Let $H \to G$ be an inclusion of abelian groups, and let $R$ be a ${\Bbb Z}[H]$-algebra. Assume that the assembly map ${\Bbb S}[BG] \otimes_{\Bbb S} K(R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G]) \to K((R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G])[G])$ is an equivalence.
This assumption is not quite applicable to $K(R \otimes_{{\Bbb Z}[H]} ({\Bbb Z}[G] \otimes {\Bbb Z}[G]))$, where ${\Bbb Z}[G] \otimes {\Bbb Z}[G]$ is given the diagonal ${\Bbb Z}[H]$-action. Nonetheless, we can compose with the shearing isomorphism $G \oplus G \to G \oplus G, (g, g') \mapsto (g, g^{-1}g')$ to obtain an isomorphism to $K((R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G])[G])$.
I was wondering if there is a type of "shearing map" on ${\Bbb S}[BG] \otimes_{\Bbb S} K(R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G])$ mirroring this under assembly. I am particularly interested in a construction where the resulting map
$$K(R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G]) \xrightarrow{K({\rm id} \otimes \Delta)} K(R \otimes_{{\Bbb Z}[H]} ({\Bbb Z}[G] \otimes {\Bbb Z}[G])) \\ \simeq {\Bbb S}[BG] \otimes_{\Bbb S} K(R \otimes_{{\Bbb Z}[H]} {\Bbb Z}[G])$$ is not homotopic to the canonical map induced by the inclusion of the trivial group.
My groups are not finite, but if there is some clever construction involving Barratt-Priddy-Quillen-Segal (or anything else) in this case I would be happy to see it.
