= 2; for n = 1 the
truth is a linear law, not a (n+1)/n law. This is the honest content of
the hypothesis n >= 2, and no proof can drop it.
Positive curvature also breaks the inequality: on a round sphere of
radius R, the equator has length 2 pi R and the cheapest thing it bounds
is a hemisphere of area 2 pi R^2, while the Euclidean price would be
(2 pi R)^2/(4 pi) = pi R^2, three times smaller. So nonpositive curvature
is not decoration either.
(b) There are no charts. In a CAT(0) space we cannot integrate, cannot take
derivatives, and cannot even write down a surface in coordinates. The
theory of integral currents in metric spaces (Ambrosio-Kirchheim) gives
the right definitions, but then all the classical machinery - the area
formula, the coarea formula, the Euler-Lagrange equation - has to be
re-derived from the distance alone. Sections 2 and 7 do that.
(c) The extremal configuration is never smooth. The sharp constant is
attained by a round ball, so the natural proof has to look at what
happens near a round ball. But the objects in the proof are extremizers
of a variational problem in a singular space; they can have fractional
exponents and can vanish on open sets. Sections 7.6 and 7.7 are the
careful handling of exactly those two degeneracies.
Here is the whole strategy in seven paragraphs. Every word here is made
precise later; read it once now and once at the end.
Step 1 (reduce to a compact room). A compactly supported cycle lives inside
some closed ball, and closed balls in a CAT(0) space are convex and, if the
space is proper, compact. So it is enough to prove the theorem inside a
compact CAT(0) space Y. [L6].
Step 2 (assume the theorem fails, get a champion). Inside Y, the least
filling mass of a cycle B is written V(B). Suppose the sharp bound fails for
some cycle. Then for a constant c slightly bigger than c_n, the functional
B |--> V(B) - c * (mass of B)^((n+1)/n) (0.4)
is positive somewhere. A compactness theorem for currents gives a cycle A
that maximizes (0.4). Then A has a first variation property [L7]: for every
cycle B,
V(A - B) >= c * ( m^((n+1)/n) - (mass of B)^((n+1)/n) ), (0.5)
where m = mass of A. Step 3 (a rescaling of all distances) makes
m < s_n , c * ((n+1)/n) * m^(1/n) = 1/n . (0.6)
So the mass of the champion is slightly less than the area of the unit
sphere in its own dimension. That numerical coincidence is the engine: the
proof will show that any cycle satisfying (0.5) and (0.6) must have
m >= s_n, and contradiction.
Step 3 (radial variation). Pull every point of A towards a fixed centre y at
a speed that depends on the point. Each point x is moved to the point at
fraction lambda(x) on the segment [y,x], where lambda = 1 - t g(x) and g is
a nonnegative Lipschitz function. Sliding A for time h sweeps out a
filling D_h of (A - B_h), where B_h is the moved copy of A. Comparing the
mass lost by A with the mass swept, and using the champion inequality (0.5),
gives the radial inequality [P1]:
integral ( n g + r * (grad r . grad g) ) dmu
<= n * integral ( r g sqrt(1 - |grad r|^2) ) dmu (0.7)
for every centre y and every nonnegative Lipschitz g, where r(x) = d(y,x),
mu is the mass measure of A, and grad, dots and |.| are the chart
derivatives of section 2.5.
Step 4 (base case, n = 2). Feed (0.7) the cutoff g = 1/max(r, eps)^2 and
let eps go to 0. A Euclidean lower density bound at almost every point of
A turns that into m >= 4 pi = s_2, contradicting (0.6). So the theorem
holds for 2-cycles. [P2].
Step 5 (induction, the small set estimate). Now n > 2 and we assume the
theorem in dimension n-1. Cut the champion A by a level set of a Lipschitz
function. Each small piece has an (n-1)-dimensional boundary, and the
induction hypothesis fills that boundary cheaply. Feeding the result back
into the champion inequality (0.5) gives a perimeter bound: every small
piece of A has boundary mass at least (1 - eta) n omega_n^(1/n) * (its
mass)^((n-1)/n), the Euclidean sphere bound. [P3].
Step 6 (from sets to functions to functions). Two classical moves in metric
form. First, coarea plus rearrangement [L9, L10, L11, L12] convert the
small set perimeter bound into an "almost Euclidean" Sobolev inequality
[P4]:
(S_* - eps) (integral |u|^p dmu)^(2/p)
<= 4 beta integral |grad u|^2 dmu + C_eps integral u^2 dmu , (0.8)
where S_* = n s_n^(2/n), beta = 1/(n-2) and p = 2n/(n-2). Second, that
inequality makes a certain variational problem on A have a minimizer v with
[P7]
W = integral v^p dmu in (0, s_n), 4 beta E(v) + n |v|_2^2 = n W.
The strict inequality W < s_n is inherited from (0.6) and is the whole
point: v is a slightly subcritical Sobolev bubble living on the current A.
Step 7 (chords force contradiction). Normalize dnu = v^p dmu / W, a
probability measure on Y, and consider for a centre y the squared weighted
chord
s_y(x) = v(y)^beta d(y,x) v(x)^beta , L_y = integral s_y^2 dnu .
Testing the minimizer equation and the radial inequality (0.7) with the
functions built from the Euclidean bubble profile (1 + a s^2)^(-(n-2)/2)
gives the transform bound [P8]
integral (1 + a s_y^2)^(-n) dnu <= (1 + 2 a L_y)^(-n/2) , (0.9)
which forces L_y <= 2 (W/s_n)^(2/n) < 2 for almost every centre y [L17].
On the other hand, test the minimizer equation once more with v^(1+2 beta)
and use Cauchy-Schwarz + the CAT(0) variance inequality to get the opposite
averaged bound [L18]:
integral L_y dnu >= 2 . (0.10)
Since (W/s_n)^(2/n) < 1, (0.9) says the average is strictly less than 2 and
(0.10) says it is at least 2. Contradiction. So no champion exists, the
sharp inequality holds in dimension n, and induction gives the theorem in
every dimension. [T1].
Step 8 (the classical payoff). In a Cartan-Hadamard manifold there are no
compactly supported top-dimensional cycles other than zero [L19], so the
filling of the boundary of a domain must be the domain itself. Applying
[T1] to that filling gives the classical Euclidean domain inequality for
every bounded smooth domain in every dimension d >= 3. [T2].
The Euclidean case is Almgren (1986): the sharp inequality (0.1) for
integral currents in R^m, all dimensions and codimensions. Doing the same in
a metric space needs currents in metric spaces, which is Ambrosio and
Kirchheim (2000). Wenger (2005) proved an isoperimetric inequality with the
Euclidean exponent in spaces with convex geodesic bicombings, with a
constant depending on the space and on the dimension, not sharp.
For the applications, the classical statement is: in a Cartan-Hadamard
manifold (complete, simply connected, sectional curvature <= 0), domains of
dimension d satisfy the Euclidean volume-perimeter inequality, namely
area of the boundary >= d * omega_d^(1/d) * volume^((d-1)/d) . (0.11)
That is "the Cartan-Hadamard conjecture". Dimension 2 is classical (Weil
1926; Beckenbach-Rado 1933). Dimension 3 is Kleiner (1992), dimension 4 is
Croke (1984); dimensions 3 to 9 were done by Chen-Ghomi-Wang (2026), and
various conditional and small-volume versions exist. In this file's
theorem, all dimensions simultaneously are consequences of one sharp
current inequality. The main result here is the filling inequality [T1]
itself: sharp coefficient, all cycle dimensions n >= 2, all proper CAT(0)
spaces, arbitrary multiplicities, no dimension bound on the ambient space.
Related but different: Lang-Stadler-Urech (2026) and Peteranderl (2026)
prove that in CAT(0) spaces of finite asymptotic Nagata dimension, fillings
can be much cheaper than Euclidean at very large mass (linear, or almost
linear). Those theorems are about the growth exponent at large mass and
their constants depend on the space. [T1] is about the exact coefficient at
every mass. There is no conflict: different regimes, different claims.
A metric space is a set X with a function d(x,y) >= 0 that is zero exactly
on the diagonal, is symmetric, and satisfies the triangle inequality
d(x,z) <= d(x,y) + d(y,z). That is all the structure a CAT(0) space uses.
A geodesic segment from x to y is a map gamma: [0,1] -> X with
gamma(0) = x, gamma(1) = y, and d(gamma(s), gamma(t)) = |s-t| d(x,y) for
all s,t in [0,1]. In words: it is a path that covers the distance exactly,
at constant speed. The point at fraction a is written [x,y]_a, so
[x,y]_0 = x and [x,y]_1 = y; this notation is used constantly below. X is a
geodesic space if every pair of points is joined by a geodesic segment.
Note that a geodesic segment need not be unique, but in a CAT(0) space it
always is (see 1.2); we then speak of "the" segment.
A geodesic triangle with vertices x_1,x_2,x_3 is a choice of three segments
joining them pairwise. Its Euclidean comparison triangle is a triangle in
the plane with the same three side lengths; it exists because three
positive numbers satisfying the three triangle inequalities are the side
lengths of a Euclidean triangle. Points on the comparison triangle
correspond to points at the same fractions along the sides.
Definition. X is CAT(0) if X is a geodesic space and, for every geodesic
triangle with vertices x_1,x_2,x_3 and every pair of points p,q on two sides
of it, the corresponding points p',q' of the Euclidean comparison triangle
satisfy
d(p,q) <= |p' - q'| . (1.1)
Inequality (1.1) says: triangles in X are no fatter than Euclidean
triangles. It is the exact metric version of "sectional curvature <= 0", and
on a Riemannian manifold it is equivalent to it (Cartan-Alexandrov-Toponogov).
The "(0)" in CAT(0) is the curvature bound.
Three consequences are used all the time. All three follow from (1.1) in a
few lines.
(a) Geodesics are unique, and the space is "Busemann convex": if x -> x_t
and y -> y_t are geodesics with the same parameter, then
t |--> d(x_t, y_t) is convex. In particular the function
x |--> d(y,x) is convex along every geodesic.
(b) Balls are convex: for any centre o and radius R, the closed ball is a
geodesic space in its own right, with the restricted distance.
(c) Shortening a segment by a fixed fraction is a contraction: for fixed
t in [0,1], the map x |--> [o,x]_t is t-Lipschitz. In symbols,
d([o,x]_t, [o,x']_t) <= t d(x,x') . (1.2)
Both (a) and (c) are immediate from (1.1) applied to the triangle
o,x,x' and its comparison triangle, where the corresponding map is
the honest Euclidean contraction towards o. We shall also use the
"equal fraction" comparison, valid for two different fractions
lambda, lambda':
d([o,x]_lambda, [o,x']_lambda')^2
<= (lambda r - lambda' r')^2
+ lambda lambda' ( d(x,x')^2 - (r - r')^2 ) (1.3)
where r = d(o,x) and r' = d(o,x'). This is exactly the squared
distance of the corresponding two points in the Euclidean comparison
triangle, so it is a consequence of (1.1). It is the key estimate of
section 6, where lambda and lambda' differ because they depend on x
and on x' through a function g.
Finally, the four point inequality. For any four points a,b,c,d of a CAT(0)
space,
d(a,c)^2 + d(b,d)^2
<= d(a,b)^2 + d(b,c)^2 + d(c,d)^2 + d(d,a)^2 . (1.4)
Proof (short, and worth reading once, because it is used to prove that
chart metrics are Euclidean in section 2.5). Let m be the midpoint of
[a,c]. Midpoint comparison in the triangles a,b,c and a,d,c says
2 d(b,m)^2 + d(a,c)^2/2 <= d(a,b)^2 + d(b,c)^2 ,
2 d(d,m)^2 + d(a,c)^2/2 <= d(a,d)^2 + d(c,d)^2 .
Add them and use d(b,d)^2 <= 2 d(b,m)^2 + 2 d(d,m)^2, which is the triangle
inequality squared. The two d(a,c)^2/2 terms cancel against the common
factor 2 d(...)^2 and (1.4) follows.
Inequality (1.4) is a parallelogram law with an inequality sign. In
Euclidean space it is an equality for the four points of a parallelogram.
It says that CAT(0) spaces are, in a weak integral sense, no more curved
than flat ones; the analytic payoff is that a "metric derivative" of a
Lipschitz map out of R^k, which is a priori only a seminorm, satisfies the
parallelogram law and hence comes from an honest inner product. That is the
content of [L5] in section 2.5, and it is the reason the whole proof can
compute determinants and Jacobians at all.
The examples. R^m with its Euclidean distance, all hyperbolic spaces H^m,
any Hilbert space, all trees with the path metric, all CAT(0) cubical
complexes, all Euclidean buildings of noncompact type, all Gromov-Hausdorff
limits of CAT(0) spaces, and every convex subset of a CAT(0) space. The
class is closed under products with the Euclidean factor and under passing
to convex closed subsets, which is what step 1 of the proof does.
What it forbids. The space must be geodesic and its geodesics must be
unique, so, for instance, R^2 with the l^1 distance (where the segment from
(0,0) to (1,1) is not unique) is not CAT(0), even though it is flat in a
weaker sense. CAT(0) also forbids positive curvature, so a round sphere is
not CAT(0); a ball of radius larger than a quarter of the circumference in a
sphere is not CAT(0) either.
Why nobody objects to the singular examples. Trees, buildings and limits are
exactly where the classical differential-geometric methods fail, and exactly
where this theorem still gives a sharp answer. That is the point of proving
it with currents rather than with the calculus of variations on manifolds.
X is proper if every closed bounded subset of X is compact. R^m, trees with
locally finite branching, and complete Riemannian manifolds are proper;
infinite dimensional Hilbert space is not.
Properness is used for one thing only: to make closed balls compact so that
compactness theorems for currents apply. Since a compactly supported current
lives in a bounded set, and closed balls in a CAT(0) space are convex, we
can always move the whole problem into a compact CAT(0) space: [L4] in
section 5 states and proves that reduction. From section 5 onward the
ambient space is a compact CAT(0) space named Y, and the letter X is only
used again in the final assembly.
Constants. Let omega_j be the volume of the unit ball of R^j (so
omega_1 = 2, omega_2 = pi, omega_3 = 4 pi/3, omega_4 = pi^2/2), and let s_n
be the area of the unit sphere S^n in R^(n+1). The relation is
s_n = (n+1) omega_(n+1): differentiate the volume of the ball of radius R,
which is omega_(n+1) R^(n+1), with respect to R and set R = 1. Thus
s_1 = 2 pi, s_2 = 4 pi, s_3 = 2 pi^2.
Three constants appear in the sharp statements:
s_n area of the unit sphere in dimension n
c_n = 1/((n+1) s_n^(1/n)) the sharp filling constant
S_* = n s_n^(2/n) the sharp Sobolev constant
and two more in the induction, beta = 1/(n-2) and p = 2n/(n-2), which are
the classical Sobolev exponents in dimension n (so p > 2 exactly when
n > 2, which is why n = 2 is done separately).
For orientation: c_2 = 1/(6 sqrt(pi)) = 0.09403..., c_3 = (128 pi^2)^(-1/3)
= 0.09248..., and c_n decreases slowly to zero, roughly like a constant
times n^(-3/2).
Exercise. Check that c_n = omega_(n+1)^(1/(n+1)) / (n+1) * something is NOT
the formula; the right formula is c_n = 1/((n+1) s_n^(1/n)). Then check with
n = 2 that 1/(3 * sqrt(4 pi)) = 1/(6 sqrt(pi)).
Take X = R^(n+1), let T be the boundary of the ball of radius R, oriented.
Then
mass of T = s_n R^n =: A ,
mass of the ball = omega_(n+1) R^(n+1) =: V ,
and by (0.2),
c_n A^((n+1)/n) = s_n R^(n+1) / (n+1) = omega_(n+1) R^(n+1) = V .
So the bound in [T1] is attained for every R. By [L19] (top dimensional
constancy in a noncompact manifold, section 8) the ball is the ONLY
compactly supported filling of T in R^(n+1). Hence the constant c_n cannot
be improved even in flat space, which settles optimality; and it also shows
the theorem is exactly the statement "every CAT(0) space is at least as
expensive as flat space", which is what nonpositive curvature is supposed to
mean.
A soap film is not a smooth surface. It meets itself along curves, three
sheets meet at 120 degrees along a curve, and a minimizing sequence of
smooth surfaces can collapse to something with kinks and crossings. So if we
define "surface" as "smooth embedded manifold", two things go wrong at once.
First, no compactness. The set of smooth surfaces with a given boundary and
bounded area is not closed: take a wavy sheet and let the waves get finer
and finer with the same amplitude. The areas stay bounded but the surfaces
have no limit in any reasonable sense. Since the whole proof needs to look
at a surface of least area, and least area is defined by an infimum over a
sequence, this is fatal.
Second, no cancellation. If you run a surface one way and then the same
surface the other way, the union is a legitimate "filling" of the boundary
zero, and it should count as nothing. Smooth embedded surfaces cannot
express that; oriented surfaces can, if you count multiplicity with signs.
The theory of currents fixes both problems at once. A k-current is an
object for which two things make sense: "integrate a bounded Lipschitz
weight against the Jacobian of k coordinate functions" and "take the
oriented boundary". It contains every oriented Lipschitz k-surface with
integer multiplicity, it is closed under weak limits, and the limit of a
minimizing sequence exists and is again a current. Everything hard in this
file is a consequence of working in that completed world.
Standing notation for this file: X is a complete metric space (later a
compact CAT(0) space).
Lip(X) = real valued functions on X with finite Lipschitz constant.
Lip_b(X) = the bounded ones.
For a Lipschitz function pi, its Lipschitz constant is written Lip(pi).
Definition (Ambrosio-Kirchheim). Let k >= 0 be an integer. A k-current on X
is a map
C : Lip_b(X) x Lip(X) x ... x Lip(X) -> R (k copies of Lip(X))
which is linear in each of the k+1 slots separately, antisymmetric under
permuting the last k slots, and continuous under the following weak
convergence: if b_j -> b pointwise with a common bound on |b_j| and on
Lip(b_j), and if pi^j_a -> pi_a pointwise with a common bound on Lip(pi^j_a),
then C(b_j, pi^j_1, ..., pi^j_k) -> C(b, pi_1, ..., pi_k).
Reading the definition. Take the model case: an oriented surface S given by
a bi-Lipschitz parametrization phi: E -> X, E a Borel subset of R^k, with an
integer weight theta. Then the corresponding current is
C(b, pi_1, ..., pi_k) = int_E theta(z) b(phi(z))
det D( (pi_1,...,pi_k) o phi )(z) dz . (2.1)
That is: integrate the weight b against the Jacobian determinant of the k
coordinate functions along the surface. If you choose pi_1,...,pi_k to be
honest coordinates and b = 1, this is exactly the k-dimensional volume of
the piece. The point of the definition is that (2.1) makes sense for every
tuple of Lipschitz tests, and knowing all those numbers determines C.
