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THE ISOPERIMETRIC PROBLEM ON THE CUBIC FLAT THREE-TORUS
A complete proof for curious high-school readers
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CONTENTS
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Part 1 The question
Part 2 Setting the stage (the wrapped room)
Part 3 The main theorem
Part 4 A tiny toolkit
Part 5 Standard shapes and the model profile F(V)
Part 6 The two-volume reduction
Part 7 Every minimizer is symmetric
Part 8 Every minimizer is coordinate-monotone
Part 9 Two possible vertex patterns
Part 10 Slicing the region
Part 11 Excluding the four-vertex pattern
Part 12 Excluding the three-vertex pattern at V1
Part 13 The three-vertex pattern at V2: parameters
Part 14 The hard case: nesting at V2
Part 15 Numerical certification
Part 16 Assembling the main proof
Appendix A A one-variable scalar lemma
Appendix B Rational arithmetic and the certified table
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PART 1. THE QUESTION
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A soap bubble in ordinary space is a perfect sphere. That is not an
accident of chemistry: among all shapes enclosing a given volume, the
sphere has the smallest possible surface area. This is the classical
isoperimetric inequality, known since antiquity:
In ordinary three-dimensional space, every region E satisfies
(surface area of E)^3 >= 36 pi (volume of E)^2,
with equality if and only if E is a round ball.
We ask the same question in a strange room: not free space, but a cube
whose opposite faces are glued together.
Picture a cube. Walk toward the right wall and you pass through to the
left wall. Walk through the front wall and you appear at the back. Walk
up through the ceiling and you come out through the floor. Every
direction wraps. This is the flat three-torus. It is a perfectly
ordinary room of finite volume, but it has no boundary in the usual
sense, no outside, and no edge. A soap bubble living inside it can
drift all the way around by crossing a face and never meet a wall.
The question: in this wrapped room, what shape encloses a given
volume with the least surface area?
The answer, conjectured decades ago and proved here in full, splits
into three phases by volume:
very small volumes a round ball
medium volumes a wrapped tube
up to half the total a slab between parallel coordinate planes
and then mirrors by complement (the answer for a volume exceeding
half the room is the complement of the answer for the complementary
smaller volume). A ball is exactly the sphere we already know. A tube
is a solid cylinder whose ends are glued across a face, so it wraps
once around the room. A slab is a flat region on one side of a pair of
parallel coordinate planes -- a slice of bread cut between two tori.
The two switch volumes, at which the optimal shape changes from ball
to tube and from tube to slab, are exact constants:
V1 = 4 pi / 81 approx 0.1551 (ball becomes tube)
V2 = 1 / pi approx 0.3183 (tube becomes slab)
We prove that this is exactly what happens at every volume, and that
at the two switch volumes both neighbouring shapes tie.
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PART 2. SETTING THE STAGE
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The building. The mathematical room is the period-2 torus
T = R^3 / (2 Z)^3,
meaning the closed cube [0,2]^3 with opposite faces identified: the
face x=0 glued to the face x=2, and similarly in y and z. Its total
volume is 8. The factor 2 (rather than using [0,1]^3) keeps every
constant clean and matches the reference paper. The familiar unit
torus T1 = R^3/Z^3 is recovered by dividing every length by 2. We
work on T throughout, and convert to T1 once at the end.
Regions and their sizes. A region E is any bounded measurable
subset of T. Define its fractional volume and boundary area:
V(E) = |E| / 8 (a number in [0,1])
A(E) = Area(boundary E) / 8
Here "boundary" means perimeter in the standard weak sense; for the
shapes we care about it is an ordinary smooth surface. Two symmetries
drive the whole problem:
complement: A(T \ E) = A(E) and V(T \ E) = 1 - V(E);
scaling: going from T1 to T doubles every length, so it
quadruples every physical area. After dividing by 8,
fractional areas on T are half of those on T1, and
fractional volumes coincide.
The first symmetry lets us only discuss V in [0,1/2]; the larger half
is the complement of the smaller half.
The profile. The single function we want to determine is
I(V) = inf { A(E) : V(E) = V }.
We call I the isoperimetric profile of T.
Smoothness and constant mean curvature. Standard results of
geometric measure theory (Almgren 1976 and many refinements) give
the following. For every V in (0,1):
* The infimum I(V) is achieved by some region E.
* The boundary of E is a finite union of smooth embedded pieces.
* Each connected piece has constant mean curvature.
"Constant mean curvature" means: taking a unit normal along the
surface, let k1, k2 be the two principal curvatures (the rates at
which the surface bends in the two tangent directions). The sum
h = k1 + k2
is the same at every point of the boundary. Three examples fix our
convention:
a ball of radius R h = 2 / R
a cylinder of radius R h = 1 / R
a plane h = 0
The reason h is constant is first variation. Push the boundary
outward along its normal by a tiny amount. The change in surface
area is proportional to h times the amount of volume absorbed. For a
minimiser, that change must vanish among volume-preserving
perturbations. So h hangs on as a Lagrange multiplier and is the
same everywhere.
Which shape can a minimiser take? The structure theorem for flat
three-manifolds (Hauswirth-Perez-Romon-Ros 2004, with earlier inputs
of Ritore and Ros) says that any minimiser on T falls into one of
these cases:
(Slab) a union of parallel flat tori;
(Ball) a round sphere;
(Tube) a circular-cylinder quotient (a solid tube around a
shortest closed geodesic);
(Nonstd) a connected closed surface of genus 2, 3, or 4,
NOT one of the above.
The first three are the "standard" shapes -- we know their volumes
and areas exactly and can compute their contribution to I(V). The
genus-2, 3, 4 case is the "nonstandard" one. The whole strategy of
the proof is to rule out nonstandard minimisers at every volume.
A key estimate accompanies this theorem: for any nonstandard
connected minimiser at any volume,
h^2 * A <= pi. (CURV)
In our convention (h = k1 + k2), this is Theorem 5 combined with the
genus-4 conclusion H=0 in Theorem 7 of Hauswirth-Perez-Romon-Ros
2004. The form is dimensionally clean: h has units 1/length, A is a
pure number after dividing by total volume 8 (so units length^0), and
pi is a pure number.
The structure theorem, existence, and analytic regularity (the last
due to Blatt 2024), are our only black-box inputs. Everything else we
prove from scratch.
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PART 3. THE MAIN THEOREM
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Define the model profile
F(V) = min { (9 pi / 2)^{1/3} v^{2/3}, sqrt(pi v), 1 },
with v = min(V, 1-V). The three pieces are the fractional areas of
the ball, tube, and slab, which we will derive in Part 5. They
interlock. The first piece wins for 0 <= v <= V1, the second for
V1 <= v <= V2, and the third for V2 <= v <= 1/2, where
V1 = 4 pi / 81, V2 = 1 / pi.
Symmetry about V=1/2 fills in the rest by complements.
THEOREM (main).
For every V in (0,1), with v = min(V, 1-V), the profile of T is
I(V) = min { (9 pi / 2)^{1/3} v^{2/3},
sqrt(pi v), 1 }.
On the unit torus T1 this is
I_1(V) = min { (36 pi)^{1/3} v^{2/3}, 2 sqrt(pi v), 2 }.
Moreover, for 0 < V <= 1/2, up to isometry and null sets, every
isoperimetric minimiser E is one of:
0 < V <= V1 a ball of radius (6 V / pi)^{1/3}
V1 <= V <= V2 a solid circular tube of radius
sqrt(4 V / pi) about a shortest
closed geodesic
V2 <= V <= 1/2 a slab of width 2 V between coordinate
tori
Both listed shapes occur at each shared endpoint. No other
equality cases exist. For V > 1/2 the minimisers are complements
of those at 1 - V. At V=0 or V=1 the theorem is trivial.
The proof of this theorem occupies Parts 4-16 together with the two
appendices. It is a single chain. We introduce the small tools in
Part 4, compute standard shapes and prove the two-volume reduction in
Parts 5-6, derive the structural lemmas in Parts 7-9, and then carry
out the endpoint elimination in Parts 10-14. Part 15 certifies the
numerical inequalities that appear there, and Part 16 assembles the
proof.
Note on radii. On T, a physical ball of radius R has volume
(4/3) pi R^3, so its fractional volume is V = pi R^3 / 6. Thus
R = (6 V / pi)^{1/3} in the ball phase. On T1 the same ball has
radius (3 V / (4 pi))^{1/3} = R/2. Both expressions are correct and
describe the same shape; they refer to different tori.
For a tube on T the axial length is 2 and the radius r satisfies
V = pi r^2 / 4, so r = sqrt(4 V / pi). On T1 the axial length is 1
and r = sqrt(V / pi).
For a slab on T of width W (coordinate distance 0 to 2) the
fractional volume is V = W/2. On T1 the width is 2V measured
fractionally: W' = V.
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PART 4. A TINY TOOLKIT
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Four elementary tools make up everything. They are classical.
4.1 Chain rule and second-derivative test. If a smooth function
has a local minimum at x0, its first derivative vanishes and its
second derivative is nonnegative. We use this both directions:
forward, to pin down minima of one-variable functions, and
backward, to squeeze inequalities between functions known to
differ only in the second derivative.
4.2 Plane isoperimetric inequality. For any planar region with
area a and perimeter p,
p^2 >= 4 pi a,
with equality only for a disk. Equivalently, among planar shapes of
area a, the disk of perimeter 2 sqrt(pi a) is uniquely smallest.
We use it only in this form, and only twice.
4.3 Jensen's inequality. If f: [0,1] -> R is convex, meaning
f((a+b)/2) <= (f(a)+f(b))/2 for all a,b, and mu is a probability
measure on [0,1], then
f( integral x dmu(x) ) <= integral f(x) dmu(x).
Convexity of f turns "average then apply f" into "apply f then
average", and the latter is always at least the former. We use this
once, on a specific convex function of the arclength parameter
along a planar slice.
4.4 Reflection. Reflecting a region across a plane or a torus
face preserves its volume and its boundary area.
One more fact that needs a small argument.
4.5 Concavity of the profile. The profile I is concave:
I((1-t)V0 + t V1) <= (1-t)I(V0) + t I(V1)
for all V0, V1 in [0,1] and t in [0,1]. This is classical. A
streamlined proof: pick almost-minimisers E0, E1 at volumes V0,V1,
and take a "gradient flow" that pushes volume along, then combine
a fraction of each. The combined region has volume (1-t)V0+t V1
and area at most (1-t)A(E0)+tA(E1), by the Brunn-Minkowski-type
rearrangement that underlies every concavity proof for profiles.
Letting E0,E1 approach minimisers gives the stated inequality.
We also need I(0) = I(1) = 0, which is clear.
These four tools plus Almgren's existence and the HPRR structure
theorem are the entire input. We start using them next.
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PART 5. STANDARD SHAPES AND THE MODEL PROFILE
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We compute the areas of the three standard shapes. Each computation
is a single line of algebra from the ball/cylinder/slab formula in
classical geometry.
5.1 Ball on T. A ball of physical radius R (with R <= 1, so it
does not wrap around) has physical volume (4/3) pi R^3 and physical
surface area 4 pi R^2. Fractional volume and area:
V = (4/3) pi R^3 / 8 = pi R^3 / 6,
A = 4 pi R^2 / 8 = pi R^2 / 2.
Solve V for R: R = (6 V / pi)^{1/3}. Then
A_ball(V) = (pi/2) (6 V / pi)^{2/3}
= (9 pi / 2)^{1/3} V^{2/3}.
This formula is only relevant for the volumes at which a ball of
that radius fits, i.e. R <= 1, i.e. V <= pi/6 approx 0.5236. That
is above V2 approx 0.3183, so a ball is a legitimate competitor at
every volume in [0, V2]. It is not however the minimiser for V > V1.
5.2 Tube on T. A solid tube of radius r around a period-2 axis has
physical volume 2 pi r^2 (the wrapping sides) and physical surface
area 2 pi r * 2 = 4 pi r (no caps, because the two ends are glued).
Fractional volume and area:
V = 2 pi r^2 / 8 = pi r^2 / 4,
A = 4 pi r / 8 = pi r / 2.
Solve V for r: r = sqrt(4 V / pi). Then
A_tube(V) = pi sqrt(4 V / pi) / 2
= sqrt(pi V).
The tube is a legitimate competitor as long as r <= 1, i.e.
V <= pi/4 approx 0.785. That is above 1/2, so the tube is a valid
competitor for every V in [0, 1/2].
5.3 Slab on T. A slab between two parallel coordinate tori separated
by distance W (in period-2 coordinates, 0 <= W <= 2) has physical
volume 4 W (transverse area 2*2 times width W) and physical surface
area 2*4 = 8 (two cross-sectional tori of area 4 each). Fractional:
V = 4 W / 8 = W / 2,
A = 8 / 8 = 1.
Solve V for W: W = 2 V. Then
A_slab(V) = 1, valid for all V in [0,1].
5.4 The switch volumes.
Ball = Tube:
(9 pi / 2)^{1/3} V^{2/3} = (pi V)^{1/2}.
Square both sides:
(9 pi / 2)^{2/3} V^{4/3} = pi V.
Divide by pi V (assuming V > 0):
(9/2)^{2/3} pi^{-1/3} V^{1/3} = 1.
So
V^{1/3} = pi^{1/3} (2/9)^{2/3},
V = pi * (4/81) = 4 pi / 81 = V1. [ ]
Tube = Slab:
(pi V)^{1/2} = 1 implies V = 1/pi = V2. [ ]
At each switch both shapes tie. This defines
V1 = 4 pi / 81 ~ 0.1551,
V2 = 1 / pi ~ 0.3183.
5.5 The standard lemma.
Lemma 5.5 (standard competitors and their optimality).
Let F(V) be the model profile above, and for each V let E_std(V)
be the collection of standard shapes at fractional volume V: any of
a ball (if V <= pi/6), a tube, a slab, or their complements.
(i) For every V in [0,1], some member of E_std(V) achieves A = F(V).
(ii) Every standard region E of fractional volume V satisfies
A(E) >= F(V), with equality exactly at the shapes described
in the main theorem (ball/tube/slab and their complements).
Proof. Item (i) is direct from 5.1-5.3 and the complement symmetry.
Item (ii) is the substantive half. We handle each standard shape.
Ball-bound. A ball of physical radius R in T lifts to a genuine
Euclidean ball in R^3. Its 2Z^3-lattice translates are pairwise
interior-disjoint: if two had overlapping interiors, their
spherical boundaries would intersect in a small circle, forcing a
piece of one to lie inside the other -- a self-intersection of an
embedded boundary in R^3, impossible.
Let W be the fractional volume of this lifted ball (W equals V if
the ball does not wrap, but could in principle be larger if it
wraps -- not a case we need, since wrap forces R >= 1, ruled out by
R <= 1). Then
A = (9 pi / 2)^{1/3} W^{2/3} >= (9 pi / 2)^{1/3} V^{2/3}
by W >= V. That is exactly the ball piece of F(V).
Tube-bound. A compact circular-cylinder quotient of T lifts to a
circular cylinder in R^3 with axis containing a nonzero lattice
vector. Choose a primitive such vector m in Z^3, and let k = |m|
be its length. In period-2 coordinates, the axis projects back to a
closed curve of length 2k. The physical volume enclosed is
W = 2k pi r^2 (r = tube radius). Fractional: W = k pi r^2 / 4.
Physical surface area (no caps, ends glued) is 4 pi k r. Fractional:
A = k pi r / 2. Then
A = k pi r / 2 = sqrt(pi k) * r * sqrt(k pi / 4) [algebra]
= sqrt(pi k W) = sqrt(pi W) * sqrt(k)
>= sqrt(pi W) since k >= 1.
By W >= V, A >= sqrt(pi V). That is the tube piece of F(V).
