Generalized Henneberg Stable Minimal Surfaces

Resul s Ma h (2023) 78:53
Online Fi s
c
2023 The Au ho (s)
h ps://doi.o g/10.1007/s00025-022-01831-0 Resul s in Ma hema ics
Gene alized Hennebe g S able Minimal
Su aces
Da id Moya and Joaqu´ın P´e ez
Abs ac . We gene alize he classical Hennebe g minimal su ace by gi -
ing an infini e amily o comple e, fini ely b anched, non-o ien able, s able
minimal su aces in R3. These su aces can be g ouped in o sub amilies
depending on a posi i e in ege (called he complexi y), which essen ially
measu es he numbe o b anch poin s. The classical Hennebe g su ace
H1is cha ac e ized as he unique example in he sub amily o he simples
complexi y m= 1, while o m≥2 mul ipa ame e amilies a e gi en.
The isome y g oup o he mos symme ic example Hmwi h a gi en
complexi y m∈Nis ei he isomo phic o he dihed al isome y g oup
D2m+2 (i mis odd) o o Dm+1 ×Z2(i mis e en). Fu he mo e, o m
e en Hmis he unique solu ion o he Bj¨o ling p oblem o a hypocycloid
o m+ 1 cusps (i mis e en), while o modd he conjuga e minimal
su ace H∗
m o Hmis he unique solu ion o he Bj¨o ling p oblem o a
hypocycloid o 2m+ 2 cusps.
1. In oduc ion
A celeb a ed esul ob ained independen ly by do Ca mo and Peng [1], Fische -
Colb ie and Schoen [2] and Pogo elo [7] es ablishes ha i Mis a comple e
o ien able s able minimal su ace in R3, hen Mis a plane. Ros [8] p o ed ha
he same cha ac e iza ion holds wi hou assuming o ien abili y. Ne e heless,
a ple ho a o comple e s able minimal su aces in R3appea i we allow hese
s able minimal su aces o ha e b anch poin s, wi h he simples example being
he classical Hennebe g minimal su ace [3].
The class o comple e, fini ely connec ed and fini ely b anched minimal
su aces wi h fini e o al cu a u e (among which s able ones a e a pa icula
case) appea s na u ally in he ollowing si ua ion: Gi en ε0>0, I∈N∪{0}
0123456789().: V,- ol
53 Page 2 o 25 D. Moya and J. P´e ez Resul s Ma h
and H0,K
0≥0, le Λ = Λ(I,H0,ε
0,K
0) be he se o imme sions F:MX
whe e Xis a comple e Riemannian 3-mani old wi h injec i i y adius Inj(X)≥
ε0and absolu e sec ional cu a u e bounded om abo e by K0,Mis a com-
ple e su ace, Fhas cons an mean cu a u e H∈[0,H
0] and Mo se index a
mos I. The second undamen al o m |AFn|o a sequence {Fn}n⊂Λ may ail
o be uni o mly bounded, which leads o lack o compac ness o Λ. Ne e he-
less, he in e es ing ambien geome y o he imme sions Fncan be p o en o
be well o ganized locally a ound a mos Ipoin s p1,n,...,p
k,n ∈Mn(k≤I)
whe e |AFn| akes on a bi a ily la ge local maximum alues. A ound any o
hese poin s pi,n, one can pe o m a blow-up analysis and find a limi o (a sub-
sequence o ) expansions λnFno he Fn( ha is, we iew Fnas an imme sion
wi h cons an mean cu a u e Hn/λnin he scaled ambien mani old λnXn
o a sequence {λn}n⊂R+ ending o ∞). This limi is a comple e imme sed
minimal su ace :ΣR3wi h fini e o al cu a u e, passing h ough he
o igin

0∈R3. Recall ha such an has fini ely many ends, each o which is
a mul i- alued g aph o fini e mul iplici y (spinning) s∈N, o e he ex e io
o a disk in he angen plane a infini e o a ha end. Thus, a bi a ily
small almos pe ec ly o med copies o la ge compac po ions o (Σ) can
be ep oduced in Fn(Mn) a ound Fn(pi,n) o nsufficien ly la ge. Comple e,
fini ely-connec ed and fini ely-b anched minimal su aces wi h fini e o al cu -
a u e in R3appea na u ally when conside ing clus e ing phenomena in his
amewo k: I may occu ha diffe en blow-up limi s o he Fna ound pi,n a
diffe en scales λ1,n >λ
2,n wi h λ1,n/λ2,n →∞as n→∞, p oduce diffe en
limi s j:Σ
jR3,j=1,2, wi h Index( 1) + Index( 2)≤I; in his case, all
he geome y o 1(Σ1) collapses a ound

