It is well-known that given a Banach space $X$, the set of compact operators (let's denote it by $K(X)$) on $X$ forms a both-sided ideal in $L(X)$, the ring of bounded linear operators on $X$. My question is

  1. Is there any natural interpretation of the quotient ring $L(X)/K(X)$?

Here "natural" is probably too vague, so I will explain what I have in mind: Is there some canonical surjective ring homomorphism $L(X)\rightarrow R(X)$, where $R(X)$ is again some ring ($\mathbb{C}$-algebra) such that $K(X)$ is its kernel? I obviously do not mean the quotient map as such; I am looking for some description in terms of functional analysis (ideally, the ring $R(X)$ should be described in terms of functional analysis and the space $X$).

Moreover, it also holds that all the compact operators form a both-sided ideal in the $\mathbb{C}$-linear category of Banach spaces. So the general question is

  1. Is there a $\mathbb{C}$-linear functor (again, described in terms of functional analysis) from the category of Banach spaces such that its kernel consits precisely of all compact operators?

I apologize if the question is too vague, but it seems to me that this is one of the vague questions worth asking.

Thanks in advance for any help.

  • $\begingroup$ These things look like a green magnetic monopole but without the hat. $\endgroup$ – Masacroso Nov 5 '14 at 22:34
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    $\begingroup$ When $X$ is a Hilbert space, the quotient algebra in question is called the Calkin algebra. To my knowledge there is no natural ring $R(X)$ as described in the question. Since this doesn't happen for Hilbert spaces, I doubt it happens for Banach spaces in general, but perhaps there are some special cases. Also, there are analogues of the Calkin algebra called Corona algebras, which occur when one replaces $K(X),L(X)$ with a generic $C^*$-algebra $\mathcal{A}$ and its multiplier algebra, respectively. $\endgroup$ – J. Loreaux Nov 5 '14 at 23:21

The Calkin algebras $\mathscr{Q}(X) = \mathscr{B}(X)/\mathscr{K}(X)$ of Banach spaces can be very different. The prototypical example is of course $\mathscr{Q}(\ell_2)$ which is humongous. See this thread.

Maybe one should ask whether this is really the correct definition. Indeed, Banach spaces can lack the approximation property in which case $\mathscr{Q}(X)$ is not simple. This suggests that the answer for your second question should be no, as compact operators on such spaces are no longer the smallest non-zero ideal of $\mathscr{B}(X)$.

Anyhow, even for spaces $X$ with bases Calkin algebras can be very different from $\mathscr{Q}(\ell_2)$. They can be for instance one-dimensional:

S.A. Argyros and R.G. Haydon, A Hereditarily Indecomposable $\mathscr{L}_\infty$-space that solves the scalar-plus-compact problem, Acta Math. 206 (2011), no. 1, 1–54.

It can also bee commutative but infinite-dimensional by a recent result of Pavlos Motakis, Daniele Puglisi and Despoina Zisimopoulou:

P. Motakis, D. Puglisi and D. Zisimopoulou, A hierarchy of separable commutative Calkin algebras, preprint, 2014.

This suggests that the answer for your first question should also be no. And as you say, it is too vague to be answered precisely.

  • $\begingroup$ Thank you for the answer. I am confused by several things, though: 1) "... in which case $\mathscr{Q}(X)$ is not simple. This suggests that the answer for your second question should be no, as compact operators on such spaces are no longer the smallest non-zero ideal of $\mathscr{B}(X)$." Did you mean "maximal closed ideal" instead, by any chance? Otherise this does not make sense to me. 2) Also, can you comment on how the facts you mentioned seuggest the negative answer? (Not that I am disagreeing with the negative answer, I am simply curious about the reasoning/intuition behind this.) $\endgroup$ – Pavel Čoupek Nov 9 '14 at 12:29
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    $\begingroup$ What I was trying to say was that there is no criterion which would rule out a semi-simple Banach algebra as a Calkin algebra of some Banach space. So even very wild commutative Banach algebras may well be isomorphic to Calkin algebras. In other words, there are no common features of Calkin algebras maybe beyond semi-simplicity etc, hence their algebraic characterisation seems to be intractable so far. $\endgroup$ – Tomasz Kania Nov 9 '14 at 12:43
  • $\begingroup$ I meant minimal closed ideal not maximal. $\endgroup$ – Tomasz Kania Nov 9 '14 at 12:44
  • $\begingroup$ Thank you again for the answer to the second part. As for the first, part, I am still confused. I still think it should me "maximal" instead of "minimal", if you want to relate it to "the quotient being simple". $\endgroup$ – Pavel Čoupek Nov 9 '14 at 13:38
  • $\begingroup$ Yes, you are right. I fell into my own trap: for me Calkin algebras are quotients of $\mathscr{B}(X)$ by the closure of the ideal of finite-rank operators :-) $\endgroup$ – Tomasz Kania Nov 9 '14 at 13:56

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