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$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $p$ and $q$ be two distinct primes. Let $$\Gamma_0(p)= \left\{ g\in \GL_3(\mathbb{Z}):g \equiv \left(\begin{matrix} \ast &\ast&\ast \\ \ast &\ast&\ast \\0&0&\ast \end{matrix}\right) \mod p \right\}$$ Let $F$ be a normalized Hecke-Maass form of type $\nu=(\nu_1,\nu_2)$ for the congruent subgroup $\Gamma_0(p)$ of $\SL(3, \mathbb{Z})$ with trivial nebentypus. Let $g\in S^\ast_k(q)$ be a newform of level $q$ on $\SL(2, \mathbb{Z})$ with trivial nebentypus. Let $L(s,F \otimes g)$ be the Rankin-Selberg convolution $L$-function which is given by $L(s,F \otimes g)=\sum_{n.m\ge 1}\lambda_F(n,m)\lambda_g(n)(nm^2)^{-s}$. Let $$\Lambda(s,F\otimes g)=Q(F\otimes g)^{s/2} L_\infty(s, F \otimes g)L(s, F \otimes g)$$ be the completed $L$-function, where $Q(F\otimes g)$ is the conductor of $L(s, F\otimes g)$. Then we have the functional equation \begin{equation} \Lambda(s, F\otimes g)=\epsilon(s, F \otimes g) \overline{\Lambda(F\otimes g,1-\bar{s})}. \end{equation}

My question is: whether or not the root number $\epsilon(s, F \otimes g)$ contains the coefficients $\lambda_g(n)$ of $g$? How we compute the local factors $\epsilon_v(s, F \otimes g)$ at places $v\le \infty$, especially at the places $v=\infty$ and $v=p,q$?

Note that in this paper (https://arxiv.org/pdf/1207.3421v3), the $\GL_2\times \GL_2$ Rankin-Selberg $L$-functions for different levels has been given in Section 2.3. A good detailed analysis has also been given at MO in this post Root number of the Rankin-Selberg convolution of two newforms. However, for $\GL_3\times \GL_2 $, there seem no a clear description, as far as I am concerned. So if any expert leans some information on this question, please show a guide or certain references.

Many thanks in advance.

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    $\begingroup$ The only self-dual $\mathrm{GL}_3$ Hecke-Maass forms on $\Gamma_0(p)$ are oldforms from $\mathrm{SL}_3(\mathbb{Z})$. Did you mean $\Gamma_0(p^2)$? Those are the ones arising as the adjoint lift of $\mathrm{GL}_2$ Hecke-Maass forms of level $p$. $\endgroup$
    – Peter Humphries
    Commented Dec 8 at 17:22
  • $\begingroup$ In any case, you might gain some intuition for thinking about what would happen if you replaced $F$ with the $\mathrm{GL}_3$ minimal parabolic Eisenstein series of level $1$ and trivial spectral parameters, so that $\Lambda(s,F \otimes g) = \Lambda(s,g)^3$ and $\epsilon(s,F \otimes g) = \epsilon(s,g)^3$. So of course the epsilon factor will not be independent of $g$. $\endgroup$
    – Peter Humphries
    Commented Dec 8 at 17:23
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    $\begingroup$ Sorry, my fault. The forms I consider are just the $GL_3$ Hecke-Maass forms of level $p$. remedied. $\endgroup$
    – hofnumber
    Commented Dec 8 at 17:27
  • $\begingroup$ @PeterHumphries Yes. However, in that situation, it seems that $\varepsilon(s,g)=\pm 1$? It appears that $\varepsilon(s,g)$ equals some complex number times $\lambda_g(q)\sqrt{q}$, which finally is independent of $g$. $\endgroup$
    – hofnumber
    Commented Dec 8 at 17:33
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    $\begingroup$ ? $\lambda_g(q) \sqrt{q}$ is not independent of $g$. $\endgroup$
    – Peter Humphries
    Commented Dec 8 at 18:50

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