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Let $\pi$ be an automorphic representation of $GL_2$ over a number field. What can I say concerning the order of the pole at $1$ of the $L$-function $L(s, \pi)$? Can we say more about $L(s, \mathrm{Sym}^2 \pi)$ or other constructions depending on $\pi$?

In the case where only holomorphic cusp forms appear, there is no pole. However, what about possible Maass forms?

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    $\begingroup$ The Maass case is the same as the holomorphic case: $L(s,\pi)$ has no poles, and $L(s,\mathrm{sym}^2 \pi)$ has a pole iff $\pi$ is dihedral with nontrivial central character. The first fact is in chapter 5 of Iwaniec and Kowalski. The second is outlined for holomorphic cusp forms here: mathoverflow.net/questions/178654/…, and the same idea holds for dihedral Maass forms with nontrivial central character. $\endgroup$ – Peter Humphries Nov 3 '17 at 9:48
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    $\begingroup$ @PeterHumphries: Please give your comment as a response, so that this question can be closed. $\endgroup$ – GH from MO Nov 3 '17 at 14:33
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Let $\pi$ be a cuspidal automorphic representation of $\mathrm{GL}_n(\mathbb{A}_F)$, where $F$ is a number field. Then $L(s,\pi)$ is entire and has a zero-free region. The latter is essentially Theorem 5.10 of Iwaniec and Kowalski, while the former is Corollary 13.8 of Godement-Jacquet.

If $\pi$ is not cuspidal, then of course this is not necessarily the case: $\zeta_F(s)$ is the $L$-function of the trivial representation of $\mathrm{GL}_1(\mathbb{A}_F)$, and this has a simple pole at $s = 1$. Of course, if $\pi$ is the isobaric sum $\pi_1 \boxplus \cdots \boxplus \pi_r$ of automorphic representations, then $L(s,\pi) = L(s,\pi_1) \cdots L(s,\pi_r)$, and if each $\pi_j$ is cuspidal (and in particular not the trivial representation), this is entire.

For $n = 2$ and cuspidal $\pi$, $\operatorname{sym}^2 \pi$ is a cuspidal automorphic representation of $\mathrm{GL}_3(\mathbb{A}_F)$ (and consequently $L(s,\operatorname{sym}^2 \pi)$ is entire) iff $\pi$ is not dihedral; note that $\operatorname{sym}^2 \pi \boxplus \omega_{\pi} = \pi \otimes \pi$, with $\omega_{\pi}$ the central character of $\pi$.

If $\pi$ is dihedral, then it is the automorphic induction $\pi(\psi)$ from $\mathrm{GL}_1(\mathbb{A}_E) = \mathbb{A}_E^{\times}$ to $\mathrm{GL}_2(\mathbb{A}_F)$ of a Hecke character $\psi$ of $\mathbb{A}_E^{\times}$, where $E/F$ is a quadratic extension to which one can associate a quadratic Hecke character $\omega_{E/F}$ of $\mathbb{A}_F^{\times}$ via class field theory. Furthermore, $\pi(\psi)$ is cuspidal iff $\psi$ does not factor through the norm map $N_{\mathbb{A}_E / \mathbb{A}_F} : \mathbb{A}_E^{\times} \to \mathbb{A}_F^{\times}$. The central character $\omega_{\pi(\psi)}$ of $\pi(\psi)$ is $\psi|_{\mathbb{A}_F^{\times}} \omega_{E/F}$ (noting that the restriction of a Hecke character of $\mathbb{A}_E^{\times}$ to $\mathbb{A}_F^{\times}$ is a Hecke character of $\mathbb{A}_F^{\times}$).

So for dihedral $\pi = \pi(\psi)$, we have that $\operatorname{sym}^2 \pi = \pi(\psi^2) \boxplus \psi|_{\mathbb{A}_F^{\times}}$, so that $L(s,\operatorname{sym}^2 \pi) = L(s,\psi^2) L(s,\psi|_{\mathbb{A}_F^{\times}})$, where $\pi(\psi^2)$ is the automorphic induction of the Hecke character $\psi^2$ of $\mathbb{A}_E^{\times}$. In particular, if $\psi|_{\mathbb{A}_F^{\times}} = 1$ (so that $\omega_{\pi(\psi)} = \omega_{E/F}$), then $L(s,\operatorname{sym}^2 \pi) = L(s,\psi^2) \zeta_F(s)$, which has a simple pole at $s = 1$. On the other hand, if $\psi^2$ is the trivial character, then $L(s,\psi^2) = \zeta_E(s) = \zeta_F(s) L(s,\omega_{E/F})$, which has a simple pole at $s = 1$; note that $\psi^2 = 1$ implies that $\pi(\psi)$ is noncuspidal, and in fact $\pi(\psi) = \omega_1 \boxplus \omega_2$ for some real (quadratic or trivial) Hecke characters $\omega_1,\omega_2$ of $\mathbb{A}_F^{\times}$ satisfying $\omega_1 \omega_2 = \omega_{E/F}$ (for $F = \mathbb{Q}$, this means that $\psi$ is a genus character of the quadratic field $E$).

Note also that $\operatorname{ad} \pi \boxplus 1 = \pi \otimes \widetilde{\pi}$ and $\operatorname{ad} \pi = \operatorname{sym}^2 \pi \otimes \omega_{\pi}^{-1}$. Thus for dihedral $\pi = \pi(\psi)$, $\operatorname{ad} \pi = (\pi(\psi^2) \otimes \psi^{-1}|_{\mathbb{A}_F^{\times}}) \boxplus \omega_{E/F}$.

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