an introduction to ii1 factors

AN INTRODUCTION TO II1 FACTORS

CYRIL HOUDAYER

Contents

1. Finite von Neumann algebras 21.1. Basics on von Neumann algebras 21.2. Finite von Neumann algebras 31.3. Orbit equivalence relations 62. Hilbert bimodules. Completely positive maps 132.1. Generalities 132.2. Dictionary between Hilbert bimodules and u.c.p. maps 152.3. Popas intertwining techniques 173. Approximation properties 223.1. Amenability 223.2. Property Gamma 273.3. Haagerup property 304. Rigidity of II1 factors 324.1. Rigid inclusions of von Neumann algebras 324.2. Applications to rigidity of II1 factors 344.3. Uniqueness of Cartan subalgebras 37Appendix A. Polar decomposition of a vector 39Appendix B. Von Neumanns dimension theory 39Appendix C. Ultraproducts 40Appendix D. On the geometry of two projections 41References 42

2000 Mathematics Subject Classification. 46L07; 46L10; 46L54.Key words and phrases. Finite von Neumann algebras; Equivalence relations; Hilbert bimod-

ules; Amenability; Haagerup property; Relative property (T); Cartan subalgebras.Research partially supported by ANR grant AGORA.

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2 CYRIL HOUDAYER

1. Finite von Neumann algebras

1.1. Basics on von Neumann algebras. Let H be a separable complex Hilbertspace. We shall denote by , the inner product on H that we assumed to belinear in the first variable (and conjugate linear in the second one). Let B(H) bethe algebra of all bounded linear maps T : H H. This is a Banach algebra forthe uniform norm:

T = sup1

T.

We moreover have ST ST, S, T B(H). The algebra B(H) isnaturally endowed with -operation called the adjonction defined as follows:

T , = , T,, H.We have (T ) = T , T = T and

T T = TT = T2.Thus, B(H) is a C-algebra. We can define several weaker topologies on B(H) aswell, in the following way. Let (Ti) be a net of operators in B(H).

Topology Ti 0norm Ti 0

ultra--strong n(Tii2 + T i n2) 0, (n) `2 H-strong Ti2 + T i 2 0, H

ultrastrongn Tin2 0, (n) `2 H

strong Ti 0, Hultraweak

nTin, n 0, (n), (n) `2 H

weak Ti, 0, , HFor a non-empty subset S B(H), define the commutant of S in B(H) by

S := {T B(H) : xT = Tx,x S}One can then define inductively S = (S), S(3) = (S), S(k+1) = (S(k)), for allk 1. It is easy to see that

S SS(k) = S(k+2),k 1.

Definition 1.1. Let M B(H) be a unital -subalgebra. We say that M is a vonNeumann algebra if M = M .

Theorem 1.2 (Von Neumanns Bicommutant Theorem). Let M B(H) be aunital -subalgebra. The following are equivalent:

(1) M = M .(2) M is strongly closed.(3) M is weakly closed.

Proof. We only sketch the proof.(1) = (2) is clear since commutants are always strongly closed.(2) = (1). Let x M . Let

V(x, 1, . . . , n, ) := {y B(H) : xi yi < ,i = 1, . . . , n}be a strong neighborhood of x in B(H). Let K = `2nH and observe that B(K) =Mn(C) B(H). Let = (1, . . . , n) K. Define V = (1 M) K. Since

AN INTRODUCTION TO II1 FACTORS 3

M is strongly closed, V is a closed subspace of K. Denote by PV B(K) thecorresponding orthogonal projection. Since (1 a)PV = PV (1 a), a M , itfollows that 1x commutes with PV , since x M . Thus (1x) V and we canfind y M such that (1 x) = (1 y), so in particular y N(x, 1, . . . , n, ).Then M is contained in the strong closure of M and hence M = M .

The fact that (2) and (3) are equivalent follows from Hahn-Banach SeparationTheorem (M is convex since it is a vector subspace!).

1.2. Finite von Neumann algebras. A von Neumann algebra M is said to befinite, if every isometry v M is a unitary, i.e.

vv = 1 = vv = 1,v M.One can show that a von Neumann algebra is finite if and only if it has a faithfulnormal tracial state : M C:

is a positive linear functional with (1) = 1. is faithful, i.e. x M , (xx) = 0 = x = 0. is normal, i.e. is weakly continuous on (M)1, the unit ball of M with

respect to the uniform norm . is a trace, i.e. x, y M , (xy) = (yx).

An infinite dimensional finite von Neumann algebra M with trivial center, i.e.M M = C, is called a II1 factor.

The most simple examples of finite von Neumann algebras are the following:(1) Abelian von Neumann algebra. Let (X,) be a standard probability

space. Represent L(X,) on the Hilbert space L2(X,) by multiplication

(f)(x) = f(x)(x),f L(x, ), L2(X,).The von Neumann algebra M = L(X,) comes equipped with the trace givenby integration against the probability measure , i.e. =

d.(2) Group von Neumann algebra. Let be a countable discrete group. The

left regular representation : U(`2()) is defined as followsst = st.

The von Neumann algebra of is then defined by

L() = {s : s }.The canonical trace on L() is = e, e. One checks that L() is a II1 factorif and only if the group has infinite conjugacy classes (icc), that is, t 6= e, theset {sts1 : s } is infinite.Exercise 1.3. Let T = [Tst]s,t B(`2()), with Tst = Tt, s. Show thatT L() if and only if T is constant down the diagonals, i.e. Tst = Txy wheneverst1 = xy1.

Assume that is abelian. Then the dual is a second countable compactabelian group. Write F : `2() L2() for the Fourier transform which is definedby F(s) = 7 s, . We get a canonical identification

L() = FL()F.(3) Group measure space construction. Let y (X,) be a probabil-

ity measure preserving (p.m.p.) action. Define an action : y L(X) by

4 CYRIL HOUDAYER

(s(F ))(x) = F (s1x), F L(X). We still denote by : U(L2(X))

the corresponding Koopman representation. We regard L(X) = L(X) 1 B(L2(X) `2()). The unitaries us = s s U(L2(X) `2()), for s ,satisfy the following covariance relation:

usFus = s(F ).

Observe that by Fells absorption principle, the unitary representation (us)s issimply a multiple of the left regular representation. The crossed product von Neu-mann algebra is then defined by

L(X)o =

{finite

asus : as L(X)} B(L2(X) `2()).

The trace is given by

(s

asus) =X

aed.

Recall that the action is said to be free if

({x X : sx = x}) = 0,e 6= s .It is moreover said to be ergodic if

A = A = (A)(1 (A)) = 0,A X.One can check that the action is free if and only if L(X) is maximal abelian inL(X)o. In that case, we say that L(X) is a Cartan subalgebra, i.e. L(X) L(X) o is maximal abelian and regular. Moreover, L(X) o is a II1 factorif and only if the action is ergodic.

Observe that when the probability space X = {pt} is a point, then the group vonNeumann algebra and the group measure space construction coincide, i.e. L(X)o = L().

Let M be a finite von Neumann algebra and fix a faithful normal trace. Weendow M with the following sesquilinear form

x, y = (yx),x, y M.Denote by L2(M, ) or simply by L2(M) the completion of M with respect to, . The corresponding 2-norm on M is defined by x2 =

(xx). Write

M 3 x x1 = x L2(M) for the natural embedding. Note that the unit vector 1is cyclic (i.e. M 1 is dense in L2(M)) and separating (i.e. x1 = 0 = x = 0) for M .For every x, y M ,

xy22 = (yxxy) (yxxy) x2y22,

so that we can represent M in a standard way on L2(M) by

pi(x)y = xy,x, y M.This is the so-called GNS-representation. Observe that pi : M B(L2(M)) is anormal -representation and pi(x) = x. Abusing notation, we identify pi(x)


with x M and regard M B(L2(M)). Let J : L2(M) 3 x1 7 x1 L2(M) bethe canonical antiunitary.

Theorem 1.4. JMJ = M .

Proof. We first prove JMJ M . Let x, y, a, b M . We haveJxJya, b = xay, b = (bxay) = (ybxa)

= xa, by = xJa, Jyb = yJxJa, b,so that JxJy = yJxJ .

Claim. The faithful normal state x 7 x1, 1 is a trace on M .Let x M . We first show that Jx1 = x1. Indeed, for every a M , we have

Jx1, a1 = Ja1, x1 = xa1, 1= ax1, 1 = x1, a1.

Let now x, y M . We havexy1, 1 = y1, x1 = y1, Jx1 = x1, Jy1

= x1, y1 = yx1, 1.Denote the trace x 7 x1, 1 on M by . Define the canonical antiunitary

K on L2(M , ) = M 1 = L2(M) by Kx1 = x1, x M . The first part ofthe proof yields KM K M = M . Since K and J coincide on M 1, which isdense in L2(M), it follows that K = J . Therefore, we have JM J M and soJMJ = M .

This Theorem shows in particular that the commutant of the (left) group vonNeumann algebra L() inside B(`2()) is the right von Neumann algebra R(),that is, the von Neumann algebra generated by the right regular representation ofthe group .

Exercise 1.5. Show that the strong operator topology on (M)1 is given by thenorm 2. Thus, strong and -strong topologies coincide on (M)1 since x2 =x2, x M .

Let B M be a von Neumann subalgebra. One can show that there existsa unique -preserving faithful normal conditional expectation EB : M B (seeSection 2 for details)1. The map EB : M B is unital completely positive andmoreover satisfies

EB(b1xb2) = b1EB(x)b2,x M,b1, b2 B.We say that EB is B-B bimodular. If we denote by eB : L2(M) L2(B) theorthogonal projection, we have eB(x1) = EB(x)1, for every x M .Proposition 1.6 (Fourier coefficients). Let y (X,) be a p.m.p. action. let A =L(X) and M = L(X)o . Every x M has a unique Fourier decomposition

x =s

xsus,

1These notes are nonlinear!

6 CYRIL HOUDAYER

with xs = EA(xus). The convergence holds for the 2-norm.2 Moreover, x22 =s xs22.

Proof. Define the unitary U : L2(M) L2(X) `2() by the formula

U

(finite

asus

)=finite

as s.

Then U 1U = 1Xe is a cyclic separating vector for M represented on the Hilbertspace L2(X)`2(). Abusing notation, we shall identify L2(M) with L2(X)`2().Under this identification eA is the orthogonal projection L2(X) `2() L2(X)Ce. Moreover useAus is the orthogonal projection L

2(X) `2() L2(X)Csand thus

s useAu

s = 1. Let x M . Regarding x(1X e) L2(X) `2(),

we know that there exists as L2(X) such that

x(1X e) =s

as s and x22 =sas22.

Then we have

as s = useAusx(1X e)= useAusxeA(1X e)= usEA(usx)(1X e)= EA(xus)(1X s).

It follows that as = EA(xus). Therefore, we have x =sEA(xu

s)us and the

convergence holds for the 2-norm. Moreover, x22 =s EA(xus)22.