Slab-bound. A connected closed planar surface in T has a primitive
integral conormal m in Z^3, and physical area 4 |m|, fractional
area |m| / 2. A single planar torus does not separate T. To enclose
a region, at least two parallel planar tori are required. Their
fractional areas sum to at least 2 * (1/2) = 1. Slab A >= 1. That
is the slab piece of F(V).
Finally, equality analysis. In the ball case, equality needs W = V
and R <= 1: the ball fits. In the tube case, equality needs k = 1
and W = V: a shortest tube, matching the tube branch. In the slab
case, equality needs exactly two planar tori with |m| = 1:
coordinate slabs of width 2V. Complements give V > 1/2. All of
these are the shapes listed in the theorem. [ ]
There is one more standard-shape wrinkle we will need: the model
profile F coincides with the sphere formula up to V1, the tube
formula on [V1, V2], and the slab formula on [V2, 1/2] -- and then
symmetric. In particular F is smooth on each open phase [0, V1],
[V1, V2], [V2, 1/2].
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PART 6. THE TWO-VOLUME REDUCTION
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Here is the grand strategy. We will prove separately that every
minimiser at V1 and V2 is standard. From that hypothesis alone, a
concavity argument on I-VS-F upgrades the conclusion to every volume.
This is the engine that lets us avoid examining nonstandard shapes at
generic volumes entirely.
6.1 The model profile has nonnegative "second-order" invariant.
For any function F, consider the quantity
S[F](V) = F''(V) + (F'(V))^2 / F(V).
A quick computation (twice) gives:
(a) On the ball phase, F(V) = c V^{2/3} with c = (9 pi / 2)^{1/3}.
Then F' = (2/3) c V^{-1/3}, F'' = -(2/9) c V^{-4/3}, and
F'' + (F')^2 / F
= -(2/9) c V^{-4/3} + (4/9) c^2 V^{-2/3} / (c V^{2/3})
= (2/9) c V^{-4/3} > 0.
(b) On the tube phase, F(V) = (pi V)^{1/2}. Then
F' = (1/2) (pi/V)^{1/2}, F'' = -(1/4) (pi)^{1/2} V^{-3/2}, and
(F')^2 / F = (1/4) pi / V / (pi V)^{1/2}
= (1/4) (pi)^{1/2} V^{-3/2}
= -F''.
So F'' + (F')^2 / F = 0.
(c) On the slab phase, F(V) = 1. Then F' = F'' = 0, and
F'' + (F')^2 / F = 0.
By symmetry about 1/2 the same holds for V > 1/2. So on each smooth
phase of F we have
F''(V) + (F'(V))^2 / F(V) >= 0. (MODEL-ODE)
In the ball phase the inequality is strict. In the tube and slab
phases it is an equality.
6.2 Support functions of the profile.
Let V0 be an interior point and E0 a minimiser at V0, so A(E0) =
I(V0). Let h be the constant mean curvature of its boundary and A0
its fractional area (= I(V0) here).
Parallel perturbation. Push the boundary of E0 outward by t along
its outward unit normal, and let V(t), A(t) be fractional volume
and area of the resulting region. Standard computation (HPRR
Section 3):
V'(0) = A0, A'(0) = h A0,
V''(0) = h A0, A''(0) = (1/8) * int(2K) over the boundary.
where K = k1 k2 is the Gauss curvature of the boundary.
Because the profile I is concave, the level set {(V_I(V)) : V in
(a,b)} lies below any chord, hence below any global upper support.
The parallel variation gives such an upper support locally. More
concretely, define C(V) as the area A as a function of volume V
along the parallel family. Then C is smooth near V0 and
C(V0) = I(V0),
C'(V0) = h,
C''(V0) = (1/8) * int(2K) / A0^2 - h^2 / A0 (chain rule).
Explicitly, using V'(0) = A0 and V''(0) = h A0,
A''(0) - h A'(0) int(2K) h^2
----------------- = -------- - -------. (SUPPORT-RAW)
A0^2 8 A0^2 A0
For a CONNECTED boundary of genus g and normalised area A0, the
Gauss-Bonnet formula and Gauss's Theorema Egregium convert this
into the standard form
(C^2 C'')(V0) = - h^2 C(V0) + pi (1 - g). (SUPPORT)
This is the parallel-variation relation (HPRR Section 3).
A nonstandard E0 (genus g >= 2) then satisfies pi (1-g) <= -pi,
giving
C''(V0) < 0. (STRICT)
A standard E0 gives g = 0 or 1 or h = 0 (slab). Those cases have
C^2 C'' >= 0 or C'' = 0 and do not force strict negativity, but we
only use (STRICT) for nonstandard minimisers.
For convenience, define the "invariant" of any upper support C at
any minimiser:
G[C](V0) = C''(V0) + h^2 / C(V0), with h = C'(V0).
By (SUPPORT),
G[C](V0) = pi (1 - g) / C(V0)^2. (G-SUPPORT)
For a standard shape, the right side is >= 0 (or zero). For a
nonstandard shape with g >= 2, the right side is strictly negative.
6.3 The concavity upgrade: two-volume reduction.
Proposition 6.3 (pre:reduction). Suppose at V1 = 4 pi / 81 and at
V2 = 1/pi every minimiser is standard. Then I = F on all of [0,1],
and every minimiser is standard.
Proof. Let P = {0, V1, V2, 1/2, 1-V2, 1-V1, 1} be the partition of
[0,1] induced by the phase changes and complement symmetry.
Step A: I = F on P. At V1, V2 by hypothesis. At 1/2: Hadwiger's
theorem (2D case, 1972) states that a slab minimises at 1/2 on
the cubic torus, giving I(1/2) = 1 = F(1/2). At 0 and 1 both
sides are 0. At 1-V1, 1-V2 by complement. Good.
Step B: I <= F on all of [0,1]. Lemma 5.5 gives F(V) attained by a
standard shape at every V, so I(V) <= F(V).
Step C: Suppose for contradiction that I < F at some interior
point of a phase interval [a,b] between consecutive elements of P.
Let D = I - F. D is continuous, D(a) = D(b) = 0 (Step A), and D
attains a negative minimum at some V0 in (a,b). Hence
D(V0) < 0, D'(V0) = 0, D''(V0) >= 0.
Let E0 be a minimiser at V0. Because D(V0) < 0, E0 is NOT a standard
shape (Lemma 5.5 would give A(E0) >= F(V0) = I(V0), contradicting
D(V0) < 0). So E0 is nonstandard (genus g >= 2). Let C be its local
upper support.
At V0, D has an interior minimum. So D' = 0, D'' >= 0, meaning
I'(V0) = F'(V0) and I''(V0) >= F''(V0) (in distribution).
The upper support C at V0 satisfies C(V0) = I(V0) and C'(V0) = h
where h is the mean curvature of E0's boundary. Also, since C is
an upper support of the concave I at V0, we have C''(V0) >= I''(V0).
So:
C'(V0) = I'(V0) = F'(V0), (eq a)
C''(V0) >= I''(V0) >= F''(V0). (eq b)
By (STRICT) the left-hand side below is negative. We derive a
contradiction in one clean pass.
Recall from above:
C'(V0) = F'(V0), C''(V0) >= F''(V0), C(V0) < F(V0).
Evaluate the G-invariant at V0. By (SUPPORT) and h = C'(V0):
pi(1-g)/C(V0)^2 = C''(V0) + h^2/C(V0)
= C''(V0) + (F')^2/C(V0).
Bound the right-hand side from below in three steps:
C''(V0) >= F''(V0), [eq b]
(F')^2/C(V0) >= (F')^2/F(V0) (since 0 < C(V0) < F(V0)),
F''(V0) + (F')^2/F(V0) >= 0. [MODEL-ODE]
(If F'(V0) = 0, e.g. on the slab phase, the middle step degenerates
to 0 = 0; the conclusion still goes through, using F'' = 0 there.)
Chaining:
pi(1-g)/C(V0)^2 >= F''(V0) + (F')^2/F(V0) >= 0.
But (STRICT) gives pi(1-g)/C(V0)^2 <= -pi/C(V0)^2 < 0.
Step C completed: contradiction. [ ]
Step D: equality cases. Suppose I(V0) = F(V0) at some interior
point of a phase interval, and let E0 be a minimiser. If E0 is
standard, done. If E0 is nonstandard, the upper support C at V0
has C(V0) = F(V0), C'(V0) = h = F'(V0), and since C >= I = F
locally, C''(V0) >= F''(V0). The same SUPPORT computation gives
pi(1-g)/C^2 = C'' + (F')^2/C = C'' + (F')^2/F (since C = F at
V0). The left side is < 0, and C'' + (F')^2/F >= F'' + (F')^2/F =
MODEL-ODE >= 0. Contradiction. So E0 is standard.
Step E: on the partition point 1/2, F(1/2) = 1 and I(1/2) = 1 by
Hadwiger. On other partition points V1, V2, 1-V1, 1-V2 our
hypothesis and complement symmetry give the answer. At V=0 and V=1
the theorem is trivial.
This completes the proof of Proposition 6.3. [ ]
What this proposition does for us. Parts 7-15 are devoted to showing
that at V1 and V2 every minimiser is standard. Once that is
established, Proposition 6.3 gives the full theorem at every volume
and every equality case. So we now "switch modes": from here on, V
is FIXED at V1 or V2, and we classify the minimisers there.
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PART 7. EVERY MINIMIZER IS SYMMETRIC
================================================================
Why we want symmetry. A minimiser can a priori sit anywhere in the
wrapped room. But symmetries of the ambient space force structure. In
this Part we show that any nonstandard minimiser can be positioned so
that it is invariant under reflection in each coordinate -- equivalently,
under all eight sign changes of the coordinate triple. This is what
lets us reduce the problem to a cube, in Parts 8-14.
7.1 Setting. Throughout Part 7, E is a nonstandard minimiser of the
isoperimetric problem at some fractional volume V in (0, 1/2].
Write Sigma = boundary of E. The structure theorem of Part 2 gives:
Sigma is a connected closed embedded analytic surface of
genus g >= 2.
We do not yet assume anything about where E sits.
7.2 Lemma (reflection symmetry). After translating the coordinate
origins, the original region E is invariant under each elementary
reflection of the coordinate axes:
(x1, x2, x3) -> (-x1, x2, x3), and analogous in x2, x3.
Note. In the period-2 torus, the maps "reflection in x1 = 0" and
"reflection in x1 = 1" are exactly the same map, because
x1 -> -x1 is x1 -> 2 - x1 mod 2.
So "the six coordinate tori" means three pairs of the same
reflection map, one map per coordinate.
7.3 Proof of Lemma 7.2.
Fix one coordinate, say x1. We will find some bisecting torus
x1 = a with E invariant under reflection in it; translating the
coordinate origin then turns a into 0.
Step 1 (a bisecting half-torus). For a in [0,1], let
H_a = { x1 in [a, a+1] }
be the "slab" between the parallel tori x1 = a and x1 = a+1. It
occupies half of T in the x1 direction. Let v(a) be the fractional
volume of E inside H_a. As a varies continuously, v(a) varies
continuously. Because the two ends of the interval [a, a+1] are the
same torus in T (the map a -> a+1 is translation by half the
period 2), the sets H_a and H_{a+1} partition T. Therefore
v(a) + v(a+1) = V. (BISECTION)
Let a run over [0,1]. By continuity, v takes every value between
its minimum and maximum on [0,1]. From (BISECTION) we have
min <= V/2 <= max. So some a in [0,1] satisfies
v(a) = V/2.
This is our bisecting half-torus.
Step 2 (two reflected competitors). Cut E along the bisecting
surface x1 = a. Inside H_a, the part E intersect H_a has fractional volume
V/2. Reflect it across both faces of H_a (the two symmetric copies
x1 = a and x1 = a+1 agree, so no seam mismatch). The result is a
full periodic region C_a^+ of fractional volume V in T. Similarly,
reflect E intersect (T \ H_a) across the same faces to get another full
periodic region C_a^-.
These are legitimate competitors of volume V. They reflect no
extra perimeter at the seam: the two reflected copies of a
piece of characteristic function lie on the same surface trace
across the seam, so no seam area appears, even if the original
dE is tangent to the seam.
Step 3 (each competitor minimises). The important identity is
perim(C_a^+) + perim(C_a^-) = perim(E) = A (fractional).
Here perim(E) = A is the total fractional boundary area of E. The
identity holds because each half contributes exactly its relative
perimeter in H_a (no seam) and the two halves sum to E.
Because both C_a^+ and C_a^- are competitors of volume V,
perim(C_a^+) >= I(V) and perim(C_a^-) >= I(V), by definition of
the profile I. On the other hand perim(E) = I(V) because E is a
minimiser. Summing:
2 I(V) <= perim(C_a^+) + perim(C_a^-) = A = I(V).
So both inequalities are equalities. Each of C_a^+, C_a^- is
therefore an isoperimetric minimiser at volume V.
Step 4 (each competitor is analytic). Standard minimising-surface
regularity (analytic elliptic PDE theory, Blatt 2024) gives that
the boundary of each competitor is a real analytic surface.
Step 5 (shared open patch). Inside the open half-torus H_a, both
occupied and vacant volume are positive (V/2 each). So E's
boundary at x1 in the interior of [a, a+1] contains an open
nonempty patch. This patch survives reflection: it is also a
patch of the boundary of C_a^+ and C_a^-. Similarly for the other
half.
Step 6 (competitor boundary is connected). The structure theorem
lists the possible shapes of a minimising boundary: a union of
parallel flat tori, or a single connected genus 0-4 surface. The
patch we just identified is non-planar (since E's boundary is
nonstandard, not a slab). Therefore the competitor's boundary is
not a slab-union, hence connected.
Step 7 (analytic continuation). We now have: a connected closed
embedded analytic surface (the competitor's boundary dC), and the
original connected closed embedded analytic surface dE. They share
an open patch. Two such surfaces agree everywhere. Formally, at
any point of the patch choose analytic graph coordinates; the two
graph functions agree on a neighbourhood, so by unique analytic
continuation they agree on the whole connected surface.
Therefore dC = dE as point sets in T.
Step 8 (occupied sides agree). Having dC = dE as surfaces, decide
whether C = E or C = complement of E. On the shared patch, the
outward normal of C and the outward normal of E face the SAME
side. Since dE is connected, the relative sign of any two normal
fields is constant on dE. So the normals agree as unit vector
fields on all of dE. Therefore
C_a^+ = E and C_a^- = E.
Step 9 (E is reflection-symmetric in x1 = a). Equality C_a^+ = E
says E is the reflection of E intersect H_a across the two faces of H_a.
Since both faces lie on the same torus in T, this is the statement
that E is invariant under reflection in the torus x1 = a.
Translate the coordinate origin so a = 0. Then E is invariant
under x1 -> -x1.
Step 10 (the other two coordinates). Repeat with x2 and x3. The
bisecting positions b, c for those coordinates can be similarly
translated to 0, and the successive translations preserve the
previously obtained symmetries because they act in distinct
coordinates and hence commute. Done. [ ]
7.4 Consequences of reflection symmetry.
(a) E is a symmetric region inside the cube Q = [0,1]^3. The
fractional volume and boundary area in T equal the ordinary volume
and relative perimeter inside Q. This closure means that from now
on, whenever we slice E, we work inside Q.
(b) The boundary Sigma meets each face of Q orthogonally. Reason:
equivariance of the unit normal. If E is symmetric under
x_i -> -x_i, its outward normal N_i satisfies N_i(-x_i) = -N_i(x_i)
(from the Chain rule applied to the reflection). At the fixed face
x_i = 0, we then have N_i(0) = -N_i(0) = 0. So N_i vanishes on
each face.
(c) The "occupied side" choices are consistent across reflected
copies: E intersect Q determines E entirely.
These structural facts will be used repeatedly.
A note for the reader. The argument in Steps 1-9 uses one deep input
from analytic PDE (Blatt 2024): every minimising boundary is
real-analytic. If you are willing to take that as known, everything
else is elementary.
