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ELECTROMAGNETIC TAILS AND THE KERR-NEWMAN PENROSE INEQUALITY
a complete counterexample, written from scratch
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CONTENTS
0. How to read this
1. The claim, in plain words
2. The setting: exactly which data are allowed
3. Step 1. Fields that already solve most of the equations
4. Step 2. One scalar equation for the shape of space
5. Step 3. Solving that equation, with uniform control
6. Step 4. The horizon, and every surface that might be smaller
7. Step 5. Mass, charge, and angular momentum at infinity
8. Step 6. The inequality fails; equality does not force Kerr-Newman
9. What this does and does not say
A. Glossary
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0. HOW TO READ THIS
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This is a complete proof, start to finish. It is not a summary and it
skips nothing. But it is written for a reader who knows calculus and
vectors and nothing else about general relativity. Every symbol is
defined where it first appears, and there is a glossary at the end.
There are only six ideas in the whole paper, and they happen in order:
(1) Write down electromagnetic fields and a "shear" tensor with a
simple shape, chosen so that three of Einstein's four constraint
equations are satisfied identically. No solving required. This is
Step 1.
(2) Show that the one remaining equation -- the Hamiltonian
constraint -- collapses to a single scalar equation for a single
unknown function u. This is Step 2.
(3) Solve that scalar equation by an explicit fixed-point argument,
and get estimates that do not degrade as the parameters vary.
This is Step 3.
(4) Check that the inner boundary sphere is a marginally trapped
surface, that no other surface is trapped, and that no enclosing
surface has smaller area. The area part uses a clever closed
2-form. This is Step 4.
(5) Compute mass, charge and angular momentum at infinity. This is
Step 5.
(6) Feed those numbers into the conjectured inequality and watch it
fail. Then tune one parameter to make it fail by exactly zero,
and observe that the resulting data cannot possibly come from a
Kerr-Newman black hole. This is Step 6.
Throughout, "s" is an amplitude in [0,1] and "L" is a large radius.
The construction is a two-parameter family. The art is in showing that
every estimate is uniform in s and L, because the conclusion needs
s = 1 and L -> infinity simultaneously.
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1. THE CLAIM, IN PLAIN WORDS
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1.1 What the conjectured inequality says
Take a snapshot of a spacetime containing a black hole. The snapshot
is a 3-dimensional space with a metric g, a "second fundamental form"
K describing how the snapshot bends inside spacetime, and electric and
magnetic vector fields E and B. Three numbers summarize the snapshot
from far away:
m = the total mass (ADM energy),
J = the total angular momentum,
Q = the total charge, Q = sqrt(Qe^2 + Qb^2).
And one number summarizes the hole itself:
A = the area of its surface (more precisely, the smallest area of
any surface that encloses the hole).
The Kerr-Newman black hole is the exact solution of Einstein's
equations describing a charged, rotating hole. Its mass formula, turned
around, predicts
m^2 >= A/(16 pi) + Q^2/2 + pi (Q^4 + 4 J^2)/A, (1.1)
together with the side condition
A >= 4 pi sqrt(Q^4 + 4 J^2). (1.2)
Inequality (1.1) is the Kerr-Newman Penrose inequality. It is the
charged rotating version of Penrose's 1973 proposal: if a black hole
forms by gravitational collapse, its surface area cannot be smaller
than the area of the (unique) Kerr-Newman hole with the same mass,
charge and angular momentum.
Two comments that matter enormously later.
First, the inequality has two branches. Think of (1.1) as a quadratic
in A. For fixed m and J the right side is smallest at
A = 4 pi sqrt(Q^4+4J^2) and grows for larger A. The "physical branch"
is the increasing one, which is exactly (1.2). On the other branch the
right side would decrease as A increases, which contradicts what
happens when you add mass far away. Every statement below is about the
physical branch, and our examples sit strictly on it.
Second, in the presence of electromagnetic fields, which J you mean is
a real choice. The "bare gravitational" angular momentum is the flux
of K alone through a large sphere. The "total conserved" angular
momentum includes an extra electromagnetic boundary term. These two
differ in general. Their difference is a boundary term
J_EM(r) = (1/4 pi) * integral over S_r of h * (flux of E), (1.3)
where h is the magnetic scalar potential. The identification of the
two, i.e. the assertion that J_EM vanishes at infinity, needs the
electromagnetic fields to fall off faster than the hypotheses below
allow.
1.2 What this paper proves
THEOREM 1.1. In the class of smooth axisymmetric source-free
electrovacuum exteriors spelled out in Section 2, both of the
following happen.
(a) There are data with
Qe = Qb = 0, A > 8 pi |J|, m^2 < A/(16 pi) + 4 pi J^2/A.
(b) There are data with
Qe = Qb = 0, A > 8 pi |J|, m^2 = A/(16 pi) + 4 pi J^2/A,
for which no smooth spacelike embedding into the Kerr-Newman
spacetime with parameters (m, J, Qe, Qb) reproduces the given
metric, second fundamental form and electromagnetic fields.
All examples are maximal (trace of K is zero) on the region
Omega = { x in R^3 : |x| >= 1 },
with connected boundary S = { |x| = 1 }. Both electromagnetic fields
are nonzero.
Part (a) says: the inequality (1.1) is false under the stated
hypotheses, and it fails even though every horizon condition and every
area condition you could ask for is satisfied. Part (b) says: even
when the inequality holds with equality, you cannot conclude the data
come from a Kerr-Newman hole. So the usual equality case of the
Penrose argument is not available either.
The proof uses no Penrose inequality, no rigidity theorem, and no
positive mass theorem. It is a construction plus a computation.
1.3 The family, and the numbers
Everything comes from one family with two dials: an amplitude
s in [0,1] and a large transition radius L. For this family
J = -4 s^2 / 15, (1.4)
m = 2 + O(s^2 / L), (1.5)
A = 64 pi + O(s^2 / L^2). (1.6)
Read (1.6) first: at s = 0 the space is exactly Schwarzschild with
mass 2, whose horizon area is 16 pi m^2 = 64 pi, and the dials only
nudge it. Read (1.4) next: the angular momentum is order s^2 and does
not care about L.
Now compute the defect, meaning the right side of (1.1) subtracted
from the left:
D(s,L) = m^2 - A/(16 pi) - 4 pi J^2/A. (1.7)
At s = 1 with L -> infinity: m^2 -> 4, A/(16 pi) -> 4, and
4 pi J^2/A -> 4 pi (16/225) / (64 pi) = 1/225. Hence
lim over L of D(1,L) = -1/225 < 0. (1.8)
So for all large enough finite L, (1.1) genuinely fails. That is
part (a).
At very small s the same quantity is positive, because adding a little
mass beats adding a little area and a little spin. So D(s,L) changes
sign as s runs from 0 to 1, and by continuity there is an s with
D(s,L) = 0 exactly. That is the equality example of part (b). It has
Q = 0, J nonzero, and nonzero electromagnetic fields. But a
Kerr-Newman black hole with Q = 0 has zero electromagnetic field on
every slice, because its field is linear in the charges. So no
Kerr-Newman realization exists. That contradiction is the whole of
part (b).
1.4 Where the loophole is
The trick is not a trick. It is exactly one thing: the hypotheses ask
only that the electromagnetic fields decay like r^{-2}. They do not
ask for the Coulomb form, which is the leading term r^{-2} with a
spherically symmetric coefficient. Our fields have r^{-2} tails whose
coefficient varies with angle. Such tails carry angular momentum in
the electromagnetic field, hence a nonzero J_EM at infinity, and hence
they are not the fields of any charged point source. Meanwhile their
energy density falls off like r^{-4}, which is integrable but only
barely: the total energy carried by the tail goes to zero like 1/L as
L grows, while it is spread over a huge volume.
So the honest one-sentence summary is: allow non-Coulomb r^{-2} tails,
define J as the bare gravitational flux, and the inequality fails;
demand Coulomb tails, or use the conserved total J, and these examples
say nothing.
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2. THE SETTING: EXACTLY WHICH DATA ARE ALLOWED
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Everything in this paper is about a single snapshot of spacetime, not
the full four-dimensional story. That is how the Penrose inequality is
usually phrased and it is what makes the problem finite-dimensional
enough to attack.
2.1 The players
Work in three spatial dimensions, with Newton's constant and the speed
of light both set to 1, and with zero cosmological constant. On an
oriented 3-dimensional Riemannian manifold (Omega, g) we have:
g a Riemannian metric (the shape of space),
K a symmetric covariant 2-tensor (how space bends in time),
tau the trace of K with respect to g, tau = tr_g K,
E, B electric and magnetic vector fields.
The cross product of two vectors is defined by the volume form:
(X x_g Y)^flat(Z) = dV_g(X, Y, Z) for all vectors Z. (2.1)
Here ^flat turns a vector into a covector using g; in index language
(X x_g Y)^flat(Z) = dV_g(X, Y, Z) is just the volume of the
parallelepiped they span, with sign.
2.2 The four constraint equations
The snapshot is not arbitrary. It must satisfy the source-free
electrovacuum constraints:
R_g + tau^2 - |K|_g^2 = 2 ( |E|_g^2 + |B|_g^2 ), (2.2a)
div_g (K - tau g) = 2 (E x_g B)^flat, (2.2b)
div_g E = 0, div_g B = 0. (2.2c)
Equation (2.2a) is the Hamiltonian constraint: it says the curvature
of space is produced by the energy of the electromagnetic field and of
the bending K. Equation (2.2b) is the momentum constraint. The two
equations in (2.2c) are Maxwell's equations for a source-free field.
Two densities come out of this and get used later:
mu = ( |E|_g^2 + |B|_g^2 ) / (8 pi), (2.3)
J_con = (E x_g B)^flat / (4 pi). (2.4)
mu is energy density. J_con is angular momentum density of the
electromagnetic field. Do not confuse J_con with the scalar total
angular momentum J defined at infinity in (2.9); they are different
objects and this paper exists because they can disagree at infinity.
2.3 The domain
The region we study is
Omega = { x in R^3 : |x| >= 1 }, S = { |x| = 1 }, (2.5)
so Omega is the outside of the unit ball, S is the unit sphere, and
"r = |x|" always means Euclidean distance from the origin. We use
(r, theta, phi) for spherical coordinates with theta the colatitude and
eta = d/dphi (2.6)
the generator of rotation about the third axis. The full class of
exteriors allowed by the theorem is more general than (2.5) -- any
connected complete exterior diffeomorphic to the complement of an open
ball, with one asymptotically flat end -- but the point of the
construction is that the example lives on the simplest possible one.
For the general class we require:
- all data smooth, including up to the boundary S;
- Omega connected, complete as a metric space with its boundary
included;
- exactly one asymptotically flat end;
- a smooth effective U(1) action preserving g, K, E, B and S, whose
generator eta has period 2 pi and agrees at infinity with the
standard rotation about the third coordinate axis.
"Effective" means the action is not secretly trivial: a rotation
fixing every point of Omega is the identity, i.e. angle 0 mod 2 pi.
2.4 Decay at infinity
In asymptotic Euclidean coordinates x, with r = |x|, require
g_ij - delta_ij = O_4(r^{-1}),
K_ij = O_3(r^{-3}),
E^i, B^i = O_3(r^{-2}). (2.7)
The notation O_k(r^{-a}) means: for each Cartesian multi-index of
order j <= k, the corresponding derivative is bounded by C r^{-a-j},
for a constant C that does not depend on x. So O_4 on g means the
metric, its first four derivatives, all decay. O_3 on E means the
field and three derivatives decay, but with only r^{-2}.
That last exponent is the crux of the paper. A Coulomb field
Q/(4 pi r^2) * xhat does decay like r^{-2}, but so does a field like
r^{-2} a(theta, phi) xhat for any smooth angular function a. The
hypotheses in (2.7) cannot tell these apart. Coulomb asymptotics would
need a separate assumption, and we deliberately do not make one.
We also require that the densities mu and |J_con|_g are integrable.
Since |E|, |B| = O(r^{-2}) gives mu = O(r^{-4}), and the volume form
in three dimensions is O(r^2) dr, integrability at infinity is exactly
marginal: r^2 * r^{-4} dr = r^{-2} dr is integrable, and no better.
2.5 The five numbers measured at infinity
Let S_R = { |x| = R }. Let n_delta be the Euclidean outward unit normal
to S_R and nu_R its g-unit outward normal. Repeated indices are summed
and d_j means d/dx^j. The asymptotic quantities are
E_ADM = (1/16 pi) lim_{R->inf} int_{S_R}
( d_j g_ij - d_i g_jj ) n_delta^i dA_delta, (2.8a)
(P_ADM)_i = (1/8 pi) lim_{R->inf} int_{S_R}
( K_ij - tau g_ij ) n_delta^j dA_delta, (2.8b)
J = (1/8 pi) lim_{R->inf} int_{S_R}
( K_ij - tau g_ij ) nu_R^i eta^j dA_g, (2.9)
Qe = (1/4 pi) lim_{R->inf} int_{S_R} g(E, nu_R) dA_g, (2.10a)
Qb = (1/4 pi) lim_{R->inf} int_{S_R} g(B, nu_R) dA_g. (2.10b)
All five limits are required to exist and be finite. We impose
P_ADM = 0, E_ADM > 0, and set
m = E_ADM, Q = sqrt( Qe^2 + Qb^2 ). (2.11)
Note what (2.9) does and does not contain: it is the flux of the
gravitational tensor K through a large sphere in the rotational
direction eta. There is no electromagnetic surface term. This is
exactly the "bare gravitational angular momentum" of Section 1, and it
is the J appearing in Theorem 1.1. That choice is the whole content of
the word "precise formulation" in the introduction.
2.6 Trapped surfaces
Let Sigma be a two-sided surface with unit normal nu. The convention
is that nu points toward the end, i.e. outward. Put
H_Sigma = div_Sigma nu,
tr_Sigma K = ( g^ij - nu^i nu^j ) K_ij,
theta_+(Sigma) = H_Sigma + tr_Sigma K. (2.12)
H_Sigma is the mean curvature of Sigma measured with outward normal,
so for a Euclidean sphere of radius R it is +2/R. theta_+ is the
"future outward expansion". A surface with theta_+ = 0 is called a
future marginally outer trapped surface, or future MOTS. Intuitively
a MOTS is a surface where outgoing light is neither expanding nor
shrinking; the boundary of a black hole in a snapshot is exactly such
a surface. A surface with theta_+ <= 0 everywhere is "outer trapped",
meaning outgoing light actually converges.
The boundary S is required to be a future MOTS:
theta_+(S) = 0. (2.13)
No condition is imposed on the inward expansion. This asymmetry is
deliberate: the results are about the outer boundary of the trapped
region, not about its inside.
We also require exact outermostness:
no compact smooth embedded enclosing surface lying entirely in the
interior of Omega, possibly disconnected, has theta_+ <= 0
everywhere. (2.14)
"Enclosing" here means the surface separates S from infinity: it is
the boundary of a region that contains the far end of Omega. So S is
not merely trapped; it is the outermost trapped thing there is.
2.7 Enclosing cuts and the area condition
The area in (1.1) is not the area of a particular surface chosen by
hand; it has to be an infimum over all surfaces that do the job. The
definition used here, carefully, is:
An enclosing cut is Gamma = dD, the full intrinsic boundary of a
connected smooth codimension-zero submanifold D of Omega, closed as
a subset of Omega, whose interior lies in the interior of Omega and
which contains the entire sufficiently distant part of the end. The
cut is smooth, embedded, two-sided; it may be disconnected and it
need not be symmetric or trapped. The full intrinsic boundary is
counted, including any portion that coincides with S. (2.15)
Two sanity checks. First, taking D = Omega is allowed, and then
Gamma = S, so S itself is always an enclosing cut and the infimum is
never bigger than Area_g(S). Second, allowing D to touch S is
essential: the interesting competing cuts are allowed to hug the
existing boundary and then wander off, and any definition that
excluded them would make the area artificially small.
The required area condition is
Area_g(Gamma) >= A := Area_g(S) > 0 for every enclosing cut Gamma.
(2.16)
So A is by definition the minimum enclosing area, and it is attained,
by S itself. This is the strongest possible reading of "A is the
surface area of the hole", and our examples satisfy it.
