A gap theorem for free boundary minimal surfaces in geodesic balls of hyperbolic space and hemisphere

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by  Haizhong Li

A GAP THEOREM FOR FREE BOUNDARY MINIMAL
SURFACES IN GEODESIC BALLS OF HYPERBOLIC
7 SPACE AND HEMISPHERE
1
0
2 HAIZHONGLI AND CHANGWEI XIONG

n
a
J

8
Abstract. Inthispaperweprovideapinchingconditionforthecharac-
1
terization ofthetotallygeodesicdiskandtherotationalannulusamong

minimalsurfaceswithfreeboundaryingeodesicballsofthree-dimensional
]
G hyperbolic space and hemisphere. The pinching condition involves the
D length of the second fundamental form, the support function of the
surface, and a naturalpotential function in hyperbolicspace and hemi-
.
h sphere.
t
a
m 1. Introduction
[
To characterize simple but important geometric objects by various stan-

1 dards has been a focus of attention for a long time. For example, F. J.
v
Almgren [1] proved that the equator is the only minimal surface in S3 of
1
8 genus 0 (up to rigid motions in S3). The Lawson conjecture confirmed by S.
9 Brendle[5]states thattheCliffordtorusistheonlyembeddedminimaltorus
4 in S3. The Pinkall-Sterling conjecture proved by B. Andrews and H. Li [3]
0
. saysthattherotationaltorusistheonlyembeddedtoruswithconstantmean
1 curvature in S3. The Willmore conjecture due to Marques and Neves [17]
0
7 showsthattheCliffordtorusistheuniqueminimizeroftheWillmoreenergy
1 among surfaces in S3 of genus at least 1. In another direction, the equator
:
iv and the Clifford tori Cm,n−m := Sm( mn)×Sn−m( n−nm), 1 ≤ m ≤ n−1,
X in Sn+1 can be characterized by the following gap tqheorem:
p
r
a Theorem 1 (Chern-do Carmo-Kobayashi [8], Lawson [14], Simons [24] ).
Let Σn be a closed minimal hypersurface in the unit sphere Sn+1. Asssume
that its second fundamental form A satisfies
A2 n. (1)
| | ≤
Then
(1) either A2 0 and Σn is an equator;
| | ≡
(2) or A2 n and Σn is one of Clifford tori Cm,n−m, 1 m n 1.
| | ≡ ≤ ≤ −
Date: January 19, 2017.
2010 Mathematics Subject Classification. 53C42, 53C20.
Keywordsandphrases. gaptheorem,minimalsurface,freeboundary,hyperbolicspace,
hemisphere.
1
2 HAIZHONGLIANDCHANGWEIXIONG
In the setting where a minimal surface lies in a 3-dimensional Euclidean
unit ball B3 with free boundary, the flat equatorial disk and the critical
catenoid play analogous roles as the equator and the Clifford torus in S3.
Here the critical catenoid is a piece of a catenoid in R3 which intersects
∂B3 orthogonally (free boundary). Due to their simple topology, these two
examplesinB3havegotextensivelystudiedtheseyears. Wereferthereaders
to [2] for a nice account of the works concerning the flat equatorial disk and
the critical catenoid. Or see e.g. [4,10–13,19,21,23]. In particular, very
recently Lucas Ambrozio and Ivaldo Nunes have obtained a gap theorem
which is similar to Theorem 1 in some sense.
Theorem 2 (Ambrozio-Nunes [2]). Let Σ be a compact free boundary min-
imal surface in Euclidean unit ball B3. Assume that for all points x Σ,
∈
A2(x) x,N(x) 2 2, (2)
| | h i ≤
where N(x) denotes a unit normal vector at the point x Σ and A denotes
∈
the second fundamental form of Σ. Then
(1) either A2 x,N(x) 2 0 and Σ is a flat equatorial disk;
| | h i ≡
(2) or A2(p) p,N(p) 2 = 2 at some point p Σ and Σ is a critical
| | h i ∈
catenoid.
Inthispaperweaimatprovingasimilargaptheoreminthe3-dimensional
hyperbolicspaceH3 andhemisphereS3. Tothatendwefirstintroducesome
+
notations.
