Foliations on double-twisted products

0.1 MB·

by  André Gomes

FOLIATIONS ON DOUBLE-TWISTED PRODUCTS
ANDRE´ GOMES
1
1
0 Abstract. In this paper we present a certain class of geodesic vector fields
2 of the double-twisted product R×(f,g)R. Some examples of totally geodesic
n foliationsaregiven.
a
J
9 1. Introduction
2
The conceptofdouble-twistedproductwas introducedby Pongeand Reckziegel

(see [5]) in the context of the semi-Riemannian Geometry.
]
G
Definition 1. Let (M ,g ) and (M ,g ) be (pseudo)Riemannian manifolds. Let
1 1 2 2
D λ : M M R (i = 1,2) be a positive and differentiable function. Consider
i 1 2
× −→
. the canonical projections π : M M M for i = 1,2. Then the double-
h i 1 2 i
× −→
at twistedproductM1×(λ1,λ2)M2 of(M1,g1)and(M2,g2)isthedifferentiablemanifold
M M equipped with the (pseudo)Riemannian metric g defined by
m 1× 2
g(X,Y)=λ g (dπ (X),dπ (Y))+λ g (dπ (X),dπ (Y))
[ 1 1 1 1 2 2 2 2

for all vector fields X and Y of M M .
1 1× 2
v
0 This definition generalizes the R. L. Bishop’s notion of an umbilic product (de-
3 noted by M M ), which B. Y. Chen calls a twisted product and which is the
1 λ 2
7 ×
double twisted product M M (see [1] and [2]). If in this situation λ only
5 1×(1,λ) 2
depends on the points of M , then M M is a warped product by definition.
1. 1 1×λ 2
0
If M and N are complete Riemannian manifolds, then the double-twisted prod-
1
uct M N is not, in general,a complete Riemannian manifold. But, for our
1 ×(λ1,λ2)
: purposes,the followingLemma is enough(the proofis similarofthat ofLemma40
v
in [4])
i
X
Lemma1. SupposethatM andN arecompleteRiemannianmanifolds. Giventwo
r differentiable functions λ ,λ : M N [1, ), then the double-twisted product
a 1 2
× −→ ∞
M N is a complete Riemannian manifold.
×(λ1,λ2)
Proof. Denote byπ :M N M andσ :M N N the naturalprojections.
× −→ × −→
Denote by and d the arc lenght and the distance in M N, respectively.
L ×(λ1,λ2)
We use the metric completeness criterion from the Hopf-Rinow theorem. Note
first that if v is tangent to M N then, since λ 1 and λ > 0, we
×(λ1,λ2) 1 ≥ 2
have v,v dπ(v),dπ(v) . Hence (α) (π α) for any curve segment α in
h i ≥ h i L ≥ L ◦
M N. Analogously, (α) (σ α). Then we have the inequalities
×(λ1,λ2) L ≥L ◦
d(x,y) d(π(x),π(y)) and d(x,y) d(σ(x),σ(y)), x,y M N
≥ ≥ ∀ ∈ ×(λ1,λ2)
Both inequalities implies that if ((p ,q )) is a Cauchy sequence in the double-
k k k N
twisted product M ×(λ1,λ2) N, then (pk)k∈∈N and (qk)k∈N are Cauchy sequences,
respectively,in(M,g1)and(N,g2). AsM andN arecomplete,(pk)k N and(qk)k N
are convergent and then ((p ,q )) is convergent in M N∈(cid:3) ∈
k k k∈N ×(λ1,λ2)
1
2 ANDRE´ GOMES
2. Preliminaries
Some computations are necessary in order to find the examples. We denote by
∂ and ∂ the canonical vector fields of R2. The new metric in R2 is given by
x y
∂ ,∂ =[f(x,y)]2
h x xi(x,y)

