1 factors generated by regular representations of the group B 0 N

1. Regular representations

Let us consider the group ˜G=B^N of all upper-triangular real matrices of infinite order with units on the diagonal

G˜=B^N={I+x|x= X

1≤k<n

x_knE_kn}, and its subgroup

G=B^N₀ ={I+x∈B^N| x is finite},

where E_kn is an infinite-dimensional matrix with 1 at the place k, n ∈ N and zeros elsewhere, x= (x_kn)_k<n isfinite means thatx_kn= 0 for all (k, n) except for a finite number of indices k, n∈N.















1 x₁₂ x₁₃ x₁₄ · · · 0 1 x23 x24 · · · 0 0 1 x₃₄ · · · 0 0 0 1 · · · . ..















Obviously,B₀^N= lim−→nB(n,R) is the inductive limit of the groupB(n,R) of real upper-triangular matrices with units on the principal diagonal

B(n,R) ={I+ X

1≤k<r≤n

xkrEkr|xkr ∈R}

with respect to the embeddingB(n,R)3x7→x+E_n+1n+1 ∈B(n+ 1,R).

We define the Gaussian measure µ_b on the group B^N in the following way

(24) dµ_b(x) = O

1≤k<n

(b_kn/π)^1/2exp(−b_knx²_kn)dx_kn=O

k<n

dµ_bkn(x_kn), whereb= (bkn)k<n is some set of positive numbers.

Let us denote by R and L the right and the left action of the group B^N on itself: R_s(t) = ts⁻¹, L_s(t) = st, s, t ∈ B^N and by Φ : B^N 7→

B^N, Φ(I+x) := (I +x)⁻¹ the inverse mapping. It is known [Kos92] that Lemma 2.1. µ^R_b^t ∼µ_b ∀t∈B₀^N for any set b= (b_kn)_k<n.

21

Lemma 2.2. µ^L_b^t ∼ µ_b ∀t ∈ B₀^N if and only if S_kn^L(b) < ∞, ∀k < n, where

S_kn^L (b) =

∞

X

m=n+1

b_km b_nm.

Lemma 2.3. µ^L_b^t ⊥µ_b ∀t∈B₀^N\{e} ⇔S_kn^L (b) =∞ ∀k < n.

Lemma 2.4. [Kos00] If E(b) = P

k<nS_kn^L(b)(bkn)⁻¹ < ∞, then µ^Φ_b ∼ µ_b.

Lemma 2.5. [Kos00]The measureµb onB^N isB₀^N ergodic with respect to the right action.

Let α : G → Aut(X) be a measurable action of a group G on the measurable space X. We recall that a measure µ on the space X is G-ergodic if f(αt(x)) = f(x) ∀t ∈ G implies f(x) = const µ a.e. for all functions f ∈L¹(X, µ).

Remark 2.6. [KZ00] Ifµ^Φ_b ∼µ_b then µ^L_b^t ∼µ_b ∀t∈B₀^N.

Proof. This follows from the fact that the inversion Φ interchanges the right and the left action: R_t◦Φ = Φ◦L_t∀t ∈ B^N. Indeed, if we denote µ^f(·) =µ(f⁻¹(·)) we have (µ^f)^g =µ^f◦g. Hence

µ_b∼µ^R_b^t ∼(µ^R_b^t)^Φ=µ^R_b^t◦Φ =µ^Φ◦L_b ^t = (µ^Φ_b)^Lt ∼µ^L_b^t.

Ifµ^R_b^t ∼µ_b and µ^L_b^t ∼µ_b ∀t∈B₀^N, one can define in a natural way (see [Kos92]), an analogue of the rightT^R,b and leftT^L,bregular representation of the groupB^N₀ in the Hilbert space H_b=L²(B^N, dµ_b)

T^R,b, T^L,b:B^N₀ →U(H_b =L2(B^N, dµb)), (T_t^R,bf)(x) = (dµ_b(xt)/dµ_b(x))^1/2f(xt), (T_s^L,bf)(x) = (dµ_b(s⁻¹x)/dµ_b(x))^1/2f(s⁻¹x).

2. Von Neumann algebras generated by the regular representations

Let A^R,b = (T_t^R,b|t∈B₀^N)⁰⁰ (resp. A^L,b= (Ts^L,b |s∈B₀^N)⁰⁰) be the von Neumann algebras generated by the right T^R,b (resp. the left T^L,b) regular representation of the groupB^N₀.

Theorem2.7. [Kos00]If E(b)<∞ thenµ^Φ_b ∼µb. In this case the left regular representation is well defined and the commutation theorem holds:

(25) (A^R,b)⁰=A^L,b.

Moreover, the operator Jµb given by

(26) (J_µbf)(x) = (dµ_b(x⁻¹)/dµ_b(x))^1/2f(x⁻¹)

2. VON NEUMANN ALGEBRAS GENERATED BY THE REGULAR REPRESENTATIONS23

is an intertwining operator:

T_t^L,b=JµbT_t^R,bJµb, t∈B₀^N and JµbA^R,bJµb=A^L,b.

Ifµ^L_b^t ⊥µ_b ∀t∈B₀^N\{e} one can’t define the left regular representation of the groupB^N₀. Moreover the following theorem holds ([Kos92])

Theorem 2.8. The right regular representation T^R,b :B₀^N → U(H_b) is irreducible if and only if µ^L_b^s ⊥µ_b ∀s∈B₀^N\{0}.

Corollary 2.9. The von Neumann algebra A^R,b is a type I∞ factor if µ^L_b^s ⊥µ_b ∀s∈B^N₀\{0}.

Let us assume now that µ^L_b^t ∼ µb ∀t ∈ B₀^N\{e}. In this case the right regular representation and the left regular representation of the group B₀^N are well defined.

In this Chapter we shall prove that if E(b) < ∞, the von Neumann algebras A^R,b and A^L,b are always factors. This is implied by the triviality of the centralizer of our algebras w.r.t. the vector 1∈L²(B^N, dµ), which is the main step in proving the type III₁ property.

Until now there existed sufficient conditions on the measure µb, for the factor property ([KZ00]). We give a short review.

Since T_t^L,b∈(A^R,b)⁰ ∀t∈B₀^N, we have A^L,b⊂(A^R,b)⁰, hence (27) A^R,b∩(A^R,b)⁰⊂(A^L,b)⁰∩(A^R,b)⁰= (A^R,b∪A^L,b)⁰. The last relation shows thatA^R,b is factor if the representation

B₀^N×B₀^N3(t, s)7→T_t^R,bT_s^L,b∈U(H_b) is irreducible.

