The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
We consider smooth oriented hypersurfaces in 2-step nilpotent Lie groups with a left invariant metric. We derive an expression for the Laplacian of the Gauss map for such hypersurfaces in the general case and in some particular cases. In the case of CMC-hypersurface in the 2m+1-dimensional Heisenber...
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irk-123456789-1065912016-10-01T03:02:21Z The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups Petrov, Ye.V. We consider smooth oriented hypersurfaces in 2-step nilpotent Lie groups with a left invariant metric. We derive an expression for the Laplacian of the Gauss map for such hypersurfaces in the general case and in some particular cases. In the case of CMC-hypersurface in the 2m+1-dimensional Heisenberg group we also give necessary and su cient conditions for the Gauss map to be harmonic and prove that for m = 1 all CMC-surfaces with the harmonic Gauss map are cylinders . 2006 Article The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups / Ye.V. Petrov // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 2. — С. 186-206. — Бібліогр.: 14 назв. — англ. 1812-9471 http://dspace.nbuv.gov.ua/handle/123456789/106591 en Журнал математической физики, анализа, геометрии Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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We consider smooth oriented hypersurfaces in 2-step nilpotent Lie groups with a left invariant metric. We derive an expression for the Laplacian of the Gauss map for such hypersurfaces in the general case and in some particular cases. In the case of CMC-hypersurface in the 2m+1-dimensional Heisenberg group we also give necessary and su cient conditions for the Gauss map to be harmonic and prove that for m = 1 all CMC-surfaces with the harmonic Gauss map are cylinders . |
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Article |
| author |
Petrov, Ye.V. |
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Petrov, Ye.V. The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups Журнал математической физики, анализа, геометрии |
| author_facet |
Petrov, Ye.V. |
| author_sort |
Petrov, Ye.V. |
| title |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups |
| title_short |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups |
| title_full |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups |
| title_fullStr |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups |
| title_full_unstemmed |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups |
| title_sort |
gauss map of hypersurfaces in 2-step nilpotent lie groups |
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
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2006 |
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http://dspace.nbuv.gov.ua/handle/123456789/106591 |
| citation_txt |
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups / Ye.V. Petrov // Журнал математической физики, анализа, геометрии. — 2006. — Т. 2, № 2. — С. 186-206. — Бібліогр.: 14 назв. — англ. |
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Журнал математической физики, анализа, геометрии |
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AT petrovyev thegaussmapofhypersurfacesin2stepnilpotentliegroups AT petrovyev gaussmapofhypersurfacesin2stepnilpotentliegroups |
| first_indexed |
2025-07-07T18:44:29Z |
| last_indexed |
2025-07-07T18:44:29Z |
| _version_ |
1837014840298700800 |
| fulltext |
Journal of Mathematical Physics, Analysis, Geometry
2006, vol. 2, No. 2, pp. 186�206
The Gauss Map of Hypersurfaces in 2-Step Nilpotent
Lie Groups
Ye.V. Petrov
Department of Mechanics and Mathematics, V.N. Karazin Kharkov National University
4 Svobody Sq., Kharkov, 61077, Ukraine
E-mail:petrov@univer.kharkov.ua
Received August 31, 2005
We consider smooth oriented hypersurfaces in 2-step nilpotent Lie groups
with a left invariant metric. We derive an expression for the Laplacian of the
Gauss map for such hypersurfaces in the general case and in some particular
cases. In the case of CMC-hypersurface in the 2m+1-dimensional Heisenberg
group we also give necessary and su�cient conditions for the Gauss map to
be harmonic and prove that for m = 1 all CMC-surfaces with the harmonic
Gauss map are �cylinders�.
Key words: 2-step nilpotent Lie group, Heisenberg group, left invariant
metric, Gauss map, harmonic map, minimal submanifold, constant mean
curvature.
Mathematics Subject Classi�cation 2000: 53C40 (primary); 53C42, 53C43,
22E25 (secondary).
It is proved in [12] that the Gauss map of a smooth n-dimensional oriented
hypersurface in Rn+1 is harmonic if and only if the hypersurface is of a constant
mean curvature (CMC). The same is proved for the cases of S3, which is a Lie
group and thus has a natural de�nition of the Gauss map [9], and, in di�erent
settings, of H3 [10]. A generalization of this proposition to the case of Lie groups
with a bi-invariant metric (this class of Lie groups includes, for example, Abelian
groups Rn+1 and S
3 �
= SU(2)) is proved in [6]. In this paper we use the methods
of [6] for an investigation of the Gauss map of a hypersurface in some 2-step
nilpotent Lie group with a left invariant metric. The theory of such groups is
highly developed (see, for example, [3] and [4]).
The paper is organized as follows. After some preliminary information
(Sect. 1.), in Sect. 2 we obtain an expression for the Laplacian of the Gauss map
The work is partially supported by the Foundation of Fundamental Researches of Ukraine,
grant No. 01.07/00132.
c
Ye.V. Petrov, 2006
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
of a hypersurface in a 2-step nilpotent Lie group (Th. 1). Using this expression we
prove some facts concerning relations between harmonic properties of the Gauss
map and the mean curvature of the hypersurface (see Sect. 3.), in particular, a
su�cient condition for the stability of CMC-hypersurfaces (Prop. 6). In Section 4
we consider the cases of Heisenberg type groups and Heisenberg groups. We show
that the harmonicity of the Gauss map of a hypersurface in such groups is, in
general, not equivalent to the constancy of the mean curvature. Also we obtain
necessary and su�cient conditions for this equivalence in the particular case of
Heisenberg groups (Prop. 7).
The author is grateful to Prof. L.A. Masal'tsev for his interest to this work.
The author would also thank Prof. Yu.A. Nikolayevsky and Prof. A.L. Yampolsky
for careful reading of manuscript.
