ELEMENTARY REPRESENTATIONS OF THE GROUP

B_{0}^{ℕ} OF FINITE UPPER-TRIANGULAR MATRICES. I

A. V. Kosyak

Received: 03.11.2000

Abstract. We define the so called ”elementary representations” T_{p}^{R,μ}, p ∈ ℕ of the group B_{0}^{ℕ} of finite upper-triangular matrices infinite in one direction, using quasiinvariant measures on some homogenuous spaces and give the criterion of irreducibility and equivalence of constructed representations. We give also the criterion of irreducibility of tensor product of finite and infinite number of ”elementary representations”. We show that the so-called regular representations of the group B_{0}^{ℕ}, studied in [1], [2] are infinite tensor product of the ”elementary representations”.

1. G− action, quasi-invariant measures and representations.

The following construction of the unitary representations of a topological group G is well known. Let we have some measurable space X with the probability measure μ on which the group G acts i.e. we have a group homomorphism α : G → Aut(X):

1) α_{e}(x) = x, ∀x ∈ X where e ∈ G is a trivial element;

2) α_{t_{1}}(α_{t_{2}}(x)) = α_{t_{1}t_{2}}(x), ∀t_{1}, t_{2} ∈ G, x ∈ X.

Let μ^{α_{t}}, t ∈ G are imagies of the measure μ with respect to the action α i.e. μ^{α_{t}}(Δ) = μ(α_{t^{−1}}(Δ)). If μ^{α_{t}} ∼ μ, ∀t ∈ G one can define the unitary representation π^{α,μ} : G → U(L^{2}(X, dμ)) of the group G

(1)  (π_{t}^{α,μ}f)(x) = (dμ^{α_{t}}(x)/dμ(x))^{1/2}f(α_{t^{−1}}(x)), f ∈ L^{2}(X, dμ).

2. Regular representations.

An analog of the regular representation for any infinite-dimensional group G, using G- quasi-invariant measures on some complitions G̃ of such a group is defined in [1], [2], [4]. The regular representation of the infinite-dimensional groups - current groups were studied firstly in [5,6,7,8].

Let B_{0}^{ℕ} be the group of finite real upper-triangular matrices with unities on the principal diagonal, B^{ℕ} be the group of all such matrices (not necessary finite):

B_{0}^{ℕ} = {I + x = I + ∑_{k<n} x_{kn}E_{kn} | x is finite},
B^{ℕ} = {I + x = I + ∑_{k<n} x_{kn}E_{kn} | x is arbitrary},

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 22E65.

Key words and phrases. Elementary representations, regular representations, infinite-dimen-sional groups.

where E_{kn}, k, n ∈ ℕ, are matrix units of infinite order. 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}. Let μ be some probability measure on the group B^{ℕ}. If μ^{R_{t}} ∼ μ and μ^{L_{t}} ∼ μ_{b} ∀t ∈ B_{0}^{ℕ} one can define in a natural way (see [1],[2]) an analog of the right T^{R,μ} and the left T^{L,μ} regular representations of the group B_{0}^{ℕ} in the space H_{μ} = L^{2}(B^{ℕ}, dμ): T^{R,μ}, T^{L,μ} : B_{0}^{ℕ} → U(H_{μ}),

(2)  (T_{t}^{R,μ}f)(x) = (dμ(xt)/dμ(x))^{1/2}f(xt),
(3)  (T_{t}^{L,μ}f)(x) = (dμ(t^{−1}x)/dμ(x))^{1/2}f(t^{−1}x).

Obviously [T_{t}^{R,μ}, T_{s}^{L,μ}] = 0, ∀t, s ∈ B_{0}^{ℕ} hence the right regular representation T^{R,μ} is reducible if μ^{L_{s}} ∼ μ for some s ∈ B_{0}^{ℕ}\e or the measure μ is not B_{0}^{ℕ}- right ergodic . Let the measure μ on B^{ℕ} is B_{0}^{ℕ}- quasi-invariant with respect to the right action: μ^{R_{t}} ∼ μ ∀t ∈ B_{0}^{ℕ}.

Conjecture 1. (Ismagilov, 1985.) The right regular representation T^{R,μ} : B_{0}^{ℕ} → U(H_{μ}) is irreducible if and only if

1) μ^{L_{s}} ⊥ μ ∀s ∈ B_{0}^{ℕ}\e;

2) the measure μ is B_{0}^{ℕ}- right ergodic.

This conjecture was proved for any Gaussian product measure μ_{b} in [1,2] and for any product measure μ = ⊗_{k<n}μ_{kn} under some technical assumption in [3]. In [4] criterion was proved for groups of the interval and circle diffeomorphisms.

Let μ be arbitrary B_{0}^{ℕ}- quasi-invariant measure on B^{ℕ}. For the generators A_{kn}^{R,μ} (A_{kn}^{L,μ}) of the one-parameter groups I + tE_{kn}, t ∈ ℝ^{1}, k < n, corresponding to the right T^{R,μ} (respectively left T^{L,μ}) regular representation we have the following formulas:

(4)  A_{kn}^{R,μ} = \frac{d}{dt}T_{I+tE_{kn}}^{R,μ} |_{t=0} = ∑_{r=1}^{k−1} x_{rk}D_{rn}(μ) + D_{kn}(μ),

(5)  A_{kn}^{L,μ} = \frac{d}{dt}T_{I+tE_{kn}}^{L,μ} |_{t=0} = −(D_{kn}(μ) + ∑_{m=n+1}^{∞} x_{nm}D_{km}(μ)),

where D_{kn}(μ) = \frac{∂}{∂x_{kn}} + \frac{d}{dt}(\frac{dμ(x(I + tE_{kn}))}{dμ(x)})^{1/2}|_{t=0} .

