Representation Theory, Part 1: Irreducibles and Maschke’s Theorem

Good luck to everyone taking the December TST tomorrow!

The goal of this post is to give the reader a taste of representation theory, a la Math 55a. In theory, this post should be accessible to anyone with a knowledge of group actions and abstract vector spaces.

Fix a ground field {k} (for all vector spaces). In this post I will introduce the concept of representations and irreducible representations. Using these basic definitions I will establish Maschke’s Theorem, which tells us that irreducibles and indecomposables are the same thing.

1. Definition and Examples

Let {G} be a group.

Definition A representation of {G} consists of a pair {\rho = (V, \cdot_\rho)} where {V} is a vector space over {k} and {\cdot_\rho} is a (left) group action of {G} on {V} which is linear in {V}. If {V} is finite-dimensional then the dimension of {\rho} is just the dimension of {V}.

Explicitly the conditions on {\cdot_\rho} are that

\displaystyle \begin{aligned} 1 \cdot_\rho v &= v \\ g_1 \cdot_\rho (g_2 \cdot_\rho v) &= (g_1g_2) \cdot_\rho v \\ g_1 \cdot_\rho (v_1 + v_2) &= g \cdot_\rho v_1 + g \cdot_\rho v_2 \\ g_1 \cdot_\rho (cv) &= c(g \cdot_\rho v). \end{aligned}

Note that another equivalent phrasing is that {\rho} is a homomorphism from {G} to the general linear group {\text{GL}(V)}; however, we will not use this phrasing.

By abuse of notation, we occasionally refer to {\rho} by just its underlying vector space {V} in the case that {\cdot_\rho} is clear from context. We may also abbreviate {g \cdot_\rho v} as just {g \cdot v}.

A simple example of a nontrivial representation is the following.

Example If {V = k^{n}} and {G = S_n}, then an example of an action is {\rho = (V, \cdot_\rho)} is simply

\displaystyle \sigma \cdot_\rho \left<x_1, \dots, x_n\right> = \left<x_{\sigma(1)}, \dots, x_{\sigma(n)}\right>

meaning we permute the basis elements of {V}. We denote this representation by {\mathrm{refl}_n}.

Let us give another useful example.

Definition Let {X} be a set acted on by {G}. We define the vector space

\displaystyle \mathrm{Fun}(X) \overset{\text{def}}{=} \left\{ \text{maps } X \rightarrow k \right\}

with the standard addition of functions.

Example We define a representation on {\mathrm{Fun}(X)} by the following action: every {f \in \mathrm{Fun}(X)} gets sent to a {g \cdot f \in \mathrm{Fun}(X)} by

\displaystyle (g \cdot_{\mathrm{Fun}(X)} f)(x) = f\left( g^{-1} \cdot_X x \right).

By abuse of notation we will let {\mathrm{Fun}(X)} refer both to the vector space and the corresponding representation.

Now that we have two nontrivial examples, we also give a trivial example.

Definition Let {G} be a group. We define the trivial representation {\text{triv}_G}, or just {\text{triv}}, as the representation {\text{triv}_G = (k, \cdot_\text{triv})}, where

\displaystyle g \cdot_\text{triv} a = a

for every {a \in k}. In other words, {G} acts trivially on {k}.

2. Homomorphisms of Representations

First, as a good budding algebraist (not really) I should define how these representations talk to each other.

Definition Let {\rho_1 = (V_1, \cdot_{\rho_1})} and {\rho_2 = (V_2, \cdot_{\rho_2})} be representations of the same group {G}. A homomorphism of {G}-representations is a linear map {T : V_1 \rightarrow V_2} which respects the {G}-action: for any {g \in G} and {v \in V},

\displaystyle g \cdot_{\rho_2} T(v) = T\left( g \cdot_{\rho_1} v \right).

The set of all these homomorphisms is written {\mathrm{Hom}_G(\rho_1, \rho_2)}, which is itself a vector space over {k}.

(Digression: For those of you that know category theory, you might realize by now that representations correspond to functors from a category {\mathcal G} (corresponding to the group {G}) into {\textbf{Vect}_k} and that homomorphisms of representations are just natural transformations.)

To see an example of this definition in action, we give the following as an exercise.

Proposition 1 Let {\rho = \left( V, \cdot_\rho \right)}. We define the {G}-invariant space {\rho^G \subseteq V} to be

\displaystyle \rho^G \overset{\text{def}}{=} \left\{ v \in V \mid g \cdot_\rho v = v \; \forall g \in G \right\}.

