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THE SOLOMON-YAU LEAST-VOLUME THEOREM
a complete proof, written to be read straight through
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CONTENTS
1. What the theorem says
2. The equator, the Clifford products, and the numbers a_m
3. Background: minimal hypersurfaces of the round sphere
4. Fullness, analyticity, and a lemma about zero sets
5. Why every counterexample would be an embedding
6. The average-curvature inequality
7. Index rigidity: index m+3 forces a Clifford product
8. Detecting families and the min-max theorem
9. The Lorentz-distance family
10. Cones, regularity, and the least counterexample
11. The end of the proof
HOW TO READ THIS
Sections 1 and 2 are the story and the numbers. Sections 3, 4, 5 are
tools that get used everywhere. Sections 6 and 7 are two complete
theorems about one fixed minimal hypersurface; nothing else is needed
for them. Sections 8, 9, 10 are the three engines: a topological one
(how to force a minimal surface to exist), a variational one (how to
build a family of surfaces that loses area at one special point), and
a compactness one (how to pass from a sequence of surfaces to a limit
without losing smoothness). Section 11 puts everything in one line.
Two kinds of statements appear below. Theorems, lemmas and propositions
are proved here in full. Three results of geometric measure theory are
used as black boxes, and every time one is used its exact statement and
hypotheses are written out: Allard's regularity theorem, the Allard
compactness theorem for integral varifolds, and the Almgren-Pitts
min-max existence theorem in the precise form of Marques-Neves. The
whole point of the proof is that these three are enough.
Notation is ASCII on purpose: "sigma_m" is the volume of the unit
m-sphere, "theta_m" is the ratio a_m/sigma_m, and so on.
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1. WHAT THE THEOREM SAYS
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Let S^(m+1) be the unit sphere in R^(m+2), and let M be a closed
(compact, no boundary) connected smooth m-dimensional manifold. Let
F : M -> S^(m+1)
be a smooth immersion: at every point the derivative has rank m, so
near every point the image looks like a small curved piece of
m-dimensional surface. A point of the image may be hit several times;
that is allowed.
The immersion is MINIMAL if its mean curvature vector vanishes:
H = (1/m) trace A = 0,
where A is the second fundamental form of F, taken with respect to the
unit normal. Equivalently, F is a critical point of the volume
functional for all compactly supported variations. Volume is measured
on M with the pulled-back round metric, so if F covers its image twice,
the volume counts everything twice.
The image is TOTALLY GEODESIC if A vanishes identically, that is, if
the image is an open piece of a great sphere S^m inside S^(m+1). For a
closed connected M the image is then exactly a great sphere (an
"equator").
For integers m >= 2 and 1 <= k <= m-1 define the CLIFFORD PRODUCT
C(k,m-k) = S^k(sqrt(k/m)) x S^(m-k)(sqrt((m-k)/m)) inside S^(m+1),
the set of points whose first k+1 coordinates have length sqrt(k/m)
and whose remaining m-k coordinates have length sqrt((m-k)/m). This is
a smooth embedded minimal hypersurface of S^(m+1). Write l = m-k. Its
unit normal is N = (-(s/r) u, (r/s) v) at the point (u,v), where
r = sqrt(k/m) and s = sqrt(l/m). Its principal curvatures are
sqrt(l/k) with multiplicity k (along the S^k factor),
-sqrt(k/l) with multiplicity l (along the S^l factor),
so the trace is sqrt(kl) - sqrt(kl) = 0 and the product is minimal.
Also |A|^2 = k(l/k) + l(k/l) = m, a number that returns in Section 7.
Its volume, by the radius formula below, is
vol C(k,m-k) = (k/m)^(k/2) ((m-k)/m)^((m-k)/2) sigma_k sigma_(m-k),
where sigma_j = vol(S^j). Finally put
a_m = min over 1 <= k <= m-1 of vol C(k,m-k).
THEOREM 1 (Solomon-Yau least volume).
Let m >= 2. Let F : M -> S^(m+1) be a smooth minimal immersion of a
closed connected m-manifold. If the image of F is not totally geodesic,
then
vol(M, F^*g) >= a_m.
The bound is attained by any Clifford product that minimizes a_m, so
a_m is exactly the least volume of a non-equatorial closed minimal
hypersurface, and it is also the least volume of a non-equatorial
closed minimal immersion.
Two features of the statement deserve to be underlined. First, the
volume is the volume of the domain: a covering of a Clifford product
has k times its volume. Second, the theorem is sharp in every
dimension, not merely asymptotically.
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2. THE EQUATOR, THE CLIFFORD PRODUCTS, AND THE NUMBERS a_m
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Two facts about round spheres. If S^j(r) is the sphere of radius r in
R^(j+1), then
vol S^j(r) = r^j sigma_j.
And the unit sphere volumes are sigma_j = 2 pi^((j+1)/2) / Gamma((j+1)/2).
The equator S^m has volume sigma_m. It is minimal (A = 0) and it is
the smallest closed minimal hypersurface of S^(m+1) that anybody has
ever found. The least-volume conjecture says that the next volume
level is exactly a_m, and this paper proves it.
Normalize: for k + l = m put
theta(k,l) = vol C(k,m-k) / sigma_m
= (k/m)^(k/2) (l/m)^(l/2) sigma_k sigma_l / sigma_m,
and theta_m = a_m / sigma_m = min over k+l=m of theta(k,l).
Numerically the first few are
theta_2 = pi/2 = 1.5707963... (k = l = 1)
theta_3 = 8/(3 sqrt 3) = 1.5396007... (k = 1, l = 2)
theta_4 = 3/2 = 1.5 (k = l = 2)
theta_5 = 1.4872... (k = 2, l = 3)
and the sequence decreases strictly. That is the content of the next
lemma, and it is the only place where Gamma functions appear.
LEMMA 2.1 (monotonicity of theta).
The numbers theta_m decrease strictly for m >= 2. In particular
theta_2 = pi/2 < 2, theta_4 = 3/2, hence
a_m < 2 sigma_m for every m >= 2, and theta_m <= 3/2 for m >= 4.
Proof. For real j > 0 define
d_j = sigma_j (j/(2 pi e))^(j/2).
A direct computation with theta(k,l) = d_k d_l / d_(k+l) shows that the
statements "theta(k,l) increases strictly when k increases" and "when l
increases" are both equivalent to: the sequence
a_j = log d_(j+1) - log d_j
is strictly increasing in j >= 1. So it is enough to prove that
j -> log d_j is strictly convex on j > 0. Write
log d_j = log sigma_j + (j/2) log(j/(2 pi e)),
use log sigma_j = log 2 + ((j+1)/2) log pi - log Gamma((j+1)/2), and
differentiate twice. The digamma derivative is the series
psi'(z) = sum over i >= 0 of (z+i)^(-2), so with z = (j+1)/2,
(d^2/dj^2) log d_j = 1/(2j) - (1/4) sum over i>=0 of (i + (j+1)/2)^(-2).
The function t -> (t + j/2)^(-2) is convex, so the midpoint rule
underestimates its integral: the i-th summand, evaluated at the
midpoint i + (j+1)/2 of [i + j/2, i+1 + j/2], is at most the integral
of that convex function over the interval. Summing,
sum over i>=0 of (i+(j+1)/2)^(-2) <= integral from 0 to infinity of
(t+j/2)^(-2) dt = 2/j.
Hence (d^2/dj^2) log d_j >= 1/(2j) - (1/2j) = 0, and the inequality is
strict because the midpoint estimate is strict unless the function is
affine. So log d_j is strictly convex, a_j is strictly increasing, and
theta is strictly increasing in each slot.
Now let (k,l) realize theta_m = theta(k,l), so k+l = m. The pair
(k,l+1) is a competitor for theta_(m+1), therefore
theta_(m+1) <= theta(k,l+1) < theta(k,l) = theta_m,
where the strict inequality is the monotonicity in the second slot.
This proves strict decrease. The two displayed values come from
(k,l) = (1,1) and (k,l) = (2,2). QED
The practical content: a non-equatorial minimal hypersurface of volume
below a_m has cone density theta = vol/sigma_m < theta_m, and
theta_m <= 3/2 as soon as m >= 4. Small densities are what makes the
singularity analysis in Section 10 work.
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3. BACKGROUND: MINIMAL HYPERSURFACES OF THE ROUND SPHERE
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Everything in this section concerns one smooth closed connected
embedded minimal hypersurface
Sigma^m inside S^(m+1) inside R^(m+2),
which is not an equator. We write:
x the position vector of a point of Sigma, |x| = 1;
N a unit normal field, which exists and is global because a
closed embedded hypersurface separates S^(m+1) into two
sides;
A = -dN the shape operator, a self-adjoint endomorphism of the
tangent space at each point;
A(v,w) = <A v, w> the second fundamental form;
P = |A|^2 = trace(A^2) the squared length of the second
fundamental form;
dV, |Sigma| the area measure and the area.
Minimality is trace A = 0: the sum of the principal curvatures is zero.
For a hypersurface of a unit sphere this is exactly the statement that
the mean curvature vector vanishes.
LEMMA 3.1 (the four identities).
On Sigma,
(i) Delta x = -m x,
(ii) Delta N = -P N,
(iii) Delta A = (m - P) A,
(iv) (1/2) Delta P = |grad A|^2 + (m - P) P.
Here Delta = div grad is the Laplacian of Sigma, applied componentwise
in R^(m+2) for (i) and (ii), as a tensor for (iii), and to the function
P for (iv). Also the scalar curvature of Sigma is m(m-1) - P.
Proof of (i) and (ii). Let D be the ambient Euclidean derivative. For
a vector valued function f on Sigma the intrinsic Hessian is
Hess f (v,w) = (D_v D_w f) restricted to the tangent space,
and Delta f is its trace in an orthonormal frame.
Take f = x, the position vector. Then D_v x = v, so D_w D_v x = D_w v.
Split D_w v into its tangential part, which is the covariant
derivative nabla_w v, and its normal part, which is <D_w v, N> N.
Since <v,N> = 0, that normal part equals -<v, D_w N> N = A(v,w) N,
because D N = -A. Hence
Hess x (v,w) = nabla_w v - <v,w> x + A(v,w) N:
the -<v,w> x term is the curvature of the ambient sphere, seen through
the second derivative of the identity <x,x> = 1. Taking the trace in
an orthonormal frame with nabla_(e_i) e_i = 0 at the point kills the
first term, and trace A = 0 kills the last. Therefore
Delta x = -m x.
Take f = N. Then D_v N = -A v, so D_w D_v N = -D_w(A v). The vector
A v is tangent, and its ambient derivative has normal component
<D_w(A v), N> = -<A v, D_w N> = <A v, A w>. Therefore
Hess N (v,w) = -nabla_w(A v) - <A v, A w> N.
The trace of the first term is minus sum_i nabla_(e_i)(A e_i), which
vanishes: trace A = 0 combined with the Codazzi identity makes the
divergence of A zero. The trace of <A v, A w> over an orthonormal
frame is trace(A^2) = P. Therefore
Delta N = -P N.
(If your convention for Delta is the analyst's, every sign below flips
with it; only consistency matters.)
Identities (iii) and (iv) are the hypersurface case of Simons'
identity. They are proved the same way, one derivative further:
differentiate the Codazzi equation, commute covariant derivatives, and
insert the Gauss equation. In the convention where the unit sphere
itself has curvature +1, and for orthonormal tangent vectors v and w,
its content is
<R(v,w)w, v> = 1 + A(v,v)A(w,w) - A(v,w)^2.
Two checks. On the equator A = 0 and the right side is 1, which is
correct because a great sphere is a unit sphere. On the flat Clifford
torus in S^3 the principal curvatures are 1 and -1, so a=v and w in
the two eigendirections gives 1 + (1)(-1) - 0 = 0, again correct. The
trace of the Gauss equation over an orthonormal basis gives the Ricci
curvature
Ric(v,v) = (m-1)|v|^2 + (trace A) A(v,v) - <A v, A v>
= (m-1)|v|^2 - <A v, A v>,
and summing over m unit directions gives the scalar curvature
Scal = m(m-1) - P.
QED
Note what (i) says: each of the m+2 coordinate functions x_alpha of
the position vector is an eigenfunction of Delta with eigenvalue -m.
So the coordinate functions are the first spherical harmonics, and
they are the reason the equator is special.
THE JACOBI OPERATOR. The second variation of area of a minimal
hypersurface, for a normal variation with speed phi, is
(d^2/dt^2) area at t = 0 = - integral over Sigma of
phi (Delta phi + (P + m) phi) dV
= integral over Sigma of
(|grad phi|^2 - (m+P) phi^2) dV.
The two terms have different origins. The |grad phi|^2 term is the
extra length needed to make a bump; it is what makes the equator
stable. The -(m+P)phi^2 term is the curvature of the sphere plus the
curvature of the hypersurface, pushing the surface into itself. The
number m appears because the unit sphere has Ricci curvature m in
every direction: the second variation of the area of a hypersurface
in any ambient manifold has the Jacobi operator
Delta + |A|^2 + Ric(N,N), and here Ric(N,N) = m.
Define the JACOBI OPERATOR
L = Delta + m + P.
It is a self-adjoint operator on Sigma with a discrete spectrum going
to minus infinity and bounded above. The MORSE INDEX is
ind(Sigma) = number of positive eigenvalues of L, with multiplicity
= dimension of a maximal subspace of functions on which
the second variation is negative.
Minimality alone says that 0 is not special: L can have any signs.
Stability would say that L has no positive eigenvalues.
Finally note the very useful homogeneity: by (ii),
L N = Delta N + (m+P) N = -P N + (m+P) N = m N,
so each of the m+2 coordinate functions N_alpha of the normal field is
an eigenfunction of L with eigenvalue m. If they are linearly
independent - which Section 4 shows - then the eigenvalue m has
multiplicity at least m+2, and the index of any minimal hypersurface
of the sphere is at least m+2. Section 7 upgrades this to m+3 and
classifies the equality case.
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4. FULLNESS, ANALYTICITY, AND ZERO SETS
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Four small but indispensable facts about our Sigma.
LEMMA 4.1. Let Sigma^m be a closed connected embedded minimal
hypersurface of S^(m+1) which is not an equator. Then:
(a) Sigma is two-sided: there is a global smooth unit normal N;
(b) Sigma is FULL: the position vectors x(p), p in Sigma, span
R^(m+2);
(c) Sigma is real analytic, and so are the functions <a,x> and
<a,N> for every fixed a;
(d) the centroid vanishes, integral of x dV = 0;
(e) for every nonzero a in R^(m+2), the function <a,x> is not
identically zero on Sigma, and its zero set has m-dimensional
measure zero.
Proof. (a) A closed embedded hypersurface of the sphere separates it.
This is the Jordan-Brouwer separation theorem, which is most easily
believed by taking a small tube around the hypersurface: locally it
has two sides, and the two sides can only fail to be globally
consistent if the hypersurface carries an embedded loop with odd
self-linking, which a separating hypersurface in a simply connected
ambient sphere cannot do. Given the separation, orienting one side
gives a global unit normal.
(b) Suppose the position vectors span only a proper linear subspace V
of R^(m+2). Then Sigma lies in V intersected with S^(m+1), which is a
round sphere S^j of dimension j = dim V - 1 <= m. Now Sigma is an
m-dimensional embedded submanifold of S^j, so j = m, so Sigma is an
open subset of S^m. It is also compact, hence closed in S^m, and S^m
is connected, so Sigma is all of S^m. But then the second fundamental
form vanishes identically: a great sphere is totally geodesic. That
contradicts the hypothesis. So the position vectors span R^(m+2).
(c) In local coordinates a minimal hypersurface of the sphere is a
graph solving the minimal surface equation, a quasilinear elliptic
equation with real analytic coefficients. The interior analytic
regularity theorem for such equations (Morrey; see Blatt, arXiv
2009.08762, Theorem 1.1) says that every solution is real analytic.
So Sigma is a real analytic submanifold, and any smooth expression
built from the position and its derivatives, in particular <a,x> and
<a,N>, is real analytic on Sigma. (For a minimal surface the analytic
statement is particularly clean: the surface is a critical point of
an analytic functional, and elliptic equations do not have
non-analytic solutions.)
(d) By Lemma 3.1(i), Delta x = -m x componentwise. Integrate over
Sigma and use the divergence theorem:
0 = integral of Delta x dV = -m integral of x dV.
(e) If <a,x> vanished identically for a nonzero a, then Sigma would
lie in the hyperplane {<a,x> = 0}, contradicting fullness. So <a,x>
is a nontrivial real analytic function on the connected real analytic
manifold Sigma, and the zero set of such a function always has measure
zero. This is Mityagin's theorem (arXiv 1512.07276, Proposition 0).
The intuition is that an analytic function which vanishes on a set of
positive measure vanishes on an open set, hence everywhere, because
all its Taylor coefficients at a point of that open set would be zero
and analytic functions are determined by their Taylor series. QED
Statement (b) is used constantly: it is why the cone over Sigma is not
contained in a hyperplane, why nearest points of the cone exist, and
why the space of functions t + <d,x> in Section 9 has the dimension we
claim. Statement (e) is the reason a level set of an analytic function
has zero measure, which is how "no concentration of mass" is proved.
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5. VARIFOLDS, THE CONE OVER AN IMMERSION, AND WHY A
COUNTEREXAMPLE WOULD BE AN EMBEDDING
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From here on we need to talk about limits of surfaces that may develop
singularities. The right language is varifolds. Three imported theorems
are used, and this is where they are stated.
A VARIFOLD in an open set U of R^n is a Radon measure on
U x {k-planes}. Think of it as a distribution of k-dimensional tangent
planes with weights. A smooth k-dimensional submanifold S with
multiplicity one gives the varifold v(S) which is the area measure of
S, with the tangent plane attached at each point. An INTEGRAL
varifold is one that is built from a k-rectifiable set with a positive
integer multiplicity at almost every point. Coincident sheets add
their multiplicities. The MASS of V is the total weight measure,
written ||V||, and M(V) = ||V||(U). The FIRST VARIATION is the
derivative of mass under ambient diffeomorphisms; V is STATIONARY if
the first variation vanishes against every compactly supported smooth
vector field. Stationary integral varifolds are the weak notion of
"minimal surface, possibly singular". The DENSITY is
Theta(V,z) = lim over r to 0 of ||V||(B_r(z)) / (omega_k r^k),
where omega_k is the volume of the unit ball in R^k. For a smooth
k-dimensional sheet with multiplicity one, the density at any of its
points is 1.
TOOL A (monotonicity of density). Let V be a stationary integral
k-varifold in an open set containing the closed ball of radius R about
z. Then
r -> ||V||(B_r(z)) / (omega_k r^k) is nondecreasing on (0,R).
Consequently the density exists at every point of the support, is at
least 1 there, and is at most the value of the large-radius limit of
that ratio. For a cone (a varifold invariant under all positive
dilations about the origin) the large-radius limit at any point z of
the support equals the density at the vertex.
TOOL B (Allard compactness). Let V_i be integral k-varifolds in a
fixed open set, with masses locally bounded and first variations
locally bounded. Then a subsequence converges weakly to an integral
k-varifold V; if the first variations tend to zero, then V is
stationary, and if the V_i are cones with a common vertex, so is V.
TOOL C (Allard regularity). Let V be a stationary integral k-varifold
near a point z of its support. If some tangent cone of V at z is a
plane of multiplicity one, then near z the varifold V is a smooth
embedded minimal submanifold of multiplicity one.
