A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
M OHAMMED G UEDIRI
On the geodesic connectedness of simply connected Lorentz surfaces
Annales de la faculté des sciences de Toulouse 6
esérie, tome 6, n
o3 (1997), p. 499-510
<http://www.numdam.org/item?id=AFST_1997_6_6_3_499_0>
© Université Paul Sabatier, 1997, tous droits réservés.
L’accès aux archives de la revue « Annales de la faculté des sciences de Toulouse » (http://picard.ups-tlse.fr/~annales/) implique l’accord avec les conditions générales d’utilisation (http://www.numdam.org/conditions).
Toute utilisation commerciale ou impression systématique est constitu- tive d’une infraction pénale. Toute copie ou impression de ce fichier doit contenir la présente mention de copyright.
Article numérisé dans le cadre du programme Numérisation de documents anciens mathématiques
http://www.numdam.org/
- 499 -
On the geodesic connectedness of
simply connected Lorentz surfaces(*)
MOHAMMED
GUEDIRI(1)
Annales de la Faculté des Sciences de Toulouse Vol. VI, n° 3, 1997
R~SUM~. - Nous montrons que les surfaces lorentziennes simplement
connexes globalement hyperboliques sont geodesiquement connexes. On
a 1’equivalence pour ces surfaces simplement connexes cocompactes.
ABSTRACT. - We discuss geodesic connectedness for simply connected
Lorentz surfaces, and show that it holds for those which are globally hyperbolic. In the compact case, we establish the equivalence for the
universal covering.
1. Introduction
It is well known
that, according
toHopf-Rinow’s theorem,
anycomplete,
connected Riemannian manifold is
geodesically connected;
thatis,
any two of itspoints
can bejoined by
a(minimizing) geodesic.
By contrast,
for Lorentzmanifolds, geodesic completeness
does notimply geodesic
connectedness(Sect. 2).
Throughout
this paper S denotes asmooth,
connected andsimply
connected surface.
If g
is an indefinite metric onS,
then we say that(S, g)
is a Lorentzian surface.
Two Lorentzian surfaces
(S, g)
and(S’, g’)
are said to be conformal if there is adiffeomorphism j~ :
? 2014~ 5~ such that~~
=/ .
~ wheref
is anon-vanishing
smooth function on S.So,
aconformally
flat surfacewill,
inthis paper, mean a Lorentzian surface which is
globally
conformal to a flatone.
(*) Reçu le 16 octobre 1995
(1) Department of Mathematics, College of Sciences, King Saud University, P.O. Box 2455, Riyadh 11451 (Saudi Arabia)
Globally hyperbolic
spacesLet denote
by J+ (p)
the set of allpoints q
E M which are related top
by
future-directednonspacelike
curves.Similarly
we define J-(p)
withpast-directed
instead future-directed.If q E J+(p)
we say that q iscausally
related to p and we write p q. .
DEFINITION . Given a Lorentzian
manifold (M, g),
we say that(M, g)
is
globally hyperbolic provided
that it isstrongly
causal andfor
each p q the setJ+ {p)
nJ- {q)
iscompact.
Recall that
(M, g)
is said to bestrongly
causal if for each p E Mnonspacelike
curves that startarbitrarily
close to p and leave some fixedneighborhood
cannot returnarbitrarily
close to p.We
point
out that Claim 1 in Section 3 shows that allsimply
connectedLorentzian surfaces are
strongly causal,
and therefore every Lorentzian sur- face which is conformal to the Minkowskiplane
isglobally hyperbolic.
The converse is not true as can be seen on the universel
covering
of(T2 , dx dy
+d y2 ) where f
is a smooth function which vanishes some- where withoutbeing identically
null.(For
more aboutglobally hyperbolic
spaces, we refer to
[1]
and[4]).
