On the local factor of Rankin-Selberg L-functions

Question

I have a puzzle on the local factors of Rankin-Selberg $L$-functions. Consider two newforms on $\text{GL}_2$. Let $f$ be a newform of square-free level $N$, and $g$ a newform of trivial level. As usual, we denote by $\alpha_f(p),\beta_f(p)$ (resp. $\alpha_g(p),\beta_g(p)$) the local parameters of the attached $L$-function $L(f, s)$ (resp. $L(g, s)$) at a prime $p$. It is well-known that $$L(f\times g, s)=\sum_{n=1}^\infty \frac{\lambda_{f\times g}(n)}{n^s}=\zeta^{(N)}(2s)\sum_{n=1}^\infty \frac{\lambda_{f}(n)\lambda_{ g}(n)}{n^s},$$where $\zeta^{(N)}(2s)=\prod_{p\nmid N}(1-p^{-2s})^{-1}$. In addition, as far as I know, the local factors for $L(f\times g, s)$ which admits a Euler product $L(f\times g, s)=\prod_{p\nmid N}L_p(f\times g, s)$$\prod_{p\mid N}L^\prime_p(f\times g, s)$ is given by \begin{align}L_p(f\times g, s)=& \left (1-\frac{\alpha_f(p)\alpha_g(p)}{p^s} \right)^{-1} \left (1-\frac{\beta_f(p)\alpha_g(p)}{p^s} \right)^{-1} \\ &\left (1-\frac{\beta_f(p)\beta_g(p)}{p^s} \right)^{-1} \left (1-\frac{\beta_f(p)\beta_g(p)}{p^s} \right)^{-1};\end{align} while, for $p|N$, $$L^\prime_p(f\times g, s)=\left (1-\frac{\lambda_f(p)\alpha_g(p)}{p^s} \right)^{-1} \left (1-\frac{\lambda_f(p)\beta_g(p)}{p^s} \right)^{-1}. $$

My question is how about the exact local factors for the $L$-function $$L(\pi, s)=\sum_{n=1}^\infty \frac{\lambda^2_{f\times g}(n)}{n^s}?$$If one writes $L(\pi, s)=\prod_{p\nmid N}L_p(\pi, s)\prod_{p\mid N}L^\prime_p(\pi, s)$. How to write the two local factors $L_p(\pi, s)$ and $L^\prime_p(\pi, s)$ by means of the local parameters?

If any expert here leans some knowledge about this question, please help to show some guides or certain relevant references. Many many thanks.

Thanks in advance.

EDIT: According to David Loeffler's comment, the notation of this $L$-function is a bit misleading. So, probably one considers $\pi^\prime=f\times g$ and its associated Rankin-Selberg $L$-function $$L(\pi^\prime\times \widetilde{\pi^\prime},s)=\sum_{n=1}^\infty \frac{\lambda_{\pi^\prime\times \widetilde{\pi^\prime}}(n)}{n^s}$$instead, where $\widetilde{\pi^\prime}$ corresponds to the dual of ${\pi^\prime}$. This is of the Riemann type (i.e., satisfies certain functional equation of the Riemann-type, and has analytic continuation to the whole complex plane, where it is holomorphic except possibly for a pole at $s=1$), and particularly admits a factorization $$L(\pi^\prime\times \widetilde{\pi^\prime},s)=\prod_{p\nmid N}L_p(\pi^\prime\times \widetilde{\pi^\prime},s) \prod_{p\mid N}L^\prime_p(\pi^\prime\times \widetilde{\pi^\prime},s).$$So, my question reduces to how to determine the local factors of $L_p(\pi^\prime\times \widetilde{\pi^\prime},s) $ and $L^\prime_p(\pi^\prime\times \widetilde{\pi^\prime},s) $ by means of the local parameters of $f,g$?

Much much obliged for any comments/answers from the so many experts here in MO-website. Thanks in advance.

