Quick Guide to Character Theory (II): Main Theory

Reminder: throughout this series, G is a finite group and K is a field. All K-vector spaces are assumed to be finite-dimensional over K.

G4. Maschke’s Theorem

If $W\subseteq V$ is a K[G]-submodule, it turns out V is isomorphic to the direct sum of W and some other submodule W’.

Maschke’s Theorem. Suppose char(K) does not divide #G (e.g. char(K) = 0). If $W\subseteq V$ is a K[G]-submodule, then there’s a submodule $W'\subseteq V$ such that:

$W + W' = V, \quad W\cap W' = 0.$

This gives $V\cong W\oplus W'$ .

As a consequence, if $W\subseteq V$ is a nontrivial submodule, then one can always decompose $V = W\oplus W'$ ; since V is finite-dimensional, this process must eventually terminate.

Definition. A K[G]-module V≠0 is said to be simple if it has no submodule other than 0 and itself. The corresponding representation is then said to be irreducible.

Thanks to Maschke’s theorem, every K[G]-module is a direct sum of simple modules.

Proof of Maschke’s Theorem.

Let X be a K-subspace of V such that $W+X = V$ and $W\cap X = 0$ , which is always possible by linear algebra. Take the projection map $p:V = W\oplus X \to W.$ Now define the linear map:

$\begin{aligned}q: V\to V, \quad q(v) = \frac 1 {|G|} \sum_{g\in G} g(p(g^{-1}v)).\end{aligned}$

We claim: q is K[G]-linear; it suffices to prove that q(hv) = h·q(v) for all $h\in G, v\in V.$ But

$\begin{aligned} h\cdot q(v) = \frac 1 {|G|} \sum_{g\in G} hg(p(g^{-1}v)) \stackrel{y=hg}{=} \sum_{y\in G} y(p(y^{-1}hv)) = q(hv).\end{aligned}$

Next claim: if $w\in W$ then q(w)=w. Indeed since W is a K[G]-submodule of V, $g^{-1}w\in W$ so $p(g^{-1}w) = g^{-1}w$ for every $g\in G.$ Hence q(w) is 1/|G| times a sum of |G| copies of w.

To complete the proof, we’ll show that W’ = ker(q) satisfies the conditions of the theorem:

Since q is K[G]-linear, its kernel W’ is a K[G]-submodule of V.
If $v\in W'\cap W$ , then q(v) = 0 since $v\in\text{ker}(q);$ on the other hand, since $v\in W$ we have q(v)=v so v=0. Thus, $W'\cap W = 0.$
For any $v\in V$ , write $v = (v-q(v)) + q(v)$ . Since $p(V)\subseteq W,$ we have $q(V)\subseteq W$ also. So $q(v)\in W$ and we get $q(q(v)) = q(v) \implies v-q(v)\in \text{ker}(q).$ Thus $v\in W + W'.$ ♦

Example

Consider $G=S_3$ acting on $V = K^3$ by permuting the coordinates (see example 3 in G1). The 1-dimensional subspace W spanned by (1, 1, 1) is clearly a G-invariant subspace. Applying the construction in the above proof gives us the complementary submodule $W' = \{(x,y,z)\in K^3: x+y+z=0\}.$

G5. Schur’s Lemma

Assume char(K) = 0. By Maschke’s theorem, every K[G]-module is a direct sum of simple modules. The following result is handy:

Schur’s Lemma. Let V and W be simple K[G]-modules.

If f : V → W is a K[G]-linear map, then f=0 or f is an isomorphism.

Suppose K = C. If f : V → V is a K[G]-linear map, then it’s a scalar multiple of the identity.

Proof.

Suppose f : V → W is non-zero. Its kernel is a submodule of V so it’s either 0 or V; since f≠0, we have ker(f) = 0. Likewise, its image is a submodule of W so it’s either 0 or W; since f≠0, we have im(f) = W. Thus f is bijective.

Let λ be an eigenvalue of f with eigenvector v, i.e. f(v) = λv. Then ker(f – λ) is non-zero so it must be whole of V, i.e. f = λ·1_V. ♦

Another way of looking at Schur’s lemma: for K = C and simple C[G]-modules V, W,

$\dim_{\mathbf{C}} \text{Hom}_{\mathbf{C}[G]}(V, W) = \begin{cases} 0, \quad &\text{if } V\not\cong W,\\ 1,\quad &\text{if } V\cong W.\end{cases}$

From now onwards, we let K = C.

G6. The Space of Fixed Vectors.

For any C[G]-module V, consider the space:

$V^G := \{v\in V: gv = v \text{ for every } g\in G\},$

the space of all vectors fixed by G. The following result is critical.

