The decomposition map is independent of the stable lattice

By [A1], after replacing $(L,L^{\mathrm{\prime }})$ by $({\pi }^{r}L,L^{\mathrm{\prime }})$ if necessary, we may assume $L\subseteq L^{\mathrm{\prime }}$. Multiplication by the scalar ${\pi }^r$ does not change the class of the reduction in the Grothendieck group, because ${\pi }^{r}L\mathrm{/}\mathfrak{m}{\pi }^{r}L\cong L\mathrm{/}\mathfrak{m}L$ as $kG$-modules.

Put $A:=L^{\mathrm{\prime }}\mathrm{/}L$. Reduction of the inclusion $L\hookrightarrow L^{\mathrm{\prime }}$ has kernel $(L\cap \pi L^{\mathrm{\prime }})\mathrm{/}\pi L$ and cokernel $L^{\mathrm{\prime }}\mathrm{/}(L+\pi L^{\mathrm{\prime }})$. Multiplication by $\pi$ identifies the kernel with $A[\pi ]:=\{a\in A:\pi a=0\}$, while the cokernel is $A\mathrm{/}\pi A$. Thus $[L\mathrm{/}\pi L]-[L^{\mathrm{\prime }}\mathrm{/}\pi L^{\mathrm{\prime }}]=[A[\pi ]]-[A\mathrm{/}\pi A]$ in $R_k(G)$.

Multiplication by $\pi$ on the finite-length $\mathcal{O}G$-module $A$ gives exact sequences $0\to A[\pi ]\to A\to \pi A\to 0,0\to \pi A\to A\to A\mathrm{/}\pi A\to 0.$ Their Grothendieck-group identities give $[A[\pi ]]=[A\mathrm{/}\pi A]$. Both end terms are annihilated by $\pi$, so this equality is an equality in $R_k(G)$. Step 2.1 therefore gives $[L\mathrm{/}\mathfrak{m}L]=[L^{\mathrm{\prime }}\mathrm{/}\mathfrak{m}{L}^{\mathrm{\prime }}]$.

Hence the Grothendieck-group class in [F1] is independent of the stable lattice.