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Let $G$ be a group of permutations of a finite set $X$. By the augmentation submodule of $\mathbb CX$, I mean the set of vectors whose coefficients sum to $0$. It is easy to show via character theory that the augmentation submodule is irreducible iff $G$ is doubly transitive. Does anybody know a proof using only the definitions of doubly transitive and irreducible representation and avoiding character theory and the orthogonality relations? It is known (I learned this from Peter Cameron, but don't know a good reference) that replacing $\mathbb C$ by $\mathbb R$ characterizes double homogeneity. So the field being algebraically closed is somehow important.

The motivation for this question comes from trying to understand the relationship between double transitivity for transformation monoids and irreducibility of the augmentation submodule. Character theory for monoids is harder to apply and so an answer to my question may provide some insight.

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Doubly transitive implies irreducible is elementary; there's a very straightforward argument that involves doing obvious things to any nonzero vector. I don't know about the other direction. –  Qiaochu Yuan Jun 23 '11 at 20:23
    
I agree doubly transitive implies irreducible is elementary, although the proof I have in mind at first sight fails for monoids. Basically, if you have a vector of the form $x-y$, you are done. If you don't, you can use transitivity of point stabilizers to kill off an appropriate element from the support of your vector. Proceed by induction on the support. This doesn't work for monoids because point stabilizers can crush elements into the point it stabilizes, although perhaps the idea is salvageable? –  Benjamin Steinberg Jun 23 '11 at 20:57
    
Yes, that's the proof I had in mind. Do you know if either direction is even true for monoids? –  Qiaochu Yuan Jun 24 '11 at 1:41
    
@Qiaochu, no I do not. But I can prove that monoids with either property have a lot in common. For instance, in either case if the monoid is not a group, then it contains all constant maps. From this, one obtains that the analogue of the permutation module is a projective indecomposable module with simple top the trivial module. In both cases, if $e$ is a nonconstant idempotent of smallest possible image, then the associated maximal subgroup is doubly transitive on eX (where X is the set the monoid acts on). I can't remember if I could prove the augmentation is indecomposable. –  Benjamin Steinberg Jun 24 '11 at 2:37
    
Also, in both cases in the above comment the transformation monoid is primitive. –  Benjamin Steinberg Jun 24 '11 at 2:39
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I am not sure where you want to draw the line about use of character theory. This equivalence is a statement about the endomorphism ring of the permutation module $\mathbb{C}X$, and doesn't really require use of characters or orthogonality relations. It does require Schur's lemma, which crucially uses algebraic closure. Using Schur's Lemma, the endomorphism ring of the permutation module is $2$-dimensional if and only if the action of $G$ on $X$ is transitive and the augmentation submodule is irreducible. On the other hand, if the acton of $G$ on $X$ is transitive and $H$ is a point stabilizer, it is just a question of looking at matrices to see that the dimension of the endmorphism ring is the number of $(H,H)$-double cosets. (These things can be proved by Frobenius reciprocity, but they can be seen directly in this case).

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Let me rephrase your argument. The centralizer algebra has basis in bijection with orbitals (this is easy). Doubly transitive means there are exactly 2 orbitals. The centralizer algebra has dimension 2 iff the augmentation is irreducible. Ok. In some sense, this is the same as the character proof since the dimension of the centralizer algebra is the inner product of the character with itself. I guess I should have instead asked for a proof that doesn't use the semisimplicity of the group algebra explicitly if possible. Maybe there is no such proof. Centralizers for monoid reps don't work. –  Benjamin Steinberg Jun 23 '11 at 22:09
    
OK, thanks. Now I understand the issues that worry you a little better. –  Geoff Robinson Jun 23 '11 at 23:04
    
@Geoff, my dream I guess is a proof which given a non-doubly transitive action "constructs" an invariant subspace. For example, if the permutation group is not primitive, then the subspace spanned by the differences of elements in some fixed block of the system of imprimitivity is an invariant subspace. This shows irreducible augmentation implies primitive. This argument works over Q. I don't expect something this simple since it is necessary to use algebraically cloded –  Benjamin Steinberg Jun 24 '11 at 1:09
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