Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This is settled in $p$-modular representation theory of finite groups and finite dim. algebras. So we let $k$ be an algebraically closed field of characteristic $p>0$ and $G$ a finite group. A module with trivial source (TS module for short) is an indecomposable direct summand of the induced module $k_Q^G$ where $Q$ is any $p$-subgroup of $G$. Note that $k_Q^G$ is the tensor product $k_Q\otimes kG$ .

Projective indecomposable $kG$-modules (PIMs for short) can be shown to have trivial source (namely as direct summands of $k_1^G$). My question is if they just ''happen'' to have trivial source as well, or if PIMs and other modules with trivial source have even more structure in common.

To put the question from another point of view: In other finite dimensional algebras over $k$, can we find analogues of modules with trivial source in the sense that they are connected to projective modules in a similar (to me unknown) way as they are in group rings?

share|improve this question
add comment

1 Answer

I wouldn't say that projective indecomposable modules "just 'happen' to have trivial source". In fact there is a bijection between the isomorphism classes of indecomposable $kG$-modules with trivial source and vertex $P$ and the isomorphism classes of projective indecomposable $k[N_G(P)/P]$-modules. This bijection is called Brauer correspondence and is given by $$M \mapsto \operatorname{Br}_P(M) := \frac{M^P}{\sum_Q tr_{Q}^P (M^Q)}$$ $(Q < P)$ 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 can be found in Theorem 3.2(3) 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

(note that the indecomposable p-permutation modules are just the indecomposable modules with trivial source).

As a slight generalization, if $A$ is any finite dim. augmented $k$-algebra you may define $M$ to have trivial source if it is isomorphic to an indecomposable direct summand of $A\otimes_B k$ for some subalgebra $B \le A$. Then finite dimensional projective indecomposable $A$-modules have trivial source (choose $B=k$).

share|improve this answer
    
I like the last paragraph of your answer in particular. Thanks. –  Natalie Jun 9 '12 at 12:30
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.