LetM3bea3-dimensionalwarpedproductRiemannianmanifold[0,R∞)
S2 equipped with the metric (see [15]) ×
g = dr2+λ(r)2gS2, (3)
where λ(r) and R∞ are given by
r, R∞ = + ,
∞
λ(r) = sinhr, R∞ = + ,
∞
sinr, R∞ = π/2.
That is, M3 is Euclidean spaceR3, hyperbolic space H3 or hemisphere S3.
 +
ForanysurfaceΣinM3,therearetwoimportantfunctions,i.e. thepotential
′
functionλ andthesupportfunction X,N . HereX = λ(r)∂ isthenatural
r
h i
conformal vector field on M3 and N is a unit normal of the surface. For
example, these two functions appear in the well-known Minkowski formulas.
Let B = [0,R) S2 M3 bethe geodesic ball in M3 with radius R, and
R
× ⊂
Σ2 be a compact free boundary minimal surface in B . Now we can state
R
our main result as follows:
Theorem 3. Let Σ be a compact free boundary minimal surface in the
geodesic ball BR M3 with radius R < R∞. Assume that for all points x
⊂
in Σ,
A2 N(x),X 2
| | h i 2, (4)
(λ′)2 ≤
A GAP THEOREM IN HYPERBOLIC SPACE AND HEMISPHERE 3
where N(x) denotes a unit normal vector at the point x Σ and A denotes
∈
the second fundamental form of Σ. Then
(1) either A2 N(x),X 2 0 and Σ is a totally geodesic disk;
| | h i ≡
A2 N(p),X 2
(2) or | | h i = 2 at some point p Σ and Σ is a rotational
(λ′)2 ∈
annulus.
The proof of Theorem 3 proceeds parallel to that of Theorem 2. Firstly
wedealwiththehyperboliccase, thatis,M3 = H3. Itisourkey observation
thatthelengthsquareofthenaturalconformalvector X candefineaconvex
function on the surface Σ under the pinching condition (4), which is the
content of Lemma4. Thenwedividethecondition (4)into two cases: either
it is a strict inequality on Σ ∂Σ, or it is an equality for some p Σ ∂Σ.
In the latter case we can use t\he Minkowski space model for H3 to∈prov\e the
surface is rotational, which is another difference from that in [2]. Secondly
for the spherical case M3 = S3, after a little bit more work, we are able to
+
find a convex function on Σ under (4), see Lemma 8. Then the remaining
is similar to that in hyperbolic case.
This paper is organised as follows. In Section 2 we check that in hyper-
bolic case the rotational minimal annulus with free boundary in B H3
R
indeed satisfies the condition (4), where the Poincar´e disk model for ⊂H3 is
convenient. Then in Section 3 as mentioned above we prove Theorem 3
for hyperbolic case. Finally in Section 4 we provide the key Lemma 8 in
spherical case to finish the proof of Theorem 3 for spherical case.
Acknowledgment. The authors would like to thank Professor Ben An-
drews for his interest in this work. The second author is also grateful to
Professor Jaigyoung Choe for discussions on minimal surfaces in hyperbolic
spaceattheworkshoponNonlinearandGeometricPartialDifferentialEqua-
tions at Kioloa Campus of ANU, 2016, Australia. The first author was sup-
portedby NSFC grant No.11671224. Thesecond authorwas supportedby a
postdoctoral fellowship funded via ARC Laureate Fellowship FL150100126.
2. The two examples for hyperbolic case
This section is devoted to examples in hyperbolic space H3 which satisfy
thecondition (4). Obviouslythetotally geodesic disksatisfies (4). Whilefor
the rotational minimal annulus with free boundary in B H3, we verify
R
⊂
(4) as follows.
First we use the Minkowski space model for H3 R4, that is,
⊂ 1
H3 = y R4 : y4 = (y1)2+(y2)2+(y3)2+1 , (5)
{ ∈ 1 }
with the metric induced from thepLorentz inner product , in R4. Then
h· ·i1 1
as in [9,20], the rotational catenoid we consider can be parametrized as the
4 HAIZHONGLIANDCHANGWEIXIONG
map F :R S1 H3 given by
a
× →
1 1
F (s,θ)= acosh2s cosθ, acosh2s sinθ,
a
− 2 − 2
r r
1 1
acosh2s+ sinhφ(s), acosh2s+ coshφ(s) ,
2 2
r r !
where φ(s) is the integral
1 s 1
φ(s)= a2 dt, (6)
r − 4Z0 (acosh2t+ 1) acosh2t 1
2 − 2
q
and a > 1 is a constant. Note that F is rotational around the plane
2 a
generated by coordinate vectors ∂ and ∂ , and s is the arclength parameter
3 4
of the generating curve.