2.8 What is deliberately not assumed
To be completely explicit about the loophole, since everything else is
bookkeeping. The hypotheses are (2.2), (2.5)-(2.7), (2.13), (2.14),
(2.16) and the symmetry. They do not include, and the theorem is
therefore allowed to violate:
- Coulomb asymptotics: E = Qe xhat / (4 pi r^2) + O(r^{-3}), and
similarly for B. We only have E = O(r^{-2}).
- The vanishing of the electromagnetic angular momentum boundary
term at infinity. We do not assume J_EM(infinity) = 0.
- The identification of the bare gravitational J with any conserved
total angular momentum.
- Any stationarity, any curl-free condition on E or B, any
multipole expansion, any first moment of the source.
Each of those extra assumptions is a real, meaningful hypothesis, and
under any of them the conclusions can change. Section 9 comes back to
this.
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3. STEP 1. FIELDS THAT ALREADY SOLVE MOST OF THE EQUATIONS
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3.1 What we want out of a "seed"
We need smooth data on Omega satisfying (2.2). Rather than solve all
four equations at once, we first write down a "seed": a flat-space
electric field E~, a flat-space magnetic field B~, and a flat-space
symmetric tensor sigma, all vanishing inside a ball of radius L, with
the three properties
div_delta E~ = div_delta B~ = 0,
tr_delta sigma = 0,
div_delta sigma = 2 (E~ x_delta B~)^flat. (3.1)
Then in Step 2 we rescale them conformally. The rescaling is designed
precisely so that (3.1) continues to hold for the physical fields and
the physical K. That leaves only the Hamiltonian constraint to solve,
which is one scalar equation. This is the standard conformal method,
run in reverse: instead of solving for the fields, we choose them.
The cutoff. Fix once and for all a smooth function chi: R -> [0,1]
with
chi(t) = 0 for t <= 1, chi(t) = 1 for t >= 2. (3.2)
An explicit choice avoids any hand-waving about existence:
zeta(t) = exp(-1/t) for t > 0, zeta(t) = 0 for t <= 0,
chi(t) = zeta(t-1) / ( zeta(t-1) + zeta(2-t) ). (3.3)
Both zeta terms are smooth and nonnegative, and never both zero, so
chi is smooth with the required values. For s in [0,1] and L > 4 set
c(r) = chi(r / L). (3.4)
Then c is smooth, monotone in spirit though not in general, and
c(r) = 0 for r <= L, c(r) = 1 for r >= 2L. (3.5)
Constants C, C_j from here on depend only on chi and the derivative
order j, never on s or L. This uniformity is what makes the whole
construction work; it is worth saying out loud every time it appears.
3.2 The seeds
Define, in the spherical coordinates of Section 2,
f = s c(r) sin^2 theta,
h = s c(r) sin^2 theta cos theta, (3.6)
and let E~ and B~ be the vector fields determined by
iota_{E~} dV_delta = d( f d phi ),
iota_{B~} dV_delta = d( h d phi ). (3.7)
Here dV_delta is the Euclidean volume form and iota is contraction:
the map "vector field -> contraction with dV" is an isomorphism of
2-forms and vector fields, so E~ and B~ are uniquely defined by (3.7).
Finally set
sigma = -( s^2 c(r)^2 sin^4 theta / r^2 )
( dr (x) d phi + d phi (x) dr ). (3.8)
A word on why these particular formulas, since they look arbitrary.
The scalar f is the electric potential of a field whose strength grows
toward the equator. The scalar h is the same thing weighted by
cos theta, which makes it odd under reflection across the equatorial
plane. That oddness is what produces angular momentum in the
electromagnetic field. The tensor sigma is the unique rotation-invariant
trace-free tensor with only mixed radial-azimuthal components, and its
size s^2 c^2 sin^4 theta / r^2 is chosen so that its divergence matches
the cross product of E~ and B~. Nothing is accidental; but the reader
is entitled to check everything, and that is Lemma 3.1.
3.3 The seed lemma
LEMMA 3.1 (Seed identities and bounds).
The fields E~, B~ and tensor sigma defined by (3.6)-(3.8) are smooth
on Omega, invariant under rotation about the third axis, and vanish on
{ 1 <= r <= L }. They satisfy
div_delta E~ = div_delta B~ = 0,
tr_delta sigma = 0,
div_delta sigma = 2 dV_delta( E~, B~, . ), (3.9)
and for every Cartesian multi-index alpha of order j >= 0,
|d^alpha E~|_delta + |d^alpha B~|_delta
<= C_j s r^{-2-j} * 1_{ r >= L }, (3.10a)
|d^alpha sigma|_delta
<= C_j s^2 r^{-3-j} * 1_{ r >= L }. (3.10b)
Moreover div_delta sigma is supported in { L <= r <= 2L } and
|d^alpha (div_delta sigma)|_delta
<= C_j s^2 r^{-4-j} * 1_{ L <= r <= 2L }. (3.11)
Here 1_{...} is the indicator function of a set.
3.4 Proof of Lemma 3.1
Everything is checked first in Cartesian coordinates, which is where
smoothness across the axis is visible, and then in spherical
coordinates, which is where the divergence identities are visible.
(a) Cartesian expressions. Put
X = (x^1, x^2, x^3), T = (-x^2, x^1, 0), w = (x^1)^2 + (x^2)^2.
Then d phi = (x^1 dx^2 - x^2 dx^1)/w and sin^2 theta = w/r^2, so
f d phi = (s c / r^2)( x^1 dx^2 - x^2 dx^1 ),
h d phi = (s c x^3 / r^3)( x^1 dx^2 - x^2 dx^1 ), (3.12)
sigma = -( s^2 c^2 w / r^7 )
( X^flat (x) T^flat + T^flat (x) X^flat ).
Every denominator here is a power of r, and r >= 1 on Omega, so all
three expressions are smooth everywhere in Omega including the axis
w = 0. That is the only reason for writing them this way: the
spherical formulas (3.6)-(3.8) contain sin theta implicitly and one
might worry at theta = 0, pi. There is no singularity, because
(X, T) are polynomial.
(b) Rotation invariance. The vectors X and T are respectively
invariant under rotations about the third axis and rotate as a vector.
The scalars w and c(r) are invariant. Hence all of X, T, w, c, r are
built only from r, w, x^3, and so all of (3.12) is invariant. In
spherical coordinates the same statement is: nothing depends on phi.
(c) Trace-freeness. In Cartesian coordinates, X . T = -x^1 x^2 +
x^2 x^1 = 0, and tr(U (x) V + V (x) U) = 2 (U . V) for covectors, so
tr_delta sigma = 2 s^2 c^2 w (X . T)/r^7 = 0.
(d) Vanishing inside r <= L. c(r) = 0 there by (3.5), and every
formula in (3.6)-(3.8) is a positive power of c. This is also why
div_delta sigma has the compact support claimed in (3.11): it is a
derivative of sigma by parts, so it lives where c' can be nonzero,
namely L <= r <= 2L.
(e) The fields are divergence-free. A contraction iota_E dV is closed
exactly when div E = 0, and d d = 0 gives
d( iota_{E~} dV_delta ) = d d ( f d phi ) = 0,
so div_delta E~ = 0. The same line with h gives div_delta B~ = 0.
That is the entire content of the first two identities in (3.9).
(f) The component formulas. We need E~ and B~ explicitly. Write the
Euclidean volume form in spherical coordinates,
dV_delta = r^2 sin theta dr ^ d theta ^ d phi, (3.13)
and expand the contraction of dV_delta with E = E^r d_r + E^theta d_theta
+ E^phi d_phi:
iota_E dV_delta = r^2 sin theta ( E^r d theta ^ d phi
- E^theta dr ^ d phi
+ E^phi dr ^ d theta ). (3.14)
On the other hand, for any scalar g, d(g d phi) = g_r dr ^ d phi
+ g_theta d theta ^ d phi. Comparing coefficients in (3.14) gives
E^r = g_theta / ( r^2 sin theta ), E^theta = - g_r / ( r^2 sin theta ),
E^phi = 0. (3.15)
Apply (3.15) to g = f, using f_theta = 2 s c sin theta cos theta and
f_r = s c' sin^2 theta:
E~^r = 2 s c cos theta / r^2,
E~^theta = - s c' sin theta / r^2,
E~^phi = 0. (3.16a)
Apply (3.15) to g = h, using h = s c sin^2 theta cos theta, so
h_theta = s c sin theta (3 cos^2 theta - 1) and
h_r = s c' sin^2 theta cos theta:
B~^r = s c (3 cos^2 theta - 1) / r^2,
B~^theta = - s c' sin theta cos theta / r^2,
B~^phi = 0. (3.16b)
Two warnings about these formulas, because they are easy to misread.
First, E~^r and E~^theta are the components in the COORDINATE frame
(d_r, d_theta, d_phi), not in an orthonormal frame. In the
orthonormal frame the polar component is r times the displayed
polar component. The polar components therefore carry a factor
r or r^2 differently from the radial ones. All actual computations
below use the coordinate frame consistently.
Second, the components in (3.16) contain both c and c'. The c'
terms are supported only in the shell L <= r <= 2L and are what
makes the divergence of sigma possible. For r >= 2L the cutoff is
one, c' = 0, and the fields become purely radial:
E~ = (2 s cos theta / r^2) d_r,
B~ = (s (3 cos^2 theta - 1) / r^2) d_r. (3.17)
These are the "tails" of the title. They have size r^{-2} but their
coefficients depend on theta. A Coulomb field would be
Qe xhat/(4 pi r^2), whose coefficient is constant on the sphere.
(g) The divergence of sigma. Write the single independent component as
a = sigma_{r phi} = sigma_{phi r} = - s^2 c^2 sin^4 theta / r^2,
and all other components of sigma vanish. (3.18)
Use the coordinate formula for the divergence of a symmetric
2-tensor,
(div_delta sigma)_i = ( 1/sqrt(det delta) ) d_j ( sqrt(det delta)
sigma_i^j ) - (1/2) sigma^{jk} d_i delta_jk,
(3.19)
with sqrt(det delta) = r^2 sin theta. The second term of (3.19)
vanishes identically: delta is diagonal in these coordinates, so
d_i delta_jk = 0 for every i, j, k. (Said differently, the Christoffel
symbols of the Euclidean metric vanish in Cartesian coordinates and
give no contribution here because the coordinate frame we are using
is orthogonal with diagonal metric.)
The tensor sigma has, in index-upper form, only one nonzero entry:
sigma_phi^r = delta^{rr} sigma_{phi r} = a, while
sigma_phi^phi = delta^{phi phi} sigma_{phi phi} = 0. So sigma_phi^j is
nonzero only for j = r, and equals a. Therefore
(div_delta sigma)_r = (1/(r^2 sin theta)) d_phi ( a / sin theta )
= 0, (3.20a)
(div_delta sigma)_theta = 0, (3.20b)
(div_delta sigma)_phi = (1/(r^2 sin theta)) d_r ( r^2 sin theta a )
= a_r + 2a/r. (3.20c)
The first two are zero because nothing depends on phi. In the third,
substitute a from (3.18). The c^2/r^2 part contributes
d/dr ( c^2 / r^2 ) = 2 c c' / r^2 - 2 c^2 / r^3, (3.21)
so
a_r + 2a/r = - s^2 sin^4 theta [ 2 c c'/ r^2 - 2c^2/r^3
+ 2c^2/r^3 ]
= - 2 s^2 c c' sin^4 theta / r^2. (3.22)
The two c^2/r^3 terms cancel exactly. Only the cutoff derivative
survives, which is why div_delta sigma is supported in L <= r <= 2L.
(h) The cross product. By definition (2.1) and (3.13),
dV_delta(E~, B~, d_phi) = r^2 sin theta ( E~^r B~^theta - E~^theta B~^r ).
Insert (3.16):
E~^r B~^theta = (2 s c cos theta / r^2)(- s c' sin theta cos theta / r^2)
= - 2 s^2 c c' sin theta cos^2 theta / r^4,
E~^theta B~^r = (- s c' sin theta / r^2)( s c (3 cos^2 theta - 1) / r^2)
= - s^2 c c' sin theta (3 cos^2 theta - 1) / r^4.
Subtracting the second from the first,
E~^r B~^theta - E~^theta B~^r
= ( s^2 c c' sin theta / r^4 )
[ - 2 cos^2 theta + 3 cos^2 theta - 1 ]
= ( s^2 c c' sin theta / r^4 )( cos^2 theta - 1 )
= - s^2 c c' sin^3 theta / r^4. (3.23)
Multiplying by r^2 sin theta,
dV_delta(E~, B~, d_phi) = - s^2 c c' sin^4 theta / r^2, (3.24)
and both other components vanish because E~ and B~ have no phi
component and because nothing depends on phi. Comparing (3.24) with
(3.22) gives exactly
(div_delta sigma)_phi = 2 dV_delta( E~, B~, d_phi ),
and (3.20a,b) say both sides vanish for the other two indices. That
is the third identity in (3.9). The point to notice: sigma had to
carry a factor s^2 to match the product E~ B~ of two fields of size s,
and its radial power had to be r^{-3} so that d_r brings it to r^{-4}.
(i) Derivative bounds. Two ingredients.
First, the Cartesian derivatives of the cutoff satisfy, for every
multi-index alpha of order j >= 1,
|d^alpha c| <= C_j r^{-j} * 1_{ L <= r <= 2L }. (3.25)
Indeed d^alpha c(x) = L^{-j} (d^alpha chi)(x/L), which is nonzero
only where x/L lies between 1 and 2, i.e. L <= r <= 2L, and there
it is bounded by C_j L^{-j}; and on that annulus r and L are
comparable.
Second, the uncut primitives in (3.12) are smooth homogeneous
functions. The coefficient of f d phi is (s c/r^2)(x^1 dx^2 - x^2
dx^1); the factor multiplying c is a smooth homogeneous function of
degree -1 in x. Differentiating it j times produces a smooth
homogeneous function of degree -1-j. Exterior differentiation
preserves that, so by (3.25)
|d^alpha E~|_delta <= C_j s r^{-2-j} 1_{r >= L}, (3.26)
where the power -2 records the size of the field for r >= 2L and
the derivative loss is absorbed into C_j. The same argument for h d
phi gives the bound on B~. For sigma, the uncut Cartesian
coefficients are smooth homogeneous of degree -3: w has degree 2,
X^flat (x) T^flat has degree 1+1 = 2, and r^{-7} has degree -7, and
2 + 2 - 7 = -3. So j derivatives produce degree -3-j, and
|d^alpha sigma|_delta <= C_j s^2 r^{-3-j} 1_{r >= L}. (3.27)
Finally div_delta sigma is built from a_r, that is from one
derivative of sigma together with the c' of (3.22), and it is
supported on the shell by (i). This gives (3.11). All bounds are
uniform in s in [0,1] and in L > 4, since only the fixed cutoff
chi entered, and the jets of chi at r = L and r = 2L are as smooth
as chi is.
That completes Lemma 3.1. Nothing in it used the equations of motion;
it is pure computation on flat space. The whole proof of the theorem
is, in a sense, an afterlife of these three formulas.
----------------------------------------------------------------
4. STEP 2. ONE SCALAR EQUATION FOR THE SHAPE OF SPACE
----------------------------------------------------------------
4.1 The rescaling
We now turn the flat seeds into curved, physical data. Pick a smooth
positive function u on Omega and define
g = u^4 delta, (4.1a)
K = u^{-2} sigma, (4.1b)
E = u^{-6} E~, (4.1c)
B = u^{-6} B~. (4.1d)
Three different powers of u, and each has a job. The power 4 in the
metric is the standard conformal weight in three dimensions: it is
exactly the exponent for which the conformal Laplacian identity below
holds. The power -2 in K is chosen so that K has zero trace with
respect to g (check: g^{ij}K_ij = u^{-4}delta^{ij} u^{-2}sigma_ij =
u^{-6} tr_delta sigma = 0). The power -6 in E, B is chosen so that the
Maxwell flux forms are u-independent, as we check in a moment.
The function u is exactly what Einstein's equation will determine. It
is the only unknown left.
4.2 What the rescaling does to the constraints
LEMMA 4.1 (Conformal electrovacuum equations).
