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A $*$-homomorphism $f:A\to B$ between C*-algebras is called non-degenerate if $f(A)B=B$.

I guess that I can prove that a non-degenerate *-homomorphism always induces a map on state spaces $f^\ast:S(B)\to S(A)$ such that $f^\ast(\phi)=\phi \circ f$?

  1. Is it correct that non-degenerate *-homomorphisms are the only *-homomorphisms such that $\phi \circ f\in S(A)$ for all $\phi \in S(B)$?

  2. This leads to the question as to whether positive linear functionals separate points from closed subspaces.

Thanks!

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The answer to 1 is "yes", and 2 is not quite well-defined, to my mind.

I'm going to follow Takesaki, Chapter III, Section 4, but this is all standard stuff. Given a C*-algebra $A$, a closed subspace $V$ of $A^*$ is left-invariant if $a\mu\in V$ for each $a\in A,\mu\in V$, where $(a\mu)(b) = \mu(ba)$ for $b\in A$. Then we have the following (Corollary~4.4 in the book):

Claim: $V$ is the closed left-invariant subspace generated by $V \cap A^*_+$.

In particular, $V$ certainly contains states. If $I\subseteq A$ is a right ideal then the annihilator $I^\perp = \{\mu\in A^* : \mu(a)=0 \ (a\in I) \}$ is left-invariant, and so

Claim: For each proper closed right ideal $I\subseteq A$ there is a state $\mu\in A^*$ which annihilates $I$.

So, if $f:A\rightarrow B$ is a degenerate $*$-homomorphism, then let $I$ be the closed linear span of $f(A)B$. By assumption $I$ is not equal to $B$, and obviously $I$ is a right ideal. So, there is some state $\mu\in B^*$ annihilating $I$, in particular, with $\mu\circ f=0$.

(If $B$ is non-unital, I guess I need to argue using approximate identities to show that $f(A) \subseteq I$. Also, using Cohen-Hewitt factorisation it follows that $I = \{ f(a) b : a\in A, b \in B \}$, no linear span or closure needed.)

So, for question 2. As every member of $A^*$ is the linear combination of (at most) 4 states, $S(A)$ certainly separates points of $A$. But if by "positive linear functionals separate points from closed subspaces" we mean that we want our functional to annihilate the subspace, then it's very easy to come up with counter-examples.

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  • $\begingroup$ could you please tell me about such a counter-example $\endgroup$ – Philip Oct 17 '16 at 20:31
  • $\begingroup$ ok sorry. got it. $\endgroup$ – Philip Oct 17 '16 at 20:47

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