IRREDUCIBILITY OF THE REGULAR GAUSSIAN

REPRESENTATIONS OF THE GROUP B_{0}^{ℤ}

A. V. Kosyak

Received: 09.04.2001

Abstract. We define the analog of the regular representations of the group B_{0}^{ℤ} of finite upper-triangular matrices infinite in both directions corresponding to quasiinvariant Gaussian product-measures on the group of all upper-triangular matrices and give the sufficent conditions of irreducibility of constructed representations. In [1], [2] the criterion was proved for an arbitrary Gaussian product-measures on the group B_{0}^{ℕ} of finite upper-triangular matrices infinite in one direction and in [3] for groups of the interval and circle diffeomorphisms.

Apparently, an analog of the regular representation of infinite-dimensional groups (current group) appears firstly in [4], [5], [6] (see also [7]). 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]. Let B_{0}^{ℤ} be the group of finite upper-triangular matrices with units on the principal diagonal infinite in both directions, B^{ℤ} be the group of all such matrices (not necessary finite):

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

B^{ℤ} = {I + x = I + ∑_{k,n∈ℤ,k<n} x_{kn}E_{kn} | x is arbitrary},

where E_{kn}, k, n ∈ ℤ are matrix units of infinite order and ℤ is the set of entire numbers. On the group B^{ℤ} we define the Gaussian measure μ_{b} as follows:

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

where b = (b_{kn})_{k<n} is a set of positive numbers.

Let us denote by R and L the right and the left action of the group B^{ℤ} on itself: R_{s}(t) = ts^{−1}, L_{s}(t) = st, t, s ∈ B^{ℤ}. Denote by μ_{b}^{R_{t}} and μ_{b}^{L_{t}}, t ∈ B_{0}^{ℤ} the imagies of the measure μ_{b} with respect to the right and the left action of the group B_{0}^{ℤ} on B^{ℤ}.

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 22E65.

Key words and phrases. Regular representation, infinite-dimensional group, Gaussian measure.

If μ_{b}^{R_{t}} ∼ μ_{b} and μ_{b}^{L_{t}} ∼ μ_{b}, ∀t ∈ B_{0}^{ℤ} one can define (see [2]) an analog of the right and the left regular representation of the group B_{0}^{ℤ},

T^{R,μ_{b}}, T^{L,μ_{b}} : B_{0}^{ℤ} → U(H_{μ_{b}} = L^{2}(B^{ℤ}, dμ_{b})),

corresponding to a B_{0}^{ℤ}− quasi-invariant measure μ_{b} on the group B^{ℤ} :

(1)  (T_{t}^{R,μ_{b}}f)(x) = (dμ_{b}(xt)/dμ_{b}(x))^{1/2}f(xt),
(2)  (T_{s}^{L,μ_{b}}f)(x) = (dμ_{b}(s^{−1}x)/dμ_{b}(x))^{1/2}f(s^{−1}x).

Lemma 1. μ_{b}^{R_{t}} ∼ μ_{b} ∀t ∈ B_{0}^{ℤ} ⇔

S_{kn}^{R}(b) = ∑_{r=−∞}^{k−1} \frac{b_{rn}}{b_{rk}} < ∞ ∀ k, n ∈ ℤ, k < n.

Lemma 2. μ_{b}^{L_{t}} ∼ μ_{b} ∀t ∈ B_{0}^{ℤ} ⇔

S_{kn}^{L}(b) = ∑_{m=n+1}^{∞} \frac{b_{km}}{b_{nm}} < ∞, ∀k, n ∈ ℤ, k < n.

Theorem 1. Let conditions

(3)  sup_{n,n>k} \frac{S_{kn}^{R}(b)}{b_{kn}} = C_{k} < ∞ ∀ k ∈ ℤ

are valid. Than the representation T^{R,μ_{b}} is irreducible if and only if:

1) no left actions are admissible for the measure μ_{b}

μ_{b}^{L_{t}} ⊥ μ_{b}, ∀t ∈ B_{0}^{ℤ}\e;

2) μ_{b} is ergodic with respect to the right action of the group B_{0}^{ℤ}.

Remark. The statement of the theorem without the condition (3) was expressed by R. S. Ismagilov as the conjecture for Gaussian centered product-measures and for the group B_{0}^{ℕ} of finite upper-triangular matrices infinite in one direction and was proved in [1], [2].

