Recently, I read a notes about Sakellaridis and Venkatesh conjecture. It mentions a technique called "unfolding" and gives an example:

Let X=A\G, X'=N\G, where G=PGL(2), A={ $\left[\begin{array}{cc} * & 0 \\ 0 & 1 \end{array}\right]$ }, N={ $\left[\begin{array}{cc} 1 & * \\ 0 & 1 \end{array}\right]$ }. Assume $\xi$ is a character of $N$, and $\xi(\left[\begin{array}{cc} 1 & x \\ 0 & 1 \end{array}\right])= \psi(x)$, where $\psi$ is a nontrivial character of a p-adic local field. Then we will have $L^2(X) \cong L^2(X',\xi)$ given by $\phi \mapsto \int_N\phi(ux')\xi(u)^{-1} $ and $\phi' \mapsto \int_A\phi'(ax)$.

It seems a direct computation, but I failed doing so. Can anyone give me a detailed calculation or tell me why the name is "unfolding"? Thank you all the time.

up vote 1 down vote accepted

It is not difficult to see why the maps are inverse to one another, but I am not sure why it is called unfolding. Let $\phi\in L^2(X)$, i.e. $\phi$ is left $A$-invariant. We want to show that $$\int_{A}\int_{N} \phi(nag)\overline{\xi(n)}dnd^\times a=\phi(g).$$ Writing $n=\begin{pmatrix}1&x\\&1\end{pmatrix}$ and $a=\begin{pmatrix}y&\\&1\end{pmatrix}$ and then doing a change of variable $n\mapsto ana^{-1}$ in the $N$-integral we obtain the integral equals to $$\int_A\int_N \phi(ang)\overline{\psi(xy)}dydx.$$ Noting that $\phi$ is left $A$-invarinat and $\int_y \psi(xy)dy =\delta_{x=0}$ distributionally the claim follows.

The other direction is similar. If $\phi$ is left $N$-equivariant with $\xi$, then $$\int_A\int_N\psi(ang)\overline{\psi(n)}dnd^\times a=\int_A\phi(ag)y^{-1}\int_N\psi((y-1)x)dx.$$ The claim follows from that the last integral is $\delta_{y=1}$.

  • I understand what you say now, but $\int_F\psi(xy)dy=\delta_{x=0}$ does not hold, right? Then it is natural to use Fourier formula to get the answer. – Cooler Panda Oct 5 at 14:24
  • It holds distributionally, i.e. thinking both sides as distributions on $F$. – Subhajit Jana Oct 5 at 18:50

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