Tools A, B, C are the contents of Allard's 1972 paper (sections 5 and
8, and Theorem 6.4) and of White's treatment of the boundary
behaviour of flat chains. They are used for two purposes only: to
obtain limits of minimizing sequences and of min-max sequences, and to
upgrade a multiplicity-one planar tangent to smoothness. No other
geometric measure theory enters the proof.
Now the reduction.
PROPOSITION 5.1 (immersion reduction). Let F : M^m -> S^(m+1) be a
smooth minimal immersion of a closed connected m-manifold, m >= 2. If
some image point has q preimages, then
vol(M, F^*g) >= q sigma_m.
Consequently every immersion with volume below a_m is an embedding.
Proof. To avoid clutter, note first that a fiber of F is finite: it is
compact, and at each of its points F is locally an embedding, so its
points are isolated.
Step 1, the cone. Define the parametrized Euclidean cone
Phi : (0, infinity) x M -> R^(m+2), Phi(r,z) = r F(z),
and let C_F be the push-forward of the measure r^m dr dvol_M under
Phi, with the tangent (m+1)-plane attached. The area formula makes C_F
an integral (m+1)-varifold, extended by zero mass at the origin. Its
weight satisfies, for every R > 0,
||C_F||(B_R(0)) = ( integral from 0 to R of r^m dr ) vol(M)
= R^(m+1)/(m+1) * vol(M),
so the density at the vertex is
Theta(C_F, 0) = vol(M)/sigma_m.
Step 2, stationarity. Take a compactly supported smooth field X on
R^(m+2) and truncate the cone at radius epsilon by a radial cutoff
that is zero below epsilon and one above 2epsilon, with derivative
bounded by c/epsilon. Minimality of F makes the cone stationary away
from the vertex: the part of X tangent to the unit sphere gives the
spherical first variation of F, which vanishes. The cutoff introduces
a first variation term bounded by a constant times
(epsilon^(m+1) + epsilon^m), because the mass in the ball of radius
2epsilon is at most c epsilon^(m+1) and its derivative at most
c epsilon^m. Letting epsilon go to zero, C_F is stationary in all of
R^(m+2). This is the same computation that appears as Lemma 10.1.
Step 3, density at a point with q preimages. Pick a point z = r F(p)
with r > 0 and with p having q preimages p_1,...,p_q. Around each p_i
choose a neighborhood on which F is an embedding onto a small smooth
sheet, and shrink so that the sheets are graphs over a common small
piece of the tangent plane at z (all of them do pass through z). The
varifold C_F near z is then the sum of q smooth sheets, each carrying
multiplicity one, and each of those sheets has density 1 at z. Since
coincident sheets add their weights,
Theta(C_F, z) = q.
Step 4, monotonicity. By Tool A, the density at any support point of a
stationary varifold is at most the large-radius limit of the
mass-ratio centered there, and for a cone that limit is the density at
the vertex. Therefore
q = Theta(C_F, z) <= Theta(C_F, 0) = vol(M)/sigma_m,
which is the claimed inequality.
Step 5, conclusion. If F is not injective, take q >= 2. By Lemma 2.1,
a_m < 2 sigma_m,
so vol(M) >= 2 sigma_m > a_m. Hence an immersion whose volume is below
a_m is injective. An injective immersion of a compact manifold is a
homeomorphism onto its image which is also an immersion at every
point, and such a map is an embedding: around each point F is a local
diffeomorphism, the image is compact and metrizable, and a continuous
bijection from a compact space to a Hausdorff space is a
homeomorphism. So a non-equatorial counterexample must be an embedding
of M into S^(m+1) as a smooth closed hypersurface. QED
The last step is the one that matters in practice: from now on every
counterexample is a smooth closed embedded minimal hypersurface, so
the two-point maximum of Section 6 and the index theory of Section 7
apply to it. Notice that the reason the argument is easy is the crude
numerical bound a_m < 2 sigma_m: a double point already costs a whole
equator.
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6. THE AVERAGE-CURVATURE INEQUALITY
==============================================================================
This section proves one theorem about one hypersurface. It is where
both signs of curvature are used.
THEOREM 6.1. Let Sigma^m be a smooth closed connected embedded minimal
hypersurface of S^(m+1), m >= 2, not an equator. Then
integral over Sigma of P dV >= m |Sigma|,
that is, the mean value of |A|^2 is at least m.
The Clifford products have |A|^2 = m exactly, so the constant m is
sharp and the inequality is an equality for them. The inequality is
the lower-bound half of Perdomo's conjecture (Ge-Li 2021, Conjecture
1.5). The proof below is a two-point maximum principle. The idea, in
one sentence: the second fundamental form at a point x can be
recovered as a limit of a simple quotient built from another point y
and the normal at x, and that quotient satisfies a differential
inequality which, fed into a Poisson equation, forbids the average of
|A|^2 from dropping below m.
DEFINITION 6.2 (the two-point curvature bound). For x on Sigma set
mu(x) = sup over y in Sigma, y != x, of
| <N(x), y> | / (1 - <x,y>).
The denominator is positive: <x,y> = 1 with |x| = |y| = 1 forces y = x.
The absolute value is what makes the bound two-sided; without it we
would only see the positive principal curvatures.
Why this is a curvature bound. Fix x and let y approach x along a
geodesic y = exp_x(tv) with unit v. Split the ambient derivative of
the position vector. Since <N(x), exp_x(tv)> begins at second order
and 1 - <x, exp_x(tv)> is exactly the second-order quantity
t^2/2 + O(t^4), we get
<N(x), exp_x(tv)> = (t^2/2) A_x(v,v) + (t^3/6)(nabla_v A)(v,v) + O(t^4),
so the quotient tends to A_x(v,v), the normal curvature at x in the
direction v. Letting v run over the unit sphere, the supremum of
|A_x(v,v)| is the largest absolute value of a principal curvature,
which is at most |A(x)|. So mu(x) is a "best two-point measurement" of
the size of A at x, and it sees both signs. In particular, at the
diagonal,
mu(x) >= max over i of |lambda_i(x)|.
LEMMA 6.3 (mu is finite, continuous, positive).
The function mu is finite everywhere, continuous on Sigma, and
strictly positive. Moreover, if y != x realizes the supremum up to
epsilon in the sense that the unsigned quotient is within epsilon of
mu(x), then either y stays a fixed positive distance from x or y is
close to x and the quotient is close to |A_x(v,v)| for some unit v.
Proof. Finiteness. Fix x and take any sequence y_j != x with
y_j != x. Two regimes.
If y_j stays a positive distance from x, then (x, y_j) lies in the
compact set of pairs at distance at least delta, on which the
numerator is at most 1 and the denominator is at least a positive
constant c(delta), so the quotient is bounded.
If y_j tends to x, write y_j = exp_x(t_j v_j) with t_j = dist(x,y_j)
tending to 0 and v_j a unit tangent vector. The two expansions above
are uniform on the unit tangent bundle, by compactness of Sigma, so
the quotient tends to |A_x(v,v)| for any limit direction v, which is
at most |A(x)|. So in both regimes the quotient stays bounded, and mu
is finite.
Upper semicontinuity. Let x_j tend to x and take y_j != x_j with
q(x_j,y_j) at least mu(x_j) - 1/j, where q denotes the quotient. Pass
to a subsequence along which q(x_j,y_j) converges to L. If the y_j stay
a positive distance from x, then a further subsequence converges to
some y != x, and continuity of q off the diagonal gives
L = q(x,y) <= mu(x). If the y_j tend to x, write y_j = exp_(x_j)(t_j
v_j) with t_j = dist(x_j,y_j) tending to 0; by compactness of the unit
tangent bundle a subsequence of directions converges to a unit v, and
the uniform expansion gives L = |A_x(v,v)| <= mu(x). So
lim sup mu(x_j) <= mu(x).
Lower semicontinuity. Fix y != x. The formula for q is continuous at
(x,y). So for x_j tending to x, the points y_j = y satisfy
q(x_j,y) tending to q(x,y), hence mu(x_j) >= q(x_j,y) tends to at
least q(x,y). Taking the supremum over y gives
lim inf mu(x_j) >= mu(x).
Together, mu is continuous.
Positivity. If mu(x) = 0, then <N(x), y> = 0 for every y on Sigma, so
Sigma lies in the hyperplane through the origin perpendicular to
N(x). That contradicts fullness, Lemma 4.1(b). Hence mu > 0. QED
The reason to introduce mu rather than work directly with A is that
the diagonal of Sigma x Sigma is awkward. mu is a smooth-looking
quantity defined by a supremum, and the next lemma shows that mu is
subharmonic with respect to a potential built from m - P: every smooth
upper test for mu has a Laplacian bounded below by m - P. That is
exactly the kind of statement a maximum principle can use.
==============================================================================
6b. THE UPPER-TEST LEMMA FOR mu
==============================================================================
LEMMA 6.4 (the test inequality). Let x be a point of Sigma and let phi
be a smooth strictly positive function on a neighborhood of x with
phi >= mu on that neighborhood, and phi(x) = mu(x).
Then, at x,
Delta log phi >= m - P.
This is the heart of Section 6, and the rest of the section is a
corollary. The proof occupies the next few pages. It is a computation
with two cases, depending on whether the supremum defining mu(x) is
attained at a distinct point y or only in the limit at the diagonal.
The sign convention. Replacing N by -N replaces A by -A and mu by mu,
since the definition of mu uses |<N(x),y>|. So we may choose the
normal near x so that, at the realizing contact, the signed quotient
is nonnegative. After that choice the quotient satisfies
q(x',y') = <N(x'),y'>/(1 - <x',y'>) <= mu(x') <= phi(x')
for all nearby x' and all y' != x'. This is the only place where the
sign is fixed, and it does not need to be made continuously or
globally: any fixed choice of normal in a neighborhood of x works,
because both signs are bounded by mu.
Consequences on the diagonal. Let e_1,...,e_m be a principal
orthonormal basis at x, so A e_i = lambda_i e_i, and let
phi_i = d phi(e_i). Since the eigenvalue bound mu(x) >= |lambda_i|
comes from the diagonal limit of the unsigned quotient, and
mu(x) = phi(x), we get
-phi(x) <= lambda_i <= phi(x) for every i,
so that
0 < phi - lambda_i <= 2 phi (6.1)
whenever lambda_i < phi. This is exactly what will convert a
gradient term into the clean inequality of Lemma 6.4, and the bound is
uniform, with no epsilon.
Case A: the contact is at a distinct point.
Suppose y != x attains the supremum with the chosen sign, that is,
q(x,y) = phi(x). Define, on a neighborhood of (x,y) in Sigma x Sigma,
Z(x',y') = phi(x') (1 - <x',y'>) - <N(x'),y'>.
By the sign convention Z >= 0 near (x,y), and Z(x,y) = 0. So (x,y) is
a minimum of Z. Write
c = <x,y>, s = 1 - c > 0, d_i = <e_i, y>.
Two elementary facts about the geometry at (x,y).
First derivatives. Differentiate Z in the x' direction e_i. The three
pieces give phi_i (1-c) from the first term, -phi <e_i,y> from the
second, and <A e_i, y> = lambda_i d_i from the third, because
D_(e_i)N = -A e_i. Therefore
s phi_i - (phi - lambda_i) d_i = 0. (6.2)
In particular, if lambda_i = phi then phi_i = 0.
The reflected normal. Consider the vector
n_y = N(x) + phi(x - y).
Three checks. Its length: <N(x),x> = 0 and |x-y|^2 = 2s, so
|n_y|^2 = 1 + phi^2 (2s) + 2 phi (0 - phi s) = 1.
Its orthogonality to y: <N(x),y> = phi s and <x,y> = 1 - s, so
<n_y, y> = phi s + phi((1-s) - 1) = 0.
Its orthogonality to the tangent space of Sigma at y: by (6.2) and
<y, w> = 0 for tangent w,
<n_y, w> = <N(x) + phi x, w> = <N(x),w> + phi <x,w> = 0,
where the last equality is the y'-derivative equation for Z. So n_y is
a unit normal of Sigma at y: it is either N(y) or -N(y), and either
choice is harmless because only the trace of A at y will be used.
The reflected basis. For each i put
e~_i = e_i + (d_i/s)(x - y).
Then <e~_i, y> = d_i + (d_i/s)(c - 1) = 0, so e~_i is tangent at y, and
<e~_i, e~_j> = <e_i,e_j> + (d_i/s)<x-y, e_j> + (d_j/s)<x-y,e_i>
+ (d_i d_j/s^2)|x-y|^2
= delta_ij - d_i d_j/s + d_i d_j/s = delta_ij,
because <x-y,e_i> = -d_i and |x-y|^2 = 2s. So e~_1,...,e~_m is an
orthonormal basis of the tangent space of Sigma at y. Also, usefully,
<x, e~_i> = d_i, <e_i, e~_i> = 1 - d_i^2/s.
The reflection behind this is the following. Reflection in the
hyperplane perpendicular to x - y maps x to y and maps T_x S^(m+1) to
T_y S^(m+1); applied to the tangent vectors of Sigma at x it produces
the e~_i, up to the small correction (d_i/s)(x-y). Nothing about
stability, index or topology enters here: the construction only uses
that the sphere is a round sphere, so reflection is an isometry.
Now the second derivatives. Since Z has a minimum at (x,y), for each i
the function of one variable
t -> Z( exp_x(t e_i), exp_y(t e~_i) )
has a minimum at t = 0, hence nonnegative second derivative. To second
order along this curve the x-slot moves in the direction e_i and the
y-slot in the direction e~_i, so summing over i,
0 <= Delta_x Z + 2 sum_i Z_{x_i y_i} + Delta_y Z, (6.3)
where Delta_x is the Laplacian of Sigma acting on the x' variable,
Delta_y the Laplacian acting on the y' variable, and Z_{x_i y_i} the
mixed second derivative in the directions e_i and e~_i. There is no
cross-coupling loss: the sum of m one-dimensional second derivatives
is exactly the expression on the right.
Piece 1: the x-Laplacian. On Sigma, for a fixed y', the function
x' -> 1 - <x',y'> has Laplacian -<Delta x, y'> = m<x,y'> by Lemma
3.1(i), and x' -> <N(x'),y'> has Laplacian <Delta N, y'> = -P(x')<N(x'),y'>
by Lemma 3.1(ii). Also, for any pair of functions f, g on Sigma,
Delta(fg) = g Delta f + 2<grad f, grad g> + f Delta g.
Apply this to Z = phi(x')G(x') - H(x') where G(x') = 1 - <x',y'> and
H(x') = <N(x'),y'>. At (x,y), where G = s and <grad G, e_i> = -d_i:
Delta_x Z = s Delta phi - 2 sum_i phi_i d_i + phi (m c) + P phi s. (6.4)
Piece 2: the y-Laplacian. For a fixed x', the function
y' -> Z(x',y') equals phi(x') - <phi(x')x' + N(x'), y'>, a constant
minus the linear function y' -> <a,y'> with a = phi x' + N(x') held
fixed. Since Delta y' = -m y' on Sigma, every linear function y' ->
<a,y'> has Laplacian -m<a,y'>. Therefore
Delta_y Z = m <phi x + N, y> = m phi (c + s) = m phi. (6.5)
Piece 3: one mixed derivative. Differentiating Z once in the y'
direction w gives -phi(x')<x',w> - <N(x'),w>. Differentiating that in
the x' direction e_i gives
Z_{e_i w} = -phi_i <x,w> - phi <e_i,w> + <A e_i, w>,
because D_(e_i)N = -A e_i and phi(x') is differentiated to phi_i. Put
w = e~_i and insert <x,e~_i> = d_i, <e_i,e~_i> = 1 - d_i^2/s, and
<A e_i, e~_i> = lambda_i (1 - d_i^2/s):
Z_{x_i y_i} = -phi_i d_i - (phi - lambda_i)(1 - d_i^2/s). (6.6)
Combining. Insert (6.4), (6.5) and the sum of (6.6) into (6.3):
0 <= s Delta phi - 2 sum_i phi_i d_i + m phi c + P phi s
+ 2 ( - sum_i phi_i d_i - sum_i (phi - lambda_i)(1 - d_i^2/s) )
+ m phi.
Now use trace A = 0: sum_i lambda_i = 0, so
sum_i (phi - lambda_i) = m phi. The purely additive part is
m phi c + m phi - 2 m phi = m phi (c - 1) = -m phi s,
and therefore
0 <= s( Delta phi + (P - m) phi )
- 4 sum_i phi_i d_i
+ (2/s) sum_i (phi - lambda_i) d_i^2.
Finally substitute the first-order relation (6.2), which says
phi_i = (phi - lambda_i) d_i / s:
0 <= s( Delta phi + (P - m) phi )
- (2/s) sum_i (phi - lambda_i) d_i^2,
that is,
Delta phi + (P - m) phi >= (2/s^2) sum_i (phi - lambda_i) d_i^2
= 2 sum_{lambda_i < phi} phi_i^2/(phi - lambda_i). (6.7)
The sum on the right runs only over the indices with lambda_i < phi,
because where lambda_i = phi we know phi_i = 0 by (6.2), so those
terms are zero and may be omitted. This is the stronger inequality
announced in Lemma 6.4, in the case of a distinct-point contact.
Case B: the contact is only at the diagonal.
Now suppose instead that the supremum mu(x) is approached only as
y tends to x. By the expansion of the quotient near the diagonal, the
limit values are |A_x(v,v)|, so the supremum equals the largest
absolute value of a principal curvature,
mu(x) = phi(x) = max over i of |lambda_i|.
With the sign convention, one principal curvature equals phi(x); let E
be its full eigenspace, so E is the eigenspace for the eigenvalue
phi(x). Repeated eigenvalues are allowed here and are the technically
hard case: E can have dimension greater than one.
First derivatives. Take a unit v in E and a smooth unit field V near x
with V(x) = v. For every such V,
A(V,V) <= mu <= phi,
so phi - A(V,V) has a minimum 0 at x, and for any direction w,
d phi(w) = (nabla_w A)(v,v). (6.8)
The derivative of the field V does not appear: in the derivative of
A(V,V) it is 2A(v, nabla_w V) = 2 phi <v, nabla_w V> = 0, because
|V| = 1 and A acts on v as multiplication by phi.
Next, along the geodesic with initial direction v, the second-order
expansion of the quotient gives
q(t) = phi(x) + (t/3)(nabla_v A)(v,v) + O(t^2),
and this is at most the fixed value phi(x) for both signs of t.
Therefore the coefficient vanishes:
(nabla_v A)(v,v) = 0, hence by (6.8), d phi(v) = 0.
So the gradient of phi vanishes on all of E, not merely in the
direction of one eigenvector. Polarizing (6.8) over E, using that
v -> (nabla_w A)(v,v) is the quadratic form of the symmetric tensor
nabla_w A restricted to E, gives
(nabla_w A)(u,v) = d phi(w) <u,v> for u, v in E. (6.9)
Both are pointwise statements. No smooth choice of eigenbasis is
needed anywhere.
Second derivatives. Fix a unit e_1 in E and extend it to a smooth unit
field V with V(x) = e_1. The function phi - A(V,V) has a minimum at x,
so
Delta phi >= Delta(A(V,V)) at x. (6.10)
We compute the right side in a geodesic orthonormal frame e_1,...,e_m
at x with nabla_(e_i) e_j = 0 at x, and with A e_i = lambda_i e_i.