The main purpose of this work is to
study
thegeodesic
connectedness ofsimply
connected surfaces. Our first and main result states that asimply
connected Lorentzian surface isgeodesically
connected if it isglobally hyperbolic,
andspecially
when it is conformal to the Minkowski2-plane Ef ,
where this latter space denotes
JR2
with the metric dxdy.
We remark that
simply connected, geodesically
connected surfaces arenot all
globally hyperbolic (and
so are not all conformal toConsider the
strip
shown indiagram
2 with the metric dxdy. Of course,
this
region
isgeodesically
connected sincegeodesics
areexactly straight
lines.
However,
it is neitherglobally hyperbolic
nor conformal toEf by
thefact that each
boundary point
isapproached by
two nullgeodesics.
In the
compact
case, we prove that the universalcovering
of a Lorentzian2-torus
1f2
isgeodesically
connected if andonly
if it isglobally hyperbolic.
Completeness
is also obtained forLorentzian, conformally
flat 2-torus(a
proof
isgiven
in[2]).
2. Generalities 2.1
Example.
A nongeodesically connected,
complete
Lorentzian manifoldLet
Rn+11
denote the usual vectorspace Rn+1
with the indefinite metric 9 - +d~1
+ ... + .In
II81 +1,
we consider thehyperquadric
Clearly, g
induces a Lorentzian metric onAlso,
it is well known[4]
that,
with thismetric, S 1
is ageodesically complete
Lorentzian manifold of constant curvature 1.The
geodesics
ofS 1 are just
the intersections ofS1
withplanes through
the
origin of Rn+11 which meet S1.
The
hyperquadric S 1
is called either the de Sitter space or the unitpseudo-sphere.
By
contrast with Riemannian case, the de Sitter spaceS1
is nevergeodesically
connected since italways
containsnon-antipodal points
p, q for which there exists nogeodesic joining
p and q(see [4, Prop. 5.38]).
2.2. The two null foliations of a Lorentz surface
Let
(,5’, g)
be asimply
connected Lorentzian surface. Then the inex- tendible nullgeodesics
of(,S, g}
may bepartitioned
into two transversal families~’1
and:F2
such that each of them covers Ssimply;
thatis,
eachpoint
p E ,S’ lies onexactly
one nullgeodesic
of eachfamily. Thus, ~’1
and~2
determine two transversal foliationsby
the nullgeodesics
of(,S’, g).
Wecall them the null
foliations
of the Lorentzian surface(S, g).
For
example,
we remark thatSf
is a ruled surfacegenerated by
eachof null foliations.
Also,
each leaf of01
meets any leaf of~’2
except one. This fact can be well understood if we work with the universalcovering
ofSf.
This space isrepresented by
thestrip
in with the Lorentzian metric
ds2
=sec2 s( -d02
+ds2).
Thus,
the universalcovering Si
of the de Sitter spaceSi
isconformally
flat. In contrast, it is not conformal to
Ef
since eachboundary point
isapproached by
two nullgeodesics (compare
withdiagram 2).
Of course,Si
is not
geodesically
connected sinceSi
is not.Together,
the two null foliations define agrid
of nullgeodesics
on ,S’ calledthe null
grid. Locally
there is a system of(local)
coordinates x, y on 5’ forwhich g
=f (x, y)
dxdy
for somenon-vanishing
functionf.
.Thus, locally,
the nullgrid
can be made as the net of level lines for the abovesystem
of coordinates.A basic fact of Lorentz
geometry
is that adiffeomorphism
is conformal if andonly
if the induced map on thetangent
bundle preserves the null-cones.This means
that,
for Lorentzsurfaces,
adiffeomorphism
is conformal iff it preserves the nullgrids.
Diagrams
1 to 4 show the nullgrids
for the metric dxdy
on four open subsets of theplane I~81.
Thestrip
shown indiagram
1 is conformal toEf,
but none of those in
diagrams
2 to 4 is conformal to This fact is related to the behaviours of~’1
and~’2.
