Global decomposition of a Lorentzian manifold as a Generalized Robertson-Walker space

0.26 MB·English

by  Manuel Gutierrez

GLOBAL DECOMPOSITION OF A LORENTZIAN MANIFOLD
AS A GENERALIZED ROBERTSON-WALKER SPACE
7
0
0 MANUELGUTIERREZANDBENJAMINOLEA
2

n
Abstract. Generalized Robertson-Walker spaces constitute a quite impor-
a
tant family in Lorentzian geometry, therefore it is an interesting question to
J
know whether a Lorentzian manifoldcan be decomposed in such a way. The
2 existence of a suitable vector field guaranties the local decomposition of the
manifold, but the global one is adelicate topological problem. Inthis paper,
] wegiveconditions onthecurvaturewhichensureaglobaldecompositionand
G
applythemtoseveralsituationswherelocaldecompositionappearsnaturally.
D Finally,westudytheuniquenessquestion,obtainingthatthedeSitterspaces
are essentially the only complete Lorentzian manifolds with more than one
.
h decomposition. Moreover, we show that the FriedmannCosmological Models
t admitanuniqueGeneralizedRobertson-Walker decomposition, evenlocally.
a
m
[

1
v 1. Introduction
7
If I R is anopen interval, f C (I) and (L,g ) a Riemannian manifold, the
6 ∞ 0
⊂ ∈
0 product manifold I L furnished with the metric g = dt2 +f(t)2g0 is called a
× −
1 Generalized Robertson-Walker space, GRW space in short, and is denoted I L.
f
0 These spaces were introduced in [1] and they have been widely studied since×then.
7
It is well known that if a Lorentzian manifold admits a timelike, irrotational and
0
conformal vector field, then it is locally a GRW space, [9, 15], but the presence
/
h of such a vector field does not allow us to decide if the decomposition is global,
t
a even in relatively simple manifolds, see example 2.6. This local decomposition ap-
m pearsinmanysituations,butthe analysisofthe globaloneinvolvesratherdelicate
: topological considerations ([2], page 118).
v
In this paper we study purely geometric conditions to achieve a global decom-
i
X position as a GRW space. The key is to analyze the periodicity of the norm of the
r vectorfield alongits integralcurves,whichis relatedto some signpropertiesofthe
a
Ricci curvature and to certain inequalities involving the lightlike sectional curva-
ture. This allows us to give a global version of Karcher’s theorem (an analogue to
Schur’s lemma).
WeapplytheaboveideastoobtainglobalGRWdecompositionsfromlocalones.
For example, any Robertson-Walker space is a perfect fluid, so it is natural to find
outunderwhatconditionsaperfectfluidisaRobertson-Walkerspace. Forthis,the
pressureand density functions must be related by a second order differentialequa-
tionandthetheoremobtainedisofasingularityflavor,i.e. themanifoldisglobally
2000 Mathematics Subject Classification. Primary53C50;Secondary 53C80.
Keywords andphrases. GeneralizedRobertson-Walkerspace,irrotationalandconformalvec-
torfield,isometricdecomposition, lightlikesectional curvature.
ThispaperwassupportedinpartbyMEYC-FEDERGrantMTM2004-06262.
1
2 MANUELGUTIERREZANDBENJAMINOLEA
aRobertson-Walkerspaceor itis incomplete. Othersituation is suggestedby pho-
tonsurfacesinGeneralRelativity, whicharetimelike totallyumbilic submanifolds,
[3]. In a local GRW space, photon surfaces inherit the local GRW structure of the
ambient space. We study how far they are in fact global GRW spaces.
Finally, we investigate the GRW decomposition uniqueness obtaining that the
de Sitter spaces are the unique complete Lorentzianmanifolds with more than one
(with non constant warping function) GRW decomposition. On the other hand,
in a non necessarily complete manifold, if the ligthlike sectional curvature is not
zero for any degenerate planes at a point p, then there is at most one local GRW
structure in a neighborhood of p.
The non complete case is interesting in Cosmology. In fact, Friedmann spaces
are incomplete GRW spaces with a distinguished family of comoving observers
which physically representsthe averageevolutionof the Galaxies in the spacetime.
From a mathematical point of view, comoving observers are the integral curves
of the unitary of a timelike irrotational and conformal vector field which gives
the global decomposition as a GRW space. Since the ligthlike sectional curvature
of the Friedmann spaces is always positive, this decomposition is unique, even
locally,whichmeansthattheFriedmannmathematicalrepresentationoftheGalaxy
evolution is unambiguous.
