Quick Guide to Character Theory (I): Foundation

Character theory is one of the most beautiful topics in undergraduate mathematics; the objective is to study the structure of a finite group G by letting it act on vector spaces. Earlier, we had already seen some interesting results (e.g. proof of the Sylow theorems) by letting G act on finite sets. Since linear algebra has much more structure, one might expect an even deeper theory.

Some prerequisites for understanding this set of notes:

basic group theory, up to group quotients and homomorphisms;
linear algebra, including tensor product of vector spaces;
elementary theory of (left) modules over non-commutative rings, including up to module quotients and homomorphisms.

[ We’ve yet to cover module theory and linear algebra; hopefully this will be rectified in the future. ]

At one point, one also needs to take the tensor product $S\otimes_R M$ , where M is an R-module and $R\subseteq S$ are (non-commutative) rings. But this is a rather minor aspect, and we’ll also describe the explicit construction so the reader can just accept some of the results at face value for now.

Throughout this document, G denotes a finite group and all linear algebra is performed over a field K. As time passes by, we’ll restrict ourselves to fields of characteristic 0, and then finally to the complex field C (or any of your favourite algebraically closed fields of characteristic 0). Also, all vector spaces over K are assumed to be of finite dimension.

Let’s begin.

G1. Group Representations and Examples

We define:

Definition. A representation of a group G is a group homomorphism $\rho : G \to GL(V),$ where V is a finite-dimensional vector space over field K.

If we fix a basis for V, then this is tantamount to giving a group homomorphism $G \to GL_n(K)$ , where n = dim(V). Thus, each element of G now corresponds to an n × n matrix with entries in K such that product in G corresponds to product of matrices.

[ Note: throughout all notes on this site, matrix representation for a linear map is obtained via $M\cdot v$ , where M is a matrix and v is a column vector. Thus, if dim(V)=m, dim(W)=n and T : V → W, then the underlying matrix has m columns and n rows, i.e. n × m. ]

Just like the case of group actions, we can think of a group representation as providing a map:

$G\times V\to V, \quad (g, v) \mapsto (\rho(g))(v)$

which is conveniently denoted g·v instead. This satisfies $e\cdot v = v$ and $(g_1 g_2)\cdot v = g_1\cdot (g_2\cdot v)$ for all group elements $g_1, g_2\in G$ and $v\in V.$ Under this notation, one also says G acts on V.

Examples

Let dim(V)=1 and G act trivially on it. Thus $\rho:G \to K^*=GL_1(V)$ takes every g to 1. We call this the trivial representation.
Suppose $G = S_n$ is the full symmetric group. Let dim(V)=1 and let G act on it via $\rho(g) = \text{sgn}(g),$ where sgn(g) = +1 if g is an even permutation and -1 if it’s odd. This is called the alternating representation. Note that it’s only available for $S_n$ and not for any old group.
Let $G = S_n$ again, and dim(V)=n be spanned by the basis $e_1, \ldots, e_n.$ Now $g\in S_n$ acts on V by taking $e_i \mapsto e_{g(i)}.$ E.g. if n = 3, the representation is:

$e \mapsto \begin{pmatrix} 1&0&0\\ 0&1&0 \\ 0&0&1\end{pmatrix},\ (1, 2) \mapsto \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\end{pmatrix},\ (1,3) \mapsto \begin{pmatrix} 0&0&1 \\ 0&1&0\\ 1&0&0\end{pmatrix},$

$(2, 3)\mapsto \begin{pmatrix} 1&0&0\\ 0&0&1\\ 0&1&0\end{pmatrix}, \ (1,2,3) \mapsto \begin{pmatrix} 0&0&1 \\ 1&0&0 \\ 0&1&0\end{pmatrix},\ (1,3,2)\mapsto \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 1&0&0\end{pmatrix}.$

Let $G = \{e, a, a^2\}$ be a cyclic group of order 3 and dim(V)=2. A representation of G is given by: $\rho(a) = \begin{pmatrix} -1 & 1\\ -1 & 0\end{pmatrix}.$ Since this matrix is of order 3, the map is well-defined.

