MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider finite-dimensional (for-simplicity) $\star$-algebras, that is, unital associative algebras over the complex numbers equipped with an antilinear antiautomorphism $\star$.

A state on a $\star$-algebra $A$ is a linear mapping $\psi: A \to \mathbb{C}$ satisfying

(i) $\psi(1) = 1$ [normalization]

(ii) $\psi(a^*) = \overline{\psi(a)}$ [reality]

(iii) $\psi(a^*a) \geq 0$ [positivity]

Denote $s(A)$ the set of all states on $A$.

Consider the following 2 ways to construct a mapping between states on different algebras:

(1) Consider $A$, $B$ $\star$-algebras and $f: A \to B$ a $\star$-homomorphism. Then we have $f^{-1}: s(B) \to s(A)$ defined by $f^{-1}(\psi)(a) := \psi(f(a))$.

(2) Consider $A$, $B$ $\star$-algebras and $\phi$ a state on $B$. Then we have $t_\phi: s(A) \to s(A \otimes B)$ defined by $t_\phi(\psi)(a \otimes b) = \psi(a) \phi(b)$.

Now, suppose we compose any number of mappings of the above 2 kinds. We get for any $\star$-algebras $A$, $B$, a certain class of mappings $s(A) \to s(B)$. I.e., these are mappings obtained by composing mapping of the above 2 kinds while inserting any number of intermediate algebras. Denote this set of mappings $\mathrm{Mor}(s(A), s(B))$.

Given a $\star$-algebra $A$, the positive cone of $A$ is the set of all linear combinations of elements of the form $a^*a$ with positive coefficients. Denote it $p(A)$.

Consider $A$, $B$ $\star$-algebras and $L: A \to B$ a linear mapping (not necessarily a homomorphism!) preserving 1 and $\star$. Suppose $L$ is positive in the sense that it maps $p(A)$ to $p(B)$. Then $L$ induces $L^{-1}: s(B) \to s(A)$ defined by $L^{-1}(\psi)(a) = \psi(L(a))$.

The Question:

Is $L^{-1}$ guaranteed to be in $\mathrm{Mor}(s(A), s(B))$?

share|cite|improve this question
This would be a lot easier to read if you wrote your question in LaTeX. – MTS Oct 7 '11 at 22:46
I've now LaTeX'd the question. I did not change anything else; although I was tempted to reorder (i), (iii), (ii) :) – José Figueroa-O'Farrill Oct 7 '11 at 23:15
I don't even understand how $L^{-1}$ is well defined if $L$ is not a homomorphism: If $L(1) = a = a^* \neq 1$ then $L^{-1}(\psi)(1) = \psi(a) \neq 1$ for some $\psi$. ' – Reimundo Heluani Oct 8 '11 at 2:08
Good point Heluani, I fixed it now – Squark Oct 8 '11 at 9:45
Thx José Figueroa-O'Farrill, I fixed the numbering – Squark Oct 8 '11 at 9:52
up vote 3 down vote accepted

Let me first stick to more conventional notation and, given a positive operator $f:C\rightarrow A$, denote by $f^\ast$ the map that you call $f^{-1}$. If $B=A\otimes C$ and $f$ is defined by $f(a\otimes c) = a\phi(c)$ with $\phi$ a state on $C$, then you get a special case of what is called a conditional expectation, and $f^\ast$ coincides with your $t_\phi$ above. If you are willing to enlarge $\mathrm{Mor}(s(A),s(B))$ to contain such maps too, then Stinespring dilation theorem tells you that any completely positive map, being the composition of an algebra morphism followed by a conditional expectation, will induce an admissible morphism on the state spaces.

share|cite|improve this answer
Well, conditional expectation is not my t_phi since conditional expectation maps states on A (x) C to states on A whereas t_phi maps states on A to states on A (x) C. Also we need the additional condition of complete positivity. – Squark Oct 20 '11 at 20:51
However, maybe this result is what I really need – Squark Oct 20 '11 at 20:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.