Let $\Bbb{F}_q$ be a finite field, let $\delta \in \Bbb{F}_q$ be a non-square, let $\Bbb{F}_{q^2} = \Bbb{F}_q\big( \sqrt{\delta} \big)$ be the corresponding quadratic extension, and let ${\frak{H}}_q:= \Bbb{F}_{q^2} - \Bbb{F}_q$ be the corresponding finite "upper" half-plane. Introduce the following notion of distance in ${\frak{H}}_q$, namely for $z, w \in {\frak{H}}_q$ define

\begin{equation} d(z,w):= \ 4 \, {\text{N}\big( z -w\big) \over {\big( z - \overline{z} \big) \big(w - \overline{w}\big) } } \end{equation}

where $\overline{z} := z^q$ and $\text{N}(z) := zz^q$. This distance is symmetric, $\Bbb{F}_q$-valued, and is invariant under the left-action of $\text{GL}_2 \big(\Bbb{F}_q \big)$ on ${\frak{H}}_q$ by linear fractional transformations. Furthermore, upon choosing a fixed increment of distance $r \in \Bbb{F}_q$, one can define an unoriented graph structure ${\frak{H}}_q^{(r)}$on ${\frak{H}}_q$ by declaring that a pair $z, w \in {\frak{H}}_q$ are joined by an edge if $d \big(z,w\big) = r$.

In view of an earlier question I asked about the Dirichlet problem for the real upper half-plane $\frak{H}_{\Bbb{R}}$ which lurks in the back of my mind, I have the following question:

Is there a way to re-attach $\Bbb{F}_q$ to ${\frak{H}}_q^{(r)}$ as a boundary --- i.e. to endow $\Bbb{F}_{q^2}$ with an unoriented graph structure $\Bbb{F}_{q^2}^{ \, (r)}$ so that ${\frak{H}}_q^{(r)}$ sits inside $\Bbb{F}_{q^2}^{ \, (r)}$ as an induced subgraph --- so that for any $t \in \Bbb{F}_q$ the solution $u: \Bbb{F}_{q^2} \longrightarrow \Bbb{F}_q$ to the discrete Dirichlet problem $ \big( \Delta_r u \big) \big|_{{\frak{H}}_q} = 0$ with $u \big|_{\Bbb{F}_q} = \delta_{t}$ (delta-function notation) is

\begin{equation} u(z) \ = \ {1 \over {2 \sqrt{\delta}} } \, {z - \overline{z} \over {\text{N}\big( g_t \cdot z \big) }} \end{equation}

where $g_t$ is the unipotent matrix $\begin{pmatrix} 1 & {\scriptstyle -}t \\ 0 & 1 \end{pmatrix}$ and where $\Delta_r$ is the Laplacian operator for $\Bbb{F}_{q^2}$ treated as an unoriented graph with boundary $\Bbb{F}_q$ ? This expression for $u$ is direct reformulation of the Poisson kernel in the finite field context. Indeed, if we write $z$ as $x + \sqrt{\delta}y$ then

\begin{equation} u(z) \ = \ {y \over {(x-t)^2 - \delta y^2}}\end{equation}

Correction/Revision: Clearly the proposed formula for $u$ as written above only makes sense as a solution to the $\Bbb{F}_q$-valued Dirichlet problem, as opposed to a complex-valued case which principally interest me. Concerning the complex-valued case, let me restate the question in the following terms: Is it possible to re-attach $\Bbb{F}_q$ as a boundary so that for $1 \leq k \leq q$ the functions

\begin{equation} c_\chi^k (z) \ = \ \chi^k \Bigg({ z - \sqrt{\delta} \over { z + \sqrt{\delta}}} \Bigg) \ \chi \Big(z + \sqrt{\delta} \Big) \end{equation}

(where $\chi: \Bbb{F}_{q^2}^* \longrightarrow \Bbb{C}^*$ is a (non-decomposable) multiplicative character) form a basis of the space of all possible solutions to the corresponding discrete Dirichlet problem ? Clearly the manner in which $\Bbb{F}_q$ is re-attached should recognise this multiplicative character.

