# Root number of the Rankin-Selberg convolution of two newforms

Let $p$ and $q$ be two distinct primes. Let $f\in \mathcal{S}_k^{\ast}(pq,\psi)$ be a holomorphic newform of level $pq$, nebentypus $\psi$, and weight $k$, where $\psi = \chi_p \chi_{0(q)}$, with $\chi_p$ a primitive character modulo $p$ and $\chi_{0(q)}$ the principal character modulo $q$ (i.e. $\psi$ is an imprimitive character induced from the primitive character $\chi_p$). Let $g$ be another holomorphic newform of level $q$, weight $k_g$, and with trivial nebentypus.

Let $L(s,f \otimes g)$ be the Rankin-Selberg convolution $L$-function. 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 complete $L$-function, where $Q(f\otimes g)$ is the conductor of $L(s, f\otimes g)$. Then we have the functional equation $$\Lambda(s, f\otimes g)=\epsilon(f \otimes g) \overline{\Lambda(f\otimes g,1-\bar{s})}.$$

My question is:

What are the root number $\epsilon(f\otimes g)$ and the conductor $Q(f\otimes g)$ in this case (the conductor equals $p^2 q^2$, I think)? Can anyone calculate the local $\epsilon_v$-factor at the place $v \mid q$ for me? I know a good reference is http://tan.epfl.ch/files/content/sites/tan/files/PhMICHELfiles/RSfinal.pdf, but most of their statements are for the levels of $f$ and $g$ to be co-prime.

You need to do this via a local argument. A good reference for local components of $\mathrm{GL}_2 \times \mathrm{GL}_2$ automorphic representations is Gelbart and Jacquet, "A relation between automorphic representations of $\mathrm{GL}(2)$ and $\mathrm{GL}(3)$". For just the $\mathrm{GL}_2$ theory, see Schmidt, "Some remarks on local newforms for $\mathrm{GL}(2)$".

• At all primes $v \nmid pq$, the local epsilon factors and conductor exponents are trivial.
• The local component of $f$ at $p$ is a principal series representation $\pi_{f,p} = \omega_{f,p,1} \boxplus \omega_{f,p,2}$, where $\omega_{f,p,1}, \omega_{f,p,2}$ are character of $\mathbb{Q}_p^{\times}$ of conductor exponent $c(\omega_{f,p,1}) = 1$ and $c(\omega_{f,p,2}) = 0$, while the local component of $g$ is a spherical principal series representation $\pi_{g,p} = \omega_{g,p,1} \boxplus \omega_{g,p,2}$ with both characters unramified, so that $c(\omega_{g,p,1}) = c(\omega_{g,p,2}) = 0$. Then $\pi_{f,p} \otimes \pi_{g,p} = \omega_{f,p,1} \omega_{g,p,1} \boxplus \omega_{f,p,1} \omega_{g,p,2} \boxplus \omega_{f,p,2} \omega_{g,p,1} \boxplus \omega_{f,p,2} \omega_{g,p,2}.$ The conductor exponent is $c(\pi_{f,p} \otimes \pi_{g,p}) = c(\omega_{f,p,1} \omega_{g,p,1}) + c(\omega_{f,p,1} \omega_{g,p,2}) + c(\omega_{f,p,2} \omega_{g,p,1}) + c(\omega_{f,p,2} \omega_{g,p,2}),$ which is $1 + 1 + 0 + 0 = 2.$ The epsilon factor $\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p)$ ostensibly depends on an additive character $\psi_p$ of $\mathbb{Q}_p$, which we may choose to be unramified (i.e. $c(\psi_p) = 0$), though the global epsilon factor is independent of this. Anyway, $\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p)$ is equal to $\epsilon_p(s,\omega_{f,p,1} \omega_{g,p,1},\psi_p) \epsilon_p(s,\omega_{f,p,1} \omega_{g,p,2},\psi_p) \epsilon_p(s,\omega_{f,p,2} \omega_{g,p,1},\psi_p) \epsilon_p(s,\omega_{f,p,2} \omega_{g,p,2},\psi_p),$ which is $\left(\omega_{g,p,1}(p) \epsilon_p(s,\omega_{f,p,1},\psi_p)\right) \cdot \left(\omega_{g,p,2}(p) \epsilon_p(s,\omega_{f,p,1},\psi_p)\right) \cdot 1 \cdot 1 = \epsilon_p(s,\pi_{f,p},\psi_p)^2.$ (See Proposition 1.4 of Gelbart and Jacquet and equations (4) and (6) of Schmidt and use the fact that $g$ has principal nebentypus means that the central character $\omega_{\pi_{g,p}} = \omega_{g,p,1} \omega_{g,p,2}$ of $\pi_{g,p}$ is trivial, so that $\omega_{g,p,1}(p) \omega_{g,p,2}(p) = 1$.) As the conductor exponent is $2$, $\epsilon_p(s,\pi_{f,p} \otimes \pi_{g,p},\psi_p) = \epsilon_p\left(\frac{1}{2},\pi_{f,p} \otimes \pi_{g,p},\psi_p\right) p^{-2\left(\frac{1}{2} - s\right)}.$
• The local components of $f$ and $g$ at $q$ are special representations $\pi_{f,q} = \omega_{f,q} \mathrm{St}_q$ $\pi_{g,q} = \omega_{g,q} \mathrm{St}_q$ with $\omega_{f,q}, \omega_{g,q}$ characters of $\mathbb{Q}_q^{\times}$ that are unramified and either trivial or quadratic. The conductor exponent of $c(\pi_{f,q} \otimes \pi_{g,q})$ is $2$, as $\pi_{f,q} \otimes \pi_{g,q}$ is the isobaric sum of an unramified character of $\mathrm{GL}_1(\mathbb{Q}_p)$ and the Steinberg representation of $\mathrm{GL}_3(\mathbb{Q}_p)$, with the conductor exponent of the former being $0$ and the latter being $2$. By Proposition 1.4 of Gelbart and Jacquet and equation (11) of Schmidt, the epsilon factor $\epsilon_q(s,\pi_{f,q} \otimes \pi_{g,q},\psi_q)$ is $\epsilon_q\left(s + \frac{1}{2},\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q\right) \epsilon_q\left(s - \frac{1}{2},\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q\right) = \epsilon_q(s,\omega_{g,q} \omega_{f,q} \mathrm{St}_q, \psi_q)^2.$ One can further show that this is equal to $\epsilon_q(s,\pi_{g,q}, \psi_q)^2 = \epsilon_q(s,\pi_{f,q}, \psi_q)^2 = p^{-2(\frac{1}{2} - s)}$ (the reference for this is Section 11.12 of Goldfeld and Hundley's book combined with Schmidt's paper).
• Finally, the local component of $f$ at $\infty$ is the discrete series representation $\pi_{f,\infty} = D_{k-1}$ of weight $k$, while $\pi_{g,\infty} = D_{k_g - 1}$. Then $\pi_{f,\infty} \otimes \pi_{g,\infty} = D_{|k - k_g|} \boxplus D_{k + k_g}$, and so $\epsilon_{\infty}(s, \pi_{f,\infty} \otimes \pi_{g,\infty},\psi_{\infty}) = \epsilon_{\infty}(s,D_{|k - k_g|},\psi_{\infty}) \epsilon_{\infty}(s,D_{k + k_g},\psi_{\infty}),$ which is $i^{|k - k_g| + 1} i^{k + k_g + 1} = (-1)^{\max\{k,k_g\}+1}.$ The best reference for this is Knapp, "Local Langlands Correspondence: The Archimedean Case".

