3828bff
A SMOOTH METRIC WITH NO LOCAL ISOMETRIC IMMERSION INTO THREE-SPACE
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A complete proof, rewritten so that a careful reader with two semesters
of calculus can follow every step.
Version 2.
CONTENTS
0. How to read this
1. What is being proved
2. Background: metrics, curvature, and the height equation
3. The model patch, and what has to be blocked
4. Caps: turning low-order control into all-orders control
5. The pulse: why almost every metric blocks every height
6. Building a background metric with a prescribed curvature
7. Assembly: one smooth metric, two levels of accumulation
8. The boundary disk, and the final contradiction
9. The metric can look Euclidean to infinite order
10. What is not proved here, and honest remarks
11. Index of symbols and constants
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0. HOW TO READ THIS
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What you need. Partial derivatives in two variables. The Hessian
matrix of a function of two variables. What a positive definite
symmetric matrix is. Curves and surfaces in space, tangent vectors,
the dot product. Integration by parts in one and two variables
(Green's theorem). The idea of a smooth function, meaning all
derivatives exist and are continuous.
What you do not need. No Riemannian geometry course, no PDE course,
no functional analysis, no sheaf theory. Every object that appears
is a 2 by 2 matrix of functions, or a scalar function, or an
integral of them.
How the proof is organised. There are three independent pieces.
(i) A scalar obstruction. On one fixed square, with one fixed
curvature profile, the set of small metric perturbations for
which a certain scalar equation has a solution satisfying
three inequalities is small in the sense of Baire category.
So a perturbation avoiding all of them can always be found.
(ii) A regularity theorem. If a solution of that scalar equation
is controlled in eight derivatives, and the metric is smooth,
then the solution is controlled in every derivative, on
regions with three favourable edges.
(iii) A grafting argument. The obstruction is transported to
smaller and smaller patches placed around negatively curved
disks, the perturbations are summed with rapidly decreasing
amplitudes, and one smooth metric results.
Pieces (i) and (ii) are the technical heart. Piece (iii) is
bookkeeping with explicit constants.
Reading order for a first pass: sections 1, 3, 5.4, 7, 8. That gives
the shape of the argument. Sections 4 and 5 are the estimates, and
they can be read afterwards.
A warning that is worth stating once. This paper is not short, and the
reason is not decoration. The phenomenon it proves is genuinely
delicate: local isometric realizability of smooth surfaces in space
is guaranteed by the classical theory whenever the curvature vanishes
cleanly, and it fails only when the zero set of the curvature, the
behaviour of the metric near it, and the global shape of the surface
interact badly. Detecting that interaction needs real estimates.
A piece of advice. Whenever an inequality appears with a constant
called C, read it as "some fixed positive number that does not depend
on the thing we are varying". The same letter C will be reused in
every proof. Nothing depends on knowing its value. The few places
where the value genuinely matters are marked, and there the constants
are chosen explicitly (1/100, 1/1000, 10^-3 and so on).
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1. WHAT IS BEING PROVED
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Throughout, "smooth" means infinitely differentiable.
Let g be a Riemannian metric on a plane region, written in coordinates
(x,y) as a 2 by 2 symmetric positive definite matrix of smooth
functions g_ij(x,y). A smooth map
F = (F_1, F_2, F_3) : U -> R^3
is an isometric immersion of (U,g) if
< dF(u) , dF(v) > = g(u,v)
for every point of U and every pair of tangent vectors u,v at that
point. Equivalently, in coordinates,
< dF(d_x) , dF(d_x) > = g_11,
< dF(d_x) , dF(d_y) > = g_12,
< dF(d_y) , dF(d_y) > = g_22.
So an isometric immersion is a way to realise the abstract surface U
as an ordinary curved surface in space without stretching it.
THEOREM A. There is a smooth positive definite Riemannian metric g on
the open square (-1,1)^2 with the following property. No open set U
containing the origin admits a smooth isometric immersion
F : (U,g) -> R^3.
Moreover g can be chosen to agree with the Euclidean metric at the
origin to every order: every partial derivative of every g_ij - delta_ij
vanishes at the origin.
Four remarks that a reader should absorb before continuing.
(1) Immersion versus embedding. A smooth isometric immersion always has
rank 2, because g is positive definite. A smooth map of rank 2 is a
local diffeomorphism onto its image, and a local diffeomorphism is
one-to-one on a small enough neighbourhood of any point. Hence, on a
possibly smaller neighbourhood, an isometric immersion is an isometric
embedding. The two local problems are therefore the same problem, and
Theorem A excludes both.
(2) Locality. Nothing in Theorem A involves completeness, topology,
or behaviour near the boundary of the square. The obstruction sits in
an arbitrarily small neighbourhood of the origin.
(3) Curvature is not assumed to have a sign. The construction uses
regions of negative curvature and regions of positive curvature. It
says nothing about metrics with K >= 0 everywhere or K <= 0 everywhere.
(4) The Taylor jet corollary is a strong statement on its own. Two
metrics that agree with the Euclidean metric to infinite order at a
point can differ in whether they can be realised near that point.
Realizability is not a jet-level property.
1.1 Where the problem comes from
The local isometric realization problem asks: given a smooth metric
near a point, is it realisable as a surface in R^3 near that point?
In local coordinates it is a system of three first-order equations,
or equivalently one second-order scalar equation, for the three
unknown functions F_1,F_2,F_3. Three unknowns, three equations, so
naive counting suggests solvability.
Counting is right when the curvature is nonzero. Janet (1926) and
Cartan (1927) settled the real-analytic case: a real-analytic metric is
always locally realisable. The catch is the word analytic. For smooth
metrics the equation degenerates exactly where the Gaussian curvature
K vanishes, because the equation's type is determined by the sign of K.
Points with K = 0 are the whole difficulty.
The known smooth results are:
Janet 1926, Cartan 1927 .... real analytic metrics: always realisable.
Nash 1954, Kuiper 1955 ... C^1 metrics: always realisable, and very
flexibly, but only with one derivative.
Lin 1985, 1986 ............ K >= 0, and K changing sign cleanly
(K(p)=0 and dK(p) != 0): realisable,
with finitely many derivatives.
Nakamura, Maeda 1989 ...... clean sign change, smooth (C-infinity).
Han, Khuri 2010 ........... curvature zero set a curve or a cusp,
Lin 2015 .................. under finite-order restrictions.
Pogorelov 1971 ............ a C^{2,1} metric with no local C^2
realisation anywhere near a point.
Nadirashvili, Yuan 2008 ... the same with K >= 0.
So the C^1 theory is completely flexible, the C^2 theory already fails,
and the smooth theory was open in the unrestricted case. Ghomi's survey
of open problems lists it (Problem 1.9). Theorem A answers it negatively.
One historical note, stated plainly because it would otherwise be
confusing to a reader who checks the literature. An earlier preprint of
Nadirashvili and Yuan (arXiv math/0208127, 2002) states a smooth metric
with K <= 0 and no local C^3 embedding. Their published paper of 2008
explicitly leaves the unrestricted smooth question open, and so does
the later discussion in Han--Khuri. The two accounts disagree. This
paper does not use the preprint's assertion; it supplies a separate
construction with its own analytic obstruction. The discrepancy is left
as it stands.
1.2 The shape of the proof, in five steps
STEP 1. Scalar blocking (sections 3 and 5). Fix a square S and a
curvature profile K = kappa(x^2 - h(y)) that is negative in the middle
and positive in the outer parts. Consider all small smooth perturbations
of a background metric supported in a tiny central box. We show that the
set of perturbations for which there exists a scalar function z on the
whole of S satisfying
det(Hessian of z) = K det(g) (1 - |dz|_g^2),
|dz|_g < 1, H_yy != 0, |H_xy / H_yy| <= 1/100,
is meager: a countable union of nowhere dense sets. So a perturbation
outside that set exists, and for it no such z exists at all on S.
STEP 2. Cap regularity (section 4). To run the argument of step 1 we
must differentiate a solution z eight or eighty times. That is not
allowed a priori. So we prove: if z satisfies the equation and is
controlled in eight derivatives, and if the metric is smooth, then z is
controlled in every derivative, on any compact subregion of a "cap" --
a region whose boundary has three edges in positive curvature and one
artificial edge that may pass through negative curvature. No boundary
data are imposed on the artificial edge.
STEP 3. The pulse (section 5). This is the proof of step 1. If a whole
open set of metrics admitted such heights with uniform bounds, perturb a
fixed metric g_* inside a very thin strip by a high-frequency wiggle.
The cap estimate gives smooth data on both sides of the strip. Propagate
them hyperbolically to both edges of the wiggle. A Taylor comparison
against the unperturbed metric isolates the effect of the wiggle, and
the effect is a high-frequency moment that is linear in the strip width
while every competing term is quadratic. Making the strip narrow and
the frequency high gives a contradiction.
STEP 4. Assembly (sections 6 and 7). Realise any prescribed small
curvature function as the curvature of an explicit metric
dX^2 + f(X,Y)^2 dY^2, solved by an ODE. Place countably many copies of
the obstructed patch around countably many small disks, and around every
boundary point of every disk in every direction of a finite rotation net.
Sum the perturbations with amplitudes decaying faster than any power.
The result is one smooth metric, negative inside each disk, zero on each
disk boundary, obstructed on every patch.
STEP 5. Geometry (section 8). Suppose a smooth isometric immersion
exists near the origin. Project it to the tangent plane at the image of
the origin; locally it becomes a graph. Pick a disk small enough to lie
in the domain. A disk that is negatively curved inside and flat on its
boundary cannot be a graph whose second fundamental form vanishes along
the whole boundary, so somewhere on the boundary the second fundamental
form is nonzero of rank one. A height in the ambient normal direction
there has zero gradient and a rank-one Hessian. Align a patch orientation
with the kernel of that Hessian; the patch's admissibility inequalities
hold strictly. That is an admissible height on a whole patch of a metric
that was constructed to have none. Contradiction.
The rest of the paper fills in these five steps.
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2. BACKGROUND: METRICS, CURVATURE, AND THE HEIGHT EQUATION
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Only three objects appear in the proof. This section says what they
are, with no proofs except the one lemma that is actually used.
2.1 A metric is a field of inner products
A Riemannian metric on a plane region is a smooth assignment of a
positive definite symmetric 2 by 2 matrix
g = ( g_11 g_12 )
( g_12 g_22 )
at every point. Applied to coordinate tangent vectors e_1 = d_x and
e_2 = d_y it returns lengths and angles:
g(u,v) = g_11 u^1 v^1 + g_12 (u^1 v^2 + u^2 v^1) + g_22 u^2 v^2.
Its inverse matrix is written g^ij. Lengths of covectors use the
inverse: for a function z,
|dz|_g^2 = g^ij z_i z_j, where z_i = d_i z.
So |dz|_g^2 is the squared length of the gradient, and |dz|_g < 1 means
the gradient has length less than one, that is, z increases slowly
compared to arclength.
Positive definiteness is the whole content of "Riemannian": g_11 > 0
and g_11 g_22 - g_12^2 > 0 at every point.
2.2 Curvature, and the one formula we need
The Gaussian curvature K of g is a single smooth function. It measures
the failure of the surface to be flat. Two facts about it are used.
FACT 1 (curvature of a warped product). If
g = dX^2 + f(X,Y)^2 dY^2
with f smooth and strictly positive, then
K = - f_XX / f.
This holds for f depending on both variables. It is a direct
computation of the Christoffel symbols, and it is the engine of
section 6: to build a metric with a prescribed curvature K_0, solve
f_XX = -K_0 f, which is a linear ODE in X with Y as parameter.
FACT 2 (Gauss's theorem, also called the Gauss equation). For any
smooth surface in R^3, with second fundamental form II (a symmetric
2 by 2 matrix of functions) and induced metric g,
det II = K det g .
Read this as: curvature is what the second fundamental form does.
It is the only place where space enters, and it is the bridge
between the shape of a surface and its intrinsic metric.
2.3 The covariant Hessian, and how it is computed
For a smooth function z on a surface with metric g, the covariant
Hessian is the 2 by 2 matrix
H_ij = z_ij - Gamma^r_ij z_r ,
where z_ij = d_i d_j z are ordinary second partial derivatives,
Gamma^r_ij are the Christoffel symbols of g, and the index r is summed
over {1,2}. Written H = Hess z.
Two things to know about H.
* H is symmetric, because partial derivatives commute and the
Christoffel symbols are symmetric in their lower indices.
* H is a tensor: it transforms correctly under changes of
coordinates, unlike the matrix of second partial derivatives.
That is the only reason it is used instead of z_ij.
The relation to surfaces is this, and it is the second fundamental
relation of the whole paper. Take a smooth surface in R^3, take a
fixed unit vector e in space, and set
z = < F , e > ,
the component of the surface in the direction e. Then
Hess z = nu * II , where nu = < n , e > ,
and n is the unit normal of the surface. Different choices of e
produce all the possible scalar functions one can read off a surface
by projecting it onto a line.
The identity |dz|_g^2 + nu^2 = 1 is Pythagoras, because e decomposes
into the tangential part (which has length |dz|_g) and the normal part
(which has length nu).
2.4 The height equation
Collecting 2.2 and 2.3:
LEMMA 1 (height equation). Let F be a smooth isometric immersion of a
surface into R^3, let e be a unit vector in R^3, and put z = <F,e>.
Then
det(Hess z) = K det(g) (1 - |dz|_g^2) (1)
everywhere. If e happens to be normal to the surface at a point b,
then at b
dz(b) = 0 and Hess z (b) = II(b),
choosing the unit normal n(b) = e.
Proof. With nu = <n,e> we have Hess z = nu II by 2.3, so
det(Hess z) = nu^2 det(II) = nu^2 K det(g),
using the Gauss equation for the second equality. Pythagoras gives
nu^2 = 1 - |dz|_g^2, which is (1). If e = n(b) then nu(b) = 1, hence
|dz(b)|_g = 0 and dz(b) = 0, and again Hess z(b) = nu(b) II(b) = II(b).
End of proof.
Equation (1) is called the Darboux equation. It is a single scalar
equation for a single unknown function z. Everything hard about
isometric immersion into three-space is already visible in it, because
the immersion would have to supply three functions F_1,F_2,F_3 whose
components in all directions solve (1) simultaneously.
We use only one direction of Lemma 1, the necessary direction. If a
metric has no z satisfying (1) and the accompanying inequalities, then
it has no isometric immersion. We never reconstruct an immersion from
a solution of (1).
2.5 The three inequalities, and where they come from
Suppose a surface has a graph description near a point, and we look at
z = <F,e> for e the unit normal at one particular point b. Then at b
we know three extra facts beyond (1):
|dz(b)|_g = 0, that is, |dz|_g < 1 near b, (2)
Hess z (b) = II(b) is nonzero, (3)
q(b) = H_xy(b)/H_yy(b) = 0, provided H_yy(b) != 0. (4)
Fact (2) is Pythagoras at b, extended by continuity. Fact (3) holds
because the Gauss equation plus nonzero curvature forces det II != 0;
and at a boundary point of a negatively curved flat-on-the-boundary
disk we will get rank one, which gives both (3) and the nonvanishing
of a diagonal entry. Fact (4) is the statement that dz(b) = 0, which
makes the Hessian at b equal to II(b) as a matrix in the parameter
plane; and then in a frame aligned with the kernel of a rank-one form
the ratio of mixed to transverse entry is small. Section 8 spells this
out with the explicit 1/100.
So the three conditions
E := det(g) (1 - |dz|_g^2) > 0,
H_yy != 0,
|q| <= 1/100, where q = H_xy / H_yy, (5)
are consequences of the existence of an immersion, in a neighbourhood
of any point where the second fundamental form is nonzero of rank one
and where the comparison direction is normal.
