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Background: Let $M_{f_i}, i=1,2$ be two modular motives associated to cusp forms $f_i \in S_{w_i}(\Gamma_0(N_i))$ of weight $w_i$ and level $N_i$ respectively. The Rankin-Selberg convolution associates an L-series $L(f_1\otimes f_2,s)$ to this pair of modular forms. In the framework of automorphic motives a natural question is whether the L-series $L(M,s)$ of a motive $M$ can be represented in terms of modular submotives $M_{f_i}$ as $L(M,s) \stackrel{?}{=} L(f_1\otimes f_2,s)$.

Question: Is there a (practical) test that can be applied to a given (motivic) L-series as to whether it admits a Rankin-Selberg product representation?

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  • $\begingroup$ Clearly, there are some necessary conditions that follow from the structure of the Euler factors, but what do you mean by a "practical test"? What is given and what operations can be performed? $\endgroup$ Commented Aug 20, 2010 at 2:36
  • $\begingroup$ Dear Laie, I am putting this here to direct your attention to my comment below David Hansen's answer, in case you don't otherwise notice it. $\endgroup$
    – Emerton
    Commented Aug 22, 2010 at 4:54

2 Answers 2

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If $L(s,M)$ is an irreducible degree 4 motivic L-function, and $L(s,\mathrm{sym}^2(M))$ has a pole, then either $M=f_1 \otimes f_2$ for a pair of distinct classical modular forms $f_1, f_2$, or $M=\mathrm{Asai}(f)$ for $f$ cuspidal on $GL_2(K)$, with $K/\mathbb{Q}$ quadratic. You can rule out the Asai case if $L(s,\mathrm{sym}^2(M)\otimes \chi)$ is entire for any nontrivial quadratic character (in particular, entire for $\chi$ the character of $K$). If by "practical" you mean "local", then I think the answer is no.

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  • $\begingroup$ Thanks for the reply, which is useful. What would be references in this direction? Practical for me means that if I compute by some method or another the L-function of a motive, or somebody gives me such an L-function, then as a person innocent in these matters I don't know whether it can be expressed in terms of the RS product, so I'm looking for tests that could decide this. Perhaps computable would have been the better word. $\endgroup$
    – Laie
    Commented Aug 20, 2010 at 16:21
  • $\begingroup$ Well, self-dual cuspidal representations of $GL_4$ are either orthogonal, in which case they come from $GO_4$, or symplectic, in which case they come from $GSp_4$. These cases are determined by whether the symmetric or exterior square L-function has a pole, respectively. Then the possible split forms of $GO_4$ are $P(\mathrm{Res}_{K/\mathbb{Q}}GL_2)$ for $K/\mathbb{Q}$ a quadratic etale algebra. I don't know a good reference for my actual statements, but you can extract them from what I just said + Asai's paper (do a "by hand" computation of the Euler factors of $\mathrm{sym^2 Asai}(f)$). $\endgroup$ Commented Aug 20, 2010 at 17:36
  • $\begingroup$ Dear Laie, Do you mean compute formally (say as an Euler product), in which case you can look at the Euler factors; or do you mean compute as a fuction of $s$? (So that you just have a whole bunch of numbers, obtained by evaluating your function at various values of $s$.) $\endgroup$
    – Emerton
    Commented Aug 22, 2010 at 4:52
  • $\begingroup$ In the above comment I am referring to your statement about "comput[ing] by some method or another the L-function of a motive". $\endgroup$
    – Emerton
    Commented Aug 22, 2010 at 4:53
  • $\begingroup$ The strategy is the former, i.e. I compute the L-function of the motive as an Euler product, arriving at a formal series. The goal now is to see whether this L-series can somehow or another be reduced to simpler building blocks, and the RS convolution is one of those gadgets that sometimes allow to do that. Occasionally the geometry associated with the motive provides guidance, but often it does not, hence the question for some kind of test on the L-function itself that one can actually perform. $\endgroup$
    – Laie
    Commented Aug 22, 2010 at 20:02
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Dear Laie, If the $f_i$ correspond to the motives $M_{f_i}$, then $L(f_1\otimes f_2,s)$ is the $L$-function of the product $M_{f_1}\times M_{f_2}$. (If you like, this is an interpretation of the Kunneth formula in cohomology.)

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