Let us denote byA^R,L,bthe the von Neumann algebras generated by the right T^R,b and the left T^L,b regular representations of the groupB₀^N:

A^R,L,b= (T_t^R,b, T_s^L,b|t, s∈B₀^N)⁰⁰= (A^R,b∪A^L,b)⁰⁰. Let us denote

(28) S_kn^R,L(b) =

∞

X

m=n+1

b_km

S_nm^L (b), k < n.

Theorem 2.10. [KZ00] The representation

B₀^N×B₀^N3(t, s)7→T_t^R,bT_s^L,b∈U(Hb) is irreducible if S_kn^R,L(b) =∞, ∀k < n.

Corollary2.11. The von Neumann algebraA^R,bis a factor ifS_kn^R,L(b) =

∞ ∀k < n.

3. Modular operator

In this section we review the construction of the modular operator for locally compact groups and extend it to the case of inductive limits of such groups. First we establish that the constant function 1 is cyclic and separat-ing forA^R,b andA^L,b(assumingE(b)<∞). We prove this by showing that setA^R,b1 contains all the polynomials, in the variablesx_kn, k < n∈N, which is dense in H_b. Using the intertwining operator J_b, we show that the same holds for the commutant ofA^R,b, which implies the separating property.

Lemma2.12. Assume thatE(b)<∞. Then the function1∈L²(B^N, µ_b) is cyclic and separating for A^R,b.

Proof. (1) First we prove the cyclic property. Consider the one-parameter groups in B₀^N,

(29) Gkn(t) :={1 +tEkn, t∈R}.

The corresponding one parameter groups T_kn^R,b(t) := {T_u^R,b;u ∈ G_kn(t)} have generators (see [Kos92], here for convenience, we omit the superscript b)

(30) A^R_kn=

k−1

X

r=1

x_rkD_rn+D_kn, where Dpq =∂pq−bpqxpq and∂pq = _∂x^∂

pq. Suppose that f ∈L²(B^N, dµ_b) and (31) (f, T_t^R,b1) =

Z

f(x)T_t^R,b1(x)dµ_b(x) = 0, ∀t∈B₀^N.

We want to prove that f = 0, which implies that the linear span of the set {T_t^R,b1;t ∈B^N₀} is dense in L²(B^N, dµ_b), since we chose f arbitrarily. We shall prove that (31) implies

(32) (f, P) = 0,

where P(x) are polynomials of finite order in the variables x_kn. Since the setP of polynomialsP is dense inL²(B^N, dµ_b) (for exam-ple by the fact that the Hermite Polynomials (23) spanL²(B^N, dµ_b)), this proves thatf = 0.

Now we shall prove the above property. First of all from (31) follows that

(f,Y T_k^R,b

ini(t_i,αj)1) = (f,Y e⁽

P_mi

j=1t_kiniαj)A^R_kini

1) = 0,

for some finite product of T_kn^R,b(t), where the index α_j varies ac-cording to the multiplicity of i. SinceT_kn^R,b(t) are strongly continu-ous one-parameter groups with generators A^R_kn and P(⊃ {1}) is a common dense domain of these generators ([Kos01]), we conclude

3. MODULAR OPERATOR 25

(after taking derivatives in all the parameters and setting them to 0) that

(f,

p

Y

i=1

(A^R_kini)^mi1) = 0, for all finite products of different A^R_k

ini.

Now we show that the set hA^R_kn;k < n∈Ni1, where hA^R_kn;k <

n ∈ Ni is the algebraic linear span generated by the generators A^R_kn, contains all the polynomials of finite order in the independent variablesxkn. We make use of multiple nested inductions as follows.

The main induction is the following:

(a) We prove that the span of {x^α_1n¹

1...x^α_1n^k

k; 1≤k, αi ∈N, i= 1..k},

where all the indices ni are mutually different, is contained in hA^R_kn;k < n∈Ni1.

(b) The second step is to prove that the span of {x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k; 1≤k, l, α_i, β_j ∈N, i= 1..k, j = 1..l}, is contained in hA^R_kn;k < n∈Ni1

(c) Now comes the induction step. Assume that the span of {x^γ_p−1s¹

1...x^γ_p−1s^r _r...x^β_2m¹ ₁...x^β_2m^l _lx^α_1n¹₁...x^α_1n^k_k; 1≤k, l, r, αµ, βν, γη ∈N, µ= 1..k, ν = 1..l, η= 1..r}

is contained in hA^R_kn;k < n ∈ Ni1 for some p > 2. Then we prove that the span of

{x^δ_ps¹₁...x^δ_ps^u_u...x^β_2m¹₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k;

1≤k, l, u, α_µ, β_ν, δ_η ∈N, µ= 1..k, ν = 1..l, η= 1..u}

is also contained in hA^R_kn;k < n∈Ni1. This implies that any polynomial of finite order is in the latter set.

(1a) We prove the first step. Again we use induction, this time on the number of factors in the monomials.

• x^α_1n¹

1 ∈ hA^R_kn;k < n∈Ni1. Indeed (again we use induction), A^R_1n11 =−b_1n1x1n1.

Furthermore, assume that Span{x^α_1n¹⁻¹

1 } ⊂ hA^R_kn;k < n∈Ni1, then also the desired property holds, since

A^R_1n1x^α_1n¹⁻¹

1 =D_1n1x^α_1n¹⁻¹

1 = (α₁−1)x^α_1n¹⁻²

1 −b_1n1x^α_1n¹

1.

• The induction step is as follows. Assume thatx^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁ ∈ hA^R_kn;k < n ∈ Ni1. The following equations show (using in-duction on αk), that also

x^α_1n^k

kx^α_1n^k−1

k−1...x^α_1n¹

1 ∈ hA^R_kn;k < n∈Ni1 : A^R_1n

kx^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁ = −b_1nkx_1nkx^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁, A^R_1n

kx^α_1n^k−1

k x^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁ = (α_k−1)x^α_1n^k−2

k ...x^α_1n¹₁

−b_1nkx^α_1n^k_kx^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁ (1b) Now we continue with the second step of the main induction.

• Consider the following equation:

A^R_2m11 = (x₁₂D_1m1 +D_2m1) 1 =−b_1m1x₁₂x_1m1 −b_2m1x_2m1.

Since the first term is in hA^R_kn;k < n∈ Ni1, so is the second term. It is also easy to see that

x^β_2m¹₁x^α_1n^k

kx^α_1n^k−1_k−1x^α_1n^k−2_k−2...x^α_1n¹₁ ∈ hA^R_kn;k < n∈Ni1 and

x^β_2m^l−1

l−1...x^β_2m¹

1x^α_1n^k

k...x^α_1n¹

1 ∈ hA^R_kn;k < n∈Ni1

for some l. This is proved in the same way as the previous case.

