A NNALES DE LA FACULTÉ DES SCIENCES DE T OULOUSE
V LADIMIR G OL ’ DSHTEIN
M ARC T ROYANOV
The L
pq-Cohomology of SOL
Annales de la faculté des sciences de Toulouse 6
esérie, tome 7, n
o4 (1998), p. 687-698
<http://www.numdam.org/item?id=AFST_1998_6_7_4_687_0>
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- 687 -
The Lpq-Cohomology of SOL(*)
VLADIMIR
GOL’DSHTEIN(1)
and MARCTROYANOV(2)
Annales de la Faculte des Sciences de Toulouse Vol. VII, nO 4, 1998
On prouve un résultat de non-annulation de la cohomologie
non reduite du groupe de Lie SOL.
ABSTRACT. - We prove a non vanishing result for the unreduced Lpq- cohomology of the Lie group SOL.
1. Introduction
SOL is the three dimensional Lie group of 3 x 3 real matrices of the form
This is a solvable and unimodular group; it is
diffeomorphic
toR3 (with
coordinates x, y,
z).
A left invariant Riemannian metric isds2
=dx2
+e2z dy2
+dz2 ;
its volume measure isgiven by
dxdy
dz and is bi-invariant.For more information about the
geometry
of this group, see[9].
Let us recall the definition of the unreduced
Lpq-cohomology
groups, let(M, ds2)
be acomplete
connected Riemannian manifold of dimension n. We( *)
regu le 12 septembre 1997, accepte le 19 juin 1998
(1) ) Departement of Mathematics, Ben Gurion University of The Negev, P.O. Box 653, ,
IL-84105 Beer Sheva (Israel)
E-mail : vladimir@bgumail.bgu.ac.il
The first author was supported by U. S.-Israeli Binational Science Foundation, grant No 94-00732
(2) Departement de Mathématiques, E.P.F.L., CH-1015 Lausanne
(Switzerland)
E-mail : troyanov@math.epfl.ch
note
LP(M, Ak)
the space of differential forms ofdegree k
with measurable coefficients on M such thatWe denote
by Zp (M)
the set of differential forms inLp(M, Ak)
which areclosed in the sense of current and
by B;q(M)
the set of forms (J EAk)
such that there exists a
form §
ELq(M, Ak-l)
withd~
= 6. The unreducedLpq-cohomology
of(M, ds2)
isby
definition thequotient
Other papers
dealing
withLpq cohomology
are[2], [3], [8]
and[10].
Thegoal
of this paper is to prove thefollowing
result about the unreducedLpq- cohomology
of SOL. .THEOREM 1.- We have = oo
for
every 1 p, q oo.2.
Auxiliary
resultsThe main
ingredient
in theproof
of Theorem 1 is the nextproposition (which
is a kind ofduality argument
inLpq-cohomology).
PROPOSITION 2.1.- Let a E and suppose that
for
every e >0,
there exists a
form
such that
where a > 0 is
independent of
E(here 1/q
+ =1/p
+ =1~.
Thena
~ (in particular, ~ 0 ).
For the
proof,
we will need thefollowing integration-by-part
lemma(for
differential forms of class
C1,
this lemma is due toGaffney ~1~ ) .
LEMMA 2.1. Let M be a
complete
Riemannianmanifold.
Let~Q
Ebe such that
dQ
Eand 03B3
E nbe such that
dy
E where1/p
+ =1/q
+ = 1. .Then
dy and 03B3
Ad/3
areintegrable
andProof.
-By
Holder’sinequality,
the formsdy
A AdQ
and y A(3
allbelong
toL1.
For smooth forms,Q
withcompact support,
the lemma is trueby
definition of the weak exterior differentialAssume first that
,Q
is smooth with noncompact support
and satisfies the conditions of the lemma. On acomplete
Riemannian manifoldM,
wecan constuct a sequence
~az ~
of smooth functions with compact support such that -~ 1uniformly
on everycompact subset, 0 ~ ~i (x)
1 and1 for all x e M. The forms have compact support, thus the lemma holds for each
A; (3.
