Submanifolds with the Harmonic Gauss Map in Lie Groups
In this paper we find a criterion for the Gauss map of an immersed smooth submanifold in some Lie group with left invariant metric to be harmonic. Using the obtained expression we prove some necessary and sufficient conditions for the harmonicity of this map in the case of totally geodesic submanifo...
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Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України
2008
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| Cite this: | Submanifolds with the Harmonic Gauss Map in Lie Groups / Ye.V. Petrov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 278-293. — Бібліогр.: 10 назв. — англ. |
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| citation_txt | Submanifolds with the Harmonic Gauss Map in Lie Groups / Ye.V. Petrov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 278-293. — Бібліогр.: 10 назв. — англ. |
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| description | In this paper we find a criterion for the Gauss map of an immersed smooth submanifold in some Lie group with left invariant metric to be harmonic. Using the obtained expression we prove some necessary and sufficient conditions for the harmonicity of this map in the case of totally geodesic submanifolds in Lie groups admitting biinvariant metrics. We show that, depending on the structure of the tangent space of a submanifold, the Gauss map can be harmonic in all biinvariant metrics or nonharmonic in some metric. For 2-step nilpotent groups we prove that the Gauss map of a geodesic is harmonic if and only if it is constant.
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Journal of Mathematical Physics, Analysis, Geometry
2008, vol. 4, No. 2, pp. 278�293
Submanifolds with the Harmonic Gauss Map
in Lie Groups
Ye.V. Petrov
Department of Mechanics and Mathematics, V.N. Karazin Kharkiv National University
4 Svobody Sq., Kharkiv, 61077, Ukraine
E-mail:petrov@univer.kharkov.ua
Received March 7, 2007
In this paper we �nd a criterion for the Gauss map of an immersed smooth
submanifold in some Lie group with left invariant metric to be harmonic.
Using the obtained expression we prove some necessary and su�cient condi-
tions for the harmonicity of this map in the case of totally geodesic submani-
folds in Lie groups admitting biinvariant metrics. We show that, depending
on the structure of the tangent space of a submanifold, the Gauss map can
be harmonic in all biinvariant metrics or nonharmonic in some metric.
For 2-step nilpotent groups we prove that the Gauss map of a geodesic is
harmonic if and only if it is constant.
Key words: left invariant metric, biinvariant metric, Gauss map, har-
monic map, 2-step nilpotent group, totally geodesic submanifold.
Mathematics Subject Classi�cation 2000: 53C42 (primary); 53C43, 22E25,
22E46 (secondary).
1. Introduction
It is proved in [9] that the Gauss map of a submanifold in the Euclidean
space is harmonic if and only if the mean curvature �eld of this submanifold is
parallel. There is a natural generalization of the Gauss map to the submanifolds
in Lie groups: for each point of a submanifold the tangent space at this point
is translated to the identity element of the group (for the precise statement see
Sect. 2). Let the Lie group be endowed with some left invariant metric. As it is
proved in [4], when this metric is biinvariant and the submanifold is hypersurface,
the Gauss map is harmonic if and only if the mean curvature is constant. Our
aim is to consider more general case of a submanifold in some Lie group with
arbitrary codimension.
The author was partially supported by N.I. Akhiezer Foundation and the Foundation of Fun-
damental Research of Ukraine (Project No. GP/F13/0019 for young scientists).
c
Ye.V. Petrov, 2008
Submanifolds with the Harmonic Gauss Map in Lie Groups
The paper is organized as follows. In Section 2 we obtain the harmonicity
criterion for the Gauss map of a submanifold in some Lie group with a left invari-
ant metric. This criterion is given in the terms of the second fundamental form
of the immersion and the left invariant Riemannian connection on the Lie group
(Th. 1).
In Section 3 we consider submanifolds in Lie groups with biinvariant metric.
Let us introduce some notation. Let N be a Lie group with biinvariant metric,
N be the Lie algebra of N ,M be a smooth immersed totally geodesic submanifold
in N . Taking if necessary the left translation ofM assume that e 2M (see Sect. 3
for details). The tangent space TeM is a Lie triple system in N . Denote by N the
Lie subalgebra TeM + [TeM;TeM ] of N . By W denote the orthogonal projection
of TeM to the semisimple Lie subalgebra N
0
= [N ;N ].
The subspace W = W \ [W;W] is an ideal (here and further by ideals we
mean the ideals in N ). Denote by V the orthogonal complement in N
0
to W. Let
V =
L
16l6m
Sl be some direct orthogonal decomposition of V into simple ideals.
Using Theorem 1 we prove
Theorem 3. Let M be a smooth immersed totally geodesic submanifold in
a Lie group N with biinvariant metric. Then:
(i) if the restriction of the metric to V is a negative multiple of the Killing form
(in particular, if V is simple), then the Gauss map of M in this metric is
harmonic;
(ii) if W \ V =
L
16l6m
Wl, where Wl � Sl is a proper Lie triple system in Sl,
i.e., Wl 6= 0 and Wl 6= Sl for each 1 6 l 6 m (in particular, if V = 0), then
the Gauss map of M is harmonic in any biinvariant metric on N ;
(iii) if the condition of (ii) is not satis�ed, then there is a biinvariant metric on
N such that the Gauss map of M is not harmonic.
In the paper [8] we considered hypersurfaces in 2-step nilpotent Lie groups
and found conditions for the Gauss maps of these hypersurfaces to be harmonic.
In particular, we showed that, unlike in the case of groups with biinvariant met-
ric, this harmonicity is not equivalent to the constancy of the mean curvature.
