Rant: Matrices Are Not Arrays of Numbers

The following is an excerpt from a current work of mine. I thought I’d share it here, as some people have told me they enjoyed it.

As I’ll stress repeatedly, a matrix represents a linear map between two vector spaces. Writing it in the form of an ${m \times n}$ matrix is merely a very convenient way to see the map concretely. But it obfuscates the fact that this map is, well, a map, not an array of numbers.

If you took high school precalculus, you’ll see everything done in terms of matrices. To any typical high school student, a matrix is an array of numbers. No one is sure what exactly these numbers represent, but they’re told how to magically multiply these arrays to get more arrays. They’re told that the matrix

$\displaystyle \left( \begin{array}{cccc} 1 & 0 & \dots & 0 \\ 0 & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \dots & 1 \\ \end{array} \right)$

is an “identity matrix”, because when you multiply by another matrix it doesn’t change. Then they’re told that the determinant is some magical combination of these numbers formed by this weird multiplication rule. No one knows what this determinant does, other than the fact that ${\det(AB) = \det A \det B}$, and something about areas and row operations and Cramer’s rule.

Then you go into linear algebra in college, and you do more magic with these arrays of numbers. You’re told that two matrices ${T_1}$ and ${T_2}$ are similar if

$\displaystyle T_2 = ST_1S^{-1}$

for some invertible matrix ${S}$. You’re told that the trace of a matrix ${\text{Tr } T}$ is the sum of the diagonal entries. Somehow this doesn’t change if you look at a similar matrix, but you’re not sure why. Then you define the characteristic polynomial as

$\displaystyle p_T = \det (XI - T).$

Somehow this also doesn’t change if you take a similar matrix, but now you really don’t know why. And then you have the Cayley-Hamilton Theorem in all its black magic: ${p_T(T)}$ is the zero map. Out of curiosity you Google the proof, and you find some ad-hoc procedure which still leaves you with no idea why it’s true.

This is terrible. Who gives a — about ${T_2 = ST_1S^{-1}}$? Only if you know that the matrices are linear maps does this make sense: ${T_2}$ is just ${T_1}$ rewritten with a different choice of basis.

In my eyes, this mess is evil. Linear algebra is the study of linear maps, but it is taught as the study of arrays of numbers, and no one knows what these numbers mean. And for a good reason: the numbers are meaningless. They are a highly convenient way of encoding the matrix, but they are not the main objects of study, any more than the dates of events are the main objects of study in history.

When I took Math 55a as a freshman at Harvard, I got the exact opposite treatment: we did all of linear algebra without writing down a single matrix. During all this time I was quite confused. What’s wrong with a basis? I didn’t appreciate until later that this approach was the morally correct way to treat the subject: it made it clear what was happening.

Throughout this project, I’ve tried to strike a balance between these two approaches, using matrices to illustrate the maps and to simplify proofs, but writing theorems and definitions in their morally correct form. I hope that this has both the advantage of giving the “right” definitions while being concrete enough to be digested. But I would just like to say for the record that, if I had to pick between the high school approach and the 55a approach, I would pick 55a in a heartbeat.

Representation Theory, Part 4: The Finite Regular Representation

Good luck to everyone taking the January TST for the IMO 2015 tomorrow!

Now that we have products of irreducibles under our belt, I’ll talk about the finite regular representation and use it to derive the following two results about irreducibles.

1. The number of (isomorphsim classes) of irreducibles ${\rho_\alpha}$ is equal to the number of conjugacy classes of ${G}$.
2. We have ${ \left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2 }$.

These will actually follow as corollaries from the complete decomposition of the finite regular representation.

In what follows ${k}$ is an algebraically closed field, ${G}$ is a finite group, and the characteristic of ${k}$ does not divide ${\left\lvert G \right\rvert}$. As a reminder, here are the representations we’ve already seen in the order we met them, plus two new ones we’ll introduce properly below.

$\displaystyle \begin{array}{|l|lll|} \hline \text{Representation} & \text{Group} & \text{Space} & \text{Action} \\ \hline \rho & V & G & G \rightarrow g \cdot_\rho V \\ \text{Fun}(X) & G & \text{Fun}(X) & (g \cdot f)(x) = f(g^{-1} \cdot x) \\ \text{triv}_G & G & k & g \cdot a = a \\ \rho_1 \oplus \rho_2 & G & V_1 \oplus V_2 & g \cdot (v_1 + v_2) = (g \cdot_{\rho_1} v_1) + (g \cdot_{\rho_2} v_2) \\ \rho_1 \boxtimes \rho_2 & G_1 \times G_2 & V_1 \otimes V_2 & (g_1, g_2) \cdot (v_1 \otimes v_2) \\ &&& = (g_1 \cdot_{\rho_1} v_1) \otimes (g_2 \cdot_{\rho_2} v_2) \\ \text{Res}^G_H(\rho) & H & V & h \cdot v = h \cdot_\rho v\\ \rho_1 \otimes \rho_2 & G & V_1 \otimes V_2 & g \cdot (v_1 \otimes v_2) = (g \cdot_{\rho_1} v_1) \otimes (g \cdot_{\rho_2} v_2) \\ \text{Reg}(G) & G \times G & \text{Fun}(G) & (g_1, g_2) \cdot f(g) = f(g_2 g g_1^{-1}) \\ \rho^\vee & V^\vee & G & (g \cdot \xi)(v) = \xi(g^{-1} \cdot_\rho v) \\ \hline \end{array}$