For an arbitrary product measure μ = ⊗_{k<n}μ_{kn} we have

D_{kn}(μ) = \frac{∂}{∂x_{kn}} + \frac{∂}{∂x_{kn}} ( ln μ_{kn}^{1/2}(x_{kn})),

where dμ_{kn}(x) = μ_{kn}(x)dx, x ∈ ℝ^{1}. Let us denote by

M_{kn}(p) = ∫_{ℝ^{1}} x^{p}μ_{kn}(x)dx, M̃_{kn}(p) = ((i^{−1}D_{kn}(μ))^{p}1, 1)_{L^{2}(ℝ^{1}, dμ_{kn})}, p ∈ ℕ.

For an arbitrary product measure μ = ⊗_{k<n}μ_{kn} we have [3]

Lemma 1. μ^{R_{t}} ∼ μ ∀t ∈ B_{0}^{ℕ}.

ELEMENTARY REPRESENTATIONS OF GROUP B_{0}^{ℕ} … 3

Lemma 2. μ^{L_{t}} ∼ μ ∀t ∈ B_{0}^{ℕ} ⇔

S_{kn}^{L}(μ) = ∑_{m=n+1}^{∞} M̃_{km}(2)M_{nm}(2) < ∞ ∀k < n.

Lemma 3. μ_{b}^{L_{t}} ⊥ μ ∀t ∈ B_{0}^{ℕ}\e ⇔ S_{kn}^{L}(μ) = ∞ ∀k < n.

Let us define the Gaussian measure μ_{b} on the group B^{ℕ} in the following way:

dμ_{b}(x) = ⊗_{k<n}(b_{kn}/π)^{1/2} exp(−b_{kn}x_{kn}^{2})dx_{kn} = ⊗_{k<n}dμ_{b_{kn}}(x_{kn}),

where b = (b_{kn})_{k<n} is some set of the positive numbers. In this case we have (see for example [9], formulas (6) and (7))

D_{kn}(μ_{b}) = \frac{∂}{∂x_{kn}} − b_{kn}x_{kn},

(6)  M_{kn}(2) = \frac{1}{2b_{kn}} , M_{kn}(4) = \frac{3}{(2b_{kn})^{2}} , M_{kn}(2m) = \frac{(2m − 1)!!}{(2b_{kn})^{m}},

(7) M̃_{kn}(2) = \frac{b_{kn}}{2} , M̃_{kn}(4) = 3 (\frac{b_{kn}}{2} )^{2} , M̃_{kn}(2m) = (2m − 1)!! (\frac{b_{kn}}{2} )^{m} .

3. Elementary representations.

Let us consider subgroups X_{p}, p ∈ ℕ and X^{{p}} in the group B^{ℕ}, where {p} is finite or infinite sequence {p} = (p_{1}, p_{2}, ...) of the natural numbers p_{n} ∈ ℕ, p_{1} < p_{2} < ...

X_{p} = {I + x ∈ B^{ℕ} | I + x = I + ∑_{n=p+1}^{∞} x_{pn}E_{pn}},

X^{{p}} = Π_{p_{k}∈{p}}X_{p_{k}} = {I + x ∈ B^{ℕ} | I + x = I + ∑_{p_{k}∈{p}} ∑_{n=p_{k}+1}^{∞} x_{p_{k}n}E_{p_{k}n}}.

Obviously the right action of the group B_{0}^{ℕ} is well defined on the groups X_{p} and X^{{p}} .

For B_{0}^{ℕ}- right quasi-invariant measure μ on X_{p} (respectively X^{{p}}) we define representation T_{p}^{R,μ} (respectively T^{R,μ,{p}}) by formulas:

(T_{p,t}^{R,μ}f)(x) = (dμ(xt)/dμ(x))^{1/2}f(xt), f ∈ H_{p}(μ) := L^{2}(X_{p}, dμ),
(T_{t}^{R,μ,{p}}f)(x) = (dμ(xt)/dμ(x))^{1/2}f(xt), f ∈ H^{{p}}(μ) := L^{2}(X^{{p}}, dμ).

In the case of {p} = (1, 2, ..., q) we will use the special notations:

X^{q} = X^{{p}}, H^{q}(μ) = H^{{p}}(μ), T^{R,μ,q} = T^{R,μ,{p}}.

In the cases {p} = (p_{n})_{n=1}^{∞}, p_{n} = n we obtaine the right regular representation T^{R,μ}. Note that the defined representations T_{p}^{R,μ}, T^{R,μ,{p}} and T^{R,μ} are particular cases of (1) with X = X_{p}, X = X^{{p}} and X = B^{ℕ} for α = R.

Definition 1. We will call the representation T_{p}^{R,μ} by the elementary representations.

Our aim is to prove the following theorem for Gaussian product measures μ = μ_{b} = ⊗_{k<n}μ_{b_{kn}} and its projections μ^{{p}} = ⊗_{p_{k}∈{p}}μ_{p_{k}} on the subgroups X^{{p}}, where μ_{p} = ⊗_{n=p+1}^{∞}μ_{b_{pn}}.

Theorem 1.