As it was shown in [2], totally geodesic submanifolds in such groups either have
the Gauss map of maximal rank or they are open subsets of subgroups (and
consequently have the constant Gauss map). In the latter case the structure of
subalgebras corresponding to such subgroups can be explicitly described (this de-
scription implies, in particular, that there are not totally geodesic hypersurfaces
in 2-step nilpotent Lie groups, see [2]). Using our criterion, in Section 4 we prove
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 279
Ye.V. Petrov
that the Gauss map of a geodesic in a 2-step nilpotent Lie group is harmonic if
and only if it is constant (Prop. 4).
The author would thank Prof. A.A. Borisenko and prof. L.A. Masal'tsev for
their attention paid to this work. Also, the author is grateful to the reviewer for
the essential improvement of the results and presentation of the paper.
2. The Harmonicity Criterion
Suppose M is a smooth manifold, dimM = n, M ! N is an immersion of
M in some Lie group N with a left invariant metric, dimN = n + q. For some
point p of M let Y1; : : : ; Yn and Yn+1; : : : ; Yn+q be orthonormal frames of tangent
space TpM � TpN and of normal space NpM � TpN , respectively. Also by Ya,
1 6 a 6 n+ q, denote the corresponding left invariant �elds on N .
Denote the left invariant metric on N (and also the corresponding inner pro-
duct on its Lie algebra) by h�; �i, the Riemannian connection of this metric by
r, its curvature tensor by R(�; �)�, and the normal connection of the immersion
M ! N by r?.
Let E1; : : : ; En+q be the vector �elds de�ned on some neighborhood U of p such
that Ei(p) = Yi, E1; : : : ; En and En+1; : : : ; En+q are orthonormal frames of the
tangent and the normal bundles of M on U , respectively, and (rEi
Ej)
T
(p) = 0,
for all 1 6 i; j 6 n. Then the mean curvature �eld H of the immersion is de�ned
on U by
H = 1
n
P
16i6n
(rEi
Ei)
? : (1)
Here (�)T and (�)? are the projections to the tangent bundle TM and the normal
bundle NM , respectively.
For 1 6 i; j 6 n, n+1 6 � 6 n+q by b�
ij
= hrEi
Ej; E�i denote the coe�cients
of the second fundamental form of the immersion on U with respect to the frame
E1; : : : ; En+q. Suppose that on U for 1 6 a 6 n+ q
Ea =
P
16b6n+q
Ab
aYb: (2)
Here fAb
ag16a;b6n+q are functions on U . Obviously, Ab
a(p) = Æab, where Æab is the
Kronecker symbol.
Let � be the Laplacian �M of the induced metric onM . The de�nition of the
Laplacian and the conditions (rEi
Ej)
T
(p) = 0 imply that for functions f and g
de�ned on U
�f(p) =
P
16i6n
EiEi(f); (3)
�(fg)(p) = g(p)�f(p) + 2
P
16i6n
Ei(f)Ei(g) + f(p)�g(p): (4)
280 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
Let � be the Gauss map of M :
�: M ! G(n; q); �(p) = dLp�1(TpM): (5)
Here G(n; q) is the Grassmannian of n-dimensional subspaces in the n+ q-dimen-
sional vector space, a point p is identi�ed with its image under the immersion, Lg
is the left translation by g 2M , dF is the di�erential of a map F .
Recall that if (M1; g1) and (M2; g2) are smooth Riemannian manifolds, then
for any � 2 C1(M1;M2) the energy of � is
E(�) = 1
2
R
M1
P
16i6m
g2(d�(Ei); d�(Ei))dVM ;
where m = dimM1, E1; : : : ; Em is the orthonormal frame on M1, dVM is the
volume form of g1. The critical points of the functional � 7! E(�) are called
harmonic maps from M1 to M2. We say that a map is harmonic at some point if
the corresponding Euler�Lagrange equations are satis�ed at this point (i.e., the
so-called tension �eld vanishes, see, for example, [10]).
Theorem 1. The map � is harmonic at p if and only ifP
16i6n
hR(Yj ; Yi)Yi; Y�i �
P
16i6n
hr(rYi
Yi)
Yj ; Y�i+ h[nH; Yj ]; Y�i
+2
P
16i;k6n
b�
ik
hrYi
Yk; Yji+ 2
P
16i6n;n+16
6n+q
b
ij
hrYi
Y
; Y�i
�
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i+
P
16i6n
h(rYi
Yj)
? ; (rYi
Y�)
?i = 0
(6)
for all 1 6 j 6 n, n+ 1 6 � 6 n+ q.
P r o o f. The Grassmannian has the structure of the symmetric space
G(n; q) = O(n + q)= (O(n)�O(q)). There is an embedding of this space in the
space of symmetric matrices of order n+ q considered with the obvious Euclidean
metric ([5]). This embedding is induced by the map A 7! AEAt, where A 2
O(n+ q), At is A transposed, and
E =
� q
n+q
In 0
0 n
n+q
Iq
!
:
Here In and Iq are the identity matrices of order n and q, respectively. The image
of � on U corresponds to the matrix A = (Ab
a
)16a;b6n+q, where A
b
a
are the func-
tions from (2). The composition of � and the embedding give the map de�ned
on U by0BBBB@
� q
n+q
In +
P
n+16
6n+q
A
j
A
k
! P
n+16
6n+q
A
j
A
�
!
P
n+16
6n+q
A�
A
k
!
n
n+q
Iq �
P
16l6n
A�
l
A
�
l
!
1CCCCA ; (7)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 281
Ye.V. Petrov
where 1 6 j; k 6 n, n+ 1 6 �; � 6 n+ q. Di�erentiate Ea with respect to Ei on
U for 1 6 a 6 n+ q, 1 6 i 6 n:
rEi
Ea =
P
16b6n+q
Ei(A
b
a
)Yb +
P
16b6n+q
Ab
a
rEi
Yb: (8)
In particular, at p
rYi
Ea =
P
16b6n+q
Ei(A
b
a
)Yb +rYi
Ya: (9)
Note that Ei(A
b
a
) = �Ei(A
a
b
) (this can be derived from (9) or simply from the
fact that so(n+ q) is the algebra of skew-symmetric matrices).
According to Th. (2.22) in Ch. 4 of [10], the criterion of the harmonicity of �
is the set of equations
��a
b
�
P
16i6n
B(d�(Ei); d�(Ei))
!
a
b
= 0: (10)
Here 1 6 a 6 b 6 n + q, �a
b
are the coordinate functions of the embed-
ding, and B is the second fundamental form of the embedding�. The �elds
f @
@A
j
�
g16j6n;n+16�6n+q form the frame of TG(n; q) on the image of U (note that
@
@A
�
j
= � @
@A
j
�
). Denote by Ca
b
for 1 6 a 6 b 6 n+q the matrix with entry 1 at the
intersection of the a�th row and b�th column and with other entries equal to 0.
The di�erential of the embedding at p maps the �eld @
@A
j
�
to the vector C
j
�. It fol-
lows that we can take as a frame of the normal space of the Grassmannian at the
image of this point the vectors Ci
j
, 1 6 i 6 j 6 n and C�
�
, n+1 6 � 6 � 6 n+ q.
The expressions (7) imply on U for 1 6 l 6 m 6 n, n+ 1 6
6 � 6 n+ q�
@
@A
j
�
�
l
m
= ÆljA
m
� + ÆmjA
l
�;
�
@
@A
j
�
�
�
= Æ
�A
�
j
+ ��A
j
:
Di�erentiate these equations:
B
�
@
@A
j
�
; @
@A
k
�
�
=
P
16l6m6n
@
@A
k
�
�
@
@A
j
�
�l
m
C l
m
+
P
n+16
6�6n+q
@
@Ak
�
�
@
@A
j
�
�
�
C
� = Æ��(1 + Æjk)C
j
k
� Æjk(1 + Æ��)C
�
�
:
for 1 6 j 6 k 6 n, n+ 1 6 � 6 � 6 n+ q. Also note that
d�(Ei) =
P
16j6n;n+16�6n+q
Ei(A
j
�)
@
@A
j
�
:
�Actually, the sign of B in [10] is di�erent because the Laplacian in this book is de�ned with
the opposite sign.
282 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
This implies that at p for 1 6 j 6 k 6 n the expressions in (10) take the form
�
� q
n+q
Æjk +
P
n+16
6n+q
A
j
A
k
!
�
P
16i6n
B(d�(Ei); d�(Ei))
!
j
k
= 2
P
16i6n;n+16
6n+q
Ei(A
j
)Ei(A
k
)� 2
P
16i6n;n+16
6n+q
Ei(A
j
)Ei(A
k
) = 0:
Here the equation (4) was used. Similarly, for n+ 1 6 � 6 � 6 n+ q obtain
�
n
n+q
�� �
P
16l6n
A�
l
A
�
l
!
�
P
16i6n
B(d�(Ei); d�(Ei))
!
�
�
= �2
P
16i;l6n
Ei(A
�
l
)Ei(A
�
l
) + 2
P
16i;l6n
Ei(A
�
l
)Ei(A
�
l
) = 0:
It follows that the conditions (10) at p become
�
P
n+16
6n+q
A
j
A
�
!
= 0
for 1 6 j 6 n; n+ 1 6 � 6 n+ q. The di�erentiation gives
�A
j
� + 2
P
16i6n;n+16
6n+q
Ei(A
j
)Ei(A
�
) = 0;
1 6 j 6 n; n+ 1 6 � 6 n+ q:
(11)
Di�erentiate (8) with respect to Ei on U for n+ 1 6 a = � 6 n+ q:
rEi
rEi
E� =
P
16j6n
EiEi(A
j
�)Yj +
P
n+16�6n+q
EiEi(A
�
�)Y�
+2
P
16j6n
Ei(A
j
�)rEi
Yj + 2
P
n+16�6n+q
Ei(A
�
�)rEi
Y�
+
P
16j6n
A
j
�rEi
rEi
Yj +
P
n+16�6n+q
A
�
�rEi
rEi
Y�:
(12)
Take the inner product of (12) with Yj at p:
EiEi(A
j
�) = hrEi
rEi
E�; Yji � 2
P
16k6n
Ei(A
k
�
)hrEi
Yk; Yji
�2
P
n+16
6n+q
Ei(A
�)hrEi
Y
; Yji � hrEi
rEi
Y�; Yji:
Therefore (11) takes the formP
16i6n
hrEi
rEi
E�; Yji � 2
P
16i;k6n
Ei(A
k
�
)hrEi
Yk; Yji
�2
P
16i6n;n+16
6n+q
Ei(A
�)hrEi
Y
; Yji �
P
16i6n
hrEi
rEi
Y�; Yji
+2
P
16i6n;n+16
6n+q
Ei(A
j
)Ei(A
�
) = 0:
(13)
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 283
Ye.V. Petrov
Here (3) was used. The de�nition (1) of the mean curvature �eld implies for
1 6 j 6 n, n+ 1 6 � 6 n+ q at p
hrYj
(nH); Y�i =
P
16i6n
hrEj
�
(rEi
Ei)
?
�
; E�i =
P
16i6n
hrEj
rEi
Ei; E�i
�
P
16i6n
hrEj
�
(rEi
Ei)
T
�
; E�i =
P
16i6n
hR(Ej ; Ei)Ei +rEi
rEj
Ei
+r[Ej;Ei]Ei; E�i �
P
16i6n
Ejh(rEi
Ei)
T ; E�i+
P
16i6n
h(rEi
Ei)
T ;rEj
E�i
=
P
16i6n
hR(Yj ; Yi)Yi; Y�i+
P
16i6n
hrEi
rEj
Ei; E�i =
P
16i6n
hR(Yj ; Yi)Yi; Y�i
+
P
16i6n
hrEi
[Ej ; Ei]; E�i+
P
16i6n
hrEi
rEi
Ej; E�i =
P
16i6n
hR(Yj ; Yi)Yi; Y�i
+
P
16i6n
Eih[Ej ; Ei]; E�i �
P
16i6n
h[Ej ; Ei];rEi
E�i+
P
16i6n
hrEi
rEi
Ej; E�i
=
P
16i6n
hR(Yj ; Yi)Yi; Y�i+
P
16i6n
hrEi
rEi
Ej ; E�i:
In the third equality the de�nition of the curvature tensor was used. The fourth
equality follows from the Frobenius theorem, the condition (rEi
Ej)
T
(p) = 0, and
its consequence
[Ek; Ei](p) = ([Ek; Ei])
T
(p) = (rEk
Ei �rEi
Ek)
T
(p) = 0:
Di�erentiate two times the expression hEj ; E�i = 0 with respect to Ei:
hrEi
rEi
Ej ; E�i+ 2hrEi
Ej ;rEi
E�i+ hEj ;rEi
rEi
E�i = 0:
This equation and (9) imply
hrYj
(nH); Y�i =
P
16i6n
hR(Yj ; Yi)Yi; Y�i � 2
P
16i6n
hrEi
Ej;rEi
E�i
�
P
16i6n
hEj ;rEi
rEi
E�i =
P
16i6n
hR(Yj; Yi)Yi; Y�i
�2
P
16i6n;n+16
6n+q
b
ij
(Ei(A
�) + hrYi
Y�; Y
i)�
P
16i6n
hEj ;rEi
rEi
E�i:
From (9) and the condition (rEi
Ej)
T
(p) = 0 obtain
b
ij
= Ei(A
j
) + hrYi
Yj; Y
i; (14)
0 = Ei(A
k
j
) + hrYi
Yj ; Yki: (15)
Hence at p
hrYj
(nH); Y�i =
P
16i6n
hR(Yj; Yi)Yi; Y�i
+2
P
16i6n;n+16
6n+q
b
ij
hrYi
Y
; Y�i
�2
P
16i6n;n+16
6n+q
hrYi
Yj; Y
iEi(A
�)
�
P
16i6n
hEj ;rEi
rEi
E�i � 2
P
16i6n;n+16
6n+q
Ei(A
j
)Ei(A
�):
(16)
284 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
The equation (14) impliesP
16i;k6n
Ei(A
k
�
)hrEi
Yk; Yji = �
P
16i;k6n
Ei(A
�
k
)hrEi
Yk; Yji
= �
P
16i;k6n
b�
ik
hrYi
Yk; Yji+
P
16i;k6n
hrYi
Yk; Y�ihrYi
Yk; Yji:
Note that for each pair of left invariant �elds X and Y the product hX;Y i is
constant, hence,
hrZX;Y i = Z (hX;Y i)� hX;rZY i = �hX;rZY i (17)
for every vector Z. This and the fact that the frame is orthonormal implyP
16i;k6n
hrYi
Yk; Y�ihrYi
Yk; Yji =
P
16i;k6n
hrYi
Yj; YkihrYi
Y�; Yki
=
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i:
Thus, P
16i;k6n
Ei(A
k
�)hrEi
Yk; Yji = �
P
16i;k6n
b�
ik
hrYi
Yk; Yji
+
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i:
(18)
Substituting (16) in (13) and taking into account (18) derive the conditionsP
16i6n
hR(Yj ; Yi)Yi; Y�i � hrYj
(nH); Y�i+ 2
P
16i;k6n
b�
ik
hrYi
Yk; Yji
+2
P
16i6n;n+16
6n+q
b
ij
hrYi
Y
; Y�i �
P
16i6n
hrEi
rEi
Y�; Yji
�2
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i = 0:
(19)
At p for 1 6 i; j 6 n, n+ 1 6 � 6 n+ q obtain
hrEi
rEi
Y�; Yji = hrEi
P
16a6n+q
Aa
i
rYaY�
!
; Yji =
P
16k6n
Ei(A
k
i
)hrYk
Y�; Yji
+
P
n+16
6n+q
Ei(A
i
)hrY
Y�; Yji+ hrYi
rYi
Y�; Yji:
Substitute into this (14) and (15) and use the de�nition of the mean curvature
hnH;E
i =
P
16i6n
b
ii
. Then use (17):
P
16i6n
hrEi
rEi
Y�; Yji = �
P
16i6n;16a6n+q
hrYi
Yi; YaihrYaY�; Yji
+
P
n+16
6n+q
hnH; Y
ihrY
Y�; Yji+
P
16i6n
hrYi
rYi
Y�; Yji
=
P
16i6n;16a6n+q
hrYi
Yi; YaihrYaYj ; Y�i
�
P
n+16
6n+q
hnH; Y
ihrY
Yj; Y�i �
P
16i6n
hrYi
Yj;rYi
Y�i:
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 285
Ye.V. Petrov
The frame is orthonormal, henceP
16i6n
hrEi
rEi
Y�; Yji =
P
16i6n
hr(rYi
Yi)
Yj ; Y�i
�hr(nH)Yj; Y�i �
P
16i6n
hrYi
Yj;rYi
Y�i:
(20)
Substitute (20) in (19) and obtain (6).
Note that the Gauss map of a Lie subgroup is constant, therefore harmonic.
If N is the Euclidean space En+q, then the curvature tensor vanishes. For any
vector �eld X and for the left invariant (i.e., constant) Y the derivatives rXY also
vanish. This yields that the conditions (6) take the form hrYj
(nH); Y�i = 0 for
1 6 j 6 n, n+1 6 � 6 n+ q, i.e., r?H = 0, and we obtain the above-mentioned
classical result of [9].
