Analysefonctionnelle/.Fwwcfr'0Ha/Analysis
A
one
parameter
group
of
automorphisms
of if
(F00)(x)B(H)
scaling
the trace
FlorinRÂDULESCU
Abstract — Weprovein thispaperthe existenceofaone realparameter, continuous group a,of
automorphismsofthe von Neumannalgebra S£(Foe)®B(H), scaling the trace by(. Hère
if
(Foe)isthevonNeumannalgebra associatedtoafreenonabeliangroupwith infmitely manygenerators,
while B(H) is the algebra ofail boundedlinear operators on a separable, infinité dimensional
Hilbertspace H,andxisthecanonical trace.
In particular it follows that there exists a type III, factor having a core isomorphic to J5f(Foe)®B(H), obtainedby takingthe crossproduct of.S?(FM)®B(H) by thegivengroup of
automorphisms.
Un groupe
à un paramètre d'automorphismes
demodule
R*sur l'algèbre du
groupelibre
à uneinfinité
degénérateurs
Résumé— Nous montrons l'existence d'un groupe a,,àunparamètre teR%,d'automorphismesde l'algèbreSC(FJ®B(H),telque x(O,(X))=tx(x), t>0,
xeif
(FJ®B(H).IciSC(FJ
est l'algèbrede vonNeumann associéàungroupelibre noncommutatifavec une infinité degénérateurs, tandisque
B(H) est l'algèbre des opérateurs bornéssur un espace hilberlien separable H et % est la trace
canonique.
En particulier, il s'ensuitqu'ilexisteunfacteurM,detypeIIIj telquel'algèbredes élémentslaissés
fixésparlegroupe modulaire (associéàunetrace généralisée sur M) soit isomorpheàS£(Foe)®B(H)
[i.e.M estleproduitcroisé deS£(Foe)®B(H)parlegroupeàunparamètred'automorphismes].
Versionfrançaiseabrégée—Dans cette Note nous utilisons une description de l'algèbre
de von Neumann .S?(F00)®B(H)à l'aide des travaux de Voiculescu [8] sur les algèbres de
von Neumann associées à des groupes libres.
On montrera que ^f(Foe)®B(H) est engendrée par une algèbre abélienne diffuse A
=
L°°(R) (avecla trace donnée par la mesure de Lesbegue) et par des parties finiespXp,
où p est une projectionfinie dans A, et X un élément symétrique non-borné. En outre, la
trace sur toute sous-algèbre engendrée parpL°°(U)et
pXp
(où l'unité p est une projection finiedans A) est uniquement déterminée parles valeurs qu'elle prend sur />L°°(R) et X (lesdeuxalgèbres sont indépendantes par rapportàla trace, ausens des probabilités non
commu-tatives de [7]).
De façon intuitive, si Y est le générateur de A, on peut dire que .5?(Foe)®B(H) est
l'algèbredes quantités mesurablesassociéeà deux observablesY, X, la deuxième ne
permet-tant de mesurer que des quantités qui sont finies par rapport à Y (i.e. de la forme
pXp,
où p est un idempotent de mesure finie, dans l'algèbre de Y).
De plus, pour toute projection finiep dans l'algèbre de Y, les observables
pYp
etpXp
sont indépendantes pour la mesure(au sens où la mesure est uniquement déterminée par sesvaleurs sur les algèbres engendrées par
pXp
et pYp).Les résultats de [7], permettent de retrouver le comportement de X, Y par rapportà la
trace, en considérant X comme une matrice aléatoire, quand Y est contenu dans l'algèbre des matrices diagonales. Si l'on considèrela limite de la représentation matricielle,la trace
est maintenant calculée comme la composée de l'espérance conditionnelleavec la trace (non normalisée) sur les matrices.
Note présentée par Alain CONNES.
Tout automorphisme de A donne un automorphisme de
if
(F00)®B(H) qui laisseX fixe.En particulier on prouve le résultat suivant:
THÉORÈME. — II existe un groupe à un paramètre (oc,)IeK*+ d'automorphismes de
5£ (Foe)®B(H), tel que les a, multiplientpar t la tracecanonique x sur
if
(Foe)®B (H) [i.e.T(a,00)
=
fT(x),*ei?(FJ®B(H),
t>0].