Second Revision:

What I have in mind here are two (of three) models for the Weil representation of $\text{GL}_2\big(\Bbb{F}_q \big)$ which I want to relate via a discrete Dirichlet problem for the finite upper half-plane: One model arises as a $(q-1)$ dimensional subspace of $\Bbb{C}\big[ {\frak{H}}_q \big]$ while the other is supported on $\Bbb{C}\big[ \Bbb{F}_q^* \big]$. In both models the formulae for the action of $\text{GL}_2\big(\Bbb{F}_q \big)$ involve a choice of indecomposable multiplicative character (albeit different) $\chi, \acute{\chi}: \Bbb{F}_{q^2}^* \longrightarrow \Bbb{C}^*$. Let us denote by $\sigma_\chi: \text{GL}_2\big(\Bbb{F}_q \big) \longrightarrow \Bbb{C}\big[ {\frak{H}}_q \big]$ and $\omega_{\acute{\chi}}: \text{GL}_2\big(\Bbb{F}_q \big) \longrightarrow \Bbb{C}\big[ \Bbb{F}_q^* \big]$ the first and second models respectively then for $u: {\frak{H}}_q \longrightarrow \Bbb{C}$ and $z \in {\frak{H}}_q$ the action in the first model is given by

\begin{equation} \Big( \sigma_\chi(g) \, u \Big)(z) \ = \ u \Bigg( {az + b \over {cz+d}} \Bigg) \, \chi \Big(cz + d \Big) \end{equation}

\begin{equation} \text{with} \quad g \in \text{GL}_2\big(\Bbb{F}_q \big) \quad \text{and} \quad g^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation}

See, for example, the book of Audrey Terras for a treatment of the second model.

Now, rather than attaching $\Bbb{F}_q$ to ${\frak{H}}_q^{(r)}$ let us instead attach $\Bbb{F}_q^*$ to ${\frak{H}}_q^{(r)}$ and let this attachment recognise $\chi$ in some manner (to be determined). In this way $\Bbb{F}_{q^2}^*$ is endowed with the structure of an unoriented graph containing ${\frak{H}}_q^{(r)}$ as an induced subgraph.

Now given any boundary condition $f: \Bbb{F}_q^* \longrightarrow \Bbb{C}$ there exists a unique complex-valued function $u_f: \Bbb{F}_{q^2}^* \longrightarrow \Bbb{C}$ such that $u_f \big|_{\Bbb{F}_q^*} = f$ and $\big( \Delta_r u_f \big) \big|_{ {\frak{H}}_q } = 0$ where

\begin{equation} \big( \Delta_r u_f \big)(z) \ = \ u_f(z) \ - \ {1 \over {\text{val}(z)}} \ \sum_{z \, \sim \, w} \ u_f(w) \end{equation}

for any $z \in \Bbb{F}_{q^2}^*$ and where the sum is taken over all $w \in \Bbb{F}_{q^2}^*$ connected to $z$ by an edge and where $\text{val}(z)$ is the valency of $z$. Such a solution $u_f$ produces a "flux" on the boundary equal to $\big( \Delta_r u_f \big)\big|_{\Bbb{F}_q^*}$. The Dirichlet-to-Neumann map is the operator $\Lambda: \Bbb{C}\big[ \Bbb{F}_q^* \big] \longrightarrow \Bbb{C}\big[ \Bbb{F}_q^* \big]$ sending $f$ to $\big( \Delta_r u_f \big)\big|_{\Bbb{F}_q^*}$.

The restriction $\Delta_{{\frak{H}}_q}$ of the Laplacian $\Delta_r$ to ${\frak{H}}_q$ can be expressed as

\begin{equation} \big( \Delta_{{\frak{H}}_q} u \big)(z) \ = \ u(z) \ - \ {1 \over {q+1}} \ \sum_{d(z,w) = r} \ u(w) \end{equation}

and is $\text{GL}_2\big(\Bbb{F}_q \big)$-invariant in the sense that

\begin{equation} \Delta_{{\frak{H}}_q} \Big( \sigma(g) \, u \Big) \ = \ \sigma(g) \, \Big( \Delta_{{\frak{H}}_q} u \Big)\end{equation}


\begin{equation} \Big( \sigma(g) \, u \Big)(z) \ = \ u \Bigg( {az + b \over {cz+d}} \Bigg) \quad \text{for} \quad g \in \text{GL}_2\big(\Bbb{F}_q \big) \quad \text{with} \quad g^{-1} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{equation}

Consequently, for $u: {\frak{H}}_q \longrightarrow \Bbb{C}$, we have $\Delta_{{\frak{H}}_q}u = 0$ if and only if $\Delta_{{\frak{H}}_q} \big( \sigma(g) u \big) = 0$ for all $g \in \text{GL}_2\big(\Bbb{F}_q \big)$.

Question: Can the attachment of $\Bbb{F}_q^*$ to ${\frak{H}}_q^{(r)}$ be done in such a manner so that for $g \in \text{GL}_2\big(\Bbb{F}_q \big)$ the operators $\omega_\chi(g)$ and $\sigma_\chi(g)$ are related by a formula of the form

\begin{equation} \omega_{\acute{\chi}}(g) f \ = \ \Lambda \Big( \sigma_\chi(g) \, u_f \Big) \end{equation}




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