$SL_2(\mathbb{F})$, decomposing $\mathbb{C}\{X\}$ into irreducible $G$-representations and dimensions Let $\mathbb{F}$ be a finite field with $q$ elements and $H = \mathbb{F}^\times$, the multiplicative group of $\mathbb{F}$. It is known that $H$ is a cyclic group of order $q - 1$, so $\widehat{H} = \text{Hom}(H, \mathbb{C}^*)$ is also a cyclic group of order $q - 1$.
Let  $G  = SL_2(\mathbb{F})$. The group $G$ acts linearly on the $2$-dimensional vector space $\mathbb{F}^2$ and fixes the origin $0 \in \mathbb{F}^2$. Hence, $G$ acts on $X := \mathbb{F}^2 \setminus \{0\}$. This is a finite set, so we have the $G$-representation in the $\mathbb{C}$-vector space $\mathbb{C}\{X\}$. The space $\text{End}_G\,\mathbb{C}\{X\} := \text{Hom}_G(\mathbb{C}\{X\}, \mathbb{C}\{X\})$, of $G$-intertwiners, is a $\mathbb{C}$-algebra with multiplication defined as a composition of intertwiners.
We also let $H$ act on $X$ by scalar multiplication $H \ni z: x \mapsto z \cdot x$. For each $\chi \in \widehat{H}$, we define a subspace of $\mathbb{C}\{X\}$ as follows:$$\mathbb{C}\{X\}^\chi := \{f \in \mathbb{F}\{X\} : f(z \cdot x) = \chi(z) \cdot f(x) \text{ for all }z \in \mathbb{F}^\times\}.$$What is the decomposition of $\mathbb{C}\{X\}$ into irreducible $G$-representations, and what are the dimensions of these simple representations?
 A: This is fairly simple to solve and does not need many of things you have introduced in the question. I'll assume $q$ is odd, the case of $q$ even is similar. One can compute the character of the permutation representation $\mathbb{C}[X]$ by counting fixed points. We only need to do this for a representative from each conjugacy class.
Let us denote by $\mu_n \leqslant \mathbb{F}^{\times}$ the subgroup $\{x \in \mathbb{F}^{\times} \mid x^n = 1\}$ of $n$th roots of unity. Let us define elements
\begin{gather*}
\mathbf{u}(\varepsilon,c) = \begin{bmatrix}
\varepsilon & c\\
0 & \varepsilon
\end{bmatrix} \in \mathrm{SL}_2(q), \quad \varepsilon \in \{\pm 1\}, c \in \mathbb{F}\\
\mathbf{d}(a) = \begin{bmatrix}
a & 0\\
0 & a^{-1}
\end{bmatrix} \in \mathrm{SL}_2(q^2), \quad a \in \mathbb{F}_{q^2}^{\times}.
\end{gather*}
For any $\xi \in \mu_{q+1}$ let us also denote by $\mathbf{d}'(\xi) \in \mathrm{SL}_2(q)$ a semisimple element with eigenvalues $\{\xi,\xi^{-1}\}$. Note that there exists an element $g \in \mathrm{GL}_2(q)$ such that $g\mathbf{d}'(\xi)g^{-1} = \mathbf{d}(\xi)$. A complete list of class representatives for $\mathrm{SL}_2(q)$ is then given by
\begin{equation*}
\mathbf{d}(\varepsilon), \mathbf{u}(\varepsilon,c), \mathbf{d}(a), \mathbf{d}'(\xi)
\end{equation*}
where $\varepsilon \in \{\pm 1\}$, $c \in \mathbb{F}_q$, $a \in \mu_{q-1}$, $\xi \in \mu_{q+1}$. Let $X_{q^2}$ be the set $(\mathbb{F}_{q^2}\oplus\mathbb{F}_{q^2})\setminus\{0\}$ then we clearly have $X \subseteq X_{q^2}$. An easy calculation shows that, for any $1 \neq \xi \in \mathbb{F}_{q^2}$, we have $\mathbf{d}(\xi)$ has no fixed points on $X_{q^2}$. This implies that for $\xi \in \mu_{q+1}$ we have $\mathbf{d}'(\xi)$ has no fixed points on $X$ because it is conjugate to an element with no fixed points. Hence one easily obtains that the character of the permutation module $\mathbb{C}[X]$ is given by
$\mathbf{d}(1)$: $q^2-1$, $\mathbf{d}(-1)$: 0, $\mathbf{u}(1,c)$: $q-1$, $\mathbf{u}(-1,c)$: 0, $\mathbf{d}(a)$: 0, $\mathbf{d}'(\xi)$: 0
With this one actually sees that this is the induced character $\mathrm{Ind}_U^G(1_U)$, where $U$ is the subgroup of uni-upper triangular matrices (a Sylow $p$-subgroup of $G$) and $1_U$ is the trivial character. The decomposition of this character is well known. See 3.2.13 of Bonnafé's "Representations of $\mathrm{SL}_2(q)$".