Answer 1

If we write $$L_p( f\times g, s) =\prod_{i=1}^4 \left(1 - \frac{m_i}{p^s} \right)^{-1}$$ for $p\nmid N$ and $L_p ( f\times g, s) =\prod_{i=1}^2 \left(1 - \frac{m_i}{p^s} \right)^{-1}$ then the Rankin-Selberg of this $L$-function with its dual, normalized so that the pole is at $s=1$, has local factor at $p\nmid N$ given by $$\prod_{i=1}^4 \prod_{j=1}^4 \left(1 - \frac{m_i}{m_j p^{s}}\right)^{-1}$$ and local factor at $p\mid N$ given by $$\prod_{i=1}^2 \prod_{j=1}^2 \left(1 - \frac{m_i}{m_j p^{s}}\right)^{-1}\left(1 - \frac{m_i}{m_j p^{1+s}}\right)^{-1}$$ if the nebentypus is trivial at $p$ or $$\prod_{i=1}^2 \prod_{j=1}^2 \left(1 - \frac{m_i}{m_j p^{s}}\right)^{-2}$$ if the nebentypus is nontrivial at $p$.

Plugging in the formulas (essentially what you have written) for $m_i$ in terms of the $\alpha$s and $\beta$s will give you a formula for these local factors in terms of $\alpha$s and $\beta$s, but this one is too long for me to write.

As with most questions of this type the easiest thing to do it is by Galois representations. A newform gives a Galois representation at each place $p$ by the local Langlands correspondence, and even a global Galois representation if it's holomorphic. The local $L$-function of a Galois representation is a product of $ \left(1 - \frac{\lambda_i}{p^s} \right)^{-1}$ where $\lambda_i$ are the eigenvalues of Frobenius on the inertia invariant parts. Rankin-Selberg coresponds to tensor product of Galois representations and Rankin-Selberg with the dual corresponds to tensor product with the dual representation. (Sometimes one must normalize the dual in a different way but here we want to take the usual dual to put the pole at $s=1$ instead of somewhere else). When the inertia is trivial, i.e. at $p\nmid N$, the tensor product has the effect of multiplying all the eigenvalues of Frobenius on one representation by the eigenvalues of Frobenius on the other to get the eigenvalues of Frobenius on the other, and dualizing has the effect of inverting the eigenvalues, so we get the $m_i/m_j$ formula.

When both representations are ramified, things are more complicated since the inertia invariants of the tensor product are not equal to the tensor product of the inertia invariants. In particular, this applies when we tensor a ramified representation with its dual. However, it's not too hard to calculate in each case that appears: When the nebentypus is trivial, the Galois representation associated to $f$ has inertia acting by a $2\times 2$ unipotent Jordan block, so the tensor product with $g$ has two $2\times 2$ unipotent Jordan blocks, and the same is true of the dual, except the Frobenius eigenvalues on the invariant part of the $2\times 2$ Jordan blocks on the dual are obtained by dividing the inverses of the original Frobenius eigenvalues by $p$ (because of how Frobenius acts on inertia). The tensor product of two $2\times 2$ unipotent Jordan blocks is the sum of a $3\times 3$ Jordan block and a $1\times 1$ Jordan block. The $3\times 3$ block corresponds to the product of the Frobenius eigenvalues on both sides, which is the product of an original Frobenius eigenvalue and an inverse of an original Frobenius eigenvalue divided by $p$, while the $1\times 1$ block corresponds to this times $p$, which is the product of an original Frobenius eigenvalue and an original Frobenius eigenvalue.

When the Nebentypus is trivial, the Galois representation associated to $f$ is a sum of two summands with different eigenvalues of inertia, each isomorphic to a twist of the other (since all one-dimensional representations are isomorphic to a twist of each other). When we tensor with the unramified representation $g$ the property is preserved. Then when we tensor with the dual, the inertia invariant part is the sum of the tensor products of the inertia invariant parts (which give products of Frobenius eigenvalues with their inverses) with the tensor product of the twists of the inertia invariant parts (which give products of Frobenius eigenvalues with their inverses again because the twists cancel).