1) Representation T^{R,μ,{p}} is the tensor product of the representations T_{p_{k}}^{R,μ_{p_{k}}},

p_{k} ∈ {p}
(8)  T^{R,μ,{p}} = ⊗_{p_{k}∈{p}}T_{p_{k}}^{R,μ_{p_{k}}};

2) representation T^{R,μ,{p}} is irreducible if and only if S_{p_{k}p_{n}}^{L}(μ) = ∞ ∀p_{k} <

p_{n}; p_{k}, p_{n} ∈ {p};

3) T^{R,μ,{p}} ∼ T^{R,μ′,{p′}} if and only if μ ∼ μ′ and {p} = {p′};

4) the tensor product of two irreducible representations T^{R,μ,{p}} ⊗ T^{R,μ′,{p′}} is irreducible if and only if {p} ∩ {p′} = {∅}, where {p} ∩ {p′} = {p_{k} | p_{k} ∈

{p}} ∩ {p′_{k} | p′_{k} ∈ {p′}} and S_{p_{k}p′_{n}}^{L}(μ ⊗ μ′) = ∞, p_{k} ∈ {p}, p′_{n} ∈ {p′}.

Since T_{p}^{R,μ} (respectively T^{R,μ,q} ) is the restrictions of the representation T^{R,μ} to the subspace H_{p}(μ) = L^{2}(X_{p}, dμ_{p}) (respectively H^{q}(μ) = L^{2}(X^{q}, dμ^{q})) of the space H_{μ} = L^{2}(B^{ℕ}, dμ), we have

(9)  A_{p,kn}^{R,μ} = \{ \begin{matrix}0, & if k < p, \\ D_{pn}(μ), & if p = k < n, \\ x_{pk}D_{pn}(μ), & if p < k < n, \end{matrix}

(10)  A_{kn}^{R,μ,q} = \{ \begin{matrix}∑_{r=1}^{k−1} x_{rk}D_{rn}(μ) + D_{kn}(μ), & if k ≤ q, k < n, \\ ∑_{r=1}^{q} x_{rk}D_{rn}(μ), & if q < k < n. \end{matrix}

4. Irreducibility and equivalence.

For Gaussian measure μ = μ_{b} and its projections μ_{p} = ⊗_{n=p+1}^{∞}μ_{b_{pn}} holds

Theorem 2.

1) Representation T_{p}^{R,μ} is irreducible if and only if the measure μ = μ_{p} on the space X_{p} is B_{0}^{ℕ}− right-ergodic;

2) two irreducible representations T_{p}^{R,μ} and T_{p′}^{R,μ′} are equivalent if and only if μ ∼ μ′ and p = p′.

Proof. 1). Let us consider the commutative set of generators {A_{p,pn}^{R,μ}}_{n=p+1}^{∞}. By formulas (9) we have A_{p,pn}^{R,μ} = D_{pn}(μ). Since the family of operators 𝔻_{p}(μ) = {D_{pn}(μ)}_{n=p+1}^{∞} has common simple spectrum in the space H_{p}(μ) = L^{2}(X_{p}, dμ) any bounded operator A in the space H_{p}(μ) commuting with this family is some essentially bounded function of this family

A = a(𝔻_{p}(μ)) = a(D_{pp+1}(μ), D_{pp+2}(μ), ..., D_{pn}(μ), ...).

Since

[D_{pn}(μ), A_{p,nm}^{R,μ}] = [D_{pn}(μ), x_{pn}D_{pm}(μ)] = D_{pm}(μ), p < n < m,

and

[D_{pn}(μ)^{s}, x_{pn}D_{pm}(μ)] = sD_{pn}(μ)^{s−1}D_{pm}(μ), p < n < m,

so

(11) [a(𝔻_{p}(μ)), A_{p,nm}^{R,μ}] = [a(𝔻_{p}(μ)), x_{pn}D_{pm}(μ)] = \frac{∂a(𝔻_{p}(μ))}{∂D_{pn}(μ)}D_{pm}(μ),

hence the commutation [a(𝔻_{p}(μ)), A_{p,nm}^{μ}] = 0 means

a(D_{pp+1}(μ), ..., D_{pn}(μ) + tD_{pm}(μ), ..., D_{pm}(μ), ...)
= a(D_{pp+1}(μ), ..., D_{pn}(μ), ..., D_{pm}(μ), ...), ∀t ∈ ℝ^{1}, p < n < m.

By ergodicity of the measure μ the function

a = a(D_{pp+1}(μ), D_{pp+2}(μ), ..., D_{pn}(μ), ...)

is const and operator A is trivial A = λI. 2). Sufficiency is obvious. Let T_{p}^{R,μ} ∼ T_{p′}^{R,μ′}, we prove that p = p′ and μ ∼ μ′. Let us assume that p < p′ and consider the restrictions T |_{G} of the representations T = T_{p}^{R,μ} and T_{p′}^{R,μ′} on the subgroup G = X_{p,0} = {I + x ∈ B_{0}^{ℕ} | I + x ∈ X_{p}}. The spectral measure 𝔼_{p}^{μ} of the restriction T_{p}^{R,μ} |_{X_{p,0}} is the spectral measure of the commutative family of self-adjoint operators i^{−1}𝔻_{p}(μ) = {i^{−1}D_{pn}(μ)}_{p=n+1}^{∞} and the spectral measure 𝔼_{p′}^{μ′} of T_{p′}^{R,μ′} |_{X_{p,0}} is trivial (see (9)), so p = p′. In this case

the spectral measures 𝔼_{p}^{μ} and 𝔼_{p}^{μ′} are equivalent, so μ ∼ μ′. □

5. Tensor product of the finite numbers of the elementary representations and irreducibility.

We prove the Theorem 1 part 1) and 2) for {p} = (1, 2, ..., q). For another finite {p} the proof is the same.

Theorem 3.