Sincewe can
apply Lebesgue’s
dominated convergence theorem. Thus we haveFinally,
for any,Q
E withdQ
E we can construct asequence
,~~
of smooth forms such that,Q~
-~,Q
inLP-topology
and ~dQ
in
LP-topology (see Corollary
1 of[4]).
Thus the samelimiting
process proves the lemma in all itsgenerality.
0Proof of Proposition
~.1.Suppose
that a =dQ
for somej~
ELq(M,
We haveby
Lemma1,
Using
Holder’sinequality,
weget
This is
impossible
since E > 0 isarbitrary. 0
Proposition
2.1 can becompleted
in thefollowing
way.LEMMA 2.2.- Let ai, a2, ... ar E
Zp (M)
and suppose that we canfind pairwise disjoint
setsSi
C M such thatfor
every ~ > 0 there exists= E n with
supp{ai )
Usupp (’Yi )
Csi
and such that
II d03B3i I)
q, E andM
y2 A ai > a where a > 0 isindependent of
E and i. Then~al~, [a2],
... ,[ar]
arelinearly independent
elementsof
Proof.-
Chooseai
ER,
i =1, ... ,
r, and set a = and y = Theassumption
on thesupports
of these formsimplies
that
This sum vanishes if and
only
if allÀi
= 0.Since y
ELP’
nLq’
andwe can deduce from
Proposition
1 that~ ~i (a~ ~
0 E unlessall
A;
= 0. DFor all
x0 ~ R
the surfaceis a
totally geodesic
surface isometric to thehyperbolic plane
Inparticular
a functionf :
SOL -~ 1~8 which is invariant under all x-translations{i.e., f
=f(y, z))
can be seen as a function on thehyperbolic plane.
LEMMA 2.3. - There exists two non
negative
smoothfunctions f
and gon
IHI2
^_rHx0
such that:~1~ f (y, z)
=g(y, z)
= 0ifz:S;
0 or~y~
>1;
(2) d f
anddg
EA1) for
any l r oo;(~~
thesupport of df A dg
is contained in{(y,x~ 1
, 0 z1}
t~4~ f ~~2 df
ndg = 1~
;(5) 8f/8y
and8g/8y
E and8f/8x, 8g/8z
havecompact support.
Remark. - The forms
d f
anddg
cannot havecompact support,
other-wise, by
Stokestheorem,
we would haveProof.
- Choose nonnegative
smooth functionshl, h2
and k : II8 --~ ~8 with thefollowing properties:
(i)
= 0 if~y~ ~ 1~ >_
0 and ~ 0 for all y;(ii)
the function has nonempty support;
(iii) k~(z) >
0 for all z, furthermoreWe set
f(y, z)
:=hl(y)k(z)
andg(y, z)
:=h2(y)k(z). Property (1)
of thelemma is dear. We prove
(3) (i.e.,
thatd f
for any 1r oo),
wehave
The first term dz has compact support, and the second term
dy
has its support in the infiniterectangle Q
=1 ,
z >0~.
Choose D oo such that D on H. We have
thus,
since the element of area ofIHI2
is dA = e~dy dz,
we havefrom which one
gets d f
E Lr. .Now observe that
hence the
property (3)
follows from the construction ofh,l, h2
and k.Property (4)
isonly
anormalisation,
andproperty (5)
is easy to check. 0The
following
is avanishing
result for some kind of"anisotropic weighted capacity" .
LEMMA 2.4.- Given any numbers 6 and
q’
such that 1q’
oo and0 6
(1/2)(q’ - 1),
we can construct afamily of Lipschitz functions 03C8t
=z),
t >1,
onR2
such that:where the constant C =
C(6)
isindependent of
t.Proof.
- We first choose some number s > 0 solarge
that(s
+ 1 -q’s)/2s
-b and setp(x, z)
:=x2
+Izl2s.
We now define03C8t : R2 ~ R2 by
We will prove that
where the constant C is
independent
of t.It will be convenient to introduce new variables X = and Z =
(we assume z > 0).
We havethus p = t
(X2
+Z2).
Let us setWt (X, Z)
:=x),
thenIn
particular, Wt
isindependent
of t(and
will henceforth be written asw)
and its
support
is the annulus A ={(X, Z) ~ 1 (X2
+Z2) 2}.