Then there is a natural bijection of vector spaces {\mathrm{Hom}_G(\text{triv}_G, \rho) \simeq \rho^G}.

Proof: Let {\rho = (V, \cdot_\rho)}. The set {\mathrm{Hom}_G(\text{triv}_G, \rho)} consists of maps {T : k \rightarrow V} with

\displaystyle g \cdot_\rho T(a) = T(g \cdot_{\text{triv}} a) = T(a)

for every {a \in k}. Since {T : k \rightarrow V} is linear, it is uniquely defined by {T(1)} (since {T(a) = a T(1)} in general). So {g \cdot_\rho T(1) = T(1)}, i.e. {T(1) \in \rho^G}, is necessary and sufficient. Thus the bijection is just {T \mapsto T(1)}. \Box

This proposition will come up again at the end of Part 4.

3. Subrepresentations, Irreducibles, and Maschke’s Theorem

Now suppose I’ve got a representation {\rho = (V, \cdot_\rho)}.

Definition Suppose we have a subspace {W \subseteq V} which is {\rho}-invariant, meaning that {g \cdot_\rho w \in W} for every {w \in W} and {g \in G}. Then we can construct a representation of {G} on {W} by restricting the action to {W}:

\displaystyle \rho' = \left( W, \cdot_\rho|_W \right).

In that case the resulting {\rho} is called a subrepresentation of {V}.

Every {\rho} has an obvious subrepresentation, namely {\rho} itself, as well as a stupid subrepresenation on the zero-dimensional vector space {\{0\}}. But it’s the case that some representations have interesting subrepresentations.

Example Consider the representation {\mathrm{refl} = (k^n, \cdot)} of {S_n} on {k^n} defined in the first section. For all {n \ge 2}, {\mathrm{refl}} is not irreducible.

Proof: Consider the subspace {W \subset k^n} given by

\displaystyle W = \left\{ (x_1, x_2, \dots, x_n) \mid x_1 + \dots + x_n = 0. \right\}

then {W} is invariant under {\mathrm{refl}}, so we have a subrepresentation of {\mathrm{refl}}, which we’ll denote {\mathrm{refl}_0}. \Box

This motivates the ideas of irreducibles.

Definition A representation {\rho} is irreducible if it has no nontrivial subrepresentations.

Of course the first thing we ask is whether any representation decomposes as a product of irreducible representations. But what does it mean to compose two representations, anyways? It’s just the “natural” definition with the direct sum.

Definition Let {\rho_1 = (W_1, \cdot_{\rho_1})} and {\rho_2 = (W_2, \cdot_{\rho_2})} be representations and suppose we have {V = W_1 \oplus W_2}. Then we define the representation {\rho = \rho_1 \oplus \rho_2} by {\rho = (V, \cdot_\rho)} where

\displaystyle g \cdot_\rho (w_1 + w_2) = (g \cdot_{\rho_1} w_1) + (g \cdot_{\rho_2} w_2).

Just like every integer decomposes into prime factors, we hope that every representation decomposes into irreducibles. But this is too much to hope for.

Example Let {G = S_2}, let {k = \mathbb F_2} be the finite field of order {2} (aka {{\mathbb Z}/2{\mathbb Z}}), and consider {\mathrm{refl} = (k^2, \cdot)}, which is not irreducible. However, we claim that we cannot write {\mathrm{refl} = \rho_1 \oplus \rho_2} for any nontrivial {\rho_1} and {\rho_2}.

Proof: This is a good concrete exercise.

Assume not, and let {V_1} and {V_2} be the underlying vector spaces of {\rho_1} and {\rho_2}. By nontriviality, {\dim V_1 = \dim V_2 = 1}, and in particular we have that as sets, {\left\lvert V_1 \right\rvert = \left\lvert V_2 \right\rvert = 2}. Take the only nonzero elements {(a,b) \in V_1} and {(c,d) \in V_2}. Since {V_1} is invariant under {\mathrm{refl}}, {(b,a) \in V_1}, so {(a,b) = (b,a) \implies (a,b) = (1,1) \in V_1}. Similarly, {(1,1) \in V_2}, which is impossible. \Box

So we hoped for perhaps too much. However, with seemingly trivial modifications we can make the above example work.