∂ ,∂ =0
 h x yi(x,y)
∂ ,∂ =[g(x,y)]2
 h y yi(x,y)
where (x,y) R2 and f,g : R2 (0,+ ) are functions of class . The
∞
∈ −−→ ∞ C
real plane equipped with this (Riemannian) metric is the double twisted product of
R andRanddenotedbyR R. Wefindnowitsconnection andcurvatureR.
(f,g)
× ∇
By metric compatibility we know that
2ff = ∂ ∂ ,∂ =2 ∂ , ∂
x xh x xi h x ∇∂x xi
2ff = ∂ ∂ ,∂ =2 ∂ , ∂
y yh x xi x ∇∂y x
2gg = ∂ ∂ ,∂ =2 ∂ , ∂
x xh y yi h(cid:10) y ∇∂x yi(cid:11)
2gg = ∂ ∂ ,∂ =2 ∂ , ∂
y yh y yi y ∇∂y y
0 = ∂ ∂ ,∂ = ∂ ,∂ + ∂ , ∂
xh x yi h∇(cid:10)∂x x yi(cid:11) h x ∇∂x yi
0 = ∂ ∂ ,∂ = ∂ ,∂ + ∂ , ∂
yh x yi ∇∂y x y x ∇∂y y
∂ = ∂
∇∂x y ∇∂y x (cid:10) (cid:11) (cid:10) (cid:11)
where in the last equation we used [∂ ,∂ ]=0.
x y
Using the equations above, we find the Riemannian connection of R R:
(f,g)
∇ ×
1 1
∂ = ∂ ,∂ ∂ + ∂ ,∂ ∂
∇∂x x f2 h∇∂x x xi x g2 h∇∂x x yi y
f ff
x y
= ∂ ∂
f x− g2 y
1 1
∂ = ∂ ,∂ ∂ + ∂ ,∂ ∂
∇∂y y f2 ∇∂y y x x g2 ∇∂y y y y
gx(cid:10)g gy (cid:11) (cid:10) (cid:11)
= ∂ + ∂
− f2 x g y
1 1
∂ = ∂ ,∂ ∂ + ∂ ,∂ ∂
∇∂y x f2 ∇∂y x x x g2 ∇∂y x y y
f (cid:10) g (cid:11) (cid:10) (cid:11)
y x
= ∂ + ∂
x y
f g
FOLIATIONS ON DOUBLE-TWISTED PRODUCTS 3
Moreover,by definition we have
R(∂ ,∂ )∂ = ∂ ∂ ∂
x y x ∇∂y∇∂x x−∇∂x∇∂y x−∇[∂x,∂y] x
= ∂ ∂
∇∂y∇∂x x−∇∂x∇∂y x
f ff f g
x y y x
= ( ∂ ∂ ) ( ∂ + ∂ )
∇∂y f x− g2 y −∇∂x f x g y
f f ff ff
x x y y
= ∂ +( ) ∂ ∂ ( ) ∂
f ∇∂y x f y x− g2 ∇∂y y− g2 y y
f f g g
y y x x
( ∂ +( ) ∂ + ∂ +( ) ∂ )
− f ∇∂x x f x x g ∇∂x y g x y
f f g f ff g g g ff
x y x x y x y y
= ( ∂ + ∂ )+( ) ∂ ( ∂ + ∂ ) ( ) ∂
f f x g y f y x− g2 − f2 x g y − g2 y y
f f ff f g f g g
y x y y x y x x
( ∂ ∂ )+( ) ∂ + ( ∂ + ∂ )+( ) ∂
− f f x− g2 y f x x g f x g y g x y
(cid:20) (cid:21)
f2 g2 f g ff g ff g
= y x + x x y y ( y) ( x) ∂