La classification [1] de A. Connes des facteurs de type
III
montre que l'existence d'unautomorphisme de
if
(Foe)®B(H), multipliant la trace par t est équivalente à l'existence d'un facteur M de typeIII,, tel que l'algèbre des éléments fixés par le groupe modulaire (associé à une trace généralisée sur M) soit isomorphe au facteur £C(Fco)®B(FL)[/e(0,
1)].Comme nous avons par ailleurs trouvé un groupe (a,),eM»+ d'automorphismes de
if
(Foe)®B(H) nousavons montré (à l'aide du travail de M. Takesaki[6]) qu'il existe aussi un facteur M de typeIIIj,
tel que l'algèbre des éléments fixés par le groupe modulaire (associé à une trace généralisée sur M) soit isomorphe au facteurif
(Foe)®B(H).COROLLAIRE. — 77existe unfacteur M de type III1 telque l'algèbredes éléments invariants
par le groupe modulaire (associée à une trace généraliséesur M) soit isomorphe au facteur »2'(Foo)®B(H). Ici M est leproduit croisé de <£(F00)®B(H) parle groupe d'automorphismes
(ai)t
6 K +•
Inthis paperwe find
a
new description forthe vonNeumannalgebra «S?(Foe)®B(H)which is a conséquence
of
Voiculescu'srandom matrix picture for the von Neumann algebras associatedtofree groups. We will provethat
if
(Foe)®B(H) is generated by a diffuse abelian algebraA=
{Y}"=
L°°(IR) and the finite partspXp,
of
an unboundedsymétrie élément X [which is
not
affiliated toif
(F00)®B(H)]. Hèrep
runs over the finite projections in A (offinite Lebesgue measure).From an heuristical point
of
view, we may saythat
we are given two symmetric observables X, Y andthat
if
(Foe)®B(H) is the algebraof
measurable observables, where only finite"parts"
pXp
(with respecttoY; i. e.p
is afinite measure projectionin thealgebragenerated by Y) may be measuredoutof
X. In additionthepositionof
the observables is sothat
for anyfinite trace projectionp
eA,p
Yp
andp
Xp
behave freely withrespectto the measure (in the senséof
the noncommutativeprobability introduced by D. Voiculescu in[8]).The matrixpicture
of
D. Voiculescuforfreeproductsshowsthat
wemay assume,that
asymptotically,Xis a matrix
of
very large size with independent random entries, while Yconsistsof
constant diagonalmatrices. Themeasure is then computedby composing the expectation value with the (nonnormalized)matrixtrace.The mainresults
that
weobtainin this wayarethe following:THEOREM. — There exists a one parameter continuous group (a,)(eK*+
of
automor-phisms
of
i?(Foe)®B(H)
such that ex, scales the trace x onif
(F00)®B(H) by t [i.e.x(a,(x))
=
tx(x),xeif
(FJ®B(H),
t>0].
By Connes's classification
of
typeIII
factors [1] it is well knownthat
any suchautomorphisma, on if(F0O)®B(H) scaling trace by
te(0,
1), gives a typeIII, factor with a core isomorphictoif
(F00)®B(H).The fact
that
we hâve a continuousone parameter groupof
automorphismsshows (by the workof
M. Takesaki[6])that
we may now in addition find a typeIIIj
factor havingacore isomorphictoif
(Foe)®B(H).COROLLARY. — There exists
a
typeïïï1factor
M havinga
core [1] isomorphic toif
(Foe)®B(H). Equivalently M is thecrossedproducto/if
(Foe)®B(H)with the actionof
Uinduced by the oneparametergroupof
automorphisms.In particularwe reprove in this way
that
the fundamental groupof
if
(Foe) is R+/0 [equivalentlythat
for any selfadjoint nonnull idempotent e in i?(Fa>)>e&ÇF^e
^£C(Fm)].This coverspartlythe main resuit in [9].
The results suggests
that
apossible connection could be realized betweenthe noncom-mutativeprobability theoryrecently introduced by D. Voiculescuand thework([3], [4], [5], [2])on algebraicquantumfieldtheory.The (verynatural)question about thepossible existence
of
atype III1 factorhavinga
core isomorphic toif
(Foe)®B(H) was first raised to me by M. Takesaki. We are indebtedtoD. Voiculescuforseveral usefulremarks.We will first construct
a
semifinite algebrase
which we show to be isomorphic toif
(Foe)®B(H) and then showthat se
has a one parameter groupof
automorphisms scalingtraceby t, te
U.The idea istoconsider firstthesubalgebra
J/0
of
thefreeproductL00(R)*C[X](X ishère an undetermined variable, assumed to be selfadjoint) generated by L00(U) and the
finite parts
pXp
of
X, wherep
runs over the setof
ail finite Lebesgue measureprojections in Loe(U).