1) Representation T^{R,μ,q} is the tensor product of the representations T_{p}^{R,μ_{p}}, 1 ≤ p ≤ q :

(12)  T^{R,μ,q} = ⊗_{p=1}^{q}T_{p}^{R,μ_{p}};

2) representation T^{R,μ,q} is irreducible if and only if S_{kn}^{L}(μ) < ∞, k < n ≤ q.

Proof. We will show that by generators A_{kn}^{R,μ,q}, k < n it is possible to approximate the operators of multiplications by independent variables x_{kn}, k < n ≤ q and the set of operators D_{kn}(μ), k < n, k ≤ q. Indeed, according to (10) we have:

A_{1n}^{R,μ,q} = D_{1n}(μ), 1 < n, A_{2n}^{R,μ,q} = x_{12}D_{1n}(μ) + D_{2n}(μ), 2 < n,
A_{3n}^{R,μ,q} = x_{13}D_{1n}(μ) + x_{23}D_{2n}(μ) + D_{3n}(μ), 3 < n,

A_{qn}^{R,μ,q} = ∑_{r=1}^{q−1} x_{rq}D_{rn}(μ) + D_{qn}(μ), q < n, A_{kn}^{R,μ,q} = ∑_{r=1}^{q} x_{rk}D_{rn}(μ), if p < k < n.

The proof is the same as in [2]. It is based on the Lemma 4–7.

Let us denote by 𝔄^{R,μ,q} the von-Neumann algebra, generated by the representation T^{R,μ,q}: 𝔄^{R,μ,q} = (T_{t}^{R,μ,q} | t ∈ B_{0}^{ℕ})′′. Let also ⟨f_{n} | n = 1, 2, ...⟩ be the closure of the linear space, generated by the set of vectors {f_{n}}_{n=1}^{∞} in a Hilbert space H.

Definition 2. Recall [10] that a non necessary bounded self-adjoint operator A in the Hilbert space H is affiliated to the von-Neumann algebra M of operators in this Hilbert space H (notation A η M) if exp(itA) ∈ M ∀t ∈ ℝ^{1}.

Lemma 4. x_{12} η 𝔄^{R,μ,q} if S_{12}^{L}(μ) = ∞. Moreover x_{12}1 ∈ ⟨A_{1n}^{R,μ,q}A_{2n}^{R,μ,q}1 | 2 < n⟩, if and only if S_{12}^{L}(μ) = ∞. In this case we have also D_{2n}(μ) η 𝔄^{R,μ,q}, 2 < n.

Proof. Let us estimate using [2], p. 25; [3], formula (11) or [11], p. 79,

δ_{12,3,m} = min_{{t_{n}}} ∥( ∑_{n=3}^{m} t_{n}A_{1n}^{R,μ,q}A_{2n}^{R,μ,q} + x_{12})1∥_{H_{μ}}^{2}.

Since A_{1n}^{R,μ,q}A_{2n}^{R,μ,q} =x_{12}D_{1n}(μ)^{2} + D_{1n}(μ)D_{2n}(μ), we have if ∑_{n=3}^{m} t_{n}M̃_{1n}(2)=1

δ_{12,3,m} = min_{{t_{n}}} ∥ [∑_{n=3}^{m} t_{n}A_{1n}^{R,μ,q}A_{2n}^{R,μ,q} + x_{12}] 1∥_{H_{μ}}^{2}

= min_{{t_{n}}} ∑_{n=3}^{m} t_{n}^{2} [M_{12}(2)(M̃_{1n}(4) − M̃_{1n}^{2}(2)) + M̃_{1n}(2)M̃_{2n}(2)]

= (∑_{n=3}^{m} \frac{M̃_{1n}(2)^{2}}{M_{12}(2)(M̃_{1n}(4) − M̃_{1n}^{2}(2)) + M̃_{1n}(2)M̃_{2n}(2)})^{−1} = (Σ_{12,3,m})^{−1} ,

where (see (6), (7))

Σ_{12,3,m} = ∑_{n=3}^{m} \frac{1}{M_{12}(2)(\frac{M̃_{1n}(4)}{M̃_{1n}^{2}(2)} − 1) + \frac{M̃_{2n}(2)}{M̃_{1n}(2)}} = ∑_{n=3}^{m} \frac{1}{M_{12}(2)(3 − 1) + \frac{b_{2n}}{b_{1n}}} ≍ ∑_{n=3}^{m} \frac{b_{1n}}{b_{2n}} .

So lim_{m→∞} δ_{12,3,m} = 0 if and only if S_{12}^{L}(μ) = \frac{1}{4} ∑_{m=3}^{∞} \frac{b_{1n}}{b_{2n}} = ∞. By Theorem VIII,

25 of [12] we have also convergence

A_{m_{1},m_{2}} = ∑_{m_{1}}^{m_{2}} t_{n}i^{−1}A_{1n}^{R,μ,q}i^{−1}A_{2n}^{R,μ,q} → x_{12}, if m_{2} → ∞

in the strong resolvent sence (see [1,2]). So if S_{12}^{L}(μ) = ∞ we have proved that x_{12} η 𝔄^{R,μ,q} and D_{2n}(μ) η 𝔄^{R,μ,q}, 2 < n. □

ELEMENTARY REPRESENTATIONS OF GROUP B_{0}^{ℕ} … 7

Lemma 5. x_{12}, x_{13}, x_{23} η 𝔄^{R,μ,q} if S_{kn}^{L}(μ) = ∞, k < n ≤ 3. In this case we have also D_{3n}(μ) η 𝔄^{R,μ,q}, 3 < n. Proof. Let us estimate

δ_{13,4,m} = min_{{t_{n}}} ∥( ∑_{n=4}^{m} t_{n}A_{1n}^{R,μ,q}A_{3n}^{R,μ,q} + x_{13})1∥_{H_{μ}}^{2}.