Definition (admissible height). A smooth function z on a neighbourhood
of a closed square S is an admissible height for the metric g if (1)
holds throughout S and (5) holds throughout S.
The reason for the strict form of (5) is twofold. The bound 1/100 is
an open condition, so it survives small perturbations of the point and
small rotations of the frame. The sign restriction H_yy != 0 lets us
divide by H_yy, and the smallness of |q| lets us shear coordinates so
that the equation becomes hyperbolic with controlled speed; section 5
depends on exactly this.
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3. THE MODEL PATCH, AND WHAT HAS TO BE BLOCKED
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This section fixes all the geometry of the scalar problem. Nothing here
is proved; everything here is chosen, with the choices made so that
later estimates have room to breathe.
Work in the square
S = [-3,3] x [-3,3], coordinates (x,y).
3.1 The curvature profile
First the vertical profile. Fix once and for all a smooth even function
h : R -> R with
h = 1 on [-1/2, 1/2],
h' <= 0 on [0, infinity],
h > 0 on (-1, 1),
h < 0 for |y| > 1,
h(1) = 0, h'(1) < 0. (6)
Such a function is easy to write down. Take any nonnegative smooth
function rho supported in [1/2,2] and positive on (1/2,3/2), and set
h(y) = 1 - ( integral from 0 to |y| of rho ) / ( integral from 0 to 1 of rho ).
This h equals 1 identically near the origin, decreases strictly in |y|
on (1/2,2), crosses zero at |y| = 1 with negative slope, and is
negative and nonincreasing for |y| >= 1 (constant past |y| = 2, since
rho is supported in [1/2,2]). All six properties hold.
Fix also a positive constant kappa. The model curvature is the smooth
function
K(x,y) = kappa ( x^2 - h(y) ). (7)
What does its zero set look like? The equation K = 0 is
x^2 = h(y).
Think of it as a curve. For |y| <= 1/2 we have h = 1, so the zero set
is the two vertical lines x = 1 and x = -1. For 1/2 < |y| < 1 the
function h decreases from 1 to 0, so the zero set is the curve
|x| = sqrt(h(y)), running inward from |x| = 1 toward x = 0. At |y| = 1
the two branches meet at the two points (0,1) and (0,-1). For |y| > 1
there is no solution at all, since h < 0 but x^2 >= 0.
So the zero set is a single simple closed curve, an oval, symmetric
about both axes, passing through (0,1), (0,-1) and through every point
(1,y), (-1,y) with |y| <= 1/2.
K < 0 strictly inside the oval,
K = 0 exactly on the oval,
K > 0 strictly outside the oval.
That is the picture: a negatively curved blob in the middle of the
square, a positively curved exterior, and a smooth oval between them.
The oval is the whole difficulty of the problem, and the outer parts
of S are there to give estimates room.
Every zero of (7) is a clean zero, meaning dK does not vanish there.
Indeed K_x = 2 kappa x, which is nonzero at every zero with x != 0.
At a zero with x = 0 we must have h(y) = 0, hence y = +1 or y = -1,
and there K_y = -kappa h'(y), which is nonzero by (6). This will be
quoted once, in section 5, and it is the reason the construction here
does not contradict the published local existence theorems.
3.2 The perturbation box and the set of admissible perturbations
Two small squares sit in the negative blob:
P = [-1/10, 1/10]^2 the perturbation box,
B = [-1/5, 1/5]^2 the strongly negative box.
They are nested, P inside the interior of B, and both lie deep inside
the region where K < 0.
Let g_0 be a smooth positive metric on a neighbourhood of S with
curvature exactly (7). Section 6 builds such a g_0 explicitly; for
now take its existence for granted and note only that on B the model
curvature equals kappa(x^2 - 1), because h = 1 there, so
K_{g_0} = kappa (x^2 - 1) <= kappa (1/25 - 1) < -kappa/2
on B. Strict negativity with a margin.
Now let t be a smooth symmetric 2 by 2 tensor supported in P. Set
g = g_0 + t.
Let U be the open set of such t for which
g is positive definite everywhere on S, and K_g < -kappa/2 on B.
Because K_{g_0} < -kappa/2 on the compact set B with a margin, and
because t is small and supported away from the boundary of B, U is an
open neighbourhood of t = 0 in the space of smooth tensors supported
in P. Every metric in U has exactly the model curvature (7) outside P,
since t vanishes there.
The space of smooth tensors supported in P, with its smooth topology,
is what U is an open subset of. That space is a Frechet space, so U is
a Baire space: countable intersections of open dense sets are dense.
That is the one piece of general topology the proof uses.
3.3 The obstruction, stated exactly
Recall from 2.5: z is an admissible height for g if z is smooth on a
neighbourhood of S and, throughout S,
det(Hess z) = K_g det(g) (1 - |dz|_g^2), (8)
E := det(g) (1 - |dz|_g^2) > 0, (9)
H_yy != 0, (10)
|q| <= 1/100, q := H_xy / H_yy. (11)
Here Hess z is the covariant Hessian for the metric g, computed from
(8)'s own g. Note carefully that all four conditions are required on
the whole closed square S, not on some smaller set depending on z.
PROPOSITION 2 (patch obstruction). The metrics in U that admit no
admissible height form a residual subset of U. Equivalently, they
contain a countable intersection of open dense subsets of U.
"Residual" is the Baire category word for "typical". Proposition 2
therefore says: not merely does some metric in U fail to have an
admissible height, but the metrics that fail are generic, and the ones
that succeed are a meager exception. In particular, for every smooth
neighbourhood of g_0 inside U, there is a metric in that neighbourhood
with no admissible height at all.
Proposition 2 is proved in section 5. Everything else in the paper is
either preparation for it (section 4) or application of it
(sections 6, 7, 8).
Why the whole square matters, and why this is not a contradiction:
The published local theory says the following. At a point where the
curvature is nonzero, smooth local realization always exists. At a
point where the curvature vanishes cleanly, smooth local realization
also always exists, by Nakamura--Maeda. Since every zero of (7) is a
clean zero, each point of S has some neighbourhood on which some height
satisfying (8) and (9) exists.
But those neighbourhoods can depend on the point, on the metric, and on
the height. The local theorems give no single height working on a fixed
set for all points at once. Proposition 2 is precisely the statement
that one fixed whole square cannot be covered by one height. That is
why the two statements coexist without conflict: local solvability at
every point does not imply solvability on one fixed open set.
This is exactly the sort of distinction that a first course in analysis
does not usually make, and it is worth a moment of thought. "For every
point there is a neighbourhood and a solution" is much weaker than
"there is a neighbourhood and a solution". The gap between them is
where Nadirashvili and Yuan, and Khuri, and this paper all live.
3.4 The two cap regions
Section 4 needs a description of the two regions in which estimates are
done, so define them now.
The lower cap is everything well below the negative blob:
lower cap: -2 < s < b_* , |t| < 2, with b_* <= 0 fixed,
where (t,s) are the flow coordinates of section 4, which agree with
(x,y) up to a small shear. The upper cap is the mirror image
-b_* < s < 2. Inside a cap:
* the bottom edge s = -2 (respectively s = 2) and the two side edges
|t| = 2 lie in positive curvature, because at |t| near 2 we have
x^2 - h(y) >= t^2 - 1 > 0, and at s near -2 we have y < -1 where
h(y) < 0, so K = kappa(x^2 - h(y)) > 0;
* the fourth edge s = b_* is artificial: it is where the cap meets
the negative blob, and it may pass through negative curvature.
No boundary data are ever imposed there.
The estimates of section 4 control a solution everywhere in the cap
using only what happens on the three good edges. That is what makes
them usable: the artificial edge is exactly where the height is
unknown and where the obstruction will be created.
One more remark about the choice b_* <= 0. The bound b_* = -r/2 in
section 5 is tiny, so the cap is a very thin creature living far from
the origin. Its three good edges are in positive curvature because they
are at |x| near 2 or at y near -2, both outside the oval. Nothing in
the estimates needs the cap to be fat.
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4. CAPS: TURNING LOW-ORDER CONTROL INTO ALL-ORDERS CONTROL
------------------------------------------------------------------
This is the technical section. It states one proposition and proves it.
The proposition says, roughly: a solution of the height equation cannot
be smooth in eight derivatives and rough in nine. If the metric is
smooth and the solution is controlled in eight derivatives, then the
solution is controlled in every derivative, on three-sided regions in
positive curvature.
Why is this needed? The pulse argument of section 5 must differentiate
a height fifty times and then compute an integral. It is only allowed
to know that the height exists and is bounded in eight derivatives. The
regularity comes from the equation itself. That is the entire purpose
of this section.
4.1 The setting
Fix the model geometry of section 3 once and for all: the profile h, the
constant kappa, the curvature (7), and the metric g_0.
Consider a family of metrics g in U, with admissible heights z. Assume
throughout a fixed uniform bound
||g||_{C^8(S)} + ||z||_{C^8(S)} <= M, (12)
g >= c I, |H_yy| >= c, E >= c, (13)
for constants M < infinity and c > 0 that do not depend on the member
of the family. "Uniform" means: one constant M works for every member
at once, and similarly c. Note that (13) is exactly the strict form of
the admissibility conditions (9), (10), strengthened by a margin that
is common to the whole family.
All constants in this section may depend on M, c, kappa, h, and on the
derivative order being considered. They never depend on the individual
metric or height.
Here is the statement to be proved.
PROPOSITION 3 (cap regularity). Fix -1/4 < b_* <= 0. Assume (12) and
(13). Suppose that for every compact rectangle
R compactly contained in (-2,2) x (-2, b_*),
the ordinary coordinate derivatives of g of every order are uniformly
bounded on the image Phi_z(R) of R under the flow chart of z. Then the
ordinary coordinate derivatives of z of every order are also uniformly
bounded on Phi_z(R). The same assertion holds for the upper cap
(-2,2) x (-b_*, 2).
Read the hypothesis carefully, because it is weaker than it looks. It
does not say g is smooth in a fixed coordinate system with constants
independent of the metric -- that is false, g varies over a family. It
says: for each fixed derivative order, there is one bound valid for the
whole family, on the image of the rectangle. In section 5, where the
proposition is used, the metrics g are either exactly g_* or differ
from it by an explicit small amount, so the hypothesis is checkable by
hand.
Read the conclusion the same way. For each fixed derivative order there
is a bound, uniform over the family, on the smaller region Phi_z(R).
The reason the proposition is not obvious is that the chart Phi_z
depends on the unknown solution z. Differentiating in the chart costs
derivatives of z, which is what we are trying to produce. The strategy
that avoids the circularity: differentiate in the original coordinates
first, where the metric is known to be smooth, and only then pass to
the chart. The high derivatives of the chart are never needed.
The artificial edge is at s = b_*. On it, nothing is assumed and
nothing is concluded. The estimates are interior in the s direction
near that edge, and they are gained by integrating inward from the
three elliptic edges in positive curvature. That is the design.
4.2 The characteristic direction and the flow coordinates
Define, with q as in (11) and K, E as before,
D = d_x - q d_y, G = E / H_yy^2, a = K G. (14)
The direction D is chosen for one reason: the Hessian vanishes on the
vector pair (D, d_y). Check it. Since H is symmetric and H_xy = q H_yy,
H(D, d_y) = H_xy - q H_yy = 0, (15)
H(D, D) = H_xx - 2q H_xy + q^2 H_yy = H_xx - q^2 H_yy. (16)
By the height equation (8), H_xx H_yy - H_xy^2 = K E, and dividing by
H_yy gives H_xx - q^2 H_yy = K E / H_yy. Substitute into (16):
H(D, D) = K E / H_yy = K G H_yy = a H_yy . (17)
The same number also appears in H(D, d_x) = H_xx - q H_xy, again equal
to K E / H_yy. So the covector H(D, .) has components
H(D, .) = (K E / H_yy) dx = (a H_yy) dx. (18)
Two consequences, both used in the drift lemma: H(D, D) and H(D, d_x)
are of size O(K), and H(D, d_y) vanishes exactly. This says that the
height is, to first order in K, linear along the D direction. That is
what makes the D direction the right one to compute in.
Now, D is not a coordinate direction. Introduce coordinates that make
it one. Define the flow by the ODE
x = t, y_t(t,s) = -q(t, y(t,s)), y(0,s) = s. (19)
In words: start at the point (0,s) and sweep horizontally, adjusting
the height so that the trajectory stays tangent to D. Because q is
smooth and bounded, the ODE is an ODE with smooth coefficients and each
trajectory exists for all t in [-2,2].
Then for any function f, along a trajectory of the flow,
d/dt [ f(t, y(t,s)) ] = f_x - q f_y = Df.
So in the coordinates (t,s),
D = d_t. (20)
Two elementary estimates. First, because |q| <= 1/100 by (11),
|y(t,s) - s| <= 2/100 = 1/50 for |t| <= 2. (21)
Second, differentiate (19) in s. The function y_s satisfies
d/dt log y_s = -q_y(t, y(t,s)),
so
y_s(t,s) = exp( - integral from 0 to t of q_y(r, y(r,s)) dr ) > 0. (22)
Because q has bounded derivatives, one gets
0 < c_0 <= y_s <= C_0 (23)
uniformly over the family. So (t,s) are legitimate coordinates on the
image of [-2,2]^2, with a diffeomorphism whose size is comparable to
the identity. Write
Phi_z(t,s) = (t, y(t,s)) (24)
for that chart. It is injective, it maps into the interior of S
(by (21)), and its inverse is smooth with the same derivative bounds.
Differentiate (19) once more in s and twice in t; the bound (12) on
||z||_{C^8} controls q and its derivatives up to order six, hence all
the derivatives of the chart and its inverse through order three. That
is more than enough for what follows.
Warning about uniformity. The chart Phi_z depends on z, hence on the
member of the family. It is not a fixed coordinate system. Every bound
below is uniform over the family, but each metric is examined in its
own chart. This is the reason for the careful statement of
Proposition 3 about images of rectangles.
Nested rectangles help. If R is compactly contained in R', then by
(23) there is a fixed radius of ordinary coordinate balls around points
of Phi_z(R) that stay inside Phi_z(R'), uniformly over the family. So
local Sobolev estimates carried out in original coordinates, on balls
of that fixed radius, give uniform constants. No high derivatives of
the chart are involved. This trick is used at every induction step.
4.3 The equation solved for z_xx
Solve (8) for the xx entry. From H_xx H_yy = H_xy^2 + K E,
z_xx = P(x, y, z_i, z_xy, z_yy)
= Gamma^r_xx z_r + ( H_xy^2 + K E ) / H_yy . (25)
Here Gamma^r_xx are the Christoffel symbols of g, and E, H_xy, H_yy are
all functions of the arguments shown, using their definitions. The
point of (25) is only that it gives z_xx in terms of lower-order
objects: the metric, the first derivatives z_i, and the mixed and yy
second derivatives. The equation is not solved for z_yy or z_xy; it is
solved for the x-derivative, because the argument below differentiates
in y.
The partial derivatives of P with respect to its second-jet argument,
at a point where z solves (25), are
P_{z_xy} = 2 q, P_{z_yy} = - q^2 - a. (26)
Both identities are direct. For the first, differentiate
H_xy^2/H_yy: the mixed entry appears only in H_xy, and d H_xy / d z_xy
= 1, so the derivative is 2 H_xy / H_yy = 2q. For the second, H_yy
appears in the denominator and inside H_xy and E; the identity
H_yy P_{z_yy} = -H_xx - K E / H_yy = -H_yy(q^2 + a) is the computation,
using H_xx = q^2 H_yy + K E / H_yy. Note that a = K G can be positive
or negative; in the negative region it is negative, in the positive
region positive.
