Fundamentals of the Theory of Operator Algebras Vol.4: by Richard V. Kadison, John R. Ringrose

By Richard V. Kadison, John R. Ringrose

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Additional info for Fundamentals of the Theory of Operator Algebras Vol.4: Special Topics-Advanced Theory, an Exercise Approach (Pure and Applied Mathematics (Academic Press), Volume 100)

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In this way, we construct a sequence {E,} with the desired properties. we may assume that CE = CF = I . With this assumption and with the notation of (i), if PEn j! P F for each nonzero central subprojection P of CF in R, then F 5 E , from the comparison theorem. If F 5 En for each n, then E,-1 - E , has a subprojection equivalent t o F for each n. In this case, R contains an infinite orthogonal family of projections equivalent t o F , which contradicts the assumption that R is finite. Thus, for some n and some non-zero central projection P , PEn 4 P F ; so that P E , will serve as G .

Now El E A, El 5 E , and F1 5 F . 26. If F - F1 # 0 , then E - El # 0 ; and F - F1 has a non-zero central carrier contained in the central carrier of E - E l . From (ii), there is a non-zero projection EO in A such that Eo 5 E - El and EO FO 5 F - F1. But then ( { E o , E c } , ( F ~ , F c }is) an element of S properly larger than ({E,},{F,}) - contradicting the maximal property of ({E,}, {F,}). It follows that El F1 = F. 6, there are n orthogonal equivalent projections F1,. . ,F, in R with sum I since R has no central portion of type I.

Thus s(g) = 0 if the reduced form of g contains any of az, a s , . .. 42, d1 is a maximal abelian subalgebra of L3,. 41, L , E d1. Of course, d1 C n:=lR,. Thus d1 = nr=lR,. 45. c. c. groups. (ii) Let G be 3 2 @ n. Show that LG is not isomorphic to C F ~ . * Solution. (i) Let el and e2 be the respective identity elements of G1 and Gz. c. groups and (g1,92) # (e1,e2). Say 91 # el. If g{ and gr are distinct conjugates of gl in GI, then (gi,g2) and (g;,gz) are distinct conjugates of (gl,g2) in G1$Gz.

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