• Now suppose that x^β_2m^l−1

l ...x^β_2m¹₁x^α_1n^k

k...x^α_1n¹₁ ∈ hA^R_kn;k < n∈Ni1 Then

A^R_2m

lx^β_2m^l−1

l−1...x^β_2m¹

1x^α_1n^k

k...x^α_1n¹

1

=x12(∂1m_l−b1m_lx1m_l)x^β_2m^l−1_l−1...x^β_2m¹₁x^α_1n^k

k...x^α_1n¹₁ +(∂2ml−b2mlx2ml)x^β_2m^l−1_l−1...x^β_2m¹₁x^α_1n^k

k...x^α_1n¹₁.

The first term in the right hand side contains only monomials of order lower than β_l in the variable x2m_l and in the second term we see the monomials

x^β_2m^l

l...x^β_2m¹₁x^α_1n^k

k...x^α_1n¹₁.

(1c) Finally we turn to the main induction step. So assume that Span{x^γ_p−1s¹

1...x^γ_p−1s^r _r...x^β_2m¹ ₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k; 1≤k, l, r, α_µ, β_ν, γ_η ∈N, µ= 1..k, ν = 1..l, η= 1..r}

is contained in hA^R_kn;k < n∈Ni1 for some p >2.

3. MODULAR OPERATOR 27

• Again we see that A^R_pu1x^γ_p−1s¹

1...x^γ_p−1s^r

r...x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k

= (Pp−1

i=1 xip(∂iu1−biu1xiu1))x^γ_p−1s¹ ₁...x^γ_p−1s^r _r...x^β_2m¹₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k

+(∂pu1 −bpu1xpu1)x^γ_p−1s¹ ₁...x^γ_p−1s^r _r...x^β_2m¹₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k.

The summation in the first term gives rise only to monomials containing xin for i < p, which are in hA^R_kn;k < n ∈ Ni1 by the induction hypothesis. Thus the latter set also contains the second term and hence

x_pu1x^γ_p−1s¹

1...x^γ_p−1s^r

r...x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k. The same holds for

x^δ_pu¹₁x^γ_p−1s¹ ₁...x^γ_p−1s^r _r...x^β_2m¹ ₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k.

• Finally suppose that also

x^δ_pu^v−1_v ...x^δ_pu¹₁x^γ_p−1s¹ ₁...x^γ_p−1s^r _r...x^β_2m¹₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k

is in hA^R_kn;k < n∈Ni1. We calculate A^R_puvx^δ_pu^v−1_v ...x^δ_pu¹₁x^γ_p−1s¹

1...x^γ_p−1s^r

r...x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k

= (Pp−1

i=1x_ip(∂_iuv−b_iuvx_iuv))x^δ_pu^v−1_v ...x^δ_pu¹₁x^γ_p−1s¹

1...x^γ_p−1s^r _r...x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k

+(∂puv−bpuvxpuv)x^δ_pu^v−1_v ...x^δ_pu¹₁x^γ_p−1s¹ ₁...x^γ_p−1s^r _r...x^β_2m¹₁...x^β_2m^l

lx^α_1n¹₁...x^α_1n^k

k. Again, the summation in the first term gives rise to a polyno-mial of order smaller than δv in the xpuv variable, which by the last induction hypothesis is contained in our span of gen-erators acting on 1. So does the first monomial in the second term (after expanding the brackets). The last monomial gives us the final statement:

x^δ_pu^v_v...x^δ_pu¹₁x^γ_p−1s¹

1...x^γ_p−1s^r

r...x^β_2m¹

1...x^β_2m^l

lx^α_1n¹

1...x^α_1n^k

k ∈ hA^R_kn;k < n∈Ni1, for any of the parameters p, δi, γi, βi, αi∈Nand v, r, l, k∈N. It follows that the set hA^R_kn, k < ni acting on 1 generates the set P of all possible polynomials in the independent variables xkn. Thus equation (32) holds for the function f. P is dense in L²(B^N, dµ_b) and hencefmust be equal to 0. Sincef ∈L²(B^N, dµ_b) was arbitrary and the equation (31) holds for all t∈B₀^N, the span of {T_t^R,b1;t ∈ B^N₀} must be dense in L²(B^N, dµ_b) and hence 1 is cyclic for A^R,b.

(2) Now we turn to the separating property. In this case we have to prove that 1 is cyclic for (A^R,b)⁰ = A^L,b. Thus, again consider f ∈L²(B^N, dµ_b) and assume

(33) (f, b1) = 0,∀b∈A^L,b.

Recall that E(b) < ∞ implies the existence of the intertwining operator J, which is anti-unitary. Then the following calculation holds:

(f, T_t^R,b1) = (J T_t^R,b1, J f)

= Z s

dµ_b(x⁻¹) dµb(x)

s

dµ_b((xt)⁻¹) dµb(x⁻¹)

s

dµ_b(x⁻¹)

dµb(x) f(x⁻¹)dµb(x)

= Z s

dµ_b(t⁻¹x⁻¹)

dµ_b(x⁻¹) f(x⁻¹)dµ(x⁻¹).

If we replacex⁻¹ byx in the above integral we obtain (f, T_t^R,b1) = (f, T_t^L,b1) for all t∈B₀^N. From (1) we know that (f, T_t^R,b1) = 0 for all t∈B₀^N implies thatf = 0. Hence (f, T_t^L,b1) = 0 for allt∈B₀^N also implies thatf = 0 and hence 1 is cyclic forA^L,b, since we chose f arbitrarily.

We recall how to find the modular operator and the operator of canonical conjugation for the von Neumann algebraA^ρ_G, generated by the right regular representationρof a locally compact Lie groupG. Lethbe a right invariant Haar measure onGand

ρ, λ:G7→U(L²(G, h))

be the right and the left regular representations of the group Gdefined by (ρ_tf)(x) =f(xt), (λ_tf)(x) = (dh(t⁻¹x)/dh(x))^−1/2f(t⁻¹x).

To define the right Hilbert algebra onGwe can proceed as follows. LetM(G) be algebra of all probability measures onGwith convolution determined by

Z

f dµ∗ν = Z Z

f(st)dµ(s)dν(t).

We define the homomorphism M(G)3µ7→ρ^µ=

Z

G

ρtdµ(t)∈B(L²(G, h)).

We have ρ^µρ^ν =ρ^µ∗ν, indeed ρ^µρ^ν =

Z

G

ρtdµ(t) Z

G

ρsdν(s) = Z

G

Z

G

ρtsdµ(t)ν(s) = Z

G

ρtd(µ∗ν)(t) =ρ^µ∗ν.