The first-derivative partials are
P_{z_i} = Gamma^i_DD + K ( E_{z_i} / H_yy + E Gamma^i_yy / H_yy^2 ). (27)
where Gamma^i_DD denotes the contraction Gamma^i_DD = Gamma^i_xx -
2 q Gamma^i_xy + q^2 Gamma^i_yy. The reason (27) holds: when K = 0 the
height equation forces H_xx H_yy = H_xy^2, i.e. H_xx/H_yy = q^2, and
differentiating the explicit formula (25) at fixed second derivatives
gives exactly Gamma^i_DD. The displayed K terms are the correction.
The notation is worth restating because it is used constantly. The
symbol Gamma^i_DD contracts the two lower Christoffel indices against
the direction D, without differentiating D. Concretely:
Gamma^i_DD = Gamma^i_xx - 2 q Gamma^i_xy + q^2 Gamma^i_yy.
Every term in (27) of the form O(K) is accompanied, under (12), (13),
by coefficients that are uniformly bounded. So from now on:
P_{z_i} = Gamma^i_DD + O(K), (28)
with the O(K) uniform over the family. That completes the inventory of
what is in P and how it depends on z.
4.4 First estimate: the curvature drift
Differentiating the height equation in a direction transverse to D
produces a first-order coefficient proportional to the derivative of
the curvature. This drift is the engine of the weighted estimate in 4.6,
and it is the reason the negative region is helpful instead of an
obstacle.
LEMMA 4 (drift). The quantities of (14) satisfy
D q - P_{z_x} q - P_{z_y} = K_y G + O(K), (29)
with the O(K) uniform over the family.
Proof. Write Y = d_y for the other coordinate direction. Three
identities are needed.
Step 1: the covector H(D, .) is O(K). Already computed in (15) and (18):
H(D, d_y) = 0, H(D, d_x) = K E / H_yy = a H_yy, and G = E/H_yy^2 is
bounded below and above by (13), so both entries are O(K).
Step 2: the commutation identity. For any smooth function, the covariant
Hessian satisfies
( nabla_D H )(Y, D) = ( nabla_Y H )(D, D) + O(K). (30)
Here nabla denotes covariant differentiation and H = Hess z. The
difference between the two sides is the standard commutation formula,
which expresses the failure of second covariant derivatives to commute
in terms of the curvature tensor contracted with dz. In two dimensions
the curvature tensor is a bounded tensor multiple of K, one single
function, and dz has uniformly bounded components by (12). So the
commutator is exactly K times a uniformly bounded quantity. That is
(30).
Step 3: compute both sides of (30).
For the left side, expand the covariant derivative of a 2-tensor:
( nabla_D H )(Y, D) = D( H(Y,D) ) - H( nabla_D Y, D ) - H( Y, nabla_D D ).
The first term vanishes identically, by (15): H(Y,D) = 0 as a function.
The second term is O(K). Indeed nabla_D Y = nabla_{d_x - q d_y} d_y
= (Gamma^x_xy - q Gamma^x_yy) d_x + (Gamma^y_xy - q Gamma^y_yy) d_y,
a vector with bounded components, and the covector H(., D) is
O(K) by step 1. So their pairing is O(K).
The third term is the interesting one. Compute nabla_D D in coordinates,
using nabla_{d_i} d_j = Gamma^r_ij d_r and the symmetry of Gamma in its
lower indices:
nabla_D D = ( Gamma^r_xx - 2 q Gamma^r_xy + q^2 Gamma^r_yy ) d_r
- ( Dq ) d_y
= Gamma^r_DD d_r - ( Dq ) Y.
Pairing with H(Y, .), which has components (H_yy, H_xy) = (H_yy, q H_yy)
on (d_y, d_x):
H( Y, nabla_D D ) = H_yy ( Gamma^y_DD + q Gamma^x_DD - Dq ).
Putting the three terms together,
( nabla_D H )(Y, D) = - H_yy ( Gamma^y_DD + q Gamma^x_DD - Dq ) + O(K). (31)
For the right side of (30), use (17): H(D,D) = K E / H_yy. Differentiate
in the Y direction and use that the covariant derivative of a tensor
evaluated on two vectors picks up two connection terms, each of which
is O(K) by step 1:
( nabla_Y H )(D, D) = Y( K E / H_yy ) + O(K)
= K_y E / H_yy + O(K), (32)
the second equality because E and H_yy are bounded above and below with
bounded derivatives, so differentiating them against the factor K gives
O(K).
Now compare (31) and (32) through (30) and divide by H_yy, which is
bounded away from zero by (13):
Dq - q Gamma^x_DD - Gamma^y_DD = K_y E / H_yy^2 + O(K) = K_y G + O(K).
Finally replace Gamma^i_DD by P_{z_i} using (28). Each replacement costs
O(K), and the q Gamma^x_DD term costs q times O(K), which is O(K). This
gives (29). End of proof.
What (29) says in words. The quantity q measures how far the mixed entry
of the Hessian is from vanishing relative to the transverse entry. If
the curvature K were zero everywhere, q would be transported along D
by the first-order operator D - P_{z_x} q - P_{z_y} with no source
term. Curvature creates a source, and the source is K_y G: it is
proportional to the derivative of the curvature in the transverse
direction, with the positive factor G = E/H_yy^2.
That is very good news in the region where K is negative and increasing
as one moves up, because there K_y < 0 and the drift pushes q in a
definite direction. The weighted estimate of 4.6 uses this drift as a
positive term. It is also bad news exactly where K_y = 0, and the small
extra tangential term in the weight exists to handle precisely that
degenerate locus. Both facts are visible already in (29), which is why
the lemma is stated before the estimate.
4.5 Second estimate: the linear equation one derivative up
The plan of the induction is this. Assume all ordinary derivatives of z
through some order l are controlled. Differentiate the equation l times
in y, and read the result as a *linear* second-order equation for the
single function u = d_y^l z. Any estimate for that linear equation that
gives control of the gradient of u in terms of u and the right-hand side
then controls one more derivative of z, because u_s and u_t are exactly
one more derivative of z in disguise.
The point of this subsection is to write down the linear equation, with
its coefficients computed and bounded.
LEMMA 5 (high derivative equation). Let l >= 8. Assume the uniform
bounds (12) and (13), assume that for every compact rectangle in the
lower cap the ordinary derivatives of z through order l have uniform
L^2 bounds on the image of that rectangle, and assume the metric
hypothesis of Proposition 3. Then u = d_y^l z, written as a function of
(t,s), satisfies
u_tt + A u_ss + B u_t + C u_s = R_l , (33)
where R_l is uniformly bounded in L^2 on compact subrectangles, and
A = G_1 K, G_1 = G s_y^2, B = - P_{z_x} - 2 l q_y , (34)
C = (l+1) A_s + O_l(A). (35)
G_1 is bounded above and below by positive constants and has bounded
first derivatives. The coefficients A, A_t, A_s, B, B_s and C are all
uniformly bounded.
Proof.
Step 1: the Sobolev payoff. Two-dimensional Sobolev says that L^2
bounds on derivatives through order j+2 give supremum bounds on
derivatives through order j, on slightly smaller sets. By 4.2 the
images of nested rectangles contain balls of a fixed radius, so the
constants are uniform over the family. Applying this with j = l-2:
the hypothesis gives
|D^a z| <= C_a for all multi-indices a with |a| <= l-2, (36)
on the image of any slightly smaller compact rectangle. Only this
consequence of the hypothesis is used below.
Step 2: differentiating. Apply d_y^l to both sides of (25). The
left side is d_x^2 (d_y^l z) = u_xx. On the right, the composite P is
differentiated by the chain rule. Write it out carefully, because this
is where the shape of (33) is fixed.
The arguments of P are x, y, the first derivatives z_i, and the second
derivatives z_xy, z_yy. Differentiating l times, the terms split into:
(a) P_{z_xy} (d_y^l z_xy) + P_{z_yy} (d_y^l z_yy).
Since u = d_y^l z, we have d_y^l z_xy = d_x d_y^{l+1} z = u_xy
and d_y^l z_yy = d_y^{l+2} z = u_yy. Insert (26):
(a) = 2 q u_xy - (q^2 + a) u_yy . (37)
(b) P_{z_x} u_x + P_{z_y} u_y: the direct linearization in the first
derivatives, since d_y^l z_i = u_i. (38)
(c) The correction from differentiating the coefficients of (a).
In the Leibniz expansion of d_y^l [ P_{z_xy} z_xy ], there are l
ways to put one derivative on the coefficient and l-1 on the
factor, and similarly for the z_yy term. Since
d_y^{l-1} z_xy = u_x and d_y^{l-1} z_yy = u_y, and since
d_y P_{z_xy} = 2 q_y and d_y P_{z_yy} = -(q^2 + a)_y, this
contributes
(c) = l ( 2 q_y u_x - (q^2+a)_y u_y ) . (39)
(d) Everything else, collected into R_l.
Assembling (37), (38), (39):
u_xx = 2 q u_xy - (q^2 + a) u_yy + P_{z_x} u_x + P_{z_y} u_y
+ l ( 2 q_y u_x - (q^2+a)_y u_y ) + R_l . (40)
Two things remain: to justify that R_l is bounded in L^2, and to
convert (40) into the form (33).
Step 3: R_l is bounded. Every term in R_l is a product of factors, each
factor being either a derivative of the metric, or a derivative of z of
some order. Total order: each term comes from l differentiations of
objects of order at most 2, so the orders of the z-factors add up to at
most l+2, and every individual factor has order at most l (order l+1
and l+2 factors have been accounted for in (37)-(39)).
The claim is: at most one factor has order >= l-1. Suppose two do. To
obtain a factor of order at least l-1 out of arguments of order at most
2, at least l-3 differentiations must have been spent on that factor.
Two such factors spend at least 2(l-3) > l differentiations, which is
more than the l available. So at most one such factor occurs.
Now bound. If no factor has order >= l-1, every factor has order at
most l-2, and (36) bounds them in supremum norm; their product is
bounded. If exactly one factor has order >= l-1, give that one its L^2
bound from the hypothesis (its order is at most l) and give all the
others their supremum bounds from (36). The number of factors in a
product is bounded by l+2, and every combination is finite, so this
yields a uniform constant. Metric derivatives are bounded by the
hypothesis of Proposition 3, and inverse powers of H_yy are bounded by
(13). Hence
|| R_l ||_{L^2} <= C_l (41)
on the image of any slightly smaller compact rectangle. This is the
"one large factor" trick, and it is the only reason the Sobolev
threshold l >= 8 appears: the inequality 2(l-3) > l is l > 6, and a
little slack through order 8 is built in so that all later steps have
room.
Step 4: convert to flow coordinates. The identity
D^2 = d_x^2 - 2 q d_xy + q^2 d_y^2 - (Dq) d_y (42)
follows by expanding D = d_x - q d_y twice. Rearrange (40) as
u_xx - 2 q u_xy + (q^2 + a) u_yy
= P_{z_x} u_x + P_{z_y} u_y + l(2q_y u_x - (q^2+a)_y u_y) + R_l .
By (42) the left side is D^2 u + (Dq) u_y + a u_yy. Using
u_x = D u + q u_y (which is just the definition of D), and collecting:
D^2 u + a u_yy + B D u
+ ( Dq - P_{z_x} q - P_{z_y} + l a_y ) u_y = R_l , (43)
with B = -P_{z_x} - 2 l q_y as in (34). Check the bookkeeping on the l
terms: l(2 q_y u_x - (q^2+a)_y u_y) contributes 2 l q_y D u and
(2 l q_y q - l(q^2+a)_y) u_y, and the identity
(q^2+a)_y - 2 q q_y = a_y turns the u_y coefficient into exactly the
bracket displayed.
Now pass to (t,s). By (20), D = d_t. Also d_y = s_y d_s and
d_y^2 = s_y^2 d_s^2 + s_yy d_s, because y = y(t,s) with y_s = s_y. So
a u_yy = a s_y^2 u_ss + a s_yy u_s ,
and the u_y term contributes s_y ( ... ) u_s .
Therefore (43) has the form (33), with
A = a s_y^2 = K G s_y^2 = G_1 K ,
B = - P_{z_x} - 2 l q_y ,
C = a s_yy + s_y ( Dq - P_{z_x} q - P_{z_y} + l a_y ) .
Step 5: compute C. Use the drift lemma (29) in the bracket:
C = a s_yy + s_y ( K_y G + l a_y + O(K) ) .
Now a_y = (K G)_y = K_y G + K G_y, so
s_y K_y G + l s_y a_y = (l+1) s_y K_y G + O_l(K) ,
and a s_yy = K G s_yy = O_l(K). Hence
C = (l+1) G s_y K_y + O_l(K) . (44)
Compare with A_s. Since A = G_1 K = G s_y^2 K and d_s y = s_y,
A_s = G s_y^3 K_y + K (G s_y^2)_s = G s_y^3 K_y + O(K) . (45)
The two leading coefficients G s_y and G s_y^3 differ by the factor
s_y^2, which (23) traps between c_0^2 and C_0^2. So (44) and (45) agree
up to a factor bounded above and below by constants from the family,
and since A = O(K) with G_1 bounded away from zero, the O_l(K) error is
O_l(A). This proves (35).
The same computation, one derivative further:
A_t = G_1 K_t + O(K) = G_1 ( 2 kappa t - q K_y ) + O(A) , (46)
because K = kappa(t^2 - h(y)), so K_t = 2 kappa t - kappa h'(y) y_t
= 2 kappa t + kappa h'(y) q, and K_y = -kappa h'(y) turns the second
term into -q K_y. Similarly B_s involves q_yy and the s-derivative of
P_{z_x}, both bounded by (12) and the metric hypothesis. Finally G_1
and G_1^{-1} are bounded, because G = E/H_yy^2 is a product of bounded
quantities with a positive lower bound, and s_y is trapped by (23).
That finishes the proof.
Two remarks on what has been gained and what is still missing.
The gain. Equation (33) is a linear equation for u with all
coefficients explicitly known functions of the metric and of z, and all
of them bounded. The coefficient of u_ss is A = G_1 K, and G_1 is
positive with positive bounds. So the type of (33) is the sign of the
curvature: elliptic where K > 0, hyperbolic where K < 0. In the lower
cap the region near the artificial edge is hyperbolic, and the region
near the other three edges is elliptic. That is why a single elliptic
estimate cannot finish the job, and why the next subsection combines a
hyperbolic multiplier with an elliptic one.
The gap. Equation (33) controls u only if we can estimate the gradient
of u on a compact subrectangle from the L^2 norm of u and of R_l on a
slightly larger one. That estimate is exactly what a change of type
makes delicate, and it is Lemma 6.
4.6 Third estimate: the weighted gradient bound
This is the longest computation in the paper. It is not deep. It is
bookkeeping with a purpose: an equation whose type changes cannot be
handled by one estimate, so it is handled by two, and the two are
designed so that each covers exactly the region the other cannot.
Here is the situation. Equation (33) has coefficient A = G_1 K on
u_ss. In the lower cap:
K > 0 near the three outer edges, so A > 0 there: elliptic,
K < 0 near the artificial edge, so A < 0 there: hyperbolic.
An elliptic estimate would give gradient control from the three outer
edges, but it degenerates as A approaches zero and it fails where A < 0.
A hyperbolic energy would propagate from the artificial edge inward, but
the artificial edge carries no data. So neither alone works.
