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Jul 30, 2021 at 18:06 comment added Stopple @PeterHumphries Why not make this an answer, so the question does not go 'unanswered'?
Jul 29, 2021 at 23:46 comment added hofnumber @PeterHumphries Many thanks to your comments and reference!
Jul 29, 2021 at 21:17 comment added Peter Humphries There should be a bound of the form $L(1,\mathrm{sym}^2 f \otimes \mathrm{sym}^2 g) \ll_{\varepsilon} P^{\varepsilon}$ by an application of the main result of Xiannan Li's Ph.D. thesis.
Jul 29, 2021 at 21:16 comment added Peter Humphries Yes, there is no pole at $s = 1$. Since $f$ has level $1$ and $g$ has level $P$ (assuming that $g$ has trivial nebentypus), their symmetric square lifts are cuspidal; they are also not dual to each other since $\mathrm{sym}^2 f$ has level $1$ whereas $\mathrm{sym}^2 g$ has level $P$. The Rankin-Selberg convolution of two cuspidal automorphic forms can only have a pole at $s = 1$ if they are dual to each other.
Jul 29, 2021 at 19:08 comment added hofnumber But I really wanna know if somewhere said about $L(1, \text{sym}^2 f \times \text{sym}^2 g)$? If $f\neq g$, the property of $L(1, \text{sym}^2 f \times \text{sym}^2 g)$ has ever been studied before?
Jul 29, 2021 at 19:05 comment added hofnumber @ F.C. @GH from MO Really sorry if any offense. I will improve the post editing.
Jul 29, 2021 at 18:58 history edited LSpice CC BY-SA 4.0
`\DeclareMathOperator`; grammar
Jul 29, 2021 at 18:48 comment added GH from MO I deleted the opening phrase "guys" as inappropriate. MathOverflow is not a casual chat site, but a scientific forum.
Jul 29, 2021 at 18:47 history edited GH from MO CC BY-SA 4.0
deleted 8 characters in body
Jul 29, 2021 at 18:41 comment added F. C. Hello. Note that some ladies here may possibly not appreciate to be named by "guys".
Jul 29, 2021 at 17:05 history asked hofnumber CC BY-SA 4.0