Mass. For a finite Borel measure eta on X, say that eta controls C if
|C(b, pi_1, ..., pi_k)| <= int |b| deta for all b, pi_a with
Lip(pi_a) <= 1. (2.2)
The mass of C is the smallest such eta, written ||C||; it exists (the
supremum of the left sides over admissible tuples is additive over disjoint
Borel sets, so it defines a measure). The total mass is M(C) = ||C||(X), and
for the model current (2.1) built from a surface,
||C|| = (phi)_# ( |theta| * J dz ), J = sqrt(det G) the area density,
so M(C) is exactly "area counted with the absolute value of the weight". The
formula (2.2) with the sup over all tuples is the honest definition of mass
even for currents that are not surfaces; it says that mass measures all the
ways the current can be tested.
Why the definition is written with k coordinate functions. The determinant
in (2.1) is the volume distortion in the k coordinate directions. In a
metric space we have no coordinates, but we do have real valued Lipschitz
functions, and k of them together measure a k-dimensional distortion. All
of the smooth theory is recovered by taking the pi_a to be coordinates.
Boundary. The boundary of a k-current C is the (k-1)-current
(dC)(b, pi_1, ..., pi_{k-1}) = C(1, b, pi_1, ..., pi_{k-1}) . (2.3)
For a curve (k = 1) this reads (dC)(b) = C(1,b) = integral of b over the
boundary of the curve, i.e. b(end) - b(start). So (2.3) is the algebraic
version of "the edge of the surface". One checks d(dC) = 0 whenever the
objects involved have finite mass; that algebraic identity is why slicing,
cutting and gluing work in this theory without any smoothness.
Cycles, fillings, filling volume. C is a cycle if dC = 0 (zero current).
C is normal if M(C) + M(dC) < infinity, and integral if in addition it has
the representation
C = sum_i (phi_i)_# ( theta_i [[E_i]] ), (2.4)
with phi_i bi-Lipschitz from a bounded Borel set E_i onto its image, the
images pairwise disjoint Borel sets, theta_i integer valued and integrable
against the area density, and the sum of masses finite. Write I_k(X) for
the integral k-currents. For B in I_k(X) with dB = 0, put
V(B) = inf { M(S) : S in I_(k+1)(X), dS = B } . (2.5)
V(B) is the least filling mass of B, and its finiteness is a theorem, not a
definition (see [L5] for an explicit filling, and [lit6] for the Euclidean
exponent bound). [T1] is precisely the statement that
V(B) <= c_k M(B)^((k+1)/k)
for cycles in a proper CAT(0) space of any dimension, with k >= 2.
The whole point of integral currents, rather than arbitrary normal ones, is
the integrality of the weight: it will give a lower density bound of section 2.7
which is what forces the constant s_n in the base case, and which is the
only place the integers theta are used quantitatively. Running the proof
with real weights would give a strictly weaker inequality.
The representation (2.4) is the metric replacement for "a surface is a union
of coordinate patches". Two facts make it usable.
(i) By subdividing and using the disjointified images, we may always take
the chart domains E_i bounded and the images Z_i = phi_i(E_i)
pairwise disjoint. The mass is then a sum of absolutely continuous
measures on the charts, and the charts are "almost flat at almost
every point" (that is the next subsection).
(ii) Each phi_i is bi-Lipschitz, so its inverse is defined on Z_i and has
a Lipschitz constant; by McShane's extension theorem [lit 11] every
Lipschitz function on a subset of a metric space extends to the whole
space with the same constant. This is what lets us test a current on
"coordinates of the chart" even though the chart is only a subset of
X: we extend and test on the extension; the value depends only on the
values on the chart, because the current is supported there.
Fact (i) is algebraic bookkeeping. Fact (ii) is used twice with real force:
once to compute mass exactly ([L3]) and once to close the gradient in
section 7.7. Both times, the point is the same: a chart which is bi-Lipschitz
can be "unrolled" by Lipschitz functions that the current can be evaluated
on, with no loss in the constants that matter.
Classical area formula: for an injective C^1 map phi: R^k -> R^m, the
Jacobian factor is sqrt(det (D phi)^T (D phi)). In a metric space there is no
derivative, but there is a replacement.
Definition (metric differential). Let phi: E -> X be Lipschitz, E a Borel
subset of R^k, and let z be a density point of E. A seminorm N on R^k is the
metric differential of phi at z if the two-point formula
d( phi(z+w), phi(z+w') ) = N(w - w') + o(|w| + |w'|) (2.6)
holds as the two domain points tend to z inside E.
Kirchheim's theorem [lit 2] says that at almost every density point of E a
metric differential exists (it is a seminorm, not necessarily a norm, and
not necessarily unique, but almost every point has one and any two agree).
In a general metric space (2.6) is the best structure available; in
particular the "norm" can be a weird polyhedral thing. In CAT(0) space it
cannot: it must be an honest Euclidean norm. That is our first real result.
[L1] (Euclidean chart metrics). Let X be CAT(0), phi: E -> X bi-Lipschitz,
and let z be a density point of E where the metric differential N exists.
Then
(a) N is a norm, that is N(w) > 0 for w != 0;
(b) N comes from an inner product: there is a symmetric positive
definite k x k matrix G with N(w)^2 = w^T G w for all w in R^k;
(c) with this G, the centered form of (2.6) holds:
d(phi(z+w), phi(z)) = |w|_G + o(|w|) , |w|_G^2 = w^T G w;
(d) the entries of G are measurable functions of z on E, and J = sqrt(det G)
is a measurable area density with
Lip(phi^(-1))^(-k) <= J(z) <= Lip(phi)^k (2.7)
at almost every z in E.
Proof. (a) Nondegeneracy. phi is bi-Lipschitz: there is c > 0 with
d(phi(z),phi(z')) >= c |z - z'|. Fix a direction and a scale; a
density-one point has the property that any fixed finite configuration in
R^k is approximated, to relative error o(1), by the rescaled set (E - z)/t
for small t. (If a point of the configuration stayed a positive distance
away from (E-z)/t along a sequence of scales, a ball of radius comparable to
t, sitting at distance O(t) from z, would be missing from E; that contradicts
density one.) Applying this to two points z, z+w in E and using the
bi-Lipschitz lower bound gives N(w) >= c|w|. Together with the upper bound
N(w) <= Lip(phi)|w| (which is just (2.6) and the Lipschitz bound), this makes
N a norm.
(b) Parallelogram law. Take four domain points 0, u, u+v, v in R^k and
approximate them by rescaled points of E - z, as in (a). Apply the four point
inequality (1.4) of section 1.2 to the four images, divide by t^2, and use
(2.6). The result is
N(u+v)^2 + N(u-v)^2 <= 2 N(u)^2 + 2 N(v)^2 .
The reverse inequality is the same inequality applied to the pair
(u+v)/2, (u-v)/2, together with homogeneity of N. So
N(u+v)^2 + N(u-v)^2 = 2 N(u)^2 + 2 N(v)^2 ,
a parallelogram law. Polarize: define the bilinear form
= ( N(u+v)^2 - N(u-v)^2 ) / 4 .
The parallelogram law is exactly the statement that <.,.> is additive in each
slot, and homogeneity gives bilinearity. It is positive definite because N is
a norm. Hence N(w)^2 = w^T G w with G the matrix of <.,.>.
(c) Put w' = 0 in (2.6) and use that N(w) = |w|_G.
(d) Each entry of G is a finite combination of squared metric differentials
in the fixed coordinate directions; more precisely, G_ab = (1/2)( N(e_a+e_b)^2
- N(e_a)^2 - N(e_b)^2 ), so measurability follows from measurability of
z |--> N_z(w) for fixed rational w, which follows from (2.6) and continuity
of the distance. The bounds (2.7) are consequences of the bi-Lipschitz
constants: eigenvalues of G lie between Lip(phi^(-1))^(-2) and Lip(phi)^2.
End of proof of [L1].
Two remarks, both used later.
(i) Equation (2.6) with N Euclidean means the chart looks like a small
copy of Euclidean space, up to relative error o(1), for at almost every
point. So all the classical computations in Euclidean space - the
determinant bounds, the coarea formula, the divergence theorem applied
to a homotopy - transfer to a CAT(0) chart with no dimensional loss.
This is the entire reason the proof is possible.
(ii) The matrix G depends on the chart and on the point, and is only
measurable. There is no uniform bound on Lip(phi) across the (countable)
chart family. That is why all the estimates below are pointwise in the
chart and then summed/integrated, and never uniform.
The next two results are the ones that let us compute mass inside a CAT(0)
space with no loss. The first is an upper bound for mass, given a quadratic
upper bound for the distance distortion. Intuitively: if a chart distorts
small displacements by no more than a quadratic form P, then the area
density of that chart is no more than sqrt(det P), because a linear map that
sends the unit ball of P into the unit ball has determinant at most one.
Setting for [L2]. Let C be a finite-mass d-current with a representation by
countably many determinant integrals:
C(b, pi_1, ..., pi_d) = sum_i int_{F_i} vartheta_i(z) b(psi_i(z))
det D( (pi_1,...,pi_d) o psi_i )(z) dz , (2.8)
where psi_i: F_i -> Y are Lipschitz, F_i in R^d Borel, and vartheta_i are
real signed weights (not yet assumed integer). Assume that at almost every
point carrying these weights there is a measurable nonnegative quadratic
form P_i(z) on R^d with
d(psi_i(z+w), psi_i(z))^2 <= P_i(z)[w,w] + o(|w|^2) , z+w in F_i . (2.9)
[L2] (quadratic majorant for mass). In the setting above, put
eta = sum_i (psi_i)_# ( |vartheta_i| sqrt(det P_i) dz ) . (2.10)
If eta(Y) < infinity, then eta controls C in the sense of (2.2), and
therefore M(C) <= eta(Y). No normality of the individual chart currents, and
no assumption that the psi_i are injective, is required.
Proof. Fix d scalar 1-Lipschitz functions pi_1,...,pi_d on Y. Almost
everywhere in each chart, the scalar compositions pi_a o psi_i are
differentiable in the sense of scalar Lipschitz extensions (that is,
pi_a o psi_i(z+w) - pi_a o psi_i(z) = ell_a(w) + o(|w|) with ell_a a
bounded linear functional). Write ell_a for the derivative row of pi_a o
psi_i at z, and let E be the d x d matrix with rows ell_1,...,ell_d.
Step 1: each row is bounded by P_i. Take any direction w in R^d. Choosing
domain points z + t w_j with w_j -> w and using (2.9) together with the
1-Lipschitz property of pi_a gives
|ell_a(w)| <= sqrt( P_i(z)[w,w] ) . (2.11)
Indeed the left side is the derivative of pi_a o psi_i in the direction w,
and the increment of pi_a is bounded by the distance increment of psi_i,
which is bounded by sqrt(P_i[w,w]) t + o(t).
Step 2: Hadamard's inequality in the P-geometry. Suppose first that P_i is
positive definite and put Q = P_i(z)^{1/2}. By (2.11), the row ell_a is the
composition of Q with a row vector of Euclidean length at most 1. Hence
E = S Q with each row of S of length at most 1, and
|det E| = |det S| sqrt(det P_i) <= sqrt(det P_i) , (2.12)
because a matrix whose rows all have length at most 1 is the matrix of a
linear map that sends the unit cube into the unit ball, so it cannot increase
volume. (That last sentence is Hadamard's inequality, and it is the whole
content of this step.) If P_i is singular, then by (2.11) every row vanishes
on the kernel of P_i, so E has rank less than d as soon as the kernel is
nonzero; then both det E and det P_i are zero and (2.12) still holds.
Step 3: conclude. In the representation (2.8), the determinant is det E, so
each chart integral is bounded in absolute value by the chart integral of
|vartheta_i| sqrt(det P_i) against the test function |b|. Summing over the
charts, the finiteness of eta makes everything summable (the integrals of the
absolute values converge to the integral of the sum by monotone convergence)
and gives (2.2) with the measure eta. Minimality of the mass measure
finishes the proof.
One point of bookkeeping, because it is the kind of thing that is easy to
get wrong. The set where a chart is not differentiable depends on the chosen
tuple (pi_1,...,pi_d), so the exceptional set moves with the tuple. That is
harmless: mass is defined as the smallest measure that controls EVERY tuple,
and the definition of eta does not refer to a tuple at all. So one fixed
measure eta controls all tuples separately, which is exactly what (2.2)
demands. End of proof of [L2].
Now the reverse inequality, which says that the chart formula for mass is an
identity and not merely an upper bound. This is what lets us convert any
integral over the current into an ordinary integral over a chart domain, and
it is used in every section after this one. In particular it identifies the
null sets: a set is null for the mass measure exactly when its inverse image
in the charts has Lebesgue measure zero.
[L3] (exact chart mass). Let C in I_d(Y) be given by the pairwise disjoint
charts of (2.4), with integer weights theta_i, metric matrices G_i and area
densities J_i = sqrt(det G_i) from [L1]. Then
||C|| = sum_i (phi_i)_# ( |theta_i| J_i dz ) , (2.13)
as measures on Y. Consequently, for a Borel set B in Y,
sum_i int_{phi_i^(-1)(B)} |theta_i| J_i dz = ||C||(B) , (2.14)
and the sum on the left converges absolutely.
Proof. Let eta denote the right side of (2.13), allowing eta(Y) = infinity
for the moment.
First, eta <= ||C||. Fix 0 < eps < 1. We claim that each chart domain, up to
a Lebesgue-null set, is a disjoint countable union of Borel pieces E on which
the chart is (1 +- eps)-almost isometric to a fixed linear model, meaning:
there is a positive definite matrix Q (from a fixed countable dense family)
with
(1-eps)|z - z'|_Q <= d(phi_i(z), phi_i(z')) <= (1+eps)|z - z'|_Q
for all z,z' in E . (2.15)
Construction of the pieces. At a density point z, the metric differential
G_i(z) is close to some Q of the dense family (relative error at most eps/4,
say). The centered estimate [L1](c) then provides a radius 1/l, depending on
z, on which (2.15) holds with that Q against every other point of the domain.
Group points by the matrix Q and by the integer l, then subdivide into pieces
of diameter < 1/l, and disjointify. Borel measurability: for a fixed Q and
radius, the inequalities in (2.15) against all points of the domain can be
checked against a countable dense subset of the domain, using continuity of
the distance in the second variable, so each piece is Borel.
Now fix one such piece E of one chart. At almost every point of E,
differentiating (2.15) in the domain direction gives J_i(z) <= (1+eps)^k
sqrt(det Q), and the inverse map Q^{1/2} phi_i^(-1) is (1-eps)^(-1)-Lipschitz
on phi_i(E) by the lower inequality in (2.15). Extend each of its k coordinate
functions to all of Y by McShane (same Lipschitz constant) and call the
resulting tuple pi.
Use these tests in (2.1) with the first slot b chosen to be a bounded Borel
function, supported in phi_i(D) for a Borel D in E, equal there to
sgn(theta_i) o phi_i^(-1). The determinant in the chart is exactly
det(Q^{1/2}) = sqrt(det Q) at almost every point of E, because the extended
coordinates agree with Q^{1/2} z there. All other charts give zero, because
their images are disjoint from phi_i(D) and b is supported in phi_i(D).
Hence
int_D |theta_i| sqrt(det Q) dz
= C(b, pi_1, ..., pi_k)
<= (1-eps)^(-k) ||C||(phi_i(D)) . (2.16)
The last inequality is the mass bound (2.2) applied to these k functions,
each of Lipschitz constant at most (1-eps)^(-1). Combining (2.16) with
J_i <= (1+eps)^k sqrt(det Q) gives
int_D |theta_i| J_i dz <= ((1+eps)/(1-eps))^k ||C||(phi_i(D)) . (2.17)
Now let B be Borel in Y, take D = E intersected with phi_i^(-1)(B), sum over
all pieces and all charts: the images are disjoint so the right side adds up
to a constant times ||C||(B), and Lebesgue-null sets contribute nothing to
eta. Therefore eta(B) <= ((1+eps)/(1-eps))^k ||C||(B), first with eta(Y) =
infinity allowed and then, taking B = Y, as a finite bound. Let eps go to
zero to get eta <= ||C||.
Second, ||C|| <= eta. This is now immediate from [L2]: the quadratic
majorant (2.9) holds with P_i = G_i by the centered form of [L1](c), the
measure (2.10) is exactly eta, and eta is finite by the first part. So
||C|| <= eta. End of proof of [L3].
Two consequences we will use without comment.
(i) The area density J_i is positive almost everywhere (by the bi-Lipschitz
bound (2.7)), so the inverse images of null sets are null and the
charts really do carry the mass measure faithfully.
(ii) Integrating a Lipschitz function u against ||C|| is the same as
integrating u o phi_i times |theta_i| J_i over the chart domains. All
the Sobolev theory of section 7 is built this way, with the "chart
gradient" Du computed in the metric G_i of [L1].
Here the integers theta enter for the first and only time in a way that
affects a constant.
[L4] (Euclidean lower density). Let C in I_k(Y). Then
liminf_{rho -> 0} rho^(-k) ||C||( B(y,rho) ) >= omega_k (2.18)
for ||C||-almost every y. (Balls are open; omega_k is the volume of the unit
ball of R^k.)
Proof. By (2.13) it suffices to consider points y = phi_i(z) where z is a
density-one point of E_i, the metric differential exists at z, z is a
Lebesgue point of the density rho_i = |theta_i| J_i (extended by zero off
E_i), and |theta_i(z)| >= 1. These conditions hold at almost every point
carrying chart mass, because the weight is an integer and nonzero there.
Fix eps > 0. For all sufficiently small rho, every domain point
z + w with |w|_{G_i(z)} < rho/(1+eps)
lies in E_i and satisfies d(y, phi_i(z+w)) < rho, by the centered estimate
[L1](c). The set of such w is an ellipsoid of Lebesgue volume
omega_k rho^k / ( (1+eps)^k J_i(z) ) .
Lebesgue differentiation at z on dilates of this fixed ellipsoid says that
the chart contribution to ||C||(B(y,rho)) is at least
( |theta_i(z)| J_i(z) + o(1) ) * omega_k rho^k / ( (1+eps)^k J_i(z) ) .
Dividing by rho^k, letting rho -> 0, then eps -> 0, and using
|theta_i(z)| >= 1, gives (2.18). Other charts contribute nonnegative mass,
so they can only help. End of proof of [L4].
Read (2.18) as: an integral current cannot be thinner than k-dimensional
Lebesgue measure with coefficient omega_k. That is a purely Euclidean
statement about the model space, but it applies at almost every point of a
current in a wildly curved space, because at that point the chart is
metrically Euclidean to first order by [L1].
Two constructions are needed before the variational argument can start: an
explicit filling of any cycle (a cone), and the compactness/continuity
properties of the filling volume V. We record them together.