For every smooth positive u on Omega, the data (4.1) satisfy
tau = tr_g K = 0, (4.2a)
div_g E = div_g B = 0, (4.2b)
div_g K = 2 (E x_g B)^flat_g. (4.2c)
They satisfy the electrovacuum Hamiltonian constraint (2.2a) if and
only if
- Delta_delta u = p u^{-7} + ell u^{-3} on Omega, (4.3)
where
p = (1/8) |sigma|_delta^2
= s^4 c^4 sin^6 theta / (4 r^6), (4.4a)
ell = (1/4)( |E~|_delta^2 + |B~|_delta^2 )
= ( s^2 / (4 r^4) ) [ c^2 ( 9 t^4 - 2 t^2 + 1 )
+ (r c')^2 ( 1 - t^4 ) ], (4.4b)
with t = cos theta. Both p and ell are nonnegative, smooth, and
rotation-invariant. For every Cartesian multi-index alpha of order
j >= 0,
|d^alpha p| <= C_j s^4 r^{-6-j} 1_{r >= L},
|d^alpha ell| <= C_j s^2 r^{-4-j} 1_{r >= L}. (4.5)
4.3 Proof of Lemma 4.1
(a) Volume forms and flux forms. In three dimensions, g = u^4 delta
gives det g = u^{12} det delta, so
dV_g = u^6 dV_delta. (4.6)
Consequently the contractions satisfy, for every vector Z,
iota_{u^{-6}E~} dV_g = iota_{E~} ( dV_g / u^6 ) = iota_{E~} dV_delta.
(4.7)
That is the whole purpose of the exponent -6: raising the volume form
by u^6 and lowering the field by u^{-6} cancel. So the physical flux
2-forms are literally the flat ones,
iota_E dV_g = d(f d phi), iota_B dV_g = d(h d phi), (4.8)
and since exact forms are closed, div_g E = div_g B = 0. That is
(4.2b), and it never used u at all.
(b) The cross product. For any vector Z, using dV_g = u^6 dV_delta and
E = u^{-6}E~, B = u^{-6}B~,
(E x_g B)^flat_g(Z) = dV_g(E, B, Z)
= u^6 u^{-12} dV_delta(E~, B~, Z)
= u^{-6} dV_delta(E~, B~, Z). (4.9)
So the electromagnetic momentum source also rescales by exactly u^{-6}.
(c) Trace of K. Already done in 4.1:
tr_g K = g^{ij} K_ij = u^{-4} delta^{ij} u^{-2} sigma_ij
= u^{-6} tr_delta sigma = 0. (4.10)
(d) The divergence of K. This is the only genuinely computational
step, so it is done in Cartesian coordinates with full indices.
Write g = e^{2b} delta with b = 2 log u, so g_{ij} = e^{2b}delta_ij
and K = e^{-b}sigma. Let
D^k_ij = Gamma^k_ij(g) - Gamma^k_ij(delta) (4.11)
be the difference of the Christoffel symbols. Since delta has
vanishing Christoffels, D is the full Christoffel symbol of g, and a
direct computation gives
D^k_ij = delta^k_i b_j + delta^k_j b_i - delta_ij b^k, (4.12)
where b_i = d_i b and indices on the right are raised with delta. Two
identities will be used, and both are one line:
delta^{jk} D^l_kj = b^l + b^l - 3 b^l = - b^l. (4.13)
delta^{jk} D^l_ki sigma_lj
= b_i sigma^j_j - b^l sigma_il
= b^j sigma_ij - b^l sigma_il
= 0. (4.14)
In (4.13) the 3 comes from delta^jk delta_kj = 3, and in (4.14) the
first equality is the expansion of D^l_ki, while the last is just the
renaming j <-> l; the middle vanishing uses tr_delta sigma = 0 from
Lemma 3.1. This is where trace-freeness of the seed pays for itself.
Now the covariant derivative of K. Since K_ij = e^{-b} sigma_ij,
nabla_k K_ij = e^{-b} [ d_k sigma_ij - sigma_ij b_k
- D^l_ki sigma_lj - D^l_kj sigma_il ]. (4.15)
Contract with g^{jk} = e^{-2b} delta^{jk} and use (4.13), (4.14):
(div_g K)_i = e^{-3b} delta^{jk} [ d_k sigma_ij - sigma_ij b_k ]
+ e^{-3b} [ delta^{jk} ( -D^l_ki sigma_lj
- D^l_kj sigma_il ) ]
= e^{-3b} [ delta^{jk}d_k sigma_ij - b^j sigma_ij
+ b^l sigma_il ]
= e^{-3b} delta^{jk} d_k sigma_ij
= u^{-6} (div_delta sigma)_i. (4.16)
Read the two middle terms: -b^j sigma_ij and +b^l sigma_il are the
same number and they cancel. That cancellation is the reason the
power -2 was chosen for K; any other power would leave behind a term
proportional to b times sigma and the construction would fail.
Combining (4.16) with Lemma 3.1 and (4.9) gives
div_g K = u^{-6} div_delta sigma
= u^{-6} * 2 dV_delta(E~, B~, .)
= 2 (E x_g B)^flat_g, (4.17)
which is (4.2c). So three of the four constraints hold for EVERY
positive u, with no equation to solve.
(e) The Hamiltonian constraint. The scalings of the individual norms
are worth writing out once. Since g_{ij} = u^{4}delta_ij:
|K|_g^2 = g^{ik}g^{jl}K_ijK_kl = u^{-8}u^{-8}u^{-4}|sigma|^2
= u^{-12}|sigma|_delta^2, (4.18a)
|E|_g^2 = g_ij E^i E^j = u^4 u^{-12}|E~|_delta^2
= u^{-8}|E~|_delta^2, (4.18b)
|B|_g^2 = u^{-8}|B~|_delta^2. (4.18c)
For the curvature we use the conformal formula in three dimensions,
which is standard: for g = e^{2b}delta,
R_g = e^{-2b} ( - 4 Delta_delta b - 2 |grad_delta b|^2 ). (4.19)
The formula is classical; two checks pin it down. If b is constant
both sides are zero, as they must be since a constant rescaling of
flat space is flat. And if b = 2 log u one computes Delta_delta b
= 2 Delta u/u - 2|grad u|^2/u^2 and |grad b|^2 = 4|grad u|^2/u^2, so
the bracket becomes -8 Delta_delta u / u exactly, with the two
|grad u|^2 terms cancelling. Hence
R_g = - 8 u^{-5} Delta_delta u. (4.20)
Now put everything into the Hamiltonian constraint (2.2a). With
tau = 0 from (4.10),
-8 u^{-5} Delta_delta u - u^{-12} |sigma|^2
= 2 u^{-8} ( |E~|^2 + |B~|^2 ). (4.21)
Multiply by -u^5/8:
Delta_delta u + (1/8) u^{-7} |sigma|^2
= - (1/4) u^{-3} ( |E~|^2 + |B~|^2 ). (4.22)
Move the first-order term to the right and read off the coefficients:
- Delta_delta u = ( |sigma|^2/8 ) u^{-7}
+ ( (|E~|^2+|B~|^2)/4 ) u^{-3}. (4.23)
That is (4.3) with p and ell as in (4.4). Notice how the exponents
-7 and -3 are forced: -7 = -12 + 5 from the trace of K squared, and
-3 = -8 + 5 from the electromagnetic energy. They are not choices.
(f) The explicit coefficients. Using (3.18) and the fact that
|sigma|_delta^2 = 2 a^2/(r^2 sin^2 theta) for a tensor with only
sigma_{r phi} = sigma_{phi r} = a,
|sigma|^2 = 2 s^4 c^4 sin^8 theta / ( r^4 * r^2 sin^2 theta )
= 2 s^4 c^4 sin^6 theta / r^6,
p = |sigma|^2/8 = s^4 c^4 sin^6 theta / (4 r^6). (4.24)
For the electromagnetic part, insert (3.16) into the coordinate-frame
norm |E~|^2 = (E~^r)^2 + r^2 (E~^theta)^2 (the r^2 because the theta
direction is orthogonal to d_r with length r):
|E~|^2 = 4 s^2 c^2 cos^2 theta / r^4 + s^2 c'^2 sin^2 theta / r^2,
|B~|^2 = s^2 c^2 (3 cos^2 theta - 1)^2 / r^4
+ s^2 c'^2 sin^2 theta cos^2 theta / r^2. (4.25)
Adding, and writing (rc')^2 = r^2 c'^2 to collect the c' terms,
|E~|^2 + |B~|^2
= (s^2/r^4) [ c^2 ( 4 cos^2 theta + (3 cos^2 theta - 1)^2 ) ]
+ (s^2 (rc')^2 / r^4) [ sin^2 theta (1 + cos^2 theta) ]. (4.26)
Now expand in t = cos theta. Since (3t^2-1)^2 + 4t^2 = 9t^4 - 6t^2 + 1
+ 4t^2 = 9t^4 - 2t^2 + 1, and since sin^2 theta (1 + cos^2 theta) =
(1-t^2)(1+t^2) = 1 - t^4,
|E~|^2 + |B~|^2 = (s^2/r^4) [ c^2 (9t^4 - 2t^2 + 1)
+ (rc')^2 (1 - t^4) ], (4.27)
and dividing by 4 gives (4.4b).
(g) Bounds. Both p and ell are explicit products of bounded factors
s, c, c' and powers of r and sin theta, and both vanish for r <= L
with the derivatives of c controlled by (3.25). Thus the derivative
bounds (4.5) follow directly from Lemma 3.1 or, equivalently, are just
the statement that p and ell are squared norms of quantities already
estimated in (3.10). Nonnegativity is manifest: p is a squared norm,
and ell is a sum of two squared norms. Rotation invariance is manifest
because nothing depends on phi.
That completes Lemma 4.1.
4.4 The boundary condition, and why u_0 = 1 + 1/r
Only one job remains: solve (4.3). But solving it on Omega requires a
boundary condition at r = 1, and the right condition comes from
demanding that S be a marginally outer trapped surface.
We will see in Section 6 that the coordinate sphere S_r has future
outward expansion
theta_+(S_r) = ( 1/(u^6 r^2) ) d_r ( r^2 u^4 ). (4.28)
So theta_+(S) = 0 forces
d_r ( r^2 u^4 ) = 0 at r = 1, i.e. u_r + u/2 = 0 at r = 1. (4.29)
That is the Robin condition. It is worth noticing immediately that
u_0(r) = 1 + 1/r (4.30)
satisfies it: u_0'(1) + u_0(1)/2 = -1 + 1 = 0. It also tends to 1 at
infinity. It is harmonic in flat space away from the origin:
- Delta_delta u_0 = 0 for r > 0, since 1/r is the fundamental solution
in three dimensions. So u_0 is the "free" background: it gives the
Schwarzschild-like metric g = u_0^4 delta with mass 2 and horizon at
r = 1, if the source terms p and ell were absent.
Therefore we write
u = u_0 + v (4.31)
with v the correction caused by the source. The equation for v is
- Delta_delta v = p u^{-7} + ell u^{-3} on Omega,
v_r + v/2 = 0 at r = 1,
v -> 0 as r -> infinity, (4.32)
because u_0 satisfies the homogeneous Robin condition and is harmonic.
Section 5 solves (4.32) and proves estimates uniform in s and L.
----------------------------------------------------------------
5. STEP 3. SOLVING THAT EQUATION, WITH UNIFORM CONTROL
----------------------------------------------------------------
5.1 What has to be proved, and why it is delicate
The equation to solve is (4.32):
- Delta_delta v = H, H = p u^{-7} + ell u^{-3}, u = u_0 + v.
Read it as a fixed point problem: H depends on v, and v is the Newton
potential of H. The point of the construction is the geometry of the
source.
The source lives only at radii r >= L, so it is a distance at least
L - 1 away from the boundary sphere S. A Newton potential at distance
D from a source of total strength M is of size M/D, not M. Meanwhile
the total strength of the source is small: from (4.5), the energy
density is O(s^2 r^{-4}) and the volume is O(r^2 dr), so
integral of |H| over R^3 is O( s^2 / L ). (5.1)
Dividing by the distance L to the boundary gives
size of v near S is O( s^2 / L^2 ). (5.2)
That is much smaller than 1 for large L, and it is this L^{-2} that
lets us keep u close to u_0 everywhere and control every surface in
Section 6.
Two remarks on why one cannot cheat here.
First, the energy tail is only marginally integrable: r^{-4} against
a volume r^2 dr gives r^{-2}dr, which converges but leaves a long
tail. There is no first moment of H: the integral of |y| |H| dy
diverges logarithmically. So multipole expansion is not available and
the decay of the Newton potential is not the usual one. Everything
has to be done with the L^1 norm of H and with explicit splitting of
the integral.
Second, the boundary condition is Robin, not Dirichlet. A Newton
potential does not satisfy it. The device that fixes this is Kelvin
reflection, and it is exact, not approximate: it produces a harmonic
function near the origin that exactly cancels the radial derivative
at r = 1.
5.2 The Newton potential and Kelvin reflection
LEMMA 5.1 (Newton potential and Robin reflection).
Let L > 4 and let H be continuous on R^3 with
|H(y)| <= a |y|^{-4} 1_{ |y| >= L }. (5.3)
Define
N_H(x) = (1/(4 pi)) * integral over R^3 of H(y)/|x-y| dy. (5.4)
Then N_H is well defined and C^1, satisfies
- Delta N_H = H (distributionally, and classically off supp H),
(5.5)
and is smooth and harmonic on { |x| < L }. For j = 0, 1 it obeys
|D^j N_H(x)| <= C_j a L^{-2-j} if |x| <= L/2,
|D^j N_H(x)| <= C_j a ( r^{-2-j} + L^{-1} r^{-1-j} ) if r >= L/2,
(5.6)
where r = |x|. The first estimate in fact holds for every j >= 0.
Define the reflected potential on Omega by
(R H)(x) = N_H(x) + r^{-1} N_H( x / r^2 ). (5.7)
Then R H preserves nonnegativity, satisfies -Delta(R H) = H
distributionally, satisfies the Robin condition
(d_r + 1/2)(R H) = 0 on S, (5.8)
and obeys the bounds
||R H||_inf <= C a L^{-2},
|R H| + r |d_r R H| <= C a L^{-2},
|d_r^2 R H| <= C a L^{-2} for 1 <= r <= 2. (5.9)
Finally, the homogeneous Robin problem on Omega has no nonzero
solution that is smooth, C^1 up to S, and O(r^{-1}) at infinity.
5.3 Proof of Lemma 5.1
(a) Size of the source. Throughout,
integral of |H| <= 4 pi a / L. (5.10)
Indeed the integral of |y|^{-4} over the outside of the ball of radius
L is 4 pi times the integral of rho^{-2} d rho from L to infinity,
which is 4 pi / L.
(b) Existence and C^1 regularity. The kernel |x-y|^{-1} is locally
integrable in three dimensions, and so is its gradient |x-y|^{-2}:
the integral of rho^{-2} times the area 4 pi rho^2 d rho converges at
0. Splitting the integral into a small ball around the singularity
and its complement, and bounding the singular part by a constant
times the radius of the ball, shows that N_H and its first derivatives
are continuous. So N_H is C^1.
(c) The distributional equation. The function G(x) = 1/(4 pi |x|) is
the fundamental solution of the Laplacian in three dimensions, so
-Delta G = delta_0. Convolving with H and applying Fubini (everything
is integrable by (5.10)) gives (5.5).
(d) Bounds away from the source. Fix x with r = |x| <= L/2. Then
|x-y| >= |y| - |x| >= |y|/2, and source separation lets us
differentiate under the integral any number of times:
Using that the derivative of order j of the kernel is bounded by
C_j |x-y|^{-1-j}, and that |x-y| >= |y|/2 here,
|D^j N_H(x)|
<= C_j a * integral over rho >= L of rho^{-4} rho^{-1-j} rho^2 d rho
= C_j a * integral over rho >= L of rho^{-3-j} d rho
= C_j a L^{-2-j} / (2+j). (5.11)
That is the first line of (5.6), valid for all j. The factor L^{-2}
and not L^{-1} is the important gain: the potential at distance
comparable to L from a source of total strength a/L is of size a/L^2.
Fix now r >= L/2 and split the integral at |x - y| = r/2. Two pieces.
Inner piece: where |x-y| <= r/2. There |y| >= r/2, and the
derivative of the kernel of order j is bounded by C|x-y|^{-1-j}, so
|inner| <= C a r^{-4} * integral over |z| <= r/2 of |z|^{-1-j} dz
<= C a r^{-4} r^{2-j}
= C a r^{-2-j}. (5.12)
(The integral of |z|^{-1-j} over the ball of radius r/2 is
C r^{2-j} for j = 0, 1, which is where the dimension 3 enters.)