The de�nition of the second fundamental form and the fact that the frame is
orthonormal allow us to rewrite (6) in the formP
16i6n
hR(Yj ; Yi)Yi; Y�i �
P
16i6n
hr(rYi
Yi)
Yj ; Y�i+ h[nH; Yj ]; Y�i
�2
P
16i6n
hr
(rYi
Yj)
TEi; Y
�i � 2
P
16i6n;
h(rYi
Ej)
? ; (rYi
Y�)
?i
�
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i+
P
16i6n
h(rYi
Yj)
? ; (rYi
Y�)
?i = 0:
(21)
Note that these expressions do not depend on the particular choice of E1; : : : ; En.
The summands in (6) that do not include the coe�cients of the second fun-
damental form and the mean curvature �eld can be rewritten:P
16i6n
hR(Yj ; Yi)Yi; Y�i �
P
16i6n
hr(rYi
Yi)
Yj ; Y�i
�
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i+
P
16i6n
h(rYi
Yj)
? ; (rYi
Y�)
?i
=
P
16i6n
hrYj
rYi
Yi �r(rYi
Yi)
Yj �rYi
rYj
Yi �r[Yj;Yi]Yi
+rYi
�
rYj
Yi + [Yi; Yj]
�
T
�rYi
�
rYj
Yi + [Yi; Yj ]
�?
; Y�i
=
P
16i6n
h[Yj ;rYi
Yi]�rYi
rYj
Yi �r[Yj ;Yi]Yi +rYi
[Yj ; Yi]
+rYi
�
rYj
Yi + 2[Yi; Yj ]
�T
�rYi
�
rYj
Yi
�?
; Y�i
=
P
16i6n
h[Yj;rYi
Yi] + [Yi; [Yj; Yi]] + 2rYi
�
([Yi; Yj ])
T �
�
rYj
Yi
�?�
; Y�i:
In particular, a totally geodesic submanifold M has the harmonic Gauss map at
p if and only ifP
16i6n
�
[Yj;rYi
Yi] + [Yi; [Yj ; Yi]] + 2rYi
�
([Yi; Yj ])
T �
�
rYj
Yi
�?��?
= 0 (22)
for all 1 6 j 6 n.
286 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
3. Lie Groups with Biinvariant Metric
In this section we consider a Lie group N with some biinvariant metric.
The conditions from Th. 1 in this particular case are relatively simple:
Proposition 2. The Gauss map of a smooth submanifold M in the Lie group
N with biinvariant metric h�; �i is harmonic at a point p 2M if and only if in the
above notation
h[nH; Yj ]; Y�i+
P
16i6n;n+16
6n+q
b
ij
h[Yi; Y
]; Y�i
+1
2
P
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i = 0
(23)
for 1 6 j 6 n, n+ 1 6 � 6 n+ q.
P r o o f. Recall that the left invariant metric h�; �i is biinvariant if and only if
h[X;Y ]; Zi = hX; [Y;Z]i for all left invariant X, Y , and Z. Also, rXY = 1
2
[X;Y ].
In particular, rXY = �rYX and rXX = 0. This, together with the symmetry
of the second fundamental form, implies
P
16i;k6n
b�
ik
hrYi
Yk; Yji = 0. The curvature
tensor is de�ned by the equation R(X;Y )Z = �1
4
[[X;Y ]; Z]. Thus,
�
P
16i6n
h(rYi
Yj)
T ; (rYi
Y�)
T i+
P
16i6n
h(rYi
Yj)
? ; (rYi
Y�)
?i
= �1
4
P
16i6n
h[Yi; Yj ]T ; [Yi; Y�]T i+
1
4
P
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i
= �1
4
P
16i6n
h[Yi; Yj ]; [Yi; Y�]i+
1
2
P
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i
=
P
16i6n
hR(Yi; Yj); Yi); Y�i+
1
2
P
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i:
Substitute this in (6) and obtain (23).
If q = 1 (i.e., M is a hypersurface), then h[Yi; Yn+1]; Yn+1i = hYi; [Yn+1; Yn+1]i
vanishes for all 1 6 i 6 n, i.e., [Yi; Yn+1]
? = 0. It follows that (23) gives the
conditions Yj(nH) = 0, where H is the mean curvature function. This implies
the result from [4] cited in the introduction.
Denote by N the Lie algebra of N . It is well known (see, for example, [7,
Lem. 7.5]), that N is compact, i.e., N = Z � N 0, where the direct sum is or-
thogonal, Z is abelian, and N 0 = [N ;N ] is semisimple with the negative de�nite
Killing form.
Let M be totally geodesic submanifold of N , : M ! N the corresponding
immersion, and p an arbitrary point ofM . Consider the immersion 0 = L (p)�1 Æ
: M ! N . The image 0(p) coincides with the identity element e of the group.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 287
Ye.V. Petrov
The Gauss map of this immersion maps each point r 2M to the subspace �0(r) =
dL 0(p)�1 Æ d 0(TrM) = dL (p)�1 Æ d (TrM) = �(r), i.e., the Gauss maps of two
immersions are the same. Left translations are isometries of N , hence 0 is also
totally geodesic. Thus we can assume without loss of generality that (p) = e.
Then the tangent space TeM is a Lie triple system in N (see, for example, [6,
Th. 4.3 of Ch. XI]). The subspace N = TeM + [TeM;TeM ] is a compact Lie
subalgebra, therefore it has an orthogonal direct decomposition N = Z�N
0
with
abelian Z and semisimple N
0
= [N ;N ]. Take the decomposition Ya = Xa + Za
for 1 6 a 6 n + q, where Xa 2 N
0
, Za 2 Z. Then for 1 6 a; b 6 n + q the Lie
bracket [Ya; Yb] = [Xa;Xb]. Denote by W the subspace spanned by X1; : : : ;Xn
(i.e., the orthogonal projection of TeM to N
0
). It is a Lie triple system in N
0
, and
N
0
=W + [W;W]. The intersection W =W \ [W;W] is an ideal (from here by
ideals we mean the ideals in N ). The Lie algebra N
0
is semisimple, consequently
the orthogonal complement V toW is an ideal and it equals the orthogonal direct
sum
L
16l6m
Sl of simple ideals Sl.