Exercise 1.7. Let y (X,) be a p.m.p. action. Let A = L(X) and M =L(X)o .

(1) Show that y X is free if and only if A = A M (A is maximal abelian).(2) Under the assumption that y X is free, show that M is a II1 factor if

and only if y X is ergodic.(3) Assume that is icc. Show that M is a II1 factor if and only if y X is

ergodic.

Exercise 1.8. A von Neumann algebra M is diffuse if it has no minimal projection.

(1) Let N M be an inclusion of von Neumann algebras and let e N be aprojection. Show that e(N M)e = (eNe) eMe.

(2) Let M be a finite von Neumann algebra. Show that M is diffuse if and onlyif there exists a sequence of unitaries un U(M) such that un 0 weakly.

For more on C-algebras and finite von Neumann algebras, we refer to the ex-cellent book by Brown and Ozawa [2].

1.3. Orbit equivalence relations.

2The convergence does not hold in the strong topology.


1.3.1. Basic facts on measured equivalence relations. In this note, (X,) will de-note a nonatomic standard Borel probability space. A countable Borel equivalencerelation R is an equivalence relation defined on the space X X which satisfies:

(1) R X X is a Borel subset.(2) For any x X, the class or orbit of x denoted by [x]R := {y X : (x, y) R} is countable.

We shall denote by [R] the full group of the equivalence relation R, i.e. [R]consists of all Borel isomorphisms : X X such that graph() R. The set ofall partial Borel isomorphisms : dom() range() such that graph() R willbe denoted by [[R]]. If is a countable group and (g, x) gx is a Borel action of on X, then the equivalence relation given by

(x, y) R( y X) g , y = gxis a countable Borel equivalence relation on X. Conversely, we have the following:

Theorem 1.9 (Feldman & Moore, [11]). If R is a countable Borel equivalencerelation on X, then there exist a countable group and a Borel action of on Xsuch that R = R( y X). Moreover, and the action can be chosen such that

(x, y) R g , g2 = 1 and y = gx.Given a countable Borel equivalence relation R on X, we say that is R-

invariant if = , [R],

where (U) = (1(U)), for any Borel U X. The following proposition isuseful and easy to prove:

Proposition 1.10. Let R be a countable Borel equivalence relation defined on X.The following are equivalent:

(1) is R-invariant.(2) is -invariant whenever is a countable group acting in a Borel way on

X such that R = R( y X).(3) is -invariant for some countable group acting in a Borel way on X

such that R = R( y X).(4) [[R]], (dom()) = (range()).

For any U X, we define the R-saturation of U by[U ]R =

xU

[x]R

= {y X : x U , (x, y) R}.We have U [U ]R and [U ]R is a Borel subset of X. We say that U X is R-invariant if [U ]R = U (up to null sets). The equivalence relation R is said to beergodic if any R-invariant Borel subset U X is null or co-null.Exercise 1.11. Let R be a measured equivalence relation for which (almost) everyorbit is infinite. Show that there exists a sequence (gn) in [R] such that g0 = IdXand R = n graph(gn). In other words, we can write the equivalence relation R asa countable disjoint union of graphs of elements in the full group [R].

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Important Convention. In the rest of this paper, when we write an equivalencerelationR defined on (X,), we always mean a countable Borel equivalence relationsuch that the measure is R-invariant. When we write U X, we always mean aBorel subset of X. From now on, we will neglect null sets, i.e. whenever a propertyis true for every x X, we mean for -almost every x X. From now on, we willalways assume that (almost) every orbit of R is infinite, that is, R is a type II1equivalence relation.

We define now a Borel measure on R. For W R a Borel subset, we defineWx = {y X : (x, y) W} and Wy = {x X : (x, y) W}. We define on R by

(W) =X

|Wx|d(x),W R.

Lemma 1.12. Since is assumed to be R-invariant, we have:X

|Wx|d(x) =X

|Wy|d(y),W R.

Proof. From Exercise 1.11, we know thatR = n graph(gn), for some gn [R]. LetW R be a Borel subset. Then W = n(graph(gn) W) and graph(gn) W =graph(n) for n [[R]]. Thus, we can write W as a countable disjoint unionof graphs of [[R]]. Consequently, we just have to prove the equality whenW = graph(), for [[R]]. For any : dom() range() [[R]], we have

(dom()) =X

| graph()x|d(x)

(range()) =X

| graph()y|d(y).

Since (dom()) = (range()), we are done.

We shall denote by D := {(x, x) : x X} R the diagonal. We have (D) = 1.For i = 1, 2 let Ri be a countable Borel equivalence relation on (Xi, i), andassumed that i is Ri-invariant. We say that R1 and R2 are isomorphic anddenote R1 ' R2 if there exists a Borel isomorphism : X1 X2, such that1 = 2 and

(x, y) R1 ((x),(y)) R2.We will be using the following important fact. For : X1 X2 a Borel isomor-phism such that 1 = 2, we associate pi : L(X1) L(X2) defined by(piF )(x) = F (1x), for any F L(X1), and any x X1. Moreover the map pi is an onto isomorphism (see [28, Proposition XIII.1.2]).

1.3.2. Construction of the von Neumann algebra of R. We define the left regularrepresentation of the equivalence relation R on the Hilbert space L2(R, ). For : dom() range() [[R]] and L2(R, ), set

(u)(x, y) = 1range()(x)(1(x), y),(x, y) R.In other words, (u)(x, y) = (1(x), y) if x range() and 0 otherwise. Thevon Neumann algebra L(X) acts in the following way. If F L(X) and L2(R, ), we have

(F)(x, y) = F (x)(x, y),(x, y) R.


For F L(X) and [[R]], define F L(X) byF(x) = 1range()(x)F (1(x)).

In other words, F(x) = F (1(x)) if x range() and 0 otherwise.Note that, we can also define the right regular representation of R on L2(R, )

in the following way: for : dom() range() [[R]] and L2(R, ), set(v)(x, y) = 1range()(y)(x, 1(y)),(x, y) R.

Exercise 1.13. Show that for any , [[R]], we have(1) uu = u.(2) u = u1 .(3) uu = 1dom() and uu

= 1range().

(4) uF = Fu, for any F L(X).Definition 1.14 (Feldman & Moore, [12]). The von Neumann algebra L(R) B(L2(R, )) of the equivalence relation R is then defined as follows:

L(R) := W {u : [[R]]}.Likewise, we define R(R) = W {v : [[R]]}. It is trivial to check that

L(R) R(R). Define the canonical anti-unitary J : L2(R, ) L2(R, ) by(J)(x, y) = (y, x), for any L2(R, ), for any (x, y) R.Proposition 1.15. Denote by 0 = 1D L2(R, ) the characteristic functioncorresponding to the diagonal D R.

(1) 0 is a cyclic separating vector for L(R).(2) The vector state = 0, 0 is a faithful normal trace on L(R). In partic-

ular, L(R) is a finite von Neumann algebra.(3) For any [[R]], JuJ = v. In particular, L(R) = R(R).

Proof. (1) Write 0 = 1D, where D R is the diagonal. For [[R]], we have1graph() = u10. Recall that R can be written as a countable disjoint unionof graphs of Borel isomorphisms R = n graph(gn). Take any W R. ThenW = nWn, with Wn =W graph(gn). Since Wn is the graph of a partial Borelisomorphism, it follows that 1W L(R)0. Consequently, 0 is cyclic for L(R).Exactly in the same way, 0 is cyclic for R(R), hence separating for L(R).

(2) It suffices to prove that (uu) = (uu), for every , [[R]]. Let, [[R]]. We have

(uu) =X

1{x=()1(x)}(x)d(x)

=X

1{x=11(x)}(x)d(x)

=X

1{x=11(x)}(x)d(x) (x = 1(x))

=X

1{x=()1(x)}(x)d(x)

= (uu).

(3) A straightforward computation shows that JuJ = v, for any [[R]].It follows that JL(R)J = R(R). By the general theory of finite von Neumannalgebras (see Proposition 1.4), we get L(R) = R(R).

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Denote A = L(X) and M = L(R). Observe that A M . We know that thereexists a unique -preserving faithful normal conditional expectation EA : M A.In order to know EA, it is sufficient to compute EA(u) for any [[R]]. Denoteby eD : L2(R) L2(D, 0) the orthogonal projection.Proposition 1.16. We have

(1) EA(ug) = 1{x=g1x}, g [R].(2) eD(x0) = EA(x)0, x L(R).

Proof. In order to prove (1) and (2), it suffices to show that EA(u) = 1{x=1(x)},for every [[R]]. Let , [[R]]. Write f = 1{x=1(x)} L(X). Then, wehave

(EA(u)u) =X

EA(u)(x)1{x=1(x)}(x)d(x)

=X

EA(u1{x=1(x)})(x)d(x)

= (EA(u1{x=1(x)})) = (u1{x=1(x)})

=X

1{x=1(x)}(x)f(x)d(x)

=X

f(x)1{x=1(x)}(x)d(x) = (fu).

Thus, ((EA(u) f)x) = 0, for any x M . Consequently, EA(u) = f =1{x=1(x)}.

Proposition 1.17. Let (gn) be a sequence in [R] such that g0 = IdX and R =n graph(g

1n ). Then any x L(R) can be uniquely written

x =n

anugn ,

where an A.Proof. Since R = n graph(g1n ), we have

L2(R, ) =n

L2(graph(g1n ), n),

where n is the restriction of to graph(g1n ). Let x L(R). Define an =EA(xugn). Recall that 1graph(g1n ) = ugn0. Denote by eD : L

2(R, ) L2(D, 0)the orthogonal projection. It is easy to check that ugneDu

gn is the orthogonal

projection L2(R, ) L2(graph(g1n ), n). We haveugneDu

gn(x0) = ugn(eDu

gnx0)

= ugnEA(ugnx)0

= ugnEA(ugnx)u

gnugn0

= EA(xugn)ugn0.

Therefore x0 =nEA(xu

gn)ugn0 in L

2(R, ) and so x = nEA(xugn)ugnwhere the convergence holds for the 2-norm.

The above proposition yields in particular L(R) = (L(X) {ug : g [R]}).Proposition 1.18. Denote M = L(R) and A = L(X). Then


(1) A = A M , i.e. A M is a maximal abelian -subalgebra.(2) NM (A) = M , i.e. A M is regular.

Proof. (1) Let u U(A M). As before we may write u = n anugn for somegn [R]. Fix F L(X). Since uF = Fu, we get an(Fgn F ) = 0, for any n.Thus, for any x supp(an), F (g1n (x)) = F (x). Using a previous remark, we getg1n (x) = x, for any x supp(an). Thus, u =

n an1supp(an) A.

(2) It is trivial once we noticed that M = (A {ug : g [R]}). From this proposition, it follows that Z(M) = M M AM = A. Moreover,

for any U X, we have the followingU = [U ]R gU = U ,g [R].

