(Representation Theory) Notes on Representations

Let \(G\) be a finite group, let \(F\) be a field (algebraically closed, char. 0).

A representation of \(G\) is a vector space \(V\) together with a group homomorphism \[\rho:G \to \operatorname{GL}(V).\]

If we choose a basis \(\mathcal{B}\) of \(V\), we obtain from \(\rho\) an assignment \(g \mapsto [\rho(g)]_{\mathcal{B}}\) which is a group homomorphism \(G \to \operatorname{GL}_n(F)\).

First example

Consider the finite cyclic group \(G = \mathbb{Z}/n\mathbb{Z}\) for a natural number \(n > 0\).

Suppose that \(M\) is an \(m \times m\) matrix for which \(M^n = \mathbf{I}\). Using \(M\), we get a representation \[\rho_M:\mathbb{Z}/n\mathbb{Z} \to \operatorname{GL}_m(F)\] by the rule \[\rho_M(i + n \mathbb{Z}) = M^i.\]

One might consider the question: If \(M\) and \(M'\) are two \(m \times m\) matrices for which \(M^n = (M')^n = \mathbf{I}\), when are the representations \(\rho_M\) and \(\rho_{M'}\) “the same”??

Consider the following \(3\times 3\) matrices, each of which satisfies \(M^3 = \mathbf{I}\):

\[\begin{equation*} \begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}, \begin{pmatrix} 1 & 0 & 0 \\ 0 & \zeta & 0 \\ 0 & 0 & \zeta^2 \end{pmatrix}, \begin{pmatrix} 0 & 0 & -2 \\ 3/2 & 0 & 0 \\ 0 & -1/3 & 0 \end{pmatrix}. \end{equation*}\] where \(\zeta\) is a root of \(\dfrac{T^3-1}{T-1} = T^2 + T + 1\). ¹

In fact, all three of these matrices have eigenvalues \(1,\zeta,\zeta^2\) each with multiplicity 1.

Example: two commuting matrices

Suppose that \(M,N\) are \(m \times m\) matrices for which \(M^n = \mathbf{I}\), \(N^k = \mathbf{I}\) and \(MN = NM\).

We get a representation \[\rho:\mathbb{Z}/n\mathbb{Z} \times \mathbb{Z}/k\mathbb{Z} \to \operatorname{GL}_m(F)\] by the rule \[\rho(i+n\mathbb{Z},j+k\mathbb{Z}) = M^iN^j.\]

Remark

Let \(\mu\) be an eigenvalue of the matrix \(M\), and let \(V_\mu \subset F^m\) be the corresponding eigenspace.

Since \(M\) and \(N\) commute, a homework exercise shows that \(V_\mu\) is invariant under the action of \(N\).

Then \[N_{\mid V_\mu}:V_\mu \to V_\mu\] is a linear transformation satisfying \((N_{\mid V_\mu})^k = \operatorname{id}\).

In particular, we may choose a basis of \(V_\mu\) of eigenvectors for \(N_{\mid V_\mu}\)

In this way, we see that \(V = F^m\) has a basis consisting of vectors which are simultaneously eigenvectors for \(N\) and for \(M\).

Construction of representations from group actions

Suppose that the (finite) group \(G\) acts on the (finite) set \(\Omega\). We are going to use this action to define a representation of \(G\).

We consider the vector space \(F[\Omega]\) of all \(F\)-valued functions on \(\Omega\).

For \(g \in G\), we consider the linear transformation \[\rho(g):F[\Omega] \to F[\Omega]\] defined for \(f \in F[\Omega]\) by the rule \[\rho(g)(f)(\omega) = f(g^{-1}\omega).\]

Proposition:

(a). \(\rho(g)\) is an \(F\)-linear map.

(b). \(\rho(gh) = \rho(g)\rho(h)\) for \(g,h \in G\).

(c). \(\rho(g):F[\Omega] \to F[\Omega]\) is invertible.

Proof of (b):

Let \(f \in F[\Omega]\). For \(\tau \in \Omega\) we have \[\rho(gh)f(\tau) = f((gh)^{-1}\tau) = f(h^{-1}g^{-1}\tau) = \rho(h)f(g^{-1}\tau) = \rho(g)\rho(h)f(\tau).\]

Since this holds for all \(\tau\), conclude that \[\rho(gh)f = \rho(g)\rho(h)f.\]

Since this holds for all \(f\), conclude that \(\rho(gh) = \rho(g)\rho(h)\) as required.