Let Φ be the inverse mapping on the group B^{ℤ}:Φ(I +x)=(I +x)^{−1}, (I +x) ∈ B^{ℤ}. Denote by μ_{b}^{Φ} the imagie of the measure μ_{b} with respect to the inverse mapping Φ. Lemmas 1–3 are proved in fact in [1,2].

Lemma 3. For k, n ∈ ℤ, k < n we have μ_{b}^{L_{I+tE_{kn}}} ⊥ μ_{b} ∀t ∈ ℝ^{1}\0 ⇔ S_{kn}^{L}(b) = ∞.

Let us denote for the measure μ_{b}

E_{m}(b) = ∑_{k<n≤m} \frac{S_{kn}^{R}(b)}{b_{kn}}, m ∈ ℤ.

IRREDUCIBILITY OF REGULAR REPRESENTATIONS OF GROUP B_{0}^{ℤ} 3

Lemma 4. The measure μ_{b} is B_{0}^{ℤ}− right ergodic if E_{m}(b) < ∞, ∀m ∈ ℤ.

For any m ∈ ℤ let us define the subgroups B^{m} and B_{(m)} of the group B^{ℤ} as follows:

B^{m} = {I + x ∈ B^{ℤ} | I + x = I + ∑_{k<n≤m} x_{kn}E_{kn}},

B_{(m)} = {I + x ∈ B^{ℤ} | I + x = I + ∑_{k<n,m<n} x_{kn}E_{kn}}.

Obviously B^{ℤ} is a semi-direct product of two subgroups B_{(m)} and B^{m} (B_{(m)} is a normal subgroup in B^{ℤ}) for any m ∈ ℤ: B^{ℤ} = B_{(m)}▷< B^{m}. Let μ_{b}^{m} and μ_{b,(m)} be the projections of the measure μ_{b} onto subgroups B^{m} and B_{(m)} respectively. We have

μ_{b}^{m} = ⊗_{k<n≤m}μ_{kn}, μ_{b,(m)} = ⊗_{k<n,m<n}μ_{kn}.

Using [8, Theorem 1] we have

Lemma 5. If E_{m}(b) < ∞, then (μ_{b}^{m})^{Φ} ∼ μ_{b}^{m}, m ∈ ℤ.

Let us denote by B_{0}^{m} = B^{m} ⋂ B_{0}^{ℤ}.

Corollary 1. If E_{m}(b) < ∞, then the measure μ_{b}^{m} on the group B^{m} is B_{0}^{m} rightergodic, m ∈ ℤ.

Proof. If E_{m}(b) < ∞ then by Lemma 5 we have (μ_{b}^{m})^{Φ} ∼ μ_{b}^{m}. By [2, Lemma 1.3] the measure μ_{b}^{m} on the group B^{m} (being the product measure) is B_{0}^{m} left-ergodic. So the proof follows from the fact that the inverse mapping Φ interchanges the right R and the left L actions. Indeed (xt)^{−1} = t^{−1}x^{−1}, so Φ(R_{t^{−1}}(x)) = L_{t}(Φ(x)), x ∈ B^{m}, t ∈ B_{0}^{m}. This means that

(4)  Φ ∘ R_{t^{−1}} = L_{t} ∘ Φ, t ∈ B_{0}^{m}.

Let B^{m} = X_{1} ⋃ X_{2} with X_{1} ⋂ X_{2} = ∅, μ_{b}^{m}(X_{1}) > 0, μ_{b}^{m}(X_{2}) > 0 and

(5)  R_{t}(X_{1}) ⊂ X_{1}, R_{t}(X_{2}) ⊂ X_{2} ∀t ∈ B_{0}^{m}.

The ergodicity of the measure μ_{b}^{m} with respect to the right action of the group B_{0}^{m} means that μ_{b}^{m}(X_{1}) = 1 or μ_{b}^{m}(X_{1}) = 1. Using (4) and (5) we conclude that Φ(X_{1}) and Φ(X_{2}) are B_{0}^{m} left-invariant:

L_{t}(Φ(X_{i})) = Φ(R_{t^{−1}}(X_{i})) ⊂ Φ(X_{i}), i = 1, 2.

