C OMPOSITIO M ATHEMATICA
A. J. E LLIS
Central decompositions for compact convex sets
Compositio Mathematica, tome 30, n
o3 (1975), p. 211-219
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211
CENTRAL
DECOMPOSITIONS
FOR COMPACT CONVEX SETSA. J. Ellis Noordhoff International Publishing
Printed in the Netherlands
1. Introduction
In this paper we continue the
investigation, begun
in[8],
into facialdecompositions
forcompact
convex sets K. Inparticular
westudy
conditions on K under which the
Bishop decomposition
determinesA(K),
or at least determines the centre of
A(K) ;
thespecial
case when K is thestate space of a unital
C*-algebra
is’investigated
in this connection. In the final section we prove a result for functionalgebras
which is relatedto facial
decompositions,
and we alsogive
asimple geometrical proof
of the Hoffman-Wermer theorem.
We are indebted to several mathematicians for discussions
concerning
the contents of this paper, and in
particular
to E. G. Effros and E.St~rmer.
2.
Terminology
andpreliminaries
Let K be a
compact
convex subset of alocally
convex Hausdorff space and letA(K)
denote the Banach space of all continuous real-valued affine functions onK,
endowed with the supremum norm. The set of extremepoints
of K will be denotedby ôK,
and its closureby
ôK. The sets ofconstancy
in ôK for the central functions inA(K)
form adecomposition {E03B1}
ofôK,
such thatEa
=8 Fa
for some closedsplit
faceF03B1
of K(cf. [1]).
It was shown
[8]
that thedisjoint
facesFa always
cover 8K and thatthey
determine
A(K)
in the sense thatThe
family {F03B1}
is called theSilov decomposition
for K.212
A subset E of ôK is a set of
antisymmetry
iff
is constant on E wheneverj E 4 j K )
andf IcoE belongs
to the centreof A(coE). (Here,
coE denotestlie closed convex hull of
E.) A(K)
is said to beantisymmetric
if 8K isa set of
antisymmetry.
It was shown[8]
that each x in êKbelongs
to amaximal set of
antisymmetry Ee
and that eachEp
has the formôfo,
where
(Fo)
is afamily
ofpairwise-disjoint
closedsplit
faces of K. Thefamily (Fo)
is called theBishop decomposition
for K.The essential set for
A(K)
is the smallest closed subset E of 8 K such thatA(K)18K
contains theideal {f ~ CR(K) : f (x)
=0, x~E}.
The setFE
= coE is the smallest closed(split)
face F of K such that whenever B and C are closed subsetsof ôK,
with B 2 F n ôK and C n F =p,
thencoB and coC are
split
faces of K(cf. [8]).
The annihilator inA(K)
ofFE
is the essential ideal of
A(K).
K is said to
satisfy Stprmer’ s
axiom if the closed convex hull of anarbitrary family
of closedsplit
faces of K isagain
asplit
face of K. Forx in ôK let
Fx
denote the smallest closedsplit
face of Kcontaining
x,and let U denote the
family
of all continuous affinebijections
a : K - Ksuch that
a(Fx)
=Fx
for all x in aK. Then K is said to admitsufficiently
many inner
automorphisms
ifFx
= co(ax :
a EUl
for all x in 8K.3
In
[8, Example 10]
anexample
wasgiven
ofasimplex
K for which theBishop decomposition
fails to determineA(K). However, using
anadaptation
of aproof
ofGlicksberg [10],
weproved
that if êK is closed then theBishop decomposition
does determineA(K) ;
we now extend thatresult.
THEOREM
(1): If
theBishop decomposition for
K covers 8K then thedecomposition
determinesA(K).
PROOF :
Let y
be an extremepoint
of the unit ball of the space of all Radon measures on êK which annihilateA(K)IaK.
As in[8,
Theorem8]
the result will follow if we can show that the
support
Dof J1
is containedin some
F /3 .
Suppose
that nosingle F 13
contains D. Let G be the smallest closedsplit
face of Kcontaining
all setsF 13
which intersect D. Then aG is nota set of
antisymmetry,
and so there exists a central functionf
inA(G)
which is not constant. Since the restriction of
f
toF 13
is central inA(F 13)’
f
must be constant on eachF03B2.
Foreach g
inA(G)
there exists an h inA(G)
such thath(x)
=f (x)g(x)
for all x in èG. If ybelongs
to D theny lies is some
F 13
contained inG,
and sincef
is constant onF 13
we haveh(y)
=f(y)g(y).
