This paper calculates the index and nullity of Lawson surfaces 1, showing they are linearized isolated with no exceptional Jacobi fields, and provides explicit formulas for these spectral invariants for all genus g.
Contribution
It provides explicit formulas for the index and nullity of Lawson surfaces 1, demonstrating their linearized stability and isolation for all genus g.
Findings
01
Index of 1 is 2g+3 for all g2.
02
Nullity of 1 is 6 for all g2.
03
1 surfaces have no exceptional Jacobi fields, indicating linearized isolation.
Abstract
We prove that the Lawson surface ξg,1 in Lawson's original notation, which has genus g and can be viewed as a desingularization of two orthogonal great two-spheres in the round three-sphere S3, has index 2g+3 and nullity 6 for any genus g≥2. In particular ξg,1 has no exceptional Jacobi fields, which means that it cannot `flap its wings' at the linearized level and is C1-isolated.
∂+Ωij:=qiqjqj+1∪qi+1qjqj+1 and ∂−Ωij:=qiqi+1qj∪qiqi+1qj+1,
∂+Ωij:=qiqjqj+1∪qi+1qjqj+1 and ∂−Ωij:=qiqi+1qj∪qiqi+1qj+1,
∂Ωij=∂+Ωij∪∂−Ωij and Qij=∂+Ωij∩∂−Ωij.
∂Ωij=∂+Ωij∪∂−Ωij and Qij=∂+Ωij∩∂−Ωij.
M=M[C,m]:=i+j∈2Z⋃Dij
M=M[C,m]:=i+j∈2Z⋃Dij
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Full text
The index and nullity of the Lawson surfaces ξg,1
Nikolaos Kapouleas
Department of Mathematics, Brown University, Providence, RI 02912
We prove that the Lawson surface ξg,1
in Lawson’s original notation,
which has genus g and can be viewed as a desingularization of two orthogonal great two-spheres in the round three-sphere S3,
has index 2g+3 and nullity 6 for any genus g≥2.
In particular ξg,1 has no exceptional Jacobi fields,
which means that it cannot “flap its wings” at the linearized level
and is C1-isolated.
1. Introduction
The general framework and brief discussion of the results
Determining the index and nullity of complete or closed minimal surfaces is a difficult problem
which has been fully solved only in a few cases;
see for example [nayatani1992, nayatani1993, morabito].
The index plays an important role in min-max theory [neves2014];
this provides partial motivation for our result.
In this article we prove Theorem 6.21,
which determines (for the first time) the index and the nullity of the Lawson surfaces ξg,1 [Lawson] with g≥2.
These are the Lawson surfaces which have genus g and can be viewed as desingularizations of two orthogonal great two-spheres in the round three-sphere S3
in the sense of [alm20]*Definition 1.3.
The index determined is consistent with (but larger than) a lower bound established by Choe [choe1990].
We prove that the nullity is 6 and so there are no exceptional Jacobi fields,
which means by Corollary 6.23 that these surfaces cannot “flap their wings” at the linearized level
and are C1-isolated.
This provides a partial answer to questions asked in [alm20]*Section 4.2.
The ideas of our proof originate with work of NK on the approximate kernel for Scherk surfaces [compact, alm20].
Our approach requires a detailed understanding of the elementary geometry of S3 and of the surfaces involved,
especially their symmetries.
The proof makes heavy use also of Alexandrov reflection in the style of Schoen’s [schoen1983].
The Courant nodal theorem [courant] and an argument of Montiel-Ros [Ros] play essential roles as well.
In ongoing work we hope to extend this result to determine the index and nullity
of all Lawson surfaces
ξm−1,k−1 in Lawson’s original notation, with m≥k≥3.
Another interesting problem, which could not be posed until the determination of the index of the Lawson surfaces,
is motivated by the characterization of the Clifford torus by Fischer-Colbrie (unpublished) and (independently) by Urbano [urbano]
as the only closed minimal surface in S3, besides the great sphere, which has index ≤5,
and also by some recent results for minimal surfaces in R3 [davi, davi2]:
the problem is to classify all closed minimal surfaces in S3 which have index ≤7
(the index of the Lawson surface of genus two),
or more generally ≤2g+3 for small g
(the index of the Lawson surface ξg,1).
Notation and conventions
We denote by S3⊂R4 the unit 3-dimensional sphere.
Notation 1.1*.*
For any
A⊂S3⊂R4
we denote by Span(A) the span of A as a subspace of R4
and by S(A):=Span(A)∩S3.
∎
Given now a vector subspace V of the Euclidean space R4,
we denote by V⊥ its orthogonal complement in R4,
and we define the reflection in R4 with respect to V,
RV:R4→R4, by
[TABLE]
where ΠV and ΠV⊥ are the orthogonal projections of R4 onto V and V⊥ respectively.
Alternatively
RV:R4→R4 is the linear map which restricts to the identity on V
and minus the identity on V⊥.
Clearly the fixed point set of RV is V.
Definition 1.3** (Reflections RA).**
Given any
A⊂S3⊂R4,
we define A⊥:=(Span(A))⊥∩S3
and
RA:S3→S3 to be the restriction to S3 of RSpan(A).
Occasionally we will use simplified notation:
for example for A as before and p∈S3 we may write S(A,p) and RA,p instead of S(A∪{p}) and RA∪{p}
respectively.
∎
Note that the set of fixed points of RA above is S(A) as in notation 1.1,
which is S3,
or a great two-sphere,
or a great circle,
or the set of two antipodal points,
or the empty set,
depending on the dimension of Span(A).
Following now the notation in [choe:hoppe], we have the following.
Definition 1.4** (The cone construction).**
For p,q∈S3 which are not antipodal we denote
the minimizing geodesic segment joining them by pq.
For A,B⊂S3 such that no point of A is antipodal to a point of B
we define the cone of A and B in S3 by
[TABLE]
If A or B contains only one point we write the point instead of A or B respectively;
we have then p\mbox{\times\times}q=\overline{pq}
for any
p,q∈S3 which are not antipodal.
More generally,
given linearly independent p1,⋯,pk∈S3,
we define inductively for k≥3\overline{p_{1}\cdots p_{k}}:=p_{k}\mbox{\times\times}\overline{p_{1}\cdots p_{k-1}}.
∎
If G is a group acting on a set B and if A is a subset of B,
then we refer to the subgroup
[TABLE]
as the stabilizer of A in G.
When A is a subset of the round 3-sphere, we will set
[TABLE]
In the next definition we find it convenient to work with piecewise-smooth functions on a domain in a surface.
By this we mean that each such function is continuous on the domain,
the domain can be subdivided into domains by a finite union of piecewise-smooth embedded curves,
and on the closure of each of these domains the function is smooth.
We use Cpw∞(U) to denote the space of piecewise-smooth functions on a domain U.
Definition 1.7** (Eigenvalues).**
We assume given a compact domain U in a smooth surface equipped with a Riemannian metric g,
a smooth function f on U,
and a linear space of piecewise-smooth functions V′⊂Cpw∞(U) which is invariant under the Schrödinger operator
L=Δg+f
defined on U.
We define
λi(V′,L) to be the ith eigenvalue,
where we are counting in non-decreasing order and with multiplicity.
(Note also that we follow the conventions which make the eigenvalues of the Laplacian on a closed surface nonnegative.)
Moreover for λ∈R we denote by
#<λ(V′,L),
#=λ(V′,L),
and
#≤λ(V′,L),
the number of eigenvalues λi(V′,L) which are <λ, or =λ, or ≤λ, respectively.
We also define the index of L on V′,
Ind(V′,L):=#<0(V′,L),
and
the nullity of L on V′,
Null(V′,L):=#=0(V′,L).
Finally note that we may omit L from the notation when it can be inferred from the context.
∎
Definition 1.8** (Eigenvalue equivalence).**
Suppose L, U, and V′ are as in 1.7
and L′, U′, and V′′ satisfy correspondingly the same conditions.
We define V′∼L,L′V′′—or V′∼V′′ if the operators are understood from the context—to mean
that there is a linear isomorphism F:V′→V′′ such that the following holds:
∀f′∈V′, f′ is an eigenfunction with respect to L if and only if
F(f′) is an eigenfunction with respect to L′ of the same eigenvalue as f.
We say then that L on V′ and L′ on V′′ are eigenvalue equivalent.
∎
Note that clearly if 1.8 holds,
then ∀i∈N we have λi(V′,L)=λi(V′′,L′).
In this article we will say that a function satisfies the Dirichlet condition on a curve if it vanishes there and the Neumann condition if
its derivative along the normal to the curve vanishes.
Definition 1.9** (Eigenvalues for mixed Dirichlet and Neumann boundary conditions).**
*Suppose L and U are as in 1.7 and moreover
the boundary ∂U is piecewise-smooth and can be decomposed as ∂U=∂DU∪∂NU—note that
∂DU, ∂NU can be empty.
We define then the following for i∈N and λ∈R:
(i)
Cpw∞[U;∂DU,∂NU] to be the space of piecewise-smooth functions on U which satisfy
the Dirichlet condition on ∂DU and the Neumann condition on ∂NU;
(ii)
λi[L,U;∂DU,∂NU]:=λi(L,Cpw∞[U;∂DU,∂NU]);
(iii)
#<λ[L,U;∂DU,∂NU]:=#<λ(L,Cpw∞[U;∂DU,∂NU])
and similarly for “=λ” and “≤λ”.
∎*
Acknowledgments
The authors would like to thank Richard Schoen for his continuous support and interest in the results of this article
and Otis Chodosh for bringing this problem to their attention.
NK was partially supported by NSF grant DMS-1405537.
NK would like to thank also for their hospitality and support the Institute for Advanced Study at Princeton during the 2018 Fall term,
and the University of California, Irvine, during Spring 2019.
Finally we would like to thank Robert Kusner and Richard Schoen for pointing out that our theorem
implies isolatedness for the surfaces involved.
2. Basic spherical geometry
Rotations along or about great circles
Note that by 1.3,
C⊥ is the great circle furthest from a given great circle C in S3.
(Note that the points of C⊥ are at distance π/2 in S3 from C and any point of S3∖C⊥
is at distance <π/2 from C).
Equivalently C⊥ is the set
of poles of great hemispheres with equator C;
therefore C and C⊥ are linked.
The group
GsymC∪C⊥
contains
GsymC=GsymC⊥
(which includes arbitrary rotation or reflection in the two circles)
and includes also
orthogonal transformations exchanging
C with C⊥.
Definition 2.1** (Rotations RCϕ, RϕC and Killing fields KC, KC).**
*Given a great circle C⊂S3, ϕ∈R,
and an orientation chosen on the totally orthogonal circle C⊥,
we define the following:
(i)
the rotation about C by angle ϕ
is the element RCϕ of SO(4) preserving C pointwise
and rotating the totally orthogonal circle C⊥ along itself by angle ϕ
(in accordance with its chosen orientation);
(ii)
the Killing field KC on S3
and the normalized Killing field KC on S3∖C
are given by
21KCp:=∂ϕ∂ϕ=0RCϕ(p)∀p∈S3
and
KCp:=∣KC∣p∣KC∣p∀p∈S3∖C.*
*Assuming now an orientation chosen on C we define the following:
(iii)
the rotation along C by angle ϕ is RϕC:=RC⊥ϕ;
(iv)
the Killing field KC:=KC⊥ on S3
and the normalized Killing field KC:=KC⊥ on S3∖C⊥.
∎*
Note that RϕC=RC⊥ϕ resembles a translation along C,
while in the vicinity of C⊥ it is a rotation.
Note also that KC is defined to be a rotational Killing field around C,
vanishing on C and equal to the unit velocity on C⊥.
