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I heard that the Langlands functoriality conjecture implies the Ramanujan conjecture for $GL(2)$. especially for the Maass form.

There are various versions of the Langlands functoriality concerning to which groups are associated.

I am wondering which version of the Langlands functorial conjecture could prove the Ramanujan conjecture for $GL(n)$ completely?

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1 Answer 1

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Let $F$ be a number field, let $\mathbb{A}_F$ be the ring of adeles of $F$, and let $\mathcal{A}(n)$ be the set of cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_F)$ with unitary central character. For $\pi\in\mathcal{A}(n)$, I will express the generalized Ramanujan conjecture (GRC) for $\pi$ as the conjectural bound

$$|\lambda_{\pi}(\mathfrak{a})|\ll_{n,\epsilon}\mathrm{N}\mathfrak{a}^{\epsilon}$$

for all $\epsilon>0$, where $\lambda_{\pi}(\mathfrak{a})$ is the Hecke eigenvalue at $\mathfrak{a}$. One way to approach GRC is to study the moments

$$\sum_{\mathrm{N}\mathfrak{a}\leq x}|\lambda_{\pi}(\mathfrak{a})|^{2k},$$

where $k\geq 1$ is a natural number. When $k=1$, this is bounded from above by

$$\sum_{\mathrm{N}\mathfrak{a}\leq x}\lambda_{\pi\times\widetilde{\pi}}(\mathfrak{a}),$$

where $\pi\times\pi'$ denotes the Rankin-Selberg convolution. The work of Jacquet, Piatetski-Shapiro, and Shalika establishes the basic properties of the Rankin-Selberg $L$-function $L(s,\pi\times\widetilde{\pi})$. These properties (particularly the absolute convergence of the Euler product that defines $L(s,\pi\times\widetilde{\pi})$ in the region $\mathrm{Re}(s)>1$) imply the first nontrivial bound: $|\lambda_{\pi}(\mathfrak{a})|\ll_{n,\epsilon}\mathrm{N}\mathfrak{a}^{1/2+\epsilon}$.

To study $k=2$, the pertinent Dirichlet series (which is conjecturally an $L$-function with an analytic continuation and functional equation) is $L(s,\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi})$. For $k=3$, we need $L(s,\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi}\times\pi\times\widetilde{\pi})$. Hopefully the pattern is clear. We expect each Dirichlet series in this sequence converges absolutely for $\mathrm{Re}(s)>1$. That would suffice to prove GRC (take $k$ to be sufficiently large in terms of $\epsilon$).

Such a region of absolute convergence follows immediately if all of these Dirichlet series are in fact products of $L$-functions of cuspidal automorphic representations (as above). Equivalently, each of these higher-order convolutions are isobaric sums of cuspidal automorphic representations. One way to express the Langlands functoriality conjecture for $\mathrm{GL}_n$, at least as it pertains to the question, is that if $\pi\in\mathcal{A}(n)$ and $\pi'\in\mathcal{A}(n')$, then there exists an isobaric sum of cuspidal automorphic representations, say $\pi\boxtimes\pi'$, such that

$$L(s,\pi\times\pi')=L(s,\pi\boxtimes\pi').$$

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  • $\begingroup$ @Thank you very much for such great answer! It seems that the generalized Ramanuzan conjecture follows from the existence of such isobaric sum of cuspidal representation. Is it not proved yet? $\endgroup$
    – Monty
    Commented Mar 1, 2023 at 2:18
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    $\begingroup$ What you need is that if $\pi$ is a (non-dihedral) cuspidal automorphic representation of $GL_2$, then the $n$-th symmetric power lifting $Sym^n \pi$ exists as a cuspidal automorphic representation of $GL_{n+1}$. This is known for a few small $n$ (maybe up to $n = 7$ or something like that) but absolutely out of reach for arbitrary $n$. $\endgroup$ Commented Mar 1, 2023 at 2:38
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    $\begingroup$ @DavidLoeffler For non-dihedral Hecke-Maass forms, we only know the automorphy up to $n=4$. But that is enough to establish the absolute convergence of the Dirichlet series for $L(s,\mathrm{Sym}^n\pi)$ for $\mathrm{Re}(s)>1$ when $n\leq 8$. $\endgroup$
    – 2734364041
    Commented Mar 1, 2023 at 2:57
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    $\begingroup$ To make the comment by @DavidLoeffler concrete (when the central character is trivial and the base field is $\mathbb{Q}$), I note that the factorization that I mention above is $\zeta(s) L(s,\mathrm{Sym}^2 \pi)$ when $k=1$, $\zeta(s)^2 L(s,\mathrm{Sym}^2\pi)^3 L(s,\mathrm{Sym}^4\pi)$ when $k=2$, and $\zeta(s)^5 L(s,\mathrm{Sym}^2\pi)^9 L(s,\mathrm{Sym}^4\pi)^5 L(s,\mathrm{Sym}^6\pi)$ when $k=3$. The formulas get complicated quickly as $k$ grows, but you'll need the first $2k$ symmetric powers to access the $2k$-th moment. $\endgroup$
    – 2734364041
    Commented Mar 1, 2023 at 3:17
  • $\begingroup$ @DavidLoeffler, Thank you very much for such wonderful answers. I understand it. $\endgroup$
    – Monty
    Commented Mar 1, 2023 at 8:20

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