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488 H. Schultz / Journal of Functional Analysis 236 (2006) 457–489<br />

Lemma 7.5. For n ∈ N and 1 i n <strong>de</strong>fine fi : C n → C n+1 by<br />

fi(z1,...,zn) = (z1,...,zn,zi).<br />

Then for n commuting operators T1,...,Tn ∈ M, one has that<br />

μT1,...,Tn,Ti<br />

Proof. Given Borel sets B1,...,Bn+1 ⊆ C we must show that<br />

C<strong>le</strong>arly,<br />

μT1,...,Tn,Ti (B1 ×···×Bn+1) = μT1,...,Tn<br />

= fi(μT1,...,Tn ). (7.13)<br />

f −1<br />

i (B1 ×···×Bn+1) . (7.14)<br />

f −1<br />

i (B1 ×···×Bn+1) = B1 ×···×(Bi ∩ Bn+1) ×···×Bn<br />

so that the right-hand si<strong>de</strong> of (7.14) is<br />

τ PT1 (B1) ∧···∧PTi (Bi<br />

<br />

∩ Bn+1) ∧···∧PTn (Bn)<br />

= τ PT1 (B1) ∧···∧PTi (Bi) ∧ PTi (Bn+1) ∧···∧PTn (Bn) .<br />

But this is exactly the <strong>le</strong>ft-hand si<strong>de</strong> of (7.14) and we are done. ✷<br />

We will not give the proof of Theorem 7.1 in full generality but rather, by way of an examp<strong>le</strong>,<br />

illustrate how it goes. Consi<strong>de</strong>r for instance 3 commuting operators T1,T2,T3 ∈ M and the<br />

polynomial q ∈ C[z1,z2,z3] given by<br />

At first <strong>de</strong>fine φ1 : C 3 → C 5 by<br />

q(z1,z2,z3) = 1 + 2z 2 2 + z1z2z3. (7.15)<br />

φ1(z1,z2,z3) = (z2,z2,z1,z2,z3).<br />

By repeated use of Lemmas 7.4 and 7.5 we find that<br />

Next <strong>de</strong>fine φ2 : C 5 → C 2 by<br />

μT2,T2,T1,T2,T3<br />

= φ1(μT1,T2,T3 ).<br />

φ2(z1,...,z5) = (2z1z2,z3z4z5),<br />

and by repeated use of (7.9) and Lemma 7.4 conclu<strong>de</strong> that<br />

With φ3 : C 2 → C given by<br />

μ 2T 2 2 ,T1T2T3 = (φ2 ◦ φ1)(μT1,T2,T3 ).<br />

φ3(z1,z2) = z1 + z2

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