Lemma 2.2** (Orbits).**
For KC as in 2.1, the orbits of KC (that is its flowlines) are planar circles
and ∀p∈C each orbit intersects the closed hemisphere C^{\perp}\mbox{\times\times}\mathsf{p} exactly once.
Moreover the intersection (when nontrivial) is orthogonal.
Proof.
This is straightforward to check already in R4 with the
hemisphere C^{\perp}\mbox{\times\times}\mathsf{p} replaced by the half-three-plane containing p and with boundary Span(C⊥).
By restricting then to S3 the result follows.
∎
This lemma allows us to define a projection which effectively identifies the space of orbits in discussion
with a closed hemisphere:
Definition 2.3** (Projections by rotations).**
For C and p as in 2.2
we define the smooth map
\Pi^{C}_{\mathsf{p}}:\mathbb{S}^{3}\to C^{\perp}\mbox{\times\times}\mathsf{p} by requiring ΠpCx to be the intersection of C^{\perp}\mbox{\times\times}\mathsf{p} with the orbit of KC containing x,
for any x∈S3.
Definition 2.4** (Graphical sets).**
A set A⊂S3 is called graphical with respect to KC (with C as above)
if each orbit of KC intersects A at most once.
If moreover A is a submanifold and there are no orbits of KC which are tangent to A,
then A is called
strongly graphical with respect to KC.
The geometry of totally orthogonal circles
We fix now some C and C⊥ as above, and orientations on both.
(Of course, after choosing an orientation on C,
choosing an orientation on C⊥ is equivalent to choosing an orientation on S3.)
We define ∀ϕ∈R the points
[TABLE]
where p0,p0 are arbitrarily fixed points on C and C⊥ respectively.
Note that we will routinely omit [C] when understood from the context.
Using 1.1 we further define ∀ϕ∈R the great spheres
[TABLE]
and ∀ϕ,ϕ′∈R
the great circles
[TABLE]
Definition 2.8** (Coordinates on R4).**
Given C as above and points as in 2.5,
we define coordinates (x1,x2,x3,x4) on R4⊃S3 by requiring that
[TABLE]
Lemma 2.9** (Basic geometry related to C and C⊥).**
*The following hold ∀ϕ,ϕ′,ϕ1,ϕ1′,ϕ2,ϕ2′∈R.
(i) pϕ+π=−pϕ and
pϕ+π=−pϕ.
Similarly
Σϕ+π=Σϕ and
Σϕ+π=Σϕ.
(ii)
Cϕϕ′∩C={pϕ,pϕ+π}
and
Cϕϕ′∩C⊥={pϕ′,pϕ′+π}
with orthogonal intersections.
Moreover
Cϕϕ′=pϕpϕ′∪pϕ′pϕ+π∪pϕ+πpϕ′+π∪pϕ′+πpϕ.
(iii)
C\mbox{\times\times}\mathsf{p}^{\phi} and C^{\perp}\mbox{\times\times}\mathsf{p}_{\phi}
are closed great hemispheres
with boundary C and C⊥ and poles pϕ and pϕ respectively.
(iv)
\Sigma_{\phi}=(C^{\perp}\mbox{\times\times}\mathsf{p}_{\phi})\cup(C^{\perp}\mbox{\times\times}\mathsf{p}_{\phi+\pi})
and
\Sigma^{\phi}=(C\mbox{\times\times}\mathsf{p}^{\phi})\cup(C\mbox{\times\times}\mathsf{p}^{\phi+\pi}).
(v)
Σϕ∩C⊥={pϕ,pϕ+π}
and
Σϕ∩C={pϕ,pϕ+π}
with orthogonal intersections.
(vi)
Cϕϕ′=Σϕ∩Σϕ′
with orthogonal intersection.
(vii)
(Cϕϕ′)⊥=Cϕ+π/2ϕ′+π/2.
(viii)
Σϕ∩Σϕ′=C unless ϕ=ϕ′(modπ)
in which case
Σϕ=Σϕ′.
Similarly
Σϕ∩Σϕ′=C⊥ unless ϕ=ϕ′(modπ)
in which case
Σϕ=Σϕ′.
In both cases the intersection angle is ϕ′−ϕ(modπ).
(ix)
Cϕ1ϕ1′∩Cϕ2ϕ2′=∅
unless
ϕ1=ϕ2(modπ)
or
ϕ1′=ϕ2′(modπ).
If both conditions hold then
Cϕ1ϕ1′=Cϕ2ϕ2′.
If only the first condition holds then
Cϕ1ϕ1′∩Cϕ2ϕ2′={pϕ1,pϕ1+π}
with intersection angle equal to ϕ2′−ϕ1′(modπ).
If only the second condition holds then
Cϕ1ϕ1′∩Cϕ2ϕ2′={pϕ2,pϕ2+π}
with intersection angle equal to ϕ2−ϕ1(modπ).*
Proof.
It is straightforward to verify all these statements by using the coordinates defined in
2.8.
∎
Definition 2.10** (Symmetries of Killing fields).**
We call a Killing field Keven (odd) under an isometry R if it satisfies
R∗∘K=K∘R
(R∗∘K=−K∘R).
∎
Lemma 2.11** (Some symmetries of Killing fields).**
*The following hold ∀ϕ,ϕ′∈R.
(i)
KC is
odd under RΣϕ
and
RCϕϕ′
and even under RΣϕ.
(ii)
KC⊥ is
odd under RΣϕ
and
RCϕϕ′
and even under RΣϕ.
(iii)
KCϕϕ′ is odd under
RΣϕ and RΣϕ′
and even under
RΣϕ+π/2 and RΣϕ′+π/2.
Moreover
Σϕ+π/2 and Σϕ′+π/2
are preserved under the flow of
KCϕϕ′ and contain the fixed points ±pϕ′∈Σϕ+π/2 and ±pϕ∈Σϕ′+π/2
and the geodesic orbit
Cϕ+π/2ϕ′+π/2=Σϕ+π/2∩Σϕ′+π/2.*
Proof.
For any great circle C′ we have that KC′ is even (odd) with respect to a reflection R
if and only if R(C′⊥)=C′⊥ and R respects (reverses) the orientation of C′⊥.
Applying this it is straightforward to confirm the lemma.
∎
3. Tessellations of S3
Lawson tessellations
Our purpose is to study the Lawson surfaces ξm−1,1 [Lawson],
which have genus g=m−1 and can be viewed as desingularizations of Σπ/4∪Σ−π/4,
where m≥3, m∈N.
With this goal it is helpful to introduce the notation
[TABLE]
Note that we have then 2m points qi for i∈Z subdividing C into 2m equal arcs of length π/m each,
and 4 points qj for j∈Z subdividing C⊥ into 4 arcs of length π/2 each.
qi+21 is the midpoint of qiqi+1 for each i∈Z and
qj+21 is the midpoint of qjqj+1 for each j∈Z.
We define now ∀i,j∈Z compact domains
Ωi,Ωj,Ωij by
[TABLE]
Clearly we have then the decompositions with disjoint interiors
[TABLE]
Note that
[TABLE]
Moreover we have
[TABLE]
Lemma 3.6** (Properties of Ωij).**
∀i,j∈Z*, Ωij is a spherical tetrahedron and satisfies the following.
(i)
Its faces are the spherical triangles
qiqjqj+1,
qi+1qjqj+1,
qiqi+1qj,
and
qiqi+1qj+1.
(ii)
Its dihedral angles are all π/2 except for the one along
qjqj+1
which is π/m.
(iii)
It is bisected by the spherical triangles
qi+21qjqj+1
and
qiqi+1qj+21
and its symmetries are given by (RS3 is the identity map on S3)*
[TABLE]
(iv) It is convex in the sense that xy⊂Ωij∀x,y∈Ωij.
Proof.
It is straightforward to check all these statements by using the definitions and for (iii) that m>2.
∎
3.6.iii
motivates us to define ∀i,j∈Z,
by modifying 3.2,
compact domains
Ωi±,Ωj±,Ωi±j± by
[TABLE]
We have then various decompositions with disjoint interiors, for example
[TABLE]
Note also that
[TABLE]
Moreover
all four tetrahedra Ωi±j± have
pimπpj2π=qi+21qj+21
as a common edge and adjacent ones have
common faces given by
Ωi−j+∩Ωi−j−=qi+21qiqj+21,
Ωi+j+∩Ωi+j−=qi+21qi+1qj+21,
Ωi−j−∩Ωi+j−=qi+21qj+21qj,
and
Ωi−j+∩Ωi+j+=qi+21qj+21qj+1.
Subdividing S3 with mutually orthogonal two-spheres
Note that by 2.9.vi,viii
Σ0, Σπ/2, Σ0, and Σπ/2
form a system of four mutually orthogonal two-spheres in S3.
We will later study the subdivisions these two-spheres effect on S3 and the Lawson surfaces.
To this end we define
Ω∗∗±∗,
Ω∗∗∗±,
Ω±∗∗∗,
and
Ω∗±∗∗,
to be the closures of the connected components into which S3 is subdivided by the removal of
Σ0, Σπ/2, Σ0, or Σπ/2 respectively,
chosen so that
[TABLE]
To further subdivide we replace ∗’s by ± signs to denote the corresponding intersections
of these domains; for example we have
[TABLE]
Clearly we have
[TABLE]
Lemma 3.14** (Elementary geometry of Ω++++).**
Ω++++* is the spherical tetrahedron
p0pπ/2p0pπ/2
and satisfies the following.
(i)
Its faces are the spherical triangles
pπ/2pπ/2p0⊂Σπ/2,
pπ/2pπ/2p0⊂Σπ/2,
p0pπ/2p0⊂Σ0,
p0pπ/2p0⊂Σ0.
All angles of all faces are π/2.
(ii)
All its edges have length π/2 and its dihedral angles are all π/2.
(iii)
Its symmetry group is isomorphic to the symmetric group on its vertices.
Ω++++ is bisected by six spherical triangles including
pπ/4p0pπ/2
and
p0pπ/2pπ/4
and its symmetries include
RΣπ/4,
RΣπ/4,
and
RCπ/4π/4.*
Proof.
It is straightforward to verify all these statements by using the definitions.
∎
Lemma 3.15** (Some decompositions).**
*We have the following.
(i)
Ω∗∗++=∪i=02m−1(Ωi0+∪Ωi1−).
(ii)
Ω+∗++=Ω++++∪RΣπ/2Ω++++=Ω0+0+∪Ω0+1−∪(∪i=1m−1(Ωi0+∪Ωi1−))∪Ωm−0+∪Ωm−1−.
(iii)
Ω++++=Ω0+0+∪Ω0+1−∪⎩⎨⎧(∪i=12m−1(Ωi0+∪Ωi1−))∪Ω2m−0+∪Ω2m−1−,(∪i=12m−1(Ωi0+∪Ωi1−)), if m∈2Z, if m∈2Z+1.*
Proof.
It is straightforward to verify all these statements by using the definitions.
∎
The coordinate Killing fields
Using the coordinates defined in 2.8,
we endow R4 with its standard orientation dx1∧dx2∧dx3∧dx4,
and we endow the six coordinate 2-planes with the orientations
[TABLE]
Note that these orientations have been chosen so that one obtains the orientation
of R4 upon taking the wedge product (in either order)
of the orientation forms of a pair of orthogonally complementary 2-planes.
In turn we orient each coordinate unit circle by taking the interior product
of its outward unit normal with the orientation form of the 2-plane it spans.
These choices are consistent with the convention that for any oriented great circle C′
we orient C′⊥ so that the wedge product of
the two corresponding 2-plane orientations will yield the standard orientation on R4.
Thus
[TABLE]
Lemma 3.18** (KCϕϕ′ on Ω+±++ for ϕ,ϕ′∈{0,π/2}).**
All claims follow easily from (3.17) and Definition 2.1.