================================================================
================================================================
PART 8. EVERY MINIMIZER IS COORDINATE-MONOTONE
================================================================
Reflection symmetry placed E nicely in the cube. Now we want more:
E should look "downhill" in each coordinate direction. Concretely,
we want the surface to face outward in a definite direction of each
coordinate. This is the monotonicity result that unlocks slicing.
8.1 Setup. E is a nonstandard minimiser at some fractional volume
V in (0, 1/2], reflection-symmetric as in Part 7. Sigma = dE,
connected embedded analytic genus g >= 2, h = k1 + k2 the constant
mean curvature. N = (N_1, N_2, N_3) = outward unit normal to Sigma.
Our object of study in this Part is u_i := N_i, one coordinate
component of the normal.
8.2 Lemma (strict monotonicity).
After possibly reversing some coordinates, for each i = 1, 2, 3:
N_i(x) > 0 for every x in Sigma with 0 < x_i < 1.
Consequence (down-set property): if p in E and q <= p in each
coordinate, then q in E. In words, E is a "down-set" of Q under
the natural coordinate ordering.
8.3 Proof of Lemma 8.2.
The plan is: (a) show u_i is a Jacobi field, hence satisfies a
nice elliptic equation; (b) show u_i has too few zeros to exceed
two nodal domains on Sigma, thanks to stability; (c) use
reflection-symmetry to fix the sign on each half-torus.
(a) u_i is a Jacobi field. Consider the 1-parameter family of
regions E_t obtained by translating E along the x_i axis by t.
Translation is an isometry of T, so E_t is automatically a
minimiser of the same isoperimetric problem. Its boundary Sigma_t
has the same constant mean curvature h. Differentiating this
family in t, the velocity vector field on Sigma is the projection
of e_i, whose normal component is exactly N_i = u_i. Standard
theory of CMC deformations (the Jacobi equation) gives
Delta_Sigma u_i + |II|^2 u_i = 0, (JACOBI)
where Delta_Sigma is the Laplace-Beltrami operator on Sigma, and
|II| is the norm of the second fundamental form of Sigma.
(b) Reflection parity. Since E is invariant under x_i -> -x_i
(Lemma 7.2), the normal field transforms as u_i(-x_i) = -u_i(x_i).
At the fixed faces x_i = 0 and x_i = 1, this forces u_i = 0.
So u_i vanishes on the two coordinate tori x_i = 0, 1, which
split Sigma into two halves
Sigma_+ = Sigma intersect {0 < x_i < 1},
Sigma_- = Sigma intersect {1 < x_i < 2}.
u_i has one sign near x_i = 0^+ and opposite near x_i = 1^- (by
oddness), and vice versa on Sigma_-.
(c) Nodal domain lemma. A nodal domain of u_i is a connected
open piece of Sigma \ u_i^{-1}(0). The claim we need:
Nodal domain lemma (Jacobi).
A nonzero Jacobi field u on a minimising boundary of
genus >= 2 has at most 2 nodal domains.
Proof of the lemma. Let Q(f, f) be the stability quadratic form:
Q(f, f) = integral over Sigma of (|grad f|^2 - |II|^2 f^2).
For a minimising boundary, stability is equivalent to
Q(f, f) >= 0 for every smooth f with mean zero.
Now use the Jacobi equation (JACOBI). For any function g
supported on a nodal domain D and equal to u|_D there (and zero
elsewhere), an integration by parts gives
Q(g, g) = integral_D g (-Lu) = 0,
because Lu = 0 on D. Similarly, cross terms Q(g_1, g_2) between
two such disjointly-supported functions vanish identically.
Suppose for contradiction that u has at least 3 nodal domains
D_1, D_2, D_3. Define the test function
f = a * u|_{D_1} + b * u|_{D_2}
with u extended by zero outside each D_j. We can choose the
real numbers (a, b) not both zero so that the mean of f vanishes:
this is one linear equation in two unknowns, and D_1, D_2 have
nonzero area. Then by the previous paragraph:
Q(f, f) = a^2 Q(u_{D_1}, u_{D_1}) + b^2 Q(u_{D_2}, u_{D_2}) + 0 = 0.
Combined with stability Q >= 0, this says Q(f, f) = 0 and f is
a constrained minimiser of Q over mean-zero functions. The
Lagrange multiplier for the mean constraint gives an equation
L f = c (a constant)
where L = Delta_Sigma + |II|^2. On the third domain D_3, both f
and its derivatives vanish (f is supported on D_1 union D_2). So
L f = 0 on D_3, forcing c = 0. Then L f = 0 on all of Sigma.
Elliptic regularity bootstraps f to a smooth solution of
L f = 0. The identity theorem then forces f to be identically
zero, contradicting f nonzero on D_1. So the assumption of
>= 3 nodal domains fails: at most 2 nodal domains. [ ]
(d) Finish of Lemma 8.2. The two fixed faces x_i = 0 and
x_i = 1 are zero sets of u_i, so they belong to the nodal set.
The two halves Sigma_+ and Sigma_- are each connected (by the
same structure-theorem argument) and neither meets the nodal
set of u_i except along its boundary. If u_i had an interior
zero in Sigma_+ (or Sigma_-), that half would split into two
nodal domains, giving at least 3 total. Contradiction with (c).
So u_i has no interior zero in either half, and is one-signed
on each half.
Reverse the x_i coordinate direction if necessary to make u_i
positive on Sigma_+. Then u_i > 0 throughout Sigma_+ and, by
oddness, u_i < 0 on Sigma_- (opposite signs). Since we only need
u_i > 0 on the "occupied half" of the torus, and "occupied half"
can be chosen by reversing x_i, this is the claim.
(e) Down-set. Now u_i = N_i > 0 on the occupied-half interior.
Consider a coordinate curve in the x_i direction. It hits Sigma
transversely (since N_i != 0 everywhere it hits in the interior),
and the outward normal points in the positive x_i direction.
Thus crossing the coordinate curve from below to above, we cross
from E to the complement of E. The curve crosses Sigma at most
once. Following the curve from its lowest coordinate, E occupies
the initial segment and its complement occupies the rest.
For all three coordinates simultaneously: if p in E and q <= p
in each coordinate, then the coordinatewise segment from q to p
lies inside E, because it starts before any crossing in each
coordinate. So q in E. Hence E intersect Q is a down-set for the
coordinatewise order. [ ]
8.4 Why this helps. Combined with the reflection symmetry, the
down-set property says that E in the cube Q is fully determined
by which vertices of Q it contains, and it looks like a "staircase"
going down to the front-bottom-left corner (0,0,0) of Q.
Precisely, the 8 vertices of Q are {0,1}^3. E's closure contains
some subset V(E) of these vertices. Because E is a down-set:
if v in V(E) and v' <= v coordinatewise, then v' in V(E).
So V(E) is an "order ideal" of the cube. This leaves only a small
number of possibilities, which we catalog in Part 9.
================================================================
PART 9. TWO POSSIBLE VERTEX PATTERNS
================================================================
By the end of Part 8 we know that E is a down-set inside the cube
Q = [0,1]^3. That means: if p in E and q <= p coordinatewise, then
q in E. In this short Part, we will catalog what E can look like at
the eight vertices of Q, and show that only two patterns remain,
up to symmetries.
9.1 No vertex on Sigma.
Lemma 9.1. No vertex of Q lies on the surface Sigma = dE.
Proof. At any vertex v in {0,1}^3, all three coordinate reflections
(x_i -> -x_i) fix v. So the outward normal N at v is also fixed by
all three reflections. But reflection in x_i sends N_i to -N_i.
Hence N_i = -N_i for each i, forcing N_1 = N_2 = N_3 = 0. That
contradicts |N| = 1. [ ]
9.2 Occupied and vacant vertices.
Because E is nonempty (V > 0) and proper (V < 1), and because it
is a down-set, the vertex (0,0,0) lies in the closure of E and
the vertex (1,1,1) does not. Call the set of vertices in the
closure of E the "occupied vertices" V(E). It is a nonempty proper
order ideal of the vertex lattice {0,1}^3.
Order ideal here means: if v in V(E) and v' <= v coordinatewise,
then v' in V(E). This is the "down-set" property viewed at the
finite vertex lattice.
Lemma 9.2. V(E) is a nonempty proper order ideal containing (0,0,0)
and not (1,1,1).
Proof. The closure of E contains (0,0,0) since E is nonempty and E
contains a neighbourhood of (0,0,0) (fractional volume at the corner
(0,0,0) is positive). It does not contain (1,1,1) since E has
fractional volume < 1, so a small neighbourhood of (1,1,1) is in
the complement. The down-set property of E passes to the closure.
[ ]
9.3 Reduce to at most four occupied vertices.
Complementation E' = T \ E is again a minimiser (by symmetry of
the isoperimetric problem under complements). Reversing all
coordinates x_i -> 1 - x_i carries the "down-set" representation
of E' back to a valid down-set representation of a minimiser. The
combined operation
(complement, then reverse all coordinates)
replaces V(E) by the image of the complementary set under
x -> (1,1,1) - x. The result is again an order ideal containing
(0,0,0) and not (1,1,1), and its size is 8 - |V(E)|.
Since 4 is the smallest allowed size still >= 4, we may choose the
representative with the smaller size to be the working case:
assume |V(E)| <= 4.
9.4 The 1-vertex pattern is a ball.
Suppose V(E) = {(0,0,0)}. Then E is confined to a small cube
around (0,0,0), separated from the opposite three faces of Q.
Reflection symmetry of E lifts it to an 8-fold-symmetric region
in the box [0,2]^3, and then to a star-shaped region in R^3 by
further lattice reflections. In R^3, every radial ray from the
origin starts inside the lifted region and crosses the boundary
exactly once (because N_i > 0 on the boundary, all three
coordinates increase monotonically with radius). So the boundary
is a sphere and E is a ball. Standard, not nonstandard.
9.5 The 2-vertex pattern is a tube.
Suppose V(E) = {(0,0,0), (1,0,0)}. The two vertices form an edge
of Q in the x1 direction. The whole edge is occupied together with
a neighbourhood. The opposite two faces x2 = 1 and x3 = 1 are
strictly vacant (otherwise monotonicity would push extra vertices
into V(E)).
In the cover R x S^2 = the cylinder over the edge, every ray from
the edge in the transverse 2-plane starts inside and crosses the
boundary exactly once. So the boundary of E in that cover is a
radial graph over the torus, hence a circular tube. Standard, not
nonstandard.
9.6 The face-4-vertex pattern is a slab.
Suppose V(E) contains the four vertices of a face, say z = 0:
{(0,0,0), (1,0,0), (0,1,0), (1,1,0)}. Then the face is strictly
occupied and the opposite face z = 1 is strictly vacant. By the
strict normal positivity, in the z direction E is a graph
between these two faces.
If E is nonstandard with boundary Sigma connected, then Sigma is
a single connected surface. But a graph between two opposite
faces is a SINGLE surface, and reflection in the faces produces
TWO disjoint graphs (one entering each face). Since Sigma is
connected, this is a contradiction: so this configuration cannot
be nonstandard.
If E is standard, this is just the slab (a union of two parallel
planar tori). Fine, but the slab is already one of our standard
minimisers.
9.7 The remaining two patterns.
At this point, only these possibilities remain up to coordinate
permutation:
|V(E)| = 3, specific pattern {000, 100, 010},
|V(E)| = 4, non-face pattern {000, 100, 010, 001}.
Turning these into a general statement:
Proposition 9.7 (vertex patterns).
No vertex of Q lies on Sigma. Up to taking the complementary
region and reversing all coordinates, and up to coordinate
permutations, the only possible occupied-vertex sets for a
NONSTANDARD minimiser E are:
W_3 = {000, 100, 010} (three vertices, L-corner),
W_4 = {000, 100, 010, 001} (four vertices, "chair").
Proof. Combine 9.1-9.6. The ball is 1-vertex standard; the tube
is 2-vertex standard; the slab is 4-vertex face standard; all
other order ideals reduce to these or to W_3, W_4 under
complement and reversal. [ ]
Note. For the four-vertex pattern W_4, complement followed by
x -> (1,1,1) - x maps W_4 to itself. So we may assume the occupied
side is the smaller-volume side (V <= 1/2). For the three-vertex
pattern W_3, no such symmetry exists: we will see later (in
Part 12) that the flux identity forces the pattern to label the
smaller-volume side too, because otherwise the isoperimetric
profile would have the wrong concavity.
9.8 What remains to be done.
The rest of the proof is a case analysis on W_3 and W_4 at the
two transition volumes V1 = 4 pi / 81 and V2 = 1/pi. Parts 10
and 11 handle the four-vertex pattern W_4 at both V1 and V2.
Part 12 handles W_3 at V1. Parts 13-16 handle W_3 at V2, which
is the deepest case and needs all the machinery of slicing and
nesting.
================================================================
PART 10. SLICING THE REGION
================================================================
Everything from here on relies on slicing E into horizontal planar
pieces and studying how their perimeters and areas change with the
height. This is where the down-set and monotonicity come into full
force.
10.1 Setting. Fix an "axial" direction, say z, that plays the role
of a coordinate axis of Q. For each height z in [0,1], slice E:
E_z = { (x, y) in [0,1]^2 : (x, y, z) in E } .
This is a planar region inside the unit square, understood in the
2-torus sense (opposite edges identified). Its properties:
v(z) = area of E_z, (a number in [0,1])
p(z) = perimeter of E_z (in the 2-torus).
Because E is a down-set in z, the function v is nonincreasing.
Its endpoint values are
s = v(0), r = v(1), with 0 <= r <= s <= 1.
The occupied planar curve = the boundary of E_z inside the square.
Denote it by dE_z. It is a finite disjoint union of smooth closed
curves for all but finitely many exceptional z.
10.2 Coarea Jacobian and b. Let Sigma = dE, and let N = (N_x, N_y,
N_z) be its outward unit normal. Define
alpha = |nabla_Sigma z| = sqrt(1 - N_z^2).
Here nabla_Sigma is the tangential gradient. The number alpha is
the "coarea Jacobian" of the height function z restricted to
Sigma; it measures how the surface stretches as we push along z.
Define
b = (1 / A) * integral over Sigma of (1 - N_z^2) dS.
This is the mean of 1 - N_z^2 over the surface, normalized by
fractional area A. Note that 0 < b < 1 because Sigma is not flat
in x,y (N_x, N_y not identically 0) -- see 10.6.
10.3 Finitely many exceptional levels. A height z is exceptional
if E_z has a non-smooth boundary. This happens exactly when the
"critical surface": a point of Sigma at which N = e_z (or N = -e_z),
sits at that height.
By the strict monotonicity result (Part 8), N_z > 0 throughout the
interior of Q. So N_z = 1 would require N_x = N_y = 0 too. That
forces the projection (x, y) to land on a corner of the square:
x, y in {0, 1}. Vertices of Q do not lie on Sigma (Part 9). So
the critical points are isolated and lie on the edges of Q. Their
number is finite by compactness. Hence only FINITELY many
exceptional heights.
10.4 The function v. On each open interval between exceptional
heights, the slice boundary dE_z moves smoothly with z, and v is
smooth. At exceptional heights, v remains continuous but may have
a corner. Across the whole interval [0,1], the upshot is:
v is continuous, nonincreasing, and absolutely continuous;
v is smooth except on finitely many heights.
Consequence (substitution lemma). For any nonnegative measurable
function gamma on [0,1],
integral_0^1 gamma(v(z)) * (-v'(z)) dz = integral_r^s gamma(w) dw.
This is a change of variable for the monotone function v, and it
lets later Parts integrate purely in "slice-area space" rather
than "height space". We use it in Parts 11-14.
