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arXiv:0803.2214v1 [math.DG] 14 Mar 2008
The Gauss Map of Hypersurfaes in
2
Step Nilpotent
Lie Groups
E. V. Petrov
V. N. Karazin Kharkiv National University
4 Svobody sq., Kharkiv, 61077, Ukraine
Email: petrovuniver.kharkov.ua
In this paper we onsider smo oth oriented hypersurfaes in
2
step nilp otent Lie
groups with a left invariant metri and derive an expression for the Laplaian
of the Gauss map for suh hypersurfaes in the general ase and in some
partiular ases. In the ase of CMChypersurfae in the
2m+ 1
dimensional
Heisenberg group we also derive neessary and suient onditions for the
Gauss map to be harmoni and prove that for
m= 1
all CMCsurfaes with
the harmoni Gauss map are ylinders.
2000 Mathematis Subjet Classiation.
Primary 53C40. Seondary 53C42,
53C43, 22E25.
Keywords.
2step nilpotent Lie group, Heisenberg group, left invariant metri,
Gauss map, harmoni map, minimal submanifold, onstant mean urvature.
It is proved in [12℄ that the Gauss map of a smooth
n
dimensional oriented
hypersurfae in
Rn+1
is harmoni if and only if the hypersurfae is of a
onstant mean urvature (CMC). The same is proved for the ases of
S3
,
whih is a Lie group and thus has a natural denition of the Gauss map [9℄,
and, in dierent settings, of
H3
[10℄. A generalization of this proposition
to the ase of Lie groups with a biinvariant metri (this lass of Lie groups
inludes, for example, abelian groups
Rn+1
and
S3∼
=SU (2)
) is proved in [6℄.
In this paper we use methods of [6℄ for an investigation of the Gauss map of a
hypersurfae in some
2
step nilpotent Lie group with a left invariant metri.
The theory of suh groups is highly developed (see, for example, [3℄ and [4℄).
The paper is organized as follows. After some preliminary information
(setion 1), in setion 2 we obtain an expression for the Laplaian of the
Gauss map of a hypersurfae in a
2
step nilpotent Lie group (Theorem 1).
1
Using this expression we prove some fats onerning relations between har
moni properties of the Gauss map and the mean urvature of the hyper
surfae (see setion 3), in partiular, a suient ondition for the stability
of CMChypersurfaes (Proposition 6). In setion 4 we onsider the ases of
Heisenberg type groups and Heisenberg groups. We show the harmoniity of
the Gauss map of a hypersurfae in suh groups is, in general, not equiva
lent to the onstany of the mean urvature. Also we obtain neessary and
suient onditions for this equivalene in the partiular ase of Heisenberg
groups (Prop osition 7).
The author is grateful to prof. L. A. Masal'tsev for onstant attention to
this work. The author would also thank prof. Yu. A. Nikolayevsky and prof.
A. L. Yampolsky for many useful advies onerning language and style.
1 Preliminaries
Let us reall some basi denitions and fats about the stability of on
stant mean urvature hypersurfaes in Riemannian manifolds. Suppose
M
is
a smooth
n
dimensional manifold immersed in a smo oth
n+ 1
dimensional
Riemannian manifold as a CMChypersurfae. Denote by
η
a unit normal
vetor eld of
M
. Let
D⊂M
be a ompat domain. The
index form
of
D
is a quadrati form
Q(·,·)
on
C∞(D)
dened by the equation
Q(w, w) = −Z
D
wLw dVM,
(1)
where
dVM
is the volume form of the indued metri on
M
,
L
is the
Jaobi
operator
∆M+Ric(η, η) + kBk2
,
Ric(·,·)
is the Rii tensor of the ambient
manifold,
kBk
is the norm of the seond fundamental form of the immersion,
and
∆M
is the Laplaian of the indued metri (see, for example, [2℄).
Let
M
be a minimal hypersurfae (a hypersurfae of a nonzero onstant
mean urvature, respetively). A ompat domain
D⊂M
is alled
stable
if
Q(w, w)>0
for every funtion
w∈C∞(D)
vanishing on
∂D
(for every
w∈C∞(D)
vanishing on
∂D
and with
R
D
w dVM= 0
, respetively). The
hypersurfae
M
is
stable
if every ompat domain
D⊂M
is stable, and is
unstable
otherwise (see, for example, [1℄). It is proved in [7, Theorem 1℄ that
if the
Jaobi equation
Lw = 0
admits a solution
w
stritly positive on
M
,
then
M
is stable.
2
Let
(M, g)
be a smo oth Riemannian manifold. Denote by
∆M
the Lapla
ian of
g
. For eah
φ∈C∞(M, S n)
denote by
∆Mφ
the vetor
(∆Mφ1,...,
∆Mφn+1)
, where
(φ1,...,φn+1)
is the oordinate funtions of
φ
for the stan
dard embedding of a unit sphere
Sn֒→Rn+1
. It is well known that the
harmoniity of
φ
is equivalent to the equation
∆Mφ= 2e(φ)φ
, where
e(φ)
is
the energy density funtion of
φ
(see [14, p. 140, Corollary (2.24)℄).
Suppose
M
is an oriented hypersurfae in a
n+ 1
dimensional Lie group
N
with a left invariant Riemannian metri. Fix the unit normal vetor eld
η
of
M
with resp et to the orientation. Let
p
be a point of
M
. Denote
by
La
the left translation by
a∈N
, and let
dLa
be the dierential of this
map. We an onsider
p
as an element of
N
if we identify this point with
its image under the immersion. Let
G
be the map of
M
to
Sn⊂ N
suh
that
G(p) = (dLp)−1(η(p))
for all
p∈N
, where
N
is the Lie algebra of
N
.
We all
G
the
Gauss map
of
M
. It is proved in [6℄ that if a metri of
N
is
biinvariant (see [11℄ on a struture of suh Lie groups), then the Gauss map
is harmoni if and only if the mean urvature of
M
is onstant.