Theorem 3. Let M be a smooth immersed totally geodesic submanifold in
a Lie group N with biinvariant metric. Then:
(i) if the restriction of the metric to V is a negative multiple of the Killing form
(in particular, if V is simple), then the Gauss map of M in this metric is
harmonic;
(ii) if W \ V =
L
16l6m
Wl, where Wl � Sl is a proper Lie triple system in Sl,
i.e., Wl 6= 0 and Wl 6= Sl for each 1 6 l 6 m (in particular, if V = 0), then
the Gauss map of M is harmonic in any biinvariant metric on N ;
(iii) if the condition of (ii) is not satis�ed, then there is a biinvariant metric on
N such that the Gauss map of M is not harmonic.
P r o o f. The conditions (23) for 1 6 j 6 n, n+1 6 � 6 n+ q take the formP
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i = 0: (24)
Also note that P
16i6n
h[Yi; Yj ]T ; [Yi; Y�]T i+
P
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i
=
P
16i6n
h[Yi; Yj ]; [Yi; Y�]i =
P
16i6n
h[[Yi; Yj]; Yi]; Y�i = 0
since the tangent space TeM is a Lie triple system. Hence the conditions (24) are
equivalent to P
16i6n
h[Yi; Yj ]T ; [Yi; Y�]T i = 0: (25)
288 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
Note that (24) and (25) can also be obtained directly from (22) using the fact
that TeM is a Lie triple system and the expression for the invariant Riemannian
connection.
The ideal W is semisimple, hence W = [W ;W] and
W = [W;W ] = [W ; [W ;W ]] � [W; [W;W]] = [TeM; [TeM;TeM ]] � TeM:
This implies that we can choose a frame of TeM such that Yi = Xi 2 W for
1 6 i 6 n1, where 0 6 n1 6 n, and Yi = Xi + Zi with Xi 2 fW = W \ V and
Z 2 Z for n1 + 1 6 i 6 n. For 1 6 j 6 n1 the equations in (24) becomeP
16i6n
h[Yi; Yj ]?; [Yi; Y�]?i =
P
16i6n1
h[Yi; Yj]?; [Yi; Y�]?i = 0
because [Yi; Yj ] 2 TeM for 1 6 i 6 n1. This yields that for to show a harmonicity
or non-harmonicity of the Gauss map at the point it su�ces to check (24) or (25)
for n1 + 1 6 j 6 n.
The subspace fW is a Lie triple system in a semisimple Lie algebra V, and
V = fW+[fW;fW ]. Moreover, fW\ [fW;fW ] = 0 because V is a direct complement to
W\ [W;W]. For each 1 6 l 6 m the restriction of the inner product to Sl is equal
to the Killing form multiplied by a negative constant: hX;Y i = �l Tr(adX Æad Y )
for X;Y 2 Sl, �l < 0 (See [7, Lem. 7.6]). Here by adX we mean the restriction of
the adjoint representation operator to the corresponding simple ideal. Denote by
Pl the orthogonal projection to Sl, then hX;Y i =
P
16l6m
�l Tr(adPl(X)Æad Pl(Y ))
for X;Y 2 V.
For each 1 6 l 6 m the operator Pl is a Lie algebra homomorphism, therefore
Pl([fW ;fW ]) = [Pl(fW); Pl(fW)] and
Sl = Pl(V) = Pl(fW + [fW ;fW]) = Pl(fW) + [Pl(fW); Pl(fW)]:
The intersection Pl(fW) \ [Pl(fW); Pl(fW)] is an ideal in simple Sl. Hence either
Pl(fW)\[Pl(fW); Pl(fW)] = 0 or Sl = Pl(fW) = [Pl(fW); Pl(fW)]. In the �rst case the
operators adX for X 2 Pl(fW) map Pl(fW) to [Pl(fW); Pl(fW)] and [Pl(fW); Pl(fW)]
to Pl(fW). The operators ad Y for Y 2 [Pl(fW); Pl(fW)] map the subspaces Pl(fW)
and [Pl(fW); Pl(fW)] to themselves. It follows that hPl(fW); [Pl(fW); Pl(fW)]i = 0.
If the restriction of the metric to V is a negative multiple of the Killing form
(the case of (i)), then the same argument shows that hfW ; [fW;fW ]i = 0.
Consider the case Pl(fW) \ [Pl(fW); Pl(fW)] = 0 for all 1 6 l 6 m. We proved
that hPl(fW); [Pl(fW); Pl(fW)]i = 0 for all l, thus hfW ; [fW ;fW ]i = 0. For each
1 6 l 6 m denote Pl(fW) by Wl. Then hWl; [fW ;fW ]i = hWl; [Wl;Wl]i = 0,
hence Wl is contained in the orthogonal complement of [fW;fW ], i.e., in fW; and
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 289
Ye.V. Petrov
fW =
L
16l6m
Wl. Subspaces Wl are Lie triple systems, Wl 6= 0 because in the
opposite case [Wl;Wl] = 0 and Sl = 0. It contradicts the fact that Sl is simple.
If Wl = Sl, then Wl = [Wl;Wl] because Sl is simple, that is a contradiction.
