# Is the Brauer correspondence injective ?

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.

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 $k[N_G(P)/P])$.

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I'd also love to know in what context you are studying this paper. – Natalie Jun 9 '12 at 12:27
@Natalie: I learned about Brauer correspondence when looking into Wildon's paper on modular representations of the symmetric group. In order to get some more background for this section I googled for "Brauer correspondence" and found Broué's paper. – Ralph Jun 14 '12 at 19:02

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).

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@Florian: I think Green correspondence can also be used to show surjectivity: We can view an indecomposable projective $k[N_G(P)/P]$-module $N$ as $k[N_G(P)]$-module with vertex $P$ and trivial source. So $N$ is a Green correspondent $g(M)$ of some indecomposable $kG$-module $M$ with vertex $P$. But since Green correspondence respects sources, $M$ has trivial source. Hence we obtain $Br_P(M)=g(M)=N$. – Ralph Jun 14 '12 at 19:04
Since I want to save this html document and it fits here, let me include a proof that an indec. proj. $k[N/P]$-module $F$ with $N=N_G(P)$ has as $kN$-module vertex $P$ and trivial source: Since $F$ is an indec. direct summand of the permutation module $k[N/P]=kN\otimes_{kP}k$ it has trivial source (Benson: Repr. and Coh. I, 3.11.2) and is $kP$-projective. Thus its vertex $Q$ is conjugated to a subgroup of $P$. Moreover, $\operatorname{res}^N_P F$ is a direkt summand of $$\operatorname{res}^N_P \operatorname{ind}^N_Q k = \oplus_g \operatorname{ind}^P_{P\cap gQg^{-1}}k=: \oplus_g L_g.$$ – Ralph Jun 14 '12 at 19:05
(cont.) But $P$ acts trivially on $F$, so $\operatorname{res}^N_P F$ is a direkt sum of copies of $k$. Since $P$ is a $p$-group, $L_g$ is indec. by a theorem of Green. Hence there is a $g \in N$ with $k=L_g$ and from $\dim L_g=(P:P \cap gQg^{-1})$ we conclude $P\le gQg {-1}$ which implies $Q=g^{-1}Pg$. – Ralph Jun 14 '12 at 19:06

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.

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@Mark: Thanks for the nice example. Also the section on the Brauer correspondence in your paper gives a very good introduction in this topic (and an elegant proof for the bijectivity of the Brauer correspondence). – Ralph Jun 14 '12 at 19:03

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 $M_{i} \cong M_{j}.$

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Is it true that "projective indecomposable $kN_G(P)/P$-module are precisely the indecomposable $kN_G(P)$ modules with vertex $P$" as you say? What about the indecomposable $kN_G(P)$-modules, which do not have $P$ in their kernel? Even though $P$ is normal in $N_G(P)$, we just know that irreducible $kN_G(P)$-modules have $P$ in their kernel (Feit p102) but not indec. modules in general. – Natalie Jun 8 '12 at 20:40
@Natalie: I missed out the important words "with trivial source", which I inteded to be there. I have re-edited- thanks for pointing that out – Geoff Robinson Jun 8 '12 at 20:51
The Brauer morpism as defined by Brou\'e in that paper is only defined for trivial source modules in any case. – Geoff Robinson Jun 8 '12 at 20:57
@Geoff: Thanks for your answer. It was equally helpfull as Florian's answer and because I can't accept both, I accepted the first one. I hope you don't mind. – Ralph Jun 14 '12 at 19:04
@Ralph: Of course I don't mind. – Geoff Robinson Jun 17 '12 at 17:23

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?).

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But the question was about modules with vertex $P$ and trivial source wasn't it? It is certainly true that things can go wrong if you move away from indecomposable (trivial source) modules with vertex $P$. – Geoff Robinson Jun 8 '12 at 20:58
In the description of the Brauer correspondence (B.C.) in the question, $M$ is required to have vertex $P$ (as stated by Geoff): "... with trivial source and vertex $P$ ...". Then, of course, $\operatorname{Br}_P(M) \neq 0$, i.e. the B.C. misses $0$ as projective indecomposable $k[N_G(P)/P]$-module. This might have led to confusion. I will edit the question accordingly when I have worked through the answers. Anyway, thanks for the hint and the references. – Ralph Jun 8 '12 at 21:09