## Baum-Connes-like “conjecture” for $l^p$-spaces

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Let $G$ be a (discrete) group. For the Baum-Connes conjecture, one looks at the reduced group $C^{\star}$-algebra: Look at the Hilbert space $l^2(G)$ and the representation of $G$ on this Hilbert space given by left multiplication. The norm-closure of the resulting $\mathbb{C}G$-representation in $B(l^2(G))$ is the reduced group $C^*$-algebra.

For any $p \geq 1$, we can pretty much do the same: Look at $l^p(G)$. We still have a representation of $G$ on $B(l^p(G))$ by left multiplication and hence obtain a kind of reduced Banach group algebra for $l^p$; lets call it $B^p(G)$.

There also should be an assembly map

$K_*(E_{Fin}G) \rightarrow K_*(B^p(G))$

as in the Baum-Connes conjecture. For $p = 1$, we have $B^1(G) = l^1(G)$ and we obtain the Bost assembly map. I have no reason to believe that for arbitrary p such an assembly map might be an isomorphism, but was wondering whether such group Banach algebras, and maybe even the assembly maps, have been considered anywhere in the literature.

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I have no answer to your question but would just like to point out that what you are calling $B^p(G)$ is known in the harmonic-analysis literature as $PF_p(G)$, the algebra of $p$-pseudofunctions. These are not very well understood, for instance if $F_2$ denotes the free group on two generators and $p\notin\{1,2,\infty\}$, then I don't think it is even known if $PF_p(F_2)$ can contain non-trivial idempotents - so we don't have an `$\ell^p$-Kadison-Kaplansky', which makes me guess that getting Baum-Connes results in this setting would be substantially harder than in the $p=2$ case. – Yemon Choi Nov 27 at 23:36

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BTW. if you do find, or know of, a proof of the $\ell^p$-Kadison-Kaplansky for the free group, please let me know :) – Yemon Choi Nov 27 at 23:48

## 2 Answers

It is likely that in cases where I proved Baum-Connes without coefficients (i.e. reductive groups over local fields and some discrete groups with RD), some variant of the Schwartz or Jolissaint algebra will be dense and stable under functional calculus in the algebra you call $B^p(G)$. This would imply the BC conjecture for it. In the case of Schwartz algebras, some arguments like this (with almost the right L^p estimates) are given in the last section of my paper in Inventiones. The big limitation of this $L^p$ variant of the Baum-Connes conjecture is that when you consider coefficients in a $G$-$C^*$-algebra A, $L^p(G,A)$ can be defined only in a naive way and cannot be a $A$-Hilbert module as in the case where p=2.

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Many thanks for the suggestions. One timid remark: functional calculus in $PF_p(F_2)$ may be tricky, we know there is a self-adjoint element of the complex group ring of $F_2$ whose spectrum in the algebra $\ell^1(F_2)$ has non-empty interior, although I admit this does not rule out some Jolissaint-type algebra working – Yemon Choi Dec 2 at 19:05

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Thanks for your answer. I do not really understand the last remark, though: Why should I expect/want $L^p(G,A)$ to be a Hilbert module? Even for $A = \mathbb{\IC}$ this does not work - or am i misunderstanding something? – Fabian Lenhardt Dec 5 at 9:28

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There is a version of $KK$-theory for Banach algebras, which was developed by Lafforgue. There also is a paper titled Banach KK-theory and the Baum-Connes conjecture, which is probably relevant for this question. I think this is a survey of K-théorie bivariante pour les algèbres de Banach et conjecture de Baum-Connes.

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Lafforgue's work is only directly useful if one deals with unconditional completions of $\mathbb{C}G$ to a Banach algebra: the norm on $\mathbb{C}G$ of, say, $\sum a_g g$ should only depend on $\right\vert a_g \left\vert$, not on the precise value. I don't think that for $p \neq 1$, any of the algebras I have learned are named $PF_p(G)$ are unconditional. – Fabian Lenhardt Nov 28 at 13:48
@Fabian: I also suspect that they are not unconditional in the sense you describe, even for G the group of integers – Yemon Choi Nov 28 at 16:11
@Yemon, Fabian: You are right, I should have read the question more carefully. – Ulrich Pennig Nov 28 at 17:37
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