The passage from the integral to the rational group ring in algebraic $K$-theory
Georg Lehner
TL;DR
The paper resolves a standing question by showing that the map $\widetilde{K_0 }\mathbb{Z}G \to \widetilde{K_0 }\mathbb{Q}G$ need not be trivial, even under Farrell–Jones-type assumptions. It develops an assembly-based, representation-theoretic framework using the Whitehead spectrum $\mathbf{Wh}(R;G)$ and the singular-character group $\text{SC}(G)$ to compute the image via $\ker$-groups tied to $\widetilde{K_0 }\mathbb{Q}G$, $\text{SC}(G)$, and $K_{-1}\mathbb{Z}G$. The authors supply explicit finite- and virtually cyclic-case calculations and culminate with a concrete counterexample $G = QD_{32} *_{Q_{16}} QD_{32}$ in which the image contains a nontrivial $\mathbb{Z}/2$-summand, demonstrated with detailed group-ring data (Schur indices, Wedderburn decomposition) and GAP calculations. This work links deep representation-theoretic invariants to assembly maps in algebraic $K$-theory, highlighting that integral obstructions can persist in infinite groups and providing a practical method to quantify them. The results have implications for Wall obstructions and Bass-type conjectures, illustrating how subtle torsion can arise in $K_0$ comparisons between $\mathbb{Z}G$ and $\mathbb{Q}G$ via the singular-character framework.
Abstract
An open question is whether the map $\widetilde{K_0 }\mathbb{Z} G \rightarrow \widetilde{K_0 }\mathbb{Q} G$ in reduced $K$-theory from the integral to the rational group ring is trivial for any group $G$. We will show that this is false, with a counterexample given by the group $QD_{32} *_{Q_{16}} QD_{32}$. We will also show how to compute the image of the map $\widetilde{K_0 }\mathbb{Z} G \rightarrow \widetilde{K_0 }\mathbb{Q} G$ using representation theoretic means, assuming $G$ satisfies the Farrell-Jones conjecture.