Timeline for Siegel domains and cuspidal functions
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2014 at 19:02 | comment | added | paul garrett | If it is also moderate growth or $L^2$ in addition, then it will be of rapid decay in Siegel sets... but not in the part "below" the Siegel set. E.g., just on the upper half-plane, trying to integrate a Bessel function against $dx\,dy/y^2$ on the strip $0\le x\le 1$ is a problem. On the Siegel set it's fine, on the whole strip, not quite. | |
| Mar 2, 2014 at 17:12 | comment | added | prochet | and if we assume that it is an eigenfunction and also right invariant by $K$, it might still not be $L^{1}$? | |
| Mar 2, 2014 at 14:15 | comment | added | paul garrett | In some contexts, "cuspidal" might merely mean satisfying the Gelfand condition(s) $\int_{N_k\\N_{\mathbb A}} f(ng(\;dn=0$, without an eigenfunction condition, so I think one should be careful about what is presumed. And, no, $f$ will be well-behaved in a Siegel set, but not "below" a Siegel set, since that "below" part contains infinitely-many copies of a Siegel set (or fundamental domain). | |
| Mar 2, 2014 at 9:27 | comment | added | prochet | as the mirabolic contains the unipotent radical $N$ of $GL_{n}$, the definition is the usual one for a cuspidal function. An other weaker question would be, do we have that $f\in L^{1}(GL_{n}(\mathbb{A})/P_{n}(F)$? | |
| Mar 1, 2014 at 23:21 | history | answered | paul garrett | CC BY-SA 3.0 |