Outer piece: where |x-y| >= r/2. There the kernel and its first
derivatives are bounded by C r^{-1-j}, and by (5.10)
|outer| <= C r^{-1-j} * (4 pi a / L) = C a L^{-1} r^{-1-j}.
(5.13)
Adding (5.12) and (5.13) gives the second line of (5.6).
(e) The reflected operator. Write I(x) = x/r^2 for the inversion in
the unit sphere. It maps Omega to the unit ball, where N_H is smooth
and harmonic. The Kelvin identity in three dimensions,
Delta ( r^{-1} N_H(I(x)) ) = r^{-5} ( Delta N_H )( I(x) ), (5.14)
is a direct computation (it is the n = 3 case of the standard formula
for the behaviour of the Laplacian under inversion). Its right side
vanishes on Omega, because I(x) has length 1/r <= 1 < L and N_H is
harmonic there. Hence the reflected term is harmonic on Omega, and
R H satisfies -Delta(R H) = H there.
Positivity: both |x-y|^{-1} and its reflected counterpart are positive
kernels, so H >= 0 implies R H >= 0.
The Robin condition. Let omega be a unit vector and evaluate along the
ray x = r omega. At r = 1 the reflected term contributes N_H(omega)
with the same argument as the direct term, and direct differentiation
gives
(R H)(omega) = 2 N_H(omega),
d_r (R H)(omega) = d_r N_H(omega) - d_r N_H(omega) - N_H(omega)
= - N_H(omega). (5.15)
So d_r(R H) + (1/2)(R H) = -N_H + N_H = 0 at r = 1. Note that this
holds with full angular dependence, not merely on average: the two
pieces of (5.7) have matching values and matching tangential
derivatives on S, and their radial derivatives combine to give (5.15).
That is exactly what a Robin condition requires.
The bounds (5.9). For 1 <= r <= L/2, use the first line of (5.6) for
both terms: the direct argument is within L/2 of the origin, and the
reflected argument has length 1/r <= 1 < L/2. This gives the first two
lines of (5.9) with |d_r R H| also bounded by C a L^{-2}. For r >= L/2,
the second line of (5.6) gives |N_H(x)| <= C a (r^{-2} + L^{-1}r^{-1})
<= C a L^{-2} since r >= L/2, and r|d_r N_H| <= C a (r^{-2} +
L^{-1}r^{-1}) <= C a L^{-2} by the same computation multiplied by r.
The reflected term also contributes at most C a L^{-2} times 1 and
C a L^{-2} times r^{-1} for its radial derivative, because the chain
rule from I(x) brings a factor r^{-2}. Adding gives (5.9) for the
first two lines. For the second derivative on 1 <= r <= 2, all
arguments stay in 1/2 <= |.| <= 2, where N_H is smooth and all its
derivatives are bounded by C_j a L^{-2-j} <= C a L^{-2} by (5.11), so
the same differentiation gives the third line.
(f) Uniqueness of the homogeneous problem. Let w be smooth and
harmonic on Omega, C^1 up to S, with w = O(r^{-1}) and
w_r + (1/2)w = 0 on S. Extend it into the ball by Kelvin reflection:
w_hat(x) = r^{-1} w(I(x)) for 0 < r < 1. (5.16)
At r = 1 the values agree, and the tangential derivatives agree
because the tangential part of I is the identity on S. The radial
derivative of the extension at S is the derivative of r^{-1}w(omega/r)
with respect to r as r approaches 1 from below, which equals
-w(omega) - w_r(omega), i.e. -w + w/2 = -w/2 = w_r, using the Robin
condition. So the extension is C^1 across S. Since it is harmonic on
both sides and C^1 across the interface, it is distributionally
harmonic on the punctured space.
Moreover it is bounded near the origin: near 0, w_hat(x) =
r^{-1}w(I(x)) with |I(x)| = 1/r, so |w_hat| <= r^{-1} C r = C. And it
is O(r^{-1}) at infinity because w is.
Now let eps > 0 and R > 0, and apply the maximum principle to w_hat
and to -w_hat on the annulus eps <= |x| <= R. On the inner sphere the
bound is C (times 1 = eps/eps), and on the outer sphere it is C/R. The
function C eps/|x| + C/R is harmonic on the annulus and dominates the
absolute value on both boundary spheres, so by the maximum principle
|w_hat(x)| <= C eps/|x| + C/R for eps <= |x| <= R. (5.17)
Let eps tend to 0 and then R to infinity at a fixed x different from
0. The right side tends to 0, so w_hat(x) = 0. Hence w = 0.
That completes Lemma 5.1. The reader should hold on to two facts from
it, because they are the whole engine of Section 5: the L^{-2} gain in
(5.11), and the exactness of the Robin cancellation in (5.15).
5.4 Deriving regularity from the source
The next lemma says: the smoother and smaller the source, the smoother
and smaller the potential, one derivative at a time, with a constant
that involves only the L^1 mass of the source and not its moments.
LEMMA 5.2 (Localized derivative bootstrap).
Let H satisfy (5.3), and let j >= 0. If H is C^j and
|D^i H(y)| <= a_j |y|^{-4-i} 1_{ |y| >= L } for 0 <= i <= j,
(5.18)
then N_H is C^{j+1} on R^3 and the bounds (5.6) hold through order
j+1, with C_j a_j in place of C_j a. In particular no first moment of
H is used anywhere.
5.5 Proof of Lemma 5.2
Fix a point x_0 and a smooth compactly supported cutoff zeta that is
one near x_0. Keep zeta fixed while differentiating in x. Let
G(x) = 1/(4 pi |x|). For a multi-index alpha with |alpha| = j+1, write
alpha = beta + e_i with |beta| = j, and integrate by parts in the
distributional formula. Since zeta H has compact support, no boundary
term appears, and near x_0
D^alpha N_H(x) = integral of d_i G(x-y) * D_y^beta ( zeta H )(y) dy
+ integral of D^alpha G(x-y) * (1 - zeta(y)) H(y) dy.
(5.19)
The first integral is continuous: d_i G is locally integrable, its
singularity is of order |x-y|^{-2}, and the small-ball argument from
step (b) above applies. The second integral is smooth near x_0 because
(x, y) -> 1 - zeta(y) keeps the kernel away from its singularity on
supp(1-zeta).
Now the estimate at |x_0| = r >= L/2. Take zeta supported in the ball
of radius r/2 about x_0 and equal to one on the ball of radius r/4,
with |D^i zeta| <= C_i r^{-i}. On that support |y| >= r/2, so by
(5.18) every Leibniz term of D^beta(zeta H) is bounded by
C_j a_j r^{-4-j}. Then the first integral of (5.19) at x_0 is bounded
by
C_j a_j r^{-4-j} * integral over |z| <= r/2 of |z|^{-2} dz
<= C_j a_j r^{-4-j} * C r = C_j a_j r^{-3-j}, (5.20)
and the second by
C_j r^{-2-j} * integral of |H| <= C_j r^{-2-j} * 4 pi a / L
= C_j a_j L^{-1} r^{-2-j}. (5.21)
These are the two terms of the second line of (5.6) at order j+1.
Applying the same formula at lower orders gives the lower orders. On
|x| <= L/2 the source is separated, differentiation under the integral
is legitimate, and (5.11) gives every order at once.
5.6 Why the source has small L^1 mass but no first moment
One calculation from Lemma 4.1 that gets used constantly, so it is
worth isolating. Using the explicit coefficients (4.4) and the bound
u >= 1 (proved with the fixed point, but already clear since u_0 >= 1
and v >= 0):
0 <= H = p u^{-7} + ell u^{-3}
<= C s^4 r^{-6} + C s^2 r^{-4}
<= C s^2 r^{-4} for r >= L > 4, s <= 1. (5.22)
So H obeys (5.3) with a = C s^2. Consequently
0 <= integral of H <= C s^2 / L, (5.23)
and every estimate of Lemma 5.1 applies with a = C s^2. In particular
R H is bounded by C s^2 L^{-2}, which is the smallness of (5.2).
Note that H is NOT of the Coulomb form and does not have a finite
first moment: the integral of |y| |H| diverges like log. This is a
statement about the tail, since H ~ s^2 r^{-4} times an angular
function, and it is why there is no multipole expansion in this paper.
5.7 The fixed point, and the uniform estimates
PROPOSITION 5.3 (Uniform conformal factor).
There is L_0 > 4, depending only on the cutoff chi, such that for every
L >= L_0 and every s in [0,1] the boundary value problem (4.32) has a
smooth axisymmetric solution u = u_0 + v on Omega satisfying
- Delta_delta u = p u^{-7} + ell u^{-3} on Omega,
u_r + (1/2) u = 0 on S,
u -> 1 as r -> infinity. (5.24)
The correction v obeys v >= 0 and v = O(r^{-1}), and with this
normalization the solution is unique. With M = 3, independent of s
and L,
1 <= u_0 <= u <= M. (5.25)
Uniformly in s in [0,1] and L >= L_0,
|v| + r |v_r| <= C s^2 L^{-2} for 1 <= r <= L, (5.26a)
|v| + r |v_r| <= C s^2 / (L r) for r >= L, (5.26b)
|v_rr| <= C s^2 L^{-2} for 1 <= r <= 2. (5.26c)
For every integer k >= 0, all Cartesian derivatives satisfy
|D^k v| <= C_k ( s^2 / L ) r^{-1-k}, (5.27a)
|D^k (u - 1)| <= C_k r^{-1-k}. (5.27b)
For fixed L, the map s -> v_s is continuous in the supremum norm, and
so are the maps
s -> integral over R^3 of H_s dx, s -> N_s(0), (5.28)
where H_s = p u_s^{-7} + ell u_s^{-3}, extended by zero inside the
unit ball, and N_s is its Newton potential. Moreover
0 <= integral of H_s <= C s^2 / L,
0 <= N_s(0) <= C s^2 / L^2. (5.29)
At s = 0 the solution is exactly u = u_0.
The two regimes in (5.26) are worth reading out loud. For r <= L the
correction is of size s^2/L^2, because the source is at distance at
least L/2. For r >= L the correction is of size s^2/(Lr), because the
source now sits at distance r and behaves like a total strength s^2/L
at that distance. The two formulas agree at r = L. In particular
|v_r| <= C s^2 L^{-2} everywhere, which is what the horizon analysis
in Section 6 needs.
5.8 Proof of Proposition 5.3
Step 1: the map. Let C_b(Omega) be the continuous bounded functions
with the supremum norm, and let
K = { v in C_b(Omega) : v >= 0 }. (5.30)
K is a closed subset of a complete metric space, hence complete. For
v in K define
H_v = p (u_0 + v)^{-7} + ell (u_0 + v)^{-3},
T_s(v)(x) = N_{H_v}(x) + r^{-1} N_{H_v}(x/r^2). (5.31)
Extend H_v by zero inside the unit ball. This extension is
continuous because H_v vanishes for r <= L (both p and ell do), and
L < 1 is impossible since L > 4.
Step 2: T_s maps K into K. Since u_0 >= 1 and v >= 0, the base
u_0 + v >= 1 on all of Omega. Using the bounds (4.5) and r >= L > 4,
0 <= H_v <= C s^4 r^{-6} + C s^2 r^{-4} <= C s^2 r^{-4}
for r >= L, (5.32)
because s <= 1 and r >= 1. So (5.3) holds with a = C s^2, and
Lemma 5.1 gives nonnegativity and
||T_s(v)||_inf <= C s^2 L^{-2}. (5.33)
Step 3: T_s is a contraction on K. Both summands of H_v are
decreasing functions of their argument, so for v, w in K, by the mean
value theorem applied to x -> x^{-7} and x -> x^{-3} on the interval
[1, infinity),
|H_v - H_w| <= [ 7 p (u_0+v)^{-8} + 3 ell (u_0+v)^{-4} ]
* |v - w|
<= C ( s^4 r^{-6} + s^2 r^{-4} ) |v - w|
<= C s^2 r^{-4} |v - w| for r >= L. (5.34)
Apply Lemma 5.1 with a = C s^2 ||v - w||_inf to the difference. The
key point is that H_v - H_w is not nonnegative, but the operator R is
bounded acting on functions obeying (5.3) and its bounds used only
|H|: the kernel |x-y|^{-1} plus its reflection is applied to
H_v - H_w, and the estimate (5.9) was derived from (5.10), which uses
|H| only. Hence
||T_s(v) - T_s(w)||_inf <= C L^{-2} ||v - w||_inf. (5.35)
(For a reader who wants this spelled out: for any H obeying (5.3) with
constant a, one has ||R H||_inf <= C a L^{-2}, and one applies this to
a = C s^2 ||v-w||. No sign is used.)
Choose L_0 > 4 so large that C L_0^{-2} <= 1/2 and C L_0^{-2} <= 1.
Then for all L >= L_0, T_s maps K into itself with contraction
constant 1/2, and T_s maps K into the smaller set
{ ||v||_inf <= C s^2 L^{-2} }, which is contained in { ||v|| <= 1 }.
Step 4: existence and basic properties. By the contraction theorem,
for each s there is a unique v in K with T_s(v) = v. Set u = u_0 + v.
By Lemma 5.1, T_s(v) satisfies the Robin condition and
- Delta_delta v = - Delta_delta T_s(v) = H_v = p u^{-7}
+ ell u^{-3}, (5.36)
distributionally and, once u is smooth, classically. Also u_0 is
harmonic and satisfies the homogeneous Robin condition, so u
satisfies (5.24). Since u_0 >= 1 and v >= 0,
1 <= u <= 1 + 1/r + 1 <= 3 for r >= 1, (5.37)
which is (5.25) with M = 3. At s = 0 we have p = ell = 0 identically,
so H_v = 0 and T_0(v) = 0 for every v; the unique fixed point is
v = 0 and u = u_0.
Step 5: axisymmetry. The coefficients p and ell depend only on r and
theta, hence are invariant under rotations about the third axis. The
operator T_s, built from the Laplacian and the reflected inversion, is
rotation-invariant as well, and it has a unique fixed point in K.
Therefore the fixed point is invariant. In coordinates: whatever
depends on phi could be averaged over the rotation group and would
still be a fixed point, hence must equal the fixed point.
Step 6: radial estimates (5.26). Apply Lemma 5.1 with H = H_v and
a = C s^2, which is legitimate by (5.32). For 1 <= r <= L/2 the first
line of (5.6) gives |N_v| + r|d_r N_v| <= C s^2 L^{-2}; for
L/2 <= r <= L the second line gives |N_v| + r|d_r N_v|
<= C s^2 (r^{-2} + L^{-1} r^{-1}) <= C s^2 L^{-2}; and for r >= L,
|N_v| <= C s^2 (r^{-2} + L^{-1}r^{-1}) <= C s^2 / (L r),
r |d_r N_v| <= C s^2 (r^{-2} + L^{-1}r^{-1}) <= C s^2 / (L r).
The reflected term contributes at most C s^2 L^{-2} to both |v| and
r|v_r| (its radial derivative brings an extra factor r^{-2} from the
inversion, and its argument stays inside the unit ball). Adding gives
(5.26a) and (5.26b). For (5.26c), on 1 <= r <= 2 both arguments of the
reflected potential lie in the ball of radius 1, where N_v is smooth
with all derivatives bounded by C_j s^2 L^{-2-j}, so differentiating
twice gives at most C s^2 L^{-2}.
Step 7: smoothness and full decay (5.27). Claim: for every k >= 0,
|D^k v| <= C_k (s^2/L) r^{-1-k} on Omega. (5.38)
Suppose (5.38) holds for all orders i <= k. Then because
u = u_0 + v with u_0 = 1 + 1/r,
|D^i (u - 1)| <= C_i r^{-1-i} for i <= k, (5.39)
using D^i(1/r) <= C_i r^{-1-i} and s^2/L <= 1. Since u >= 1, the
chain rule applied to x -> x^{-7} and x -> x^{-3}, together with
(4.5) and (5.39), gives
|D^i H_v| <= C_i s^2 r^{-4-i} 1_{r >= L} for i <= k. (5.40)
Also H_v is C^k, because u is C^k by the induction hypothesis (v is a
Newton potential and by Lemma 5.2 gains one derivative at a time).