Indeed, assume that U = [U ]R and fix g [R]. For any x U , since (x, gx) R,then gx U . Conversely, assume that gU = U , for any g [R]. Recall that thereexists a countable group and a p.m.p. action of on X such that R = R( y X).If (x, y) R, with x U , there exists g such that y = gx. But then y U .Thus we obtain:

Proposition 1.19. L(R) is a factor if and only if R is ergodic.Then we summarize what we did so far in the following theorem:

Theorem 1.20. Let (X,) be a nonatomic probability space. Let R be an ergodiccountable Borel equivalence relation on X such that is R-invariant. Then L(R)is a II1 factor and L(X) L(R) is a Cartan subalgebra.1.3.3. The full group of R and consequences. Denote A = L(X) and M = L(R).We prove the following:

Theorem 1.21. We have

[R] = NM (A)/U(A).Proof. Let u NM (A). As before, we may write u =

n anugn , for some an A,

gn [R]. We know that there exists a Borel isomorphism : X X such that = and uFu = F, for any F A. Thus, uF = Fu and so an(FgnF) =0, for any n and any F A. Hence, for any x supp(an), for any F A

F (g1n (x)) = F (1(x)).

Denote Y =n supp(an). The Borel subset Y is co-null. Indeed, for all n N, we

have1X\Y anugn = 0

and so 1X\Y u = 0. Thus, (X\Y ) = 0. This finally proves that [R].We have constructed a group morphism

:NM (A) [R]

u 7 which is onto since = u. Moreover, u ker() if and only if u AM . Thus,ker() = U(A). This completes the proof. Corollary 1.22. We have NM (A) = U(A)o [R].

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Proof. We already know that the sequence

1 U(A) NM (A) [R] 1is exact. It moreover splits with the following section

s :[R] NM (A)g 7 ug .

Theorem 1.23. For i = 1, 2, let Ri be a measured equivalence relation on (Xi, i).Denote Ai = L(Xi) and Mi = L(Ri). Then

R1 ' R2 (A1 M1) ' (A2 M2).Proof. = First assume that R1 ' R2. Then there exists a Borel isomorphism : X1 X2 such that 1 = 2 and for any (x, y) X1 X1, (x, y) R1 iff((x),(y)) R2. We define a unitary as follows:

U :L2(R1, 1) L2(R2, 2)

7 ((x, y) 7 (1(x),1(y))) .For any g [R1], F L(X1) and any L2(R2, 2), we have

(UugU)(x, y) = (g11(x), y)

(UFU)(x, y) = F (1(x))(x, y).

Thus, UugU = ug1 and UFU = F. Consequently, = Ad(U) : M1 M2is an onto -isomorphism such that (A1) = A2.= Assume now that there exists an onto -isomorphism : M1 M2 such

that (A1) = A2. We know that there exists a Borel isomorphism : X1 X2such that 1 = 2 and (F ) = F, for any F A1. Fix g [R1]. Since ugnormalizes A1 inside M1, it follows that (ug) normalizes A2 inside M2. Thus thereexist h [R2] and v U(A2) such that (ug) = uhv.

Fix F A1. From the one hand, we know that ugFug = Fg. Thus we obtain(ugFug) = (Fg) = (Fg). On the other hand, we have

(ugFug) = (ug)(F )(ug)

= uhvFvuh= uhFuh= (F)h.

Consequently, (Fg) = (F)h, and so g11 = 1h1 on X2. Equivalently,g = h on X1. For any x X1, ((x),g(x)) = ((x), h(x)) R2, sinceh [R2].

Let now (x, y) R1. We know that there exists g [R1] such that y = gx.Thus, ((x),(y)) = ((x),g(x)) R2. Reasoning exactly the same way with1, we obtain that is an isomorphism of equivalence relations.

1.3.4. Group actions and their orbit equivalence relations. Given a p.m.p. action y (X,), one can associate the orbit equivalence relation R( y X) defined by

(x, y) R( y X) s , y = sx.When the action y X is free, the map

(X, counting) 3 (s, x) 7 (x, sx) (R( y X), )


is a p.m.p. Borel isomorphism.

Exercise 1.24. Let y (X,) be a free p.m.p. action. Show that the vonNeumann algebra of the orbit equivalence relation L(R( y X)) and the groupmeasure space construction L(X)o are -isomorphic.Definition 1.25. Let y (X,) and y (Y, ) be two free p.m.p. actions. Weshall say that

(1) y (X,) and y (Y, ) are conjugate if there exist a p.m.p. Borelisomorphism : (X,) ' (Y, ) and a group isomorphism : ' suchthat (sx) = (s)(x), s ,x X.

(2) y (X,) and y (Y, ) are orbit equivalent (abbreviated OE) if thereexist a p.m.p. Borel isomorphism : (X,) ' (Y, ) such that (x) =(x), x X.

(3) y (X,) and y (Y, ) areW -equivalent (abbreviated WE) if L(X)o ' L(Y )o .

Let A = L(X) L(X) o = M and B = L(Y ) L(Y ) o = N .Observe that Theorem 1.23 yields

y (X,) OE y (Y, ) R( y X) ' R( y Y ) (A M) ' (B N)

Therefore the following implications are true

conjugacy = orbit equivalence = W-equivalence.

2. Hilbert bimodules. Completely positive maps

2.1. Generalities.

2.1.1. Hilbert bimodules. The discovery of the appropriate notion of representationsfor von Neumann algebras, as so-called correspondences or bimodules, is due toConnes [5].

Definition 2.1. Let M,N be finite von Neumann algebras. A Hilbert space His said to be an M -N -bimodule if it comes equipped with two commuting normal-representations pi : M B(H) and : Nop B(H). We shall intuitively writex y = pi(x)(yop), H,x M, y N .

We shall see that an M -M bimodule H is the analog of a unitary group repre-sentation pi : U(Hpi).Example 2.2. The following are important examples of bimodules:

(1) The identity bimodule L2(M) with x y = xy.(2) The coarse bimodule L2(M) L2(N) with x ( ) y = (x) (y). It

can be checked that as M -N -bimodules,

L2(M) L2(N) ' HS(L2(M), L2(N))where HS(L2(M), L2(N)) denotes the M -N -bimodule of Hilbert-Schmidtoperators on from L2(M) to L2(N).

(3) For any -preserving automorphism Aut(M), we regard L2(M) withthe following structure: x y = x(y).

14 CYRIL HOUDAYER

(4) Let B M be a von Neumann subalgebra and denote by eB : L2(M) L2(B) the orthogonal projection. Consider the basic construction M, eBwhich is the von Neumann subalgebra of B(L2(M)) generated byM and eB .We endow M, eB with the following semifinite trace: Tr(xeBy) = (xy),for all x, y M (see Subsection 2.3). Then L2(M, eB,Tr) is naturally en-dowed with a structure of M -M -bimodule. Moreover, as M -M -bimodules,

L2(M, eB,Tr) ' L2(M)B L2(M)where B denotes Connes fusion tensor product (see [5, Appendix V.B]).

2.1.2. Unital completely positive maps. Let (M, M ), (N, N ) be finite von Neumannalgebras endowed with a fixed faithful normal trace. A linear map : M N issaid to be completely positive if the maps

n = Idn : Mn(C)M Mn(C)Nare positive for every n 1. We shall say that is unital if (1) = 1, and trace-preserving if moreover N ((x)) = M (x), for every x M .Theorem 2.3 (Stinespring dilation). Let : M N be a (normal) u.c.p. map.Then there exist a Hilbert space H, an isometry V : L2(N) H and a (normal)-representation pi : M B(H) such that

(x) = V pi(x)V,x M.Proof. Equip H0 = M alg L2(N) with the following sesquilinear form

i

ai i,j

bj j =i,j

(bjai)i, jL2(N)

and promote it to a Hilbert space H by separation and completion. Denote by(j bj j) the vector in H which it represents. Define now V : L2(N) H by

V = (1 ). It is clear that V is an isometry, i.e. V V = 1L2(N). For everyx M , we define a bounded linear operator pi(x) on H by

pi(x)(j

bj j) = (j

xbj j).

As expected, pi : M B(H) is a (normal) -representation such that (x) =V pi(x)V , x M .

It follows that a u.c.p. map : M N satisfies for every x M ,(xx) = V pi(xx)V

= V pi(x)V V pi(x)V + V pi(x)(1 V V )pi(x)V (x)(x).

If is moreover assumed to be trace-preserving, the operator T : L2(M) L2(N)defined by

T(x) = (x),x M.is bounded and T = 1.Example 2.4. The following are important examples of -preserving u.c.p. maps:

(1) The trace : M M , the identity map Id : M M and more generallyall -automorphisms : M M which preserve the trace.


(2) LetB M be a von Neumann subalgebra. Denote by eB : L2(M) L2(B)the orthogonal projection. Denote by EB : M B the unique -preservingconditional expectation from M onto B which satisfies

EB(x) = eB(x),x M.It is easy to see that EB is indeed a u.c.p. map.

(3) Let M = L() be the von Neumann algebra of a countable group . Let : C be a normalized positive definite function, i.e. for any finiteset F the matrix ((st1))s,tF is positive. Define the corresponding -preserving u.c.p. map : L() L() by

(s

asus) =s

(s)asus.

Exercise 2.5. Let : L() L() be a -preverving u.c.p. map. Show that : C defined by (s) = ((us)us) is a positive definite function.2.2. Dictionary between Hilbert bimodules and u.c.p. maps.

2.2.1. From u.c.p. maps to Hilbert bimodules. Let : M N be a trace-preservingu.c.p. map. Equip H0 = M L2(N) with the following sesquilinear form

i

ai i,j

bj j =i,j

(bjai)i, jL2(N)

and promote it to a Hilbert space H by separation and completion. Observe thatH is a Stinespring Dilation of . Abusing notation, denote by b the vector inH which it represents. The action is given by

a(b )x = (ab) (x),for a M and x N . With the unit vector = 1 1 H, we have

ax, yH = (a)x, yL2(N)for every a M , x, y N . Since the u.c.p. map is assumed to be trace-preserving,we get

, = M and , = N .2.2.2. From Hilbert bimodules to u.c.p. maps. Let H be an M -N bimodule, witha tracial unit vector , i.e. , = M and , = N . Then the linear operatorL : L2(N) H defined by L(x) = x is bounded and L = 1. For anyx, y N , we have

x, L(y)L2(N) = x, yH= (xy)= x, yL2(N),

so that L(y) = y. Therefore L is an isometry, i.e. LL = 1. Denote by J = JN

the canonical antiunitary. Moreover, for any a M , LaL N . Indeed for anyy, z1, z2 N , we have

(LaL)(JyJ)z1, z2L2(N) = aLJyJz1, L z2H= az1y, z2H

(JyJ)(LaL)z1, z2L2(N) = xL z1, LJyJz2H= az1, z2yH.

