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Hybrid bounds for ${\rm{GL}}(4)\times {\rm{GL}}(1)$ twisted $L$-functions

Fei Hou

Abstract

Let $P,M$ be a two primes such that $(P,M)=1$. Let $Π$ be a normalized Hecke-Maaß form on ${\rm{GL}}(4)$ of level $P$, and $χ$ a primitive Dirichlet character modulo $M$. In this paper, we study the hybrid subconvexity problem for $L(s, Π\otimes χ)$ simultaneously in the level and conductor aspects. Among other things, we prove a hybrid subconvex bound, so long as $M^{1/5}<P<M^{2/5}$.

Hybrid bounds for ${\rm{GL}}(4)\times {\rm{GL}}(1)$ twisted $L$-functions

Abstract

Let be a two primes such that . Let be a normalized Hecke-Maaß form on of level , and a primitive Dirichlet character modulo . In this paper, we study the hybrid subconvexity problem for simultaneously in the level and conductor aspects. Among other things, we prove a hybrid subconvex bound, so long as .
Paper Structure (13 sections, 5 theorems, 114 equations)

This paper contains 13 sections, 5 theorems, 114 equations.

Table of Contents

  1. Introduction
  2. Prerequisites
  3. Automorphic $L$-functions and functional equations
  4. Voronoı̆ summation formula
  5. Proof of Theorem \ref{['03']}
  6. Treatment of $\mathcal{S}^{\text{Non-de.}}_{\Pi}$
  7. Bounding $\mathcal{S}^{\text{Non-de.,1}}_{\Pi}$
  8. Bounding $\mathcal{S}^{\text{Non-de.,2}}_{\Pi}$
  9. Treatment of $\mathcal{S}^{\text{Dua.}}_{\Pi}$.
  10. Treatment of $\mathcal{S}^{\text{Deg.}}_{\Pi}$
  11. The endgame
  12. Appendix A
  13. A. 1. Proof of Lemma \ref{['9945fyhjt3464lljyrdfv432']}

Key Result

Theorem 1.1

Let $P,M\ge 1$ be two primes such that $(P,M)=1$. Let $\chi$ be a primitive Dirichlet character modulo $M$. Let $\theta={\log P}/{\log M}$. Then, for any normalized Hecke-Maaß form $\Pi$ of level $P$ with trivial nebentypus, we have for any $0<\mu<1/2$, where $\mathcal{Q}=PM^4$ denotes the (analytic) conductor of $L (1/2, \Pi \otimes \chi)$, and the implied $\ll$-constant depends only on $\vareps

Theorems & Definitions (5)

  • Theorem 1.1
  • Theorem 1.2
  • Corollary 1.3
  • Lemma 2.1
  • Lemma 2.2