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Fix number field $F$, its ring of adeles $\mathbb{A}$, a "nice" algebraic group defined over $F$ (at least reductive but for my purposes I can assume simple and simply connected) and a parabolic subgroup (over $F$) $P=M\cdot U$. Fix an irreducible representation $\sigma$ of $M(\mathbb{A})$ (Either cuspidal or not, I actually care for characters) and consider a subquotient $\Pi$ (a sub, a quotient or a sub of a quotient) of the parabolic induction $Ind_{P(\mathbb{A})}^{G(\mathbb{A})} \sigma$. By Flath's theorem, if $\Pi$ is irreducible then it is factorizable in the sense that it can be written as a restirted tensor product $\otimes'_\nu \Pi_\nu$.

Here $G$ is defined over the ring of integers a.e. and hence $G(\mathcal{O}_\nu)$ is defined a.e. and hence also the notion of "spherical vector".

My question is as follows: What if $\Pi$ is not irreducible, will it still be factorizable in a similar sense? If so, is there a reference and if not can somebody suggest a counter example?

Just to make my intentions clearer, I am interested in "special values" of Eisenstein series, namely what is the representation generated by the leading term of Eisenstein series at points of holomorphicity (which won't be part of the discrete spectrum).

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It is not even true for finite direct products. Take $G = G_1 \times G_2$, $\chi_i, \chi_i'$ to be characters of $G_i$. Then let $\rho = (\chi_1 \otimes \chi_2) \oplus (\chi_1' \otimes \chi_2')$. This is 2-dimensional, and the restriction $\rho_i \simeq \chi_i \oplus \chi_i'$ of $\rho$ to each $G_i$ component is also 2-dimensional. So $\rho$ cannot be a tensor product of $\rho_1$ with $\rho_2$.

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  • $\begingroup$ But is it a parabolic induction? Maybe I should ask for something more modest, I care only for quotients and not subs. Probably your example can be cooked accordingly. Is there perhaps something that can be said for Eisenstein series? $\endgroup$
    – Matht111101111
    Commented Jun 14, 2016 at 18:11
  • $\begingroup$ @Guest111101111 Well, this argument gets you examples on $G$. You can view it as a case of parabolic induction with $M=G$. I wouldn't expect any pure factorization for Eisenstein series. Already for GL(2), the constant term of an Eisenstein series itself is not factorizable, but a sum of things which are factorizable. $\endgroup$
    – Kimball
    Commented Jun 14, 2016 at 20:36

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