Timeline for Integrals of modular forms
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 9, 2018 at 22:55 | comment | added | paul garrett | The "holomorphic" Eisenstein series are never in $L^1$, but the so-called "non-holomorphic" Eisenstein series $E_s$ are in $L^1$ for $0<\Re(s)<1$, away from poles, by knowing that they are asymptotic to their constant term(s), which are explicit and elementary: $y^s+c_sy^{1-s}$ for $c_s$ a ratio of zetas. | |
| Aug 9, 2018 at 22:52 | comment | added | Kim | It's making sense, at least. So are Eisenstein series in $L^1$? | |
| Aug 9, 2018 at 22:48 | comment | added | paul garrett | Well, since the total volume is finite, $L^2$ implies $L^1$. Holomorphic cuspforms are provably in $L^2$. Eisenstein series of all sorts are not. If "cuspform" generally includes an assumption of $L^2$-ness, then that does imply $L^1$-ness. Is this helping? | |
| Aug 9, 2018 at 22:47 | comment | added | Kim | Ah, I see. So is the integrability harder to show? | |
| Aug 9, 2018 at 22:39 | comment | added | paul garrett | $\int_{\Gamma\backslash G} f(g)\,dg=\int_{\Gamma\backslash G} f(gk)\rho(k)^{-1}\,dg$ for all $k\in K$. Replacing $g$ by $gk^{-1}$ gives $\int_{\Gamma\backslash G}f(g)\,dg=\rho(k)^{-1}\int_{\Gamma\backslash G}f(g)\,dg$ for all $k\in K$. Since $\rho$ is non-trivial, there is $k\in K$ so that $\rho(k)\not=1$, which implies that the integral is $0$. Note that this is a very general mechanism, not depending on holomorphy or all the other details here. | |
| Aug 9, 2018 at 22:24 | comment | added | Kim | I am happy to ask it in any case. Your reasoning above is a little too terse for me to follow. How do we get vanishing from $K$-equivariance again? | |
| Aug 9, 2018 at 22:16 | comment | added | paul garrett | If $2k\not=0$, then the integral is $0$, because the integrand is right $K=SO(2,\mathbb R)$ equivariant by a non-trivial character $\sim 2k$. This presumes that $f$ and $F$ are integrable... So, is this the question you really intended to ask? | |
| Aug 9, 2018 at 22:00 | history | asked | Kim | CC BY-SA 4.0 |