0∈ 2(Σ2), and e e y end o 1(Σ1)
wi h mul iplici y m≥3 p oduces a b anch poin a he o igin o 2(Σ2)o
b anching o de s−1. Fo de ails abou his clus e ing phenomenon and how
o o ganize hese blow-up limi s in hie a chies appea ing a ound {pi,n}n,see
he pape [4] by Meeks and he second au ho .
The main goal o his pape is o gene alize he classical Hennebe g min-
imal su ace H1 o an infini e amily o connec ed, 1-sided, comple e, fini ely
b anched, s able minimal su aces in R3. B anch poin s a e una oidable i we
seek o comple e, non-fla s able minimal su aces by he a o emen ioned e-
sul s [1,2,7,8]; 1-sidedness is also necessa y condi ion o s abili y (see P opo-
si ion 3below). Ou examples can be g ouped in o sub amilies depending on
he numbe o b anch poin s ( his will be encoded by an in ege m∈Ncalled
he complexi y). The mos symme ic examples Hmin each sub amily o com-
plexi y mwill be s udied in dep h (Sec . 5.3). Depending on he pa i y o m,
ei he Hmo i s conjuga e minimal su ace H∗
m(which does no gi es ise o a
1-sided su ace, see Sec . 5.4) can be iewed as he unique solu ion o a Bj¨o ling
p oblem o a plana hypocycloid (Sec . 5.7). The isome y g oup o Hmis iso-
mo phic o he dihed al g oup D2m+2 i mis odd and o he g oup Dm+1 ×Z2
i mis e en (Sec . 5.8). We will also p o e ha H1is he only elemen in he
Gene alized Hennebe g S able Minimal Su aces Page 3 o 25 53
sub amily wi h complexi y m= 1 (Theo em 11), while o m≥2, Hmcan
be de o med in mul ipa ame e amilies: P oposi ion 14 gi es an explici 1-
pa ame e amily o examples wi h complexi y m= 2, in e pola ing be ween
H2and a limi which u ns ou o be H1(Sec . 6.2.1), and he sub amily o
examples wi h complexi y m= 2 is a wo-dimensional eal analy ic mani old
a ound H2(Sec . 6.2.2).
2. 1-Sided B anched S able Minimal Su aces
We s a wi h he Weie s ass da a (g,ω) on a Riemann su ace Σ, so ha
(g,ω) sol es he pe iod p oblem and p oduces a con o mal ha monic map
X:ΣR3gi en by he classical o mula
X=Re(φ1,φ
2,φ
3)=Re1
2(1 −g2)ω, i
2(1 + g2)ω,gω.(1)
We will assume ha Xis an imme sion ou side o a locally fini e se o poin s
B⊂Σ, whe e X ails o be an imme sion (poin s o Ba e called b anch poin s
o X). Such an Xwill be called a b anched minimal imme sion. The induced
(possible b anched) me ic is gi en by
ds2=1
4(1 + |g|2)2|ω|2.(2)
The local s uc u e o Xa ound a b anch poin in Bis well-known, see
e.g. Micalle and Whi e [5, Theo em 1.4] o de ails. Gi en p∈B, he e exis s
a con o mal coo dina e (D,z) o Σ cen e ed a p(he e Dis he closed uni
disk in he plane), a diffeomo phism uo Dand a o a ion φo R3such ha
φ◦X◦uhas he o m
z→ (zq,x(z)) ∈C×R∼R3
o znea 0, whe e q∈N,q≥2, xis o class C2,andx(z)=o(|z|q). In his
se ing, he b anching o de o pis defined o be q−1∈N.
Le us assume ha Xp oduces a 1-sided b anched minimal su ace; his
means ha he e exis s an an i-holomo phic in olu ion wi hou fixed poin s
I:Σ→Σ such ha I◦φj=φj o j=1,2,3. This is equi alen o
−1/g=g◦I, I∗ω=−g2ω. (3)
In pa icula , Imus p ese e he se B.Σ/Iis a non-o ien able diffe en iable
su ace endowed wi h a con o mal class o me ics, and he ha monic map X
induces ano he ha monic map 
X:Σ/IR3such ha 
X◦π=X, whe e
π:Σ→Σ/Iis he na u al p ojec ion ( 
Xis a b anched minimal imme sion).
Recip ocally, e e y 1-sided con o mal ha monic map can be cons uc ed in his
way.
53 Page 4 o 25 D. Moya and J. P´e ez Resul s Ma h
Rema k 1. In he pa icula case ha he compac ifica ion o Σ is C,wecan
assume ha I(z)=−1/zand w i e ω= dz globally. In his se ing, he
abo e equa ions gi e
−1/g(z)=g(−1/z), ◦I=−z2g2 . (4)
De ini ion 2. Gi en a 1-sided con o mal ha monic map 
X:Σ/IR3,we
deno e by Δ, |A|2 he Laplacian and squa ed no m o he second undamen al
o m o 
X.Theindex o 
Xis defined as he numbe o nega i e eigen alues o
he ellip ic, sel -adjoin ope a o L=Δ+|A|2(Jacobi ope a o o X) defined
o e he space o compac ly suppo ed smoo h unc ions φ:Σ→Rsuch ha
φ◦I=−φ.
Xis said o be s able i i s index is ze o.
In he case 
Xis fini ely b anched, he eigen alues and eigen unc ions o
he Jacobi ope a o o Xa e well defined ia a a ia ional app oach, since he
codimension o he singula i y se Bis wo (see [9]), and s abili y also makes
sense.
The nex esul is p o en by Meeks and he second au ho in [4].
P oposi ion 3. Le X:ΣR3be comple e, non-fla , fini ely b anched mini-
mal imme sion wi h b anch locus B⊂Σ.Then:
1. [4, P oposi ion 3] I Xis s able, hen Σis non-o ien able and X(B)
con ains mo e han 1 poin .
2. [4, Rema k 3.6] Suppose ha Σis non-o ien able, Xhas fini e o al cu -
a u e and i s ex ended uno ien ed Gauss map G:P2=S2/{±1}→P2
is a diffeomo phism. Then, Xis s able.
3. The Bj¨o ling P oblem
We nex ecall he basics o he classical Bj¨o ling p oblem, o be used la e .
Le γ:I⊂R→R3be an analy ic egula cu e and ηan analy ic ec o field
along γsuch ha γ( ),η( )=0andη( )= 1 o all ∈I. The classical
esul due o E.G. Bj¨o ling asse s ha he ollowing pa ame iza ion gene a es
a minimal su ace Swhich con ains γand has ηas uni no mal ec o along
γ:
X(u, )=Reγ(w)−iw
w0η(w)×γ(w)dw,
whe e γ, ηa e analy ic ex ensions o he co esponding γ,η and w=u+i is
defined in a simply connec ed domain Ω ⊂Cwi h I⊂Ω. In pa icula , he
su ace Sis locally unique a ound γwi h his da a (i is called he solu ion o
he Bj¨o ling p oblem wi h da a γ,η).
In wha ollows, we will conside diffe en Bj¨o ling p oblems o analy ic
plana cu es γ⊂{z=0} ha ail o be egula a fini ely many poin s. The
abo e cons uc ion can be applied o each o he egula a cs o hese cu es
Gene alized Hennebe g S able Minimal Su aces Page 5 o 25 53
a e emo ing he ze os o γ. In all ou applica ions, ηwill be aken as he
(uni ) no mal ec o field o γas a plana cu e.
4. The Classical Hennebe g Su ace
The classical Hennebe g minimal su ace H1is he 1-sided, comple e, s able
minimal su ace in R3gi en by he Weie s ass da a:
g(z)=z, ω =z−4(z±i)(z±1)dz =z−4(z4−1)dz, z ∈C−{0,∞}.
(5)
H1has wo b anch poin s1a [1] = {1,−1},[i]={i, −i}∈P2=C/A, whe e
A(z)=−1/zis he an ipodal map. By P oposi ion 3,H1is s able.
H1can be con o mally pa ame e ized (up o ansla ions) by eq. (1).
A e ansla ing Xso ha X(eiπ/4)=

