Suppose $G$ is a connected real reductive Lie group, $\mathfrak{g} = \text{Lie}(G)$, and $\mathcal{Z} = \mathcal{Z}[U(\mathfrak{g})]$ the center of the universal enveloping algebra of $\mathfrak{g}$.

Let $A$ denote the space of automorphic forms on $G$, which, for the sake of argument, I'll assume are functions $f:G_\mathbb{Q}\backslash G \to \mathbb{C}$ that are $K$-finite for some maximal compact $K$, $\mathcal{Z}$-finite, and of uniform moderate growth, for each $f \in A$. For each $d \in \mathbb{N}_0$, let $A_d$ denote the subspace of $A$ consisting of those automorphic forms $f$ for which there exists $C > 0$ such that $|Xf(g)| \leq C\|g\|^d$ for all $X \in U(\mathfrak{g}_\mathbb{C})$.

Every $f \in A$ is annihilated by a finite codimensional ideal $I$ in $\mathcal{Z}$. My question is about how strong a converse exists for this statement. Specifically, given finite codimensional ideals $I_1, I_2 \subset \mathcal{Z}$, if $I_1 \subsetneq I_2$, can one always find an automorphic form $f \in A$ that is annihilated by $I_1$ but not by $I_2$? What about finding $f \in A_d$? If not, are there stronger conditions one can impose on the ideals to ensure that this holds?

don'twant to consider... ? $\endgroup$ – paul garrett Jul 22 '16 at 14:58