1. The Regular Representation

Recall that ${\mathrm{Fun}(G)}$ is the vector space of functions from ${G}$ to ${k}$, with addition being defined canonically. It has a basis of functions ${\delta_g}$ for each ${g \in G}$, where

$\displaystyle \delta_g(x) = \begin{cases} 1 & x = g \\ 0 & \text{otherwise} \end{cases}$

for every ${x \in G}$. (Throughout this post, I’ll be trying to use ${x}$ to denote inputs to a function from ${G}$ to ${k}$.)

Definition Let ${G}$ be a finite group. Then the finite regular representation, ${\mathrm{Reg}(G)}$ is a representation on ${G \times G}$ defined on the vector space ${\mathrm{Fun}(G)}$, with the following action for each ${f \in \mathrm{Fun}(G)}$ and ${(g_1, g_2) \in G \times G}$:

$\displaystyle ( g_1, g_2 ) \cdot f(x) \overset{\text{def}}{=} f(g_2 x g_1^{-1}).$

Note that this is a representation of the product ${G \times G}$, not ${G}$! (As an aside, you can also define this representation for infinite groups ${G}$ by replacing ${\mathrm{Fun}(G)}$ with ${\mathrm{Fun}_c(G)}$, the functions which are nonzero at only finitely many ${g \in G}$.)

In any case, we now can make ${\mathrm{Reg}(G)}$ into a representation of ${G}$ by this restriction, giving ${\mathrm{Res}_G^{G \times G} \left( \mathrm{Reg}(G) \right)}$, which I will abbreviate as just ${\mathrm{Reg}^\ast(G)}$ through out this post (this is not a standard notation). The action for this is

$\displaystyle (g \cdot_{\mathrm{Reg}^\ast(G)} f)(x) \overset{\text{def}}{=} \left( (g, g) \cdot_{\mathrm{Reg}(G)} f \right)(x) = f\left( g^{-1} x g \right).$

Exercise Consider the invariant subspace of ${\mathrm{Reg}^\ast(G)}$, which is

$\displaystyle \left( \mathrm{Reg}^\ast(G) \right)^G = \left\{ f : G \rightarrow V \mid f(g^{-1} x g) = f(x) \; \forall x,g \in G \right\}.$

Prove that the dimension of this space is equal to the number of conjugacy classes of ${G}$. (Look at the ${\delta_g}$ basis.)

Recall that in general, the invariant subspace ${\rho^G}$ is defined as

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

2. Dual Representations

Before I can state the main theorem of this post, I need to define the dual representation.

Recall that given a vector space ${V}$, we define the \textbf} by

$\displaystyle V^\vee \overset{\text{def}}{=} \mathrm{Hom}(V,k)$

i.e. it is the set of maps from ${V}$ to ${k}$. If ${V}$ is finite-dimensional, we can think of this as follows: if ${V}$ consists of the column vectors of length ${m}$, then ${V^\vee}$ is the row vectors of length ${m}$, which can be multiplied onto elements of ${V}$. (This analogy breaks down for ${V}$ infinite dimensional.) Recall that if ${V}$ is finite-dimensional then there is a canonical isomorphism ${V \simeq (V^\vee)^\vee}$ by the map ${v \mapsto \mathrm{ev}_v}$, where ${\mathrm{ev}_v : V^\vee \rightarrow k}$ sends ${\xi \mapsto \xi(v)}$.

Now we can define the dual representation in a similar way.

Definition Let ${\rho = (V, \cdot_\rho)}$ be a ${G}$-representation. Then we define the dual representation ${\rho^\vee}$ by

$\displaystyle \rho^\vee = \left( V^\vee, \cdot_{\rho^\vee} \right) \quad\text{where}\quad \left( g \cdot_{\rho^\vee} \xi \right)(v) = \xi \left( g^{-1} \cdot_\rho v \right).$

Lemma 1 If ${\rho}$ is finite-dimensional then ${(\rho^\vee)^\vee \simeq \rho}$ by the same isomorphism.