2. Global Generalized Robertson-Walker decompositions
Let (M,g) be a connected Lorentzian manifold with dimension n > 1 and E a
timelike,unitary,completeandirrotational(curlE =0)vectorfield. WecallΦthe
flow of E, ω the metrically equivalent one-form to E and L the orthogonal leaf
p
through p M. It follows that L ω = d i ω+i dω = 0, thus Φ is foliated,
E E E
i. e. Φ (L∈) L for all t R, and◦the map Φ◦ : R L M is a local
t p ⊂ Φt(p) ∈ × p →
diffeomorphism which is onto for all p M. Usually, we will drop the point of L
p
∈
and will write simply L. Since Φ identifies ∂ with E, we have Φ g = dt2+g ,
where g is a metric tensor on L for each t ∂tR and g =g . ∗ − t
t p ∈ 0 |Lp
Lemma 2.1. If E is a timelike, unitary, complete and irrotational vector field,
then the map Φ:(R L , dt2+g ) (M,g) is a Lorentzian covering map for all
p t
× − →
p M.
∈
Proof. Let σ : [0,1] M be a geodesic and (t ,x ) R L a point such that
0 0 p
Φ(t ,x ) = σ(0). W→e must show that there exists a l∈ift α×: [0,1] R L of σ
0 0 p
throughΦstartingat(t ,x ), [12]. Thereisageodesicα:[0,s ) →R L×,α(s)=
0 0 0 p
→ ×
(t(s),x(s)),suchthatΦ α=σ andα(0)=(t ,x )becauseΦisalocalisometry. If
0 0
◦
wesupposes <1,thereisageodesic(t (s),x (s))suchthatΦ(t (s),x (s))=σ(s)
0 1 1 1 1
with s (s ε,s +ε). Then, in the open interval (s ε,s ) it holds
0 0 0 0
∈ − −
Φ(t(s),x(s))=Φ((t (s),x (s)).
1 1
Differentiating and using that Φ is foliated, it is easy to see that t (s) t(s) =
1
c R. Therefore it exists lim α(s) and the geodesic α is extendible. − (cid:3)
∈ s→s0
Let i : L M be the inclusion map. The induced maps in homotopy i and
#
→
Φ are equal, thus i is injective and π (L) may be considered as a subgroup
# # 1
of π (M). On the other hand, Φ is injective (and thus a global isometry) if and
1
only if an integral curve of E meets the orthogonal leaves at only one value of its
parameter. This fact will be used frequently in this paper.
GLOBAL DECOMPOSITION AS A GRW SPACE 3
Lemma 2.2. Let (M,g) be a non compact Lorentzian manifold and E a timelike,
unitary and irrotational vector field with compact orthogonal leaves. Then M is
isometric to (I L, dt2+g ), where I R is an open interval and L is a compact
t
× − ⊂
Riemannian manifold.
Proof. LetAbethedomainofΦ(theflowofE),Lanorthogonalleafand(a,b) R
themaximalintervalofRsuchthat(a,b) L A. Weclaimthatitisthemaxi⊂mal
× ⊂
definition interval of each integral curve with initial value on L. Indeed, suppose
that Φ (p ) is defined in (a,b+δ) for some p L. There is a η R such that
t 0 0
∈ ∈
(−η,η)×LΦb(p0) ⊂ A. Since LΦb(p0) is compact, Φ−η2 : LΦb(p0) → LΦb−η2(p0) is
onto, and therefore a diffeomorphism. Now, for an arbitrary p L, Φ (p) can be
t
∈
defined in (a,b+η) and we obtain a contradiction.
If there were (t ,p) (a,b) L such that Φ(t ,p) L, t = 0, then M =
0 0 0
∈ × ∈ 6
Φ (L), and it would be compact. Therefore Φ : (a,b) L M is an
i∪ntj∈e[c0t,ti0v]e mt ap and we obtain the desired result. × → (cid:3)
Intheaboveproposition,ifEiscompleteandaleafiscompact,thenallleavesare
compact, since given L and L there is a parameter t R such that Φ :L L
p q t p q
and it is a diffeomorphism. In this situation (a,b)=R.∈ →
The following lemma is provedin a general form in [13], but it does not include
an explicit expression for the warping function.