By ergodicity of the measure μ_{b}^{m} with respect to the left action of the group B_{0}^{m} we conclude that (μ_{b}^{m})^{Φ}(X_{1}) = μ_{b}^{m}(Φ(X_{1})) = 1 or (μ_{b}^{m})^{Φ}(X_{2}) = μ_{b}^{m}(Φ(X_{2})) = 1 so μ_{b}^{m}(X_{1}) = 1 or μ_{b}^{m}(X_{2}) = 1 since (μ_{b}^{m})^{Φ} ∼ μ_{b}^{m}. □

Proof of the Lemma 4. Let now a measurable function f ∈ L^{1}(B^{ℤ}, dμ_{b}) is B_{0}^{ℤ} right-invariante i.e. f(zt) = f(z) (mod μ_{b}) ∀t ∈ B_{0}^{ℤ}. We prove that f(z) = const (mod μ_{b}). Indeed, let us denote for any m ∈ ℤ

f_{m}(x) = ∫_{B_{(m)}} f(yx) dμ_{b,(m)}(y)

where z = yx ∈ B^{ℤ} is the decomposition of the element z ∈ B^{ℤ} corresponding to the semi-direct product B^{ℤ} = B_{(m)}▷<B^{m}. Obviously

f_{m}(xt) = f_{m}(x) (mod μ_{b}^{m}) ∀t ∈ B_{0}^{m}.

By Corollary 1 we have f_{m}(xt) = const (mod μ_{b}^{m}) ∀m ∈ ℤ, hence f(x) = const (mod μ_{b}) since f_{m} → f, m → ∞ in L^{1}(B^{ℤ}, dμ_{b}). □

For the generators A_{kn}^{R,μ_{b}} (A_{kn}^{L,μ_{b}}) of the one-parameter groups I + tE_{kn}, t ∈ ℝ^{1}, k < n, corresponding to the right T^{R,μ_{b}} (respectively left T^{L,μ_{b}} if μ_{b}^{L_{t}} ∼ μ_{b} ∀t ∈ B_{0}^{ℤ}) regular representation we have the following formulas:

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

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

where D_{kn}(μ_{b}) = ∂/∂x_{kn} − b_{kn}x_{kn}.

We recall the definition and properties of the Helliger integral ([9], Chap. 2, § 2). Suppose that μ and ν are two probability measures on the measure space (X, B). Assume that λ is probability measure such that μ ≺ λ, ν ≺ λ for example λ = (μ + ν)/2. The Hellinger integral H(μ, ν) of two measures μ and ν is defined as follows:

H(μ, ν) = ∫_{X} \sqrt{\frac{dμ}{dλ}}\sqrt{\frac{dν}{dλ}} dλ.

It does not depend on λ and has the following properties:

(H1) 0 ≤ H(μ, ν) ≤ 1 (the Schwartz inequality );
(H2) H(μ, ν) = 1 ⇔ μ = ν;
(H3) H(μ, ν) = 0 ⇔ μ ⊥ ν;
(H4) μ ∼ ν ⇒ H(μ, ν) > 0.

The converse to (H4) does not hold in general.

Proof of the Lemma 2 and 3. The proof of the Lemma 1 is the same. Since the one-dimensional subgroups I + tE_{kn}, t ∈ ℝ^{1}, k, n ∈ ℤ, k < n generate the group B_{0}^{ℤ} we have

μ_{b}^{L_{t}} ∼ μ_{b}, ∀t ∈ B_{0}^{∞} ⇔ μ_{b}^{L_{I+sE_{kn}}} ∼ μ_{b}, ∀k, n ∈ ℤ, k < n, s ∈ ℝ^{1}.

IRREDUCIBILITY OF REGULAR REPRESENTATIONS OF GROUP B_{0}^{ℤ} 5

Since μ_{b} and μ_{b}^{L_{I+tE_{kn}}} are product measures and

(I + tE_{kn})(I + x) = (I + tE_{kn}) ( \begin{matrix}& ... & & & ... & & ... \\ 0 & ... & x_{kn} & x_{kn+1} & ... & x_{km} & ... \\ & ... & & & ... & & ... \\ 0 & ... & 1 & x_{nn+1} & ... & x_{nm} & ... \\ & ... & & & ... & & ... \end{matrix} )

= ( \begin{matrix}& ... & & & ... & & ... \\ 0 & ... & x_{kn} + t & x_{kn+1} + tx_{nn+1} & ... & x_{km} + tx_{nm} & ... \\ & ... & & & ... & & ... \\ 0 & ... & 1 & x_{nn+1} & ... & x_{nm} & ... \\ & ... & & & ... & & ... \end{matrix} ),

the condition μ_{b}^{L_{I+tE_{kn}}} ∼ μ_{b}, ∀k < n is equivalent by Kakutani criterion [10] (see also [11, § 16, Theorem 1]) with H(μ_{b}^{L_{I+tE_{kn}}} , μ_{b}) > 0, t ≠ 0. Moreover

μ_{b}^{L_{I+tE_{kn}}} ⊥ μ_{b} ∀t ∈ ℝ^{1} ⇔ H(μ_{b}^{L_{I+tE_{kn}}} , μ_{b}) = 0, t ≠ 0.