Proof: We want to check that the isomorphism ${V = (V^\vee)^\vee}$ by ${v \mapsto \mathrm{ev}_v}$ respects the action of ${G}$. That’s equivalent to checking

$\displaystyle \mathrm{ev}_{g \cdot_\rho v} = g \cdot_{(\rho^\vee)^\vee} \mathrm{ev}_v.$

But

$\displaystyle \mathrm{ev}_{g \cdot v}(\xi) = \xi(g \cdot_\rho v)$

and

$\displaystyle \left( g \cdot_{(\rho^\vee)^\vee} \mathrm{ev}_v \right)(\xi) = \mathrm{ev}_v(g^{-1} \cdot_{\rho^\vee} \xi) = \left( g^{-1} \cdot_{\rho^\vee} \xi \right)(v) = \xi(g \cdot_\rho v).$

So the functions are indeed equal. $\Box$

Along with that lemma, we also have the following property.

Lemma 2 For any finite-dimensional ${\rho_1}$, ${\rho_2}$ we have ${\mathrm{Hom}_G(\rho_1, \rho_2) \simeq \mathrm{Hom}_G(\rho_1 \otimes \rho_2^\vee, \text{triv}_G)}$.

Proof: Let ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and ${\rho_2 = (V_2, \cdot_{\rho_2})}$. We already know that we have an isomorphism of vector homomorphisms

$\displaystyle \mathrm{Hom}_{\textbf{Vect}}(V_1, V_2) \simeq \mathrm{Hom}_{\textbf{Vect}} (V_1 \otimes V_2^\vee, k)$

by sending each ${T \in \mathrm{Hom}_{\textbf{Vect}}(V_1, V_2)}$ to the map ${T' \in \mathrm{Hom}_{\textbf{Vect}} (V_1 \otimes V_2^\vee, k)}$ which has ${T'(v \otimes \xi) = \xi(T(v))}$. So the point is to check that ${T}$ respects the ${G}$-action if and only if ${T'}$ does. This is just a computation. $\Box$

You can deduce as a corollary the following.

Exercise Use the lemma to show ${\mathrm{Hom}_G(\rho, \text{triv}_G) \simeq \mathrm{Hom}_G(\text{triv}_G, \rho^\vee)}$.

Finally, we want to talk about when ${\rho^\vee}$ being irreducible. The main result is the following.

Lemma 3 Consider a representation ${\rho}$, not necessarily finite-dimensional. If ${\rho^\vee}$ is irreducible then so is ${\rho}$.

When ${\rho}$ is finite dimensional we have ${(\rho^\vee)^\vee \simeq \rho}$, and so it is true for finite-dimensional irreducible ${\rho}$ that ${\rho^\vee}$ is also irreducible. Interestingly, this result fails for infinite-dimensional spaces as this math.SE thread shows.

Proof: Let ${\rho = (V, \cdot_\rho)}$. Let ${W}$ be a ${\rho}$-invariant subspace of ${V}$. Then consider

$\displaystyle W^\perp = \left\{ \xi \in V^\vee : \xi(w) = 0 \right\}.$

This is a ${\rho^\vee}$-invariant subspace of ${V^\vee}$, so since ${\rho^\vee}$ is irreducible, either ${W^\perp = V^\vee}$ or ${W^\perp = \{0\}}$. You can check that these imply ${W=0}$ and ${W=V}$, respectively. $\Box$

3. Main Result

Now that we know about the product of representations and dual modules, we can state the main result of this post: the complete decomposition of ${\mathrm{Reg}(G)}$.

Theorem 4 We have an isomorphism

$\displaystyle \mathrm{Reg}(G) \simeq \bigoplus_{\alpha} \rho_\alpha \boxtimes \rho_\alpha^\vee.$

Before we can begin the proof of the theorem we need one more lemma.

Lemma 5 Let ${\pi}$ be a representation of ${G \times G}$. Then there is an isomorphism

$\displaystyle \mathrm{Hom}_{G \times G}(\pi, \mathrm{Reg}(G)) \simeq \mathrm{Hom}_G(\mathrm{Res}^{G \times G}_G(\pi), \text{triv}_G).$

Proof: Let ${\pi = (V, \cdot_\pi)}$. Given a map ${T : V \rightarrow \mathrm{Fun}(G)}$ which respects the ${G \times G}$ action, we send it to the map ${\xi_T : V \rightarrow k}$ with ${\xi_T(v) = T(v)(1)}$. Conversely, given a map ${\xi : V \rightarrow k}$ which respects the ${G}$ action, we send it to the map ${T_\xi : V \rightarrow \mathrm{Fun}(G)}$ so that ${T_\xi(v)(x) = \xi\left( (x,x^{-1}) \cdot v \right)}$.

Some very boring calculations show that the two maps are mutually inverse and respect the action. We’ll just do one of them here: let us show that ${\xi_T(v)}$ respects the ${G}$ action given that ${T}$ respects the ${G \times G}$ action. We want to prove

$\displaystyle \xi_T\left( (g,g) \cdot_\pi v \right) = g \cdot_\text{triv} \xi_T(v) = \xi_T(v).$

Using the definition of ${\xi_T}$

\displaystyle \begin{aligned} \xi_T\left( (g,g) \cdot_\pi (v) \right) &= T\left( (g,g) \cdot_\pi v \right)(1) \\ &= \left( (g,g) \cdot_{\mathrm{Fun}(G)} T(v) \right)(1) \\ &= T(v)\left( g 1 g^{-1} \right) = T(v)(1) = \xi_T(v). \end{aligned}

The remaining computations are left to a very diligent reader. $\Box$

Now let’s prove the main theorem!