Regular Representation

Example 3 above is clearly generalisable: if G acts on finite set X, then let V be a vector space with abstract basis $\{e_x : x\in X\}.$ Thus, dim(V) = #X. Now $g\in G$ acts on V by taking $e_x \to e_{g\cdot x}.$

In particular, any group G acts on itself by left multiplication, so this gives a representation of dimension #G. Explicitly, V is given an abstract basis $\{e_g : g\in G\}$ and the action of $h\in G$ is given by:

$\rho_{reg}(h) : V\to V, \quad e_g \mapsto e_{hg}.$

This is called the regular representation of group G. Note that example 3 is not the regular representation since in the regular representation of S₃, dim(V) = 3! = 6.

G2. The Group Algebra

We define:

Definition. Given field K and finite group G, the group algebra K[G] is a K-vector space with an abstract basis given by:

$\{e_g : g\in G\}$

and multiplication $K[G]\times K[G]\to K[G]$ given by $e_g \cdot e_h = e_{gh}$ and extended linearly.

Some concrete computations will make it much clearer. Suppose $G=S_3$ and K=C. Then a typical product of elements of C[G] looks like:

$\begin{aligned}&(\frac 1 2 e_{(1,2)} + \sqrt 2 e_{(1, 3,2)}) (-3 e_{(1,2)} + \sqrt 3 e_{(1,2,3)})\\ = &(-\frac 3 2 e_1 + \frac {\sqrt 3}2 e_{(2, 3)}) + (-3\sqrt 2 e_{(2,3)} +\sqrt 6 e_1)\\ = &(\frac 3 2 +\sqrt 6)e_1 + (\frac{\sqrt 3} 2 - 3\sqrt 2)e_{(2,3)}.\end{aligned}$

The following should now be clear.

Theorem. The group algebra K[G] is a ring which contains K as a subring. It is commutative if and only if G is abelian.

As a ring, we can talk about left modules over K[G]. These turn out to correspond precisely to representations of G.

Let’s do the easy direction first: suppose we’re given a left K[G]-module V. Then V is naturally a K-vector space and we obtain an action of G on V by restricting the left-module action $K[G] \times V \to V$ to the basis $\{e_g : g\in G\} \subset K[G].$ Since $e_g \cdot e_h = e_{gh}$ for any $g, h\in G,$ we get a representation of G on V.

Conversely, suppose G acts on V via K-linear maps, i.e. every $g\in G$ gives rise to a linear map $\rho(g) : V\to V.$ We’ll define a K[G]-module structure on V, by first decreeing that $e_g\in K[G]$ act on V via ρ(g), then extending linearly to the whole K[G]. Explicitly:

$\begin{aligned}K[G] \times V\to V\end{aligned}$ takes $\begin{aligned}\left(\sum_{g\in G} c_g e_g, v\right) \mapsto \sum_{g\in G} c_g \rho(g)(v) \in V.\end{aligned}$

Concrete Example

Consider example 4 from section G1, where $G = \{e, a, a^2\}$ is cyclic of order 3 and the representation takes a to $\begin{pmatrix} -1 & 1 \\ -1 & 0\end{pmatrix}.$ Now a typical element of K[G] is of the form:

$c_0 e + c_1 a + c_2 a^2$ where $c_0, c_1, c_2 \in K.$

The corresponding matrix is then:

$c_0 \rho(e) + c_1 \rho(a) + c_2 \rho(a)^2 = c_0\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix} + c_1 \begin{pmatrix} -1 & 1 \\ -1 & 0\end{pmatrix} + c_2 \begin{pmatrix} 0 & -1 \\ 1 & -1\end{pmatrix},$

or $\begin{pmatrix} c_0 - c_1 & c_1-c_2 \\ -c_1+c_2 & c_0-c_2 \end{pmatrix}$ . This represents the action of $c_0 + c_1 a + c_2 a^2 \in K[G]$ on V as a K[G]-module.

G3. Creating New Representations

We’ll look at ways to create new representations of G from existing ones.