0, he b anch poin s o H1a e mapped
by X o (0,0,±1) and a pa ame iza ion o H1in pola coo dina es z= eiθ
is gi en by
X( eiθ)=⎛
⎜
⎜
⎝
cos θ
2( −1
)−cos(3θ)
6( 3−1
3)
−sin θ
2( −1
)−sin(3θ)
6( 3−1
3)
cos(2θ)
2( 2+1
2)
⎞
⎟
⎟
⎠.(6)
Since X(eiθ)=(0,0,cos(2θ)), hen Xmaps he uni ci cle in o he e ical
segmen {(0,0, )| ∈[−1,1]}. In his way, θ∈[0,2π]→ X(eiθ) bounces
be ween he wo b anch poin s o H1(obse e ha he complemen o his
closed segmen in he x3-axis is no con ained in H1), see Fig. 1.
4.1. Isome ies o H1
I is s aigh o wa d o check ha
1. The an ipodal map A:C→C(in pola coo dina es ( , θ)→ (1/ , π +θ))
lea es he su ace in a ian . This is he deck ans o ma ion, which is
o ien a ion e e sing.
2. The map z→−z(in pola coo dina es ( , θ)→ ( , π +θ)) induces he
o a ion by angle πabou he axis x3on he su ace.
3. The in e sion o he z-plane wi h espec o he uni ci cle, z→ 1/z, (in
pola coo dina es ( , θ)→ (1/ , θ)) is he composi ion o Awi h z→−z,
and hus, i also induces a o a ion o angle πabou he x3-axis on he
su ace.
4. The conjuga ion map z→ z(in pola coo dina es ( , θ)→ ( , −θ)) in-
duces he eflec ion o Xabou he plane (x1,x
3).
5. The eflec ion abou he imagina y axis (in pola coo dina es ( , θ)→
( , π −θ)) induces he eflec ion o Xabou he plane (x2,x
3).
1B anch poin s o H1all ha e o de 1 (locally he su ace winds wice a ound he b anch
poin ); his ollows om di ec compu a ion, o om P oposi ion 21 in Whi e’ s ”Lec u es
on minimal su aces heo y”.

53 Page 6 o 25 D. Moya and J. P´e ez Resul s Ma h
Figu e 1. The Hennebe g su ace H1. A e a ansla ion,
he b anch poin s o H1a e con ained in he x3-axis. H1con-
ains wo ho izon al, o hogonal lines ha bisec he x1-and
x2-axis. Le : In e sec ion o H1wi h a ball o adius 8. Righ :
op iew o H1
6. Xmaps he hal -line { e−iπ/4| ∈(0,∞)}( espec i ely { eiπ/4|
∈(0,∞)}) injec i ely in o l1= Span(1,1,0) ( espec i ely l2= Span
(1,−1,0)). Thus, he o a ions R1,R
2o angle πabou l1,l
2a e isome-
ies o X(R1is induced by z→−iz and R2by z→ iz).
7. The map z→ iz (in pola coo dina es ( , θ)→ ( , θ +π/2)) induces he
o a ion o angle π/2 abou he x3-axis composed by a eflec ion in he
(x1,x
2)-plane.
Toge he wi h he iden i y map, he abo e isome ies o m a subg oup o
he isome y g oup Iso(H1)o H1, isomo phic o he dihed al g oup D4.
Lemma 4. These a e all he (in insic) isome ies o H1.
P oo . This is a di ec consequence o he ac ha e e y in insic isome y φ
o H1p oduces a con o mal diffeomo phism o C {0}in o i sel ha p ese es
he se o b anch poin s o H1. In pa icula φis o one o he a o emen ioned
eigh cases. 
4.2. Associa ed Family and he Conjuga e Su ace H∗
1
The flux ec o o H1a ound he o igin in C anishes (in o he wo ds, he
Weie s ass o m Φ = (φ1,φ
2,φ
3) associa ed o H1is exac ). This implies ha
all associa ed su aces {
H1(ϕ)|ϕ∈[0,2π)} o he o ien able co e 
H1=
H1(0)
o H1a e well-defined as su aces in R3( he b anched minimal imme sion