∎
Some quadrilaterals in S3
We consider now ∀i,j∈Z
the spherical quadrilateral Qij⊂∂Ωij
consisting of the four edges of the spherical tetrahedron Ωij
not contained in C or C⊥;
that is
[TABLE]
For ease of reference we define the set of vertices of Qij
(the same as the set of vertices of Ωij)
[TABLE]
Recall that by 3.7∀i,j∈Z the circle
S(qi+21,qj+21)=Ciπ/mjπ/2 is an axis of symmetry of Ωij.
It is natural then to call this circle the “axis” of Ωij
and study rotations along it as in the following lemma.
We also define
Lemma 3.23** (Ωij and rotations along its axis).**
*The following are true ∀i,j∈Z and any orbit O of
KC,
where
C:=(Ciπ/mjπ/2)⊥=Ciπ/m+π/2jπ/2+π/2.
(i) (RCtΩij)∩Ωij=∅ for t∈(−3π/2,−π/2)∪(π/2,3π/2).
Moreover either
(RC±π/2Ωij)∩Ωij={qi+21} or
(RC±π/2Ωij)∩Ωij={qj+21}
(depending or the orientation of Ciπ/mjπ/2 and the sign).
(ii)
For each Ωi±j± either O∩Ωi±j±=O∩Ωij or O∩Ωi±j±=∅.
(iii)
If O∩Ωij=∅,
then (recall 3.21)
O∩∂±Ωij={x±} for some x±∈∂±Ωij.
Moreover O∩Ωij is
a connected arc (possibly a single point) whose endpoints are x+ and x−.
(iv)
If O∩Qij=∅, then x+=x−∈Qij
and
O∩Ωij={x+}.
(v)
If O∩(Ωij∖Q\mspace0.8mu/ij)=∅,
then O intersects each face of Ωij containing x+ (x−) transversely.
(vi)
\Pi_{i}^{j}(\Omega_{i}^{j})\subset C_{i\pi/m+\pi/2}^{j\pi/2+\pi/2}\mbox{\times\times}\mathsf{p}_{i\pi/m}
is homeomorphic to a closed disc with boundary
Πij(Qij),
where
Πij:=Πpiπ/mCiπ/mjπ/2
is defined as in 2.3
(recall also piπ/m=qi+21).*
Proof.
We can clearly assume without loss of generality that i=j=0.
To prove (i) note that H:=\mathsf{p}^{0}\mbox{\times\times}C^{\pi/2}_{\pi/2} and H^{\prime}:=\mathsf{p}_{0}\mbox{\times\times}C^{\pi/2}_{\pi/2} are orthogonal
closed hemispheres with common boundary Cπ/2π/2,
intersecting C00 orthogonally at p0 and p0 respectively,
and satisfying
RCπ/2(H′)=H.
Moreover q0q1⊂H and q0q1⊂H′ with both geodesic segments avoiding the boundary Cπ/2π/2.
Since two orthogonal hyperplanes separate R4 into four convex connected components,
(i) follows easily.
Because each of the bisecting spheres Σ0 and Σ0
is preserved by the family RtC00,
the orbits of KC00 cannot cross either sphere, proving (ii).
Before turning to the remaining items
we first show that no orbit of KC00 intersects any face of Ω00
tangentially, except at a vertex.
By the symmetries it suffices to prove that orbits
intersect p0q1q1⊂Σπ/2m
and p0q1q1⊂Σπ/4 transversely (if at all)
except at q1 (the orbit through which is tangential to Σπ/2m)
and q1 (the orbit through which is tangential to Σπ/4).
Of course the spheres Σπ/2m and Σπ/4 are minimal surfaces
and neither contains C00,
so the Killing field KC00
induces a nontrivial Jacobi field on each of them.
A point where an orbit meets one of these spheres tangentially is a zero
of the corresponding Jacobi field,
but we know these nontrivial Jacobi fields are simply first harmonics,
each of whose nodal sets consists of a single great circle.
Clearly the reflection RΣπ/2 (RΣπ/2) preserves
the sides of Σπ/2m (Σπ/4)
and reverses each orbit of KC00.
Thus orbits can meet Σπ/2m (Σπ/4) tangentially
only along Cπ/2mπ/2 (Cπ/2π/4),
which intersects q1q1 only at q1 (q1),
establishing the asserted transversality.
Next we argue that no orbit of KC00 intersects
any face of Ω00 at more than one point.
Again (by the symmetries) it suffices to
show that every orbit intersects each of the faces
p0q1q1⊂Σπ/2m
and p0q1q1⊂Σπ/4
at most once.
To see this first note that the orbits of KC00 in R4⊃S3
are planar circles, so if one intersects a great 2-sphere at more than one point,
then the intersection must be either a great circle (the entire orbit) or a pair of points.
In the first case the 2-sphere so intersected must contain C00,
but neither the sphere Σπ/4⊃p0q1q1
nor the sphere Σπ/2m⊃p0q1q1
contains C00, and so the orbits of KC00 must meet these spheres at most twice.
However, the reflection RΣπ/2 (RΣπ/2)
preserves both Σ0 (Σπ/4)
and each orbit (as a set) of KC00,
so that if an orbit intersects Σ0 (Σπ/4) in two points,
these points must lie on opposite sides of Σπ/2 (Σπ/2).
Since in fact Ω00 crosses neither sphere of symmetry,
we see that any orbit meets each face at most once, as claimed.
Now we are ready to prove (iii), (iv), and (v).
By the symmetries
it suffices to consider an orbit O intersecting Ω0+0+.
Of course by (i) O is not contained in Ω0+0+
and obviously by (ii) O can enter (or exit) Ω0+0+ only through
p0q1q1 or p0q1q1,
but by the preceding paragraph it intersects each at most once.
Since q1q1 lies on both these triangles,
it follows that any orbit O meeting q1q1
intersects Ω00 at only one point.
If on the other hand O misses q1q1,
then, by the transversality above,
it must intersect the interior of Ω00,
so in this case it must cross p0q1q1∪p0q1q1 at least twice,
meaning, by the above, that in fact O must intersect each of these triangles exactly once.
This completes the proof of (iii), (iv), and (v).
For (vi) set Π:=Π00.
Since the quadrilateral Q00 is itself a closed curve missing
Cπ/2π/2=Π−1(Cπ/2π/2),
its image Q′:=Π(Q00) under Π
is likewise a closed curve missing Cπ/2π/2.
By item (iv) (and the embeddedness of Q00)
it follows that Q′ is an embedded closed curve in the interior of C_{\pi/2}^{\pi/2}\mbox{\times\times}\mathsf{p}_{0},
so that \left(C_{\pi/2}^{\pi/2}\mbox{\times\times}\mathsf{p}_{0}\right)\backslash Q^{\prime}
has two connected components, one the disc bounded by Q′
and the other the annulus bounded by Q′ and Cπ/2π/2.
Call the closure of the disc D′.
Since the hemisphere
\Pi^{-1}\left(\overline{\mathsf{p}_{0}\mathsf{p}_{\pi/2}}\right)=C_{0}^{0}\mbox{\times\times}\mathsf{p}_{\pi/2}\subset\Sigma^{0}
intersects Q00 only at q1,
we see that the geodesic arc p0pπ/2
intersects Q′ exactly once (at q1),
and so we conclude that p0∈D′.
A second application of item (iv)
ensures that Π(Ω00\Q00)
misses Q′,
but Ω00\Q00 is connected and includes p0,
so we have Π(Ω00)⊂D′.
Last, note that D′′:=q0q1q0∪q0q1q1
is a disc in Ω00 whose boundary is Q00 and thereby mapped by Π
homeomorphically onto Q′=∂D′.
It follows (by degree theory) that Π(D′′)=D′,
and so of course Π(Ω00)=D′ as well.
∎
4. The Lawson surfaces
Definition, uniqueness, and symmetries
Note that the surfaces we define below are the surfaces called ξm−1,1 in [Lawson].
Recall that these surfaces can be viewed as desingularizations of two orthogonal great two-spheres.
In this article we do not consider any other Lawson surfaces and when we refer to Lawson surfaces we mean these surfaces only.
The surfaces defined in the next theorem are positioned so that they can be viewed as desingularizations of Σπ/4∪Σ−π/4 along C.
Note also that we restrict our attention to the case m≥3 because the surfaces produced
otherwise are the great sphere (m=1) and the Clifford torus (m=2).
Theorem 4.1** (Lawson 1970 [Lawson]).**
Given an integer m≥3
there is a unique compact connected
minimal surface Dij⊂Ωij with
∂Dij=Qij (recall (3.2) and (3.19)).
Moreover Dij is a disc, minimizing area among such discs,
and
[TABLE]
is an embedded connected closed (so two-sided) smooth minimal surface of genus m−1.
Proof.
The theorem except for the uniqueness part but including
the existence of a minimizing disc Dij is proved in [Lawson].
Although the uniqueness is also claimed in [Lawson],
the subsequent literature
(for example [choe:soret:2009]) does not assume uniqueness known.
We provide now a simple proof of uniqueness.
Suppose D′ij is another connected minimal surface in Ωij with boundary Qij.
By 3.23RCiπ/m+π/2jπ/2+π/2tDij cannot intersect
D′ij for any t∈(−π,0)∪(0,π) because
otherwise we can consider the sup or inf of such t’s which we call t′.
For t′ then we would have tangential contact on one side in the interior.
By the maximum principle
[schoen1983]*Lemma 1
this would imply
equality of the surfaces and the boundaries, a contradiction.
By 3.23
the orbits which are close enough to Qij∖Q\mspace0.8mu/ij and intersect Ωij
also intersect Dij and D′ij.
Since there are no intersections for t=0 above, we conclude that
Dij and D′ij agree on a neighborhood of Qij∖Q\mspace0.8mu/ij and therefore by analytic continuation they are identical.
∎
Corollary 4.2** (Symmetries of the Lawson discs).**
∀i,j∈Z* Dij inherits the symmetries of Ωij: it is preserved as a set by
RΣiπ/m=RΣti+21=RC⊥,qi+21,
RΣjπ/m=RΣtj+21=RC,qj+21,
and
the composition of those
RCiπ/mjπ/2.
Moreover it has no more symmetries.*
Proof.
That the symmetries of Ωij are symmetries of Dij follows from the uniqueness of Dij discussed in 4.1.
Any symmetry of Dij has to be a symmetry of its boundary and then of its vertices, and hence of Ωij as well.
By 3.6.iii this completes the proof.
∎
Lemma 4.3** (Generating symmetries of the Lawson surfaces).**
*For M=M[C,m] as in 4.1 we have the following symmetries,
which generate GsymM.
(i)
∀i,j∈Z we have
RΣjπ/2,RΣiπ/m∈GsymM.
Moreover
the collection of the great two-spheres of symmetry of M is
{Σjπ/2}j∈Z∪{Σiπ/m}i∈Z and contains m+2 spheres.
(ii)
∀i,j∈Z we have
RC(2i−1)2mπ(2j−1)4π=Rqi,qj∈GsymM.
Moreover
the collection of great circles contained in M is
{C(2i−1)2mπ(2j−1)4π=S(qi,qj)}i,j∈Z
and contains 2m great circles.*
Furthermore if ν:M→S3 is a unit normal smoothly chosen on M, then
ν is even under the symmetries in (i)
(that is for such a symmetry R we have ν∘R=R∗∘ν)
and odd under the symmetries in (ii)
(that is for such a symmetry R we have ν∘R=−R∗∘ν).
Proof.
Set Q:={Qij}i+j∈2Z
and Ω:={Ωij}i+j∈2Z.
It is easy to see (keeping in mind that m>2)
that an element of O(4) permutes Q
if and only if it permutes Ω.
By the uniqueness assertion of Theorem 4.1
any element of O(4) permuting Q
is then a symmetry of M.