Proof: We have that ${\mathrm{Reg}(G)}$ is the sum of finite-dimensional irreducibles ${\rho_\alpha \boxtimes \rho_\beta}$, meaning

$\displaystyle \mathrm{Reg}(G) = \bigoplus_{\alpha, \beta} \left( \rho_\alpha \boxtimes \rho_\beta \right) \otimes \mathrm{Hom}_{G \times G}\left( \rho_\alpha \boxtimes \rho_\beta, \mathrm{Reg}(G) \right).$

But using our lemmas, we have that

$\displaystyle \mathrm{Hom}_{G \times G}\left( \rho_\alpha \boxtimes \rho_\beta, \mathrm{Reg}(G) \right) \simeq \mathrm{Hom}_G(\rho_\alpha \otimes \rho_\beta, \text{triv}_G) \simeq \mathrm{Hom}_G(\rho_\alpha, \rho_\beta^\vee).$

We know that ${\rho_\beta^\vee}$ is also irreducible, since ${\rho_\beta}$ is (and we’re in a finite-dimensional situation). So

$\displaystyle \mathrm{Hom}_G\left( \rho_\alpha, \rho_\beta^\vee \right) \simeq \begin{cases} k & \rho_\beta^\vee = \rho_\alpha \\ \{0\} & \text{otherwise}. \end{cases}$

Thus we deduce

$\displaystyle \mathrm{Reg}(G) \simeq \bigoplus_{\alpha} \left( \rho_\alpha \boxtimes \rho_\alpha^\vee \right) \otimes k \simeq \bigoplus_{\alpha} \left( \rho_\alpha \boxtimes \rho_\alpha^\vee \right)$

and we’re done. $\Box$

4. Corollaries

Recall that ${\mathrm{Fun}(G)}$, the space underlying ${\mathrm{Reg}(G)}$, has a basis with size ${\left\lvert G \right\rvert}$. Hence by comparing the dimensions of the isomorphsims, we obtain the following corollary.

Theorem 6 We have ${\left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2}$.

Moreover, by restriction to ${G}$ we can obtain the corollary

$\displaystyle \mathrm{Reg}^\ast(G) \simeq \bigoplus_\alpha \mathrm{Res}_{G}^{G \times G} \left( \rho_\alpha \otimes \rho_\alpha^\vee \right) = \bigoplus_\alpha \rho_\alpha \otimes \rho_\alpha^\vee.$

Now let us look at the ${G}$-invariant spaces in this decomposition. We claim that

$\displaystyle \left( \rho_\alpha \otimes \rho_\alpha^\vee \right)^G \simeq k.$

Indeed, {Proposition 1} in {Part 1} tells us that we have a bijection of vector spaces

$\displaystyle \left( \rho_\alpha \otimes \rho_\alpha^\vee \right)^G \simeq \mathrm{Hom}_G(\text{triv}_G, \rho_\alpha \otimes \rho_\alpha^\vee).$

Then we can write

\displaystyle \begin{aligned} \mathrm{Hom}_G(\text{triv}_G, \rho_\alpha \otimes \rho_\alpha^\vee) &\simeq \mathrm{Hom}_G\left(\text{triv}_G, \left( \rho_\alpha^\vee \otimes \rho_\alpha \right)^\vee \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha^\vee \otimes \rho_\alpha, \text{triv}_G \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha \otimes \rho_\alpha^\vee, \text{triv}_G \right) \\ &\simeq \mathrm{Hom}_G\left(\rho_\alpha, \rho_\alpha \right) \\ &\simeq k \end{aligned}

by the lemma, where we have also used Schur’s Lemma at the last step. So that means the dimension of the invariant space ${(\mathrm{Reg}^\ast (G))^G}$ is just the number of irreducibles.

But we already showed that the invariant space of ${(\mathrm{Reg}^\ast (G))^G}$ has dimension equal to the conjugacy classes of ${G}$. Thus we conclude the second result.

Theorem 7 The number of conjugacy classes of ${G}$ equals the number of irreducible representations of ${G}$.

Hooray!

Time permitting I might talk about the irreducibles of ${S_n}$ in subsequent posts. No promises here though.

Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.

Represenation Theory, Part 3: Products of Representations

Happy New Year to all! A quick reminder that ${2015 = 5 \cdot 13 \cdot 31}$.

This post will set the stage by examining products of two representations. In particular, I’ll characterize all the irreducibles of ${G_1 \times G_2}$ in terms of those for ${G_1}$ and ${G_2}$. This will set the stage for our discussion of the finite regular representation in Part 4.

In what follows ${k}$ is an algebraically closed field, ${G}$ is a finite group, and the characteristic of ${k}$ does not divide ${\left\lvert G \right\rvert}$.