Since

A_{1n}^{R,μ,q}A_{3n}^{R,μ,q} = x_{13}D_{1n}^{2}(μ) + x_{23}D_{1n}(μ)D_{2n}(μ) + D_{1n}(μ)D_{3n}(μ),
D_{2n}(μ)A_{3n}^{R,μ,q} = x_{13}D_{1n}(μ)D_{2n}(μ) + x_{23}D_{2n}^{2}(μ) + D_{2n}(μ)D_{3n}(μ),

hence we have if ∑_{n=4}^{m} t_{n}M̃_{1n}(2) = 1,

δ_{13,4,m} = min_{{t_{n}}} ∥ [∑_{n=4}^{m} t_{n}A_{1n}^{R,μ,q}A_{3n}^{R,μ,q} + x_{13}] 1∥_{H_{μ}}^{2}

= ∑_{n=4}^{m} t_{n}^{2} [M_{13}(2)(M̃_{1n}(4) − M̃_{1n}^{2}(2)) + M_{23}(2)M̃_{1n}(2)M̃_{2n}(2) + M̃_{1n}(2)M̃_{3n}(2)]

= ( ∑_{n=4}^{m} \frac{M̃_{1n}^{2}(2)}{M_{13}(2)(M̃_{1n}(4) − M̃_{1n}^{2}(2)) + M_{23}(2)M̃_{1n}(2)M̃_{2n}(2) + M̃_{1n}(2)M̃_{3n}(2)})^{−1}

=: (Σ_{13,4,m})^{−1}. So lim_{m→∞} δ_{13,4,m} = 0 if and only if Σ_{13,4,∞} = ∞, where

Σ_{13,4,∞} ≍ ∑_{n=4}^{∞} \frac{1}{(\frac{M̃_{1n}(4)}{M̃_{1n}^{2}(2)} − 1) + \frac{M̃_{2n}(2)}{M̃_{1n}(2)} + \frac{M̃_{3n}(2)}{M̃_{1n}(2)}}

= ∑_{n=4}^{∞} \frac{1}{3 − 1 + \frac{b_{2n}}{b_{1n}} + \frac{b_{3n}}{b_{1n}}} ≍ ∑_{n=4}^{∞} \frac{b_{1n}}{b_{2n} + b_{3n}} =: σ_{13}(μ).

Using (10) we have if ∑_{n=4}^{m} t_{n}M̃_{2n}(2) = 1,

δ_{23,4,m} = min_{{t_{n}}} ∥ [∑_{n=4}^{m} t_{n}A_{1n}^{R,μ,q}A_{3n}^{R,μ,q} + x_{23}] 1∥_{H_{μ}}^{2}

≍ ( ∑_{n=4}^{m} \frac{M̃_{2n}^{2}(2)}{M̃_{1n}(2)M̃_{2n}(2) + (M̃_{2n}(4) − M̃_{2n}^{2}(2)) + M̃_{2n}(2)M̃_{3n}(2)})^{−1}

=: (Σ_{23,4,m})^{−1}. So lim_{m→∞} δ_{23,4,m} = 0 if and only if Σ_{23,4,∞} = ∞ where

Σ_{23,4,∞} = ∑_{n=4}^{∞} \frac{1}{\frac{M̃_{1n}(2)}{M̃_{2n}(2)} + (\frac{M̃_{2n}(4)}{M̃_{2n}^{2}(2)} − 1) + \frac{M̃_{3n}(2)}{M̃_{2n}(2)}}

= ∑_{n=4}^{∞} \frac{1}{\frac{b_{1n}}{b_{2n}} + 3 − 1 + \frac{b_{3n}}{b_{2n}}} ≍ ∑_{n=4}^{∞} \frac{b_{2n}}{b_{1n} + b_{3n}} =: σ_{23}(μ).

8 A. V. KOSYAK

By Lemma 7 one of the series σ_{13}(μ) or σ_{23}(μ) is divergent. Let σ_{13}(μ) = ∞ so x_{13} η 𝔄^{R,μ,q} and

x_{23} ∈ ⟨D_{2n}(μ)(A_{3n}^{R,μ,q} − x_{13}D_{1n}(μ))1 | 4 < n⟩

⇔ ∑_{n=4}^{∞} \frac{1}{(\frac{M̃_{2n}(4)}{M̃_{2n}^{2}(2)} − 1) + \frac{M̃_{3n}(2)}{M̃_{2n}(2)}} ≍ ∑_{n=4}^{∞} \frac{M̃_{2n}(2)}{M̃_{3n}(2)} = ∑_{n=4}^{∞} \frac{b_{2n}}{b_{3n}} = 4S_{23}^{L}(μ) = ∞,

so x_{23} η 𝔄^{R,μ,q}. If σ_{23}(μ) = ∞ we have x_{23} η 𝔄^{R,μ,q} and x_{13} ∈ ⟨D_{1n}(μ)(A_{3n}^{R,μ,q} − x_{23}D_{2n}(μ))1 | 4 < n⟩

⇔ ∑_{n=4}^{∞} \frac{1}{(\frac{M̃_{1n}(4)}{M̃_{1n}^{2}(2)} − 1) + \frac{M̃_{3n}(2)}{M̃_{1n}(2)}} ≍ ∑_{n=4}^{∞} \frac{M̃_{1n}(2)}{M̃_{3n}(2)} = ∑_{n=4}^{∞} \frac{b_{1n}}{b_{3n}} = 4S_{13}^{L}(μ) = ∞.