Conversely,
since M is disjoint from the interior of every Ωij with i+j∈2Z+1,
every element of GsymM
must permute Q.
Now write G for the subgroup of O(4)
generated by all the orthogonal transformations named in the statement of the lemma.
It is immediately verified from definitions
1.3, 2.1, 2.6, 2.7, 3.2, and 3.19
that every element of G indeed permutes
Q,
confirming that G⊆GsymM.
In fact it is clear that G acts transitively on Ω,
so in order to show that GsymM⊆G
it suffices to show that any orthogonal transformation
preserving Ω00 as a set belongs to G,
but this is evident from 3.7.
Thus GsymM=G.
The counts of the spheres and circles named in (i) and (ii)
are obvious from (2.6) and (2.7) alone.
It is also obvious from the definitions that every circle
in item (ii) of the lemma
indeed lies on M,
and furthermore for this very reason
reflection through such a circle
must reverse ν.
Note that for each j∈Z the symmetry RΣjπ/2
fixes C pointwise.
In particular RΣjπ/2
fixes the point q1∈C∩M,
but q1
lies on each of the circles of symmetry
Cπ/2mπ/4 and Cπ/2m−π/4
orthogonally intersecting C there,
and so ν(q1) points along C
and is thereby preserved by RΣjπ/2.
A similar argument shows that ν is likewise preserved by every RΣiπ/m
with i∈Z.
The only assertions left to prove are
that M is invariant under no spheres of symmetry
other than those enumerated in (i)
and that M contains no circles of symmetry other than those enumerated in (ii)
(since the reflection principle [Lawson]*Proposition 3.1
then ensures that M contains no other great circles at all).
Accordingly suppose that S is such a sphere or circle of symmetry.
so that RS∈GsymM.
As explained above,
RS therefore permutes Q,
but because m>2, this requires in particular that RS preserve C (and so C⊥ too) as a set.
If S is a great sphere, it must consequently intersect either C or C⊥ orthogonally
(containing the other), but to permute Q it can then be only one of the spheres listed in (i)
(a sphere bisecting some Ωij, since reflection through a sphere containing a face of an Ωij
takes the corresponding Qij to a quadrilateral outside of Q).
If instead S is a great circle, in order to preserve C as a set it
must (a) coincide with C, (b) coincide with C⊥, or (c) intersect C (and so also C⊥) orthogonally.
Clearly neither C nor C⊥ is contained in M, since, for example, neither
q0q1 nor q0q1 is contained in ∂D00.
In case (c), in order to permute Q, S can be only one of the circles
listed in (ii) (a circle containing an edge of a quadrilateral in Q)
or one of the circles of intersection of a pair of spheres of symmetry
(a circle bisecting the edges on C and C⊥ of some Qij, not necessarily having i+j even),
but none of these latter circles is contained in M,
since, for example, for all i,j∈Zqi+21∈∂Dij.
∎
Corollary 4.4** (Umbilics on the Lawson surfaces).**
For M=M[C,m] as in 4.1 we have only four umbilics, q1, q2, q3, and q4,
of degree (as in [Lawson]) m−2 each.
Proof.
By the symmetries it is clear that each of these point is an umbilic of degree m−2.
By a result of Lawson [Lawson]*Proposition 1.5
the total degree of the umbilics is 4g−4=4m−8 and so there can be no other.
∎
Corollary 4.5** (The unit normal on the geodesic segments qiq1).**
By appropriate choice of the unit normal
ν:M→S3 smoothly defined on M we have ∀i∈Z
[TABLE]
Proof.
A unit vector normal to a great circle C′⊂S3
lies on the circle C′⊥.
By the symmetries (Lemma 4.3)
the unit normal ν on M∩C must point along C,
while on M∩C⊥ it must point along C⊥.
Thus ν(q1)=±q1+m/2 and ν(q1)=±q0.
Assume that
ν(q1)=q1+m/2.
Using Lemma 4.3 again,
it suffices to complete the proof for i=1.
Since M is disjoint from the interior of Ω10 (and Ω01),
we conclude that along all of q1q1
the normal ν cannot cross either Σπ/2m or Σπ/4
and more specifically, by our choice of ν(q1), must point into Ω10.
It follows that ν(q1)=q0
and ν(x)⋅ν(q1)≥0 and ν(x)⋅ν(q1)≥0
for all x∈q1q1,
completing the proof.
∎
Although it is not needed in this article,
we include the following lemma to offer a fuller picture of the symmetry group.
Lemma 4.6** (Further symmetries of the Lawson surfaces).**
*For M=M[C,m] as in 4.1 we have the following symmetries.
(i) A great circle C′⊂M is a circle of symmetry for M
if and only if (a) C′=C, (b) C′ is one of the 2m circles
Ciπ/mjπ/2 having i,j∈Z,
or (c) m is even and C′=C⊥.
Each such RC′ preserves ν.
(ii) The antipodal map R∅ belongs to GsymM if and only if m is even,
in which case R∅ preserves ν.
(iii) A point x∈S3 is a point of symmetry for M
(Rx∈GsymM) if and only if
(a) x is one of the 2m points piπ/m with i∈Z
or (b) x is one of the 4 points pjπ/4 with j∈2Z+m.
Each such Rx reverses ν.
(iv) For every i∈Z the map
Rqi∘RCπ/2∈GsymM and reverses ν.*
Proof.
It is easy to see that for any i∈Z
both Rqi and RCπ/2
exchange the sets {Ωij}i+j∈2Z
and {Ωij}i+j∈2Z+1,
so that the composite acts as a permutation on each of these sets
and therefore (as explained in the proof of Lemma 4.3)
belongs to GsymM;
it is also easy to see that the composite reverses the normal at qi,
completing the proof of (iv).
Item (ii) follows from (i),
since R∅=RCRC⊥.
The fact that the circles listed in (i)
exhaust all circles of symmetry not lying on M
follows from the final paragraph of the proof of Lemma 4.3.
The rest of (i) is easily proven using Lemma 4.3 itself
(and the group structure of O(4)) as follows.
Clearly RΣ′∘RΣ′′=RΣ′∩Σ′′
for any two great spheres Σ′ and Σ′′ intersecting orthogonally.
On the other hand, according to Lemma 4.3,
the great 2-spheres of symmetry of M are precisely the spheres Σiπ/m and Σjπ/2
for i,j∈Z, so in particular Σπ/2 is a sphere of symmetry precisely when m is even.
Together, the preceding two sentences complete the proof of (i).
To prove (iii)
first note that for any point x∈S3
the set x⊥
is the round 2-sphere centered at ±x,
and
moreover
Rx=R−x=−Rx⊥=RΣx∘RCx,
where Σx is any great sphere through ±x and Cx is any
great circle orthogonally intersecting Σx at ±x.
In particular Rx∈GsymM precisely when Rx⊥
takes M to −M.
Since −Dij=Di+mj+2,
we have −M=⋃i+j≡mmod2Dij.
It is clear from Lemma 4.3 that GsymM preserves C as a set
(since each generator obviously does so). Thus in order for x to be a point
of symmetry of M (whatever the parity of m) x must lie on either C or C⊥.
Since C is itself a great circle of symmetry,
any point of symmetry lying on C must also lie on a sphere of symmetry intersecting C orthogonally.
Thus the set of points of symmetry lying on C
is simply {piπ/m}i∈Z.
To identify the points of symmetry on C⊥
we observe by Lemma 4.3
that GsymM preserves the set {qj}j∈Z,
which means that a point of symmetry on C⊥ must lie in
{qj}j∈21Z={pjπ/4}j∈Z.
It is easy to see that RΣjπ/4+π/2
takes M to −M precisely when j−m∈2Z,
which completes the proof.
∎
Graphical properties
Lemma 4.7** (Graphical property and subdivisions of Dij).**
∀i,j∈Z* the following hold.
(i)
Dij is graphical—with its interior strongly graphical—with respect to
KCiπ/mjπ/2=KCiπ/m+π/2jπ/2+π/2 (recall 2.4)
and each orbit which intersects Ωij intersects Dij as well.
(ii)
Each of Di±j:=Dij∩Ωi±j, Dij±:=Dij∩Ωij±, and Di±j±:=Dij∩Ωi±j±,
is homeomorphic to a closed disc.*
Proof.
To prove (i) we first prove that Dij is graphical.
This follows by the same argument
as in the second paragraph of the proof of 4.1 but with D′ij replaced by Dij.
Consider now the Jacobi field ν⋅KCiπ/m+π/2jπ/2+π/2,
which clearly by the graphical property and appropriate choice of ν is ≥0 on Dij and hence by the maximum principle is >0 on the interior of Dij.
This implies that the interior of Dij is strongly graphical.
Next we
recall the projection map
[TABLE]
defined in 3.23.vi.
Let D′:=Πij(Ωij),
which by 3.23.vi is homeomorphic to a closed disc with ∂D′=Πij(Qij).
Clearly then Πij(Dij)⊂D′.
Since ∂Dij=Qij we have also
Πij(∂Dij)=∂D′,
and therefore
Πij(Dij)=D′,
which completes the proof of (i).
Furthermore, as shown above, Dij is graphical with respect to KCiπ/mjπ/2,
so the restriction Πij∣Dij is one-to-one.
We conclude that Πij takes Dij homeomorphically onto
D′.
The proof of (ii) is then completed by the fact that Πij clearly respects the symmetries of Ωij.
∎
By the definitions when i+j∈2Z we have
M∩Ωi±j±=Di±j±;
otherwise we have
M∩Ωi±j±=∅.
By 4.7
each Di±j± is an embedded minimal disc.
To study ∂Di±j± and the intersections with two-spheres of symmetry we define
the intersections of Dij and Di±j± with the bisecting two-spheres as follows.
[TABLE]
Lemma 4.10** (The α and β curves).**
∀i,j∈Z* the following hold.
(i)
Dij intersects pimπpj2π=qi+21qj+21
at a single point which we will call xij.
(ii)
The sets αij−, αij+, βi−j, βi+j, αij, and βij
are connected curves with
∂αij−={qj,xij},
∂αij+={qj+1,xij},
∂βi−j={qi,xij},
∂βi+j={qi+1,xij},
∂αij={qj,qj+1},
and
∂βij={qi,qi+1}.
As in the previous proof we consider Πij,
which is a homeomorphism from Dij onto D′ and moreover respects the symmetries of Ωij.
Using the various definitions it is then straightforward to complete the proof.
∎
Lemma 4.11** (Graphical properties of Di±j±).**
∀i,j∈Z* the following hold.
(i)
The interior of Dij is contained in the interior of Ωij and the conormal of Dij at a point of Qij∖Q\mspace0.8mu/ij
is transverse to each face of Ωij containing the point.
(ii)
Di±j (as in 4.7(ii)) is graphical with respect to KC⊥ and strongly graphical in its interior.
(iii)
Dij± (as in 4.7(ii)) is graphical with respect to KC and strongly graphical in its interior.*
Proof.
(i) follows easily by the maximum principle
[schoen1983]*Lemma 1.
The proofs of (ii) and (iii) are based on Alexandrov reflection in the style of [schoen1983].
Clearly ΠqiCΩij=qiqjqj+1
by 2.3 and 3.6.
For t∈[0,π/m] we define (recall 3.1)
[TABLE]
We clearly have then Dij=Di,tj∪Di:tj and Di,tj∩Di:tj=Dij∩qi,tqjqj+1.
Clearly
Ωij∖Dij consists of two connected components,
which in this proof we call U1 and U2,
where
U1 is chosen to be the component which contains the interior of
qiqjqj+1.
We define
[TABLE]
(i) implies that Di,tj is graphical for t small enough,
and therefore
t∈/T for t small enough.
We conclude that t′:=infT>0.