H1(ϕ) has Weie s ass da a gϕ=g,ωϕ=eiϕωand i is isome ic o 
H1,in
pa icula i has he same b anch locus as 
H1).
Gene alized Hennebe g S able Minimal Su aces Page 7 o 25 53
Figu e 2. The as oid γ4( ed) and he ou ays ob ained
by in e sec ing H∗
1wi h he (x1,x
2)-plane (blue) (Colo figu e
online)
None o he su aces 
H1(ϕ) excep o ϕ= 0 descends o he non-
o ien able quo ien P2 {[0]}, because he second equa ion in (3) is no p e-
se ed i we exchange ωby eiϕω,ϕ∈(0,2π). In pa icula , none o hese
associa ed su aces a e cong uen o H1.
The conjuga e su ace H∗
1:= 
H1(π/2) is symme ic by eflec ion in he
(x1,x
2)-plane. The in e sec ion be ween H∗
1and {z=0}consis s o he as oid
γ4pa ame e ized by
→ γ4( )=⎛
⎜
⎝−sin(θ)+sin(3θ)
3
−cos(θ)−cos(3θ)
3
0
⎞
⎟
⎠,
oge he wi h ou ays s a ing a he cusps o he as oid in he di ec ion o
hei posi ion ec o s, see Fig. 2.
In pa icula , H∗
1is he solu ion o he Bj¨o ling p oblem o he cu e γ4
and he choice o uni no mal field he no mal ec o o γ4as a plana cu e,
see also Rema k 8below.
5. Gene alized Hennebe g Su aces
We will nex sea ch o a 1-sided, comple e, s able minimal su ace in X:Σ
R3wi h Σ = C E,Efini e and g(z)=z. Hence, I(z)=−1/z,
X=X/I:Σ/
IR3is s able and (4) w i es
(−1/z)=−z4 (z).(7)
53 Page 8 o 25 D. Moya and J. P´e ez Resul s Ma h
5.1. Gene al o m o
We ake a gene al a ional unc ion
(z)= c
zm+3 M
j=1(z−aj)
N
j=1(z−bj),(8)
whe e c, aj,b
j∈C∗,m∈N,M,N ∈N∪{0}a e o be de e mined.
Rema k 5. 1. Henne be g’s su ace H1has (z)=z−4(z4−1), hence c=1,
m=1,N=0,M=4,{aj}={±1,±i}.
2. The ze os o he induced he me ic (2) (b anch poin s o he su ace)
occu p ecisely a he poin s aj; he ends occu a 0,∞and a he poin s
bj(in pa icula , bo h amilies {aj}j,{bj}jmus hen come in pai s o
an ipodal poin s, see also (12) below).
3. A consequence o he las obse a ion is ha when he abo e o a ions
in R3o ou su aces (p o ided ha he Weie s ass da a close pe iods)
a e no allowed unless he axis o o a ion is e ical.
Imposing (7) o(8)wege
c(−1)m−1+M−Nz3+m−M+NM
j=1(1 + ajz)
N
j=1(1 + bjz)= (−1/z)=−z4 (z)
=−c
zm−1M
j=1(z−aj)
N
j=1(z−bj),
hus
c(−1)m+M−Nz2+2m−M+N
M

j=1
(1 + ajz)
N

j=1
(z−bj)=c
M

j=1
(z−aj)
N

j=1
(1 −bjz),
(9)
om whe e we deduce ha
2+2m−M+N=0,(10)
in pa icula M−Nis e en. Subs i u ing z=0in(9)wege
c(−1)m
N

j=1
bj=c
M

j=1
aj.(11)
Using (11), we can ew i e (9) as an equali y be ween monic polynomials in
z:
M

j=1 1
aj
+zN

j=1
(z−bj)=
M

j=1
(z−aj)
N

j=1 1
bj
+z,
Gene alized Hennebe g S able Minimal Su aces Page 9 o 25 53
om whe e we deduce ha
{a1,...,a
M}={−1/a1,...,−1/aM},{b1,...,b
N}={−1/b1,...,−1/bN}.
(12)
ha is, M,N a e e en, he aj( esp. bj) a e gi en by M/2( esp.N/2) pai s
o an ipodal poin s in C∗.Now(10) and (11) gi e espec i ely:
1+m−
M+
N=0,(13)
−c
N/2

j=1
bj
bj
=c
M/2

j=1
aj
aj
.(14)
5.2. Sol ing he Pe iod P oblem in he One-Ended Case: Complexi y
F om (3) and (8) we see ha he poin s whe e ds2can blow up a e z=
0,b
1,...,b
Nand i s an ipodal poin s. In o de o keep he compu a ions simple,
we will assume he e a e no bj’s, i.e. N= 0 (o equi alen ly M/2=m+ 1),
which educes he pe iod p oblem o imposing
γ
g2ω=γ
ω, Re γ
gω =0,
whe e γ={|z|=1}, o equi alen ly,
Res0(g2 )=−Res0( ),Im Res0(g )=0.(15)
We can simpli y (8) o
(z)= c
zm+3
m+1

j=1
(z−aj)z+1
aj,(16)
which sa isfies (7) ( his is he condi ion o descend o he quo ien as a 1-sided
su ace, p o ided ha he pe iod p oblem (15) is sol ed) i and only i (14)
holds, which in his case educes o
−c
c=
m+1

j=1
aj
aj
.(17)
We call
P(z):=
m+1

j=1
(z−aj)z+1
aj=
2m+2

h=0
Ahzh.(18)
53 Page 16 o 25 D. Moya and J. P´e ez Resul s Ma h
hence
A1=R( 1)e2iθ2+R( 2)eiθ2,(27)
A2=−1+e2iθ2−R( 1)R( 2)eiθ2,(28)
A3=−R( 1)+R( 2)eiθ2.(29)
W i ing c=eiβ,weha e
cA1+cA3=R( 1)e−i(β+2θ2)−eiβ+R( 2)e−i(β+θ2)−ei(β+θ2)
=R( 1)e−iθ2e−i(β+θ2)−ei(β+θ2)−2R( 2)sinh(i(β+θ2))
=−2e−iθ2R( 1)sinh(i(β+θ2)) −2iR( 2)sin(β+θ2)
=−2iR( 1)e−iθ2+R( 2)sin(β+θ2),(30)
cA2=−eiβ 1+e2iθ2+R( 1)R( 2)ei(β+θ2)
=−ei(β+θ2)e−iθ2+eiθ2+R( 1)R( 2)ei(β+θ2)
=−[2 cosh(iθ2)−R( 1)R( 2)] ei(β+θ2)
=−[2 cos θ2−R( 1)R( 2)] ei(β+θ2).(31)
A lis (c, a1,a
2) sol es he pe iod p oblem i and only i he igh -hand-side
o (30) anishes and he igh -hand-side o (31) is eal.
The hi d equa ion in (26) educes o
e2i(β+θ2)=−1.(32)
Theo em 11. The Hennebe g su ace H1is he only su ace wi h m=1 ha
sol es he pe iod p oblem and descends o a 1-sided quo ien .
P oo . By he abo e a gumen s, he igh -hand-side o (30) anishes, he igh -
hand-side o (31) is eal and (32) holds.
(32) implies ha sin(β+θ2)=±1. Since he igh -hand-side o (30)
anishes, we ha e
R( 1)e−iθ2+R( 2)=0.(33)
We ha e wo possibili ies:
• 1=1.Thus(33) implies 2= 1. F om, (32)weha eβ+θ2≡π/2modπ
and om (31)weha ecosθ2=0, husθ2=π/2o θ2=3π/2. This gi es
he lis s (1,1,i), (−1,1,i), (1,1,−i) and (−1,1,−i). All o hem gi e aise
o he Hennebe g su ace.
• 1= 1. This implies e−iθ2=−R( 2)
R( 1), which is eal. Hence e−iθ2=±1.
As he unc ion → R( ) is injec i e, his implies 1= 2and θ2=πo
2=1/ 1and θ2= 0. Since he igh -hand-side o (31) is eal and (32)
holds, 2 cos θ2−R( 1)R( 2) = 0. Bu in bo h cases 2 cos θ2−R( 1)R( 2)
does no anish. Hence his possibili y canno occu .