A. Direct Sum

If R is a ring, then the direct sum of two R-modules is another one. In particular, this holds for R = K[G] as well. Specifically, if $\rho_1 : G \to GL(V_1)$ and $\rho_2 : G\to GL(V_2)$ are both representations, then the direct sum $V := V_1\oplus V_2$ gives:

$\rho : G \to GL(V_1 \oplus V_2), \quad g\cdot (x, y) := (g\cdot x, g\cdot y).$

If we pick bases of V₁ and V₂, then the resulting basis of $V=V_1 \oplus V_2$ gives the matrix of g : V → V as

B. Submodules and Quotients

Generally, if M is a left R-module and $N\subseteq M$ a submodule, we get a quotient module M/N. When V is a left K[G]-module, a submodule $W\subseteq V$ is said to be a G-invariant subspace. Clearly, this is a vector subspace; also, for each $g\in G,$ the action of g on V results in $g(W) \subseteq W.$ Conversely, if W is a vector subspace of V which is invariant under all $g\in G,$ then it is a K[G]-submodule.

If we pick a basis of W and extend it to V, then the matrix representation of $g\in G$ is:

C. Tensor Product

If V and W are K-vector spaces, we can take their tensor product over K: $X = V\otimes W$ . Explicitly, if $\{e_i\}_{i\in I}$ is a basis of V and $\{f_j\}_{j \in J}$ a basis of W, then $\{e_i\otimes f_j\}_{(i,j)\in I\times J}$ gives a basis of the tensor product X.

Given $g\in G$ , since the action is linear on both V and W, this induces a linear map

$\phi_g : V\otimes W\to V\otimes W, \quad v\otimes w \mapsto (gv)\otimes (gw).$

Note that $\phi_g \circ \phi_{g'} = \phi_{gg'}$ ; indeed, on elements $v\otimes w$ this is easily seen to be true:

$\phi_g(\phi_{g'}(v\otimes w)) = \phi_g((g'v)\otimes (g'w)) = g(g'v)\otimes g(g'w) = (gg')v\otimes (gg')w = \phi_{gg'}(v\otimes w).$

Since the set of all such elements spans $V\otimes W,$ the result follows. In terms of matrix representation, we get:

D. Space of Linear Functions

Suppose V and W are K[G]-modules. The space of all K-linear maps $X := \text{Hom}_K(V, W)$ is also a K[G]-module. To define the action of G on X, let’s imagine a K-linear map f : V → W written in the form of a huge lookup table (v, f(v)) such that each v occurs exactly once on the left. Now let G act on the entire table by replacing (v, f(v)) with the pair (g·v, g·f(v)). Unwinding the definition, we see that G acts on X via:

$g : \text{Hom}_K(V,W) \to \text{Hom}_K(V,W), \quad f \mapsto (g\circ f \circ g^{-1} : V\to W).$

Note that in the composition gfg^-1, the left g acts on W while the right g^-1 acts on V.

E. Dual Space.

A special case of the above is when W = K with the trivial representation. The resulting Hom_K(V, K) is known in linear algebra as the dual space V*. The above definition then gives us an action of G on V* via: $f \mapsto f\circ g^{-1}\in V^*.$

Let’s do some sanity check here. From linear algebra, there’s a canonical isomorphism:

$V^*\otimes W \cong \text{Hom}_K(V, W), \quad (f\otimes w) \mapsto (v \mapsto f(v)w).$

If both V and W are K[G]-modules, then there appears to be two different ways to define a G-action on Hom_K(V, K). Fortunately, both ways are identical; this can be checked by letting $g\in G$ act on the element $f\otimes w$ on the left and the map $v\mapsto f(v)w$ on the right.

On the left, we get $(f\circ g^{-1})\otimes (gw)$ .
On the right, we get the composition $v\overset{g^{-1}}\longrightarrow g^{-1}v \rightarrow f(g^{-1}v)w \overset{g}\longrightarrow f(g^{-1}v)\cdot gw,$ which is the image of $(f\circ g^{-1})\otimes (gw)$ .

In a Nutshell

Given a finite group G and field K, we’ve defined the group algebra K[G] which is a ring containing K. This is done by using an abstract basis $\{e_g : g\in G\}$ so that the dimension of K[G] is precisely the order of G. Product is defined via $e_g \cdot e_{g'} = e_{gg'}$ and extended linearly.

There’s a one-to-one correspondence between (1) K[G]-modules, and (2) linear representations of G on K-vector spaces.

The usual operations to construct new K[G]-modules are (A) direct sums, (B) submodules and quotients, (C) tensor products, (D) Hom_K(V, W) and (E) duals.

Everything presented so far is rather generic; in fact, one could even take K as any commutative ring and there’d be no effect on the theory thus far. In the next installation, we’ll explore the structure of K[G]-modules in greater detail.

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