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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 $\mathrm{Mor}(s(A), s(B))$?

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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.

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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

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