Now we onsider the ase of nilpotent Lie groups. Let
N
be a nite
dimensional Lie algebra over
R
with a Lie braket
[·,·]
. The lower entral
series of
N
is dened indutively by
N1=N
,
Nk+1 =Nk,N
for all
positive integers
k
. The Lie algebra
N
is alled
k
step nilpotent
if
Nk6= 0
and
Nk+1 = 0
. A Lie group
N
is alled
k
step nilpotent
if its Lie algebra
N
is
k
step nilp otent.
In the sequel, we onsider a
2
step nilpotent onneted and simply on
neted Lie group
N
and its Lie algebra
N
. Let
Z
be the enter of
N
. Sine
N
is
2
step nilp otent,
06= [N,N]⊂ Z
. Supp ose that
N
is endowed with a
salar produt
h·,·i
. This salar produt indues a left invariant Riemannian
metri on
N
, whih we also denote by
h·,·i
. Let
V
be an orthogonal om
plement to
Z
in
N
with respet to
h·,·i
. Then
[V,V] = [N,N]⊂ Z
. For
eah
Z∈ Z
a linear op erator
J(Z): V → V
is well dened by
hJ(Z)X, Y i=
h[X, Y ], Zi
, where
X, Y ∈ V
are arbitrary vetors.
An important lass of
2
step nilp otent groups onsists of soalled
2m+1

dimensional
Heisenberg groups
, whih appear in some problems of quantum
and Hamiltonian mehanis [8℄. The Lie algebra of a Heisenberg group has
a basis
K1,...,Km
,
L1,...,Lm
,
Z
and the struture relations
[Ki, Lj] = δijZ, [Ki, Kj] = [Li, Lj] = [Ki, Z] = [Li, Z] = 0,16i, j 6m,
where
δij
is the Kroneker symbol. We introdue a salar produt suh that
this basis is orthonormal. The threedimensional Heisenberg group with a
3
left invariant Riemannian metri is often denoted by
Nil
and is a three
dimensional Thurston geometry. A Lie algebra
N
is of
Heisenberg type
if
J(Z)2=−hZ, ZiId V
, for every
Z∈ Z
[4℄. Its Lie group
N
is alled a
Lie group of
Heisenberg type
. This lass of groups ontains, for example,
Heisenberg groups and quaternioni Heisenberg groups [3, p. 617℄. A general
approah to the struture of
2
step nilpotent Lie algebras was developed in
the pap er [5℄.
The Riemannian onnetion asso iated with
h·,·i
is dened on left invari
ant elds by (see [3℄)
∇XY=1
2[X, Y ], X, Y ∈ V;
∇XZ=∇ZX=−1
2J(Z)X, X ∈ V, Z ∈ Z;
∇ZZ∗= 0, Z, Z∗∈ Z.
(2)
From this one an obtain for the urvature tensor
R(X, Y )X∗=1
2J([X, Y ])X∗
−1
4J([Y, X ∗])X
+1
4J([X, X∗])Y,
X, X∗, Y ∈ V;
R(X, Z)Y
R(X, Y )Z
=
=−1
4[X, J (Z)Y],
−1
4[X, J (Z)Y]
+1
4[Y, J (Z)X],
X, Y ∈ V, Z ∈ Z;
R(X, Z)Z∗
R(Z, Z∗)X
=
=−1
4J(Z)J(Z∗)X,
−1
4J(Z∗)J(Z)X
+1
4J(Z)J(Z∗)X,
X∈ V, Z, Z∗∈ Z;
R(Z, Z∗)Z∗∗ = 0, Z, Z∗, Z∗∗ ∈ Z.
(3)
And the Rii tensor is dened by
Ric(X, Y ) = 1
2
l
P
k=1hJ(Zk)2X, Y i, X, Y ∈ V;
Ric(X, Z) = 0, X ∈ V, Z ∈ Z;
Ric(Z, Z∗) = −1
4Tr(J(Z)J(Z∗)), Z, Z∗∈ Z.
(4)
Here
dim Z=l
, and
Z1,...,Zl
is an orthonormal basis for
Z
.
2 The Laplaian of the Gauss map
Suppose
dim N= dim N=n+ 1
,
dim Z=n−q+ 1
, where
n
and
q
are
positive integers,
q6n
.
4
Let
M
be a smooth oriented manifold,
dim M=n
. Suppose
M→
N
is an immersion of this manifold in
N
as a hypersurfae, and
η
is the
unit normal vetor eld of
M
in
N
. For eah point
p
of
M
, supp ose that
η(p) = Yn+1 =Xn+1 +Zn+1
, where
Xn+1 ∈ V
,
Zn+1 ∈ Z
. Throughout this
paper, we denote by
Xi, Yi, Zi
elements of
TpN
as well as the orresponding
left invariant vetor elds, whih are elements of
N
. Cho ose an orthonormal
frame
{Y1,...,Yn}
in the vetor spae
TpM⊂TpN
suh that for
16i6q−1
Yi=Xi
,
Yq=Xq−Zq
, and for
q+ 1 6i6n Yi=Zi
, where
X1,...,Xq
are elements of
V
,
Zq,...,Zn
belong to
Z
,
Xn+1 =λXq
,
Zn+1 =µZq
, where
λ>0
and
µ>0
,
Xq=Zn+1
,
Zq=Xn+1
. Let
E1,...En
be an
orthonormal frame dened on some neighborhoo d
U
of
p
suh that
Ei(p) = Yi
and
(∇EiEj)T(p) = 0
, for all
i, j = 1,...n
(suh a frame is alled geodesi
at
p
). Here we denote by
(·)T
the pro jetion to
TpM
.
