2
$\begingroup$

Let $G = PGL(2, \mathbb{R})$, $\Gamma \subset SL(2, \mathbb{Z})$ a congruence subgroup, $\chi: \Gamma \to S^1$ a character, and $(\pi, L^2_0(\Gamma \backslash G, \chi))$ be the usual Hilbert space right regular representation of $G$ consisting of $L^2$ functions on $G$ satisfying $$f(\gamma g) = \chi(\gamma)f(x)$$ and the condition (via Fourier expansion at the cusps) of being cuspidal.

I learned how to deduce the fact that this $L^2$-space decomposes into irreducibles with finite multiplicity from the fact that for $\phi \in C_c^\infty(G)$, the convolved operator $$\pi(\phi) = \int_G \phi(h)\pi(h)\, dg$$ is compact when restricted to $L^2_0(\Gamma \backslash G, \chi)$. But I am not sure how to establish the compactness of this operator when $\chi$ is not trivial when restricted to the upper-triangular unipotent part of $\Gamma$. The proof I learned in Bump's book Automorphic forms and representations, Theorem 3.2.3, assumes that $\chi|_{\Gamma \cap N} = 1$, and I am having trouble salvaging the proof when this is not true.

The key step in the proof is the estimate $\lVert \pi(\phi) f \rVert_{L^\infty} \ll \lVert f \rVert_{L^2}$. In Bump's proof (which is reproduced from Gelfand--Graev--Pjateckii-Shapiro which also makes this assumption on $\chi$) this is done by writing $$\pi(\phi) f(g) = \int_{(\Gamma \cap N)\backslash G}K(g, h)f(h)\, dh,$$ where $K(g, h) = \sum_{\gamma \in \Gamma \cap N} \chi(\gamma)\phi(g^{-1}\gamma h)$. When $\chi(\gamma) = 1$ in all those terms, Bump uses the Poisson summation formula to rewrite this as a sum of things that are easily bounded, except the value at $0$ of the Fourier transform of $n \mapsto \chi(n)\phi(g^{-1}nh)$ as a function on $N$ (here $\chi$ I treat as the unique extension from $\Gamma \cap N$ to $N$). The point is then that this term has zero contribution by the assumption that $f$ and thus $\pi(\phi)f$ is cuspidal: $$\int_{(\Gamma \cap N) \backslash N} \int_{(\Gamma \cap N)\backslash G}K(ng, h)f(h)\, dh\, dn = 0$$ means that the contribution of $\int_{(\Gamma \cap N) \backslash N} \sum_{\gamma \in \Gamma \cap N} \chi(\gamma)\phi(g^{-1}n^{-1}\gamma h)$ is zero. This only seems helpful when $\chi$ is trivial, otherwise this isn't equal to the value at $0$ of the Fourier transform above, which is $\int_{(\Gamma \cap N) \backslash N} \sum_{\gamma \in \Gamma \cap N} \chi(n^{-1}\gamma)\phi(g^{-1}n^{-1}\gamma h)\, dn$. So how does one establish the desired bound in general? Can the same technique be made to work?

$\endgroup$

0

You must log in to answer this question.