We have

(8) \begin{aligned}H & (μ_{b}^{L_{I+tE_{kn}}} , μ_{b}) \\ & = H(μ_{kn}^{L_{I+tE_{kn}}} , μ_{kn}) ∏_{m=n+1}^{∞} H((μ_{km} ⊗ μ_{nm})^{L_{I+tE_{kn}}} , (μ_{km} ⊗ μ_{nm})) \\ & = ∫_{ℝ^{1}} (\frac{dμ_{kn}(x_{kn} + t)}{dμ_{kn}(x_{kn})})^{1/2} μ_{kn}(x_{kn}) dx_{kn} \\ & × ∏_{m=n+1}^{∞} ∫_{ℝ^{2}} (\frac{dμ_{km}(x_{km} + tx_{nm})}{dμ_{nm}(x_{nm})})^{1/2} μ_{km}(x_{km})μ_{nm}(x_{nm}) dx_{km} dx_{nm} \\ & = \frac{exp ( − \frac{t^{2}b_{kn}}{4} )}{∏_{m=n+1}^{∞} (1 + \frac{t^{2}}{4} \frac{b_{km}}{b_{nm}} )^{1/2}} . \end{aligned}

So μ_{b}^{L_{t}} ∼ μ_{b} ∀t ∈ B_{0}^{ℤ} ⇔ S_{kn}^{L}(b) = ∑_{m=n+1}^{∞} \frac{b_{km}}{b_{nm}} < ∞, ∀k, n ∈ ℤ, k < n,

μ^{L_{I+tE_{kn}}} ⊥ μ, ∀t ∈ ℝ^{1}\0 ⇔ S_{kn}^{L}(b) = ∞, k, n ∈ ℤ, k < n.

□

Proof of the Theorem 1. Necessity. is obvious since in the oposite case i.e. if μ_{b}^{L_{t_{0}}} ⊥ μ_{b} for some id ≠ t_{0} ∈ B_{0}^{ℤ} we have a nontrivial operator T_{t_{0}}^{L,μ_{b}} that commute with the right regular representation: [T_{t}^{R,μ_{b}}, T_{t_{0}}^{L,μ_{b}}] = 0 ∀t ∈ B_{0}^{ℤ}. If the measure μ_{b} is not ergodic then the representation T^{R,μ_{b}} is also reducible.

Sufficiency. Let μ_{b}^{L_{t}} ⊥ μ_{b}, ∀t ∈ B_{0}^{ℤ}\e hence μ_{b}^{L_{I+tE_{kn}}} ⊥ μ_{b}, ∀t ∈ ℝ^{1}\0 ∀k, n ∈ ℤ, k < n, then by Lemma 3 S_{kn}^{L}(μ_{b}) = ∞, ∀k, n ∈ ℤ, k < n. We will show that in this case we will be able to approximate by generators A_{kn}^{R,μ_{b}}, k, n ∈ ℤ, k < n the operators of multiplication by independent variables (x_{kn})_{k<n} using the condition (3).

Let us denote by 𝔄^{R,μ_{b}}(B_{0}^{ℤ}) the von-Neumann algebra, generated by the right regular representation: 𝔄^{R,μ_{b}}(B_{0}^{ℤ}) = (T_{t}^{R,μ_{b}} | t ∈ B_{0}^{ℤ})′′. By < f_{n} | n = 1, 2, ... > we denote the closure of the linear space, generated by the set of vectors {f_{n}}_{n=1}^{∞} in a Hilbert space H.

Definition. Recall [12] 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}.

Let us denote for p = 1, 2, ...

M_{kn}(p) = (x_{kn}^{p}1, 1)_{H_{μ_{b}}} = ∫_{ℝ^{1}} x^{p}μ_{kn}(x) dx, M̃_{kn}(p) = ((i^{−1}D_{kn}(μ_{b}))^{p}1, 1)_{H_{μ_{b}}} .

The direct calculation gives us (see also [13] formula (6) and (7)) (9)

M_{kn}(2) = \frac{1}{2b_{kn}} , M_{kn}(4) = \frac{3}{(2b_{kn})^{2}} , M̃_{kn}(2) = \frac{b_{kn}}{2} , M̃_{kn}(4) = 3(\frac{b_{kn}}{2} )^{2}.

