2024-01-22--notes-RT.md

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title: (Representation Theory) Notes on Representations
date: 2024-01-22

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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.

[^1]: If $F = \mathbb{C}$ then we can take $\zeta = e^{2\pi i/3}$.

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 $$c \sum_{\omega \in \Omega} \delta_\omega.$$