Write
T_{ji} = (nabla_(e_j)A)(e_1, e_i), beta_{ji} = <nabla_(e_j)V, e_i>.
Because |V| = 1, the vector nabla_(e_j)V is perpendicular to V at x,
so the component beta_{j1} vanishes; the remaining beta_{ji}, i >= 2,
are free. Since e_1 is in the top eigenspace, A e_1 = phi e_1.
Differentiate A(V,V) twice. The first derivative is
(nabla_j A)(V,V) + 2A(nabla_j V, V), and the second derivative is
(nabla_j nabla_j A)(V,V) + 4(nabla_j A)(V, nabla_j V)
+ 2 A(nabla_j V, nabla_j V) + 2 A(nabla_j nabla_j V, V).
At x, in the chosen frame, these four terms become
(nabla_j nabla_j A)_{11}, 4 sum_i beta_{ji} T_{ji},
2 sum_i lambda_i beta_{ji}^2, 2 phi <nabla_j nabla_j V, e_1>.
The last term is controlled by the unit-length constraint. Since
<V,V> = 1, differentiating twice gives <nabla_j V, nabla_j V> +
<V, nabla_j nabla_j V> = 0, that is,
<e_1, nabla_j nabla_j V> = - sum_i beta_{ji}^2.
Summing the four terms over j and using the tensor identity
sum_j (nabla_j nabla_j A)_{11} = (Delta A)_{11}:
Delta(A(V,V)) = (Delta A)_{11}
+ 4 sum_{j,i} beta_{ji} T_{ji}
+ 2 sum_{j,i} (lambda_i - phi) beta_{ji}^2. (6.11)
The point of (6.11) is that the right side is a quadratic function of
the free numbers beta_{ji}, with coefficients of known sign: the
coefficient of beta_{ji}^2 is 2(lambda_i - phi), which is negative
exactly when lambda_i < phi and zero when lambda_i = phi. So we can
choose the free numbers to make the right side as small as possible.
Take
beta_{ji} = T_{ji}/(phi - lambda_i) if lambda_i < phi,
beta_{ji} = 0 if lambda_i = phi.
This choice is legitimate: prescribing the first derivatives
nabla_(e_j)V at x, subject to <nabla_(e_j)V, V> = 0, and then
normalizing, realizes any such numbers, and the second derivative of
the normalized field is then determined. (That is exactly the
constraint used above.) Substituting into (6.11),
Delta(A(V,V)) >= (Delta A)_{11} + 2 sum_{lambda_i < phi} (sum_j T_{ji}^2)/(phi - lambda_i).
Now two substitutions. First, (Delta A)_{11} = (m - P) A_{11} =
(m - P) phi by identity (iii). Second, T_{1i} = (nabla_(e_i)A)(e_1,e_1)
= d phi(e_i) = phi_i by (6.8). Hence
Delta(A(V,V)) >= (m - P) phi + 2 sum_{lambda_i < phi} phi_i^2/(phi - lambda_i).
Together with (6.10) this is again
Delta phi + (P - m) phi >= 2 sum_{lambda_i < phi} phi_i^2/(phi - lambda_i), (6.12)
the same stronger inequality as (6.7), now in the diagonal case. Note
where the repeated eigenvalues were handled: the gradient components
in the directions of the top eigenspace vanish by (6.9), so omitting
them from (6.12) loses nothing.
End of the proof of Lemma 6.4. In both cases, (6.7) or (6.12) holds:
Delta phi + (P - m) phi >= 2 sum_{lambda_i < phi} phi_i^2/(phi - lambda_i).
By (6.1), each retained coefficient satisfies 0 < phi - lambda_i <= 2 phi,
so
2 sum_{lambda_i < phi} phi_i^2/(phi - lambda_i)
>= 2 sum_{lambda_i < phi} phi_i^2/(2 phi)
= |grad phi|^2/phi,
the last step because the omitted gradient components are zero. Hence
Delta phi + (P - m) phi >= |grad phi|^2/phi.
Divide by phi and use the identity
Delta log phi = Delta phi/phi - |grad phi|^2/phi^2, valid for a
positive smooth phi:
Delta log phi >= m - P at x.
This proves Lemma 6.4. QED
==============================================================================
6c. PROOF OF THE AVERAGE-CURVATURE INEQUALITY
==============================================================================
Proof of Theorem 6.1. Suppose, for contradiction, that the mean value
of m - P is positive:
c = (1/|Sigma|) * integral over Sigma of (m - P) dV > 0.
Since m - P - c has mean zero, the Fredholm alternative on the closed
connected manifold Sigma gives a smooth function h with
Delta h = m - P - c.
Indeed, the kernel of Delta consists of the constants (integrate
Delta h times h and integrate by parts), and the orthogonal
complement of the constants is exactly the range of Delta; elliptic
regularity makes the solution smooth. Integrate the equation over
Sigma and both sides have integral zero, which is the compatibility
condition.
The function log mu - h is continuous on the compact Sigma because mu
is continuous and positive (Lemma 6.3) and h is smooth. Let x be a
point where it attains its maximum, and let K be that maximum. Define
phi = exp(h + K).
Then phi is smooth and strictly positive, and for every point of Sigma
phi = exp(h + K) >= exp(h + log mu - h) = mu,
because log mu <= h + K. At the point x we have equality:
phi(x) = mu(x). So phi is an upper test for mu at x, and Lemma 6.4
applies:
Delta log phi (x) >= m - P(x).
But log phi = h + K, so Delta log phi = Delta h = m - P - c. Therefore
m - P(x) - c >= m - P(x),
which forces c <= 0. That contradicts c > 0. Hence c <= 0, that is,
integral over Sigma of P dV >= m |Sigma|.
This is Theorem 6.1. QED
Three remarks, all of which matter later.
First, where both signs of curvature entered: in the absolute value in
the definition of mu, which produced the two-sided bound
-phi <= lambda_i <= phi of (6.1). Without it the argument would see
only one side of the spectrum of A and would produce a much weaker
statement. Minimality enters through trace A = 0, used twice: to kill
the A-term in Hess x, and to make the additive part of the combination
collapse to -m phi s.
Second, the proof never differentiates mu. It only uses upper tests,
which is why continuity and positivity of mu suffice and no regularity
theory for the supremum is needed. This is what makes the "diagonal"
case with repeated eigenvalues treatable at all.
Third, equality. If P = m identically, the conclusion is an equality.
Conversely the proof can be run backwards to see that equality forces
equality in the arithmetic-geometric mean step, hence all the
eigenvalues of rI - h A equal, which in the Clifford case is exactly
what happens. Section 7 only needs the inequality, so we do not push
this further.
==============================================================================
7. INDEX RIGIDITY: INDEX m+3 FORCES A CLIFFORD PRODUCT
==============================================================================
Continue with Sigma^m a smooth closed connected embedded minimal
hypersurface of S^(m+1), m >= 2, not an equator. Recall from Section 3:
L = Delta + m + P,
the Jacobi operator, with ind(Sigma) the number of positive
eigenvalues of L counted with multiplicity.
LEMMA 7.1 (the normal coordinates are independent).
The map a -> <a,N> from R^(m+2) to smooth functions on Sigma is
injective. Consequently, the eigenvalue m of L has multiplicity at
least m+2.
Proof. Suppose a is nonzero and <a,N> vanishes identically. Set
f = <a,x> and compute the Hessian of f on Sigma. Two standard facts
combine.
First, on the unit sphere itself the function <a,x> is a first
spherical harmonic:
Hess_S <a,x> (v,w) = -<v,w> <a,x> (v, w tangent to S^(m+1)).
This is the vector identity Hess_S x = -x g, contracted with a.
Second, if i : Sigma -> S^(m+1) is the inclusion with second
fundamental form II(v,w) = A(v,w) N, then for the composed function
f = <a,x> of i,
Hess_Sigma f (v,w) = Hess_S (f)(v,w) - df(II(v,w))
= -<v,w> <a,x> - A(v,w) <a,N>.
The correction term is the usual one: following a test curve in the
ambient sphere instead of inside Sigma costs the normal component
II(v,w), and that is exactly what a Hessian comparison along an
inclusion measures. Under the assumption <a,N> = 0 this collapses to
Hess f (v,w) = -f <v,w> (7.1)
at every point of Sigma. Equation (7.1) is Obata's equation.
The function f is not constant. If f were the constant c, then (7.1)
would give -c <v,w> = 0 for all tangent v, w, so c = 0, and then
Sigma would lie in the hyperplane {<a,x> = 0}, contradicting fullness
(Lemma 4.1(b)). So f is nonconstant, and it solves (7.1) on the
complete connected Riemannian manifold (Sigma,g). Obata's theorem
(Obata 1962, Theorem A) then says that (Sigma,g) is isometric to the
unit round sphere S^m. But the Gauss equation gives the scalar
curvature of Sigma,
Scal = m(m-1) - P,
whereas the unit round S^m has scalar curvature m(m-1) exactly.
Therefore P = 0 on Sigma, so A vanishes identically and Sigma is an
equator. That contradicts the hypothesis. Hence no nonzero a satisfies
<a,N> = 0.
For the consequence: by Section 3, L N = m N. If the m+2 coordinate
functions N_alpha of N were linearly dependent, some nonzero a would
satisfy sum_a a_alpha N_alpha = <a,N> = 0, which we just excluded. So
they are independent, and m is an eigenvalue of L with multiplicity at
least m+2. QED
LEMMA 7.2 (the top eigenvalue is strictly larger than m).
Let lambda_1 be the largest eigenvalue of L. Then L has a largest
eigenvalue; it is simple; it has an eigenfunction which is strictly
positive; and
lambda_1 >= m + (1/|Sigma|) * integral of P dV > m.
Proof. The operator L is self-adjoint on Sigma with a discrete
spectrum bounded above, since its quadratic form is at most
max(m + P) times the L^2 norm. So a largest eigenvalue lambda_1
exists. Let K > 1 + max(m+P) and consider the operator K - L, a
Schrodinger operator with a strictly positive quadratic form plus a
positive potential. Its principal eigenvalue is simple and has a
strictly positive eigenfunction: this is the first-eigenfunction
theorem, and it follows from the strong maximum principle applied to
the equation (K-L)u = lambda u with lambda the principal eigenvalue.
Translating back, lambda_1 is simple with a positive eigenfunction.
For the inequality, test the Rayleigh quotient with the constant
function 1. Since the largest eigenvalue maximizes the Rayleigh
quotient over all nonzero functions,
lambda_1 >= (integral of 1 . L 1) / (integral of 1)
= (1/|Sigma|) * integral of (m + P) dV
= m + (1/|Sigma|) * integral of P dV.
By Theorem 6.1 the last integral is at least m|Sigma|, and in fact the
argument below only needs it to be positive. So
lambda_1 > m. QED
PROPOSITION 7.3 (index bound).
ind(Sigma) >= m+3.
Proof. Lemma 7.1 gives m+2 independent eigenfunctions with eigenvalue
m > 0. Lemma 7.2 gives one more positive eigenvalue lambda_1, larger
than m and hence not in the span of those. All of these are positive
eigenvalues, so ind(Sigma) >= m+3. QED
The rest of Section 7 is about the equality case: which hypersurfaces
have index exactly m+3? The answer is: exactly the Clifford products,
up to rigid motions of the sphere. This is where Theorem 6.1 is used
twice, once as an inequality and once through its equality case.
==============================================================================
7b. CONFORMAL MOVES OF THE AMBIENT SPHERE AND BALANCING
==============================================================================
We need one more tool: a family of ambient conformal maps of the sphere
which can be used to "center" a hypersurface.
For a in S^(m+1) and tau >= 0 put C = cosh tau and S = sinh tau, and
define
T_{tau,a}(X) = [ X + ((C-1)<a,X> - S) a ] / (C - S<a,X>). (7.2)
The denominator is positive: C - S<a,X> >= C - S = e^(-tau) > 0.
LEMMA 7.8 (conformality and length factor).
T_{tau,a} is a smooth diffeomorphism of S^(m+1) onto itself, with
T_{0,a} = identity. For W tangent to the sphere at X,
dT_{tau,a}(W) = beta ( BW + S<a,W> T_{tau,a}(X) ), beta = (C - S<a,X>)^(-1),
where B = I + (C-1)aa^t is symmetric with B a = C a and B = I on
a^perp. Consequently |dT(W)| = beta |W|: the map is conformal with
length factor beta. The maps form the connected group generated by the
"boosts" T_{tau,a} with the orthogonal transformations of the sphere.
Proof. First check that T maps the sphere to itself. Write X =
<a,X> a + X_perp, so X_perp is perpendicular to a. Then
T(X) = (X_perp + (C<a,X> - S) a)/(C - S<a,X>).
The squared norm is
|X_perp|^2 + (C<a,X> - S)^2 = 1 - <a,X>^2 + C^2<a,X>^2 - 2CS<a,X> + S^2
= (C^2 - 1)<a,X>^2 - 2CS<a,X> + (1 + S^2),
using C^2 - S^2 = 1. On the other hand,
(C - S<a,X>)^2 = C^2 - 2CS<a,X> + S^2<a,X>^2
= 1 + S^2 + (C^2-1)<a,X>^2 - 2CS<a,X>,
which is the same expression. So |T(X)| = 1, and T maps the sphere to
itself. Since T_{-tau,a} is an inverse, T is a diffeomorphism.
For the differential, differentiate (7.2) along a curve X(t) with
X'(0) = W. The denominator differentiates to -S<a,W>, the numerator to
W + (C-1)<a,W>a, and (7.2)'s quotient rule gives
dT_{tau,a}(W) = (C-S<a,X>)^(-1) ( W + (C-1)<a,W>a )
+ (S<a,W>/(C-S<a,X>)^2) ( X + ((C-1)<a,X>-S)a ).
The bracket in the second line is exactly (C - S<a,X>) T(X). Collecting
the two terms with (C-S<a,X>)^(-1),
dT(W) = beta ( W + (C-1)<a,W> a + S<a,W> T(X) ),
which is the displayed formula because B W = W + (C-1)<a,W>a.
Now the length. This is a pleasant computation in the decomposition
X = f a + X_perp with f = <a,X> and X_perp perpendicular to a. Write a
tangent vector as W = u a + W_perp with u = <a,W>. Since W is tangent
to the sphere at X, <W,X> = 0, that is,
u f + <W_perp, X_perp> = 0. (7.3a)
Also |X_perp|^2 = 1 - f^2 and |W|^2 = u^2 + |W_perp|^2. Put D = C - S f,
so beta = 1/D, and note the identity
C + S(Cf - S)/D = (CD + SCf - S^2)/D = (C^2 - S^2)/D = 1/D, (7.3b)
which is where C^2 - S^2 = 1 gets used.
Differentiate the decomposition T = (X_perp + (Cf - S)a)/D. The
derivative of f is u, and
dT(W) = [W_perp + Cu a]/D + (Su/D^2)[X_perp + (Cf - S)a]
= (1/D)[ W_perp + (Su/D) X_perp + (u/D) a ],
with (7.3b) collapsing the a-components. Take squared lengths, using
that W_perp and X_perp are both perpendicular to a:
|dT(W)|^2 = (1/D^2)[ |W|^2 - u^2 + 2(Su/D)(-uf) + (S^2u^2/D^2)(1-f^2) ]
+ u^2/D^4,
because |W_perp|^2 = |W|^2 - u^2 and <W_perp,X_perp> = -uf by (7.3a).
So |dT(W)|^2 = |W|^2/D^2 holds if and only if
-D^2 - 2SfD + S^2(1-f^2) + 1 = 0,
which is what remains after multiplying by D^2 and dividing by u^2.
Expand with D = C - Sf: D^2 = C^2 - 2CSf + S^2f^2 and
2SfD = 2CSf - 2S^2f^2, so
-D^2 - 2SfD + S^2 - S^2f^2 + 1
= -C^2 + 2CSf - S^2f^2 - 2CSf + 2S^2f^2 + S^2 - S^2f^2 + 1
= -C^2 + S^2 + 1 = 0.
Therefore |dT(W)| = |W|/D = beta |W|, and T is conformal with factor
beta. QED
The reason to care: the area of the image T(Sigma) is
integral over Sigma of beta^m dV, because the pullback of the area form
of an m-surface under an ambient conformal map with factor beta
multiplies the area element by beta^m.
LEMMA 7.9 (balancing).
Let u be any smooth strictly positive function on Sigma. Then some T
of the form (7.2) satisfies
integral over Sigma of u T(x) dV = 0. (7.5)
Proof. Parametrize the maps by z = (tanh tau) a in the open unit ball
B of R^(m+2), with z = 0 giving the identity, and set
F(z) = ( integral of u T_z(x) dV ) / ( integral of u dV ),
a continuous map from the open ball to R^(m+2). We extend it to the
closed ball. As z tends to a boundary point a_*, that is, as tau tends to infinity
with a = a_*, divide the numerator and the denominator of (7.2) by
C = cosh tau. The numerator tends to (<a,X> - 1)a and the denominator
to 1 - <a,X>. Hence
T_{tau,a}(X) -> -a for every X on the sphere with X != a;
the convergence is uniform on compact subsets of the sphere minus the
point a, and the single exceptional point a carries no mass under
u dV. So F extends
continuously to the closed ball with
F(a_*) = -a_* for every a_* in S^(m+1).
Now suppose F has no zero. Then G(z) = F(z)/|F(z)| is a continuous map
from the closed ball to S^(m+1) whose restriction to the boundary
sphere is the antipodal map a_* -> -a_*. But the boundary restriction
of a map defined on the whole ball is null-homotopic, whereas the
antipodal map of S^(m+1) has degree (-1)^(m+2) and is therefore not
null-homotopic. Contradiction. So F has a zero z inside the open ball,
and the corresponding T satisfies (7.5). QED
Because u > 0, condition (7.5) is exactly the statement that the image
hypersurface T(Sigma) has centroid zero in the ambient Euclidean space.
So every hypersurface can be conformally moved to have centroid zero,
and the price paid is measured by the conformal factor.
LEMMA 7.10 (conformal moves strictly decrease area).
Fix a in S^(m+1) and let f = <a,x>, eta = <a,N>, beta = (C - S f)^(-1)
as in (7.2). Then for every tau >= 0,
integral over Sigma of beta^m dV <= |Sigma|, (7.6)
with strict inequality when tau > 0. Consequently also
integral over Sigma of beta^2 dV <= |Sigma|, (7.7)
again with strict inequality when tau > 0.
Proof. Every point of Sigma satisfies the orthogonal decomposition
a = f x + (grad f) + eta N,
because x is a unit normal of Sigma and the normal N is perpendicular
to it. Squaring and using |a| = 1,
|grad f|^2 = 1 - f^2 - eta^2. (7.8)
Also, differentiating beta = (C - S f)^(-1),
grad beta = S beta^2 grad f. (7.9)
The whole lemma follows from one divergence identity. By Lemma 3.1(i),
Delta f = -m f. Therefore, using (7.8) and (7.9),
div( beta^m grad f )
= beta^m Delta f + m beta^(m-1) <grad beta, grad f>
= -m f beta^m + m S beta^(m+1) |grad f|^2
= m beta^(m+1) ( S(1 - f^2 - eta^2) - f(C - S f) )
= m beta^(m+1) ( S - C f - S eta^2 ),
where the last step expands f/beta = f(C - Sf) = Cf - Sf^2 and cancels
S f^2 against S f^2. Sigma is closed, so the integral of a divergence
vanishes:
0 = integral of beta^(m+1) ( S - C f - S eta^2 ) dV,
that is,
integral of beta^(m+1) (C f - S) dV = - S * integral of beta^(m+1) eta^2 dV. (7.10)
Now differentiate (7.6)'s left side with respect to tau. Since
dC/dtau = S and dS/dtau = C,
d beta / d tau = -(C - S f)^(-2) (S - C f) = beta^2 (C f - S).