Lemma 6. For p, q ∈ ℤ, p < q we have x_{pq}1 ∈< A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}}1 | q < n >, if and only if

Σ_{pq}(b) = ∑_{n=q+1}^{∞} \frac{b_{pn}^{2}}{(b_{pn} + S_{pn}^{R}(b))(b_{qn} + S_{qn}^{R}(b))} = ∞.

Proof. Let us estimate

δ_{pq,m_{1},m_{2}} = min_{{t_{n}}} ∥( ∑_{n=m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} + x_{pq})1∥_{H_{μ_{b}}}^{2}

when ∑_{n=m_{1}}^{m_{2}} t_{n}M̃_{pn}(2) = ∑_{n=m_{1}}^{m_{2}} t_{n}b_{pn}/2 = 1. According to (9) we obtain

(10)  \begin{matrix}δ_{pq,m_{1},m_{2}} = min_{\{t_{n}\}} ∥[ ∑_{n=m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} + x_{pq}]1∥_{H_{μ_{b}}}^{2} \\ = min_{\{t_{n}\}} ∥[ ∑_{n=m_{1}}^{m_{2}} t_{n}( ∑_{r=−∞}^{p−1} x_{rp}D_{rn}(μ_{b}) + D_{pn}(μ_{b})) \\ × ( ∑_{s=−∞}^{q−1} x_{sq}D_{sn}(μ_{b}) + D_{qn}(μ_{b})) + x_{pq}]1∥_{H_{μ_{b}}}^{2} \end{matrix}

IRREDUCIBILITY OF REGULAR REPRESENTATIONS OF GROUP B_{0}^{ℤ} 7

= min_{{t_{n}}} ∥[ ∑_{n=m_{1}}^{m_{2}} t_{n}( ∑_{r=−∞}^{p−1} x_{rp}D_{rn}(μ_{b}))( ∑_{s=−∞}^{q−1} x_{sq}D_{sn}(μ_{b}) + D_{qn}(μ_{b}))

+ D_{pn}(μ_{b})( ∑_{s=−∞,s≠p}^{q−1} x_{sq}D_{sn}(μ_{b}) + D_{qn}(μ_{b})) + x_{pq}(D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))]1∥_{H_{μ_{b}}}^{2}

≍ min_{{t_{n}}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}[ ∑_{r=−∞}^{p−1} ∥x_{rp}D_{rn}(μ_{b})1∥^{2}( ∑_{s=−∞}^{q−1} ∥x_{sq}D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥D_{pn}(μ_{b})1∥^{2}( ∑_{s=−∞,s≠p}^{q−1} ∥x_{sq}D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥x_{pq}∥^{2}∥(D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))1∥_{H_{μ_{b}}}^{2} ]

= min_{{t_{n}}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}[ ∑_{r=−∞}^{p−1} M_{rp}(2)M̃_{rn}(2)( ∑_{s=−∞}^{q−1} M_{sq}(2)M̃_{sn}(b) + M̃_{qn}(2))

+ M̃_{pn}(2)( ∑_{s=−∞,s≠p}^{q−1} M_{sq}(2)M̃_{sn}(2) + M̃_{qn}(2)) + M_{pq}(2)(M̃_{pn}(4) − M̃_{pn}^{2}(2))]

= min_{\{t_{n}\}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}[\frac{1}{4} ∑_{r=−∞}^{p−1} \frac{b_{rn}}{b_{rp}} (\frac{1}{4} ∑_{s=−∞}^{q−1} \frac{b_{sn}}{b_{sq}} + \frac{b_{qn}}{2} )

+ \frac{b_{pn}}{2} (\frac{1}{4} ∑_{s=−∞,s≠p}^{q−1} \frac{b_{sn}}{b_{sq}} + \frac{b_{qn}}{2} ) + \frac{b_{pq}}{2} (\frac{3b_{pn}^{2}}{4} − \frac{b_{pn}^{2}}{4} )]

≍ min_{\{t_{n}\}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}( ∑_{r=−∞}^{p−1} \frac{b_{rn}}{b_{rp}} + b_{pn})( ∑_{s=−∞}^{q−1} \frac{b_{sn}}{b_{sq}} + b_{qn})

= min_{{t_{n}}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}(b_{pn} + S_{pn}^{R}(b))(b_{qn} + S_{qn}^{R}(b)).