Gene alized Hennebe g S able Minimal Su aces Page 17 o 25 53
6.2. Solu ions wi h Complexi y m=2
Suppose ha a lis (c=eiβ,a
1= 1,a
2= 2eiθ2,a
3= 3eiθ3)∈S1×
R+×(C∗)2is a solu ion o he pe iod p oblem wi h 1-sided quo ien and
associa ed b anched minimal imme sion X. The lis ha gi es ise o H2is
(±i, 1,e
iπ/3,e
2iπ/3).
Sol ing he pe iod p oblem wi h 1-sided quo ien is equi alen o sol ing
cA2=−cA4,Im(cA3)=0,−c
c=a2
a2
a3
a3
(34)
The hi d equa ion in (34) educes o
e2i(β+θ2+θ3)=−1.(35)
(18) can be w i en as
P(z)=z6+A5z5+A4z4+A3z3+A2z2+A1z+A0,
whe e
A2=e2i(θ2+θ3)+e2iθ2+e2iθ3−R( 1)R( 2)ei(θ2+2θ3)−R( 1)R( 3)ei(2θ2+θ3)
−R( 2)R( 3)ei(θ2+θ3),(36)
A3=2
R( 2)cosθ3+R( 3)cosθ2+R( 1)cos(θ2−θ3)
−1
2R( 1)R( 2)R( 3)ei(θ2+θ3),(37)
A4=−(1 + e2iθ2+e2iθ3)+R( 1)R( 2)eiθ2
+R( 1)R( 3)eiθ3+R( 2)R( 3)ei(θ2+θ3).(38)
Thus,
cA2+cA4=2e−i[β+2(θ2+θ3)]F, (39)
cA3=±2iG (40)
whe e
F=e2iθ3+[2cosθ2−R( 1)R( 2)] eiθ2−R( 3)R( 1)+R( 2)eiθ2eiθ3,
(41)
G=R( 2)cosθ3+R( 3)cosθ2+R( 1)cos(θ2−θ3)−1
2R( 1)R( 2)R( 3).
(42)
Rema k 12. (I) F om (42) we deduce ha Gis eal, hence he condi ion
Im(cA3) = 0 only holds i and only i G= 0. We deduce ha a lis
(c, a1,a
2,a
3) sol es he 1-sided pe iod p oblem i and only i (35) holds
and F=G=0.
(II) The exp ession (41) is symme ic in ( 2,θ
2),( 3,θ
3). This can be deduced
om he symme y o A2,A
4, o di ec ly checked by using he equali y
e2iθ = 2 cos θeiθ −1,θ∈R,(43)
53 Page 18 o 25 D. Moya and J. P´e ez Resul s Ma h
which ans o ms (41)in o
F=(1+e2iθ2+e2iθ3)−R( 1)
3