The
partial
derivatives of1/;t
may be written asThe maximum of the function z --~ on 0 z oo is achieved at
z = (s - 1),
hencefor
all z >
0. From the secondequation
in(2.2)
and(2.3),
we conclude thatWe see from the first
equation
in(2.2)
and theinequality (2.4)
thatSince
we obtain
(2.1)
withwhere the domain of
integration
is the half annulusA+
={(X, Z) ~ Z >
3. Proof of the main theorem
The
proof
is technical and will be divided in fivesteps:
we first fix somearbitrarily
e > 0.Step
1. We construct a closed2-form
a EZp(SOL)
We start
by choosing
apair
of functionsf
=f(y, z)
and g =g(y, z)
withthe
properties
of Lemma 2.3. We then choose a smooth function A : IL8 -~ R such that, _ ~ _ _
Then we set
z)
= and note thatWe
finally
defineObserve that d~ == 0 and
!~/! ~1~>0, !.c!~};
2022
03BB’(e-z :c)
hascompact support;
2022
2014-
A is bounded(since )d.c
Ady|
= 1 and~f/~y
isbounded);
2022
2014 |x||dy^dz|
is bounded(since
== ez and|x|
~ e-z onQ).
From these estimates and
0 ~ V ~ 1,
we deduceeasily
that~ ~
const e"~on Q and _
for all 1
p
oo. It follows that a EStep
2. We construct afamily of
almost closedforms
03B3t EFix 0
(1/2)(q’ - 1)
and choose a function1/;t
=z)
asin Lemma 2.4.
Define 03B3t
:=z) dg.
In order to show that it E observe that it has itssupport
contained in the boxQt
:={(x,
y,z)
E SOL1 ,
z >0, ~x~ 2t} .
Recall that
0 ~ 1/;t(z, z)
1 and the volume form of SOL isd(vol) =
dx
dy
dz. We thus haveBy
Lemma2.3,
we know thatfrom which one
gets
an estimates Inparticular
Step
3. We estimate~d03B3t~q’
We have
and
Recall that is
bounded,
has compact support anddit
has itssupport in the
region Qt.
ThusSince 0 6
(1 /2)(q’ - 1),
Lemma 2.4implies
Step 4.
We estimate theintegral of
a A 03B3tLet _
We have
(since dzAd fAdg
=0). By
Lemma2.3, df^dg ~ 0,
and since 03BB’(e-zx)
> 0we see that a
n yt
is a nonnegative
3-form. Inparticular At > f0
an yt
for every measurable subset A C SOL.
We set
Recall that then
z)
=1,
we thusget
Now set u =
e-z x,
du = e-zdx,
uo =~~ ~B/~
and u1 =e-z ~.
We haveif t is
large enough (i.e., 1).
ThusObserve that the constant
C4
ispositive
andindependent
of t(in fact, using
equation (4)
of Lemma 2.3 and(i)
of Lemma2.4,
we see thatAt -
1 asStep
5.Recapitulation
Let us summarize the
previous
estimates(3.1), (3.2)
and(3.3):
If we let t ~ oo and
apply Proposition 2.1,
weobtain 03B1 ~ By
the constructionUsing
the group of isometries T(z,
y,z)
2014~(x,
y +2k, z),
k E?L,
we canproduce
an infinitefamily
of forms ai EZp (SOL ) satisfying
thehypothesis
of Lemma 2.2. Therefore
for all 1 p, q oo. The
proof
iscomplete D
4. Final remark
The above
proof
of Theorem 1 isonly
true for unreducedcohomology.
Infact,
the work of JeffCheeger
and Mikhael Gromovgives
us thefollowing
result in the reduced case
(for
p = q =2)
.THEOREM 2.2014 The reduced
L2-cohomology of
SOL is trivial.Proof.
- The Lie group SOL admits uniform lattices(i.e.,
discretecocompact
subgroups),
see[11]
forexplicit
constructions. The result thus follows from[5], [6]
and[7]
since every lattice in SOL is amenable. 0Acknowledgments
Finally,
we thank the referee who discovered a mistake in an earlier version of Lemma 2.1.References
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