Is the Brauer correspondence injective ?

Question

Let $G$ be a finite group and $k$ a field. In modular representation theory the Brauer correspondence establishes a bijection between the isomorphism classes of indecomposable $kG$-modules with trivial source and vertex $P$ and the isomorphism classes of non-zero projective indecomposable $k[N_G(P)/P]$-modules. This bijection is induced by the Brauer map $$M \mapsto \operatorname{Br}_P(M) := \frac{M^P}{\sum_Q tr_{Q}^P (M^Q)}$$ $(Q < P)$ where $M^P$ are the $P$-invariants and $tr^P_Q: M^Q \to M^P,\;\; m \mapsto \sum_{g \in P/Q}gm$ is the trace map. This was proved in Theorem 3.2 of the paper

M. Broué: On Scott Modules and p-Permutation Modules: An Approach through the Brauer Morphism. Proc. Amer. Math. Soc. 93(1985), 401-408

and the statement can also be found in this paper of M. Wildon (end of p. 12).

However, to my understanding, Broué's proof doesn't show that if $M_1,M_2$ aren't ismorphic then $\operatorname{Br}_P(M_1), \operatorname{Br}_P(M_2)$ aren't isomorphic as well (as $k[N_G(P)/P]$-modules).

Question: Does someone know, if the Brauer correspondence $[M] \mapsto [\operatorname{Br}_P(M)]$ ($[-]$ denotes isomorphism classes) is injective ? If so, a proof or a reference which contains a proof is also welcome.

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Let me just explain why I think Broué's proof doesn't show injectivity in full strenght: He decomposes $kG \otimes_{kP}k = \bigoplus_{i=1}^n M_i$ into indecomposable $kG$-Modules whereby the ones with vertex $P$ are exactly those such that $\operatorname{Br}_P(M_i) \neq 0$. Assume this holds for $i=1,...m$. Then $$k[N_G(P)/P] = \operatorname{Br}_P(kG \otimes_Pk)= \bigoplus_{i=1}^m \operatorname{Br}_P(M_i)$$ and the $\operatorname{Br}_P(M_i),\;(i=1,...m)$ can be shown to be indecomposable. Since every indecomposable projective $k[N_G(P)/P]$-module is isomorphic to a direct summand of $k[N_G(P)/P]$, the Brauer correspondence is surely surjective.

However, to my understanding, it may happen that $M_i,M_j$ aren't isomorphic, but $\operatorname{Br}_P(M_i) \cong \operatorname{Br}_P(M_j)$ (i.e. this summand has multiplicity $>1$ in $k[N_G(P)/P])$.

Answer 1

Actually the answer seems to be in the paper of Wildon you linked. Right after Lemma 1.3.7. he shows that $Br_P$ coincides with the Green correspondence on indecomposable direct summands of $1\uparrow _P^G$ (which are exactly your modules $M_i$). Since Green corrensondence establishes a bijection between isomorphism classes of $kG$-modules with vertex $P$ and isomorphism classes of $kN_G(P)$-modules with vertex $P$, injectivity of $Br_P$ follows (surjectivity doesn't follow immediately from this argument, because we restricted the domain of definition of the Green correspondence to trivial source modules).

Answer 2

As Geoff Robinson's answer makes clear, the Brauer map is injective when restricted to indecomposable $p$-permutation $kG$-modules with vertex $P$, because it agrees with the Green correspondence. This is stated in Theorem 3.4 of the paper by Broué cited in the question. Natalie's answer mentions Thévenaz's book, which also has a proof of this fact.

If we allow non $p$-permutation modules (but still require fixed vertex $P$) then the map is not injective. For example, take $G = P = \langle g \rangle \cong C_8$ and let $k = \mathbf{F}_2$. Let $V_k$ be the unique $k$-dimensional indecomposable representation of $G$, for $1 \le k \le 8$. Then $V_3$ has a basis $v_1,v_2,v_3$ such that $v_i(g+1) = v_{i+1}$ for $i=1,2$ and $v_3(g+1) = 0$. Clearly $V_3^P = v_3$. Since

$$v_2g^2 = v_2 + v_2(1+g^2) = v_2 + v_2(1+g)^2 = v_2$$

and $\mathrm{Tr}_{\langle g^2 \rangle}^P v_2 = v_2(1+g) = v_3$, we have $\mathrm{Br}_P(V_3) = 0$. A similar argument shows that $\mathrm{Br}_P(V_5) = \mathrm{Br}_P(V_7) = 0$. But since $V_3$, $V_5$ and $V_7$ have odd dimension, they all have full vertex $P$.

There is a related example in Example 1.3.6 of the note of mine linked in the question.

Answer 3

I think that Broué was trying to give a different proof of something that was previously know to specialists via Green Correspondence. Green correspondence tells us that if we induce an indecomposable module with vertex $P$ from $N_{G}(P)$ to $G$, then there is a unique indecomposable summand with vertex $P$ (and necessarily with the same source). It also tells us that if we restrict an indecomposable module with vertex $P$ from $G$ to $N_{G}(P)$, then there is a unique indecomposable summand with vertex $P$ (again with the same source). Since the projective indecomposable $kN_{G}(P)/P$-module are precisely the indecomposable $kN_{G}(P)$ modules with vertex $P$ and trivial source, the claimed bijection holds. This accords with what is quoted in Florian's answer from Wildon's book. The projective indecomposable $kN_{G}(P)/P$-modules are the indecomposable summands of ${\rm Ind}_{P}^{N_{G}(P)}(k)$, and if we apply the Brauer map ${\rm Br}_{P}$ to the permutation module ${\rm Ind}_{P}^{G}(k)$, we get precisely ${\rm Ind}_{P}^{N_{G}(P)}(k)$.` Further comment: The point is that when ${\rm Br}_{P}(M_{i})$ is induced from $N_{G}(P)$ back up to $G$, $M_{i}$ occurs as a summand of the induced module, and it is the only summand of the induced module with vertex $P.$ Hence if ${\rm Br}_{P}(M_{i}) \cong {\rm Br}_{P}(M_{j})$, we must have $M_{i} \cong M_{j}.$

Answer 4

I carefully disagree with the above answers.

Take two $kG$-modules with trivial source (i.e. p-permutation) modules $M_1,M_2$ which are not isomorphic and whose vertices are $Q_i$ such that $Q_i$ is a proper subgroup of $P$ for $i=1,2$. Then $Br_P(M_i)=0$ for $i=1,2$ (see (1.3) in Broue's Paper you mentioned or Cor. (27.7) in Thevenaz' book "$G$-algebras and modular representation theory"). So the answer to your question is no. You might also look at M. Cabanes Paper "Brauer Morphisms between Modular Hecke Algebras", Section A.

But I think the above arguments hold if you consider modules with the same vertex (up to conjugation) such that their images under $Br$ are isomorphic (and maybe not zero?).