Using this model, we can prove for any geodesic ball B H3 centered
R
⊂
at (0,0,0,1), there exists a unique a = a(R) such that F is a minimal
a(R)
annulus with free boundary in B . In fact, it suffices to prove, for any
R
a > 1 and fixed θ (say 0), the tangent straight line F(s)+kF′(s): k R
2 { ∈ }
in R4 will intersect with the y4-axis for some s > 0. Then F(s ) is on the
1 0 0
boundaryof the geodesic ball. The proof is elementary, which we omit here.
However, under this model it is not easy to see that (4) holds on Σ. To
that end, we shall use the Poincar´e disk model for H3. That is,
3
H3 = z R3 : (zi)2 < 1 , (7)
{ ∈ }
i=1
X
with the conformal flat metric
3
4
g = ρ(z)2g = (dzi)2. (8)
0 (1 3 (zi)2)2
− i=1 Xi=1
Here ρ(z) = 2 and g = P3 (dzi)2. Now consider the rotational
1−P3i=1(zi)2 0 i=1
minimal surface Σ in H3 given by
P
z(t,θ) = (t,f(t)cosθ,f(t)sinθ), 1 a¯ t a¯ 1. (9)
− ≤ − ≤ ≤ ≤
As a surface in Euclidean space, Σ has a unit normal
′
(f , cosθ, sinθ)
N¯ = − − , (10)
1+(f′)2
and two principal curvatures p
′′
f 1
κ¯ = , κ¯ = . (11)
1 −(1+(f′)2)23 2 f(1+(f′)2)21
A GAP THEOREM IN HYPERBOLIC SPACE AND HEMISPHERE 5
Thenasasurfaceinhyperbolicspace, i.e. withaconformalmetricg = ρ2g ,
0
the surface Σ has two principal curvatures (see e.g. [16, Lemma 10.1.1])
κ¯ N¯(ρ)
i
κ = , i = 1,2, (12)
i ρ − ρ2
here N¯(ρ) means the directional derivative, namely N¯(ρ) = g (N¯, ρ).
With these preparation, since Σ is a minimal surface in H3,0we ca∇ng0derive
′′ ′
f 1 f tf
= +4 − , (13)
1+(f′)2 f 1 t2 f2
− −
with initial data
′
f(0) = c (0,1), f (0) = 0. (14)
∈
Then for a geodesic ball B H3 centered at the origin with hyperbolic
R
⊂
radiusR,wecanfindauniquec = c(R)suchthatthesolutioncurvef(t)(t
∈
[ a,a]) generates a minimal annulus with free boundary in B . Moreover,
R
−
the free boundary condition requires that a is the first zero of the function
′
f(t) tf (t).
−
Now we are in a position to verify (4). First note that (4) is equivalent to
κ ( < N,∂ >) tanhr 1, (15)
2 r
· − · ≤
where r is the geodesic distance from z = (t,f(t)cosθ,f(t)sinθ) to the
origin with respect to g. We can compute it to get
1+ z
r =ln | |g0, (16)
1 z
−| |g0
which implies
2z 2 t2+f2
tanhr = | |g0 = . (17)
1+ z 2 1+t2+f2
| |g0 p
Meanwhile,
′
z tf f
< N,∂ > = g (N¯, )= − .
r 0
|z|g0 1+(f′)2 t2+f2
In summary, we have p p
1 t2 f2 tf′ f
κ ( < N,∂ >) tanhr = − − −
2 · − r · 2f 1+(f′)2 − 1+(f′)2!
f tf′ 2 pt2+f2 p
−
× 1+(f′)2 t2+f2 × 1+t2+f2
p
(1+f2 t2 2tff′)(f tf′)
p p
= − − − . (18)
f(1+(f′)2)(1+t2+f2)
In view of the equation (13), we know that on (0,a)
f′ 0, f′′ 0, f tf′ > 0, and 1+f2 t2 2tff′ > 0.
