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I am struggling to see why Eisenstein series is well defined, and I would greatly appreciate clarification.

Let $$ E(x, \lambda) = \sum_{\delta \in P(\mathbb{Q}) \backslash G(\mathbb{Q}) } e^{\langle H_P(\delta x), \lambda + \rho_P\rangle}, $$ where $x \in G(\mathbb{A}_{\mathbb{Q}})$ and $\lambda \in X^*(M)_{\mathbb{Q}}$. I want to understand why $e^{\langle H_P(\delta x), \lambda + \rho_P\rangle}$ is well defined with $\delta \in P(\mathbb{Q}) \backslash G(\mathbb{Q})$.

$H_P: G(\mathbb{A}_{\mathbb{Q}}) \rightarrow \mathfrak{a}_M$ is defined by sending $nmk \mapsto H_M(m)$. $nmk$ is from the Iwasawa decomposition of $G(\mathbb{A}_{\mathbb{Q}})$ and $\langle H_M(m), \lambda\rangle = \log \lvert\lambda(m)\rvert$. I am struggling to see how this is well defined on $P(\mathbb{Q}) \backslash G(\mathbb{Q})$.

PS. Here $G$ is a connected reductive group defined over $\mathbb{Q}$, $P$ is a parabolic subgroup, and $P=MN$ is a Levi decomposition.

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    $\begingroup$ $\mid \lambda (m )\mid$ is the adelic modulus. Hence , for a global $rational$ element $p$ of $P$ , the product formula says that $\mid \lambda (pm)\mid =\mid \lambda (m)\mid$. This is the reason $H_P$ is left invariant under elements of $P(\mathbbb Q)$. $\endgroup$
    – Venkataramana
    Commented May 10, 2019 at 13:37
  • $\begingroup$ I thought it made sense but turns out I'm actually still a bit confused. $H_P$ is defined by taking the $M$ part, $H_M(m)$. Is it obvious that if we take $pg$, say with $g=nmk$ then the $M$ part of the Iwasawa decomposition of $pg$ is $pm$? $\endgroup$
    – Johnny T.
    Commented May 10, 2019 at 14:24
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    $\begingroup$ Ah, yes. $p=um'$ where $u$ is in the unipotent radical and $m'\in M({\mathbb Q})$. Hence $H_P(pm)= H_P(m'm)=\mid \lambda (m'm) \mid =\mid \lambda (m)\mid$. $\endgroup$
    – Venkataramana
    Commented May 10, 2019 at 14:32
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    $\begingroup$ No, for a global point, the m part involves the maximal compact, hence m need not be rational. $\endgroup$
    – Venkataramana
    Commented May 10, 2019 at 14:38
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    $\begingroup$ Global point $x\in G$ but not in $P$. $\endgroup$
    – Venkataramana
    Commented May 10, 2019 at 14:38

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