j=2
R( j)eiθj−R( 2)R( 3)ei(θ2+θ3).(44)
Lemma 13. I F=0, hen he coefficien o R( 1)in (44)is non-ze o.
P oo . Suppose R( 2)eiθ2+R( 3)eiθ3= 0. This leads o one o he ollowing wo
possibili ies: (a) eiθ2=eiθ3and R( 2)=−R( 3)o else(b)eiθ2=−eiθ3and
R( 2)=R( 3). (a) implies 3=1/ 2and hus, (44) gi es F= 1+e2iθ2(1
2
2+ 2
2).
(b) implies 2= 3and (44) gi es he same exp ession o F. In any case, we
deduce om F= 0 ha e2iθ2is eal nega i e, hence 2
2
4
2+1 =−e2iθ2= 1. This is
impossible, since he unc ion x>0→ x
1+x2has a unique maximum a x=1
wi h alue 1/2. 
The nex esul desc ibes a one-pa ame e amily o non- i ial examples
o complexi y m=2diffe en omH2.
P oposi ion 14. Suppose ha a lis (c, a1,a
2,a
3)sol es he 1-sided pe iod p ob-
lem. Then:
1. I 1=1, and a leas one o 2o 3equals one, hen (c, a1,a
2,a
3)=
(±i, 1,e
iπ/3,e
2iπ/3)and he example is H2.
2. I θ2+θ3=0(mod π), hen 2= 3o 2=1/ 3and ( 1,
2)a e gi en
by he ollowing unc ions o θ2∈(π
4,π
3]∪[2π
3,3π
4):
R( 1(θ2)) = 1
8√2 (θ2)−3
cos θ2cos(2θ2)[ (θ2) + 3 + 4 cos(2θ2)],(45)
R( 2(θ2)) = − (θ2)−3
√2,(46)
o else ( 1,
2)a e gi en by he opposi e exp essions o bo h R( 1(θ2)),R
( 2(θ2)), which exchange ( 1,
2)by (1
1,1
2). He e, is he unc ion
(θ2)=1−8 cos(2θ2)−8 cos(4θ2).(47)
P oo . I 1= 1, and a leas one o 2o 3equals one, hen (44) gi es
1+e2iθ2+e2iθ3=0and(42) gi es R( 2)cosθ3+R( 3)cosθ2= 0. Since a
leas one o 2o 3equals one, hen a leas one o R2o R3equals ze o. In
ac , bo h R2=R3= 0 (because o he wise we ge cos θ2= 0 o cos θ3=0,
which p e en s 1 + e2iθ2+e2iθ3 om cancelling), and hus, 2= 3=1.In
his se ing, 1 + e2iθ2+e2iθ3= 0 leads o (c, a1,a
2,a
3)=(±i, 1,e
iπ/3,e
2iπ/3),
which p o es i em 1.
Now assume θ2+θ3= 0. Then (44),(42) gi e espec i ely
1 + 2 cos(2θ2)−R( 2)R( 3)=R( 1)[R( 2)eiθ2+R( 3)e−iθ2],
(48)
(R( 2)+R( 3)) cos θ2+R( 1) cos(2θ2)=1
2R( 1)R( 2)R( 3).(49)
Gene alized Hennebe g S able Minimal Su aces Page 19 o 25 53
Obse e ha R( 1) canno anish by Lemma 13 (ano he eason is ha o h-
e wise, (49) gi es cos θ2= 0, and (48) gi es −1−R( 2)R( 3)=0whichis
absu d). F om (48) we deduce ha R( 2)eiθ2+R( 3)e−iθ2is eal. This implies
ha [R( 2)−R( 3)] sin θ2= 0. We claim ha sin θ2= 0; o he wise θ2≡0
(mod π) and (48),(49) gi e he sys em
3−R( 2)R( 3)=±R( 1)[R( 2)+R( 3)],
R( 1)±(R( 2)+R( 3)) = 1
2R( 1)R( 2)R( 3),
(wi h he same choice o signs), which can be easily seen no o ha e solu ions.
Thus, sin θ2= 0 hence R( 2)=R( 3)and 2= 3. In his se ing,
(48),(49) educe o
1 + 2 cos(2θ2)−R( 2)2=2R( 1)R( 2)cosθ2,(50)
2R( 2)cosθ2+R( 1) cos(2θ2)=1
2R( 1)R( 2)2.(51)
I we assume θ2+θ3=π, hen (44),(42) gi e espec i ely
1 + 2 cos(2θ2)+R( 2)R( 3)=R( 1)[R( 2)eiθ2−R( 3)e−iθ2],
(52)
(−R( 2)+R( 3)) cos θ2−R( 1) cos(2θ2)=1
2R( 1)R( 2)R( 3).(53)
Again, R( 1) can no anish due o Lemma 13.F om(52) we deduce ha
R( 2)eiθ2−R( 3)e−iθ2is eal. This implies ha [R( 2)+R( 3)] sin θ2=0.We
claim ha sin θ2= 0; o he wise θ2≡0 (mod π) and (52),(53) gi e he sys em
3+R( 2)R( 3)=±R( 1)[R( 2)−R( 3)],
−R( 1)±(−R( 2)+R( 3)) = 1
2R( 1)R( 2)R( 3),
(wi h he same choice o signs), which again has no solu ions. Thus, sin θ2=0
hence R( 2)=−R( 3)and 2=1/ 3. In his se ing, (48),(49) educe again
o (50) and (51).
The sys em (50),(51) has wo equa ions and h ee unknowns 1,
2,θ
2.
Nex we desc ibe i s solu ions. Conside he unc ion gi en by (47). Then,
(π−θ2)= (θ2), o each θ2, (θ2,0)=0= (π−θ2,0),
whe e θ2,0=1
2co −19
√32√10+95 ∼0.499841, and he domain o is
[θ2,0,π−θ2,0]+πZ. The se {θ2∈[θ2,0,π−θ2,0]| (θ2)≥3}equals A:=
[π
4,π
3]∪[2π
3,3π
4].
The unique solu ion ( 1,
2) o he sys em (50),(51) as a unc ion o θ2is
gi en by (45), (46) and he opposi e exp essions o bo h R( 1(θ2)),R( 2(θ2)),
which exchange ( 1,
2)by(1
1,1
2). 
53 Page 20 o 25 D. Moya and J. P´e ez Resul s Ma h
6.2.1. The One-Pa ame e Family o Examples in I em 2 o P oposi ion 14.
Obse e ha he map θ2∈(π
4,π
3]→ π−θ2∈[2π
3,3π
4) is a diffeomo phism.
Using he no a ion in i em 2 o P oposi ion 14, o eachθ2∈(π
4,π
3], we ha e
R( 1(π−θ2)) = −R( 1(θ2)),R( 2(π−θ2))) = R( 2(θ2)).(54)
Each o hese lis s wi h θ2∈(π
4,π
3]∪[2π
3,3π
4) sol es he 1-sided pe iod p oblem,
hence i defines a non-o ien able, b anched minimal su ace H(θ2). Fu he -
mo e, (54) implies ha
1(π−θ2)= 1
1(θ2),
2(π−θ2)= 2(θ2).(55)
We claim he su aces H(θ2)andH(π−θ2) a e cong uen . To see his, no e
ha he se o poin s {aj,−1/ajj=1,2,3} ha defines h ough (16)and
gene a es he su ace H(θ2), is:
 1,−1
1
,
2eiθ2,1
2
ei(π+θ2),
2e−iθ2,1
2
ei(π−θ2).(56)
The analogous se o poin s o he su ace H(π−θ2) is gi en h ough (55):
1
1
,− 1,− 2e−iθ2,1
2
e−iθ2,− 2e−iθ2,1
2
eiθ2,
which is up o sign he se desc ibed in (56). The e o e, he unc ion defined
by equa ion (16) and he co esponding unc ion 
defined by he same o mula
o he su ace H(π−θ2) a e ela ed by 
(−z)=− (z), o each z∈C.Using
ha ω= dz and ω=
dz define, ia he Weie s ass ep esen a ion (1),
ela ed b anched minimal imme sions X=(x1,x
2,x
3) o H(θ2)and 
X=
(x1,x2,x3) o H(π−θ2), we ge ha H(θ2)andH(π−θ2) a e cong uen .
In he sequel, we will educe ou s udy o he amily {H(θ2)|θ2∈(π
4,π
3]}.
F om (45), (46)weha e
lim
θ2→π/3−R( 1(θ2)) = lim
θ2→π/3−R( 2(θ2)) = 0,
which implies ha
lim
θ2→π/3−H(θ2)=H2.
We nex iden i y he limi (a e escaling) o he su aces H(θ2)asθ2→
π/4+. We fi s obse e ha
lim
θ2→π/4+R( 1(θ2)) = −∞,lim
θ2→π/4+R( 2(θ2)) = 0.(57)
This implies ha he b anch poin a1=a1(θ2) is ending o ze o, hence he
limi o H(θ2) when θ2→π/4+(i i exis s) canno be an example wi h
complexi y m= 2. In ui i ely, i is clea han he complexi y canno inc ease
when aking limi s (e en wi h diffe en scales), hence by Theo em 11 i is
Gene alized Hennebe g S able Minimal Su aces Page 21 o 25 53
na u al o hink ha he limi o sui able e-scalings o H(θ2) when θ2→π/4+
be H1. We nex o malize his idea.
Ano he consequence o (57) is ha he lis (c, a1,a
2,a
3)=(i, 1(θ2),
2
(θ2)eiθ2,
2(θ2)e−iθ2) con e ges as θ2→π/4+ o (c, a1,a
2,a
3)=(i, 0,e
iπ/4,
e−iπ/4). A e applying o H(θ2) a homo he y o a io 1(θ2)>0 (which
sh inks o ze o), he Weie s ass da a o he sh unk su ace 1(θ2)H(θ2)is
(g(z)=z, 1(θ2) (z)), whe e (z) is gi en by (16). Fo z∈C {0}fixed,
lim
θ2→π/4+ 1(θ2) (z)(16)
= lim
θ2→π/4+ 1(θ2)i
z5
3