16 CYRIL HOUDAYER

Define the u.c.p. map : M N by (a) = LaL. SinceN ((a)) = LaL1, 1 = a, = M (x),

it follows that is trace-preserving. We moreover have

(a)x, yL2(N) = ax, yH,We can now prove the uniqueness and existence of the trace-preserving faithful

normal conditional expectation EB : M B.Proposition 2.6. Let (M, ) be a finite von Neumann algebra with a fixed traceand let B M be a von Neumann subalgebra such that B = |B. Then there existsa unique normal faithful trace-preserving conditional expectation EB : M B.Proof. Consider the M -B-bimodule H = ML2(M)B . The vector 1 H is obviouslya unit tracial vector. Denote by EB the corresponding normal trace-preservingu.c.p. map EB : M B. Recall

x1a, 1bH = EB(x)a, bL2(B),x M,a, b B.Let x M , a, b, c, d B. We have

EB(axb)c, dL2(B) = axb1c, 1dH= x1bc, 1adH= EB(x)bc, adL2(B)= aEB(x)bc, dL2(B),

hence EB(axb) = aEB(x)b. Assume now that EB(xx) = 0. Then

0 = EB(xx)1, 1L2(B) = xx1, 1H = x22,and so x = 0. Let E : M B be another trace-preserving conditional expectation.Then

EB(x)c, dL2(B) = x1c, 1dH= (dxc)= (E(dxc))= (dE(x)c)

= E(x)c, dL2(B),hence EB(x) = E(x). Therefore EB : M B is the unique normal faithful trace-preserving conditional expectation.

2.2.3. From unitary group representations to Hilbert bimodules. Let pi : U(Hpi)be a unitary representation of a countable discrete group . Let M = L() be thegroup von Neumann algebra and denote by (us)s the canonical unitaries. Defineon Kpi = Hpi `2() the following left and right commuting multiplications: forevery Hpi and every s, t ,

us ( t) = (pis s)( t) = pis st( t) us = (1Hpi s1)( t) = ts.


It is clear that the right multiplication extends to the whole von Neumann algebraM . Observe now that the unitary representations pi and 1Hpi are unitarilyconjugate. Indeed, define U : Hpi `2() Hpi `2() by

U( t) = pit t.It is routine to check that U is a unitary and U(1Hpi s)U = pis s, for everys . Therefore, the left multiplication extends to M . We have proven:Proposition 2.7. The formulae above endow the Hilbert space Kpi with a structureof L()-L()-bimodule.

Observe that in the case pi = is the left regular representation of , the M -M -bimodule K is nothing but the coarse bimodule L2(M) L2(M).Exercise 2.8. Let : C be a normalized positive definite function. Let(pi,Hpi, ), with Hpi unit vector, be the GNS-representation of , i.e., (s) =pis, , for all s . Show that the u.c.p. map associated to the bimodule Kpisatisfies

(s

asus) =s

(s)asus.

Exercise 2.9. Prove the following dictionary between u.c.p. maps and Hilbertbimodules:

u.c.p. maps Hilbert bimodulesId : M M Identity bimodule L2(M) : M M Coarse bimodule L2(M) L2(M)

Automorphism : M M L2(M) with x y = x(y)EB : M M L2M, eB

2.3. Popas intertwining techniques.

2.3.1. The basic construction. Throughout this section, we will denote by M a finitevon Neumann algebra with a distinguished faithful normal trace . Let B M bea unital von Neumann subalgebra. Let eB : L2(M) L2(B) be the orthogonalprojection. We will denote by EB : M B the unique faithful normal -preservingconditional expectation. It satisfies the following:

EB(x) = eB(x),EB(axb) = aEB(x)b,x M,a, b B.

The basic construction M, eB is the von Neumann subalgebra of B(L2(M)) gener-ated by M and the projection eB . Observe that JeB = eBJ and eBxeB = EB(x)eB ,x M .Proposition 2.10. The following are true.

(1) M, eB = JBJ B(L2(M)).(2) The central support of eB in M, eB equals 1. In particular, the -subalgebra

span(MeBM) is a -strongly dense in M, eB. The formula eBxeB =EB(x)eB extends the conditional expectation EB : M, eB B.

(3) M, eB is endowed with a semifinite faithful normal trace defined byTr(xeBy) = (xy),x, y M

.

18 CYRIL HOUDAYER

Proof. (1) For x B, we clearly have xL2(B) L2(B) and xL2(B) L2(B),hence xeB = eBx. If x M {eB}, then

EB(x)1 = eB(x1) = xeB(1) = x1.

Therefore x = EB(x) B. It follows that B = M {eB}. Thus,JBJ = JM J, JeBJ = M, eB.

(2) The map B 3 x 7 xeB BeB is a -isomorphism. Indeed, if xeB = 0,then x = 0, for every L2(B). Since x B, it follows that x = 0. Denote byz(eB) the central support of eB in B. Then z(eB) B and z(eB)eB = eB . Hencez(eB) = 1. Thus the central support of eB = JeBJ in JBJ is equal to 1. It isclear that I = span(MeBM) is a -subalgebra of M, eB and a 2-sided ideal ofthe -algebra generated by M and eB . Thus I is a closed 2-sided ideal of M, eB.Moreover

IL2(M) = MeBL2(M) = ML2(B) M 1.Since I is nondegenerate, we get I = M, eB.

(3) Since eB has central support 1 in M, eB, one can find partial isometries(vi) in M, eB such that vi vi eB and

i viv

i = 1. It follows that

i

viL2(B) = L2(M).

Define the following normal weight Tr on M, eB byTr(x) =

i

xvi1, vi1,x M, eB+.

Assume that Tr(xx) = 0. Then xvi1 = 0, for every i. For every b B, we havexvib1 = xviJbJ 1 = JbJxvi1 = 0.

Therefore x = 0 and Tr is faithful. For every x, y M , we haveTr(xeBy) =

i

xeByvi1, vi1 =i

eByvieB 1, eBxvieB 1

=i

EB(yvi)eB 1, EB(xvi)eB 1 =i

(EB(xvi)EB(yvi))

=i

(EB(vi y)EB(vi x)) =

i

EB(vi x)eB 1, EB(vi y)eB 1

=i

eBvi xeB 1, eBvi yeB 1 =i

vivi x1, y1

= i

vivi x1, y

1 = x1, y1 = (xy).

We get that Tr is semifinite since span(MeBM) is a -strongly dense -subalgebrain M, eB. For every x, y, z, t M, eB, we have

Tr(xeByzeBt) = Tr(xEB(yz)eBt) = (xEB(yz)t)= (EB(yz)EB(tx)) = (zEB(tx)y)= Tr(zEB(tx)eBy) = Tr(zeBtxeBy).

Thus Tr is a trace. This completes the proof.


It follows from the previous proposition that

M, eB = {T B(L2(M)) : T (b) = T ()b, L2(M),b B}.Let HB be a right B-submodule of L2(M)B . Write PH : L2(M) H for theorthogonal projection. It is clear that PH M, eB. We define the von Neumanndimension of HB by dim(HB) := Tr(PH).Exercise 2.11 ([2]). Let (N,Tr) be a semifinite von Neumann algebra. Let ={x N : x2,Tr 1}. Prove that the formal inclusion L2(N,Tr) is ultraweak-weak continuous.

2.3.2. Intertwining subalgebras. In [23, 19], Popa introduced a very powerful toolto prove the unitary conjugacy of two von Neumann subalgebras of a tracial vonNeumann algebra (M, ). If A (M, ) is a (possibly non-unital) von Neumannsubalgebras, denote by 1A the unit of A.

Theorem 2.12 (Popa, [23, 19]). Let (M, ) be a finite von Neumann algebra. LetA M be a possibly non-unital von Neumann subalgebra and B M be a unitalvon Neumann subalgebra. The following are equivalent:

(1) There exist n 1, a possibly non-unital -homomorphism : AMn(C)B and a non-zero partial isometry v M1,n(C) 1AM such that

xv = v(x),x A.(2) There exists a nonzero A-B-subbimodule H of AL2(1AM)B such that

dim(HB)

20 CYRIL HOUDAYER

Since H is a left A-module, we get a (unital) -homomorphism : A pBnpsatisfying x() = ((x)) for all x A, and pL2(B)n. Define now ej L2(B)n as ej = (0, . . . , 1, . . . , 0) and = (1, . . . , n) M1,n(C) H, with j =(pej). Let j {1, . . . , n}. For any x A, write (x) = (kl(x))kl pBnp. Wehave

xj = x(pej) = ((x)pej) = (p(x)ej) = (pni=1

ij(x)ei)

=ni=1

(p(0, . . . , ij(x), . . . , 0))

=ni=1

((pei)ij(x))

=ni=1

(pei)ij(x) ( is a right B-module isomorphism)

=ni=1

iij(x).

Consequently, for every x A, x = (x). In the von Neumann algebra Mn+1 B(L2(M) L2(M)n), define

Xx =(x 00 (x)

),x A.

In the space L2(Mn+1), define

=(

0 0 0

).

Note that Xx 1An+1Mn+11An+1 , x A, and 1An+1L2(Mn+1)1An+1 . Weobtain Xx = Xx, for all x A. Denote by T the corresponding unboundedoperator affiliated with Mn+1. and write T = V |T| for its polar decomposition(see Appendix A.1). We get XxV = V Xx, for every x A, and V V 1An+1 .Write

V =(u vv w

).

It is straightforward to check that v M1,n(C) 1AM is a partial isometry fromkerw onto keru such that xv = v(x), for every x A.

(1) = (4). By contradiction, assume that there exists a sequence of unitaries(uk) be a sequence of unitaries in A such that limk EB(aukb)2 = 0 for all a, b 1AM . Then (IdnEB)(vukv)2 0. But for every k N, vukv = (uk)vv.Moreover, (uk) U(pBnp) and vv p. Thus,

(IdnEB)(vv)2 = (uk)(IdnEB)(vv)2= (IdnEB)((uk)vv)2= (IdnEB)(vukv)2 0.

We conclude that (IdnEB)(vv) = 0 and so v = 0. Contradiction.(4) = (3). We can take > 0 and K 1AM finite subset such that

maxa,bK

EB(aub)2 ,u U(A).


Note that

EB(aub)22 = (EB(aub)EB(aub))= Tr(eB(aub)eB(aub)eB).

Define now the element c =aK aeBa

in 1AM, eB+1A. Note that Tr(c) =aK (aa

)

22 CYRIL HOUDAYER

now cut down and w by one of projections (0, . . . , qi, . . . , 0) and assume n = 1from the beginning.

