By Peter L Duren; Richard Askey; Uta C Merzbach; Harold M Edwards

ISBN-10: 0821801244

ISBN-13: 9780821801246

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Since a,i(l — pi) = (1 — Pi)di = 0. We therefore get (*) Vi^PiViPi- Since J^a^ = ^ 6 j = 1, we must have y\ 4- 2/2 +1/3 = 0. It follows from (*) that r r r r , 2/2 = 5 UJS UJ*S ss Since 2/1+2/2 + 2/3 = 0, we get r + s + t = 0, r + u*s + ut = 0 and r + u>s + uj*t = 0. This means that r + UJ*S — uj(r + s) = 0 and r + UJS — uj*(r + s) = 0. Hence, r ( l — u;) = 5(0; — w*) and r ( l — u;*) = S(UJ* — a;), which gives U) — UJ* 1 - UJ UJ* — UJ 1 - UJ* and therefore either 5 = 0 or UJ* — 1 = 1 — UJ. But in the latter case we would have 2 = UJ* +UJ = —1 and so s = 0.

1 does not generalize to the noncommutative case. 2. Write Pi = 5 1 1 1 1 V2 = and p3 = - UJ 1 u> u*] 1J ' where u = — \ + *(^r), and note that each pi is a projection. Define If a is a composite point, then we may find 6 = 61 0 62 0 63 and c = c± 0 c2 0 c 3 such that a ^ 6, a ^ c, 5^6» = J^Cj = 1, all terms lie between 0 and 1 and a = (|)(6 + c). If we write bi = ^ + yi, where a^ = (§)pi, then it follows that C{ = &i 2/j. We have (1 -pMl -Pi) = {-)(l-Pi)(bi + Ci)(l -Pi) = 0, LYAPUNOV THEOREMS FOR OPERATOR ALGEBRAS 43 and, since all terms are positive, we get (1 - pi)bi(l - Pi) = (1 - Pi)ci(l - p^ = 0.

To see that ii) is true, fix a composite point a\ 0 • • • 0 a^ in Q. By the first paragraph of the proof there is a projection r in M such that if — r < b < r then se#(ai0- • -0afc+5(60(-6)0O0- • -00), a i 0 - • • 0 a f c - 5 ( 6 0 ( - 6 ) 0 O 0 - • -00)) C Q. Thus, if we put y = R-span({6 0 (-6) 0 O - - - 0 O : - r < 6 < r}), then y is lineal with respect to a\ 0 • • • 0Ojb and y is infinite-dimensional because Af is nonatomic. 4. 1 does not generalize to the noncommutative case. 2. Write Pi = 5 1 1 1 1 V2 = and p3 = - UJ 1 u> u*] 1J ' where u = — \ + *(^r), and note that each pi is a projection.