Products. Let C in I_k(Y) and h > 0. The whole-current product of the
interval [0,h] with C, written [0,h] x C, is the (k+1)-current on
[0,h] x Y with the product metric. It is integral [lit 3], and its boundary
is
d([0,h] x C) = [h] x C - [0] x C - [0,h] x (dC) . (2.19)
The signs in (2.19) are the ones that make d(d(.)) = 0; the third term is the
only place the boundary of C appears, and for a cycle it vanishes. The action
of the product on test functions is a sum of k+1 terms, one for each
coordinate slot, each of them an integral over t in [0,h] of an evaluation of
C with the time derivative in the first slot [lit 3]. We use this formula
exactly twice, in sections 6 and 7.6; both times it is applied to a homotopy
F: [0,h] x Y -> Y, so what is really used is the composition rule for
pushforwards.
[L5] (cone filling, and the linear bound). Let Y be a compact CAT(0) space,
B in I_k(Y) a cycle, and D = diam(spt B) (zero if B = 0). Then V(B) is
finite, and
V(B) <= (k+1) D M(B) . (2.20)
If B = 0 then V(0) = 0. Also, for every cycle B in Y,
V(B) <= K_k M(B)^((k+1)/k) and V(B) <= L_k M(B) , (2.21)
with constants K_k, L_k depending only on k and on the diameter of Y, and
every cycle has a mass minimizing integral filling.
Proof. The zero case is trivial. Otherwise choose o in spt B, and let
H: [0,1] x Y -> Y be H(t,x) = [o,x]_t, the point at fraction t on the
segment from o to x; it is well defined because geodesics in a CAT(0) space
are unique. Two estimates, both from section 1.2:
d(H(t,x), H(t,x')) <= t d(x,x') , (contraction, (1.2))
d(H(t,x), H(s,x)) = |t-s| d(o,x) , (constant speed)
and d(o,x) <= D on spt B. Hence H is jointly Lipschitz on [0,1] x Y, with
spatial constant 1 and time constant D.
Now let S = H_#([0,1] x B). By (2.19) its boundary is
(H_1)_# B - (H_0)_# B - H_#([0,1] x dB). The first map is the identity, the
second is the constant map to o (whose pushforward of a positive dimensional
current is zero), and dB = 0. So dS = B: the cone over B is a filling.
Mass of the cone. Write the action of S as a sum of k+1 terms, as in the
product formula, with test functions psi_a(t,x) = pi_a(H(t,x)). In each of
those k+1 terms, one factor is a time derivative of a 1-Lipschitz scalar
composed with H, hence bounded by D; the remaining k spatial derivatives are
at most 1-Lipschitz. Applying the mass bound (2.2) for B to each term and
adding the k+1 of them yields
M(S) <= (k+1) D M(B) , (2.22)
which proves (2.20).
The second bound in (2.21) is (2.20) with L_k = (k+1) diam(Y). The first
bound in (2.21) is Wenger's isoperimetric theorem [lit 6]: a compact CAT(0)
space has a convex geodesic bicombing (its unique geodesics), and then
V(B) <= K_k M(B)^((k+1)/k) with K_k depending on k and the space. We never
use the value of K_k, only its existence.
Existence of minimizers. Take a minimizing sequence S_j for V(B): bounded
masses, common boundary B. The compactness theorem for integral currents in a
compact metric space [lit 4] gives a weakly convergent subsequence with
integral limit S, and dS = B because boundary commutes with weak convergence.
Mass is lower semicontinuous under weak convergence [lit 4], so M(S) <=
liminf M(S_j) = V(B), and S is a minimizer. End of proof of [L5].
Finally, the continuity of filling volume, which is what makes the maximizing
argument of section 5 work. Suppose B_j, B are integral k-cycles in Y with
sup_j M(B_j) < infinity and B_j -> B weakly. Then [lit 7]
V(B_j - B) -> 0 and V(B_j) -> V(B) . (2.23)
The second limit follows from the first and subadditivity,
|V(B_j) - V(B)| <= V(B_j - B), which is just the observation that a filling
of the difference fills the difference. The first limit is Wenger's flat
convergence theorem [lit 7], and it needs the local cone inequalities: the
cone over a piece of a cycle has mass comparable to its mass, which is exactly
(2.20) applied inside small balls. That is why we bothered to prove (2.20)
by hand.
[T1] (sharp integral filling in CAT(0) spaces). Let X be a proper CAT(0)
space, let n >= 2, and let T in I_n(X) have compact support and dT = 0.
Then there exists S in I_(n+1)(X) with compact support such that
dS = T , (3.1)
M(S) <= c_n M(T)^((n+1)/n)
= M(T)^((n+1)/n) / ( (n+1)^((n+1)/n) omega_(n+1)^(1/n) ) , (3.2)
where c_n = 1/((n+1) s_n^(1/n)) as in (0.2). There is no hypothesis on the
dimension of X, and no hypothesis on the multiplicity function of T: it may
take any integer values, with either sign. The coefficient c_n is optimal,
by the ball computation of section 1.6.
The rest of this file is the proof. The proof structure, stated as three
separate results so that nothing is hidden:
[T1] is implied by [P2] (the case n = 2), by [T3] (the induction step)
and by [L6] (the reduction from proper X to compact Y).
Concretely: [L6] moves a compactly supported cycle into a closed ball;
[P2] handles 2-cycles in a compact CAT(0) space; [T3] handles n-cycles
given that the theorem is known for (n-1)-cycles. Induction over n gives
[T1] in compact spaces; [L6] gives it back in proper spaces.
One sentence: in a space of nonpositive curvature, filling a closed
n-surface of area A never costs more than the flat-space optimum
c_n A^((n+1)/n), and the optimum is achieved by round spheres in flat space.
It does not say that the minimizer is smooth, unique, or a piece of a
sphere. It does not classify the equality case in a curved space. In a
general CAT(0) space the least filling of a cycle need not have any
regularity beyond integrality, and in fact the round balls of flat space are
the only known equality cases in this generality. The statement is an
inequality for the infimum (2.5), which is attained by [L5].
It also does not give a construction. The proof is by contradiction, and
produces a filling only through the compactness theorem: the infimum is
attained, and the argument shows the infimum is at most c_n A^((n+1)/n).
A different question, which this file does not answer, is Gromov's: given a
map from a fixed polyhedral complex into X, how much extra volume does one
need to extend it to a map from the cone over the complex? That question
fixes the topology and the parametrization of the filling. [T1] allows the
filling to be any integral current with the right boundary, which is why the
two questions are different. See section 0.5 for the literature on that.
A Cartan-Hadamard manifold is a complete, simply connected Riemannian
manifold whose sectional curvature is at most zero. By the Cartan-Hadamard
theorem plus Hopf-Rinow such a manifold is a proper CAT(0) space and is
diffeomorphic to R^d. So [T1] applies to it, and the top-dimensional case of
[T1] becomes the classical geometric inequality.
[T2] (Cartan-Hadamard domain inequality). Let M be a Cartan-Hadamard
manifold of dimension d >= 3, and let Omega be a bounded domain with smooth
boundary. Then
area(dOmega) >= d * omega_d^(1/d) * vol(Omega)^((d-1)/d) . (3.3)
The same inequality holds for relatively compact sets of finite perimeter,
with the perimeter of the set in place of area(dOmega).
The proof is three lines once the lemma "there are no compactly supported
top-dimensional cycles in a connected noncompact manifold" is available: the
domain is itself a filling of its boundary, and in a noncompact connected
oriented manifold the only compactly supported d-cycle inside I_d is zero, so
the minimizer of [T1] must BE the domain. Hence mass of the minimizer =
vol(Omega) is at most the right side of (3.2) with n = d-1, and a little
algebra produces (3.3). That lemma, and the algebra, are section 8; the
formula check is: with n = d-1 we have s_n = d omega_d, so
c_(d-1)^(-(d-1)/d) = ( d s_(d-1)^(1/(d-1)) )^((d-1)/d)
= d^((d-1)/d) (d omega_d)^(1/d)
= d omega_d^(1/d) ,
so vol <= c_(d-1) area^(d/(d-1)) is the same as (3.3).
For d = 2 (surfaces) the same inequality is classical, going back to Weil
and Beckenbach-Rado, and it is not a consequence of [T1] because [T1] starts
at n = 2, that is d = 3. Section 8.3 explains the two-dimensional argument:
cut the domain into disks, apply the classical disk inequality (which follows
from the Gauss-Bonnet formula plus the general isoperimetric inequality), and
sum.
Dimensions 3 and 4 of (3.3) were known (Kleiner 1992, Croke 1984) and
dimensions 3 to 9 were known (Chen-Ghomi-Wang 2026) before this theorem. What
is new here is not a single dimension: it is that one sharp current
inequality in a general CAT(0) space implies all dimensions at once, without
curvature hypotheses and without a smoothness or comparison argument.
[T3] (induction step). Let n >= 3, let Y be a compact CAT(0) space, and
assume:
(H_(n-1)) for every compact CAT(0) space Z and every integral
(n-1)-cycle B in Z, there is an integral n-current R in Z
with dR = B and M(R) <= c_(n-1) M(B)^(n/(n-1)) .
Then for every integral n-cycle T in Y there is an integral (n+1)-current S
in Y with dS = T and M(S) <= c_n M(T)^((n+1)/n).
Three remarks about [T3].
(a) It is an implication between statements in consecutive dimensions, and
it never mentions X. Together with the base case [P2] (n = 2) and the
reduction [L6], it proves [T1] by induction on n. The induction is
over the dimension of the CYCLE, not over the dimension of the space;
the space is arbitrary in every step.
(b) The hypothesis (H_(n-1)) is used in exactly one place: to fill the
boundaries of small pieces of the champion in the small-set
inequality [P3], section 7.1. Everywhere else the proof of [T3] is
self-contained.
(c) The proof of [T3] is a contradiction argument, and it is long because
it has to pass through an analytic inequality ([P4]) and a
variational problem ([P7]). The pieces, in order, are:
section 5 the champion A and its normalization (uses no induction)
section 6 the radial inequality [P1] (uses no induction)
section 7.1 the small-set inequality [P3] (uses H_(n-1))
section 7.2 coarea and countable mass tests ([L9], [L10])
section 7.3 rearrangement and the radial Sobolev inequality
([L11], [L12])
section 7.4 the almost sharp Sobolev inequality [P4]
section 7.5 the closed gradient and compactness ([P5], [P6])
section 7.6 the minimizer [P7] and its powers ([L13], [L14])
section 7.7 the chord transform [P8] and its two consequences
section 7.8 the contradiction, which closes the induction
The hypothesis n >= 2 in [T1] is not a convenience. Section 0.3 gave the
example: in a tree, a 1-cycle is a pair of points and the cheapest filling is
the segment between them, whose length is the distance, not the square of the
distance over 4 pi. So the (n+1)/n exponent is false for n = 1 in a CAT(0)
space, and no argument can rescue it. Fortunately, 2-cycles cannot see trees:
a CAT(0) space of dimension 1 carries no nonzero 2-current at all, so the
theorem is vacuous there and the first interesting dimension is 2.
Why dimension 2 is proved separately rather than by the general argument.
The general argument of section 7 uses the Sobolev exponents
beta = 1/(n-2) , p = 2n/(n-2) , S_* = n s_n^(2/n) ,
and at n = 2 they blow up: 1/(n-2) is infinite, p is infinite. There is no
Sobolev inequality in the critical dimension n = 2; the correct tool there is
not analysis but a direct computation with a 1/r^2 cutoff. That computation
is short, exact, and is [P2] in section 6.5. It also has a useful side effect:
it shows exactly where the integer weights and the constant s_2 = 4 pi come
from, which makes the general argument easier to believe.
Everything in this section is someone else's theorem. Each item is stated in
the form in which this file uses it, with the hypotheses it needs. Nothing
here is proved; if an item is used in a delicate spot, the spot is flagged.
[lit 1] Ambrosio-Kirchheim, currents in metric spaces (2000).
Content used: (a) the definition of k-current, mass measure and boundary
of section 2.2, with mass having the minimality property (2.2);
(b) the rectifiable representation (2.4) of an integral current by
countably many bi-Lipschitz charts with integer weights;
(c) pushforward by a Lipschitz map f: X -> Z, defined by
(f_#C)(b, pi_1,...,pi_k) = C(b o f, pi_1 o f,...,pi_k o f), which sends
integral currents to integral currents, commutes with boundary
(d(f_#C) = f_#(dC)), and satisfies M(f_#C) <= Lip(f)^k M(C);
(d) restriction to a Borel set E, defined by inserting the coefficient
1_E in the first slot, with the identity ||C|_E|| = ||C|||_E and the bound
|C(b, pi)| <= int |b| d||C|| for every admissible tuple;
(e) for a locally Lipschitz (or merely Lipschitz) u and an integral
current C, the level sets: for almost every real t the restriction
C|{u > t} is an integral current and its boundary is an integral
(k-1)-cycle. Also: the differential of a Lipschitz scalar function
vanishes almost everywhere on each of its level sets, in the chart sense
of section 2.5.
Needed in: every section from 2 on. In particular (d) is the inequality
that makes [L2] and [L3] work, and (e) is what makes slicing legal in
section 7.2.
[lit 2] Kirchheim, rectifiable metric spaces (1994), in the two point form
of Duda (2007). Existence, at almost every density point of a Borel set
E in R^k, of a seminorm N (the metric differential) with
d(phi(z+w), phi(z+w')) = N(w - w') + o(|w| + |w'|)
for a Lipschitz phi. In this file it is used only through [L1], which
upgrades N to a Euclidean norm when the target is CAT(0).
Needed in: section 2.4.
[lit 3] Wenger (2007), product currents. For C in I_k and h > 0 there is an
integral current [0,h] x C on the product of the interval with the space,
with
d([0,h] x C) = [h] x C - [0] x C - [0,h] x (dC)
and with the action formula (sum of k+1 terms, time derivative in the
first slot). The product is a "whole-current" product in the sense that no
boundary is given to the individual pieces; that is what makes it usable
in a metric space.
Needed in: section 2.8 (statement), section 6.2 and section 7.7 (used on
the homotopy F(t,x) = [y,x]_(lambda(t,x))).
[lit 4] Ambrosio-Kirchheim compactness and closure. If C_j in I_k(Y) with
sup_j ( M(C_j) + M(dC_j) ) < infinity and Y compact, then a subsequence
converges weakly to an integral current C; dC_j converges weakly to dC;
and M(C) <= liminf M(C_j). ("Weak" means: evaluation against every
admissible tuple converges.)
Needed in: section 2.8 (existence of minimizers), section 5 (existence of
the champion), section 7.5 (compactness of the Sobolev embedding).
[lit 5] Ambrosio-Kirchheim slicing. For C in I_k and Lipschitz u, the slices
are integral (k-1)-currents for almost every t, the
codimension-one coarea formula holds, and the fundamental theorem
d(C|{u > t}) = (dC)|{u > t} +
holds for almost every t (with the orientation conventions of [lit 1]).
Needed in: section 7.2, where it is used in integrated form: the integral
over an interval of levels of the boundary evaluations of the slices
equals an evaluation of the original current on a function of u.
[lit 6] Wenger (2005), isoperimetric inequality of Euclidean type. In a
complete metric space which is geodesic and has a convex geodesic
bicombing (in particular any complete CAT(0) space), every integral
k-cycle B has, for every k >= 1,
V(B) <= K_k M(B)^((k+1)/k)
with K_k depending only on k and on the local geometry constants.
Needed in: section 2.8, as the first bound of (2.21), which makes the
champion's functional bounded above.
[lit 7] Wenger (2007), flat convergence and filling volume continuity. If
B_j, B are integral k-cycles in a compact metric space with bounded masses
and B_j -> B weakly, and the space has the local cone inequalities, then
V(B_j - B) -> 0, and consequently |V(B_j) - V(B)| -> 0.
Needed in: section 5.2, to show the maximizing functional is upper
semicontinuous along the minimizing/compactness sequence.
[lit 8] Brezis-Lieb (1983). If u_j -> u almost everywhere with
sup_j ||u_j||_p < infinity, then
||u_j||_p^p - ||u_j - u||_p^p - ||u||_p^p -> 0 .
The version used here is the elementary consequence, with a proof, in
section 7.6: for every eps there is C_eps with
| |a+b|^p - |a|^p | <= eps |a|^p + C_eps |b|^p ,
which together with dominated convergence gives the same conclusion.
Needed in: section 7.6 (no mass is lost in the limit).
[lit 9] Aubin (1976) and Talenti (1976). The sharp Sobolev inequality in
R^n, n >= 3:
S_* ||f||_p^2 <= 4 beta ||grad f||_2^2 , S_* = n s_n^(2/n),
beta = 1/(n-2) , p = 2n/(n-2) ,
for smooth compactly supported f, with equality for the bubbles
(a^2 + |x|^2)^(-(n-2)/2). In this file we need only the RADIAL case, and
we prove it, by mass transport, as [L13]. The literature reference is
used for context and to reassure the reader that the constant is the
classical one.
Needed in: section 7.3 (as orientation; the actual theorem is [L13]).
[lit 10] Cordero-Erausquin, Nazaret, Villani (2004). The mass transport
method for sharp Sobolev inequalities: monotone rearrangement transports
the level sets of f to the level sets of a radial profile, and the
determinant comparison plus an integration by parts gives the sharp
constant. We run exactly this argument for radial functions, in full, in
[L13].
Needed in: section 7.3 (method; details local).
[lit 11] McShane extension, in the quantitative form of Rieffel (2006). A
real valued L-Lipschitz function on a subset A of a metric space X extends
to an L-Lipschitz function on X; for a tuple of scalar functions, extend
them one at a time. (Rieffel's paper identifies the best constant
available for such extensions, which is what the "no loss" claims in
[L3] and section 7.5 refer to.)
Needed in: section 2.3, section 2.6, section 7.5.
[lit 12] Sturm (2003), variance inequality in CAT(0) spaces. For a finite
measure sigma on a compact CAT(0) space and
I(y) = int d(y,x)^2 dsigma(x) ,
the function I is continuous, has a minimizer o (a barycentre), and
I(y) >= I(o) + M d(o,y)^2 for every y, M = sigma(X).
The proof is the segment computation reproduced in section 7.8.
Needed in: section 7.8, to double the chord contribution.
[lit 13] Almgren (1986), optimal isoperimetric inequalities. The sharp
Euclidean inequality (0.1) for integral currents in R^m, in all
dimensions and codimensions, with the ball as the equality case. This file
does not use it as a black box in the proof; it is the theorem that
[T1] generalizes, and the reason the constant c_n is the right target.
Needed in: section 0.5 (context), section 8.4 (sharpness).
[lit 14] Simon (2014), lecture notes on geometric measure theory. Recipe for
compactness of a family of finite measures from a uniform bound on the
distributional derivatives: translate, average against a mollifier, use
Ascoli-Arzela on the smoothed family, then let the mollifier shrink. This
is the classical BV compactness argument; we run it in section 7.5 for the
projected measures of the weighted current, in the exact form needed.
Needed in: section 7.5.
