210

Revista de la Union Matematica Argentina Volumen 38, 1993.

A CLASS OF MINIMAL SUBMANIFOLDS IN A 2-STEP NILPOTENT LIE GROUP Carlos J. Ferraris

ABSTRACT

We study the existence of minimal submanifolds in a simply connected 2-step nilpotent Lie group N with a left invariant metric, the main result being that everyone-parameter subgroup of N lies in at least one 2-dimensional connected subgroup which is a minimal submanifold of N. A similar result is also obtained for a class of solvable Lie groups with a left invariant metric.

INTRODUCTION

This article deals with the existence of minimal submanifolds ina 2-step nilpotent Lie group N with a left invariant metric. In Section 1, we introduce the notation and recall some of the concepts to be used in the sequel. In Section 2, we derive the existence of minimal submanifolds uaturally associated to one-parameter subgroups of N. In Section 3, we obtain a similiar result for a class of solvable Lie groups (semi-direct product of R and a 2-step nilpotent Lie group) with a left invariant metric.

211

1. MINIMAL SUBMANIFOLDS IN A LIE GROUP

Let (G, ( , ) be a connected Lie group with a left invariant metric ( , ) and let (9, ( , ) denote its Lie algebra with the inner product ( ,). The Levi-Civita connection for ( G, ( , ) can be expressed in terms of 9 by the formula

'V X Y = 1/2([X, Y] - ad

(1) where

x- adyX),

X, Y E g,

* stands for the adjoint relative to the inner product

( , ) on g, cf. [2].

DEFINITION 1.1. A submanifold M of (G, ( , )) is called a minimal submanifold if trace 11M == 0, where 11M stands for the second fundamental form of Min (G, ( , ). Let H be a connected Lie subgroup of G and let H be its Lie sub algebra.

PROPOSITION 1.2. A connected Lie subgroup H of G is a minimal submanifold

L: 'VXiXi

of (G, ( , ) if and only if

(H, ( , ).

(L: 'VX.Xi

belongs to H, for an orthonormal basis (X;) of

is independent of the orthonormal basis (Xi».

i

PROOF: It follows from the formula X,YEg

(where 'V 1l indicates the covariant derivative in H) and the fact that trace lIn

= LIIH(Xi,X;l i

for an orthonormal basis {Xi) of (9, ( , ). The remark tha.t

L: 'VX.Xi

is independent of the orthonormal basis (Xi) follows from

the properties of the covariant derivative and the fact that if (l'j ) is another orthonormal basis, then l'j =

L: aij Xj,

where (aij) is an orthogonal matrix, and hence

i

= Lbik'VX.Xk = L'VX.Xi. i,k

i

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2. 2-STEP NILPOTENT LIE GROUPS Let N be a real finite dimensional 2-step nilpotent Lie algebra, i.e. 0 -=f. [N,N] where Z is.the center of N. If ( , ) is a positive definite inner product on

where V

c

Z,

N, then

= Z.l = orthogonal complement of Z.

For every Z E Z, there is a skew-symmetric operator j(Z) on V defined by (j(Z)V, W) = ([V, W], Z),

Z E Z, V, WE V.

Let (N, ( , ) denote the simply connected 2-step nilpotent Lie group with Lie algebra

N end left invariant metric ( , ). From formula (1) follows:

i)

\1vW =:= 1/2 [V, W], V, WE V,

ii)

\1 z Z' = 0, Z,Z' E Z,

iii)

\1vZ = \1zV = -1/2 j(Z)V, Z E Z, V E V.

THEOREM 2.1. Let (N, ( , ) be a simply connected 2-step nilpotent Lie group with a left invariant metric ( , ). Every one-parallleter tiubgl'oup of N lies in at least one 2-dimensional connected (abelian) subgroup H which is a minimal submanifold of (N,{, ).

PROOF: We consider first the generic case, i.e., let X E N, with X

=

V

+ Z,

0 -=f.

