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

Let $X$ be a compact metric space. Recall that the hyperspace $2^{X}$ is the set of all non empty compact subsets of $X$ with the Hausdorff metric. Assume that $\mathcal{C}$ is the category of all compact metric spaces. Define a functor $F:\mathcal{C}\to \mathcal{C}$ with $F(X)=2^{X}$. (F can natrually acts on morphisms).

Since the category of compact metric spaces is equivalents to the category of commutative unital separable $C^{*}$ algebras, the above functor $F$ gives us a natural and unique functor $G$ on the category $\mathcal{S}$ of all commutative unital separable $C^{*}$ algebras: \begin{equation} G:\mathcal{S}\to \mathcal{S} \end{equation} Let $\tilde{\mathcal{S}}$ be the category of all (not necessarily commutative) unital separable $C^{*}$ algebras.

My question:

What is a natural extension $\tilde{G}$ of $G$, as a functor $\tilde{G}:\tilde{\mathcal{S}} \to \tilde{\mathcal{S}}$?

If we construct a nice $\tilde{G}$ as above, then for every (non commutative) unital separable $C^{*}$ algebra $A$, $\tilde{G}(A)$ would be called the hyper algebra of $A$. After construction of such $\tilde{G}$, two possible natural question could be:

1)If $A$ is an idempotent-less $C^{*}$ algebra, is $\tilde{G}(A)$ idempotent-less, too?

2)Is $\tilde{G}$ an exact or half exact functor?

share|cite|improve this question

Your Answer


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

Browse other questions tagged or ask your own question.