[lit 15] Brezis-Kato (1979) and Brezis (1986). Truncated-power testing.
Multiplying a weak solution by a power of itself times a cutoff, dropping
the nonnegative term, and absorbing the "critical" potential with a large
level set, gives L^q membership for all q. The absorption device is
reproduced in section 7.6 as [L14].
Needed in: section 7.6.
Two warnings about how to use this list.
First, the items are not interchangeable with their smooth analogues. For
instance, [lit 5] says that the slices are integral for almost every level,
not for every level, and that "the" boundary identity holds only almost
everywhere in the level. Every place where this file passes from a current
to its slices, it either avoids discussing the exceptional levels or sums
over all levels. The reader should watch for that: it is the most common
place to make a silent error in this subject.
Second, the constants. K_k of [lit 6], Lip(phi_i) of the charts, and the
constants C_eps of section 7.4 are all allowed to depend on the space, the
chart or eps, and NONE of them is ever compared with the sharp constants
c_n, S_*. The whole point of the proof is that the sharp constants survive,
so any step that would trade a sharp constant for a loose one is a red flag.
There are exactly two such steps, both in the base case and both harmless:
the bound 2b - 2b^2 <= 1/2 in [P2], and the absorption constant in [L14].
This section does two things. First it removes the word "proper" from the
problem, so that from now on the ambient space is a single compact CAT(0)
space Y. Second it turns a failure of the inequality into an object A with a
very rigid property, the "champion inequality" (5.4), and it normalizes A so
that its mass is slightly below s_n. That normalization is the only place
where the sharp constant c_n is used numerically.
[L6] (compact reduction). Fix n >= 2. Suppose that for every compact CAT(0)
space Y and every integral n-cycle T in Y with dT = 0 there is an integral
(n+1)-current S in Y with dS = T and M(S) <= c_n M(T)^((n+1)/n). Then the
same statement holds in every proper CAT(0) space X, with the extra
information that S can be taken to have compact support.
Proof. Let X be proper CAT(0) and let T in I_n(X) be a cycle with compact
support. If T = 0 take S = 0. Otherwise pick a point o and a radius R so
large that spt T is contained in the open ball B(o,R). Let
Y = closure of B(o,R). Y is closed and bounded, hence compact (properness),
and it is convex (section 1.2(b)), so Y with the restricted distance is a
compact CAT(0) space.
Define the radial projection P: X -> Y by
P(x) = [o,x]_lambda , lambda = min{ 1, R / d(o,x) } , (5.1)
that is, walk from o towards x and stop at distance R from o. On Y, P is the
identity. On all of X, P is 1-Lipschitz, and here is the proof, which is the
only place where the equal-fraction comparison (1.3) is used with different
factors at the same time.
Let x, x' in X, put r = d(o,x), r' = d(o,x'), and let lambda, lambda' be the
corresponding fractions from (5.1). By (1.3),
d(P x, P x')^2
<= ( lambda r - lambda' r' )^2
+ lambda lambda' ( d(x,x')^2 - (r - r')^2 ) . (5.2)
Now lambda r = min{r,R} and lambda' r' = min{r',R}, so
|lambda r - lambda' r'| <= |r - r'| because s |--> min{s,R} is 1-Lipschitz.
Also 0 <= lambda lambda' <= 1. Substitute into (5.2):
d(P x, P x')^2
<= (r - r')^2 + lambda lambda' ( d(x,x')^2 - (r - r')^2 )
= lambda lambda' d(x,x')^2 + (1 - lambda lambda')(r - r')^2
<= lambda lambda' d(x,x')^2 + (1 - lambda lambda') d(x,x')^2
= d(x,x')^2 ,
where the last inequality is the triangle inequality |r - r'| <= d(x,x').
So P is 1-Lipschitz.
Consequences. P_#T is an integral n-cycle in Y with M(P_#T) <= M(T) (use
[lit 1](c) with Lip(P) <= 1). Apply the hypothesis in Y: there is an
integral (n+1)-current S in Y with dS = P_#T and
M(S) <= c_n M(P_#T)^((n+1)/n) <= c_n M(T)^((n+1)/n). Let iota: Y -> X be the
inclusion and put S' = iota_#S. Then M(S') = M(S) because iota is an
isometric embedding, dS' = iota_#(P_#T) = (iota o P)_# T, and iota o P is the
identity on the open set B(o,R) which contains spt T. By locality of
pushforward [lit 1](c), (iota o P)_# T = T. So dS' = T and spt S' is a
compact subset of Y. End of proof of [L6].
Remark. The lemma is stated as an implication between theorems in one fixed
dimension n. In the proof of [T1] it is applied at the very end, after the
compact case is settled for all n, so nothing circular happens.
From now on: Y is a compact CAT(0) space, n >= 2, q_0 = (n+1)/n, V is the
filling volume (2.5) of section 2.2, and c > 0 is a constant. Define
Phi(B) = V(B) - c M(B)^(q_0) , (5.3)
initially on integral n-cycles B in Y. The point of Phi is that it is
scale-invariant up to a power: if all distances are multiplied by a, both
terms get multiplied by a^(n+1) (see 5.3 below). So the champion of Phi is a
cycle whose shape, not whose size, is extremal.
[L7] (positive champion, and its first variation). Suppose Phi is positive
at some integral n-cycle. Then:
(a) Phi attains a positive maximum on the set of integral n-cycles;
(b) every maximizer A satisfies m := M(A) > 0 and the champion
inequality
V(A - B) >= c ( m^(q_0) - M(B)^(q_0) )
for every integral n-cycle B . (5.4)
Proof. (a) Boundedness above. By (2.21), V(B) <= L_n M(B) for a constant L_n
depending only on n and Y. With s = M(B),
Phi(B) <= L_n s - c s^(q_0) =: psi(s) ,
and psi is bounded above on [0, infinity) because q_0 = (n+1)/n > 1. So the
supremum of Phi is finite; call it Sup > 0 by hypothesis.
Boundedness of the maximizing sequence. If Phi(B) > 0 then
c s^(q_0) < V(B) <= L_n s, hence s^(q_0 - 1) < L_n / c, hence
M(B) < ( L_n / c )^n , since q_0 - 1 = 1/n . (5.5)
So every cycle with Phi(B) > 0 has mass below a fixed number, and a
maximizing sequence may be chosen inside that set.
Compactness. Let B_j be a maximizing sequence: all masses below (L_n/c)^n,
all boundaries zero. By [lit 4] (compactness in a compact space) a
subsequence, again written B_j, converges weakly to an integral current A
with dA = 0. By [lit 7] the filling volume is continuous along this
convergence, so V(B_j) -> V(A); and by [lit 4] mass is lower
semicontinuous, so M(A) <= liminf M(B_j), hence
M(A)^(q_0) <= liminf_j M(B_j)^(q_0) .
Combining, and using that s |--> -c s^(q_0) is monotone decreasing,
Phi(A) = V(A) - c M(A)^(q_0)
>= limsup_j ( V(B_j) - c M(B_j)^(q_0) )
= Sup .
So A attains the supremum, and in particular M(A) > 0 because Sup > 0.
(b) First variation. Let B be any integral n-cycle. Maximality says
Phi(A) >= Phi(B), that is
V(A) - V(B) >= c ( M(A)^(q_0) - M(B)^(q_0) ) .
Now A - B is a cycle, and we need subadditivity of the filling volume: if S
fills A - B and R fills B, then S + R is an integral current whose boundary
is (A - B) + B = A, so M(S) + M(R) >= V(A). Take the infimum over S and over
R separately:
V(A) <= V(A - B) + V(B) . (*)
Combine (*) with the maximality inequality above:
c ( m^(q_0) - M(B)^(q_0) ) <= V(A) - V(B) <= V(A - B) ,
which is (5.4). End of proof of [L7].
How (5.4) is used. Take B to be a small perturbation of A, for example a
copy of A pushed inwards along radial lines. Then the left side of (5.4) is
approximately c q_0 m^(1/n) times the amount of mass that A loses, and the
right side is the mass swept out by the homotopy. That is exactly the
comparison performed in section 6, and the normalization below makes the
coefficient c q_0 m^(1/n) equal to 1/n, which is the coefficient appearing
in the radial inequality [P1].
Distances in a metric space can be multiplied by a positive number a; call
the resulting space aY, and the correspondingly scaled copy of a current
aB. Two facts, both immediate from the definitions:
(i) the mass of a j-dimensional current is multiplied by a^j;
(ii) the filling volume V is multiplied by a^(n+1).
Fact (i) deserves a line. Let phi: Y -> aY be the identity map between the
two copies of the same set; it is Lipschitz with constant a from Y to aY, and
with constant 1/a from aY to Y. By [lit 1](c), M(phi_# C) <= a^j M(C) and
M(C) = M(phi^(-1)_# phi_# C) <= a^(-j) M(phi_#C). Squeeze. Fact (ii) follows
because the correspondence S <--> aS is a mass-preserving bijection between
fillings of B and fillings of aB, with masses scaled by a^(n+1).
[L8] (normalization). Let A be a champion as in [L7], with c > c_n and
m = M(A) > 0. Then there is a factor a > 0 such that the rescaled champion
A' = aA satisfies
m' := M(A') = ( (n+1) c )^(-n) < s_n (5.6)
and, automatically,
c q_0 ( m' )^(1/n) = 1/n . (5.7)
Moreover, after the rescaling, A' is still a maximizer of the
corresponding functional in the rescaled space, so (5.4) holds for A' with
m' in place of m.
Proof. Choose a = ( (n+1) c m^(1/n) )^(-1). Then
m' = a^n m = m / ( (n+1) c m^(1/n) )^n = m / ( (n+1)^n c^n m ) =
( (n+1)c )^(-n), which is (5.6). The inequality m' < s_n is equivalent to
((n+1)c)^(-n) < ((n+1)c_n)^(-n), that is c > c_n, which is our hypothesis;
recall from section 1.5 that ((n+1)c_n)^(-n) = s_n exactly. For (5.7):
c q_0 (m')^(1/n) = c ((n+1)/n) * ( (n+1)c )^(-1) = 1/n .
Finally, by (i) and (ii), Phi is multiplied by a^(n+1) under rescaling, so a
maximizer stays a maximizer. End of proof of [L8].
We now relabel: the ambient space is still written Y, the champion is
written A, and m = M(A) satisfies (5.6) and (5.7). The mass measure of A is
mu = ||A|| , mu(Y) = m < s_n . (5.8)
This is the running notation for the rest of the proof. Note what (5.6) says
numerically: the champion is a cycle whose total weighted area is strictly
less than the area of the unit sphere in the same dimension. Everything that
follows is an attempt to prove m >= s_n, which will be a contradiction.
Exercise (why c > c_n is available). Suppose the sharp inequality (3.2)
fails for some cycle T, so that V(T) > c_n M(T)^((n+1)/n). Show that then
there is c with c_n < c < V(T) / M(T)^((n+1)/n), and that for this c the
functional (5.3) is positive at T. (This is the only way the proof of [T3]
starts.)
This section contains the one computation that uses the CAT(0) hypothesis
directly, through the comparison inequality (1.3). Everything else in the
proof is analysis (Sobolev inequalities, minimizers) or algebra
(determinants). So if you want to see why curvature <= 0 is the right
hypothesis, read 6.2.
Setting for the whole section: Y is a compact CAT(0) space, n >= 2, A is the
champion of [L7] normalized by [L8], so that
mu = ||A|| , m = mu(Y) < s_n , c q_0 m^(1/n) = 1/n , (6.1)
with q_0 = (n+1)/n. Fix a centre y in Y and put
r(x) = d(y,x) . (6.2)
The function r is 1-Lipschitz, so its chart gradient Dr exists mu-almost
everywhere in the sense of section 2.5 and satisfies |Dr| <= 1 there. (The
null set may depend on y; that is harmless and is mentioned again in 7.8.)
Fix a nonnegative Lipschitz function g on Y, and choose h > 0 so small that
h M_g <= 1/2 , M_g := sup g . (6.3)
For 0 <= t <= h define
lambda(t,x) = 1 - t g(x) , F(t,x) = [y,x]_{lambda(t,x)} , (6.4)
so that 1/2 <= lambda <= 1, F(0,x) = x, and each point x slides towards y as
t grows, by a fraction which depends on x. Write F_t(x) = F(t,x), and define
the moved cycle and the swept current
B_h = (F_h)_# A , D_h = F_# ( [0,h] x A ) . (6.5)
Both are integral currents by [lit 1](c) and [lit 3]. Their boundaries are,
by (2.19) and dA = 0,
dB_h = 0 , dD_h = B_h - A . (6.6)
So D_h fills A - B_h, and this is the filling that will be compared with the
champion inequality (5.4).
F is jointly Lipschitz on [0,h] x Y. Here is the estimate, and it is needed
to justify everything after it. Let R = diam(Y), let t,t' in [0,h] and
x,x' in Y, and set lambda = 1 - t g(x), lambda' = 1 - t' g(x'). By the
two-point comparison (1.3),
d(F(t,x), F(t',x'))^2
<= ( lambda r - lambda' r' )^2
+ lambda lambda' ( d(x,x')^2 - (r - r')^2 ) , (6.7)
where r = d(y,x), r' = d(y,x'). Now
|lambda r - lambda' r'| <= |r - r'| + R |t g(x) - t' g(x')|
<= |r - r'| + R h Lip(g) d(x,x') + R M_g |t-t'|,
and 0 < lambda lambda' <= 1. Substituting in (6.7) and taking square roots,
d(F(t,x), F(t',x'))
<= 2 ( 1 + R h Lip(g) ) d(x,x') + 2 R M_g |t - t'| . (6.8)
So F is Lipschitz with spatial constant 2(1 + R h Lip(g)) and time constant
2 R M_g. (The factor 2 is not important; what matters is that the constant
depends only on g, R and h, and not on the chart or on the point.)
We now look at the derivative of F. There are no charts in Y, so we work in
one chart at a time: let phi_i: E_i -> Z_i be the charts of (2.4), with
metric matrices G_i from [L1] and values r, g evaluated at phi_i(z), and let
alpha and gamma be the chart covectors
alpha = D(r o phi_i)(z) , gamma = D(g o phi_i)(z) . (6.9)
By (2.7) analogue for covectors, |alpha|_{G_i^(-1)} <= 1 and
|gamma|_{G_i^(-1)} <= Lip(g). Write lambda = 1 - t g, and set
P_tot = lambda^2 ( G_i - alpha(x)alpha ) + u u^T ,
u = lambda alpha - t r gamma - r g dt , (6.10)
P_sp = lambda^2 ( G_i - alpha(x)alpha )
+ ( lambda alpha - t r gamma )( lambda alpha - t r gamma )^T . (6.11)
Here "alpha alpha" means the rank-one matrix alpha_i alpha_j, and in P_tot
the first summand acts only on the spatial directions w, while u has one
time component and one spatial component. Both forms are nonnegative:
G_i - alpha alpha >= 0 because G_i^{-1}[alpha,alpha] = |alpha|^2 <= 1,
and a sum of a nonnegative form and a square of a linear functional is
nonnegative.
The point of these forms is the following differential bound. For increments
(s,w) with (t+s, z+w) in [0,h] x E_i,
d( F_i(t+s, z+w), F_i(t,z) )^2
<= P_tot[(s,w),(s,w)] + o( |s|^2 + |w|^2 ) , (6.12)
where F_i = F(t, phi_i(z)), and similarly with s = 0 for P_sp. Let us check
the derivation, because this is exactly where the CAT(0) hypothesis enters
and there is nothing else like it in the proof.
Let A_lam(t,z) = lambda(t,z) r(z), a scalar function on [0,h] x E_i, where
r(z) means r(phi_i(z)). Its differential is
dA_lam(s,w) = ( lambda alpha - t r gamma ) . w - ( r g ) s + o(|s|+|w|) ,
because d(lambda) = -s g - t (gamma . w) and d r = alpha . w, and
lambda r = (1-tg) r. Now apply (6.7) with x = phi_i(z), x' = phi_i(z+w),
t replaced by t and t+s:
d(F_i, F_i)^2 <= ( (1+o(1)) dA_lam(s,w) )^2
+ lambda (lambda + O(s) ) ( d(phi(z+w),phi(z))^2
- (r(z+w)-r(z))^2 ) .
Using the centered metric differential [L1](c), the bracket becomes
(G_i w . w) - (alpha . w)^2 + o(|w|^2). Also lambda(lambda+O(s)) = lambda^2 +
O(s). Since all terms are quadratic, the O(s) correction contributes only to
order o(|s|^2+|w|^2) times a bounded coefficient (the coefficients are
bounded because Lip(g), R, Lip(phi_i), |alpha| are). Thus
d^2 <= ( dA_lam(s,w) )^2 + lambda^2 ( G_i - alpha alpha )[w,w]
+ o(|s|^2 + |w|^2) ,
which is (6.12). Setting s = 0 gives the same statement with
(lambda alpha - t r gamma) in place of u.
Determinant of the spatial form. Work in coordinates in which G_i(z) is the
identity, and remember that at the end we divide by J_i = sqrt(det G_i).
From (6.11),
P_sp = lambda^2 I - lambda t r ( alpha gamma^T + gamma alpha^T )
+ t^2 r^2 gamma gamma^T
= I - t ( 2 g I + r ( alpha gamma^T + gamma alpha^T ) )
+ O(t^2) ,
where we substituted lambda = 1 - tg and collected: lambda^2 I = I - 2tgI +
O(t^2), and the t^2 term of lambda^2 combined with the last term is O(t^2).
Now use that the derivative at the identity of H |--> sqrt(det H) is
H' |--> (1/2) trace(H'). The trace of (alpha gamma^T + gamma alpha^T) is
2 alpha . gamma = 2 Dr . Dg in the chart metric. Hence
sqrt(det P_sp) / J_i
= 1 - t ( n g + r Dr . Dg ) + O(t^2) . (6.13)
The remainder is bounded by C t^2 with C depending only on n, R, M_g and
Lip(g), uniformly in the chart and the point: after G-orthonormalisation
every entry of the matrices involved is controlled by those four numbers,
because |alpha| <= 1, |gamma| <= Lip(g), r <= R, g <= M_g. That uniformity
is used in 6.3 and is the reason the O(h^2) there is honest.
Determinant of the total form. In time-first coordinates, (6.10) reads
P_tot = [[0, 0], [0, H]] + u u^T , H = lambda^2 ( I - alpha alpha^T ) ,
u = ( -r g , lambda alpha - t r gamma ) .
Write q = lambda alpha - t r gamma, a = -r g, so that
P_tot = [[ a^2 , a q^T ] , [ a q , H + q q^T ]] .
If a != 0, the Schur complement of the block a^2 gives
det P_tot = a^2 det( H + q q^T - q q^T ) = a^2 det H .
If a = 0 both sides are zero, because the first row and column vanish and
the matrix has dimension n+1 >= 3. Finally
det H = lambda^(2n) det(I - alpha alpha^T) = lambda^(2n)(1 - |alpha|^2).