so x_{13} η 𝔄^{R,μ,q}. Finally we have

x_{13}, x_{23} η 𝔄^{R,μ,q} and D_{3n}(μ) η 𝔄^{R,μ,q}, 3 < n,

since D_{3n}(μ) = A_{3n}^{R,μ,q} − x_{13}D_{1n}(μ) − x_{23}D_{2n}(μ), 3 < n. □ Lemma 6. {x_{kn}}_{k<n≤q} η 𝔄^{R,μ,q} if S_{kn}^{L}(μ) = ∞, k < n ≤ q. In this case we have also D_{kn}(μ) η 𝔄^{R,μ,q}, k < n, k ≤ q. Proof. Let us estimate for some r, 1 ≤ r < q

δ_{rq,q+1,m} = min_{{t_{n}}} ∥( ∑_{n=q+1}^{m} t_{n}D_{rn}(μ)A_{qn}^{R,μ,q} + x_{rq})1∥_{H_{μ}}^{2}.

We have (see (10)): D_{rn}(μ)A_{qn}^{R,μ,q} = D_{rn}(μ)(∑_{s=1}^{q−1} x_{sq}D_{sn}(μ) + D_{qn}(μ)), hence, if ∑_{n=q+1}^{m} t_{n}M̃_{rn}(2) = 1, we have

δ_{rq,q+1,m} = min_{{t_{n}}} ∥ [ ∑_{n=q+1}^{m} t_{n}D_{rn}(μ)A_{qn}^{R,μ,q} + x_{rq}] 1∥_{H_{μ}}^{2} = (Σ_{rq,q+1,m})^{−1},

where

Σ_{rq,q+1,m}

= ∑_{n=q+1}^{m} ( ∑_{s=1,s≠r}^{q−1} \frac{M_{sq}(2)M̃_{sn}(2)}{M̃_{rn}(2)} + \frac{M̃_{qn}(2)}{M̃_{rn}(2)} + M_{rq}(2)(\frac{M̃_{rn}(4)}{M̃_{rn}^{2}(2)} − 1))^{−1}.

We have

Σ_{rq,q+1,∞} ≍ ∑_{n=q+1}^{∞} ( ∑_{s=1,s≠r}^{q−1} \frac{M̃_{sn}(2)}{M̃_{rn}(2)} + \frac{M̃_{qn}(2)}{M̃_{rn}(2)} + (\frac{M̃_{rn}(4)}{M̃_{rn}^{2}(2)} − 1))^{−1}

≍ ∑_{n=q+1}^{∞} \frac{M̃_{rn}(2)}{∑_{s=1,s≠r}^{q} M̃_{sn}(2)} = ∑_{n=q+1}^{∞} \frac{b_{rn}}{∑_{s=1,s≠r}^{q} b_{sn}} =: σ_{rq}(μ).

We need here the Lemma 2.5, p. 256 in [2]:

Lemma 7. Let for fixed q = 2, 3, ... holds S_{rq}^{L}(μ) = \frac{1}{4} ∑_{n=q+1}^{∞} \frac{b_{rn}}{b_{qn}} = ∞, 1 ≤ r < q.

Then for some r, 1 ≤ r < q we have σ_{rq}(μ) = ∞.

Let for some r_{1} we have σ_{r_{1}q}(μ) = ∞. In this case we can aproximate the variable x_{r_{1}q}. To aproximate x_{rq} for r ≠ r_{1} we use anouther combination

D_{rn}(μ)(A_{qn}^{R,μ,q} − x_{r_{1}q}D_{r_{1}n}(μ)) = D_{rn}(μ)( ∑_{s=1,s≠r_{1}}^{q−1} x_{sq}D_{sn}(μ) + D_{qn}(μ)).

We have if ∑_{n=q+1}^{m} t_{n}M̃_{rn}(2) = 1

δ_{rq,q+1,m}^{r_{1}} = min_{{t_{n}}} ∥[ ∑_{n=q+1}^{m} t_{n}D_{rn}(μ)(A_{qn}^{R,μ,q} − x_{r_{1}q}D_{r_{1}n}(μ)) + x_{rq}]1∥_{H_{μ}}^{2}

= min_{{t_{n}}} ∑_{n=q+1}^{m} t_{n}^{2}[M̃_{rn}(2)( ∑_{s=1,s≠{r,r_{1}}}^{q−1} M_{sq}(2)M̃_{sn}(2) + M̃_{qn}(2))

+ M_{rq}(2)(M̃_{rn}(4) − M̃_{rn}^{2}(2))] = (Σ_{rq,q+1,m}^{r_{1}})^{−1},

where

Σ_{rq,q+1,m}^{r_{1}}

= ∑_{n=q+1}^{m} ( ∑_{s=1,s≠\{r,r_{1}\}}^{q−1} \frac{M_{sq}(2)M̃_{sn}(2)}{M̃_{rn}(2)} + \frac{M̃_{qn}(2)}{M̃_{rn}(2)} + M_{rq}(2)(\frac{M̃_{rn}(4)}{M̃_{rn}^{2}(2)} − 1))^{−1}.