1. Preliminaries
Let us recall some basic de�nitions and facts about the stability of constant
mean curvature hypersurfaces in Riemannian manifolds. Suppose M is a smooth
n-dimensional manifold immersed in a smooth n + 1-dimensional Riemannian
manifold as a CMC-hypersurface. Denote by � a unit normal vector �eld of M .
Let D �M be a compact domain. The index form of D is a quadratic form Q(�; �)
on C
1
(D) de�ned by the equation
Q(w;w) = �
Z
D
wLw dVM ; (1)
where dVM is the volume form of the induced metric on M , L is the Jacobi
operator �M +
�
Ric(�; �) + kBk2
�
, Ric(�; �) is the Ricci tensor of the ambient
manifold, kBk is the norm of the second fundamental form of the immersion, and
�M is the Laplacian of the induced metric (see, for example, [2]).
Let M be a minimal hypersurface (a hypersurface of a nonzero constant
mean curvature, correspondingly). A compact domain D � M is called sta-
ble if Q(w;w) > 0 for every function w 2 C
1
(D) vanishing on @D (for every
w 2 C
1
(D) vanishing on @D and with
R
D
w dVM = 0). The hypersurface M is
stable if every compact domain D � M is stable, and is unstable otherwise (see,
for example, [1]). It is proved in [7, Th. 1] that if the Jacobi equation Lw = 0
admits a solution w strictly positive on M , then M is stable.
Let (M; g) be a smooth Riemannian manifold. Denote by �M the Laplacian
of g. For each � 2 C
1
(M;S
n
) denote by �M� the vector (�M�1; : : : ; �M�n+1),
where (�1; : : : ; �n+1) is the coordinate functions of � for the standard embedding
of a unit sphere S
n
,! R
n+1 . It is well known that the harmonicity of � is
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 187
Ye.V. Petrov
equivalent to the equation �M� = 2e(�)�, where e(�) is the energy density
function of � (see [14, p. 140, Cor. (2.24)]).
Suppose M is an oriented hypersurface in an n+ 1-dimensional Lie group N
with a left invariant Riemannian metric. Fix the unit normal vector �eld � of
M with respect to the orientation. Let p be a point of M . Denote by La the
left translation by a 2 N , and let dLa be the di�erential of this map. We can
consider 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 such that G(p) = (dLp)
�1
(�(p))
for all p 2 N , where N is the Lie algebra of N . We call G the Gauss map ofM . It
is proved in [6] that if a metric of N is bi-invariant (see [11] on a structure of such
Lie groups), then the Gauss map is harmonic if and only if the mean curvature
of M is constant.
Now we consider the case of nilpotent Lie groups. Let N be a �nite dimen-
sional Lie algebra over R with a Lie bracket [�; �]. The lower central series of N is
de�ned inductively by N 1
= N , N k+1
=
�
N k
;N
�
for all positive integers k. The
Lie algebra N is called k-step nilpotent if N k 6= 0 and N k+1
= 0. A Lie group N
is called k-step nilpotent if its Lie algebra N is k-step nilpotent.
In the sequel, we consider a 2-step nilpotent connected and simply connected
Lie group N and its Lie algebra N . Let Z be the center of N . Since N is 2-step
nilpotent, 0 6= [N ;N ] � Z. Suppose that N is endowed with a scalar product
h�; �i. This scalar product induces a left invariant Riemannian metric on N , which
we also denote by h�; �i. Let V be an orthogonal complement to Z in N with
respect to h�; �i. Then [V;V] = [N ;N ] � Z. For each Z 2 Z a linear operator
J(Z) : V ! V is well de�ned by hJ(Z)X;Y i = h[X;Y ]; Zi, where X;Y 2 V are
arbitrary vectors.
An important class of 2-step nilpotent groups consists of the so-called 2m+1-
dimensional Heisenberg groups, which appear in some problems of quantum and
Hamiltonian mechanics [8]. The Lie algebra of a Heisenberg group has a basis
K1; : : : ;Km, L1; : : : ; Lm, Z and the structure relations
[Ki; Lj ] = ÆijZ; [Ki;Kj ] = [Li; Lj ] = [Ki; Z] = [Li; Z] = 0; 1 6 i; j 6 m;
where Æij is the Kronecker symbol. We introduce a scalar product such that
this basis is orthonormal. The three-dimensional Heisenberg group with a left
invariant Riemannian metric is often denoted by Nil and is a three-dimensional
Thurston geometry. A Lie algebraN is ofHeisenberg type if J(Z)2 = �hZ;Zi Id jV ,
for every Z 2 Z [4]. Its Lie group N is called a Lie group of Heisenberg type. This
class of groups contains, for example, Heisenberg groups and quaternionic Heisen-
berg groups [3, p. 617]. A general approach to the structure of 2-step nilpotent
Lie algebras was developed in the paper [5].
188 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
The Riemannian connection associated with h�; �i is de�ned on left invariant
�elds by (see [3])
rXY =
1
2
[X;Y ]; X; Y 2 V;
rXZ = rZX = �1
2
J(Z)X; X 2 V; Z 2 Z;
rZZ
�
= 0; Z; Z
� 2 Z:
(2)
From this one can obtain for the curvature tensor
R(X;Y )X
�
=
1
2
J([X;Y ])X
�
�1
4
J([Y;X
�
])X
+
1
4
J([X;X
�
])Y;
X;X
�
; Y 2 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 2 V; Z 2 Z;
R(X;Z)Z
�
R(Z;Z
�
)X
=
=
�1
4
J(Z)J(Z
�
)X;
�1
4
J(Z
�
)J(Z)X
+
1
4
J(Z)J(Z
�
)X;
X 2 V; Z; Z� 2 Z;
R(Z;Z
�
)Z
��
= 0; Z; Z
�
; Z
�� 2 Z:
(3)
And the Ricci tensor is de�ned by
Ric(X;Y ) =
1
2
lP
k=1
hJ(Zk)2X;Y i; X; Y 2 V;
Ric(X;Z) = 0; X 2 V; Z 2 Z;
Ric(Z;Z
�
) = �1
4
Tr(J(Z)J(Z
�
)); Z; Z
� 2 Z:
(4)
Here dimZ = l, and Z1; : : : ; Zl is an orthonormal basis for Z.