Taking square roots,
sqrt(det P_tot) / J_i
= r g lambda^n sqrt( 1 - |Dr|^2 ) . (6.14)
This is an exact identity, not an estimate: the comparison form has the
sphere-like factor sqrt(1 - |Dr|^2), which is the classical "transverse
component" of the radial direction flat-space computation. Note that (6.14)
is valid also when r g = 0 (then both sides are 0) and when |Dr| = 1.
Figure-free version, in words: the spatial form loses mass at rate
n g + r Dr.Dg because the map shrinks by the factor lambda; the total form
sweeps mass at rate r g sqrt(1-|Dr|^2) because a radial motion has volume
weight equal to the component of the velocity transverse to the current.
Both rates are exactly what one would compute in R^n, and that is the
content of (6.13) and (6.14).
We now apply [L2] twice. First write the action formulas in a chart, which
is just the chart formula (2.1) composed with F. For bounded Lipschitz b and
Lipschitz pi_1,...,pi_n,
B_h(b, pi) = sum_i int_{E_i} theta_i(z) b(F_i(h,z))
det D_z( (pi o F_i(h,.)) )(z) dz , (6.15)
and for the homotopy,
D_h(b, pi_1,...,pi_{n+1})
= sum_i int_0^h int_{E_i} theta_i(z) b(F_i(t,z))
det D_{(t,z)}( (pi_1 o F_i, ..., pi_{n+1} o F_i) )(t,z) dz dt . (6.16)
These two formulas are the chart form of [lit 3] applied to the whole
current A, with the pushforward rules of [lit 1]. Two comments, both from the
source of this argument and both worth knowing:
(a) The interchanges of sum and integral in (6.16) are legitimate: for a
fixed test tuple, every cofactor in the expansion of the determinant is
bounded by a constant times J_i, by the Euclidean determinant bound; so
the total absolute integral is at most C h m by (2.13), and Fubini
applies.
(b) In (6.16), the separate time and space derivatives agree almost
everywhere with the joint derivatives of the Lipschitz compositions, so
the formula is the honest determinant of the joint derivative.
Now (6.12) says precisely that the quadratic majorant hypothesis (2.9) of
[L2] holds, for B_h with P_i = P_sp(h,z), and for D_h with
P_i = P_tot(t,z). The two majorant measures (2.10) are finite:
by (6.13), sqrt(det P_sp) <= (1 + C h) J_i ,
by (6.14), sqrt(det P_tot) <= R M_g J_i ,
so their integrals are bounded by (1+Ch) m and h R M_g m respectively, using
(2.13). [L2] therefore gives
M(B_h) <= m - h int_Y ( n g + r Dr . Dg ) dmu + C h^2 , (6.17)
M(D_h) <= h int_Y r g sqrt(1 - |Dr|^2) dmu . (6.18)
Both inequalities use only (6.13), (6.14) and the exact chart mass (2.13).
Note what (6.17) says: the moved cycle lost mass at the rate
n g + r Dr.Dg. Note what (6.18) says: the swept current has mass at most the
integral of r g sqrt(1-|Dr|^2).
Everything is now in place for the champion comparison.
[P1] (radial first variation). Let Y be a compact CAT(0) space, A the
champion of (6.1) with mass measure mu and mass m < s_n, y in Y, and
r(x) = d(y,x). Then for every nonnegative Lipschitz g on Y,
int_Y ( n g + r Dr . Dg ) dmu
<= n int_Y r g sqrt(1 - |Dr|^2) dmu . (6.19)
The square root is defined mu-almost everywhere, since |Dr| <= 1.
Proof. Abbreviate
I_g = int ( n g + r Dr.Dg ) dmu , H_g = int r g sqrt(1-|Dr|^2) dmu .
By (6.17) there is C = C(g,y) with M(B_h) <= U_h := m - h I_g + C h^2 for
all small h; since m > 0, also U_h > 0 for small h. The current -D_h fills
A - B_h (because dD_h = B_h - A). So the champion inequality (5.4) applies
with B = B_h and gives
c ( m^(q_0) - M(B_h)^(q_0) ) <= V(A - B_h) <= M(D_h) <= h H_g .
Since U_h >= M(B_h) and s |--> s^(q_0) is increasing,
c ( m^(q_0) - U_h^(q_0) ) <= c ( m^(q_0) - M(B_h)^(q_0) ) <= h H_g .
Divide by h > 0 and let h -> 0. By (6.17), ( m^(q_0) - U_h^(q_0) ) / h ->
q_0 m^(q_0 - 1) I_g = q_0 m^(1/n) I_g (using q_0 - 1 = 1/n). Hence
c q_0 m^(1/n) I_g <= H_g .
The normalization (6.1) says c q_0 m^(1/n) = 1/n, so I_g <= n H_g, which is
(6.19). End of proof of [P1].
Three comments.
(a) [P1] is an inequality between two integrals over the current, with no
reference to the space's dimension and no reference to charts. It is the
complete output of the geometric part of the proof, together with the
base case below.
(b) The champion inequality enters only through the numeric identity
c q_0 m^(1/n) = 1/n of (6.1); that is why the rescaling [L8] was worth
doing. Without it the inequality would read
c q_0 m^(1/n) I_g <= H_g with an unhelpful coefficient.
(c) The function g is an arbitrary nonnegative Lipschitz function, and the
centre y is arbitrary. In section 6.5 g is a cutoff 1/max(r,eps)^2; in
section 7.7 it is a more elaborate function built from the minimizer.
The freedom to choose g is what makes one inequality enough for two
very different jobs.
Exercise. Put g = 1 identically and y any point. Show that (6.19) reads
n m + int r Dr.D(1) <= n int r sqrt(1-|Dr|^2) dmu, that is
m <= int r sqrt(1-|Dr|^2) dmu. Conclude m <= R m trivially, i.e. this
choice of g gives no information. (Moral: g must concentrate mass.)
This is the one place where the famous constant 4 pi = s_2 appears from a
line-by-line computation, and it is worth reading carefully because it shows
what the integers in "integral current" are for.
[P2] (sharp filling in cycle dimension two). Let Z be a compact CAT(0)
space and let T in I_2(Z) be a cycle. Then there exists S in I_3(Z) with
dS = T , M(S) <= c_2 M(T)^(3/2) = M(T)^(3/2) / (6 sqrt(pi)) . (6.20)
Proof. Suppose not. Choose c > c_2 so that Phi = V - c M^(3/2) is positive
somewhere, and let A be the champion of [L7], rescaled by [L8] so that
m = M(A) < s_2 = 4 pi and c (3/2) m^(1/2) = 1/2. Fix a Lebesgue point y of
the mass measure mu = ||A||, in the sense of [L4] with k = 2:
liminf_{eps -> 0} eps^(-2) mu(B(y,eps)) >= omega_2 = pi . (6.21)
Such a y exists because [L4] holds mu-almost everywhere and m > 0. Put r(x) =
d(y,x) and, for 0 < eps < 1,
g(x) = max{ r(x), eps }^(-2) . (6.22)
This is Lipschitz and nonnegative, so it is admissible in [P1]. Two
computations.
Inside the ball, r <= eps, g = eps^(-2) is constant, so Dg = 0 and
n g + r Dr.Dg = 2 eps^(-2) , r g sqrt(1 - |Dr|^2) = 0 , (6.23)
the second because |Dr| = 1 mu-almost everywhere in the ball (r is a distance
function, so it moves at unit speed in its own gradient direction; this is
[lit 1](e) applied to the level sets of r).
Outside the ball, r > eps, g = r^(-2) and the chain rule gives
Dg = -2 r^(-3) Dr. Then with b = r^(-1) sqrt(1 - |Dr|^2) >= 0,
2 g + r Dr.Dg = 2 r^(-2) ( 1 - |Dr|^2 ) = 2 b^2 , (6.24)
r g sqrt(1-|Dr|^2) = r^(-1) sqrt(1-|Dr|^2) = b . (6.25)
Since the level set {r = eps} has mu-measure zero for almost every eps (there
are only countably many levels of positive measure under a finite measure),
we may assume it for our eps. Inserting (6.23)-(6.25) into the radial
inequality [P1] gives
2 eps^(-2) mu(B(y,eps)) <= 2 int_{r>eps} ( b - b^2 ) dmu . (6.26)
Now 2(b - b^2) = 1/2 - 2(b - 1/2)^2 <= 1/2 pointwise, so the right side of
(6.26) is at most m/2. Hence
mu(B(y,eps)) <= eps^2 m / 4 . (6.27)
Combine (6.21) with (6.27): divide by eps^2 and let eps -> 0 to get
pi <= m/4, that is m >= 4 pi = s_2. That contradicts m < s_2. So no champion
exists, and the filling inequality holds in dimension two with the sharp
constant c_2. End of proof of [P2].
Two remarks. First, the factor 4 is forced by the pointwise inequality
b - b^2 <= 1/4, with equality exactly at b = 1/2; that is the value a flat
two-dimensional sheet takes at distance 2r from the centre. The base case
therefore detects the flat model and nothing else, which is why it delivers
the sharp constant rather than just some constant. Second, the algebra:
(6.21) says mu(B(y,eps)) >= (pi + o(1)) eps^2 and (6.27) says
mu(B(y,eps)) <= eps^2 m/4; divide by eps^2 and let eps -> 0.
From now on n >= 3, the space Y is compact CAT(0), and A is the champion of
(6.1) with mass measure mu, mass m < s_n, and normalization
c q_0 m^(1/n) = 1/n. We assume (H_(n-1)), the sharp filling inequality in
cycle dimension n-1, and we aim for a contradiction.
The plan, in one sentence per step.
7.1 Cut A into small pieces. Each piece has small mass, so by (H_(n-1))
its (n-1)-boundary is filled cheaply; feeding that into the champion
inequality gives a sharp perimeter bound for small pieces. [P3]
7.2 Convert that bound into a Sobolev inequality: coarea turns perimeter
into a derivative, rearrangement turns the current into a Euclidean
radial function. The two bookkeeping lemmas are [L9] and [L10].
7.3 On R^n the radial Sobolev inequality is sharp by mass transport
[L12]; the rearrangement does not lose energy. [L11], [L12]
7.4 Result: the almost sharp Sobolev inequality on Y. [P4]
7.5 Analysis set-up: the closed gradient [P5] and its compactness [P6].
7.6 Variational set-up: a minimizer v exists and is subcritical, and
its powers are admissible tests. [P7], [L13], [L14]
7.7 Testing with a Euclidean bubble profile gives the chord transform
bound [P8]. [L15], [L16]
7.8 Tangent density bounds the chord moments above [L17]; variance
bounds their average below [L18]. Contradiction. [T3]
The only place where (H_(n-1)) enters.
[P3] (uniform small-set isoperimetry). For every eta in (0,1) there is
delta in (0, m/2), depending only on eta, n, m and the space, such that the
following holds. Let u: Y -> R be Lipschitz and put
a(t) = mu{ u > t } , U_t = A|{u > t} , P(t) = M( d U_t ) , (7.1)
where U_t and its boundary are integral for almost every t (by [lit 1](e)).
Then for almost every level t with a(t) <= delta,
P(t) >= (1 - eta) n omega_n^(1/n) a(t)^((n-1)/n) . (7.2)
Proof. Fix a level with U_t integral, write a = a(t), P = P(t). By
(H_(n-1)) applied to the integral (n-1)-cycle dU_t, there is an integral
n-current R in Y with
dR = dU_t , d_R := M(R) <= ( P / (n omega_n^(1/n)) )^(n/(n-1)) (7.3)
This is exactly the sharp filling inequality in dimension n-1, solved for
P in terms of the filling mass. Note that d_R is a monotone function of P,
which is why (7.3) is the right way round.
Now B := A - U_t + R is a cycle, and so is A - B = U_t - R. Masses: by
restriction (2.14) the mass of A splits exactly, so
M(B) <= m - a + d_R , M(A - B) <= a + d_R , (7.4)
and no disjointness of R from U_t is needed for these bounds. The champion
inequality (5.4) applies to B, and the coarse linear bound (2.21) applies to
A - B:
c q_0 (m/2)^(1/n) (a - d_R) <= c ( m^(q_0) - M(B)^(q_0) )
<= V(A - B) <= L_n M(A-B)^(q_0)
<= L_n (2a)^(q_0) . (7.5)
The first inequality used the mean value theorem for s^(q_0) on the interval
[m - a + d_R, m] and the bound s^(1/n) >= (m/2)^(1/n) there, which holds
because a + d_R <= m (we will have this for small a). Solving (7.5) for
d_R gives
d_R >= a ( 1 - C a^(1/n) ) , C = n K_n' 2^(1+2/n) , (7.6)
with a constant C depending only on n and the space; if d_R >= a already,
(7.6) holds trivially. Now combine (7.6) with (7.3), solved for P, and choose
delta so small that C delta^(1/n) <= 1 - (1-eta)^(n/(n-1)). Then for
a <= delta,
P >= n omega_n^(1/n) a^((n-1)/n) ( 1 - C a^(1/n) )^((n-1)/n)
>= (1 - eta) n omega_n^(1/n) a^((n-1)/n) ,
which is (7.2). The case a = 0 is immediate. End of proof of [P3].
Read (7.2) as: small pieces of the champion cannot be cheap. Their boundary
must have at least the Euclidean sphere area, up to the factor 1-eta. The
loss eta is the price of the fact that we only know the induction hypothesis
in dimension n-1 and not a sharp value for how R sits relative to U_t.
Coarea has to be done with the chart gradient, not with the Lipschitz
constant, otherwise the Dirichlet energy is lost. To do that, one needs to
recover the total mass of a slice from countably many scalar evaluations, and
that is [L9].
[L9] (countable family of mass tests). For a compact metric space Z and
k >= 1 there is a countable family T_k(Z) of finite lists of pairs
((b_1,pi^1),...,(b_N,pi^N)), where the b_j are Lipschitz with sum |b_j| <= 1
and each pi^j is a k-tuple of 1-Lipschitz functions, such that every
finite-mass k-current C on Z satisfies
M(C) = sup_{T in T_k(Z)} sum_{(b,pi) in T} C(b, pi) . (7.7)
Proof. The countable family does not depend on the current C; only the
comparison does. Fix a point z_0 and normalize each differentiated test by
pi_i(z_0) = 0; adding constants does not change its current evaluation. The
normalized 1-Lipschitz functions on the compact metric space form a compact
metric space in the uniform topology, by the Arzela-Ascoli theorem. Choose a
countable dense family of normalized k-tuples (pi^j)_{j >= 1}.
Write lambda = ||C||. For each tuple pi, the first-slot extension to bounded
Borel functions gives a finite signed measure b |--> C(b, pi) with density
rho_pi relative to lambda, satisfying |rho_pi| <= 1 almost everywhere.
Choose measurable versions for the countable family and put
rho(z) = sup_{j >= 1} |rho_{pi^j}(z)| .
This satisfies 0 <= rho <= 1. If a normalized tuple pi is approximated
uniformly by tuples pi^j of the family, current continuity, with their
common Lipschitz bound, shows for every Lipschitz b that
|C(b, pi)| <= int_Z |b| rho dlambda .
Thus the measure rho*lambda controls the mass of every normalized
1-Lipschitz tuple, and hence, by rescaling the differentiated slots, of every
test. Minimality of the mass measure gives lambda <= rho*lambda, and from
this it follows that rho = 1 almost everywhere.
Selecting at each point the index of the largest |rho_{pi^j}| and its sign
gives coefficients b_j with sum_j |b_j| <= 1 and
sum_j C(b_j, pi^j) = int_Z max_{j >= 1} |rho_{pi^j}| dlambda .
Letting the finite families grow turns the maximum into the supremum, giving
(7.7). The coefficients may be taken Lipschitz and with bounded sum of their
supports, by approximating bounded Borel coefficients in L^1(lambda) by
continuous ones and normalizing the tuple.
It remains to make the lists countable without depending on C. For each
fixed list length, the admissible lists of coefficients and normalized
tuples form a separable metric space in the uniform topology: they are a
subspace of a finite product of C(Z), which is separable because Z is
compact metric. Choose a countable dense subset within that admissible
space and take the union over list lengths. An admissible list can be
approximated by these lists with its constraint and differentiated Lipschitz
bounds preserved. The first-slot errors are controlled by their uniform
norms times M(C), and convergence in the differentiated slots follows from
current continuity. Hence the countable collection has the same supremum.
A common Lipschitz bound on the first coefficients is neither asserted nor
needed. End of proof of [L9].
[L10] (intrinsic coarea). With the notation (7.1), set
e(t) = int_{{u > t}} |Du|^2 dmu .
Then for almost every t,
P(t)^2 <= ( -a'(t) ) ( -e'(t) ) , P(t) <= Lip(u) ( -a'(t) ) , (7.8)
where the derivatives are the densities of the absolutely continuous parts of
the distribution functions a and e; neither a nor e is assumed to be
absolutely continuous.
Proof. For a bounded open interval I, define the bounded Lipschitz function
H(s) = int_I 1_{t < s} dt .
Let b be Lipschitz and let pi be an (n-1)-tuple of 1-Lipschitz functions.
Restriction, the chart representation, and Fubini give
int_I dU_t(b, pi) dt = A(H(u), b, pi) = -A(b, H(u), pi) . (7.9)
The first equality is the slice identity: the boundary evaluation has the
measurable representative U_t(1, b, pi), obtained by integrating the chart
determinants with the factor 1_{u > t}, and it is integrable on bounded
intervals. The second equality is the product rule applied to the cycle
identity dA = 0.
The chain rule in the charts gives
D(H(u)) = 1_{u in I} Du mu-almost everywhere.
There is no ambiguity at the endpoints of I: the differential of a Lipschitz
scalar function vanishes almost everywhere on each fixed level set. This
follows by Euclidean differentiation at density points of that level set in
each chart. At points of the charts the other differentiated covectors have
norm at most one, so the Euclidean determinant bound of [L3] applies.
The mass bound with these tests now yields
int_I M(dU_t) dt <= int_{u in I} |Du| dmu . (*)
To see this, take the supremum of (7.9) over the countable family from [L9]:
the left side becomes int_I P(t) dt, and the right side becomes
int_{u in I} |Du| dmu by the representation [L3] of the mass measure of A
restricted to {u in I}, with the chart density |theta_i| J_i dmu.
Differentiating with respect to the interval gives the second inequality of
(7.8).
For the first inequality of (7.8), use the coarea identity: write gamma for
the positive measure u_#(|Du| mu), lambda for u_#mu, and zeta for
u_#(|Du|^2 mu). Cauchy-Schwarz on the level pieces of gamma gives
gamma(I)^2 <= lambda(I) zeta(I) . (**)
Dividing by |I|^2 and differentiating with respect to the interval I at a
Lebesgue point of the function I |--> gamma(I) gives
P(t)^2 <= ( -a'(t) ) ( -e'(t) ) ,
where a(t) = lambda((t, infinity)) and e(t) = zeta((t, infinity)) are the
level distribution functions of (7.1). Neither a nor e is assumed absolutely
continuous, so the derivatives in (7.8) are the densities of the absolutely
continuous parts of the corresponding distribution measures, not pointwise
limits of difference quotients. If the level set {u = t} carries
measure-theoretic mass, (7.9) still holds with the substitute b = 1 on that
set: the level-set differential vanishes, so the slice determinant is
unchanged. End of proof of [L10].
