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Divisibility theorems for group representations II October 14, 2009

Posted by Akhil Mathew in algebra, representation theory.
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So last time we proved that the dimensions of an irreducible representation divide the index of the center. Now to generalize this to an arbitrary abelian normal subgroup.

There are first a few basic background results that I need to talk about. 

Induction  

Given a group {G} and a subgroup {H} (in fact, this can be generalized to a non-monomorphic map {H \rightarrow G}), a representation of {G} yields by restriction a representation of {H}. One obtains a functor {\mathrm{Res}^G_H: Rep(G) \rightarrow Rep(H)}. This functor has an adjoint, denoted by {\mathrm{Ind}_H^G: Rep(H) \rightarrow Rep(G)}.

It can be described explicitly using tensor products in a simple manner. Given an inclusion {H \rightarrow G}, there is a map {\mathbb{C}[H] \rightarrow \mathbb{C}[G]} from which one gets the restriction functor {Rep(G) \simeq \mathbf{Mod}(\mathbb{C}[G]) \rightarrow \mathbf{Mod}(\mathbb{C}[H]) \simeq Rep(H)}. As is well-known for rings {A \rightarrow B}, the adjoint to this functor is given by the tensor product:

\displaystyle \mathrm{Ind}_H^G(M) := \mathbb{C}[G] \otimes_{\mathbb{C}[H]} M.

It follows that if {N = \mathrm{Ind}_H^G(M)}, then {M} is a {H}-submodule of {N}. Also, if {G = \bigcup_i g_i H} is a coset decomposition, then

\displaystyle M = \bigoplus g_i N

as vector spaces. In particular, {\dim N = (G:H ) \dim N}. Conversely, if the above decomposition holds then {M = \mathrm{Ind}_H^G(N)}.

The adjointness relation is usually called Frobenius reciprocity, and is written as

\displaystyle \hom_G(\mathrm{Ind}_H^G(X), Y) \simeq \hom_H(X, \mathrm{Res}^G_H(Y)), \quad \forall X \in Rep(H), Y \in Rep(G). 

Restriction  

First, we discuss the restriction to abelian normal subgroups via a lemma, which will enable us to induct on {|G|}

Lemma 1 Let {G} be a group, {A} an abelian normal subgroup. Let {M} be a simple {G}-representation. Then there are two possibilities: 

  1. {A} acts by scalar matrices on {M}.
  2. There is a subgroup {H} with {A \subset H \subsetneq G} and

    \displaystyle M = \mathrm{Ind}_H^G(M')

    for some {M' \subset M}

 

A more general version of this is in Serre or Curtis-Reiner, in the latter attributed to Clifford.  The more general result drops the abelian hypothesis and replaces 1 above by “the restriction to A is isotypic, i.e. a direct sum of isomorphic simple objects.”

So, by restriction {M} is an {A}-module. By the previous post, we can write as {A}-modules

\displaystyle M = \bigoplus_{\chi \in \hom(A, \mathbb{C}^*)} M_{\chi},

where

\displaystyle M_{\chi} := \{ m \in M: am = \chi(a)m, \forall a \in A \}.

Perhaps {M_{\chi}} ought to be thought of as a weight space, as in the theory of semisimple Lie algebras, when {A} is replaced by a Cartan subalgebra. It turns out that in this case, the elements of {G} permute the spaces {M_{\chi}}: if {m \in M_{\chi}}, then {gm \in M_{\chi_g}} where {\chi_g(x) = \chi(g^{-1}xg)} because of the “fundamental calculation” (Fulton and Harris’ phrase)

\displaystyle xgm = g (g^{-1} x g )m = g \chi_g(x)m, \quad x \in H.

Now pick {\chi} with {M_{\chi} \neq 0}. If {M_{\chi}=M} then we are in the first case. Otherwise, take {H} to be the stabilizer of {M_{\chi}} and the result follows.  

Finally, the theorem   

Theorem 2 Let {G} be a finite group and {A} an abelian normal subgroup. Then each simple representation of {G} has dimension at most {|G|/|A|} 

Induction on {G}. Assume the theorem proved for smaller groups.

First of all, let’s make some reductions. Assume that {V} is a simple representation of {G} which is faithful, because otherwise we could replace {G} by the image in {Aut(V)} to get a representation of a quotient {G'} of {G}. The image {A'} of {A} in that quotient satisfies {(G':A') \mid (G:A)}, so we reduce to the faithful case.

Next, I claim that we may assume that {A} does not act by scalars on {V}. If it did, then {A} would be contained in the center of {G} (by faithfulness), in which case the theorem is already proved.

By the lemma, this means there is a subgroup {H \subsetneq G} containing {A} with {V = \mathrm{Ind}_H^G(W)} for some (necessarily simple) {W \in Rep(H)}. Then by the inductive hypothesis

\displaystyle \dim V = (G:H) \dim W \mid (G:H)(H:A),

proving the theorem. 

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Comments»

1. Todd Trimble - October 17, 2009

Hi Akhil; please excuse my contacting you this way. We really appreciate the additions you’ve made to the nLab, and we hope that you’ll also set up a user account at the nForum. That way you’ll be able to log any changes or new entries you’ve created, and people who have subscribed to the RSS feed will get an immediate heads-up and check out what you’ve done (without having to hit ‘Recently Revised’, which had given us problems in the past).

It’s pretty straightforward; just go here and hit Account to get started. You might also consider subscribing to the RSS feed.

To any others reading this: the nLab is a wiki originally created by hosts and patrons of the n-Category Cafe blog to record working notes on all aspects of mathematics. The home page describes pretty well what we’re about. We always welcome new contributors; please do not be intimidated by the content. Contributors come from a wide variety of backgrounds and with varying levels of expertise.

2. Akhil Mathew - October 17, 2009

@Todd: Thanks, I just set up the nForum account and logged the changes I made yesterday.


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