≥ ≥ − − −
6 HAIZHONGLIANDCHANGWEIXIONG
Furthermore, it is straightforward to check the numerator in (18) is decreas-
ing, while the denominator is increasing. As a consequence, κ < N,∂ >
2 r
− ·
tanhr is decreasing from 1 to 0. This finishes the verification of (4).
·
3. Proof of Theorem 3 for hyperbolic case
In this section we assume M3 = H3. Firstly we prove a useful lemma.
Lemma 4. Let Σ2 be a free boundary minimal surface in B H3. Let ϕ
R
⊂
be the function defined by
ϕ(x) = λ(r(x))2, x Σ. (19)
∈
Then we have
(a) Σϕ(x) = 2λλ′∂ for all x ∂Σ.
r
∇ ∈
A2 N(x),X 2
(b) For each x Σ, when | | h i 2, the matrix Hess ϕ(x)
∈ (λ′)2 ≤ Σ
is positive semi-definite.
Proof. Note that X = λ(r)∂ is a conformal vector field, i.e. for any vector
r
field v TM3 we have (see e.g. [6,15])
∈
′
D (λ(r)∂ )= λ(r)v. (20)
v r
Here D is the connection of ambient space M3. And we will use = Σ to
∇ ∇
denote the connection of Σ.
Note that ϕ(x) = X 2. Let e ,e T Σ be an orthonormal basis for
1 2 x
| | ∈
T Σ. Then for (a),
x
ϕ(x) = (e X 2)e
i i
∇ | |
= 2 D X,X e
h ei i i
′
= 2 λe ,λ∂ e
i r i
h i
= 2λλ′∂T,
r
where ()T denotes the orthogonal projection onto T Σ. Now along ∂Σ, the
x
· ′
vector ∂ is the conormal of ∂Σ in Σ. So we have ϕ(x) = 2λλ∂ for all
r r
∇
x ∂Σ, which proves (a).
∈
For (b), taking any Y,Z T Σ, we have
x
∈
Hess ϕ(x)(Y,Z) = YZ(X 2) ( Z)(X 2)
Σ Y
| | − ∇ | |
′ ′
= 2Y λZ,X 2 λ Z,X
Y
h i− h ∇ i
= 2λ′′ Y,∂ Z,X +2 λ′D Z,X +2(λ′)2 Z,Y 2 λ′ Z,X
r Y Y
h ih i h i h i− h ∇ i
= 2λλ′′ Y,∂ Z,∂ +2(λ′)2 Z,Y +2λ′ A(Y),Z N(x),X
r r
h ih i h i h ih i
1
= 2λλ′′ Y,∂ Z,∂ +2(λ′)2 Y + A(Y) N(x),X ,Z . (21)
h rih ri λ′ h i
(cid:28) (cid:29)
A GAP THEOREM IN HYPERBOLIC SPACE AND HEMISPHERE 7
′′
Note that λ (r) = sinhr 0 and the second fundamental form A has
≥
A2 N(x),X 2
eigenvalues A/√2. Therefore when | | h i 2, the matrix
±| | (λ′)2 ≤
Hess ϕ(x) is positive semi-definite.
Σ
(cid:3)
Then as in [2], we can obtain the following proposition. The proof for it
is the same as in [2]. We omit the details here.
Proposition 5. Let Σ be a compact free boundary minimal surface in B
R
H3. Define ⊂
= p Σ : r(p)= minr(x) . (22)
C { ∈ x∈Σ }
A2 N(x),X 2
If | | h i 2 on Σ, then
(λ′)2 ≤
(a) either contains a single point p Σ ∂Σ, in which case Σ must be
C ∈ \
a totally geodesic disk;
(b) or is a simple closed geodesic in Σ ∂Σ and Σ is homeomorphic
C \
to an annulus.
Moreover, reasoning as in [2], we have a corollary.
Corollary 6. Let Σ be a compact free boundary minimal surface in B
R
A2 N(x),X 2 ⊂
H3. If | | h i < 2 on Σ, then Σ is a totally geodesic disk.