Hence, using (7.10) in the second step,
d/dtau ( integral of beta^m dV )
= m * integral of beta^(m-1) beta^2 (C f - S) dV
= m * integral of beta^(m+1) (C f - S) dV
= - m S * integral of beta^(m+1) eta^2 dV. (7.11)
This is an exact identity, not an inequality. Now: eta = <a,N> is not
identically zero, by Lemma 7.1, since a is a unit vector. So the right
side of (7.11) is strictly negative for every tau > 0. Since
beta = 1 when tau = 0, integrating from 0 to tau gives (7.6) with
strict inequality.
For (7.7), apply Holder's inequality with exponents m/2 and
m/(m-2), and m/(m-2) = infinity when m = 2:
integral of beta^2
<= ( integral of beta^m )^(2/m) * ( integral of 1 )^(1 - 2/m)
<= |Sigma|^(2/m) * |Sigma|^(1 - 2/m) = |Sigma|.
When m = 2 the first inequality is an equality and the second one is
(7.6) itself, which is strict for tau > 0; for m > 2 the second
inequality is strict as well. QED
Interpretation. The map T_{tau,a} pulls the round metric of the
ambient sphere back to a conformally rescaled metric on Sigma, and
(7.6) says the resulting m-volume is strictly smaller than |Sigma| for
every nontrivial conformal move. So the round position is the unique
maximum of area in this family of conformal moves, and the loss is
strict. That strictness is what turns the inequality of the
average-curvature theorem into an equality statement about area, in
the next two lemmas.
==============================================================================
7c. THE EQUALITY CASE: INDEX m+3 FORCES A CLIFFORD PRODUCT
==============================================================================
PROPOSITION 7.6. Let Sigma^m be a smooth closed connected embedded
minimal hypersurface of S^(m+1), m >= 2, not an equator, with
ind(Sigma) = m+3. Then, up to an orthogonal change of coordinates in
R^(m+2), the image of Sigma is the Clifford product
S^k(sqrt(k/m)) x S^(m-k)(sqrt((m-k)/m))
for some k with 1 <= k <= m-1. In particular its volume is at least a_m.
Proof. Step 1, the spectral consequence. By Lemmas 7.1 and 7.2 the
positive spectrum of L consists of the simple eigenvalue lambda_1 > m,
with a strictly positive eigenfunction u, together with the eigenvalue
m with multiplicity m+2, and nothing else; otherwise the index would
exceed m+3. Therefore
the eigenvalues of L on u^perp are all at most m. (7.12)
Consequently, for every smooth function phi with
integral over Sigma of u phi dV = 0,
we have, expanding phi in the orthonormal basis of eigenfunctions of L
lying in u^perp,
integral of phi L phi dV <= m * integral of phi^2 dV. (7.13)
Step 2, apply this to a conformally moved hypersurface. Let T be the
map given by Lemma 7.9 for the weight u, so that
integral of u T(x) dV = 0 in R^(m+2),
and write v = T(x) for the moved position vector, with components
v_1,...,v_{m+2}. Each component satisfies the orthogonality in (7.13),
so summing (7.13) over the m+2 components,
integral of <v, L v> dV <= m * integral of |v|^2 dV = m|Sigma|, (7.14)
because |v| = 1. Now compute the left side. Since L = Delta + m + P and
|v|^2 = 1,
<v, L v> = sum_a v_a Delta v_a + (m+P) |v|^2
= - |grad v|^2 + (m+P),
where |grad v|^2 = sum_a |grad v_a|^2. The map T is conformal with
factor beta, so the pullback of the round metric of the ambient sphere
is beta^2 times the metric of Sigma, and therefore
|grad v|^2 = m beta^2. (7.15)
(For an isometric embedding of an m-manifold into a Euclidean space the
sum of the squared gradients of the m+2 coordinate functions equals m;
undoing the conformal factor beta multiplies that sum by beta^2.)
Substituting (7.15) into (7.14),
m|Sigma| + integral of P dV - m * integral of beta^2 dV <= m|Sigma|,
that is,
integral of P dV <= m * integral of beta^2 dV. (7.16)
Step 3, close the chain. Theorem 6.1 says integral of P >= m|Sigma|,
and Lemma 7.10 says integral of beta^2 <= |Sigma|. Together with (7.16),
m|Sigma| <= integral of P <= m * integral of beta^2 <= m|Sigma|. (7.17)
So all three quantities are equal. Three consequences.
First, integral of beta^2 = |Sigma|, and by the strict part of Lemma 7.10
this forces tau = 0, that is, T is the identity. So v = x.
Second, integral of P = m|Sigma|, so the average-curvature inequality
of Theorem 6.1 is an equality.
Third, and most importantly, each individual inequality of (7.13)
must be an equality. Indeed, with v = x, the left side of (7.14)
equals m|Sigma| + integral of P - m|Sigma| = integral of P = m|Sigma|,
which is also the right side; and each summand is at most m|Sigma| by
(7.13). So every defect
d_a = m * integral of x_a^2 dV - integral of x_a L x_a dV
is nonnegative and their sum is zero. Hence d_a = 0 for every a.
Step 4, deduce P = m. Equality in (7.13) for phi = x_a, whose expansion
in eigenfunctions has all eigenvalues at most m, means that the
component of x_a in each eigenspace with eigenvalue strictly less than
m vanishes. So x_a lies in the eigenspace for the eigenvalue m:
L x_a = m x_a.
On the other hand, by Lemma 3.1(i),
L x_a = Delta x_a + (m + P) x_a = -m x_a + (m+P) x_a = P x_a.
Comparing, P x_a = m x_a for each a, that is, P x = m x as vectors.
Taking the inner product with x and using |x| = 1,
P = m everywhere on Sigma. (7.18)
Step 5, A is parallel. Insert (7.18) into Simons' identity (iv) of
Lemma 3.1:
(1/2) Delta P = |grad A|^2 + (m - P) P = |grad A|^2 + 0.
The left side vanishes because P is constant, so |grad A|^2 = 0 and
therefore
grad A = 0 (7.19)
at every point. The second fundamental form is parallel. QED (for the
first part)
What remains is a classification: which minimal hypersurfaces of the
sphere have parallel, nonvanishing second fundamental form? The answer
is a short and satisfying computation, and it is the classical
classification of Simons.
==============================================================================
7d. SIMONS' CLASSIFICATION: PARALLEL A MEANS CLIFFORD
==============================================================================
Continue the proof of Proposition 7.6. We have P = m and grad A = 0.
Step 6, constant eigenvalues and parallel eigendistributions. Since A is
parallel, its eigenvalues along Sigma are constant: differentiate
A V_i = lambda_i V_i in a direction w, with V_i a local unit
eigenfield, and use grad A = 0 to get A(nabla_w V_i) = lambda_i
nabla_w V_i. So the covariant derivative of an eigenfield stays in the
same eigenspace, and the eigenvalues cannot vary along a connected
Sigma. Write the distinct eigenvalues as lambda, with eigenspaces
E_lambda, and note that each E_lambda is a parallel subbundle.
Step 7, the curvature preserves the eigenspaces. The Ricci identity for
a self-adjoint endomorphism A reads, in the notation of Simon's paper,
nabla_X nabla_Y A - nabla_Y nabla_X A = R(X,Y) A - A R(X,Y),
where the right side is the commutator of R(X,Y) with A acting on
tangent vectors. Both left side terms vanish because A is parallel, so
R(X,Y) A = A R(X,Y) for all tangent X, Y. (7.20)
Therefore R(X,Y) maps every eigenspace E_lambda into itself. Now take
unit U in E_lambda and unit W in E_kappa with lambda != kappa. Since W
lies in E_kappa and R(U,W) preserves E_kappa, the vector R(U,W)W lies
in E_kappa, and it is therefore orthogonal to U:
<R(U,W)W, U> = 0. (7.21)
Step 8, the Gauss equation determines the eigenvalues. Insert (7.21)
into the Gauss equation of Lemma 3.1, which for orthonormal U, W and
principal directions reads
<R(U,W)W, U> = 1 + A(U,U)A(W,W) - A(U,W)^2.
Since A(U,W) = 0 for eigenvectors with different eigenvalues,
0 = 1 + lambda kappa, that is lambda kappa = -1, (7.22)
for every pair of distinct eigenvalues. There cannot be three distinct
eigenvalues lambda, kappa, mu: (7.22) applied twice would give
kappa = mu = -1/lambda. So there are at most two. There cannot be
exactly one either, because A is not zero (Sigma is not an equator) and
trace A = 0. Hence there are exactly two eigenvalues, of opposite
signs, satisfying lambda kappa = -1.
Step 9, the two values. Let k be the multiplicity of the positive
eigenvalue lambda and l = m-k the multiplicity of kappa. Since
trace A = 0,
k lambda + l kappa = 0, kappa = -1/lambda,
so k lambda = l/lambda, lambda^2 = l/k, and
lambda = sqrt(l/k), kappa = -sqrt(k/l). (7.23)
Both k and l are positive, so 1 <= k <= m-1.
Step 10, the hypersurface is a product of spheres. Consider the two
affine subspaces of R^(m+2)
V_+ = E_lambda(x) + R( lambda N(x) - x ), dimension k+1,
V_- = E_kappa(x) + R( kappa N(x) - x ), dimension l+1.
Their dimensions sum to m+2, and they are orthogonal: since x and N
are unit vectors with <x,N> = 0,
< lambda N - x, kappa N - x > = lambda kappa + 1 = 0
by (7.22), and E_lambda, E_kappa are tangent, hence orthogonal to both
x and N, and orthogonal to each other. So V_+ and V_- span R^(m+2)
orthogonally at each point. In particular x has a component in each.
One must check that V_+ and V_- do not move with x; this is the point
where parallelism of A is used. Let V be a local section of E_lambda
and w a tangent vector. Because D_w x = w and D_w N = -A w,
D_w V = (nabla_w V) - <V,w> x + A(w,V) N
= (nabla_w V) - <V,w> x + lambda <V,w> N,
where the second equality uses A(w,V) = lambda <w,V>, and where the
term -<V,w>x comes from differentiating <V,x> = 0. Since
nabla_w V lies in E_lambda, this shows D_w V lies in V_+. Similarly
D_w (lambda N - x) = -(lambda A + I) w,
which lies in E_lambda when w is in E_lambda, because
lambda A + I = (lambda^2 + 1)I there, and which vanishes when w is in
E_kappa, because lambda A + I = lambda kappa + 1 = 0 there. So the
ambient derivative of every generator of V_+ stays in V_+, and the same
holds for V_- by the symmetric computation with kappa. Since Sigma is
connected, V_+ and V_- are therefore fixed linear subspaces of R^(m+2),
independent of the point x.
Step 11, the constant lengths of the projections. Let pi_+ and pi_-
denote the orthogonal projections onto these fixed subspaces. Since
E_lambda and E_kappa are tangent, and x and N are perpendicular to
them, the V_+-component of x is its component along lambda N - x:
pi_+ x = < x, y > y, y = (lambda N - x)/sqrt(1 + lambda^2),
where |y| = 1 because <lambda N - x, lambda N - x> = lambda^2 + 1.
And <x, lambda N - x> = -1, so <x,y> = -1/sqrt(1+lambda^2) and
|pi_+ x| = 1/sqrt(1+lambda^2) = 1/sqrt(1 + l/k) = sqrt(k/m). (7.24)
In the same way, |pi_- x| = sqrt(l/m).
Step 12, conclusion. Equations (7.24) say the image of Sigma lies in
the intersection of the unit sphere with the two quadrics
|pi_+ X| = sqrt(k/m) and |pi_- X| = sqrt(l/m), and that intersection is
exactly the Clifford product of Step 9. Both sets are smooth closed
connected m-dimensional submanifolds, so Sigma is an open subset of the
Clifford product; being compact, it is also closed; the Clifford
product is connected because k, l >= 1; hence Sigma is the whole
product. Its volume is therefore
(k/m)^(k/2) ((m-k)/m)^((m-k)/2) sigma_k sigma_(m-k) >= a_m,
the inequality being the definition of a_m as a minimum. This
completes the proof of Proposition 7.6. QED
The classification is exact, and it is the classical rigidity theorem of
Simons: every closed embedded minimal hypersurface of the unit sphere
with identically parallel second fundamental form is an equator or a
Clifford product.
COROLLARY 7.7 (volume below a_m raises the index).
Let Sigma^m be as in Proposition 7.6 and suppose its volume is less
than a_m. Then
ind(Sigma) > m+3.
Proof. By Proposition 7.3, ind(Sigma) >= m+3. Suppose ind(Sigma) = m+3.
By Proposition 7.6 the image of Sigma is a Clifford product, whose
volume is at least a_m; that contradicts the assumption vol < a_m. So
ind(Sigma) > m+3. QED
This corollary is the reason index appears in the proof of the volume
theorem. A counterexample of least volume will have index strictly
greater than m+3, and Section 9 will convert each index direction
beyond p = m+3 into a strict decrease of area.
==============================================================================
8. DETECTION: FORCING A HALF-VOLUME, CENTERED SURFACE
==============================================================================
Sections 8, 9, 10 are the three engines of the proof. This section is
the topological one, and its output is a lower bound: any family of
surfaces that "winds around" in the right way must contain a surface of
perimeter at least a fixed constant gamma_m, which is strictly bigger
than the perimeter of an equator.
Throughout, m >= 2, n = m+1 is the dimension of the ambient sphere,
p = m+3 is the dimension of the parameter space (why p = m+3 and not
something smaller will become clear in Section 9), and all chains have
coefficients in Z_2 = {0,1}.
8a. Chains, fillings, and the covering of the cycle space
A set E of finite perimeter in S^n determines a mod-2 chain [E], and
its boundary cycle d[E] is an integral (n-1 = m)-cycle with
mass(d[E]) = Per(E),
the perimeter being computed in the sphere. Conversely, every cycle in
the flat path component of the zero cycle bounds at least one set of
finite perimeter. The proof is the isoperimetric choice: for two cycles
T, T' that are flat-close, there is a filling R with dR = T - T' and
mass(R) <= c * flatdist(T, T'), (8.1)
with c a constant depending only on the sphere; this is the
small-filling theorem of Marques-Neves (2017), section 3.1. Chaining
finitely many such fillings along a flat-continuous path from the zero
cycle to T produces one filling of T. Write Z for this path component;
its elements are the flat m-boundaries in S^n.
LEMMA 8.1 (the filling cover).
The boundary map
pi : { sets of finite perimeter in S^n, mod null sets }
-> Z, pi(E) = d[E],
is a two-sheeted covering map. Its deck involution is complementation
E -> E^c = S^n minus E.
Proof. It is a map to Z, because [E] is a chain and d[E] is a flat
cycle, and it is onto, as just recalled. Two sets have the same
boundary iff E and F differ by a set with zero perimeter, since
d[E] - d[F] = d[E Δ F] with Δ the symmetric difference. A set of
finite perimeter in a connected manifold with zero perimeter has
constant characteristic function: its distributional gradient vanishes,
so it is constant almost everywhere, and the only constants are 0 and
1. Hence F = E or F = E^c almost everywhere. So every fiber of pi has
at most two points, and those are complementary sets.
It remains to see that pi is a local homeomorphism onto Z. Metrize the
fillings by d(E,F) = volume(E Δ F), for which
flatdist(d[E], d[F]) <= d(E,F), so pi is continuous. Fix T = d[E] and
choose delta > 0 so small that 2 delta < the small-filling threshold in
(8.1) and 4 c delta < sigma_(m+1). For a cycle T' that is flat
delta-close to T, the small filling R(T,T') is uniquely determined:
two choices would differ by a set with zero boundary, hence by nothing
or by the whole sphere, and their total mass is below sigma_(m+1).
Define
[E_{T'}] = [E] + R(T,T').
Then E_{T'} is a filling of T' of finite perimeter, and for T', T''
both delta-close to T, the cycle R(T,T') + R(T,T'') + R(T',T'') has
mass at most 4c delta < sigma_(m+1) and zero boundary, so it is zero
and
d(E_{T'}, E_{T''}) <= c * flatdist(T', T'').
So T' -> E_{T'} is a continuous branch of the inverse, and the same
with E^c gives the other branch. The two branches are disjoint open
sets in the preimage, distinguished by whether volume(E_{T'} Δ E) is below
or above sigma_(m+1)/2. QED
The covering pi has exactly two sheets, so it defines a cohomology
class
lambda in H^1(Z; Z_2),
by the rule: lambda of a loop in Z equals 1 when the lifted filling
path ends at the complement of where it started, and 0 otherwise. This
is the standard correspondence between two-sheeted covers, cohomology
with Z_2 coefficients, and homomorphisms from the fundamental group to
Z_2. Concretely, lambda is the obstruction to choosing fillings
continuously and consistently around loops.
DEFINITION 8.2 (detecting).
Let X be a finite CW complex and Phi : X -> Z a flat-continuous map.
Phi is DETECTING if
(Phi^* lambda)^p is not zero in H^p(X; Z_2), (8.2)
where p = m+3. Phi has NO CONCENTRATION OF MASS if
lim over r to 0 of sup over x in X and z in S^n of
||Phi(x)||(B_r(z)) = 0, (8.3)
that is, if the mass of the cycles Phi(x) is uniformly small in small
balls, uniformly in x.
Detection is invariant under flat homotopies, since homotopic maps
induce the same map in cohomology.
The natural example is a projective family, and the next lemma records
exactly what has to be checked for the family of Section 9.
LEMMA 8.3 (projective criterion).
Let Q be the closed unit ball of R^p, and consider maps
q -> E_q from Q to sets of finite perimeter such that
q -> volume(E_q Δ E_{q'}) is small when q' is close to q,
E_{-q} = E_q^c for every q on the boundary sphere of Q.
Then q -> d[E_q] descends to a map on Q/(q ~ -q) = RP^p, and it is
flat continuous there.
If in addition E_q depends on q through a family of level sets whose
fillings satisfy the hypotheses above and which are continuous in
volume, then the descended map is detecting.
Proof. Two complements have the same boundary, d[E_q^c] = d[E_q],
because a chain and its complement differ by the fundamental class of
the closed sphere, which is a cycle. So the formula defines a map on
the quotient, and flat continuity follows from (8.1), since flat
distance is bounded by the volume of a symmetric difference.
For detection, take any q on the boundary sphere of Q. The straight
segment s -> s q, s in [-1,1], joins -q to q, and the two endpoints are
identified in the quotient: so this segment becomes a closed loop in
RP^p, based at the class of q. Along this loop, the filling goes from
E_{-q} = E_q^c at s = -1 to E_q at s = 1. The projected cycles
d[E_{sq}] therefore form a loop in Z whose lift in the filling cover
starts at E_q^c and ends at E_q, which is the other sheet over the same
cycle. So lambda of this loop is 1. Therefore Phi^*lambda is a nonzero
element of H^1(RP^p; Z_2), and since
H^*(RP^p ; Z_2) = Z_2[alpha]/(alpha^(p+1)), |alpha| = 1,
with alpha the nonzero class in degree one (Hatcher 2002, Theorem
3.19), Phi^*lambda = alpha. Hence
(Phi^* lambda)^p = alpha^p is not zero in H^p(RP^p; Z_2) = Z_2.