Lemma 2.3. Let I R be an open interval, L a manifold and g a metric on I L
such that the canon⊂ical foliations are orthogonal. If E = ∂ is timelike, unit×ary
∂t
and orthogonally conformal, then g is the twisted product dt2+f(t,q)2g where
0
f(t,q) = exp( t divE(s,q)ds) and g = g . If moreover d−ivE is proportional to
E, then g is tRh0e wanr−p1ed product 0dt2 +|Lf(t)2g where f∇(t) = exp( t divE(s,q)ds),
− 0 0 n 1
being q a fixed point of L. R −
Proof. Since E is orthogonally conformal, L g(X,Y) = 2ag(X,Y) for all X,Y
E
∈
E , where a = divE. Fix q L and identify it with (0,q). If we take v,w T L
⊥ n 1 ∈ ∈ q
and call h(t)=g(Φ− (v),Φ (w)) then h(t)=2a(Φ (q))h(t). Therefore
t∗q t∗q ′ t
t divE(s,q)
g((0 ,v ),(0 ,w ))=exp 2 ds g(v,w).
t q t q
(cid:18) Z n 1 (cid:19)
0 −
If E is proportional to E then divE is constant on L and the lemma follows. (cid:3)
A timelike, irrotational and conformal vector field is characterized by the equa-
tion U =aX forallvectorfieldX,whereaiscertainfunction. Wecallλ= U
X
∇ | |
and E = U from now on.
λ
Lemma 2.4. Let U be a timelike, irrotational and conformal vector field. Then
(1) U = E(λ)X for all vector field X and λ is constant trough the orthog-
X
∇
onal leaves of U.
(2) If h is a constant function through the orthogonal leaves of U, then E(h) is
constant through the orthogonal leaves of U.
(3) E is irrotational, orthogonally conformal and divE is proportional to E.
∇
Proof. It is a straightforwardcomputation. (cid:3)
Corollary 2.5. Let (M,g) be a Lorentzian manifold and U a timelike, irrota-
tional and conformal vector field with complete unitary E. Then the map Φ :
4 MANUELGUTIERREZANDBENJAMINOLEA
(R L , dt2+f(t)2g ) (M,g) is a Lorentzian covering for all p M, where Φ
p 0
× − → ∈
is the flow of E and f(t)= λ(Φt(p)).
λ(p)
Proof. Apply lemmas 2.1, 2.3 and 2.4. (cid:3)
However, the global GRW decomposition is not guaranteed by the existence of
such a vector field, as the following example shows.
Example 2.6. Take M =S1 R2 furnished with the Lorentzian metric
×
g =(1+Re(ei(t−x)))dt2+2(3+Re(ei(t−x)))dtdx+
(1+Re(ei(t x)))dx2+(4+2Re(ei(t x)))dy2.
− −
ThevectorfieldU =f(t,x)(∂ ∂ ), wheref(t,x)= 2+Re(ei(t x)),is timelike,
∂t−∂x −
irrotational and conformal. In fact, take the orthogopnal basis X ,X ,X where
1 2 3
{ }
X = ∂ ∂ , X = ∂ + ∂ and X = ∂ . Since [X ,X ] = 0, U = fX
1 ∂t − ∂x 2 ∂t ∂x 3 ∂y i j 1
and X (f) = X (f) = 0 we can use the Koszul formula to show that U =
2 3 ∇Xi
X (f)X for i = 1,2,3. Now, γ(t) = (eit, t,0) is an integral curve of X and
1 i 1
L= (eix,x,y):(x,y) R2 is an orthogona−l leaf of U. The curve γ intersects L
{ ∈ }
in two different point for t=0 and t=2π. Hence M is not a global GRW space.
Now we obtain a global decomposition result which will be used later.
Proposition 2.7. Let M be a Lorentzian manifold and U a timelike, irrotational,
conformal and non parallel vector field with complete unitary. Suppose one of the
following is true
(1) divU 0 (or divU 0).
≥ ≤
(2) Ric(U) 0.
≤
(3) U 0.
△| |≤
Then M is isometric to a GRW space.
Proof. Take L an orthogonal leaf and γ(t) an integral curve of E with γ(0) L.
∈
If γ returned to L then λ(γ(t)) would be non constant and periodic, since Φ is
t
foliatedandλisconstantthroughL. Thisisacontradictionifwesupposecaseone
or two.
Nowwesuppose λ 0. Adirectcomputationgivesus λ=E(E(λ))+(n
△ ≤ −△ −
1)E(λ)2,thusifz(t)=ln λ(γ(t)) weget0 z +(n 1)z2. Sinceλ(γ(t))isperiodic
λ λ(γ(0)) ≤ ′′ − ′
and non constant there exists t < t such that z (t ) = z (t ) = 0 and z (t) = 0
0 1 ′ 0 ′ 1 ′
6
for all t (t ,t ). Suppose that z >0 in (t ,t ) (the case z <0 is similar). Then
0 1 ′ 0 1 ′
∈
for all t (t +ε,t ), where ε is small enough, we get
0 1
∈
t z t
′′
− (n 1).
Z z2 ≤Z −
t0+ε ′ t0+ε
Thus 1 (n 1)(t t ε)+ 1 and we get a contradiction taking
z′(t) ≤ − − 0 − z′(t0+ε)
t t .
