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Setup: Write $G = \text{SL}_2(\mathbf{R})$ and $\Gamma = \text{SL}_2(\mathbf{Z})$.

Let $f$ be a modular form on $\mathbf{H}$ of weight $2k$, so that $$f(gz) = f(z) (cz + d)^{2k} \qquad \text{for} \qquad g = \left( \begin{array}\ a & b \\ c & d \end{array} \right) \in \Gamma$$ There is an associated lift of $f$ to a function $F$ on $\Gamma \backslash G$ given by setting $$F(g) = f\left(\frac{ai + b}{ci + d}\right)(ci + d)^{-2k} \qquad \text{for} \qquad g = \left( \begin{array}\ a & b \\ c & d \end{array} \right) \in G$$ $$$$ Question: What is the value of the integral $$\int_{\Gamma \backslash G} F \, d\mu$$ where $\mu$ denotes the Haar probability measure on $\Gamma \backslash G$?

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    $\begingroup$ 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? $\endgroup$
    – paul garrett
    Aug 9, 2018 at 22:16
  • $\begingroup$ 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? $\endgroup$
    – Kim
    Aug 9, 2018 at 22:24
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    $\begingroup$ $\int_{\Gamma\backslash G} f(g)\,dg=\int_{\Gamma\backslash G} f(gk)\rho(k)^{-1}\,dg$ 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. $\endgroup$
    – paul garrett
    Aug 9, 2018 at 22:39
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    $\begingroup$ 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? $\endgroup$
    – paul garrett
    Aug 9, 2018 at 22:48
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    $\begingroup$ 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. $\endgroup$
    – paul garrett
    Aug 9, 2018 at 22:55

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