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Notes on the Von Neumann Algebra of a Group

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Notes on the Von Neumann Algebra of a Group Notes on the von Neumann algebra of a group Ioannis Emmanouil Contents 1 The strong and weak operator topologies on B(H) 1 2 The von Neumann algebra of a group 4 3 The center of N G 6 4 The center-valued trace on N G 9 5 Exercises 19 0 1 The strong and weak operator topologies on B(H) Let H be a Hilbert space. Besides the operator norm topology, the algebra B(H) can be also endowed with the strong operator topology (SOT). The latter is the locally convex topology which is induced by the family of semi-norms (Q»)»2H, where Q»(a) = ka(»)k for all » 2 H and a 2 B(H). Hence, a net of operators (a¸)¸ in B(H) is SOT-convergent to 0 if and only if lim¸ a¸(») = 0 for all » 2 H. The weak operator topology (WOT) on B(H) is the locally convex topology which is induced by the family of semi-norms (P»;´)»;´2H, where P»;´(a) = j<a(»); ´ >j for all »; ´ 2 H and a 2 B(H). In other words, a net of operators (a¸)¸ in B(H) is WOT- convergent to 0 if and only if lim¸ <a¸(»); ´ > = 0 for all »; ´ 2 H. Remarks 1.1 (i) Let (a¸)¸ be a net of operators on H. Then, we have k¢k¡ lim a¸ = 0 =) SOT ¡ lim a¸ = 0 =) WOT ¡ lim a¸ = 0: ¸ ¸ ¸ If the Hilbert space H is not ¯nite dimensional, none of the implications above can be reversed (cf. Exercise 5.1). (ii) For any a 2 B(H) we consider the left (resp. right) multiplication operator La : B(H) ¡! B(H) (resp. Ra : B(H) ¡! B(H)); which is de¯ned by letting La(b) = ab (resp. Ra(b) = ba) for all b 2 B(H). It is easily seen that the operators La and Ra are WOT-continuous. On the other hand, if the Hilbert space H is not ¯nite dimensional, then the multiplication in B(H) is not (jointly) WOT-continuous (cf. Exercise 5.1). (iii) The adjoint operator ( )¤ : B(H) ¡! B(H); which is de¯ned by a 7! a¤, a 2 B(H), is WOT-continuous. Proposition 1.2 Let (a¸)¸ be a bounded net of operators on H. Then, the following con- ditions are equivalent: (i) WOT ¡ lim¸ a¸ = 0. (ii) There is an orthonormal basis (ei)i of the Hilbert space H, such that for all i; j we have lim¸ <a¸(ei); ej > = 0. (iii) There is a subset B ⊆ H, whose closed linear span is H, such that for all »; ´ 2 B we have lim¸ <a¸(»); ´ > = 0. (iv) There is a dense subset X ⊆ H, such that lim¸ <a¸(»); ´ > = 0 for all »; ´ 2 X. Proof. It is clear that (i)!(ii)!(iii), whereas the implication (iii)!(iv) follows by letting X be the (algebraic) linear span of B. It only remains to show that (iv)!(i). To that end, assume that M > 0 is such that ka¸ k · M for all ¸ and consider two vectors »; ´ 2 H. For any positive ² we may choose two vectors »0; ´0 2 X, such that k» ¡»0 k < ² and k´ ¡´0 k < ². Since 0 0 0 0 0 <a¸(»); ´ > ¡ <a¸(» ); ´ > = <a¸(» ¡ » ); ´ > + <a¸(» ); ´ ¡ ´ >; 1 it follows that 0 0 0 0 0 j<a¸(»); ´ > ¡ <a¸(» ); ´ >j · j<a¸(» ¡ » ); ´ >j + j<a¸(» ); ´ ¡ ´ >j 0 · ka¸ k ¢ k» ¡ » k ¢ k´ k + 0 0 ka¸ k ¢ k» k ¢ k´ ¡ ´ k · M²(k» k + k´ k +²): 0 0 0 0 Since lim¸ < a¸(» ); ´ > = 0, we may choose ¸0 such that j< a¸(» ); ´ >j < ² for all ¸ ¸ ¸0. Then, j< a¸(»); ´ >j < ²(1 + M(k » k + k ´ k +²)) for all ¸ ¸ ¸0 and hence lim¸ < a¸(»); ´ > = 0, as needed. 2 Theorem 1.3 Let H be a separable Hilbert space, r a positive real number and B(H)r = fa 2 B(H): kak · rg the closed r-ball of B(H). Then, the topological space (B(H)r; WOT) is compact and metrizable. Proof. In order to prove compactness, we consider for any »; ´ 2 H the closed disc D»;´ = fz 2 C : jz j · r k» k ¢ k´ kg ⊆ C Q and the product space »;´2H D»;´. In view of Tychono®'s theorem, the latter space is compact. We now de¯ne the map Y f : B(H) ¡! D ; r »;´2H »;´ by letting f(a) = (<a(»); ´ >)»;´ for all a 2 B(H)r. It is clear that f is a homeomorphism of (B(H) ; WOT) onto its image. Therefore, the compactness of (B(H) ; WOT) will follow, as r Q r soon as we prove that the image im f of f is closed in »;´2H D»;´. To that end, let (z»;´)»;´ be an element in the closure of im f. Then, the family (z»;´)»;´ is easily seen to be linear in » and quasi-linear in ´, whereas j z»;´ j · r k » k ¢ k ´ k for all »; ´. Hence, there is a vector a» 2 H with ka» k · r k» k, such that z»;´ = <a»; ´ > for all »; ´ 2 H. Using the linearity of the family (z»;´)»;´ in the ¯rst variable, it follows that there is an operator a 2 B(H)r, such that a» = a(») for all » 2 H. Then, (z»;´)»;´ = f(a) 2 im f, as needed. 