(I follow up this question from MO, since it appears to get no real interest in there)

Let $F$ be a non-archimedean local field and $E$ a quadratic extension on $F$, $\chi$ a quasi-character of $E^\star$ and $\psi$ a positive character of $E^\star$. I would like to understand why the usual Rankin-Selberg zeta integrals converge, and this question reduces to the convergence of

$$\int_{E^\star} \chi(a) \psi(a) |a|^s d^\times a$$

this convergence seems to be true in a half-plane $Re(s)>s_0$ for some $s_0 \in \mathbb{R}$, depending only on $\chi$ and $\psi$.

Why is that true? What do we know about characters of $E^\star$ to be able to conclude that straigthforwardly? And why, when converges, this integral gives a polynomial in $q^{—s}$ where $q$ is the cardinality of the residue field?

  • $\begingroup$ what's $E{}{}{}$ ? $\endgroup$ – mercio Jun 21 '17 at 9:14
  • 1
    $\begingroup$ @mercio Sorry, $E$ is a quadratic extension of $F$, and everything is in a local setting. $\endgroup$ – Wolker Jun 21 '17 at 16:41
  • $\begingroup$ So $\chi$ is a character of $F$, not $E$? Because then $\chi(a)$ for $a \in E^{\times} \setminus F^{\times}$ would make no sense... $\endgroup$ – Peter Humphries Jun 21 '17 at 17:19
  • $\begingroup$ @PeterHumphries The OP meant $\int_{E^*} \chi(a) 1_{|a| \le 1} |a|^s d^\times a$ where $\chi$ is completely multiplicative, right ? I find it is $\displaystyle= \frac{\int_{\mathcal{O}^\times} \chi(x) d^\times x}{1-\chi(\pi) q^{-s}}$ where $\mathcal{O}/(\pi) = \mathbf{F}_q$ $\endgroup$ – reuns Jun 26 '17 at 17:38

First off, the fact that $E$ is a quadratic extension of $F$ is a red herring; it's completely irrelevant.

So why does this integral converge? Well, we may write \[E^{\times} = \left\{ \varpi_E^k x \colon k \in \mathbb{Z}, \, x \in \mathcal{O}_E^{\times}\right\},\] where $\varpi_E$ is a uniformiser for $E$ satisfying $\varpi_E \mathcal{O}_E = \mathfrak{p}$, the prime ideal of $\mathcal{O}_E$, and $|\varpi_E|_E^{-1} = q = \# \mathcal{O}_E / \mathfrak{p}$, the cardinality of the residue field.

It follows that \[\int_{E^{\times}} \chi(a) \psi(a) |a|_E^s \, d^{\times} a = \sum_{k=-\infty}^{\infty} \int_{\varpi_E^k \mathcal{O}_E^{\times}} \chi(a) \psi(a) |a|_E^s \, d^{\times} a = \sum_{k = -\infty}^{\infty} q^{-ks} \int_{\varpi_E^k \mathcal{O}_E^{\times}} \chi(a) \psi(a) \, d^{\times} a,\] as $|a|_E = q^{-k}$ for $a \in \varpi_E^k \mathcal{O}_E^{\times}$.

If $\chi$ is unramified, so that $\chi(\varpi_E^k x) = \chi(\varpi_E)^k$ for $x \in \mathcal{O}_E^{\times}$, then the inner integral is equal to \[\chi(\varpi_E)^k q^k \left(\int_{\varpi_E^k \mathcal{O}_E} \psi(a) \, da - \int_{\varpi_E^{k + 1} \mathcal{O}_E} \psi(a) \, da\right)\] as $d^{\times} a = \frac{da}{|a|_E}$ and $\varpi_E^k \mathcal{O}_E^{\times} = \varpi_E^k \mathcal{O}_E \setminus \varpi_E^{k + 1} \mathcal{O}_E$. The first integral vanishes unless $k \geq c(\psi)$, where $c(\psi)$ denotes the conductor exponent of $\psi$, in which case $\psi$ is trivial on $\varpi_E^k \mathcal{O}_E$, and so the integral is equal to $q^{-k}$ (which is the volume of this space with respect to this measure); similarly with the second integral vanishing unless $k + 1 \geq c(\psi)$, in which case the integral is equal to $q^{-k - 1}$. So the whole thing is equal to $\chi(\varpi_E)^k (1 - q^{-1})$ if $k \geq c(\psi)$, $-\chi(\varpi_E)^k q^{-1}$ if $k = c(\psi) - 1$, and $0$ if $k \leq c(\psi) - 2$.

So when $\chi$ is unramified, the integral is equal to \[- \chi(\varpi_E)^{c(\psi) - 1} q^{-(c(\psi) - 1)s - 1} + (1 - q^{-1}) \sum_{k = c(\psi)}^{\infty} \chi(\varpi_E)^k q^{-ks}.\] This geometric series converges if and only if $|\chi(\varpi_E) q^{-s}| < 1$: i.e. if $\Re(s) > \log |\chi(\varpi_E)|$, in which case it is equal to \[\frac{\chi(\varpi_E)^{c(\psi)} q^{-c(\psi)s}}{1 - \chi(\varpi_E) q^{-s}},\] and the whole expression simplifies to \[\chi(\varpi_E)^{c(\psi)} q^{-c(\psi)s} \frac{1 - \chi(\varpi_E)^{-1} q^{-(1 - s)}}{1 - \chi(\varpi_E) q^{-s}}.\]

Finally, we claim that if $\chi$ is ramified, then $\int_{\varpi_E^k \mathcal{O}_E^{\times}} \chi(a) \psi(a) \, d^{\times} a$ vanishes for all but one value of $k$. A proof is given in Lemma 1.1.1 of this paper of Ralf Schmidt. From this, we see that this integral is equal to a monomial, and so converges for all $s \in \mathbb{C}$.

  • $\begingroup$ What are the properties of $\chi,\psi$ ? They are not multiplicative for $|a| \le p^n$ and $\chi \psi$ vanishes for $|a| > p^n$ ? $\endgroup$ – reuns Jun 27 '17 at 22:41
  • $\begingroup$ @user1952009, no. $\chi$ is multiplicative and satisfies $\chi(1 + x) = 1$ for all $x \in \mathfrak{p}^{c(\chi)}$. $\psi$ is additive and satisfies $\psi(x) = 1$ for $x \in \mathfrak{p}^{c(\psi)}$. They are both nonvanishing everywhere. $\endgroup$ – Peter Humphries Jun 28 '17 at 17:09

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