It follows that Wl 6= Sl. This is the case of (ii). It is easy to see also that the
condition in (ii) implies Pl(fW) \ [Pl(fW); Pl(fW)] = 0 for all 1 6 l 6 m. In fact,
if fW =
L
16l6m
Wl with Wl � Sl, then Pl(fW) = Wl, [Pl(fW); Pl(fW)] = [Wl;Wl],
therefore the case Sl = Pl(fW) = [Pl(fW); Pl(fW)] is excluded by the condition
Wl 6= Sl.
Assume that hfW ; [fW ;fW]i = 0. Take any n1 + 1 6 j 6 n. For 1 6 i 6 n1
[Yi; Yj] = 0 and for n1 + 1 6 i 6 n
[Yi; Yj ]
T =
P
16k6n
h[Yi; Yj ]; YkiYk =
P
n1+16k6n
h[Xi;Xj ];XkiYk = 0
because Xi 2 fW for n1 + 1 6 i 6 n. This yields that (25) is satis�ed. We proved
(i) and (ii).
Finally, in the case (iii) there is 1 6 l0 6 m such that Sl0 = Pl0(
fW) =
[Pl0(
fW); Pl0(
fW)]. Consider the new metric h�; �i0 such that it is equal to h�; �i on
the orthogonal complement to V and
hX;Y i0 =
P
16l6m
(�Tr(adPl(X) Æ adPl(Y )))� �2 Tr(adPl0(X) Æ adPl0(Y ))
for X;Y 2 V, where � 6= 0. It is a biinvariant metric. Denote �Tr(adPl0(X) Æ
adPl0(Y )) by hX;Y i
00.
The ideal Sl0 is not contained in fW because in the opposite case it is contained
also in fW\ [fW ;fW] = 0, that is a contradiction. It follows that there is a vector Y
orthogonal to fW such that Pl0(Y ) 6= 0. Then Sl0 = Pl0(
fW) implies that there is
a vector X 2 fW such that Pl0(X) = Pl0(Y ). Note that Y is orthogonal to TeM .
We can consider that the norm of Y equals 1. Choose the orthonormal frames of
TeM and NeM such that Yj0 = X + Zj0 for some n1 + 1 6 j0 6 n and Y�0 = Y
for some n + 1 6 �0 6 n + q. Then the discussion above implies that for any
n1 + 1 6 i 6 n in the new metric
[Yi; Yj0 ]
T =
X
16k6n
h[Yi; Yj0 ]; Yki
0Yk =
X
n1+16k6n
h[Xi;X];Xki
0Yk
= �2
X
n1+16k6n
h[Pl0(Xi); Pl0(X)]; Pl0 (Xk)i
00Yk
= ��2
X
n1+16k6n
h[Pl0(Xi); Pl0(Xk)]; Pl0(X)i00Yk:
290 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
There is some n1 + 1 6 i 6 n such that this expression does not vanish because
Sl0 = [Pl0(
fW); Pl0(
fW)]. Similarly,
[Yi; Y�0 ]
T = ��2
P
n1+16k6n
h[Pl0(Xi); Pl0(Xk)]; Pl0(Y )i00Yk:
The expression in (25) for j = j0 and � = �0 thus becomes
�4
P
n1+16i;k6n
(h[Pl0(Xi); Pl0(Xk)]; Pl0(X)i00)2 6= 0:
Therefore the Gauss map is not harmonic.
A Lie triple system U is reducible if U = U1�U2, where U1 and U2 are nonzero
Lie triple systems such that [U1;U2] = 0, and is irreducible otherwise (see, for
example, App. 1 of [3]). Theorem 3 then implies that if fW is irreducible and V is
not simple, then there is a biinvariant metric on N such that the Gauss map of
M is not harmonic.
Consider an example. Let N be so(3)� so(3) with the orthogonal basis con-
sisting of the vectors e1; e2; e3; f1; f2; f3 with the nonzero brackets
[e1; e2] = �[e2; e1] = e3; [e2; e3] = �[e3; e2] = e1; [e3; e1] = �[e1; e3] = e2;
[f1; f2] = �[f2; f1] = f3; [f2; f3] = �[f3; f2] = f1; [f3; f1] = �[f1; f3] = f2:
Let W be the subspace spanned by e1 + f1, e2 � f2, and e3 + f3. Let M be
exp(W), hence TeM = W. The bracket [W;W] is spanned by e1 � f1, e2 + f2,
and e3 � f3. It is easy to see that W is a Lie triple system. In our notation,
N = N = N
0
= V = W + [W;W]. The intersection W = W \ [W;W] vanishes,
therefore fW = W. Choose a metric such that hei; eji = Æij and hfi; fji = Æija
2,
where 0 < a 6= 1, then W and [W;W] are not orthogonal. The orthonormal
frames of the tangent and the normal spaces of M can be chosen in the following
way:
Y1 =
1p
1+a2
(e1 + f1); Y2 =
1p
1+a2
(e2 � f2); Y3 =
1p
1+a2
(e3 + f3);
Y4 =
p
1+a2
a
�
e1 �
1
a2
f1
�
; Y5 =
p
1+a2
a
�
e2 +
1
a2
f2
�
; Y6 =
p
1+a2
a
�
e3 �
1
a2
f3
�
:
Compute (25), e.g., for j = 1 and � = 4:P
16i63
h[Yi; Y1]T ; [Yi; Y4]T i
= 1
a(1+a2)
h(�e3 + f3)
T ;
�
�e3 �
1
a2
f3
�T
i+ 1
a(1+a2)
h(e2 + f2)
T ;
�
e2 �
1
a2
f2
�T
i
=
�2(�1+a2)
a(1+a2)2
+
2(1�a2)
a(1+a2)2
6= 0:
It follows that the Gauss map is not harmonic.
Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 291
Ye.V. Petrov
4. 2-Step Nilpotent Groups and Geodesics
Recall that a Lie group N is 2-step nilpotent if and only if its Lie algebra
N is 2-step nilpotent, i.e., [N ;N ] 6= 0, [[N ;N ];N ] = 0. In other words,
0 6= [N ;N ] � Z, where Z is the center of N . Consider a 2-step nilpotent Lie
group N with left invariant metric induced by an inner product h�; �i on N as
above. Denote by V the orthogonal complement to Z in N . For each Z 2 Z
de�ne a linear operator J(Z) : V ! V by hJ(Z)X;Y i = h[X;Y ]; Zi for all X, Y
from V. All J(Z) are skew-symmetric. The group N and the Lie algebra N are
called nonsingular if for each Z 6= 0 the operator J(Z) is nondegenerate.
The left invariant Riemannian connection is de�ned by (see [1])
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:
(26)
Let us study whether the Gauss map of a totally geodesic submanifold M
in N is harmonic. It was proved in [2, Th. (4.2)] that if N is simply connected
and nonsingular, then a totally geodesic submanifold either have the Gauss map
of maximal rank at any point or it is a left translation of some open subset
in a totally geodesic subgroup. The latter case takes place for many classes of
submanifolds, for example, for all totally geodesic M such that dimM > dimZ
in 2-step nilpotent groups N with dimN > 3 (see [2, Cor. (5.6)]). The structure
of the corresponding subgroups (or their Lie algebras) is also described in [2] (and
allows to prove, for example, that there are no totally geodesic hypersurfaces in
nonsingular 2-step nilpotent Lie groups, see [2, Cor. (5.8)]). Anyway, in this case
the Gauss map is constant, thus harmonic. Therefore it su�ces to consider the
case of the Gauss map with maximal rank. For n = dimM = 1, i.e., for geodesics,
the answer can be found in the next statement:
Proposition 4. A smooth geodesic in a 2-step nilpotent group has the har-
monic Gauss map if and only if it is a left translation of some one-parameter
subgroup.
P r o o f. The "if" part is clear, let us prove the "only if" one. Taking if
necessary a left translation we can think that our geodesic contains the identity e
of N (similarly to the discussion in the previous section). Decompose its tangent
vector at e as X + Z, where X 2 V and Z 2 Z. The condition (22) with n = 1,
j = 1, and Y1 = X + Z becomes
0 =
�
[X + Z;�J(Z)X] + 2rX+Z (J(Z)X)
?
�?
=
�
[J(Z)X;X] + [X;J(Z)X] � J(Z)2X
�?
= �
�
J(Z)2X
�?
:
292 Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2
Submanifolds with the Harmonic Gauss Map in Lie Groups
Here we used the de�nition of the 2-step nilpotent Lie algebra, the equations
(26), and the fact that hJ(Z)X;X + Zi = hJ(Z)X;Xi = 0 because J(Z) is
skew-symmetric, therefore (J(Z)X)
?
= J(Z)X. The conditions mean J(Z)2X =
�(X + Z), where � 2 R. Thus �Z = 0, hence Z = 0 or � = 0, in any case
J(Z)2X = 0. This yields 0 = hJ(Z)2X;Xi = �jJ(Z)Xj2, therefore J(Z)X
vanishes. Then Proposition (3.5) of [1] implies that the geodesic is de�ned by the
formula exp(t(X + Z)). This gives us the desired result.
Actually, the proof implies that the geodesic is a left translation of one-
parameter subgroup if the Gauss map is harmonic only at some point. Anyway,
it follows that for n = 1 the Gauss maps of maximal rank are not harmonic. It is
interesting to check whether the similar statement is true for other values of n.
References
[1] P.B. Eberlein, Geometry of 2-Step Nilpotent Groups with a Left Invariant Metric.
� Ann. Sci. �Ecole Norm. Sup. 27 (1994), 611�660.
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Journal of Mathematical Physics, Analysis, Geometry, 2008, vol. 4, No. 2 293
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| id | nasplib_isofts_kiev_ua-123456789-106507 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1812-9471 |
| language | English |
| last_indexed | 2025-12-07T18:54:42Z |
| publishDate | 2008 |
| publisher | Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| record_format | dspace |
| spelling | Petrov, Ye.V. 2016-09-29T18:09:48Z 2016-09-29T18:09:48Z 2008 Submanifolds with the Harmonic Gauss Map in Lie Groups / Ye.V. Petrov // Журнал математической физики, анализа, геометрии. — 2008. — Т. 4, № 2. — С. 278-293. — Бібліогр.: 10 назв. — англ. 1812-9471 https://nasplib.isofts.kiev.ua/handle/123456789/106507 In this paper we find a criterion for the Gauss map of an immersed smooth submanifold in some Lie group with left invariant metric to be harmonic. Using the obtained expression we prove some necessary and sufficient conditions for the harmonicity of this map in the case of totally geodesic submanifolds in Lie groups admitting biinvariant metrics. We show that, depending on the structure of the tangent space of a submanifold, the Gauss map can be harmonic in all biinvariant metrics or nonharmonic in some metric. For 2-step nilpotent groups we prove that the Gauss map of a geodesic is harmonic if and only if it is constant. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии Submanifolds with the Harmonic Gauss Map in Lie Groups Article published earlier |
| spellingShingle | Submanifolds with the Harmonic Gauss Map in Lie Groups Petrov, Ye.V. |
| title | Submanifolds with the Harmonic Gauss Map in Lie Groups |
| title_full | Submanifolds with the Harmonic Gauss Map in Lie Groups |
| title_fullStr | Submanifolds with the Harmonic Gauss Map in Lie Groups |
| title_full_unstemmed | Submanifolds with the Harmonic Gauss Map in Lie Groups |
| title_short | Submanifolds with the Harmonic Gauss Map in Lie Groups |
| title_sort | submanifolds with the harmonic gauss map in lie groups |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/106507 |
| work_keys_str_mv | AT petrovyev submanifoldswiththeharmonicgaussmapinliegroups |