3. MODULAR OPERATOR 29

Let us consider a subalgebra Mh(G) := (ν ∈M(G) |ν ∼h) of the algebra M(G) In the case when µ ∈ M_h(G) we can associate with the measure µ its Rodon-Nikodim derivative dν(t)/dh(t) = f(t). When f ∈ C₀^∞(G) or f ∈L¹(G) we can write

ρ^f = Z

G

f(t)ρtdh(t),

hence we can replace the algebra M_h(G) by its subalgebra identified with algebra of functions C₀^∞(G) or L¹(G, h) with convolutions. If we replace the Haar measure h with some measure µ∈ Mh(G) we obtain the isomor-phic imageT^R,µ of the right regular representationρin the spaceL²(G, µ):

T_t^R,µ = U ρ_tU⁻¹ where U : L²(G, h) 7→ L²(G, µ) defined by (U f)(x) = dh(x)

dµ(x)

1/2

f(x). We have

(T_t^R,µf)(x) =

dµ(xt) dµ(x)

1/2

f(xt), and

T^f = Z

G

f(t)T_t^R,µdµ(t).

We have (see [Con94], p.462) (we shall writeT_t instead ofT_t^R,µ ) S(T^f) := (T^f)^∗ =

Z

G

f(t)T_t⁻¹dµ(t) = Z

G

f(t)T_t⁻¹ dµ(t)

dµ(t⁻¹)dµ(t⁻¹) Z

G

dµ(t⁻¹)

dµ(t) f(t⁻¹)Ttdµ(t).

Hence

(Sf)(t) = dµ(t⁻¹)

dµ(t) f(t⁻¹).

To calculate S^∗ we use the fact that S is anti-linear so (Sf, g) = (S^∗g, f).

We have

(Sf, g) = Z

G

dµ(t⁻¹)

dµ(t) f(t⁻¹)g(t)dµ(t) = Z

G

f(t⁻¹)g(t)dµ(t⁻¹) = Z

G

g(t⁻¹)f(t)dµ(t) = (S^∗g, f),

hence (S^∗g)(t) = g(t⁻¹). Finally the modular operator ∆ defined by ∆ = S^∗S has the following form (∆f)(t) = _dµ(t^dµ(t)−1)f(t). Indeed we have

f(t)7→^S dµ(t⁻¹)

dµ(t) f(t⁻¹) ^S

∗

7→ dµ(t) dµ(t⁻¹)f(t).

Finally, sinceJ =S∆^−1/2 (see [Con94] p.462) we get f(t)^∆

−1/2

7→

dµ(t⁻¹) dµ(t)

1/2

f(t)7→^J dµ(t⁻¹) dµ(t)

dµ(t) dµ(t⁻¹)

1/2

f(t⁻¹)

=

dµ(t⁻¹) dµ(t)

1/2

f(t⁻¹).

Hence

(J f)(t) =

dµ(t⁻¹) dµ(t)

1/2

f(t⁻¹), and (∆f)(t) = dµ(t) dµ(t⁻¹)f(t).

To prove that J T_t^R,µJ =T_t^L,µ we get f(t)7→^J

dµ(x⁻¹) dµ(x)

1/2

f(x⁻¹)^T

R,µ

7→t

dµ(xt) dµ(x)

1/2

dµ((xt)⁻¹) dµ(xt)

1/2

f((xt)⁻¹) = dµ(t⁻¹x⁻¹)

dµ(x)

1/2

f(t⁻¹x⁻¹)7→^J

dµ(x⁻¹) dµ(x)

1/2

dµ(t⁻¹x) dµ(x⁻¹)

1/2

f(t⁻¹x) = dµ(t⁻¹x)

dµ(x) 1/2

f(t⁻¹x) = (T_t^L,µf)(x).

Remark 2.13. The representation T^R,b is the inductive limit of the representations T^R,µmb of the group B(m,R) where the measure µ^m_b is the projection of the measureµbonto subgroupB(m,R). Obviouslyµ^m_b is equiv-alent with the Haar measureh_m on B(m,R).

Hence, we conclude that the modular operator of (A^R,b,1) is defined by

(34) ∆(f)(x) = dµ_b(x)

dµ_b(x⁻¹)f(x).

4. Examples

We still need to verify whether the von Neumann algebrasA^R,bandA^L,b exist. Here we give an example of a measure µ_b for which the conditions S_kn^L(b)<∞, for all k < n∈Nand E(b)<∞ are fulfilled.

In the example 1 below for the particular caseb_kn= (a_k)ⁿ we give some sufficient conditions on the sequence an.

Example 1. Let us takeb_kn= (a_k)ⁿ, k, n∈N. We have

(35) S_kn^L(b) =

∞

X

m=n+1

ak

an

m

= ak

an

n+1 ∞

X

m=0

ak

an

m

= ak

an

n+1

1 1−_a^ak

n

<∞.

ForE(b) holds:

E(b) =

∞

X

k=1

∞

X

n=k+1

S_kn^L(b) b_kn =

∞

X

k=1

∞

X

n=k+1

ak

an

n+1

1 1−_a^ak

n

1 aⁿ_k =

∞

X

k=1

ak

∞

X

n=k+1

1 an

n+1 1 1−^a_a^k

n

<

∞

X

k=1

a_k 1−_a^ak

k+1

∞

X

n=k+1

1 an

n+1

5. THE TYPE III1 FACTORS 31

<

∞

X

k=1

a_k 1−_a^ak

k+1

∞

X

n=k+1

1 a_k+1

n+1

=

∞

X

k=1

a_k 1−_a^ak

k+1

1 a_k+1

k+2

1 1−_a¹

k+1

=

∞

X

k=1

ak

ak+1

1−_a^ak

k+1

2

1 a_k+1

k+1

.

Example 2. Let us take b_kn = (a_k)ⁿ, k, n ∈N where a_k =s^k, k ∈N withs >1.

Conditions (35) hold for a_k =s^k. We have S_kn^L (b) =

ak

an

n+1

1 1−^a_a^k

n

= 1

s^n−k n+1

1 1− ¹

s^n−k

∼ 1

s^n−k n+1

<∞ Using the latter equivalence we conclude thatE(b)<∞.Indeed we have

E(b) =

∞

X

k=1

∞

X

n=k+1

S_kn^L (b) b_kn ∼

∞

X

k=1

∞

X

n=k+1

1 s^(n−k)(n+1)

1 s^kn

=

∞

X

k=1

s^k

∞

X

n=k+1

1 sⁿ⁽ⁿ⁻¹⁾ =

∞

X

k=1

s^k

∞

X

n=0

1 s(n+k+1)(n+k+2)

<

∞

X

k=1

s^k 1 s^(k+1)(k+2)

∞

X

n=0

1

sⁿ = 1 1−¹_s

∞

X

k=1

1

s^(k+1)2+1 <∞.

