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Let $\Bbb{F}_q$ be a finite field with $q$ elements, let $\Bbb{F}_{q^2}$ be its quadratic extension, and consider the finite "upper" half space ${\frak{H}}_q := \Bbb{F}_{q^2} - \Bbb{F}_q$. Apeing a construction found in one of S. Lang's immortal works one can show for $g \in \text{GL}_2(\Bbb{F}_q)$ and $f: {\frak{H}}_q \longrightarrow \Bbb{C}$ that $\sigma_\chi$ given by

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

defines a $q(q-1)$ dimensional representation of $\text{GL}_2(\Bbb{F}_q)$ on the vector space $\Bbb{C}\big[ {\frak{H}}_q \big]$ consisting of all complex-valued function on ${\frak{H}}_q$ where

\begin{equation} g^{-1} \ = \ \begin{pmatrix} a & b \\ c & d\end{pmatrix} \end{equation}

and where $\chi: \Bbb{F}_{q^2}^* \longrightarrow \Bbb{C}^*$ is a fixed (indecomposable) multiplicative character.

Is it possilbe to identify the Weil representation of $\text{GL}_2(\Bbb{F}_q)$ corresponding to $\chi$ as a direct summand of $\Bbb{C}\big[ {\frak{H}}_q \big]$ and if so, with which multiplicity does it occur ?

yours,

Ines

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

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  • $\begingroup$ Dear Dror, Thank you ! Comprehensive :) In the last line, what does the notation $\chi^{\theta}$ mean ? Am I correct in understanding that $\sigma_{\chi}$ is a direct sum of Weil representations (each with multiplicity one) --- just that $W(\chi)$ is not one of them ? This comes as a surprise to me because I always thought that Weil representatiosn do not arise through inductions --- which exercise 4 asserts. best, Ines $\endgroup$ – Ines Institoris Apr 11 '15 at 18:52
  • $\begingroup$ @Ines: You're welcome :P The latex is being a bit weird. It's not a superscript, it's just a multiplication of characters, $\chi\cdot\theta^{i(q-1)}$. Indeed the above Weil representations are components of $\sigma_\chi$. This shouldn't be too much of a surprise, since every representation of a finite group is a component of an induction, as the regular representation is such. Moreover, I would say that induction from $T$ is not only the formally correct choice to get $W(\chi)$, but also the intuitive one. But, yeah, intuition comes after working with these things for a long while. $\endgroup$ – Dror Speiser Apr 11 '15 at 19:36
  • $\begingroup$ Sure, like everything else they occur as summands of an induction --- but individually they are not induced representations. In view of your exercise 6 there are no $\text{GL}_(\Bbb{F}_q)$-invariants in this upper-half plane model --- so are there no finite analogues of modular forms? Ines $\endgroup$ – Ines Institoris Apr 12 '15 at 15:48
  • $\begingroup$ I think you are confusing some ideas, and doing the exercises should help with understanding what is going on. Exercise 4 does not claim that the Weil representations are inductions. As for modular forms, what would a function in $\sigma_\chi$ look like if it was $GL_2(\mathbb{F}_q)$-invariant? And in the global case, which are the $SL_2(\mathbb{Z})$-invariant modular forms? These are good elementary questions. $\endgroup$ – Dror Speiser Apr 12 '15 at 16:13

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