Let $\chi$ be a character of $\mathbb{Q}^{\times}\setminus\mathbb{A}^{\times}$. We define an induced representation of $Mp_2(\mathbb{A})\simeq SL_2(\mathbb{A})\times \mathbb{C}^1$, $$I(s,\chi) := \{\Phi(g,s),\;s.t.\;\Phi((n(b)m(a)g,z),s) = \chi(a)|a|^{s+1}\Phi((g,z),s)\},$$ where $n(b)$ is a unipotent element of $SL_2(\mathbb{A})$ and $m(a)$ is the diagonal matrix $\mathrm{diag}(a,a^{-1})$ with $a\in\mathbb{A}^{\times}$. Given a factorizable section $\Phi(s) = \otimes\Phi(s)_p$ we can construct an Eisenstein series $$E(g,s,\Phi) = \sum_{\gamma\in P(\mathbb{Q})\setminus SL_2(\mathbb{Q})}\Phi(\gamma g,s).$$ Let $\Phi_{\infty}^{l}(s)$ be the normalized eigenfunction of $I_{\infty}(s,\chi)$, i.e. $$\Phi^l_{\infty}(gk,s) = \nu_{l}(k)\Phi^{l}_{\infty}(g,s),$$ where $\nu_l(k)$ is a character of $\widetilde{SO}(2)$ of weight $l$. In "Eisenstein series for $SL_2$" thm. 2.4 (https://link.springer.com/article/10.1007/s11425-010-4097-1), the authors gave an expression for the constant term of the Eisenstein series $E(\tau,s,\Phi^l\otimes\Phi_f) := v^{-l}E(g_{\tau},s,\Phi^l\otimes\Phi_f)$. In order to simplify the expression I will suppose that $\Phi_f(s) = \otimes_{p\nmid\infty}\Phi_p(s),$ with $\Phi_p(s)$ the $SL_2(\mathcal{O}_p)-$fixed vector. With this input the constant term is given by $$v^{(s+1-l)/2}\Phi_f(1)+4\pi^2(-i)^{l}v^{l-2s}\frac{2^{-s}\Gamma(s)^2}{\Gamma((s+1+l)/2)^2\Gamma((s+1-l)/2)^2}\frac{\zeta(2s)}{\zeta(2s+1)}.$$ In "Eisenstein series for $SL_2$" cor. 2.4 the authors proved that the Eisenstein series is holomorphic in $s$ except in $l = 1/2$ and $s = 1/2$ without proof. How can they deduce this from the expression of the constant term? How is the case $s = 1/2$ treated? At first appearence the constant term of the Eisenstein series is not holomorphic at $s = 1/2$ except for $l = \pm 3/2$, where the Gamma function cancels the pole of the zeta function. Is there some analytic continuation process behind? If the answer is affirmative, is there any way to compute explicitly the constant term of the Eisenstein series?
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2$\begingroup$ You misread Corollary 2.5: they first multiply the Eisenstein series by $\zeta(2s + 1)$ to get a "completed" Eisenstein series $E^{\ast}$, then conclude that $E^{\ast}$ (not $E$!) is holomorphic except possibly at $s = 1/2$. The key point is that the completed zeta function $\pi^{-s/2} \Gamma(s/2) \zeta(s)$ has poles only at $s = 0$ and $s = 1$, and otherwise is holomorphic; in particular, every term in the Fourier-Whittaker expansion in Theorem 2.4 is holomorphic. $\endgroup$– Peter HumphriesCommented Sep 29, 2021 at 20:17
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