Lemma 5.2 then gives N_v is C^{k+1} with
|D^{k+1} N_v(x)| <= C_k s^2 L^{-2-k} if |x| <= L/2,
|D^{k+1} N_v(x)| <= C_k s^2 ( r^{-3-k}
+ L^{-1} r^{-2-k} ) if r >= L/2. (5.41)
For the reflected term, the argument x/r^2 always lies in the closed
unit ball, where all derivatives of N_v are bounded by C_k s^2 L^{-2-k}
by the first line of (5.41). Differentiating r^{-1} N_v(x/r^2)
therefore gives, for every order k+1,
|D^{k+1} ( r^{-1} N_v(x/r^2) )|
<= C_k s^2 L^{-2} r^{-2-k}. (5.42)
Each of the three bounds (5.41)-(5.42) is at most
C_k (s^2/L) r^{-2-k} in its own region, and the region matters:
- if r <= L/2, the quantity s^2 L^{-2-k} is bounded by
s^2 L^{-1} r^{-1-k} times 2^{1+k} (because r <= L/2), and
s^2 L^{-2} r^{-2-k} <= s^2 L^{-1} r^{-2-k} <= s^2 L^{-1} r^{-1-k};
- if r >= L/2, then r^{-3-k} <= 2 r^{-2-k}/L, and L^{-1}r^{-2-k}
is already in the desired form; and s^2 L^{-2} r^{-2-k} <=
s^2 L^{-1} r^{-2-k}.
So |D^{k+1} v| <= C_k (s^2/L) r^{-2-k}, which is (5.38) at order
k+1. The base cases k = 0 and k = 1 are exactly (5.26). By induction,
(5.38) holds for all k, and (5.27b) follows from
u - 1 = 1/r + v.
Finally, since every derivative of v is bounded and decays, u is
smooth on Omega and (5.36) holds pointwise, not merely
distributionally. Note that the smoothness came from the gain of one
derivative at a time in Lemma 5.2, not from a bootstrap on the
equation itself.
Step 8: uniqueness. Let u' = u_0 + v' be another smooth solution of
(5.24) with v' >= 0 and v' = O(r^{-1}). Then -Delta v' = H_{v'} and
v' satisfies the homogeneous Robin condition. So v' - R H_{v'} is
harmonic, satisfies the homogeneous Robin condition, and is O(r^{-1}).
By the last part of Lemma 5.1, v' - R H_{v'} = 0, that is,
v' = R H_{v'} = T_s(v'). (5.43)
Also v' is bounded and nonnegative, so v' lies in K. Since T_s is a
contraction with constant 1/2 on K, it has only one fixed point, and
v' = v.
Step 9: continuity in s and the source integrals. The coefficients
obey p_s = s^4 p_1 and ell_s = s^2 ell_1. Hence for v in K and s, t
in [0,1],
|H_{s,v} - H_{t,v}| <= |s^4 - t^4| p_1 (u_0+v)^{-7}
+ |s^2 - t^2| ell_1 (u_0+v)^{-3}
<= C |s - t| r^{-4} 1_{r >= L}, (5.44)
using s, t <= 1 and u_0 + v >= 1. So by Lemma 5.1,
||T_s(v) - T_t(v)||_inf <= C L^{-2} |s - t|. (5.45)
Write v_s - v_t = T_s(v_s) - T_t(v_t) and split:
||v_s - v_t|| <= ||T_s(v_s) - T_s(v_t)|| + ||T_s(v_t) - T_t(v_t)||
<= (1/2) ||v_s - v_t|| + C L^{-2} |s - t|, (5.46)
so ||v_s - v_t|| <= 2 C L^{-2} |s - t|. That is sup-norm continuity,
which also gives continuity of u_s restricted to S, hence of the
horizon area (Section 6).
For the integrals, (5.44) with v = v_s and t = s gives the pointwise
bound |H_s(y) - H_t(y)| <= C |s-t| |y|^{-4}, which is integrable over
r >= L and also integrable against the weight |y|^{-1} since
|y|^{-5} is integrable on the outside of a ball in three dimensions.
Dominated convergence gives continuity of both maps in (5.28).
Finally the upper bounds in (5.29) follow from (5.23) and from
|N_s(0)| <= (1/(4 pi)) * integral of |H_s(y)| / |y| dy
<= (1/(4 pi)) (1/L) * integral of |H_s| <= C s^2 / L^2,
because every |y| in the support satisfies |y| >= L. Nonnegativity of
both quantities is clear since H_s >= 0.
That completes Proposition 5.3. The two properties that Section 6
consumes are: u is trapped between 1 and 3, and the correction v
together with its first two radial derivatives is smaller than
C s^2 L^{-2}.
----------------------------------------------------------------
6. STEP 4. THE HORIZON, AND EVERY SURFACE THAT MIGHT BE SMALLER
----------------------------------------------------------------
6.1 The geometric statement
Everything in this section is uniform in s in [0,1] and L >= L_0, and
"L sufficiently large" means possibly one further increase of the
threshold L_0 beyond the one used in Section 5. There are finitely many
such increases and none of them depends on s.
PROPOSITION 6.1 (Geometry).
For all sufficiently large L, uniformly in s, the data constructed in
Sections 3-5 are smooth source-free electrovacuum data on
Omega = { |x| >= 1 } with boundary S = { r = 1 }, with tr_g K = 0. The
metric is complete with its boundary included, has exactly one
asymptotically flat end, and satisfies
g_ij - delta_ij = O_4(r^{-1}),
K_ij = O_3(r^{-3}),
E^i, B^i = O_3(r^{-2}). (6.1)
The rotation about the third axis preserves g, K, E, B and S, and the
densities mu and |J_con|_g are integrable. With the normal toward the
end, S is a future MOTS, and no compact smooth embedded enclosing
surface in the interior of Omega has theta_+ <= 0 everywhere. Every
enclosing cut has Area_g(Gamma) >= Area_g(S). Finally, with a constant
C_A independent of s and L,
64 pi <= A := Area_g(S) <= 64 pi + C_A s^2 / L^2, (6.2)
and A(s,L) is continuous in s for fixed L.
The three substantive claims are: S is a MOTS, nothing else is
trapped, and S minimizes enclosing area. They are proved in three
separate lemmas so that each can be read on its own.
6.2 Proof of the smoothness, completeness and integrality claims
(a) Smoothness and symmetry. Sections 3-5 give smooth g, K, E, B on
Omega, smooth up to S, and invariant under rotations about the third
axis (the seeds by construction, u by the uniqueness of the fixed
point). The three non-Hamiltonian constraints and the Hamiltonian
constraint are Lemma 4.1 and Proposition 5.3. The action is effective:
a rotation that fixes every point of Omega is the identity.
(b) Completeness. Since 1 <= u <= M,
delta <= g <= M^4 delta (6.3)
as quadratic forms. The closed exterior with the Euclidean intrinsic
length metric is complete: a Cauchy sequence has a Euclidean limit in
Omega, and local boundary charts show that intrinsic length converges
too. Because (6.3) compares lengths in both directions, a curve's
g-length and its delta-length are comparable, so the g-length metric is
complete as well, with S included.
(c) Decay. From u - 1 = 1/r + v and (5.27b),
|D^k(u - 1)| <= C_k r^{-1-k} for every k. Since u^4 = 1 +
4(u-1) + 6(u-1)^2 + ... and u - 1 = O(r^{-1}), the product rule gives
u^4 - 1 = O_4(r^{-1}), hence g_ij - delta_ij = O_4(r^{-1}). Similarly
K = u^{-2}sigma with sigma bounded by (3.10b) and u^{-2} smooth with
decaying derivatives gives K = O_3(r^{-3}), and E = u^{-6}E~ gives
E = O_3(r^{-2}), likewise B. That is (6.1).
(d) Integrability. From (4.18) and (3.10),
mu = (|E|_g^2 + |B|_g^2)/(8 pi) = O(r^{-4}),
and mu is supported on r >= L. Its integral converges since
r^{-4} r^2 dr = r^{-2}dr. For J_con = (E x_g B)^flat_g/(4 pi): the
seeds E~ and B~ are radial for r >= 2L by (3.17), hence parallel, so
E~ x B~ = 0 there. Therefore J_con is supported in the compact annulus
L <= r <= 2L and is bounded, hence integrable.
That is all of Proposition 6.1 except the three lemmas.
6.3 The coordinate spheres, and the boundary as a MOTS
LEMMA 6.2 (Spheres).
For L sufficiently large, the coordinate sphere S_r has future outward
expansion
theta_+(S_r) = 2 W / ( r u^3 ), W = u + 2 r u_r, (6.4)
which vanishes at r = 1 and is strictly positive for every r > 1.
Uniformly in s and L,
W_r >= 1/8 and W >= (r-1)/8 for 1 <= r <= 2,
W >= 1/4 for r >= 2. (6.5)
6.4 Proof of Lemma 6.2
(a) K vanishes on the sphere's tangent space. The tangent space of S_r
is spanned by d_theta and d_phi. The tensor sigma has only the mixed
components sigma_{r phi} = sigma_{phi r} (3.18), so
K(d_theta, d_theta) = u^{-2} sigma_{theta theta} = 0,
K(d_theta, d_phi) = 0, K(d_phi, d_phi) = 0. (6.6)
The same holds on the axis theta = 0, pi by continuity. Hence
tr_{S_r} K = 0 and theta_+(S_r) = H_{S_r}. (6.7)
This is a real simplification and it is the reason the construction is
workable: the only thing to compute is the mean curvature of the
coordinate spheres.
(b) The normal and the mean curvature. The g-unit normal to S_r
pointing toward the end is
nu_r = u^{-2} d_r, (6.8)
because |dr|_g^2 = g^{rr} = u^{-4}. It is a unit vector field on
Omega, so by the definition H = div_{S_r} nu_r and the identity
div_g nu = div_{S_r} nu + g(nabla_nu nu, nu), together with
g(nabla_nu nu, nu) = (1/2) nu(|nu|_g^2) = 0, we get
H_{S_r} = div_g( u^{-2} d_r ). (6.9)
Now use the coordinate divergence formula in the metric g, for which
sqrt(det g) = u^6 r^2 sin theta and the vector u^{-2}d_r has only the
component u^{-2} in the r direction:
H_{S_r} = ( 1/(u^6 r^2 sin theta) ) d_r ( u^6 r^2 sin theta u^{-2} )
= ( 1/(u^6 r^2) ) d_r ( r^2 u^4 )
= ( 2 r u^4 + 4 r^2 u^3 u_r ) / ( u^6 r^2 )
= 2 ( u + 2 r u_r ) / ( r u^3 ). (6.10)
Together with (6.7) that is (6.4). Note that this formula is valid
with full angular dependence: the normal nu_r has no theta or phi
component, so no angular derivative of u enters.
(c) The Robin condition kills the boundary expansion. At r = 1 the
Robin condition (4.29) gives u_r + u/2 = 0, so
W(1, omega) = u(1, omega) + 2 u_r(1, omega) = 0. (6.11)
Since u >= 1 > 0, (6.4) then gives theta_+(S) = 0. So S is a future
MOTS. And since u >= 1, the sign of theta_+(S_r) is the sign of W.
(d) W is positive off the boundary. Insert u = u_0 + v with
u_0 = 1 + 1/r, so u_0' = -1/r^2:
W = 1 - 1/r + v + 2 r v_r. (6.12)
For r >= 2, use (5.26b),
|v| + r|v_r| <= C s^2/(L r) <= C s^2/(2L), so
W >= 1 - 1/2 - ( |v| + 2 r |v_r| ) >= 1/2 - C s^2 / L, (6.13)
which is at least 1/4 once L >= 4 C (s <= 1).
For 1 <= r <= 2, differentiate (6.12) in r. With u_0' = -1/r^2 and
u_0'' = 2/r^3,
W_r = 3 u_r + 2 r u_{rr}
= 1/r^2 + 3 v_r + 2 r v_{rr}. (6.14)
Since 1/r^2 >= 1/4 and, by (5.26a) and (5.26c),
|3v_r| + |2rv_{rr}| <= C s^2 L^{-2},
W_r >= 1/4 - C s^2 / L^2 (6.15)
on 1 <= r <= 2, which is at least 1/8 once L is large.
Finally, W(1,omega) = 0 by (6.11), so integrating (6.15) along the ray
gives W(r,omega) >= (r-1)/8 for 1 <= r <= 2. That is (6.5) and it
proves W > 0 for r > 1, hence theta_+(S_r) > 0 for r > 1.
Two comments on why this is stronger than it looks.
First, the expansion of the coordinate spheres is positive already
at r = 1 + epsilon, not merely far away. That is what makes the
barrier argument of Lemma 6.3 work at the point of maximum radius,
which could be arbitrarily close to S.
Second, the positivity is angularly uniform. The constant 1/8 does
not depend on theta, on s or on L. The reason is again the L^{-2}
smallness from Proposition 5.3: the correction v and its radial
derivatives are far smaller than the background 1 - 1/r near the
boundary.
6.5 Nothing else is trapped
We must exclude every other candidate horizon. The next lemma does
that, and it uses only a maximum principle for the function r on the
competing surface. The trick is to evaluate the comparison at a point
where r is largest, so that the surface is tangent to one of the
coordinate spheres and the section's computation of theta_+ applies.
Throughout, "compact smooth embedded surface" has its standard meaning:
a compact smooth embedded 2-manifold without boundary. (If one instead
allows surfaces with edges, the class of surfaces is larger and the
hypothesis that none is trapped is correspondingly stronger; that
stronger statement is not claimed here. Everything below is about the
boundaryless surfaces of (2.14).)
LEMMA 6.3 (No outer trapped enclosing surface).
For L sufficiently large, no compact smooth embedded enclosing surface
Sigma contained in the interior of Omega satisfies
theta_+(Sigma) <= 0 everywhere.
6.6 Proof of Lemma 6.3
(a) The maximum of radius. Suppose such a Sigma exists. Since
Sigma lies in the interior of Omega and is compact, the function
r restricted to Sigma attains its maximum at some point p, and
R := r(p) > 1. (6.16)
The value R is the maximum radius reached by any point of Sigma.
(b) Tangency at the maximum. At p the function r restricted to Sigma
has a critical point, so the differential dr vanishes on the tangent
space T_p Sigma. Hence
T_p Sigma <= ker( dr ) = T_p S_R. (6.17)
Both spaces have dimension 2, so they are equal:
T_p Sigma = T_p S_R. (6.18)
In particular, at p, the normal to Sigma is the normal to the
coordinate sphere of radius R, and the only ambiguity is its sign.
(c) The sign. The surface Sigma encloses S, meaning it is the boundary
of a region containing the end, so its normal pointing toward the end
is the one that points in the direction of increasing r at the point
of largest r. More explicitly, the ray { p + t d_r : t > 0 } leaves the
region enclosed by Sigma and never returns, so the outward normal at p
is
nu_Sigma(p) = u(p)^{-2} d_r. (6.19)
(d) The Laplacian comparison. On the surface Sigma, consider the
function r restricted to Sigma. It attains a maximum at p, so
Delta_Sigma r (p) <= 0. (6.20)
(For a smooth function on a surface, at an interior local maximum the
surface Laplacian is nonpositive; apply the maximum principle to r on
a small disc around p on Sigma.)
On the other hand, decompose the ambient Hessian of r into its parts
tangent and normal to Sigma. For any surface with unit normal nu and
mean curvature H_Sigma = div_Sigma nu,
Delta_Sigma r = tr_{T_p Sigma} ( grad^2 r ) - H_Sigma nu(r). (6.21)
By (6.18) and (6.19), nu_Sigma(r) = dr(u^{-2}d_r) = u^{-2}, and the
tangent trace is the tangent trace for the sphere S_R,
tr_{T_p S_R}( grad^2 r ) = |grad r|_g * H_{S_R}(p). (6.22)
The identity behind (6.22) is the standard one for level sets: if F is
a function with nonvanishing gradient, then for tangent vectors X, Y
to a level set of F,
grad^2 F (X, Y) = |grad F| * A_F(X, Y),
where A_F is the second fundamental form of the level set with respect
to the unit normal grad F/|grad F|. Taking the trace on both sides
gives (6.22). Here |grad r|_g = u^{-2} > 0.
