# Rankin-Selberg convolution and product of degrees

As I'm kinda obsessed with the Selberg class and because of the general converse conjecture, I'm still trying to get a rough idea of what automorphic representations and their L-functions as well as the properties thereof are. So, letting $n$ and $n'$ be two distinct positive integers, $\pi$ (respectively $\pi'$) an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ (respectively of $\operatorname{GL}_{n'}(\mathbb{A}_{\mathbb{Q}})$), is it true that the Rankin-Selberg convolution $\pi\times\pi'$ of $\pi$ and $\pi'$ is an automorphic representation of $\operatorname{GL}_{n.n'}(\mathbb{A}_{\mathbb{Q}})$?

• This is conjectured to be true, by "Langlands functoriality", but proven in very, very few cases. – paul garrett Jan 24 '15 at 17:27
• @paulgarrett: You should turn your comment into answer, so that this question can be closed. Perhaps you can mention the pairs $(n,n')$ when we know the statement. – GH from MO Jan 24 '15 at 18:12
• I totally agree with GH from MO. – Sylvain JULIEN Jan 24 '15 at 18:39
• @Paul Garrett: is it proven to be equivalent to the Ramanujan conjecture or not? – Sylvain JULIEN Feb 6 at 21:14
• So far as I know, we could have the Ramanujan conjecture without assertions about modularity. Also, the basic way to deduce Ramanujan is just from holomorphy of symmetric powers (as observed by Serre long ago). Modularity would just be one way to get this... More complicated chains of inference based on functoriality and modularity are of course possible... – paul garrett Feb 6 at 21:20

I first disclaim being up-to-date on the precise issue in the question! Given that:

The only truly interesting example I know to have been definitely worked out, that exactly fits the question is Ramakrishnan's result from 2000 (Annals) which proves that the Rankin-Selberg convolutions for $GL_2\times GL_2$ (when there's no pole!) are standard cuspidal $L$-functions for $GL_4$, as would be expected from Langlands Functoriality. (The case that there's a pole is the case that the convolution is of a cuspform/cuspidal-repn and its contragredient, so that the convolution factors as the symmetric square and a zeta.)

The rather boring case of $GL_1\times GL_2$ has been known for a long time...

The Cogdell-Jacquet-PiatetskiShapiro approach to modularity by converse theorems already required Ramakrishnan to treat some quite delicate aspects of the triple convolution on $GL_2$, that is, $(GL_2\times GL_2)\times GL_2$, since to prove modularity on $GL_n$ we need to know good behavior of convolutions $... \times GL_m$ with $m\le n-2$ (although it is conjectured, I think, that less would suffice).

Thus, the question amounts to asking about good behavior of triple convolutions $GL_m\times GL_n \times GL_q$ with $q$ up to $mn-2$ (with the current state of knowledge, or at least mine at last notice, about converse theorems).

So, even though Henry Kim and collaborators have exhaustively examined/exploited the Langlands-Shahidi method to see which such $L$-functions are obtained (via exceptional groups, etc), already to show that a convolution on $GL_2\times GL_3$ (without poles) came from $GL_6$, one would need $GL_2\times GL_3\times GL_q$ with $q=1,2,3,4$. I very vaguely remember seeing something (Henry Kim?) about $GL_2\times GL_3\times GL_3$ or maybe $GL_2\times GL_3\times GL_4$ or so, but I've not been to re-locate it.

In any case, there is a very finite supply of Langlands-Shahidi and/or Rankin-Selberg integral repns of the relevant triples known, with no real prospect of improvement in the near future. E.g, $GL_2\times GL_4\times GL_q$ with $q=1,2,3,4,5,6$ are surely not all known, so modularity for $GL_2\times GL_4$ would not be proven by current converse theorems.

(If someone can find those few triple cases, it would be helpful!)

• Do already existing techniques allow to hope one may prove a convolution on $GL_{m}\times GL_{n}$ where $m$ and $n$ have the same radical, comes from $GL_{m.n}$ or is it totally out of reach ? – Sylvain JULIEN Jan 28 '19 at 21:57
• I think that is out of reach... – paul garrett Jan 29 '19 at 0:43