It feels like it should be. In a sense it's really almost there! But alas, it's not.

Set $G=GL_2(\mathbb{F}_q)$. Fix $\epsilon\in\mathfrak{H}_q$, such that $\epsilon^2\in\mathbb{F}_q^\times$. We define the usual action of $G$ on $\mathfrak{H}_q$ by:

$$\left( \begin{array}{ccc}
a & b \\
c & d \end{array} \right)(z)=\frac{az+b}{cz+d}$$

and also the "automorphic factor":

$$j_\chi(\left( \begin{array}{ccc}
a & b \\
c & d \end{array} \right), z)=\chi(cz+d)$$

**Exercise 0.0 (you're probably familiar with this):** Show that the above is an action, i.e. $gg'(z)=g(g'(z))$

**Exercise 0.1** Show that $j_\chi$ is a cocycle in the sense that: $j_\chi(gg',z)=j_\chi(g,g'(z))j_\chi(g',z)$

Let $T$ be the stabiliser of $\epsilon\in\mathfrak{H}_q$.

**Exercise 1:** Show that $T$ is the subgroup consisting of the matrices
$$\left( \begin{array}{ccc}
a & b\epsilon^2 \\
b & a \end{array} \right),\ \ (a,b)\not=0$$

**Exercise 2:** Show that
$$d:\left( \begin{array}{ccc}
a & b\epsilon^2 \\
b & a \end{array} \right)\mapsto a+\epsilon b$$
is an isomorphism $T\cong\mathbb{F}_{q^2}^\times$.

Define $\varphi:\mathbb{C}[\mathfrak{H}_q]\rightarrow\mathbb{C}[G]$ as:

$$\varphi(f)(g)=f(g(\epsilon))j_\chi(g,\epsilon).$$

Endow $\mathbb{C}[G]$ with an action of $G$ on the left, by:
$$(gf)(h):=f(g^{-1}h).$$

**Exercise 3:** Show that $\varphi$ is an intertwining operator, i.e: $g\varphi(f)=\varphi(\sigma_\chi(g)f)$.

**Exercise 3$\frac{1}{2}$:** Show that $\varphi$ is injective.

Define $\text{Ind}_T^G\ \chi\circ d$ to be the subspace of $\mathbb{C}[G]$ of functions:
$$\{ F\in\mathbb{C}[G]\ |\ F(gt)=F(g)\chi(d(t)), \text{for all } g\in G, t\in T \}$$

**Exercise 4:** Show that $\varphi(\mathbb{C}[\mathfrak{H}_q])=\text{Ind}_T^G\ \chi\circ d$.

So in order to answer the question, we must compute $(W(\chi), \text{Ind}_T^G\ \chi\circ d)$. By Frobenius reciprocity, this is equal to $(Res_T\ W(\chi), \chi\circ d)_T$

Given the character of $Res_T\ W(\chi)$, this is a very easy computation. It is well known that:
$$Tr_{W(\chi)}(\left( \begin{array}{ccc}
a & 0 \\
0 & a \end{array} \right))=(q-1)\chi(a),\ \ a\not=0$$
and
$$Tr_{W(\chi)}(\left( \begin{array}{ccc}
a & b\epsilon^2 \\
b & a \end{array} \right))=-\chi(a+\epsilon b)-\chi(a-\epsilon b),\ \ b\not=0$$

So, finally:

**Exercise 5:** Show that $(W(\chi), \sigma_\chi)=0$.

If you've come this far, it is now also possible to prove:

**Exercise 6:** Let $\theta\in\widehat{\mathbb{F}_{q^2}^\times}$ be a generator. Show that $(W(\chi\theta^{i(q-1)}), \sigma_\chi)=1$ for $1\le i\le q$.

And we get:
$$\sigma_\chi=\bigoplus_{1\le i\le q} W(\chi\theta^{i(q-1)})$$