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How does one prove that an Eisenstein series (adelically formulated as in the book of Moeglin-Waldspurger) is not identically zero? Namely how does one prove that the sum $\sum_{\gamma\in P(k)\backslash G(k)}\varphi(\gamma g)$ is not identically zero, provided the sum converges? The book by M-W does not prove it, and neither could I find a reference.

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1 Answer 1

The space $P(k) \backslash G(k)$ has a discrete set of representatives in $ U(A) M(k) \backslash G(A)$. See for example the discussion after Lemma 3.3 in Gelbart-Jacquet "Forms of GL(2) from the analytic view point" for $G=GL(2)$ . I am sure somewhere in Moeglin-Waldspurger a similar lemma is quoted/proved somewhere for the more general $G$. Arthur has something similar certainly for $G$ reductive, but I remember that M-W consider also more generally finite covers etc. The main concern of these lemmas is actually the absolute convergence, but also they also provide the non-triviality.

A suitable set of representatives can be given via the Bruhat decomposition. Pick a compact set $K \subset U(A) M(k) \backslash G(A)$ containing only one representative and having non-empty interior. Let $\varphi :U(A) M(k) \backslash G(A) \rightarrow \mathbb{C}$ be a function which is positive, smooth, compactly supported on this set $K$, non-vanishing on the interior. Then your sum will not be zero only for $\gamma^{-1} g$ for $ g \in K$. This works equally well if you work modulo the center.

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Thanks for your quick reply. But how can one know that such $\varphi$ is indeed in the space of automorphic forms giving rise to the Eisenstein series? My understanding is that $\varphi$ has to be an automorphic form on the space $U(\mathbb{A})M(k)\backslash G(\mathbb{A})$ in the notation of Moegling−Waldspurger. But how is your $\varphi$ in this space? –  Windi Apr 2 '13 at 16:24
    
Sorry, wrong space of functions. The argument is the same though. I will have to look up whether my first sentence needs more explanation tomorrow. The claim there might be nontrivial and a theorem. –  Marc Palm Apr 2 '13 at 19:05

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