2. The Laplacian of the Gauss Map
Suppose dimN = dimN = n + 1, dimZ = n � q + 1, where n and q are
positive integers, q 6 n.
Let M be a smooth oriented manifold, dimM = n. Suppose M ! N is an
immersion of this manifold in N as a hypersurface, and � is the unit normal vector
�eld ofM in N . For each point p ofM , suppose that �(p) = Yn+1 = Xn+1+Zn+1,
where Xn+1 2 V, Zn+1 2 Z. Throughout this paper, we denote by Xi; Yi; Zi
elements of TpN as well as the corresponding left invariant vector �elds, which
are elements of N . Choose an orthonormal frame fY1; : : : ; Yng in the vector
space TpM � TpN such that for 1 6 i 6 q � 1 Yi = Xi, Yq = Xq � Zq, and for
q+1 6 i 6 n Yi = Zi, where X1; : : : ;Xq are elements of V, Zq; : : : ; Zn belong to Z,
Xn+1 = �Xq, Zn+1 = �Zq, where � > 0 and � > 0, jXqj = jZn+1j, jZqj = jXn+1j.
Let E1; : : : En be an orthonormal frame de�ned on some neighborhood U of p
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 189
Ye.V. Petrov
such that Ei(p) = Yi and (rEi
Ej)
T
(p) = 0, for all i; j = 1; : : : n (such a frame is
called geodesic at p). Here we denote by (�)T the projection to TpM .
We can rewrite (4) in the following form:
Ric(X;Y ) =
1
2
n+1P
k=q
hJ(Zk)2X;Y i; X; Y 2 V;
Ric(X;Z) = 0; X 2 V; Z 2 Z;
Ric(Z;Z
�
) = �1
4
P
16k6q; k=n+1
hJ(Z)J(Z�)Xk; Xki; Z; Z
� 2 Z:
(5)
In particular, for all X;Y 2 VP
16i6q; i=n+1
hJ([X;Xi])Xi; Y i
=
P
16i6q; i=n+1
n+1P
j=q
h[X;Xi]; Zjih[Xi; Y ]; Zji
= �
n+1P
j=q
P
16i6q; i=n+1
hJ(Zj)X;XiihJ(Zj)Y;Xii
=
n+1P
j=q
hJ(Zj)2X;Y i = 2Ric(X;Y ):
(6)
For 1 6 i; j 6 n, denote by bij = hrEi
Ej; �i the coe�cients of the second
fundamental form of the immersion, by kBk the norm of this form, and by H the
mean curvature of the immersion on U . Since the frame is orthonormal on U ,
H =
1
n
nP
i=1
bii;
kBk2 =
P
16i;j6n
(bij)
2
:
(7)
Suppose that on U
� =
n+1X
j=1
ajYj;
where fajgn+1
j=1 are some functions on U . It is clear that aj(p) = Æj n+1. Then the
Gauss map G : U ! S
n � R
n+1 takes the form
G =
n+1X
j=1
ajYj(e):
In particular, G(p) = Yn+1(e). Denote by � the Laplacian �M of the induced
metric on M .
190 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
Theorem 1. Let M be a smooth oriented manifold immersed in a 2-step nilpo-
tent Lie group N as a hypersurface and G be the Gauss map of M . Then, in the
above notation
�G(p) =
qP
k=1
�Yk(nH) +
q�1P
j=1
hJ([Xk;Xj ])Xj ;Xn+1i
+ 4hR(Xk; Zn+1)Zn+1;Xn+1i � 2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi;Xki
+2
qP
i=1
biq(p)hJ(Zq)Xi; Xki+ nH(p)hJ(Zn+1)Xn+1;Xki
�
Yk(e)
+
nP
k=q+1
�
� Yk(nH)
�
Yk(e)
+
q�1P
j=1
hJ([Xn+1;Xj ])Xj ;Xn+1i+ 4hR(Xn+1; Zn+1)Zn+1;Xn+1i
� 2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi;Xn+1i+ 2
qP
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 function nH with respect to the
vector �eld Yk.
P r o o f. Since the frame E1; : : : ; En is geodesic at p, the Laplacian at this
point has the form
�G(p) =
n+1X
j=1
nX
i=1
EiEi(aj)Yj(e): (9)
For 1 6 i 6 n we have on U
rEi
� =
n+1X
j=1
Ei(aj)Yj +
n+1X
j=1
ajrEi
Yj; (10)
rEi
rEi
� =
n+1X
j=1
EiEi(aj)Yj + 2
n+1X
j=1
Ei(aj)rEi
Yj +
n+1X
j=1
ajrEi
rEi
Yj: (11)
Considering this expression at p and taking its scalar product with Yk for 1 6
k 6 n+ 1, we get
hrEi
rEi
�; Yki = EiEi(ak) + 2
n+1X
j=1
Ei(aj)hrEi
Yj; Yki+ hrEi
rEi
Yn+1; Yki:
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 191
Ye.V. Petrov
Then for scalar coe�cients in (9) we have
nP
i=1
EiEi(ak) =
nP
i=1
hrEi
rEi
�; Yki
�2
n+1P
j=1
nP
i=1
Ei(aj)hrEi
Yj ; Yki �
nP
i=1
hrEi
rEi
Yn+1; Yki:
(12)
For 1 6 k 6 n, the �rst expression in (7) and the de�nition of a second
fundamental form imply at p
Yk(nH) = Ek
nX
i=1
hrEi
Ei; �i
!