So the descended map is detecting. QED
Remark on the shape of Q. The parameter space is a ball times an
interval, and only the boundary of the ball is subject to the
antipodal identification. The loop used above is diametral in the ball,
and the interval direction is not involved. Nothing in the argument
needs the parameter complex to have dimension exactly p; what matters
is that the detection class survives, which is a statement about the
cohomology ring of the parameter space and not about the geometry of
the family.
8b. The detection gap
The point of detection is that it forbids the family from being made of
small surfaces. Here is the precise statement.
LEMMA 8.4 (detection gap).
There is a constant gamma_m > sigma_m, depending only on m, such that
every detecting map Phi : X -> Z satisfies
sup over x in X of mass(Phi(x)) >= gamma_m. (8.4)
Proof. Step 1, the constrained isoperimetric problem. Consider the
class of sets of finite perimeter that have exactly half the volume of
the sphere and centroid zero:
A_m = { E : volume(E) = sigma_(m+1)/2 and integral over E of
y dVol(y) = 0 }. (8.5)
It is nonempty: the union of two antipodal spherical caps, with the
common radius chosen so that the total volume is half of sigma_(m+1),
has centroid zero by symmetry. Define
gamma_m = inf { Per(E) : E in A_m }. (8.6)
This infimum is attained. Take a minimizing sequence E_j. Their
perimeters are bounded, so compactness for sets of finite perimeter on
the compact sphere gives a subsequence converging in L^1 to a set E of
finite perimeter. Both the volume and every centroid coordinate are
continuous under L^1 convergence, because the sphere is bounded, so E
still has half volume and zero centroid; and the perimeter is lower
semicontinuous, so Per(E) <= liminf Per(E_j) = gamma_m. Hence E is a
minimizer, and gamma_m = Per(E) for that E.
Step 2, the strict inequality. The isoperimetric inequality on the
sphere says that a set of half volume has perimeter at least sigma_m,
with equality only for a hemisphere (Boge-lein, Duzaar, Fusco 2017,
section 2.3, equation (2.2)). A hemisphere has nonzero centroid:
slicing by the coordinate along its axis,
integral over {y . a > 0} of y dVol(y) = omega_m a, (8.7)
where omega_m = sigma_(m-1)/m is the volume of the unit m-ball. (Check:
slice the sphere at height t along a; the slice is a round sphere of
radius sqrt(1-t^2) and dimension m-1, and the t-integral of t times its
volume gives sigma_(m-1)/m. For m = 1 the hemisphere is a semicircle
and the integral is 2 = omega_1; for m = 2 it is pi = omega_2.) So the
minimizer of (8.6), which must have zero centroid, is not a hemisphere,
and therefore
gamma_m > sigma_m. (8.8)
This is the only numerical input of the section, and it is a strict
inequality, which is what makes the min-max output land strictly above
the equator.
Step 3, detection forces a member of A_m. Now let Phi : X -> Z be
detecting. Pull the covering of Lemma 8.1 back to X: the total space is
X~ = { (x, E) : x in X, E of finite perimeter, d[E] = Phi(x) },
with the obvious topology, and the projection X~ -> X is a two-sheeted
covering with deck transformation (x,E) -> (x,E^c). On X~ define
G(x,E) = ( volume(E) - sigma_(m+1)/2 , integral over E of y dVol(y) )
a continuous map into R x R^(m+2) = R^p. Because a set and its
complement have complementary volumes and complementary centroids
(the total centroid of the sphere is zero),
G(x, E^c) = - G(x, E). (8.9)
Suppose G has no zero. Then N = G/|G| maps X~ continuously to S^(p-1)
and satisfies N(x,E^c) = -N(x,E), so it descends to a continuous map
f : X -> RP^(p-1).
Moreover the covering X~ -> X is exactly the pullback under f of the
antipodal covering S^(p-1) -> RP^(p-1): fiberwise, the two points E and
E^c over x are sent to the two points N and -N over f(x), and this
identification is continuous. Two-sheeted coverings are classified by
their cohomology class in H^1(., Z_2), so
Phi^* lambda = f^* alpha_(p-1),
where alpha_(p-1) is the nonzero degree-one class of RP^(p-1). But
alpha_(p-1)^p = 0 in H^p(RP^(p-1); Z_2), because the cohomology ring of
RP^(p-1) is truncated at degree p-1. Hence
(Phi^* lambda)^p = f^* (alpha_(p-1)^p) = 0,
which contradicts the detecting property (8.2). Therefore G has a zero:
there are x in X and a filling E of the cycle Phi(x) with
volume(E) = sigma_(m+1)/2, centroid of E = 0,
that is, E belongs to A_m. For such an E, by the definition (8.6),
mass(Phi(x)) = Per(E) >= gamma_m.
So sup over x of mass(Phi(x)) >= gamma_m, and (8.4) is proved. QED
What this says in words: a family of surfaces that wraps around the
cycle space in a detecting way cannot have all its members of area below
gamma_m. The reason is that somewhere in the family the enclosed region
is exactly half of the sphere and balanced, and such a region cannot be
a hemisphere, so its boundary has perimeter strictly bigger than an
equator.
8c. The min-max theorem
The last piece of this section is the engine that converts a detecting
family into a minimal surface. Its proof is a long story - Almgren's
existence theory, Pitts' regularity, and the Marques-Neves
discretization - and none of that is repeated here. What matters is
that the hypotheses are exactly the ones Lemma 8.3 and Lemma 8.4
provide.
First, a bookkeeping lemma, so that the family of Section 9, which
lives on the projective space, satisfies the combinatorial hypothesis
of the min-max theorem.
LEMMA 8.5 (cubical replacement).
Let Phi : RP^p -> Z be a detecting map without concentration of mass
and with sup mass D. Then there is a finite cubical complex K and a
detecting map Psi : K -> Z without concentration of mass such that
sup over K of mass(Psi) = D. (8.10)
Proof. Embed RP^p into the space of real symmetric (p+1) x (p+1)
matrices by [u] -> u u^t with |u| = 1. Every such matrix has a simple
top eigenvalue 1, separated from the eigenvalues 0, and the map sending
a symmetric matrix with simple top eigenvalue to its top eigenline is
continuous on a neighbourhood U of the image. That map is a retraction
r : U -> RP^p. Choose a cubical grid in a large cube fine enough that
the union K of all closed cubes meeting the image of RP^p, together
with their faces, lies inside U. Then K is finite, and both
i : RP^p -> K (the inclusion) and r : K -> RP^p satisfy r o i = id.
Consequently r^* is injective on cohomology, so Psi = Phi o r detects
whenever Phi does, and r(K) = RP^p, so the set of cycles attained by
Psi is exactly the set attained by Phi. Both assertions follow. QED
The dimension of K may exceed p. That is harmless: in the theorem
below, the parameter complex enters only through its finiteness and its
detection class.
THEOREM 8.6 (Almgren-Pitts min-max, Marques-Neves form).
Let m >= 2 and p = m+3. Let X be a finite cubical complex and let
Phi : X -> Z be a flat-continuous detecting map (Definition 8.2) with
no concentration of mass and
D_0 = sup over X of mass(Phi(x)) < infinity.
Then there exists a stationary integral m-varifold V in S^(m+1) with
gamma_m <= mass(V) <= D_0. (8.11)
If 2 <= m <= 6, V may be chosen to be a finite sum of smooth closed
embedded minimal hypersurfaces of S^(m+1) with positive integer
multiplicities.
Sketch of the imported content, so that the shape of the argument is
visible. A discrete map on the vertices of a cubical subdivision has a
fineness, the largest mass of the difference of its values on adjacent
vertices. One considers the class of all homotopy sequences of discrete
maps that are flat homotopic to Phi with fineness tending to zero, and
the width L of that class, the infimum over sequences of the limsup of
the maximum vertex mass. Then:
(a) The discretization theorem (Marques-Neves 2017, Theorem 3.9)
produces a sequence in the class with fineness tending to zero and
limsup of the maximum vertex mass at most D_0. This uses exactly
the no-concentration hypothesis, which allows a fine
triangulation of the parameter space without losing mass.
(b) The Almgren extension theorem (ibid., Theorem 3.10) turns a fine
discrete map into a flat-continuous one, with a controlled
increase of mass proportional to the fineness. Hence the
extensions are detecting for all large indices, and the detection
gap Lemma 8.4 gives L >= gamma_m.
(c) Taking the given Phi as one competitor gives L <= D_0.
(d) The Almgren-Pitts existence theorem (ibid., Theorem 2.14) gives a
stationary integral varifold of mass L, almost minimizing in
annuli. That theorem has no restriction on the dimension of the
parameter complex.
(e) The regularity statement (ibid., Theorem 2.11, using Schoen-Simon
1981) upgrades that varifold to a sum of smooth closed embedded
minimal hypersurfaces when 2 <= m <= 6.
The two inequalities gamma_m <= L <= D_0 are the whole content. QED
Two remarks.
First, when m >= 7 the theorem still gives a stationary integral
varifold, and it may be singular. Section 10 handles that by an
induction on dimension which shows that a varifold below the Clifford
threshold is smooth anyway. So the dimension restriction in (e) is
temporary.
Second, there is no index requirement hidden in the hypotheses. The
parameter complex may be of any dimension; detection is a statement
about H^p, not about the index of the outcome. The index theory of
Section 7 is used only in Section 9, to deform the family itself. This
matters: the classical min-max results with index bounds assume
parameter complexes of dimension equal to the index one wants, whereas
here the deformation is performed directly on the family and the
parameter dimension stays p = m+3.
The engine is now assembled. Given a family that detects and has no
concentration, min-max returns a stationary integral hypersurface of
mass at least gamma_m and at most the largest mass in the family. To
prove the volume theorem, the plan is therefore:
take a counterexample Sigma of least area H;
construct a detecting family whose largest mass is at most area(Sigma);
use its index to deform the family slightly, strictly reducing the
largest mass to some D_1 < H;
run min-max: get a stationary integral varifold of mass in
[gamma_m, D_1) which is therefore smaller than H;
show it is a smooth hypersurface, hence a counterexample smaller
than H, a contradiction.
Sections 9 and 10 supply the second and fifth steps. Section 11 puts the
five steps together.
==============================================================================
9. THE LORENTZ-DISTANCE FAMILY
==============================================================================
This is the construction that turns an index direction into a decrease
of area. Throughout, Sigma^m is a fixed smooth closed connected embedded
minimal hypersurface of S^(m+1), m >= 2, not an equator. We write
|Sigma| its area, dmu its area measure, N its normal,
A = -dN its shape operator, C(Sigma) the Euclidean cone over it
in R^(m+2), that is, all points s x with s >= 0, x in Sigma.
The idea in one paragraph. Take the cone over Sigma. Its signed
distance function phi is a 1-Lipschitz function, homogeneous of degree
one, and it encodes Sigma as its zero level. Now look at the graph of
phi in a space with one extra "time" coordinate, and apply Lorentz
boosts, which preserve the quadratic form |z|^2 - h^2. Intersecting the
boosted graph with the unit spatial sphere produces, for each boost and
each level, a hypersurface of the sphere. The miracle is that the area
of every one of those hypersurfaces can be bounded by an integral over
the fixed source Sigma, because the boost formulas give an explicit
map from Sigma to the level set whose Jacobian is controlled by the
nonnegativity of r I - hA coming from the nearest-point condition.
9a. Signed distance to the cone
Let phi : R^(m+2) -> R be the signed Euclidean distance to C(Sigma),
positive on the side of the cone pointed to by N and vanishing on the
cone. It is homogeneous of degree one, phi(s z) = s phi(z) for s > 0,
and it is 1-Lipschitz.
Why 1-Lipschitz. For two points on the same side of the cone the claim
is the triangle inequality for distance to a closed set. For two points
on opposite sides, the segment joining them meets the cone, so the sum
of their distances to the cone is at most the length of that segment.
LEMMA 9.1 (cone contact).
There is a constant K < 1 with
|phi(z)| <= K |z| for every z in R^(m+2). (9.1)
For z != 0 let h = phi(z). Every closest point of C(Sigma) to z has
the form r_z x with x in Sigma and
r_z = sqrt(|z|^2 - h^2) > 0, z = r_z x + h N(x),
r_z I - h A(x) >= 0, (9.2)
the last inequality being an inequality of self-adjoint
endomorphisms of the tangent space T_x Sigma.
Proof. Define, for z in R^(m+2),
H(z) = max over x in Sigma of <z, x>.
First, H(z) > 0 for every z != 0. If H(z) <= 0 then <z,x> <= 0 at every
point of Sigma, and integrating against the area measure,
0 = <z, integral of x dmu> = integral of <z,x> dmu <= 0,
with equality only if <z,x> = 0 on a set of full measure, hence
everywhere since it is real analytic (Lemma 4.1(e)). That would put
Sigma in the hyperplane perpendicular to z, contradicting fullness
(Lemma 4.1(b)). So H(z) > 0.
Second, the distance to the cone. A point of the cone has the form
s x with s >= 0 and x in Sigma. For fixed x,
|z - s x|^2 = |z|^2 - 2 s <z,x> + s^2,
which is minimized over s >= 0 at s = max(<z,x>, 0), with value
|z|^2 - <z,x>^2. Therefore
dist(z, C(Sigma))^2 = |z|^2 - H(z)^2. (9.3)
By compactness of the unit sphere and continuity of H,
c = min over |z| = 1 of H(z) > 0,
and (9.3) together with homogeneity gives dist(z,C)^2 <=
(1 - c^2)|z|^2. This proves (9.1) with K = sqrt(1 - c^2) < 1, and it
also shows that the vertex is never a closest point of the cone to a
nonzero z, since H(z) > 0.
Third, the contact formulas. Let x maximize <z, .>, which exists by
compactness. The nearest cone point is s x with s = <z,x> = H(z) > 0.
The residual z - H(z)x is perpendicular to the tangent space of Sigma
at x, because x maximizes the restriction of the linear function <z,.>
to Sigma. Hence z - H(z)x is a multiple of N(x), say lambda N(x) with
|lambda| = dist(z,C) = |h|. Comparing with the definition of the signed
distance, lambda = h: the segment from z to its closest cone point does
not cross the cone, so z lies on the side of the cone determined by the
sign of h, which is the side of N(x). Therefore
z = r_z x + h N(x), r_z = H(z) = sqrt(|z|^2 - h^2).
Fourth, the second-order condition. Since x maximizes the restriction
of <z,.> to Sigma, the Hessian of that restriction at x is negative
semidefinite. By the computation in Lemma 3.1, with z = r_z x + h N(x),
Hess (<z, .>)(v,w) = -<z,x><v,w> + <z,N>A(v,w) = -r_z<v,w> + h A(v,w).
So -r_z I + h A <= 0 as a quadratic form on T_x Sigma, which is (9.2).
The argument uses only that x is a maximizing point, so it applies to
each closest point separately, and there may be several. QED
Two consequences worth naming. First, phi satisfies a strict
spacelike bound (9.1): the graph of phi over the unit sphere can never
be vertical. Second, at every contact the matrix r I - h A has
nonnegative eigenvalues, and its trace is exactly mr, because trace A =
0. That trace is fixed, and the arithmetic-geometric mean inequality
will convert it into a bound on the Jacobian.
9b. Lorentz boosts and the global graph
Parametrize the boost parameters by the open unit ball of R^(m+2). For
v = 0 take the identity. For v != 0 write v = tanh(tau) a with tau > 0
and |a| = 1, and set
C = cosh tau, S = sinh tau, B = I + (C - 1) a a^t. (9.4)
So B acts as multiplication by C on the line R a and as the identity on
a-perpendicular. All three quantities are smooth functions of
v, because
S a = C v and (C - 1) a a^t = C^2 v v^t/(C + 1),
which involve only v and |v|.
The LORENTZ BOOST with parameter v sends a point (y, b) of
R^(m+2) x R to the pair (z, h) given by
z = B y + b S a, h = C b + S <a, y>. (9.5)
Its defining property is the identity
|z|^2 - h^2 = |y|^2 - b^2. (9.6)
Indeed, expanding |By + bSa|^2 - (Cb + S<a,y>)^2 and using
B^2 = I + S^2 a a^t, <By, a> = C<a,y>, and C^2 - S^2 = 1, all the cross
terms cancel. So (9.5) is a Lorentz transformation: it preserves the
quadratic form of signature (m+2, 1), and in particular it preserves
the cone {|z|^2 - h^2 = 0} and the two "spacelike" regions.
The graph of phi is the set of pairs (z, h) with h = phi(z). Boosting
it produces a new surface, and intersecting that surface with the unit
spatial sphere gives a hypersurface of the sphere. The next lemma says
that every such intersection can be described as a graph over the whole
sphere.
LEMMA 9.2 (global graph and compactification).
Fix v in the open unit ball. Then the set
{ (y, b) : C b + S <a, y> = phi(B y + b S a) } (9.7)
is the graph of a continuous function b = psi_v(y) defined on all of
R^(m+2), and psi_v depends continuously on (v, y). On the unit sphere
one has |psi_v| < 1, and the functions extend jointly continuously to
the closed ball of R^(m+2) times the unit sphere, with boundary values
psi_a(y) = - <a, y> for |a| = 1. (9.8)
Proof. Fix v and y and consider the one-variable function
F(b) = C b + S <a, y> - phi(B y + b S a).
It is continuous. When b_2 > b_1, the two arguments of phi differ by
(b_2 - b_1) S a, a vector of length (b_2 - b_1) S, so the 1-Lipschitz
property gives |phi(z_2) - phi(z_1)| <= (b_2 - b_1) S and
F(b_2) - F(b_1) >= C(b_2 - b_1) - S(b_2 - b_1) = (C - S)(b_2 - b_1) > 0.
So F is strictly increasing, with slope at least C - S = e^(-tau) > 0.
For b > 0 use the bound (9.1) and |B y| <= C |y|:
F(b) >= C b - S |y| - K |B y + b S a|
>= C b - S |y| - K (C |y| + S b)
= (C - K S) b - (S + K C) |y|,
and C - K S > 0 because K < 1 and S < C. So F tends to +infinity as b
tends to +infinity, and the same estimate applied to -b shows it tends
to -infinity as b tends to -infinity. A continuous strictly increasing
function with those limits has exactly one zero. That zero is
psi_v(y), and (9.7) is its graph.
Continuity in (v, y) is now immediate from the uniform slope bound: the
zero of a family of strictly increasing functions with slope at least
C - S > 0, depending continuously on a parameter, depends continuously
on that parameter; and C - S is bounded below by a positive constant on
compact subsets of the ball. So psi_v is continuous jointly.
Now restrict to |y| = 1. By (9.6) and (9.1),
1 - b^2 = |y|^2 - b^2 = |z|^2 - h^2 >= (1 - K^2) |z|^2, (9.9)
so if |b| were 1 then z = 0 and h = 0, which is impossible because
|y| = 1 and the boost (9.5) is invertible. Hence |psi_v(y)| < 1 for
|y| = 1, and the level sets of psi_v meet the sphere transversally.
For the boundary values, divide (9.5) by C and use S a = C v to see
that
b + <v, y> = h/C.