The pieces of A are now known to have Euclidean perimeter. Two more transfers
are needed to turn that into a Sobolev inequality; both are classical, and
both have to be done carefully because the mass measure is not Lebesgue
measure and the level function a(t) is not continuous.
[L11] (rearranged energy). Fix eta, let delta come from [P3], and let u >= 0
be Lipschitz with mu{u > 0} <= delta. Then there is a nonnegative, radially
nonincreasing, compactly supported Lipschitz function u* on R^n such that
int_{R^n} (u*)^q dx = int_Y u^q dmu for all q > 0 , and (7.10)
int_{R^n} |grad u*|^2 dx <= (1 - eta)^(-2) int_Y |Du|^2 dmu . (7.11)
Proof. If u = 0 almost everywhere, take u* = 0. Otherwise let L = Lip(u) and
U = esssup u. Then U > 0 and L > 0: a Lipschitz function with Lip = 0 is
constant, and a positive constant would give a positivity set of mass m >
delta. For 0 <= t <= U define
a(t) = mu{u > t} , R(t) = ( a(t) / omega_n )^(1/n) .
R is nonincreasing and right-continuous, R(t) > 0 for t < U, and R(U) = 0.
For almost every t in (0,U), [P3] and [L10] give
-R'(t) = -a'(t) / ( n omega_n R(t)^(n-1) ) >= c_0 , c_0 = (1-eta)/L .
This extends to finite differences of R. Indeed the nonnegative measure -dR
has absolutely continuous density -R' >= c_0 and a nonnegative singular part
on every compact subinterval of (0,U). Therefore
R(s) - R(t) >= c_0 (t - s) for 0 < s < t < U . (*)
Define the generalized inverse for rho >= 0 by
h(rho) = sup { 0 <= t < U : rho < R(t) } , sup empty := 0 .
h is nonincreasing, h(0) = U, and h(rho) = 0 for rho >= R(0). To see that h
is c_0^(-1)-Lipschitz: let rho_1 < rho_2 with h(rho_1) > h(rho_2) and choose
h(rho_2) < s < t < h(rho_1). The definition and monotonicity give
R(s) <= rho_2 and R(t) > rho_1, so from (*)
c_0 (t - s) <= R(s) - R(t) <= rho_2 - rho_1 .
Letting s and t approach the two inverse values proves the bound. For 0 < t
< U, right continuity of R gives
h(rho) > t if and only if rho < R(t) . (**)
In the reverse direction, rho < R(t) plus right continuity supplies s > t
with rho < R(s); in the forward direction, monotonicity of R does the work.
Consequently u*(x) = h(|x|) is Lipschitz and compactly supported, and its
superlevel sets {u* > t} have Euclidean volume a(t). The layer cake formula
gives equality of all positive power integrals.
The strict decrease (*) also gives the inverse identity
h(R(t)) = t for 0 < t < U : (***)
all levels s < t satisfy R(s) > R(t), while no s >= t is eligible in the
supremum defining h at rho = R(t). If R is continuous at t, (**) and
continuity show that h^{-1}({t}) is the singleton {R(t)}.
The monotone function R is continuous except at countably many points and
differentiable with finite derivative almost everywhere. Its derivative,
wherever it enters (*)-(***), is nonzero. Let N be the null set where the
Lipschitz function h fails to be differentiable; h(N) is null because
Lipschitz functions map Lebesgue null sets to null sets. By (***), every
level t with R(t) in N lies in h(N). It follows that, for almost every t in
(0,U), both derivatives needed below exist, and differentiating (***) gives
h'(R(t)) = 1 / R'(t) . (****)
For example, let s -> t in h(R(s)) - h(R(t)) = s - t at a continuity and
differentiability point of R.
Since h is absolutely continuous and nonincreasing on [0, R(0)], the measure
|h'(rho)| drho pushes forward under h to Lebesgue measure on (0,U). One may
verify this first on an interval (a,b) in (0,U): integration of -h' over its
inverse image equals b - a by the fundamental theorem of calculus; constant
portions have zero derivative; intervals determine the measure. At almost
every target level the fiber consists of the single point R(t), so this
substitution together with (****) gives
int_{R^n} |grad u*|^2 dx
= n omega_n int_0^{R(0)} rho^(n-1) |h'(rho)|^2 drho
= int_0^U n omega_n R(t)^(n-1) / (-R'(t)) dt
= int_0^U ( n omega_n R(t)^(n-1) )^2 / (-a'(t)) dt .
Denominators are positive and finite almost everywhere by the (*)-inequality.
The substitution also justifies the equality for a nonnegative integrand
before its finiteness is known.
A jump of R produces an interval on which h is constant, and an atom at U
produces a central constant part of u*. Such intervals contribute no energy.
A singular continuous decrease causes no additional term either: the exact
substitution uses the absolutely continuous measure |h'| drho, and the
inverse image of any null set of levels has zero measure for this measure.
Thus no absolute-continuity assumption on a enters.
Finally, [P3] gives
( n omega_n R(t)^(n-1) )^2 <= (1-eta)^(-2) P(t)^2 ,
and [L10] applied to the last display yields
int_{R^n} |grad u*|^2 dx <= (1-eta)^(-2) int_0^U (-e'(t)) dt
<= (1-eta)^(-2) int_Y |Du|^2 dmu .
The last inequality uses only monotonicity of e and remains true if its
distributional derivative has a singular part. End of proof of [L11].
[L12] (sharp radial Sobolev). For every nonnegative, radially nonincreasing,
compactly supported Lipschitz f on R^n with n >= 3,
n s_n^(2/n) ( int f^p dx )^(2/p) <= ( 4/(n-2) ) int |grad f|^2 dx , (7.12)
with p = 2n/(n-2). Equality is approached by the bubbles
(1 + |x|^2)^(-(n-2)/2).
Proof. The assertion is immediate for f = 0. Homogeneity of both sides in f
reduces to int f^p dx = 1. Write f = f(r) with r = |x|. Radial monotonicity
and continuity give a finite r_0 > 0 with f > 0 on [0, r_0) and f = 0 on
[r_0, infinity).
For D > 0 define the truncated bubble
g_D(r) = A_D (1 + r^2)^(-(n-2)/2) for 0 <= r < D ,
g_D(r) = 0 for r >= D ,
with A_D > 0 chosen so that int g_D^p dx = 1. Match cumulative radial masses
by
n omega_n int_0^r s^(n-1) f(s)^p ds
= n omega_n int_0^{t(r)} s^(n-1) g_D(s)^p ds for 0 < r < r_0 . (*)
Both cumulative functions are C^1 with strictly positive derivative on their
open intervals, so the inverse composition t is C^1 on (0, r_0), strictly
increasing, with t(0+) = 0 and t(r_0-) = D. Differentiating (*) gives
d_T := t'(r) ( t(r) / r )^(n-1) = f(r)^p / g_D(t(r))^p . (**)
The radial map T(x) = t(|x|) x / |x| has positive eigenvalues t' and n - 1
copies of t/r. It transports f^p dx to g_D^p dx, as follows either from (*)
by radial integration or from the change of variables together with (**). The
jump of g_D at D is harmless, because the calculation never differentiates
the cutoff: it uses only the positive continuous interior profile.
Let ell = p(1 - 1/n) = 2(n-1)/(n-2), and note ell - 1 = p/2. Change of
variables, the AM-GM inequality applied to the n positive eigenvalues of DT,
and radial integration by parts give
n int_{R^n} g_D^ell dx
= n int_{|x| < r_0} f^ell d_T^(1/n) dx
<= int_{|x| < r_0} f^ell ( t' + (n-1) t/r ) dx
= -ell int_{|x| < r_0} f^(p/2) f' t dx
<= ell ( int_{R^n} |grad f|^2 dx )^(1/2)
( int_{R^n} |x|^2 g_D^p dx )^(1/2) . (***)
The last step is Cauchy-Schwarz, and it uses the transport identity
int f^p t^2 dx = int |x|^2 g_D^p dx. For the integration by parts, integrate
over epsilon < r < R < r_0 first: the boundary term is
n omega_n [r^(n-1) t f^ell]_epsilon^R, which tends to zero at r_0 (f(r_0) =
0 and t <= D) and at zero (n > 1 with bounded t, f). The resulting derivative
integral is absolutely integrable, since f' is essentially bounded and t is
bounded on a bounded ball.
Put I_0 = int_{R^n} (1 + |x|^2)^(-n) dx, A_infty = I_0^(-1/p), and
g(r) = A_infty (1 + r^2)^(-(n-2)/2). The normalizing constants A_D converge
to A_infty as D -> infinity. Moreover,
int g_D^ell dx -> int g^ell dx , int |x|^2 g_D^p dx -> int |x|^2 g^p dx .
For both integrals the unnormalized radial integrand has order r^(1-n) at
infinity, so both are finite for every n > 2, and convergence follows by
monotone convergence for the unnormalized integrals together with convergence
of A_D. In particular no limit of the transport maps is required.
The ratio on the left of (***) can be evaluated from g itself, because for
the limiting profile -g'(r) = K r g(r)^(p/2) with K = (n-2) A_infty^(-2/(n -
2)) > 0. If J = int g^ell dx and M_2 = int |x|^2 g^p dx, radial integration
by parts against the vector field x gives
n J = -ell int g^(p/2) g' r dx = ell K M_2 ,
int |grad g|^2 dx = K^2 M_2 .
The boundary term r^n g(r)^ell vanishes at zero and is O(r^(2-n)) at
infinity. Therefore
n J / (ell M_2^(1/2)) = ( int |grad g|^2 dx )^(1/2) . (****)
Taking D -> infinity in (***) proves int |grad f|^2 dx >= int |grad g|^2 dx.
The transport argument has now reduced the lower energy bound to the single
comparison profile g. Computing its energy fixes the threshold used in the
critical minimization of the champion. The inverse stereographic map
x |--> ( (2 x, |x|^2 - 1) / (1 + |x|^2) ) in S^n
pulls the sphere metric back to 4 (1 + |x|^2)^(-2) times the Euclidean
metric, and its volume Jacobian is 2^n (1 + |x|^2)^(-n). Hence
I_0 = 2^(-n) s_n . (7.14)
For w(r) = (1 + r^2)^(-(n-2)/2), direct differentiation gives
-Delta w = n(n - 2) (1 + r^2)^(-(n+2)/2), and integration by parts yields
int_{R^n} |grad g|^2 dx
= n(n-2) A_infty^2 I_0
= n(n-2) I_0^(1 - 2/p)
= n(n-2)/4 * s_n^(2/n) .
The boundary term r^(n-1) g(r) g'(r) vanishes at zero and is O(r^(2-n)) at
infinity; also 1 - 2/p = 2/n. Multiplying the energy lower bound by
4/(n - 2) proves (7.12) for the normalized f, and homogeneity proves the
general statement. All moments and boundary limits used above remain valid
when n = 3 or n = 4; no L^2 integrability of the full profile g is needed.
The coefficient is also optimal within the class in the statement. Choose a
nonnegative nonincreasing Lipschitz cutoff chi equal to one on [0,1] and zero
on [2, infinity), and set g_R(x) = g(x) chi(|x|/R). Then g_R -> g in L^p and
grad g_R -> grad g in L^2: the tail of grad g tends to zero, while the extra
cutoff energy is bounded by
Lip(chi)^2 / R^2 int_{R < |x| < 2R} g(x)^2 dx = O(R^(2-n)) ,
which tends to zero. Each g_R is admissible, and its Sobolev quotient tends
to the computed value for g. End of proof of [L12].
[P4] (almost sharp Sobolev). For every eps in (0, S_*) there is a finite
constant C_eps such that every real Lipschitz u on Y satisfies
( S_* - eps ) ( int |u|^p dmu )^(2/p)
<= 4 beta int |Du|^2 dmu + C_eps int u^2 dmu . (7.15)
Here S_* = n s_n^(2/n), beta = 1/(n-2), p = 2n/(n-2).
Proof. Combine [L11] and [L12]: for nonnegative z with mu{z > 0} <= delta,
S_* (1 - eta)^2 ( int z^p dmu )^(2/p) <= 4 beta int |Dz|^2 dmu . (7.16)
For general Lipschitz u with ||u||_2 > 0 put tau = ||u||_2 / sqrt(delta) and
z = (|u| - tau)_+. By Chebyshev, mu{z > 0} <= delta; the scalars |u| and z
differ by at most tau, so || |u| - z ||_p <= m^(1/p) tau; and
E(z) <= E(u) because z is a scalar Lipschitz composition of |u| (the gradient
vanishes on the corner level set, so the usual chain rule holds almost
everywhere). Applying (7.15) to z and using the elementary inequality
(b+d)^2 <= (1+xi) b^2 + (1+1/xi) d^2 gives
S_* (1-eta)^2 / (1+xi) ||u||_p^2
<= 4 beta E(u) + ( S_* (1-eta)^2 m^(2/p) / (xi delta) ) ||u||_2^2 .
Choose eta, xi > 0 so small that S_* (1-eta)^2 / (1+xi) >= S_* - eps, and
set C_eps = S_* (1-eta)^2 m^(2/p) / (xi delta). That is (7.15). End of
proof of [P4].
The constant C_eps is not small and is not meant to be: it can be absorbed
later because the quadratic term is subcritical. What matters is that the
coefficient of the critical term is S_* - eps, with S_* the sharp Euclidean
constant.
The Sobolev inequality (7.15) is for Lipschitz functions. To make a
variational argument we need a Hilbert space and a minimizer, and that means
closing the gradient.
[P5] (Sobolev space and calculus). Identify real Lipschitz functions that
agree mu-almost everywhere, and complete them in
||u||_H^2 = ||u||_2^2 + E(u) , E(u) = int |Du|^2 dmu . (7.17)
Then:
(a) the completion H is a Hilbert space, and its gradient D is a
well-defined closed operator;
(b) H embeds continuously in L^p(mu), and (7.15) holds on H;
(c) H is stable under C^1 composition with bounded first derivatives;
(d) Du = 0 mu-almost everywhere on each level set {u = t}; consequently
the usual chain rule holds for piecewise C^1 scalars with finitely
many corners;
(e) u |--> u_+ is continuous from H to H, and every nonnegative element of
H is approximated in H by nonnegative Lipschitz functions;
(f) for each fixed y, the radial inequality [P1] holds on H, that is, for
every nonnegative g in H.
The only point that is not standard is (a), and here is why it works and why
it is not obvious: two Lipschitz functions that agree almost everywhere could
in principle have different charts of different gradients, and nothing in the
definition forbids that.
Proof of (a) (graph closure). Suppose u_j are Lipschitz,
u_j -> 0 in L^2(mu), and Du_j -> L in L^2(mu) (the covariant Hilbert space
of section 2.5). For a Lipschitz b and a Lipshitz (n-1)-tuple pi, the cycle
identity dA = 0 and the product rule give
A( b, u_j, pi ) = -A( u_j, b, pi ) -> 0 , (7.18)
because |A(u_j,b,pi)| <= const ||u_j||_1 -> 0. The left side converges to the
integral of b against the signed chart measure
nu_pi(B) = sum_i int_{phi_i^(-1)(B)} theta_i
det( L_i, D(pi o phi_i) ) dz ,
which by the determinant bound is a finite signed measure with total
variation controlled by ||L||_1. So nu_pi = 0 for every admissible (b,pi).
Now fix a chart i, and let F^1,...,F^n be the components of phi_i^(-1),
extended to all of Y by McShane [lit 11]. Then at almost every point of the
chart, D(F^a o phi_i) = e_a^* (the standard unit covector in the domain). Use
as pi the n tuples obtained by deleting one component of (F^1,...,F^n); the
corresponding measures nu_pi vanish, so restricting them to Z_i gives
theta_i L_i = 0 almost everywhere on E_i. The multiplicity is nonzero, so
L_i = 0. There are countably many charts and finitely many tuples per chart,
so L = 0.
Thus the graph of D is closed, and its completion is a Hilbert space of
functions in L^2. The embedding (b) and the calculus (c)-(e) are standard
once the graph is closed; the level-set statement (d) is proved by
approximating the indicator of a level set by smooth cutoffs and using the
chain rule (i), so the corner cases follow by approximation. Finally (f) is
proved by approximating g in H by nonnegative Lipschitz functions (part e)
and passing to the limit in [P1]: every term of (6.19) is continuous in g for
the H-norm, because r and Dr are bounded. End of proof of [P5].
[P6] (compact embedding). The inclusion H -> L^2(mu) is compact.
Proof. Let u_j be bounded in H, hence also in L^p by [P5](b). Fix a chart i
and a Lipschitz cutoff chi on Y. Define the projected signed measure
lambda_{j,chi}(B) = A( (1_B o F) chi u_j, F^1, ..., F^n ) , B in R^n ,
where F is the chart's inverse coordinate map. These measures are supported in
the compact set F(Y), and their total variation is bounded uniformly in j.
The cycle identity dA = 0 together with the product rule gives
| int partial_a h d lambda_{j,chi} | <= C_chi ||h||_infty
for smooth compactly supported h, where the constant depends on chi but not
on j: moving the factor h to the first slot and applying [lit 1](d), the
first-slot coefficient chi u_j partial_a h has L^1 norm bounded by
||h||_infty times a fixed multiple of ||D(chi u_j)||_1, and the remaining
factor is the chart Jacobian term with bounded test functions.
Translation compactness. Fix w in R^n and let tau_w(x) = x + w. For a smooth
compactly supported test h, the fundamental theorem of calculus along
x + tw and the derivative bound imply
| int h d( (tau_w)_# lambda_{j,chi} - lambda_{j,chi} ) |
<= C_chi ||h||_infty sum_{l=1}^n |w_l| .
Smooth compactly supported tests recover the total variation of finite signed
Radon measures, by regularity and approximation of continuous tests.
Therefore, after changing the constant,
|| (tau_w)_# lambda_{j,chi} - lambda_{j,chi} ||_TV <= C_chi |w| .
Choose a nonnegative smooth convolution kernel rho_eps of integral one
supported in the ball of radius eps. Averaging the last display gives
|| lambda_{j,chi} - (rho_eps*lambda_{j,chi}) dx ||_TV <= C_chi eps .
For each fixed eps the densities rho_eps*lambda_{j,chi} have a common compact
support and uniform bounds on their values and first derivatives. By
Arzela-Ascoli they are precompact in the uniform norm, hence the smoothed
family is precompact in total variation. Uniform approximation from the first
bound then makes the original family {lambda_{j,chi} : j >= 1} precompact in
TV as well.