j=1
(z−aj)z+1
aj
=i
z5z−eiπ/4z+eiπ/4z−e−iπ/4z+e−iπ/4lim
θ2→π/4+(z− 1(θ2)) ( 1(θ2)z+1)
=i
z4z−eiπ/4z+eiπ/4z−e−iπ/4z+e−iπ/4:= 
(z).
Plugging he Weie s ass da a (g(z)=z
dz)in o(1), we ob ain a pa ame iza-
ion o he limi su ace o 1(θ2)H(θ2)asθ2→π/4+in pola coo dina es
z= eiθ:

X( eiθ)=⎛
⎜
⎜
⎝
−sin θ
2( −1
)+sin(3θ)
6( 3−1
3)
−cos θ
2( −1
)−cos(3θ)
6( 3−1
3)
−cos θsin θ( 2+1
2)
⎞
⎟
⎟
⎠.(58)
We claim ha his pa ame iza ion gene a es he Hennebe g su ace H1.To
see his, obse e ha i we fi s pe o m he change o a iables θ=
θ+π/4
and hen o a e he su ace an angle o −π
4a ound he x3-axis, we ge
⎛
⎜
⎝
cos π
4sin π
40
−sin π
4cos π
40
001
⎞
⎟
⎠·
X( ei(
θ+π
4))=−⎛
⎜
⎜
⎝
cos 
θ
2( −1
)−cos(3
θ)
6( 3−1
3)
−sin 
θ
2( −1
)−sin(3
θ)
6( 3−1
3)
cos(2
θ)
2( 2+1
2)
⎞
⎟
⎟
⎠,
which is, up o a sign, he pa ame iza ion gi en in (6) o H1(see Fig. 5 o
images o he su ace H(θ2) o h ee diffe en alues o θ2∈(π
4,π
3]).
6.2.2. A ound H2 he Space o Examples wi h Complexi y m=2is Two-
Dimensional. I em 2 o P oposi ion 14 defines a non-compac amily o non-
o ien able, b anched minimal su aces {H(θ2)|θ2∈(π
4,π
3]}inside he moduli
space o examples wi h complexi y m= 2. Appa en ly, he space o solu ions
o his complexi y has eal dimension 2 ( he a iables a e 1,
2,
3,θ
2,θ
3,
F= 0 is a complex condi ion and G= 0 is a eal condi ion). We can ensu e
his a leas a ound H2 ia he implici unc ion heo em ( his is consis en
wi h i em 2 o P oposi ion 14, since i imposes he ex a condi ion θ2+θ3=0
mod π), as we will show nex .
Conside he (smoo h) pe iod map gi en by
P:(R+)3×R2−→ R3≡C×R
(( 1,
2),( 3,θ
2,θ
3)) −→ (F( 1,
2,
3,θ
2,θ
3),G( 1,
2,
3,θ
2,θ
3)),