1
→Therefore, in the three cases, the covering map Φ:R L M is injective and
corollary 2.5 finishes the proof. × → (cid:3)
Remark2.8. Itisclearthatwecandropthehypothesis U nonparallel intheabove
proposition taking strict inequalities. On the other hand, if a GRW space R L
f
×
verifies Ric(∂ ) 0 then f is constant and ∂ is parallel, [15]. Hence we can not
∂t ≥ ∂t
suppose Ric(U) 0 instead of Ric(U) 0 in the above proposition.
≥ ≤
GLOBAL DECOMPOSITION AS A GRW SPACE 5
In [11] it was proven that a complete Lorentzian manifold with non negative
constantscalar curvature which admits a timelike, irrotational,conformaland non
parallelvector field is isometric to a globalGRW space. Now we can getdecompo-
sition theorems on Lorentzian manifolds using Ricci curvature hypothesis.
Theorem 2.9. Let M be a complete and non compact Lorentzian manifold with
n 3 and U a timelike, irrotational, conformal and non parallel vector field. If
Ri≥c(v) 0 for all v U then M is isometric to a GRW space R L. Moreover,
⊥
≥ ∈ ×
if U is bounded from above then L is compact.
| |
Proof. If λ 0 then we can apply proposition 2.7. Suppose that there exists a
△ ≤
pointp M with λ(p)>0andcallLtheleafthroughp.Takez Landv T L
z
∈ △ ∈ ∈
anunitaryvector. SinceM islocallyawarpedproduct,adirectcomputationgives
us
1 E(λ)2
Ric (v)=Ric (v) E(E(λ))+(n 2) .
L M
− λ(cid:18) − λ (cid:19)
Then RicL(v) ≥ −λ1 (cid:16)E(E(λ))+(n−1)E(λλ)2(cid:17) = △λλ(z). But λ and △λ are con-
stant through the orthogonal leaves, thus RicL(v) ≥ △λλ(p) > 0 for all unitary
v TL. Using the completeness of M and the theorem of Myers we conclude that
∈
L is compact. Lemmas 2.2 and 2.3 gives us that M is isometric to a GRW space.
Suppose now that λ is bounded fromabove. It is sufficientto showthat there is
a point p M such that λ(p) > 0. Suppose that λ 0. Since E is complete,
∈ △ △ ≤
X = λnE is complete too. Indeed, if γ is an integral curve of E and β = γ h of
◦
X,thenthefunctionhverifiestheequationh(t)=λn(γ(h(t)))whichhasmaximal
′
solution defined on the whole R due to the boundness of λ. Now
E(λ)2 1
λ E(E(λ)) n = X(X(λ)).
△ ≥− − λ −λ2n
If β : R M is an integral curve of X and y(t) = λ(β(t)) then 0 y and it is
′′
bounded→from above. Therefore y is constant and U would be parall≤el. (cid:3)
In the above theorem, if dimM = 2 then 0 Ric(v) = E(E(λ)) = Ric(U) for
≤ λ −
all unitary v U and proposition 2.7 gives us the global GRW decomposition.
⊥
∈
IfRic(u) 0foralltimelikevectoruthenitissaidthatthetimelikeconvergence
≥
condition holds (TCC) and if Ric(u) 0 for all lightlike vector u then it is said
≥
that the null convergence condition holds (NCC).
Theorem 2.10. Let M be a complete and non compact Lorentzian manifold with
n 3 and U a timelike, irrotational, conformal and non parallel vector field. If M
≥
satisfies the NCC then it is isometric to a GRW space.
Proof. Let h : M R be given by h = E(E(λ)) E(λ)2. We consider two
→ − λ
possibilities: there exists p M with h(p)>0 or h(q) 0 for all q M.
Assumethesecondonea∈ndtakeγ :R M aninteg≤ralcurveofE∈.Thefunction
→
z(t) = ln λ(γ(t)) verifies z 0. Therefore γ can not return to L and M is
λ(γ(0)) ′′ ≤ γ(0)
isometric to a GRW space (corollary 2.5).
Suppose now that there exists p M such that h(p) > 0 and take L the leaf
∈
through p. If v T L is an unitary vector then v+E is lightlike and a straight-
z z
∈
forwardcomputation gives us
n 2 E(λ)2
0 Ric (v+E )=Ric (v)+ − E(E(λ)) .
M z L
≤ λ (cid:18) λ − (cid:19)
6 MANUELGUTIERREZANDBENJAMINOLEA
Thenweget n 2h(z) Ric (v). Butλ andh areconstantthroughthe orthogonal
λ(−z) ≤ L
leaves. Hence 0< n 2h(p) Ric (v) for all unitary v TL and we can conclude
λ(−p) ≤ L ∈
as in theorem 2.9. (cid:3)
Corollary 2.11. Let (L,g ) be a non compact and complete Riemann manifold
0
and M =S1 L endowed with a warped product metric dt2+f(t)2g . If the NCC
0
× −
holds then f is constant.