Lemma. We have:

$\begin{aligned}\dim(V^G) = \frac 1 {|G|}\sum_{g\in G} \text{tr}(g|_V),\end{aligned}$

the average of the traces of $g\in G$ acting on V.

Proof.

Take the averaging map (which is C-linear):

$\begin{aligned}p:V\to V, \quad p(v) = \frac 1{|G|} \sum_{g\in G} gv.\end{aligned}$

We claim: for any $v\in V$ , we have $p(v)\in V^G$ . Indeed, for any $h\in G$ , we have

$\begin{aligned}h\cdot p(v) = \frac 1 {|G|}\sum_{g\in G} hg(v)\stackrel{y=hg}{=} \frac 1 {|G|}\sum_{y\in G} y(v) = p(v).\end{aligned}$

On the other hand, if $w\in V^G$ then gw = w, which gives us p(w) = w. Just as in the proof of Maschke’s theorem, this shows $\text{ker}(p)\cap V^G = 0$ and $\text{ker}(p) + V^G = V.$ If we pick a basis of ker(p) and of V^G, then the matrix for p is:

So $\dim(V^G) = \text{tr}(p) = \frac 1 {|G|}\sum_{g\in G}\text{tr}(g|_V).$ ♦

G7. Introducing Characters

Since the trace of g‘s action is important, let’s define:

Definition. Let $\rho :G\to GL(V)$ be a representation of G. Its character is the function

$\chi:G\to\mathbf{C}, \quad \chi(g) = \text{tr}(\rho(g)).$

Looking at V as a C[G]-module, we have $\chi_V(g) = \text{tr}(g|_V).$

From G6, we see that $\dim(V^G) = \frac 1 {|G|}\sum_{g\in G} \chi_V(g).$ Suppose V and W are C[G]-modules. Let’s consider the various constructions in G3.

Direct sum : $\chi_{V\oplus W}(g) = \chi_V(g) + \chi_W(g)$ .
Tensor product : $\chi_{V\otimes W}(g) =\chi_V(g) \chi_W(g)$ .
Dual : $\chi_{V^*}(g) = \overline{\chi_V(g)}$ .

The first two equalities are easy. For the third, let’s pick a basis $B=\{e_1, \ldots, e_n\}$ of V and consider the dual basis $B^*=\{e_1^*, \ldots, e_n^*\}.$ If M is the matrix for g : V → V with respect to B, the corresponding matrix for g : V* → V* with respect to B* is given by $(M^{-1})^t.$ Hence:

$\chi_{V^*}(g) = \chi_V(g^{-1}).$

Since each g : V → V has finite order, it is diagonalisable (via expressing g in its Jordan canonical form) and every eigenvalue is a root of unity. With respect to a basis of eigenvectors, g is diagonal with entries $(\lambda_i : i=1,\ldots, n).$ Thus g^-1 is diagonal with entries $(\lambda_i^{-1} = \overline\lambda_i: i=1,\ldots, n)$ and $\text{tr}(g^{-1}) = \overline{\text{tr}(g)}$ as desired.

This gives:

Hom : $\chi_{\text{Hom}_\mathbf{C}(V,W)}(g) = \chi_{V^*\otimes W}(g) = \chi_{V^*}(g)\chi_W(g) = \overline{\chi_V(g)}\chi_W(g).$

Now consider Hom_C(V, W)^G. A C-linear map f : V → W is fixed by G if and only if:

$g\circ f \circ g^{-1} = f \iff \forall v\in V, g(f(g^{-1}v)) = f(v) \iff \forall v\in V, g(f(v)) = f(g(v))$

if and only if f is C[G]-linear. Thus $\text{Hom}_{\mathbf{C}}(V, W)^G = \text{Hom}_{\mathbf{C}[G]}(V, W)$ and:

$\begin{aligned}\dim_\mathbf{C} \text{Hom}_{\mathbf{C}[G]}(V, W) = \frac 1 {|G|} \sum_{g\in G} \overline{\chi_V(g)}\chi_W(g).\end{aligned}$

Together with Schur’s lemma, we have:

Corollary (Orthogonality of Characters).

