Linear representations of finite groups

Linear transformations
Let $$\mathcal{V}$$ and $$\mathcal{W}$$ be two vector spaces over the same field $$\mathbb{K}$$, and let $$T: \mathcal{V} \to \mathcal{W}$$ be a mapping between the two vector spaces. If $$T$$ fulfills



T(a \vec{v}_1 + b \vec{v}_2) = a T(\vec{v}_1) + b T(\vec{v}_2) \quad \forall \, a, b \in \mathbb{K} \;\text{and} \;\vec{v}_1, \vec{v}_2 \in \mathcal{V}, $$ then $$T$$ is said to be a linear transformation between the two vector spaces. The group of all such linear transformations when $$\mathcal{W} = \mathcal{V}$$ is called the general linear group of $$\mathcal{V}$$ and denoted $$GL(\mathcal{V})$$.

If $$\mathcal{V}$$ is finite dimensional with dimensionality $$n$$, then any element of $$GL(\mathcal{V})$$ is isomorphic to a matrix in $$GL(n, \mathbb{K})$$. Choosing a basis $$\{ \vec{e}_1,\ldots,\vec{e}_n \}$$ for $$\mathcal{V}$$, the effect of a linear representation $$T$$ is given by its effect on the basis vectors:



T \vec{e}_i = \sum_{j=1}^{n} \vec{e}_j T_{j i} $$.

Invariant subspaces
Let $$\mathcal{T}$$ be a linear transformation on $$\mathcal{V}$$. If $$\mathcal{U} \subset \mathcal{V}$$ is a subspace which is unaltered by $$\mathcal{T}$$, i.e.



\vec{u}^\prime = \mathcal{T} \vec{u} \in \mathcal{U} \quad \forall \vec{u} \in \mathcal{U} $$,

then $$\mathcal{U}$$ is said to be an invariant subspace (under $$\mathcal{T}$$).

Example
Let $$\mathcal{V} = \mathbb{R}^3$$ and let the linear transformation $$\mathcal{T}$$ pick out the $$x$$-component of any vector $$\vec{v} \in \mathbb{R}^3$$:



\mathcal{T} \vec{v} = \mathcal{T} \begin{pmatrix} v_x \\ v_y \\ v_z \end{pmatrix} = \begin{pmatrix} v_x \\ 0 \\ 0 \end{pmatrix} $$.

Then the subspace $$\mathcal{U} = \{ (u, 0, 0)^T | u \in \mathbb{R} \}$$ is an invariant subspace.

Linear group representations
A linear representation of a group $$G$$ is a mapping from the group to linear transformations on a vector space $$\mathcal{V}$$, $$\rho : G \to GL(\mathcal{V})$$, such that group multiplication is preserved:


 * $$ \rho(g_1 g_2) = \rho(g_1) \rho(g_2) \quad \forall g_1, g_2 \in G$$.

Note that to the left the multiplication is in the group $$G$$, while to the right the multiplication the combination of successive linear transformations in $$GL(\mathcal{V})$$.

Example
Let $$G$$ be any group and represent all of its elements by the unit element 1 (of $$\mathbb{K}$$). This is allways a representation (check it!), and it is for obvious reasons called the trivial representation.

Irreducible representations
If $$\rho$$ is a linear representation of $$G$$ on $$\mathcal{V}$$, we say that a subspace $$\mathcal{U} \subset \mathcal{V}$$ is an invariant subspace under the representation $$\rho$$ if


 * $$ \rho(g) \mathcal{U} \subset \mathcal{U} $$

for all group elements $$g \in G$$.

If $$\{ 0 \}$$ and $$\mathcal{V}$$ are the only invariant subspaces of $$\mathcal{V}$$ (under $$\rho$$), then the representation $$\rho$$ is said to be irreducible.

The irreducible representation can be thought of as the building blocks of which one can construct general representations of the group.

Example
Our previous example, where all group elements were represented by the unit element 1, is an irreducible representation. Since any vector multiplied by unity equals itself, each unique vector defines its own subspace under this representation.