V E V and 0 -=f. Z E Z. In this case, we take H = span {V, Z} and we prove that the

connected subgroup H = exp(H), which

contaill~

the one-parameter. subgroup exp tX,

is a minimal submanifold of (N, ( , ). In fact, by applying formulas (i) and (ii) to the orthonomlal basis {V/IIVII' Z/llzll} of (H, ( , ) we obtain

\1 V/ 11 1'1I V/IIVII

+ \1 z/llzIIZ/llzll =

O.

Hence H = exp(H) is a minimal submanifold of (N, ( , ) by Proposition 1.2. In the case

V = 0 or Z = 0 we just take a non zero vector ei ther in V or Z, to obtain a 2-dimensional minimal subgroup H of (N, ( , ). This concludes the proof of the theorem.

213

REMARK. We observe that if H

o -#

= exp(1t)

= span {V, Z}, 0 -# V i.e., j2(Z) = -IIZII2 Id.,

for 1t

Z E Z and (N, ( , ) is of Heisenberg type,

Z E Z, cf. [3] and [5], or more generally nonsing-ular adx: N

--+

E V and for every

Z is surjective for

every X E N - Z} then th~ subgroup H is a minimal submanifold which is neither totally geodesic ('V X Y E 1t, -for every X, Y E 1t) nor fiat (the sectional curvature

K(V, Z) = 0). In fact,

0-# 'VvZ =

...,.1/2j(~)V

tf. 1t

and a simple calculation shows that

K(V,Z) =

~ IU(Z)VW

if V and Z are unit vectors.

The property of Theorem 2.1 is not shared for every 3-step nilpotent Lie group with a left invariant metric as it is shown by the following example. Let N be the Lie algebra of matrices

}f~q~

Xl

YI

0 0 0

X2

~l

l'

0 0

3

Xl, X2, X3, YI , Y2, Z

,

E

R}

0

N is a 3-step nilpotent Lie algebra aud if we set

X,

Y,

-

we have

[! !l ' !l ' ~ [! !l ' ~ [~ ~l ' ,~ [~ ~l ~d ~ [~ ~l ~

L 0 0, 0 o· 0 0 0

0 0

0 0 0 0

1 0 0 0

_

){2

y,

- 0

=

[0~

0 0 0 0

0 0 0 0

0 1 0 0

0 0 0 0

X,

z

0 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

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We give to N the inner procitlct ( , ) such that the vectors {Xl, X 2 , X 3 , Yl , Yz , Z} form an orthonormal basis. Let (N, ( , )) denote the simply connected 3-step nilpotent Lie group with Lie algebra N anclleft invariant metric ( , ) We show there is a vector inN for which the corresponding one-parameter subgroup cannot lie in any 2-dimensional minimal subgroup of (N, ( , ). We consider V = 1/ J3 (X3 +1'2+Z) and we shall prove that the one-parameter subgroup exp tV cannot lie in any 2-dimensional minimal subgroup H of (N, ( , ). In fact, let us assume that H parameter subgroup exp and [U, Vl

=

nT,

= exp(H)

is a 2-dimensional subgroup containing the one-

then H must cCllltaill aunit vector U such that (U, V) = 0

0, the latter is due to the fact that H must be abelian since it is a

2-climensional nilpotent Lie algebra. We shall show that H cannot be minimal. Let

smce

this implies

02

= 0 and 1'1

On the other hand

and hence

Therefore

and

+n

j

= O.

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From now on we indicate a = a3 and (3 = (32, wi"th this notation a computation using formula (1) gives us

+ "VvV =

"VuU

+ (3)]Xl + [1/3 + (a~ + a(3)]X2 - al(a + (3)X3 + [1/3 - a(a + (3)]Yl + + al(a + (3)Y2 .

[1/3 - (3(a

We have two cases to be considered, namely.

CASE 1: al= O. We claim that "V uU

+ "V V V =I O.

In fact, if it were equal to zero,

then 1/3 - (3(a

+ (3) =

1/3 + a(3 = 1/3 - a(a

+ (3) = o.

From this follows easily that a 2 = (32 =2/3. On the other hand

and by substituting the values of a 2 and (32, we would have that the .fact that U is a unit vector. It is also clear that "V uU

IIUI1 2 =

+ "V v V

2, contradicting

¢. 'H.

