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Let $\Gamma$ be a countable (discrete) group (in what follows, make additional assumptions as you wish). Let $C^*_r(\Gamma)$ and $W^*_r(\Gamma)$ be the reduced $C^*$-algebra respectively the reduced von-Neumann-algebra of $\Gamma$, in other words, the closure of $\mathbb{C}[\Gamma] \subset B(\ell^2(\Gamma))$ with respect to the norm respectively the strong operator topology.

Now extension of scalars gives a functor $$ C^*_r(\Gamma)\text{-}{\mathrm{HMod}} \longrightarrow W^*_r(\Gamma)\text{-}{\mathrm{HMod}},$$ where for a $C^*$-algebra $A$, I write $A\text{-}\mathrm{HMod}$ for the category of finitely generated projective Hilbert-$A$-modules.

Q: What can be said about this functor?

More specifically, it is clear that free modules are sent to free modules, but in general, the preimage of free modules need not be free. Can we characterize Hilbert-$C^*_r(\Gamma)$-modules that are sent to free Hilbert-$W^*_r(\Gamma)$-modules?

More coarsely: The functor above induces a map in K-theory $$K_0(C^*_r(\Gamma)) \longrightarrow K_0(W^*_r(\Gamma)).$$ Q: What can be said about this map?


For example, if $\Gamma = \mathbb{Z}^d$, then Fourier transform induces isomorphisms $C^*_r(\Gamma) \cong C^0(T^d)$ and $W^*_r(\Gamma) = L^\infty(T^d)$, where $T^d = \mathbb{R}/\mathbb{Z}$. Finitely generated $C^0(T^d)$-modules are given by continuous sections of finite rank vector bundles $\mathcal{E}$ over $T^d$ and all modules are of this type.

Now if $\Gamma^0(T^d, \mathcal{E})$ is such a modules, its extension of scalars is the set of $L^\infty$-sections of $\mathcal{E}$. This is a free module, of course, since any vector bundle has a trivialization of non-continuous sections.

Hence all modules over $C^*_r(\mathbb{Z}^d)$ become free after extension of scalars, and the map is the zero map after passing to reduced K-theory.

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    $\begingroup$ @YemonChoi von Neumann algebras have trivial $K_1$ but, for example $K_0(M)\cong \mathbb{R}$ when $M$ is a $II_1$ factor $\endgroup$ Nov 6, 2018 at 4:00
  • $\begingroup$ @CalebEckhardt D'oh! I had B(H) and other properly infinite examples in my head, of course things with trace would be different $\endgroup$
    – Yemon Choi
    Nov 6, 2018 at 12:50
  • $\begingroup$ In the $II_1$ factor case it reduces to the image of the trace : $\tau(K_0(C_r^*(\Gamma)) \subset \mathbb R$. It is known that in some case the image is $\mathbb Z$. Take for exemple $\Gamma = F_n$ the free group. In full generallity, for any ICC amenable group i don't know. $\endgroup$ Nov 14, 2018 at 13:49

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