We an rewrite (4) in the following form
Ric(X, Y ) = 1
2
n+1
P
k=qhJ(Zk)2X, Y i, X, Y ∈ V;
Ric(X, Z) = 0, X ∈ V, Z ∈ Z;
Ric(Z, Z∗) = −1
4P
16k6q, k=n+1hJ(Z)J(Z∗)Xk, Xki, Z, Z∗∈ Z.
(5)
In partiular, for all
X, Y ∈ V
P
16i6q, i=n+1hJ([X, Xi])Xi, Y i
=P
16i6q, i=n+1
n+1
P
j=qh[X, Xi], Zjih[Xi, Y ], Zji
=−
n+1
P
j=qP
16i6q, i=n+1hJ(Zj)X, XiihJ(Zj)Y , Xii
=
n+1
P
j=qhJ(Zj)2X, Y i= 2 Ric(X, Y ).
(6)
For
16i, j 6n
, denote by
bij =h∇EiEj, ηi
the oeients of the seond
fundamental form of the immersion, by
kBk
the norm of this form, and by
H
the mean urvature of the immersion on
U
. Sine the frame is orthonormal
over
U
,
H=1
n
n
P
i=1
bii,
kBk2=P
16i,j6n
(bij )2.
(7)
5
Suppose that on
U
η=
n+1
X
j=1
ajYj,
where
{aj}n+1
j=1
are some funtions on
U
. It is lear that
aj(p) = δj n+1
. Then
the Gauss map
G:U→Sn⊂Rn+1
takes the form
G=
n+1
X
j=1
ajYj(e).
In partiular,
G(p) = Yn+1(e)
. Denote by
∆
the Laplaian
∆M
of the indued
metri on
M
.
Theorem 1.
Let
M
be a smooth oriented manifold immersed in a 2step
nilpotent Lie group
N
as a hypersurfae and
G
be the Gauss map of
M
.
Then, in the above notation
∆G(p) =
q
P
k=1 −Yk(nH) +
q−1
P
j=1hJ([Xk, Xj])Xj, Xn+1i
+ 4hR(Xk, Zn+1)Zn+1, Xn+1i − 2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Xki
+2
q
P
i=1
biq (p)hJ(Zq)Xi, Xki+nH(p)hJ(Zn+1)Xn+1, XkiYk(e)
+
n
P
k=q+1 −Yk(nH)Yk(e)
+ q−1
P
j=1hJ([Xn+1, Xj])Xj, Xn+1i+ 4hR(Xn+1, Zn+1)Zn+1, Xn+1i
−2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Xn+1i+ 2
q
P
i=1
biq (p)hJ(Zq)Xi, Xn+1i
−kBk2(p)−Ric(Yn+1, Yn+1 )Yn+1(e).
(8)
Here
Yk(nH)
denotes the derivative of the funtion
nH
with respet to
the vetor eld
Yk
.
6
P r o o f. Sine the frame
E1,...,En
is geo desi at
p
, the Laplaian at this
point has the form
∆G(p) =
n+1
X
j=1
n
X
i=1
EiEi(aj)Yj(e).
(9)
For
16i6n
we have on
U
∇Eiη=
n+1
X
j=1
Ei(aj)Yj+
n+1
X
j=1
aj∇EiYj,
(10)
∇Ei∇Eiη=
n+1
X
j=1
EiEi(aj)Yj+ 2
n+1
X
j=1
Ei(aj)∇EiYj+
n+1
X
j=1
aj∇Ei∇EiYj.
(11)
Considering this expression at
p
and taking its salar produt with
Yk
for
16k6n+ 1
, we get
h∇Ei∇Eiη, Yki=EiEi(ak) + 2
n+1
X
j=1
Ei(aj)h∇EiYj, Yki+h∇Ei∇EiYn+1, Yki.
Then for salar oeients in (9) we have
n
P
i=1
EiEi(ak) =
n
P
i=1h∇Ei∇Eiη, Yki
−2
n+1
P
j=1
n
P
i=1
Ei(aj)h∇EiYj, Yki −
n
P
i=1h∇Ei∇EiYn+1, Yki.
(12)
For
16k6n
, the rst expression in (7) and the denition of a seond
fundamental form imply at
p
Yk(nH) = Ek n
X
i=1 h∇EiEi, ηi!=
n
X
i=1 h∇Ek∇EiEi, ηi
=
n
X
i=1 hR(Ek, Ei)Ei, ηi+h∇Ei∇EkEi, ηi+h∇[Ek,Ei]Ei, ηi!
=
n
X
i=1 hR(Yk, Yi)Yi, Yn+1i+h∇Ei∇EkEi, ηi!.
7
The seond equality in the equation above follows from the fat that the
projetion
(∇EiEi)T= 0
at
p
and the vetor
∇Ekη
is tangent to
M
. The
fourth equality is a onsequene of
[Ek, Ei] = ([Ek, Ei])T= (∇EkEi− ∇EiEk)T= 0
at
p
. Sine
h[Ek, Ei], ηi= 0
on
U
, and
[Ek, Ei](p) = 0
, at
p
we have
0 = h∇Ei[Ek, Ei], ηi=h∇Ei∇EkEi− ∇Ek∇EiEi, ηi,
for
16i6n
, hene
Yk(nH) =
n
X
i=1 hR(Yk, Yi)Yi, Yn+1i+h∇Ei∇EiEk, ηi!.
(13)
Dierentiating
hEk, ηi= 0
two times with resp et to
Ei
(here we put
16k, i 6n
) and using
h∇EiEk,∇Eiηi(p) = 0
, we derive from (13)
n
P
i=1h∇Ei∇Eiη, Yki=−
n
P
i=1h∇Ei∇EiEk, ηi
=−Yk(nH) +
n
P
i=1hR(Yk, Yi)Yi, Yn+1i=−Yk(nH) + Ric(Yk, Yn+1).