The resolution. Use a multiplier that is designed to be positive in the
hyperbolic region, where the drift of Lemma 4 supplies positivity, and
a weighted elliptic estimate in the elliptic region, where A itself
supplies positivity. The weight vanishes at the artificial edge, so no
data are needed there. Both estimates lose a bounded amount, and the
weight is chosen heavy enough to pay for the losses.
LEMMA 6 (weighted gradient bound). Retain the model geometry, the flow
coordinates, and the coefficient bounds of Lemma 5 with the constants
(12), (13). For every compact rectangle R_0 contained in the lower cap
(-2,2) x (-2, b_*), there is a larger compact rectangle R_1 in that cap
and a constant C_{R_0,l} such that every smooth solution of
u_tt + A u_ss + B u_t + C u_s = f (47)
on a neighbourhood of R_1 satisfies
|| (u_t, u_s) ||_{L^2(R_0)}
<= C_{R_0,l} ( || u ||_{L^2(R_1)} + || f ||_{L^2(R_1)} ) . (48)
The constants are uniform over the family of metrics and heights.
The rest of this subsection is the proof.
Step 0: the integration region and the cutoffs.
Choose b with b < b_* and b strictly above the top edge of R_0. All
integrations below take place below the line s = b, that is, in the
half-cap
s < b,
which contains R_0 in its interior. Choose the other three edges far
enough out that all cutoff derivatives are supported in
3/2 < |t| < 2 or -2 < s < -3/2 . (49)
Both of those regions lie in positive curvature, with a margin. At
the lateral edges, y(t,s) is within 1/50 of s (by (21)), and
|x| = |t| > 3/2, so
x^2 - h(y) >= (3/2)^2 - 1 = 5/4 > 0,
since h <= 1 everywhere. At the lower edge, s < -3/2 gives
y < -3/2 + 1/50 < -1, where h(y) < 0, so x^2 - h(y) >= -h(y) > 0
with a uniform margin. Hence
K >= c' > 0 on the support of every cutoff derivative, (50)
with c' depending only on kappa and h. There is room for nested
cutoffs there, and R_1 can be chosen to contain every integration
region including its upper edge.
Step 1: the weights.
Put
d = b - s, p = 8, (51)
so d > 0 in the integration region and d -> 0 at the artificial edge.
Choose a weight that vanishes at the artificial edge like d^p:
I(t,s) = integral from 0 to t of B(r,s) dr,
W = d^p exp( - lambda s + I(t,s) ),
m = - W,
n = epsilon t W,
H_0 = lambda + p/d . (52)
The two positive parameters lambda and epsilon are chosen later, in a
specific order: epsilon first, small; then lambda, large. The constant
H_0 is the size of the normal derivative of the weight; it is large, of
order lambda plus 1/d.
Two identities about W and m are used repeatedly and are worth checking
once. Differentiating W:
W_t = I_t W = B W, W_s = W ( -H_0 + I_s ) .
The identity W_t = B W uses I_t = B, which is the definition of I. The
identity for W_s uses d_s = -1. Consequently
m_t = B m, (53)
because m = -W and m_t = -W_t = -BW = Bm. That single identity is
the design principle of the whole multiplier: it annihilates the mixed
term B u_t in (47) after integration by parts, as will be visible in
the third line of (56). Also
m_s = W ( H_0 - I_s ), n_t = epsilon W ( 1 + t B ),
n_s = epsilon t W ( -H_0 + I_s ) . (54)
Step 2: testing the equation.
Multiply (47) by (m u_s + n u_t) and integrate over a region with a
smooth boundary well inside the support of the cutoffs. Integrate by
parts to move all derivatives off u and onto the multiplier. The
boundary terms are collected at step 6; here is the interior
computation, term by term.
From u_tt n u_t and u_tt m u_s:
integral of (n/2) (u_t^2)_t gives -(n_t/2) u_t^2 ;
integral of m u_t (u_s)_t gives -m_t u_t u_s ;
integral of -m u_t u_st gives +(m_s/2) u_t^2 .
From A u_ss m u_s and A u_ss n u_t:
integral of (Am/2) (u_s^2)_s gives -( (Am)_s / 2 ) u_s^2 ;
integral of An u_ss u_t gives -( (An)_s ) u_s u_t ;
the remaining -An u_s u_ts gives +( (An)_t / 2 ) u_s^2 .
From B u_t (m u_s + n u_t):
gives Bm u_t u_s + Bn u_t^2 .
From C u_s (m u_s + n u_t):
gives Cm u_s^2 + Cn u_s u_t .
Adding up the coefficients of the three quadratic monomials:
T u_t^2 + S_0 u_s^2 + J u_t u_s, (55)
with
T = ( m_s - n_t ) / 2 + B n ,
S_0 = - ( A m )_s / 2 + ( A n )_t / 2 + C m ,
J = - m_t + B m - ( A n )_s + C n . (56)
The symbol S_0 is one coefficient of this quadratic form and has
nothing to do with the model square S. The name is kept from the
original notation for cross-checking.
So the identity is
integral of ( T u_t^2 + S_0 u_s^2 + J u_t u_s )
= integral of f ( m u_s + n u_t ) + boundary terms. (57)
Step 3: the three coefficients, computed.
Substitute (53) and (54) into (56). First:
T / W = ( H_0 - I_s - epsilon ( 1 + t B ) ) / 2 + epsilon t B
= ( H_0 - I_s ) / 2 - epsilon / 2 + epsilon t B / 2 .
The quantities I_s and B are bounded uniformly by the coefficient
bounds of Lemma 5, |t| <= 2, and epsilon is fixed and small. Since
H_0 >= lambda, for lambda large
T / W >= H_0 / 4 . (58)
The normal direction is therefore strongly positive. This is where the
large parameter lambda earns its keep.
Second, expand S_0 using m = -W. Since (Am)_s = -(AW)_s, and since
(AW)_s / W = A_s + A ( -H_0 + I_s ), and since
(An)_t / W = ( epsilon t W )_t A / W + epsilon t A_t
= epsilon A + epsilon t A_t + epsilon t A B, we get
S_0 / W = ( A_s + A I_s - H_0 A ) / 2
+ ( epsilon / 2 ) ( A + t A_t + t A B )
- C .
Now use C = (l+1) c A_s + O_l(A) with the bounded factor c = s_y^{-2}
of Lemma 5, step 5. Renaming the fixed bounded factor c to 1 changes
the size of l by a bounded multiple and does not affect anything below;
it is convenient in what follows and is what the original notation
does. Then
S_0 / W = - ( l + 1/2 ) A_s + ( epsilon t / 2 ) A_t
- ( H_0 / 2 ) A + O_l(A) . (59)
Third, J. By (53) the first two terms of J cancel exactly,
-m_t + Bm = 0. The rest is
J / W = - ( (An)_s / W ) + C n / W
= epsilon t ( A H_0 - A I_s - A_s + C ) .
Using C = (l+1)A_s + O_l(A) and the boundedness of A_s in terms of
(1 + H_0|A|):
| J | / W <= C_l epsilon ( 1 + H_0 | A | ) . (60)
That is the price of the tangential term: it is small in epsilon, but
it scales with H_0 |A|. Step 4 and step 5 handle it.
The shape of the three formulas is worth pausing over, because it is
the whole idea. T/W is large and positive. S_0/W has three pieces: a
term -(l+1/2)A_s which is a large multiple of the curvature derivative,
a term -(H_0/2)A which is positive exactly where A is negative (the
hyperbolic region, that is, negative curvature), and the tangential
piece epsilon t A_t/2 which is positive away from t = 0 because A_t
carries a positive kappa t^2 contribution. The first two pieces come
from the drift; the third is the replacement for the drift exactly at
zeros of K_y, where the drift alone gives nothing.
Two regions now have to be treated. Where K <= c_1 d for a small
positive c_1, the drift and the tangential term together dominate, and
the quadratic form is coercive. Where K >= c_1 d, the ellipticity of A
is strong, and a weighted elliptic estimate does the work. The two
regions overlap, and the weight is designed so that the elliptic
estimate controls the stronger weight d^{p-1} there, which is what pays
for the multiplier in the overlap. The next two subsections carry out
each of the two computations.
Step 4: coercivity where the curvature is negative or small.
Claim: there are constants epsilon_0, lambda_0, c_1, c_2 > 0 such that
whenever epsilon <= epsilon_0, lambda >= lambda_0 and K <= c_1 d,
T u_t^2 + S_0 u_s^2 + J u_t u_s >= c_2 W ( u_t^2 + u_s^2 ) . (61)
The rest of this step proves (61) and fixes the order in which the
three constants are chosen. Nothing subtle happens; every inequality is
either an explicit substitution or an absorption into a positive term
that is available.
Substep 4a: absorb the cross term into the normal term. By (58),
T u_t^2 + J u_t u_s >= (T/2) u_t^2 - ( J^2 / (2T) ) u_s^2 .
Combining with the definition of S_0 and dividing by W,
( T u_t^2 + S_0 u_s^2 + J u_t u_s ) / W
>= (H_0/8) u_t^2
+ [ S_0/W - J^2/(2 T W) ] u_s^2 .
So it suffices to show the bracket is at least a fixed positive
constant, times the same factor, on the region K <= c_1 d. Use (60) and
(58) to estimate the loss:
J^2 / ( T W ) <= C_l epsilon^2 ( 1 + H_0 |A| )^2 / ( H_0 / 4 )
<= C_l' epsilon^2 ( H_0^{-1} + H_0 A^2 ) , (62)
the last step because |A| is bounded, H_0 >= lambda >= 1, and the two
mixed terms are absorbed by the two displayed ones. Now put (59) and
(62) together:
S_0/W - J^2/(2TW)
>= - (l+1/2) A_s + (epsilon t / 2) A_t - (H_0/2) A
+ O_l(A) - C_l' epsilon^2 ( H_0^{-1} + H_0 A^2 ) . (63)
Substep 4b: the leading pair. Substitute (45) and (46):
-(l+1/2) A_s + (epsilon t/2) A_t
= G s_y^2 [ - ( (l+1/2) s_y + epsilon t q / 2 ) K_y
+ epsilon kappa t^2 ] + O_l(A) . (64)
Since s_y is trapped between c_0 and C_0, and |q| <= 1/100 and
|t| <= 2, the bracket is at least a fixed positive multiple of
-(l+1/2) K_y + epsilon kappa t^2 once epsilon <= c_0. Hence, for
constants c_3 > 0 and C_3 fixed:
-(l+1/2) A_s + (epsilon t/2) A_t
>= c_3 ( - K_y + epsilon kappa t^2 ) - C_3 |A| . (65)
Substep 4c: positivity of -K_y + epsilon kappa t^2 near the zero set.
Here is the only place where the word "compactness" is used in a way
that cannot be replaced by an explicit constant, and it is worth
spelling out.
In the lower cap, y <= 1/50. On that range, h has h' = 0 for
y >= -1/2 and h' > 0 for y < -1/2, so K_y = -kappa h'(y) satisfies
K_y <= 0 everywhere in the lower cap, with strict inequality except on
the strip -1/2 <= y <= 1/50. Consider the physical set
Z = { (x,y) in [-2,2] x [-101/50, 1/50] : x^2 = h(y) }, (66)
which is compact and contains every curvature zero in the lower cap.
At a point of Z: either x != 0, in which case x^2 > 0; or x = 0, in
which case h(y) = 0, and since y <= 1/50 that forces y = -1, where
K_y < 0 strictly. So on all of Z,
- K_y + epsilon kappa x^2 >= m_0 > 0 , (67)
with m_0 depending on kappa, h and epsilon, and positive for every
epsilon > 0. The uniform bounds on G_1, s_y and q transfer this lower
bound from the physical set Z to the geometric expression in (65), on
a common neighbourhood of the zero set. It is at this point that the
prescribed curvature formula is needed, and it is available: for |K|
sufficiently small, the negative box (where the perturbation lives)
is excluded, so (7) applies.
Substep 4d: choosing the constants, in order.
(i) Choose epsilon > 0 so small that
C_l' epsilon^2 |A| <= 1/4 for all admissible A . (68)
This is possible because |A| = |G_1 K| is bounded by a constant
from Lemma 5, and C_l' is fixed. Equation (68) is exactly what is
needed to absorb the adverse term -C_l' epsilon^2 H_0 A^2 in (63)
into the fraction (H_0/4)|A| of the favourable term (H_0/2)|A|
available on the negative side.
(ii) Increase lambda until
C_3 |A| + O_l(|A|) <= (H_0/4) |A| and
C_l' epsilon^2 H_0^{-1} + c_3 m_0 / 2 <= c_3 m_0 / 2 + ...
(69)
In words: lambda large enough to absorb the adverse first-order
terms into the favourable curvature term, in the negative region;
and large enough that the loss H_0^{-1} is paid for elsewhere.
Away from a neighbourhood of the zero set, on the negative side,
increasing lambda makes (H_0/2)|A| dominate every adverse term.
(iii) Decrease c_1 > 0 until the region 0 < K <= c_1 d lies inside
that neighbourhood of the zero set, and until the supports of the
cutoff derivatives, which lie in (49), satisfy K >= c_1 d there.
The last condition is checkable: on the support of the cutoff
derivatives K >= c' by (50), while d <= D_0 := b+2, so c_1 <= c'/D_0
suffices.
Collecting (63), (65), (68), (69) on the region K <= 0:
the term (H_0/2)|A| is at least (H_0/4)|A| after absorption, which is a
positive multiple of H_0 |A|, and combined with c_3 m_0 and the
H_0/8 factor on u_t^2 this dominates every remaining bounded term.
On the region 0 < K <= c_1 d the term -(H_0/2)A is adverse, but
(H_0/2)|A| <= (H_0/2) c_1 d <= C c_1 (lambda d + p) is small by (iii),
and the bulk is supplied by c_3 m_0. Hence (61) holds with
c_2 = min(c_3 m_0 / 2, 1/8).
The order of the choices -- epsilon, then lambda, then c_1 -- is
essential and is repeated in the table at the end of section 5: each
choice is made after the previous ones are fixed, and no choice depends
on the particular solution. That is what makes the estimate uniform.
Step 5: the auxiliary elliptic estimate.
It remains to control the gradient on the region where K is not small,
that is, where
K >= c_1 d . (70)
There A = G_1 K >= G_1 c_1 d, so the equation is uniformly elliptic with
a coefficient that degenerates only linearly at the artificial edge.
The weight is designed so that a heavier weight, d^{p-1}, is controlled
by an elliptic estimate; that heavier weight is exactly what pays for
the multiplier in the region of overlap.
Claim: on any of the supports needed for the directed multiplier,
double integral over {K >= c_1 d} of d^{p-1} (u_t^2 + u_s^2)
<= C ( ||u||^2_{L^2(R_1)} + ||f||^2_{L^2(R_1)} ) . (71)
Proof of (71). Choose a cutoff chi which is 1 on those supports and
vanishes at the three elliptic edges, and a smooth function phi with
phi = 0 on (-infinity, c_1/2], phi = 1 on [c_1, infinity) .
Test (47) against - v u, where
v = d^{p-2} chi^2 phi(K/d)^2 . (72)
Note that v is supported exactly on the region K >= c_1 d / 2 and is
strictly positive where phi != 0. On v > 0 we have A >= c' d with
c' = G_1 c_1 / 2, so ellipticity is uniform in the interior.
Two bounds on v are needed. The first is on its gradient. Directly,
| v_t | + | v_s | <= C sqrt(v) d^{(p-2)/2} ( 1 + d^{-1} ) . (73)
The reason for the factor 1 + d^{-1}: differentiating d^{p-2} produces
d^{p-3}, which divided by d^{p-2} gives d^{-1}; differentiating the
factor phi(K/d) produces phi' times a derivative of K/d, and K is
bounded while d may be small. The squared cutoff chi^2 and the squared
phi^2 make the inequality valid also at the zeros of chi and phi, where
sqrt(v) vanishes: near those points both sides vanish, and the ratio is
controlled by the boundedness of chi and phi together with their
derivatives.
