6
$\begingroup$

I think that there is a metric on the huge space of all $C^{*}$ algebras. What is the explicit

definition of this metric?may you introduce me a reference?

Moreover is the restriction of this metric to commutative $C^{*}$ algebras gives us a discrete metric? by discrete I mean "every commutative $C^{*}$ algebra has a neighborhood, with respect to this metric, which contains no other commutative $C^{*}$ algebra"

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ What makes you think that such a metric exists? $\endgroup$
    – Eric Wofsey
    Commented Jan 1, 2014 at 20:34
  • $\begingroup$ I heard this from a participant, when I was participating in a workshop on NCG.(about 9 years ago) $\endgroup$
    – Ali Taghavi
    Commented Jan 1, 2014 at 20:44
  • 1
    $\begingroup$ I also heard this from an speaker who gave a talk on NCG about 6 years ago. I remember he said that this metric is defined by A. Connes. In that time I did not peruse the details. But now I am interested to know about this metric $\endgroup$
    – Ali Taghavi
    Commented Jan 1, 2014 at 20:52

2 Answers 2

Reset to default
14
$\begingroup$

Probably the most standard metric is Banach-Mazur distance, and there is indeed a theorem due to Amir which says that if the Banach-Mazur distance between $C(K)$ and $C(L)$ is less than $2$ then $K$ and $L$ are homeomorphic. There's also something called the Kadets distance which is basically a linearized Gromov-Hausdorff distance. I don't know what the Kadets distance says about commutative C*-algebras.

Adding an edit to call attention to Caleb Eckhardt's comment. Caleb point out that CB Banach-Mazur distance is a better candidate (I agree) and cites this paper which shows that if two C*-algebras have sufficiently small CB Banach-Mazur distance, and one of them is nuclear (in particular, commutative) then they are isomorphic.

Improve this answer
$\endgroup$
2
  • 7
    $\begingroup$ The completely bounded Banach-Mazur distance (which is just Banach-Mazur for commutative C*-algebras) would probably be more useful for noncommutative C*-algebras. Ricard and Roydor recently showed a (completely bounded) analogue of the Amir-Cambern theorem for nuclear C*-algebras arxiv.org/abs/1108.1970 $\endgroup$
    – Caleb Eckhardt
    Commented Jan 2, 2014 at 14:43
  • $\begingroup$ NiK,Caleb thank you very much for your valuable answer and comment $\endgroup$
    – Ali Taghavi
    Commented Jan 2, 2014 at 21:57
3
$\begingroup$

Another possibility is the Kadison-Kastler metric on C*-subalgebras of $B(H)$, which is just the Hausdorff distance between their unit balls. This paper gives references to a number of results about stability under perturbation, which include the case where one of the algebras is separable and abelian; this paper generalizes to the case where one of the algebras is separable and nuclear.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .