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I met the following relation (if my understanding is correct) of local $\gamma$-factors when I was reading Hiraga-Ichino-Ikeda's paper "Formal Degrees and Adjoint $\gamma$-Factors": Let $F$ be a $p$-adic field with residue field order $q$. Let $G=GL(n,F)$ and $\pi$ be a discrete series of $G$, $\tilde{\pi}$ its contragredient. Then $$Res_{s=0}\gamma (s,\pi\times \tilde{\pi},\psi)=(\log q)^{-1}(1-q^{-1})\gamma(0,\pi,Ad,\psi)^{-1}.$$ Here the $\gamma$-factor is defined in the usual sense according to local Langlands correspondence ($L$-parameters, Artin $L$-functions, and local root number/$\epsilon$-factors), and $Ad$ means the adjoint representation of the $L$-group on its own Lie algebra.

I saw this relation in that paper, at the end of the proof of the lemma 4.1 (if my understanding is correct). It seems that the authors regarded this as a very direct fact and gave no more explanation. But I have no idea how left hand side is related to the adjoint $\gamma$-factor of $\pi$. Does this follow from some routine properties or calculations of local Langlands correspondence? Thanks a lot in advance for any help or suggestions!

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The equality they used in the paper should be $$\mathrm{Res}_{s=0}\gamma(s,\pi\times\widetilde{\pi},\psi)^{-1}=(\log q)^{-1}(1-q^{-1})\gamma(0,\pi,Ad,\psi)^{-1}.$$

Notice that the difference between L-factors $L(s,\pi\times\widetilde{\pi})$ and $L(s,\pi,Ad)$ is $L(s,triv)$, where $triv$ stands for the trivial representation. Hence we get $$\gamma(s,\pi,Ad,\psi)\gamma(s,\pi\times\widetilde{\pi},\psi)^{-1}=L(s,triv)L(1-s,triv)^{-1}.$$ Directly from the definition of local L-factors we have: $$L(1,triv)=(1-q^{-1})^{-1},~\mathrm{Res}_{s=0}L(s,triv)=\mathrm{Res}_{s=0}\frac{1}{1-q^{-s}}=(\log q)^{-1}.$$

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  • $\begingroup$ Thanks a lot. I guess the point here is that I don’t know the relation $\gamma(s,\pi,Ad,\psi)\gamma(s,\pi\times\widetilde{\pi},\psi)^{-1}=L(s,triv)L(1-s,triv)^{-1}.$ Does it come from some concrete computations of the Rankin-Selberg theory? $\endgroup$
    – youknowwho
    Commented Apr 10, 2023 at 16:01
  • $\begingroup$ @tooweaktolearnmathematics No, this step is just from the definition of $\gamma$-factors. The quotient turns out to be the $\gamma$-factor of the trivial representation and the $\epsilon$-factor becomes trivial. $\endgroup$
    – Erica
    Commented Apr 10, 2023 at 16:54

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