Rankin-Selberg convolution : normalization issues I was told several times on MO that if  $ F : s\mapsto\sum_{n>0}\frac{a_{n}}{n^{s}}$ and  $ G : s\mapsto\sum_{n>0}\frac{b_{n}}{n^{s}} $ for  $ \Re(s)>1 $ are L-functions, then provided the Rankin-Selberg convolution  $F\otimes G $ is itself an L-function, it is not, generally speaking, equal to  $s\mapsto\sum_{n>0}\frac{a_{n}b_{n}}{n^{s}}$ for  $\Re(s)>1 $ . 
Is it only due to the fact that the operation of mapping  $ (a_{n},b_{n}) $ to  $ a_{n}b_{n} $ doesn't commute with the normalization operation that is required to define an L-function or are there other reasons ? 
 A: The issue stems to an unfortunate coincidence for small values of $n$, namely that the Rankin-Selberg $L$-function is, in some sense, "close to" being the Dirichlet series you write when both $F$ and $G$ have degree $1$ or $2$ Euler factors. But for $n \geq 3$, this is no longer the case.
The Rankin-Selberg product $\pi \otimes \pi'$ of two automorphic representations $\pi,\pi'$ of $\mathrm{GL}_n(\mathbb{A})$ and $\mathrm{GL}_m(\mathbb{A})$ is something that is defined locally as follows.
N.B. The $L$-function of an automorphic representation is the canonical example of an element of the Selberg class, and it is conjectured that there is a bijection between these two sets of $L$-functions.
Write $L(s,\pi) = \prod_p L(s,\pi_p)$ and $L(s,\pi') = \prod_p L(s,\pi_p')$, where the Euler factors are of the form
\[L(s,\pi_p) = \prod_{j = 1}^{n} \frac{1}{1 - \alpha_{\pi,j}(p) p^{-s}}, \quad L(s,\pi_p') = \prod_{k = 1}^{m} \frac{1}{1 - \alpha_{\pi',k}(p) p^{-s}}.\]
We can write
\[L(s,\pi) = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n)}{n^s}\]
where $\lambda_{\pi}(n)$ is the multiplicative function satisfying
\[\lambda_{\pi}(p^r) = \sum_{\substack{r_1,\ldots,r_n = 0 \\ r_1 + \cdots + r_n = r}}^{r} \alpha_{\pi,1}(p)^{r_1} \cdots \alpha_{\pi,n}(p)^{r_n},\]
and similarly for $L(s,\pi')$.
Then the Rankin-Selberg $L$-function is essentially equal to $L(s,\pi \otimes \pi') = \prod_p L(s,\pi_p \otimes \pi_p')$ with
\[L(s,\pi_p \otimes \pi_p') = \prod_{j = 1}^{n} \prod_{k = 1}^{m} \frac{1}{1 - \alpha_{\pi,j}(p) \alpha_{\pi',k}(p) p^{-s}}.\]
(This is not quite correct at primes dividing the levels of $\pi$ and $\pi'$.)
For $n = m = 1$, $\pi$ and $\pi'$ are just Dirichlet characters $\chi$ and $\psi$, so that $\alpha_{\pi,1}(p) = \chi(p)$, $\alpha_{\pi',1}(p) = \psi(p)$, $\lambda_{\pi}(n) = \chi(n)$, and $\lambda_{\pi'}(n) = \psi(n)$. In this case, it is indeed the case that
\[L(s,\pi \otimes \pi') = \sum_{n = 1}^{\infty} \frac{\chi(n) \psi(n)}{n^s} = \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}.\]
For $n = m = 2$, things are a little more complicated. In this case, $\alpha_{\pi,1}(p) \alpha_{\pi,2}(p)$ is equal to $\chi(p)$ for some Dirichlet character $\chi$, and similarly $\alpha_{\pi',1}(p) \alpha_{\pi',2}(p) = \psi(p)$ for some Dirichlet character $\psi$. (More precisely, $\pi$ and $\pi'$ are associated to modular forms, and $\chi$ and $\psi$ are their nebentypen.) Then $L(s,\pi \otimes \pi')$ is not equal to $\sum_{n = 1}^{\infty} \lambda_{\pi}(n) \lambda_{\pi'}(n) n^{-s}$, but it is surprisingly close to being so:
\[L(s,\pi \otimes \pi') = L(2s,\chi \psi) \sum_{n = 1}^{\infty} \frac{\lambda_{\pi}(n) \lambda_{\pi'}(n)}{n^s}.\]
(Again, this is not quite true; one has to correct the Euler factors at the bad primes.)
However, if $n$ or $m$ are at least $3$, then $\sum_{n = 1}^{\infty} \lambda_{\pi}(n) \lambda_{\pi'}(n) n^{-s}$ is, in some sense, very far from being equal to $L(s,\pi \otimes \pi')$. You can check this manually by comparing the Euler products of each of these two Dirichlet series.

I hope this answers your question somewhat. I have no idea what you actually mean by "Is it only due to the fact that the operation of mapping $(a_n,b_n)$ to $a_n b_n$ doesn't commute with the normalization operation that is required to define an L-function or are there other reasons ?"
