6.1. Traces and the right regular representation 213(JS J)'. With these simple observations and the technical calculations of the
previous lemma, we can demonstrate the result of Murray and von Neumann
mentioned earlier.Theorem 6.1.4. If T is a tracial state on A, then 7r 7 (A)" = 1f~P(A^0 P)' and
(hence) 1r 7 (A)' = 7r~P (A^0 P)".Proof. Since 7r 7 (A)" c 1f~P(A^0 P)', we only have to observe the opposite
inclusion. But J7r 7 (A)J = 7r~P(A^0 P) and hence1f~P(AOP)' = J(7rT(A)')J.
Thus we will be finished once we observe that J(7r 7 (A)')J and 7r 7 (A)' are
commuting sets of operators. For any x, y, z E 7r 7 (A)', we have by Lemma
6.1.3 thatJxJy(zi) = Jxz*y*i = yzx*i = yJxz*i = yJxJ(zi).
Since i is 7r 7 (A)'-cyclic and z was arbitrary, it follows that JxJy = yJxJ.
DExercises
Exercise 6.1.1. A bijective, adjoint-preserving, linear map 1r: A -+ B
is called an anti-isomorphism if 7r(ab) = 7r(b)7r(a). Prove that A is anti-
isomorphic to itself if and only if it is isomorphic to A op.
Exercise 6.1.2. The conjugate algebra of A, denoted A= {a : a EA}, is
just A as an involutive ring but it has the conjugate vector space structure
- i.e., .\a = A.a for all A.· E CC and a E A. Show that A op is isomorphic to A
via the adjoint map.
Exercise 6.1.3. Let A= M2(CC). Give an example of a state l.{J on A such
that the obvious definition of the right regular representation fails (e.g.,
arrange that 1f~P(a)b = ba is not even well-defined).
Exercise 6.1.4. For a group r, let r^0 P denote the opposite group - i.e.,
r^0 P = r as sets but multiplication in r^0 P gets reversed (g·h = hg). Show that
g 1--t g-^1 gives an isomorphism r -+ r^0 P. Prove that C (r)^0 P :::. C (r^0 P) ~
C(r) and C~(r)^0 P ~ C~(r^0 P) '::::'. C~(r). In other words, group C-algebras
are always anti-isomorphic to themselves.
Exercise 6.1.5. Show that the right regular representation, as defined in
this section, is really the same (after identifying group C* -algebras with their
opposite algebras) as the usual right regular representation of a group.