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 ?

• I guess you meant "by <b>multiplying</b> by the missing Euler factor" ? – Sylvain Julien Nov 11 '18 at 20:43
• How come html code doesn't seem to work ? – Sylvain Julien Nov 11 '18 at 20:45
• @reuns, this is not the reason why this Dirichlet series is not the Rankin-Selberg $L$-function. – Peter Humphries Nov 12 '18 at 13:08
• @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 $z$ – reuns Nov 13 '18 at 15:12

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 ?"

• Thank you very much, Peter. I guess "nebentypen" is the plural of "nebentypus" ? Is it a German word ? – Sylvain Julien Nov 12 '18 at 14:02
• Yep. It's also just called the central character of a modular form. – Peter Humphries Nov 12 '18 at 14:06
• 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 $L_{\sigma\otimes\sigma'}(s,\pi_{p}\otimes\pi'_{p})=\prod_{j=1}^{n}\prod_{k=1}^{m}\frac{1}{1-\sigma(\alpha_{\pi,j}(p))\sigma'(\alpha_{\pi',k}(p))p^{-s}}$? – Sylvain Julien Nov 13 '18 at 11:04
• 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. – Peter Humphries Nov 13 '18 at 11:06
• Yes, that's precisely my goal, to preserve the L-function. My question was more about the correctness of the notations I used. – Sylvain Julien Nov 13 '18 at 12:11