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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 ?

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  • $\begingroup$ I guess you meant "by <b>multiplying</b> by the missing Euler factor" ? $\endgroup$ – Sylvain Julien Nov 11 '18 at 20:43
  • $\begingroup$ How come html code doesn't seem to work ? $\endgroup$ – Sylvain Julien Nov 11 '18 at 20:45
  • $\begingroup$ @reuns, this is not the reason why this Dirichlet series is not the Rankin-Selberg $L$-function. $\endgroup$ – Peter Humphries Nov 12 '18 at 13:08
  • $\begingroup$ @PeterHumphries Tks I confused $\zeta(2s)E_s(z)=\sum_{n,m} Im( nz+m)^s$ with $E_s(z)=\sum_{(n,m)=1} Im( nz+m)^s$ needed for the Rankin-Selberg convolution. It is $\zeta(2s)E_s(z)$ having a functional equation $s \to 1-s$ while both are modular in $z$ $\endgroup$ – reuns Nov 13 '18 at 15:12
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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 ?"

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  • $\begingroup$ Thank you very much, Peter. I guess "nebentypen" is the plural of "nebentypus" ? Is it a German word ? $\endgroup$ – Sylvain Julien Nov 12 '18 at 14:02
  • $\begingroup$ Yep. It's also just called the central character of a modular form. $\endgroup$ – Peter Humphries Nov 12 '18 at 14:06
  • $\begingroup$ Suppose that for a given $ p $ $ \sigma $ is a permutation of the $ n $ numbers $ \alpha_{\pi,j}(p) $ and $ \sigma' $ a permutation of the $ m $ numbers $ \alpha_{\pi',k}(p) $, and write $ L_{\sigma}(s,\pi_{p})=\prod_{j=1}^{n}\frac{1}{1-\sigma(\alpha_{\pi,j}(p))p^{-s}} $ and similarly $ L_{\sigma'}(s,\pi'_{p}) $ . Can we write $L_{\sigma\otimes\sigma'}(s,\pi_{p}\otimes\pi'_{p})=\prod_{j=1}^{n}\prod_{k=1}^{m}\frac{1}{1-\sigma(\alpha_{\pi,j}(p))\sigma'(\alpha_{\pi',k}(p))p^{-s}} $? $\endgroup$ – Sylvain Julien Nov 13 '18 at 11:04
  • $\begingroup$ The local $L$-function$L(s,\pi_p)$ is independent of the ordering of the Satake parameters $\alpha_{\pi,1}(p),\ldots,\alpha_{\pi,n}(p)$, so permutations of them leave the $L$-function unchanged. $\endgroup$ – Peter Humphries Nov 13 '18 at 11:06
  • $\begingroup$ Yes, that's precisely my goal, to preserve the L-function. My question was more about the correctness of the notations I used. $\endgroup$ – Sylvain Julien Nov 13 '18 at 12:11

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