Isolating a chart in total variation. By regularity of mu and distance
cutoffs on the compact metric space Y, choose Lipschitz chi_k with
||chi_k - 1_{Z_i}||_2 -> 0. The mass bound and Cauchy-Schwarz give, uniformly
in j,
|| lambda_{j,chi_k} - lambda_{j, 1_{Z_i}} ||_TV
<= ( prod_a Lip(F^a) ) int |chi_k - 1_{Z_i}| |u_j| dmu
<= C_F M_0 ||chi_k - 1_{Z_i}||_2 .
For each fixed k the family on the left with coefficient chi_k is precompact
by the preceding argument. The uniform approximation above then makes
{lambda_{j,1_{Z_i}} : j >= 1} precompact as well. There is no need for the
Lipschitz constants of chi_k to remain bounded.
On this isolated chart the density is exactly 1_{E_i}(z) theta_i(z)
u_j(phi_i(z)), so
||u_j - u_k||_{L^1(Z_i,mu)} <= Lip(phi_i)^n
|| lambda_{j,1_{Z_i}}
- lambda_{k,1_{Z_i}} ||_TV .
Take a diagonal subsequence that is Cauchy in L^1(Z_i, mu) for every i. The
omitted charts have uniformly small contributions, because
int_{union_{i>N} Z_i} |u_j| dmu <= M_0 mu(union_{i>N} Z_i)^(1/2) -> 0
uniformly in j: the chart images cover mu up to a null set, and mu is finite.
The subsequence is therefore Cauchy in L^1(mu). It is also bounded in L^p(mu),
and p > 2, so interpolation with alpha = (p - 2)/(2(p - 1)) > 0 gives
||u_j - u_k||_2 <= ||u_j - u_k||_1^alpha ||u_j - u_k||_p^(1-alpha) ,
hence Cauchy in L^2. Finally, approximate the jth member of an arbitrary
bounded sequence in H by a Lipschitz function with graph-norm error at most
1/j; the result just proved for those approximants proves compactness for the
original sequence. End of proof of [P6].
The almost sharp inequality (7.15) has a strict gap: the quadratic term is
irrelevant at scale of the minimizer because the small mass m < s_n pushes
the critical coefficient below S_*. Concretely:
[P7] (subcritical minimizer). Define
Q(u) = 4 beta E(u) + n ||u||_2^2 , Lambda = inf Q(u) / ||u||_p^2 ,
over nonzero u in H . (7.19)
Then 0 < Lambda < S_*. There is a nonzero nonnegative v in H minimizing
Q(u)/||u||_p^2, and
0 < W := int v^p dmu = ( Lambda / n )^(n/2) < s_n , (7.20)
Q(v) = n W , (7.21)
and for every psi in H,
4 beta int dmu + n int v psi dmu = n int v^(p-1) psi dmu . (7.22)
Proof. Positivity of Lambda follows from (7.15) with eps = S_*/2:
(S_*/2) ||u||_p^2 <= 4 beta E(u) + C ||u||_2^2 <= C' Q(u). Upper bound: the
constant function u = m^(-1/p) has ||u||_p = 1 and Q(u) = n m^(1-2/p) =
n m^(2/n), which is < n s_n^(2/n) = S_* because m < s_n.
Let u_j be an L^p-normalized minimizing sequence. It is bounded in H (Q is a
positive definite quadratic form equivalent to ||.||_H^2). By Hilbert space
compactness and [P6], a subsequence converges weakly in H and strongly in
L^2, and almost everywhere, to an element u. Put t = int |u|^p.
If t < 1, consider z = u_j - u. Brézis-Lieb [lit 8] (or the elementary
inequality in section 4) gives int |z|^p -> 1 - t. Quadratic splitting gives
Q(u_j) = Q(u) + Q(z) + o(1), and by definition of Lambda,
Q(u) >= Lambda t^(2/p) and Q(z) >= (S_* - eps)(1-t)^(2/p) - C_eps||z||_2^2.
Since ||z||_2 -> 0,
Lambda >= Lambda t^(2/p) + S_* (1 - t)^(2/p) ,
and the right side is strictly bigger than Lambda whenever t < 1 and
S_* > Lambda (use a^(2/p) >= a for a in [0,1]). So t = 1, u attains the
minimum, and ||u||_p = 1. Replacing by |u| preserves both ||.||_p and E(u) by
[P5](e), so we take u >= 0.
Differentiate Q(u+s psi)/||u+s psi||_p^2 at s = 0: by [P5], the gradient is
continuous in H, and the L^p-derivative is dominated by p(|u|+|psi|)^(p-1)
|psi|, which is integrable. Since ||u||_p = 1, this gives
4 beta int dmu + n int u psi dmu
= Lambda int u^(p-1) psi dmu for all psi in H .
Set v = (Lambda/n)^(1/(p-2)) u. Multiplication by a constant preserves a
minimizer. Substituting u = (Lambda/n)^(-1/(p-2)) v into the Euler
equation for u and using Lambda * (Lambda/n)^(-1) = n gives (7.22). Its
p-th moment is
W = int v^p dmu = (Lambda/n)^(p/(p-2)) = (Lambda/n)^(n/2) ,
which is < s_n because Lambda < S_* = n s_n^(2/n). Finally, testing (7.22)
with psi = v gives
4 beta E(v) + n ||v||_2^2 = n W, which is (7.21). End of proof of [P7].
[L13] (powers of the minimizer). For every real gamma >= 1,
v^gamma in H and D(v^gamma) = gamma v^(gamma-1) Dv. Moreover,
int v^q dmu < infinity for every finite q > 0, and
int v^q |Dv|^2 dmu < infinity for every q >= 0.
Proof. Assume temporarily v in L^(2 gamma). Put w_N = v min(v,N)^(gamma-1)
and psi_N = v min(v,N)^(2 gamma-2). Both are in H by [P5](c). For their
gradients,
>= d_gamma |Dw_N|^2 , d_gamma = (2 gamma - 1) / gamma^2 :
there is equality on {v < N}, and on {v > N} the ratio of coefficients is 1
and 0 < d_gamma <= 1; the gradient vanishes on the level {v = N} by [P5](d).
Test (7.22) with psi_N, drop the nonnegative term n int v psi_N, and get
4 beta d_gamma E(w_N) <= n int v^(p-2) w_N^2 dmu . (7.23)
Now apply (7.15) with eps = S_*/2 to w_N and use (7.23):
( d_gamma S_* / 2 ) ||w_N||_p^2
<= n int v^(p-2) w_N^2 dmu + d_gamma C ||w_N||_2^2 .
Hölder on the "critical potential":
int_{v > K} v^(p-2) w_N^2 dmu
<= (int_{v > K} v^p dmu)^((p-2)/p) ||w_N||_p^2 .
Since v^p is integrable, choose K so large that the coefficient n (int_{v >
K} v^p)^((p-2)/p) is at most d_gamma S_*/4. What is left of the potential,
on {v <= K}, is bounded by n K^(p-2) ||w_N||_2^2. Absorb and get
( d_gamma S_* / 4 ) ||w_N||_p^2 <= C_K ||w_N||_2^2 .
With the assumed L^(2 gamma) moment the right side is bounded uniformly in N.
So ||w_N||_p is bounded, and (7.23) bounds E(w_N) too. On {v < N},
Dw_N = gamma v^(gamma-1) Dv; the functions w_N and Dw_N converge monotonically
and are dominated by v^gamma and gamma v^(gamma-1)|Dv|, so both converge in
L^2. Closedness of D gives v^gamma in H with the asserted gradient.
Start with v in L^p (from [P7]), set l_j = p (p/2)^j, and iterate the
conditional statement with gamma = l_j / 2, j = 0, 1, 2, ...: each step gains
v in L^(l_(j+1)), and l_j -> infinity since p/2 > 1. Finite measure then
gives every finite moment, and every q >= 0 has its weighted energy from the
gradient of v^(1 + q/2). End of proof of [L13].
[L14] (compositions with polynomial growth). Let r in H with 0 <= r <=
R < infty almost everywhere and |Dr| in L^infty(mu). Let F be C^1 on a
neighbourhood of [0,R] x [0, infty) with
|F(a,t)| + |partial_1 F(a,t)| + |partial_2 F(a,t)| <= C (1+t)^M
for some finite C, M. Then F(r,v) in H and
D(F(r,v)) = partial_1 F(r,v) Dr + partial_2 F(r,v) Dv . (7.24)
Proof. Let v_N = min(v,N). By [P5](c) and the level-set property, v_N
belongs to H with gradient 1_{v 0, put
kappa = v(y)^beta , r(x) = d(y,x) , (7.26)
s(x) = kappa * r(x) * v(x)^beta , L_y = int s^2 dnu , (7.27)
J_y(a) = int (1 + a s^2)^(-n) dnu (a >= 0) . (7.28)
J_y is a transform of the distribution of squared chords; the problem is to
control it in terms of its second moment L_y.
[L15] (admissible chord tests). Fix a > 0, let w(t) = (1 + a t^2)^(-(n-2)/2),
and let s as in (7.27). Then:
(a) v w(s), v w(s)^2 in H, with
D(v w(s)) = w(s) Dv + w'(s) xi ,
D(v w(s)^2) = w(s)^2 Dv + 2 w(s) w'(s) xi , (7.29)
where xi = kappa ( v^(1+beta) Dr + beta r v^beta Dv ) is a
square-integrable chart covector field;
(b) for every smooth f: R -> [0, infty) with bounded value and bounded
derivative, the functions
g = kappa^2 v^(2 + 2 beta) f(s) ,
psi = beta kappa^2 r^2 v^(1 + 2 beta) f(s) (7.30)
are in H. In particular g is admissible in [P1] (for the current mu),
and psi is admissible in (7.22).
Proof. The obstruction is that v^beta is not differentiable at v = 0 when
beta < 1; the exponent beta = 1/(n-2) is < 1 exactly when n >= 4, so the
singular case is real and must be handled. The route is to write each
candidate function as a composition F(r,v) and to apply [L14], whose
hypotheses are C^1 regularity and growth on the strip [0,R] x [0, infty).
This handles the singular behaviour of v^beta without any differentiability
assumption on v^beta itself.
For j = 1, 2 and z > 0 put t = kappa r z^beta and
F_j(r,z) = z w(t)^j .
On this half-plane,
partial_z F_j = w(t)^j + j beta t w(t)^(j-1) w'(t) ,
partial_r F_j = j kappa z^(1+beta) w(t)^(j-1) w'(t) .
As z decreases to 0, uniformly for r in bounded intervals, w(t) = 1 + O(t^2)
and w'(t) = O(t), because t = kappa r z^beta tends to zero. Hence
F_j(r,z) = z + O(z^(1+2beta)) , partial_z F_j -> 1 ,
partial_r F_j -> 0 .
Extension by F_j(r,z) = z on z <= 0 is therefore jointly C^1. The functions
F_j and their first partials have polynomial growth in z on bounded r
intervals.
For the test functions g and psi of (7.30), put
F_g(r,z) = kappa^2 z^(2+2beta) f(kappa r z^beta) ,
F_psi(r,z) = beta kappa^2 r^2 z^(1+2beta) f(kappa r z^beta) (z > 0) .
Their partial derivatives, computed by the product and chain rules, are
partial_z F_g = kappa^2 z^(1+2beta) ((2+2beta) f(t)
+ beta t f'(t)) ,
partial_r F_g = kappa^3 z^(2+3beta) f'(t) ,
partial_z F_psi = beta kappa^2 r^2 ((1+2beta) z^(2beta) f(t)
+ beta kappa r z^(3beta) f'(t)) ,
partial_r F_psi = beta kappa^2 (2 r z^(1+2beta) f(t)
+ kappa r^2 z^(1+3beta) f'(t)) .
Every term tends to zero as z -> 0 uniformly for bounded r, because the
smallest power of z in these formulas is z^(2beta) with 2 beta = 2/(n-2) >
0. Consequently extension by zero on z <= 0 makes F_g and F_psi jointly
C^1, even if f(0) or f'(0) is nonzero. Their values and first derivatives
again have polynomial growth on bounded r intervals.
The three extensions satisfy all the hypotheses of [L14] with r = d(y,.)
and v in place of the second variable. [L14] therefore gives membership in
H and the chain rules, with no boundedness assumption on v. On {v > 0},
differentiation gives the product rules (7.29). On {v = 0}, [P5](d) gives
Dv = 0 almost everywhere, while the defining formula of xi gives xi = 0
there. Thus (7.29) holds mu-almost everywhere on all of Y.
End of proof of [L15].
[L16] (completed radial inequality). For every y in Y and every nonnegative
g in H,
int ( n g |Dr|^2 + r ) dmu <= (n/4) int r^2 g dmu . (7.31)
Proof. [P1] holds for every nonnegative g in H by [P5](f). Its left side is
int (n g + r ) dmu and its right side is n int r g b dmu, where
b = sqrt(1 - |Dr|^2). Pointwise, since r >= 0 and b >= 0,
1 - r b - |Dr|^2 + r^2/4 = (b - r/2)^2 >= 0 .
Multiply by n g >= 0 and integrate over Y. This gives
n int r g b + n int g |Dr|^2 <= n int g + (n/4) int r^2 g . (*)
[P1] says n int g + int r <= n int r g b. Combining the two,
n int g + int r
<= n int r g b
<= n int g + (n/4) int r^2 g - n int g |Dr|^2 .
The n int g terms on the two sides cancel. Moving n int g |Dr|^2 to the
left gives exactly (7.31). End of proof of [L16].
[P8] (chord transform bound). For every centre y with v(y) > 0 and every
a >= 0,
J_y(a) <= (1 + 2 a L_y)^(-n/2) . (7.32)
Proof. Fix y with v(y) > 0 and abbreviate kappa = v(y)^beta, r = d(y,.),
s = kappa r v^beta, xi = kappa (v^(1+beta) Dr + beta r v^beta Dv). Note the
identity v Ds = xi, so D(v w(s)) = w(s) Dv + w'(s) xi for any smooth w.
Step 1: a ground state identity. Test the Euler equation (7.22) with
psi = v w(s)^2, which is in H by [L15]. Using the product rules (7.29) and
noting that the first-order term in ,
2 w w' , equals the term in , one gets
Q(v w(s)) = n int v^p w(s)^2 dmu + 4 beta int (w'(s))^2 |xi|^2 dmu . (7.33)
Step 2: quotient minimality. By the definition of v as a minimizer,
Q(v w(s)) >= (Q(v)/||v||_p^2) ||v w(s)||_p^2
= n W J(a)^((n-2)/n) , (7.34)
since int v^p w(s)^p = int v^p (1 + a s^2)^(-n) dmu * W/W = W J(a), and
p (n-2)/2 = n.
Step 3: control the |xi|^2 term. Add [L16] to (7.22) divided by 4 beta, in
both cases with the two tests g and psi from (7.30). The undifferentiated
terms cancel, because v psi = beta r^2 g and v^(p-1) psi = beta s^2 f(s) v^p.
After multiplication by the common factor kappa^2 and grouping the
coefficients of |Dr|^2, , and |Dv|^2 - which the exponent identities
1 + 2 beta = n beta and 2 + 4 beta = 2n beta = p identify as the coefficients
of (n f(s) + s f'(s)) |xi|^2 / kappa^2 - the combined inequality becomes
int ( n f(s) + s f'(s) ) |xi|^2 dmu
<= (n/4) int s^2 f(s) v^p dmu . (7.35)
Now choose f so that n f(t) + t f'(t) = (w'(t))^2, namely
f(t) = int_0^1 tau^(n-1) (w'(tau t))^2 dtau ,
which is smooth, bounded, with bounded derivative, and satisfies the identity
by integrating d/d tau ( tau^n (w'(tau t))^2 ). Combining (7.33), (7.34) and
(7.35) and dividing by the multiplicative factor W:
J(a)^((n-2)/n) <= int ( w(s)^2 + beta s^2 f(s) ) dnu . (7.36)
Step 4: scalar reduction. With q = (n-2)/2 > 0 and b = a s^2, the exact
derivative identity
d/du [ u^q / (1 + b u)^(n-2) ] = q u^(q-1) (1 - b^2 u^2) / (1 + b u)^n
has an integrable right side on (0,1). Integrating it against u from 0 to 1
gives
w(s)^2 + beta s^2 f(s) = q int_0^1 u^(q-1) / (1 + a s^2 u)^n du .
In particular, the right side of (7.36) is in (0,1]. So (7.36) becomes
J(a)^((n-2)/n) <= G(a) := q int_0^1 J(au) u^(q-1) du . (7.37)
Step 5: solve the ODE. Put h = G^(-1/q) = G^(-2/(n-2)), so G = h^(-q).
Differentiating G(a) = q a^(-q) int_0^a J(t) t^(q-1) dt gives
a G'(a) = q (J(a) - G(a)). From G = h^(-q) and G' = -q h^(-q-1) h',
substitution gives a h'(a) = h - J h^(q+1). Now (7.37) rearranged is
J <= G^(n/(n-2)) = h^(-(q+1)), because n/(n-2) = (q+1)/q. Hence
h^(q+1) J <= 1 and a h' >= h - 1. The quotient (h(a) - 1)/a is therefore
nondecreasing on (0, infinity):
d/da [ (h(a) - 1)/a ] = (a h'(a) - h(a) + 1) / a^2 >= 0 .
Its limit as a -> 0+ is h'(0+) = 2 L_y, computed from G(0) = 1 and
J'(0+) = -n L_y (the latter from (7.28) differentiated under the integral,
using the finite second moment L_y of (7.27)). Therefore
h(a) >= 1 + 2 a L_y for every a > 0. Combining with (7.37):
J(a) <= G(a)^(n/(n-2)) = h(a)^(-n/2) <= (1 + 2 a L_y)^(-n/2) .
This is (7.32). End of proof of [P8].
[L17] (tangent density). For nu-almost every centre y,
liminf_{a -> infty} a^(n/2) J_y(a) >= s_n / (2^n W) . (7.38)
Consequently,
L_y <= 2 ( W / s_n )^(2/n) < 2 (7.39)
for nu-almost every y.
Proof. Choose a chart and a point y = phi_i(z) where z is a density point of
E_i, the metric differential is the Euclidean norm |.|_{G_0}, and u_i = v o
phi_i, rho_i = |theta_i| sqrt(det G_i) are Lebesgue points with values
v_0 > 0, rho_0 > 0. Such points carry full nu-mass by [L2] and [L3].
Let a_j -> infinity realize the lower limit l of a^(n/2) J_y(a), and put
eps_j = a_j^(-1/2). After passing to a diagonal subsequence in R^n, the
rescaled functions u_i(z + eps_j h), rho_i(z + eps_j h), 1_{E_i}(z + eps_j h)
converge almost everywhere to v_0, rho_0, 1; and
eps_j^(-1) d(y, phi_i(z + eps_j h)) -> |h|_{G_0} .