If
one requiresthatp Xp
is semicircular (withthecorresponding normalization conditions coming from Proposition2.3 in [7]) andfree with the algebrapLx(U),
one gets in this way a well defined trace x on sé0. We réfèreto [7] for the définitions regardingnoncommutativeprobability.We let
se
be the von Neumannalgebra coming from G.N.S. construction forsé0 andx. Clearly anyautomorphism
a
of
L00(U),scalingtraceby / inducesan automorphismoc0ofstf0 (mapping Xinto t~1,2X). Moreover, by proposition2.3 in [7],<x0also scales
trace byt.
It
remainsto showthat
se
isisomorphictoif
(Foe)®B(H). Thisis done exactly by the same technics as the one used in theproof
ofTheorem 3.2 in[7].Werecall first
a
fewdéfinitions from [7], [8]. Let (A, cp) be a C*-algebra withunit1equiped with
a
trace cp. A familyof
subalgebras 1eA,£
A(ieï)
iscalledafree familyof
subalgebrasif
q>(a1a2.. .a„)
=
0 whenever0,-eA,-.withij^ij+1,
l^j-^n—l,
(p(aj)=
0,7=1,
2...n. A family
of
subsets(n;),-eI is called freeif
the familyof
the subalgebrasgenerated by theCîtand 1 is free. Moreoverafamily(/j)i6lis free
if
the family{f,
1}"is free. A free family (fi)i is called semicircularif
f
areselfadjointandif
thedistributionsof
theyj-,ieï,
are given bythesemicirclelawIn particular it follows
that
the von Neumannalgebra generated by an infinité free semicircular family C/î),eS is isomorphic to if(FcardS). To observe that, one has to consider unitaries F;that
generate the same abelian (diffuse) von Neumannalgebra asf,
for eachieS.
It
followsthat
the family (F,)ieS is also free andhencethat
the vonNeumannalgebra generated by them isisomorphicto
i?
(Fcards) (see[7]). BycardS wedénote the cardinality
of
S.CONSTRUCTION OF THE ALGEBRA
JJ.
-
Let A=
Loe(IR) with the trace <ï> commingfrom the canonical Lesbegue measure on U and let
Lf(U)
be the subalgebraL2(R)
H
L°°(R). Let X be an arbitrary variable and let sé0 be the subalgebraof
the freealgebraicproductgeneratedby
Lf
(U) and{pXp\p
projectioninLf(U)}.
srf0 has
a
canonical involutivestructureif
we requirefor X tobe selfadjoint.We now definea traceT onsé0 which is uniquely determined subjectto thefollowing
conditions:
(1) xrestrictedto
Lf
(R)=
O.(2)
pXp
andp
L00(R) are free with respecttotherestrictionof
x topsé0p
(3) <!)(p)~ll2pXpis semicircular with respectto thetrace(TQ?)-1)T
onpsé^p.
In the conditions(2) and (3) abovep
runs over the setof
ail finitetraceprojections in L00(R).Theexistenceand the unicity
of
x followsfrom the factthat p Xp
andp
L00(R) aregenerating in
psi'0p
an algebra stfp (with unit p) which isa
copyof
the algebraic free productwhile on thislastalgebra (withunit p) conditions(2), (3)are defininga uniquenormalized trace xp= (x(p))_1x [7]. Since s#0 is the union
of
the algebras sev, after ail finite projectionsp
inLf
(R), weobtainin this waya trace onsrf0.The fact
that
the trace is well defined followsfrom proposition2.3 [7] which assertsthat
wheneverD is an abelian algebra, free to the algebra generated by a semicircular élément Z, then for any nonnull projectionqeD,
the algebras generated by qT> andqZq
are free, while %(q)~112qZq is semicircular with respect to the normalized trace(x(<7))_1X(Xistne trace onthe algebra generated byD andZ).
To check
that
the traceis well defined we hâve to take two nonnullprojectionsp, q inLf
(R). We may assumethat
q^P-
The tracexp= (T(/?)_1)T defined by the condi-tions (2), (3) on the subalgebra sé°pof
s/0
(with unit p) is sothat
x(p)"ll2pXp
is semicircularandsothat
L00(U)p andpXp
arefree.But the propositionrecalled before (from[7]) shows
that
in qstf°pq,(xp(q))-ll2q((x(p))-ll2pXp)q=(x(q))-ll2qXp
is semicircular and
that
(T(q))~1,2qXq and qL°° (R) are free with respect to the traceAs for
p
let srf\ be the subalgebraof
s40 (with unit q) generated by Loe(R) q andqXq.