=
nX
i=1
hrEk
rEi
Ei; �i
=
nX
i=1
hR(Ek; Ei)Ei; �i+ hrEi
rEk
Ei; �i+ hr[Ek;Ei]Ei; �i
!
=
nX
i=1
hR(Yk; Yi)Yi; Yn+1i+ hrEi
rEk
Ei; �i
!
:
The second equality in the equation above follows from the fact that the projection
(rEi
Ei)
T
= 0 at p and the vector rEk
� is tangent to M . The fourth equality is
a consequence of
[Ek; Ei] = ([Ek; Ei])
T
= (rEk
Ei �rEi
Ek)
T
= 0
at p. Since h[Ek; Ei]; �i = 0 on U , and [Ek; Ei](p) = 0, at p we have
0 = hrEi
[Ek; Ei]; �i = hrEi
rEk
Ei �rEk
rEi
Ei; �i;
for 1 6 i 6 n, hence
Yk(nH) =
nX
i=1
hR(Yk; Yi)Yi; Yn+1i+ hrEi
rEi
Ek; �i
!
: (13)
Di�erentiating hEk; �i = 0 two times with respect to Ei (here we put 1 6
k; i 6 n) and using hrEi
Ek;rEi
�i(p) = 0, we derive from (13)
nP
i=1
hrEi
rEi
�; Yki = �
nP
i=1
hrEi
rEi
Ek; �i
= �Yk(nH) +
nP
i=1
hR(Yk; Yi)Yi; Yn+1i = �Yk(nH) + Ric(Yk; Yn+1):
(14)
192 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
For 1 6 i 6 n, di�erentiating h�; �i = 1 two times with respect to Ei, we get
2hrEi
rEi
�; �i+ 2hrEi
�;rEi
�i = 0. This equation and the second expression in
(7) imply at p
nP
i=1
hrEi
rEi
�; Yn+1i = �
nP
i=1
hrEi
�;rEi
�i
= �
nP
i=1
nP
j=1
hrEi
�;EjihrEi
�;Eji
= �
P
16i;j6n
hrEi
Ej ; �i2 = �kBk2(p):
(15)
Consider the scalar products hrEi
�; Yki at the point p, for 1 6 i 6 n, 1 6 k 6
n+ 1. As j�j = jYn+1j = 1, we obtain from (10)
0 = hrEi
�; �i(p) = hrEi
�; Yn+1i = Ei(an+1) + hrEi
Yn+1; Yn+1i = Ei(an+1):
For 1 6 k 6 n, hEk; �i = 0 implies
bik(p) = hrEi
Ek; �i(p) = �hrEi
�;Eki(p) = �hrEi
�; Yki
= �Ei(ak)� hrEi
Yn+1; Yki:
Hence at p we have
�2
n+1X
j=1
nX
i=1
Ei(aj)hrEi
Yj; Yki
= 2
nX
j=1
nX
i=1
bij(p) + hrYi
Yn+1; Yji
!
hrYi
Yj ; Yki:
It follows from (2) that for 1 6 i; j 6 q the expression bij(p)hrXi
Xj ; Yki is skew-
symmetric with respect to i; j; hence the sum of such terms with respect to i and
j vanishes. Sum up other expressions using the symmetry rXZ = rZX for all
X 2 V, Z 2 Z and the symmetry of the second fundamental form. We obtain
�2
n+1P
j=1
nP
i=1
Ei(aj)hrEi
Yj; Yki = �2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi; Yki
+2
qP
i=1
biq(p)hJ(Zq)Xi; Yki+ 2
P
16i;j6n
hrYi
Yn+1; YjihrYi
Yj; Yki:
(16)
Now we can complete the proof of the theorem using the following technical
lemmas.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 193
Ye.V. Petrov
Lemma 2. The last summand on the right hand side of (16) is equal to
2
P
16i;j6n
hrYi
Yn+1; YjihrYi
Yj ; Yki
=
2
66666666664
2Ric(Yk; Yn+1) + 4hR(Xk; Zn+1)Zn+1;Xn+1i; 1 6 k 6 q � 1;
2Ric(Xq;Xn+1) + 2Ric(Zq; Zn+1)
+4hR(Xq; Zn+1)Zn+1;Xn+1i
�4hR(Xn+1; Zq)Zn+1;Xn+1i;
k = q;
�2Ric(Yk; Yn+1)
+4hR(Xn+1; Zk)Zn+1;Xn+1i;
q + 1 6 k 6 n;
2Ric(Xn+1;Xn+1)� 2Ric(Zn+1; Zn+1)
+8hR(Xn+1; Zn+1)Zn+1;Xn+1i;
k = n+ 1:
(17)
Lemma 3. The last summand on the right hand side of (12) can be reduced
to the form
�
nP
i=1
hrEi
rEi
Yn+1; Yki
=
2
6666664
nH(p)hJ(Zn+1)Xn+1;Xki �Ric(Yk; Yn+1); 1 6 k 6 q � 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 6 k 6 n;
�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 combine (12) with (16), (17), (18), and (6), we get (8).
P r o o f o f L e m m a 2. For 1 6 k 6 q� 1 from the expressions for the
Riemannian connection we get
nX
i=1
nX
j=1
hrYi
Yn+1; YjihrYi
Yj ;Xki =
q�1X
i=1
h
1
2
[Xi;Xn+1];�Zqi
+ h�
1
2
J(Zn+1)Xi;Xqi
!
h
1
2
J(Zq)Xi;Xki
+
q�1X
i=1
nX
j=q+1
h
1
2
[Xi;Xn+1]; Zjih�
1
2
J(Zj)Xi;Xki
+
q�1X
j=1
h�
1
2
J(Zn+1)Xq +
1
2
J(Zq)Xn+1; Xjih
1
2
J(Zq)Xj ;Xki
194 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
+
h
1
2
J(Zq)Xn+1 �
1
2
J(Zn+1)Xq;Xqi+ h
1
2
[Xq; Xn+1];�Zqi
!