If t′<2mπ, then by the definition of t′,
RC⊥,qi,t′Di,t′j
and
Di:t′j have a point of one-sided interior or boundary tangential contact.
By the maximum principle
[schoen1983]*Lemma 1 and analytic continuation this implies that
RC⊥,qi,t′Di,t′j
and
Di:t′j are identical contradicting the symmetries of Qij (alternatively 4.3).
We conclude that t′≥2mπ and hence
[TABLE]
Using this we prove now that
Di−j
is graphical with respect to KC⊥:
Otherwise there would be an orbit which would contain two points y1=y2 with
yi∈Dij∩qi,tiqjqj+1
for i=1,2, where 0<t1<t2≤2mπ.
y2 is then a point of interior one-sided tangential contact of
RC⊥,qi,t∗Di,t∗j
and Dij, where t∗=2t1+t2∈(0,2mπ).
This implies that
RC⊥,qi,t∗
is a symmetry of Dij, and hence of ∂Dij=Qij, which is a contradiction.
To prove that it is strongly graphical in the interior we argue as in the proof of 4.7.
By symmetry we conclude the statement for
Di+j.
This completes the proof of (ii).
The proof of (iii) is similar with the roles of C and C⊥ exchanged.
∎
We define [m:2]:=0 if m∈2Z and [m:2]:=1 otherwise.
Lemma 4.12** (Some intersections of M with great two-spheres).**
That the circles are contained in the intersections in (i) and (iii) follows from the definition of M in 4.1
and the reverse inclusions follow from 4.11.i
completing the proof of (i) and (iii).
By 2.9.iv we have
\Sigma_{i\frac{\pi}{m}\,}=(C^{\perp}\mbox{\times\times}\mathsf{p}_{i\frac{\pi}{m}\,})\cup(C^{\perp}\mbox{\times\times}\mathsf{p}_{i\frac{\pi}{m}+\pi\,})
and
\Sigma^{j\frac{\pi}{2}\,}=(C\mbox{\times\times}\mathsf{p}^{j\frac{\pi}{2}\,})\cup(C\mbox{\times\times}\mathsf{p}^{j\frac{\pi}{2}+\pi\,}).
By 4.9 and 4.1 we have
M\cap(C^{\perp}\mbox{\times\times}\mathsf{p}_{i\frac{\pi}{m}\,})=\bigcup_{j\in 2\mathbb{Z}-i}\alpha_{i}^{j}
and
M\cap(C\mbox{\times\times}\mathsf{p}^{j\frac{\pi}{2}\,})=\bigcup_{i\in 2\mathbb{Z}-j}\beta_{i}^{j} .
Using 3.1 we complete the proof of (ii) and (iv).
∎
Subdividing the Lawson surfaces with mutually orthogonal two-spheres
Definition 4.13**.**
For M=M[C,m] as in 4.1
we define M±±±±:=M∩Ω±±±±,
where instead of ± we could also have ∗
(recall 3.12).
For example
M+−−∗:=M∩Ω+−−∗.
∎
Lemma 4.14** (Description of M∗∗++).**
*The following hold.
(i)
M∗∗±± is homeomorphic to a closed disc and
M∗∗++=∪i=0m−1(D2i0+∪D2i+11−).
(ii)
∂M∗∗++=(Σ0∩M∗∗++)∪(Σπ/2∩M∗∗++)
and is homeomorphic to a circle.
(iii)
Σ0∩M∗∗++=∪i=0m−1β2i0,
and so consists of m connected components,
each homeomorphic to a closed interval.
(iv)
Σπ/2∩M∗∗++=∪i=0m−1β2i+11,
and so consists of m connected components,
each homeomorphic to a closed interval.
(v)
Σ0∩M∗∗++ is homeomorphic to a closed interval and
Σ0∩M∗∗++={α00+∪αm0+ if m∈2Z,α00+∪αm1− if m∈2Z+1.
(vi)
Σπ/2∩M∗∗++ is homeomorphic to a closed interval and
Σπ/2∩M∗∗++=⎩⎨⎧αm/20+∪α3m/20+q1q2m+1∪q1q23m+1αm/21−∪α3m/21− if m∈4Z, if m∈2Z+1, if m∈4Z+2.*
(i)
M++++=D0+0+∪⎩⎨⎧∪i=12m−1Di[i:2]±∪D2m−0+∪i=12m−1Di[i:2]±∪i=12m−1Di[i:2]±∪D2m−1− if m∈4Z, if m∈2Z+1, if m∈4Z+2,
where the ± signs are + for i even and − for i odd.
Therefore M++++ is homeomorphic to a closed disc.
(ii)
∂M++++ is homeomorphic to a circle.
Moreover we can write
∂M++++=γ1∪γ2∪γ3, where
γ1:=γ1−∪γ1+,
γ1−:=Σ0∩M++++,
γ1+:=Σπ/2∩M++++,
γ2:=Σ0∩M++++, and
γ3:=Σπ/2∩M++++,
and each of γ1,γ2,γ3 is homeomorphic to a closed interval.
(iii)
γ1=γ1−∪γ1+=β0+0∪⎩⎨⎧∪i=12m−1βi[i:2]∪β2m−0∪i=12m−1βi[i:2]∪i=12m−1βi[i:2]∪β2m−1 if m∈4Z, if m∈2Z+1, if m∈4Z+2,
(iv)
γ1−=Σ0∩M++++=p0pπ/2p0∩M++++=β0+0∪⎩⎨⎧∪i=14m−4β2i0∪β2m−0∪i=14m−1β2i0∪i=14m−2β2i0∪i=14m−3β2i0 if m∈4Z, if m∈4Z+1, if m∈4Z+2, if m∈4Z+3.
(v)
γ1+=Σπ/2∩M++++=pπ/2pπ/2p0∩M++++=⎩⎨⎧∪i=1m/4β2i−11∪i=14m−1β2i−11∪i=14m−2β2i−11∪β2m−1∪i=14m+1β2i−11 if m∈4Z, if m∈4Z+1, if m∈4Z+2, if m∈4Z+3.
(vi)
γ2=Σ0∩M++++=p0pπ/2p0∩M++++=α00+.
(vii)
γ3=Σπ/2∩M++++=pπ/2pπ/2p0∩M++++=⎩⎨⎧αm/20+q1q2m+1=pπ/4pπ/2αm/21− if m∈4Z, if m∈2Z+1, if m∈4Z+2,*
(i)
M+∗++=D0+0+∪{∪i=1m−1Di[i:2]±∪Dm−0+∪i=1m−1Di[i:2]±∪Dm−1− if m∈2Z, if m∈2Z+1,
where the ± signs are + for i even and − for i odd.
Therefore M+∗++ is homeomorphic to a closed disc.
(ii)
∂M+∗++ is homeomorphic to a circle.
Moreover we can write
∂M+∗++=γ4∪γ5, where
γ4:=γ4−∪γ4+,
γ4−:=Σ0∩M+∗++,
γ4+:=Σπ/2∩M+∗++,
γ5:=Σ0∩M+∗++,
and each of γ4,γ5 is homeomorphic to a closed interval.
(iii)
γ4=γ4−∪γ4+=β0+0∪{∪i=1m−1βi[i:2]∪βm−0∪i=1m−1βi[i:2]∪βm−1 if m∈2Z, if m∈2Z+1,
(iv)
γ4−=Σ0∩M+∗++=β0+0∪⎩⎨⎧∪i=12m−2β2i0∪βm−0∪i=12m−1β2i0 if m∈2Z, if m∈2Z+1.
(v)
γ4+=Σπ/2∩M+∗++={∪i=1m/2β2i−11∪i=12m−1β2i−11∪βm−1 if m∈2Z, if m∈2Z+1.
(vi)
γ5=Σ0∩M+∗++=Σ0∩M∗∗++={α00+∪αm0+ if m∈2Z,α00+∪αm1− if m∈2Z+1.
(vii)
Σπ/2∩M+∗++=Σπ/2∩M++++=⎩⎨⎧αm/20+q1q2m+1=pπ/4pπ/2αm/21− if m∈4Z, if m∈2Z+1, if m∈4Z+2,*
p2π=q2m+1* and for m even p4π=q4m+21.
Moreover the following hold.
(i)
If m∈4Z then RΣπ/4=Rq4m+21,C⊥ is a symmetry of M++++ preserving the unit normal.
(ii)
If m∈4Z+2 then RCπ/4π/4=Rq4m+2,q1 is a symmetry of M++++ reversing the unit normal.
(iii)
If m∈2Z+1 then RCπ/2π/4=Rq2m+1,q1 is a symmetry of M+∗++
reversing the unit normal and exchanging M++++ with M+−++.*
Proof.
All items follow easily from 3.14.iii, 4.3, 4.13, and 4.15.
∎
5. Jacobi fields
As is well known the linearized operator for the mean curvature on
M=M[C,m] (recall 4.1)
is given by
[TABLE]
where ∣A∣2 is the square of the length of the second fundamental form of the surface.
We recall next the following standard definition.
Definition 5.2** (Jacobi fields on M=M[C,m]).**
We call a function J∈C∞(M) a Jacobi field on M=M[C,m] if it satisfies LJ=0.
∎
It is well known that Killing fields induce Jacobi fields as in the following definition.
Definition 5.3** (Jacobi fields JC′).**
We adopt now for the rest of this article the same choice for the unit normal ν
on the Lawson surface M=M[C,m] as in 4.5.
Given then a great circle C′ in S3 and assuming an orientation on C′⊥,
we define the Jacobi fieldJC′⊥=JC′⊥[C,m]=JC′=JC′[C,m]∈C∞(M[C,m]) by
JC′⊥=JC′:=KC′⋅ν
(recall 2.1).
∎
Note that multiplying a Jacobi field by −1
changes neither its nodal set nor any other significant properties,
and so the orientation of C′⊥ and direction of ν
do not play a fundamental role.
Definition 5.4** (Exceptional and non-exceptional Jacobi fields).**
We call a Jacobi field on Mnon-exceptional if it is induced by a Killing field;
otherwise we call it exceptional.
∎
We proceed to study some non-exceptional Jacobi fields which as we will see in 5.9 form a basis.
It is useful to introduce first the following notation.
Notation 5.5*.*
We define 0⊥:=π/2 and (π/2)⊥:=0.
∎
Lemma 5.6** (Symmetries of Jacobi fields).**
∀i,j∈Z* we have the following.
(i)
JC is odd under RΣjπ/2=RC,qj+21
and even under RΣiπ/m=RC⊥,qi+21
and
RC(2i−1)2mπ(2j−1)4π=Rqj,qi.
(ii)
JC⊥ is odd under RΣiπ/m=RC⊥,qi+21
and even under RΣjπ/2=RC,qj+21
and
RC(2i−1)2mπ(2j−1)4π=Rqj,qi.
(iii)
If m∈2Z and ϕ,ϕ′∈{0,π/2},
then JCϕϕ′ is odd under RΣϕ and RΣϕ′
and even under RΣϕ⊥ and RΣϕ⊥′.
(iv)
If m∈2Z+1, then the symmetries in (iii) hold except for the ones associated with RΣπ/2.*
Proof.
All items follow from 2.11 and 4.3.
Note that RΣπ/2 preserves M only when m is even.
∎
Lemma 5.7** (Action of some symmetries on some Jacobi fields).**
*The following hold.
(i)
If m∈4Z, then
JC00∘RΣπ/4=−JCπ/20 and
JCπ/2π/2∘RΣπ/4=JC0π/2.
(ii)
If m∈4Z+2, then
JC00∘RCπ/4π/4=JCπ/2π/2 and
JCπ/20∘RCπ/4π/4=−JC0π/2.
