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notes/20090409-weinstein1

Jared Weinstein, UW, 20090409 -- Jacquet Langlands

Fourier transofrms on finite rings

$R$ finite $\mathbf{F}_q$-algebra

$\psi:R\to\mathbf{C}^*$ and $\psi(xy)=\psi(yx)$

$\pi:R^*\to{\rm Aut}(V)$ irreducible repn.

For $y\in R$,

$$\hat{\pi}(y) = \frac{1}{\sqrt{\#R}}\sum_{x\in R^*} \psi(xy) \pi(x) \in {\rm End}(V).$$

In some cases $\hat{\pi}$ is supported on $R^*$.

If $y\in R^*$, then $\hat{\pi}(y) = \hat{\pi}(1) \pi(y^{-1})$ and $\hat{\pi}(y) = \pi(y^{-1})\hat{\pi}(1)$, so $\hat{\pi}(1) = \varepsilon(\pi, \psi)$ is a scalar.

Example: $R = \mathbf{F}_q$, $\psi:\mathbf{F}_q \to \mathbf{C}^*$. $\pi:R^*\to\mathbf{C}^*$ nontrivial.

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l.7 ...frac{1}{\sqrt{q}} \sum_{x\in\mathbf{F}_q^*$
                                                   \pi(x) \psi(x).$$
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Example: $R=M_2(\mathbf{F}_q)$, $\psi(x) =\psi(tr(x))$.

$$\pi:GL_2(\mathbf{F}_q) \to GL_N(\mathbf{C})$$

Macdonald, 1980s for $GL_n$

Example:

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l.7 ...M_2({\mathbf{F}_q) \times \mathbf{F}_{q^2}$
                                                  
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.

$$\psi(x,y) = \psi(tr(x) - tr(y))$$

$$R_0^* = GL_2(\mathbf{F}_q) \times \mathbf{F}_{q^2}^*$$

Give representation sing cohomology.

$$X_0: (xy^q - x^qy)^{q-1} = 1$$

The action is by linear fractional transformations

[a,b; c,d] * (x,y) = (a*x + b*y, c*x + d*y)

Also, $\alpha \in F_q^*$ acts by $\alpha(x,y) = (\alpha^{-1} x, \alpha^{-1} y)$.


2013-05-11 18:34