## 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 (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 TV → 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

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

1. 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 RK[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:

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 fV → W written in the form of a huge lookup table (vf(v)) such that each v occurs exactly once on the left. Now let G act on the entire table by replacing (vf(v)) with the pair (g·vg·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 WK with the trivial representation. The resulting HomK(VK) 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 HomK(VK). 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) HomK(VW) 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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