The Proposition shows that \[\rho:G \to \operatorname{GL}(F[\Omega])\] is a representation of \(G\).

Here is an alternate perspective on \(F[\Omega]\) and the action of \(G\).

For \(\omega \in \Omega\), consider the Dirac function \(\delta_\omega \in F[\Omega]\) defined by \[\delta_\omega(\tau) = \left\{ \begin{matrix} 1 & \omega = \tau \\ 0 & \text{else} \end{matrix} \right.\]

Proposition:

The collection \(\delta_\omega\) forms a basis for the \(F\)-vector space \(F[\Omega]\).

Sketch:

For \(f \in F[\Omega]\), we have \[f = \sum_{\omega \in \Omega} f(\omega) \delta_\omega.\] This shows that the \(\delta_\omega\) span \(F[\Omega]\).

Now, suppose \(0 = \sum_{\omega \in \Omega} a_\omega \delta_\omega\) for \(a_\omega \in F\) for each \(\omega \in \Omega\).

To prove linear independence, we must argue that \(a_\omega = 0\) for every \(\omega\).

Set \(f = \sum_{\omega \in \Omega} a_\omega \delta_\omega\). For \(\tau \in \Omega\), the function \(f\) satisfies \[f(\tau) = \sum_{\omega \in \Omega} a_\omega \delta_\omega(\tau) = a_\tau.\]

Since \(f = 0\) - i.e. “\(f\) is the function which is takes value zero at every argument” – we conclude that \(a_\tau = 0\) for each \(\tau\). This proves linear independence.

Let’s consider the action of \(G\) on these basis vectors.

Proposition:

For \(g \in G\) and \(\omega \in \Omega\), we have \[\rho(g)\delta_\omega = \delta_{g \omega}.\]

Proof:

For \(\tau \in \Omega\) we have \[\rho(g) \delta_\omega(\tau) = \delta_\omega(g^{-1}\tau) = \left\{ \begin{matrix} 1 & \omega = g^{-1}\tau \\ 0 & \text{else} \end{matrix} \right. = \left\{ \begin{matrix} 1 & g\omega = \tau \\ 0 & \text{else} \end{matrix} \right.\]

This shows that \(\rho(g)\delta_\omega(\tau) = \delta_{g\omega}(\tau)\); since this holds for all \(\tau\) we conclude \(\rho(g)\delta_\omega = \delta_{g\omega}\) as required.

Remark:

Viewing \(F[\Omega]\) as the vector space with basis \(\delta_\omega\) gives us a perhaps simpler proof that \(\rho(gh) = \rho(g)\rho(h)\).

Indeed, for \(\omega \in \Omega\) we have \[\rho(gh)\delta_\omega = \delta_{gh \omega} = \rho(g)\delta_{h\omega} = \rho(g)(\rho(h)\delta_\omega) = \rho(g)\rho(h)\delta_\omega.\]

Homomorphisms

Let \((\rho,V)\) and \((\psi,W)\) be representations of \(G\). a linear mapping \(\Phi:V \to W\) is a homomorphism of \(G\)-representations provided that for every \(v \in V\) and for every \(g \in G\) we have \[\Phi(\rho(g)v) = \psi(g) \Phi(v).\]

The \(G\)-representations \((\rho,V)\) and \((\psi,W)\) are isomorphic is there is a homomorphism of \(G\)-representations which is invertible.

Example

For any \(G\) one has the so-called trivial representation \(\mathbb{1}.\) This representation has vector space \(V = F\) (it is 1 dimensional!) and the mapping \[G \to \operatorname{GL}(V) = \operatorname{GL}_1(F) = F^\times\] is just the trivial homomorphism \(g \mapsto 1\).

One often writes \(\mathbb{1}\) for the underlying vector space \(F\).

Let \(G\) act on the (finite) set \(\Omega\). Then there is a homomorphism of \(G\)-representations \[\mathbb{1} \to F[\Omega].\] Viewing \(\mathbb{1} = F\), the scalar \(c\) is mapped to the constant function with value \(c\).

Put another way, the scalar \(c\) is mapped to the vector \[c \sum_{\omega \in \Omega} \delta_\omega.\]