5. The type III1 factors

Let A^R,b and A^L,b be the von Neumann algebras defined in Section 2 and assume that E(b) <∞. In this section we prove that A^L,b (and hence A^R,b) are type III₁factors, with no further assumptions. The main step is to prove that the fixed point algebra ofA^L,bw.r.t. the modular group is trivial.

From the last section follows that the stateφ(.) = (1, .1) is a faithful normal state on A^L,b (and A^R,b). Note that in this case, M_φ =M^σφ, whereσ_φ is the modular group ofφ and M_φ is the centralizer of Mw.r.t. φ ([BR02], Prop. 5.3.28). We want to prove thatM_φ=C.1implies thatM is of type III1.

Lemma 2.14 ([Bau95]). Let M be a von Neumann algebra onH andη a cyclic and separating vector for M, with σ its associated modular group.

Assume that σ is inner, i.e. σ_t(a) = U(t)aU(t)^∗, with a one-parameter group U(t)∈ M for all t∈R. Then U(t)∈ M^σ for all t∈R.

Proof. The stateφ(a) := (η, aη) is invariant w.r.t. σ(follows for exam-ple from the modular condition, see Chapter 1). Thus (η, U(t)aU(−t)η) = (η, aη). Since U(t)∈ Mwe can replace aby aU(t) and obtain

(η, U(t)aη) = (η, aU(t)η), a∈ M.

Since M_φ=M^σ, we have proven the result.

Assume now thatM^σ =C.1. First of all we note thatC_M⊆ M_φ=C1, which follows from the definition. Thus in this case M is a factor. It also follows that if M is semi-finite and hence σ is inner (Theorem 1.11 from Chapter 1), that σt(a) = a, since σ is implemented by a scaling operator.

Thus C1 = M^σ = M. Hence a non-trivial von Neumann algebra with an ergodic modular automorphism group must be type III factor. In fact, by the following theorem, it is a type III₁ factor!

Theorem2.15. [Bau95]Let Mbe a von Neumann algebra (M 6=C.1) and φ a faithful normal state. Assume that the centralizer of φ is trivial, i.e.

M_φ:={a∈ M;φ(ab) =φ(ba),∀b∈ M}=C.1.

Then Mis a type III1 factor.

Proof. The strategy of the proof is to exclude the other cases. Recall the definition (11) of Connes spectrum Γ(M).

(1) We assert that Γ ={1}is impossible. IfSp∆\{0}={1}then either Sp∆ ={1} orSp∆ ={1,0}. Sp∆ ={1}means that ∆ =1, which implies M = M_φ =C.1, and this is not the case. Sp∆ = {1,0}

means that 0 is an isolated point of Sp∆, therefore the eigenpro-jection of the point 0 does not vanish, hence there is a vectore6= 0 with ∆e= 0 which is impossible since ∆ is invertible.

(2) We assert that Γ =λ^Z,0< λ <1 is impossible. On the contrary, let Sp∆\{0}=λ^Z. This implies, among other things, that 0< λ <1 is an isolated point of Sp∆. Then lnλis an isolated point of Spln ∆.

Now we use some results from spectral analysis of automorphism groups (see Appendix B). Since Spln ∆ = Spσ^φ, we have lnλ ∈ Spσ^φ and lnλ is isolated. Now, by Corollary B.2 from Appendix B, there is an 06=a∈ M, such that

(36) ∆^ita∆^−it=λ^ita.

Now we use the KMS condition (see Chapter 1) to obtain a con-tradiction. Let b∈ M and define the functions

F_a,b(t) = φ(σ^φ_t(a)b) =λ^itφ(ab), F_a,b(t+i) = φ(bσ^φ_t(a)) =λ^it(ba).

On the other hand, Fa,b(t+i) =λ^i(t+i)φ(ab). It follows that

(37) φ(ab) =λφ(ba).

Now we use (36), in the adjoint form:

∆^ita^∗∆^−it=λ^−ita^∗.

This implies, using the automorphism property,

∆^ita^∗a∆^−it =a^∗a,

5. THE TYPE III1 FACTORS 33

i.e. a^∗a∈C.1 and evena^∗a=µ1,µ >0, becausea6= 0. The same argument holds for aa^∗, which yields aa^∗ = κ1, κ > 0. Now, we can normalize asuch thata^∗a=1, thenaa^∗ is a projection, hence aa^∗=1follows and ais unitary. But if we putb:=a^∗ in equation (37), we obtain

φ(aa^∗) =λφ(a^∗a), or λ= 1 which is a contradiction.

From Section 3 we know that the state φ(.) := (1, .1) (here we consider it on A^L,b) is faithful, since 1 is cyclic and separating forA^R,b andA^L,b and that its modular group is defined by

σ_φ(a) := Ad∆^it(a), a∈A^L,b,

∆(f)(x) = dµb(x)

dµ_b(x⁻¹)f(x), f ∈L²(B^N, dµb).

Now we state the first main theorem.

Theorem 2.16. Consider the von Neumann algebra A^L,b generated by the left regular representation T^L,b of B₀^N. Assume that E(b) < ∞. Let φ(a) = (1, a1) be the faithful normal state associated to the cyclic and sepa-rating vector1and∆the corresponding modular operator. ThenA^L,b_φ =C.1 and hence A^L,b is a type III₁ factor. The same holds forA^R,b.

We need some intermediate results.

Lemma 2.17. Let g be a multiplication on L²(B^N, dµb) by a measurable functiong onB^N, then

(T_t^R,bgT_t^R,b−1f)(x) =g(xt)f(x), ∀x∈B^N, t∈B₀^N, f ∈L²(B^N, dµb).

Proof. Letf ∈L²(B^N, dµ_b). The following calculation holds:

(T_t^R,bgT_t^R,b−1f)(x) = s

dµb(xt)

dµ_b(x)(gT_t^R,b−1f)(xt)

= s

dµb(xt) dµ_b(x)g(xt)

s

dµb(x) dµ_b(xt)f(x)

= g(xt)f(x).

Proposition2.18. LetMbe a von Neumann algebra onH_b =L²(B^N, dµ_b).

If e^isxkn ∈ M⁰, k < n, T_t^R,b ∈ M⁰,∀t ∈ B₀^N, s ∈ R and the measure µ_b is ergodic, then M=C.1.

Proof. From Chapter 1, Proposition 1.16 we know that if H_b were L²(R, dµ_b), then the result would hold. The space L²(B^N, dµ_b) is isomor-phic to L²(R^∞, dµb) = ⊗_k<n∈NL²(R¹, dµbkn). Since the variables xkn are independent, the condition e^ixkns ∈ M⁰, for all k < n and s ∈ R, means thatL^∞(R, dµ_bkn)⊂ M⁰ for all k < n. This implies that the von Neumann algebra generated by (L^∞(R, dµ_bkn))_k<n is contained in M⁰. The latter is isomorphic to L^∞(B^N, dµ_b), which is maximally abelian. Hence, M ⊂ L^∞(B^N, dµb)⁰ =L^∞(B^N, dµb). Moreover, since we assume that T_t^R,b ∈ M⁰ for all t ∈B₀^N, all functions in M are B₀^N-right invariant, by Lemma 2.17.