1 In order to prove metrizability, we ¯x an orthonormal basis (en)n=0 of the separable Hilbert space H and de¯ne for any a; b 2 B(H)r X 1 dr(a; b) = j<(b ¡ a)(en); em >j : n;m 2n+m It is easily seen that dr is a metric on B(H)r, which induces, in view of Proposition 1.2, the weak operator topology on B(H)r. 2 Our next goal is to prove a result of von Neumann, describing the closure of unital self- adjoint subalgebras of B(H) in the weak and strong operator topologies in purely algebraic terms. To that end, we consider for any subset X ⊆ B(H) the commutant X0 = fa 2 B(H): ax = xa for all x 2 Xg: The bicommutant X00 of X is the commutant of X0. It is clear that X ⊆ X00. Lemma 1.4 For any X ⊆ B(H) the commutant X0 is WOT-closed. 2 Proof. For any operator x 2 B(H) we consider the linear endomorphisms Lx and Rx of B(H), which are given by left and right multiplication with x respectively. Then, X0 = T x2X ker (Lx ¡ Rx) and hence the result follows from Remark 1.1(ii). 2 If n is a positive integer and X ⊆ B(H) a set of operators, we shall consider the set n X ¢ In = fxIn : x 2 Xg ⊆ Mn(B(H)) 'B(H ). Then, the following two properties are easily veri¯ed (cf. Exercise 5.2): 0 n 0 (i) The commutant (X ¢ In) of X ¢ In in Mn(B(H)) 'B(H ) is the set Mn(X ) of matrices with entries in the commutant X0 of X in B(H). 00 n 00 (ii) The bicommutant (X ¢ In) of X ¢ In in Mn(B(H)) 'B(H ) is the set X ¢ In, where X00 is the bicommutant of X in B(H). Lemma 1.5 Let A be a self-adjoint subalgebra of B(H) and V ⊆ H a closed A-invariant subspace. Then: (i) The orthogonal complement V ? is A-invariant. (ii) If p is the orthogonal projection onto V , then p 2 A0. (iii) The subspace V is A00-invariant. Proof. (i) Let » 2 V ? and a 2 A. Then, for any vector ´ 2 V we have a¤(´) 2 AV ⊆ V and hence <a(»); ´ > = <»; a¤(´)> = 0. Therefore, it follows that a(») 2 V ?. (ii) We ¯x an operator a 2 A and note that the subspaces V and V ? are a-invariant, in view of our assumption and (i) above. It follows easily from this that the operators ap and pa coincide on both V and V ?. Hence, ap = pa. (iii) Let » 2 V , a00 2 A00 and consider the orthogonal projection p onto V . In view of (ii) above, we have a00p = pa00 and hence a00(») = a00p(») = pa00(») 2 V , as needed. 2 We are now ready to state and prove von Neumann's theorem. Theorem 1.6 (von Neumann bicommutant theorem) Let A ⊆ B(H) be a self-adjoint sub- algebra containing the identity operator. Then, ASOT = AW OT = A00, where we denote by ASOT (resp. AW OT ) the SOT-closure (resp. WOT-closure) of A in B(H). Proof. It is clear that ASOT ⊆ AW OT . Since A ⊆ A00, it follows from Lemma 1.4 that AW OT ⊆ A00. Hence, it only remains to show that A00 ⊆ ASOT . In order to verify this, we consider an operator a00 2 A00, a positive real number ², a positive integer n and vectors »1;:::;»n 2 H. We have to show that the SOT-neighborhood 00 00 N²;»1;:::;»n (a ) = fa 2 B(H): k(a ¡ a )»i k < ² for all i = 1; : : : ; ng 00 of a intersects A non-trivially. To that end, we consider the self-adjoint subalgebra A¢In ⊆ n Mn(B(H)) acting on the Hilbert space H by left multiplication and the closed subspace n V = f(a(»1); : : : ; a(»n)) : a 2 Ag ⊆ H : It is clear that V is A ¢ In-invariant. Invoking Lemma 1.5(iii) and the discussion preceding 00 it, we conclude that the subspace V is left invariant under the action of the operator a In 2 00 00 Mn(B(H)).
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