So defining the global epsilon factor $\epsilon(s,\pi_f \otimes \pi_g) = \epsilon_{\infty}(s, \pi_{f,\infty} \otimes \pi_{g,\infty},\psi_{\infty}) \prod_{p'} \epsilon_{p'}(s, \pi_{f,p'} \otimes \pi_{g,p'},\psi_{p'}),$ where $p'$ runs over all primes, the functional equation for the completed Rankin-Selberg $L$-function $\Lambda(s, \pi_f \otimes \pi_g)$ is $\Lambda(s, \pi_f \otimes \pi_g) = \epsilon(s,\pi_f \otimes \pi_g) \Lambda(1 - s, \widetilde{\pi}_f \otimes \widetilde{\pi}_g).$ By the above discussion, $\epsilon(s,\pi_f \otimes \pi_g) = (pq)^2 (-1)^{\max\{k,k_g\}+1} \epsilon_p\left(\frac{1}{2},\pi_{f,p},\psi_p\right)^2.$ Finally, we note that $\epsilon_p\left(\frac{1}{2},\pi_{f,p},\psi_p\right) = \eta_f(p)$, the pseudo-eigenvalue of $f$ corresponding to the Atkin-Lehner operator $W_p$; see equation (7.6) of Templier, "Voronoï Summation for $\mathrm{GL}(2)$", and equation (7.8) shows that $\eta_f(p) = \overline{\lambda_f}(p) \tau(\chi_p) p^{-1/2}$ (note that $|\lambda_f(p)| = 1$).

• Can you possibly tell me how do the local root numbers $\epsilon_p(\frac{1}{2},\pi_{f,p},\psi_p)$ and $\epsilon_q(\frac{1}{2},\pi_{g,q},\psi_q)$ exactly look like, in terms of the character $\psi$ and the pseudo-eigenvalues of $f$ and $g$? – lin Apr 26 '17 at 19:54
• @Lin, see the latest edit. Let me know if you have any other questions. – Peter Humphries Apr 26 '17 at 23:06
• Fantastic! This is exactly what I want. Many thanks for your time! – lin Apr 26 '17 at 23:53
• May I ask you one more question? Here the root number $(-1)^{\max\{k,k_g\}+1} \eta_f(p)^2$ turns out to be independent of $\psi(q)$. If we assume the form $g$ to be of level 1 (instead of level $q$), is the root number becoming $(-1)^{\max\{k,k_g\}+1} \eta_f(p)^2 \psi(q)$ instead? – lin Jul 14 '17 at 19:01
• @lin, yes, I believe that's correct (and in fact is covered in the article that you linked in your question). – Peter Humphries Jul 17 '17 at 15:48