Question

In $C^\ast$-algebras we use $K(H)$, the algebra of compact operators on a separable Hilbert space, for stabilization of a $C^\ast$-algebra, i.e. $S(A):=A\otimes K(H)$. Is there any similar stabilization functor in Banach algebras? What is the substitute of $K(H)$?

Answer 1

This is not an answer, but too long for a comment: Even for $C^*$-algebras there is more than one method of stabilization. The critical feature of the compact operators is that $K(H) \otimes K(H) \cong K(H)$ for an infinite dimensional Hilbert space and that $K_*(K(H)) \cong K_*(\mathbb{C})$. But this also holds true for other $C^*$-algebras like for example the infinite Cuntz algebra $\mathcal{O}_{\infty}$ or the Jiang-Su algebra $\mathcal{Z}$. In general - again for the $C^*$-case - you might want to search for strongly self-absorbing $C^*$-algebras.

There is the following theorem:

If $A$ is a simple, separable and nuclear $C^*$-algebra, then $A \cong A \otimes \mathcal{O}_{\infty}$ if and only if $A$ is purely infinite.

And there is a famous theorem by Kirchberg, which you may read like this:

Assume $A$ and $B$ are simple, separable, nuclear and $\mathcal{O}_{\infty} \otimes K(H)$-stable $C^*$-algebras (i.e. $A \otimes \mathcal{O}_{\infty} \otimes K(H) \cong A$ and likewise for $B$), then $A$ and $B$ are isomorphic if and only if they are $KK$-equivalent.

If you are looking for a setup like this, you may want to search for Banach algebras, which have the $K$-theory of $\mathbb{C}$ and are isomorphic to the tensor product of two copies of themselves. Or one might be able to generalize the notion of strongly self-absorbing to Banach algebras.

That was a lot about $C^*$-algebras and not much about the Banach case. Sorry about that.