By the ergodicity of the measure, they are constantµ_b.a.e.

LetM:=A^L,b_φ . Then

M⁰ = (A^L,b∩ {∆^is;s∈R}⁰)⁰ ={T_t^R,b,∆^is;t∈B₀^N, s∈R}⁰⁰. Lemma 2.19. Let Q_knf(x) :=x_knf(x), where f ∈L²(B^N, dµ_b). Then

Q_kn η M⁰, ∀k < n∈N,

which is equivalent to e^iQkns∈ M⁰ for all s∈R(by Chapter 1, Section 2).

Proof. We shall give two possible methods to prove the Lemma. How-ever, only one of them will lead to a rigorous proof.

Some useful formulas[Kos02]

Let us denote byX⁻¹ the inverse matrix to the upper triangular matrix X=I+x=I+P

k<nx_knE_kn∈B^N

X⁻¹ = (I+x)⁻¹ =I+X

k<n

x⁻¹_knE_kn∈B^N. We have by definitionX⁻¹X =XX⁻¹ =I hence

(38) XX⁻¹

kn=

n

X

r=k

X_krX_rn⁻¹ =δ_kn=

n

X

r=k

X_kr⁻¹X_rn= X⁻¹X

kn, k≤n, hence

x⁻¹_kn+

n−1

X

r=k+1

xkrx⁻¹_rn +xkn= 0 =xkn+

n−1

X

r=k+1

xkrx⁻¹_rn +x⁻¹_kn, k < n, and

(39) x⁻¹_kn =−x_kn−

n−1

X

r=k+1

xkrx⁻¹_rn =−x_kn−

n−1

X

r=k+1

x⁻¹_krxrn. Fromx⁻¹_kk+1 =−x_kk+1, by induction follows

(40)

x⁻¹_kn =−x_kn+

n−k−1

X

r=1

(−1)^r+1 X

k≤i₁<i2<...<ir≤n

xki1xi1i2...xirn, k < n−1.

5. THE TYPE III1 FACTORS 35

Remark 2.20. Using (40) we see that x⁻¹_kn depends only on xrs with k≤r < s≤n.

We have

(41) x⁻¹_kn +xkn=−

n−1

X

r=k+1

xkrx⁻¹_rn, x⁻¹_kn−xkn= 2xkn−

n−1

X

r=k+1

xkrx⁻¹_rn. Let us denote

w_kn:=w_kn(x) := (x_kn+x⁻¹_kn)(x_kn−x⁻¹_kn).

Using (63) we get

(42) ∆(x) = dµ_b(x)

dµb(x⁻¹) = exp



 X

k,n∈N, k<n

b_kn (x⁻¹_kn)²−x²_kn



.

−ln ∆(x) = X

k,n∈N, k<n

bkn

x²_kn−(x⁻¹_kn)²

= X

k,n∈N, k<n

bkn(xkn+x⁻¹_kn)(xkn−x⁻¹_kn) X

k,n∈N, k<n

b_kn(x_kn+x⁻¹_kn)[2x_kn−(x_kn+x⁻¹_kn)] = X

k,n∈N, k<n

b_knw_kn(x).

Method 1: Consider the one-parameter groups inB₀^N, (43) G_kn(t) :={1 +tE_kn, t∈R}.

Recall that the corresponding one parameter groups T_kn^R,b(t) := {T_u^R,b;u ∈ Gkn(t)} have generators

(44) A^R_kn=

k−1

X

r=1

x_rkD_rn+D_kn, whereDpq = _∂x^∂

pq−b_pqxpq. Hence we have two types of generators, affiliated to M⁰ at our disposal, namely A^R_kn and iln ∆. Using certain commutator relations involving these operators, we shall obtain the independent variables x_kn. This suggests that these variables are also affiliated toM⁰, since this would certainly be true if A^R_kn and ln ∆ were bounded. However, since we are dealing with unbounded operators, a more rigorous argument is needed in order to prove the Lemma.

First, we study the commutators ofA^R_kn on ln ∆. In order to do this we need to know the action of D_pq on x⁻¹_kn.

Lemma 2.21. We have (45)

[Dpq, x⁻¹_kn] =

−x⁻¹_kpx⁻¹_qn −δ_kpx⁻¹_qn −δ_qnx⁻¹_kp −δ_kpδ_qn, if k≤p < q ≤n,

0, otherwise.

Proof. We prove (45) by induction inn−k, such thatk≤p < q≤n.

Since x⁻¹_kn =−x_kn, when n−k= 1,

[Dpq, xkn] =−δ_pkδqn

in this case. Next we consider the induction step.

Let us suppose that (45) holds for all (p, q) withr ≤p < q ≤n, k < r.

We prove that then (45) holds also for (p, q), k ≤p < q ≤n. Indeed we have

[D_pq, x⁻¹_kn] =−[D_pq, x_kn+

n−1

X

r=k+1

x_krx⁻¹_rn] =−δ_kpδ_pn−δ_kpx⁻¹_kn−

n−1

X

r=k+1

x_kr[D_sq, x⁻¹_rn] By induction, the last term equals to

n−1

X

r=k+1

x_kr(x⁻¹_rpx⁻¹_qn +δprx⁻¹_qn +δqnx⁻¹_jp +δprδqn) Hence, by (39) [Dpq, x_kn⁻¹] =

−δ_kpδpn−δ_kpx⁻¹_kn−(x⁻¹_kpx⁻¹_qn +x_kpx⁻¹_qn) +x_kpx⁻¹_qn−x⁻¹_kpδqn−x_kpδqn+x_kpδqn,

which reduces to the desired result.

Using (45) we get (46)

[D_pq, x_kn+x⁻¹_kn] =

−x⁻¹_kpx⁻¹_qn −δ_kpx⁻¹_qn −δqnx⁻¹_kp, ifk≤p < q≤n,

0, otherwise.

Using (46) we have [D_pq,(x_kn+x⁻¹_kn)(x_kn−x⁻¹_kn)] = (47)







2x⁻¹_kpx⁻¹_qnx⁻¹_kn+ 2δkpx⁻¹_qnx⁻¹_kn+ 2δqnx⁻¹_kpx⁻¹_kn, ifk≤p < q ≤n, (p, q)6= (k, n), 2(x_kn+x⁻¹_kn), if (p, q) = (k, n),

0, otherwise.