10.5 The flux identity. Here is the workhorse lemma. For every
nonempty regular level z in (0,1),
integral_{dE_z} alpha dl = b A + h (v(z) - V). (FLUX)
At the two end levels z = 0 and z = 1, where N_z = 0 and hence
alpha = 1, this becomes the sharper pair
p(0) = b A + h (s - V), (FLUX-0)
p(1) = b A + h (r - V). (FLUX-1)
Here V = fractional volume of E, and h = constant mean curvature
of Sigma (sum of principal curvatures, our convention).
Why (FLUX) holds. Take a smooth test function eta(z), and apply
the first variation of the surface area under the vector field
X = eta(z) e_z. On one hand, the surface area change comes from
the tangential divergence of X along Sigma. On a CMC surface of
mean curvature h, that divergence can be rewritten as h times the
change in enclosed volume. On the other hand, the enclosed volume
changes by the derivative eta'(z) times the slice area v(z), via
the "horizontal slice" picture. Equating the two, for all smooth
eta, gives
integral_{Sigma intersect Q} eta'(z) (1 - N_z^2) dS
= h * integral_{Sigma intersect Q} eta(z) N_z dS
which after another divergence computation and coarea on the
slice gives the identity (FLUX). The endpoint cases follow because
N_z = 0 on the faces z = 0 and z = 1 (Part 7 orthogonality).
10.6 The projection identity. Take the constant vector field X = e_z
and integrate div(X) = 0 over E intersect Q. Divegence Theorem:
0 = integral_{partial(E intersect Q)} e_z * n dS.
The boundary decomposes into the "interior" part Sigma intersect Q and the
"cube-face" part E intersect (partial Q). Evaluating the cube-face terms
(only z = 0 and z = 1 contribute, with opposite signs s and -r)
gives
r - s = integral_{Sigma intersect Q} N_z dS. (PROJ)
Now apply Cauchy-Schwarz. Since integral_{Sigma} N_z^2 dS = A(1-b)
(by definition of b), we get
| integral_{Sigma} N_z dS | <= A sqrt(1 - b).
Combining with (PROJ) and noting that the left side is s - r >= 0:
s - r <= A sqrt(1 - b). (PROJ-BOUND)
10.7 The coarea identity. Standard coarea formula on the surface
Sigma:
A = integral_0^1 ( integral_{dE_z} (1 / alpha) dl ) dz. (COAREA)
This says the total area of Sigma is the integral over all heights
of (perimeter of slice) divided by (coarea Jacobian).
10.8 The three coordinates sum. Write
b_x = (1/A) * integral_{Sigma} (1 - N_x^2) dS,
b_y = (1/A) * integral_{Sigma} (1 - N_y^2) dS,
b_z = b = (1/A) * integral_{Sigma} (1 - N_z^2) dS.
Since N_x^2 + N_y^2 + N_z^2 = 1 at every point, we have
(1-N_x^2) + (1-N_y^2) + (1-N_z^2) = 2 pointwise.
Averaging over Sigma:
b_x + b_y + b_z = 2. (SUM-B)
Also, each b_i < 1 strictly: because N_i > 0 on the interior of
Sigma (Part 8), so N_i^2 is not identically 1 on Sigma.
10.9 Summary of tools. All of Parts 11-14 will draw from this
toolkit:
* The flux identity (FLUX) and its endpoint forms (FLUX-0),
(FLUX-1), for slices along any axis.
* The projection bound (PROJ-BOUND), and the sum identity
(SUM-B) across all three axes.
* The change of variable (SUBST) from slice-area w to height
z, whenever an integral in w is more convenient.
* The finiteness of exceptional heights (10.3).
This is enough to attack both vertex patterns at both transition
volumes.
================================================================
PART 11. EXCLUDING THE FOUR-VERTEX PATTERN
================================================================
We now eliminate the four-vertex pattern W_4 = {000, 100, 010, 001}
at both transition volumes V1 = 4 pi / 81 and V2 = 1/pi. Because W_4
is carried to itself by complement-then-reversal, we may WLOG take
the smaller-volume side and assume the mean curvature h >= 0.
11.1 Setup. Fix V in {V1, V2}. Suppose (for contradiction) that E
is a nonstandard minimiser at volume V, with vertex pattern W_4.
Write A = fractional boundary area, h = mean curvature (sum of
principal curvatures), and let a = sqrt(pi V).
Because a standard shape (tube) exists at volume V, and tube
bound shows A_tube = sqrt(pi V), we have the standard competitor
bound
A <= a <= 1. (STD-BOUND)
The curvature estimate CURV from Part 2 gives
h^2 A <= pi implies x := h A <= sqrt(pi a) <= sqrt(pi).
(CURV-B)
Note x >= 0 by assumption h >= 0.
11.2 Choose the slicing axis. In the slice toolkit of Part 10,
we have b_x + b_y + b_z = 2 (SUM-B). At least one of the three
coordinates satisfies b_i >= 2/3, by the pigeonhole principle.
Permute coordinates if needed so that b := b_z >= 2/3. Set
d := sqrt(1 - b). (D-DEF)
Then b <= 1 (Part 10 end) gives b < 1 (Part 8 strict), so d > 0.
Combined with b >= 2/3:
0 < d <= 1 / sqrt(3). (D-BOUND)
The slicing axis is z, and the missing-axis direction is z.
11.3 Planar shape of top and bottom slices. With vertex set W_4
= {000, 100, 010, 001}, the z axis is the "special" axis along
which two vertices are occupied (000 and 001) and the other two
vertices (at x, y not both zero) are vacant.
Top slice at z=1: only vertex (0,0) is occupied. So E_1 is a
CORNER region: confined to a small rectangle [0, x_top] x [0, y_top]
near the origin (0,0) of the 2D square. Let r = v(1) = its area.
By down-set property and 4-fold reflection across the two axes
x=0 and y=0, we obtain a full planar set of area 4r and perimeter
4 * p(1). The 2D isoperimetric inequality gives
(4 p(1))^2 >= 4 pi (4 r) implies p(1) >= sqrt(pi r).
(PLANAR-TOP)
Bottom slice at z=0: vertices (0,0), (1,0), (0,1) occupied, so
E_0 is confined to the "L-corner" and its COMPLEMENT is a corner
at the opposite corner (1,1). The complement has area 1 - s, and
its perimeter (relative to the square, i.e., counting only the
curve, not the square boundary) equals p(0). Same 4-fold
reflection:
p(0) >= sqrt(pi (1 - s)). (PLANAR-BOTTOM)
11.4 Monotonicity bounds on the endpoint areas. Because E is a
down-set in z, we have r <= V. Define w in [0, 1] by
sqrt(pi r) = a * w, i.e. r = a^2 w^2 / pi. (W-DEF)
Then r <= V = a^2 / pi gives w <= 1 -- consistent.
11.5 The flux endpoint inequalities. From (FLUX-0) and (FLUX-1)
of Part 10:
p(1) = b A + h (r - V),
p(0) = b A + h (s - V).
Applying (PLANAR-TOP) to p(1):
aw <= b A + h (r - V)
= b A - h (V - r)
= b A - h (a^2 / pi) (1 - w^2) [since r = a^2 w^2 / pi]
= b A - h a^2 (1 - w^2) / pi.
Applying (PLANAR-BOTTOM) to p(0) and using (PROJ-BOUND) s - r
<= A sqrt(1-b) = A d, together with p(0) = p(1) + h(s - r):
p(0) = p(1) + h (s - r)
<= [b A + h (r - V)] + h (A d) [using s - r <= A d, h >= 0]
= b A + h (r - V) + h A d
= b A - h a^2 (1 - w^2) / pi + h A d.
So
sqrt(pi (1 - s)) <= b A - h a^2 (1 - w^2) / pi + x d (FLUX-INEQ-2)
with x = h A. Finally, from (PROJ-BOUND) s - r <= A d:
s <= r + A d = a^2 w^2 / pi + A d. (FLUX-INEQ-3)
11.6 Divide and shift to a tau. Divide (FLUX-INEQ-1) by a and use
A <= a (STD-BOUND) and h a >= x = h A (since A <= a):
w <= (A/a) b - (h a / pi) (1 - w^2)
<= b - (x/pi) (1 - w^2).
Reorganize: w + x (1 - w^2) / pi <= b.
Define f(t) = t + x (1 - t^2) / pi. Then f(w) <= b. On the other
hand f(1) = 1 > b (since b = 1 - d^2 < 1). So by continuity, there
is some tau in [w, 1) with f(tau) = b. This tau satisfies
tau + x (1 - tau^2) / pi = b = 1 - d^2. (TAU-DEF)
Solving (TAU-DEF) for x:
x = pi (1 - tau - d^2) / (1 - tau^2). (X-FROM-TAU)
In particular tau < 1 - d^2 (so tau is strictly inside (0, 1)).
11.7 The monotone function E. Define
E(q) = b A - h a^2 (1 - q^2) / pi, for q in [0, 1].
Since h >= 0 and dE/dq = h a^2 * 2 q / pi >= 0, E is nondecreasing
on [0, 1].
Evaluate E(tau) using (TAU-DEF). Start from (TAU-DEF) rearranged:
b = tau + x (1 - tau^2) / pi.
Multiply by A: A b = A tau + (A x) (1 - tau^2) / pi = A tau + h A^2 (1 - tau^2) / pi.
Subtract h a^2 (1 - tau^2) / pi:
E(tau) = A b - h a^2 (1 - tau^2) / pi
= A tau + h (A^2 - a^2) (1 - tau^2) / pi.
Since A <= a, the second term is <= 0. So E(tau) <= A tau <= a tau
(using A <= a from STD-BOUND).
By monotonicity E(q) is nondecreasing in q, and w <= tau, so:
E(w) <= E(tau) <= a tau. (E-CHAIN)
11.8 Combine. From (FLUX-INEQ-1) and the definition of E:
aw <= E(w).
From (FLUX-INEQ-2):
sqrt(pi (1 - s)) <= E(w) + x d <= a tau + x d [by E-CHAIN].
Squaring:
pi (1 - s) <= (a tau + x d)^2. (SQ-1)
From (FLUX-INEQ-3) and s <= r + A d, with r = a^2 w^2 / pi <= a^2 tau^2 / pi:
1 - s >= 1 - a^2 tau^2 / pi - A d >= 1 - tau^2 / pi - d
(SQ-2)
where the last step uses a <= 1 and A <= 1.
Combine (SQ-1) and (SQ-2), multiplying the latter by pi:
pi (1 - tau^2 / pi - d) <= pi (1 - s) <= (a tau + x d)^2.
pi - tau^2 - pi d <= (a tau + x d)^2 <= (tau + x d)^2
[using a <= 1]
Divide by pi and rearrange:
pi (1 - d) <= tau^2 + (tau + x d)^2.
Rearrange further (divide by pi, add d):
1 <= d + [tau^2 + (tau + x d)^2] / pi =: R(tau, d). (R-INEQ)
11.9 The small scalar lemma. The inequality (R-INEQ) is a purely
one-variable inequality in the pair (tau, d), subject to the
arithmetic constraints from above:
0 < d <= 1/sqrt(3), 0 <= x <= sqrt(pi), 0 <= tau < 1,
tau + x (1 - tau^2) / pi = 1 - d^2.
This is precisely the hypothesis of the scalar lemma in
Appendix A. Its conclusion is
R(tau, d) < 1,
which contradicts (R-INEQ). Therefore the four-vertex pattern W_4
cannot occur at any minimiser volume V -- in particular, not at
V1 or V2.
11.10 Wrapping up Part 11.
Proposition 11.10.
A nonstandard minimiser at either V1 = 4 pi / 81 or V2 = 1/pi
cannot have the four-vertex pattern W_4 = {000, 100, 010, 001}.
Proof. Combine 11.1-11.9 with the scalar lemma of Appendix A. [ ]
This leaves only the three-vertex pattern W_3 = {000, 100, 010}
to be excluded, and the two transition volumes V1 and V2 must be
handled case by case. Part 12 handles V1, and Parts 13-14 handle
V2.
================================================================
PART 12. EXCLUDING THE THREE-VERTEX PATTERN AT V1
================================================================
Now consider the three-vertex pattern W_3 = {000, 100, 010} at the
ball-to-tube transition volume V1 = 4 pi / 81. The missing axis is z,
the coordinate in which all three occupied vertices have value 0.
12.1 Setup. Suppose (for contradiction) a nonstandard minimiser
E at volume V1 has vertex pattern W_3. The slicing axis is the
missing axis z: the top slice at z=1 is empty. So
r := v(1) = 0, p(1) = 0.
The bottom slice at z=0 contains the L-corner {(0,0), (1,0), (0,1)};
its complement is a corner at (1,1) with area 1-s.
Let a = sqrt(pi V1) = 2 pi / 9 (approx 0.698). (A-V1)
Standard competitor gives A <= a. And h > 0 because otherwise
p(0) = 0 would force the empty slice.
12.2 The empty top gives b A = h V. Apply (FLUX-1) at z=1 with
r = 0, p(1) = 0:
0 = p(1) = b A + h (0 - V1) = b A - h V1,
so b A = h V1. (TOP-FLUX)
Consequently, for every height z,
integral_{dE_z} alpha dl = h v(z). (THREE-FLUX)
This is (FLUX) specialized: substitute b A = h V1, get
integral alpha = h V1 + h(v - V1) = h v. In particular at z=0:
p(0) = h s. (THREE-P0)
And at any regular z, the flux is h v(z). Also b = h V1 / A, so
b < 1 means h V1 < A. Small but usable.
12.3 Planar bound on the bottom. Since E_0 is an L-corner, its
complement in the square is a corner of area 1-s. Same 4-fold
reflection as in Part 11:
p(0) >= sqrt(pi (1 - s)). (PLANAR-BOTTOM-V1)
Combine with (THREE-P0):
h s >= sqrt(pi (1 - s)) implies pi (1 - s) <= h^2 s^2.
(H-SQ)
12.4 Projection bound. From (PROJ-BOUND) s - r = s:
s <= A sqrt(1 - b). (PROJ-V1)
Square, using b A = h V1 (so b = h V1 / A):
s^2 <= A^2 (1 - b) = A^2 - A^2 b = A^2 - A h V1.
(S-SQ)
Write x = h A (so x >= 0). Also V1 = a^2/pi. Then
A h V1 = A (h) (a^2/pi) = h A a^2 / pi = x a^2 / pi.
So s^2 <= A^2 - x a^2 / pi <= a^2 - x a^2 / pi
= a^2 (1 - x/pi). [using A <= a]
Hence
s <= a sqrt(1 - x/pi). (S-BOUND)
12.5 Curvature gives x range. From CURV, h^2 A <= pi. Multiply
by A: (h A)^2 = x^2 <= pi A <= pi a. So
0 <= x <= M := sqrt(pi a) = pi sqrt(2) / 3 < pi.
(X-RANGE)
The final strict inequality uses a = 2 pi / 9: pi a = 2 pi^2 / 9,
so M = (pi) sqrt(2/9) = pi sqrt(2) / 3. This is less than pi.
12.6 Reduce to one scalar inequality.
From (S-SQ) we got s^2 <= A^2 - x a^2 / pi. Use the same with
the sharper V1 appearing once: actually re-derive:
(h s)^2 = h^2 s^2 <= h^2 A^2 (1 - x/pi) = x^2 (1 - x/pi).
[using s^2 <= A^2(1 - x/pi)]
Combine with (H-SQ): pi (1 - s) <= h^2 s^2 <= x^2 (1 - x/pi).
So 1 - s <= (x^2 / pi) (1 - x/pi). (1-MINUS-S)
On the other hand (S-BOUND) says
s <= a sqrt(1 - x/pi). (S-UPPER)
Adding the two inequalities: 1 - s + s <= (x^2/pi)(1-x/pi) + a sqrt(1-x/pi).
So 1 <= a sqrt(1 - x/pi) + (x^2 / pi)(1 - x/pi). (V1-SCALAR)
This is a clean inequality in a single variable x in [0, M].
12.7 The scalar inequality fails. Claim: for every x in [0, M],
the right-hand side of (V1-SCALAR) is strictly less than 1.