"g2 − g2 fg − g3 − g2 y − g x# y
and then
f2 g2 f g ff g ff g
R(∂ ,∂ )∂ ,∂ =g2 y x + x x y y ( y) ( x)
h x y x yi "g2 − g2 fg − g3 − g2 y − g x#
Analogously, we have
R(∂ ,∂ )∂ = ∂ ∂ ∂
x y y ∇∂y∇∂x y−∇∂x∇∂y y−∇[∂x,∂y] y
= ∂ ∂
∇∂y∇∂x y−∇∂x∇∂y y
f g g g g
y x x y
= ( ∂ + ∂ ) ( ∂ + ∂ )
∇∂y f x g y −∇∂x − f2 x g y
f f g g
y y x x
= ∂ +( ) ∂ + ∂ +( ) ∂
f ∇∂y x f y x g ∇∂y y g y y
gg gg g g
x x y y
+ ∂ +( ) ∂ ∂ ( ) ∂
f2 ∇∂x x f2 x x− g ∇∂x y− g x y
f f g f g g g g g
y y x y x x y x
= ( ∂ + ∂ )+( ) ∂ + ( ∂ + ∂ )+( ) ∂
f f x g y f y x g − f2 x g y g y y
gg f ff gg g f g g
x x y x y y x y
+ ( ∂ ∂ )+( ) ∂ ( ∂ + ∂ ) ( ) ∂
f2 f x− g2 y f2 x x− g f x g y − g x y
(cid:20) (cid:21)
f2 g2 gf g g f f gg
= y x + x x y y +( y) +( x) ∂
"f2 − f2 f3 − fg f y f2 x# x
3. Geodesic vector fields
A codimension-one foliation on a Riemannian manifold M is totally geodesic
F
if their leaves are totally geodesic submanifolds of M. If an unit vector field
U =a(x,y)∂ +b(x,y)∂
(x,y) x y
4 ANDRE´ GOMES
defined on R R is geodesic, then its integral curves determines a totally
(f,g)
geodesic foliati×on on R R. The vector field U is geodesic U =0. But
(f,g) U
× ⇔ ∇
U= a U +b U
∇U ∇∂x ∇∂y
= a( (a∂ +b∂ ))+b (a∂ +b∂ )
∇∂x x y ∇∂y x y
= a(a∇∂x∂x+ax∂x+b∇(cid:0)∂x∂y+bx∂y)+b(cid:1)a∇∂y∂x+ay∂x+b∇∂y∂y+by∂y
f ff f g
x y y (cid:0) x (cid:1)
= a a( ∂ ∂ )+a ∂ +b( ∂ + ∂ )+b ∂
f x− g2 y x x f x g y x y
(cid:18) (cid:19)
f g gg g
y x x y
+b a( ∂ + ∂ )+a ∂ +b( ∂ + ∂ )+b ∂
f x g y y x − f2 x g y y y
(cid:18) (cid:19)
f g g f
= a2 x b2 x +2ab y +aa +ba ∂
f − f2 f x y x
(cid:18) (cid:19)
ff g g
+ a2 y +b2 y +2ab x +ab +bb ∂
− g2 g g x y y
(cid:18) (cid:19)
Therefore, the condition U =0 is equivalent to the following system
U
∇
a2fx b2gxg +2abfy +aa +ba =0
f − f2 f x y

