Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

This is kind of vague question, but I'll try to make it more precise.

$T$ is a compact operator on a Hilbert Space, $H$, if $\overline{T(D)}$ is compact in $H$, where of course, $D$ is the closed unit ball in $H$. So we use the underlying Hilbert space in order to define these operators. Now $B(H)$ is a $C^*$-algebra. Can we abstractly define/characterize compact operators only in terms of $C^*$-notions, without appealing to the underlying Hilbert space? Stated otherwise, is it possible to define, say "compact" elements in a $C^*$-algebra $A$, which agree with the usual notion if we take $A=B(H)$?

share|cite|improve this question
$H$ is not a C*-algebra. – Jonas Meyer Jun 17 '12 at 17:13
@JonasMeyer I assume the OP meant $K(H)$ – user16299 Jun 17 '12 at 20:52
Yeah, sorry I have edited the question. – Nirakar Neo Jun 18 '12 at 10:21
If $H$ is separable, then $K(H)$ is the only two-sided closed proper ideal of $B(H)$. This theorem is due to Calkin. – Norbert Jun 18 '12 at 11:14
up vote 3 down vote accepted

I am not familiar with such a concept, but you could call a projection $p$ in a $C^*$-algebra $A$ finite if $pAp$ is finite-dimensional; and then consider the closed ideal generated by the finite projections.

I don't know if this is useful, however.

share|cite|improve this answer
I think this works. – Nirakar Neo Jun 18 '12 at 10:34

(With the disclaimer that I am not an operator algebraist, so there may be technical details that are not quite right.)

There is a notion of "compact element" for a semifinite von Neumann algebra, which is roughly speaking one on which there is a faithful tracial weight whose ideal of definition is ultraweakly dense (at least in the case of von Neumann algebras whose predual is separable). The idea is similar to that described by Harald; you call a projection in the von Neumann algebra finite if its trace is a finite number, and then you define the ideal of compact elements to be the norm-closed ideal generated by the finite projections.

When your von Neumann algebra is $B(H)$, with the trace being the usual one, then a projection is finite in the sense above if and only if it has finite rank, and so one recovers $K(H)$.

share|cite|improve this answer
Umm...but I couldn't get that middle paragraph, as I haven't studied von Neumann algebras. But I'll try to understand the answer once I get some understanding. Thanks. :) – Nirakar Neo Jun 18 '12 at 10:36

this is a late reply, but here goes:

Yes, there is a very general notion of "compact operators" for Hilbert $C^*$-modules (these are analogous to Hilbert spaces; however their inner products lie in $C^*$-algebras and they have an additional module structure), and a $C^*$-algebra is a special case of a Hilbert $C^*$-module.

Assuming that a definition exists (I won't go into details here), denote the set of compact operators on a $C^*$-algebra $A$ as $K(A)$.

With such a definition, one discovers that there is an isomorphism $A\cong K(A)$, where $A$ is considered as a Hilbert $C^*$-module over itself.

Further, if $A$ is unital, all elements in the multiplier algebra of $A$ are compact according to this definition.

For more details I refer you to Chapter 1 and 2 of the book by E.C.Lance, titled "Hilbert $C^*$-modules: A toolkit for Operator Algebraists".

share|cite|improve this answer

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.