# $L^2(X) \cong L^2(X',\xi)$

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.

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