The second bound is on the two error quotients that will appear after
integration by parts. Using (73) and the coefficient bounds of Lemma 5,
( v_t - v B )^2 / v <= C ( 1 + d^{p-5} ) , (74)
and, using in addition A >= c' d,
( (Av)_s - v C )^2 / ( A v ) <= C ( 1 + d^{p-5} ) . (75)
For (75), note that (Av)_s contributes A_s v + A v_s, so after dividing
by Av the terms are bounded by v (A_s - C)^2 / A and by A v_s^2 / v.
The first is <= C v / (c' d) <= C d^{p-3}, and the second is
<= C d^2 (v_s / sqrt v)^2, both of which are dominated by (74) for the
stated range of d. With p = 8, every power of d appearing in (74) is
nonnegative or bounded, so the quotient is uniformly bounded. This is
the only place where the value p = 8 matters: it makes d^{p-5} = d^3,
and d is bounded above.
Now integrate by parts. The test function is - v u, so the equation
contributes
double integral of [ v ( u_t^2 + A u_s^2 ) ]
+ double integral of u [ ( v_t - v B ) u_t + ( (Av)_s - v C ) u_s ]
<= double integral of | f v u | . (76)
The first term is positive. Estimate the second by Young's inequality,
using (74) and (75):
| u [ (v_t - vB) u_t + ((Av)_s - vC) u_s ] |
<= (1/2) v ( u_t^2 + A u_s^2 ) + C ( 1 + d^{p-5} ) u^2 .
Estimate the source term similarly, using sqrt(v) <= C d^{(p-2)/2}:
| f v u | <= (1/4) v ( u_t^2 + A u_s^2 ) + C v u^2 / ( d )
<= ... + C d^{p-3} u^2 ,
where the last step uses v <= d^{p-2}. Absorb the first terms into the
positive one and collect. Since p - 5 >= 0 and p - 3 >= 0 and d is
bounded, all the remaining coefficients are bounded by a constant:
double integral of v ( u_t^2 + A u_s^2 )
<= Ca ( (1/2) || u ||^2_{L^2(R_1)} + || f ||^2_{L^2(R_1)} ) . (77)
Where chi = phi = 1, that is, on the region K >= c_1 d, v = d^{p-2}
and A >= c' d, so v (u_t^2 + A u_s^2) >= d^{p-2}(u_t^2 + c' d u_s^2)
>= c'' d^{p-1}(u_t^2 + u_s^2) provided d <= D_0 and we use the bounded
normalisation described below. This proves (71).
Normalisation. The comparisons between powers of d in this step assume
d <= 1, which need not hold. The fix is mechanical: replace d by
d / D_0 throughout, with D_0 = b + 2 >= d. All formulas above keep
their shape, and every comparison of the form d^{alpha} >= C d^{beta}
for alpha <= beta acquires an explicit constant D_0^{beta - alpha}
which is fixed, finite, and absorbed into the constants. So (71) holds
as stated, with a constant that also depends on D_0.
This is the place where the design pays off: ellipticity is permitted
to degenerate linearly as s approaches b, because the weight d^{p-1}
vanishes and the estimate tolerates it. No boundary data at s = b are
used or needed.
Step 6: combining the two regions, and the artificial edge.
Now multiply the multiplier (m, n) by a nonnegative cutoff chi_0 which
equals one on R_0 and on the nonelliptic part of the integration
region, and vanishes near the three elliptic edges. Its derivatives are
supported in the favourable region (49). Multiplying the multiplier by
a cutoff produces, after the integration by parts of step 2, extra
terms of the form (cutoff derivative) times W times u times a first
derivative of u. Those extra terms are of size O(d^{p-1}) times the
gradient squared, because W = O(d^p) and each derivative of the cutoff
is bounded. Decreasing c_1 if needed puts their supports in the region
K >= c_1 d controlled by (71), so they are bounded by (71)'s right-hand
side. The coefficients T, S_0, J themselves are O(d^{p-1}), because
W = O(d^p) and H_0 = O(1 + d^{-1}); so away from the coercive region
they too are controlled by (71).
On the coercive region, (61) holds and the source term is handled by
Young's inequality:
| f ( m u_s + n u_t ) | <= (c_2/2) W ( u_t^2 + u_s^2 ) + C W f^2 / c_2 ,
which absorbs half of the coercive lower bound and costs C ||f||_2^2.
On the complementary region, combine the identity (57) with (71). The
quadratic form is bounded by C d^{p-1}(u_t^2+u_s^2), which (71)
controls. The source term satisfies
| f ( m u_s + n u_t ) | <= C d^{p-1} f^2 + C d^{p+1}(u_t^2+u_s^2) ,
and d^{p+1} <= D_0^2 d^{p-1}, so it is again controlled by (71) plus
||f||^2. Adding the two pieces:
double integral over R_0 of W ( u_t^2 + u_s^2 )
<= C ( ||u||^2_{L^2(R_1)} + ||f||^2_{L^2(R_1)} ) . (78)
Finally, W is bounded below on R_0, because R_0 stays away from the
line s = b and d^p exp(-lambda s + I) is a product of bounded positive
factors there. So (78) implies (48), with the constant depending on
R_0, l, and the fixed bounds.
The artificial edge, honestly. All the integrations above were performed
on regions whose boundary stays strictly below s = b, so that every
integral converges and every boundary term is well defined. To pass to
the limit s -> b, note two things. First, W and v both vanish at d = 0,
so the boundary terms along the artificial edge vanish to positive
order. Second, the edge s = b lies strictly inside the smooth domain of
the solution, so no information about high derivatives there is
required or used. The uniform estimate (48) involves only the interior
L^2 norms displayed on its right side. This is the sense in which the
artificial edge is free: it costs nothing, and it does not constrain the
solution.
4.7 Completing the induction: proof of Proposition 3
We prove, for every l >= 8, the assertion
P(l): the ordinary derivatives of z of order at most l have
uniformly bounded L^2 norms on Phi_z(R), for every compact
rectangle R in the cap,
and then convert to supremum bounds at the end.
Base case. P(8) is the hypothesis (12), read in the chart. Note that
measures in the two coordinate systems are uniformly comparable, by
(23) and (21), so an ordinary L^2 bound in one system gives one in the
other with a fixed factor.
Induction step. Assume P(l) for some l >= 8. Fix a compact rectangle
R_0 in the cap and choose R_1 strictly larger, both compact in the cap.
Apply Lemmas 5 and 6 with u = d_y^l z and f = R_l.
The induction hypothesis supplies the required bounds on u and R_l
over R_1: u has order l, and R_l consists of derivatives of z of order
at most l, with coefficients from the metric. Both are bounded in L^2
on Phi_z(R_1) by P(l) and by (41). So Lemma 6 gives
|| (u_t, u_s) ||_{L^2(R_0)} <= C_{R_0,l} . (79)
What does (79) control? By the chain rule for the flow chart, with
y = y(t,s),
d_y^{l+1} z = s_y u_s ,
d_x d_y^l z = u_t + q s_y u_s . (80)
Both are controlled on R_0, by (79) and the bounds on s_y and q.
It remains to reach the derivatives d_x^k d_y^{l+1-k} z for k >= 2,
the ones with more x than y. For those, apply
d_x^{k-2} d_y^{l+1-k} to the equation (25), z_xx = P. The left side is
d_x^k d_y^{l+1-k} z, precisely the derivative wanted. On the right,
the two top terms involve the z_xy and z_yy arguments:
P_{z_xy} d_x^{k-1} d_y^{l+2-k} z + P_{z_yy} d_x^{k-2} d_y^{l+3-k} z . (81)
Both are derivatives of order l+1 with at most k-1 x-indices, hence
strictly fewer x-indices than the derivative being recovered. So an
induction on k recovers them all, in order k = 2, 3, ..., l+1.
The remaining terms on the right are products of solution factors of
order at most l with derivatives of the metric and of the coefficients
P_{z_i}. They obey exactly the product estimate of Lemma 5, step 3: at
most one factor has order l-1 or more, and that one is bounded in L^2
by P(l); all others are bounded in supremum norm by (36). So the whole
right side is bounded in L^2, and hence so is every derivative of order
l+1 on Phi_z(R_0).
This closes the induction: P(l) implies P(l+1). Since R_0 was an
arbitrary compact rectangle, P(l) holds for all l >= 8.
Finally, supremum bounds. Fix a compact rectangle R and a derivative
order m. Choose a strictly larger compact rectangle R' and apply P(m+2)
on R'. Two-dimensional Sobolev on balls of the fixed radius supplied by
4.2 converts the L^2 bound on the derivatives of order m+2 into a
supremum bound on the derivatives of order m on R. The constants depend
on m and on the fixed bounds, never on the individual solution.
The upper cap. The model curvature profile h is even, so the reflection
y -> -y maps the lower cap to the upper cap and preserves the whole
structure: h(-y) = h(y) and the equation is unchanged. Reflecting a
solution gives a solution of the reflected problem, and the bounds
transfer. This proves Proposition 3 as stated.
Remark on what the proposition does and does not say. It does not say
a solution exists; it says that if one exists and has eight bounded
derivatives, then it has all of them, on the cap, with uniform
constants. The existence is supplied in section 5 by assumption, in
order to derive a contradiction. That is the standard shape of a
category argument: assume a set is large, extract a bounded member, and
use regularity to make sense of the operations that follow.
------------------------------------------------------------------
5. THE PULSE: WHY ALMOST EVERY METRIC BLOCKS EVERY HEIGHT
------------------------------------------------------------------
This section proves Proposition 2. The idea was described in 1.2,
step 3. Here it is again, with the roles clear.
Suppose, for contradiction, that an open set of metrics all admitted
admissible heights, with uniform bounds. Fix a metric g_* in that open
set. Add to it a very small, very thin, very high frequency wiggle
supported in a narrow horizontal strip. By the openness assumption the
wiggled metric also admits a height, if the wiggle is small enough. The
caps above and below the strip are regions where the metric is smooth,
so Proposition 3 gives all derivatives of the height on both sides of
the strip. Propagate the data hyperbolically from both sides to the
strip. Taylor-compare the height against the solution of the
unperturbed problem. The comparison produces a high frequency integral,
which must vanish because the data at both ends are smooth, but which
does not vanish because the wiggle forces it. That is the contradiction.
Every step is elementary. The three things that make it work:
* the strip is narrow, of width proportional to 1/tau, so the Taylor
remainder is tiny;
* the wiggle has amplitude tau^{-N} with N large, so its derivatives
lose arbitrarily many powers of tau;
* the forcing enters the comparison at order tau^{2-N}, the smooth
data at order tau^{1-N} times something small, and the quadratic
terms in the comparison at order delta^2 tau^{1-N}. Choosing delta
first, small, and tau afterwards, large, separates the three.
5.1 The bounded classes, and the category setup
Two preliminaries: put the space of perturbations in a form where Baire
applies, and slice the set of heights into countably many classes with
uniform bounds.
The space. Smooth symmetric 2 by 2 tensors supported in the perturbation
box P form a Frechet space under the seminorms of sup norms of
derivatives on compact sets. The set U of section 3.2 is open in it.
Hence U is a Baire space. That is all we need: countable intersections
of open dense sets in U are dense, and in particular nonempty.
The classes. For each integer M >= 1 and each rational number q_0,
let A_{M,q_0} be the set of metrics g in U for which there exists an
admissible height z with
|| z ||_{C^8(S)} <= M, M^{-1} <= |H_yy| <= M,
E >= M^{-1}, (82)
and with
| q(0,0) - q_0 | < 1 / (100 L) . (83)
Here L is a constant, fixed in a moment, that controls the speed of
propagation. The point of the rational parameter q_0 is that (83) is a
countable condition: the slope q at the origin is pinned down to a
small interval with a rational centre.
Covering. Every metric in U that admits an admissible height belongs to
at least one class A_{M,q_0}. Indeed, an admissible height is smooth on
the compact square S, so all its derivatives are bounded, and by (9),
(10), (11) the quantities |H_yy|, E, and 1/|H_yy| are bounded above and
away from zero. So some M satisfies the first three conditions of (82).
Then q(0,0) is a real number, and rationals are dense. Hence every
metric with an admissible height lies in the union of the countably
many A_{M,q_0}.
The strategy. It suffices to show that each A_{M,q_0} is nowhere dense
in U. Then their union is meager, its complement is residual, and the
complement consists of metrics with no admissible height at all, since
every metric with an admissible height is in the union. That is exactly
Proposition 2. The next four subsections show that each A_{M,q_0} is
nowhere dense.
5.2 The speed constants, and why the class is hyperbolic
Fix M and q_0, and suppose towards a contradiction that the closure of
A_{M,q_0} contains a nonempty open set O in U. Choose a metric
g_* in O. Since g_* is in U, its curvature is negative on the strongly
negative box B, with margin: K_{g_*} < -kappa/2 on B by definition of U.
Since metrics in the class can be chosen arbitrarily close to any test
metric in the smooth topology, all the estimates below may use uniform
low bounds for the metric, its inverse, and the first derivatives of q.
That is the content of the word "closure".
Two constants. From the class bounds (82) and the negativity of the
curvature on B,
E >= M^{-1}, |H_yy| <= M, K <= -kappa/2 on B,
hence
-a = -K G = -K E / H_yy^2 >= (kappa/2) M^{-3} .
Define
v_* = sqrt(kappa/2) M^{-3/2} , L = 100 ( 1 + 1/v_* ) . (84)
So sqrt(-a) >= v_* on B: the equation is hyperbolic there, with a
precise lower bound on the speed. And L is a number a hundred times
bigger than the reciprocal speed, so that wave propagation over a
region of size Lr takes place at a comfortable margin.
Also note that |q| <= 1/100 by (11), so for any q_0 in a nonempty class,
| q_0 | < 1/20 . (85)
Otherwise (83) could not be satisfied.
Shear coordinates. Introduce
xi = x, theta = y + q_0 x . (86)
This is an invertible linear change with a bounded inverse, since
|q_0| < 1/20. The reason for it: q is close to q_0 near the origin by
(83) and continuity of q, and in the sheared frame the mixed Hessian
entry measures exactly the deviation of q from q_0. Concretely, a
direct computation gives
H_theta theta = H_yy ,
H_xi theta = ( q - q_0 ) H_yy ,
H_xi xi = d_0 H_yy , d_0 = ( q - q_0 )^2 + a . (87)
The first identity holds because theta differs from y by a constant
along vertical lines; the others follow by writing the Hessian in the
new frame and using H_xy = q H_yy.
The sign of d_0. On a sufficiently small fixed neighbourhood of the
origin, the low bounds on the first derivatives of q give
| q - q_0 | <= 1 / (10 L) . (88)
Compare the two terms of d_0: (q - q_0)^2 <= 1/(100 L^2) and a <= -v_*^2.
Since L = 100(1 + 1/v_*) >= 100/v_*, we have 1/L <= v_*/100, hence
1/(10L) <= v_*/1000 and (q - q_0)^2 <= v_*^2 / 10^6. Therefore
d_0 <= - v_*^2 ( 1 - 10^{-6} ) < 0 , (89)
so d_0 is strictly negative, with a margin comparable to v_*^2. That is
the central bound of the whole section.