CASE 2: al =I 0: We have two subcases: i) ii)

+ (3 = 0, and then it follows immediately that "VuU + "VvV ¢.'H. a + (3 =I 0, then "VuU + "V v V =I o. We claim that "VuU + "V v V ¢. 'H,

a

if it were not

so, then there would exist constants A and B such that "VuU

+ "VvV =

AU

+ BV.

Now by looking at the components in the X 3 , Y2 and Z directions we would have

,

A = B = 0, contradicting the fact that "V u U + "V v V

=I O. This concludes our assertion.

REMARK. It seems quite reasonable to conjecture that the property of Theorem 2.1 characterizes thf; 2-step nilpotent Lie groups in the class of nilpotent Lie groups with a left invariant metric.

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3. SOLVABLE LIE GROUPS Let (N, ( , )) be a 2-step nilpotent Lie algebr'a with an inner product ( , ) as in Section 2.

We define

by

,\ E R, ,\

adAV

= ,\V,

adAZ

= 2,\Z,

V E V, Z E Z,

i= O.

S becomes a solvable Lie algebra and we define an inner product on S. by assuming that

IIAII =

1 and A is orthogonal do

N.

Let (S, ( , )) denote the simply connected solvable Lie group 'with Lie algebra Sand left invariant metric ( , ). We have the following formulas for the covariant derivative

\7 AB = O. \7 v A

= -/\V;

\7AV = 0; \7 zA

\7AZ = 0;

= -2'\Z;

\7vW

= 1/2 [V, ~'V] +,\ (V, W) A;

\7 zZ'

= 2,\ (Z, Z') A;

\7 v Z

= \7zV = -1/2j(Z)V

where B ERA, V, 'IV E V, Z, Z' E Z. For more details on the geometry of (S, ( , )), see [1] and [4].

THEOREM 3.1. Let (S, ( , )) be as above. Every one-parameter subgroup of S lies in at least one 3-dimensional connected subgroup H which is a minimal submanifold of

(S, ( , )).

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PROOF: We just consider the generic case, that is, X = a A + V + Z, a

= span {A., V, Z} and H = exp(H), By (2) to the orthonormal basis {A, VIIIVII, Z/IIZII} of (H, ( , )) we get

and 0

=f

Z E Z. We take

7-{

=1=

0, 0

=1=

V EV

appling formulas

\7 AA + \7v/IIVIIVIIIVII + \7 z/llzllZ/IIZII = = >. (VIIIVII, VIIIVII) A + 2>' (Z/IIZII, Z/IIZII) A = = 3>'A E H.

This shows that H = exp(H) is a minimal submanifold of (S, ( , )) by Proposition 1.2.

= \7 Z V =

REMARK. If (N, ( , )) is of Heisenberg type or nonsingular then the subgroup H

exp(H) of Theorem 3.1 is minimal without being totally geodesic, in fact, -1/2j(Z)V

i

H.

ACKNOWLEDGMENT. The author wishes to thank the referee for pointing out a fact that simplified the proof of Theorems 2.1 and 3.1.

218

REFERENCES [1] Boggino, J., Generalized Heisenberg groups and solvmanifolds naturally associated, Rend. Sem. Mat. Univ. Politec. Torino 43 (1985), 529-547. [2] Cheeger, J. and Ebin, D., Compa'l'ison' Theo'l'ems in Riemannian Geometry, Nort.h Holland, Amsterdam, 1975. [3] Eberlein, P., Geometry of 2-step nilpotent groups (Preprint). [4] Ferraris, C.J., Geometry of Cm'not solvmanifolds (to appear in Rend. Sem. Mat. Univ. Politec, Torino). [5] Kaplan, A., Riemannian manifolds attached to Clifford modules, Geom. Dedicata 11 (1981), 127--136. Carlos J. Ferrari::; CFM - Depto. de I\1atemchica Universidade Federal de Santa Catarina Campus Universit1:1.rio- Trinclade CEP 88049 Florianopolis, SC Brasil

Recibido en abril de 1993. Version modificada en junio de 1993.