Write e = ww A (since A pMp = A) and f = ww (A) qMq. Byspatiality, we have

f((A) qMq)f = ((A)f) fMf = (wAw) fMf = wAw,which is abelian. Let Q := (A) qMq, which is a finite von Neumann algebra.Since Bq Q is maximal abelian and f Q is an abelian projection, [29, LemmaC.2] yields a partial isometry u Q such that uu = f and uQu Bq. Definenow v = wu. We get

vAv = uwAwu = uf((A) qMq)fu Bq.Moreover vv = wuuw = wfw = e A. Since vAv and Bvv are both maximalabelian, we get vAv = Bvv.

We can even go further if we moreover assume that A,B M are both Cartansubalgebras and M is a II1 factor (see [19, Theorem A.1]).

Theorem 2.14 (Popa, [19]). Let M be a II1 factor. Let A,B M be Cartansubalgebras. The following are equivalent:

(1) A M B.(2) There exists u U(M) such that uAu = B.

Proof. We only need to prove (1) = (2). By Proposition 2.13, there exists anonzero partial isometry v M such that vv A, vv B and vAv = Bvv.Since A is diffuse, we may shrink vv A so that (vv) = 1/n, for some n N.Write p1 = vv, q1 = vv and take projections p2, . . . , pn A, q2, . . . , qn Bsuch that (pi) = (qj) = 1/n. Since NM (A) = NM (B) = M and M is a II1factor, a classical exhaustion argument gives partial isometries ui, wj M such thatp1 = ui ui, pi = uiu

i , u

iAui = Au

i ui, uiAu

i = Auiu

i and likewise q1 = wjw

j ,

qj = wjwj , wjBwj = Bw

jwj , wjBw

j = Bwjw

j . Define

u =ni=1

uivwi.

It is now routine to check that u U(M) and uAu = B.

3. Approximation properties

3.1. Amenability.

3.1.1. Noncommutative Lp spaces. We refer to [27, Chapter IX] for the details ofthe following facts on noncommutative Lp spaces. Let (N ,Tr) be a semifinitevon Neumann algebra endowed with a faithful, normal, semifinite trace Tr. For1 p < , we define the Lp-norm on N by xp = Tr(|x|p)1/p. By completing{x N : xp


which satisfies xyr xpyq, x, y. The Banach space L1(N ) is identified withthe predual of N under the duality

L1(N )N 3 (, x) 7 Tr(x) C.The Banach space L2(N ) si identified with the GNS-Hilbert space L2(N ,Tr). El-ements in Lp(N ) can be regarded as closed operators on L2(N ) which are affil-iated with N and hence in addition to the above-mentioned product, there arewell-defined notions of positivity, square root, etc... We shall use the generalizedPowers-Strmer Inequality (see [27, Theorem IX.1.2]):

|| ||22 2 21 + 2 2,, L2(N ).The Hilbert space L2(N ) is an N -N bimodule such that xy, = Tr(xy),x, y N , , L2(N ). We also recall the following formulae. Let fa be thecharacteristic function of the interval (

a,+). For any , L2(N )+, we have

(see [7, Proposition 1.1] and [27, Theorem IX.2.14]) 0

fa()22 da = 22, 0

fa() fa()22 da 2 + 2.

Let H be a complex separable Hilbert space and let (N ,Tr) = (B(H),Tr),where Tr is the canonical trace on B(H). Then, L1(B(H)) can be identified withthe space of trace-class operators on H, denoted by S1(H) in the sequel. In thesame way, L2(B(H)) can be identified with the space of Hilbert-Schmidt operatorson H, denoted by S2(H) in the sequel.

Let (M, ) be a finite von Neumann algebra, denote by H = L2(M, ) its L2-space with respect to the finite trace . The Hilbert space H is endowed with acanonical anti-unitary J defined by Jx = x, x M . In the sequel, we shall simplydenote H 3 7 H. We regard M B(H) through the GNS-construction.The following linear map

U :H H S2(H)

k

k k 7k

, kk

defines a unitary. We shall identify S2(H) and H H through this unitary U .Moreover, for any , H, for any x M ,

U(x ) = , x = x, = xU( )U( x) = , (x) = , x = x, = U( )x.

Thus, U preserves the M -M -bimodule structure: S2(L2(M)) with its bimodulestructure, as a two-sided ideal of B(L2(M)), is identified with the so-called coarsecorrespondence L2(M) L2(M).

Finally, note that the symbol Lim will be used for a state on `(N) or moregenerally on `(I) with I directed set.

3.1.2. Amenable finite von Neumann algebras. Recall that a countable discretegroup is amenable if one the following equivalent conditions holds:

There exists a -invariant state : `() C, i.e. (sf) = (f), for alls , f `().

24 CYRIL HOUDAYER

There exists a sequence of almost invariant unit vectors n `2(), i.e.limn sn n2 = 0, for all s .

Let (M, ) be a finite von Neumann algebra with separable predual. Denote H =L2(M, ). We regard M B(H) through the GNS-construction. A state onB(H) is said to be M -central if Ad(u) = , u U(M).Theorem 3.1 (Connes, [7]). Let M be a finite von Neumann algebra. The followingare equivalent:

(1) There exists a conditional expectation E : B(H)M .(2) There exists an M -central state on B(H) such that |M = .(3) There exists a net of unit vectors (n) in S2(H) such that xn, n (x),x M , and [n, u]2 0, u U(M).

Proof. (1) = (2). Let E be a conditional expectation from B(H) onto M . Denote = E. Then x B(H), u M , one has

(uxu) = (E(uxu)) = (uE(x)u) = (E(x)) = (x).

Thus, the state is M -central and |M = .(2) = (3). Let be an M -central state on B(H) such that |M = . Take a

net (n) of positive norm-one elements in S1(H) such that Tr(n) converges to pointwise. Then x B(H), u U(M), one has

limn

Tr((n Ad(u)n)x) = limn

Tr(nx) limn

Tr(unux)

= limn

Tr(nx) limn

Tr(nuxu)

= (x) (Ad(u)(x)) = 0by assumption. It follows that the net (n Ad(u)n) in S1(H) converges to 0 inthe weak topology. By the Hahn-Banach Separation Theorem, one may assume,by passing to finite convex combinations, that the net (n Ad(u)n) in S1(H)converges to 0 in norm. Thus, [u, n]1 0, u U(M). Define the unit vectorsn =

1/2n S2(H). Using the Powers-Strmer Inequality, u U(M),

[u, n]22 = unu n22 unu n1= [u, n]1 .

This implies that [u, n]2 0, u U(M). Moreover, x M ,limnxn, n = lim

nTr(xnn) = lim

nTr(x2n)

= limn

Tr(nx) = (x) = (x).

This proves (3).(3) = (1). Assume that there exists a net of unit vectors (n) in S2(H) such

that xn, n (x), x M , and [n, u]2 0, u U(M). Note that we alsohave [n, u]2 0, u U(M). Write n = wnn for the polar decomposition,with n = (nn)

1/2 0, n2 = 1, and wn partial isometry in B(H). Thus, n,x B(H),

xn, n = Tr(xnn) = Tr(x2n) = xn, n.


Consequently, limnxn, n = (x), x M . Moreover, using the Powers-StrmerInequality, n, u U(M),

[u, n]22 = unu n22 u2nu 2n1= u(nn) (nn)u1 (un nu)n1 + n(un nu)1 un nu2 + un nu2.

Therefore, limn [u, n]2 = 0. So, we may moreover assume that n 0, n. Thus,c M ,

limnnc22 = lim

nnc, nc = lim

nTr(nccn)

= limn

Tr(2ncc) = (cc) = c22, .

Define the following state

(x) = Limnxn, n,x B(H).

Note that |M = . Moreover, b, c M , y, z B(H), since [n, bc]2 0, weget

|(bcyz)| = |Limn

Tr(2nbcyz)|= |Lim

nTr(nbcnyz)|

= |Limn

Tr((znb)(cny))| lim sup

nznb2cny2

lim supnnb2cn2yz

= b2,c2,yz.Take now x B(H), a M and write a = v|a| its polar decomposition in M .Thus, we have a = (v|a|1/2)|a|1/2 and

|(ax)| v|a|1/22,|a|1/22,x |a|1/22,|a|1/22,x= a1,x.

It follows that |(ax)| a1,x, a M , x B(H). By the dualityM = L1(M), we know that there exists (x) M , such that (a(x)) = (ax),a M , x B(H). It is straightforward to check that : B(H) M is aconditional expectation.

The state in (2) of Theorem 3.1 is called a hypertrace. Condition (3) saysthat the identity bimodule L2(M) is weakly contained in the coarse bimoduleL2(M) L2(M), that we usually denote by L2(M) weak L2(M) L2(M). Thisis the analog of the notion of amenability in the case of groups since the identitybimodule plays the role of the trivial representation and the identity bimodule playsthe one of the left regular representation. For this reason, a finite von Neumannalgebra M which satisfies one of the equivalent conditions of Theorem 3.1 is said tobe amenable. Note that more generally for any M -M -bimodule H, L2(M) weak H

26 CYRIL HOUDAYER

if there exists a net (n) of unit vectors in H such that limnn, n = pointwiseand limn [u, n] = 0.Proposition 3.2. Let be a countable discrete group. Then, is amenable if andonly if the group von Neumann algebra L() is amenable.

Proof. Assume that is amenable. Denote by the left regular representation of on `2(). Then 1 , i.e. there exists a net of unit vectors (n) in `2() suchthat limn gn n2 = 0, g G. We shall identify `2() with the L2-space ofL(). Denote the unit vector n = n n `2() `2() = S2(`2()). Moreover,we have for all g ,

ugn nug2 = ugn n n nug2 ugn n n n2 + n n n nug2 gn n2 + g1n n2.

Therefore, limn ugn nug2 = 0, g . Thus Condition (3) in Theorem 3.1 issatisfied, and L() is amenable.

Assume that L() is amenable. Let H = `2(). By Condition (3) in Theorem3.1, we know that there exists a sequence of unit vectors (n) in H H suchthat for any g , limn ugn nug2 = 0. Define the unitary representationpi : U(H H) by

pi(g) = ugug.It is straightforward to see that pi is a mutiple of and that 1 pi. Consequently,1 and is amenable. Exercise 3.3. Let M be a diffuse amenable finite von Neumann algebra. Showthat a hypertrace given by Theorem 3.1 can never be normal on B(H).

Recall that a finite von Neumann algebra M is said to be approximately finitedimensional (AFD) if there exists an increasing sequence of finite dimensional unital-subalgebras Qn M such that

nQn is ultraweakly dense in M . Murray and

von Neumann showed in their seminal work the uniqueness of the AFD II1 factor.The following is easy to prove.

Proposition 3.4. Let M be a finite AFD von Neumann algebra. Then M isamenable.

Proof. Let Qn M be a sequence of finite dimensional unital -subalgebras suchthat

nQn is ultraweakly dense in M . Denote by n the (probability) Haar mea-

sure on the compact group U(Qn). Fix (N)\N a free ultrafilter. For everyx B(L2(M)) define

E(x) = limn

U(Qn)

uxudn(u).

