Return to 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?

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?

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 $$ M(P):= M^P/\left( J(\mathcal{O}M^P + \sum_{Q<P}\mathrm{tr}_Q^P(M^Q)\right) $$

where $tr_Q^P:M^Q \to M^P$ 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?

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?

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 $$ M(P):= M^P/\left( J(\mathcal{O})M^P + \sum_{Q<P}\mathrm{tr}_Q^P(M^Q)\right) $$

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

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

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?

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 $$ M(P):= M^P/\left( J(\mathcal{O}M^P + \sum_{Q<P}\mathrm{tr}_Q^P(M^Q)\right) $$

where $tr_Q^P:M^Q \to M^P$ 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?