(14)
For
16i6n
, dierentiating
hη, ηi= 1
two times with respet to
Ei
, we get
2h∇Ei∇Eiη, ηi+2h∇Eiη, ∇Eiηi= 0
. This equation and the seond expression
in (7) imply at
p
n
P
i=1h∇Ei∇Eiη, Yn+1i=−
n
P
i=1h∇Eiη, ∇Eiηi
=−
n
P
i=1
n
P
j=1h∇Eiη, Ejih∇Eiη, Eji
=−P
16i,j6nh∇EiEj, ηi2=−kBk2(p).
(15)
Consider the salar produts
h∇Eiη, Yki
at the p oint
p
, for
16i6n
,
16k6n+ 1
. As
η=Yn+1= 1
, we obtain from (10)
0 = h∇Eiη, ηi(p) = h∇Eiη, Yn+1 i=Ei(an+1) + h∇EiYn+1, Yn+1i=Ei(an+1).
For
16k6n
,
hEk, ηi= 0
imply
8
bik(p) = h∇EiEk, ηi(p) = −h∇Eiη, Eki(p) = −h∇Eiη, Yki
=−Ei(ak)− h∇EiYn+1, Yki.
Hene at
p
we have
−2
n+1
X
j=1
n
X
i=1
Ei(aj)h∇EiYj, Yki
= 2
n
X
j=1
n
X
i=1 bij (p) + h∇YiYn+1, Yji!h∇YiYj, Yki.
It follows from (2) that for
16i, j 6q
the expression
bij (p)h∇XiXj, Yki
is skewsymmetri with resp et to
i, j
; hene the sum of suh terms with
respet to
i
and
j
vanishes. Sum up other expressions using the symmetry
∇XZ=∇ZX
for all
X∈ V
,
Z∈ Z
and the symmetry of the seond
fundamental form. We obtain
−2
n+1
P
j=1
n
P
i=1
Ei(aj)h∇EiYj, Yki=−2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Yki
+2
q
P
i=1
biq(p)hJ(Zq)Xi, Yki+ 2 P
16i,j6nh∇YiYn+1, Yjih∇YiYj, Yki.
(16)
Now we an omplete the pro of of the theorem using the following tehnial
lemmas.
Lemma 2.
The last summand on the right hand side of
(16)
is equal to
2P
16i,j6nh∇YiYn+1, Yjih∇YiYj, Yki
=
2 Ric(Yk, Yn+1) + 4hR(Xk, Zn+1)Zn+1, Xn+1i,16k6q−1;
2 Ric(Xq, Xn+1) + 2 Ric(Zq, Zn+1)
+4hR(Xq, Zn+1)Zn+1, Xn+1i
−4hR(Xn+1, Zq)Zn+1, Xn+1i,
k=q;
−2 Ric(Yk, Yn+1)
+4hR(Xn+1, Zk)Zn+1, Xn+1i,q+ 1 6k6n;
2 Ric(Xn+1, Xn+1)−2 Ric(Zn+1, Zn+1)
+8hR(Xn+1, Zn+1)Zn+1, Xn+1i,k=n+ 1.
(17)
9
Lemma 3.
The last summand on the right hand side of
(12)
an be redued
to the form
−
n
P
i=1h∇Ei∇EiYn+1, Yki
=
nH(p)hJ(Zn+1)Xn+1, Xki − Ric(Yk, Yn+1),16k6q−1;
−Ric(Xq, Xn+1)−Ric(Zq, Zn+1 )
+4hR(Xn+1, Zq)Zn+1, Xn+1i,k=q;
Ric(Yk, Yn+1)−4hR(Xn+1, Zk)Zn+1, Xn+1i, q + 1 6k6n;
−Ric(Xn+1, Xn+1) + Ric(Zn+1, Zn+1)
−4hR(Xn+1, Zn+1)Zn+1, Xn+1i,k=n+ 1.
(18)
Now, if we ombine (12) with (16), (17), (18), and (6), we get (8).
P r o o f o f L e m m a 2. For
16k6q−1
from the expressions for the
Riemannian onnetion we get
n
X
i=1
n
X
j=1 h∇YiYn+1, Yjih∇YiYj, Xki=
q−1
X
i=1 h1
2[Xi, Xn+1],−Zqi
+h−1
2J(Zn+1)Xi, Xqi!h1
2J(Zq)Xi, Xki
+
q−1
X
i=1
n
X
j=q+1h1
2[Xi, Xn+1], Zjih−1
2J(Zj)Xi, Xki
+
q−1
X
j=1 h−1
2J(Zn+1)Xq+1
2J(Zq)Xn+1, Xjih1
2J(Zq)Xj, Xki
+ h1
2J(Zq)Xn+1 −1
2J(Zn+1)Xq, Xqi+h1
2[Xq, Xn+1],−Zqi!hJ(Zq)Xq, Xki
+
n
X
j=q+1h1
2[Xq, Xn+1], Zjih−1
2J(Zj)Xq, Xki
+
n
X
i=q+1
q−1
X
j=1 h−1
2J(Zi)Xn+1, Xjih−1
2J(Zi)Xj, Xki
10
+
n
X
i=q+1h−1
2J(Zi)Xn+1, Xqih−1
2J(Zi)Xq, Xki.
The skewsymmetry of
J
implies
hJ(Z)Xq, Xn+1i=−hJ(Z)Xn+1, Xqi= 0
for all
Z∈ Z
, and
[Xq, Xn+1] = 0
. Hene we an rewrite the above expression
in the form
n
X
i=1
n
X
j=1 h∇YiYn+1, Yjih∇YiYj, Xki
=1
2
n
X
j=qX
16i6q, i=n+1h−J(Zj)Xn+1, XiihJ(Zj)Xk, Xii
=−1
2
n
X
j=qhJ(Zj)Xn+1, J (Zj)Xki=1
2
n
X
j=qhJ(Zj)2Xn+1, Xki
= Ric(Xk, Xn+1) + 2hR(Xk, Zn+1)Zn+1, Xn+1i.