It is clear that E : B(L2(M))M is a conditional expectation. Denote by J thecanonical antiunitary. Thus E : B(L2(M))M defined by E(x) = JE(JxJ)J isa conditional expectation.

The converse is Connes fundamental result.

Theorem 3.5 (Connes, [7]). Let M be a finite von Neumann algebra. Then M isAFD if and only if M is amenable. There exists a unique amenable II1 factor. Inparticular, all icc amenable countable discrete groups give the same II1 factor.


Exercise 3.6. Let y P be a trace-preserving action of an amenable group onan amenable finite von Neumann algebra P . Show that M = P o is amenable.

3.2. Property Gamma.

Definition 3.7. A II1 factor M is said to have the property Gamma if for every > 0 and every x1, . . . , xn (M)1, there exists u U(M) such that (u) = 0 and[u, xi]2 < , for all 1 i n.Theorem 3.8 (Connes, [8]). Let M be a II1 factor. The following are equivalent:

(1) M does not have property Gamma.(2) There exists a non-trivial central sequence in M , i.e. M M 6= C.(3) M M is diffuse.

Proof. For a free ultafilter , we will denote by pi : `(N,M)M the quotientmap. For (1) = (2), fix a free ultrafilter . Since M does not have propertyGamma, there exists a sequence of unitaries un U(M) such that (un) = 0 andlimn [un, x]2 = 0, for all x M . Define u = pi((un)) U(M M). We havethat (u) = 0 and so M M 6= C. For (2) = (3), let e M M be a nonzeroprojection such that e 6= 1. Write = (e) (0, 1). We may find a sequence ofprojections en which represents e such pi((en)) = e and (en) = , for all n N.

Observe that (en) is a central sequence and since M is a II1 factor, we have thaten weakly (by weak compactness of the unit ball (M)1). Thus we construct asubsequence ekn which satisfies for every n N,

[ej , ekn ]2 0,a > 0, limn fa(|n|)|n|2 > c.

We may then choose a subsequence (kn) such that fn(|kn |)|kn |2 c, for all n N. Then with n = 1fn(|kn |)|kn |2 fn(|kn |)|kn |, we still have that limn xn nx2 = 0, for all x M . Observe that for all a > 0,

n fa(n)n afa(n)

and so (fa(n)) 1/a.

Let F U(M) be a finite subset and > 0. Let = 4/(4|F |). Choose n Nlarge enough such that 1/

n < and = n satisfies

uF uu22 < .

It is clear that fa(uu) = ufa()u. Hence, we get 0

fa()22da = 22 = 1and

uF

0

fa() ufa()u22da uF uu2 + uu2

2(|F |uF uu22)1/2

< 2|F |

= 2|F |

0

fa()22da.Therefore there exists a n such that

uFfa() ufa()u22 < 2

|F |fa()22.

Letting e = fa(), we have (e) (fa()) < and ue eu2 < e2, for allu F . The claim is proven.

The next claim uses a maximality argument.

Claim. For every finite subset F U(M), for every > 0, there exists a projectione M such that (e) = 1/2 and [u, e]2 < , for all u F .

Once the claim is proven, we are done. Indeed, we can construct a sequence ofprojections en M such that (en) = 1/2 and limn [x, en]2 = 0, for all x M .We the get a sequence of unitaries un = 2en 1, with (un) = 0 and such thatlimn [x, un]2 = 0, for all x M . Hence, M has property Gamma. It only remainsto prove the claim.


Let u1, . . . , uk U(M) and > 0. Let I be the set of families i = (E,U1, . . . , Uk)such that:

E M is a projection such that (E) 1/2. Each Uj U(M) is a unitary commuting with E. Uj uj1 (E), for all 1 j k.

We define a partial ordering on I in the following way: we write i i ifE E and Uj U j1 (E E),1 j k.

It is easy to see that if i i and i i, then i i. The set (I,) is moreoverinductive. By Zorns Lemma, there exists a maximal element i = (E,U1, . . . , Uk) I. Assume that (E) < 1/2. We will deduce a contradiction. Let > 0 suchthat (E) + < 1/2 and 4 . Let vj = (1 E)Uj = Uj(1 E) U(N) whereN = (1E)M(1E). By the previous claim (with N), we know that there exists aprojection e N such that N (e) < and [vj , e]2,N e2,N , for all 1 j k.By Proposition D.1, there exists wj U(N) such that wjvjevjwj = e andwj (1 E)1,N

2vjevj e1,N 4vjevj e2,N e2,N 4N (e).

Let E = E + e M . It is a projection stricly larger than E and such that(E) 1/2. Let U j = UjE + wjvj . It is easy to see that U j U(M) and U j iscommuting with E. Moreover, wjvj vj1 4(e), U j Uj1 4(e). Since4 and Uj uj1 (E) by assumption, we get

U j uj1 (E + e) = (E),1 j k.The element i = (E, U 1, . . . , U

k) satisfies i i and i 6= i, which contradicts the

maximality of i I. Therefore, we have that (E) = 1/2. We have thus shown foreach > 0, the existence of a projection E M such that (E) = 1/2, [E,Uj ] = 0and Uj uj1 . We finally get

[uj , E]2 = [uj Uj , e]2 2uj Uj2 2(2)1/2.As > 0 is arbitrary, we are done.

(1) = (3). Assume that K(L2(M)) C(M,M ) 6= {0}. Since C(M,M )is a simple C-algebra, we get K(L2(M)) C(M,M ). Let (xn) be a (uni-formly bounded) central sequence in M . We get that [y, xn] 0 -strongly for ally C(M,M ). Denote by PC : L2(M) C the orthogonal projection. SinceK(L2(M)) C(M,M ), we get [PC, xn] 0 -strongly. We have that

limnxn (xn)12 = lim

n(xnPC PCxn)1 = 0,

and so (xn) is trivial.(3) = (2). Assume that it is impossible to find a sequence of unit vectors n

like in (2). Then then exists > 0, a finite subset F U(M) such thatmaxuF uu2 2, L2(M).

We may assume that F = F . Define the self-adjoint operator

T =1|F |

uF

uJuJ C(M,M ).

We have T 1, T 1 = 1, so that 1 is an eigenvector for the eigenvalue 1. Weshow that T 1 is invertible on L2(M) C1, so that there is a spectral gap at 1

30 CYRIL HOUDAYER

for the operator T . Otherwise, since T is selfadjoint, there would be a sequence ofunit vectors k L2(M)C1 such that limk (T 1)k2 = 0. We have

(1 T )k2 = 1|F |uF

(1 uJuJ)k2.

Recall that a Hilbert space is uniformly convex. For all k N, we have thatuF

1|F | k uJuJk

2 = 2 2


Example 3.12. The following groups have the Haagerup property: amenablegroups, free groups and more generally all groups which act properly on a tree.The Haagerup property is moreover stable under taking subgroups, amenable ex-tentions, free products, wreath products.

We refer to the book by P.A. Cherix, M. Cowling, P. Jolissaint, P. Julg and A.Valette [3] for a comprehensive of groups with the Haagerup property

Definition 3.13 (Choda, [4]). Let (N, ) be a finite von Neumann algebra endowedwith a fixed faithful normal trace. We say that N has the Haagerup property ifthere exists a sequence n : N N of -preserving ucp maps which satisfies:

limn n(x) x2 = 0, for all x N . Whenever wk (N)1 is a sequence such that wk 0 weakly, then we have

limk n(wk)2 = 0, for all n N.Proposition 3.14 (Choda, [4]). Let be a countable discrete group. Then hasthe Haagerup property if and only if L() has the Haagerup property.

Proof. Assume that has the Haagerup property. Then there exists a sequencen : C such that limn n = 1 pointwise and n c0(), for all n N. Wemay assume that n(e) = 1, for all n N. Define n : L() L() by

n(s

asus) = n(s)asus.

It is straightforward to check that n is a sequence of -preserving ucp maps whichsatisfies conditions of Definition 3.13.

Conversely, assume that L() has the Haagerup property. Let n be a sequenceof -preverving ucp maps given by Definition 3.13. Define n(s) = (n(us)us),for all s . Then n is a sequence of positive definite functions that does thejob.

Theorem 3.15 (Haagerup, [14]). The free groups Fn have the Haagerup property.

Proof. Denote by Fn 3 g 7 |g| R+ the natural length function. We will showthat for all 0 < < 1, the function defined by

Fn 3 g 7 |g|

is a positive definite function on Fn. Since c0(Fn) and lim1 = 1, we getthat Fn has the Haagerup property.

We give a proof using Popas free malleable deformation [21, 24]. We may assumethat n < . Let M = L(Fn) and M = L(Fn Fn), where Fn is a copy of Fnwhich is free from Fn. Denote by a1, . . . , an the canonical generators of L(Fn)(resp. b1, . . . , bn the ones of L(Fn)). For 1 k n, let

hk =11pi log(bk),

where log denotes the principal branch of the logarithm. We get that hk is selfad-joint and bk = exp(

1pihk). For t R, definebtk := exp(

1pithk) U(L(Fn)).

32 CYRIL HOUDAYER

It is straightforward to check that

(t) := (btk) = 11

exp(1pitx)dx = sin(pit)

pit

Define the following -automorphism t : M M byt(ak) = akbtk and t(bk) = bk.

Since the unitaries {a1, . . . , an, a1bt1, . . . , akbtk} generate M and are -free from eachother, we check easily that (t) is a one-paramater family of trace-preserving -automorphisms. Write N = L(Fn) and N = 1(N). We see that N N = Mand N is -free from N . Consequently, we get M = N N and (t) satisfies1(x 1) = 1 x, for all x N . Moreover, we have

(EN t)(ug) = (t)2|g|ug,g Fn,Since EN t is u.c.p., it follows that t : Fn C defined by t(g) = ((EN t)(ug)ug) = (t)

2|g| is a positive definite function. We are done.

4. Rigidity of II1 factors

4.1. Rigid inclusions of von Neumann algebras. The notion of property (T)for a II1 factor was introduced by Connes and Jones in their seminal work [9]. Itsrelative version for a inclusion of finite von Neumann algebras is due to Popa [19].

Definition 4.1 (Popa, [19]). Let (M, ) be a finite von Neumann algebra with afixed trace and B M be a subalgebra. The inclusion is said to be rigid if for every > 0, there exist > 0 and a finite subset F M such that for every -preservingu.c.p. map : M M , we have

supxF(x) x2 = sup

x(B)1(x) x2 .

The von Neumann algebra M has property (T) if the identity inclusion M Mis rigid. Note that we can relax the assumptions in Definition 4.1. Indeed, let : M M be a completely positive map such that (1) 1 and . Then, : M M defined by

(x) = (x) +( )(x)( )(1) (1 (1))

is a -preserving u.c.p. map.