(iii)
If m∈2Z+1, then
JC00∘RCπ/2π/4=JC0π/2
and
JCπ/2π/2∘RCπ/2π/4=JCπ/20.*
Proof.
All items follow easily from 4.17 and the definitions, using in particular the orientation conventions specified in 3.17.
∎
Lemma 5.8** (Gradient of Jacobi fields at q1).**
If ϕ,ϕ′∈{0,π/2},
then
JCϕϕ′(q1)=0
and the gradient
∇q1JCϕϕ′ at q1
is nonzero and tangential to Σϕ⊥.
Proof.
By 4.4M has high-order contact with Σπ/4 at q1,
so we can consider the corresponding Jacobi field on Σπ/4 instead.
That Jacobi field is clearly a first harmonic of Σπ/4
and the result follows without calculation by the symmetries.
∎
Lemma 5.9** (Non-exceptional Jacobi fields).**
JC,
JC⊥,
and
JCϕϕ′
for
ϕ,ϕ′∈{0,π/2}
form a basis of the space of non-exceptional Jacobi fields on M=M[C,m].
Proof.
Since the space of Killing fields has dimension six,
it is enough to prove that the Jacobi fields under consideration are linearly independent.
By symmetrizing and antisymmetrizing with respect to
RΣ0, RΣπ/2, and (for the last four) RΣ0,
we can kill all of them by 5.6 except for a single Jacobi field arbitrarily chosen in advance.
This reduces the proof to proving that each of the six Jacobi fields does not vanish identically.
Clearly JC(q1)=0, JC⊥(q1)=0, and for the rest we consider the gradient
at q1 and we appeal to 5.8. This completes the proof.
An alternative proof is that
the map K↦K⋅ν from the space of Killing fields
to the space of Jacobi fields is injective:
if not,
there would exist a nontrivial Killing field everywhere tangential to M,
meaning M would have a one-parameter family of symmetries.
By Lemma 4.3, however, GsymM is discrete, completing the proof.
∎
Lemma 5.10** (Some Jacobi fields on geodesic segments).**
*We have the following.
(i)
For i∈(2Z+1)∩[1,(m+1)/2]
and
i∈(2Z)∩[(m+1)/2,m]
we have
JC00≥0 on
qiq1⊂M+∗++.
(ii)
For i∈(2Z)∩[1,(m+1)/2]
and
i∈(2Z+1)∩[(m+1)/2,m]
we have
JC0π/2≥0 on
qiq1⊂M+∗++.
(iii)
For i∈(2Z)∩[1,(m+1)/2]
we have
JCπ/20≤0 on
qiq1⊂M++++
and
for i∈(2Z+1)∩[(m+1)/2,m]
we have
JCπ/20≥0 on
qiq1⊂M+−++.
(iv)
For i∈(2Z+1)∩[1,(m+1)/2]
we have
JCπ/2π/2≥0 on
qiq1⊂M++++
and
for i∈(2Z)∩[(m+1)/2,m]
we have
JCπ/2π/2≤0 on
qiq1⊂M+−++.*
Meanwhile, according to Lemma 4.5,
at each point on qiq1 the unit normal ν
is a convex combination of
ν(qi)=p2m2i−1π+(−1)i+12π
and ν(q1)=p−π/4.
Note that
[TABLE]
so on M+∗++
[TABLE]
and
[TABLE]
On the other hand, obviously
[TABLE]
while of course
[TABLE]
All items now follow from the convexity and the signs of the inner products recorded above.
∎
We define now a kind of discrete derivative ∂\mspace0.8mu/ for functions on M by appropriately adapting to the current
situation the discrete derivative defined in [alm20]*(8.13), page 319:
Definition 5.17** (T and the discrete derivative ∂\mspace0.8mu/).**
We define an isometry T:S3→S3 by T:=Rq1,q1∘RΣ0 and
a linear map ∂\mspace0.8mu/:C∞(M)→C∞(M) by ∂\mspace0.8mu/f:=2sin(π/m)1(f∘T−f∘T−1)∀f∈C∞(M).
∎
Lemma 5.18** (Elementary properties of T and ∂\mspace0.8mu/).**
∂\mspace0.8mu/* as in 5.17 is well defined and
T preserves C, C⊥, and M=M[C,m],
and on M satisfies T∗−1∘ν∘T=−ν.
Moreover T=Rπ/mC∘RΣπ/4 and so T rotates C along itself by angle π/m and reflects C⊥ to itself
while fixing q1=pπ/4 and q3=−pπ/4.*
Proof.
The first statement about T follows from 4.3.
It follows then that ∂\mspace0.8mu/ is well defined.
Using the definitions it is easy to check that T maps p0,pπ/2,p0,pπ/2 to
pπ/m,pπ/2+π/m,pπ/2,p0
respectively.
This implies the last statement and completes the proof.
∎
Lemma 5.19** (Discrete derivatives of some Jacobi fields).**
*The following hold.
(i) ∂\mspace0.8mu/JC00=JCπ/2π/2 and ∂\mspace0.8mu/JCπ/2π/2=−JC00.
(ii) ∂\mspace0.8mu/JC0π/2=−JCπ/20 and ∂\mspace0.8mu/JCπ/20=JC0π/2.*
Proof.
Note that
if J=K⋅ν is a Jacobi field induced by a Killing field K,
then
J∘T=(K∘T)⋅(ν∘T)=(T∗−1∘K∘T)⋅(T∗−1∘ν∘T)=−(T∗−1∘K∘T)⋅ν
and similarly
J∘T−1=−(T∗∘K∘T−1)⋅ν,
so we have
where in this proof we simplify the notation by taking c:=cosmπ and s:=sinmπ.
It is easy to calculate then by referring to 3.17 that
[TABLE]
If we exchange T and T−1 in the left hand sides we obtain the same expressions but with “s” replaced by “−s”.
Subtracting, applying 5.20, and referring to 3.17 again, we conclude the proof.
∎
6. Eigenfunctions on the Lawson surfaces
In this section we study the index and nullity of the linear operator L on M=M[C,m] defined in 5.1.
L is the only operator we consider in this section and so we often omit it in order to simplify the notation,
especially in the notation of 1.9.
We start by defining
[TABLE]
where the ± signs are taken correspondingly, the first one referring to RΣ0 and the second one to RΣπ/2.
We clearly have
[TABLE]
where we use ⊕L to mean “direct sum” not only in the sense of linear spaces,
but also to mean that the summands are invariant under L,
and therefore the same decomposition holds for the corresponding eigenspaces.
(i) follows from the definitions,
4.14, and 5.6, where the linear isomorphism is given by restriction to M∗∗++ in one direction
and its inverse by extension by appropriate reflections.
For (ii) recall first
that Lemma 5.6 implies that
JC is nonnegative on M∗∗++ by 4.11.iii and the symmetries.
JC is nontrivial by 5.9 and therefore,
as a consequence of Courant’s nodal theorem B.1,
there are no other eigenfuctions in V−− of the same or lower eigenvalue as the eigenvalue of JC,
which is zero.
The result follows.
∎
To study
V++ now we define
[TABLE]
where in the second equation the ± signs are taken correspondingly.
Note that
[TABLE]
On the other hand V++ is not the direct sum of
V+++
and
V−++.
(i)
JC⊥∈V−+++ and
V−++∼Cpw∞[D0+0+∪D1−1−;α11−∪α00+,β0+0∪β1−1].
(ii)
λ1(V−++)=0<λ2(V−++).*
Proof.
JC⊥∈V−+++ follows from 5.6.ii and the definitions.
Recall now
that
D0+0+∪D1−1− is homeomorphic to a closed disc and its boundary is β0+0∪β1−1∪α11−∪α00+.
We clearly have (i) then,
where the linear isomorphism is given by restriction in one direction and its inverse by extending using reflections.
On D0+0+JC⊥ is nonnegative by 4.11.ii and nontrivial by 5.9.
By 5.6.ii it is then nonnegative on D1−1− as well.
As a consequence of
Courant’s nodal theorem B.1JC⊥ corresponds then to the lowest eigenvalue and the proof is complete.
∎
(i) follows easily by the symmetries with linear isomorphisms being restrictions in one direction and inverses given by extensions using
even or odd reflections appropriately.
By 6.5 to prove (ii) it is enough to prove
[TABLE]
Let ϕ1 be an eigenfunction corresponding to the eigenvalue λ1(V+−++).
Since D00 is a minimizing disc, it is weakly stable, and so λ1(V+−++)≥0.
The first inequality in 6.8 will follow if we prove λ1(V+−++)=0.
By Courant’s nodal theorem B.1ϕ1 cannot change sign on D00 and so by the maximum principle
(without loss of generality) ϕ1>0 on the interior of D00 and ηϕ1<0 on
∂D00∖Q\mspace0.8mu/00=Q00∖Q\mspace0.8mu/00,
where η is the outward unit conormal derivative of D00 at ∂D00∖Q\mspace0.8mu/00.
By applying Green’s second identity to ϕ1 and JCπ/2π/2 we conclude
[TABLE]
If λ1(V+−++)=0, then the left hand side vanishes.
Since ϕ1 satisfies the Dirichlet condition on ∂D00,
we conclude
[TABLE]
By 4.7.i JCπ/2π/2 does not change sign on D00.
Since ηϕ1<0 on ∂D00∖Q\mspace0.8mu/00, we conclude that JCπ/2π/2=0 on ∂D00.
This contradicts 5.8 (alternatively it implies odd symmetries which together with 5.6
contradict the nontriviality of JCπ/2π/2) and therefore we conclude the first inequality in 6.8.
The positivity of the zeroth-order term of L
implies the second inequality in 6.8.
Suppose now that λ2(V++++)≤0 and let ϕ2 be a corresponding eigenfunction.
Then ϕ2 satisfies Neumann conditions on ∂D0+0+
and moreover by Courant’s nodal theorem B.1 will have two nodal domains on D0+0+.
It follows from [Cheng]*Theorem 2.5 that the nodal set ϕ2−1({0})
contains (at least) a piecewise C2 embedded circle or segment
whose endpoints (if it has any) lie on ∂D0+0+
but which is otherwise disjoint from ∂D0+0+.
In particular this nodal curve separates D0+0+
into two components and misses the interior of at least one of the three sides—β0+0,
α00+, and q1q1—of ∂D0+0+.
We call the missed side γ.
By domain monotonicity we conclude that
λ1[D0+0+;γ,∂D0+0+∖γ]<0.
If γ=q1q1, this would contradict the first inequality in 6.8, which already has been proved.
If γ=α00+, this would contradict 6.6.
Finally if γ=β0+0, this would contradict 6.3, and the proof is complete.
∎
Proposition 6.9**.**
We have Ind(V++)=2m−1
and
Null(V++)=1.
Proof.
By considering M∗∗++ and subdividing along the curves of intersection with Σiπ/m
and imposing the Dirichlet or the Neumann conditions appropriately,
the result follows from A.1 by using 6.6 and 6.7.
∎
To study
V+− and V−+ now we define
[TABLE]
Note that
[TABLE]
Lemma 6.12** (The sign of some Jacobi fields).**
*The following hold.
(i)
If m≥3, then
JC00≥0 and JC0π/2≥0 on M+∗++.
(ii)
If in addition m is even, then
JCπ/2π/2≥0 and JCπ/20≤0
on M++++.*
Proof.
*Step 1:
We prove that
∀i∈Z∩[1,(m+1)/2]
we have JC00≥0 on qiq1—equivalently
JC00≥0 on all geodesic segments contained in M++++. *
If i is odd or i=2m+1,
we already know this by 5.10.i.
We can assume then that m≥4 because for m=3 step 1 is proved.
By 5.10.iv and 5.19.i we have for
i∈(2Z+1)∩[1,(m+1)/2] that
JC00∘T≥JC00∘T−1 on
qiq1.