This implies the rst equality in (17).
For
q+ 1 6k6n
we have
n
X
i=1
n
X
j=1 h∇YiYn+1, Yjih∇YiYj, Zki=
q−1
X
i=1
q−1
X
j=1 h−1
2J(Zn+1)Xi, Xjih1
2[Xi, Xj], Zki
+1
2
q−1
X
i=1 h−1
2J(Zn+1)Xi, Xqi+h1
2[Xi, Xn+1],−Zqi!h1
2[Xi, Xq], Zki
+
q−1
X
j=1 h−1
2J(Zn+1)Xq+1
2J(Zq)Xn+1, Xjih1
2[Xq, Xj], Zki
=1
4
q
X
i=1
q
X
j=1 h−J(Zn+1)Xi, Xjih[Xi, Xj], Zki
=1
4X
16i6q, i=n+1 X
16j6q, j =n+1h−J(Zn+1)Xi, XjihJ(Zk)Xi, Xji
−1
2X
16i6q, i=n+1h−J(Zn+1)Xn+1, XiihJ(Zk)Xn+1 , Xii
11
=1
4X
16i6q, i=n+1h−J(Zn+1)Xi, J(Zk)Xii − 1
2h−J(Zn+1)Xn+1, J(Zk)Xn+1i
=1
4X
16i6q, i=n+1hJ(Zk)J(Zn+1)Xi, Xii − 1
2hJ(Zk)J(Zn+1)Xn+1, Xn+1i
=−Ric(Zk, Zn+1) + 2hR(Xn+1, Zk)Zn+1, Xn+1i.
This ompletes the proof of (17) and of the lemma, as
Yq=Xq−Zq
, and
Yn+1 =Xn+1 +Zn+1
.
P r o o f o f L e m m a 3. Let on
U
Ei=
n+1
X
j=1
cij Yj,
(19)
where
cij
,
16i6n
,
16j6n+ 1
are salar funtions on
U
. Note that
Ei(p) = Yi
, so
cij (p) = δij
. Using (19), we get
∇EiYn+1 =
n+1
P
j=1
cij∇YjYn+1 =1
2
q
P
j=1
cij [Xj, Xn+1]−J(Zn+1)Xj
+1
2ciqJ(Zq)Xn+1 −1
2
n
P
j=q+1
cij J(Zj)Xn+1 −ci n+1J(Zn+1)Xn+1.
(20)
Also, for
16k6n
at
p
we have
∇EkEi=
n+1
X
j=1 Yk(cij)Yj+cij (p)∇YkYj!=
n+1
X
j=1
Yk(cij)Yj+∇YkYi.
(21)
In partiular, at
p
bki(p) = h∇EkEi, ηi(p) = Yk(ci n+1) + h∇YkYi, Yn+1i.
(22)
Considering (21) for
k=i
, pro jeting both sides of it to
TpM
, and using the
properties of the geo desi frame, we get
0 =
n
X
j=1
Yi(cij)Yj+ (∇YiYi)T.
12
For
16i6q−1
and
q+ 1 6i6n∇YiYi= 0
, and
∇YqYq=J(Zq)Xq=
∇YqYqT
, sine
hJ(Zq)Xq, Xn+1 +Zn+1i= 0
. Then, for
16j6q
we obtain
Yi(cij ) =
0,16i6q−1;
−hJ(Zq)Xq, Yji, i =q;
0, q + 1 6i6n.
We an dedue from (22) and the above onsiderations that for
16i6n
bii(p) = Yi(ci n+1)
. Dierentiate (20) with respet to
Ei
at
p
. For
16i6q−1
we get
∇Ei∇EiYn+1 =−Yi(ci n+1)J(Zn+1)Xn+1 +1
2∇Xi [Xi, Xn+1]−J(Zn+1)Xi!
=−bii(p)J(Zn+1)Xn+1 −1
4J([Xi, Xn+1])Xi−1
4[Xi, J(Zn+1)Xi].
For
i=q
we have
∇Eq∇EqYn+1 =−Yq(ci n+1)J(Zn+1)Xn+1 +
n
X
j=1
Yq(cqj )∇YjYn+1
+1
2∇Yq [Xq, Xn+1]−J(Zn+1)Xq+J(Zq)Xn+1!=−bqq (p)J(Zn+1 )Xn+1
−1
2
q−1
X
j=1 hJ(Zq)Xq, Xji [Xj, Xn+1]−J(Zn+1)Xj!
−1
4[Xq, J(Zn+1)Xq] + 1
4[Xq, J(Zq)Xn+1]−1
4J(Zq)J(Zn+1)Xq+1
4J(Zq)2Xn+1.
For
q+ 1 6i6n
we obtain
∇Ei∇EiYn+1 =−Yi(ci n+1)J(Zn+1)Xn+1 −1
2∇Zi(J(Zi)Xn+1)
=−bii(p)J(Zn+1)Xn+1 +1
4J(Zi)2Xn+1.
Summing up these expressions, we get for
16k6q−1
−
n
X
i=1 h∇Ei∇EiYn+1, Xki=nH(p)hJ(Zn+1)Xn+1, Xki
13
+1
4
q−1
X
i=1 hJ([Xi, Xn+1])Xi, Xki+1
2
q−1
X
j=1 hJ(Zq)Xq, Xjih−J(Zn+1)Xj, Xki
+1
4hJ(Zq)J(Zn+1)Xq, Xki − 1
4hJ(Zq)2Xn+1, Xki − 1
4
n
X
i=q+1hJ(Zi)2Xn+1, Xki
=nH(p)hJ(Zn+1)Xn+1, Xki − 1
2X
16i6q, i=n+1hJ([Xn+1, Xi])Xi, Xki
=nH(p)hJ(Zn+1)Xn+1, Xki − Ric(Xk, Xn+1).
