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Let $\pi$ be any automorphic Maass form on $\text{GL}_m$ of level $N$, say. Assume that the associated $L$-function $L(s,\pi)$ satisfies some good conditions; for example, it satisfies the functional equation of the Riemann type, and has the analytic continuation to the whole complex plane, where it is holomorphic except possibly for a pole at $s = 1$.

I wonder if we have the lower bound at the edge of the critical strip $\Re s=1$ such that $$ |L(1+it,\pi)| \gg ( N (1+|t|) )^{-\varepsilon} ?$$ for any $\varepsilon>0$. And I also ask if there exists a better bound compared with the above estimate?

I think a good reference should be Li's paper-Upper bounds on L-functions at the edge of the critical strip on IMRN; see https://www.et-fine.com/10.1093/imrn/rnp148. However, it appears that there is no a direct conclusion??

If any expert knows some knowledge upon this, please give some valuable comments or certain relevant references.

Thanks in advance.

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  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented May 24 at 22:43

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This is not known. The best known lower bound is due to Brumley (2013), to be found in an Appendix to a paper by Lapid, which gives that $$|L(1+it,\pi)| \gg_{\pi_{\infty},\varepsilon} ( N (1+|t|) )^{-m-\varepsilon}.$$ Note that the implied constant depends on the archimedean parameters of $\pi$ (as necessary). Alternatively, if $N$ is replaced by the analytic conductor of $\pi$, then the implied constant will only depend on $m$ and $\varepsilon$. A small improvement in the $t$-aspect was achieved by Qiao Zhang (2022).

On the other hand, if $\pi$ is not the determinant twist of a self-dual cuspidal irreducible representation, then the sought lower bound (with exponent $-\varepsilon$) is certainly true. I don't know if this is written anywhere, but it should follow from the work of Brumley (2019) published as an Appendix to a paper by Humphries.

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  • $\begingroup$ Many thanks for explaining the history of this direction and the development of this topic with new references. But what's astonishing is that the lower bounds for $L(1+it, \pi)$ at $\Re s=1$ and $L(1, \pi)$ at $s=1$ are totally different! Thanks. $\endgroup$
    – user528074
    Commented May 24 at 22:07
  • $\begingroup$ @user528074 1. I added the dependence of the implied constant on $\pi_\infty$ (as necessary). 2. It is not clear what you mean by "totally different". The best known lower bound for $L(1,\pi)$ is just a special case of the best known lower bound for $L(1+it,\pi)$. In fact the two problems are equivalent, because $$L(1+it,\pi)=L(1,\pi\otimes|\det|^{it}).$$ 3. f you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Commented May 24 at 22:56

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