Letxq=(x(q)_1)xbe the tracedefined by conditions (2), (3) on ,s/° [Le. sothat
(T(q))~1!2qXqissemicircularandfree with qL°°(R)]. Clearlysé°q
£
qstfpqandwe only hâve to checkthat
xpq restricted tosé\
coincides with xq [as this will implythat
xprestricted to
sé\
is(0(^)/0
(p))xg]. Butthis follows from the factthat
both traces are makingqXq
semicircular and free with respect to qL™(R). Thus the restrictionof
0
(p)xptoséq coincideswith4>(q)xqandhencexis welldefined.The algebra
se
is now obtainedby takingthe G.N.S. constructionassociated tos/0
and x and by considering the weak closure
of
stf0 in this représentation. In this way we obtain a semifinite faithfull tracex onse
which extendsthe given trace. A systemof
generators forse
is L°°(R) and\pXp\p
projectionin L*(R)}. [Notethat
nowLoe(R) is
a
subalgebraof se
(containing the unitof
se) andthat
the restrictionof
x toL00(R) is$.]
PROPOSITION. — With the notations above wehâve
J^
=
if
(F00)®B (H).Proof. — Let
eu
e2,. . . be
a
partitionof
the unity in LM(R) with projectionsof
equal finite trace. Let vt,
ieH
be the partialisometriesfrom the polar décompositionof
eiXe1=
vibi,ieN,
i^>2. The behaviourof
Xwhencutted offby finiteprojectionsin L°°(R) [Le. properties (2), (3) above] and by theproof of
Theorem 3.2 in [7], showsthat
vfvi=
e1;vtvf=
et and moreoverthat
the family(4) {e1Lco(U)e1}U{vfReXvi,
vfïmXvtliJeN,2Si^j}U{bi, a,|ieN,
z^2} is free. Hère at is a semicircular generator for vfLao(U)vi. On the other hand, by constructionse
is generated as avonNeumannalgebra by L00(M)and{pXp\p
projection in Lj?(R)}.Hence it follows
that
the family (4) is indeed a Systemof
generators for e1se
ex and thusthat
psép
is isomorphic toif
(Foe) for any projectionpeLf(U).
Thusse
is afactorisomorphicto
if
(F00)®B(H).Using this new description
of
if
(Foe)®B(H)(^sé)
we are going to showthat
any isomorphism
of
L°°(R) induces an automorphismof
if
(Foe)®B(H)^^/.
The automorphismis first defined algebraicallyonsé0 and thenextended bycontinuity to se.PROPOSITION. — With the notations above let 0 be an automorphism scaling trace by
t>0
o/L°°(U). Dénote byS the automorphisminducedon L00(R)*C[X],mapping L00(U)onto L00(R) by 0andX into
t~m
X.ThenS induces (by restriction) an automorphism
of
sé0, also denoted by S, which scalesthe trace by t andthus an automorphismS
of se
scaling trace by t.In addition the depencence8-»Sispoint-continuous.
Proof. — Clearly we only hâve to prove
that
S scales trace by t.It
is sufficienttoshow
that
S restrictedto
psé0p
scales the trace by t. We may also assumethat
x(p)
=
$(p)=l.
Letq=Q(p).
Let
eu
e2,. .
.,
enbe any nnonnullprojections inp
L°°(R). Dénotebyxp=
(x(p))~Jx,
xq= (x(#))-1x the trace on
psép,
respectivelyonqstfq,
given by the conditions(2),(3). [Inparticularx restrictedto
p
se
p
isx^while therestrictionof
x toqse
qisx(q)xq.]Moreover
forany
iu
. .
.,
ime{1,...,«}
since 0 scalestrace onL™(U)by t.Thus
Because
of
property(3)of
x,t~1/2qXq
is semicircularandfreewith respectto Lco(U)qin the algebra
qsé
q (with trace xqand unitq).The relation(5) and the fact
that
the trace is uniquely determined on a free product[8]
if
it
is determined on his factors showsthat
we may continue the previous equality withWe hâve used hèrethefact
that p Xp
is semicircularandfree withp
Loe(U)in se'p,with respectto the tracex . ThusSscales the trace onsé0 andhenceforth on se.Thelasttwopropositionsgivethe
proof
of
our mainresuit.Noteremisele5mars1992,acceptéele 15avril1992.
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