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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}\DeclareMathOperator\sym{sym}$I encountered a question on the poles of $\GL_3\times \GL_3$ $L$-function, which needs the knowledge of the experts here.

Let $f$ be an $\SL_2$ newform of level $P$ (such as a primitive holomorphic form) and $g$ a newform with trivial level. Does any expert here know whether or not the $L$-function $L(s, \sym^2 f \times \sym^2 g)$ has a pole at $s=1$? If not, how about the upper-bound of the $L$-value at $s=1$? I guess $L(1, \sym^2 f \times \sym^2 g)\ll P^\varepsilon$, but I don't know any ideas of the proof.

If any experts here know something about this question, please give a guide sharing some of your valuable comments or give some references.

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    $\begingroup$ Hello. Note that some ladies here may possibly not appreciate to be named by "guys". $\endgroup$
    – F. C.
    Commented Jul 29, 2021 at 18:41
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    $\begingroup$ I deleted the opening phrase "guys" as inappropriate. MathOverflow is not a casual chat site, but a scientific forum. $\endgroup$
    – GH from MO
    Commented Jul 29, 2021 at 18:48
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    $\begingroup$ @ F.C. @GH from MO Really sorry if any offense. I will improve the post editing. $\endgroup$
    – hofnumber
    Commented Jul 29, 2021 at 19:05
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    $\begingroup$ 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. $\endgroup$
    – Peter Humphries
    Commented Jul 29, 2021 at 21:16
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    $\begingroup$ 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. $\endgroup$
    – Peter Humphries
    Commented Jul 29, 2021 at 21:17

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