What goes wrong with the Brauer construction for a module over a complete DVR?

Question

Let $G$ be a finite group, $\mathcal{O}$ a complete discrete valuation ring, and $F$ the residue field. The Brauer construction for a fintely-generated $\mathcal{O}G$-module $M$ with respect to a $p$-subgroup $P$ is defined to be the $F[N_G(P)/P]$-module $$ M(P):= M^P/\left( J(\mathcal{O})M^P + \sum_{Q<P}\operatorname{tr}_Q^P(M^Q)\right) $$

where $\operatorname{tr}_Q^P:M^Q \to M^P$ is the relative trace map. The Brauer construction is an additive functor which fits into the following diagram which commutes up to natural isomorphism:

$$\require{AMScd}\begin{CD} {}_G\mathbf{set} @>(-)^P>> {}_{N_G(P)/P}\mathbf{set} \\ @V\mathcal O[-]VV @VV F[-]V \\ {}_{\mathcal OG}\mathbf{triv} @>>(-)(P)> {}_{F[N_G(P)/P]}\mathbf{triv} \end{CD}$$

where $\mathbf{triv}$ is the category of $p$-permutation modules.

I remember reading somewhere that the Brauer construction "does not work" over $\mathcal{O}$. I think they meant that there is no functor making the diagram commute if you replace the $F$ by $\mathcal{O}$. Does anyone know what goes wrong?

Answer 1

I presume you want $\mathcal{O}$ to have characteristic zero and $F$ to have characteristic $p$.

Consider the case where $p=2$, $G=C_2$ and $P=G$. If $\mathcal{O}$ is the trivial $\mathcal{O}G$-module, then there are module homomorphisms $\mathcal{O}\to\mathcal{O}G\to\mathcal{O}$ composing to multiplication by $2$.

If there were a functorial Brauer construction defined over $\mathcal{O}$ then this would induce maps $\mathcal{O}(P)\to\mathcal{O}G(P)\to\mathcal{O}(P)$ composing to multiplication by $2$. But $\mathcal{O}(P)$ would be $\mathcal{O}$ and $\mathcal{O}G(P)$ would be $0$, so this is impossible.