Here we use the equation
J(Zq)J(Zn+1)Xq=J(Zn+1)2Xn+1
, whih follows
from the onstrution of the frame. Thus we obtain the rst expression in
(18).
For
q+ 1 6k6n
we have
−
n
X
i=1 h∇Ei∇EiYn+1, Zki=1
4
q−1
X
i=1 h[Xi, J(Zn+1)Xi], Zki
+1
2
q−1
X
j=1 hJ(Zq)Xq, Xjih[Xj, Xn+1], Zki
+1
4h[Xq, J(Zn+1)Xq], Zki − 1
4h[Xq, J(Zq)Xn+1], Zki
=−1
4X
16i6q, i=n+1hJ(Zk)J(Zn+1)Xi, Xii+hJ(Zk)J(Zn+1)Xn+1, Xn+1i
= Ric(Zk, Zn+1)−4hR(Xn+1 , Zk)Zn+1, Xn+1 i.
In the above alulation we used the fat that
J(Zq)Xq=J(Zn+1)Xn+1
and
[Xq, J(Zq)Xn+1] = [Xn+1 , J(Zn+1)Xn+1]
. As
Yq=Xq−Zq
and
Yn+1 =
Xn+1 +Zn+1
, we get the last three equalities in (18).
3 Mean urvature and harmoniity
Consider the tangent bundle
T N
and the distribution in
T N
formed
by left invariant vetor elds from
Z
. Sine
Z
is an abelian ideal, we an
integrate this distribution and obtain a foliation. Denote this foliation by
FZ
. Let G b e harmoni. Sine by (8), in this ase
Yk(nH) = 0
for all
q+ 1 6k6n
, we have
14
Corollary 4.
If the Gauss map of
M
is harmoni, then for eah leaf
M′
of
FZ
the mean urvature of the immersion is onstant on
M∩M′
.
Now we obtain some analogues of the results for Lie groups with bi
invariant metris that were stated in [6℄.
Let
ν
be a vetor eld on
M
dened by
ν(p) = Yq
, for
p∈M
. In other
words, we obtain
ν(p)
rotating the unit normal vetor
η(p)
by the angle
π
2
in
the 2plane ontaining
η(p)
and orthogonal to both
dLp(V)
and
dLp(Z)
.
Proposition 5.
Let
M
be a ompat smooth oriented hypersurfae in a
2
step
nilpotent Lie group
N
. Assume that
(i). the mean urvature of
M
is onstant on the integral urves of
ν
;
(ii). the Gauss map of
M
is harmoni;
(iii).
kBk2+ Ric(η, η)>0
on
M
and
kBk2+ Ric(η, η)>0
in some point of
M
;
(iv). the set of points
p∈M
suh that
η(p)/∈dLp(V)
is dense in
M
.
Then
G(M)
is ontained in a losed hemisphere of
Sn
if and only if
G(M)
is ontained in a great sphere of
Sn
.
P r o o f. One of the impliations in the proposition is obvious. Suppose that
some losed hemisphere of
Sn
ontains
G(M)
, i.e., there exists a unit vetor
v∈Rn+1
suh that for all
p∈MhG(p), vi
is nonp ositive. Consider a smo oth
funtion
f=hG, vi
on
M
. The oeient of
Yq(e)
in (8) vanishes. For all
points from some dense set of
M
we have
Xq6= 0
and thus
Xn+1 =Xn+1
XqXq
.
This, together with
Yq(nH) = 0
, implies that the oeient of
Yn+1(e)
is
equal to
−kBk2−Ric(η, η)
on the dense subset of
M
and hene on the whole
M
beause both the o eient and
−kBk2−Ric(η, η)
are ontinuous. Taking
the salar produt of (8) with
v
, we obtain
∆f=−kBk2+ Ric(η, η)f>0.
Then
f
is a subharmoni funtion on the ompat manifold
M
. Thus
f
is
onstant, and
kBk2+ Ric(η, η)f=−∆f= 0
. From the hypothesis, this
implies
f= 0
, hene
G(M)
is ontained in the equator
v⊥
. This ompletes
the pro of.
15
Proposition 6.
Suppose that a smooth oriented hypersurfae
M
in a
2

step nilpotent Lie group
N
is CMC, its Gauss map is harmoni, for all
p
from some dense set of
M
the normal vetor
η(p)/∈dLp(V)
, and
G(M)
is
ontained in an open hemisphere of
Sn
. Then
M
is stable.
P r o o f. From the hypothesis, there exists
v∈Rn+1
suh that for all
p∈M
hG(p), vi>0
. As in the proof of Proposition 5, onsider a salar funtion
w(p) = hG(p), vi
on
M
. This funtion is smooth and positive. As above, (8)
implies the Jaobi equation
(∆ + kBk2+ Ric(η, η)) w= 0
. Now [7, Theorem
1℄ implies the stability of
M
.
4 Groups of Heisenberg type
Let
N
be a group of Heisenberg type. Then from (5), for all
X, Y ∈ V
,
Ric(X, Y ) = 1
2
n+1
X
k=qhJ(Zk)2X, Y i=−1
2(n+ 1 −q)hX, Y i.
Also, we an rewrite the oeients in (8) for
16k6q
and for
k=n+ 1
in the form
q−1
X
j=1 hJ([Xk, Xj])Xj, Xn+1i+ 4hR(Xk, Zn+1)Zn+1, Xn+1i
=
0,16k6q−1;
Zn+1Xn+1q−n−1 + Zn+1 2, k =q;
Xn+12q−n−1 + Zn+12, k =n+ 1.
Moreover,
Ric(Zn+1, Zn+1) = −1
4Tr J(Zn+1)2=q
4Zn+12,
and thus
Ric(Yn+1, Yn+1) = q
4Zn+12−1
2(n+ 1 −q)Xn+12.