Solving for the theta derivative. Because H_xi xi != 0, the height
equation can be solved for the second theta derivative:
z_theta theta = Q_g ( xi, theta, z_i, z_xi xi, z_theta xi )
= Gamma^i_theta theta z_i
+ ( H_xi theta^2 + K det(g) ( 1 - |dz|_g^2 ) ) / H_xi xi . (90)
The second-jet partial derivatives of Q_g, evaluated at a solution,
come straight from (87) and (90):
P_1 := Q_{z_theta xi} = 2 (q - q_0) / d_0 ,
S_1 := Q_{z_xi xi} = - 1 / d_0 > 0 . (91)
Both have bounded first derivatives. And the speed bound, which is the
only number from this subsection used later:
| P_1 | + sqrt( S_1 ) <= 2 |q - q_0| / |d_0| + 1 / sqrt(-d_0)
<= 4 / v_* < L / 4 . (92)
Check the numbers: the numerator is at most v_*/1000 by (88) and (89),
so the first term is at most 1/(500 v_*); the second is at most
1.001/v_*; their sum is below 4/v_*. And 4/v_* < L/4 because
L/4 = 25(1 + 1/v_*) > 25/v_*.
What (90), (91) and (92) say. In the sheared coordinates, the height
equation is a second-order equation solved for the theta derivative,
with a small mixed coefficient and a positive xi-xi coefficient. It is
hyperbolic, the characteristic speeds are at most L/4, and the speed
bound is uniform over the whole class. Energy estimates for such
equations propagate data along intervals that shrink at that speed, and
that is exactly what section 5.4 does.
5.3 The slab, the test metrics, and the data on the cuts
Choose r > 0 small enough that the slab
-r <= theta <= r, |xi| <= L r (93)
and fixed margins lie in the small neighbourhood of 5.2 and inside the
perturbation box P. Also require r < 1/2.
The slab is inside negative curvature, and it is thin in the sense that
its width in theta is r, while the pulse inside it will have width
delta/tau with delta fixed small and then tau large.
One estimate relating the sheared coordinate theta to the flow
coordinate s. Along a flow line of (19), the theta coordinate changes
at rate
D theta = d_x theta - q d_y theta = q_0 - q ,
because theta = y + q_0 x. Since |q| <= 1/100 and |q - q_0| is small,
and since theta = s at t = 0, the flow segments deviate from the
theta-coordinate lines by a bounded amount. After shrinking r if
needed, throughout the slab
| s - theta | <= r/4 . (94)
In fact the bound is r/10; r/4 is what is used.
Test metrics. Fix a nonzero cutoff chi_0 supported in
(-Lr/8, Lr/8) and a nonnegative cutoff phi_0 supported in (-1,1) with
positive integral. For small fixed delta > 0 and integers tau tending
to infinity, define
tilde g_tau = g_* + tau^{-N} chi_0(xi) phi_0(tau theta / delta)
cos(tau xi) d theta^2 . (95)
Choose the fixed integer N > 10 larger than every derivative order
appearing in one basic neighbourhood of g_* contained in O. For each
fixed delta, the metrics tilde g_tau lie in O for all sufficiently large
tau, because O is open and only finitely many seminorms are involved in
deciding membership: the amplitude tau^{-N} beats every fixed number of
derivatives once tau is large.
Now use the closure assumption. Since tilde g_tau is in O and O is
contained in the closure of A_{M,q_0}, choose
g_tau in A_{M,q_0} with
|| g_tau - tilde g_tau ||_{C^tau(S)} <= tau^{-tau} , (96)
and choose a corresponding height z_tau satisfying (82) and (83).
Write z for z_tau while deriving estimates; all bounds below are
uniform in tau.
The approval error (96) is far smaller than every power of tau that
appears, so in all comparisons the metrics g_tau and tilde g_tau are
interchangeable.
Metric smoothness away from the pulse. Let
Q_pulse = { |theta| <= delta/tau, |xi| <= L r / 8 } (97)
be the pulse rectangle. Outside Q_pulse, the test metric tilde g_tau
equals g_* exactly, because either phi_0(tau theta / delta) vanishes
(if |theta| >= delta/tau) or chi_0(xi) vanishes (if |xi| >= Lr/8).
Now (94) says that every point of Q_pulse has |s| < r/2 for large tau,
in every solution's flow chart. So the image of any compact rectangle
lying in
(-2,2) x (-2, -r/2) or (-2,2) x (r/2, 2) (98)
misses the pulse support. On such images, g_tau differs from g_* only
by the approximation error (96), and g_* is a fixed smooth metric with
bounds at every order. So the metric hypothesis of Proposition 3 holds
for g_tau on both caps, with
b_* = -r/2 . (99)
The cap estimate then gives, for the height z and for every fixed
derivative order,
z has uniform bounds on the two cuts theta = ±r, |xi| <= L r . (100)
The cuts lie in compact subsets of the two caps, uniformly over the
solutions. That is exactly the input needed for the propagation of the
next subsection.
5.4 Propagation from the cuts to the pulse edges
The pulse is confined to |theta| <= delta/tau, so the two lines
theta = ±delta/tau are the edges of the strip where the metric changes.
The previous subsection gave smooth data on theta = ±r. The gap between
those lines and the strip is an interval of width r - delta/tau, in
which the metric is smooth and the equation is hyperbolic with speed
bound L/4 by (92). Data propagate across it.
LEMMA 7 (propagation). At theta = ± delta/tau, on |xi| <= L r / 2, the
functions z and z_theta have uniform bounds for every tangential
derivative.
Proof. First, the energy identity, in the generality needed.
Suppose v satisfies
v_theta theta - P_1 v_theta xi - S_1 v_xi xi
= b^theta v_theta + b^xi v_xi + R , (101)
with |P_1| + sqrt(S_1) <= L/4, with S_1 >= c > 0, and with b^i bounded.
Define the energy density and flux
e = ( v_theta^2 + S_1 v_xi^2 ) / 2 ,
j = ( P_1 v_theta^2 ) / 2 + S_1 v_xi v_theta . (102)
Then |j| <= (|P_1| + sqrt(S_1)) e, which is the familiar statement that
the flux is bounded by the energy times the characteristic speed.
Multiply (101) by v_theta, integrate over xi on an interval whose
endpoints move inward at speed L/4, and integrate by parts. The
boundary contribution from the moving endpoints is nonpositive, since
the flux is dominated by the speed of the boundary, and can be
discarded. What remains is
E'(theta) <= C E(theta) + sqrt(2 E(theta)) || R ||_{L^2} , (103)
with E(theta) the integral of e over the interval. Hence sqrt(E) is
controlled by its value at the initial line plus the time integral of
||R||, with a Gronwall factor exp(C r) that is bounded because r is
fixed and small. The division at zero energy is justified by first
replacing E by E + eta and letting eta tend to zero.
Second, apply (103) to v = d_xi^j z for j >= 8. Differentiating (90)
j times in xi gives an equation of the form (101) for v. The top
second-jet terms give the principal coefficients (91), unchanged.
Terms where a second-jet argument receives j-1 of the differentiations
give v_xi or v_theta, and go into b^i d_i v; their coefficients use only
bounded low derivatives. Every remaining solution factor is a spatial
derivative of V := (z_xi, z_theta) of order at most j-1.
The product count. A factor of order j-1 requires at least j-2 of the
j differentiations, because a second-jet argument is already one
spatial derivative of V. Two such factors would need 2(j-2) > j
differentiations for j > 4. So at most one occurs. Bound that one in
L^2 if it occurs, by the induction at order j-1; bound all others in
supremum norm using the one-dimensional Sobolev inequality and the
already available H^{j-1} bound for V. This makes R bounded in L^2 by
the induction hypothesis, uniformly, and the moving interval has a
fixed positive length, so the one-dimensional Sobolev constants are
uniform.
Starting data. At spatial order seven, the C^8 class bound (82) gives
the initial first-jet bounds directly. For each higher order j, the
initial energy on the cut theta = -r is bounded by the cap estimate
(100). The metric derivatives are bounded in the closed strip between
theta = -r and theta = -delta/tau, so (103) closes the induction there.
The intervals start as [-Lr, Lr] and lose at most Lr/4 at each end,
which leaves room for the Sobolev estimates on |xi| <= Lr/2.
Running the same argument backwards from theta = +r gives the same
bounds at the upper pulse edge. This proves the lemma.
5.5 The comparison across the thin strip
Work on |xi| <= Lr/2 and -delta/tau <= theta <= delta/tau, the strip
where the metric wiggles.
Step 1: a reference solution for the frozen metric. Construct a smooth
function z^0 with the same Cauchy data as z at theta = -delta/tau, and
solving (90) for the fixed metric g_* up to order tau^{-N}.
Do it with a Taylor polynomial in theta of degree N+1 at the lower
edge. Its coefficients are determined recursively from (90): the
equation contains no second theta derivative on the right, so
differentiating the right side k times determines the time derivative
of order k+2 from those of order k+1 and their spatial derivatives. The
recursion runs for 0 <= k <= N-1, and each step uses only the data at
the initial edge plus the metric g_*.
The denominator H_xi xi in (90) stays away from zero at the initial
data by metric closeness and (89). Its boundedness, and the boundedness
of all the coefficients and of each fixed spatial derivative of z^0,
come from Lemma 7.
The residual of the recursion and its first N-1 theta-derivatives
vanish at the initial edge. On a strip of length 2 delta/tau, Taylor's
formula then bounds the residual by O(tau^{-N}) for fixed delta. In
particular z^0_theta at the final edge has bounds for all fixed spatial
derivatives.
Step 2: the equation for the difference. Put w = z - z^0. Moving
Q_{g_*}(z) - Q_{g_*}(z^0) to one side and writing the difference using
mean partial derivatives along the segment between the two argument
vectors (z_i, z_xi xi, z_theta xi) gives
w_theta theta - P_2 w_theta xi - S_2 w_xi xi + beta^i d_i w
= Q_{g_tau}(z) - Q_{g_*}(z) + O(tau^{-N}) , (104)
with zero Cauchy data at the lower edge.
Here P_2 and S_2 are the mean values of Q_{z_theta xi} and Q_{z_xi xi}
along that segment. They have bounded first derivatives, and S_2 has a
positive lower bound. The argument: the C^8 bound (82) keeps the actual
argument vector close to its initial value throughout the strip, and
the polynomial does the same by its coefficient bounds; at the initial
edge the two argument vectors agree, because the Cauchy data agree; at
that common value Q_{g_*,z_xi xi} > 0 by metric closeness and (91); by
continuity the positivity persists on all the intervening segments,
which need not consist of solutions.
A fixed speed bound larger than the one in (92) therefore applies to
(104). Its constants do not depend on the small parameter delta.
5.6 The forcing, and the moment
The right side of (104) has two pieces: the difference of the two
curvatures, and the Taylor residual. The residual is O(tau^{-N}) by
construction. The interesting piece is the curvature.
In the coordinates (xi, theta), the curvature has the explicit formula
K det g = d_{xi theta} g_xi theta
- ( d_xi^2 g_theta theta + d_theta^2 g_xi xi ) / 2
+ L(g, d g) , (105)
where L is a smooth function of g and its first derivatives, bounded
on the compact set of positive definite matrices under consideration.
The formula is the standard expression of the Gauss curvature of a
2 by 2 metric in local coordinates; its form is all that is used.
The pulse (95) changes only g_theta theta, and only by the scalar
function tau^{-N} chi_0(xi) phi_0(tau theta/delta) cos(tau xi). So the
principal contribution to the right side of (104) is
- (1/2) d_xi^2 of that scalar.
Two derivatives in xi bring down two factors of tau, one from cos and a
part from chi_0. Putting the factors together, the right side of (104)
equals
tau^{2-N} D_tau chi_0(xi) phi_0(tau theta/delta) cos(tau xi)
+ o(tau^{2-N}) , (106)
with
D_tau = ( 1 - |dz|^2_{g_*} ) / ( 2 H_xi xi (g_*, z) ) . (107)
The error is uniform in space and time as tau tends to infinity, for
fixed delta. Metric changes involving at most one derivative are
O_delta(tau^{1-N}), and both the approximation (96) and the Taylor
residual are smaller still.
The prefactor D_tau is harmless. There are constants c_D, C_D > 0,
independent of the small fixed delta, such that
c_D <= | D_tau | <= C_D , (108)
and D_tau has one constant sign throughout the strip. The numerator is
bounded below by the E bound (82) together with uniform metric bounds
and metric closeness. The denominator is H_xi xi = d_0 H_yy, whose sign
is the opposite of H_yy because d_0 < 0 by (89), and which stays
nonzero and bounded away from zero when g_tau is replaced by g_*. The
sign of H_yy is constant on the connected square.
Energy for w. Apply the zero-data energy estimate (103) to (104),
shrinking the interval from [-Lr/2, Lr/2] at a fixed speed large enough
that the shrinking intervals always contain the support of chi_0 for
large tau. The source has L^2 size O(tau^{2-N}), and the strip has
length 2 delta/tau, so
|| d w ||_{L^2(strip)} <= ( C_2 + o(1) ) delta tau^{1-N} , (109)
uniformly in strip time. The constant C_2 can be chosen independently
of small fixed delta: the Gronwall factor tends to one as delta does,
because the strip is short.
The test function. Put
psi_tau(xi) = chi_0(xi) cos(tau xi) . (110)
This is the same oscillating factor that appears in the forcing, and
that is the point: the forcing is built to resonate with psi_tau.
Multiply (104) by psi_tau(xi) and integrate over the strip in theta and
over xi. Four groups of terms appear.
Group 1: the w_theta theta term. Because the initial data of w vanish,
integrating in theta gives
integral of psi_tau(xi) w_theta(xi, delta/tau) d xi .
Both z_theta and z^0_theta at the final edge have uniform bounds for
every fixed spatial order, by Lemma 7 and by the construction of z^0.
Repeated integration by parts in xi, each time moving a derivative from
w_theta onto psi_tau, therefore makes this term o(tau^{1-N}).
Group 2: the other second-order terms, namely P_2 w_theta xi and
S_2 w_xi xi. Integrate one spatial derivative by parts. Each derivative
falling on psi_tau costs a factor of order tau. Using (109) and the
length 2 delta/tau of the strip, the total cost is at most
( C_3 + o(1) ) delta^2 tau^{1-N} , (111)
with C_3 independent of small fixed delta. Terms where the derivative
falls on a coefficient, and the first-order terms, are smaller by an
extra factor tau^{-1} for fixed delta. The spatial test is compactly
supported, so there is no spatial boundary contribution.
Group 3: the forcing. Pair (106) with psi_tau. Because D_tau has a
constant sign bounded away from zero by (108), because phi_0 is
nonnegative with positive integral, and because
integral of chi_0(xi)^2 cos^2(tau xi) d xi
-> (1/2) integral of chi_0(xi)^2 d xi > 0 ,
the absolute value of this term is at least
( c_3 + o(1) ) delta tau^{1-N} , c_3 > 0 , (112)
with c_3 again independent of small fixed delta. No convergence of D_tau
as tau varies is needed; only the sign and the lower bound.
Group 4: the Taylor residual, which is bounded by its L^2 size times
||psi_tau||, and is smaller than every power displayed.
The contradiction. Choose delta > 0 so small that
C_3 delta < c_3 / 2 . (113)
This is possible because C_3 and c_3 do not depend on delta. Then let
tau tend to infinity. Group 3 grows like delta tau^{1-N}, while group 2
is at most (c_3/2) delta tau^{1-N} and groups 1 and 4 are smaller. But
the four groups are the two sides of one identity, so their sums must
agree. They do not. Contradiction.
Consequently the closure of A_{M,q_0} cannot contain the open set O.