We can not suppose the less restrictive TCC condition in the above theorem
because U would be parallel, see remark 2.8. But the TCC condition can be used
to obtain a GRW decomposition in a necessarily incomplete manifold.
Proposition2.12. LetM beaLorentzian manifold andU atimelike, irrotational,
conformal andnon parallel vectorfieldwith compact orthogonal leaves. IfRic(U)
0 then M is isometric to a GRW space (a,b) L where (a,b)=R. ≥
× 6
Proof. We canproveinthe samewayasinlemma 2.2thatthe integralcurveswith
initial value on an orthogonalleaf L all have the same maximal definition interval,
say (a,b). Since E(E(λ)) 0 and λ is not constant it follows that (a,b) = R.
≤ 6
Then, necessarily Φ : (a,b) L M is a diffeomorphism and lemma 2.3 finished
the proof. × → (cid:3)
TheClosedFriedmannCosmologicalModelisanexampleofamanifoldsatisfying
the hypotheses of the above proposition.
Causalityhypotheses arefrequently usedinLorentziangeometry,besides curva-
turehypotheses. SincetheinjectivityofΦdependsonthebehaviorofsometimelike
curves it seems natural to impose a causality condition to reach global decompo-
sitions. However, a hard condition as being globally hyperbolic is not sufficient to
obtain a global product, as the following example shows.
Example 2.13. (Compare with proposition 2 in [9]) Take M˜ =R2 with the metric
dt2+f(t)2dx2, where f(t) = 4+sin(2πt). Call Γ the isometry group generated
−
by Ψ(t,x) = (t+1,x+1) and Π : M˜ M = M˜/Γ the projection. The vector
→
field U =Π (f ∂ ) is timelike, irrotational and conformal. The manifold M verifies
∂t
∗
any causality condition. In fact, Π( (t,x):t=x ) is a Cauchy hypersurface. But
{ }
M does not split as a GRW, since any integral curve of E = U intersects its
U
orthogonal leaves at an infinite number of points. | |
With the chronologicalhypothesis we can obtain certain information about the
fundamental group of the manifold. First we need the following lemma.
Lemma 2.14. Let M be a Lorentzian manifold and U a timelike, irrotational and
conformal vector field with complete unitary E. If there exists t > 0 and p M
0
∈
such that Φ (p) L , being Φ the flow of E, then
t0 ∈ p
(1) Φ :M M is an isometry.
t0 →
(2) If M is chronological, then the isometry group Γ generated by Φ is iso-
t0
morphic to Z and acts on M in a properly discontinuous manner.
Proof. (1) Since U is irrotational and conformal it follows that L g(X,Y) =
E
2E(λ)g(X,Y) for all X,Y E . If we take v,w T M with v,w E and
λ ∈ ⊥ ∈ q ∈ ⊥
call h(t) = g(Φ (v),Φ (w)), then the above equation can be written as h(t) =
t∗q t∗q ′
GLOBAL DECOMPOSITION AS A GRW SPACE 7
2λ′(t)h(t), where λ(t)=λ(Φ (p)), and thus h(t)= λ(t) 2g(v,w). Since λ is con-
λ(t) t λ(0)
(cid:16) (cid:17)
stant through the orthogonal leaves, we get g(Φ (v),Φ (w)) = g(v,w). Now,
t0∗q t0∗q
using that Φ is foliated, it is easy to conclude that Φ :M M is an isometry.
t t0 →
(2) Suppose that the manifold is future oriented by the vector field E. We will
construct for each q M an open set U of q such that Φ (U) U = for all
∈ nt0 ∩ ∅
n Z 0 . Recall that Φ:R L M is a local isometry with f(t)= λ(Φq(t))
∈ −{ } ×f q → λ(q)
(corollary 2.5). Take k =max f(t)2/t [ t ,t ] and the ball W =B(q, ε )
{ ∈ − 0 0 } 2√k ⊂
L , where ε < t is small enough to Φ : ( ε,ε) W U be an isometry and
q 0 f
− × →
W a normal neighborhood of q. Given Φ (z) U, with z = exp (v) W and
s q
∈ ∈
s ( ε,ε), we can construct future timelike curves for t [0,1]
∈ − ∈
ε
α(t) = Φ (t 1),exp ((1 t)v)
(cid:16)2 − q − (cid:17)
ε
β(t) = Φ t,exp (tv)
2 q
(cid:16) (cid:17)
from Φ ε(z) to q and from q to Φε(z) respectively.