Theorem 4.2 (Popa, [19]). Let B M be an inclusion of finite von Neumannalgebras and let be a fixed trace on M . The following are equivalent:

(1) The inclusion B M is rigid;(2) For every > 0, there exist > 0 and a finite subset F M such that

for any M -M bimodule H and any tracial unit vector H for whichx x , x F , there exists a B-central vector H such that .

Proof. We prove both directions.(1) = (2). Let 0 < < 1. Let F M be a finite subset and > 0 given by

Condition (1). Let H be an M -M -bimodule and a tracial unit vector H suchthat x x , x F . Let : M M be the -preserving u.c.p. map


associated with (H, ). Recall that ax, yH = (a)x, yL2(N), a, x, y M .Then we have

(x) x22 = (x)22 + x22 2

34 CYRIL HOUDAYER

Example 4.4. Here are a few classical examples.

(1) The example by excellence of a pair with relative property (T) is (Z2 oSL(2,Z),Z2).

(2) For any n 3, SL(n,Z) has property (T).(3) For any property (T) group and any group H, the pair (H ,) has

relative property (T).

We refer to the book by Bekka, de la Harpe and Valette [1] for a comprehensivelist of groups with (relative) property (T).

Theorem 4.5 (Popa, [19]). Let be an inclusion of countable groups. Thefollowing are equivalent:

(1) The pair (,) has the relative property (T);(2) The inclusion L() L() is rigid.

Proof. Write B = L() L() = M .(1) = (2). Assume that the pair (,) has the relative property (T). Fix > 0.

We know that there exist > 0 and a finite subset F such that for any unitaryrepresentation pi : U(Hpi) which has a (pi(F ), )-invariant unit vector , thereexists a non-zero pi()-invariant vector such that . Let now H be aM -M -bimodule and a unit tracial vector H such that us us , s .Define the unitary representation pi : U(H) by pis() = usus. The vector is then (pi(F ), )-invariant. Thus there exists a pi()-invariant vector H suchthat . It is then clear that is B-central.

(1) = (2). Assume that the inclusion B M is rigid. Let = 1/2. Weknow that there exist > 0 and a finite subset F M such that for any M -M -bimodule H and any tracial vector H for which x x , x F , thereexists a B-central vector H such that . Let pi : U(Hpi) be aunitary representation. Consider the M -M -bimodule Kpi = Hpi `2() associatedwith pi. Take a sequence (n) Hpi of almost invariant unit vectors and set n =n e. It is then clear that (n) Kpi is a sequence of tracial vectors for whichlimn xn nx = 0, x M . For n N large enough, we have xn nx ,x F . Write = n. Therefore there exists a nonzero B-central vector Kpisuch that 1/2. Regard `2(, Hpi) and write =

s s s, where

s Hpi. We have

e 2 +

s{e}s2 = 2 1/4.

Since Hpi is a unit vector, we have that e 6= 0. Since e is moreover pi()-invariant, the proof is complete.

The previous theorem shows in particular that the inclusion L(T2) L(T2)oSL(2,Z) is rigid.

4.2. Applications to rigidity of II1 factors. We use now the tools we introducedin the previous sections to get structural properties for property (T) II1 factors.Most of the proofs are based on a separability vs property (T) argument thatgoes back to Connes [6].


4.2.1. Symmetry groups of property (T) factors are countable. For a II1 factor M ,we endow the group Aut(M) with the topology of pointwise 2-convergence: fora sequence (n) in Aut(M), we have

n Id n(x) x2 0,x M.Note that Aut(M) is a polish group. We shall denote by Inn(M) Aut(M) thesubgroup of inner automorphisms. For any u U(M), we shall write Adu(x) =uxu, x M . We denote by Out(M) the quotient group Aut(M)/ Inn(M).

For the hyperfinite II1 factor R, the outer automorphisms group Out(R) ishuge. Indeed, one can embed any second countable locally compact group Gin Out(R). For property (T) II1 factors, the situation is dramatically different.

Theorem 4.6 (Connes, [6]). Let M be a property (T) II1 factor. Then Out(M) iscountable.

Proof. We show that Inn(M) Aut(M) is an open subgroup. Thus, it follows thatOut(M) = Aut(M)/ Inn(M) is a Hausdorff discrete group, and by separability,Out(M) is necessarily countable. Fix = 1/2.

By property (T) of M , we know that there exists > 0 and F M finite subsetsuch that for every u.c.p. -preserving map : M M , we have

supxF(x) x2 < = sup

x(M)1(x) x2 < 1/2.

Let V,F = { Aut(M) : (x) x2 < ,x F} be an open neighborhoodof Id in Aut(M). Let V,F . Since (x) x2 < , x F , we know that(x) x2 1/2, for every x (M)1. Observe that (u)u 12 1/2, forevery u U(M). Consider

C = cow{(u)u : u U(M)}the weak closure of the convex hull of all the (u)us, for u U(M). Observe thatC (M)1 is closed in L2(M).

Denote by a C the unique element of minimum 2-norm. It follows thata 12 1/2. We get a 6= 0. Observe that for every u U(M), (u)au C and(u)au2 = a2. By uniqueness, we get (u)au = a, for every u U(M). Sowe have aau = uaa, for every u U(M). It follows that aa = R+ since Mis a factor. Therefore v = a/

U(M) and = Adv.

Observe that a property (T) factor cannot have property Gamma. Recall thatfor a II1 factor M , the fundamental group of M is defined as follows:

F(M) = {(p)/(q) : pMp ' qMq}.Murray & von Neumann showed that the unique AFD II1 factor R has full fun-damental group, i.e. F(M) = R+. There is an alternative way of defining thefundamental group of M . Denote by M = MB(`2) the corresponding II fac-tor with semifinite trace Tr given by Tr = TrB(`2). For any Aut(M),there exists a unique > 0 such that Tr = . We shall denote this by mod().Moreover, the map mod : Aut(M) R+ is a group homomorphism. It is theneasy to check that

F(M) = {mod() : Aut(M)}.Using property (T), Connes [6] gave the first example of II1 factor with countable

fundamental group.

36 CYRIL HOUDAYER

Theorem 4.7 (Connes, [6]). Let M be a property (T) II1 factor. Then F(M) iscountable.

Proof. We construct a one-to-one map : F(M) Out(MM). Since Mhas property (T), MM has property (T) as well (cf Exercise 4.3) and thenOut(MM) is countable by Theorem 4.6. Therefore, F(M) follows countable.Claim. Let N be a II1 factor. Let Aut(N) such that mod() = 1. Thenthere exist u U(N) and Aut(N) such that

= Adu ( IdB(`2)).Therefore the group homomorphism

{ Aut(N) : mod() = 1} 3 7 [] Out(N)is well-defined.

Denote by (eij) B(`2) the canonical matrix unit such that Tr(eii) = 1, i N. Let Aut(N) such that mod() = 1. Write fij = (1 eij). SinceTr(f00) = Tr(1 e00) = 1, f00 and 1 e00 are equivalent projections in the factorN, so that there exists a partial isometry v N such that vv = f00 andvv = 1 e00. Define u =

fj0v(1 e0j). It is routine to check that u U(N)

and u(1 eij)u = fij , which finishes the proof of the claim.Let t F(M) and choose t Aut(M) such that mod(t) = t. Since tt1

Aut((MM)) has modulus 1, the claim yields a unique t = [t] Out(MM),so that the map

(1) F(M) 3 t 7 t Out(MM)is well-defined. If s 6= t, 1s t is outer, and so is (s s1)1(t t1). Thuss 6= t. Since the map (1) is one-to-one, we are done. 4.2.2. Connes rigidity conjecture. Connes in the late 70s conjectured the following:any countable icc property (T) groups ,,

' L() ' L().Popa suggested the following strenghthening of Connes rigidity conjecture:

Conjecture 4.8 (Popa, [22]). If is an icc property (T) group and is a group,then any -isomorphism : L() ' L()t forces t = 1 and there exist a groupisomorphism : ' and a character Hom(,T) such that

(s

asus) =s

(s)asu(s).

In particular,

F(L()) = {1} and Out(L()) = Out()Hom(,T).Popa observed in [20] that Ozawas original result [18] could be used to prove

Connes rigidity conjecture up to countable classes.

Theorem 4.9 (Ozawa, [18]). For icc countable property (T) groups, the map L() is countable-to-one.


Proof. The proof is an analog of the one of Theorem 2 in [18]. We prove theresult by contradiction and assume that there are uncountably many pairwise non-isomorphic icc property (T) groups (i)iI which give the same II1 factor M . ByShaloms result [25], we know that each property (T) group is the quotient of afinitely presented property (T) group. Since there are only countably many finitelypresented groups, we may assume that all the is are quotients of the same property(T) group .

We regard M B(L2(M)) represented in its standard form and M = L(i) forevery i I, so that i U(M). Denote by pii : U(M) a group homomorphismsuch that pii() = i. Fix = 1/4. Since is a property (T) group, there exist0 < < 1 and a finite subset E such that for any unitary representationpi : U(H) and any (pi(E), )-invariant unit vector H, there exists a pi()-invariant vector H such that 1/4.

Since the finite von Neumann `(E,M) is 2-separable, there exist i 6= j I such that maxsE pii(s) pij(s)2 . Let J be the canonical antiunitarydefined on L2(M) and define the unitary representation pi : U(M) by pi(s) =pii(s)Jpij(s)J . We have

pi(s)1 1L2(M) = pii(s)pij(s) 12 .Since the vector 1 is (pi(E), )-invariant, it follows that there exists a pi()-invariantvector L2(M) such that 1L2(M) 1/4. Thus, for every s we have

pii(s) pij(s)2 = pi(s)1 1L2(M)= pi(s)(1 ) (1 )L2(M) 1/2.

Exactly as in the proof of Theorem 4.6, denote by a the unique element of minimum 2-norm in the weakly closed convex set C = cow{pii(s)pij(s) : s }. Sincea 12 < 1, it follows that a 6= 0. Observe that pii(s)apij(s) C andpii(s)apij(s)2 = a2. By uniqueness, we have pii(s)apij(s) = a. Moreover,aapij(s) = pij(s)aa, for every s . Since M is a factor and pij() = M , wehave aa = R+. Thus v = a/

U(M) and pii(s) = vpij(s)v, for every

s . It follows in particular that i and j are isomorphic, contradiction.

4.3. Uniqueness of Cartan subalgebras.

Theorem 4.10 (Popa, [19]). Let y B be a trace-preserving action of a countablegroup with the Haagerup property on a finite von Neumann algebra B. Denoteby M = Bo the crossed product. Let A M be a rigid von Neumann subalgebra.Then A M B.Proof. Since has the Haagerup property, let n : C be a sequence of positivedefinite functions such that limn n = 1 pointwise and n C0(), for all n N.Define n : M M the sequence of -preserving ucp maps as follows:

n(s

asus) =s

n(s)asus.