We have used the following equality

((D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))1, (D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))1)
= (D_{pn}^{2}(μ_{b})1, D_{pn}^{2}(μ_{b})1) + 2M̃_{pn}(2)(D_{pn}^{2}(μ_{b})1, 1) + M̃_{pn}^{2}(2)
= (D_{pn}^{4}(μ_{b})1, 1) − 2M̃_{pn}^{2}(2) + M̃_{pn}^{2}(2) = M̃_{pn}(4) − M̃_{pn}^{2}(2).

To estimate the last but one expression we use the following equality (see [2], p. 25)

(11)  min_{\{t_{n}\}} \{ ∑_{n=1}^{m} t_{n}^{2}a_{n} | ∑_{n=1}^{m} t_{n}b_{n} = 1\} = \frac{1}{∑_{n=1}^{m} \frac{b_{n}^{2}}{a_{n}}} ,

see also [14, § 26 bibliography, p. 79]:

min_{\{t_{n}\}} \{ ∑_{n=1}^{m} x_{n}^{2}c_{n} | ∑_{n=1}^{m} x_{n} = 1\} = \frac{1}{∑_{n=1}^{m} \frac{1}{c_{n}}} .

We have

δ_{pq,m_{1},m_{2}} = min_{{t_{n}}} ∥( ∑_{n=m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} + x_{pq})1∥_{H_{μ_{b}}}^{2} ≍ (Σ_{pq,m_{1},m_{2}})^{−1},

where

Σ_{pq,m_{1},m_{2}} = ∑_{n=m_{1}}^{m_{2}} \frac{b_{pn}^{2}}{(b_{pn} + S_{pn}^{R}(b))(b_{qn} + S_{qn}^{R}(b))}.

So lim_{m→∞} δ_{pq,m_{1},m} = 0 if and only if Σ_{pq,q+1,∞} = Σ_{pq}(b) = ∞. □

Lemma 7. x_{pq} η 𝔄^{R,μ_{b}}(B_{0}^{ℤ}), ∀p, q ∈ ℤ, p < q if the conditions (3) true and no left actions are admissible for the measure μ_{b}.

Proof. It suffices to show that we have convergence

A_{m_{1},m_{2}} = ∑_{m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} → −x_{pq}, if m_{2} → ∞

in the strong resolvent sence. This means that

exp(itA_{m_{1},m_{2}}) → exp(-itx_{pq}) ∀t ∈ ℝ^{1}.

By [15, Theorem VIII, 25] it suffices to show the convergence A_{m_{1},m_{2}}f → −x_{pq}f for any f ∈ D, where D is a common essential domain for all the operators A_{m_{1},m_{2}} and A = x_{pq}. In fact we show that it suffices to show the convergence A_{m_{1},m_{2}}f → −x_{pq}f only for the identity vector f = 1. For the role of D we choose a dense set consisting of finite linear combinations of arbitrary finite monomials

x_{q}^{α} = Π_{k≤q,k<n}x_{kn}^{α_{kn}}, α_{kn} = 0, 1, ... .

Obviously D is the common essential domain for the operators A_{m_{1},m_{2}} and A, since D consists of analytic vectors for the operators A_{m_{1},m_{2}} and A. The function f ∈ D is cylindrical, so for some k_{1}, m_{1}, k_{1} < q < m_{1}, this function f depends only on the variables x_{kn} for k_{1} ≤ k ≤ q, n < m_{1} and we have f = f^{m_{1}} ⊗ 1 with respect to the tensor product

H_{μ_{b}} = H^{m_{1}} ⊗ H_{(m_{1})},
where H^{m_{1}} = L^{2}(B^{m_{1}}, dμ_{b}^{m_{1}}), H_{(m_{1})} = L^{2}(B_{(m_{1})}, dμ_{b,(m_{1})}).

IRREDUCIBILITY OF REGULAR REPRESENTATIONS OF GROUP B_{0}^{ℤ} 9

Since the operator A_{m_{1},m_{2}} does not act on the variables x_{kn}, −∞ < k ≤ q < n < m_{1} we have using (10)

min_{{t_{n}}} ∥( ∑_{n=m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} + x_{pq})f^{m_{1}} ⊗ 1∥_{H_{μ_{b}}}^{2}

= min_{{t_{n}}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}[ ∑_{r=−∞}^{p−1} ∥x_{rp}f^{m_{1}}∥^{2}∥D_{rn}(μ_{b})1∥^{2}

× ( ∑_{s=−∞}^{q−1} ∥x_{sq}f^{m_{1}}∥^{2}∥D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥D_{pn}(μ_{b})1∥^{2}( ∑_{s=−∞,s≠p}^{q−1} ∥x_{sq}f^{m_{1}}∥^{2}∥D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥x_{pq}f^{m_{1}}∥^{2}∥(D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))1∥_{H_{μ_{b}}}^{2} ]