If x,−y2 U with y = Φ (x)2and n N, then using the above curves, we
∈ nt0 ∈
can construct a broken closed timelike curve, in contradiction with the chronology
hypothesis. Thus, Φ (U) U = for all n Z 0 .
nt0 ∩ ∅ ∈ −{ }
Take the Riemannian metric g = g + 2ω ω. The group Γ is a group of
R
⊗
isometries for g too. Since g is Riemannian, the existence of the above open
R R
set U for each q M is sufficient to show that the action of Γ in M is properly
discontinuous. ∈ (cid:3)
Theorem 2.15. Let M be a chronological Lorentzian manifold and U a timelike,
irrotational, conformal and non parallel vector field with complete unitary. Then
M is a global GRW space or there is a Lorentzian covering map Ψ: M S1 N
and π (M)=Z π (L), where L is an orthogonal leaf of U. → ×
1 1
⊕
Proof. Suppose M is not a GRW space. Then there is p M such that divU >0
p
∈
(proposition 2.7) and the integral curve Φ (t) of E returns to L, the leaf of E
p ⊥
through p. We fix this p throughout the proof. Take t = inf t > 0 : Φ (t) L
0 p
{ ∈ }
and call λ(t) = λ(Φ (t)). Since λ(0) = E(λ) = 1divU > 0 and λ is constant
p ′ p n p
throughtheorthogonalleaves,wehavet >0anditisaminimum. Indeed,suppose
0
itisnotandtakeasequencet witht <t ,Φ (t ) Landlim t =t . Since
n 0 n p n n n 0
λ(t ) = λ(0) for all n N, it follows λ(t ) > 0, b∈ut this is a c→on∞tradiction with
′ n ′ ′ 0
λ(t )=λ(0) for all n ∈N and therefore Φ (t ) L. Applying the above lemma, Γ
n p 0
∈ ∈
acts on M in a properly discontinuous manner.
Take the quotient manifold M/Γ and the canonical projection P : M M/Γ.
→
SinceΦ preservesthevectorfieldU,thereisatimelike,irrotationalandconformal
t0
vector field W on M/Γ such that P (U)=W. If we call F = W , then its integral
W
∗ | |
curvesareP(Φ (x)). ThusitiseasytocheckthattheintegralcurvesofF intersect
t
theleafofF givenbyN =P(L)atonlyonepointandtheyaret -periodic. Then,
⊥ 0
fromtheLorentziancovering(R N, dt2+f2g ) M/Γgivenincorollary2.5we
N
canconstructanisometry(S1 ×N, −dt2+f2g ) →M/Γ. Withthis identification,
N
× − →
P :M S1 N is given by P(Φt(x))=(e2πitt0,P(x)), for all x L and t R.
Now→,π (M×)isisomorphictothekernelofthemapΘ:π (S1 ∈N) Γgi∈venby
1 1
Θ([γ]) = Φ where Φ (p) = γ˜(1) and γ˜ is a lift of γ through×P : M→ S1 N
startingatnpt,0[7]. Identinfty0ing Γandπ (S1)withZ, themapΘverifiesΘ(→n,[0])×=n
1
8 MANUELGUTIERREZANDBENJAMINOLEA
and Θ(0,[γ]) = n, where Φ (p) = γ˜(1) and γ˜ is a lift of γ. Since the isometry
nt0
Φ :M M canbe restricted to an isometry Φ :L L, it is easy to show that
thte0 restr→ictionofP to L,P :L N is anormalt0cover→ingmapwith Γ Z asdeck
0
transformationgroup,henceπ (→N)=Z π (L),[7]. Noticethatthis id≃entification
1 1
⊕
works as follows: If α˜ is a curve in L from p to Φ (p) then [P α˜] π (N)
t0 0 ◦ ∈ 1
is identified with (1,0) Z π (L) and (0,[β]) Z π (L) is identified with
1 1
∈ × ∈ ×
[P β] π (N). Taking into account all these identifications we can consider
Θ0:◦Z ∈Z 1π (L) Z. It is easy to show theat Θ(1,0,0) = Θ(0,1,0) = 1 and
e⊕ ⊕ 1 →
Θ(0,0,[β])=0 for [β] π (L) . Therefore
1
∈
e KerΘe= (n, n,[β]):n Z,[β] π1(L) Z π1(L).
{ − ∈ ∈ }≃ ⊕
(cid:3)
e e
RecallthatM isisometrictoaGRWifandonlyifπ (L)=π (M). Thistheorem
1 1
says that if M is chronologicalonly two possibilities can occur: π (M)=π (L) or
1 1
π (M)=Z π (L).