1. Products of Representations

First, I need to tell you how to take the product of two representations.

Definition Let ${G_1}$ and ${G_2}$ be groups. Given a ${G_1}$ representation ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and a ${G_2}$ representation ${\rho_2 = (V_2, \cdot_{\rho_2})}$, we define

$\displaystyle \rho_1 \boxtimes \rho_2 \overset{\text{def}}{=} \left( V_1 \otimes V_2, \cdot \right)$

as a representation of ${G_1 \times G_2}$ on ${V_1 \otimes V_2}$. The action is given by

$\displaystyle (g_1, g_2) \cdot (v_1 \otimes v_2) = \left( g_1 \cdot_{\rho_1} v_1 \right) \otimes (g_2 \cdot_{\rho_2} v_2).$

In the special case ${G_1 = G_2 = G}$, we can also restrict ${\rho_1 \boxtimes \rho_2}$ to a representation of ${G}$. Note that we can interpret ${G}$ itself as a subgroup of ${G \times G}$ by just looking along the diagonal: there’s an obvious isomorphism

$\displaystyle G \sim \left\{ (g,g) \mid g \in G \right\}.$

So, let me set up the general definition.

Definition Let ${\mathcal G}$ be a group, and let ${\mathcal H}$ be a subgroup of ${\mathcal G}$. Then for any representation ${\rho = (V, \cdot_\rho)}$ of ${\mathcal G}$, we let

$\displaystyle \mathrm{Res}^{\mathcal G}_{\mathcal H} (\rho)$

denote the representation of ${\mathcal H}$ on ${V}$ by the same action.

This notation might look intimidating, but it’s not really saying anything, and I include the notation just to be pedantic. All we’re doing is taking a representation and restricting which elements of the group are acting on it.

We now apply this to get ${\rho_1 \otimes \rho_2}$ out of ${\rho_1 \boxtimes \rho_2}$.

Definition Let ${\rho_1 = (V_1, \cdot_{\rho_1})}$ and ${\rho_2 = (V_2, \cdot_{\rho_2})}$ be representations of ${G}$. Then we define

$\displaystyle \rho_1 \otimes \rho_2 \overset{\text{def}}{=} \mathrm{Res}^{G \times G}_G \left( \rho_1 \boxtimes \rho_2 \right)$

meaning ${\rho_1 \otimes \rho_2}$ has vector space ${V_1 \otimes V_2}$ and action ${g \cdot (v_1 \otimes v_2) = (g \cdot_{\rho_1} v_1) \otimes (g \cdot_{\rho_2} v_2)}$.

This tensor product obeys some nice properties, for example the following.

Lemma 1 Given representations ${\rho}$, ${\rho_1}$, ${\rho_2}$ we have

$\displaystyle \rho \otimes \left( \rho_1 \oplus \rho_2 \right) \simeq \left( \rho \otimes \rho_1 \right) \oplus \left( \rho \otimes \rho_2 \right).$

Proof: There’s an obvious isomorphism between the underlying vector spaces, and that isomorphism respects the action of ${G}$. $\Box$

To summarize all the above, here is a table of the representations we’ve seen, in the order we met them.

$\displaystyle \begin{array}{|l|lll|} \hline \text{Representation} & \text{Group} & \text{Space} & \text{Action} \\ \hline \rho & V & G & g \cdot_\rho v \\ \text{Fun}(X) & G & \text{Fun}(X) & (g \cdot f)(x) = f(g^{-1} \cdot x) \\ \text{triv}_G & G & k & g \cdot a = a \\ \rho_1 \oplus \rho_2 & G & V_1 \oplus V_2 & g \cdot (v_1 + v_2) = (g \cdot_{\rho_1} v_1) + (g \cdot_{\rho_2} v_2) \\ \rho_1 \boxtimes \rho_2 & G_1 \times G_2 & V_1 \otimes V_2 & (g_1, g_2) \cdot (v_1 \otimes v_2) \\ &&& = (g_1 \cdot_{\rho_1} v_1) \otimes (g_2 \cdot_{\rho_2} v_2) \\ \text{Res}^G_H(\rho) & H & V & h \cdot v = h \cdot_\rho v\\ \rho_1 \otimes \rho_2 & G & V_1 \otimes V_2 & g \cdot (v_1 \otimes v_2) = (g \cdot_{\rho_1} v_1) \otimes (g \cdot_{\rho_2} v_2) \\ \hline \end{array}$

2. Revisiting Schur and Maschke

Defining a tensor product of representations gives us another way to express ${\rho^{\oplus n}}$, as follows. By an abuse of notation, given a vector space ${k^m}$ we can define an associated ${G}$-representation ${k^m = (k^m, \cdot_{k^m})}$ on it by the trivial action, i.e. ${g \cdot_{k^m} v = v}$ for ${v \in k^m}$. A special case of this is using ${k}$ to represent ${\text{triv}_G}$. With this abuse of notation, we have the following lemma.