Since T±1(qiq1)=qi±1q1 by 5.18,
this means that JC00 on qiq1⊂M++++ is increasing with increasing even i.
Arguing inductively on even i it is enough to prove then that
JC00≥0 on
q2q1.
Taking i=1 in the inequality in the previous paragraph we establish that
JC00∘T≥JC00∘T−1 on
q1q1.
By 5.17 we have
T−1=RΣ0 on q1q1,
so by 5.6.iii-iv we know that
JC00∘T−1=−JC00 on
q1q1.
Combining we obtain
JC00∘T+JC00≥0 on q1q1.
We consider now the symmetrization φ:=JC00∘RΣπ/m+JC00 of JC00 on D11−⊂M++++.
Recall that
∂D11−=q1q1∪q2q1∪β11.
Since by 5.18T=RΣπ/m on q1q1,
by the last inequality above we have φ≥0 on
q1q1∪q2q1⊂∂D11−.
By 5.6.iii-iv φ satisfies the Neumann condition on the remaining boundary β11.
If we assume that φ attains negative values on D11−, then by domain monotonicity,
and since Lφ=0,
we obtain a contradiction to λ1(V+−++)>0 in 6.7.ii by using 6.7.i.
Hence φ≥0 on D11− and since φ=2JC00 on α11−=D11−∩Σπ/m (by 4.9),
we conclude that JC00≥0 on α11−⊂D11−.
We consider now the domain Φ:=D1+1−∪D20+.
Clearly Φ is homeomorphic to a disc and has ∂Φ=β1+1∪β20∪q3q1∪α11−
and q2q1⊂Φ.
We postpone the case m=4 for later and we assume that m≥5 so that
q3q1⊂Φ⊂M++++.
We already know then
(by the preceding paragraphs and 5.6)
that
JC00≥0 on q3q1∪α11− and satisfies the Dirichlet condition on β20 and the Neumann condition on β1+1.
In order to apply now A.1
we subdivide Φ along
q2q1 into
D1+1− and D20+.
We clearly have
Cpw∞[L,D1+1−;α11−,q2q1∪β1+1]∼V−+++
and
λ1[L,D20+;β20∪q3q1,q2q1]>λ1[L,D20+;β20,q2q1∪q3q1]≥λ1(V−−).
By referring to 6.6 and 6.3 we conclude that
#<0[L,D1+1−;α11−,q2q1∪β1+1]=0
and
#≤0[L,D20+;β20∪q3q1,q2q1]=0.
By A.1 then we conclude that
λ1[L,Φ;β20∪q3q1∪α11−,β1+1]>0,
which by domain monotonicity would contradict an assumption that JC00 takes negative values on Φ.
We conclude that
JC00≥0 on q2q1⊂Φ
which completes step 1 under the assumption m≥5.
We consider now the case m=4.
Note that
D1+1−∪D2−0+ is homeomorphic to a disc and can be subdivided into
D1+1− and D2−0+ by q2q1.
Recall that
∂D1+1−=α11−∪β1+1∪q2q1 and
∂D2−0+=α20+∪β2−0∪q2q1.
We have λ1[D1+1−;α11−,β1+1∪q2q1]=0 by 6.6 and
λ1[D2−0+;β2−0,α20+∪q2q1]=0 by 6.3.
Applying A.1 we conclude that
λ1[D1+1−∪D2−0+;α11−∪β2−0,β1+1∪α20+]≥0.
Since for m=4 we have α20+⊂Σπ/2,
it follows by 5.6 that
JC00 satisfies the same boundary conditions except on α11−,
where we proved above that it is ≥0.
Moreover
JC00 cannot vanish identically on α11− by 5.8.
By domain monotonicity we obtain a contradiction on the assumption that
JC00 is not nonnegative on
D1+1−∪D2−0+.
We conclude
JC00≥0 on q2q1⊂D1+1−∪D2−0+ and step 1 is complete in all cases.
*Step 2:
We prove that
∀i∈Z∩[1,(m+1)/2]
we have JC0π/2≥0 on qiq1—equivalently
JC0π/2≥0 on all geodesic segments contained in M++++. *
Unlike the case of
JC00 we now know this by 5.10.ii when i is even.
Using the discrete derivative as before,
we have
by 5.10.iii and 5.19.ii
for
i∈(2Z)∩[1,(m+1)/2] that
JC0π/2∘T≥JC0π/2∘T−1 on
qiq1.
Arguing inductively on odd i,
it is enough to prove then that
JC0π/2≥0 on
q1q1.
For this we consider the domain Φ′:=D0+0+∪D11−.
Clearly Φ′ is isometric to Φ in the previous step (in fact Φ=T(Φ′)) and
has ∂Φ′=β0+0∪β11∪q2q1∪α00+
and q1q1⊂Φ′.
Similarly to the previous step,
we already know that
JC0π/2≥0 on q2q1 and satisfies the Dirichlet condition on β11∪α00+ and the Neumann condition on β0+0.
Arguing then as in the previous step,
we conclude that JC0π/2≥0 on Φ′ and hence on
q1q1⊂Φ′,
which completes step 2.
*Step 3:
We prove that JC00≥0 and JC0π/2≥0 on M++++. *
Recall from 4.15 that M++++ can be subdivided along the geodesic segments it contains
into Di[i:2]±’s,
D0+0+, and D2m−0+ (for m∈4Z) or
D2m−1− (for m∈4Z+2).
We already know that on the geodesic segments in the boundaries of these regions
we have JC00,JC0π/2≥0,
while on the rest of each boundary JC00 and JC0π/2
satisfy Dirichlet or Neumann conditions.
We also know, using 6.3 and
6.7,
that if we impose the Dirichlet condition on each geodesic segment
and leave the remaining boundary conditions unchanged,
then the corresponding lowest eigenvalue on each region
(obtained by subdividing along the geodesic segments)
is strictly positive.
If we assume then that
JC00 or JC0π/2 attains negative values,
we will have a contradiction by domain monotonicity.
This completes step 3.
Step 4:
We complete the proof of the lemma.
For m even (i) follows from step 3 and the even symmetry with respect to RΣπ/2,
as asserted in 5.6.iii,
which exchanges M++++ with M+−++.
For m odd (i) follows from step 3 and
by using 4.17.iii and the identity
JC00∘RCπ/2π/4=JC0π/2
from 5.7.iii.
Finally (ii) follows from step 3 and (for m∈4Z)
4.17.i and 5.7.i
or (for m∈4Z+2)
4.17.ii and 5.7.ii.
∎
(i)
JC00∈V−−+
and
V−−+∼Cpw∞[M+∗++;γ4−∪γ5,γ4+].
(ii)
JC0π/2∈V−+−
and
V−+−∼Cpw∞[M+∗++;γ4+∪γ5,γ4−].
(iii)
Ind(V−−+)=Ind(V−+−)=0
and
Null(V−−+)=Null(V−+−)=1.*
Proof.
(i) and (ii) follow easily from the definitions and the symmetries in 5.6,
with the linear isomorphisms between the spaces given by restriction to M+∗++
and their inverses by extending using the appropriate reflections.
(iii) follows then from 6.12.i and B.1.
∎
We proceed to study now
V+−+ and V++−.
One would like to decompose these spaces further,
but unfortunately it is clear how to do this only when m is even.
If m is even, we define
[TABLE]
where the upper circles can be +− or −+ (on both sides) and the lower circle + or − (on both sides).
We have then for m even that
[TABLE]
Although we will not use the following lemma, we state it for completeness of exposition—compare also with 5.7.
Lemma 6.16** (Some eigenvalue equivalences).**
*The following hold.
(i)
If m∈4Z, then
V−+−+∼V+−−+
and
V−++−∼V+−+−.
(ii)
If m∈4Z+2, then
V−+−+∼V+−+−,
and
V−++−∼V+−−+.
(iii)
If m∈2Z+1, then
V−−+∼V−+−
and
V+−+∼V++−.*
Proof.
All items follow easily from 4.17 and the definitions.
∎
(i)
JCπ/20∈V+−+
and
V+−+∼Cpw∞[M+∗++;γ4−,γ4+∪γ5].
Moreover if m is even, we have
JCπ/20∈V+−−+
and
V+−−+∼Cpw∞[M++++;γ1−∪γ3,γ1+∪γ2].
(ii)
JCπ/2π/2∈V++−
and
V++−∼Cpw∞[M+∗++;γ4+,γ4−∪γ5].
Moreover if m is even, we have
JCπ/2π/2∈V+−+−
and
V+−+−∼Cpw∞[M++++;γ1+∪γ3,γ1−∪γ2].
(iii)
Ind(V+−+)=Ind(V++−)=1
and
Null(V+−+)=Null(V++−)=1.*
Proof.
As in the proof of 6.13,
(i) and (ii) follow easily from the definitions and the symmetries in 5.6,
with the linear isomorphisms between the spaces given by restriction to M+∗++
and their inverses by extending using the appropriate reflections.
To prove (iii) now we provide different arguments depending on whether m is even or odd,
the even case being easier because of the extra symmetry we can employ.
We assume first that m is even.
By (i), (ii), 6.12.ii, and B.1 we conclude that λ1(V+−+−)=λ1(V+−−+)=0,
λ2(V+−+−)>0, and λ2(V+−−+)>0.
Replacing the Dirichlet condition with the Neumann condition reduces the eigenvalues and therefore
λ1(V+++−)<λ1(V+−+−)=0
and
λ1(V++−+)<λ1(V+−−+)=0.
By Courant’s nodal theorem B.1 and arguing as in the proof of 6.7 using
[Cheng]*Theorem 2.5, we conclude that
the eigenfunction corresponding to
λ2(V+++−)
must contain a separating nodal curve in M++++
which does not intersect at least one of γ1, γ2, or γ3 defined as in 4.15.ii.
There is a nodal domain then in M++++ which does not intersect
at least one of γ1, γ2, or γ3.
If it does not intersect γ1, by extending to M++++
and by using domain monotonicity we conclude that
λ2(V+++−)>λ1(V−−).
If it does not intersect γ2, using domain monotonicity we conclude that
λ2(V+++−)>λ1(V−++−).
If it does not intersect γ3, using domain monotonicity we conclude that
λ2(V+++−)>λ1(V+−+−).
Since
λ1(V−−)=0,
λ1(V−++−)=0,
and
λ1(V+−+−)=0
by 6.3, 6.13, and the above,
we conclude that
0<λ2(V+++−).
Arguing similarly we conclude that
0<λ2(V++−+).
The above together with the decompositions (by 6.15)
[TABLE]
imply (iii) in the case that m is even.
Suppose now that m is odd.
Recall 4.16.
By “cutting through” q1q1 and α11−
we obtain the decomposition
M+∗++=D0+0+∪D1−1−∪M′,
where
[TABLE]
with the signs as in 4.16.i.
By 4.7.ii D0+0+, D1−1−, and M′ are each homeomorphic to a disc and
we have
∂D0+0+=β0+0∪q1q1∪α00+,
∂D1−1−=β1−1∪q1q1∪α11−,
and
∂M′=γ4′∪α11−∪αm1−,
where
γ4′:=γ4−′∪γ4+′,
γ4−′=γ4−∩M′=γ4−∖β0+0,
and
γ4+′=γ4+∩M′=γ4+∖β1−1.
The advantage of M′ over M+∗++ is that M′ has an extra symmetry,
Rq2m+1,C⊥=RΣ2π+2mπ,
which preserves each of γ4−′ and γ4+′.
To exploit this we define
[TABLE]
We clearly have then the decomposition W=W+⊕LW−.
We claim now that
[TABLE]
To prove the claim it is enough to prove that λ1(W−)>0 and λ2(W+)>0.
