Irreducible representations of inhomogeneous linear transformations over $\mathbb{F}_q$

Question

Let $\mathbb{F}_q$ be a finite field with $q$ elements and let $G$ be the group of inhomogeneous linear transformations over $\mathbb{F}_q$, that is $x\mapsto ax+b$ for some $a\in \mathbb{F}_q^*$ and $b\in\mathbb{F}_q$. Now I have to find all the irreducible representations of $G$. The hint in the book is that we should look at $V=\{f\colon \mathbb{F}_q\rightarrow \mathbb{C}\colon \sum_{x\in \mathbb{F}_q}f(x)=0\}$.

The dimension of this vector space is $q-1$ since we can choose the first $q-1$ 'coordinates' freely. First I need to show that this is indeed a representation, thus if $f\in V$ then also $\rho(g)f\in V$ where the action is defined by $\rho(g) f(x)=f(g(x))=f(ax+b)$. So I need to show that $\sum_{x\in \mathbb{F}_q}f(ax+b)=0$. Since the function $g\colon x\mapsto ax+b$ is bijective, this equality indeed holds. Now I have to show that this is irreducible. The order of $G$ is $q(q-1)$ and $(q-1)^2=q^2-2q+1$, so after this we have to find other representations such that the sum of squares formula is fulfilled. Can you give me hints on showing that this representation is irreducible? Thanks.

Answer 1

Choose a basis $\mathcal{B}=\{f_a-f_0\}_{a\neq0}$ for $V$, where $f_a$ is defined as follows: $f_a(a)=1$ and $f_a(b)=0$ for every $b\neq a$.

Suppose $W$ is a $G$-subspace of $V$. First, note that $(u,v)\sum_a c_a f_a =\sum_a c_{ua+v}f_a$. It follows that $W$ must contain a vector where $c_0\neq 0$. Now, choose an arbitrary generator $\alpha$ of $\mathbb{F}_q^*$, since $(\alpha,0)^{q-1}=(1,0)$ and the polynomial $x^{q-1}-1$ over $\mathbb{C}$ has no multiple roots, there exists a basis for $W$ in which every element is an eigenvector of $(\alpha,0)$. By the earlier observation, the basis must include at least one vector with $c_0\neq 0$. Moreover, the equality $$ (\alpha,0)\left(\sum_{i=0}^{q-2}c_{\alpha^i}f_{\alpha^i}+c_0f_0\right)=\sum_{i=0}^{q-2}c_{\alpha^{i+1}}f_{\alpha^i}+c_0f_0 $$ shows that there must exist an eigenvector in the basis with eigenvalue $1$, taking the form $c_{\alpha}\left(\sum_{a}f_{a}-q f_0\right)$. Next, consider the action of $(1,0)-(1,-b)$ for every element $b\in\mathbb{F}_q$. This transforms the vector into $qc_\alpha(f_b-f_0)$. Therefore, $W=V$ and it follows that $V$ is irreducible.

There are $q$ conjugation classes in total: $\{(1,0)\}$, $\{(1,b)\mid b\neq 0\}$ and $\{(a,b)\mid a=a_0\}$ for each fixed $a_0\neq 1$. The remaining $q-1$ representations must be $1$ dimensional to satisfy the sum of squares formula.

The commutator group of $G$ consists of all transformations of the form $x\mapsto x+b$. The Abelianization of $G$ is isomorphic to the subgroup of scaling transformations, i.e. $\mathbb{F}_q^*$. Choose a generator $\alpha$ of $\mathbb{F}_q^*$, then the character of $\alpha$ can be expressed as $\exp\left(i\frac{2k\pi}{q-1}\right)$, where $0\le k\le q-2$. Each such character corresponds to a distinct $1$ dimensional representation of $\mathbb{F}_q^*$.