Lemma 2 Let ${M}$ be an ${m}$-dimensional vector space over ${k}$. Then ${\rho^{\oplus m} \simeq \rho \otimes M}$.

Proof: It reduces to checking that ${\rho \otimes k \overset{\text{def}}{=} \rho \otimes \text{triv}_G}$ is isomorphic to ${\rho}$, which is evident. We can then proceed by induction: ${\rho \otimes (k \oplus k^{t-1}) \simeq (\rho \otimes k) \oplus (\rho \otimes k^{t-1})}$. $\Box$

So, we can actually rewrite Maschke’s and Schur’s Theorem as one. Instead of

$\displaystyle \rho \simeq \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha} \quad\text{where}\quad n_\alpha = \dim \mathrm{Hom}_G(\rho, \rho_\alpha)$

$\displaystyle \bigoplus_\alpha \rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \simeq \rho.$

Now we’re going to explicitly write down the isomorphism between these maps. It suffices to write down the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \rightarrow \rho_\alpha^{\oplus n_\alpha}}$, and then take the sum over each of the ${\alpha}$‘s. But

$\displaystyle \mathrm{Hom}_G(\rho, \rho_\alpha) \simeq \mathrm{Hom}_G(\rho_\alpha^{\oplus n_\alpha}, \rho_\alpha) \simeq \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)^{\oplus n_\alpha}.$

So to write the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)^{\oplus n_\alpha} \rightarrow \rho_\alpha^{\oplus n_\alpha}}$, we just have to write down the isomorphism ${\rho_\alpha \otimes \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha) \rightarrow \rho_\alpha}$,

Schur’s Lemma tells us that ${\mathrm{Hom}_G(\rho_\alpha, \rho_\alpha) \simeq k}$; i.e. every ${\xi \in \mathrm{Hom}_G(\rho_\alpha, \rho_\alpha)}$ just corresponds to multiplying ${v}$ by some constant. So this case is easy: the map

$\displaystyle v \otimes \xi \mapsto \xi(v)$

works nicely. And since all we’ve done is break over a bunch of direct sums, the isomorphism propagates all the way up, resulting in the following theorem.

Theorem 3 (Maschke and Schur) For any finite-dimensional ${\rho}$, the homomorphism of ${G}$ representations

$\displaystyle \bigoplus_\alpha \rho_\alpha \otimes \mathrm{Hom}_G(\rho, \rho_\alpha) \rightarrow \rho$

given by sending every simple tensor via

$\displaystyle v \otimes \xi \mapsto \xi(v)$

is an isomorphism.

Note that it’s much easier to write the map from left to right than vice-versa, even though the inverse map does exist (since it’s an isomorphism). (Tip: as a general rule of thumb, always map out of the direct sum.)

3. Characterizing the ${G_1 \times G_2}$ irreducibles

Now we are in a position to state the main theorem for this post, which shows that the irreducibles we defined above are very well behaved.

Theorem 4 Let ${G_1}$ and ${G_2}$ be finite groups. Then a finite-dimensional representation ${\rho}$ of ${G_1 \times G_2}$ is irreducible if and only if it is of the form

$\displaystyle \rho_1 \boxtimes \rho_2$

where ${\rho_1}$ and ${\rho_2}$ are irreducible representations of ${G_1}$ and ${G_2}$, respectively.

Proof: First, suppose ${\rho = (V, \cdot_\rho)}$ is an irreducible representation of ${G_1 \times G_2}$. Set

$\displaystyle \rho^1 \overset{\text{def}}{=} \mathrm{Res}^{G_1 \times G_2}_{G_1} (\rho).$

Then by Maschke’s Theorem, we may write ${\rho^1}$ as a direct sum of the irreducibles

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \otimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho^1$

with the map ${v \otimes \xi \mapsto \xi(v)}$ being the isomorphism. Now we can put a ${G_2}$ representation structure on ${\mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1)}$ by

$\displaystyle (g_2 \cdot f)(g) = g_2 \cdot_{\rho} (f(g)).$

It is easy to check that this is indeed a ${G_2}$ representation. Thus it makes sense to talk about the ${G_1 \times G_2}$ representation

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \boxtimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1).$

We claim that the isomorphism for ${\rho^1}$ as a ${G_1}$ representation now lifts to an isomorphism of ${G_1 \times G_2}$ representations. That is, we claim that

$\displaystyle \bigoplus_\alpha \rho_\alpha^1 \boxtimes \mathrm{Hom}_{G_1} (\rho_\alpha^1, \rho^1) \simeq \rho$

by the same isomorphism as for ${\rho^1}$. To see this, we only have to check that the isomorphism ${v \otimes \xi \mapsto \xi(v)}$ commutes with the action of ${g_2 \in G_2}$. But this is obvious, since ${g_2 \cdot (v \otimes \xi) = v \otimes (g_2 \cdot \xi) \mapsto (g_2 \cdot \xi)(v)}$.

