How do endomorphisms of Banach space modulo compact operators look like?

Question

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

2. 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.

Answer 1

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.