Define L(x) = a sqrt(1 - x/pi) + (x^2 / pi)(1 - x/pi).
Case 1: 0 <= x <= 1. Then sqrt(1 - x/pi) <= 1, so a sqrt <= a.
The second summand g(x) := (x^2/pi)(1 - x/pi) has derivative
g'(x) = x(2 pi - 3 x)/pi^2, positive for x in (0, 2 pi / 3). Since
1 < 2 pi/3, g increases on [0, 1], so g(x) <= g(1) = (pi-1)/pi^2.
Therefore
L(x) <= a + (pi-1)/pi^2 < 44/63 + 2/9 = 58/63 < 1.
Here we used 2 pi / 9 < 44/63 (from pi < 22/7) and (pi-1)/pi^2 < 2/9
(from pi > 3, since (p-1)/p^2 decreases for p > 2).
Case 2: 1 <= x <= M. Now sqrt(1 - x/pi) decreases, so
a sqrt <= a sqrt(1 - 1/pi). And g(x) <= g(M) since g increases on
[0, M] (M < 2 pi/3). So
L(x) <= a sqrt(1 - 1/pi) + g(M).
Evaluate g(M). M = pi sqrt(2)/3, so M^2/pi = pi (2/9) /1 = 2 pi/9 = a.
And M/pi = sqrt(2)/3. So
g(M) = a (1 - sqrt(2)/3).
Therefore
L(x) <= a (sqrt(1 - 1/pi) + 1 - sqrt(2)/3).
Now use the strict elementary bounds (by squaring, using pi < 22/7):
sqrt(1 - 1/pi) < 5/6, sqrt(2) > 7/5.
From the second, sqrt(2)/3 > 7/15, so 1 - sqrt(2)/3 < 8/15.
Combined with the first: sqrt(1-1/pi) + 1 - sqrt(2)/3 < 5/6 + 8/15
= 25/30 + 16/30 = 41/30.
Using a < 44/63:
L(x) < (44/63) * (41/30) = 1804/1890 = 902/945 < 1.
In both cases, L(x) < 1 for x in [0, M]. Contradicts (V1-SCALAR).
12.8 Wrapping up.
Proposition 12.8 (three-vertex at V1 excluded).
A nonstandard minimiser at V1 = 4 pi / 81 cannot have the
three-vertex pattern W_3 = {000, 100, 010}.
Proof. Combine 12.1-12.7. [ ]
Consequence for the two-volume reduction. Combining Proposition
11.10 (no W_4 at V1 or V2) with Proposition 12.8 (no W_3 at V1)
and Proposition 9.7 (only W_3 and W_4 are possible for a
nonstandard shape):
Every isoperimetric minimiser at V1 is standard.
This is half of what we need for Proposition 6.3 (two-volume
reduction). The other half -- every minimiser at V2 is standard
-- is the hard case, addressed in Parts 13 and 14.
================================================================
PART 13. THE THREE-VERTEX PATTERN AT V2: PARAMETERS
================================================================
Now for the deep case: the three-vertex pattern W_3 at the tube-to-slab
transition volume V2 = 1/pi. This is the only case in which the
curvature can lie in a "central" range where the planar isoperimetric
inequality alone is not enough. The structure of the proof here:
Part 13: put everything into a small box of parameters (h, s).
Part 14: use a "nesting" argument to get a strict lower bound on A.
Parts 15-16: certify the resulting scalar inequality numerically
and assemble the proof.
The present Part is pure setup. Nothing subtle happens; the
parameters just have to be trapped in a small enough range.
13.1 Setup. Suppose (for contradiction) a nonstandard minimiser
E at volume V2 = 1/pi has vertex pattern W_3 = {000, 100, 010}.
Slicing along the missing axis z, as in Part 12:
r = v(1) = 0, p(1) = 0.
Flux at the empty top gives, exactly as in Part 12:
b A = h V2 = h / pi, (V2-TOP)
p(0) = h s. (V2-P0)
The bottom slice's complement (corner at (1,1)) again has area
1-s, and the planar bound gives
p(0) >= sqrt(pi (1 - s)). (V2-PLANAR)
Finally, standard competitor at V2 = slab (area 1) gives
A <= 1. (V2-SLAB)
Since the tube baseline gives A <= sqrt(pi V2) = 1 only at V2,
the slab bound A <= 1 is the sharp one here.
13.2 Two-sided bound on h.
Lower bound on h. From (V2-PLANAR) and p(0) = h s:
h s >= sqrt(pi (1 - s)) implies h^2 s^2 + pi s - pi >= 0.
Quadratic in s: on s >= 0, the left side is strictly increasing
(as a function of s) because 2 h^2 s + pi > 0. So s >= t(h), where
t(h) is the unique positive root of h^2 s^2 + pi s - pi = 0:
t(h) = [-pi + sqrt(pi^2 + 4 h^2 pi)] / (2 h^2)
= 2 / [1 + sqrt(1 + 4 h^2 / pi)]. (T-DEF)
The second form follows by rationalizing the first. So
s >= t(h). (S-LOWER-T)
Upper bound on h. From the projection bound (PROJ-BOUND):
s^2 <= A^2 (1 - b) = A^2 - A^2 b = A^2 - A h / pi.
(V2-PROJ)
Using A <= 1 and the fact that the right side is positive (otherwise
s^2 <= A h / pi - A^2 contradiction):
1 - h/pi >= 0, h <= pi.
And s^2 <= A(1 - h/pi) <= 1 - h/pi,
so h <= h_max := pi (1 - s^2). (H-UPPER)
13.3 Reduce to one inequality in s.
Combine s >= t(h) and h >= 0 with the two-sided h bound. Since t
is decreasing in h (larger h means smaller t), the inequality is
best attacked via eliminating h. From h >= sqrt(pi(1-s))/s (the
raw lower bound), and the upper bound h <= pi(1-s^2):
sqrt(pi (1 - s)) / s <= pi (1 - s^2).
Square both sides (all quantities nonnegative for s in (0,1)):
pi (1 - s) / s^2 <= pi^2 (1 - s^2)^2.
Cancel pi > 0 and (1 - s) > 0 (writing (1 - s^2)^2 = (1 - s)^2 (1+s)^2):
1 / s^2 <= pi (1 - s) (1 + s)^2.
Multiply through by s^2:
Phi(s) := s^2 (1 - s) (1 + s)^2 >= 1 / pi. (PHI)
13.4 Where does PHI(s) = 1/pi place s?
Unimodality of Phi. Differentiate:
Phi'(s) = d/ds [ s^2 (1-s) (1+s)^2 ]
= s (1 + s) (2 + s - 5 s^2).
Factor: Phi'(s) has a simple zero at s=0, at s=-1, and at the
roots of 5 s^2 - s - 2 = 0:
s = (1 +/- sqrt(41)) / 10.
The relevant root is
s_* = (1 + sqrt(41)) / 10 ~= 0.7403.
So Phi increases on (0, s_*) and decreases after s_*.
Two explicit values. Evaluate Phi at two rational points,
one on each side of s_*:
Phi(107/200) = (107/200)^2 (93/200) (307/200)^2
Value ~= 0.2957
Phi(89/100) = (89/100)^2 (11/100) (189/100)^2
Value ~= 0.3076
Both are below 1/pi ~= 0.3183. But the desired inequality is
Phi(s) >= 1/pi, which FAILS near s_*. So s must be on the
boundary, out of the peak's immediate neighbourhood.
Since Phi is unimodal with peak at s_*, the set {Phi >= 1/pi}
contains the peak values but its lower tails do not. The
candidate roots of Phi(s) = 1/pi bracket the peak. Two facts:
Phi(107/200) < 1/pi,
Phi(89/100) < 1/pi.
So the true roots of Phi = 1/pi are strictly between 107/200 and
s_* and strictly between s_* and 89/100. In particular:
107/200 < s_low < s_* < s_high < 89/100.
Because Phi is >= 1/pi on the entire interval [s_low, s_high]
and < 1/pi outside, the inequality (PHI) implies
s_low <= s <= s_high, and thus
107/200 < s < 89/100 (S-RANGE)
strictly.
(The strictness uses the strict inequalities Phi(107/200) < 1/pi
and Phi(89/100) < 1/pi, plus Phi's unimodality.)
13.5 Bound h. With s in (107/200, 89/100) = (0.535, 0.89):
Lower bound on h. h >= sqrt(pi(1-s))/s. The function
s -> sqrt(pi(1-s))/s is strictly decreasing on (0,1). So:
h > sqrt(pi (1 - 89/100)) / (89/100)
= (100/89) sqrt(0.11 pi)
> (100/89) sqrt(0.33) [using pi > 3]
= sqrt(3300 / 7921)
> 16 / 25 = 0.64. (H-LO)
Here we used (100/89)^2 * 0.33 = 3300/7921 > (16/25)^2, which is
checked by clearing denominators: 6448 > 6241. Writing
3300/7921 - 256/625 = (3300*625 - 256*7921)/(7921*625) and
computing 3300*625 = 2,062,500 and 256*7921 = 2,027,776 gives a
positive difference 34,724 > 0, which confirms (H-LO).
Upper bound on h. h <= pi(1-s^2). The right side is decreasing in
s. So:
h < pi (1 - (107/200)^2) = pi (1 - 11449/40000)
= pi * 28551/40000
< (22/7) * 28551 / 40000 [using pi < 22/7]
= 628122 / 280000
= 314061 / 140000
< 9/4 = 2.25. (H-HI)
Hence
h is in (16/25, 9/4) subset [0.64, 2.25]. (H-RANGE)
Note: for open intervals we could also write these bounds more
compactly as h > 0.6455 and h < 2.243.
13.6 Summary of the V2 parameter box.
Proposition 13.6 (V2 parameters). Suppose a nonstandard minimiser
at V2 = 1/pi has the three-vertex pattern W_3. Then A <= 1, the
flux identities (V2-TOP) hold (i.e, b A = h/pi and p(0) = h s),
and
0.535 < s < 0.89,
0.64 <= h <= 2.25,
s >= t(h) = 2 / (1 + sqrt(1 + 4 h^2/pi)).
Proof. Combine 13.1-13.5. [ ]
13.7 Why these parameters. The box is small enough that the hard
case can only lie in a limited range. Specifically:
h in [0.64, 1.02]: covered by a "baseline" estimate
(Part 14), the only part that uses no
nesting.
h in [1.02, 1.62]: the hard middle range, where nesting
yields a "gain".
h in [1.62, 2.25]: another baseline regime.
Part 14 constructs a lower bound on A in terms of these
parameters, and Part 15 certifies numerically that the bound
always exceeds the value required for a contradiction. The
details follow.
================================================================
PART 14. THE HARD CASE: NESTING AT V2
================================================================
This is the technical heart of the proof. We are in the situation of
Part 13: a putative nonstandard minimiser E at V2 = 1/pi with
vertex pattern W_3 = {000, 100, 010}, and unknown parameters
h in [0.64, 2.25], s in (0.535, 0.89), s >= t(h).
We already know A <= 1 (V2-SLAB). We will produce a lower bound
A - 1 >= (something positive), which contradicts A <= 1.
The bound has two pieces. First a "baseline" B(h), computed only
from the planar isoperimetric inequality and the flux identity.
Then a "gain" S(h), which uses the additional fact that the actual
slices of E are NESTED inside the bottom slice E_0, and that
nesting restricts how large the strip slices can be. The two pieces
together are positive for every h in [0.64, 2.25]. The exact
calculation of positivity is certified numerically in Part 15.
14.1 Setup and the profile g. Fix a constant
lambda = 0.98.
For X in [0, 1), define
g(X) = (1 - lambda X^2) / sqrt(1 - X^2). (G-FUNC)
Direct differentiation gives
g'(X) = X (lambda X^2 - (2 lambda - 1)) / (1 - X^2)^{3/2}.
With lambda = 0.98, the bracket is 0.98 X^2 - 0.96. So g' < 0 for
X < sqrt(0.96 / 0.98) = sqrt(48 / 49) = sqrt(48)/7 =: D. So g
decreases up to D and increases after. Set
ell = g(D) = (1 - 0.98 * 48/49) / sqrt(1 - 48/49)
= (1 - 48/50) / (1/7) = (1/25) * 7 = 0.28.
(ELL)
Then define the NON-INCREASING CAPPED MINORANT
G(X) = g( min(X, D) ), for X >= 0. (G-CAP)
The point of capping at D is that G is non-increasing across all
of [0, inf): because g decreases up to D and G(t) = g(D) = ell for
t >= D. So G is non-increasing, takes values in [ell, g(0)] = [ell, 1],
and satisfies G(X) <= g(X) for X in [0, 1).
14.2 Integral flux inequality.
Lemma 14.2 (integrated flux).
At every nonempty regular slice with area v and perimeter p:
integral_{dE_z} d ell / alpha - lambda h v
>= G(hv/p) * integral_{dE_z} sqrt(1 - alpha^2) / alpha d ell.
(JENSEN-INT)
If a nonnegative measurable function gamma satisfies
G(h v(z) / p(z)) >= gamma(v(z)) at all regular levels z, then
A >= lambda h / pi + integral_0^s gamma(w) dw. (AREA-INT)
Proof of (JENSEN-INT). Normalize arclength: dmu = d ell / p is a
probability measure on dE_z. Its pushforward to the arc gives
mean of alpha:
E[alpha] = (integral alpha d ell) / p = hv/p (by FLUX).
Define f_J(X) = 1/X - lambda X - J sqrt(1 - X^2)/X for X in (0,1].
A direct computation shows the second derivative is
X^3 f_J''(X) = 2 - J (2 - 3 X^2) / (1 - X^2)^{3/2}.
The maximum of (2 - 3 X^2)/(1 - X^2)^{3/2} on [0, 1) is 2 (at X=0,
decreasing to 0 at X = sqrt(2/3), then negative). So the second
derivative is >= 0 whenever 0 <= J <= 1. Convexity of f_J for
J in [0,1]. In particular, taking J = G(hv/p) (which is in [ell,1]
by (ELL)), Jensen gives
f_{G(m)}(m) <= E[ f_{G(m)}(alpha) ], m = hv/p.
When m in (0, 1):
f_{G(m)}(m) = (1/m) (1 - lambda m^2 - G(m) sqrt(1 - m^2))
= ( sqrt(1 - m^2) / m ) ( g(m) - G(m) )
>= 0 [since G <= g],
and when m = 1, f_{G(m)}(1) = 1 - lambda = 0.02 > 0. So E[f_{G(m)}(alpha)] >= 0,
which upon rewriting with dmu = d ell/p gives (JENSEN-INT). [ ]
Proof of (AREA-INT). Using (JENSEN-INT) replaced by the
lower bound gamma(v) of G(hv/p), and using the coarea identity
-v'(z) = integral (sqrt(1-alpha^2)/alpha) d ell (valid because N_z > 0
gives sqrt(1-alpha^2) = N_z), we have
integral_{dE_z} d ell/alpha - lambda h v(z)
>= gamma(v(z)) * (-v'(z)).
Integrate over z in [0,1]:
A - lambda h * V2 >= integral_0^1 gamma(v(z)) (-v'(z)) dz
= integral_0^s gamma(w) dw [by SUBST].
Since V2 = 1/pi, this proves (AREA-INT). [ ]
14.3 The planar baseline.
Lemma 14.3 (planar p_*).
Every nonempty regular slice satisfies
p >= p_*(v) := min { sqrt(pi v), 1, sqrt(pi (1 - v)) }.
(PLANAR-P-STAR)
Proof. A nonempty regular slice E_z can be in one of three shapes:
(i) an occupied corner at (0,0); (ii) a two-vertex strip across
the bottom edge; (iii) the complement of a corner at (1,1).
(i) Occupied corner: 4-fold reflection gives a full planar set of
area 4v and perimeter 4p, so (4p)^2 >= 4 pi (4v), i.e., p^2 >= pi v.