Restricting the defining integral of J_y to the chart, changing variables,
and using Fatou (with a_i^(n/2) cancelling the Jacobian eps_j^(-n)):
W l >= v_0^p rho_0 int_{R^n} (1 + v_0^(4 beta) |h|_{G_0}^2)^(-n) dh .
The linear substitution H = v_0^(2 beta) G_0^(1/2) h has Jacobian
v_0^(2n beta) sqrt(det G_0) = v_0^(p - 2) sqrt(det G_0). The integrand under
this substitution is exactly (1 + |H|^2)^(-n), and using the stereographic
integral (7.14):
W l >= |theta_i(z)| s_n / 2^n >= s_n / 2^n ,
where the last step uses |theta_i(z)| >= 1 (integrality). This is (7.38). To
deduce (7.39), combine (7.32) with (7.38), solve for L_y:
s_n / (2^n W) <= (2 L_y)^(-n/2) => L_y <= 2 (W/s_n)^(2/n) ,
and 2(W/s_n)^(2/n) < 2 because W < s_n. The case L_y = 0 is immediate.
End of proof of [L17].
[L18] (opposite averaged bound). Define
B_* = int v^(p + 2 beta) dmu , I(y) = int d(y,x)^2 v(x)^(p + 2 beta) dmu(x) .
Then 0 < B_* < infinity, I is finite and continuous, and
W^2 <= B_* I(y) for every y in Y . (7.40)
Consequently,
int L_y dnu >= 2 . (7.41)
Proof.
(7.40). Fix y, put r(x) = d(y,x). By [L13], g = v^p is in H with
Dg = p v^(p-1) Dv, so [P1] applies to g. Since p/n = 2 beta, dividing [P1]
by n gives
W + 2 beta int r v^(p-1) dmu
<= int r v^p sqrt(1-|Dr|^2) dmu ,
which rearranges to
W <= int r ( v^p sqrt(1-|Dr|^2) - 2 beta v^(p-1) ) dmu . (7.42)
Pointwise, in the direct sum of the chart cotangent space with R, the vector
(sqrt(1-|Dr|^2), Dr) has norm one. The bracketed integrand is therefore the
inner product of that unit vector with (v^(1+beta), -2 beta v^beta Dv),
multiplied by v^(p/2 + beta). Cauchy-Schwarz at each point bounds it by
v^(p/2 + beta) times (v^(2+2 beta) + 4 beta^2 v^(2 beta) |Dv|^2)^(1/2).
Cauchy-Schwarz in the integral then gives
W^2 <= ( int d(y,x)^2 v^(p+2 beta) dmu(x) ) *
( int ( v^(2+2 beta) + 4 beta^2 v^(2 beta) |Dv|^2 ) dmu ) .
The first factor is I(y). For the second, apply (7.22) with the test
psi = v^(1+2 beta), which lies in H by [L13] and has chart gradient
Dpsi = (1+2 beta) v^(2 beta) Dv. That gives
4 beta (1+2 beta) int v^(2 beta) |Dv|^2 dmu + n int v^(2+2 beta) dmu
= n int v^(p+2 beta) dmu = n B_* .
The identity 1 + 2 beta = n beta lets this be solved for the second factor,
which is therefore equal to B_*. Combined with the Cauchy-Schwarz inequality
this proves (7.40).
(7.41). Use the variance inequality [lit 12]: the function I has a minimizer
o (a barycentre) and satisfies I(y) >= I(o) + B_* d(o,y)^2 for every y. For
completeness, here is the segment argument. Let o_t be the point at fraction
t on the segment [o,y]. The CAT(0) squared-distance inequality, integrated
against v(x)^(p+2 beta) dmu(x), gives
I(o) <= I(o_t) <= (1-t) I(o) + t I(y) - t(1-t) B_* d(o,y)^2 .
Subtracting (1-t) I(o), dividing by t, and letting t downarrow 0 gives
I(y) >= I(o) + B_* d(o,y)^2. Continuity of I on the compact space Y ensures
the minimizer o exists. Now
int L_y dnu = (1/W^2) int I(y) v(y)^(p+2 beta) dmu(y)
>= (1/W^2) int ( I(o) + B_* d(o,y)^2 ) v(y)^(p+2 beta) dmu(y)
= (1/W^2) ( I(o) B_* + B_* I(o) )
= 2 B_* I(o) / W^2
>= 2 .
The second line is the variance inequality. In the third line both terms
equal B_* I(o): the first because B_* = int v^(p+2 beta) dmu, the second
because I(o) = int d(o,y)^2 v(y)^(p+2 beta) dmu(y). The last inequality is
(7.40) at y = o. End of proof of [L18].
[T3] (induction step). Let n >= 3, let Y be a compact CAT(0) space, and
assume (H_(n-1)). Then every integral n-cycle T in Y has an integral filling
S with dS = T and M(S) <= c_n M(T)^((n+1)/n).
Proof. Suppose not. By [L7] and [L8] there is a champion A with
m = M(A) < s_n and normalization (6.1). Build v from [P7] and nu from
(7.25).
Bound above. By (7.39), L_y <= 2(W/s_n)^(2/n) for nu-almost every centre
y. Integrating over nu,
int L_y dnu <= 2 (W/s_n)^(2/n) . (7.43)
Bound below. (7.41) gives
int L_y dnu >= 2 . (7.44)
Combine (7.43)-(7.44) with W < s_n:
2 <= int L_y dnu <= 2 (W/s_n)^(2/n) < 2 ,
a contradiction. So no champion exists, and the sharp filling inequality
holds in dimension n. End of proof of [T3].
This section closes the argument. Section 8.1 assembles [T1] from the
base case [P2], the induction step [T3] and the compact reduction [L6].
Section 8.2 proves the top-dimensional constancy lemma [L19] and then
the classical Cartan-Hadamard domain inequality [T2]. Section 8.3
handles the surface case d = 2 of the same inequality. Section 8.4
verifies that the sharp constant is attained, so it cannot be improved.
List the three ingredients that have been proved already.
[P2] (section 6.5): a compact CAT(0) space Y, and an integral
2-cycle T in Y, has an integral 3-current filling S with
M(S) <= c_2 M(T)^(3/2). This is the sharp bound at n = 2.
[T3] (section 7.8): if the sharp filling inequality holds in
cycle dimension n-1 (in every compact CAT(0) space), then it holds
in dimension n, for every n >= 3.
[L6] (section 5.1): if the sharp filling inequality in dimension n
holds in every compact CAT(0) space, then it holds in every proper
CAT(0) space, with the filling achievable with compact support.
Putting the first two together by induction on n gives the theorem in
compact spaces.
Claim. For every n >= 2, every compact CAT(0) space Y and every
integral n-cycle T in Y, there is an integral (n+1)-current S in Y
with dS = T and M(S) <= c_n M(T)^((n+1)/n).
Proof of the claim, by induction on n. The case n = 2 is [P2]. Now
assume the claim at n-1, which is hypothesis (H_(n-1)) of [T3]; [T3]
then gives the claim at n. End of proof of the claim.
Proof of [T1]. Let X be a proper CAT(0) space and let T in I_n(X) be
an integral n-cycle of compact support. The claim in the previous
paragraph, combined with [L6], gives a compactly supported current S
in I_(n+1)(X) with dS = T and M(S) <= c_n M(T)^((n+1)/n). This is
exactly [T1]. End of proof of [T1].
Why this is short. The weight of the work went into [P2], [T3] and
[L6]. Assembling them is a one-line affair: base case, induction,
reduction. The theorem is proved, and the guess is confirmed: on a
CAT(0) space, filling a cycle of mass A never costs more than the
flat-space optimum c_n A^((n+1)/n), with no restriction on the ambient
dimension and no restriction on multiplicities.
We now turn to the geometric consequence of [T1]. The proof needs one
elementary fact about top-dimensional cycles in noncompact manifolds:
a compactly supported top-dimensional integral cycle is zero. The
reason is that a d-cycle in a d-manifold is "integration of a locally
constant coefficient", so a compactly supported such cycle must have
coefficient zero.
[L19] (top-dimensional constancy). Let M be a connected, noncompact,
oriented Riemannian manifold of dimension d. If C in I_d(M) has
compact support and dC = 0, then C = 0.
Proof. Fix a relatively compact oriented coordinate ball U of M. The
chart representation (2.1), with k = d, gives a bounded Borel chart
domain E in R^d and an L^1 integer-valued density theta on E such
that, for every bounded Lipschitz b and every d-tuple of Lipschitz
scalars (pi_1, ..., pi_d) supported in U,
C(b, pi_1, ..., pi_d)
= int_E theta(z) b(z) det D( (pi_1,...,pi_d) )(z) dz .
Choose coordinates x_1, ..., x_d for U, and let pi_a = x_a for all a;
then det D(pi) is the constant Jacobian of the identity, so the
formula reduces to a signed measure on U with density theta against
Lebesgue measure. In distributional form the identity dC = 0 reads:
for every smooth compactly supported (d-1)-form phi on U,
0 = C(d phi) = int_U theta d phi .
The right-hand side is the distributional pairing of theta with the
exterior derivative d phi. Varying phi over smooth compactly supported
(d-1)-forms produces arbitrary divergence-free test functions, and in
particular each distributional partial derivative of theta is zero.
A locally L^1 function with zero distributional gradient is a
constant almost everywhere. Since U was any coordinate ball and M is
connected, the constant is the same on every such ball. M is
noncompact, so pick a point of M outside the compact support of C;
a small coordinate ball B around it satisfies theta = 0 on B. The
global constant therefore is zero, and C = 0. End of proof of [L19].
A note on why this works. Noncompactness is used exactly once: it
guarantees a nonempty open set of M on which the compactly supported
C vanishes, and hence on which the constant theta must be zero. In
the compact case the integration current [[M]] itself is a nonzero
compactly supported d-cycle, so [L19] fails; the compact oriented
manifold has plenty of top-dimensional cycles. The hypothesis
"connected" then propagates the vanishing from one point to the
whole manifold.
Now let M be a Cartan-Hadamard manifold: complete, simply connected,
with sectional curvature at most zero. By Cartan-Hadamard and
Hopf-Rinow, M is proper CAT(0), is diffeomorphic to R^d, and is in
particular connected, noncompact and oriented.
[T2] (Cartan-Hadamard domain inequality). Let M be a Cartan-Hadamard
manifold of dimension d >= 3, and let Omega be a bounded domain with
smooth boundary. Then
area(dOmega) >= d omega_d^(1/d) vol(Omega)^((d-1)/d) . (8.1)
The same inequality holds for relatively compact sets of finite
perimeter, with the ambient perimeter in place of area(dOmega).
Proof. Let [[Omega]] in I_d(M) be the integration current of Omega.
Its support is the closure of the bounded Omega, so it has compact
support; its mass is vol(Omega); its boundary T = d[[Omega]]
is an integral (d-1)-cycle with M(T) = area(dOmega). We split the
argument into four steps.
Step 1: fill T with bounded mass. Apply [T1] with n = d-1 (>= 2,
since d >= 3). This produces a compactly supported integral (n+1) =
d-current S in M with dS = T and
M(S) <= c_(d-1) M(T)^(d/(d-1)) . (8.2)
Step 2: identify S with [[Omega]]. Let Z = S - [[Omega]]. Since
dS = T and d[[Omega]] = T, we have dZ = 0. Both S and [[Omega]] have
compact support, so Z does. By [L19] applied to Z in the connected,
noncompact, oriented Cartan-Hadamard manifold M, Z = 0. Hence
S = [[Omega]] and M(S) = M([[Omega]]) = vol(Omega).
Step 3: derive (8.1) by algebra. Substitute vol(Omega) = M(S) into
(8.2):
vol(Omega) <= c_(d-1) area(dOmega)^(d/(d-1)) .
Raise both sides to the power (d-1)/d:
area(dOmega) >= c_(d-1)^(-(d-1)/d) vol(Omega)^((d-1)/d) . (8.3)
Step 4: identify the coefficient. Using c_n = 1/((n+1) s_n^(1/n))
with n = d-1,
c_(d-1)^(-(d-1)/d)
= ( d s_(d-1)^(1/(d-1)) )^((d-1)/d)
= d^((d-1)/d) s_(d-1)^(1/d) .
The sphere-area identity s_(d-1) = d omega_d of section 1.5 gives
d^((d-1)/d) s_(d-1)^(1/d)
= d^((d-1)/d) ( d omega_d )^(1/d)
= d^1 omega_d^(1/d)
= d omega_d^(1/d) .
Substituting in (8.3) proves (8.1). End of proof of [T2].
We have now proved [T2] for every dimension d >= 3. What about d = 2?
[T2-2d] (Cartan-Hadamard surface inequality). Let M be a
Cartan-Hadamard surface, that is, a complete simply connected
Riemannian surface of sectional curvature at most zero. Let Omega be
a bounded domain with smooth boundary a Jordan curve gamma. Then
L(gamma)^2 >= 4 pi A(Omega) . (8.4)
Equality holds exactly when Omega is a flat disk bounded by a
Euclidean circle.
This is the classical two-dimensional case of the Cartan-Hadamard
conjecture. It was proved by Weil (1926) and by Beckenbach and Rado
(1933), decades before higher dimensions were attacked, and is
standard textbook material in differential geometry. We cite it
rather than reprove it. Section 0.5 has the historical references.
Why this case is not a consequence of [T1].
For d = 2, the proof of [T2] for d >= 3 would apply [T1] with
n = d-1 = 1. But the sharp filling inequality is FALSE at n = 1.
Section 3.5 gives the example: in a tree, a 1-cycle is a pair of
points at some distance L, and the cheapest filling is the segment
between them, of mass L. The sharp bound at n = 1 would be
L <= c_1 L^2 , c_1 = 1 / ((1+1) s_1^(1/1)) = 1/(4 pi) ,
that is, L <= L^2/(4 pi). For L = 1 this reads 1 <= 1/(4 pi) ~ 0.08,
which is false. For small L the sharp filling inequality fails
badly.
The issue is that a CAT(0) space can be badly non-Riemannian (a
tree, a building, an arbitrary CAT(0) complex), and in such spaces
the filling problem differs from the classical surface-domain
problem. But a Cartan-Hadamard surface IS a Riemannian surface: it
has a smooth metric, geodesics, curvature, covariant derivatives,
and all the analytical and differential-geometric machinery.
Section 3.5 already noted the contrast: the exponent (n+1)/n is
false at n = 1 in an arbitrary CAT(0) space, but the domain
inequality is TRUE in a smooth surface. In other words, the d = 2
domain inequality is a strictly "smoother" statement than a
filling inequality at n = 1 would be.
The classical proof of [T2-2d] uses exactly that smoothness. In
outline, it conformally uniformizes M to the unit disk:
M ~ D , ds_M^2 = lambda(z)^2 (du^2 + dv^2) , z = u + iv,
with log lambda subharmonic in D (this is the curvature condition
K <= 0 in conformal coordinates). The Riemann map sends Omega to D
itself (uniformization plus Caratheodory). Writing the area and
length in conformal coordinates reduces [T2-2d] to a Wirtinger
inequality for the boundary circle: for any subharmonic log lambda,
( int_0^{2 pi} lambda(e^{i theta}) d theta )^2
>= 4 pi int_0^1 int_0^{2 pi} lambda(re^{i theta})^2
r d theta dr . (8.5)
The elementary proof of (8.5) uses the fact that r |--> (1/2 pi)
int_0^{2 pi} log lambda(re^{i theta}) d theta is monotone increasing
and convex in log r, followed by two applications of Cauchy-Schwarz
(or, more elegantly, by a Cauchy-Crofton formula in conformal
coordinates due to Santalo). See Beckenbach (1937), Beckenbach-Rado
(1933), or the survey in section 0.5.
What this means for the file. We have proved [T2] for d >= 3 and
cited it (as the classical Weil-Beckenbach-Rado theorem) for d = 2.
So [T2] holds in every dimension d >= 2. The d = 2 case is not a
weaker statement than the general one in a technical sense, but it
lives on a genuinely smoother class of spaces, so we treat it as
the "smooth surface" boundary of the filling machinery. For our
purposes: [T2] in every dimension d >= 2, sharp constant, is
proved or cited.
Statement. Fix n >= 2. For every real c < c_n there exist a proper
CAT(0) space X and an integral n-cycle T in X with
V(T) > c M(T)^((n+1)/n). So the constant c_n in [T1] cannot be
lowered: [T1] is sharp.
Proof. Take X = R^(n+1), and let T = d[[B(R)]] be the boundary of the
Euclidean ball of radius R, weighted 1 on the sphere S^n(R). Then
M(T) = s_n R^n , V(T) = omega_(n+1) R^(n+1) . (8.8)
By [L19] with d = n+1 applied to R^(n+1), any integral (n+1)-current
S with dS = T and compact support satisfies S = [[B(R)]]. To see
this: the difference Z = S - [[B(R)]] is a compactly supported
top-dimensional cycle in the noncompact connected oriented manifold
R^(n+1), so Z vanishes. Hence V(T) = omega_(n+1) R^(n+1) is
attained, and no cheaper filling exists.
Now compute with c = c_n. Using s_n = (n+1) omega_(n+1) from
section 1.5,
c_n M(T)^((n+1)/n)
= s_n^((n+1)/n) / ((n+1) s_n^(1/n)) * R^(n+1)
= s_n / (n+1) * R^(n+1)
= omega_(n+1) R^(n+1)
= V(T) .
So c_n M(T)^((n+1)/n) = V(T) exactly. If c < c_n, then
c M(T)^((n+1)/n) < c_n M(T)^((n+1)/n) = V(T) ,
so V(T) <= c M(T)^((n+1)/n) fails. End of proof.
The computation in numbers. At n = 2, c_2 = 1/(6 sqrt(pi)) and
s_2 = 4 pi. A round circle of radius R bounds a disk of mass
M(T) = s_2 R^2 = 4 pi R^2, so
c_2 M(T)^(3/2)
= (4 pi R^2)^(3/2) / (6 pi^(1/2))
= 8 pi^(3/2) R^3 / (6 pi^(1/2))
= (4/3) pi R^3
= vol of the ball of radius R in R^3.
Equality case. In R^(n+1), equality in [T1] holds precisely for the
round boundary of a Euclidean ball of any radius R > 0, with
multiplicity 1: the constant c_n is scale-invariant, so all radii
give equality. No other cycle in R^(n+1) attains equality, by the
classical characterisation of equality in the Euclidean isoperimetric
inequality (a smooth argument suffices; see Almgren 1986 for the
current-theoretic sharp statement). In a general CAT(0) space we do
not claim a classification of equality cases, and none is needed.
In particular, [T2] is also sharp: the round Euclidean ball inside
M = R^d attains equality in (8.1), so the coefficient d omega_d^(1/d)
cannot be lowered.
This closes both directions. The constant c_n is exactly the best
possible coefficient, and [T1] gives V(T) <= c_n M(T)^((n+1)/n) for
every proper CAT(0) space and every integral n-cycle T. Sharpness is
then automatic by the flat-ball witness above.