We have

Σ_{rq,q+1,∞}^{r_{1}} ≍ ∑_{n=q+1}^{∞} ( ∑_{s=1,s≠\{r,r_{1}\}}^{q−1} \frac{M̃_{sn}(2)}{M̃_{rn}(2)} + \frac{M̃_{qn}(2)}{M̃_{rn}(2)} + (\frac{M̃_{rn}(4)}{M̃_{rn}^{2}(2)} − 1))^{−1}

≍ ∑_{n=q+1}^{∞} \frac{M̃_{rn}(2)}{∑_{s=1,s≠\{r,r_{1}\}}^{q} M̃_{sn}(2)} = ∑_{n=q+1}^{∞} \frac{b_{rn}}{∑_{s=1,s≠\{r,r_{1}\}}^{q} b_{sn}} =: σ_{rq}^{r_{1}}(μ).

By Lemma 7 for some r ≠ r_{1}, 1 ≤ r < q we have σ_{r_{2}q}^{r_{1}}(μ) = ∞. Then x_{r_{2}q} η 𝔄^{R,μ,q} and so on. At the end we have

{x_{rq}}_{r<q} η 𝔄^{R,μ,q} and {D_{qn}(μ)}_{q<n} η 𝔄^{R,μ,q}

since D_{qn}(μ) = A_{qn}^{R,μ,q} − ∑_{r=1}^{q} x_{rq}D_{rn}(μ) η 𝔄^{R,μ,q}, q < n. □

Finally we have {x_{kn}}_{k<n≤q} η 𝔄^{R,μ,q}, {D_{kn}(μ)}_{k<n,k≤q} η 𝔄^{R,μ,q}, so the commutant (𝔄^{R,μ,q})′ of the von Neumann algebra 𝔄^{R,μ,q} coinsides with essentially bounded functions from the family of operators 𝔻^{q}(μ) = {D_{kn}(μ)}_{k≤q<n}.

Let now a bounded operator A ∈ L(H^{q}(μ)) commute with T_{t}^{R,μ,q}, t ∈ B_{0}^{ℕ} then this operator A is an operator of multiplication in the space H^{q}(μ) by some essentially bounded function: A = a({D_{kn}(μ)}_{k≤q<n}). Since for 1 ≤ m ≤ q < k < n, we have

[D_{mk}(μ), A_{kn}^{R,μ,q}] = [D_{mk}(μ), ∑_{r=1}^{q} x_{rk}D_{rn}(μ)] = D_{mn}(μ),

using (11) the commutation [a(𝔻^{q}(μ)), A_{kn}^{R,μ,q}] = 0 ∀q < k < n, gives us

0 = [a(𝔻^{q}(μ)), A_{kn}^{R,μ,q}] = ∑_{r=1}^{q} \frac{∂a(𝔻^{q}(μ))}{∂D_{rk}(μ)}D_{rn}(μ)

and the last relation means the equality

a ( \begin{matrix}D_{1q+1}(μ) & … & D_{1k}(μ) + tD_{1n}(μ) & … & D_{1n}(μ) & … \\ & … & & … & & … \\ D_{qq+1}(μ) & … & D_{qk}(μ) + tD_{qn}(μ) & … & D_{qn}(μ) & … \end{matrix} )

= a ( \begin{matrix}D_{1q+1}(μ) & … & D_{1k}(μ) & … & D_{1n}(μ) & … \\ & … & & … & & … \\ D_{1q+1}(μ) & … & D_{qk}(μ) & … & D_{qn}(μ) & … \end{matrix} ) ∀t ∈ ℝ^{1}, ∀q < k < n.

This means that the function a({D_{kn}(μ)}_{k≤q<n}) is const a(𝔻^{q}(μ)) =const. □

6. Regular representations as infinite tensor product of the ”elementary representations”.

We prove the Theorem 1 part 1)–3) for {p} = (p_{n})_{n=1}^{∞}, p_{n} = n. For another infinite {p} the proof is the same.

Theorem 4.

1) Representation T^{R,μ} is the infinite tensor product of the representations T_{p}^{R,μ_{p}}, 1 ≤ p < ∞ :

(13)  T^{R,μ} = ⊗_{p=1}^{∞}T_{p}^{R,μ_{p}};

2) representation T^{R,μ} is irreducible if and only if S_{kn}^{L}(μ) < ∞, ∀k < n;

3) two irreducible representations T^{R,μ} and T^{R,μ′} are equivalent if and only if μ ∼ μ′.

Proof. The irreducibility and equivalency are proved in [1,2]. The representation (13) follows from (4) and (10). □

Proof of the Theorem 1 part 3) for general {p}. Sufficiency is obvious. Necessity is based on the Theorem 4 part 3) and Theorem 2 part 2). Let T^{R,μ,{p}} ∼ T^{R,μ′,{p′}}, where {p} = (p_{1}, p_{2}, ...), {p′} = (p′_{1}, p′_{2}, ...). We prove that {p} = {p′} and μ ∼ μ′.

Let us assume that p_{1} < p′_{1} and consider the spectral measures 𝔼_{p_{1}}^{μ_{p_{1}}} and 𝔼_{p_{1}}^{μ′_{p_{1}}} of the restrictions of the representations T^{R,μ,{p}} and T^{R,μ′,{p′}} on the subgroup X_{p_{1},0} (see the proof of the Theorem 2 part 2)). The spectral measure 𝔼_{p_{1}}^{μ_{p_{1}}} is the spectral measure of the commutative family of self-adjoint operators i^{−1}𝔻_{p_{1}}(μ) = {i^{−1}D_{p_{1}n}(μ)}_{n=p_{1}+1}^{∞} and the spectral measure 𝔼_{p_{1}}^{μ′_{p_{1}}} is trivial (see (9) and (10)). It contradicts with T^{R,μ,{p}} ∼ T^{R,μ′,{p′}}, so p_{1} = p′_{1}. In this case the spectral

measures 𝔼_{p_{1}}^{μ_{p_{1}}} and 𝔼_{p_{1}}^{μ′_{p_{1}}} are equivalent, so μ_{p_{1}} ∼ μ′_{p_{1}} and T_{p_{1}}^{R,μ_{p_{1}}} ∼ T_{p_{1}}^{R,μ′_{p_{1}}} . Since by formula (13) we have