≍ min_{{t_{n}}} ∑_{n=m_{1}}^{m_{2}} t_{n}^{2}[ ∑_{r=−∞}^{p−1} ∥x_{rp}∥^{2}∥D_{rn}(μ_{b})1∥^{2}

× ( ∑_{s=−∞}^{q−1} ∥x_{sq}∥^{2}∥D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥D_{pn}(μ_{b})1∥^{2}( ∑_{s=−∞,s≠p}^{q−1} ∥x_{sq}∥^{2}∥D_{sn}(μ_{b})1∥^{2} + ∥D_{qn}(μ_{b})1∥^{2})

+ ∥x_{pq}∥^{2}∥(D_{pn}^{2}(μ_{b}) + M̃_{pn}(2))1∥_{H_{μ_{b}}}^{2} ]

= min_{{t_{n}}} ∥( ∑_{n=m_{1}}^{m_{2}} t_{n}A_{pn}^{R,μ_{b}}A_{qn}^{R,μ_{b}} + x_{pq})1∥_{H_{μ_{b}}}^{2} = δ_{pq,m_{1},m} ≍ (Σ_{pq,m_{1},m_{2}})^{−1}.

Using conditions (3) we have

Σ_{pq,m_{1},m_{2}} = ∑_{n=m_{1}}^{m_{2}} \frac{b_{pn}}{b_{qn}(1 + \frac{S_{pn}^{R}(b)}{b_{pn}})(1 + \frac{S_{qn}^{R}(b)}{b_{qn}})}

≥ ∑_{n=m_{1}}^{m_{2}} \frac{b_{pn}}{b_{qn}(1 + C_{p})(1 + C_{q})} ≍ ∑_{n=m_{1}}^{m_{2}} \frac{b_{pn}}{b_{qn}} .

So lim_{m→∞} δ_{pq,m_{1},m} = 0 if S_{pq}^{L}(b) = ∑_{n=q+1}^{∞} \frac{b_{pn}}{b_{qn}} = ∞. □

Let now a bounded operator A ∈ L(H_{μ_{b}}) commutes with T_{t}^{R,μ_{b}}, t ∈ B_{0}^{ℤ} i.e. [A, T_{t}^{R,μ_{b}}] = 0 ∀t ∈ B_{0}^{ℤ} so the operator A is an operator of multiplication in the space H_{μ_{b}} by some function: A = a(x). Since [a(x), T_{t}^{R,μ_{b}}] = 0 ∀t ∈ B_{0}^{ℤ}, the function a(x) is B_{0}^{ℤ}-right invariant: a(xt) = a(x) ∀t ∈ B_{0}^{ℤ} so by ergodicity of the measure μ_{b} we have a(x) = const. □

Corollary 2. Let condition (3) and 2) of the Teorem 1 are true for the measure μ_{b}. In this case three following conditions are equivalent:

i) the representation T^{R,μ_{b}} is irreducible, ii) μ_{b}^{L_{t}} ⊥ μ_{b}, ∀t ∈ B^{ℤ}\e, iii) S_{kn}^{L}(b) = ∞, ∀k, n ∈ ℤ, k < n.

Proof. By the Theorem 1 and Lemma 3 we have: ii) ⇒ iii) ⇒ i) so

i) ⇒ ii) ⇒ iii) ⇒ i).
□

We show that conditions of the Theorem 1 are compatible i.e. that there exist b = (b_{kn})_{k<n} such that for measure μ_{b} we have:

0) μ_{b}^{R_{t}} ∼ μ_{b}∀t ∈ B_{0}^{ℤ} ⟺^{Lemma 2} S_{kn}^{R}(b) < ∞ ∀ k, n ∈ ℤ, k < n;

1) μ_{b}^{L_{t}} ⊥ μ_{b}∀t ∈ B_{0}^{ℤ}\e ⟺^{Corollary 2} S_{kn}^{L}(b) = ∞ ∀k, n ∈ ℤ, k < n;

2) the measure μ_{b} is ergodic with respect to the right action of the group B_{0}^{ℤ};

3) sup_{n,n>k} \frac{S_{kn}^{R}(b)}{b_{kn}} = C_{k} < ∞ ∀ k ∈ ℤ.