If V and W are simple C[G]-modules, then:

$\begin{aligned}\frac 1{|G|}\sum_{g\in G} \overline{\chi_V(g)}\chi_W(g) = \dim_{\mathbf{C}} \text{Hom}_{\mathbf{C}[G]}(V,W) =\begin{cases} 1, \ \text{if } V\cong W,\\ 0,\ \text{if } V\not\cong W.\end{cases}.\end{aligned}$

G8. Class Functions

Since tr(AB) = tr(BA) for square matrices of the same size (in fact, it’s even true for rectangular matrices, as long as AB and BA are both square), we have

$\chi_V(h^{-1}gh) = \text{tr}(h^{-1}\cdot gh|_V) = \text{tr}(gh\cdot h^{-1}|V) = \text{tr}(g|_V) = \chi_V(g).$

Definition. A function $\psi : G\to \mathbf{C}$ is called a class function if $\psi(h^{-1}gh) = \psi(g)$ for any $g,h\in G.$

Thus, the character of a representation is a class function. Define an inner product between class functions $\psi, \phi:G\to \mathbf{C}$ as:

$\begin{aligned}\left<\psi, \phi\right> := \frac 1 {|G|} \sum_{g\in G} \psi(g)\overline{\phi(g)}.\end{aligned}$

From G7, the set of characters of simple C[G]-modules (or irreducible representations) forms an orthonormal set. In fact, we can say more:

Theorem. The set of irreducible characters forms an orthonormal basis for the space of all class functions.

Proof.

It suffices to show that the irreducible characters span the space of class functions. Otherwise, there’s a class function f : G → C which is orthogonal to all irreducible characters. Take the element $\sigma := \frac 1 {|G|}\sum_{g\in G} f(g)g \in \mathbf{C}[G].$ For any simple C[G]-module V, σ acts on V as a C-linear map. Now,

$\begin{aligned}h\cdot\sigma(v) = \frac 1 {|G|}\sum_{g\in G} f(g)hg(v) = \frac 1 {|G|} \sum_{g\in G} f(g) hgh^{-1}(hv) \stackrel{y=hgh^{-1}}{=}\frac 1 {|G|}\sum_{y\in G} f(y)yhv = \sigma(hv)\end{aligned}$

so it is in fact C[G]-linear and must be a scalar multiple of the identity. To compute this identity, we take the trace:

$\begin{aligned} c = \frac 1 {\dim V}\text{tr}\left(\frac 1 {|G|}\sum_{g\in G} f(g) g\right) = \frac 1 {|G|\dim V} \sum_{g\in G} f(g)\chi_V(g) = \frac 1 {\dim V}\left<f, \overline\chi_V\right>.\end{aligned}$

But f is orthogonal to each $\overline\chi_V = \chi_{V^*}$ so c=0. Thus σ=0 on an irreducible representation V and hence on any representation as well. In particular, σ=0 on C[G] itself so 0 = σ·1 = σ. ♦

Corollary. The number of irreducible representations of G is the number of its conjugancy classes m.

The m × m table comprising of values $\chi_V(g)$ is called a character table.

Note.

On an intuitive level, characters thus provide an excellent way to analyse non-abelian groups. Indeed, if G is “highly non-abelian” (e.g. simple), it has fewer conjugancy classes and thus a smaller character table. On the other hand, abelian groups have the largest possible character table since every conjugancy class has size 1.

G9. Degrees of Irreducible Characters

If V is a C[G]-module, its dimension as a C-space is called its degree.

Let’s take the obvious C[G]-module, i.e. C[G] itself (on the representation side, this gives rise to the regular representation we saw in example 3, G1). Writing C[G] as a direct sum of irreducible representations $V_1^{d_1}\oplus V_2^{d_2} \oplus \ldots \oplus V_m^{d_m}$ , the corresponding character gives:

$\chi_{reg} = d_1\chi_1 + d_2\chi_2 + \ldots + d_m \chi_m$ , where each $d_i \ge 0$ .

To compute each d_i, we take the inner product $d_i = \left<\chi_{reg}, \chi_i\right>$ . But $\chi_{reg}$ is easy to compute: it takes e to |G| and all other $g\in G$ to zero (since if g≠e, the action of g on G via left-multiplication has no fixed point). Thus, $d_i = \frac 1 {|G|} |G|\cdot \chi_i(1)=\chi_i(1) =\dim_{\mathbf{C}} V_i.$

We have thus proven:

Theorem. If V is an irreducible representation of G, the number of times it occurs in the regular representation is dim(V).

Corollary. $|G| = \sum_{i=1}^m d_i^2$ . [ Follows from $\chi_{reg}(1) = \sum_{i=1}^m d_i \chi_i(1).$ ]

Corollary. If G is abelian, then each $d_i = 1$ . [ This follows from: size of character table is |G| × |G|. ]

Example of a Character Table

Here’s the character table of the symmetric group S₄. It’s 5 × 5 since the number of conjugancy classes of S₄ is p(4) = 5. The size of each conjugancy class is written in square brackets [s].

In the next installation, we’ll see how this table is obtained.