Suppose $f$ and $g$ are weight $k$ eigenforms for the Hecke operators, normalized, with Fourier coefficients $a_{n}$ and $b_{n}$ respectively. I wanted to see if I could derive a Euler product for $$L(s,f \times \overline{g}) = \sum_{n \ge 0}\frac{a_{n}\overline{b_{n}}}{n^{s}}$$ without the use of the Satake parameters. By this I mean I do not assume $$a_{p} = \alpha_{p}+\overline{\alpha_{p}} \quad \text{and} \quad a_{p^{n}} = \frac{\alpha_{p}^{n+1}-\overline{\alpha_{p}}^{n+1}}{\alpha_{p}-\overline{\alpha_{p}}}$$ My approach so far: Using the multiplicativity of the coefficients I want to find a closed form expression for the $p$-th part $$\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}.$$ Using the recurrence relation $$a_{p^{n}} = (a_{p^{n-1}}a_{p}-p^{k-1}a_{p^{n-2}})$$ (and similarily for $\overline{b}_{n}$) I was able to compute:
\begin{align*} \sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}} &= 1+\frac{a_{p}\overline{b_{p}}}{p^{w}}+\sum_{n \ge 2}\frac{(a_{p^{n-1}}a_{p}-p^{k-1}a_{p^{n-2}})(\overline{b_{p^{n-1}}}\overline{b_{p}}-p^{k-1}\overline{b_{p^{n-2}}})}{p^{nw}} \\ &= 1+\frac{a_{p}\overline{b_{p}}}{p^{w}}+\frac{a_{p}\overline{b_{p}}}{p^{w}}\sum_{n \ge 1}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}+\frac{p^{2(k-1)}}{p^{2w}}\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}-p^{k-1}\sum_{n \ge 2}\frac{a_{p^{n-1}}a_{p}\overline{b_{p^{n-2}}}+a_{p^{n-2}}\overline{b_{p^{n-1}}}\overline{b_{p}}}{p^{nw}} \\ &= 1+\left(\frac{a_{p}\overline{b_{p}}}{p^{w}}+\frac{p^{2(k-1)}}{p^{2w}}\right)\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}-\frac{p^{k-1}}{p^{2w}}\sum_{n \ge 0}\frac{a_{p^{n-1}}a_{p}\overline{b_{p^{n-2}}}}{p^{nw}}-\frac{p^{k-1}}{p^{2w}}\sum_{n \ge 0}\frac{a_{p^{n-2}}\overline{b_{p^{n-1}}}\overline{b_{p}}}{p^{nw}} \\ &= 1+\left(\frac{a_{p}\overline{b_{p}}}{p^{w}}+\frac{p^{2(k-1)}}{p^{2w}}\right)\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}-\frac{p^{k-1}}{p^{2w}}\sum_{n \ge 0}\frac{(a_{p^{n}}+p^{k-1}a_{p^{n-2}})\overline{b_{p^{n-2}}}}{p^{nw}} \\ &-\frac{p^{k-1}}{p^{2w}}\sum_{n \ge 2}\frac{a_{p^{n-2}}(\overline{b_{p^{n}}}+p^{k-1}\overline{b_{p^{n-2}}})}{p^{nw}} \\ &= 1+\left(\frac{a_{p}\overline{b_{p}}}{p^{w}}+\frac{p^{2(k-1)}}{p^{2w}}-\frac{2p^{3(k-1)}}{p^{4w}}\right)\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n}}}}{p^{nw}}-\frac{p^{k-1}}{p^{2w}}\sum_{n \ge 0}\frac{a_{p^{n}}\overline{b_{p^{n-2}}}+\overline{b_{p^{n}}}a_{p^{n-2}}}{p^{nw}}. \end{align*}
But I can't express the last sum in terms of $a_{n}b_{n}$ using the recurrence. If I could do this, the original sum could be isolated and I would be done. Is there something subtle happening or might have I overlooked a detail? I've seen the computation of the Euler product for $L(s,f \otimes \overline{g})$ using the description of the Fourier coefficients using the Ramanujan conjecture so I'm not sure why this isn't working out.
(annoying calculation, no idea why I accepted to do it)
The recurrence relation is different at the $p$ dividing the level, so perhaps you assume level $1$. If so then
$$L(s,f)=\sum_{n\ge 1} a_nn^{-s}=\prod_p \frac1{1-a_pp^{-s}+p^{k-1-s}}$$ $$= \prod_p \frac1{(1-\alpha_p p^{-s})(1-\beta_p p^{-s})} =\prod_p (\frac{\frac{\alpha_p}{\alpha_p-\beta_p}}{1-\alpha_p p^{-s}}- \frac{\frac{\beta_p}{\alpha_p-\beta_p}}{1-\beta_p p^{-s}}) $$ ie. $$a_{p^k} = \frac{\alpha_p^{k+1}-\beta_p^{k+1}}{\alpha_p-\beta_p}$$ And simimarly for $$\overline{b_{p^k}} = \frac{\alpha_p'^{k+1}-\beta_p'^{k+1}}{\alpha_p'-\beta_p'}$$ so that
$$\sum_{n\ge 1} a_n \overline{b_n} n^{-s}=\prod_p \sum_{k\ge 0} a_{p^k} \overline{b_{p^k}}p^{-sk}$$
$$=\prod_p\sum_{k\ge 0} \frac{\alpha_p^{k+1}\alpha_p'^{k+1}+ \beta_p^{k+1}\beta_p'^{k+1}- \alpha_p^{k+1}\beta_p'^{k+1}-\alpha_p'^{k+1}\beta_p^{k+1}}{(\alpha_p-\beta_p)(\alpha_p'-\beta_p')}p^{-sk}$$
$$ = \prod_p \frac1{(\alpha_p-\beta_p)(\alpha_p'-\beta_p')} (\frac{\alpha_p\alpha_p'}{1-\alpha_p \alpha_p' p^{-s}}+\frac{\beta_p\beta_p'}{1-\beta_p \beta_p' p^{-s}} -\frac{\alpha_p\beta_p'}{1-\alpha_p \beta_p' p^{-s}} -\frac{\alpha_p'\beta_p}{1-\alpha_p' \beta_p p^{-s}}) $$ $$ = \prod_p \frac{1-\alpha_p\beta_p\alpha_p'\beta_p' p^{-2s}}{(1-\alpha_p\alpha_p'p^{-s})(1-\beta_p\beta_p'p^{-s})(1-\alpha_p\beta_p'p^{-s})(1-\alpha_p'\beta_pp^{-s})}$$ $$ = \prod_p \frac{1-p^{2k-2} p^{-2s}}{1 - p^{-s}a_p\overline{b_p}+ p^{-4s} p^{4(k-1)} - p^{-3s}p^{2(k-1)}a_p\overline{b_p} + p^{-2s}p^{k-1}(a_p^2 + \overline{b_p}^2-2p^{k-1}) }$$ Note that replacing the numerator by $1$ gives the tensor product of automorphic representations $L(s,\pi_f \otimes \overline{\pi_g})$.