1 1
⊕
3. Perfect Fluids
A Lorentzian four dimensional manifold M is called a perfect fluid if there is
an unitary timelike vector field E and ρ,η C (M) (the energy and pressure)
∞
∈
such that the stress-energy tensor is T =(ρ+η)ω ω+ηg, or equivalently Ric=
⊗
(ρ+η)ω ω+ 1(ρ η)g. It is well known that any Robertson-Walker space is a
⊗ 2 −
perfect fluid, [12]. More specifically, if we denote ρ and η the derivative along E,
′ ′
we have
Lemma 3.1. A GRW spacetime is a perfect fluid if and only if it is a RW space.
Moreover, if ρ+η =0 then
6
2
ρ′ ′ 1 ρ′ 1
= + (ρ+3η).
(cid:18)ρ+η(cid:19) 3(cid:18)ρ+η(cid:19) 2
Proof. It is a straightforwardcomputation. (cid:3)
AnaturalquestioniswhatconditionsonaperfectfluidmakeitaglobalRobertson-
Walker space, [5, 6]. Using the decomposition theorems of the previous section we
can give an answer to this question. First we need a preliminary lemma.
Lemma 3.2. Let M be a complete Lorentzian manifold and E a timelike, unitary,
irrotational and orthogonally conformal vector field with divE proportional to E.
Then there exists an irrotational and conformal vector fi∇eld U such that E = U .
U
| |
Proof. If Π : M˜ M is the universal covering of M and E˜ the lift of E then M˜
decomposes as R→ L˜ where E˜ = ∂ . Let ψ(t,p) = (A(t,p),B(t,p)) be a deck
×f ∂t
transformation. Since it is an isometry and preserves E˜, it is easy to show that
ψ(t,p)=(t+k,B(p))and f(t+k) =aforallt R,being aandk certainconstants.
f(t) ∈
Since M˜ is complete, [15]
k
= ∞f(t)dt= ∞ an f(t)dt
∞ Z Z
0 nX=0 0
0 k
∞
= f(t)dt= a−n−1 f(t)dt.
∞ Z Z
−∞ nX=0 0
GLOBAL DECOMPOSITION AS A GRW SPACE 9
Thusa=1andf(t+k)=f(t)forallt R. ThevectorfieldU˜ =f ∂ isirrotational
∈ ∂t
andconformalandit is preservedby the decktransformations. Thus there exists a
vector field U on M irrotationaland conformal such that E = U . (cid:3)
U
| |
Example 3.3. The above lemma does not hold if M is not complete. Consider
M˜ =R R2, with the metric g = dt2+et(dx2+dy2), and Γ the isometry group
× −
generated by φ(t,(x,y)) = (t+1,e−12(x,y)). Then there exists a vector field E on
M =M˜/Γwhichverifiesthehypotheses ofthelemmabutthereisnotU irrotational
and conformal with E = UU . Otherwise, the lift to M˜ of U would be e2t ∂∂t, which
is not invariant under the|a|ction of Γ.
A perfect fluid is called barotropic if it satisfies an equation of state η = η(ρ).
ObservethatinaRWperfectfluidR Lbothfunctionsη andρcanbe expressed
f
×
in terms of the warping function f, [12], and therefore it is barotropic in the open
set dρ = 0. The following theorem shows that this condition together with the
dt 6
relationbetweentheenergyandthepressureobtainedinlemma3.1giverisetothe
global decomposition as a RW space.
Theorem 3.4. Let M be a non compact spacetime with a barotropic perfect fluid
(E,ρ,η) such that E is geodesic, dη = 0, d2η = 0, ρ+η = 0 and ρ > 0 is not
dρ 6 d2ρ 6 6
constant. If
ρ′ ′ 1 ρ′ 2 1
= + (ρ+3η),
(cid:18)ρ+η(cid:19) 3(cid:18)ρ+η(cid:19) 2
then either M is incomplete or a global Robertson-Walker space.
Proof. WefirstshowthatE isirrotational,thatis,dω =0. Fromtheforceequation
g(X, η)= (ρ+η) E forallX E wegetdη =hw,beinghcertainfunction,
E ⊥
∇ − ∇ ∈
and
X(h)=X(g(E, η))=g(E, η)=g(X, η)=0.