_Assuming
that 0f 1,
define Radon measures li 1, ,u2 on8K by
for all u in
CR(aK). Then /1
=03BC1 +
112and ~03BC1~ + ~03BC2~
= 1. For each g inA(K)
thereexists, by
the aboveargument together
with the fact that Gis a closed
split
face ofK,
an h inA(K)
such thath(y)
=f (y)g(y)
for ally in D. Therefore
so that 03BC1, and
similarly
,u2, annihilatesA(K)IîK. Since y
is extremeit follows that /11 = /1 = 03BC2, and
that f
is constant on D.If
f(y) = À
for all y inD,
let H = co{x~~G : f(x) = 03BB}.
Then H is asplit
face ofG,
and hence asplit
face of K. Moreover H contains all the facesFp
which intersectD,
so that H = G. But thenf
is constant onG,
and this contradiction
completes
theproof.
A consequence of Theorem 1 is that
if every point
of 8Kbelongs
to someclosed
split antisymmetric
face of K then theBishop decomposition
determines
A(K).
Thefollowing example
shows that the converse of Theorem 1 is false even for metrizablesimplexes.
EXAMPLE: Let K be a
compact
metrizablesimplex
such that K = K(cf. [14]).
DefineK
such thatwhere L is a fixed two-dimensional
subspace
ofA(K) containing
theconstants
and ~f~
=sup {~fn~ :n ~ 01.
ThenK
is the closed convex hull ofcountably
manycopies Kn
ofK, together
with a closed linesegment
I.The
Bishop decomposition
forK
consists of the facesKn
and the extremepoints
ofI,
andaK
is the union of theK, together
with I. Hence theBishop decomposition
fails to coverêk. However,
iff ~ CR(aK)
is suchthat
11 K,,
EA(Kn)
for each n it isstraightforward
to check that1
is affineon
I,
and that1
is the restriction of somef
inA(K);
therefore theBishop decomposition
determinesK. Finally,
since K is asimplex A(K)
has theRiesz
interpolation property,
and iff l , f2,
,g 1, 22 EA(k)
with214
for some e >
0,
it is easy to constructcoordinately
anh ~ A(K)
withfi, f2 ~ h ~ g1,
g2 . It follows thatK
is asimplex.
It has been
pointed
out to us that if K is the state space of the C*-algebra
in[16,
p.136] then,
in a similar manner, it can be shown that theBishop decomposition
determinesA(K)
but does not cover aK. We donot know any necessary and sufficient conditions for the
Bishop
de-composition
to determineA(K).
Every
closed linearsubspace
L ofCR(ÊK)
which containsA(K)IAK
isisometrically order-isomorphic
to a spaceA(H),
for somecompact
convexset H. The next result shows that there
always
exists a smallest such space L for which theBishop decomposition
for N determinesA(H).
THEOREM
(2): If K
denotes the state spaceof
the ordered Banach space L ={f
ECR(ôK) : f l(Fp
noK)EA(Fp)I(Fp
naK), 03B2},
then theBishop decomposition for K
determinesA(K).
PROOF : Since the
Choquet boundary êk
for L containsîK,
theSilov
boundaries
~K
for L and îK forA(K)loK
coincide. Let x ~ ~K and let C be the maximalA(K)-antisymmetric
subset ofak
which contains x. Ifx ~
F 13
and if G is the smallest closedsplit
face ofK
which contains~F03B2,
then we will show that
A(G)
isantisymmetric.
In
fact,
letf
be central inA(G),
so thatf
extends to a function inA(K).
Then,
ifg ~ A(F03B2)
=A(K)IF 13’
there exists an h inA(k)
such thath(y)
=f(y)g(y)
for all y in~F03B2.
The definition ofL, together
with the factthat
F 13 belongs
to theBishop decomposition
forK,
nowimplies
thatf
is constant on
oF p. Since f
is central inA(G)
and G is the smallest closedsplit
face ofK containing ~F03B2,
we conclude thatf
is constant onG,
therefore C contains~F03B2.
If
f
ECR(~K)
and iff 1 C
EA(co C)j C (where
co C denotes the closedconvex hull of C
in K,
and hencebelongs
to theBishop decomposition
for
K)
thenf|C
has an extensionf E L.
HencefloF 13
has an extensionbelonging
toA(F 13).
Since x ~ ~K was chosenarbitrarily
it follows that theBishop decomposition
forK
determinesA(K).
For the
simplex
K in[8, Example 10]
the setK
is a Bauersimplex,
identifiable with the base of the
positive
conein t 1 .
In thisexample
thecentres of
A(K)
andA(K)
aredistinct,
and the essential ideals ofA(K)
and
A(k)
are also distinct. If K(not necessarily
asimplex)
satisfies certainconditions,
then thefollowing
result shows that these distinctions do not occur.THEOREM
(3):
Let Ksatisfy Stormer’s
axiom and admitsiffficiently
many inner
automorphisms.