Indeed, if k≤p < q≤n, (p, q)6= (k, n) we have

[Dpq,(xkn+x⁻¹_kn)(xkn−x⁻¹_kn)] = [Dpq,(xkn+x⁻¹_kn)(2xkn−(xkn+x⁻¹_kn))]

= [Dpq,(xkn+x⁻¹_kn)](2xkn−(xkn+x⁻¹_kn))−(xkn+x⁻¹_kn)[Dpq,(xkn+x⁻¹_kn)] =

−2x⁻¹_kn[D_pq,(x_kn+x⁻¹_kn)]⁽⁴⁶⁾= 2x⁻¹_kpx⁻¹_qnx⁻¹_kn+ 2δ_kpx⁻¹_qnx⁻¹_kn + 2δ_qnx⁻¹_kpx⁻¹_kn. Lemma 2.22. We have

(48) [A^R_pq, wkn] =



















0, if k < n≤p, 2xkpxkq, if n=q, k ≤p−1,

0, if 1≤k≤m−1, m+ 1< n, 2x⁻¹_pnx⁻¹_qn, if k=p, n≥q+ 1,

0, if q ≤k < n.

2(x_pq+x⁻¹_pq), if k=p, q=n.

5. THE TYPE III1 FACTORS 37

hence

(49) −[A^R_pq,ln ∆(x)] = 2

p−1

X

r=1

brqxrpxrq+ 2

∞

X

n=q+1

bpnx⁻¹_pnx⁻¹_qn+ 2(xpq+x⁻¹_pq).

Proof. Since

A^R_pq=

p−1

X

r=1

x_rpD_rq+D_pq

and wkn, k < n≤p does not depend on xrq, 1 ≤r ≤q we conclude that [A^R_pq, w_kn] = 0 fork < n≤pand q ≤k < n.

Letn=q, since [D_rq, w_kq] = 0 for 1< r < k we get [A^R_pq, w_kq] =

p−1

X

r=k

x_rp[D_rq, w_kq] + [D_pq, w_kq] =

2 x_kp(x_kq+x⁻¹_kq) +

p−1

X

r=k+1

x_rpx⁻¹_krx⁻¹_kq +x⁻¹_kpx⁻¹_kq

!

=

2 xkpxkq+ xkp+

p−1

X

r=k+1

x⁻¹_krxrp+x⁻¹_kp

! x⁻¹_kq

!

(39)= 2xkpxkq. Similarly, for 1≤k≤p−1, q < n we get

[A^R_pq, w_kn] =

p−1

X

r=k

xrp[Drq, w_kn] + [Dpq, w_kn] =

2 x_kpx⁻¹_qn +

p−1

X

r=k+1

x_rpx⁻¹_krx⁻¹_qn +x⁻¹_kpx⁻¹_qn

!

2 xkp+

p−1

X

r=k+1

xrpx⁻¹_kr +x⁻¹_kp

!

x⁻¹_qn ⁽³⁹⁾= 0.

Ifk=p and n≥q+ 1 we have as before

[A^R_pq, w_pn] = [D_pq, w_pn]⁽⁴⁷⁾= 2x⁻¹_pnx⁻¹_qn. Finally if (p, q) = (k, n),

[A^R_pq, w_pq] = 2(x_pq+x⁻¹_pq)

For p = m, q = m+ 1 the last term of equation (49) vanishes, and we obtain the following formula.

(50) −[A^R_mm+1,ln ∆] = 2

m−1

X

r=1

b_rm+1x_rmx_rm+1+ 2

∞

X

n=m+2

b_mnx⁻¹_mnx⁻¹_m+1n.

Next, we shall use induction to obtain the independent variables. First, we want to act with A^R_m−1,m+1 and A^R_m−1m on the above formula. From (45) we conclude that the terms involving components ofx⁻¹ vanish. The action of A^R_pq onx_kn is easily computed:

(51) [A^R_pq, x_kn] =

p−1

X

r=1

xrp[Drq, x_kn] + [Dpq, x_kn] = (x_kp+δ_kp)δqn. We get the following formulas:

(52)

[A^R_m−1m+1,[A^R_mm+1,ln ∆]] =

−2Pm−2

r=1 b_rm+1xrm−1x_rm−2bm−1m+1xm−1m, [A^R_m−1m,[A^R_mm+1,ln ∆]] =

−2Pm−2

r=1 b_rm+1xrm−1x_rm+1−2bm−1m+1xm−1m+1

To getx1m we compute

[A^R_1m−1,[A^R_m−1m+1,[A^R_mm+1,ln ∆]]] =−2b_1m+1x_1m.

Next we consider the induction step. Suppose that we have obtained all variablesx_rn, r < p < m−1 for somep. The variablesxm−1m andxm−1m+1

are deduced from (52). To get x_pm+1, p < m−1 we compute [A^R_pm−1,[A^R_m−1m+1,[A^R_mm+1,ln ∆]]] =−2

p−1

X

r=1

b_rm+1x_rpxrm−1−2b_pm+1x_pm+1. This method looks nice at the first glance, and it would lead to a rigorous proof if we were to find a common invariant domain for the operators A^R_kn and ln ∆. Indeed, suppose that D is such a domain and A and B are two operators onD, affiliated toM⁰. Than, by Definition 1.4, uD⊆D,u∈ M and

u[A, B]u^∗ξ = (uAu^∗uBu^∗−uBu^∗uAu^∗)ξ

= (uAu^∗B−uBu^∗A)ξ

= [A, B]ξ,

sinceAξ, Bξ∈D. Hence the closure of [A, B] is also affiliated to M⁰. A possible candidate for a common invariant domain for A^R_pq and ln ∆ is the set of functions

hA^R_pq,ln ∆;p < q∈Ni1,

whereh.idenotes the linear algebraic span (we already know that the above set contains the set P of polynomials). However, proving that the above functions are indeed inL²(B^N, dµb) is a very difficult task and hence another approach seems to be more plausible. The second method, which we describe below, will give us the solution.

5. THE TYPE III1 FACTORS 39

Method 2: In this method we circumvent the problem of dealing with unbounded operators, by directly working with elements in M⁰. By certain combinations of elements in M⁰ we would like to obtain the one-parameter groups generated by the multiplication operators Q_kn, k < n ∈ N. First we consider a special case, which leads to a better understanding of the situation. Consider the restriction of the group B^N to the first two lines.

