 By A. van Daele

ISBN-10: 0521219752

ISBN-13: 9780521219754

The speculation of von Neumann algebras has passed through fast improvement because the paintings of Tonita, Takesaki and Conner. those notes, according to lectures given on the collage of Newcastle upon Tyne, offer an advent to the topic and display the real position of the idea of crossed items. half I bargains with basic non-stop crossed items and proves the commutation theorem and the duality theorem. half II discusses the constitution of kind III von Neumann algebras and considers crossed items with modular activities. proscribing the remedy to the case of o-finite von Neumann algebras permits the writer to paintings with devoted common states.

Best mathematics books

Additional info for Continuous Crossed Products and Type III Von Neumann Algebras

Sample text

Al Notation. By L2(X, 3C) we denote the set of JC-valued functions on X such that (i) W-), 77) is measurable for all 77 e 3C where ( , ) denotes the scalar product in 3C. (ii) there is a separable subspace X o of X such that c(s) E JCo for all s X. (iii) f (s))ds < Remark that it follows from conditions (i) and (ii) that (( ),( ) ) is measurable so that condition (iii) makes sense. Indeed if { en) n=l, 00 is an orthogonal basis in JC0 then 00 (k(s), (s)) = Z n=1 k (g(s), en) 12 and for each n we have that (( ), en) is measurable.

11 Theorem. M ®a G = {x E M 9) 03(L2(G))10t(x) = x,Vt E G ). We have shown already that 0t(x) = x for all t E G and x E M Oa G. Conversely take a x E M 0 ((L2(G)) such that 0t(x) = R. We will now show that x E M 9a G. Fix a function f E Cc(G) and normalize it so that f f(t)dt = 1 and define OK as in the previous lemma for any compact set K in G by OK(s) = fK f(st)dt. By the same lemma m(OK) = fK Ot(m(f))dt. Put Proof. xK = f 9t(m(OK)x m(f))dt. Then from lemma 3. 9 we know that xK E M Oa G. We will now show that for any h E Cc(G) we have xKm(h) - x m(h) and m(h)iK - m(h)x.

Now in section 3 we found that M ® G were precisely the fixed points in M 0 G (L2(G)) under 0. It is fairly straightforward to check that M ®a G is the image under the isomorphism of the subalgebra ira(M ®a G) of (M ®a G) %6. So it follows that na(M 0a G) are the fixed points in (M ®a G) ®a G for the bidual action a. In fact this is a general result: 4. 12 Proposition. {x E M ® G I a (x) = x for all p E G1= {W(x)Ix EM). Proof. We obtained already that ap(e(x)) = n(x). Now if E M ®a G and ap(x) = x, then if we assume a spatial x EM0iB(L2(G)) n {as ®ps)' n {1 ®vp)'.