X E
− ∇ ∇ ∇ ∇ ∇
Hence 0=dh ω+hdω =hdω. Thus, to show dω =0 it is enough to consider a
p
∧
critical point p of η. Call η(t) =η(γ(t)) and ρ(t)=ρ(γ(t)) where γ is the integral
curve of E with γ(0) = p. Then η = 0 if and only if ρ = 0. If ρ(t) = 0 for all
′ ′ ′
t ( ε,ε)then ρ+3η=0 in contradictionwith d2η =0. Thus there is a sequence
∈ − d2ρ 6
t converging to 0 with ρ(t )=0 and therefore dω =0.
n ′ n p
6
Now, the endomorphism A : E E given by A(X) = E is self-adjoint
⊥ ⊥ X
→ ∇
andastraightforwardcomputationgiveusRic(E)= E(divE) A 2. Fromthis
− −k k
and the energy equation divE = E(ρ), we get
−ρ+η
2
A 2 = ρ′ ′ 1(ρ+3η)= 1 ρ′ = 1(trace A)2.
k k (cid:18)ρ+η(cid:19) − 2 3(cid:18)ρ+η(cid:19) 3
Then A = divEid and E is orthogonally conformal. It follows from the energy
3
equation that divE is constant through the orthogonal leaves, hence divE is
∇
proportional to E. Suppose M is complete, then lemma 3.2 says that there exists
U irrotational and conformal such that E = U . If U were parallel then M would
U
| |
be locally a direct product, which implies that ρ is constant, [12].
Now, using the expression of the Ricci tensor of a perfect fluid at the beginning
of this section, if ρ+η < 0 apply theorem 2.9 and if ρ+η > 0 apply theorem
2.10. (cid:3)
10 MANUELGUTIERREZANDBENJAMINOLEA
4. Lightlike sectional curvature of a Lorentzian manifold
Let M be a Lorentzian manifold and E a timelike and unitary vector field. We
can define a curvature for degenerate planes as follows, [8]. Take Π a degenerate
planeandabasis u,v ,whereuistheuniquelightlikevectorinΠwithg(u,E)=1.
{ }
We define the lightlike sectional curvature of Π as
g(R(v,u,u),v)
(Π)= .
E
K g(v,v)
This sectional curvature depends on the choice of the unitary timelike vector
fieldE, butits signdoes notchangeif wechooseanothervectorfield. In fact,if E
′
is another unitary timelike vector field then KE′(Π)= g(u,1E′)2KE(Π) where u∈Π
is the unique lightlike vector such that g(u,E) = 1. Thus, it makes sense to say
positive lightlike sectional curvature or negative lightlike sectional curvature.
If E is geodesic, n >3 and is a never zero point function then M is locally
E
a Robertson-Walker space wheKre E is identified with ∂ , [9, 10]. Therefore E is
∂t
irrotational,orthogonallyconformaland divE isproportionaltoE anditfollows
fromlemmas2.1and2.3thatM iscovere∇dbyaRWspaceR L. Asanapplication
×
of the decomposition results presented above,we obtain conditions on the lightlike
sectional curvature which ensure the global decomposition of M as a Robertson-
Walker space. First, we give some relations between lightlike sectional curvature
and Ricci curvature.
Lemma 4.1. Let M be a Lorentzian manifold and E a timelike and unitary vector
field. Ifuisalightlikevectorwithg(u,E)=1,thenRic(u)= in=−12KE(span(u,ei))
where e1,...,en 2 is an orthonormal basis of u⊥ E⊥. P
{ − } ∩
Lemma 4.2. Let M = I L be a GRW space and u a lightlike vector such that
f
×
g(u, ∂ ) = 1. If u = ∂ +w, w TL, and Π = span(u,v) is a degenerate plane
∂t −∂t ∈
then K∂∂t(Π)= KL(span(v,fw2))+f′2−f′′f, where KL is the sectional curvature of L. If
moreover L has constant sectional curvature then Ric(u)=(n 2) (Π).
∂
− K∂t
Proof. It is a straightforward calculation. If L has constant sectional curvature,
then isapointfunctionandtheabovelemmagivesusRic(u)=(n 2) (Π).
∂ ∂
K∂t − K∂t (cid:3)
Lemma 4.3. Let M = R L be a GRW space. If M is lightlike complete and
f
×
Ric(u)>0 for all lightlike vector u, then Ric (v)>0 for all v TL.
L
∈
Proof. Suppose there is v TL unitary such that Ric (v) 0. Then
L
∈ ≤
∂ n 2 f 2
′
0<Ric( +v)=RicL(v)+ − f′′ .
∂t f (cid:18) f − (cid:19)
If we call y = lnf then y′′ = f′′ff−2f′2 < 0. We can suppose y′(0) > 0. Now,
0 ey 0 ey′(0)t+y(0) < and we conclude from [15] that M is lightlike
Rin−c∞omple≤te.R−∞ ∞ (cid:3)
Proposition4.4. Let M bea non compact and complete Lorentzian manifold with
n>3 and E a timelike, unitary and geodesic vector field. If the lightlike sectional
curvature is a positive point function, then M is a RW space R L with L of
E
K ×
constant positive sectional curvature.