That is, let X⁽²⁾ be the space X⁽²⁾:={1 +x;x=

∞

X

n=2

x1nE1n+

∞

X

n=3

x2nE2n}

X⁽²⁾=

1 x12 x13 x14 · · · 0 1 x23 x24 · · ·

We restrict the action of B₀^N to X⁽²⁾ and keep the previous notation for the restricted operatorsT_t^R,b. We get the following generators of the groups {T_t^R,b;t∈G_kn(t)}:

A^R,2_1n = D1n=∂1n−b1nx1n, A^R,2_2n = x₁₂D_1n+D_2n, A^R,2_kn = x1kD1n+x2kD2n. Define the restriction of the measure µb on X⁽²⁾ by

µ²_b := O

k=1,2;n>k

µb_kn. The formulas for the inverse matrix are the following:

x⁻¹_1n = −x_1n+x₁₂x_2n, x⁻¹_2n = −x_2n.

Furthermore, the modular operator has the following form. Note that the terms with w₁₂and w_kn fork >1 vanish in this case.

(53) ln ∆(x) =

∞

X

n=3

b_1n(x⁻¹_1n−x_1n)(x⁻¹_1n+x_1n) =

∞

X

n=3

b_1n((x₁₂)²(x_2n)²−2x₁₂x_1nx_2n).

LetA^L,b₂ be the von Neumann algebra onL²(X⁽²⁾, dµ²_b) defined by:

A^L,b₂ :={T_t^L,b;t∈B₀^N}⁰⁰.

We remind that also in this case the centralizer of A^L,b₂ w.r.t. the vector 1 is given by

M₂ ={T_t^R,b,∆^is;t∈B₀^N, s∈R}⁰.

Now we want to prove that the one-parameter groups generated by the multiplication operatorsQ₁₂, Q₁₃, ... andQ₂₃, Q₂₄, ..., are contained in M⁰₂.

First we try to obtain Q12. Let X⁰ = XG23(s), where X = 1 +x. We calculate the following expression, where we use Lemma 2.17:

(54) ∆^it(x)T₂₃^R,b(s)∆^−it(x)T₂₃^R,b(−s) = ∆^it(x)∆^−it(x⁰) =eit(ln ∆(x)−ln ∆(x⁰))

Note that only x13 and x23 change under the above transformation. We formally compute ln ∆(x⁰):

ln ∆(x⁰) = b₁₃ (x⁰₁₂)²(x⁰₂₃)²−2x⁰₁₂x⁰₁₃x⁰₂₃

+

∞

X

n=4

b_1n (x₁₂)²(x_2n)²−2x₁₂x_1nx_2n

= b13{(x₁₂)²(x23)²+ 2s(x12)²x23+s²(x12)²

−2x₁₂x₁₃x₂₃−2s((x₁₂)²x₂₃+x₁₂x₂₃)−2s²(x₁₂)²}

+

∞

X

n=4

b_1n (x₁₂)²(x_2n)²−2x₁₂x_1nx_2n

= −b₁₃(2sx₁₂x₁₃+s²(x₁₂)²) + ln ∆(x).

Hence, expression (54) becomes

U₂₃(s, t)(x) :=e^{it{b13(2sx12x13+s2(x12)2)}}

Remark 2.23. It is important to say that the exponentials of infinite sums such as the ones above are well defined, since functional calculus in multiple variables applies to the set of commuting operators{Q_kn}. Indeed, the above formulas with infinite sums (and all formulas in this dissertation, involving sums of functions in the variablesx_kn) consist of terms, which act as multiplication operators on mutually independent Hilbert spaces in the decomposition

L²(B^N, dµ_b) =O

k<n

L²(R¹, dµ_bkn).

Hence we do not need to verify any convergence properties.

It follows that the multiplication operator of the function U23(s, t)(x) is inM⁰₂ for all s, t∈R. Let nowX⁰=XG₁₃(s⁰). To obtainx₁₂we compute:

T₁₃^R,b(s⁰)U₂₃(1,1)(x)T₁₃^R,b(−s⁰) = e^ib13(^2x12x013+(x12)²)

= e^ib13(^2x12x13+(x12)²+2s⁰x12)

Upon multiplying the above expression byU23(1,−1) we obtain the one pa-rameter groupe^2b13s0Q12,s⁰ ∈R. Similarly, we can obtain the one-parameter

5. THE TYPE III1 FACTORS 41

groups generated by the operators Q1m, m > 2. In order to do this we compute

Umm+1(s, t) = ∆^it(x)T_mm+1^R,b (s)∆^−it(x)T_mm+1^R,b (−s)

instead ofU₂₃. Again only the term of ln ∆(x), (53) withn=m+ 1 changes under the above transformation and we extract

(55)

Umm+1(s, t)(x) :=e^{it{b1m+1(2sx1mx1m+1+s2(x1m)2)+b2m+1(2sx2mx2m+1+s2(x2m)2)}}

and hence we obtain x1m analogously to x12, by transforming the x1m+1

term and subtracting the initial part.

In order to obtain the operatorsQ2m, we operate in a similar way. Again consider formula (55). Now substituteX⁰ =XG_2m+1(s) forX and subtract the initial part:

(56)

Umm+1(1,−1)T_2m+1^R,b (s⁰)Umm+1(1,1)(x)T_2m+1^R,b (−s⁰) = e^{2is0(b1m+1x1mx12+b2m+1x2m)}.

We already proved that the one-parameter groups generated by the self-adjoint operators Q_1n are in M⁰₂. From Chapter 1 we know that this is equivalent toQ_1nηM⁰₂. According to Theorem 1.5 in Chapter 1, this implies that also closure of products and sums of these operators are affiliated to the abelian von Neumann algebra generated by Q_1n, which is contained in M⁰₂. Equation (56) then implies that also Q2m η M⁰₂. Hence we obtained Q_kn η M⁰₂ fork= 1,2,n > k.

Now we generalize the above discussion to the case (B^N₀, B^N).

Lemma 2.24. Fix anm∈Nand let

Umm+1(s, t) := ∆^itT_mm+1^R,b (s)∆^−itT_mm+1^R,b (−s)∈ M⁰ for s, t∈R.

Then (U_mm+1(s, t)f)(x) = (57)

exp(itPm−1

k=1 b_km+1 2sx_kmx_km+1+s²(x_km)² +itP∞

n=m+2bmn 2sx⁻¹_mnx⁻¹_m+1n−s²(x⁻¹_m+1n)² )f(x) Furthermore, the following identities hold for all s⁰∈Rand p < m:

(58)

Umm+1(1,−1)T_1m+1^R (s⁰)Umm+1(1,1)T_1m+1^R (−s⁰)f

(x) = exp(2is⁰b1m+1x1m)f(x),

(59)

Umm+1(1,−1)T_pm+1^R (s⁰)Umm+1(1,1)T_pm+1^R (−s⁰)f (x) = exp

2is⁰(Pp−1

r=1brm+1xrmxrp+bpm+1xpm)

f(x).

Im Dokument Type III von Neumann algebras in the Theory of Infinite-dimensional Groups (Seite 31-55)