Calkin algebra

Assume the Axiom of Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)). Let $X$ be an infinite-dimensional complex Banach space and let $\mathcal{B}(X)$ be the Banach algebra of bounded operators (If $Y$ is Banach then $\mathcal{B}(X,Y)$ is Banach, The operator norm as the least bound and as the unit-sphere or unit-ball supremum). Let $\mathcal{K}(X)\subseteq \mathcal{B}(X)$ be the set of compact operators (Compact linear operator). Then:

• $\mathcal{K}(X)$ is a linear subspace of $\mathcal{B}(X)$ (Linear combinations of compact operators are compact);
• it is a two-sided ideal: if $K$ is compact and $A,B$ are bounded then $AK$ and $KB$ are compact (Compositions with a compact operator are compact);
• it is closed in the operator norm: a norm limit of compact operators is compact (Norm limit of compact operators is compact, which is stated under Countable Choice and requires the target to be Banach, as here). More explicitly, if $T$ lies in the norm closure, Countable Choice selects $K_n\in \mathcal{K}(X)$ with $\mathrm{\parallel }{K}_n-T\mathrm{\parallel }<1\mathrm{/}n$ for $n\ge 1$; the theorem makes $T$ compact;
• and it is proper: the identity $I_X$ is not compact precisely because $X$ is infinite-dimensional (On an infinite-dimensional normed space, the identity operator is not compact).

The Calkin algebra of $X$ is the quotient algebra

$\mathcal{C}(X):=\mathcal{B}(X)\mathrm{/}\mathcal{K}(X),$

with the quotient vector-space structure, the quotient norm $\mathrm{\parallel }A+\mathcal{K}(X)\mathrm{\parallel }:=\inf \{\mathrm{\parallel }A-K\mathrm{\parallel }:K\in \mathcal{K}(X)\}$, and the multiplication $(A+\mathcal{K})(B+\mathcal{K}):=AB+\mathcal{K}$. Multiplication is well-defined because $\mathcal{K}(X)$ is a two-sided ideal, and it is submultiplicative: for $K_1,K_2\in \mathcal{K}$ and representatives $A,B$, $(A+K_1)(B+K_2)=AB+(A{K}_2+K_{1}B+K_1{K}_2)$ with the bracket in $\mathcal{K}$, so taking infima gives $\mathrm{\parallel }(A+\mathcal{K})(B+\mathcal{K})\mathrm{\parallel }\le \mathrm{\parallel }A+\mathcal{K}\mathrm{\parallel }\mathrm{\parallel }B+\mathcal{K}\mathrm{\parallel }$. The quotient is complete for the quotient norm by A quotient of a Banach space by a closed subspace is Banach. Its unit is $1:=I_X+\mathcal{K}(X)$.

• The unit has norm one and is nonzero. The quotient norm satisfies $\mathrm{\parallel }{I}_X+\mathcal{K}\mathrm{\parallel }\le \mathrm{\parallel }{I}_X\mathrm{\parallel }=1$. The quotient norm is definite because $\mathcal{K}$ is closed, and $I_X\mathrm{\notin }\mathcal{K}$, so $c=\mathrm{\parallel }{I}_X+\mathcal{K}\mathrm{\parallel }>0$. The unit is idempotent and the quotient norm is submultiplicative, giving $c\le c^2$. Division by $c$ yields $c\ge 1$, hence $c=1$.

• Finite-dimensional $X$ is excluded, not normalized away. If $\dim X<\mathrm{\infty }$ then every bounded operator has finite-dimensional range and is compact (Bounded finite rank operators are compact), $\mathcal{K}(X)=\mathcal{B}(X)$ and the quotient is the zero algebra, which carries no unit in the sense of Unital Banach algebra. The definition therefore restricts to infinite-dimensional $X$; the finite-dimensional case is the zero quotient and is not called a Calkin algebra here.

• Where Countable Choice is spent. It is used in the norm-limit compactness theorem, in selecting the approximating sequence above to turn sequential closure into norm closure, and in quotient completeness. The ideal formulas, identity noncompactness and the unit-norm argument introduce no further choice beyond those supplied closed-quotient facts.

• The Calkin algebra forgets compact perturbations. Two operators have the same coset exactly when they differ by a compact operator, so $\mathcal{C}(X)$ records the "Fredholm part" of $\mathcal{B}(X)$; this is what makes the Atkinson theorem a statement about invertibility in $\mathcal{C}(X)$ (Atkinson in Calkin algebra language).