5
$\begingroup$

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 \begin{equation} \Lambda(s, f\otimes g)=\epsilon(f \otimes g) \overline{\Lambda(f\otimes g,1-\bar{s})}. \end{equation}

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.

$\endgroup$
5
$\begingroup$

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 $|\lambda_f(p)| = 1$).

$\endgroup$
  • $\begingroup$ 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 $g$? $\endgroup$ – lin Apr 26 '17 at 19:54
  • $\begingroup$ @Lin, see the latest edit. Let me know if you have any other questions. $\endgroup$ – Peter Humphries Apr 26 '17 at 23:06
  • $\begingroup$ Fantastic! This is exactly what I want. Many thanks for your time! $\endgroup$ – lin Apr 26 '17 at 23:53
  • $\begingroup$ 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? $\endgroup$ – lin Jul 14 '17 at 19:01
  • 1
    $\begingroup$ @lin, yes, I believe that's correct (and in fact is covered in the article that you linked in your question). $\endgroup$ – Peter Humphries Jul 17 '17 at 15:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.