ThenThe essential ideals
of A(K)
andA(K)
coincideand,
inparticular,
K is aBauer
simplex if
andonly if K
is a Bauersimplex.
PROOF : For each x E aK denote
by Jx
theprimitive
idealso that
Jx
E PrimA(K). By [1,
Theorem II6.30]
the map xJx
isopen from the relative
topology
of aK to the hull-kerneltopology
ofPrim
A(K).
Now iff
ECR(oK)
is constant on eachFp
naK,
thenf
isconstant on each
aFx
sinceôFx
is a setof antisymmetry
forA(K).
Thereforethe
equation
defines a continuous function
f
on PrimA(K),
and hence Alfsen and Andersen’s version of the Dauns-Hofmann theorem[1,
Theorem II7.20]
shows that
f|~K belongs
to(the
restrictionof)
the centre ofA(K).
If hbelongs
toA(K)
then the definition ofA(k)
shows thatf h. belongs
toA(K)IêK,
and thereforef belongs
also to(the
restrictionof)
the centreof
A(K).
If 1 is central in
A(K)
then l iscertainly
constant on eachF03B2.
If l’ iscentral in
A(K)
thenl’|(F03B2 ~ ~K)
is central inA(Fp);
in fact ifpEA(Fp)
then p =
qlF 13
for some q inA(K),
and so there exists ans ~ A(K)
withs = l’ · q
onoK,
that issl(Fp
n~K) ~ A(F03B2)|(F03B2
naK)
and s = l’ · p onaffi. Therefore,
since the centre ofA(F 13)
istrivial,
l’ is constant onF 13
n ôK. Thiscompletes
theproof
of the first statement of the theorem.If
f belongs
to the essential ideal ofA(k)
thenf
is in the centre ofA(K),
and hencebelongs
toA(K) by
the aboveargument.
Therefore the essential ideals ofA(K)
andA(K)
coincide. Since K andK
are Bauersimplexes
if andonly
if their essential ideals coincide withA(K)
andA(K)
respectively,
the final statement is evident.COROLLARY
(4): If
K is the state spaceof
a unitalC*-algebra A,
then theBishop decomposition for
K de termines the centreof
d.PROOF:
A(K)
can be identified withd h,
the set of hermitian elementsof A,
and K satisfiesStormer’s
axiom and admitssufficiently
many inner216
automorphisms.
The centre ofA(K)
isnaturally
identifiable with the hermitian elements in the centre ofA,
and hence the result follows.(For
references see
[1].)
When K is the state space of .91 the closed
split
faces of K are theinvariant
faces,
andthey comprise
the annihilators in K of the closed two-sided ideals of A.Using
these facts it is not difficult to see that theSilov decomposition {F03B1}
for K consists of the annihilators of the closed two-sided ideals{I03B1}
in A which aregenerated by
the maximal ideals of the centre of A. TheBishop decomposition {F03B2}
for Kcorresponds
tothe
family {I03B2}
of closed two-sided ideals in,.ç/ which are minimalsubject
to the
property
thatA/I03B2
has trivial centre.In
general
we have been unable to decide whether theBishop
de-composition
determines A. Thisis,
of course, so if aK is closed and it is also true if .91 is aweakly
centralalgebra.
For an
arbitrary compact
convex set K we say thatA(K)
isweakly
central
if J1
=J2
wheneverJ1
andJ2
are maximal near-lattice ideals inA(K)
such thatJ1
n Z =J2 n Z,
where Z denotes the centre ofA(K) (cf. [5]).
THEOREM
(5): If A(K)
isweakly
central then theBishop
andSilov decompositions for
K coincide. Inparticular,
theBishop decomposition
de termines
A(K).
PROOF : Let
Fa
be a face in theSilov decomposition
for K and suppose that the centre ofA(Fa)
is non-trivial. Then there existnon-empty disjoint
closedsplit
faces G and H ofFa
andhence, by
Zorn’slemma,
there existdisjoint
minimalsplit
facesG 1 and H 1
of K contained inFa .
But then I =
(G1)
and J =(H1)
are maximal near-lattice ideals inA(K)
with I n Z =(Fa)1
n Z = J nZ,
since the functions in Z areconstant on
Fa .
Therefore we obtain I = J andG 1
=H1,
and this contra- diction shows thatA(Fa)
has trivial centre. Since theBishop decomposi-
tion is at least as fine as the
Silov decomposition
the twodecompositions
must coincide.
COROLLARY
(6):
Let .91 be aweakly
central unitalC*-algebra, for example
aW*-algebra,
with state space K andBishop decomposition {I}.