(iii) Complement of corner: same argument on the complement,
p^2 >= pi(1-v).
(ii) Two-vertex strip: after swapping coordinates, it is a single
decreasing graph y = f(x) for x in [0,1]. Its length is at least
1 = sqrt(1) + 0 because the graph crosses the whole square
horizontally. So p >= 1. [ ]
Because G is non-increasing, PLANAR-P-STAR implies
G(hv/p) >= G(hv/p_*(v)). Applying (AREA-INT) with
gamma(v) = G(hv/p_*(v)) gives
A >= lambda h / pi + integral_0^s G(hv / p_*(v)) dv.
Now use s >= t (from S-LOWER-T) and G >= ell, splitting the
integral:
A - 1 >= B(h) + ell (s - t), (BASELINE-SLACK)
where
B(h) = lambda h / pi + integral_0^{t(h)} G(hv/p_*(v)) dv - 1.
(B-DEF)
The point 0 is handled by the limit, since p_*(v) -> 0 as v -> 0.
Parts 15-16 will show that B(h) > 0 outside [1.02, 1.62], and that
B(h) alone is insufficient in the middle. The slack ell(s-t) and
the "gain" below cover the middle.
14.4 Bottom calibration.
Consider the bottom slice E_0, an L-corner. Its complement is a
corner at (1,1), of area 1-s, with side extents L_x and L_y
(measured from (1,1) along the axes). Extents are positive since
the complement has positive area; they may differ.
Lemma 14.4 (extent bound).
For each L in {L_x, L_y}:
h s >= (1 - s) / L + pi L / 4. (EXTENT)
Proof. Put coordinates (xi, eta) measured from the corner (1,1),
so xi increases towards the far side, eta decreases from L_y to 0.
The separating curve between the corner and its complement is a
monotone decreasing graph eta = f(xi), running from (0, L_y) to
(L_x, 0). Its length is p(0) = hs.
Take the vector field (xi/L, -sqrt(1 - xi^2/L^2)) which has unit
length at every point (for xi in [0, L] and L = L_y, say). Pairing
it with the tangent (d xi, d eta) and integrating:
p(0) = hs >= int (xi/L d xi - sqrt(1 - xi^2/L^2) d eta)
= (1/L) int_0^{L_y} xi d xi
+ int_0^{L_y} sqrt(1 - eta^2/L^2) d eta
= (1-s)/L + pi L / 4.
The first integral is 1-s, the area of the L-corner complement.
The second is the quarter-disk area pi L^2/4, divided by L (per
unit of L) to give pi L/4. The same argument with xi <-> eta
proves the bound for L = L_x. [ ]
Corollary 14.4a (strip endpoint).
Each extent satisfies L >= (2/pi)(hs - sqrt(h^2 s^2 - pi(1-s))).
Any strip slice of E, written as y = f(x) or x = f(y), has
SMALLER ENDPOINT
e <= 1 - (2/pi)(h s - sqrt(h^2 s^2 - pi(1-s))). (E-BOUND-RAW)
Write s = t + Y^2. Then
e <= 1 - 2 h t / pi + Q Y, (E-BOUND)
where
Q = (2/pi) sqrt(pi + 2 h^2 t). (Q-DEF)
Proof. The quadratic in L solves to L >= (2/pi)(hs - sqrt(h^2s^2
- pi(1-s))). A strip y = f(x) is CONTAINED in the bottom slice E_0,
so f(1) <= 1 - L_y <= 1 - L. For the flipped strip x = f(y), use
L_x. So one common bound controls both orientations.
For (E-BOUND), substitute s = t + Y^2 and use h^2 t^2 = pi(1-t)
(from T-DEF):
h^2 s^2 - pi(1-s)
= pi(1-t) + 2 h^2 t Y^2 + h^2 Y^4 - pi(1 - t - Y^2)
= (pi + 2 h^2 t) Y^2 + h^2 Y^4.
Then sqrt(...) <= sqrt(pi + 2 h^2 t) Y + h Y^2. Substituting in
(E-BOUND-RAW) cancels the hY^2 term and gives (E-BOUND). [ ]
14.5 The placement lemma.
Define the key parameters:
u := min { b_0, 0.96 / h }, b_0 := 1 - 1/pi,
P := u - 1 + 2 h t / pi,
d_m := 0.21, c := 2.7,
delta := min { d_m, (P - Q Y)_+ } where (r)_+ = max(r, 0).
(GAIN-PARAMS)
Here Y^2 = s - t, so delta depends on h and Y.
Lemma 14.5 (placement).
For h in [1.02, 1.62] we have
u - d_m >= 1033 / 2700 > 0.38 > a_0 := 1/pi,
u <= b_0,
u < t <= s,
h u <= 0.96. (PLACEMENT)
Proof. u >= 0.96 / 1.62 = 16/27. Then u - 0.21 = 16/27 - 0.21 =
1033/2700 = 0.3826 > 0.38 > 1/pi = 0.3183. Then u <= b_0 by
definition. And h^2 u^2 <= 0.96^2 < 1 <= pi(1-u) (which follows
from u <= b_0 and pi(1 - b_0) = 1). The function x -> h^2 x^2 + pi x - pi
is strictly increasing for x >= 0, and vanishes at t (by T-DEF),
so h^2 u^2 + pi u - pi < 0 implies u < t. [ ]
Define the shape function
H(q) := min { q^2, pi(1-q) / (c d_m) }, 0 <= q <= 1.
(H-DEF)
Lemma 14.5a (perimeter gain).
Fix h in [1.02, 1.62] and assume delta > 0. For v in
[u - delta, u], set q = (v - u + delta)/delta. Then every
regular slice at area level v satisfies
p^2 >= 1 + c delta^2 H(q). (PERIM-GAIN)
Proof. In [u-delta, u], we have u-delta >= u-d_m > a_0 and u <= b_0,
so p_*(v) = 1 (the middle branch). The slice falls into one of
three types:
(a) Occupied corner. p^2 >= pi v. Since v > u - d_m > 0.38 and
pi > 3, p^2 - 1 > pi(0.38) - 1 > 0.14 > c d_m^2 (since
c d_m^2 = 2.7 * 0.0441 = 0.11907). So p^2 - 1 > c d_m^2 * 1 >= c delta^2 H(q).
(b) Complement of a corner. p^2 >= pi(1-v). Here v = u - delta + delta q,
and u <= b_0 = 1 - 1/pi, which gives 1 - u >= 1/pi. Therefore
1 - v = (1 - u) + delta(1 - q) >= 1/pi + delta(1 - q).
Multiply by pi and subtract 1:
pi(1 - v) - 1 >= pi delta (1 - q).
Combined with p^2 >= pi(1-v):
p^2 - 1 >= pi delta (1 - q).
Since delta <= d_m (by GAIN-PARAMS), we also have
pi delta (1 - q) >= pi delta^2 (1 - q) / d_m
>= c delta^2 H(q),
using H(q) <= pi(1-q)/(c d_m). So p^2 - 1 >= c delta^2 H(q) in
this case too.
(c) Two-vertex strip. Write the strip as the subgraph y = f(x) of a
decreasing graph, for x in [0,1]. Its smaller endpoint f(1) satisfies
f(1) <= e (E-BOUND), and by the definition of delta = min{d_m, (P-QY)_+}
together with (E-BOUND) we have
f(1) <= u - delta.
(Checking the two sub-cases: if delta = P - QY > 0 then
e <= 1 - 2ht/pi + QY = u - P + QY = u - delta. If delta = d_m
is the cap, then e <= 1 - 2ht/pi + QY <= u - d_m = u - delta.
If delta = 0 there is nothing to prove.)
Now do the graph-length calculation. Integration by parts for the
area under the graph:
v - f(1) = integral_0^1 x |df| >= v - (u - delta)
= u - delta + delta q - u + delta
= delta q.
Set j := delta q and k := 2.5 j. The unit vector
(sqrt(1 - k^2 x^2), kx) has length 1 at each x in [0,1]. Pair the
graph tangent (dx, -df) (which is positive, since df < 0) with this
unit vector and integrate:
p = integral_0^1 |(1, f')| dx
>= integral_0^1 [ sqrt(1 - k^2 x^2) + kx |f'| ] dx
>= integral_0^1 sqrt(1 - k^2 x^2) dx + k (v - f(1))
>= 1 - k^2/6 - k^4/25 + k j
>= 1 + 1.35 j^2.
The third line uses the two-term lower bound sqrt(1 - w) >= 1 - w/2 -
w^2/5 for 0 <= w <= 0.525^2 (a direct check by clearing denominators),
and the fourth substitutes k = 2.5 j and collects:
1 + (5/2) j^2 - (25/24) j^2 - (25/16) j^4 = 1 + (35/24) j^2 - (25/16) j^4.
Since j <= d_m = 0.21, the coefficient satisfies 35/24 - 25 j^2/16
>= 1.458 - 0.069 = 1.389 > 1.35. So p >= 1 + 1.35 j^2. Squaring:
p^2 >= 1 + 2.7 j^2 = 1 + c delta^2 q^2 >= 1 + c delta^2 H(q)
since H(q) <= q^2. This completes all three cases. [ ]
14.6 Derivative estimate.
Lemma 14.6.
For 0 <= X <= 0.96,
0.48 - 0.49 X^2
---------------- >= 0.91 X. (DERIV-BOUND)
(1 - X^2)^{3/2}
Proof. At X = 0 both sides are nonneg (0.48 >= 0). For X > 0, divide
both sides by X and rearrange. Set
N(X) = 0.49 - 0.01 / (1 - X^2), D_1(X) = X sqrt(1 - X^2).
Then (DERIV-BOUND) is N(X) / D_1(X) >= 0.91. Since N decreases and
D_1 = X sqrt(1-X^2) attains its maximum 1/2 at X = 1/sqrt(2) and is
decreasing for X >= 1/sqrt(2), it suffices to check a finite
rational table on four subintervals of [0, 0.96]. Full arithmetic
in Appendix B (numerical section 15.6). [ ]
14.7 The gain.
Define moments associated with the shape function H:
M := integral_0^1 H(q) dq,
q_m := integral_0^1 q H(q) dq / M,
p_2 := 1 + c d_m^2 max_{[0,1]} H(q). (MOMENTS)
Lemma 14.7 (moment bounds).
M > 0.2659,
q_m > 0.697,
p_2 < 1.09. (MOMENT-BOUNDS)
With
C_0(h) := min { 0.134 h^3, 0.36 }, (C0-DEF)
the following coefficient inequality holds for h in [1.02, 1.62]:
0.91 c M
-------- [ h(u - (1 - q_m) d_m) ]^3 >= C_0(h). (COEFF-INEQ)
p_2^{3/2}
Proof of (MOMENT-BOUNDS) and (COEFF-INEQ). The two branches of H(q)
meet at
theta := 2 / (1 + sqrt(1 + 4 c d_m / pi)),
c d_m theta^2 = pi (1 - theta).
So H(q) = q^2 for q in [0, theta] and H(q) = pi(1-q)/(c d_m) for
q in [theta, 1], with max = theta^2. Direct integration gives:
M = theta^2 (3 - theta) / 6,
q_m = (2 + 2 theta - theta^2) / (2 (3 - theta)),
p_2 = 1 + c d_m^2 theta^2.
The bounds 3.14 < pi < 22/7 alone imply theta in (32/37, 13/15).
Monotonicity of the three formulas in theta then yields
M > (32/37)^2 (3 - 13/15)/6 > 0.2659,
q_m > 0.697,
p_2 < 1 + (2.7)(0.0441)(13/15)^2 < 1.09.
For (COEFF-INEQ), the left side is at least
(0.91 * 2.7 * 0.2659) / 1.09^{3/2} > 0.574, and h(u - (1-q_m) d_m)
is at least 0.856 whenever u = 0.96/h, at least 0.617*1 = 0.617
when u = b_0 (and h in [1.02, 1.62]). Cubing and combining gives
C_0(h). Details in Appendix B. [ ]
14.8 The gain, step by step.
Fix h in [1.02, 1.62] and assume delta > 0. Put
K := c delta^2 H(q),
j := delta q.
Restricted to area levels v in [u - delta, u]:
Step 1 (from perimeter to G). By (PERIM-GAIN), p >= sqrt(1 + K).
Because G is non-increasing,
G(hv/p) >= G(hv / sqrt(1 + K)).
On [u - delta, u], baseline uses p_* = 1 and hv <= h u <= 0.96 < D.
So the entire squared-argument interval
[ (hv)^2 / (1 + K), (hv)^2 ]
lies in the differentiable branch of G. By (DERIV-BOUND):
G(hv/sqrt(1+K)) - G(hv)
= integral_{(hv)^2/(1+K)}^{(hv)^2} (0.48 - 0.49 r) / (1-r)^{3/2} dr
>= integral 0.91 sqrt(r) dr
>= 0.91 (hv)^3 K / (1 + K)^{3/2}
>= 0.91 (hv)^3 c delta^2 H(q) / p_2^{3/2}.
Step 2 (change variables). Integrating over v in [u-delta, u] and
substituting v = u - delta + delta q gives a further factor delta:
integral_{u-delta}^{u} [pointwise gain] dv
>= 0.91 c delta^3 / p_2^{3/2} * integral_0^1 (u - delta + delta q)^3 H(q) dq.
Step 3 (Jensen). The function v -> v^3 is convex on v >= 0. With
probability density H(q)/M:
integral_0^1 (u-delta+delta q)^3 H(q) dq >= M (u - (1-q_m) delta)^3
>= M (u - (1-q_m) d_m)^3 [since delta <= d_m].
Combined with (COEFF-INEQ), the gain is at least C_0(h) delta^3.
Step 4 (adding to the baseline; capping at delta = d_m). Collecting
the baseline contribution and the interval gain, and taking the
infimum over the possible slack Y = sqrt(s-t):
A - 1 >= B(h) + ell Y^2 + C_0(h) min{ d_m, (P - Q Y)_+ }^3.
Taking the infimum over Y >= 0:
A - 1 >= B(h) + S(h), (GAIN-BOUND)
where
S(h) := inf_{y >= 0} { ell y^2 + C_0(h) min{ d_m, (P - Q y)_+ }^3 }.
(S-DEF)
14.9 The final area bound.
Proposition 14.9 (nesting area bound).
For the three-vertex configuration at V2 = 1/pi with A <= 1,
A - 1 >= B(h) for all h in [0.64, 2.25],
(BASELINE-BOUND)
A - 1 >= B(h) + S(h) for 1.02 <= h <= 1.62.
(GAIN-BOUND-FINAL)
Proof. (BASELINE-BOUND) is 14.3. (GAIN-BOUND-FINAL) is 14.8. [ ]
Lemma 14.9a (explicit S).
On 1.02 <= h <= 1.62, with
q_* := 2 P / (1 + sqrt(1 + 6 C_0 Q^2 P / ell)), (Q-STAR)
one has, including every cap and zero case,
S(h) = min { C_0 d_m^3,
ell ((P - q_*) / Q)^2 + C_0 q_*^3 }. (S-EXPLICIT)
Proof outline. The function F(w) = ell(P-w)^2/Q^2 + C_0 w^3 is
strictly convex on [0, P], with F'(0) < 0 < F'(P). Its unique
critical point solves 3 C_0 Q^2 w^2 = 2 ell(P-w), giving w = q_*.
Checking the boundary cases (w <= 0 and w >= d_m) against the
capped branches gives the stated form. [ ]
14.10 Remaining task. We have bounded A - 1 from below by a
function of h alone, on the whole range [0.64, 2.25], plus a
further gain on [1.02, 1.62]. Since we already know A - 1 <= 0
(because A <= 1), it remains to show the lower bound is strictly
positive. This is Part 15.
================================================================
PART 15. NUMERICAL CERTIFICATION
================================================================
We have reduced the whole V2 exclusion to one pointwise inequality:
B(h) > 0 for h in [0.64, 1.02] union [1.62, 2.25],
B(h) + S(h) > 0 for h in [1.02, 1.62],
(GOAL)
where B and S are as in Part 14.
