Skip to main content
3 events
when toggle format what by license comment
Sep 9, 2022 at 16:04 comment added Tian An I meant that one should always work with $L^2(G(\mathbb Q)\backslash G(\mathbb A)^1)$ or something like $L^2(Z(\mathbb A)G(\mathbb Q)\backslash G(\mathbb A))$ for the reasons you mentioned.
Sep 8, 2022 at 21:37 comment added D_S So $L^2(G(\mathbb Q) \backslash G(\mathbb A))$ is really as the notation suggests, and consists of all measureable functions on $G(\mathbb A)$ which are $G(\mathbb Q)$-invariant and satisfy $\int\limits_{G(\mathbb Q) \backslash G(\mathbb A)} |f(g)|^2 dg < \infty$? This seems strange, since $G(\mathbb Q) \backslash G(\mathbb A)$ doesn't usually have finite volume.
Sep 8, 2022 at 3:20 history answered Tian An CC BY-SA 4.0