Substituting (6.21) and (6.22) into (6.20), and using nu_Sigma(r)
= u^{-2} = |grad r|_g:
0 >= Delta_Sigma r(p)
= |grad r| H_{S_R}(p) - H_Sigma(p) |grad r|
= u^{-2} ( H_{S_R}(p) - H_Sigma(p) ). (6.23)
Dividing by u^{-2} > 0 and using Lemma 6.2, which says that
H_{S_R} = theta_+(S_R) > 0 for R > 1:
H_Sigma(p) >= H_{S_R}(p) > 0. (6.24)
(e) The trace of K on Sigma. By (6.18), T_p Sigma = T_p S_R, and by
(6.6) the tensor K vanishes on that plane. So
tr_Sigma K (p) = 0. (6.25)
(f) Contradiction. Combining (6.24) and (6.25),
theta_+(Sigma)(p) = H_Sigma(p) + tr_Sigma K(p)
= H_Sigma(p)
>= H_{S_R}(p)
> 0. (6.26)
So Sigma has a point where its future outward expansion is strictly
positive, which contradicts the hypothesis theta_+(Sigma) <= 0
everywhere. Therefore no such Sigma exists.
The logic is worth restating in one breath: if a surface were trapped
everywhere, then in particular at its outermost point, but at an
outermost point a trapped surface would have to bend inward at least
as hard as the coordinate sphere through that point, and the
coordinate spheres bend outward strictly. Hence no trapped surface
exists.
6.7 Minimal area: the calibration
Now the area condition (2.16). Recall that an enclosing cut is the
full intrinsic boundary Gamma = dD of a connected region D that
contains the end. The claim is that every such Gamma has area at least
that of S.
The tool is a closed 2-form whose "size" is at most one and whose
integral over any enclosing cut is exactly -A. Closed 2-forms with
this property are usually called calibrations; the argument is the
two-dimensional analogue of showing that a surface has minimal area by
exhibiting a closed form of comass one whose integral is the area of
the candidate.
LEMMA 6.4 (Calibrated area).
For L sufficiently large, every enclosing cut Gamma in the sense of
(2.15) satisfies
Area_g(Gamma) >= Area_g(S) = A. (6.27)
Moreover 64 pi <= A <= 64 pi + C_A s^2/L^2, and A(s,L) is continuous
in s for fixed L.
6.8 Proof of Lemma 6.4
(a) The form. Let pi: Omega -> S^2 be the radial projection
pi(r, omega) = omega, and let dA_{S^2} denote the unit round area form
on the sphere. Define
beta = pi^* ( u(1, omega)^4 dA_{S^2} ). (6.28)
Here u(1,omega) is the value of u on the boundary sphere; it is a
smooth function of the angular coordinate omega, so the inner form is
a smooth 2-form on S^2. Since every 2-form on a 2-manifold is closed,
d beta = pi^*( d( u(1,.)^4 dA_{S^2} ) ) = 0. (6.29)
So beta is a smooth closed 2-form on Omega. Note that it is
independent of r except through the argument of pi: beta restricts to
the same angular form on every sphere.
(b) Its comass. Recall that the comass of a 2-form with respect to a
metric is the supremum of |beta(e_1, e_2)| over pairs (e_1, e_2) that
are orthonormal for that metric. We compute it for g.
Since beta is the pullback by pi of a form on S^2, and pi is constant
along rays, beta annihilates radial vectors: beta(d_r, X) = 0 for all
X. So the supremum is over orthonormal pairs in the angular plane.
Choose the oriented pair
e_theta = u^{-2} r^{-1} d_theta, e_phi = u^{-2} (r sin theta)^{-1} d_phi,
(6.30)
which is g-orthonormal because |d_theta|_g = u^2 r and
|d_phi|_g = u^2 r sin theta. Then
beta(e_theta, e_phi)
= u^{-4} (r^2 sin theta)^{-1} * u(1,omega)^4 * sin theta
= u(1,omega)^4 / ( r^2 u(r,omega)^4 ). (6.31)
This is the value on any oriented orthonormal angular pair, up to
sign, so
||beta||_{*,g}(r, omega) = u(1,omega)^4 / ( r^2 u(r,omega)^4 ). (6.32)
Claim: this is at most 1. Indeed the inequality
u(1,omega)^4 <= r^2 u(r,omega)^4 is equivalent to
u(1,omega)^2 <= r u(r,omega)^2, (6.33)
that is, to the statement that r u^2 is nondecreasing in r along each
ray. And
d_r ( r u^2 ) = u^2 + 2 r u u_r = u ( u + 2 r u_r ) = u W, (6.34)
which is nonnegative for r >= 1 by Lemma 6.2 (W >= 0, with equality
only at r = 1). Hence r u(r,omega)^2 >= u(1,omega)^2 and
||beta||_{*,g} <= 1 everywhere on Omega. (6.35)
This is the whole point of the construction of u: the Robin condition
makes r u^2 stationary at r = 1 in the right direction, so the radial
projection form is a calibration rather than merely a closed form.
(c) The integral over any enclosing cut. Let D be as in (2.15) and
Gamma = dD its full intrinsic boundary. Choose R so large that
Gamma is contained in { r <= R } and D contains the whole region
{ r >= R }. Put D_R = D intersect { r <= R }, a compact region with
piecewise smooth boundary. Then D_R has no corners: near Gamma the
truncation does nothing (Gamma lies inside r < R), and near S_R the
truncation is a regular level-set cut in the interior of D. So
dD_R = Gamma union S_R (6.36)
as an oriented boundary, where the orientation of Gamma is the
intrinsic one coming from D and the orientation of S_R is the one
making it the boundary of the part of D inside the ball.
Since beta is closed, Stokes gives
0 = integral over D_R of d beta
= integral over Gamma of beta + integral over S_R of beta. (6.37)
On S_R the form beta restricts to u(1,omega)^4 dA_{S^2}, because pi is
a diffeomorphism from S_R onto S^2 and the pullback of the angular
form is exactly the angular part. Hence
integral over S_R of beta = integral over S^2 of u(1,omega)^4 dA
= A, (6.38)
the last equality being the definition of the area of S (see (6.40)
below). So
integral over Gamma of beta = - A. (6.39)
(d) The comparison. Now bound the left side using the comass. For a
2-form and a surface, the integral is bounded by the integral of the
comass times the area, provided the surface is two-sided with a chosen
orientation, and the bound holds for each component separately:
A = | integral over Gamma of beta |
<= integral over Gamma of ||beta||_{*,g} dA_g
<= integral over Gamma of 1 dA_g
= Area_g(Gamma). (6.40)
Note that Gamma is allowed to be disconnected, to be non-symmetric, and
to have a piece lying on S itself; none of that was used. The comass
bound is pointwise in the tangent plane, so it controls every tangent
plane equally well.
That is (6.27). Since D = Omega is admissible and gives Gamma = S, the
infimum of the areas of enclosing cuts equals A and is attained.
(e) The value of A. On S, the induced metric is u(1,omega)^4 times the
unit round metric, so
A = integral over S^2 of u(1,omega)^4 dA_{S^2}. (6.41)
By (5.26a) at r = 1, |v(1,omega)| <= C s^2 L^{-2}, and by (5.25)
u <= M. Since u(1,.) = 2 + v(1,.),
0 <= A - 64 pi
= integral over S^2 of ( (2 + v)^4 - 16 ) dA
<= 4 M^3 * integral over S^2 of v dA
<= C_A s^2 / L^2, (6.42)
using the mean value theorem for the fourth power on the interval
[2, 2 + C s^2/L^2] and the fact that v >= 0. At s = 0, v = 0 and
A = 16 * 4 pi = 64 pi exactly.
Continuity in s: by Step 9 of Proposition 5.3, v_s -> v_t uniformly on
Omega, hence u_s(1,.) -> u_t(1,.) uniformly on S^2, hence the integral
(6.41) converges. That is the continuity statement.
(f) Conclusion. Lemma 6.4 gives (6.27), (6.2) and continuity. Together
with Lemma 6.2 (S is a MOTS, and the coordinate spheres have positive
expansion) and Lemma 6.3 (no other surface is trapped), this completes
Proposition 6.1 and hence all the geometric requirements listed in
Section 2.
----------------------------------------------------------------
7. STEP 5. MASS, CHARGE, AND ANGULAR MOMENTUM AT INFINITY
----------------------------------------------------------------
7.1 The statement
Everything in this section holds for L above the threshold of
Proposition 6.1, uniformly in s in [0,1]. Write H and N for the source
and its Newton potential at the fixed point, so u = u_0 + v with
v = N + (reflected).
PROPOSITION 7.1 (Asymptotic quantities).
The data have finite asymptotic quantities with
P_ADM = 0, Qe = 0, Qb = 0, (7.1)
and
m = E_ADM = 2 + 2 N(0) + (1/(2 pi)) * integral of H dx
= 2 + O(s^2/L). (7.2)
In particular m >= 2. If M is a uniform upper bound for u (one may take
M = 3), then
m - 2 >= s^2 / (3 M^3 L). (7.3)
The angular momentum of (2.9) is
J = - 4 s^2 / 15. (7.4)
For s > 0 both electromagnetic fields are nonzero.
Note what is being claimed and what is not. The formula (7.2) for m is
EXACT, not asymptotic; the O(s^2/L) refers only to the size of the
correction. Similarly (7.4) is exact for every L. The two estimates
(7.2) and (7.3) together say that the correction to the mass is
positive, of size between s^2/(const * L) and C s^2/L, and that this
is much larger than the correction to the area, which is O(s^2/L^2).
That ratio is what produces the violation in Section 8.
7.2 Proof: momentum and charge
(a) Momentum. In (2.8b) the integrand is K_ij n^j since tau = 0, and
by (6.1) K = O_3(r^{-3}). The area of S_R is 4 pi R^2, so the integral
is bounded by C R^{-3} * 4 pi R^2 = O(R^{-1}), which tends to 0.
Therefore P_ADM = 0.
(b) Charges. By (4.8) the physical electric flux form is exactly the
flat one,
iota_E dV_g = d( f d phi ). (7.5)
The electric charge (2.10a) is the integral of the flux form over the
closed surface S_R, and the integral of an exact 2-form over a closed
surface is zero by Stokes. So Qe = 0 for every R, with no limit
needed. The same line with h in place of f gives Qb = 0.
For a reader who wants the pedestrian version: parameterize S_R by
(theta, phi) and note the form (7.5) has no dr component, so only
f_theta d theta ^ d phi survives. Integrating over phi gives 2 pi times
the integral from theta = 0 to theta = pi, which is
2 pi ( f(R, pi) - f(R, 0) ) = 0 because f contains the factor
sin^2 theta. Either way, the charge vanishes identically.
7.3 Proof: the mass formula
(a) Reducing the ADM integral. Substituting g_ij = u^4 delta_ij into
(2.8a) gives, for every component,
d_j g_ij - d_i g_jj = d_i(u^4) - 3 d_i(u^4) = - 2 d_i(u^4), (7.6)
because g is a multiple of the identity. Since n_delta = d_r on S_R,
E_ADM = - (1/(8 pi)) lim over R of integral over S_R of d_r(u^4) dA_delta
= - (1/(2 pi)) lim over R of integral over S_R of u^3 u_r dA_delta.
(7.7)
(b) Replacing u^3 by 1. By (6.1), u - 1 = O(R^{-1}) and
u_r = O(R^{-2}) on S_R. Hence
| u^3 u_r - u_r | <= C (u - 1) |u_r| <= C R^{-3} (7.8)
and the integral of the error over S_R is O(R^{-3} R^2) = O(R^{-1}).
Therefore
E_ADM = - (1/(2 pi)) lim over R of integral over S_R of u_r dA_delta
+ O(R^{-1}). (7.9)
This is the step where the exactness of the mass formula comes from:
the correction to u^3 is integrable and its contribution vanishes.
(c) Splitting off the background. Write u = u_0 + v with u_0 = 1 + 1/r.
Then u_0' = -1/r^2, so the integral over S_R of u_0' dA_delta is
(-1/R^2)(4 pi R^2) = -4 pi. Hence
integral over S_R of u_r dA_delta = - 4 pi
+ integral over S_R of v_r dA_delta. (7.10)
(d) The flux of the correction. Write v = N + phi with
phi(x) = r^{-1} N(x/r^2), the reflected term. Two calculations.
First, the ordinary Newton potential. Since -Delta N = H,
integral over S_R of d_r N dA_delta
= integral over B_R of Delta N dx
= - integral over B_R of H dx. (7.11)
(The first equality is the divergence theorem; the second is the
pointwise equation.) Note that the right side tends to -integral of H
as R grows, since H is integrable.
Second, the reflected term. On the ray x = R omega,
phi(R omega) = R^{-1} N(omega/R),
d_R phi(R omega) = - R^{-2} N(omega/R)
- R^{-3} (d_r N)(omega/R). (7.12)
The second identity is the chain rule, using
d/dR N(omega/R) = -(1/R^2)(d_r N)(omega/R). Since the source
vanishes inside r < L, the function N is smooth and harmonic at the
origin, so N(omega/R) = N(0) + O(R^{-1}) and the second term of
(7.12) is O(R^{-3}). Hence
d_R phi = - N(0) R^{-2} + O(R^{-3}),
integral over S_R of d_R phi dA_delta
= - 4 pi N(0) + O(R^{-1}). (7.13)
(e) Assembling. Insert (7.11) and (7.13) into (7.10), then into (7.9):
E_ADM = - (1/(2 pi)) [ - 4 pi - integral of H - 4 pi N(0) ]
= 2 + (1/(2 pi)) * integral of H + 2 N(0). (7.14)
That is (7.2). All terms are nonnegative because H >= 0 and
N(0) >= 0, so m >= 2, and m = E_ADM by (2.11).
Note also that no first moment of H and no multipole expansion of N was
used. The only facts were the pointwise equation, integrability of H,
and smoothness of N at the origin.
(f) Size of the correction. From (5.29), the integral of H is at most
C s^2/L and N(0) is at most C s^2/L^2. So m = 2 + O(s^2/L).
(g) Positive lower bound. Here is a place where the angular structure
of the tails is used. For r >= 2L the cutoff is one, so by (3.17) the
electric field is E~ = (2 s cos theta / r^2) d_r. Using only the
electric contribution to ell, and u <= M,
H = p u^{-7} + ell u^{-3}
>= ell u^{-3}
>= ( s^2 cos^2 theta / r^4 ) / M^3 for r >= 2L. (7.15)
The inequality ell >= s^2 cos^2 theta / r^4 is a one-line check on
(4.4b) with c = 1, c' = 0: it says
(1/4)(9t^4 - 2t^2 + 1) >= t^2 for t = cos theta, and
9t^4 - 6t^2 + 1 = (3t^2 - 1)^2 >= 0.
Integrating (7.15) over the outside of the sphere of radius 2L:
integral of H >= (s^2/M^3) * (2/3) * 2 pi * (1/(2L))
= 2 pi s^2 / (3 M^3 L), (7.16)
using the angular integral of cos^2 theta sin theta from 0 to pi, which
is 2/3, and the radial integral of r^{-2} from 2L to infinity, which is
1/(2L). Then (7.2) gives
m - 2 >= (1/(2 pi)) * (2 pi s^2/(3 M^3 L)) = s^2 / (3 M^3 L). (7.17)
That is (7.3). The bound is positive for s > 0 and it is the mass that
makes the inequality fail.
7.4 Proof: angular momentum
(a) The integrand simplifies completely. In (2.9) we have tau = 0,
K = u^{-2} sigma, and the g-unit normal nu_R = u^{-2} d_r. Therefore
K(nu_R, eta) = u^{-4} sigma(d_r, d_phi) = u^{-4} a, (7.18)
where a = sigma_{r phi} = -s^2 c^2 sin^4 theta / r^2 by (3.18). And
dA_g = u^4 dA_delta by (4.6). The two conformal factors cancel exactly:
K(nu_R, eta) dA_g = a dA_delta. (7.19)
This is the same cancellation as for the flux forms, and it is the
reason this construction gives a clean number for J rather than an
integral that depends on u.
(b) The integral. Insert a and dA_delta = R^2 sin theta d theta d phi
on S_R:
J_grav(R) := (1/(8 pi)) integral over S_R of K(nu_R, eta) dA_g
= (1/(8 pi)) integral over S_R of a dA_delta
= (1/(8 pi)) ( - s^2 c(R)^2 ) ( 2 pi ) * integral from 0 to pi of
sin^5 theta d theta
= - (s^2 c(R)^2 / 4) * (16/15)
= - 4 s^2 c(R)^2 / 15. (7.20)
(Here c(R) is constant on the sphere, so it factors out of the
integral. The angular integral is the standard one: the integral of
sin^5 theta from 0 to pi is 16/15.)
