1
$\begingroup$

Knowing the fact that standard representation arising out of permutation representation of $A_5$ over $\mathbb{C}$ is irreducible and of degree $4$. What can we conclude about the irreducibility over general field, whose characteristics does not divide the order of $A_5$. Is it irreducible ? Can we use Clifford Theory here ? How ?

$\endgroup$
17
  • 5
    $\begingroup$ You write "whose characteristic does NOT divide the order", is that a typo? I'm asking since you also added the "modular representation theory" tag. Typically modular representation theory refers to the case where the characteristic divides the group order. In characteristic not dividing the group order, the representation theory behaves as in characteristic zero, in particular your representation is irreducible. $\endgroup$ Commented Apr 6, 2021 at 9:27
  • 2
    $\begingroup$ In other words the argument that the dimension of the endomorphism algebra of a permutation module equals the number of orbits on pairs can be carried out in nonmodular characteristic without using characters by looking at hom set dimensions and then the result follows from there. $\endgroup$ Commented Apr 6, 2021 at 9:59
  • 2
    $\begingroup$ I'm suggesting a conceptual way to do this using permutation modules and centralizer algebras or endomorphism rings that would work for An for n\geq 4 and S_n. If you want just A5 by bare hands I think math stack exchange is a more appropriate site. $\endgroup$ Commented Apr 6, 2021 at 10:16
  • 2
    $\begingroup$ As a general comment, Clifford theory is only useful in the representation theory of finite groups in the presence of proper non-trivial normal subgroups., so is not so relevant for simple groups. $\endgroup$ Commented Apr 6, 2021 at 12:31
  • 2
    $\begingroup$ @GeoffRobinson Clifford theory could be relevant if the problem was somehow easier for $S_5$. $\endgroup$
    – lambda
    Commented Apr 6, 2021 at 12:48

1 Answer 1

6
$\begingroup$

Here is a fairly general answer for the characteristic zero question you ask, which I will mark as Community Wiki:

Let $G$ be any doubly transitive permutation group of degree $n$ on a set $\Omega$. Let $V$ be the underlying permutation module, viewed as the $\mathbb{Q}G$-module $\mathbb{Q}\Omega$. Then $V \cong U \oplus W$, where $U$ is the trivial $\mathbb{Q}G$-module and $W$ is $n-1$-dimensional ($U$ may be realised as the space of $\mathbb{Q}$-linear combinations of elements of $\Omega$ with all coefficients equal, and $W$ is the set of all $\mathbb{Q}$-linear combinations of elements of $\Omega$ in which the sum of coefficients appearing is zero.

It is an easy exercise that ${\rm End}_{\mathbb{Q}G}(W) \cong \mathbb{Q}$ (this uses the double transitivity of the permutation action). This implies that the representation afforded by $W$ is absolutely irreducible- that is, remains irreducible after any extension of scalars enlarging $\mathbb{Q}$ to any extension field. Since any field $\mathbb{F}$ of characteristic zero has prime subfield isomorphic to $\mathbb{Q}$, we see that $W \otimes_{\mathbb{Q}} \mathbb{F}$ is an irreducible $\mathbb{F}G$-module.

( This is, of course, all standard theory)

Later edit: In view of some of the comments, let me point out that this argument does not always generalize to prime characteristic dividing the group order. Possibly The easiest case it fails is when the prime characteristic $p$ is a divisor of $n$, the degree of the permutation representation, a case mentioned by Derek Holt in comments.

For if $\mathbb{F}$ is a field of prime characteristic $p$ which divides $n = |\Omega|$, then the permutation module $\mathbb{F}\Omega$ has two obvious submodules $U$ and $W$ defined as before, but note that $U \subseteq W$ in this case because the coefficients appearing in any element of $U$ sum to zero. Hence $W$ is not irreducible (if $n >2$ to discount a trivial exception). We also find that $\mathbb{F}\Omega/W \cong U$. Hence the permutation module $\mathbb{F}\Omega$ has at least three composition factors, at least two of which are trivial. The $n-2$-dimensional module $W/U$ is sometimes called the Green heart (after J.A. Green). It is often irreducible, but I think examples exist when it is not.

$\endgroup$
7
  • 2
    $\begingroup$ This all works in any characteristic that doesn’t divide the order of the group with the easy exercise being only a tiny bit less easy. The centralizer algebra of the permutation module has basis the characteristic matrices of orbitals (orbits on pairs). $\endgroup$ Commented Apr 6, 2021 at 13:15
  • $\begingroup$ @DerekHolt, I think this only works when $p$ doesn't divide the order of the group. 2 divides the order of PSL_3(2). $\endgroup$ Commented Apr 6, 2021 at 13:26
  • $\begingroup$ @DerekHolt, are you sure that it is a direct sum rather than a composition series? $\endgroup$ Commented Apr 6, 2021 at 13:28
  • 1
    $\begingroup$ On the contrary, I am sure it is not! Let me start again. For a counterexample when $p$ divides $|G|$, consider the permutation representation of ${\rm SL}_3(2)$ on $7$ points, which is doubly transitive. This decomposes as $T \oplus V$ with $T$ the trivial module and $V$ indecomposable of dimension $6$, but $V$ is not irreducible - it has two mutually dual composition factors, each of dimension $3$. $\endgroup$
    – Derek Holt
    Commented Apr 6, 2021 at 13:32
  • 2
    $\begingroup$ @Geoff Thanks for so much efforts in answering. Actually the concepts of double transitivity of permutation groups and absolute irreducibility were new for me yesterday. But now, I have understood the journey from double transitivity to irreducibility by using character theory and also without using it( in non-modular cases ) with the help of stablizers. Thanks Benjamin, Derek, lambda, Achim for your time and energy. $\endgroup$
    – HIMANSHU
    Commented Apr 7, 2021 at 5:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .