By Steve Wright

Based on lectures dropped at the Seminar on Operator Algebras at Oakland collage throughout the iciness semesters of 1985 and 1986, those notes are a close exposition of contemporary paintings of A. Connes and U. Haagerup which jointly represent an evidence that each one injective elements of style III1 which act on a separable Hilbert area are isomorphic. This end result disposes of the ultimate open case within the type of the separably performing injective components, and is likely one of the impressive fresh achievements within the conception of operator algebras. The notes can be of substantial curiosity to experts in operator algebras, operator thought and staff in allied parts comparable to quantum statistical mechanics and the speculation of workforce representations.

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**Uniqueness of the Injective III1 Factor**

In response to lectures brought to the Seminar on Operator Algebras at Oakland college in the course of the wintry weather semesters of 1985 and 1986, those notes are a close exposition of modern paintings of A. Connes and U. Haagerup which jointly represent an explanation that every one injective elements of variety III1 which act on a separable Hilbert area are isomorphic.

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The set of all such automorphisms is denoted by Int R. Our goat here is to prove the following theorem. Theorem 3. Let M be a type III factor, 0 6 Aut M . ,~. 1 Proof of nece~ity. Let ~ , . . -,~. 1) limll~,oO-1 - u:~u~II = O, ~, e M +. By definition, O(~j) is the representing vector in P~ of q0j 0 -1, and u* ~j u~ is the representing o vector in ph for u~ qojun. 1), limll0(~jl - ~; ¢j ~,,11 ~ = o , v j . 42 Hence we can choose N such that whence Thus 0 satisfies the condition in T h e o r e m 3 with x = u ~ 5£ 0.

J(x*x))}. ~ + ( ~ + ~ ~(x*~)~)~ < ~ + (¢~+~)~ < 2772, providede + ,k} < 1 . ~o¢~))e. (~)~,, 11~ < 20~ ~ ~(~o(~)%(x)). 1 ! Since u = u,~(x) is a partial isometry, we have the conclusion of the temma except for u 2 = 0 . To get this, set y = u(1 - uu*). (,*y) _. ). 34) with u~(x) replaced by an appropriate us(y). Since u,~(y) 2 = 0 for each a, this gives us what we want. 36). 37) _< 11:(1 - i)a(au*J + ~,~ I l u ( a ( ~ - f ) J -< tl(zx2, - x ~ ) ~ l l a~(~ - f)),'il I - a~,)el II - (1-f))~,ll + ~'itl(af J- f)~,ll, since u {1 - f, ~il = u If, (i].

E i tl(~÷f)5,11, i and c = 1 3 . 2 7 , E = e(a0) satisfy the requirements of L e m m a 3. D. The next l e m m a is the key to the proof of Theorem 3. Lemma 4. Let M, O, ~j, and e be as in Theorem 3, and let c be the constant in Lemma 3. , (i~) tlx~j-0(~j)xll ~ _< ~ r llx~,tl ~, Pro@ We ~; j = l , , n begin by noting that there exists 6 > 0 such that whenever e l , . . , an, /31, . . , /3, are nonnegative scalars with ai _> ½ /3i - /3j, i = 1 , . . , n , then Ec~ >_ 2 -4 E/3~. 2 Set 7 = rain {10 - s c -2 n - l e ~, 10 - s c -2 n -1 6}.