hJ(Zq)Xq;Xki
+
nX
j=q+1
h
1
2
[Xq;Xn+1]; Zjih�
1
2
J(Zj)Xq;Xki
+
nX
i=q+1
q�1X
j=1
h�
1
2
J(Zi)Xn+1;Xjih�
1
2
J(Zi)Xj ; Xki
+
nX
i=q+1
h�
1
2
J(Zi)Xn+1;Xqih�
1
2
J(Zi)Xq;Xki:
The skew-symmetry of J implies hJ(Z)Xq; Xn+1i = �hJ(Z)Xn+1;Xqi = 0 for all
Z 2 Z, and [Xq;Xn+1] = 0. Hence we can rewrite the above expression in the
form
nX
i=1
nX
j=1
hrYi
Yn+1; YjihrYi
Yj;Xki
=
1
2
nX
j=q
X
16i6q; i=n+1
h�J(Zj)Xn+1; XiihJ(Zj)Xk;Xii
= �
1
2
nX
j=q
hJ(Zj)Xn+1; J(Zj)Xki =
1
2
nX
j=q
hJ(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 6 k 6 n, we have
nX
i=1
nX
j=1
hrYi
Yn+1; YjihrYi
Yj ; Zki =
q�1X
i=1
q�1X
j=1
h�
1
2
J(Zn+1)Xi;Xjih
1
2
[Xi;Xj ]; Zki
+
1
2
q�1X
i=1
h�
1
2
J(Zn+1)Xi;Xqi+ h
1
2
[Xi;Xn+1];�Zqi
!
h
1
2
[Xi;Xq]; Zki
+
q�1X
j=1
h�
1
2
J(Zn+1)Xq +
1
2
J(Zq)Xn+1;Xjih
1
2
[Xq; Xj ]; Zki
=
1
4
qX
i=1
qX
j=1
h�J(Zn+1)Xi;Xjih[Xi; Xj ]; Zki
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 195
Ye.V. Petrov
=
1
4
X
16i6q; i=n+1
X
16j6q; j=n+1
h�J(Zn+1)Xi;XjihJ(Zk)Xi;Xji
�
1
2
X
16i6q; i=n+1
h�J(Zn+1)Xn+1;XiihJ(Zk)Xn+1; Xii
=
1
4
X
16i6q; i=n+1
h�J(Zn+1)Xi; J(Zk)Xii �
1
2
h�J(Zn+1)Xn+1; J(Zk)Xn+1i
=
1
4
X
16i6q; i=n+1
hJ(Zk)J(Zn+1)Xi;Xii �
1
2
hJ(Zk)J(Zn+1)Xn+1; Xn+1i
= �Ric(Zk; Zn+1) + 2hR(Xn+1; Zk)Zn+1;Xn+1i:
This completes 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+1X
j=1
cijYj; (19)
where cij , 1 6 i 6 n, 1 6 j 6 n + 1 are scalar functions on U . Note that
Ei(p) = Yi, so cij(p) = Æij . Using (19), we get
rEi
Yn+1 =
n+1P
j=1
cijrYj
Yn+1 =
1
2
qP
j=1
cij
�
[Xj ; Xn+1]� J(Zn+1)Xj
�
+
1
2
ciqJ(Zq)Xn+1 � 1
2
nP
j=q+1
cijJ(Zj)Xn+1 � ci n+1J(Zn+1)Xn+1:
(20)
Also, for 1 6 k 6 n at p we have
rEk
Ei =
n+1X
j=1
Yk(cij)Yj + cij(p)rYk
Yj
!
=
n+1X
j=1
Yk(cij)Yj +rYk
Yi: (21)
In particular, at p
bki(p) = hrEk
Ei; �i(p) = Yk(ci n+1) + hrYk
Yi; Yn+1i: (22)
Considering (21) for k = i, projecting both sides of it to TpM , and using the
properties of the geodesic frame, we get
0 =
nX
j=1
Yi(cij)Yj + (rYi
Yi)
T
:
196 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
For 1 6 i 6 q�1 and q+1 6 i 6 nrYi
Yi = 0, andrYqYq = J(Zq)Xq =
�
rYqYq
�
T
,
since hJ(Zq)Xq;Xn+1 + Zn+1i = 0. Then, for 1 6 j 6 q we obtain
Yi(cij) =
2
4 0; 1 6 i 6 q � 1;
�hJ(Zq)Xq; Yji; i = q;
0; q + 1 6 i 6 n:
We can deduce from (22) and the above considerations that for 1 6 i 6 n bii(p) =
Yi(ci n+1). Di�erentiate (20) with respect to Ei at p. For 1 6 i 6 q � 1 we get
rEi
rEi
Yn+1 = �Yi(ci n+1)J(Zn+1)Xn+1 +
1
2
rXi
[Xi;Xn+1]� J(Zn+1)Xi
!
= �bii(p)J(Zn+1)Xn+1 �
1
4
J([Xi;Xn+1])Xi �
1
4
[Xi; J(Zn+1)Xi]:
For i = q we have
rEqrEqYn+1 = �Yq(ci n+1)J(Zn+1)Xn+1 +
nX
j=1
Yq(cqj)rYj
Yn+1
+
1
2
rYq
[Xq;Xn+1]� J(Zn+1)Xq + J(Zq)Xn+1
!