Matrix representations
Since we are concerned with finite groups, i.e. groups with only a finite number of members, it suffices also to choose finite dimensional vector spaces $$\mathcal{V}$$. If we will choose a basis for the vector space $$\mathcal{V}$$, we can further regard all representations as matrix representations:


 * $$ \rho : G \to GL(n, \mathbb{K})$$,

where $$\mathbb{K}$$ is any field and $$n$$ is the dimension of $$\mathcal{V}$$. We will mostly be concerned with the fields of the real ($$\mathbb{R}$$) and complex ($$\mathbb{C}$$) numbers, in which case the entries of the representation matrices will be real or complex, respectively.

The components of the representation matrix are obtained from the effect of the representation on the basis vectors $$\vec{e}_i$$, $$i = 1,\ldots,n$$ where $$n$$ is the dimension of the vectors space $$\mathcal{V}$$:


 * $$ \rho(g) \vec{e}_i = \sum_{j=1}^{n} \vec{e}_j D^{(\rho)}_{j i}(g) $$.

These representation matrices have to obey



D^{(\rho)}_{j i}(g h) = \sum_{k=1}^{n} D^{(\rho)}_{j k}(g) D^{(\rho)}_{k i}(h) $$, which is nothing other than ordinary matrix multiplication.

Proof: Since $$\rho$$ is a representation, the two calculations

\begin{align} \rho(g h) \vec{e}_i &= \sum_{j=1}^{n} \vec{e}_j D^{(\rho)}_{j i}(g) \\ \rho(g h) \vec{e}_i &= \rho(g) \rho(h) \vec{e}_i = \rho(g) \sum_{k=1}^{n} \vec{e}_k D^{(\rho)}_{k i}(h) = \sum_{k=1}^{n} \rho(g) \vec{e}_k D^{(\rho)}_{k i}(h) \\ &= \sum_{k=1}^{n} \sum_{j=1}^{n} \vec{e}_j D^{(\rho)}_{j k}(g) D^{(\rho)}_{k i}(h) = \sum_{j=1}^{n} \vec{e}_j \left( \sum_{k=1}^{n} D^{(\rho)}_{j k}(g) D^{(\rho)}_{k i}(h) \right) \end{align} $$ must yield the same.

Schur's lemma
Let $$\rho$$ and $$\sigma$$ be irreducible representations of the group $$G$$ on $$\mathcal{V}$$ and $$\mathcal{W}$$. Assume that $$S:\mathcal{V} \to \mathcal{W}$$ is a linear transformation such that


 * $$S \rho(g) = \sigma(g) S, \quad \forall g \in G$$.

Then $$S$$ is either invertible or identically zero.

Proof: $$\text{Im}S = S(\mathcal{V}) = \{ S \vec{v} | v \in \mathcal{V} \}$$ is a subspace of $$\mathcal{W}$$ which is invariant under $$\sigma$$:



\sigma(g) S \vec{v} = S \rho(g) \vec{v} = S \vec{v}' \in \text{Im}S, \quad \forall \vec{v} \in \mathcal{V} $$.

Since $$\sigma$$ is irreducible this means that either
 * $$\text{Im} S = \{0\}$$, in which case $$S = 0$$, or
 * $$\text{Im} S = \mathcal{W}$$, in which case $$S$$ is onto.

We also have that $$\text{Ker} S = \{ \vec{v} \in \mathcal{V} | S \vec{v} = 0 \}$$ is an invariant subspace of $$\mathcal{V}$$ under $$\rho$$, since $$\rho(g) \vec{v}$$ belongs to the kernel if $$\vec{v}$$ does:


 * $$S \rho(g) \vec{v} = \sigma(g) S \vec{v} = 0, \quad \forall \vec{v} \in \text{Ker} S$$.

Therefore, since $$\rho$$ is irreducible either
 * $$\text{Ker}S = \{0\}$$, in which case $$S$$ is one-to-one, or
 * $$\text{Ker}S = \mathcal{V}$$, in which case $$S = 0$$.

Therefore, $$S$$ is either zero or invertible.