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 $K(H)$?
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$\begingroup$ @Z254R, are you pointing out deficiencies in a question? ;-) Well in any case I agree completely with you. @Vahid, what do you want a stabilization functor to do? Do you have in mind K-theory? $\endgroup$– Yemon ChoiCommented Jan 14, 2013 at 23:58
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2$\begingroup$ Is it clear which tensor product should be used for Banach algebras? (For $C^*$-algebras, one secretly enjoys nuclearity of $K(H)$...) $\endgroup$– Alain ValetteCommented Jan 15, 2013 at 17:58
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1$\begingroup$ @Vahid: not being a C-star algebraist, nor a (NC) geometer, I'm afraid I still don't understand what is meant to be special about tensoring with $K(H)$. I mean, why not tensor with $C$? What is it one gains in the C-star world by tensoring with $K(H)$? Is it important to you that the stabilization of the stabilization of A is the stabilization of A? It would help Banach-algebra people like myself if you could add some more specific sub-questions or requirements to your original question. $\endgroup$– Yemon ChoiCommented Jan 16, 2013 at 19:37
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1$\begingroup$ @Yemon: I list some instances that clarify the importance of stabilization in $C^*$-algebras: 1. Both K-theory and KK-theory are stable functors meaning $K(A)\simeq K(A\otimes K(H))$. 2. Two separable $C^*$-algebras $A$ and $B$ are Morita equivalent if and only if they are stably isomorphic, i.e. $A\otimes K(H)\simeq B\otimes K(H)$. 3. Tensoring by $K(L^2(G))$ also appears in some theorems too, for example see Takai-Takesaki duality. So, it is nice to have a similar notion in Banach algebras, for instance proving item 2 for Banach algebras would be a good start. $\endgroup$– user23860Commented Jan 17, 2013 at 5:58
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1$\begingroup$ @Yemon: You are welcome. Following Vincent Lafforgue's works, Walter Paravicini has studied Morita equivalence of Banach algebras too. $\endgroup$– user23860Commented Jan 25, 2013 at 20:58
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2 Answers
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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.
answered Jan 15, 2013 at 12:36
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$\begingroup$ Thanks Ulrich. I guess one can choose is $K(H)$ for stabilization of Banach algebras too. But I am not sure if this gives rise to similar theorems about Morita equivalence and $K$-theory of Banach algebras. So, it seems there is no obvious choose "known yet" and it is open for more investigations. $\endgroup$– user23860Commented Jan 15, 2013 at 12:44
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$\begingroup$ Out of curiosity, Ulrich: *what is it that one gains by tensoring a given $C^\ast$-algebra with $K(H)$? (not a rhetorical question; I am genuinely ignorant of, or have forgotten, the reason one does this). $\endgroup$ Commented Jan 16, 2013 at 19:38
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1$\begingroup$ Yemon, one gains, for example idempotents. The group $K_0(A)$ can be constructed looking at the projections of $A\otimes K$ (looking at $\bigcup_n M_n(A)$ also works here, but passing to the completion may be more appealing at times). One also looses after tensoring with $K$ (which can be a good thing). E.g., Brown's theorem says that if $a,b\in A$ are positive then thr C*-algebras $\overline{aAa}$ and $\overline{bAb}$ become isomorphic after tensoring with $K$ if and only if $a$ and $b$ generate the same closed two-sided ideal. $\endgroup$ Commented Jan 17, 2013 at 17:05
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$\begingroup$ Stabilization with K is also interesting from the point of view of Hilbert bimodules, since the Brown-Green-Rieffel theorem says that two C*-algebras with countable approximate identities are strongly Morita equivalent if and only if they are stably isomorphic. Apart from that we have an exact sequence $0 \to Inn(A \otimes \mathbb{K}) \to Aut(A \otimes \mathbb{K}) \to Pic(A \otimes \mathbb{K}) \to 0$. $\endgroup$ Commented Jan 17, 2013 at 18:22
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$\begingroup$ Oops, I just realized my last comment was mentioned above already. $\endgroup$ Commented Jan 17, 2013 at 18:26
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There is now a theory of $L^p$-operator algebras, defined mainly by Chris Phillips. (See for example http://arxiv.org/pdf/1309.6406.pdf.) The objects there considered are norm-closed subalgebras of $B(L^p(X,\mu))$ for $\sigma$-unital spaces $(X,\mu)$. (Note that even when $p=2$, these are not necessarily $C^*$-algebras.) Of course not every Banach algebra has this form, but they are a considerable generalization of C*-algebras (or operator algebras acting on Hilbert spaces), and one can use them as a test case for what could be a general theory for Banach algebras.
In this context, what seems to be the right analog of stabilization (tensoring with $\mathcal{K}(\mathcal{H})$ in the C*-algebra case), is tensoring with the closure of the union of the finite matrices. When $p=2$, what we get is just the compact operators, and this is also the case whenever $p>1$, but for $p=1$ one gets something smaller (which does not contain all rank one projections). One reason why this seems to be the right analog is that Takai duality seems to hold with this algebra in place of the usual $\mathcal{K}(\mathcal{H})$.
In general, tensor products of Banach algebras are not so well behaved. One takes the algebraic tensor product first, and then one should complete in a suitable norm. Even in the $C^*$-algebra case there is more than one choice, but the situation is much worse even in the $L^p$-operator algebra case. Alain already mentioned the ambiguity of tensor products in $C^*$-algebras, which is not an issue if one of the algebras is nuclear (=amenable). For $L^p$-operator algebras, it is not known whether amenability is equivalent to nuclearity, and even in the presence of both, defining a tensor product is a complicated issue.
I don't know what applications you had in mind, but you can start by looking at $L^p$-operator algebras and tensor them with the closure of the union of the finite matrices, and see whether the statements you were hoping would be true, actually hold in this case.
