By Dr. Leslie Cohn (auth.)

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Suppose that b E~. (X g(~ , b E~). Then F(b) = O. 1) F(bac) = ~(a)o~F(b) (b ~ , Proof. (H e O~). c ~7)2, a ~(I). From property 2) above, it is clear that if X e ~, F(X) = 0. Property B) implies that the kernel of F is a left ideal in 9 . Hence the kernel of F contains ~ , Let ~ ( J ) = C + ~c ~ ( J )analogously. F(c) = c ~ (c ~ ) . as claimed. 1), we show first that If c = V e 9T~, we find using 2) that F(V) = - V = V ~. Assume that F(c) = c ~ if c a ~ ( J ) ; and assume that c ~ o~(J), V c Ogq.

If XI, X2 e ~ l , then q(Xl)¢i(x2) - Q(x2)~i(xi) = ~/[Xl,X2]). Proof. We have q(X1)q(X2)I = q ( X 1 ) ( ¢ i ( X 2 ) I ) = (q(X1)¢I(X2))I + ¢I(X2)q(XI)I = (q(X1)¢l(X2))l + ¢I(XI)¢I(~)I, Hence, [q(Xl) , q(X2)]I = {q(Zl)¢i(X2) - q(X~(Xl)}i" Also, [q(Xl) , q(X2)]I = q([XI,X2])I = ¢I([XI,X2])I. Therefore, ¢i([Xl,X2])I = {~(×l)¢i(x 2) - q(z2)¢i(Xl)}I. Since ~ is an integral domaih and I # 0 by assumption, we may cancel I to obtain the desired identity. 5. Suppose that X1,X2 ~ i " Then 46 FI(X 1 ® X 2 - X 2 ® X 1 - [X1,X2]) = 0, Proof.

Proof. 1, Let IR(~:m) = ]V(R)f(~l~;X_yIm)d~. AdtBs 1 (where VB(R) = {~ e HIIts(~) I = R, It6(~) I ~ R for 6 e P+, ~ ~ 8 or - @S}). ,s, lira fVsi(R)IF(~)lexp{-(Im~+0)(H(~))}Idt81 A ... ,A). 8s I= 0 Clearly, ve may n assume that F(~) ~ ~yep+tT(~) 7, where ~Tcp+7(Ho)n 7 < d + B i(Ho) = d i. Set B~(d) = ma~ l J=l,,.. zm~v,~j> >_B~(d). (~)l"s. Adt6 I. A dt 8s ] = C'Rs-l. Adtsi i I(R) denote this integral. 5). Hence, Vsi(R ) = {[ ¢ ~Itsi(~ ) = +_2, Its(i)[ ~ R if (5# 8i }. xR . xRxR = RldZlXvol(A R s.