Diagrams 1 and 2
Diagrams 3 and 4
2.3 The conformal
boundary
Let
I,
m be two leaves of the two null foliations01
and~’2 respectively,
and let us attach ideal
boundary points /-}-,
1- and m+, m- in thepositive
and the
negative
directionsrespectively.
The conformal
boundary
of(S, g)
denotedby 8cS (and
which is due toKulkarni
[3])
will be the set ofequivalence
classesrelatively
to a certainequivalence
relation on the set of all idealboundary points. Thus,
it is invariant under conformalchanges.
A
topology
is then defined on S = S U8cS
which induces atopology
on8cS.
Ingeneral, 8cS
may becomplex
and need not be Hausdorff.(See [3]
for more
details).
Now,
it is not difficult to see that the fourregions
indiagrams
1 to 4represent
different conformal classes. Of course, not everysimply
connectedLorentzian surface is
conformally
flat. Infact,
for theregion
shown indiagram 4,
someboundary points
areapproached by infinitely
many nullgeodesics.
Note also that thisregion
may be seen as the universalcovering
of a
punctured
disc with the metric dxd y.
A remarkable
conformal
invariantLet
i(p)
be the number of allpossible
nullgeodesics
in Sending
at theboundary point
p. It is a conformal invariant of(S, g)
.In the
examples
of thediagrams above, 8cS
can be identified with the subset of alltopological boundary points
which are endpoints
of nullgeodesics living
in S.Then, i(p) ~
0 for allpoints
of8cS.
Thus,
indiagram 1, i ( p)
= 1 for allpoints
of Indiagram 2, i ( p)
= 2since each
boundary point
isapproached by exactly
two nullgeodesics,
oneof each of the two null foliations. In
diagram 3,
the sixboundary points
at outer corners are
approached along by
no nullgeodesic,
while those atthe two inner corners are
approached by exactly
two nullgeodesics.
Thismeans that
only
the latter(i.e.,
the two innercorners)
are in8cS. Thus, i(p)
= 1 except at the two inner corners on8cS
wherei(p)
= 2. Indiagram 4, i(p)
= oo for everyboundary point
on8cS corresponding
to theorigin
ofthe
punctured
disk. This is anexample
of asimply
connected Lorentzian surface which cannot beconformally
flat.The
following
result will beeasily proved, by
a closereading
of[3].
PROPOSITION 2.1.2014 Let
(S, g)
be aconnected, simply
connected Lorentzsurface
with smoothableThen,
1) (8, g) is (globally) conform ally flat iff i(p)
3for
every p in8cS;
2) (S, g)
isconformal
toE1 iff i(p)
= 1for
every p in3) (S, g)
isconformal
toS21 iff i(p)
= 2for
every p inThe
boundary points
p E such thati(p) ~
2 werecalled, by Kulkarni,
characteristic
points.
Remark. - Another conformal invariant is the number of corner
points
on
8c8. Thus, by varying
the number of horizontal and vertical linesegments making
up thepolygonal boundary
of a certainregion,
we canconstruct an infinite number of
conformally
distinctsimply
connectedLorentz surfaces. This contrasts
completely
with the three well known conformal classes ofsimply
connected Riemannian surfaces.The next result
points
out anequivalence
between the fact to be confor- mal to a subset ofEf
and certainassumptions
on the behaviours of~"i
and~’2. But,
beforestating it,
we shall need a definition.DEFINITION . We say that the
pair ~~1, ~2)
make up aproduct (which
is
always
truelocally) if
there existglobal
coordinates(x, y)
such that the metric isgiven by
g =f (x, y)
dxdy for
a certainnon-vanishing function f
on S.
PROPOSITION 2.2.2014 Under the
assumptions of proposition
we have:1) (S, g)
isconform ally fl’at iff (F1, F2)
is aproduct;
2) (S, g)
isglobally hyperbolic iff
eachleaf of 01
intersects eachleaf of
~’2
andconversely.