By “cutting through” with
Σ2π+2mπ
we have the decomposition
M′=M+′∪RΣ2π+2mπM+′,
where
[TABLE]
We have then
M+′∩RΣ2π+2mπM+′=M′∩Σ2π+2mπ=α2m+1[2m+1:2]±,
γ4±′∩M+′=γ4±∩M+′,
[TABLE]
Next we reposition M+′ by using T−2m+1 (recall 5.18) to obtain
[TABLE]
and we use
Rqm/2,C⊥=RΣ2π−2mπ to “double” M′′,
producing
[TABLE]
where α2m+1[2m+1:2]±,
which was used to subdivide M′,
has been moved and “doubled” to α00+∪αm−10+⊂∂M′′′.
We have then that a first eigenfunction in W− (corresponding to λ1(W−))
corresponds to an eigenfunction
in Cpw∞[M′′′;α00+∪αm−10+∪(γ4±∩M′′′),(γ4∓∩M′′′)]
which moreover is even under reflection with respect to
Rqm/2,C⊥=RΣ2π−2mπ,
and where the ± and ∓ signs are opposite and depend on whether m∈4Z+1 or m∈4Z+3.
Either way by 6.13 and by domain monotonicity
(since M′′′⊊M+∗++)
we conclude
that λ1(W−)>0.
We have also W+∼Cpw∞[M′′;(γ4±∩M′′),α00+∪α2m−1[2m−1:2]±∪(γ4∓∩M′′)].
Suppose φ is an eigenfunction corresponding to
λ2(W+).
By Courant’s nodal theorem B.1 and arguing as in the proof of 6.7 using
[Cheng]*Theorem 2.5, we conclude that
there is a
separating nodal curve γ which has to avoid at least one of
γ4∩M′′, α00+, or α2m−1[2m−1:2]±.
In the first case by domain monotonicity we conclude that
[TABLE]
where the last equality follows from 6.3.
In the second case we again use domain monotonicity,
but the comparison is with λ1(W−),
which we proved positive above.
In the third case we reposition M′′ and we argue as for the second case.
This completes the proof that
λ2(W+)>0 and hence of our claim
6.18.
We consider now an eigenfunction corresponding to
λ2[D1−1−;∅,β1−1∪q1q1∪α11−].
By Courant’s nodal theorem B.1 and arguing as in the proof of 6.7 using
[Cheng]*Theorem 2.5 again, we conclude that
there is a separating nodal curve which avoids at least one of
β1−1, q1q1, or α11−.
We can use domain monotonicity then to assert that
λ2[D1−1−;∅,β1−1∪q1q1∪α11−]
is > one of
λ1[D1−1−;β1−1,q1q1∪α11−],
λ1[D1−1−;q1q1,β1−1∪α11−],
or
λ1[D1−1−;α11−,β1−1∪q1q1].
By appealing to 6.3, 6.7, or 6.6 correspondingly we conclude that
[TABLE]
We can apply A.1 now to the decomposition
M+∗++=D0+0+∪D1−1−∪M′ to conclude
by referring to 6.18, 6.19, and 6.20, that
#≤0(V+−+)=#≤0[M+∗++;γ4−,γ4+∪γ5]≤2.
Since
JCπ/20∈V+−+ changes sign on M+∗++ by 5.8,
it cannot be a first eigenfunction.
Since it has eigenvalue [math], we conclude by the last inequality that
it is a second eigenfunction, which completes the proof of (iii) for V+−+.
The proof for V++− is similar.
∎
The main theorem follows.
Recall that ξg,1 in the notation of [Lawson] denotes the genus-g Lawson surface
which can be viewed as a desingularization of two orthogonal great two-spheres in the round three-sphere S3.
Theorem 6.21**.**
If g∈N and g≥2, then the index of ξg,1 is 2g+3.
Moreover ξg,1 has nullity 6 and no exceptional Jacobi fields.
Proof.
Recall that m=g+1.
Combining then 6.2, 6.11,
Propositions 6.3, 6.9, 6.13 and 6.17,
we conclude the proof.
∎
Remark 6.22* (Alternative proof for high genus).*
The Lawson surfaces of high genus can be constructed by gluing and then one obtains a detailed knowledge of their geometry.
The gluing construction is a straightforward desingularization construction for \Sigma^{\pi/4}\cup\Sigma^{-\pi/4}=\cup_{j=1}^{4}(\mathsf{q}^{j}\mbox{\times\times}C),
that is two orthogonal great two-spheres,
in the fashion of those constructions in [kapouleas:wiygul] which are for two orthogonally intersecting Clifford tori.
The surfaces constructed are modeled in the vicinity of C after the classical Scherk surface [Scherk] desingularizing two orthogonal planes in R3
and
given in appropriate Cartesian coordinates
by the equation sinhx1sinhx2=sinx3.
For each large m we can impose on the construction all the symmetries of M=M[C,m].
By the uniqueness then in 4.1 we can infer that the surface constructed is actually M=M[C,m]=ξm−1,1.
By the control the construction provides we can conclude then that for large m (equivalently large genus)
the region of the Lawson surface M=ξm−1,1 in the vicinity of C can be approximated by an appropriately scaled singly periodic Scherk surface,
which has been transplanted to S3 so that its axis covers C.
The rest of the Lawson surface approximates \cup_{j=1}^{4}(\mathsf{q}^{j}\mbox{\times\times}C) (that is the two great two-spheres being desingularized)
with a small neighborhood of C removed.
This information can be used to simplify the proofs of many intermediate results on which the proof of the main theorem is based,
thus avoiding the need for many of the arguments we have used in this article.
∎
Note now that there is a smooth family of singly periodic Scherk surfaces which can be parametrized by the angle θ∈(0,π) between
two adjacent asymptotic half planes.
The Scherk surface in 6.22 corresponds then to θ=π/2.
Since we can then prescribe θ, we say that the Scherk surfaces can “flap their wings”.
In [alm20]*Section 4.2 a heuristic argument was provided indicating that this is not the case for the Lawson surfaces
and further questions motivated by this were asked.
The non-existence of exceptional Jacobi fields as in 6.21 provides partial answers to some of those questions.
In particular it implies that each ξg,1 is isolated as in the following corollary.
Isolatedness can be proved by adapting the proof of [KMP]*Proposition 3.1 or by a more direct argument suggested to us by R. Schoen as follows.
Corollary 6.23** (No flapping and isolatedness).**
ξg,1* as in 6.21
cannot “flap its wings” at the linearized level and moreover it is isolated in the sense that there is an ϵ>0 such that
any minimal surface
within a C1ϵ-neighborhood of ξg,1
is congruent to ξg,1.*
Proof.
By “no flapping at the linearized level” we mean that there are no Jacobi fields
which are infinitesimal deformations consistent with varying the angle of intersection of the
two spheres of which the surfaces can be viewed as desingularizations.
Since there are no exceptional Jacobi fields by Theorem 6.21,
the result follows.
To prove now isolatedness
suppose M=ξg,1 is not isolated modulo congruence.
Then there exists a sequence {Mn}
of embedded
minimal surfaces none of which is congruent to M
but which C1-converge to M.
Each Mn is then the graph (via the exponential map in the normal direction)
of some function un on M, so that {un} converges to [math] in C1(M).
By appropriately rotating each Mn we may assume that eventually un
is L2(M)-orthogonal to the space of nonexceptional Jacobi fields,
at least to first order in ∥un∥C0.
Since each Mn is minimal, by elliptic regularity
the sequence {un/∥un∥C1}
is bounded in C2,α(M), so has a subsequence converging in C2(M),
thereby producing a nontrivial exceptional Jacobi field, a contradiction.
∎
Appendix A Eigenvalues and subdivisions
In this appendix following [Ros] we state two bounds on the number of eigenvalues on a domain
in terms of the number of eigenvalues on appropriate subdivisions of the domain.
More precisely suppose that we are given L, U, g, and ∂U=∂DU∪∂NU as in 1.9.
We assume further that by removing a finite union of smooth embedded one-dimensional submanifolds,
γ⊂U,
we subdivide U into n∈N connected components
whose (compact) closures we denote by Ui for i=1,…,n.
We define ∂DUi:=∂Ui∩∂DU,
∂NUi:=∂Ui∩∂NU,
and
γi:=∂Ui∩γ.
Clearly then we have the decomposition
[TABLE]
We have then the following.
Proposition A.1** (Montiel-Ros [Ros]).**
*Assuming the above and in the notation of 1.9 we have the following ∀λ∈R.
(ii)
#≤λ[L,U;∂DU,∂NU]≤#≤λ[L,U1;∂DU1,γ1∪∂NU1]+∑i=2n#<λ[L,Ui;∂DUi,γi∪∂NUi].*
Proof.
Parts (i) and (ii) generalize Lemma 12 and Lemma 13 respectively of [Ros],
whose proofs carry over here with only minor modification.
Nevertheless we sketch a proof for ease of reference.
First we introduce some general notation:
for L, U, g, and ∂U=∂DU∪∂NU as in 1.9
and λ∈R
we will write Eλ[U;∂DU,∂NU]
for the λ-eigenspace of L on U
with Dirichlet condition on ∂DU and Neumann condition on ∂NU.
We will understand Eλ[U;∂DU,∂NU]={0}
when λ is not an eigenvalue.
Now we make the same assumptions on U and its boundary as above
and fix λ∈R.
To prove (i) we define the n spaces of test functions
[TABLE]
Clearly for any ui∈Vi and uj∈Vj with i=j
we have ∇ui⊥L2(U)∇uj
and ui⊥L2(U)huj for any h∈Cpw∞(U).
Define also
[TABLE]
We claim that the L2(U)-orthogonal projection
⨁i=1nVi→V<λ is injective,
which will establish (i).
To check the injectivity suppose u∈⨁i=1nVi, so that
u=∑i=1nui
for some u1∈V1, u2∈V2, …, un∈Vn.
Then
[TABLE]
but the additional assumption u⊥L2V<λ
forces the equality case,
which then implies
u∣U1=0 and Lu=−λu everywhere.
We conclude by the unique-continuation principle [Aronszajn] that in fact u=0.
To prove
now (ii) we define the 2+n vector spaces
[TABLE]
Clearly ∏i=1nWi is a subspace of W
and the map
ι:u∈V≤λ↦(u∣U1,…,u∣Un)∈W
is injective.
Endowing W with the obvious L2 inner product and writing
π:W→∏i=1nWi
for the corresponding projection onto ∏i=1nWi,
we claim that π∘ι is injective,
implying (ii).
To check the injectivity suppose u∈V≤λ.
Then
[TABLE]
On the other hand, if ι(u)⊥∏i=1nWi,
then for each i
[TABLE]
with strict inequality when n=1, unless u∣U1=0.
Since ∑i=1n⟨u∣Ui,u∣Ui⟩L2=⟨u,u⟩L2,
we conclude that if u∈kerπ∘ι, then u is a solution to Lu=−λu
and vanishes identically on U1,
forcing u=0 by unique continuation.
∎
Appendix B The Courant nodal theorem
In this appendix we recall Courant’s nodal theorem in the form we use it.
Suppose that we are given L, U, g, and ∂U=∂DU∪∂NU as in 1.9.
Suppose moreover U is connected.
We define the number of nodal domains of an eigenfunction u of L to be the number of connected components of
U∖u−1(0).
We have then the following,
where for ease of reference we include in the theorem its corollary on the simplicity of the first eigenvalue.
Given L, U, g, and ∂U=∂DU∪∂NU as above,
let Nn for each n∈N
be the number of nodal domains of an eigenfunction corresponding to the nth eigenvalue
λn[L,U;∂DU,∂NU]
in the notation of 1.9.
We have then for n=1: N1=1 and
λ1[L,U;∂DU,∂NU]<λ2[L,U;∂DU,∂NU];
and for n>1: 2≤Nn≤n.