Thus the isomorphism holds. But ${\rho}$ is irreducible, so there can only be one nontrivial summand. Thus we derive the required decomposition of ${\rho}$.

Now for the other direction: take ${\rho_1}$ and ${\rho_2}$ irreducible. Suppose ${\rho_1 \boxtimes \rho_2}$ has a nontrivial subrepresentation of the form ${\rho_1' \boxtimes \rho_2'}$. Viewing as ${G_1}$ representation, we find that ${\rho_1'}$ is a nontrivial subrepresentation of ${\rho_1}$, and similarly for ${\rho_2}$. But ${\rho_1}$ is irreducible, hence ${\rho_1' \simeq \rho_1}$. Similarly ${\rho_2' \simeq \rho_2}$. So in fact ${\rho_1' \boxtimes \rho_2' \simeq \rho_1 \boxtimes \rho_2}$. Hence we conclude ${\rho_1 \boxtimes \rho_2}$ is irreducible. $\Box$

4. Conclusion

In particular, this means that any representation ${\rho}$ of ${G \times G}$ decomposes as

$\displaystyle \rho \simeq \bigoplus_{\alpha, \beta} \rho_\alpha \boxtimes \rho_\beta$

and we even have

$\displaystyle \mathrm{Res}_{G}^{G\times G} \rho \simeq \bigoplus_{\alpha, \beta} \rho_\alpha \otimes \rho_\beta.$

In the next post I’ll invoke this on the so-called finite regular representation to get the elegant results I promised at the end of Part 2.

Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.

Representation Theory, Part 2: Schur’s Lemma

Merry Christmas!

In the previous post I introduced the idea of an irreducible representation and showed that except in fields of low characteristic, these representations decompose completely. In this post I’ll present Schur’s Lemma at talk about what Schur and Maschke tell us about homomorphisms of representations.

1. Motivation

Fix a group ${G}$ now, and consider all isomorphism classes of finite-dimensional representations of ${G}$. We’ll denote this set by ${\mathrm{Irrep}(G)}$. 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. This begs the question: what is ${n_\alpha}$? Is it even uniquely determined by ${\rho}$?

To answer this I first need to compute ${\mathrm{Hom}_G(\rho, \pi)}$ for any two distinct irreducible representations ${\rho}$ and ${\pi}$. One case is easy.

Lemma 1 Let ${\rho}$ and ${\pi}$ be non-isomorphic irreducible representations (not necessarily finite dimensional). Then there are no nontrivial homomorphisms ${\phi : \rho \rightarrow \pi}$. In other words, ${\mathrm{Hom}_G(\rho, \pi) = \{0\}}$.

I haven’t actually told you what it means for representations to be isomorphic, but you can guess — it just means that there’s a homomorphism of ${G}$-representations between them which is also a bijection of the underlying vector spaces.

Proof: Let ${\phi : \rho_1 \rightarrow \rho_2}$ be a nonzero homomorphism. We can actually prove the following stronger results.

• If ${\rho_2}$ is irreducible then ${\phi}$ is surjective.
• If ${\rho_1}$ is irreducible then ${\phi}$ is injective.

Exercise Prove the above two results. (Hint: show that ${\text{Im } \phi}$ and ${\ker \phi}$ give rise to subrepresentations.)

Combining these two results gives the lemma because ${\phi}$ is now a bijection, and hence an isomorphism. $\Box$

2. Schur’s Lemma

Thus we only have to consider the case ${\rho \simeq \pi}$. The result which relates these is called Schur’s Lemma, but is important enough that we refer to it as a theorem.

Theorem 2 (Schur’s Lemma) Assume ${k}$ is algebraically closed. Let ${\rho}$ be a finite dimensional irreducible representation. Then ${\mathrm{Hom}_{G} (\rho, \rho)}$ consists precisely of maps of the form ${v \mapsto \lambda v}$, where ${\lambda \in k}$; the only possible maps are multiplication by a scalar. In other words,

$\displaystyle \mathrm{Hom}_{G} (\rho, \rho) \simeq k$

and ${\dim \mathrm{Hom}_G(\rho, \rho) = 1}$.

This is NOT in general true without the algebraically closed condition, as the following example shows.

Example Let ${k = {\mathbb R}}$, let ${V = {\mathbb R}^2}$, and let ${G = {\mathbb Z}_3}$ act on ${V}$ by rotating every ${\vec x \in {\mathbb R}^2}$ by ${120^{\circ}}$ around the origin, giving a representation ${\rho}$. Then ${\rho}$ is a counterexample to Schur’s Lemma.

Proof: This representation is clearly irreducible because the only point that it fixes is the origin, so there are no nontrivial subrepresentations.

We can regard now ${\rho}$ as a map in ${{\mathbb C}}$ which is multiplication by ${e^{\frac{2\pi i}{3}}}$. Then for any other complex number ${\xi}$, the map “multiplication by ${\xi}$” commutes with the map “multiplication by ${e^{\frac{2\pi i}{3}}}$”. So in fact

$\displaystyle \mathrm{Hom}_G(\rho, \rho) \simeq {\mathbb C}$

which has dimension ${2}$. $\Box$

Now we can give the proof of Schur’s Lemma.