= �bqq(p)J(Zn+1)Xn+1
�
1
2
q�1X
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
4
J(Zq)J(Zn+1)Xq +
1
4
J(Zq)
2
Xn+1:
For q + 1 6 i 6 n we obtain
rEi
rEi
Yn+1 = �Yi(ci n+1)J(Zn+1)Xn+1 �
1
2
rZi
(J(Zi)Xn+1)
= �bii(p)J(Zn+1)Xn+1 +
1
4
J(Zi)
2
Xn+1:
Summing up these expressions, we get for 1 6 k 6 q � 1
�
nX
i=1
hrEi
rEi
Yn+1;Xki = nH(p)hJ(Zn+1)Xn+1;Xki
+
1
4
q�1X
i=1
hJ([Xi;Xn+1])Xi;Xki+
1
2
q�1X
j=1
hJ(Zq)Xq;Xjih�J(Zn+1)Xj ;Xki
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 197
Ye.V. Petrov
+
1
4
hJ(Zq)J(Zn+1)Xq;Xki �
1
4
hJ(Zq)2Xn+1;Xki �
1
4
nX
i=q+1
hJ(Zi)2Xn+1;Xki
= nH(p)hJ(Zn+1)Xn+1; Xki �
1
2
X
16i6q; i=n+1
hJ([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)
2
Xn+1, which follows from
the construction of the frame. Thus we obtain the �rst expression in (18).
For q + 1 6 k 6 n we have
�
nX
i=1
hrEi
rEi
Yn+1; Zki =
1
4
q�1X
i=1
h[Xi; J(Zn+1)Xi]; Zki
+
1
2
q�1X
j=1
hJ(Zq)Xq;Xjih[Xj ;Xn+1]; Zki
+
1
4
h[Xq; J(Zn+1)Xq]; Zki �
1
4
h[Xq; J(Zq)Xn+1]; Zki
= �
1
4
X
16i6q; i=n+1
hJ(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+1i:
In the above calculation we used the fact 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 Curvature and Harmonicity
Consider the tangent bundle TN and the distribution in TN formed by left
invariant vector �elds from Z. Since Z is an Abelian ideal, we can integrate
this distribution and obtain a foliation. Denote this foliation by FZ . Let G be
harmonic. Since by (8), in this case Yk(nH) = 0 for all q + 1 6 k 6 n, we have
Corollary 4. If the Gauss map of M is harmonic, then for each leaf M 0 of
FZ the mean curvature of the immersion is constant on M \M
0.
Now we obtain some analogues of the results for Lie groups with bi-invariant
metrics that were stated in [6].
Let � be a vector �eld on M de�ned by �(p) = Yq, for p 2M . In other words,
we obtain �(p) rotating the unit normal vector �(p) by the angle �
2
in the 2-plane
containing �(p) and orthogonal to both dLp(V) and dLp(Z).
198 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
Proposition 5. Let M be a compact smooth oriented hypersurface in a 2-step
nilpotent Lie group N . Assume that:
(i) the mean curvature of M is constant on the integral curves of �;
(ii) the Gauss map of M is harmonic;
(iii) kBk2 +Ric(�; �) > 0 on M and kBk2 +Ric(�; �) > 0 at some point of M ;
(iv) the set of points p 2M such that �(p) =2 dLp(V) is dense in M .
Then G(M) is contained in a closed hemisphere of Sn if and only if G(M) is
contained in a great sphere of Sn.
P r o o f. One of the implications in the proposition is obvious. Suppose
that some closed hemisphere of Sn contains G(M), i.e., there exists a unit vector
v 2 R
n+1 such that for all p 2 M hG(p); vi is nonpositive. Consider a smooth
function f = hG; vi on M . The coe�cient of Yq(e) in (8) vanishes. For all points
from some dense set of M we have Xq 6= 0 and thus Xn+1 =
jXn+1j
jXq j Xq. This,
together with Yq(nH) = 0, implies that the coe�cient of Yn+1(e) is equal to
�kBk2 �Ric(�; �) on the dense subset of M and hence on the whole M because
both the coe�cient and �kBk2 � Ric(�; �) are continuous. Taking the scalar
product of (8) with v, we obtain
�f = �
�
kBk2 +Ric(�; �)
�
f > 0:
Then f is a subharmonic function on the compact manifoldM . Thus f is constant,
and
�
kBk2 +Ric(�; �)
�
f = ��f = 0. From the hypothesis, this implies f = 0,
hence G(M) is contained in the equator v?. This completes the proof.
Proposition 6. Suppose that a smooth oriented hypersurface M in a 2-step
nilpotent Lie group N is CMC, its Gauss map is harmonic, for all p from some
dense set of M the normal vector �(p) =2 dLp(V), and G(M) is contained in an
open hemisphere of Sn. Then M is stable.
P r o o f. From the hypothesis, there exists v 2 R
n+1 such that for all
p 2 M hG(p); vi > 0. As in the proof of Prop. 5, consider a scalar function
w(p) = hG(p); vi on M . This function is smooth and positive. As above, (8)
implies the Jacobi equation
�
�+ kBk2 +Ric(�; �)
�
w = 0. Now [7, Th. 1] implies
the stability of M .
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 199
Ye.V. Petrov
4. Groups of Heisenberg Type
Let N be a group of Heisenberg type. Then from (5), for all X;Y 2 V,
Ric(X;Y ) =
1
2
n+1X
k=q
hJ(Zk)2X;Y i = �
1
2
(n+ 1� q)hX;Y i:
Also, we can rewrite the coe�cients in (8) for 1 6 k 6 q and for k = n+ 1 in the
form
q�1X
j=1
hJ([Xk;Xj ])Xj ;Xn+1i+ 4hR(Xk; Zn+1)Zn+1;Xn+1i
=
2
664
0; 1 6 k 6 q � 1;
jZn+1j jXn+1j
�
q � n� 1 + jZn+1j2
�
; k = q;
jXn+1j2
�
q � n� 1 + jZn+1j2
�
; k = n+ 1:
Moreover,
Ric(Zn+1; Zn+1) = �
1
4
TrJ(Zn+1)
2
=
q
4
jZn+1j2 ;
and thus
Ric(Yn+1; Yn+1) =
q
4
jZn+1j2 �
1
2
(n+ 1� q) jXn+1j2 :
Equation (8) now takes the form
�G(p) =
q�1P
k=1
�Yk(nH)� 2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi;Xki
+ 2
qP
i=1
biq(p)hJ(Zq)Xi;Xki+ nH(p)hJ(Zn+1)Xn+1;Xki
�
Yk(e)
+
�
� Yq(nH) + jZn+1j jXn+1j
�
q � n� 1 + jZn+1j2
�
� 2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi;Xqi
+2
qP
i=1
biq(p)hJ(Zq)Xi;Xqi+ nH(p)hJ(Zn+1)Xn+1;Xqi
�
Yq(e)
+
nP
k=q+1
�
� Yk(nH)
�
Yk(e)
+
�2
qP
i=1
nP
j=q+1
bij(p)hJ(Zj)Xi;Xn+1i
+ 2
qP
i=1
biq(p)hJ(Zq)Xi;Xn+1i � kBk2(p)� q
4
jZn+1j2
+ jXn+1j2
�
1
2
(q � n� 1) + jZn+1j2
��
Yn+1(e):
(23)
200 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
Consider the case n = q, i.e., dimZ = 1. It is easy to see that n is then
even, n = 2m, where m is a positive integer, and N is isomorphic to the 2m+ 1-
dimensional Heisenberg group (recall that N is connected and simply connected).
