How to relate Rankin triple L-function to its Dirichlet series

Question

I have a very tricky question which may look naive to many experts here.

Let $f$ be a newform of level prime $P$, and $g,h$ two newforms of level 1, respectively. These three forms $f,g,h$ are all of trivial nebentypus.

Then how to express the Rankin triple L-function $L(s,f\times g\times h)$ as something times its Dirichlet series $\sum_{n\ge }\frac{\lambda_f(n)\lambda_g(n)\lambda_h(n)}{n^s}$? Here $\lambda_f(n),\lambda_g(n),\lambda_h(n)$ denote respectively the $n$-th normalized Fourier coefficients of the these forms.

Notice that on $GL_2$, for the Rankin-Selberg $L$-function $L(s,f\times g)$ we have $$L(s,f\times g)=\zeta ^{{N}}(s)\sum_{n\ge 1} \frac{\lambda_f(n)\lambda_g(n)}{n^s},$$ where the factor $\zeta ^{{N}}(2s)=\prod_{p\nmid N}(1-p^{-2s})^{-1}$. So I wonder if there exists some specific factor so that the triple L-function $L(s,f\times g\times h)$ can be expressed in terms of the Dirichlet series; this is my concern.

I research many places in the literature but find either the papers are geometric involving integral representations from the representation theoretic point of view, or the triple L-function is defined from $L(s,f\times g\times h) \leftrightsquigarrow L(2-s,f\times g\times h)$. I wander if one has an explicit version from the point view of the classical analytic number theory for this topic to relate $L(s,f\times g\times h) \leftrightsquigarrow L(1-s,f\times g\times h)$ having the central value at $s=1/2$, which shows an relation with the Dirichlet series $\sum_{n\ge }\frac{\lambda_f(n)\lambda_g(n)\lambda_h(n)}{n^s}$? To my knowledge, the best reference is due to S Bcherer，R Schulze-Pillot (see here), but it seems very difficult to figure out the Euler factors as shown in the chapter 3 of the paper.

If any expert leans something on the question, please give a guide. Many thanks in advance!

Answer 1

It's easier to give an elementary formula for this sum directly than in terms of the sum you give, though one can give a formula in terms of this sum by taking a ratio of Euler factors.

Let $q$ be a prime different from $P$. Then we can write $\lambda_f(q) = \alpha_f + \beta_f $ where $\alpha_f \beta_f =1$ and similarly for $g$ and $h$. Then the Euler factor of $L(s, f \times g \times h)$ is $$\frac{1}{ (1 - \alpha_f \alpha_g \alpha_h q^{-s} ) (1- \alpha_f \alpha_g \beta_h q^{-s}) (1-\alpha_f \beta_g \alpha_h q^{-s}) (1- \alpha_f \beta_g \beta_h q^{-s}) (1 - \beta_f \alpha_g \alpha_h q^{-s} ) (1- \beta_f \alpha_g \beta_h q^{-s}) (1-\beta_f \beta_g \alpha_h q^{-s}) (1- \beta_f \beta_g \beta_h q^{-s})}.$$

Similarly, at the prime $P$, we can still write $\lambda_g = \alpha_g +\beta_g$ with $\alpha_g \beta_g=1$, and similarly for $h$, but now $\lambda_f = a_f= \pm 1/\sqrt{q}$. Then the local factor is simply

$$\frac{1}{ (1 - a_f \alpha_g \alpha_h P^{-s} ) (1- a_f \alpha_g \beta_h P^{-s}) (1-a_f \beta_g \alpha_h P^{-s}) (1- a_f \beta_g \beta_h P^{-s})}.$$

Taking the product of these prime factors over all primes gives you the $L$-function.

The way I know how to prove this is by taking the local Langlands correspondence and using the definition of the local $L$-factor in terms of Galois representations, but surely there is a purely automorphic proof as well.

If you want to divide this by the naive Dirichlet series, you'll get at each prime $q$ other than $P$ an Euler factor that is the inverse of some explicit polynomial of degree $6$ in $q^{-s}$, and at the prime $P$ an Euler factor $\frac{1}{ 1- a_f^2 P^{-2s}}$. You can get these formulas by factorizing the naive Dirichlet series into a product of sums over powers of primes $q$, then multiplying those sums by the denominators appearing here and cancelling terms until you obtain a polynomial.

Edit: The "correction" Euler factor at primes away from $P$ is given by the inverse of

$$ (1 - \alpha_f \alpha_g \alpha_h q^{-s} ) (1- \alpha_f \alpha_g \beta_h q^{-s}) (1-\alpha_f \beta_g \alpha_h q^{-s}) (1- \alpha_f \beta_g \beta_h q^{-s}) (1 - \beta_f \alpha_g \alpha_h q^{-s} ) (1- \beta_f \alpha_g \beta_h q^{-s}) (1-\beta_f \beta_g \alpha_h q^{-s}) (1- \beta_f \beta_g \beta_h q^{-s}) \sum_{n=0}^{\infty} \sum_{i,j,k=0}^n \alpha_f^i \beta_f^{n-i} \alpha_g^{j} \beta_g^{n-j} \alpha_h^k \beta_h^{n-k} q^{-ns} $$

The only way I know how to evaluate this is just to multiply it out. It's possible to check that the coefficient of every term with degree in $q^{-s}$ at least $7$ will cancel, so you really get a finite formula by multiplying it out, albeit a complicated one.

The coefficient of $q^{-0s}$ is $1$ since all the terms are $1$ there.

The coefficient of $q^{-s}$ is $0$ since each product of $\alpha$s and $\beta$s is canceled by a corresponding product on the other side.

The coefficient of $q^{-2s}$ is $$-\alpha_f \beta_f \alpha_g \beta_g (\alpha_h^2 + \beta_h^2)- \alpha_f \beta_f (\alpha_g^2 + \beta_g^2) \alpha_h \beta_h - (\alpha_f^2 + \beta_f^2 ) \alpha_g \beta_g\alpha_h \beta_h - 3 \alpha_f \beta_f \alpha_g \beta_g\beta_g\alpha_h \beta_h = - \lambda_f^2 - \lambda_g^2 - \lambda_h^2 +3$$ since all the terms $$\alpha_f^i \beta_f^{2-i} \alpha_g^{j} \beta_g^{2-j} \alpha_h^k \beta_h^{2-k}$$ where none of $i,j,k$ equal $1$ give $1-1$ and cancel, all the terms where exactly one of the $i,j$ equals $1$ give $1 -2 + 1$ and cancel, all the terms where exactly two of the $i,j$ equal $1$ give $1 - 4 + 2 = -1$, and all the terms where all three of the $i,j,k$ are $1$ give $1 - 8 + 4 =-3$.

For the coefficient of $q^{-3s}$, all the terms $$\alpha_f^i \beta_f^{3-i} \alpha_g^{j} \beta_g^{3-j} \alpha_h^k \beta_h^{3-k}$$

where one of the $i,j,k$ equals $0$ or $3$ cancel, and the remaining terms all contribute $1 - 8 + 13 -4 =2$, so this coefficient is

$$ 2\alpha_f \beta_f (\alpha_f + \beta_f) \alpha_g \beta_g (\alpha_g+\beta_g) \alpha_h \beta_h (\alpha_h + \beta_h) = 2 \lambda_f \lambda_g \lambda_h.$$

Computing the next three coefficients may be beyond my stamina.