By (9.9), |h| <= (K/sqrt(1 - K^2)) sqrt(1 - b^2) <= K/sqrt(1 - K^2),
so
sup over |y| = 1 of | psi_v(y) + <v, y> | <= K/(C sqrt(1 - K^2)).
As v tends to a point a of the unit sphere, tau tends to infinity, so
C tends to infinity and the right side tends to zero. Hence the
functions psi_v converge uniformly on the unit sphere to
psi_a(y) = - <a, y>.
That is the asserted boundary value, and it completes the proof. QED
Note what the boundary value means. For |a| = 1 the level sets of
psi_a are the intersections of the sphere with the hyperplanes
<a,y> = -b, which are round spheres of dimension m-1, and the sublevel
sets are round caps:
{ y on the sphere : psi_a(y) < b } = { y : <a, y> > -b }. (9.10)
Two antipodal boundary parameters give complementary caps:
{ y : <a,y> > -b } and { y : <a,y> > b }
are complementary modulo their common boundary. That is the symmetry
which will make the family descend to the projective parameter space.
9c. The level formulas and the Jacobian bound
Now we invert the boost on a contact point and obtain an explicit map
from the source Sigma onto each level set.
Fix v in the open unit ball, so a, C, S, B are determined, and fix a
level b with |b| < 1. Put
r = sqrt(1 - b^2), (9.11)
so 0 < r <= 1, with r = 1 exactly when b = 0. For x in Sigma write
f = <a, x>, g = <a, N(x)>, D = C - S g. (9.12)
The number D is positive: C - S g >= C - S = e^(-tau) > 0, since
|g| <= 1. Define
N' = (B N(x) - S a)/D,
h_{v,b}(x) = (b + S r f)/D,
Y_{v,b}(x) = r ( B x + S f N' ) + b N'. (9.13)
Here is where these come from. Let y be a point of the unit sphere with
psi_v(y) = b, and let (z, h) be its boost, so z = By + bSa and
h = phi(z), and by (9.6)
|z|^2 - h^2 = |y|^2 - b^2 = 1 - b^2 = r^2.
By Lemma 9.1 there is a contact point x in Sigma with
z = r x + h N(x).
Apply the inverse boost, which is the boost with parameter -tau:
y = B z - S h a, b = C h - S <a, z>.
Substituting z = rx + hN(x) into the second equation,
b = C h - S r f - S h g = h(C - S g) - S r f = h D - S r f,
which is exactly h = (b + Srf)/D = h_{v,b}(x). Substituting into the
first equation and using (BN - Sa) = D N',
y = B( rx + hN ) - S h a = r Bx + h(BN - Sa) = r Bx + h D N'
= r Bx + (b + S r f) N' = Y_{v,b}(x).
So every level point is of the form Y_{v,b}(x). Two useful identities
come out of this normalization:
|N'| = 1, Y_{v,b}(x) is a point of the unit sphere,
<N', Y_{v,b}(x)> = b. (9.14)
The first is the computation |BN - Sa|^2 = D^2, which follows by
expanding and using B^2 = I + S^2 a a^t. The second follows from
<Bx, BN - Sa> = <x, B^2 N> - S<a,x> = S^2 f g - S C f = -SD f, so
<Bx, N'> = -S f, and then <Y, N'> = r(-Sf + Sf) + b = b.
Two decompositions are at work and they look exactly alike. At the
source, the contact point z is written as rx + hN(x): a multiple of the
position vector plus a multiple of the normal. At the target, put
u = Bx + S f N'.
Then u is a unit vector perpendicular to N', because
|Bx|^2 = 1 + S^2 f^2 and <Bx, N'> = -S f, so |u|^2 = 1 + S^2f^2 +
S^2f^2 - 2S^2f^2 = 1, and <u,N'> = -Sf + Sf = 0. With that notation
Y_{v,b}(x) = r u + b N', |u| = 1, u perpendicular to N',
which is the exact analogue of z = rx + hN(x): the coefficient r is the
same in both, and h has been replaced by the level b. That is Lorentz
invariance in coordinate form, and it is why the area estimate can be
carried out entirely on the source Sigma.
What we need is a differential.
LEMMA 9.3 (the differential of Y and its Jacobian).
Fix v and b as above and x in Sigma. Define, for w perpendicular to
N(x),
O(w) = B w + S <a, w> N'. (9.15)
Then |O(w)| = |w| and <O(w), N'> = 0, so O is a linear isometry from
the normal space of Sigma at x onto the plane perpendicular to N' in
R^(m+2). Moreover
dY_{v,b}(w) = O( (r I - h_{v,b}(x) A(x)) w ) (9.16)
for every w perpendicular to N(x), and consequently the m-dimensional
Jacobian of Y_{v,b} at x is
J_m Y_{v,b}(x) = | det( r I - h_{v,b}(x) A(x) ) |. (9.17)
Proof. First the isometry. Since B = I + (C-1)aa^t and B^2 = I + S^2aa^t,
|B w|^2 = |w|^2 + (C^2 - 1) <a,w>^2,
and, as computed above, <Bw, N'> = -S<a,w>. Therefore
|O(w)|^2 = |Bw|^2 + S^2<a,w>^2 + 2S<a,w><Bw,N'>
= |w|^2 + (C^2 - 1)<a,w>^2 + S^2<a,w>^2 - 2S^2<a,w>^2
= |w|^2,
because C^2 - S^2 = 1. And <O(w), N'> = -S<a,w> + S<a,w> = 0. So O is
an isometry onto its image, which is the orthogonal complement of N'.
Second the differential of N'. Differentiating (9.12) and (9.13) along
a curve with velocity w perpendicular to N, and using
dN(w) = -A w, dg(w) = -<a, Aw>,
dN'(w) = D^(-1) ( -B A w + S dg(w) N' ) = D^(-1) O( -A w ). (9.18)
Third the differential of Y. Differentiating (9.13),
dY(w) = r ( B w + S<a,w> N' + S f dN'(w) ) + b dN'(w)
= r O(w) + ( b + r S f ) dN'(w).
But b + rSf = h_{v,b}(x) D by (9.13), so with (9.18),
dY(w) = r O(w) - h O(A w) = O( (r I - h A) w ).
This is (9.16). Since O is an isometry, the Jacobian of Y at x equals
the absolute value of the determinant of rI - hA, which is (9.17). QED
Remark on the sign convention. The shape operator was defined as
A = -dN, so A(x) is a self-adjoint endomorphism of the tangent space,
and rI - hA(x) is self-adjoint as well. Its determinant is the product
of its eigenvalues, and (9.17) is the usual area formula: a linear map
multiplies m-dimensional volume by the absolute value of its
determinant once an orthonormal basis is chosen in both spaces.
Now the key estimate.
LEMMA 9.4 (the restricted area bound).
For every (v,b) in Q the set
E_{v,b} = { y on the unit sphere : psi_v(y) < b } (9.19)
has finite perimeter. For |v| < 1 and |b| < 1, put
T_{v,b} = { x in Sigma : r I - h_{v,b}(x) A(x) >= 0 }. (9.20)
Then for every open set U of the unit sphere,
Per(E_{v,b}; U) <= H^m( level set of psi_v at b, intersected with U )
<= integral over T_{v,b} intersected with
Y_{v,b}^{-1}(U) of det(r I - h_{v,b} A) dmu, (9.21)
and on all of T_{v,b},
0 <= det( r I - h_{v,b}(x) A(x) ) <= r^m <= 1. (9.22)
Proof. Take a level point y, so psi_v(y) = b, and let (z,h) be its
boost. The inversion above produced a contact point x in Sigma with
y = Y_{v,b}(x), and Lemma 9.1 says that this contact satisfies
r I - h A(x) >= 0, that is, x lies in T_{v,b}. So
the level set of psi_v at b is contained in Y_{v,b}(T_{v,b}). (9.23)
On T_{v,b} the eigenvalues nu_1,...,nu_m of the self-adjoint matrix
rI - hA are nonnegative, and their sum is
sum of nu_i = m r - h trace A = m r,
because trace A = 0. The arithmetic-geometric mean inequality gives
det( r I - h A ) = product of nu_i <= ( (sum of nu_i)/m )^m = r^m,
which is (9.22), including the nonnegativity. Now integrate the area
formula over the source. The area formula for a smooth map of an
m-manifold, with multiplicities counted in the source integral, gives
H^m( Y_{v,b}(T_{v,b}) intersected with U )
<= integral over T_{v,b} intersected with Y_{v,b}^{-1}(U) of
J_m Y_{v,b} dmu,
with equality when Y is injective on the source set. Multiplicities and
extra points of the source set only increase the right side, which is
what we want. Inserting (9.17) and (9.22) proves the second inequality
of (9.21).
For the first inequality: E_{v,b} is open, so its essential boundary is
contained in its topological boundary, which is contained in the level
set; the reduced-boundary representation of the perimeter of a set of
finite perimeter then gives
Per(E_{v,b}; U) = H^m( reduced boundary of E_{v,b} intersected with U )
<= H^m( level set intersected with U ).
That the sets have finite perimeter follows from (9.21), which bounds
the H^m measure of the level set by r^m |Sigma| < infinity, and from
the finite-essential-boundary criterion.
Finally the endpoint cases. For |v| < 1 and b = -1 the sublevel is
empty, and for b = +1 it is everything, since |psi_v| < 1 on the
sphere. For |v| = 1 the level sets are round caps by (9.10), whose
boundary is a round sphere of dimension m and radius r, with area
r^m sigma_m; and at b = +1 or -1 the cap degenerates to a point or to
the whole sphere, with zero or full volume and zero perimeter. QED
This is the estimate that makes the whole construction work: the area
of every level is bounded by an integral over the one fixed source
Sigma, with an integrand bounded by r^m. No compactness of level sets,
no uniqueness of contacts and no transversality is needed.
9d. The family, its mass bound, and no concentration
We can now assemble the family. Let
Q = (closed unit ball of R^(m+2)) x [-1, 1],
a compact convex body of dimension p = m+3, and for u = (v,b) in Q put
E_u = { y on the unit sphere : psi_v(y) < b }, Phi(u) = d[E_u].
Two remarks on the centre. At u = (0,0) the boost is the identity, so
psi_0 is the restriction of phi to the sphere; its zero level is Sigma
itself, and its negative sublevel is the side of Sigma pointed to by N.
Write Omega for that side. Then
E_{(0,0)} = Omega, Phi(0,0) = d[Omega] = Sigma,
mass(Phi(0,0)) = |Sigma|, (9.24)
with multiplicity one because Sigma is embedded.
LEMMA 9.5 (flat continuity).
The map u -> Phi(u) is flat continuous on Q. More precisely, for
u_j -> u the volumes of the symmetric differences
E_{u_j} Δ E_u tend to zero, and therefore
flatdist(Phi(u_j), Phi(u)) tends to zero.
Proof. By Lemma 9.2, psi_v depends continuously on (v,y) jointly, and
the convergence as v tends to a boundary point is uniform in y. So if
(v_j,b_j) -> (v,b), then psi_{v_j} - b_j converges uniformly on the
sphere to psi_v - b. The indicators of the sublevels therefore converge
pointwise at every point where psi_v is not equal to b, that is, off the
level set. Level sets have zero (m+1)-dimensional measure: inside the
ball they have finite m-measure by Lemma 9.4; on the boundary |v| = 1
they are round caps; at |b| = 1 they are empty or everything. So
pointwise convergence almost everywhere and dominated convergence give
volume( E_{u_j} Δ E_u ) -> 0.
Finally, the flat distance between two boundary cycles is bounded by
the volume of the symmetric difference of their fillings, which is the
elementary inequality already recorded in Lemma 8.1's proof. QED
LEMMA 9.6 (mass bounds).
For every u = (v,b) in Q, with r = sqrt(1 - b^2),
mass(Phi(u)) = Per(E_u) <= r^m |Sigma| if |v| < 1,
mass(Phi(u)) = Per(E_u) <= r^m sigma_m if |v| = 1. (9.25)
In particular mass(Phi(u)) <= |Sigma| everywhere, with equality at
u = (0,0) and at no other point of the interior of Q. Moreover for
every neighbourhood U of (0,0) in Q there is a number D(U) < |Sigma|
with
sup over u in Q minus U of mass(Phi(u)) <= D(U). (9.26)
Proof. The bounds (9.25) are Lemma 9.4 combined with the endpoint
description of the caps. For (9.26), the strict loss away from the
centre has to be proved, and that is Lemma 9.8 below; for the moment
note only that r^m |Sigma| < |Sigma| whenever |b| > 0, so the loss is
immediate except for parameters with b close to zero and v bounded away
from zero. QED (for (9.25))
The interesting assertion is really (9.26), which says that the family
has a strict maximum at the centre. It is proved in 9e.
LEMMA 9.7 (no concentration of mass).
The family Phi satisfies
lim over rho to 0 of sup over u in Q and z on the sphere of
mass( Phi(u) restricted to the ball of radius rho about z ) = 0. (9.27)
Proof. Suppose not. Then there are u_j = (v_j,b_j) in Q, points y_j on
the sphere, radii rho_j tending to zero, and delta > 0 with
mass( Phi(u_j) restricted to B_{rho_j}(y_j) ) >= delta for all j.
Pass to a subsequence with u_j -> u_* = (v_*,b_*) and y_j -> y_*.
If |b_*| = 1 then r_j -> 0 and the total masses r_j^m |Sigma| tend to
zero, contradiction. If |v_*| = 1 with |b_*| < 1, then the cycles are
round caps of dimension m and radius r_* = sqrt(1 - b_*^2): an m-sphere
of radius r in any affine (m+1)-plane has at most c_m rho^m area in a
ball of radius rho, uniformly in r and in the plane. For r <= 2 rho use
its total area; for r > 2 rho the intersection lies in a spherical cap
with angular radius at most arcsin(rho/r), and the cap area formula
gives the same bound. So the ball masses tend to zero, contradiction.
Hence |v_*| < 1. By (9.21) of Lemma 9.4, the mass of Phi(u) inside a
ball B is at most
integral over T_u intersected with Y_u^{-1}(B) of J_m Y_u dmu,
with J_m Y_u <= 1. Write I_j for the integrand on Sigma:
I_j(x) = 1 if x is in T_j and Y_j(x) is in B_{rho_j}(y_j), and 0
otherwise, times J_m Y_j. Each I_j is between 0 and 1, and
mass( Phi(u_j) restricted to B_{rho_j}(y_j) ) <= integral of I_j dmu.
Because |v_*| < 1, the maps Y_j converge smoothly on the compact Sigma
to Y_*, and the Jacobians converge uniformly. Away from the fiber
Y_*^{-1}(y_*) the shrinking ball indicator is eventually zero, so
I_j -> 0 there. On the fiber, the area formula gives
integral over Y_*^{-1}(y_*) of J_m Y_* dmu = 0,
because a single point has zero H^m measure in the target, and hence
J_m Y_* = 0 almost everywhere on the fiber. Smooth convergence then
gives I_j -> 0 almost everywhere on the fiber as well. So I_j -> 0
almost everywhere on Sigma, and dominated convergence contradicts
delta > 0. QED
Why the argument can fail at |v_*| = 1, and why that case was handled
separately. As v tends to the boundary of the ball the maps Y_{v,b}
degenerate: on the set Z_a = { x in Sigma : N(x) = a } the denominator
D = C - Sg tends to zero, and the limiting map
Y_infinity(x) = r [ x + (<a,x>/(1 - <a,N>)) (N - a) ] - b a (9.28)
is only defined off Z_a. That is harmless, because Z_a is contained in
the zero set of the real analytic function <a,x>, hence has measure
zero by Lemma 4.1(e). No rank assumption on the Gauss map is needed
anywhere: the set where the Gauss map takes a prescribed value is
source-null for a different reason, namely that on it the position
function is orthogonal to a. This is the one place where the analyticity
of Sigma saves the argument at infinite boosts.
9e. The centre is a strict maximum
We promised the strict bound (9.26). Here it is.
LEMMA 9.8 (strict loss away from the centre).
For every neighbourhood U of (0,0) in Q, there is D < |Sigma| with
sup over u in Q minus U of mass(Phi(u)) <= D. (9.29)
Consequently |Sigma| is attained as a mass exactly at u = (0,0), and
the maximum is strict away from every neighbourhood of the centre.
Proof. Suppose not. Then there are u_j = (v_j,b_j) outside U with
mass(Phi(u_j)) -> |Sigma|.
Because mass(Phi(u_j)) <= r_j^m |Sigma| with r_j = sqrt(1-b_j^2), and
r_j^m <= 1 with equality only at b_j = 0, we must have b_j -> 0. Since
u_j stays outside U and v_j cannot tend to 0 (that would put u_j inside
U for large j), a subsequence has v_j -> v_* != 0. Write t_* = |v_*|
and, for each j, t_j = |v_j|, a_j = v_j/|v_j|, so that t_j -> t_* and
a_j -> a = v_*/t_*, with t_* in (0,1].
Now choose a compact patch. The open set {A != 0} is nonempty because
Sigma is not an equator, and the real analytic function f = <a,x> cannot
vanish on all of it, by Lemma 4.1(e). So there is a compact set K_0 of
positive measure on which
|A| >= alpha > 0 and |f| >= beta > 0. (9.30)
Also K_0 does not meet Z_a = {x : N(x) = a}, because N(x) = a gives
f = <a,x> = <N(x),x> = 0. So on K_0 the denominators stay bounded away
from zero and the functions
h_j = ( b_j/C_j + t_j r_j f_j ) / ( 1 - t_j g_j )
converge uniformly to
h_* = t_* f / ( 1 - t_* <a,N> ), (9.31)
a finite function which is nonzero on K_0 by (9.30) and t_* > 0.
(Here f_j, g_j are f, <a_j,N> respectively.)
Since trace A = 0, the eigenvalues nu_1,...,nu_m of r_j I - h_j A sum to
m r_j. On T_{v_j,b_j} they are all nonnegative, so the arithmetic-
geometric mean inequality gives
det( r_j I - h_j A ) <= r_j^m,
with equality precisely when all nu_i are equal, that is, when
h_j A = 0. On K_0 the inequality is strict, uniformly: the set of pairs
(h, A) with |h| >= beta/2, |A| >= alpha, trace A = 0 is compact, and it
does not meet the equality set {(h,A) : hA = 0}. Hence there is
epsilon > 0 with
det( r_j I - h_j A ) <= ( 1 - epsilon ) r_j^m <= 1 - epsilon
on K_0, for all large j. Therefore, by (9.21),
mass(Phi(u_j)) <= integral over T_j of det(r_j I - h_j A) dmu
<= (1 - epsilon) mu(K_0) + ( |Sigma| - mu(K_0) )
= |Sigma| - epsilon mu(K_0),
which contradicts mass(Phi(u_j)) -> |Sigma|. QED
So the family has: a global bound by |Sigma|, a strict maximum at the
centre, and a uniform gap away from every neighbourhood of the centre.
All three are used below, the second and third in the deformation
(9g), the first in the estimate that leads to a counterexample.
9f. The family is detecting
The parameters live on Q = (closed unit ball) x [-1,1], which is the
closed ball of dimension p = m+3. But two boundary parameters give the
same cycle, and that is what makes the family detect.
LEMMA 9.9 (complementary boundary parameters).
Let |a| = 1 and |b| < 1. Then
psi_a(y) = - <a,y> for every y on the unit sphere, (9.32)
so that
E_{a,b} = { y : <a,y> > -b },
E_{-a,-b} = { y : <a,y> < -b }, (9.33)
which are complementary modulo the common boundary { <a,y> = -b }.
