3.4. Takesaki 's Theorem 79= elA(a)elB(b).
Hence e(x) = elA 0 elB(x) for all x EA 0 B.
To prove the second assertion, we assume A is abelian and e is a pure
state on A@a B. Thus the weak closure of 1f~(A @a B) is all of IIB(H~). But
1f~(A@aB) is contained in the C*-algebra generated by 1f~,A(A) and 1f~,B(B)
and since A is abelian and 1f~,A and 1f~,B have commuting ranges, we see
that 1f~,A(A) lives in the center of IIB(H~) - i.e., 1f~,A(A) = Cl. It follows
that elA is a character (hence pure) and the weak closure of 1f~,B(B) is all
of IIB(H~). Hence elB is also pure and we are done. D
Lernrna 3.4.4. Let <p E S(A) and 'ljJ E S(B). Then <p 0 'ljJ : A 0 B -t C is
algebraically positive - i.e., <p 0 'ljJ(x*x) ;::: 0 for all x EA 0 B.Proof. Let (1f\O, 1-{\0, v\O) and (1f'lj;, 1-{'l/J, v'l/J) be the GNS representations. By
Exercise 3.3.4 we have a representation 1f\O@ 1f'lj;: A@ B -t IIB(H\O @H'lj;)· A
calculation shows that <p 0 'ljJ ( x) is equal to ( 1f \0 @ 1f 'l/J ( x )v\O @ v'l/J, v\O @ V'lj;) for
all x EA 0 B, which implies the result. DWe now begin to show that product states are always continuous on
A @a B. The proce~s starts by considering the abelian case and building up
(via Lemma 3.4.1 and Corollary 3.4.3) to the general case. The target result
is Proposition 3.4. 7, but if reading the next two lemmas in order makes little
sense, try jumping ahead to Proposition 3.4. 7 and reading backwards.
Lernrna 3.4.5. Assume that both A and B are unital and abelian. Then for
every C*-norm II· Ila on A0B and pair of pure states <p E S(A), 'ljJ E S(B)
the linear functional <p 0 'ljJ : A 0 B -t C extends to a state on A @a B.
Proof. Let P(A) (resp. P(B)) denote the pure states on A (resp. B). Let
U = {(<p, '!/J) E P(A) x P(B): l'P 0 'l/J(x)I::::; llxlla, \fx EA 0 B}.
Then the lemma asserts that U = P(A) x P(B).^12
So assume that U is a proper subset of P(A) x P(B). A standard
calculation shows that U is a closed subset of P(A) x P(B) (in the product
of the weak-* topologies) and hence we can find open sets UA C P(A) and
UB c P(B) such that (UA x UB) nU = 0. Let 0::::; f EA and 0::::; g EB be
norm-one elements whose supports (via the Gelfand transform) live in UA
and UB, respectively. Then <p 0 'I/JU@ g) = 0 for every (<p, '!/J) EU.
This will lead to a contradiction since we can construct a pair ( <p, 'ljJ) E U
such that <p 0 'I/JU@ g) > 0. Indeed, let e E P(A@a B) be a pure state such
12If ( <p, 'lj;) E U, then \0 8 'lj; extends to a contractive linear functional on A ®a B. One then
either appeals to algebraic positivity of \0 8 'lj; to deduce that the extension is a state or to the
fact that norm-one functionals mapping the identity to 1 are necessarily positive.