 −a2fgf2y +b2ggy +2abggx +abx+bby =0
Theorem 1. Consider the following unit vector field defined on R (f,g)R
×
1 1
U := ∂ + ∂
(x,y) √2f(x,y) x √2g(x,y) y
Then the vector field U above is an unit geodesic vector field on R R iff
(f,g)
×
f =g
y x
Proof. In this case we have a = 1/√2f and b = 1/√2g. Substituting this in the
above system we obtain
1 fx 1 gxg + 1 fy + 1 ∂ ( 1 )+ 1 ∂ ( 1 )=0
2f2 · f − 2g2 · f2 fg · f √2f · ∂x √2f √2g · ∂y √2f

 −2f12 · fgf2y + 2g12 · ggy + f1g · ggx + √12f · ∂∂x(√12g)+ √12g · ∂∂y(√12g)=0
Aftersimplifications, both equations are equivalent to the equation fy =gx (cid:3)
Example. If f(x,y) := ex+y +sin2(x)+1 and g(x,y) := ex+y +1, then f = g
y x
and the following unit vector field of R R
(f,g)
×
1 1
U := ∂ + ∂
(x,y) √2(ex+y+sin2(x)+1) x √2(ex+y +1) y
is a geodesic vector field.
4. Foliations with geodesic normal vector field
We recall some results about totally geodesic foliations. Let Mn+1 be an ori-
entable Riemannian manifold and be a codimension-one foliation on M.
∞
F C −
Suppose that is transversely orientable, i.e., we may choose a differentiable unit
vector field NF X(M) normal to the leaves of .
∈ F
FOLIATIONS ON DOUBLE-TWISTED PRODUCTS 5
Given a point p M we may always choose an orthonormalframe field
∈
e ,e ,...,e ,e
1 2 n n+1
{ }
defined in a neighborhood of p and such that the vectors e ,...,e are tangent to
1 n
the leaves of and e =N. The Ricci curvature in the direction e =N is
n+1 n+1
F
n
1
Ric(N):= R(e ,N)e ,N
i i
n h i
i=1
X
The divergence of a vector field V X(M) is locally defined as
∈
n+1
div(V):= V,e
h∇ek ki
k=1
X
In [3] the Author proved the following
Theorem2. Let beacodimension-onefoliation ofacompleteRiemannianman-
F
ifold Mn and let N be an unit vector field normal to the leaves of . Then
F
1
is totally geodesic Ric(N) div( N)=0
N
F ⇐⇒ − n ∇
As a consequence we obtain
Corollary 1. Let be a codimension-one foliation of a complete Riemannian
F
manifold M and let N be a geodesic unit vector field normal to the leaves of .
F
Then is a totally geodesic foliation if, and only if, Ric(N)=0.
F
The following theorem gives us some examples of such foliations
Theorem 3. Let f = f(x) 1 and g = g(y) 1 be differentiable functions.
Consider the following unit vec≥tor field defined on≥R R
(f,g)
×
1 1
N := ∂ + ∂
(x,y) √2f(x) x √2g(y) y
Then N is a normal vector field of a totally geodesic foliation of R R.
(f,g)
×
Proof. We have f = g = 0 and then N is a geodesic vector field. Moreover, for
y x
thischoiceoffunctions,thedouble-twistedproductR Risflat(seetheexpres-
(f,g)
×
sionsofcurvatureRinthe 2)andcomplete(byLemma1). Therefore,Ric(N)=0
§
and (by Corollary 1) N is a normal vector field of a totally geodesic foliation of
R R (cid:3)
(f,g)
×
Example. If f(x):=ex+1 and g(y):=sin2(y)+1, the following vector field
1 1
N := ∂ + ∂
(x,y) √2(ex+1) x √2(sin2(y)+1) y
is a geodesic vector field and normal to the leaves of a totally geodesic foliation of
R R.
(f,g)
×
6 ANDRE´ GOMES
References
[1] Bishop,R.L.: Clairautsubmersions,DifferentialGeometry,inHonorofK.Yano,Kinokuniya,
Tokyo, pp.21-31(1972)
[2] Chen, B. Y.: Geometry of Submanifolds and its Applications, Science University of Tokyo,
(1981)
[3] Gomes, A.O.: The mean curvature of a transversely orientable foliation, Result.Math. 46,
31-36(2004)
[4] O’Neill, B.: Semi-Riemannian Geometry - with applications to Relativity, Academic Press
(1983)
[5] Ponge,R.andReckziegel,H.: Twistedproductsinpseudo-Riemanniangeometry,Geom.Ded-
icata,48,15-25,(1993)
Departmentof Mathematics
Institute of MathematicsandStatistics
University of Sa˜oPaulo
Sa˜oPaulo-05508-090-SP
Brazil
E-mail address: [email protected]