Then a
function f
inCR(ÎK)
is the restriction to îKof
an hermitianelement
of
.91if
andonly if, for
eachf3, f
coincides onIt
n aK with anhermitian element
of WIIP -
The result of
Corollary
6 can also be deduced from the results ofVesterstr~m [16].
The
Bishop decomposition
for K will determine A if itdistinguishes -Çih amongst
the Banachsubspaces
ofCR(êK).
If somesubspace
L ofCR(ôK)
such thatA(K)IAK g
L is identifiable withP4h
for some unitalC*-algebra P4 containing
.91 as asubalgebra,
then Aseparates
thepoints
of the pure state space aK of £3 so that A = B
by
Glimm’s Stone-Weierstrass theorem
[11].
In this sense theBishop decomposition
for Kalways distinguishes
Aamongst C*-algebras.
4
Let A be a function
algebra
on acompact
Hausdorff spaceX,
let S bethe state space of A and let K = co
(S ~ - iS)
with the relative w*-topology.
It was shown[8]
that theBishop decomposition
for K corre-sponds
toBishop’s decomposition
of X into maximal setsof antisymmetry
for A
[4],
and hence thedecomposition
determinesA(K);
in fact theBishop decomposition
for K consists of the sets co(Eo w - iEp) together
with Xy
and - ixy,
where{x03B3, Epl
are the maximal setsof antisymmetry
in X for A
(the Ep containing
more than onepoint),
and so theBishop decomposition
for K covers X and aK.The map 0 : A -
A(K),
where0 fiz)
= ref (z)
forf
EA,
z EK,
is areal-linear
homeomorphism (cf. [3]). Using
this map, we see thatre A = {re f : f ~ A}
andA(K)|S
areisometrically
orderisomorphic (cf. [6]).
We now use these facts to
give
asimple geometrical proof
of theHoffman-Wermer theorem
[12].
THEOREM
(7): (Hoffman-Wermer) If
A is afunction algebra
on X suchthat re A is
uniformly closed,
then A =Cc(X).
PROOF : We need to show that the
Bishop decomposition
for K consistsof
singletons. Suppose
thatKo
= co(S, u - iSo) belongs
to theBi’shop decomposition
forK,
whereSo
is a closedsplit
face of S. ThenA|(S0
nX)
is a function
algebra
onSo
n X with state spaceSo.
Since re A isuniformly
closed we have
A(K)|S
=A(S)
and henceTherefore lin
So
is w*-closed in linKo
=A(Ko)* (cf. [1,11.5])
where ’lin’denotes real-linear hull. Since
A(Ko)
isantisymmetric (S0)
is one-dimensional in
A(Ko),
so that linSo
has codimension one in linKo.
If- iSo
has two distinct extremepoints
xland
x2they
are bothsplit
faces218
of
Ko,
and so x1 and co(SO U x2)
aredisjoint
closedsplit
faces ofKo.
Hence lin x1 and lin
(80 U x2)
intersectonly
at0,
and this contradiction shows thatSo
is asingleton.
But thenKo
cannot beantisymmetric,
andthe theorem is
proved.
A result of
Fakhoury [9]
andNagel [13]
states that if K is asimplex
then the centre of
A(K)
is thelargest
closed sublattice ofCR(AK)
which iscontained in
A(K)laK. (See
also[15]
for the connections with facialtopologies
ofêK.)
If A is a non-trivial functionalgebra
and K is theassociated set defined above we will show that a
largest
closed sublattice ofCR(ÊK)
contained inA(K)IÊK always exists,
but is neverequal
to thecentre.
THEOREM
(8): If
A is afunction algebra
on X andif A1
= A nCR(X)
then
03B8(A1
+iA1)loK
is thelargest
sublatticeof CR(AK)
which is containedin
A(K)IêK.
This space coincides with the restrictionof
the centreof A(K) if
andonly if A
=Cc(X).
PROOF : Let L be a maximal sublattice of
CR(ôK)
which is contained inA(K)loK,
and letf
=8(u+iv)EL with ~f~ ~
1. Since L is a closedsubalgebra
ofCR(aK) containing
the constants, the functioncp(f)
=03B8(~(u)
+i~(v))
belongs
to L for every continuous real-valued function cpon [-1, 1].
It follows that
~(u)~re A,
and a result of Arenson[2] gives u~A1.
Similarly,
we havev E A
so that L is containedin,
and henceequal
to,the sublattice
8(A
1 +iA1)loK.
It was shown
[7]
that the centre Z ofA(K)
consists of the functions0(u + iv),
where u,v E A,
and(u - v) belongs
to the essential ideal I of A.The function
0(1) belongs
toL,
but it does notbelong
to Z unless 1e I,
that is A =Cc(X).
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(Oblatum 1-VII-1974) Department of Pure Mathematics
University College of Swansea Swansea, Wales.