T^{R,μ,{p}} = T_{p_{1}}^{R,μ_{p_{1}}} ⊗ T^{R,μ^{{p_{2}}},{p_{2}}}, T^{R,μ′,{p′}} = T_{p_{1}}^{R,μ′_{p_{1}}} ⊗ T^{R,μ′^{{p′_{2}}},{p′_{2}}},

and equivalence T^{R,μ,{p}} ∼ T^{R,μ′,{p′}} holds we conclude that T^{R,μ^{{p_{2}}},{p_{2}}} ∼ T^{R,μ′^{{p′_{2}}},{p′_{2}}} where {p_{2}} = (p_{2}, p_{3}, ...), {p′_{2}} = (p′_{2}, p′_{3}, ...) and

T^{R,μ,{p_{2}}} = ⊗_{p_{k}∈{p_{2}}}T_{p_{k}}^{R,μ_{p_{k}}} , T^{R,μ′,{p′_{2}}} = ⊗_{p_{k}∈{p′_{2}}}T_{p_{k}}^{R,μ′_{p_{k}}} .

Analogously we conclude that p_{2} = p′_{2} and μ_{p_{2}} ∼ μ′_{p_{2}}. Finally {p} = {p′} and μ_{p_{k}} ∼ μ′_{p_{k}} ∀p_{k} ∈ {p} = {p′}. For finite {p}, {p′} the proof is finished since in this case we have μ = ⊗_{p_{k}∈{p}}μ_{p_{k}} ∼ μ′ = ⊗_{p_{k}∈{p′}}μ′_{p_{k}}. In general case (for infinite {p}, {p′}) the equivalence μ_{p_{k}} ∼ μ′_{p_{k}} ∀p_{k} ∈ {p} = {p′} does not implies μ = ⊗_{p_{k}∈{p}}μ_{p_{k}} ∼ μ′ = ⊗_{p_{k}∈{p′}}μ′_{p_{k}}. In order to finish the proof we use the theorem 4 part 3) for particular case {p} = (p_{k})_{k=1}^{∞}, p_{k} = k, k ∈ ℕ. For general case {p} the proof is the same.

Proof of the Theorem 1 part 4). Sufficiency follows from Theorem 1 part 1) and 2) since in this case we have

T^{R,μ,{p}} ⊗ T^{R,μ′,{p′}} = T^{R,μ⊗μ′,{p}∪{p′}},

where {p} ∪ {p′} = {p_{k}, p′_{n} | p_{k} ∈ {p}, p′_{n} ∈ {p′}}.

Let now {p} ∩ {p′} = {p′′} := (p_{1}, ..., p_{k}). In this case we have {p} = {q} ∪ {p′′} and {p′} = {q′} ∪ {p′′}, so {p} ∪ {p′} = {q} ∪ {q′} ∪ {p′′} and we have

T^{R,μ,{p}} ⊗ T^{R,μ′,{p′}} = T^{R,μ^{{q}}⊗μ^{{p′′}}⊗μ′^{{q′}},{q}∪{p′′}∪{q′}} ⊗ T^{R,μ′^{{p′′}},{p′′}}.

So the proof that the last tensor product is reducible is similar to the proof that the following tensor product

T^{R,μ,q} ⊗ T^{R,μ′,q+k}

is reducible.

Consider the essentially bounded function a : X^{q} ∋ x ↦ a(x) ∈ ℂ^{1} and let A_{0} be the operator of multiplication in the space

H^{q}(μ) ⊗ H^{q+k}(μ′) = L^{2}(X^{q}, dμ) ⊗ L^{2}(X^{q+k}, dμ′) = L^{2}(X^{q} × X^{q+k}, dμ ⊗ μ′)

by the function a_{0} : X^{q} × X^{q+k} ∋ (x, y, z) ↦ a_{0}(x, y, z) = a(yx^{−1}) ∈ ℂ^{1}. We show that the representation T^{R,μ,q} ⊗ T^{R,μ′,q+k} commute with the operator A_{0}. Indeed, for any function f(x, y, z) ∈ L^{2}(X^{q} × X^{q+k}, dμ ⊗ μ′), using the property that for any (y, z) ∈ X^{q} × X^{k} = X^{q+k} in B^{ℕ} holds (y, z) = zy, we have

(T_{t}^{R,μ,q} ⊗ T_{t}^{R,μ′,q+k}A_{0}f)(x, zy) = (T_{t}^{R,μ,q} ⊗ T_{t}^{R,μ′,q+k}a_{0}f)(x, zy)

= (\frac{dμ(xt)}{dμ(x)})^{1/2} (\frac{dμ′(zyt)}{dμ′(zy)})^{1/2} a((yt)(xt)^{−1})f(xt, zyt)

= a(yx^{−1}) (\frac{dμ(xt)}{dμ(x)})^{1/2} (\frac{dμ′(zyt)}{dμ′(zy)})^{1/2} f(xt, zyt)

= (A_{0}(T_{t}^{R,μ,q} ⊗ T_{t}^{R,μ′,q+k})f)(x, zy).

□

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Institute of Mathematics, National Academy of Sciences of Ukraine, 3 Tereshchenkivs’ka, Kyiv, 01601, Ukraine

E-mail address: kosyak@imath.kiev.ua