Proof: Clearly any map ${v \mapsto \lambda v}$ respects the ${G}$-action.

Now consider any ${T \in \mathrm{Hom}_G(\rho, \rho)}$. Set ${\rho = (V, \cdot_\rho)}$. Here’s the key: because ${k}$ is algebraically closed, and we’re over a finite dimensional vector space ${V}$, the map ${T}$ has an eigenvalue ${\lambda}$. Hence by definition ${V}$ has a subspace ${V_\lambda}$ over which ${T}$ is just multiplication by ${\lambda}$.

But then ${V_\lambda}$ is a ${G}$-invariant subspace of ${V}$! Since ${\rho}$ is irreducible, this can only happen if ${V = V_\lambda}$. That means ${T}$ is multiplication by ${\lambda}$ for the entire space ${V}$, as desired. $\Box$

3. Computing dimensions of homomorphisms

Since we can now compute the dimension of the ${\mathrm{Hom}_G}$ of any two irreducible representations, we can compute the dimension of the ${\mathrm{Hom}_G}$ for any composition of irreducibles, as follows.

Corollary 3 We have

$\displaystyle \dim \mathrm{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) = \sum_{\alpha} n_\alpha m_\alpha$

where the direct sums run over the isomorphism classes of irreducibles.

Proof: The ${\mathrm{Hom}}$ just decomposes over each of the components as

\displaystyle \begin{aligned} \mathrm{Hom}_G \left( \bigoplus_\alpha \rho_\alpha^{\oplus n_\alpha}, \bigoplus_\beta \rho_\beta^{\oplus m_\beta} \right) &\simeq \bigoplus_{\alpha, \beta} \mathrm{Hom}_G(\rho_\alpha^{\oplus n_\alpha}, \rho_\beta^{\oplus m_\beta}) \\ &\simeq \bigoplus_{\alpha, \beta} \mathrm{Hom}_G(\rho_\alpha, \rho_\beta)^{\oplus n_\alpha m_\alpha}. \end{aligned}

Here we’re using the fact that ${\mathrm{Hom}_G(\rho_1 \oplus \rho_2, \rho) = \mathrm{Hom}_G(\rho_1, \rho) \oplus \mathrm{Hom}_G(\rho_2, \rho)}$ (obvious) and its analog. The claim follows from our lemmas now. $\Box$

As a special case of this, we can quickly derive the following.

Corollary 4 Suppose ${\rho = \bigoplus_\alpha \rho_\alpha^{n_\alpha}}$ as above. Then for any particular ${\beta}$,

$\displaystyle n_\beta = \dim \mathrm{Hom}_G(\rho, \rho_\beta).$

Proof: We have

$\displaystyle \dim \mathrm{Hom}_G(\rho, \rho_\beta) = n_\beta \mathrm{Hom}_G(\rho_\beta, \rho_\beta) = n_\beta$

as desired. $\Box$

This settles the “unique decomposition” in the affirmative. Hurrah!

It might be worth noting that we didn’t actually need Schur’s Lemma if we were solely interested in uniqueness, since without it we would have obtained

$\displaystyle n_\beta = \frac{\dim \mathrm{Hom}_G(\rho, \rho_\beta)}{\dim \mathrm{Hom}_G(\rho_\beta, \rho_\beta)}.$

However, the denominator in that expression is rather unsatisfying, don’t you think?

4. Conclusion

In summary, we have shown the following main results for finite dimensional representations of a group ${G}$.

• Maschke’s Theorem: If ${G}$ is finite and ${\text{char } k}$ does not divide ${\left\lvert G \right\rvert}$, then any finite dimensional representation is a direct sum of irreducibles. This decomposition is unique up to isomorphism.
• Schur’s Lemma: If ${k}$ is algebraically closed, then ${\mathrm{Hom}_G(\rho, \rho) \simeq k}$ for any irreducible ${\rho}$, while there are no nontrivial homomorphisms between non-isomorphic irreducibles.

In the next post I’ll talk about products of irreducibles, and use them in the fourth post to prove two very elegant results about the irreducibles, as follows.

1. The number of (isomorphsim classes) of irreducibles ${\rho_\alpha}$ is equal to the number of conjugacy classes of ${G}$.
2. We have ${ \left\lvert G \right\rvert = \sum_\alpha \left( \dim \rho_\alpha \right)^2 }$.

Thanks to Dennis Gaitsgory, who taught me this in his course Math 55a. My notes for Math 55a can be found at my website.

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

$\displaystyle \sigma \cdot_\rho \left = \left$

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 \mid a \in k \right\} \\ W_2 &= \left\{ \left \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 ${\left = \frac{1}{2} \left + \frac{1}{2} \left}$ 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.