Example 1. Let b_{kn} = a_{n}^{k}, k, n ∈ ℤ, where (a_{n})_{n∈ℤ} is some sequence of positive numbers. We have: 0)

S_{kn}^{R}(b) = ∑_{r=−∞}^{k−1} \frac{b_{rn}}{b_{rk}} = ∑_{r=−∞}^{k−1} \frac{a_{n}^{r}}{a_{k}^{r}} = ∑_{r=1−k}^{∞} (\frac{a_{k}}{a_{n}} )^{r}

= (\frac{a_{k}}{a_{n}} )^{1−k} \frac{1}{1 − a_{k}/a_{n}} < ∞ ∀ k, n ∈ ℤ, k < n ⇔ a_{n} < a_{n+1} ∀ n ∈ ℤ.

So 0) ⇔ a_{n} < a_{n+1}, ∀ n ∈ ℤ. 1)

S_{kn}^{L}(b) = ∑_{m=n+1}^{∞} \frac{b_{km}}{b_{nm}} = ∑_{m=n+1}^{∞} \frac{a_{m}^{k}}{a_{m}^{n}} = ∑_{m=n+1}^{∞} \frac{1}{a_{m}^{n−k}} = ∞ ∀ k, n ∈ ℤ, k < n

⇔ ∑_{m=0}^{∞} \frac{1}{a_{m}^{p}} = ∞ ∀p = 1, 2, ...

3)

C_{k} = sup_{n,n>k} \frac{S_{kn}^{R}(b)}{b_{kn}} = sup_{n,n>k} \frac{1}{a_{n}^{k}} (\frac{a_{k}}{a_{n}} )^{1−k} \frac{1}{1 − a_{k}/a_{n}}

= a_{k}^{1−k} sup_{n,n>k} \frac{1}{a_{n}} \frac{1}{1 − a_{k}/a_{n}} = a_{k}^{1−k} \frac{1}{a_{∞} − a_{k}} < ∞ ∀ k ∈ ℤ,

where a_{∞} = lim_{n→∞} a_{n}. 2)

E_{m}(b) = ∑_{k<n≤m} \frac{S_{kn}^{R}(b)}{b_{kn}} = ∑_{k<n≤m} \frac{a_{k}^{1−k}}{a_{n}} \frac{1}{1 − a_{k}/a_{n}} = ∑_{n=−∞}^{m} \frac{1}{a_{n}} ∑_{k=−∞}^{n−1} \frac{a_{k}^{1−k}}{1 − a_{k}/a_{n}}

≤ ∑_{n=−∞}^{m} \frac{a_{n−1}}{a_{n}} \frac{1}{1 − a_{n−1}/a_{n}} ∑_{k=−∞}^{n−1} a_{k}^{−k} ≤ ∑_{n=−∞}^{m} \frac{a_{n−1}}{a_{n} − a_{n−1}} ∑_{k=−∞}^{m−1} a_{k}^{−k} = σ_{m}^{1}σ_{m−1}^{2},

where σ_{m}^{1} = ∑_{n=−∞}^{m} \frac{a_{n−1}}{a_{n} − a_{n−1}} and σ_{m}^{2} = ∑_{k=−∞}^{m} a_{k}^{−k}, m ∈ ℤ.

So E_{m}(b) < ∞, m ∈ ℤ if σ_{0}^{1} < ∞ and σ_{0}^{2} < ∞. Let us fix some c ∈ (0, 1) and put

a_{n} = \{ \begin{matrix}c\frac{n(n − 1)}{2} & , for n < 0, \\ 1 & , for n = 0, \\ ln(n + 2) & , for n > 0. \end{matrix}

We have

σ_{0}^{2} = ∑_{k=−∞}^{0} a_{k}^{−k} = ∑_{k=0}^{∞} a_{−k}^{k} = ∑_{k=0}^{∞} c^{\frac{k^{2}(k+1)}{2}} < ∞,

σ_{0}^{1} = ∑_{n=−∞}^{0} \frac{a_{n−1}}{a_{n} − a_{n−1}} = ∑_{n=−∞}^{0} \frac{a_{n−1}}{a_{n}} \frac{1}{1 − a_{n−1}/a_{n}} = ∑_{n=0}^{∞} \frac{c^{n+1}}{1 − c^{n+1}} < ∞,

since c ∈ (0, 1) and

\frac{a_{−n−1}}{a_{−n}} = c^{\frac{(n+1)(n+2)}{2} − \frac{n(n+1)}{2}} = c^{n+1}.

Obviously, a_{n} < a_{n+1}, n ∈ ℤ and ∑_{m=0}^{∞} \frac{1}{a_{m}^{p}} = ∞ ∀p = 1, 2, ... .

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

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