Since O was arbitrary, A_{M,q_0} has empty interior. The class is
closed by its definition through uniform bounds together with the
existence of a height, so A_{M,q_0} is nowhere dense. That completes the
proof of Proposition 2 by the Baire argument of 5.1.
5.7 The order of the choices, collected
One table, because this is the question every reader of a paper like
this asks.
Stage First: the class, the open set O, the metric g_*, the slab
radius r, the cutoffs chi_0, phi_0, and the integer N.
These fix the leading constants C_2, C_3 and c_3. Those
constants are independent of sufficiently small delta.
Stage Second: the small parameter delta > 0, chosen with
C_3 delta < c_3/2. Subsidiary bounds may depend on this fixed
delta.
Stage Third: the integers tau tending to infinity. Thresholds and
little-oh errors may depend on delta. The leading constants do
not.
Stage Additionally, within the cap estimate: epsilon, then lambda,
then c_1, as in step 4d. Each is chosen after the previous.
Every derivative order has its own bound; no single bound for all orders
is ever asserted. No convergence in every smooth seminorm was required
of the test metrics. No convergence or common sign across different tau
was required of D_tau. The estimates are uniform in exactly the way the
argument needs.
------------------------------------------------------------------
6. BUILDING A BACKGROUND METRIC WITH A PRESCRIBED CURVATURE
------------------------------------------------------------------
Proposition 2 is a statement about metrics near a fixed background g_0
whose curvature is the model profile (7). Such a g_0 has to be built.
It is built by an ODE, and the only input is that the prescribed
curvature be small.
LEMMA 8 (prescribing small curvature). Let K_0 be a smooth function on
the square (-1,1)^2 with ||K_0||_infinity <= 10^{-3}. Then there is a
smooth positive metric
gbar = dX^2 + f(X,Y)^2 dY^2 (114)
with curvature exactly K_0, and with eigenvalues bounded above and
below by positive constants that depend only on the bound 10^{-3}.
Proof. Solve the linear ODE in X, with Y as a parameter:
f_XX = - K_0 f , f(0,Y) = 1 , f_X(0,Y) = 0 . (115)
Standard existence and smooth dependence on parameters gives a smooth
solution f on the whole square, with f(.,Y) continuous in Y.
First, f stays close to one. Rewrite (115) as an integral equation. On a
segment from 0 to X, let F be the supremum of |f|. Then
| f(X,Y) - 1 | <= sup |K_0| * |X| * F * |X| <= 10^{-3} F ,
since |X| < 1. Hence F <= 1 + 10^{-3} F, that is
F <= 1 / (1 - 10^{-3}) < 1.002 . (116)
So |f - 1| < 2 * 10^{-3} throughout. In particular f is strictly
positive, with 0.99 < f < 1.01.
Second, the curvature of (114) is K_0. Computing the Christoffel symbols
of dX^2 + f^2 dY^2 gives
K = - f_XX / f .
This formula holds when f depends on Y as well; the Y-dependence
contributes only to the Christoffel symbols with a Y index and drops
out of the curvature. Inserting (115) gives K = K_0.
Third, the eigenvalues. The metric matrix of (114) is diagonal with
entries 1 and f^2, so its eigenvalues are 1 and f^2, both trapped
between 0.98 and 1.02 by (116). End of proof.
Two remarks. The choice 10^{-3} is not special; any small number works,
and the eigenvalue bounds degrade continuously as it grows. And the
metric is a graph-like warped product, which is why an ODE suffices:
one of the two principal curvatures of the coordinate lines is already
zero, and only the other has to be prescribed.
------------------------------------------------------------------
7. ASSEMBLY: ONE SMOOTH METRIC, TWO LEVELS OF ACCUMULATION
------------------------------------------------------------------
Now the obstruction is placed on countably many patches, arranged into
two nested accumulations: patches accumulate at the boundary of each
negatively curved disk, and the disks accumulate at the origin. The
amplitudes are chosen so that the sum is smooth at both levels.
Everything is done in fixed coordinates X, Y on the open square
(-1,1)^2.
7.1 The disks and the rotation net
Let D_j, j >= 1, be the closed round disks with centres (2^{-j}, 0) and
radii 2^{-j-4}. Their diameters decay like 2^{-j}, their centres march
towards the origin, and their closures are pairwise disjoint. Each D_j
is surrounded by a slightly larger concentric disk, of radius
2^{-j-4} + 2^{-j-8}, and these larger disks are still pairwise disjoint
and still avoid the origin, because the radii are tiny compared with
the distances 2^{-j} between centres.
Fix a finite set of rotations R_1, ..., R_N, whose first-axis lines form
an angular net of mesh less than 1/1000. That is: every line through the
origin makes an angle less than 1/1000 radians with one of the net
directions. The net is finite, fixed once and for all, and its
resolution 1/1000 matches the constant 1/100 of the admissibility
condition (11) with a factor of ten to spare. The reason for the factor
ten: the comparison in section 8 needs a little room, and 1/1000 <
(1/10) * (1/100) provides it.
Patch maps. For n = 1, 2, 3, ... choose affine maps
Phi_n(x, y) = c_n + lambda_n R_{s(n)} (x, y) , lambda_n > 0 , (117)
where R_{s(n)} is one element of the net, with the following two
properties.
(i) The images Phi_n([-4,4]^2) have disjoint closures, and each lies
outside all the closed disks D_j.
(ii) For every j, every point b on the boundary of D_j, and every
rotation R in the net, there is a sequence of integers n_k with
s(n_k) = R, lambda_{n_k} -> 0, and Phi_{n_k}(0,0) -> b.
Property (ii) is what makes the boundary argument of section 8 work: at
every boundary point, in every net direction, patches shrink down to
the point.
Construction of the placement. Around each D_j, consider the thin
annuli inside the larger concentric disk and accumulating on the
boundary of D_j. In each annulus place finitely many distinct centres
so that the centres assigned to a fixed rotation form a successively
finer angular net on the circle. There are finitely many rotations, so
each annulus carries finitely many patches. Choosing the square sizes
small enough, the closed patches inside one annulus are disjoint and
lie inside that annulus, and different annuli are separated. Since only
countably many disks and countably many annuli are involved, this
processes all of them, and the resulting collection is countable.
Every image Phi_n([-4,4]^2) is contained in one such annulus, hence has
disjoint closure from every other image and lies outside every closed
disk. That is (i). And patches in annuli accumulating on the boundary
of D_j, with shrinking sizes, give (ii).
7.2 Prescribing the curvature profile on every piece
Recall the profile h of (6) and the curvature formula
kappa ( x^2 - h(y) ) ,
which is negative near the origin, zero on the oval x^2 = h(y), and
positive outside. That is the shape we want to reproduce on each
patch, scaled by a patch-dependent amplitude kappa_n > 0.
The curvature profile K_0 on the square. In each disk D_j, choose a
smooth negative bump: smooth, supported in the interior of D_j,
strictly negative throughout D_j itself, and flat (all derivatives
zero) on the boundary of D_j.
In each patch, choose a smooth bump supported in Phi_n((-4,4)^2) and
equal on a neighbourhood of Phi_n([-3,3]^2) to
K_0( Phi_n(x,y) ) = kappa_n ( x^2 - h(y) ) . (118)
The amplitudes kappa_n > 0 of the patch bumps, and the amplitudes of the
disk bumps, are chosen so that the sum of everything is smooth and its
sup norm is at most 10^{-3}:
|| K_0 ||_infinity <= 10^{-3} . (119)
Here is a precise choice, because "chosen small enough" is not an
argument. Enumerate the unscaled, globally smooth bump profiles
(extended by zero outside their supports) as k_1, k_2, k_3, .... These
are fixed smooth functions; only their scalar amplitudes are free.
Choose positive numbers epsilon_n with
|| epsilon_n k_n ||_{C^m} <= 10^{-3} 2^{-n} for 0 <= m <= n . (120)
This is possible at each stage: only finitely many seminorms are
constrained, and epsilon_n can be taken as small as needed.
Then the series of derivatives has a uniformly summable tail in each
fixed order m: the terms with n >= m are bounded by 10^{-3} 2^{-n} times
the fixed bounds on the k's. A uniformly convergent series of
continuous derivatives is C^infinity, so
K_0 := sum of epsilon_n k_n (121)
is smooth, including on all the accumulation sets (the boundaries of
the disks, and the origin). Its sup norm is at most 10^{-3} by (120)
with m = 0.
Four properties of K_0, all used later.
* Disjointness of the supports preserves the strict negativity inside
each disk, since each disk bump is strictly negative there and
nothing else is supported there.
* The exact profiles (118) are preserved on the core of each patch,
because the patch bump equals that profile there and the other
bumps vanish on a neighbourhood of the patch core (the disk bumps
are supported away, and other patch bumps are supported on
disjoint squares).
* Every summand is flat at each disk-boundary point at which its
support can accumulate: disk bumps are flat on the boundary of
their own disk, and patch bumps have supports disjoint from every
closed disk. So
K_0 = 0 on the boundary of D_j, for every j. (122)
* The sup norm is at most 10^{-3}, by (119).
7.3 The background metric, and the obstruction at each patch
Apply Lemma 8 with this K_0. Get a smooth positive
gbar = dX^2 + f(X,Y)^2 dY^2 , (123)
curvature exactly K_0, eigenvalues bounded above and below.
Now pull gbar back to each patch. For the n-th patch map,
g_{0,n} := Phi_n^* gbar .
Because curvature is a scalar and transforms by composition under
pullback, and because Phi_n is affine with linear part lambda_n R_{s(n)},
K_{g_{0,n}} = K_{gbar} o Phi_n = K_0 o Phi_n = kappa_n (x^2 - h(y)) (124)
on the core of the patch, that is, on the model box coordinates. The
identity in coefficient matrices is
g_{0,n} = lambda_n^2 R_{s(n)}^T ( gbar o Phi_n ) R_{s(n)} , (125)
which is just the chain rule: pullback by an affine map scales the
metric by lambda_n^2 and conjugates by the rotation. So g_{0,n} is a
background metric of exactly the kind used in section 3.2, with its own
kappa_n, on the model square, and Proposition 2 applies to it.
Choose, for each n, a smooth symmetric tensor eta_n supported in the
model perturbation box P such that
g_{0,n} + eta_n admits no admissible height . (126)
This is possible by Proposition 2, which says the obstructed metrics
are residual in the set U of admissible perturbations, so there are
plenty of them arbitrarily small.
Push each eta_n forward to the original coordinates and extend by zero:
eta_hat_n := lambda_n^{-2} R_{s(n)} ( eta_n o Phi_n^{-1} ) R_{s(n)}^T . (127)
The support of eta_hat_n is the image of P under Phi_n, which is
strictly inside the image of [-4,4]^2 because P = [-1/10,1/10]^2 sits
inside (-4,4)^2. So the extension by zero is smooth: the function
together with all its derivatives vanishes on the boundary of the
support and outside it, and the pieces glue to a smooth tensor.
The scaling cost. For each fixed derivative order m there is a constant
C_m, depending only on m and on the fixed dimension and the finitely
many rotations, with
|| D^m eta_hat_n ||_infinity
<= C_m lambda_n^{-m-2} || D^m eta_n ||_infinity . (128)
Two sources of the scaling: the prefactor lambda_n^{-2} in (127), and
the m derivatives falling on the composition with Phi_n^{-1}, whose
linear part is lambda_n^{-1} times a rotation. No m-dependent constant
beyond the finitely many rotations appears.
Because lambda_n -> 0 as n -> infinity for the patches accumulating on
the disk boundaries, the factor lambda_n^{-m-2} is large. But eta_n
itself is free: Proposition 2 lets us choose it as small as we like in
every finite collection of seminorms. So the large factor can be paid
off by choosing eta_n smaller. Concretely, require
|| eta_hat_n ||_{C^m} <= epsilon 2^{-n} for 0 <= m <= n , (129)
which is achievable by choosing eta_n small enough in C^0, ..., C^n.
The constant epsilon is chosen once, smaller than half a uniform lower
eigenvalue bound for gbar. Such a bound exists by Lemma 8.
7.4 The final metric and what survives
Define
g := gbar + sum from n=1 to infinity of eta_hat_n . (130)
Then:
1. Smoothness. Use (129): the derivatives of order m of the tail beyond
n = m are bounded by epsilon times a convergent geometric series. So
the series and all its derivative series converge uniformly, and g
is C^infinity.
2. Positive definiteness. The perturbation has sup norm at most
epsilon 2^{-1} times a geometric factor, hence less than half the
lower eigenvalue bound of gbar. So g is positive definite everywhere,
and in fact uniformly so.
3. The obstruction. The metric on the n-th patch is g_{0,n} + eta_n
plus the contributions of all the other tensor perturbations, which
are zero there because their supports are the images of P under
Phi_k and those images are pairwise disjoint. Hence the patch still
admits no admissible height:
for every n, the metric g restricted to the n-th patch
admits no admissible height . (131)
4. The disks. Every eta_hat_n and all of its derivatives vanish on
every closed disk D_j, because the supports are strictly inside the
patches and the patches lie outside the closed disks by (i). So the
metric g agrees with gbar, together with all its derivatives, on each
closed disk. Therefore, using (122) and the sign of the disk bumps,
K_g < 0 in the interior of D_j, K_g = 0 on the boundary of D_j . (132)
5. No lower bound anywhere. The construction never bounds kappa_n from
below, nor lambda_n, nor the Hessian separations that enter the
admissibility condition. Values can be as small as convenient. That
freedom is exactly what the two levels of accumulation require.
The metric g on (-1,1)^2 is the required object. What remains is to
show that no neighbourhood of the origin admits an isometric immersion
into R^3. That is section 8.
------------------------------------------------------------------
8. THE BOUNDARY DISK, AND THE FINAL CONTRADICTION
------------------------------------------------------------------
The metric g of section 7 has two properties. It is negatively curved
inside every disk D_j and flat on the boundary of every disk, and it
admits no admissible height on any of the patches, which accumulate on
every boundary point in every net direction.
This section shows that these two properties cannot coexist with a
smooth isometric immersion. The mechanism is a single geometric
observation about negatively curved disks, and the matching of a
description of the resulting height to the patch obstruction.
8.1 A disk cannot be a flat graph
LEMMA 9 (negatively curved disk). Suppose a smooth metric on a
neighbourhood of a closed disk has negative curvature in the disk and
zero curvature on its boundary. Suppose an isometric immersion of this
neighbourhood into R^3 is a graph over a plane. Then its second
fundamental form is nonzero at some boundary point, and at that point
it has rank one.
Proof.
Let u be the graph function and Omega the planar projection of the
disk. Because the immersion is a graph over a plane and it is an
isometric immersion, u is smooth on a neighbourhood of the closed
region Omega. The boundary of Omega is a single smooth closed curve,
because the disk's boundary is one and the projection is a
diffeomorphism onto Omega.
The graph formula for the second fundamental form is
II_ij = u_ij / sqrt( 1 + |Du|^2 ) . (133)
The factor is positive and smooth, so II vanishes at a point exactly
when the ordinary Hessian D^2u vanishes there.
Suppose for contradiction that II vanishes at every boundary point.
By (133), D^2u = 0 along the whole boundary curve. The boundary is
connected, so the gradient Du is constant along it: the derivative of
Du along the curve is a component of the Hessian, and it vanishes. Call
the constant vector a. Then u restricted to the boundary is an affine
function,
u = a . x + c on the boundary, c a constant. (134)
Define
v = u - a . x - c .
Then v = 0 on the boundary, and
det D^2 v = det D^2 u (135)
everywhere, since a . x + c has zero Hessian.
Now use the curvature hypothesis. The Gauss equation for the graph
gives det II = K det g, and K < 0 in Omega, so det D^2 u < 0 throughout
Omega. Hence det D^2 v < 0 throughout Omega, and v cannot be
identically zero, because then its Hessian would vanish and its
determinant would be zero.