For R >= 2L the cutoff takes the value c = 1, so the flux is constant
at -4 s^2/15 for all large R. Hence the limit exists and
J = - 4 s^2 / 15. (7.21)
Notice that J is of order s^2 and does not involve L at all. That is
because the tail is radial with an angular profile, and its angular
momentum content is a purely angular integral.
(c) The fields are nonzero. For s > 0 and r >= 2L, (3.17) gives
E~ = (2 s cos theta / r^2) d_r,
B~ = (s (3 cos^2 theta - 1)/r^2) d_r. (7.22)
The polynomial cos theta is not identically zero and the polynomial
3cos^2 theta - 1 is not identically zero, so both fields are nonzero
in the tail. The physical fields are E = u^{-6}E~ and B = u^{-6}B~ with
u positive, so they are nonzero as well.
7.5 The conserved electromagnetic flux
This subsection explains why the number (7.21) is the "bare
gravitational" angular momentum, and where the missing piece goes. It
is needed for the interpretation in Section 9, and it also verifies
that nothing is inconsistent.
PROPOSITION 7.2 (Conserved flux).
Let A_B = h d phi, the smooth magnetic potential one-form of Section 3,
and let h = A_B(eta) be its value on the rotation generator. Then
div_g ( K(., eta)^sharp_g + 2 h E ) = 0 on Omega. (7.23)
The electromagnetic boundary contribution on every coordinate sphere
is
J_EM(r) := (1/(4 pi)) integral over S_r of h * iota_E dV_g
= 4 s^2 c(r)^2 / 15. (7.24)
Consequently J_grav(r) + J_EM(r) = 0 for every r >= 1, while
J_EM(r) -> 4 s^2/15 as r -> infinity.
7.6 Proof of Proposition 7.2
(a) The divergence of the gravitational piece. Let eta be the rotation
generator and V = K(., eta)^sharp, so V^i = g^{ij} K_jl eta^l. Then
div_g V = g^{ij} (nabla_i K_jl) eta^l + g^{ij} K_jl nabla_i eta^l.
(7.25)
The first term is (div_g K)(eta). For the second, note that eta is a
Killing field, so nabla_i eta_l + nabla_l eta_i = 0. Hence
g^{ij} K_jl nabla_i eta^l = K^{il} nabla_i eta_l = 0, (7.26)
because K is symmetric and nabla eta is antisymmetric. So
div_g V = (div_g K)(eta) = 2 dV_g(E, B, eta), (7.27)
using the momentum constraint (4.2c).
(b) The electromagnetic piece. By Cartan's identity applied to the
invariant one-form A_B = h d phi and the Killing field eta with
A_B(eta) = h,
iota_eta dA_B = (iota_eta d h) d phi - d h ^ (iota_eta d phi)
= 0 - d h = - d h, (7.28)
since eta(h) = d_phi h = 0 by axisymmetry and iota_eta d phi = 1. On
the other hand dA_B = iota_B dV_g, so
iota_eta iota_B dV_g = - d h, (7.29)
which says dV_g(B, eta, X) = - X(h) for every vector X. Taking
X = E and using the cyclic symmetry of a 3-form
(dV_g(E, B, eta) = dV_g(B, eta, E)) gives
dV_g(E, B, eta) = - E(h). (7.30)
Inserting (7.30) into (7.27) and adding the divergence of 2hE, which is
2E(h) because div_g E = 0:
div_g ( V + 2 h E ) = 2(-E(h)) + 2E(h) = 0. (7.31)
That is (7.23). So the sum of the gravitational flux and the
electromagnetic boundary term is the flux of a divergence-free vector
field, hence independent of r.
(c) The electromagnetic boundary term. By (4.8), the form iota_E dV_g
equals d(f d phi). Restricting to the sphere S_r and using the
coordinate expression, only the d theta ^ d phi part contributes:
J_EM(r) = (1/(4 pi)) integral over S_r of h f_theta d theta ^ d phi
= (1/2) * integral from 0 to pi of h f_theta d theta. (7.32)
Insert h = s c sin^2 theta cos theta and f_theta = 2 s c sin theta cos
theta:
J_EM(r) = s^2 c(r)^2 * integral from 0 to pi of
sin^3 theta cos^2 theta d theta
= s^2 c(r)^2 * (4/15)
= 4 s^2 c(r)^2 / 15. (7.33)
(The angular integral is 4/15: substituting t = cos theta gives the
integral of t^2 - t^4 from -1 to 1, which is 2/3 - 2/5 = 4/15.)
(d) Consistency. Comparing (7.20) and (7.33):
J_grav(r) + J_EM(r) = - 4 s^2 c(r)^2/15 + 4 s^2 c(r)^2/15 = 0,
(7.34)
for every r >= 1, as promised by (7.31). At infinity c -> 1 and
J_EM -> 4 s^2/15, while J_grav -> -4 s^2/15. So the conserved total
flux is zero, and the two pieces separate at infinity.
7.7 Why there is no contradiction: two different J
Section 2 defined J by (2.9), which is exactly J_grav at infinity. That
is the number appearing in the inequality (1.1) as formulated in
Theorem 1.1. The conserved total, by (7.34), is
J_total = J_grav(infinity) + J_EM(infinity) = 0. (7.35)
So the data have a large bare gravitational angular momentum and zero
total conserved angular momentum. Both statements are true, and they
are statements about different quantities. The inequality with
J = 0 on the right side would be a different inequality, and this
construction says nothing about it.
There is no singular gauge choice hidden here. A_B = h d phi is a
globally smooth one-form on Omega, its weight h vanishes on the axis
where eta = 0, and the only reason h = A_B(eta) does not tend to zero
at infinity is that |eta| grows like r sin theta: the one-form decays
like r^{-1} while the vector field grows. Adding a constant C to a
scalar potential would change the boundary term only by C Qe = 0, and
the charges vanish.
Finally, why is this compatible with the divergence constraints? A
reader might expect the constraint div E = 0 to forbid an angularly
varying r^{-2} radial field, since the Coulomb field is the only
sourceless radial field with that decay. It does not, in three
dimensions with no further hypotheses:
div_delta ( r^{-2} a(omega) d_r ) = 0 (7.36)
for every smooth function a on the sphere. In coordinates the
divergence is (1/(r^2 sin theta)) d_r( r^2 sin theta r^{-2} a(omega) ),
which vanishes identically because the r-dependence cancels before the
derivative is taken. So an angularly varying r^{-2} radial field is
exactly divergence-free. It is not curl-free, and it is not the field
of a point charge, but the constraints as stated do not require either.
What the constraints do capture is the total charge, which by Stokes is
the flux of an exact form and therefore zero. And what distinguishes
these tails from Coulomb tails is precisely the angular dependence,
which is where the missing angular momentum sits.
----------------------------------------------------------------
8. STEP 6. THE INEQUALITY FAILS, AND EQUALITY DOES NOT FORCE
KERR-NEWMAN
----------------------------------------------------------------
8.1 The defect, and the strict physical branch
By (7.1) the data have Qe = Qb = 0, so Q = 0 and inequality (1.1)
reduces to
m^2 >= A/(16 pi) + 4 pi J^2 / A, (8.1)
with the side condition A >= 8 pi |J|, which is (1.2) with Q = 0.
Define the defect as the amount by which the left side exceeds the
right side:
D(s,L) = m(s,L)^2 - A(s,L)/(16 pi) - 4 pi J(s,L)^2 / A(s,L).
(8.2)
So (8.1) holds if and only if D >= 0.
Before evaluating D, here is a computation that makes both conclusions
strict. From (6.2) and (7.4),
A >= 64 pi, 8 pi |J| = 8 pi (4 s^2/15) <= 32 pi / 15. (8.3)
Since 32 pi/15 < 64 pi, we get
A >= 64 pi > 32 pi/15 >= 8 pi |J| for all s, L. (8.4)
So the side condition A >= 8 pi|J| always holds, with a lot of room to
spare, and in particular A > 8 pi|J| strictly. The data are strictly on
the physical area branch at every value of the parameters, not merely
in the limit. This matters because there is a well-known second branch
of (8.1), and a counterexample on that branch would be dismissed as
uninteresting.
Note also that the inequality in (8.4) is quantitative and uniform: the
margin is at least 64 pi - 32 pi/15, which is about 194.4, and it does
not shrink as L grows.
8.2 Part (a): the inequality is violated
LEMMA 8.1 (Violation).
For all sufficiently large L, D(1,L) < 0. Consequently the data with
s = 1 and such L satisfy every hypothesis of Section 2 and violate
(8.1), hence (1.1).
Proof. Put s = 1 and let L tend to infinity.
For the mass: by (7.2) and (5.29), m = 2 + O(1/L), so m^2 -> 4.
For the area: by (6.2), A = 64 pi + O(1/L^2), so A/(16 pi) -> 4.
For the angular momentum: by (7.4), J = -4/15 exactly, so
4 pi J^2/A -> 4 pi (16/225) / (64 pi) = 1/225. (8.5)
Therefore
lim over L of D(1,L) = 4 - 4 - 1/225 = -1/225 < 0. (8.6)
Since D(1,L) tends to a strictly negative number, there is L_1 such
that D(1,L) < -1/450, say, for all L >= L_1. Take L above the larger
of L_1 and the geometric threshold of Proposition 6.1. Those data
satisfy all the hypotheses and have D < 0.
That is conclusion (a) of Theorem 1.1. The size of the violation is
1/225 at the limit, i.e. about 0.44 percent. Small,
but strictly nonzero, and uniform in a whole tail of L.
8.3 Part (b): an equality case with no Kerr-Newman realization
LEMMA 8.2 (Equality at small amplitude).
Fix any L above the thresholds of Proposition 6.1 and of Lemma 8.1.
Then:
(i) D(s,L) is continuous in s on [0,1];
(ii) D(s,L) > 0 for all sufficiently small s > 0;
(iii) hence there is s_* in (0,1) with D(s_*,L) = 0.
The data at (s_*, L) have Qe = Qb = 0, J nonzero, both
electromagnetic fields nonzero, and A > 8 pi|J| strictly. They admit no
smooth spacelike embedding into the Kerr-Newman spacetime with
parameters (m, J, 0, 0) that induces the given metric, second
fundamental form and electromagnetic fields.
Proof. Step 1: continuity, item (i). By Step 9 of Proposition 5.3 the
map s -> v_s is continuous in the supremum norm, and by (5.28) the maps
s -> integral of H_s and s -> N_s(0) are continuous. The mass formula
(7.2) is a sum of the last two, so m is continuous in s. The area
formula (6.41) is an integral of u_s(1,.)^4 over the sphere, and
uniform convergence of u_s on S gives continuity of A. Finally J
= -4s^2/15 is polynomial in s. Hence D is continuous.
Step 2: positivity at small s, item (ii). Compute the three terms of D
separately. Throughout, M = 3 is the uniform bound (5.25).
Mass term. By m >= 2 and (7.3),
m^2 - 4 = (m-2)(m+2) >= 4 (m-2) >= 4 s^2/(3 M^3 L). (8.7)
Area term. By (6.2),
- (A - 64 pi)/(16 pi) >= - C_A s^2 / (16 pi L^2). (8.8)
Spin term. By (7.4) and A >= 64 pi,
- 4 pi J^2/A >= - 4 pi (16 s^4/225) / (64 pi) = - s^4/225. (8.9)
Adding the three inequalities and using 64 pi/(16 pi) = 4,
D(s,L) >= 4 s^2/(3 M^3 L) - C_A s^2/(16 pi L^2) - s^4/225. (8.10)
Now fix L so large that
4/(3 M^3 L) - C_A/(16 pi L^2) > 0. (8.11)
This is possible because the first term decays like 1/L and the second
like 1/L^2. Then for s > 0 small enough that
s^2 * ( 4/(3 M^3 L) - C_A/(16 pi L^2) ) > s^4/225, (8.12)
which holds whenever 0 < s^2 < 225 ( 4/(3M^3L) - C_A/(16 pi L^2) ),
the bound (8.10) is strictly positive. So D(s,L) > 0 for all small
enough positive s.
Step 3: the root, item (iii). D(1,L) < 0 by Lemma 8.1, D(s,L) > 0 for
small positive s by Step 2, and D is continuous by Step 1. The
intermediate value theorem gives s_* in (0,1) with D(s_*, L) = 0.
Step 4: properties of the root. Since s_* > 0, we have J
= -4s_*^2/15 != 0. Both fields are nonzero by Section 7.4(c). The
strict branch A > 8 pi|J| is (8.4), which holds for all parameters.
Step 5: the inequality at the root implies subextremality. At the root
D = 0, so m^2 = A/(16 pi) + 4 pi J^2/A with Q = 0. Multiply by 16 pi A:
16 pi A m^2 = A^2 + 64 pi^2 J^2. (8.13)
Now compute m^2 - |J|, again using the equality:
m^2 - |J| = A/(16 pi) + 4 pi J^2/A - |J|
= ( A^2 + 64 pi^2 J^2 - 16 pi A |J| ) / (16 pi A)
= ( A - 8 pi |J| )^2 / (16 pi A)
> 0, (8.14)
where the last step uses A > 8 pi|J| from (8.4) and A > 0. So
m^2 > |J|, i.e. |J| < m^2. (8.15)
Step 6: what (8.15) means for Kerr-Newman. The Kerr-Newman family is
parameterized by mass m, angular momentum J, and the two charges. For a
member of this family to be a black hole (rather than a naked
singularity) one needs m^2 >= |J| when Q = 0, and the extremal member
has m^2 = |J| exactly; the subextremal members have m^2 > |J|. So
(8.15) says precisely that the Kerr-Newman member with our parameters
would be a subextremal black hole.
Step 7: the electromagnetic field of that member is identically zero.
This is the crux, and it is a one-line observation about the family. In
Kerr-Newman, the electromagnetic field strength is linear in the two
charges: the dyonic potential is a sum of an electric piece
proportional to Qe and a magnetic piece proportional to Qb, and the
mass and spin enter only through the geometry (the shape of the
horizon and the dragging of frames), never as an additive contribution
to the field. So with Qe = Qb = 0,
F = 0 identically on the whole Kerr-Newman spacetime. (8.16)
(Equivalently: the Reissner-Nordstrom and Kerr-Newman metrics with
zero charge reduce to Schwarzschild and Kerr, and their Maxwell tensors
reduce to the zero tensor term by term.)
Step 8: no spacelike embedding. Suppose there were a smooth spacelike
embedding of Omega into that Kerr-Newman spacetime inducing our metric,
our K, and our E and B. The induced fields are, by the convention of
Section 2,
E^flat = ( iota_n F ) restricted to T Omega,
B^flat = ( star_g ( F restricted to T Omega ) ). (8.17)
Both expressions are linear in F, and F = 0 by (8.16). So the induced
fields vanish identically. But our constructed fields are nonzero by
Section 7.4(c). Contradiction.
Note what this step does NOT depend on: the choice of spacelike slice,
the choice of future unit normal n, the extrinsic curvature, the value
of the time function, or the metric. The vanishing of F is a property
of the ambient spacetime, so the obstruction holds for every spacelike
slice at once.
Step 9: why nothing sneaks through the extrinsic curvature. A reader
might wonder whether the embedding could be non-graph-like and somehow
recover a nonzero field. It cannot, because (8.17) is an identity: the
induced E and B are the pullback and contraction of F, and the
pullback of the zero tensor is zero. There is no slice-dependence to
exploit.
That proves Lemma 8.2, and with it conclusion (b) of Theorem 1.1.
8.4 Assembling the main theorem
PROOF OF THEOREM 1.1. Fix the cutoff chi of (3.3) once and for all.
Let L_0 be the largest of the thresholds appearing in Proposition 5.3,
Proposition 6.1 and Lemma 8.2. Fix any L >= L_0 and let s in [0,1].
Then by Proposition 5.3 the conformal factor u = u_0 + v exists,
is smooth, axisymmetric, and satisfies 1 <= u <= 3 with the estimates
(5.26)-(5.27). By Lemma 4.1 the data (4.1) satisfy the electrovacuum
constraints, and tr_g K = 0. By Proposition 6.1 they satisfy every
topological, asymptotic, symmetry, boundary, outermostness and area
condition of Section 2, and 64 pi <= A <= 64 pi + C_A s^2/L^2. By
Proposition 7.1 the asymptotic quantities exist and satisfy
P_ADM = 0, Qe = Qb = 0, m = 2 + O(s^2/L), J = -4s^2/15. And
A > 8 pi|J| strictly for every s by (8.4).