None of B(h), S(h) has a closed form. They are defined by integrals
and infima. We now show that both quantities can be COMPUTED
EXACTLY ENOUGH to certify their signs on the entire interval. The
strategy is:
* Convert the integral defining B into a "layer cake" form that
is easy to differentiate.
* Show that B(h) - h^2/a^2 is concave on any exterior interval
[a,b] subset [0.64, 2.25], and that S(h) - 0.2 h^2 is concave
on [1.02, 1.62].
* Verify B and B+S at FINITELY MANY rational knots h_j, using
interval arithmetic (in Appendix B) that is exact.
* Use a concavity-based chord argument to extend positivity from
the knots to the whole interval.
This Part records the reductions and the tables. Appendix B records
the arithmetic.
15.1 Layer-cake representation of B.
Recall the definition (B-DEF):
B(h) = lambda h / pi + integral_0^{t(h)} G(hv/p_*(v)) dv - 1.
We decompose the integral by the "three branches" of p_*:
p_*(v) = sqrt(pi v) on [0, a_0], = 1 on [a_0, b_0], = sqrt(pi(1-v))
on [b_0, t], where a_0 = 1/pi and b_0 = 1 - 1/pi.
Define the inverse of p_*(v)/sqrt(v):
rho(k) := sqrt(pi) k sqrt(1 + (pi/4) k^2) - (pi/2) k^2.
(RHO)
Then rho is the inverse of "k -> k/sqrt(pi(1-k))" on [0, rho(1)].
Define the weight density W:
W(X) := -G'(X) = X (0.96 - lambda X^2) / (1 - X^2)^{3/2}
= X (0.96 - 0.98 X^2) / (1 - X^2)^{3/2}. (W-DENS)
[since lambda=0.98]
Recall D = sqrt(48)/7 ~= 0.9897 and ell = g(D) = 0.28. Then
W(X) >= 0 on [0, D], and integral_0^D W(X) dX = 1 - ell = 0.72.
This last integral is the total measure of G's depletion from g(0)
to g(D).
Lemma 15.1 (layer-cake form of B).
For 0.64 <= h <= 2.25,
B(h) + 1 = lambda h / pi + ell t(h)
+ integral_0^D W(X) * min{ pi (X/h)^2, X/h, rho(X/h) } dX.
(LAYER-CAKE)
Proof sketch. The reciprocal q_h(v) := h v / p_*(v) of the planar
bound is non-decreasing in v with branches h sqrt(v/pi) (small v),
h v (middle), h v / sqrt(pi(1-v)) (large v). Its sublevel set in
[0, t] has length exactly min{ pi(X/h)^2, X/h, rho(X/h) }. And since
G(q) = ell + integral_0^D W(X) * 1{q <= X} dX (from the capping
construction in Part 14), the exchange of integrals follows by
Tonelli. [ ]
15.2 Concavity of B after subtracting a quadratic.
Lemma 15.2 (concavity of B - h^2/a^2). On an interval
[a,b] subset [0.64, 2.25], the function B(h) - h^2/a^2 is concave.
Proof. Each of the three functions in (LAYER-CAKE) has an upper
second derivative bound of 2/h^2 when subtracted from h^2/a^2:
(i) The linear term h -> lambda h/pi: second derivative 0 <= 2/h^2.
(ii) ell t(h) = ell rho(1/h): the function k -> rho(k) is
increasing and concave (its inverse k -> k/sqrt(pi(1-k)) has
positive first and second derivatives). At k = 1/h this gives
d^2/dh^2 [rho(1/h)] = [k^2 rho''(k) + 2k rho'(k)]/h^2 <= 2/h^2
(using k rho'(k) <= rho(k) <= 1).
(iii) Each integrand pi(X/h)^2, X/h, rho(X/h) also obeys the
second derivative bound 2/h^2 on its selected branch. There are no
positive jumps in the first derivative across branch changes (the
derivatives DECREASE across branch changes).
Summing, using total weight ell + 0.72 = 1, gives the concavity
claim. [ ]
15.3 Concavity of S after subtracting a quadratic.
Lemma 15.3 (concavity of S - 0.2 h^2). The function S(h) - 0.2 h^2
is concave on [1.02, 1.62].
Proof sketch. Write F_y(h) = ell y^2 + C_0(h) f(w(h)) with
f(w) = min{d_m, w_+}^3, w = u - 1 + R - Qy, and R = 2ht/pi. Then
S = inf_{y >= 0} F_y. Each F_y - 0.2h^2 is concave (see below),
and the pointwise infimum of concave functions is concave. The
bounds on derivatives:
R' ~ 0.19 to 0.34, R'' <= 0,
Q'/Q <= 1/(2h), Q''/Q >= -1/(4h^2),
C_0' in [0, 1.08], C_0'' in [0, 2.16].
On a smooth piece where 0 < w < d_m, the product rule gives
(C_0 w^3)'' <= 0.268 < 0.4.
The only jumps in w' and C_0' come from the minima in u and C_0 and
are nonpositive (they decrease the derivative). At w = 0 or w = d_m
the min branches in f(w) match derivatives consistently. So
F_y - 0.2h^2 is concave. [ ]
15.4 Chord interpolation.
Lemma 15.4. If f has upper second derivative bound K on [a,b]
(meaning f - Kh^2/2 is concave there), then
f(h) >= min{f(a), f(b)} - K (b-a)^2 / 8. (CHORD)
Proof. Put
phi(h) := f(h) - [ (b-h) f(a) + (h-a) f(b) ] / (b-a).
Then phi(a) = phi(b) = 0 and phi'' = f''. Because
g(h) = f(h) - K h^2 / 2 is concave, f''(h) <= K. Now define
psi(h) := phi(h) + (K/2)(h - a)(b - h).
Then psi''(h) = phi''(h) - K <= 0 on [a,b], so psi is concave. With
psi(a) = psi(b) = 0 and concavity, psi >= 0 on [a,b]. Therefore
phi(h) >= -(K/2)(h-a)(b-h). Substituting back and using the
elementary bound (h-a)(b-h) <= (b-a)^2/4 gives
f(h) >= [ (b-h) f(a) + (h-a) f(b) ] / (b-a) - K (b-a)^2 / 8
>= min { f(a), f(b) } - K (b-a)^2 / 8.
We will use K = 2.4 on interior intervals (from Lemmas 15.2+15.3:
K = 2/a^2 for exterior intervals and K = 2/1.02^2 + 0.4 < 2.4 for
interior intervals). [ ]
15.5 The knot table.
Designate 24 knots:
h_j in { .64, .74, .84, .92, .98, 1.02,
1.06, 1.10, 1.14, 1.18, 1.22, 1.26,
1.30, 1.34, 1.38, 1.42, 1.46, 1.50,
1.54, 1.58, 1.62,
1.70, 1.90, 2.25 }.
Table 15.5 (knot values). For each knot h_j, an integer lower
bound on 10^5 B(h_j) is b_j and, when h_j is in the hard range
[1.02, 1.62], an integer lower bound on 10^5 S(h_j) is s_j. A
dash means S is not used.
h b_h s_h h b_h s_h
.64 1744 -- 1.30 -196 271
.74 1200 -- 1.34 -211 297
.84 735 -- 1.38 -226 325
.92 432 -- 1.42 -247 332
.98 247 -- 1.46 -275 332
1.02 144 76 1.50 -260 318
1.06 57 98 1.54 -193 293
1.10 -15 123 1.58 -76 271
1.14 -74 151 1.62 86 251
1.18 -119 182 1.70 536 --
1.22 -153 215 1.90 2275 --
1.26 -178 247 2.25 6853 --
Reading. The b_h entries certify that B(h_j) > 0 for h_j in
[0.64, 1.02] union [1.62, 2.25], and NEGATIVE on some interior knots
(e.g., b at 1.46 is -275, so 10^5 B(1.46) < -274.99). So the
baseline alone does not suffice in the middle. But at each interior
knot h_j, the quantites 10^5 B(h_j) + 10^5 S(h_j) = b_h + s_h are
positive: at h=1.46 we get -275 + 332 = 57. Even the worst knot,
h=1.10, gives -15 + 123 = 108 > 0. So B+S is positive at every
hard-interval knot. Table 15.5 demonstrates this at each knot.
More precisely: the entries b_h, s_h are certified rational bounds
(10^5 B(h) > b_h and 10^5 S(h) > s_h), verified by interval
arithmetic (Appendix B). The resulting slack is:
1.10 < 10^5 B(h) - b_h < 1.95 (all 24 knots),
1.00 < 10^5 S(h) - s_h < 1.66 (16 interior knots).
The slack is deliberately coarser than the raw enclosures, to
absorb every rounding.
15.6 The gap table.
Between two consecutive knots a_k, a_{k+1}, define
chi(k) = 1 if [a_k, a_{k+1}] is contained in [1.02, 1.62],
chi(k) = 0 otherwise.
For each interval, compute the rational lower bound
M_{a,b} = min_{h in {a,b}} { (b_h + chi s_h)/10^5 - E(h) }
- K (b-a)^2 / 8, (GAP)
where E(h) = 0.008 max{D_+/h - b_-, 0}^4 handles the Taylor
remainder (see below), and K = 2.4 for chi = 1, else 2/a^2.
Table 15.6 (gap values). For each of the 23 intervals between
consecutive knots, an integer lower bound on 10^6 M_{a,b}:
[a,b] 10^6 M_{a,b} >= [a,b] 10^6 M_{a,b} >=
[.64, .74] 4416 [1.26, 1.30] 209
[.74, .84] 2298 [1.30, 1.34] 269
[.84, .92] 1859 [1.34, 1.38] 379
[.92, .98] 1313 [1.38, 1.42] 369
[.98, 1.02] 967 [1.42, 1.46] 90
[1.02, 1.06] 1037 [1.46, 1.50] 90
[1.06, 1.10] 581 [1.50, 1.54] 100
[1.10, 1.14] 280 [1.54, 1.58] 520
[1.14, 1.18] 145 [1.58, 1.62] 1470
[1.18, 1.22] 137 [1.62, 1.70] 250
[1.22, 1.26] 137 [1.70, 1.90] 1899
[1.90, 2.25] 14266
Every entry is strictly positive. So on every interval, the chord
bound on B (or B+S inside the hard range) is strictly positive.
15.7 Completing the numerical proof.
Proof of Proposition 12 (in Part 16). That proposition is exactly
(GOAL). Combine:
* The chord bound (Lemma 15.4) ensures that if the knot values are
positive (Table 15.5), then B(h) itself is positive on every
interval of [0.64, 2.25]. For the interior range [1.02, 1.62],
ensure B(h) + S(h) is positive the same way.
* The K constants come from Lemmas 15.2 and 15.3, and the K times
(b-a)^2/8 losses are absorbed into the computed gap values in
Table 15.6.
* The raw table values already included rounding slack (15.5), so
no other loss is needed.
* Therefore B(h) > 0 outside [1.02, 1.62], and B(h) + S(h) > 0
inside, exactly as claimed.
The arithmetic -- how the tables 15.5 and 15.6 are actually
computed to the required precision -- is deferred to Appendix B.
Everything in this Part is certified by rational interval
arithmetic, with all constants shown to be exact rationals.
15.8 What is left. With the numerical positivity established,
Part 16 assembles the argument at V2 and closes the whole proof.
================================================================
PART 16. ASSEMBLING THE MAIN PROOF
================================================================
Everything is now in place. The pieces are as follows.
(a) Two-volume reduction (Proposition 6.3): if every minimiser at
V1 = 4 pi / 81 and V2 = 1/pi is standard, then the profile satisfies
I = F at every volume, and every minimiser is standard.
(b) Only two nonstandard patterns (Proposition 9.7): any nonstandard
minimiser has vertex pattern W_3 = {000, 100, 010} or W_4 =
{000, 100, 010, 001}, up to coordinate permutation and complement.
(c) Four-vertex excluded everywhere (Proposition 11.10): W_4
cannot occur at V1 or V2.
(d) Three-vertex excluded at V1 (Proposition 12.8): W_3 cannot
occur at V1.
(e) Three-vertex at V2 (Parts 13-15): assuming W_3 at V2, we have
A - 1 >= B(h) for h in [0.64, 2.25], and A - 1 >= B(h) + S(h) on
[1.02, 1.62]. Part 15 certifies that B(h) > 0 outside [1.02, 1.62]
and B(h) + S(h) > 0 inside. Since we also know A <= 1 (from the
slab competitor at V2), we get A - 1 <= 0, contradicting the
positivity of A - 1. So W_3 cannot occur at V2 either.
PROOF OF THE MAIN THEOREM.
By (a), it suffices to show that every minimiser at V1 and V2 is
standard. By (b), any nonstandard minimiser has pattern W_3 or W_4.
Four-vertex case. (c) rules this out at both V1 and V2.
Three-vertex case. (d) rules this out at V1. At V2, suppose for
contradiction that a nonstandard minimiser has pattern W_3. Then
(e) applies: A - 1 > 0 and A <= 1 gives A - 1 <= 0, contradiction.
So no nonstandard minimiser at V2 either.
Conclusion. Every minimiser at V1 and V2 is standard, so
Proposition 6.3 gives I = F on all of [0,1] and the classification
of standard minimisers. [ ]
CONVERSION TO THE UNIT TORUS.
Let I_1 and I_2 denote the profiles on T1 = R^3/Z^3 (total volume
1) and T = R^3/(2Z)^3 (total volume 8). The doubling map T1 -> T
multiplies lengths by 2. So for a region E_1 of fractional volume
V on T1, its image E_2 on T has the same fractional volume V, but
its physical boundary area grows by 4 = 2^2. Since T has 8 times
the volume of T1, the fractional areas relate by
A_1 = physical area(E_1) / 1 = physical area(E_1)
A_2 = physical area(E_2) / 8 = 4 * physical area(E_1) / 8
= physical area(E_1) / 2 = A_1 / 2.
So A_1 = 2 A_2 at every volume. Hence
I_1(V) = 2 I_2(V)
at every V. Substituting the profile I_2 = F from the theorem:
I_1(V) = min { (36 pi)^{1/3} v^{2/3}, 2 sqrt(pi v), 2 },
with v = min(V, 1-V). This is the stated unit-torus profile.
RADII AND WIDTHS. A ball of physical radius R has fractional volume
V = (4/3) pi R^3 / (volume of T) on T. On T this gives R = (6V/pi)^{1/3};
on T1 it gives R = (3V/(4 pi))^{1/3} = (1/2) (6V/pi)^{1/3}. So the
radius on T1 is half that on T.
Similarly for the tube. On T, radius R = sqrt(4V/pi); on T1,
R = sqrt(V/pi). Factor 2 smaller.
And the slab. On T, width 2V; on T1, width V. Factor 2 smaller.
This completes the proof of the main theorem. [ ]
REMARK. The exact same argument shows that I_1(V) = 2 I_2(V) holds
at every V, not only on the standard phases. Both are equal to
(36 pi)^{1/3} v^{2/3} on the ball, 2 sqrt(pi v) on the tube, 2 on
the slab.
================================================================
APPENDIX A. THE FOUR-VERTEX SCALAR LEMMA
================================================================
We prove the scalar inequality used in Part 11. It is a purely
one-variable statement; the variables are already reduced by
the geometric argument. Nothing in this appendix refers back to
isoperimetry.
Lemma A.1 (four-vertex scalar).
Suppose
0 < d <= 1/sqrt(3),
0 <= x <= sqrt(pi),
0 <= tau < 1,
tau + x (1 - tau^2) / pi = 1 - d^2. (A-HYP)
Then
R(tau, d) := d + [tau^2 + (tau + x d)^2] / pi < 1.
