Maschke’s Theorem

Maschke’s theorem is a fundamental result that will make our project of understanding the representation theory of finite groups — and of symmetric groups in particular — far simpler. It tells us that every representation of a finite group is completely reducible.

We saw last time that in the presence of an invariant form, any reducible representation is decomposable, and so any representation with an invariant form is completely reducible. Maschke’s theorem works by showing that there is always an invariant form!

Let’s start by picking any form whatsoever. We know that we can do this by picking a basis $\{e_i\}$ of $V$ and declaring it to be orthonormal. We don’t anything fancy like Gram-Schmidt, which is used to find orthonormal bases for a given inner product. No, we just define our inner product by saying that $\langle e_i,e_j\rangle=\delta_{i,j}$ — the Kronecker delta, with value $1$ when its indices are the same and $0$ otherwise — and extend the only way we can. If we have $v=\sum v^ie_i$ and $w=\sum w^je_j$ then we find

$\displaystyle\begin{aligned}\langle v,w\rangle&=\left\langle\sum\limits_{i=1}^{\dim(V)}v^ie_i,\sum\limits_{j=1}^{\dim(V)}w^je_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\left\langle v^ie_i,w^je_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\overline{v^i}w^j\left\langle e_i,e_j\right\rangle\\&=\sum\limits_{i=1}^{\dim(V)}\sum\limits_{j=1}^{\dim(V)}\overline{v^i}w^j\delta_{i,j}\\&=\sum\limits_{i=1}^{\dim(V)}\overline{v^i}w^i\end{aligned}$

so this does uniquely define an inner product. But there’s no reason at all to believe it’s $G$ -invariant.

We will use this arbitrary form to build an invariant form by a process of averaging. For any vectors $v$ and $w$ , define

$\displaystyle\langle v,w\rangle_G=\sum\limits_{g\in G}\langle gv,gw\rangle$

Showing that this satisfies the definition of an inner product is a straightforward exercise. As for invariance, we want to show that for any $h\in G$ we have $\langle hv,hw\rangle_G=\langle v,w\rangle_G$ . Indeed:

$\displaystyle\begin{aligned}\langle hv,hw\rangle_G&=\sum\limits_{g\in G}\langle ghv,ghw\rangle\\&=\sum\limits_{k\in G}\langle kv,kw\rangle\\&=\langle v,w\rangle_G\end{aligned}$

where the essential second equality follows because as $g$ ranges over $G$ , the product $k=gh$ ranges over $G$ as well, just in a different order.

And so we conclude that if $V$ is a representation of $G$ then we can take any inner product whatsoever on $V$ and “average” it to obtain an invariant form. Then with this invariant form in hand, we know that $V$ is completely reducible.

Why doesn’t this work for our counterexample representation of $\mathbb{Z}$ ? Because the group $\mathbb{Z}$ is infinite, and so the averaging process breaks down. This approach only works for finite groups, where the average over all $g\in G$ only involves a finite sum.

September 28, 2010 - Posted by John Armstrong | Algebra, Group theory, Representation Theory

14 Comments »

That “only” part of your last sentence is quite wrong. The averaging process works any time you have a sense of integration over the group and that actually happens pretty often thanks to the measure theory (e.g. Haar measures for compact groups, Pontryagin duals for any LCA group, etc.).

The more important point I’d like to see here (and I think it hasn’t been mentioned yet) is that the underlying field of the vector space matters _a lot_. As in, any irreducible complex representation of any abelian group is one-dimensional, but this need not happen for real representations (consider two dimensional real irrep. for SO(2) and compare that with unitary representation of U(1)) even though they are compact (or even finite). Of course this can be taken much further by taking a field with positive characteristic (and I guess one could venture quite far into number theory down this road) and I hope you’ll mention some of that when discussing reps. of the symmetric group.

Last but not least, though I think I again sound quite disapproving, your review of representation theory is much appreciated and I hope you’ll continue writing. It’s just that I don’t have anything to comment on usually as it’s completely correct 🙂

Comment by Marek | September 28, 2010 | Reply
- You’re jumping ahead, but still it’s true that the averaging process described above doesn’t work for the cases you suggest. They use a different averaging process to achieve the same result, and use compactness to stand in for finiteness!
  
  And yes, the underlying field matters. But I’ve stated all along that I’m only dealing with complex representations here.
  
  Comment by John Armstrong | September 28, 2010 | Reply
  - It’s certainly not a different process, it’s the one and the same. Just consider any finite group with discrete topology, i.e. any function on the group is continuos. Then the invariant (Haar) integration with respect to this tepology is just summing over the elements of the group which is precisely what you are doing here. Finite groups are just a small part of a much bigger picture. Although it’s true that you don’t need all the measure and topologic machinery to work with finite groups, I think at least a little notion about the generalization to compact (even locally compact) groups should be stated.
    
    Oh, all right, must have slipped my mind. But I still hope you’ll point some differences at least between real and complex cases.
    
    Comment by Marek | September 28, 2010 | Reply
    - And when I come to that generalization I’ll tie back here.
      
      Comment by John Armstrong | September 28, 2010 | Reply
I think that you need to change the definition of to
_G = (1/|G|) \sum

to get the averaging process you’re describing.

Comment by Dan | September 29, 2010 | Reply
- i mean {v,w}_G = (1/|G|) \sum_{g \in G} {gv,gw} where {v,w} is the arbitrary form you defined. in particular, isn’t this the thing that sinks Maschke’s theorem in the case where the characteristic of the field divides |G|? I realize that you’re just working over the complex / characteristic zero but it seems worth mentioning in generality
  
  Comment by Dan | September 29, 2010 | Reply
  - Yes, dividing by $\lvert G\rvert$ is what can go wrong in finite characteristic. That’s why I didn’t do it here! But the great thing is that we don’t need any particular “average” inner product so long as whatever we get is $G$ -invariant.
    
    Comment by John Armstrong | September 29, 2010 | Reply
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