(λ′)2
In view of this corollary, it remains to deal with the case where the func-
A2 N(x),X 2
tion | | h i attains its maximum value 2 at some point on Σ.
(λ′)2
Proposition 7. Let Σ be a compact free boundary minimal surface in B
R
A2 N(x),X 2 A2 N,X 2 ⊂
H3. If | | h i 2 on Σ, and | | h i = 2 at some point p Σ,
(λ′)2 ≤ (λ′)2 ∈
then Σ is a rotational annulus.
The proof is similar to that of [2, Proposition 4], with some adaptation
to curved space. We include it here for completeness.
A2 N,X 2
Proof. Thefunctionϕ :Σ Rand aredefinedasbefore. Since | | h i =
→ C (λ′)2
2 at some point p Σ, then case (a) in Proposition 5 can not happen. In
∈
fact, should (a) occur, then by [25, Theorem 4.1] or [22, Theorem 2.1] (an
extension of Nitsche’s result [21]), Σ is a totally geodesic disk, a contra-
diction. Thus Σ is homeomorphic to an annulus and is a simple closed
C
geodesic γ :[0,l] Σ with arc parameter. Note that inf ϕ > 0.
Σ
Let r > 0 sat→isfy sinh2r = inf ϕ and let S H3 be the geodesic
sphere o0f radius r centered 0at the oΣrigin. Then Σr0 ⊂ x H3 : r(x) r
0 0
⊂ { ∈ ≥ }
and Σ S = γ([0,l]). Therefore T Σ = T S for all s [0,l]. Note
∩ r0 γ(s) γ(s) r0 ∈
8 HAIZHONGLIANDCHANGWEIXIONG
that γ is a geodesic in Σ. So it is also a geodesic in S . That is, γ is a great
r0
circle of S .
r0
Now we recall the Minkowski space model for H3 R4, that is,
⊂ 1
H3 = y R4 : y4 = (y1)2+(y2)2+(y3)2+1 , (23)
{ ∈ 1 }
and we set B H3 centered atp(0,0,0,1). Based on the analysis above,
R
⊂
we can assume that ∂ is orthogonal to the hyperplane which contains the
3
′
circle γ and ∂ . Then γ (s),∂ is an orthonormal basis of T Σ for all
4 3 γ(s)
{ }
s [0,l].
∈
Now as in [25, Theorem 5.1], we define for any three vectors v ,v ,v in
1 2 3
R4 their Lorentz cross product v v v as the unique vector such that
fo1r any v R4: 1 ∧ 2 ∧ 3
∈ 1
v v v ,v = det(v ,v ,v ,v), (24)
1 2 3 1 1 2 3
h ∧ ∧ i
where , is the Lorentz inner product and det is the determinant in the
1
h· ·i
canonical basis.
With this preparation, we define a function u :Σ R by
→
u(x) = φ(x) ∂ ∂ ,N(x) . (25)
3 4 1
h ∧ ∧ i
Here φ : Σ H3 R4 denotes the immersion of the surface. It is straight-
→ ⊂ 1
forward to verify that
∂u
∆u+(A2 2)u =0 on Σ, = cothR u on ∂Σ, (26)
| | − ∂ν ·
which means that u is a Jacobi function on Σ. And note that u 0 on
Σ if and only if Σ is a rotational surface in R4 around the two-dime≡nsional
1
subspace generated by ∂ and ∂ .
3 4
Now on the one hand, γ([0,l]) is contained in u−1(0), which implies at
′ d(u(γ(s)))
any point γ(s) we have u,γ = = 0. On the other hand, recall
′ h∇ i ds ′
γ (s),∂ isanorthonormalbasisofT Σandγ (s)isaprincipaldirection
3 γ(s)
{ }
of Σ. Then ∂ is the other principal direction at γ(s), which implies D N
3 ∂3
is parallel to ∂ . Therefore,
3
u,∂ = ∂ ∂ ∂ ,N(x) + φ(x) ∂ ∂ , ∂ ,N(x) φ(x)+D N(x)
h∇ 3i h 3 ∧ 3∧ 4 i1 h ∧ 3∧ 4 h 3 i1 ∂3 i1
= 0.
To sum up, u = 0 along the circle γ, which means the critical points of u
in the nodal∇set u−1(0) are not isolated. Then a result of S. Y. Cheng [7,
Theorem2.5] showsu 0onΣ. Asaconsequence, Σis arotational annulus
≡
as claimed.
(cid:3)
4. Proof of Theorem 3 for spherical case
For the problem in hemisphere, the key lemma is the following.
A GAP THEOREM IN HYPERBOLIC SPACE AND HEMISPHERE 9
Lemma 8. Let Σ be a free boundary minimal surface in B S3 with
R
⊂
R (0,π/2). Let ϕ be the function defined by
∈
ϕ(x) = λ(r(x))2, x Σ, (27)
∈
and choose Φ(s)= 1 √1 s. Then we have
− −
(a) ΣΦ(ϕ)(x) = 2Φ′(ϕ)λλ′∂ for all x ∂Σ.
r
∇ ∈
A2 N(x),X 2
(b) Foreachx Σ, when | | h i 2, thematrixHess Φ(ϕ)(x)
∈ (λ′)2 ≤ Σ
is positive semi-definite.
Proof. (a) Since ΣΦ(ϕ) = Φ′(ϕ) Σϕ, the conclusion follows immediately.
∇ ∇
(b) First we have for any Y,Z T Σ,
x
∈
Hess Φ(ϕ)(Y,Z) = YZ(Φ(ϕ)) Z(Φ(ϕ))
Σ Y
−∇
′ ′
= Y(Φ(ϕ)Zϕ) Φ(ϕ) Zϕ
Y
− ∇
′′ ′ ′
= Φ (ϕ)Yϕ Zϕ+Φ(ϕ)YZϕ Φ(ϕ) Zϕ
Y
· − ∇
= Φ′′(ϕ)4λ2(λ′)2 Y,∂ Z,∂ +Φ′(ϕ)Hess ϕ(Y,Z).
r r Σ
h ih i
Now using the computation in Lemma 4, i.e. Equality (21), we get
Hess Φ(ϕ)(Y,Z) = Φ′′(ϕ)4λ2(λ′)2 Y,∂ Z,∂
Σ r r
h ih i
1
+Φ′(ϕ) 2λλ′′ Y,∂ Z,∂ +2(λ′)2 Y + A(Y) N(x),X ,Z
h rih ri λ′ h i
(cid:18) (cid:28) (cid:29)(cid:19)
= 2 2Φ′′(ϕ)λ2(λ′)2+Φ′(ϕ)λλ′′ Y,∂ Z,∂
r r
h ih i
1
+2Φ(cid:0)′(ϕ)(λ′)2 Y + A(Y) N((cid:1)x),X ,Z .
λ′ h i
(cid:28) (cid:29)
Note that λ = sinr, λ′ = cosr, λ′′ = sinr, Φ′ = 1(1 s)−21 and Φ′′ =
− 2 −
1(1 s)−23. So we can check that 2Φ′′(ϕ)λ2(λ′)2+Φ′(ϕ)λλ′′ = 0. Then the
4 −
conclusion follows.
(cid:3)
The rest of the proof of Theorem 3 for spherical case is similar to that
for hyperbolic case, with minor modification. Just note that now Φ(ϕ) is
a convex function under the pinching condition (4). Then the remaining
goes through. Here we omit the details. Therefore we complete the proof of
Theorem 3.
Remark 9. The existence of the rotational free boundary minimal annulus
in B S3 (R < π/2) is easier now. For example, see the deep result [18]
R ⊂ +
by Davi Maximo, Ivaldo Nunes and Graham Smith. Moreover, consider the
10 HAIZHONGLIANDCHANGWEIXIONG
parametrization of Σ as F : R S1 S3 given by (see [9])
a
× →
1 1
F (s,θ)= +acos2scosθ, +acos2ssinθ,
a
2 2
r r
1 1
acos2ssinφ(s), acos2scosφ(s) ,
2 − 2 −
r r !
where φ(s) is the integral
1 s 1
φ(s) = a2 dt, (28)
r4 − Z0 (1 acos2t) 1 +acos2t
2 − 2
and 1 < a < 0 is a constant. Then after a leqngthy, but direct calculation,
−2
one can check that (4) holds for F in some geodesic ball B S3 with
a R(a)
⊂
R(a)< π/2.
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