Consequently, on the boundary |v| = 1 of Q,
Phi(-v,-b) = Phi(v,b) as cycles. (9.34)
Two warnings about what this does and does not say. The identity
psi_{-v} = -psi_v is FALSE for |v| < 1. At v = 0 the boost is the
identity, so psi_0 is the restriction of phi to the sphere, which is not
identically zero; an identity psi_0 = -psi_0 would force phi to vanish
on the whole sphere, which is absurd. The symmetry is genuinely a
degeneration phenomenon of the boundary: only for |v| = 1.
Proof. The boundary limits are Lemma 9.2: as v tends to a unit vector a,
C = cosh tau tends to infinity and
sup over |y| = 1 of | psi_v(y) + <v,y> | <= K/(C sqrt(1-K^2))
-> 0. (9.35)
So psi_v converges uniformly to the function y -> -<a,y>, which is
(9.32). Its level sets are the intersections of the sphere with the
hyperplanes <a,y> = -b, which are round (m-1)-spheres, and its
sublevel sets are the round caps (9.33). A set and its complement have
the same boundary, so the boundary cycles agree, which is (9.34). QED
Now quotient. Let
X = Q / ~, (9.36)
where q ~ q' identifies q' = -q only when |v| = 1, that is, only on
the boundary of Q. Since Q is a compact convex body containing the
origin in its interior, Q is homeomorphic to the closed ball of
dimension p, and the quotient identifying antipodal boundary points is
the real projective space RP^p.
By Lemma 9.9, Phi respects the identification, so it descends to
Phi_bar : X -> Z, Phi_bar([u]) = Phi(u). (9.37)
LEMMA 9.10 (the family is detecting).
The map Phi_bar : RP^p -> Z of (9.37) is detecting in the sense of
Definition 8.2, with no concentration of mass, and
Phi_bar([0,0]) = d[Omega], mass = |Sigma|, (9.38)
where Omega is the side of Sigma pointed to by N.
Proof. Detection. The fundamental group of RP^p is Z_2, generated by
the loop obtained from a diametral segment of Q: pick q_0 on the
boundary of Q and run
ell(t) = (1 - 2t) q_0, 0 <= t <= 1,
from q_0 to -q_0. In X this is a closed loop, and it represents the
nonzero element of pi_1(RP^p).
Follow the lifted filling along this loop. At t = 0 the filling is
E_{q_0}. At t = 1 the parameter is -q_0, so by Lemma 9.9 the filling is
the complement of E_{q_0}. The two are different sets, and the
cohomology class lambda of Lemma 8.1 takes the value 1 on this loop.
Since pi_1(RP^p) = Z_2 and Phi_bar^*lambda is evaluated on the generator
as 1,
Phi_bar^* lambda = alpha, the nonzero class in H^1(RP^p; Z_2).
Therefore
(Phi_bar^* lambda)^p = alpha^p != 0 in H^p(RP^p; Z_2),
because H^p(RP^p; Z_2) = Z_2 is generated by alpha^p. That is exactly
detection (8.2). The statement (9.38) is (9.24), and no concentration of
mass is Lemma 9.7. QED
The choice of p = m+3 is now explained: the parameter space is a ball
of dimension m+2 times an interval, and its quotient by the boundary
antipodal map is RP^(m+3), whose cohomology has a nonzero class in
degree m+3. A family on a lower-dimensional space could not carry the
m+3-fold product of a degree-one class, and detection would be
impossible.
9g. The central tube
Near the centre the family is as smooth as it looks: the level sets are
embedded copies of Sigma, not merely images of a map with folds.
LEMMA 9.11 (the central tube).
There is eta > 0 and a neighbourhood U of (0,0) in Q such that for
(v,b) in U:
(a) the level set is exactly Y_{v,b}(Sigma), and Y_{v,b} is a smooth
embedding of Sigma;
(b) every level point has a unique closest point on the cone;
(c) the tangential gradient of psi_v on the sphere is
N' - b Y_{v,b}, and its length is r = sqrt(1-b^2) > 0;
(d) E_{v,b} is a set of finite perimeter with reduced boundary
Sigma and multiplicity one, and E_{v,b} and its complement have
the same boundary cycle.
Proof. (b) is a tubular-neighbourhood statement. The cone over Sigma is
smooth away from the vertex and compact on the annulus
{ s x : x in Sigma, 1/2 <= s <= 3/2 }. Choose that annulus inside a
slightly larger annulus and use a tubular neighbourhood of the smooth
cone there: there is epsilon > 0 such that a point s x + h N(x) with
1/2 <= s <= 3/2 and |h| < epsilon has a unique closest cone point,
namely s x, and signed distance h. Shrinking epsilon, points at radii
outside the larger annulus cannot be closest points, so uniqueness holds
for the whole cone in that band.
By (9.9), a level point y with boost data (z,h) satisfies
|h| <= K/sqrt(1-K^2) * sqrt(1-b^2),
which tends to 0 as |b| tends to 1. So for |b| close enough to 1 the
bound is already below epsilon, and we then have |h| < epsilon, and then
(b) holds, so the contact point x is unique. That gives (a): every level
point is Y_{v,b}(x) for exactly one x, so the level set is the image and
the image map is injective; it is an immersion by Lemma 9.3, hence an
embedding of the compact Sigma.
For (c), differentiate the level equation. In the tube the signed
distance satisfies d phi = N . d z, so differentiating
phi(B y + b S a) = C b + S<a,y> gives
0 = d(Cb + S<a,y>) - N . d(B y + b S a)
= D ( db - N' . dy ),
with D = C - S<a,N> > 0. Hence d psi_v = N' . dy at level points, and
projecting N' to the tangent space of the sphere at y and using
<N', y> = b from (9.14),
| grad_S psi_v |^2 = |N'|^2 - <N',y>^2 = 1 - b^2 = r^2 > 0. (9.36)
So level sets are regular level sets of a smooth function with nonzero
gradient: E_{v,b} is one side of a smooth hypersurface, its topological
and reduced boundaries coincide, and the multiplicity is one because Y
is an embedding. Part (d) follows, and the cycle of a set and of its
complement are the same. QED
Note the role of each hypothesis. Minimality of Sigma entered through
trace A = 0 in Lemma 9.4, where it fixed the trace of rI - hA at mr and
allowed the arithmetic-geometric mean inequality. Embeddedness of Sigma
entered through the tube and through fullness in Lemma 4.1. Neither
compactness of the tube argument nor the area bound uses the index of
Sigma; the index is about to enter, and it enters only here.
9h. The second variation at the centre, and the index deformation
We now compute how the area of the family changes as the parameter
moves away from the centre. This is the only place where index enters
the whole paper.
LEMMA 9.12 (the area expansion at the centre).
Let A(u) = mass(Phi(u)) = Per(E_u). Then
A(u) = |Sigma| - Q(u) + O(|u|^3), u = (v,b), (9.37)
where
Q(v,b) = (1/2) [ (m|Sigma| + int P) b^2
+ 2 b * int P <v,x> dmu
+ int P <v,x>^2 dmu ]. (9.38)
The quadratic form Q is positive definite on R^(m+2) x R.
In particular, the second variation of area at the centre is negative
definite on a (m+3)-dimensional space of normal variations, namely the
space
W = { t + <d,x> : t in R, d in R^(m+2) }, (9.39)
considered as a space of functions on Sigma.
Proof. In the central tube, by Lemma 9.11, the area is
A(v,b) = integral over Sigma of det( r I - h_{v,b}(x) A(x) ) dmu. (9.40)
Expand the determinant. Since A is symmetric with eigenvalues
lambda_1,...,lambda_m summing to zero and with sum of squares P,
det( r I - h A ) = product over i of ( r - h lambda_i )
= r^m - h r^(m-1) (sum of lambda_i)
+ h^2 r^(m-2) ( sum over i<j of lambda_i lambda_j ) + O(h^3),
= r^m - (P/2) h^2 r^(m-2) + O(h^3), (9.41)
because sum lambda_i = 0 and
sum over i<j of lambda_i lambda_j
= ( (sum lambda_i)^2 - sum lambda_i^2 ) / 2 = -P/2.
Next, the scalars. With C = cosh tau = 1 + tau^2/2 + O(tau^4),
S = sinh tau = tau + O(tau^3), D = C - S g = 1 - tau g + O(tau^2), and
r = sqrt(1-b^2) = 1 - b^2/2 + O(b^4),
h_{v,b} = ( b/C + S r f ) / D = b + <v,x> + O(|u|^2), (9.42)
using S f = tau <a,x> = <v,x> to first order and 1/D = 1 + tau g +
O(tau^2). Also r^m = 1 - (m/2) b^2 + O(b^4). Substituting into (9.41)
and then (9.40),
A(v,b) = |Sigma| - (m/2) b^2 |Sigma|
- (1/2) * integral of P ( b + <v,x> )^2 dmu + O(|u|^3),
which is (9.37) with (9.38).
Positive definiteness. For fixed v, Q is a quadratic polynomial in b
with leading coefficient (1/2)(m|Sigma| + int P) > 0 and discriminant
(1/4) [ ( int P<v,x> )^2 - (m|Sigma| + int P) ( int P<v,x>^2 ) ].
By Cauchy-Schwarz, ( int P<v,x> )^2 <= ( int P )( int P<v,x>^2 ), so the
discriminant is at most (1/4)( int P<v,x>^2 )[ int P - m|Sigma| -
int P ] = -(1/4) m|Sigma| ( int P<v,x>^2 ) <= 0, with equality only if
int P<v,x>^2 = 0. Also Q(v,0) = (1/2) int P<v,x>^2, which is positive
for v != 0: the analytic function <v,x> is not identically zero by
Lemma 4.1(e), and P > 0 on a set of positive measure because Sigma is
not an equator. So Q(u) > 0 for every u != 0. QED
The space (9.39) is exactly the space of first-order normal variations
of the family at the centre: moving v in the direction d moves the
boundary in the normal direction <d,x>, and moving b moves it in the
constant normal direction 1. That is why W has dimension m+3.
Now the deformation. The idea is simple: if Sigma has one more
negative direction than W, we can push the whole family downhill in
that extra direction.
LEMMA 9.13 (removing the maximum).
Suppose ind(Sigma) > p = m+3. Then there is a detecting map
Psi : RP^p -> Z without concentration of mass with
sup over RP^p of mass(Psi) < |Sigma|. (9.43)
Proof. Step 1, find the extra direction. Let B be the symmetric
bilinear form on functions on Sigma,
B(phi, chi) = integral of phi L chi dmu
= integral of ( <grad phi, grad chi> - (m+P) phi chi ) dmu,
the second variation of area. Since ind(Sigma) > p, the positive
spectral subspace of L has dimension greater than p. Choose a
finite-dimensional subspace of it, spanned by L-eigenfunctions with
positive eigenvalues, of dimension greater than p = dim W. The p linear
conditions B(eta, w) = 0 for w in W admit a nonzero element eta of that
subspace. For this eta,
B(eta, w) = 0 for all w in W, B(eta, eta) > 0. (9.44)
Step 2, the two-parameter family. Extend eta N, a normal vector field
along Sigma, to a smooth ambient vector field on the sphere, and let
F_s be its flow. For small u and s put
E(u,s) = F_s( E_u ), A(u,s) = Per( E(u,s) ). (9.45)
By Lemma 9.11 the family E_u is smooth in u near the centre, so A is
smooth near (0,0). Its gradient vanishes at (0,0) because the centre is
a maximum of A, and its Hessian is negative definite: on the W
directions by Lemma 9.12, and in the s direction because the second
derivative in s is -B(eta, eta) < 0, while the mixed terms vanish by
(9.44). Hence there are c > 0 and delta > 0 with
A(u,s) <= |Sigma| - c( |u|^2 + s^2 ) (9.46)
whenever |u|^2 + s^2 <= delta^2. Shrink delta so that this ball lies
inside the central tube where Lemma 9.11 holds.
Step 3, a cutoff. Choose a smooth function beta of |u| with 0 <= beta
<= 1, equal to 1 for |u| <= delta/4, equal to 0 for |u| >= delta/2, and
with |grad beta| <= 4/delta. For small epsilon > 0, replace E_u near
the centre by E(u, epsilon beta(u)) and leave it alone elsewhere. The
graph
{ (u, epsilon beta(u)) : u in Q, |u| <= delta/2 }
is compact and misses (0,0), and it lies in the ball where (9.46)
holds. Its area maximum is therefore at most |Sigma| - c' for some
c' > 0.
Step 4, the global gap. Off the region |u| <= delta/2 the original
domains are unchanged, and Lemma 9.8 gives a uniform gap: there is D_1
< |Sigma| with
mass(Phi(u)) <= D_1 for u outside the delta/2-neighbourhood of 0.
So the modified family has sup mass at most max(|Sigma| - c', D_1) <
|Sigma|.
Step 5, detection and no concentration are preserved. Replacing
epsilon by t epsilon for 0 <= t <= 1 gives a flat homotopy, fixed
outside the chosen neighbourhood of the centre. Flat homotopies
preserve detection, because homotopic maps induce the same map in
cohomology. And on the modified region the boundaries belong to a
compact smooth family of embedded copies of Sigma: a finite cover by
coordinate graph patches, with uniform metric and first-derivative
bounds over the compact parameter set, bounds their masses in
radius-rho balls by a constant times rho^m. Outside that region the
original estimate of Lemma 9.7 applies unchanged. QED
We can now state the main result of this section, which collects
everything.
THEOREM 9.14 (the Lorentz-distance family).
Let Sigma^m be a smooth closed connected embedded minimal hypersurface
of S^(m+1), m >= 2, not an equator, and put p = m+3. Then there is a
detecting map
Psi : RP^p -> Z, Z the flat cycle space of S^(m+1),
without concentration of mass, such that
Phi_bar([0,0]) = d[Omega] with mass |Sigma|,
and
sup over RP^p of mass(Psi) <= |Sigma|,
with strict inequality if ind(Sigma) > p. Here Omega is the side of
Sigma pointed to by N.
Proof. Lemma 9.10 gives the family Phi_bar with the required
properties at the centre. Lemmas 9.6 and 9.8 give the mass bound and
the strict maximum, and Lemma 9.13 gives the strict global inequality
when the index exceeds p. QED
==============================================================================
10. CONES, REGULARITY, AND THE LEAST COUNTEREXAMPLE
==============================================================================
The previous section produced a detecting family with mass strictly
below the least counterexample H. Min-max then produces a stationary
integral hypersurface of mass below H. For 2 <= m <= 6 that surface is
automatically smooth, by the regularity clause of Theorem 8.6, and the
job is done. For m >= 7 the min-max output may be singular, and this
section is what handles it: it converts "stationary integral with small
mass" into "smooth embedded with multiplicity one", using the
induction hypothesis in dimensions below m.
Two things are needed. First, a compactness and regularity theory for
low-density cones, which is the induction in the dimension k of the
cone. Second, Allard's regularity theorem, which upgrades a cone with
one planar tangent onto a smooth manifold. Both are collected here.
Write, throughout, sigma_d for the volume of the unit d-sphere and
omega_d for the volume of the unit d-ball.
10a. Coning a spherical varifold
LEMMA 10.1 (the cone operation).
Let d >= 1 and let V be a stationary integral d-varifold in S^(d+1).
Let C(V) be its Euclidean cone in R^(d+2): the push-forward of
r^d dr (x) ||V|| under (r,z) -> rz, extended by zero mass at the origin.
Then:
(a) C(V) is a stationary integral (d+1)-varifold in R^(d+2);
(b) its masses are
||C(V)||(B_R(0)) = R^(d+1)/(d+1) * M(V), (10.1)
so its density at the vertex is
Theta(C(V),0) = M(V)/sigma_d; (10.2)
(c) if V is a finite sum of smooth closed embedded minimal
hypersurfaces of S^(d+1) with positive integer multiplicities,
then C(V) is cyclic modulo two.
Proof. (a) Integrality is the statement that a cone over a rectifiable
set with integer multiplicity is rectifiable with integer multiplicity;
the area formula for the map (r,z) -> rz gives it, and the mass formula
(10.1) comes out of the same computation:
||C(V)||(B_R(0)) = integral from 0 to R of r^d dr * M(V)
= R^(d+1)/(d+1) * M(V).
Stationarity away from the vertex is exactly stationarity of V, since
the cone is a product metric r^2 g_{S^(d+1)} + dr^2 and the first
variation in the radial direction is a boundary term. The boundary term
from truncating at radius epsilon is bounded by a constant times
(epsilon^(d+1) + epsilon^d), which tends to zero as epsilon tends to 0.
This is the same truncation computation as in Proposition 5.1; do it
once and use it twice. Density (10.2) is (10.1) with the definition of
density, since sigma_d = (d+1)omega_(d+1).
(c) A smooth embedded minimal hypersurface of the sphere, with
multiplicity one, gives a cone which is smooth away from the vertex, and
whose mod-2 boundary is the product of the mod-2 class of the
hypersurface with the radial interval. On every annulus the boundary
vanishes: an odd multiplicity gives a smooth mod-two product cylinder,
and an even multiplicity gives zero. Truncate the associated mod-two
chain inside radius epsilon: the difference has mass at most
M(V)epsilon^(d+1)/(d+1) and its inner boundary has mass at most
M(V)epsilon^d. Both tend to zero, so the truncated chains converge
locally flat to the full chain and their boundaries converge locally
flat to zero. Continuity of the boundary operator gives
d[C(V)] = 0 also at the vertex. QED
10b. Density, monotonicity, and the radial splitting
LEMMA 10.2 (density at every support point).
Let V be a stationary integral k-varifold in an open set U of R^n. At
every z in the support of ||V||,
Theta(V,z) >= 1. (10.3)
If V = C is a cone in all of R^n, then
1 <= Theta(C,z) <= Theta(C,0) for z in the support, (10.4)
and every tangent cone of C at such a z is nonzero and has density
Theta(C,z) at its vertex.
Proof of (10.3) and (10.4). Integrality gives, at every point of the
rectifiable set underlying V, a density which is a positive integer,
in particular at least one; and ||V||-almost every support point is
such a point. For an exceptional support point z, choose points x_i of
the rectifiable set with x_i -> z. If B_r(z) is inside the open set U
and d_i = |x_i - z| < r, monotonicity at x_i and the inclusion
B_{r-d_i}(x_i) inside B_r(z) give
||V||(B_r(z)) >= ||V||(B_{r-d_i}(x_i)) >= omega_k (r - d_i)^k.
Let i tend to infinity, then r tend to zero. This proves (10.3) also at
the points where rectifiability alone gives nothing.
For a cone, homogeneity gives ||C||(B_R(0)) = omega_k Theta(C,0) R^k
exactly. Squeeze a ball centred at z between centred balls:
B_{R-|z|}(0) inside B_R(z) inside B_{R+|z|}(0) for R > |z|.
Divide by omega_k R^k and let R tend to infinity: both bounds tend to
Theta(C,0). So the mass ratio centred at z has large-radius limit
Theta(C,0). By monotonicity that ratio is nondecreasing in R, hence
Theta(C,z) = lim over R to 0 of the ratio
<= lim over R to infinity of the ratio = Theta(C,0),
which is the upper bound in (10.4); the lower bound is (10.3).