53 Page 22 o 25 D. Moya and J. P´e ez Resul s Ma h
Figu e 5. Su aces gene a ed by he p e ious lis s
(c, a1,a
2,a
3)=(i, 1(θ2),
2(θ2)eiθ2,
2(θ2)e−iθ2) wi h θ2=1
(le ), θ2=0.83 (cen e ), θ2=0.7854 ( igh ). The limi o
1(θ2)H(θ2)asθ2→π/4+∼0.785398 is he Hennebe g su -
ace H1
whe e F,G a e gi en by (44), (42) espec i ely. Gi en ( 1,
2)∈(R+)2,le
P 1, 2:R+×R2→R3be he es ic ion o P o {( 1,
2)}×R+×R2. Then,
d(P 1, 2)( 3,θ2,θ3)≡
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
∂Re(F)
∂ 3
∂Re(F)
∂θ2
∂Re(F)
∂θ3
∂Im(F)
∂ 3
∂Im(F)
∂θ2
∂Im(F)
∂θ3
∂G
∂ 3
∂G
∂θ2
∂G
∂θ3
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
.(59)
Recall ha he lis associa ed o H2is ( 1,
2,
3,θ
2,θ
3)=(1,1,1,π/3,2π/3).
Imposing his choice o pa ame e s and compu ing he de e minan o (59)we
ge
d(P1,1)(1,π/3,2π/3) =2
√3=0.
Thus, he implici unc ion heo em gi es an open neighbo hood U⊂(R+)2
o ( 1,
2)=(1,1), an open se W⊂(R+)3×R2wi h ( 1,
2,
3,θ
2,θ
3)=
(1,1,1,π/3,2π/3) ∈Wand a smoo h map ϕ:U→R3such ha all he solu-
ions ( 1,
2,
3,θ
2,θ
3) a ound (1,1,1,π/3,2π/3) o he equa ion P( 1,
2,
3,
θ2,θ
3) = 0 a e o he o m ( 3,θ
2,θ
3)=ϕ( 1,
2). By Rema k 12(I), he lis
(c=eiβ( 1, 2),
1,
2eiθ2,
3eiθ3)
wi h β=β( 1,
2) gi en by (35) sol es he 1-sided pe iod p oblem and so,
i defines a 1-sided b anched minimal su ace. This p oduces a 2-pa ame e
Gene alized Hennebe g S able Minimal Su aces Page 23 o 25 53
de o ma ion o he su ace H2in he moduli space o examples wi h m=2
a ound H2, which in u n desc ibes he whole moduli space a ound H2.
Rema k 15. A nice consequence o he classical Leibniz o mula o he de i a-
i e o a p oduc is a ecu si e law ha gi es he coefficien s o he polynomial
P(z) defined by (18) in e ms o he coefficien s o he ela ed polynomial o
one complexi y less. To ob ain his ecu si e law, we fi s adap he no a ion
o he complexi y:
Pm+1(z):=
m+1

j=1
(z−aj)z+1
aj=
2m+2

h=0
Am+1,hzh.(60)
(19) can now be w i en
cAm+1,m =−cAm+1,m+2,Im(cAm+1,m+1)=0.(61)
We wan o find exp essions o he abo e coefficien s Am+1,m,A
m+1,m+2,
Am+1,m+1, depending only on coefficien s o he ype Am,h (i.e., o one com-
plexi y less). W i ing aj= jeiθjin pola coo dina es, obse e ha
Pm+1(z):=Pm(z)Qm+1(z),whe e Qm+1(z)=(z− m+1eiθm+1 )z+eiθm+1
m+1 .
Hence o h∈{m, m +1,m+2},
Am+1,h =1
h!P(h)
m+1(0) = 1
h!(PmQm+1)(h)(0) = 1
h!
h

k=0 h
kP(k)
m(0)Q(h−k)
m+1 (0),
whe e in he las equali y we ha e used Leibniz o mula. Since Qm+1 is a
polynomial o deg ee wo, i s de i a i es o o de h ee o mo e anish. Hence
we can educe he las sum o e ms whe e he index ksa isfies h−k≤2, i.e.,
k∈{h−2,h−1,h}and hus,
Am+1,h =1
h!h
h−2P(h−2)
m(0)Q
m+1(0) + h
h−1P(h−1)
m(0)Q
m+1(0) + h
hP(h)
m(0)Qm+1(0)
=1
h!h!
(h−2)!2 P(h−2)
m(0) ·2−hP(h−1)
m(0)R( m+1)eiθm+1 −P(h)
m(0)e2iθm+1 
=1
(h−2)! P(h−2)
m(0) −1
(h−1)! P(h−1)
m(0)R( m+1)eiθm+1 −1
h!P(h)
m(0)e2iθm+1 
=Am,h−2−Am,h−1R( m+1)eiθm+1 −Am,he2iθm+1 ,(62)
which is he desi ed ecu ence law. (62) can be used o find solu ions o (61) o
complexi y m= 3 besides he mos symme ic example H3, bu he equa ions
a e complica ed and we will no gi e hem he e.
Au ho con ibu ions Bo h au ho s ha e con ibu ed equally in he s udy.
Funding Funding o open access publishing: Uni e sidad de G anada/CBUA
This wo k is suppo ed in pa by he IMAG–Ma ia de Maez u G an CEX2020-
001105-M/AEI/10.13039/501100011033, MICINN PID2020-117868GB-I00 and
Jun a de Andaluc´ıa P18-FR-4049 and A-FQM-139-UGR18.
53 Page 24 o 25 D. Moya and J. P´e ez Resul s Ma h
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Re e ences
[1] do Ca mo, M., Peng, C.K.: S able comple e minimal su aces in R3a e planes.
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[2] Fische -Colb ie, D., Schoen, R.: The s uc u e o comple e s able minimal su aces
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[3] Hennebe g, L.: ¨
Ube solche Minimal߬ache, welche eine o gesch iebene ebene
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hechikum, Zu ich (1875)
[4] Meeks III, W.H., P´e ez, J.: Hie a chy s uc u es in fini e index CMC su aces.
Wo k in p og ess
[5] Micalle , M., Whi e, B.: The s uc u e o b anch poin s in minimal su aces and
in pseudoholomo phic cu es. Ann. Ma h. 141(1), 35–85 (1995)
[6] Odehnal, B.: On algeb aic minimal su aces. KoG 20, 61–78 (2016)
[7] Pogo elo , A.V.: On he s abili y o minimal su aces. So . Ma h. Dokl. 24, 274–
276 (1981)
[8] Ros, A.: One-sided comple e s able minimal su aces. J. Diffe . Geom. 74, 69–92
(2006)
[9] Tysk, J.: Eigen alue es ima es wi h applica ions o minimal su aces. Pac. J.
Ma h. 128, 361–366 (1987)
Da id Moya and Joaqu´ın P´e ez
Depa men o Geome y and Topology and Ins i u e o Ma hema ics (IMAG)
Uni e si y o G anada
G anada
Spain
e-mail: [email p o ec ed];
[email p o ec ed]
Gene alized Hennebe g S able Minimal Su aces Page 25 o 25 53
Recei ed: July 5, 2022.
Accep ed: Decembe 23, 2022.
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ional claims in published maps and ins i u ional affilia ions.