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I am trying to understand the functoriality conjectures of Langlands. We know that the functoriality conjectures imply that automorphic $L$-functions of a connected reductive group are equal to products of automorphic $L$-functions of general linear groups. We know that the functoriality is known for split (also quasi-split?) orthogonal groups, from the work of Cogdell $\textit{et al}$. I am particularly interested in the following question.

Let $G = SO(q)$ be a connected reductive algebraic group over $\mathbb{Q}$ where $q$ has signature $(d,1)$, i.e. $G(\mathbb{R}) = SO(d,1)$. Let $\pi$ be an irreducible cuspidal representation of $G(\mathbb{A})$. Consider the Rankin-Selberg $L$-function $L(s,\pi\times\tilde{\pi})$, where $\bar{\pi}$ is the contragradient of $\pi$. Do we know whether this $L$-function is Eulerian (i.e. can be decomposed into local $L$-factors at local places)?

Now if we assume functoriality, the above Rankin-Selberg $L$-function should be a product of some $GL(n)$ (Godement-Jacquet?) $L$-functions and hence Eulerian. Also we know that, the local factors of $GL(n)$ $L$-functions somehow correspond to the 'eigenvalues of local Hecke operators'. Then how would those Euler factors correspond to Hecke eigenvalues of a Maass form of $SO(d,1)$? Precisely if, \begin{align*} L(s,\pi\times\tilde{\pi})&=\prod_P\prod_{i=1}^n\prod_{j=1}^n(1-\alpha_i(p)p^{-s})^{-1}(1-\overline{\alpha_j(p)}p^{-s})^{-1}\\&=\prod_p(1-A(p)p^{-s}+\dots+(-1)^{n^2}p^{-n^2s})^{-1}\\&=L(s,\phi\times\bar{\phi}), \end{align*} where $\phi$ is a Maass form of $G(\mathbb{R})$, then what is the relation between $\phi$ and $A(p)$?

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  • $\begingroup$ Regarding your first question, i.e the "Eulerianity" of the L-function you consider, wouldn't this follow from the fact that the Selberg class should be closed under tensor product (i.e Rankin-Selberg convolution on the automorphic side)? $\endgroup$ May 8, 2015 at 21:41
  • $\begingroup$ How is $L(s, \pi \times \widetilde \pi)$ for $\pi$ as above defined? Likewise $L(s, \pi \times \bar \phi)$? $\endgroup$ Nov 15, 2016 at 3:02

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