In this case at p we can choose X1; : : : ;X2m+1 so that
J(Z)Xi = Xm+i; 1 6 i 6 m� 1;
J(Z)Xm =
X2m
jX2mj
=
X2m
jZ2m+1j
if Z2m+1 6= 0; or
X2m+1
jX2m+1j
if X2m+1 6= 0;
J(Z)Xm+i = �Xi; 1 6 i 6 m� 1;
J(Z)X2m = � jX2mjXm = � jZ2m+1jXm;
J(Z)X2m+1 = � jX2m+1jXm:
Choose Z2m = jX2m+1jZ and Z2m+1 = jZ2m+1jZ. Then (23) has the form
�G(p) = �
m�1P
k=1
�
Yk(2mH) + 2b2mm+k(p) jX2m+1j
�
Yk(e)
�
�
Ym(2mH) + 2mH(p) jX2m+1j
+ 2b2m 2m(p) jX2m+1j jZ2m+1j
�
Ym(e)
�
m�1P
k=1
�
Ym+k(2mH)� 2b2mk(p) jX2m+1j
�
Yk(e)
�
�
Y2m(2mH) + jX2m+1j3 jZ2m+1j
� 2b2mm(p) jX2m+1j jZ2m+1j
�
Y2m(e)
�
�
kBk2(p) + m
2
jZ2m+1j2 � 1
2
jX2m+1j2
+ jX2m+1j4 � 2b2mm(p) jX2m+1j2
�
Y2m+1(e):
(24)
Consider an example of the three-dimensional Heisenberg group Nil. In the
space R3 with Cartesian coordinates (x; y; z), de�ne vector �elds
X =
@
@x
; Y =
@
@y
+ x
@
@z
; Z =
@
@z
:
Then Span(X;Y;Z) is a Lie algebra (with the only nonzero bracket [X;Y ] = Z),
which is the Lie algebra of Nil. Introduce a scalar product in such a way that
the vectors X;Y and Z are orthonormal. Consider the following unit vector �eld:
� =
xY + Zp
1 + x2
;
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 201
Ye.V. Petrov
and vector �elds
F1 = X; F2 =
Y � xZp
1 + x2
;
which are orthogonal to �. In the notation of section , in each p F1 = X1,
F2 = X2 � Z2, � = X3 + Z3. By direct computation of covariant derivatives
it can be shown that the distribution spanned by F1 and F2 is integrable and
form the tangent bundle of some two-dimensional foliation F in Nil. From the
computation of the second fundamental form we obtain kBk2 =
(x
2�1)
2
2(1+x2)
2 , and
H = 0. Thus the leaves of this foliation are minimal surfaces. The Laplacian on
G = � is
�G =
�
F1F1 + F2F2 � (rF1
F1)
T � (rF2
F2)
T
�
G
= �
x
(1 + x2)
2
F2 �
1
(1 + x2)
2
�:
We obtain the same result considering (24) at some p. In fact,
2b22 jX3j jZ3j = 0;
jX3j3 jZ3j � 2b21 jX3j jZ3j =
x
(1 + x2)
2
;
kBk2 +
1
2
jZ3j2 �
1
2
jX3j2 + jX3j4 � 2b21 jX3j2 =
1
(1 + x2)
2
:
In particular, foliation F gives an example of a CMC-surface in Nil such that
its Gauss map is not harmonic.
Proposition 7. Suppose that M is a smooth oriented 2m-dimensional ma-
nifold immersed in the 2m + 1-dimensional Heisenberg group. If any two of the
following three claims are true, then the third one is also true:
(i) M is CMC;
(ii) the Gauss map of M is harmonic;
(iii) at every point of M , the following holds:8>>>>>><
>>>>>>:
b2mk = 0; 1 6 k 6 m� 1; m+ 1 6 k 6 2m� 1;
jZ2m+1j
�
jX2m+1j2 � 2b2mm
�
= 0;
jZ2m+1j (b1 1 + � � � + b2m�1 2m�1 + 3b2m 2m) = 0:
(25)
Here bij, 1 6 i; j 6 2m are the coe�cients of the second fundamental form
of M in the basis chosen as above.
202 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
P r o o f. If (iii) is true, then the equivalency of (i) and (ii) immediately
follows from (24). Suppose (i) and (ii) are true. Let A be a set of such points
of M that jX2m+1j 6= 0. At the points of A (24) implies the expressions in (25).
Since the distribution orthogonal to Z is nonintegrable, A is dense in M . Now
the continuity of the left hand sides of the equations (25) implies (iii).
In the case m = 1 the next theorem shows that the restrictions for M arising
from (25) are rather strict.
