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If $X$ is a compact metrisable space, a metric $d$ on $X$ can be take as an element of $C(X\times X)$ such that

(1) $ev_x\otimes ev_y (d)=d(x,y)\geq 0$ for all $x,y\in X$ (Non-negativity).

(2) $ev_x\otimes ev_y (d)=0$ iff $x=y$ (Identity of indiscernibles).

(3) $ev_x\otimes ev_y (d)=ev_y\otimes ev_x (d)$ for $x,y\in X$ (Symmetry).

(4) For all $x,y,z\in X$, $ev_x\otimes ev_y (d)\leq ev_x\otimes ev_z (d)+ev_z\otimes ev_y (d)$ (Triangle inequality).

Motivated by this, we can give the following definition of quantum compact metrisable space.

A unital nuclear $C^*$-algebra $A$ is called a quantum compact metrisable space if there exists an element $d\in A\otimes A$ such that

(1) $d\geq 0$(non-negativity).

(2) $\psi\otimes \phi(d)=0$ iff $\psi=\phi$ for $\psi,\phi\in P(A)$ where $P(A)$ is the pure state space of $A$(Identity of indiscernibles).

(3) $\psi\otimes \phi(d)=\phi\otimes \psi(d)$ for $\psi,\phi\in P(A)$(Symmetry).

(4) $\psi\otimes \phi(d)\leq \psi\otimes \varphi(d)+\varphi\otimes \phi(d)$ for all $\psi,\phi$ and $\varphi\in P(A)$(Triangle inequality).

I check that for $M_n(C)$, the $C^*$-algebra of $n$ by $n$ complex matrices and found that if a $d$ satisfies non-negativity, identity of indiscernibles and symmetry, then $d$ does not satisfy triangle inequality. So this means that $M_n(C)$ can only admit a semi-metric in this sense.

My question: is there any genuine quantum compact metrisable space(i.e. noncommutative $C^*$-algebra $A$ with such a $d\in A\otimes A$ satisfying the above properties)?

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I would suggest that you look at the work of Rieffel on compact quantum metric spaces. His point of view is not to directly generalize the metric by understanding it as an element of the tensor square $A \otimes A$ (NB: you have not specified which tensor product you use here), but rather he generalizes the Lipschitz seminorm associated to the metric.

As I understand it from Rieffel, it was known already to Kantorovich that the metric on a compact space $X$ is determined by the Lipschitz seminorm on $C(X)$: $$ L(f) = \sup_{x \neq y} \{\frac{|f(x)-f(y)|}{d(x,y)}\} $$ via the identity $$ d(x,y) = \sup\{ |f(x) - f(y)| : L(f) \le 1\}. $$

Anyway, this doesn't directly answer your question, but this has been a fruitful line of inquiry, and I suggest you look at Rieffel's papers to see if they have anything useful for you. I think "Metrics on State Spaces" is a good one to start with.

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  • $\begingroup$ Thanks for comments. Since $A$ is nuclear, there is only one $C^*$-norm on the algebraic tnensor product $A\odot A$. Actually I am working on the subject of compact quantum metric space initiated by M. Rieffel. I once discussed this question with him and he said that one obstacle for this alternate definition of quantum compact metrisable space is that one has to deal with pure states which in general are quite complicated. $\endgroup$
    – Huichi Huang
    Nov 19, 2012 at 23:07
  • $\begingroup$ To try to avoid the use of pure states in the definition of ``compact quantum metrisable space'' here, we can replace axiom (3) by $\sigma(d)=d$ where $\sigma$ is the flip operator from $A\otimes A$ to $A\otimes A$ and axiom (4) by $d_{12}≤d_{13}+d_{23}$ where we use the leg number notations, say, $d_{12}\in A\otimes A\otimes A$ stands for the image of d under the map $x\to x\otimes1$ from $A\otimes A$ to $A\otimes A\otimes A$. Is there any good way to express axiom (2) (symmetry) without using pure states? $\endgroup$
    – Huichi Huang
    Dec 5, 2012 at 1:14
  • $\begingroup$ I do not know if there is a way to do this, sorry. $\endgroup$
    – MTS
    Dec 5, 2012 at 1:48

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