It is straightforward to check that every n satisfies the following relative compact-ness property: if (wk) is a sequence in (M)1 which satisfies limk EB(awkb)2 = 0,for all a, b M , then limk n(wk)2 = 0. Indeed, write wk =

s(wk)

sus, where

38 CYRIL HOUDAYER

(wk)s = EB(wkus). Then we have n(wk) =s n(s)(wk)

sus and thus

n(wk)22 =s|n(s)|2(wk)s22.

Assume that limk (wk)s2 = 0, for all s . Fix > 0. Since n c0(),V := {s : |n(s)| 2/2} is a finite set. Then we have

n(wk)22 =

sV|n(s)|2(wk)s22 +

sV|n(s)|2(wk)s22

2/2

sV(wk)s22 +

sV|n(s)|2(wk)s22

2/2 +sV|n(s)|2(wk)s22.

We can choose k0 N large enough so thatsV |n(s)|2(wk)s22 2/2. There-

fore, we have n(wk)2 , for all k k0.Since A M is rigid, there exists n N such that n(w) w2 1/4, for all

w U(A). By contradiction, assume that A M B. Then there exists a sequencewk U(A) such that limk EB(awkb)2 = 0, for all a, b M . For k N largeenough, we get

1 = wk2 n(wk) wk2 + n(wk)2 1/4 + 1/4 = 1/2,which is absurd. Corollary 4.11 (Popa, [19]). Consider the linear action SL2(Z) y T2. Then, upto unitary conjugacy, L(T2) is the unique rigid Cartan subalgebra in L(T2) oSL2(Z).

Proof. Write M = L(T2) o SL2(Z). Let A M be rigid Cartan subalgebra.We get A M L(T2) by the previous Theorem. Since A,L(T2) M are bothCartan subalgebras of the II1 factor M , Theorem 2.14 yields u U(M) such thatuAu = L(T2).

We can now apply this last result to compute explicitely the fundamental group ofL(Z2oSL2(Z)). Gaboriau [13] showed that the group SL2(Z) has fixed price and itscost equals 13/12. This means that for every free ergodic p.m.p. SL2(Z) y X, theequivalence relation R(SL2(Z) y X) has cost 13/12. It follows from the inductionformula [13, Proposition II.6] that R(SL2(Z) y X) has trivial fundamental group.Corollary 4.12 (Popa, [19]). We have F(L(Z2 o SL2(Z))) = {1}.Proof. Let M = L(Z2 o SL2(Z)) = Ao SL2(Z). Let R be the equivalence relationinduced by the action SL2(Z) y T2. Let t 1 such that M 'M t. We can assumethat p A is a projection of trace t so that (pAp pMp) ' (At M t). It followsthat At M is a rigid Cartan subalgebra and thus there exists u U(M) suchthat uAtu = A, by Corollary 4.11. This shows that R ' Rt (see Theorem 1.23)and thus t = 1.

Popas result was the first explicit computation of a fundamental group of a II1factor that was different from R+, solving then a long-standing open problem ofKadison.


Appendix A. Polar decomposition of a vector

Let (N, ) be a finite von Neumann algebra. Since is fixed, we simply denoteL2(N, ) by L2(N). We regard N B(L2(N)). Let L2(N) such that 6= 0.Let T 0 : N L2(N) be the linear operator defined by T 0 (x) = x, for all x N .Proposition A.1. The densily defined operator T 0 is closable. Denote by T itsclosure. The operator T is affiliated with N . Write T = v|T| for its polar decom-position. Then v N and |T| is affiliated with N .

Let B N be a von Neumann subalgebra such that x = x, for all x B. Thenwe have xv = vx, for all x B.Proof. First, we prove that the operator T 0 is closable. It suffices to show that(T 0 )

is densily defined. Let y N and z N . Then,T 0 (y), z = y, z = Jy, z

= zJyJ, 1 = JyJz, 1= y, Jz = y, (J)z = y, T 0J(z).

Then T 0J (T 0 ) and so (T 0 ) is densily defined. Thus T 0 is closable and wedenote by T its closure. We prove now that T is affiliated with N . Let a, x N .On the one hand,

T 0 JaJ(x) = T 0 (xa) = xa.

On the other hand,JaJT 0 (x) = xa.

Consequently, we have JaJT 0 T 0 JaJ , for all a N . Since JNJ = N , itfollows that T is affiliated with N . Write T = v|T| for the polar decompositionof T. Since v is bounded and affiliated with N , we have that v N . Moreover,|T| is affiliated with N .

At last, let B N be a von Neumann subalgebra such that for any x B,x = x. Fix x B. It is straightforward to check that xT Tx. We also havex(T) (T)x, and so x(T)T (T)Tx. By functional calculus, it followsthat x|T| |T|x. Moreover, since N is a finite von Neumann algebra, since x Nand |T| is affiliated with N , it follows that x|T| and |T|x are closed, affiliatedwith N and consequently the equality x|T| = |T|x holds. Thus,

xv|T| = xT Tx v|T|x vx|T|.It follows that xv and vx coincide on the range of |T|, and so xv = vx. Thus,xv = vx, for every x B.

Appendix B. Von Neumanns dimension theory

Let (N, ) be a finite von Neumann algebra with a distinguished faithful normaltrace. Let H be a right Hilbert N -module, i.e. H is a complex (separable) Hilbertspace together with a normal -representation pi : Nop B(H). For any b N ,and H, we shall simply write pi(bop) = b.Proposition B.1. Let H be a right N -module. Then there exists an isometryv : H `2 L2(N) such that v(b) = v()b, for all H, b N .

40 CYRIL HOUDAYER

Proof. Let : Nop B(H (`2 L2(N))) be the -representation defined by

(yop) =(pi(yop) 0

0 1`2 yop).

Let Q = (Nop). It is clear that Q is a finite von Neumann algebra and theprojections

p =(

1 00 0

)and q =

(0 00 1 1

)belong to QB(H(`2L2(N))) which is a semifinite von Neumann algebra. Sinceq is infinite and p is finite, [26, Theorem V.1.8] yields a nonzero central projectionz Z(Q) such that zp - zq. There exists an isometry w Q such that ww = zpand ww zq. If we write

z =(z1 00 1 z2

)and w =

(a bc d

),

we get

aa+ cc = z1bb+ dd = 0aa + bb = 0cc + dd 1 z2.

Thus a = 0, b = 0, d = 0, cc = z1 and v = c : Hz1 `2 L2(N) is an isometry(since uu = z1) which moreover satifies vpi(x) = (1 x)v (since w Q). Then asimple maximality argument finishes the proof.

Since p = vv commutes with the right N -action on `2 L2(N), it follows thatp B(`2)N . Thus, as right N -modules, we have

HN ' p(`2 L2(N))N .On B(`2)N , we define the following faithful normal semifinite trace Tr (which

depends on ): for any x = [xij ]i,j (B(`2)N)+,Tr ([xij ]i,j) =

i

(xii).

We set dim(HN ) = Tr(vv). Note that the dimension ofH depends on but doesnot depend on the isometry v. Indeed take another isometry w : H `2 L2(N),satisfying w(b) = w()b, for any H, b N . Note that vw B(`2)N andww = vv = 1. Thus, we have

Tr(vv) = Tr(vwwv) = Tr(wvvw) = Tr(ww).

We define dim(HN ) := Tr(vv), for any isometry v as in Proposition B.1.

Appendix C. Ultraproducts

Let be a free ultrafilter on N and (N, ) a finite von Neumann algebra. LetI be the norm closed ideal of `(N, N) defined by

I ={

(xn) `(N, N) : limn xn2 = 0

}.

Denote by pi : `(N, N) `(N, N)/I the quotient map. The tracial ul-traproduct of N is defined as N := pi(`(N, N)) with tracial faithful state


(pi((xn))) = limn (xn). It is easy to check that N is indeed a von Neumannalgebra and is normal on N.

Exercise C.1. Let be a free ultrafilter and (N, ) a finite von Neumann algebra. Prove that every projection e N lifts to a projection (en) `(N, N)

such that limn (en) = (e). Show that if N is a II1 factor, so is N.

Regard L2(N) as a Hilbert subspace of L2(N). For a 0, denote by fa thecharacteristic function of the interval (

a,+).

Proposition C.2 ([7]). Let = (n) L2(N). Then L2(N) if and only if satisfies the following equi-integrability condition:

(2) > 0,a > 0, limn fa(|n|)|n|2 < .

Proof. Assume that L2(N) satisfies (2). Simply write = (n). Fix > 0.Then for some a > 0, one has limn fa(|n|)n2 < so that the vector = (n)with n = n(1fa)(|n|) satisfies 2 = limn nn2 and n a.Thus, L2(N).

Conversely assume that L2(N). We may assume that n2 1, for alln N. Let (0, 1). We can then find a > 0 and xn N such that nxn2 and xn a, for all n N. Up to extracting, Powers-Strmer Inequality yieldsin particular

|n| |xn|22 |n|2 |xn|21 n xn2n + xn2 3, n N.Using the same trick as in the proof of Theorem 3.9, we get

0

fb(|n|) fb(|xn|)22 |n|2 |xn|21 3.

Since Eb(|xn|) = 0 for all b > a, we get f2a(|n|)2 (3)1/2/a, for all n N.Then, for all n N,

f2a(|n|)|n|2 f2a(|n|)(|n| |xn|)2 + xnf2a(|n|)2 |n| |xn|2 + xnf2a(|n|)2 2(3)1/2.

This finishes the proof.

Appendix D. On the geometry of two projections

Proposition D.1 ([7]). Let M be a finite von Neumann algebra. If e and f areequivalent projections in M , then there exists u U(M) such that

ueu = fu|e f | = |e f |u|u 1|

2|e f |.

Proof. Recall the analysis of two projections in [26, Chapter V]. Let e = 1 eand f = 1 f . Set

e0 = e e f e ff0 = f f e f e.

42 CYRIL HOUDAYER

We know that e0 f0 = 0 and e0 e0 f0 = e0 f0. We then represent

e = e f (

1 00 0

) e f 0 0,

f = e f (c2 cscs s2

) 0 e f 0,

with c, s (e0 f0)M(e0 f0) positive elements such that c2 + s2 = e0 f0. Sincee f and e0 f0, the finiteness of M yields ef ef . Consequently, efand e f are represented by matrices:

e f =(

1 00 0

)e f =

(0 00 1

).

Therefore we come to the following situation:

e = e f (

1 00 0

)(

1 00 0

) 0,

f = e f (c2 cscs s2

)(

0 00 1

) 0.

and

|e f | = 0(s 00 s

)(

1 00 1

) 0.

We set

u = e f (c ss c

)(

0 11 0

) (e f).

It is straightforward to check that u is a unitary, ueu = f , u|e f | = |e f |u andthat

(u 1)(u 1) 2(e f)2.

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CNRS-ENS Lyon, UMPA UMR 5669, 69364 Lyon cedex 7, FranceE-mail address: [email protected]

an introduction to ii1 factors

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