“…standard proof in von Neumann algebras [15,Proposition 3.6.7] (which applies here, too), v à is taken to be any extreme point in the weak-* compact, convex set K X f$ P E yy j f $ 1 k$kgX Since f 9u$u à f $ for all unitaries u P A yy and all $ P E yy Y K is preserved under the maps $ U 3 9u$u à Y u unitary in A yy X Further, these maps clearly carry extreme points of K to extreme points of K. Thus 9uv à u à is an extreme point of K for every unitary in A yy X But then, the uniqueness part of the polar decomposition implies that there is only one extreme point in K. Whence, 9uv à u à v à for all unitaries u P A yy Y and so v à P F 1 X This, in turn, implies that for all a P A yy and all unitaries u P A yy Y we have jf jau f v à au f 9uv à a f v à ua jf juaY which means that jf j is tracial. For the converse, simply note that if jf j is tracial on A yy while v à P F 1 Y then for all $ P E yy and all unitaries u P A yy Y f 9u$u à jf jv9u$u à jf ju à v9u$ jf jv$ f $X Suppose M is a von Neumann algebra and that is an automorphism of MX In [9], Kallman termed to be free, or freely acting, in case the only solution b to the equation ab ba, for all a P MY is b 0X If one views M as a correspondence E over itself, with 9 , then is free precisely when F 1 vanishes. (We don't make any distinction, at this point, between analysis in E and analysis in E yy X Thus, Theorem 15 says that our Condition F is a correspondence-theoretic way of saying that all the powers of an autoon the simplicity of some cuntz-pimsner algebrasmorphism are freely acting.…”