Tangent cones. Rescale C about z by arbitrarily small factors. The
rescaled varifolds have uniformly bounded mass in the unit ball, by
monotonicity, and their first variations vanish, so Tool B gives a
convergent subsequence; the limit T is a stationary integral cone, a
tangent cone at z. Its density at the vertex is the limit of the mass
ratios of C centred at z, that is Theta(C,z), which is positive by
(10.3), so T is nonzero. QED
LEMMA 10.3 (splitting off the radial line).
Let k >= 2 and let C be a stationary integral k-dimensional cone in
R^(k+1). Let z be a nonzero support point and put e = z/|z|. Then every
tangent cone T of C at z has the form
T = v(R e) x W, (10.5)
the product of the line R e with multiplicity one and a nonzero
stationary integral (k-1)-dimensional cone W in e^perp = R^k, and
Theta(W,0) = Theta(T,0) = Theta(C,z). (10.6)
If C is cyclic modulo two, then so are T and W.
Proof. Write T_i = (eta_(z,rho_i))_# C -> T with
eta_(z,rho)(y) = (y - z)/rho and rho_i tending to zero. Fix a real
number b. Because C is invariant under dilation by the positive factor
1 + b rho_i, the rescaled cone T_i is invariant under the conjugate map
y -> (1 + b rho_i) y + b z. (10.7)
Let i tend to infinity: the maps converge to the translation
y -> y + bz, and invariance passes to the weak limit, so T is invariant
under every translation in the direction e. Its weight therefore
disintegrates as the product of Lebesgue measure on R e with a measure
nu on e^perp:
||T|| = L^1 (x) nu on R e (+) e^perp.
That nu is the weight of an integral varifold W. At almost every
rectifiable point of T, a blow-up of the measure is an integer multiple
of the measure of the approximate tangent plane, and those blow-ups
inherit invariance under translations in the e-direction; hence that
plane contains e. So the coordinate pi(te + w) = t has tangential
Jacobian one almost everywhere. Rectifiable coarea gives, for almost
every t, a (k-1)-rectifiable slice with the original integer
multiplicities, whose tangent plane is the intersection of the tangent
plane of T with e^perp; and product disintegration identifies the weight
of that slice, translated to e^perp, with nu. That is (10.5), with W
integral and nonzero.
Stationarity of W follows from stationarity of T tested against fields
tangent to e^perp and independent of t: the first variation of a product
splits, and the line factor contributes nothing.
The densities. By Lemma 10.2, Theta(W,0) = Theta(T,0) = Theta(C,z),
using that T is a tangent cone of C at z.
Cyclicity. Suppose d[C] = 0. By Lemma 10.1(c) applied to the rescaled
cones, each T_i is cyclic, and cyclicity passes to the limit. The mod-2
slicing identity for the boundary, applied to the coordinate pi, gives
d<[T], pi, t> = <d[T], pi, t> = 0 for almost every t,
where signs disappear modulo two. Choose one such level t at which the
rectifiable slice description above also holds. That slice is the
translate of [W] to te + e^perp, because T is the product just
established. Translating back proves d[W] = 0. QED
10c. Allard regularity from a planar tangent
The next lemma is the third and last imported theorem. It is a form of
Allard's interior regularity theorem, and its content is that a
stationary integral varifold whose one tangent cone is a simple plane
has nothing else going on locally.
TOOL C (Allard regularity, consequence form).
Let V be a stationary integral k-varifold, k >= 2, in an open set of
Euclidean space. If at a point z of its support one tangent cone is a
plane with multiplicity one, then V is, in a neighbourhood of z, a
smooth embedded minimal k-dimensional submanifold with multiplicity one.
Hypotheses, spelled out, because they are all that is used. Allard's
theorem (Allard 1972, section 8; in the formulation of Simon's lecture
notes, chapter 5, hypotheses (5.1) and theorem 5.2) requires: an
almost-everywhere lower bound of one for the density, which integrality
supplies; a bound on the generalized mean curvature in some L^p with
p > k, which stationarity supplies with mean curvature zero; and
smallness of the height and of the tangent-plane excess on a small
ball, which convergence to the multiplicity-one plane supplies, on a
rescaled ball chosen from the sequence that gives the tangent cone.
The point z stays in the support throughout. Allard then produces a
single C^(1,alpha) graph. On a connected graph the stationary integral
multiplicity is constant, and the tangent multiplicity is one, so the
constant is one. The higher regularity statement (Simon, chapter 5,
theorem 6.1) applied with mean curvature prescribed to be zero makes
the graph smooth; that statement allows arbitrary codimension. No
stability hypothesis and no upper bound on k is used.
10d. Low-density cones, by induction on the cone dimension
Here is the induction that the dimension reduction rests on. Read the
statement carefully: the dimension m of the ambient problem is fixed,
and the induction is on the dimension k of the cone.
PROPOSITION 10.5 (low-density cones).
Fix m >= 2 and assume the volume theorem in dimensions 2,...,m-1 (an
empty assumption when m = 2). Let C be a nonzero stationary integral
k-dimensional cone in R^(k+1) with
Theta(C,0) < theta_m. (10.8)
When m = 2 or m = 3, assume in addition that C is cyclic modulo two.
Then:
(i) if 1 <= k <= m, then C is a multiplicity-one hyperplane;
(ii) if k = m+1, then C is smooth and embedded with multiplicity one
away from its vertex, and its link in S^(m+1) is a smooth closed
embedded minimal hypersurface.
The proof is an induction on k. Two features deserve comment before it
starts. First, the threshold theta_m is the same for all k: a cone of
low density is a plane regardless of its dimension, as long as its
dimension does not exceed m+1. Second, the exceptional case k = m+1 is
the only place where the cone can be curved, and even there it is
smooth away from the vertex.
Proof. Throughout, m is fixed and k varies.
Step 1: base case k = 1. An integral one-dimensional cone in R^2 is a
finite union of rays with positive integer weights. Here is why. The
cone's weight measure is conical, so slabbing it radially turns it into
a finite Borel measure on the unit circle, rectifiable of dimension
zero, with integer values; local finiteness makes the total mass finite.
So write the distinct unit directions of the rays as u_1,...,u_r, with
positive integer weights q_1,...,q_r, and put N = q_1 + ... + q_r =
2 Theta(C,0). Stationarity at the origin, tested against the constant
vector field, says
sum over j of q_j u_j = 0. (10.9)
Now distinguish two cases. If m >= 4, Lemma 2.1 gives theta_m <= 3/2,
so (10.8) gives N < 3. Since N is a positive integer, N is 1 or 2.
N = 1 would make (10.9) read u_1 = 0, which is impossible for a unit
vector. So N = 2. Then there are either two rays of weight one, and
(10.9) forces them to be opposite, or one ray of weight two, which would
again violate (10.9). Either way C is a line with multiplicity one.
If m = 2 or m = 3, Lemma 2.1 gives theta_m < 2, hence N < 4, and
cyclicity of C modulo two says that the mod-two boundary coefficient at
the vertex, which is N mod 2, vanishes; so N is even, hence N = 2, and
the same conclusion follows. This proves (i) for k = 1.
Step 2: the induction step. Suppose 2 <= k <= m+1 and that (i) has been
proved for dimension k-1. Let z be a nonzero point of the support of C
and let T be any tangent cone at z. By Lemma 10.3,
T = v(R e) x W, e = z/|z|,
with W a nonzero stationary integral (k-1)-cone in e^perp = R^k, which
is cyclic modulo two whenever C is, and with
Theta(W,0) = Theta(C,z) <= Theta(C,0) < theta_m,
the inequality being (10.4). The dimension of W is k-1, which is at
most m. So the induction hypothesis (i), applied in dimension k-1 with
the same m, says that W is a multiplicity-one plane. Therefore
T = v(R e) x v(e^perp) = v(e^perp (+) R e) = v(R^k),
a plane with multiplicity one. So C has a multiplicity-one planar
tangent cone at every nonzero support point z. By Tool C, C is a smooth
embedded minimal submanifold with multiplicity one in a neighbourhood of
each such z. Since z was an arbitrary nonzero support point,
C is a smooth embedded minimal cone away from the origin,
with multiplicity one. (10.10)
Step 3: finish (i) for k <= m. By (10.10), the link Sigma of C is a
smooth closed embedded minimal hypersurface of S^k of dimension k-1.
Because C is a cone, its mass is determined by the link, and by (10.2)
applied to the link,
M(Sigma) = Theta(C,0) sigma_(k-1) < theta_m sigma_(k-1). (10.11)
Now theta is decreasing in the dimension (Lemma 2.1), and k-1 <= m-1,
so theta_m <= theta_(k-1). Hence
M(Sigma) < theta_(k-1) sigma_(k-1) = a_(k-1). (10.12)
If k-1 >= 2, the volume theorem in dimension k-1, which is part of the
induction hypothesis, now applies to the closed embedded minimal
hypersurface Sigma: a volume below a_(k-1) forces Sigma to be totally
geodesic. If k-1 = 1, then Sigma consists of two points on the circle,
and total geodesicness means that the two points are antipodal; the
same conclusion by inspection. Either way Sigma is a great sphere, and
C is a hyperplane with multiplicity one. This proves (i).
Step 4: finish (ii) for k = m+1. Nothing remains to prove: (10.10) is
exactly (ii), and the link is smooth, closed, embedded minimal. QED
10e. Consequence: small-mass min-max output is smooth
COROLLARY 10.6 (regularity of the min-max output).
Fix m >= 2 and assume the volume theorem in dimensions 2,...,m-1. Let
W be a stationary integral m-varifold in S^(m+1) with
gamma_m <= M(W) < a_m, (10.13)
and assume W is a finite sum of smooth closed embedded minimal
hypersurfaces with positive integer multiplicities when m = 2 or m = 3.
Then
W = v(S) for a smooth closed embedded minimal hypersurface S of
S^(m+1) of multiplicity one, with M(S) = M(W) < a_m.
Moreover W is not a union of equators, and S has at least one connected
component which is non-equatorial.
Proof. Consider the cone C(W) over W in R^(m+2). It is stationary,
integral and of dimension m+1, by Lemma 10.1. Its density at the vertex
is, by (10.2) and (10.13),
Theta(C(W),0) = M(W)/sigma_m < a_m/sigma_m = theta_m. (10.14)
When m = 2 or m = 3, W is a sum of smooth closed embedded minimal
hypersurfaces with integer multiplicities, so C(W) is cyclic modulo two
by Lemma 10.1(c); that is exactly the extra hypothesis of Proposition
10.5. So Proposition 10.5(ii), with k = m+1, applies and says that
C(W) is smooth and embedded with multiplicity one away from the origin.
A cone which is smooth away from the origin with multiplicity one is
the cone over a smooth closed embedded submanifold S of the sphere, and
that S has multiplicity one and is minimal, since its cone is
stationary. So W = v(S) with the stated properties, and M(S) = M(W).
Finally, a union of equators is a union of great m-spheres, hence is
the boundary of a union of hemispheres. Such a varifold has mass
k sigma_m for a positive integer k. Now k = 1 gives mass sigma_m, which
is less than gamma_m, contradicting the lower bound in (10.13); and
k >= 2 gives mass at least 2 sigma_m, which is greater than a_m by
Lemma 2.1, hence greater than M(W), contradicting (10.13). So W is not
a union of equators, and since total mass is the sum of the masses of
the components, at least one component is not an equator. That
component is a closed embedded minimal hypersurface of volume at most
M(W) < a_m. QED
==============================================================================
11. THE END OF THE PROOF
==============================================================================
We now put the pieces together. The statement to be proved, for each
fixed m >= 2, is Theorem 1. The proof is by induction on m, and the
induction hypothesis for dimension m is: the theorem holds in
dimensions 2, 3, ..., m-1.
Step 1, setup and the bound H < a_m. Suppose the theorem fails in
dimension m. Then there is a closed connected manifold M and a smooth
minimal immersion F : M -> S^(m+1) whose image is not totally geodesic
and whose volume is strictly less than a_m. Define
H = infimum of the volumes of all such counterexamples in
dimension m. (11.1)
By assumption the set is nonempty and H <= its elements' volumes; and H
is a nonnegative number. Indeed H < a_m, and by Lemma 2.1,
a_m < 2 sigma_m, so H < 2 sigma_m.
Step 2, a counterexample at volume H exists. Take a minimizing sequence
F_j : M_j -> S^(m+1). By Proposition 5.1, applied to each F_j, the
volume below 2 sigma_m forces F_j to be injective, so each image is a
smooth closed connected embedded minimal hypersurface Sigma_j with
|Sigma_j| -> H. In particular
|Sigma_j| <= 2 sigma_m for all large j.
Fix j and apply the reasoning of Step 3 and Step 5 below to the single
counterexample Sigma_j. Two facts are used, and both are proved in
those steps, so there is no circularity: a counterexample whose volume
is below a_m has index strictly larger than m+3, and Lemma 9.13 and
Theorem 8.6 then produce a stationary integral m-varifold W_j with
gamma_m <= M(W_j) < |Sigma_j|. (11.2)
Since |Sigma_j| tends to H, the masses M(W_j) are bounded by H + 1, and
each W_j is stationary and integral. Tool B gives a subsequence,
relabelled, converging weakly to a stationary integral m-varifold W
with
gamma_m <= M(W) <= H < a_m. (11.3)
The lower bound is the important one. Because gamma_m > sigma_m and
H < a_m < 2 sigma_m, the varifold W is neither a single equator (mass
sigma_m) nor a union of two or more equators (mass at least 2 sigma_m,
which already exceeds a_m > H). So W is nonzero, its mass is below a_m,
and Corollary 10.6 applies to it. The corollary says
W = v(S), S a smooth closed embedded minimal hypersurface,
M(S) = M(W) <= H < a_m, (11.4)
and since W is not a union of equators, some connected component S_0 of
S is a non-equatorial closed embedded minimal hypersurface with
|S_0| <= M(S) <= H.
By the definition (11.1) of H as an infimum over non-equatorial
counterexamples, |S_0| >= H. So |S_0| = H, and S_0 is an attaining
counterexample. Replacing Sigma by S_0, we have
Sigma a smooth closed connected embedded non-equatorial minimal
hypersurface of S^(m+1) with |Sigma| = H < a_m. (11.5)
This is Step 2. Note that the connectedness of Sigma is arranged free
of charge by taking a component at the end.
Step 3, the index is large. Let d = ind(Sigma) be the Morse index. By
Proposition 7.3,
ind(Sigma) >= m+3 = p. (11.6)
Suppose equality holds. Then Proposition 7.6 says that Sigma is, up to
orthogonal change of coordinates, a Clifford product
S^k(sqrt(k/m)) x S^(m-k)(sqrt((m-k)/m)). Its volume is therefore exactly
the volume of that Clifford product, which is at least a_m by the
definition of a_m. That contradicts |Sigma| = H < a_m. Hence
ind(Sigma) > p = m+3. (11.7)
Step 4, a detecting family with mass strictly below H. Since
ind(Sigma) > p, Lemma 9.13 applies to Sigma and produces a detecting
map
Psi : RP^p -> Z (p = m+3)
without concentration of mass, and with
sup over RP^p of mass(Psi) = D_1 < |Sigma| = H. (11.8)
Lemma 8.5 replaces the projective parameter space by a finite cubical
complex K without changing the detecting property, the absence of
concentration of mass, or the supremum D_1.
Step 5, min-max. Now apply Theorem 8.6 with X = K and Phi = Psi. It
produces a stationary integral m-varifold W_1 in S^(m+1) with
gamma_m <= M(W_1) <= D_1 < H. (11.9)
The lower bound gamma_m is the detection gap of Lemma 8.4, and the
upper bound is the mass bound of the family. If 2 <= m <= 6, the
regularity clause of Theorem 8.6 additionally says that W_1 is a finite
sum of smooth closed embedded minimal hypersurfaces with positive
integer multiplicities; that is what Corollary 10.6 needs for m = 2, 3.
Step 6, the contradiction. Apply Corollary 10.6 to W_1, with the upper
bound M(W_1) < H < a_m from our bound (11.9). It gives
W_1 = v(S_1), S_1 a smooth closed embedded minimal hypersurface,
M(S_1) = M(W_1) < H < a_m, (11.10)
with at least one connected component S_1' which is non-equatorial.
That component satisfies
|S_1'| < a_m.
So S_1' is a non-equatorial closed embedded minimal hypersurface of
S^(m+1) of volume strictly less than a_m. It is therefore a
counterexample in dimension m of volume strictly less than H, and that
contradicts the definition (11.1) of H as the infimum of such volumes.
Step 7, conclusion of the induction. The contradiction proves that no
counterexample exists in dimension m: every smooth minimal immersion of
a closed connected m-manifold into S^(m+1) whose image is not totally
geodesic has volume at least a_m. Since m >= 2 was arbitrary, and the
base case m = 2 needs no smaller dimension (the assumption "theorem in
dimensions 2,...,m-1" is empty there), the induction is complete and
Theorem 1 is proved. QED
==============================================================================
APPENDIX: WHAT IS IMPORTED, AND WHAT IS PROVED HERE
==============================================================================
Three results are used as black boxes. Everything else in this
exposition is proved, and each of these three is stated with its exact
hypotheses at the point of use.
1. Allard's interior regularity theorem, in the consequence form of
Tool C in Section 10c. Citation: Allard (1972), Section 8; Simon's
lecture notes, Chapter 5, Theorem 5.2 and Theorem 6.1.
2. Allard's compactness theorem for integral varifolds with locally
bounded mass and first variation, Tool B in Section 5. Citation:
Simon's notes, Chapter 4; White's survey.
3. The Almgren-Pitts min-max existence theorem in the Marques-Neves
formulation, Theorem 8.6. Citations: Almgren (1962); Pitts
(1981); Schoen-Simon (1981) for the regularity of minimizing
hypersurfaces; Marques-Neves (2017), Sections 2-3, for the
discrete formulation and the required extension and homotopy
statements.
Three further facts of analysis and geometry are used, each classical
and each stated where it is needed: the small-filling and isoperimetric
choice lemma of Marques-Neves (2017), Section 3.1, used in Lemma 8.1;
the isoperimetric inequality on the sphere with its hemisphere
equality case, due to Boge-lein, Duzaar and Fusco (2017), Section 2.3,
used in Lemma 8.4; and Obata's characterization of the round sphere by
the existence of a nonconstant solution of Hess f = -f g, used in
Lemma 7.1.
The two genuinely new-looking ingredients are elementary. The
two-point function mu of Section 6 and its upper-test lemma need only
the second fundamental form and a maximum principle. The Lorentz
distance family of Section 9 needs only the signed distance to a cone,
a boost, and the identity trace(A) = 0, which turns into a determinant
bound through the arithmetic-geometric mean inequality.
A reader who wants to check the logic can do it on one page:
- Section 5: a counterexample is an embedding.
- Section 6: the mean of |A|^2 is at least m.
- Section 7: index m+3 forces |A|^2 = m and then a Clifford product,
hence volume at least a_m.
- Section 8: a detecting family must contain a half-volume surface
of perimeter at least gamma_m > sigma_m, and min-max turns such a
family into a stationary integral surface.
- Section 9: a minimal hypersurface of index above m+3 produces a
detecting family whose largest member has area strictly less than
itself.
- Section 10: a stationary integral surface of small mass and small
cone density is smooth, embedded, multiplicity one.
- Section 11: if a counterexample of least volume H exists, its index
exceeds m+3, Section 9 gives a smaller family, Section 8 and 10 give
a smaller counterexample, and H is contradicted.