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Let $\Bbb A$ the ring of rational adeles and let $\pi=\bigotimes_{p\leq\infty}\pi_p$ be an automorphic (cuspidal) representation of ${\rm GL}_2(\Bbb A)$. Fix a quadratic extension $K\supset\Bbb Q$. Then there exists an automorphic representation $\pi_K$ of ${\rm GL}_2(\Bbb A\otimes_{\Bbb Q}K)$ (known as the base-change of $\pi$ to $K$) which is characterized by the identity $$ L(\pi_K,s)=L(\pi,s)L(\pi\otimes\eta_K,s) $$ of $L$-functions, where $\eta_K$ is the character of $\Bbb A^\times$ attached to $K$, namely that corresponding to the Hecke character taking value $1$ at split primes and vale $-1$ at inert primes.

So if I understand things correctly when $p$ is a prime which is unramified both for $\pi$ and $K$ the local $L_p$ factor is explicitely given by $$ L_p(\pi_K,s)=\begin{cases} \left(\frac1{1-\mu_1(p)p^{-s}}\right)^2\left(\frac1{1-\mu_2(p)p^{-s}}\right)^2 & \text{if $p$ splits in $K$}\\ \frac1{1-\mu_1(p)^2p^{-2s}}\frac1{1-\mu_2(p)^2p^{-2s}} & \text{if $p$ is inert in $K$} \end{cases}. $$ where $\pi_p\simeq\pi(\mu_1,\mu_2)$ as a principal series.

Now, I came across the formula $$ L_p(\eta_K,2s)^{-1}L_p(\pi_K,s)=\int_{\Bbb Z_p-\{0\}} \frac{\mu_1(ap)-\mu_2(ap)}{\mu_1(p)-\mu_2(p)}|a|^{s-{\frac12}}d^\times a $$ given very matter-of-factly but I am not sure how to derive it. Can anyone give a hint?

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  • $\begingroup$ has your integral at the end any equivalent in the case of Dedekind zeta functions ? (the rest does) $\endgroup$ – reuns May 24 '16 at 16:03
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    $\begingroup$ @user1952009: The Dedekind zeta of $K$ factors as $\zeta_K(s)=\zeta(s)L(\eta_K,s)$ but I see no direct link between the two situations. $\endgroup$ – AdLibitum May 24 '16 at 16:08
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This is too long to fit into the comment section - but really is one... Does the following agree with your calculations?

Doing the obvious, and evaluating (as you probably have done) the integral... Call the integral $I(s)$.

Assume that $\mu_1$ and $\mu_2$ are unramified at $p$.

$$ \begin{align} I( s ) &= \int_{ \mathbb Z_p - \{0\} } \frac { \mu_1(ap)-\mu_2(ap) } { \mu_1(p)-\mu_2(p) } |a|^{s-1/2} \, d^*a \\ &= \frac {1} { \mu_1(p)-\mu_2(p) } \left( \mu_1(p)J_{\mu_1}(s) - \mu_2(p) J_{\mu_2}(s) \right), \\ \end{align} $$

where $$ \begin{align} J_{\mu} (s) &= \int_{\mathbb Z_p - \{0\}} \mu(a)|a|^{s-1/2}\, d^*a \\ &= \sum_{k=0}^\infty \int_{p^k\mathbb Z_p^*} \mu(a)|a|^{s-1/2}\, d^*a \\ &= \sum_{k=0}^\infty \int_{\mathbb Z_p^*} \mu(p^ka)|p^ka|^{s-1/2}\, d^*a. \\ \end{align} $$

The last equality holds because $d^*a$ is multiplicative Haar measure.

Now, if $a$ a unit, $ |p^ka| = p^{-k} $, and $\mu (p^ka) = \mu(p)^k$ for $\mu$ unramfied. Therefore, the integrand(s) does not (do not) depend on $a$.

So, since $d^*a$ gives the units volume $1$, one has

$$ \begin{align} J_\mu(s) &= \sum_{k=0 }^\infty \left( \frac{\mu(p) } { p^{s-1/2}} \right)^k \cdot \int_{\mathbb Z_p^*} \,d^*a\\ &= \sum_{k=0 }^\infty \left( \frac{\mu(p) } { p^{s-1/2}} \right)^k\\ &= \frac{1 } { 1 - \frac{\mu(p) }{ p^{s-1/2} } }.\\ \end{align} $$

Therefore $$\begin{align} I(s) &= \frac {1} { \mu_1(p)-\mu_2(p)} \left( \frac{\mu_1(p) }{ 1 - \frac{\mu_1(p)}{ p^{s-1/2} } } - \frac{\mu_2(p) }{1 - \frac{\mu_2(p) }{ p^{s-1/2} } } \right) \\ &= \frac{1 }{ 1 - \frac{\mu_1(p) }{ p^{s-1/2} } } \cdot \frac{1}{1 - \frac{\mu_2(p)} { p^{s-1/2} } }. \end{align} $$

The claim is that $$ L_p(\pi_K, s) = I(s) L_p(\eta_K,2s). $$

Now, suppose $p$ is inert. Then, one should have

$$ \frac{1}{ 1-\frac{\mu_1(p)^2}{p^{2s} } } \cdot \frac{1}{ 1-\frac{\mu_2(p)^2}{p^{2s} } }\overset{?}{=} \frac{1 }{ 1 - \frac{\mu_1(p) }{ p^{s-1/2} } } \cdot \frac{1}{1 - \frac{\mu_2(p)} { p^{s-1/2} } } \cdot \frac 1 {1+ \frac{1}{p^{2s}}}. $$ I don't see how this can be - if I haven't made a mistake. Anybody care to comment?

Where did the claim appear?

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