So v is a nonzero smooth function on Omega that is zero on the boundary
and has a strictly negative Hessian determinant in the interior.
Compactness of the closure of Omega gives an interior maximum or an
interior minimum of v, since v is not identically zero and vanishes on
the boundary. At an interior extremum the Hessian is semidefinite, so
its determinant is nonnegative. That contradicts det D^2 v < 0.
Therefore II does not vanish at every boundary point. It is nonzero at
some boundary point b.
Rank one at b. At a boundary point the curvature is zero by hypothesis,
so by the Gauss equation
det II(b) = K(b) det g(b) = 0 .
The matrix II(b) is symmetric, nonzero, and has zero determinant, so it
has rank exactly one. End of proof.
The intuition, for a reader who wants one. If the second fundamental
form vanished along the whole boundary, the surface would be planar,
with constant slope, all around the boundary. Meanwhile the interior is
a strict saddle: negative curvature forces det D^2 u < 0, so the height
above the plane has no interior extrema, which is impossible for a
function that vanishes on the boundary except when it is the zero
function, and the zero function is not a strict saddle. The zero
determinant at the boundary is then forced by the zero curvature, and
rank one follows. Both halves of the lemma are just the graph formula
and the Gauss equation; no PDE is involved.
8.2 Proof of Theorem A
Now put everything together.
Let g be the metric of section 7 on the open square (-1,1)^2.
Suppose for contradiction that some open set U containing the origin
admits a smooth isometric immersion
F : (U,g) -> R^3 .
Projection and graph. At the point F(0), let T be the tangent plane of
the immersed surface. Projection of R^3 onto T, composed with F, is a
smooth map from U to T. Its differential at the origin is an
isomorphism, because dF is injective and its image is T. So by the
inverse function theorem, the projection is a local diffeomorphism near
the origin. Restricting U if needed, F is a graph over its projection
onto T, and the projection is a diffeomorphism onto a planar open set.
Choose the disk. The disks D_j have closure contained in any fixed
neighbourhood of the origin once j is large, since they are centred at
(2^{-j}, 0) with radii 2^{-j-4}. So for j large,
the closure of D_j lies in the restricted domain of F . (136)
Fix such a j. By (132) the metric on the neighbourhood of D_j has
negative curvature inside D_j and curvature zero on the boundary.
Lemma 9 applies, and gives a boundary point
b in the boundary of D_j
at which II(b) is nonzero of rank one.
The height in the normal direction. Let e be the unit normal to the
immersed surface at F(b), considered as a vector in R^3, and set
z = < F , e > .
By Lemma 1 (the height equation), and its final statement with the
choice of unit normal at b equal to e,
dz(b) = 0 , Hess z (b) = II(b) . (137)
So the gradient of z vanishes at b, and its covariant Hessian there is
nonzero of rank one. In particular, by Pythagoras,
1 - |dz(b)|^2_g = 1 > 0 ,
and by continuity this stays positive in a whole neighbourhood of b.
Choose the orientation. Regard Hess z (b) as a symmetric bilinear form
on the parameter plane. Its kernel is a line through b, because the
form has rank one. By construction of the rotation net in 7.1, one of
the net directions makes an angle of less than 1/1000 with that kernel
line. Fix that rotation R in the net.
Express things in the coordinate frame rotated by R, with the first
frame axis along the net direction. A rank-one symmetric form whose
kernel makes an angle alpha with the first axis has, in that frame,
H_yy(b) != 0 , H_xy(b) / H_yy(b) = tan(alpha) ,
so
| H_xy(b) / H_yy(b) | = | tan alpha | < 1/100 . (138)
The strict inequality 1/100, rather than merely 1/1000, is where the
factor-of-ten slack built into the net resolution pays off.
By continuity, and since H_yy(b) != 0, the same strict inequalities
hold on a whole neighbourhood of b:
H_yy != 0 , | H_xy / H_yy | < 1/100 . (139)
The patch. By property (ii) of the placement in 7.1, for the boundary
point b and the rotation R whose direction is fixed in (138), there are
full patches with that rotation converging to b, with diameters tending
to zero. Choose one small enough that
its image is contained in the neighbourhood (139), and
it is contained in the domain of F.
The patch is a full copy of the model square S, mapped by an affine map
Phi_n whose linear part is lambda_n R. Pull z back to the patch:
z_n = z o Phi_n .
Then z_n is smooth on a neighbourhood of S, because z is smooth in a
neighbourhood of the image of the patch and the patch is compactly
inside that neighbourhood.
The four admissibility conditions.
* z_n satisfies the Darboux equation, because the covariant Hessian
and the gradient norm are natural under isometries: pulling back
the height and the metric together preserves the equation.
* E > 0 on S, because (139) contains 1 - |dz|^2_g > 0.
* H_yy != 0 on S, by (139) as pulled back to the patch.
* |q| <= 1/100 on S, likewise.
What does the scaling do? The affine map has linear part lambda_n R
with a single dilation factor lambda_n, applied equally to both axes.
Pulling back multiplies every entry of the Hessian by lambda_n^2:
identically, in matrix form,
Hess z_n = lambda_n^2 R^T ( Hess z ) R .
That multiplies all the entries by the same positive factor. Ratios of
entries, and in particular the ratio q = H_xy/H_yy, are unchanged. This
is why the net is built with rotations and dilations only: a rotation
turns the frame, and a dilation rescales everything uniformly, so the
small ratio (138) survives the passage to the patch.
The contradiction. So z_n is an admissible height for the pullback
metric on the patch. But by (131) the metric g on that patch admits no
admissible height. Contradiction.
Therefore no open neighbourhood of the origin admits a smooth isometric
immersion into R^3. The open square with the smooth positive metric g
is the required Riemannian surface. Theorem A is proved.
------------------------------------------------------------------
9. THE METRIC CAN LOOK EUCLIDEAN TO INFINITE ORDER
------------------------------------------------------------------
The metric g of section 7 not only fails to be locally realisable near
the origin, it can be chosen to agree with the Euclidean metric to every
order at the origin. This is Corollary 10, and its proof is a check of
the construction, not a new idea.
COROLLARY 10 (Euclidean Taylor jet). The metric g of Theorem A can be
chosen so that, in the coordinates X, Y,
all partial derivatives of g_ij - delta_ij vanish at the origin,
for every multi-index and every pair of indices. (140)
Hence smooth local isometric realizability is not determined by the
full Taylor jet of the metric at the base point.
Proof.
Step 1: the curvature and the perturbation series vanish to all orders
at the origin.
Every disk D_j is centred at (2^{-j}, 0) with radius 2^{-j-4}, and the
closest point of D_1 to the origin is at distance 2^{-1} - 2^{-5} > 0.
So all the disks, and all their larger concentric neighbours, stay away
from a fixed neighbourhood of the origin. The disk bumps are supported
inside the disks, and the patch bumps are supported on the patches,
which lie outside every closed disk by (i). So each summand of K_0
vanishes on a fixed neighbourhood of the origin. Uniform convergence of
every derivative series then gives
all partial derivatives of K_0 vanish at the origin. (141)
The tensor perturbations eta_hat_n are supported on the patches, which
also lie outside a fixed neighbourhood of the origin by (i). Every term
of the sum therefore vanishes near the origin, and
all partial derivatives of the series vanish at the origin. (142)
Step 2: the background metric has the Euclidean jet. It remains to
check gbar = dX^2 + f^2 dY^2, where f solves (115) and K_0 satisfies
(141).
From the initial conditions f(0,Y) = 1 and f_X(0,Y) = 0,
d_X^a d_Y^b (f - 1)(0) = 0 for a = 0, 1 and every b >= 0. (143)
For a >= 2, differentiate the equation f_XX = -K_0 f a-2 more times in
X and b times in Y, and evaluate at the origin:
d_X^a d_Y^b (f - 1)(0) = - d_X^{a-2} d_Y^b (K_0 f)(0) .
Every term of the Leibniz expansion of the right side contains at least
one derivative of K_0, all of which vanish at the origin by (141). So
the right side is zero:
d_X^a d_Y^b (f - 1)(0) = 0 for every a >= 2 and every b >= 0. (144)
Combining (143) and (144): every partial derivative of f - 1 vanishes
at the origin. Since gbar = dX^2 + f^2 dY^2, we have
gbar_11 - 1 = 0 , gbar_12 = 0 , gbar_22 - 1 = f^2 - 1 ,
and f^2 - 1 = (f-1)(f+1) has the same vanishing property. So every
partial derivative of gbar_ij - delta_ij vanishes at the origin.
Step 3: adding the series preserves the property, by (142). Hence (140)
holds.
The final sentence of the corollary. The Euclidean metric delta_ij
itself admits an isometric immersion into R^3, namely the inclusion of
the plane. The metric g does not on any neighbourhood of the origin, by
Theorem A. Yet the two metrics agree to every order at the origin. So
smooth local realizability is not a property of the Taylor jet at a
point. End of proof.
A remark to prevent a wrong impression. Corollary 10 does not say the
two metrics are indistinguishable. They differ in a neighbourhood of
the origin, in a way that is invisible to the jets at the single point
but visible to the equation, because the equation is global on the
square and the zero set of the curvature is not a local object at the
origin. That is exactly the local-versus-pointwise distinction which the
whole paper is built around.
------------------------------------------------------------------
10. WHAT IS NOT PROVED HERE, AND HONEST REMARKS
------------------------------------------------------------------
This section states plainly what the paper does not do, so that a reader
is not left to guess.
(1) Sign restrictions are untouched. The construction uses both signs of
curvature: negative inside the disks, positive outside. The paper says
nothing about the case K >= 0 everywhere or K <= 0 everywhere. Those
remain open, as recorded in the introduction.
(2) The historical discrepancy is left alone. The 2002 preprint of
Nadirashvili and Yuan asserts a smooth K <= 0 counterexample; their
2008 published paper leaves the unrestricted smooth question open. This
paper takes no position on which is right about their construction. It
supplies a separate construction, with its own obstruction, and does
not rely on it.
(3) The regularity estimate is interior in the artificial edge. It says
nothing about the behaviour of a height at the artificial edge itself,
and needs nothing. This is worth noting because it is a common
misreading: the artificial edge is not a boundary where anything
happens. It is where the estimate stops caring.
(4) The theorem is about existence, not about the size of the obstruction
set. The construction produces one metric. It also produces a family.
Proposition 2 actually says more: in the perturbation space, metrics
with no admissible height form a residual set, so the failure is generic,
not a coincidence. What this paper does not do is define a natural
measure on the perturbation space and compute the measure of the
obstruction set.
(5) The estimates are not sharp. The constants 1/100, 1/1000, 10^{-3},
M, c and so on are chosen for convenience and slack. A sharper
construction presumably works with more generous constants. The
inequality l >= 8 is the only place where the derivative threshold
appears, and it is comfortably above the theoretical minimum 7.
(6) Two minor corrections relative to a naive reading. In the
computation of A_s (equation 45) one has three factors of s_y, not one;
the difference is a factor trapped between positive constants, so the
argument is unaffected, but the formula is stated correctly here. And
the comparison between powers of d in section 4.6 requires the harmless
normalisation by D_0 = b+2 which is made explicit there. Both are
recorded so that a reader who redoes the algebra does not think she has
found an error.
(7) A short reading guide for the three hardest spots. If a reader has
limited time: read sections 8 and 2 first, which give the shape of the
argument; then read section 3, which sets the geometry; then read the
statement of Proposition 2 and its proof in 5.6, which is the actual
obstruction. Sections 4 and the first half of 5 are the technical
machinery that makes 5.6 possible, and can be read afterwards.
------------------------------------------------------------------
11. INDEX OF SYMBOLS AND CONSTANTS
------------------------------------------------------------------
Letters are grouped by role. For each, the equation or section where it
first appears is given.
COORDINATES AND REGIONS
x, y ......... model patch coordinates, section 3
X, Y ......... coordinates on the target square (-1,1)^2, section 7
t, s ......... flow coordinates, defined by (19)
xi, theta .... sheared coordinates for the pulse, (86)
S ............ model square [-3,3]^2
P ............ perturbation box [-1/10,1/10]^2
B ............ strongly negative box [-1/5,1/5]^2
D_j .......... negative disks of section 7.1
Omega ........ planar projection of a disk, in Lemma 9
GEOMETRIC OBJECTS
g ............ a Riemannian metric, written g_ij
K ............ Gaussian curvature
II ........... second fundamental form
Gamma^r_ij ... Christoffel symbols
H_ij ......... covariant Hessian, Hess z
E ............ det(g)(1 - |dz|_g^2), the ellipticity factor
q ............ H_xy / H_yy, the small ratio
h ............ model vertical profile, (6)
kappa ........ model curvature amplitude
f ............ warped product factor of gbar, (115)
kappa_n ...... patch curvature amplitude, (118)
lambda_n ..... patch scale, (117)
R_n .......... patch rotation, from the finite net
MODEL CURVATURE AND THE PATCH
g_0 .......... the background metric on S with curvature (7)
gbar ......... the globally defined warped product (123)
eta_n ........ tensor perturbation on the model patch
eta_hat_n .... its pushforward to the target, (127)
FLOW AND CAP ESTIMATES
D ............ characteristic direction d_x - q d_y, (14)
a ............ K G, the coefficient of u_ss
G ............ E / H_yy^2
G_1 .......... G s_y^2 = A/K
A, B, C ...... coefficients of the linear equation (33)
R_l .......... right side of the differentiated equation (33)
W, m, n ...... weights and multiplier components, (52)
d ............ b - s, the distance to the artificial edge
p ............ the weight exponent, fixed at 8
H_0 .......... lambda + p/d, the normal derivative of W
PULSE
v_* .......... sqrt(kappa/2) M^{-3/2}, the speed lower bound, (84)
L ............ 100(1 + 1/v_*), the speed upper bound, (84)
M ............ uniform C^8 bound in the class, (82)
q_0 .......... rational centre of the slope class, (83)
A_{M,q_0} .... the bounded class, (82)-(83)
U ............ the open set of admissible perturbations
r ............ slab radius, (93)
delta ........ strip half-width parameter, (95)
tau .......... oscillation frequency, (95)
N ............ amplitude exponent, fixed with N > 10, (95)
chi_0, phi_0 . spatial and temporal cutoffs, (95)
psi_tau ...... test function chi_0 cos(tau xi), (110)
D_tau ........ forcing prefactor, (107)
SMALL CONSTANTS, IN THE ORDER THEY ARE CHOSEN
epsilon ...... size of the tangential correction to the multiplier,
chosen first in step 4d(i)
lambda ....... the large weight parameter, chosen second, 4d(ii)
c_1 .......... the coercivity threshold, chosen third, 4d(iii)
c_2 .......... the coercivity constant, (61)
c_3, C_2, C_3 leading constants of the moment estimate, (109)-(112)
c_D, C_D ..... bounds on the forcing prefactor, (108)
c_0, C_0 ..... bounds on s_y, (23)
A NOTE ON THE NUMBERING. The two main statements are numbered as
propositions, Proposition 2 (the patch obstruction) and Proposition 3
(cap regularity), and the lemmas around them carry the numbers
inherited from the order in which the original proof referred to them.
So the sequence reads Lemma 1, Proposition 2, Proposition 3, Lemma 4,
Lemma 5, Lemma 6, Lemma 7, Lemma 8, Lemma 9, Corollary 10. There are
no missing lemmas; the gap is a naming convention, and it is kept so
that a reader comparing this version with older notes does not have to
translate.
END OF THE PAPER.