For conclusion (a): take s = 1 and L >= L_0 as in Lemma 8.1. Then
D(1,L) < 0, i.e. m^2 < A/(16 pi) + 4 pi J^2/A, with Q = 0. All the
other hypotheses hold. Done.
For conclusion (b): with the same L, take s = s_* from Lemma 8.2.
Then D(s_*,L) = 0, so m^2 = A/(16 pi) + 4 pi J^2/A exactly,
with Q = 0 and A > 8 pi|J|, and no Kerr-Newman spacelike realization
exists. Done.
In both cases the examples are maximal, tr_g K = 0, on
Omega = { |x| >= 1 }, with connected boundary S = { r = 1 }, and both
electromagnetic fields are nonzero. That is everything claimed.
----------------------------------------------------------------
9. WHAT THIS DOES AND DOES NOT SAY
----------------------------------------------------------------
9.1 The honest summary
The theorem says exactly this. There is a smooth, maximal,
axisymmetric, source-free electrovacuum exterior with a connected
outermost future marginally outer trapped boundary that minimizes
enclosing area, sitting strictly on the physical area branch, for which
the inequality
m^2 >= A/(16 pi) + pi (Q^4 + 4 J^2)/A, A >= 4 pi sqrt(Q^4 + 4J^2)
fails, where J is the bare gravitational ADM flux (2.9), and where the
electromagnetic assumption is only E, B = O_3(r^{-2}).
That is the whole claim. Everything else is either hypothesis (Section
2), construction and estimates (Sections 3-6), computation (Section 7)
or arithmetic (Section 8).
9.2 The two loopholes, separately
The counterexample uses exactly two weaknesses in the formulation, and
they can be listed separately. It is important to see that neither one
alone is a trick about arithmetic; each corresponds to a specific
choice in the definition of J or in the decay assumptions.
Loophole 1: non-Coulomb r^{-2} tails. The hypotheses ask only
E, B = O_3(r^{-2}). The tails (3.17) have size r^{-2} but angularly
varying coefficients: 2 cos theta for E and 3 cos^2 theta - 1 for B.
Coulomb asymptotics would require the leading term to be a constant
multiple of the unit radial vector, i.e. a spherically symmetric
coefficient, plus O(r^{-3}) remainders. If one imposes
E = Qe xhat/(4 pi r^2) + O_3(r^{-3}),
B = Qb xhat/(4 pi r^2) + O_3(r^{-3}), (9.1)
then in particular the angular profile of the leading term is fixed,
and the boundary term (7.24) that produced the missing angular
momentum has nothing to feed on. The identification of the bare
gravitational flux with a conserved total then becomes available; that
is precisely what the Coulomb expansion in Dain-Khuri-Weinstein-Yamada
is used for. Our examples are excluded by (9.1).
Loophole 2: which J. The quantity J in (2.9) is the flux of K alone.
The conserved total, by (7.34), is
J_total = J_grav(infinity) + J_EM(infinity) = -4s^2/15 + 4s^2/15 = 0.
(9.2)
Both pieces are nonzero and they cancel at infinity exactly. If one
formulates the inequality with J_total, then for our data (9.2) gives
J = 0, and the inequality reads m^2 >= A/(16 pi) with Q = 0. For our
numbers, m^2 -> 4 and A/(16 pi) -> 4 as L grows. More precisely, write
m = 2 + d and A = 64 pi + e, with d = O(s^2/L) and e = O(s^2/L^2).
Then m^2 - A/(16 pi) = 4d + d^2 - e/(16 pi), which by (7.3) and (6.2)
is at least 4s^2/(3M^3L) - C_A s^2/(16 pi L^2), hence positive of order
s^2/L for large L. So the J_total formulation is satisfied, not
violated, by these data. This paper makes no claim about that
formulation: it is a different inequality and it needs its own proof
or disproof.
The point of Proposition 7.2 is to show that this second loophole is
not a bookkeeping accident: the sum J_grav + J_EM is the flux of a
genuinely divergence-free vector field, it is constant in r, and it
vanishes here. The data have plenty of gravitational angular momentum
and an equal and opposite amount of electromagnetic angular momentum
at infinity.
9.3 What this does not claim
To prevent misreading, an explicit list.
It does not disprove a Penrose inequality in the class (9.1). In that
class the identification of angular momenta succeeds and the present
examples are simply not admissible.
It does not disprove a Penrose inequality formulated with the
conserved total J_total of Khuri-Weinstein type. Our data satisfy
that version.
It does not disprove anything for charged configurations: the
examples have Qe = Qb = 0, so the charged terms Q^2/2 and
pi Q^4/A play no role at all. A version of the conjecture
restricted to Q != 0 is untouched.
It does not say anything about the Riemannian (time-symmetric)
Penrose inequality. At s = 0 the data are time-symmetric and the
classical result applies; at s > 0 the trace-free K is nonzero, so
the Riemannian theorem is not in play. Indeed the Riemannian
inequality cannot fail here because K = 0 is exactly the case s = 0,
where the data are Schwarzschild-like.
It does not violate positive mass: m >= 2 > 0 throughout (7.2), and
the mass of the spacetime is positive. The violation is a
discrepancy of order 1/225 in a dimensionless comparison of order 4,
not a sign error in mass.
It does not claim any failure of the area condition. On the contrary,
A is achieved as the true minimum of enclosing area (Lemma 6.4) and
the strict branch holds (8.4), both with room to spare.
It does not claim that the equality case is generic. The equality
amplitude s_* exists by continuity for each fixed large L, and the
construction gives one such family; nothing is said about uniqueness
of s_* or about other equality data.
It does not claim instability, nor any statement about evolution.
These are constraint data, not a Cauchy development.
9.4 Why the result is not an accident of small numbers
Three structural facts make the construction work, and each is visible
in the estimates.
The source sits at distance at least L from the horizon, and its
total strength is only O(s^2/L). Newton potentials at distance L
from strengths s^2/L are of size s^2/L^2, which is the smallness
that keeps 1 <= u <= 3 and keeps every surface in Section 6 under
control. The mechanism is the L^{-2} gain of (5.11), not a
miraculous cancellation.
The conformal weights were chosen so that three of the four
constraints hold identically (Lemma 4.1), and the remaining one has
the two exponents -7 and -3 forced by the trace of K squared and by
the electromagnetic energy. There is no freedom left to tune.
The angular-momentum boundary term is supported on a shell of width
L, so its integral is O(1) as an angular integral while its energy
integral is O(1/L). That ratio is what lets the defect (8.2) be
dominated by the mass term for small s and by the spin term at s = 1.
9.5 Where each hypothesis is used
For a reader who wants to audit the proof, here is the correspondence
between the assumptions of Section 2 and the places they are checked
or used.
R_g + tau^2 - |K|^2 = 2(|E|^2 + |B|^2) Lemma 4.1(e), Prop 5.3
div_g(K - tau g) = 2(E x B)^flat Lemma 4.1(d), (4.17)
div_g E = div_g B = 0 Lemma 4.1(a), (4.8)
tau = tr_g K = 0 (4.10), Prop 6.1
Omega = { |x| >= 1 }, S = { r = 1 } (2.5), Section 3
smooth up to S Lemma 3.1, Prop 5.3
complete with boundary included Section 6.2(b)
one asymptotically flat end Section 6.2(b,c)
effective U(1) with period-2pi generator Section 6.2(a)
g - delta = O_4(r^{-1}) Section 6.2(c)
K = O_3(r^{-3}), E, B = O_3(r^{-2}) Section 6.2(c)
mu and |J_con|_g integrable Section 6.2(d)
theta_+(S) = 0 Lemma 6.2(c)
no other surface with theta_+ <= 0 Lemma 6.3
every enclosing cut has area >= A Lemma 6.4
P_ADM = 0, Qe = Qb = 0 Prop 7.1
m, J finite and with the stated values Prop 7.1, 7.4
Every line is a check, not an assumption. That is why the theorem is a
construction rather than a conditional statement.
9.6 What the equality case really shows
The violation in Lemma 8.1 is small and is obtained at s = 1. The
equality case of Lemma 8.2 is the more interesting one, because it
shows that the usual use of the Penrose inequality -- equality forces
the data to be a Kerr-Newman slice -- fails here even though the
inequality holds with equality.
The reason is stripped to one line in Step 7 of the proof: a
Kerr-Newman black hole with zero charge has zero electromagnetic field,
identically, in the whole spacetime. Any embedding of any slice into
it, with any normal, any time function, any extrinsic curvature,
induces zero E and zero B. Since our data have nonzero fields, no
embedding exists. The rigidity conclusion fails not because the
geometry is strange but because the field content of the target
spacetime is empty.
Note also that this obstruction is robust in a way that a metric
computation would not be: it uses nothing about the metric at all. One
could replace the metric of our example by any other metric and the
argument would still run, as long as the charges vanish and the fields
do not.
9.7 Where one would go from here
Three natural questions, stated so that a student can see what a
research problem looks like.
Under exactly which decay hypotheses on E and B does the
identification J = J_total hold at infinity? The examples here show
that O(r^{-2}) is not enough. Something stronger, presumably
excluding angularly varying leading terms, is needed, and one would
like the sharp condition.
Is there a formulation of the inequality, with the bare
gravitational J, that survives non-Coulomb tails? Proving one would
presumably require a mass lower bound that controls the angular
structure of the tails rather than merely their L^1 mass.
Do similar tails break the equality rigidity in other families, or
is the failure here special to the zero-charge case? The mechanism
is the emptiness of F at Q = 0, so the charged case is the one
where a genuine rigidity statement might still hold.
None of these is answered here. They are what remains.
----------------------------------------------------------------
A. GLOSSARY
----------------------------------------------------------------
Alphabetical. Each entry says what the object is and, where relevant,
the formula or section number.
ADM energy, E_ADM. The total mass measured by a flux integral at
spatial infinity, formula (2.8a). Equal to m in this paper.
ADM momentum, P_ADM. The total linear momentum measured at infinity,
formula (2.8b). Zero in this paper.
angular momentum, J. The flux (2.9) of the gravitational tensor K in
the rotational direction. In this paper J = -4s^2/15 exactly (7.4).
Distinct from J_con and from J_total.
axisymmetric. Invariant under rotations about a fixed axis. Here the
third coordinate axis, with generator eta = d/dphi (2.6).
calibration. A closed differential form whose comass is at most one,
used to prove that a submanifold has minimal area: the integral over
any competitor is bounded by the comass times the area. The form beta of
(6.28) is used this way.
comass. For a 2-form omega and a metric g, the supremum of
|omega(e_1, e_2)| over g-orthonormal pairs. Formula (6.32).
conformal factor. The positive function u in g = u^4 delta, (4.1a).
Everything about the geometry is encoded in it.
conformal Laplacian identity. For g = u^4 delta in three dimensions,
R_g = -8 u^{-5} Delta_delta u, formula (4.20). It is what converts the
Hamiltonian constraint into the scalar equation (4.3).
constraint equations. The four equations (2.2) that a snapshot of
spacetime must satisfy: one Hamiltonian, one momentum (three
components), two divergence-free conditions.
Coulomb asymptotics. The requirement that the leading r^{-2} term of E
be a constant multiple of the unit radial vector (similarly for B),
plus faster-decaying remainders, (9.1). Not assumed here.
cut, enclosing. The full intrinsic boundary of a region containing the
end, definition (2.15). The area condition (2.16) is about these.
extrinsic curvature, K. The symmetric 2-tensor describing how the
snapshot bends inside spacetime. Here K = u^{-2} sigma has zero trace.
future marginally outer trapped surface, future MOTS. A surface whose
future outward expansion theta_+ vanishes, (2.12)-(2.13). The boundary
S is one.
Killing field. A vector field whose flow preserves the metric, so
nabla_i eta_j + nabla_j eta_i = 0. The rotation generator eta is one.
Used in (7.26).
Laplacian, Delta. The trace of the Hessian. Delta_delta is the flat one,
Delta_g the curved one. The sign convention here is that
-Delta_delta(1/r) = 0 and -Delta_delta N = H.
marginally outer trapped, see MOTS.
mean curvature, H_Sigma. The divergence of the unit normal along the
surface, H_Sigma = div_Sigma nu, formula (2.12). For a Euclidean
sphere of radius R it is +2/R with the outward normal.
monotonicity of r u^2. The fact that d_r(r u^2) = uW >= 0, (6.34), which
is what makes the radial-projection 2-form a calibration. Equivalent
to W >= 0 from Lemma 6.2.
MOTS, see future marginally outer trapped surface.
Newton potential, N_H. The convolution (5.4) of H with the fundamental
solution 1/(4 pi |x|) of the Laplacian in three dimensions. Solves
-Delta N_H = H.
outer expansion, theta_+. Formula (2.12): mean curvature plus the
trace of K over the surface. Positive means outgoing light expands.
Penrose inequality. The conjectured lower bound on mass in terms of the
area of a horizon, originally Penrose 1973. The Kerr-Newman version is
(1.1); the Riemannian version is the time-symmetric special case.
physical branch. The branch of the inequality on which the right side
increases with A, i.e. the side condition (1.2). Our data are strictly
on it, (8.4).
Robin condition. A boundary condition relating a function to its normal
derivative: here u_r + u/2 = 0 on S, (4.29). It makes the boundary
sphere a MOTS.
rotation generator, eta. The vector field (2.6) whose flow is rotation
about the third axis; periodic with period 2 pi, and it is a Killing
field.
scalar equation. Equation (4.3), -Delta_delta u = p u^{-7} + ell u^{-3},
the single remaining constraint after the seeds are chosen.
second fundamental form. See extrinsic curvature, K.
source, H. The right side of the scalar equation, H = p u^{-7} +
ell u^{-3}. Supported on r >= L. Its integral is O(s^2/L) but it has no
finite first moment.
tails. The r^{-2} radial fields (3.17) that survive in the region
r >= 2L, with angularly varying coefficients. The name of the paper.
trace-free. Having zero trace with respect to the metric. Both sigma
and K are trace-free, by construction and by (4.10).
U(1) action. The one-parameter group of rotations preserving the data,
with generator eta. "Effective" means no nonzero rotation acts
trivially, see Section 2.3.
volume form. The canonical 3-form dV_g induced by a metric. In the flat
case dV_delta = r^2 sin theta dr ^ d theta ^ d phi, and dV_g = u^6
dV_delta by (4.6).
W. The combination W = u + 2 r u_r, which controls everything about the
boundary: W(1,omega) = 0 is the Robin condition, theta_+(S_r) =
2W/(r u^3), and W >= 0 is the calibration inequality. Formula (6.4).
weakly future outer trapped. Having theta_+ <= 0 everywhere, the
condition excluded by Lemma 6.3 for every surface except S itself.
----------------------------------------------------------------
B. A CHECKLIST FOR A SKEPTICAL READER
----------------------------------------------------------------
If you want to verify the paper without reading the whole thing, check
these nine points in this order.
1. (3.16) and (3.24): differentiate the two components in (3.15) and
confirm the cross product equals the divergence of sigma. This is
the only place where two independent formulas must agree.
2. (4.16): expand the two Christoffel contractions and confirm they
cancel. If they had not, the power -2 in (4.1b) would be wrong.
3. (4.23): substitute u^4 delta into the Hamiltonian constraint and
confirm the exponents -7 and -3.
4. (5.15): differentiate the reflected potential along a ray at
r = 1 and confirm the Robin condition. This is the single most
important line of the elliptic theory.
5. (6.10): compute the divergence of u^{-2}d_r in the metric u^4
delta and confirm theta_+ = 2W/(r u^3).
6. (6.23): confirm the sign. The maximum principle gives
Delta_Sigma r <= 0, hence H_Sigma >= H_{S_R} > 0. A sign error
here would reverse the whole argument.
7. (6.34): confirm d_r(r u^2) = uW, and hence that the calibration
has comass at most one.
8. (7.14): confirm the two flux identities and the resulting exact
mass formula, in particular that the reflected term contributes
-4 pi N(0) with the sign shown.
9. (8.14): confirm the identity m^2 - |J| = (A - 8 pi|J|)^2/(16 pi A)
at the root, which is the only algebra needed for the rigidity
failure.
Every one of these is a short computation, and each one is done in full
in the text.
END.