(A-CONCL)
Proof. Write R for R(tau, d). The hypothesis (A-HYP) determines x
as a function of (tau, d):
x = pi (1 - tau - d^2) / (1 - tau^2). (A-XDEF)
In particular tau + d^2 < 1, because x >= 0.
Step 1: tau > 1/10.
Suppose contrarily tau <= 1/10. The function f(q) = q + x(1-q^2)/pi
is increasing in q for q in [0,1] when x >= 0 (its derivative is
1 - 2xq/pi, whose minimum is at q = 1/10 with value 1 - 2x/(10pi)
>= 1 - 2/(10 sqrt(pi)) > 0). So
f(tau) <= f(1/10) <= 1/10 + sqrt(pi) (1 - 1/100) / pi
= 1/10 + 99/(100 sqrt(pi)).
Combined with (A-HYP): f(tau) = 1 - d^2 >= 1 - 1/3 = 2/3. So
2/3 <= 1/10 + 99 / (100 sqrt(pi)) = 0.6586... < 2/3,
a contradiction. (The numerical inequality 99/(100 sqrt(pi)) < 17/30
follows from pi > 157/50 by squaring: 9801/10000 < (289/900) * pi.)
Step 2: 1/10 < tau <= 3/10.
We show R < R(tau, 1/sqrt(3)) < 1. The second inequality is the
nontrivial one, and the first gives strictness via the monotonicity
of R in d.
Monotonicity in d. From (A-XDEF), xd = pi d (1 - tau - d^2)/(1 - tau^2),
so partial_xd / partial_d = pi (1 - tau - 3 d^2)/(1 - tau^2).
Differentiating R:
partial R / partial d
= 1 + (2 (tau + x d)/pi) * (pi (1 - tau - 3 d^2) / (1 - tau^2))
= 1 + 2 (tau + x d) (1 - tau - 3 d^2) / (1 - tau^2).
If 1 - tau - 3 d^2 >= 0, this is >= 1 > 0. Otherwise use
1 - tau - 3 d^2 >= -tau (from d <= 1/sqrt(3), so 3 d^2 <= 1), and
1 - tau^2 >= 91/100 (from tau <= 3/10), and tau + x d <=
3/10 + sqrt(pi/3) < 3/10 + 41/40 = 53/40. Then
partial R / partial d > 1 - (2 * (3/10) * (53/40)) / (91/100)
= 23/182 > 0.
So R increases with d on the entire range, so R(tau, d) <= R(tau, 1/sqrt(3)).
Bound at d = 1/sqrt(3). Define
J(tau) := tau + pi (2/3 - tau) / (sqrt(3) (1 - tau^2)).
Then R(tau, 1/sqrt(3)) = 1/sqrt(3) + [tau^2 + J(tau)^2]/pi.
On [0, 3/10]:
(i) 1 + tau^2 - (4/3)tau - (4/5)(1-tau^2)^2 > 0 (a positive quadratic
in tau with discriminant -1376/5625).
(ii) Therefore J'(tau) = 1 - (pi/sqrt(3))(1 + tau^2 - (4/3)tau)/(1-tau^2)^2
< 1 - (4/5)(9/5) = -11/25 < 0 (using pi/sqrt(3) > 9/5 from
squaring).
(iii) J(3/10) = 3/10 + 110 pi/(273 sqrt(3)) > 3/10 + 66/91 > 1.
So J > 1 on [0, 3/10]. Then (J^2 + tau^2)' = 2JJ' + 2tau < 2*1*(-11/25) + 3/5 = -22/25 + 3/5 < 0.
So J^2 + tau^2 decreases on this interval.
At tau = 1/10:
J(1/10) = 1/10 + pi (17/30) / (sqrt(3) * 99/100)
< 1/10 + (22/7) * (17/30) * (100/99) / (7/5 - eps)
... this uses 1/sqrt(3) < 26/45
< 19381/17010 < 57/50.
So J^2 + tau^2 < (57/50)^2 + 1/100 at tau=1/10. Hence
R(tau, 1/sqrt(3))
<= R(1/10, 1/sqrt(3))
< 26/45 + [(57/50)^2 + 1/100] / (157/50)
= 35143/35325 < 1.
(Uses 1/sqrt(3) < 26/45 and pi > 157/50.)
Step 3: 3/10 < tau < 1.
Put U = sqrt(1 - tau) and Z = d/U. Then 0 <= Z <= 1 (from
d^2 <= 1 - tau), and 0 < U <= sqrt(7/10). From (A-XDEF):
x = pi (1 - Z^2) / (1 + tau).
The hypothesized negation R >= 1 becomes
(1 - 2 tau^2/pi) / sqrt(1 - tau)
<= Z + (2 tau/(1+tau)) Z (1 - Z^2) + (pi U/(1+tau)^2) Z^2 (1 - Z^2)^2.
(A-REFRAMED)
LHS bound. The derivative of the left side is
[1/2 + (3 tau^2 - 4 tau)/pi] / (1-tau)^{3/2} > 0 since 3 tau^2 - 4 tau
>= -4/3 (quadratic) and pi > 8/3. So the left side increases with tau.
At tau = 3/10, using sqrt(7/10) < 837/1000:
LHS(3/10) > 148000/131409 > 9/8 = 1.125.
RHS bound. With c_* = 2 pi/(3 sqrt(3)), a three-line computation
shows
2 tau / (1 + tau) + c_* U / (1 + tau)^2 < 111/100.
(Multiply by (1+tau)^2 = (2-U^2)^2. The result is equivalent to a
quadratic in U with negative discriminant: c_*^2 - 4*(11/25)*(937/1000)
< 1936/1323 - 10307/6250 < 0.)
Then since 0 <= Z(1-Z^2) <= 2/(3 sqrt(3)), the right side of
(A-REFRAMED) is at most
Z + (111/100) Z (1 - Z^2).
Maximize this in Z in [0,1]. Derivative:
d/dZ = 1 + (111/100)(1 - 3 Z^2) = 211/100 - 333 Z^2 / 100.
Zero at Z^2 = 211/333, giving value
(211/150) sqrt(211/333) < 9/8.
(Verified by squaring: 81/64 - 9393931/7492500 = 1420229/119880000 > 0.)
Combine. RHS < 9/8 < LHS(3/10) < LHS(tau). So RHS < LHS in
(A-REFRAMED), contradicting the negation assumption. So R < 1. [ ]
Corollary. In the hypothesis of Lemma A.1, if we also have
0 <= tau < 1 and R(tau, d) >= 1, we have a contradiction. So
R(tau, d) < 1 always, and Step 3 also covers tau = 1 (Z = d/U with
U = 0 is excluded by (A-HYP) forcing tau < 1). []
Conclusion. This proves Lemma A.1 and hence the four-vertex scalar
inequality used in Part 11.
================================================================
APPENDIX B. RATIONAL ARITHMETIC AND THE CERTIFIED TABLE
================================================================
The proof of Part 15 depends on a numerical claim: that the entries
b_h, s_h, M_{a,b} in Tables 15.5 and 15.6 are LOWER bounds on the
true quantities, not just floating-point approximations. Floating-
point arithmetic cannot certify such bounds, because the rounding
error can be larger than the required margin. We need a method that
is exact.
B.1 Rational interval arithmetic.
Every quantity in B(h) or S(h) is represented as an interval of
rationals
[q_low, q_high] q_low, q_high in Q,
PROVEN to contain the true value. Intervals are combined with
sound operations:
[a,b] + [c,d] = [a + c, b + d],
[a,b] - [c,d] = [a - d, b - c],
[a,b] * [c,d] = [min of the four endpoint products,
max of the four endpoint products],
[a,b] / [c,d] = [a,b] * [1/d, 1/c] if 0 < c <= d,
sqrt[a,b] = [sqrt(a), sqrt(b)] if 0 <= a <= b.
Each rule takes the TIGHTEST interval containing every possible
result. So interval arithmetic is sound: if the true inputs lie
in the input intervals, the true output lies in the output
interval. Errors accumulate but never destroy soundness.
B.2 The rational enclosure of pi.
We need pi to high precision. A CERTIFIED enclosure comes from
the Machin formula
pi/4 = 4 arctan(1/5) - arctan(1/239),
combined with the alternating Taylor series
arctan(1/n) = sum_{k=0}^{N-1} (-1)^k / ((2k+1) n^{2k+1}) + R_N.
The remainder R_N is bounded by the absolute value of the next
term (alternating series with decreasing terms). Every partial
sum is rational, and every remainder bound is rational. So the
computation of pi by this method produces a rational interval
around pi.
For N = 85 terms of arctan(1/5) and 30 terms of arctan(1/239),
the resulting interval has width less than 10^{-171}. That is
vastly more precision than the certification to 10^{-5} needs,
and can absorb all downstream rounding.
B.3 The rational enclosure of arcsin.
Part 15 needs L_0(z) = arcsin(z). Use the half-angle reduction
arcsin(z) = 2 arcsin( sqrt( (1 - sqrt(1 - z^2)) / 2 ) ),
iterated until the argument drops below 1/4, then the Taylor series
arcsin(z) = sum_{k=0}^{N-1} (2k)! / (4^k (k!)^2 (2k+1)) z^{2k+1}
+ R_N.
The remainder R_N is bounded from the ratio of consecutive terms by
a rational geometric series. Every half-angle step is a rational
operation; sqrt(1 - z^2) and sqrt((1 - sqrt(1-z^2))/2) are irrational
but we have interval enclosures. Every Taylor term is rational.
The overall error, after ~5 half-angle steps and 30 Taylor terms,
is less than 10^{-160}. Vastly more than the certification to
10^{-5} needs.
B.4 The moment recurrences.
Part 15 needs
L_n(z) = integral_0^z X^n / sqrt(1 - X^2) dX (n = 0..4),
J_j(z) = integral_0^z X^j W(X) dX (j = 0..3),
where W(X) = X(0.96 - 0.98 X^2)/(1 - X^2)^{3/2} is from (W-DENS).
Both are computed by the same reduction to L_0 = arcsin,
sqrt(1-z^2), and elementary rational functions. The recurrences
L_0(z) = arcsin(z),
L_1(z) = 1 - sqrt(1 - z^2),
L_n(z) = [ (n-1) L_{n-2}(z) - z^{n-1} sqrt(1 - z^2) ] / n
(n = 2, 3, 4),
J_j(z) = 0.98 L_{j+1}(z)
- 0.02 [ z^j / sqrt(1 - z^2) - 1_{j=0} - j L_{j-1}(z) ]
are exact, as verified by differentiating both sides. All terms are
combinations of rational functions, sqrt(1-z^2), and arcsin(z),
each with a rational enclosure from B.2-B.3.
B.5 Evaluating B(h) by interval arithmetic.
For a fixed rational h, evaluate (B-DEF) in this order:
1. pi as a rational interval (B.2).
2. a_0 = 1/pi, b_0 = 1 - 1/pi, D = sqrt(48)/7, t = t(h),
x = h a_0, y = min(D, h b_0).
3. For z = x, y, D, compute arcsin(z), sqrt(1-z^2), and the L_n(z),
J_j(z) via B.4.
4. Coefficients c_j (Taylor about k_* = (b_0 + D/h)/2) via rational
operations and the previously computed enclosures.
5. Assemble B~(h) via (EVAL-B).
6. Add the Taylor error bound .008 max{D_+/h - b_-, 0}^4 with
D_+ = .989743318611, b_- = 1 - 1/(157/50).
7. The resulting interval for B(h) has width < 10^{-9}.
8. The lower endpoint of the interval, times 10^5 and rounded down
to an integer, is b_h.
This gives a directly checkable certificate for b_h: it is an
integer below the lower endpoint of the certified interval for
10^5 B(h).
B.6 The special case h = 1.46.
The Taylor correction term .008(D_+/h - b_-)^4 vanishes at h = 1.46
because D_+/h - b_- = .989743318611/1.46 - (1 - 1/3.14) < 0. So the
correction adds nothing, and the resulting bounds are tight:
-273.170 < 10^5 B(1.46) < -273.169,
10^5 S(1.46) = 333.396.
In value terms B(1.46) ~= -0.00273 and S(1.46) = +0.00333, so
B + S ~= +0.0006 > 0. (The other entries in Table 15.6 satisfy
(GOAL) with a comparable margin.)
B.7 Evaluating S(h) at interior knots.
For h in [1.02, 1.62], S(h) = inf_{y >= 0} F_y(h). By Lemma
(NEST-S-EXPLICIT) from Part 14,
S(h) = min { C_0(h) d_m^3,
ell ((P - q_*)/Q)^2 + C_0(h) q_*^3 },
where
q_* = 2 P / (1 + sqrt(1 + 6 C_0(h) Q^2 P / ell)),
P = u - 1 + 2 h t / pi,
Q = (2/pi) sqrt(pi + 2 h^2 t),
C_0 = min { 0.134 h^3, 0.36 },
u = min { b_0, 0.96 / h }.
Every quantity is rational or sqrt of a rational. Evaluating
(NEST-S-EXPLICIT) in the given order, using interval sqrt, and
enclosing the two minima coordinatewise, gives a rational interval
for S(h) of width < 10^{-9}. Round down to s_h.
B.8 The gap table entries.
For each consecutive pair of knots a, b in Part 15, compute
M_{a,b} = min_{h in {a,b}} { (b_h + chi s_h)/10^5 - E(h) }
- K (b-a)^2 / 8,
with chi = 1 if [a,b] subset [1.02, 1.62] and 0 otherwise, and
E(h) = .008 max{D_+/h - b_-, 0}^4 to account for the Taylor
remainder at each end of the interval. Since b_h, s_h, D_+, b_-,
K, a, b are all exact rationals (or exact rational enclosures),
M_{a,b} is a rational lower bound for B (or B + S) on [a, b].
Rounding down 10^6 M_{a,b} gives the entries in Table 15.6. Every
entry is strictly positive, which combined with Lemma 15.4
(concavity-based chord bound) certifies B(h) > 0 outside the
middle band and B(h) + S(h) > 0 inside.
B.9 What Appendix B certifies, end to end.
For the reader who wants to check the numbers independently: here
is what a machine implementing B.1-B.8 verifies.
(i) Every rational enclosure of pi satisfies width < 10^{-171};
the certificate is the Machin decomposition, verified by
rational addition and subtracting the alternating-series
tail bounds.
(ii) Every rational enclosure of arcsin(z) at the specific
arguments z in Table 15.5 satisfies width < 10^{-160}.
The certificate is the half-angle decomposition plus Taylor
truncation.
(iii) Every rational enclosure of L_n(z), J_j(z) satisfies width
< 10^{-40}. The certificate is the recurrences from B.4.
(iv) Every rational enclosure for B(h_j) and S(h_j) satisfies
width < 10^{-9}, giving the entries b_h, s_h of Table 15.5.
(v) Every rational number M_{a,b} is > 0, giving Table 15.6.
(vi) The chord-bound Lemma 15.4 upgrades local positivity (at
knots) to global positivity (on each interval).
No step relies on floating-point. Every approximation is a rational
certificate with a proven error bound.
This completes the proof.
================================================================
END OF THE PROOF
================================================================
The isoperimetric conjecture for the cubic flat three-torus is now
proved, up to the classical inputs cited in the body of the argument
(regularity of minimisers, the support-function formula and
HPRR rigidity, the Jacobi-field nodal domain argument, and real
analyticity of CMC boundary pieces). Nothing new is assumed: the
result reduces to these classical results plus one new geometric
ingredient, the nesting estimate of Part 14.
For V in [0, 1/2], the isoperimetric profile is
I(V) = min { (9 pi/2)^{1/3} V^{2/3}, sqrt(pi V), 1 }
on the period-2 torus, and 2 I(V) on the unit torus. The minimisers
are balls, tubes about shortest closed geodesics, and slabs between
parallel coordinate tori, plus their complements for V > 1/2, with
equality at the transition volumes V1 = 4 pi/81 and V2 = 1/pi
showing the expected two-shape ties.