16
Equation (8) now takes the form
∆G(p) =
q−1
P
k=1 −Yk(nH)−2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Xki
+ 2
q
P
i=1
biq (p)hJ(Zq)Xi, Xki+nH(p)hJ(Zn+1)Xn+1, XkiYk(e)
+−Yq(nH) + Zn+1Xn+1q−n−1 + Zn+12
−2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Xqi
+2
q
P
i=1
biq (p)hJ(Zq)Xi, Xqi+nH(p)hJ(Zn+1)Xn+1, XqiYq(e)
+
n
P
k=q+1 −Yk(nH)Yk(e)
+ −2
q
P
i=1
n
P
j=q+1
bij (p)hJ(Zj)Xi, Xn+1i
+ 2
q
P
i=1
biq (p)hJ(Zq)Xi, Xn+1i − kBk2(p)−q
4Zn+12
+Xn+121
2(q−n−1) + Zn+12Yn+1(e).
(23)
Consider the ase
n=q
, i.e.,
dim Z= 1
. It is easy to see that
n
is then
even,
n= 2m
, where
m
is a positive integer, and
N
is isomorphi to the
2m+ 1
dimensional Heisenberg group (reall that
N
is onneted and simply
onneted).
In this ase at
p
we an hoose
X1,...,X2m+1
so that
J(Z)Xi=Xm+i,16i6m−1;
J(Z)Xm=X2m
X2m=X2m
Z2m+1
if
Z2m+1 6= 0;
or
X2m+1
X2m+1
if
X2m+1 6= 0;
J(Z)Xm+i=−Xi,16i6m−1;
J(Z)X2m=−X2mXm=−Z2m+1Xm;
J(Z)X2m+1 =−X2m+1Xm.
17
Choose
Z2m=X2m+1Z
and
Z2m+1 =Z2m+1Z
. Then (23) has the form
∆G(p) = −
m−1
P
k=1 Yk(2mH) + 2b2m m+k(p)X2m+1Yk(e)
−Ym(2mH) + 2mH(p)X2m+1
+ 2b2m2m(p)X2m+1Z2m+1Ym(e)
−
m−1
P
k=1 Ym+k(2mH)−2b2m k(p)X2m+1 Yk(e)
−Y2m(2mH) + X2m+13Z2m+1
−2b2m m(p)X2m+1Z2m+1Y2m(e)
−kBk2(p) + m
2Z2m+12−1
2X2m+12
+X2m+14−2b2m m (p)X2m+12Y2m+1(e).
(24)
Consider an example of the threedimensional Heisenberg group
Nil
. In
the spae
R3
with Cartesian oordinates
(x, y, z)
, dene vetor elds
X=∂
∂x , Y =∂
∂y +x∂
∂z , Z =∂
∂z .
Then
Span(X, Y, Z)
is a Lie algebra (with the only nonzero braket
[X, Y ] =
Z
), whih is the Lie algebra of
Nil
. Introdue a salar pro dut in suh a way
that the vetors
X, Y
and
Z
are orthonormal. Consider the following unit
vetor eld:
η=xY +Z
√1 + x2,
and vetor elds
F1=X, F2=Y−xZ
√1 + x2,
whih are orthogonal to
η
. In the notation of setion 2, in eah
p F1=X1
,
F2=X2−Z2
,
η=X3+Z3
. By diret omputation of ovariant derivatives
it an be shown that the distribution spanned by
F1
and
F2
is integrable and
form the tangent bundle of some twodimensional foliation
F
in
Nil
. From
the omputation of the seond fundamental form we obtain
kBk2=(x2−1)2
2(1+x2)2
,
18
and
H= 0
. Thus the leaves of this foliation are minimal surfaes. The
Laplaian on
G=η
is
∆G=F1F1+F2F2−(∇F1F1)T−(∇F2F2)TG
=−x
(1 + x2)2F2−1
(1 + x2)2η.
We obtain the same result onsidering (24) at some
p
. In fat,
2b22 X3Z3= 0;
X33Z3 − 2b21 X3Z3=x
(1 + x2)2;
kBk2+1
2Z32−1
2X32+X34−2b21 X32=1
(1 + x2)2.
In partiular, foliation
F
gives an example of a CMCsurfae in
Nil
suh
that its Gauss map is not harmoni.
Proposition 7.
Suppose that
M
is a smooth oriented
2m
dimensional man
ifold immersed in the
2m+ 1
dimensional Heisenberg group. If any two of
the following three laims are true, then the third one is also true.
(i).
M
is CMC;
(ii). the Gauss map of
M
is harmoni;
(iii). at every point of
M
, the fol lowing holds:
b2m k = 0,16k6m−1, m + 1 6k62m−1;
Z2m+1X2m+12−2b2m m= 0;
Z2m+1(b1 1 +···+b2m−1 2m−1+ 3b2m2m) = 0.
(25)
Here
bij
,
16i, j 62m
are the oeients of the seond fundamental
form of
M
in the basis hosen as above.
19
P r o o f. If
(iii)
is true, then the equivaleny of
(i)
and
(ii)
immediately
follows from 24. Supp ose
(i)
and
(ii)
are true. Let
A
be a set of suh p oints
of
M
that
X2m+1 6= 0
. At the points of
A
24 implies the expressions in
(25) . Sine the distribution orthogonal to
Z
is nonintegrable,
A
is dense in
M
. Now the ontinuity of the left hand sides of the equations (25) implies
(iii)
.
In the ase
m= 1
the next theorem shows that the restritions for
M
arising from (25) are rather strit.
Theorem 8.
Let
M
be a smooth oriented CMCsurfae in the Heisenberg
group
Nil
whose Gauss map is harmoni. Then
M
is a ylinder, that is,
its position vetor in the oordinates
x
,
y
,
z
has the form
r(s, t) = (f1(s), f2(s), t),
(26)
where
f1
and
f2
are some smooth funtions.
P r o o f. For eah
p∈M
denote
a(p) = X3
,
b(p) = Z3
. Then
a
and
b
are
smooth salar funtions on
M
, and
a2+b2= 1
. Consider an arbitrary p oint
p
of
M
. Choose
X1
as above and put
X2=J(Z)X1
. Denote by
T1
and
T2
the vetor elds that at eah
p∈M
are equal to
X1
and
X2
respetively.
Consider unit tangent vetor elds
F1
and
F2
, and a unit normal vetor eld
η
of
M
of the form
F1=T1, F2=bT2−aZ, η =aT2+bZ.
Denote by
κ1
and
κ2
the geodesi urvatures of the integral urves of
F1
and
F2
respetively. In other words,
∇F1F1=κ1F2,∇F1F2=−κ1F1,∇F2F1=−κ2F2,∇F2F2=κ2F1,
(27)
where
∇
is the Riemannian onnetion on
M
indued by the immersion. The
Gaussian urvature of the surfae is
K=F1(κ2) + F2(κ1)−(κ1)2−(κ2)2.
(28)
Assume that for some
p∈M a(p)6= 0
and
b(p)6= 0
. Then
ab 6= 0
on
some neighborhood
U
of
p
. Then (25) implies that on
U
the matrix of the
seond fundamental form of
M
is
3H1
2a2
1
2a2−H.
(29)
20
In partiular, the extrinsi urvature
Kext
of the surfae is
−3H2−1
4a4
.
Denote by
B
the seond fundamental form of the immersion. Then the
Codazzi equations for
M
are
(∇F1B) (F2, F1)−(∇F2B) (F1, F1) = hR(F1, F2)F1, ηi=ab;
(∇F2B) (F1, F2)−(∇F1B) (F2, F2) = hR(F2, F1)F2, ηi= 0.
Computing the ovariant derivatives of the seond fundamental form, we
obtain for
UaF1(a) + 4Hκ1−a2κ2−ab = 0,
aF2(a)−4Hκ2−a2κ1= 0.
(30)
The Gauss equation has the form
K=Kext +hR(F1, F2)F2, F1i=−3H2−1
4a4−3
4b2+1
4a2.
From (28) we obtain
F1(κ2) + F2(κ1)−(κ1)2−(κ2)2=−3H2−1
4a4−3
4b2+1
4a2.
(31)
Using (30) and the form of
F2
and
η
, we an derive
h∇F1F2, ηi=−hF2,∇F1ηi=−hF2,∇F1a
bF2+a2
b+bZi
=−F1a
bhF2, F2i − F11
bhF2, Zi − a
bhF2,∇F1F2i − 1
bhF2,∇F1Zi
=−F1a
b+aF11
b−1
bhF2,−1
2T2i=−1
b−4Hκ1
a+aκ2+b+1
2
=−1
2+4Hκ1
ab −a
bκ2;
h∇F2F1, ηi=−hF1,∇F2ηi=−hF1,∇F2a
bF2+a2
b+bZi
=−a
bhF1,∇F2F2i − 1
bhF1,∇F2Zi=−a
bκ2−1
bhT1,1
2bT1i=−a
bκ2−1
2.
21
In the above equations we used the fat that
Z
is left invariant and the
expressions (2) for the ovariant derivative. Sine
ab 6= 0
, the integrability
ondition
h[F1, F2], ηi= 0
takes the form
Hκ1= 0
. Besides, (29) imply
3H=b11 =h∇F1F1, ηi=−hF1,∇F1ηi
=−hF1,∇F1a
bF2+a2
b+bZi=−a
bhF1,∇F1F2i−1
bhF1,∇F1Zi=a
bκ1.
Thus
H=κ1= 0
. In partiular,
∇F1F1= 0
, hene
T1=F1
is a geodesi
vetor eld in the ambient manifold. Note that
T1
belongs to the distri
bution that spans the left invariant vetor elds of
V
. Considering the set
of geodesis in
Nil
(see [3, proposition (3.1), proposition (3.5)℄), we obtain
that
T1=cX +dY
, where
c, d ∈R
are some onstants, i.e.,
T1=X1
and
T2=X2
are left invariant. Note that the seond equation of (30) implies
F2(a) = F2(b) = 0
. Thus we obtain
∇F2F2=∇F2(bX2−aZ) = b∇bX2−aZ X2−a∇bX2−aZZ=−abX1.
Therefore
κ2=−ab
. It follows from this equation, from the omputations
above in this proof, and from (29) that
1
2a2=b12 =h∇F1F2, ηi=−a
bκ2−1
2=a2−1
2,
and
a2=b2=1
2
. But then
a=b=√2
2
, and the rst equation in (30) implies
aκ2+b= 0
, whih leads to a ontradition.
Thus
ab = 0
at eah point of
M
. Sine
a2+b2= 1
and
a
,
b
are ontinuous,
a= 1
or
b= 1
identially. The latter ase is impossible beause
Z⊥
is not
integrable; then the normal vetor of
M
is orthogonal to
Z
, and
F2=−Z
.
Therefore
M
is invariant under the ation of
Z
by left translations, and
M
is formed by integral urves of
Z
, whih are geo desis
(0,0, t)
. Then
M
has
the form (26).
Note that similar result for another denition of the Gauss map was
obtained in [13℄. Also, in [13℄ the equations of the CMCsurfaes of the form
(26) were obtained. Proposition 7 then implies that the Gauss maps of all
these surfaes are harmoni.
22
Referenes
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J.L. Barbosa, M.P. do Carmo, J. Eshenburg,
Stability of hypersur
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[2℄
T.H. Colding , W.P. Minikozzi II,
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[3℄
P.B. Eberlein,
Geometry of
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G.B.Fol land,
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[9℄
L.A. Masal'tsev,
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L.A. Masal'tsev,
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[11℄
J. Milnor,
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23
[12℄
E.A. Ruh, J. Vilms,
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[13℄
A. Sanini,
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H. Urakawa,
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24