Theorem 8. LetM be a smooth oriented CMC-surface in the Heisenberg group
Nil whose Gauss map is harmonic. Then M is a �cylinder�, that is, its position
vector in the coordinates x, y, z has the form
r(s; t) = (f1(s); f2(s); t); (26)
where f1 and f2 are some smooth functions.
P r o o f. For each p 2 M denote a(p) = jX3j, b(p) = jZ3j. Then a and b
are smooth scalar functions on M , and a
2
+ b
2
= 1. Consider an arbitrary point
p of M . Choose X1 as above and put X2 = J(Z)X1. Denote by T1 and T2 the
vector �elds that at each p 2 M are equal to X1 and X2, respectively. Consider
unit tangent vector �elds F1 and F2, and a unit normal vector �eld � of M of the
form
F1 = T1; F2 = bT2 � aZ; � = aT2 + bZ:
Denote by �1 and �2 the geodesic curvatures of the integral curves of F1 and F2,
respectively. In other words,
rF1
F1 = �1F2; rF1
F2 = ��1F1; rF2
F1 = ��2F2; rF2
F2 = �2F1; (27)
where r is the Riemannian connection on M induced by the immersion. The
Gaussian curvature of the surface is
K = F1(�2) + F2(�1)� (�1)
2 � (�2)
2
: (28)
Assume that for some p 2 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 second
fundamental form of M is �
3H
1
2
a
2
1
2
a
2 �H
�
: (29)
In particular, the extrinsic curvature Kext of the surface is �3H2 � 1
4
a
4.
Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2 203
Ye.V. Petrov
Denote byB the second fundamental form of the immersion. Then the Codazzi
equations for M are
(rF1
B) (F2; F1)� (rF2
B) (F1; F1) = hR(F1; F2)F1; �i = ab;
(rF2
B) (F1; F2)� (rF1
B) (F2; F2) = hR(F2; F1)F2; �i = 0:
Computing the covariant derivatives of the second fundamental form, we obtain
for U
aF1(a) + 4H�1 � a
2
�2 � ab = 0;
aF2(a)� 4H�2 � a
2
�1 = 0:
(30)
The Gauss equation has the form
K = Kext + hR(F1; F2)F2; F1i = �3H2 �
1
4
a
4 �
3
4
b
2
+
1
4
a
2
:
From (28) we obtain
F1(�2) + F2(�1)� (�1)
2 � (�2)
2
= �3H2 �
1
4
a
4 �
3
4
b
2
+
1
4
a
2
: (31)
Using (30) and the form of F2 and �, we can derive
hrF1
F2; �i = �hF2;rF1
�i = �hF2;rF1
�
a
b
F2 +
�
a
2
b
+ b
�
Z
�
i
= �F1
�
a
b
�
hF2; F2i � F1
�
1
b
�
hF2; Zi �
a
b
hF2;rF1
F2i �
1
b
hF2;rF1
Zi
= �F1
�
a
b
�
+ aF1
�
1
b
�
�
1
b
hF2;�
1
2
T2i = �
1
b
�
�
4H�1
a
+ a�2 + b
�
+
1
2
= �
1
2
+
4H�1
ab
�
a
b
�2;
hrF2
F1; �i = �hF1;rF2
�i = �hF1;rF2
�
a
b
F2 +
�
a
2
b
+ b
�
Z
�
i
= �
a
b
hF1;rF2
F2i �
1
b
hF1;rF2
Zi = �
a
b
�2 �
1
b
hT1;
1
2
bT1i = �
a
b
�2 �
1
2
:
In the above equations we used the fact that Z is left invariant and the expres-
sions (2) for the covariant derivative. Since ab 6= 0, the integrability condition
h[F1; F2]; �i = 0 takes the form H�1 = 0. Besides, (29) implies
3H = b11 = hrF1
F1; �i = �hF1;rF1
�i
= �hF1;rF1
�
a
b
F2 +
�
a
2
b
+ b
�
Z
�
i = �
a
b
hF1;rF1
F2i �
1
b
hF1;rF1
Zi =
a
b
�1:
204 Journal of Mathematical Physics, Analysis, Geometry, 2006, vol. 2, No. 2
The Gauss Map of Hypersurfaces in 2-Step Nilpotent Lie Groups
Thus H = �1 = 0. In particular, rF1
F1 = 0, hence T1 = F1 is a geodesic vector
�eld in the ambient manifold. Note that T1 belongs to the distribution that spans
the left invariant vector �elds of V. Considering the set of geodesics in Nil (see
[3, Prop. (3.1) and (3.5)]), we obtain that T1 = cX+dY , where c; d 2 R are some
constants, i.e., T1 = X1 and T2 = X2 are left invariant. Note that the second
equation of (30) implies F2(a) = F2(b) = 0. Thus we obtain
rF2
F2 = rF2
(bX2 � aZ) = brbX2�aZX2 � arbX2�aZZ = �abX1:
Therefore �2 = �ab. It follows from this equation, from the computations above
in this proof, and from (29) that
1
2
a
2
= b12 = hrF1
F2; �i = �
a
b
�2 �
1
2
= a
2 �
1
2
;
and a
2
= b
2
=
1
2
. But then a = b =
p
2
2
, and the �rst equation in (30) implies
a�2 + b = 0, which leads to a contradiction.
Thus ab = 0 at each point of M . Since a2 + b
2
= 1 and a, b are continuous,
a = 1 or b = 1 identically. The latter case is impossible since Z? is not integrable;
then the normal vector of M is orthogonal to Z, and F2 = �Z. Therefore M is
invariant under the action of Z by left translations, and M is formed by integral
curves of Z, which are geodesics (0; 0; t). Then M has the form (26).
Note that a similar result was obtained in [13] by a di�erent method. Also, in
[13] the equations of the CMC-surfaces of the form (26) were obtained. Proposi-
tion 7 then implies that the Gauss maps of all these surfaces are harmonic.
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