3. Geodesic connectedness
In this Section we shall discuss the main purpose of this paper.
3.1 General case
In the
simply
connected case, we shall prove thatgeodesic
connectedness is obtained when(,5’, g)
isglobally hyperbolic.
The idea is to prove that every inextendible
geodesic starting
at a fixedpoint
p of S may beunbounded,
thatis,
it does not remain in any compact subset of S.Therefore,
we canapply
theimplicit
function theorem to conclude that everypoint
of S can bejoined
to the fixedpoint
p.We are
going
to state our mainresult,
but first we need the next lemma.LEMMA 3.1.2014
Let y
:[0, b ~
~ S be agiven
nullgeodesic of
asimply
connected Lorentzian
surface (S, g). Then,
each othergeodesic
intersects y at most once.Proof.
-Suppose
that 03C3 is anothergeodesic
in S which intersects y twice(possibly
at the samepoint). Then, by
thetransversality
of thenull
foliations,
we can construct anon-vanishing
vector field which will be transversal to theboundary
of the disk made up of y and a, but this isobviously
absurd.Hence,
thegeodesic 03C3
mustintersect y
at most once. 0THEOREM 3.2.2014 A
simply
connected Lorentziansurface (S, g)
isgeode- sically
connectedif
it isglobally hyperbolic.
Proof.
- Let p, q be two distinctpoints
in S and let R be therectangle
made up of the null
geodesics through
p and q as infigure
1. Since(S, g)
isglobally hyperbolic
it follows that R is compact.CLAIM 1.2014
Every geodesic starting
at p andgoing
in ~Z must go out.Proof.
-Let y : 0 , b [
~S,
b ~, be an inextendiblegeodesic.
Weare
going
to showthat y
does not remain in R when t -~ b.Since R is
compact,
if y wereentirely
contained in R then there would be a sequence sn --~ b such that -~ r, with r E R.Thus,
y intersects one of the two nullgeodesics through
rinfinitely
many times.
Otherwise, y
would be extendiblepast b, contradicting
theassumption that y
is inextendiblebeyond
b.But, according
to lemma3.1,
y cannot intersect a null
geodesic
twice. 0Now,
weparametrize
thegeodesics starting
at p andgoing
in Rby
theangle
between suchgeodesics
and the fixed nullgeodesic joining
p and b asshown in
figure
1.Fig. 1
If
A(a, q, b)
denotes the reunion of thesegments ] a
, q]
and[
q, b ~ [ where
a, b are determined
by
the intersections of the nullgeodesics through
p and q, then weget
amap ~ :
:] a, /3 ~
-~A(a,
q,b)
whichassociates,
for everygeodesic starting
at p andgoing
in R its intersectionpoint
withA(a,
q,b) .
.Now,
it is obviousby
Claim 1that §
is well defined.CLAIM 2.2014 The map
~
is continuous.Proof.
- Since(S, g)
isglobally hyperbolic then, by Proposition 2.2,
there are
global
coordinates(x, y)
such that nullgeodesics
of~’1 (for example)
aregiven by
x = constant. .Assume that the null
geodesic joining b and q
isgiven by x
= c, and let(t, u)
~{~{t, u)
,y(t u))
be the two parametersfamily
ofgeodesics starting
at p and
going
in R. Thisprovides
that themap ~
isgiven by :
On the other
hand,
everygeodesic starting
at p andgoing
in TZ istransversal to
A(a,
q,b)
andconsequently,
at any(t, u)
such thatx(t, u)
= c,8~(t, u)/~t
is never vertical. But thisimplies that,
for every(t, u)
such thatx(t, u)
= c,~x(t, u)/~t #
0.Now,
for every(t, u)
such thatx(t, u)
= c, we canapply
theimplicit
function theorem to
guarantee
the existence of aneighborhood
of(t, u)
forwhich t can be
expressed
as a continuous function of u. In otherwords,
y(t(u), u)
is continuous in u andconsequently §
is continuous too. D We return now to theproof
of Theorem 3.2.Since a , ~3 ~ [ is
connectedand since a, b are attained
by geodesics starting
at p andgoing
in7Z,
themap ~ (which
iscontinuous)
will be onto andconsequently q
may bejoined
to p
by
ageodesic,
as we wished. D3.2
Compact
caseSuppose
now that S is a compact surface and recall that S could be Lorentzian if andonly
if its Euler-Poincaré characteristic vanishes.So,
the 2-torus
1f2
is theonly
orientable surfacesusceptible
to be Lorentzian.Furthermore,
we remark that:a)
the unit disk(D,
dxdy)
is notgeodesically complete;
b)
if(M, g)
is asimply
connected Lorentziansurface,
then there is asmooth conformal factor 11 :
M -~ ~ ] 0, oo [
such that isgeodesically complete (cf. [1]).