Example In the same example as above, suppose we replace {k} with any field which does not have characteristic {2}. Then {\rho} does decompose.

Proof: Consider the following two subspaces of {V = k^2}:

\displaystyle \begin{aligned} W_1 &= \left\{ \left<a, a\right> \mid a \in k \right\} \\ W_2 &= \left\{ \left<a, -a\right> \mid a \in k \right\}. \end{aligned}

It’s easy to see that both {W_1} and {W_2} are both invariant under {\rho}. Moreover, if {\text{char } k \neq 2} then we in fact have

\displaystyle V = W_1 \oplus W_2

because {\left<x,y\right> = \frac{1}{2} \left<x+y, x+y\right> + \frac{1}{2} \left<x-y, y-x\right>} for any {x,y \in k}. So if we let {\rho_1 = (W_1, \cdot_{\rho_1})} be the subrepresentation corresponding to {W_1}, and define {\rho_2} on {W_2} similarly, then we have {\rho = \rho_1 \oplus \rho_2}. \Box

Thus the only thing in the way of the counterexample was the fact that {\text{char } k = 2}. And it turns out in general this is the only obstacle, a result called Maschke’s Theorem.

Theorem 2 (Maschke’s Theorem) Suppose that {G} is a finite group, and {\text{char } k} does not divide {\left\lvert G \right\rvert}. Then every finite-dimensional representation decomposes as a direct sum of irreducibles.

Before proceeding to the proof, I’ll draw an analogy between the proof that every positive integer {m} decomposes as the product of primes. We use by strong induction on {m}; if {m} is prime we are done, and if {m} is composite there is a nontrivial divisor {d \mid m}, so we apply the inductive hypothesis to {d} and {m/d} and combine these factorizations. We want to mimic the proof above in our proof of Maschke’s Theorem, but we have a new obstacle: we have to show that somehow, we can “divide”.

So why is it that we can divide in certain situations? The idea is that we want to be able to look at an “average” of the form

\displaystyle \frac{1}{\left\lvert G \right\rvert} \sum_{g \in G} g \cdot v

because this average has the nice property of being {G}-invariant. We’ll use this to obtain our proof of Maschke’s Theorem.

Proof: We proceed by induction on the dimension of the representation {\rho}. Let {\rho = (V, \cdot_\rho)} be a representation and assume its not irreducible, so it has a nontrivial subspace {W} which is {\rho}-invariant. It suffices to prove that there exists a subspace {W' \subset V} such that {W'} is also {\rho}-invariant and {V = W \oplus W'}, because then we can apply the inductive hypothesis to the subrepresentations induced by {W} and {W'}.

Let {\pi : V \rightarrow W} be any projection of {V} onto {W}. We consider the averaging map {T : V \rightarrow V} by

\displaystyle T(v) = \frac{1}{\left\lvert G \right\rvert} \sum_{g \in G} g^{-1} \cdot_\rho \pi(g \cdot_\rho v).

We’ll use the following properties of the map.

Exercise Show that the map {T} has the following three properties.

  • For any {w \in W}, {T(w) = w}.
  • For any {v \in V}, {T(w) \in W}.
  • {T \in \mathrm{Hom}_G(\rho, \rho)}.

As with any projection map {T}, we must have {V = \ker T \oplus \text{Im } T}. But {\text{Im } T = W}. Moreover, because the map {T} is {G}-invariant, it follows that {\ker T} is {\rho}-invariant. Hence taking {W' = \ker T} completes the proof. \Box

This completes our proof of Maschke’s Theorem, telling us how all irreducibles decompose. Said another way, Maschke’s Theorem tells us that any finite-dimensional representation {\rho} can be decomposed as

\displaystyle \bigoplus_{\rho_\alpha \in \mathrm{Irrep}(G)} \rho_{\alpha}^{\oplus n_\alpha}

where {n_\alpha} is some nonnegative integer, and {\mathrm{Irrep}(G)} is the set of all (isomorphism classes of) irreducibles representations.

You may wonder whether the decomposition is unique, and if so what we can say about the “exponents” {n_\alpha}. In the next post I’ll show how to compute the exponents {n_\alpha} (which in particular gives uniqueness).

Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. Thanks also to the MOPpers at PUMaC 2014 who let me run this by them during a sleepover; several improvements were made to the original draft as a result. My notes for Math 55a can be found at my website. Thanks also to N for pointing out an error in my proof of Maschke’s Theorem.

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