Because of these two
facts,
itmight
be natural to state thefollowing conjecture :
: asimply
connectedcomplete
Lorentzsurface
isgeodesically
connected
iff
it isglobally hyperbolic.
So,
in this context we shall now prove thefollowing.
THEOREM 3.3.2014 Let
g)
be a Lorentzian 2-torus.Then,
its universalcovering
isgeodesically
connectediff
it isglobally hyperbolic.
Proof.
-By
theorem3.2,
we haveonly
to prove the converse.Suppose
that the universelcovering
is notglobally hyperbolic.
Then thereexist
causally
relatedpoints
p q such thatJ+(p)
nJ- (q)
isnoncompact.
This means,
by proposition 2.2,
that there exists at least a leafof F1
whichdoes not meet certain leaves
of :F2.
Because the surface is the universal
covering
of a torus, it must containat least one Reeb’s
component.
Now,
let x, y be twopoints
as shown infigure
2.CLAIM 3.2014 There is no
geodesic joining
x and y.Proof.
- We first assume that~’1
and~’2
are oriented as shown infigure
2. It is clear that no nullgeodesic cannot join
x and y.Also,
a timelikegeodesic starting
at x could not even cross the line I offigure 2, otherwise,
it would intersect the nullgeodesic lx or l y twice,
whichcontradicts Lemma 3.1.
Equally,
aspacelike geodesic starting
at x cannot neither reach thepoint
y, for almost the same reason.
Indeed, according
to lemma3.1,
if sucha
geodesic
meets the line I at a certainpoint z
then it would beentirely imprisoned
in theright
half-cone made up of the two nullgeodesics segments lz
= I and mzthrough
thepoint z (fig. 2).
Thus such ageodesic
could notreach the
point
y, which concludes theproof
of Theorem 3.3. DFig. 2
3.3 Final remarks
1)
It is obvious thatglobal hyperbolicity
is invariant(at
least forsurfaces)
under conformal
changes. However,
there is no reason for thatgeodesic
connectedness should be invariant
by
conformalchanges.
We can thinkabout the universel
covering
of the de Sitter space. We know that thissimply
connected surface is conformal to a subset ofEf although
it is notgeodesically
connected.2)
In[1]
and[4],
we can find a result due to Seifert which says that for aglobally hyperbolic
Lorentz manifold any twocausally
relatedpoints
may bejoined by
anonspacelike geodesic.
Acknowledgements
It is a
pleasure
to thank Professor J. Lafontaine for hisencouragements
and
stimulating conversations,
and the referee for his comments and sug-gestions.
- 510 - References
[1] BEEM (J. K.) and EHRLICH (P. E.) .2014 Global Lorentzian Geometry, Dekker, New York, 1981.
[2] CARRIÈRE (Y.) and ROZOY (L.) .2014 Complétude des métriques lorentziennes de T2 et difféomorphismes du cercle, Bul. Soc. Bras. Math. 25, n° 2
(1994),
pp. 223-235.