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Suppose $\pi$ is an unramified principal series representation of ${\rm GL}_2(F)$, where $F$ is a non-archimedean local field with integers $\mathfrak{o}$. Let $W$ be a function in its Whittaker model. If $\chi$ is a quasi-character of $F^\times$ then we define its local zeta integral as $$ Z(W, s, \chi, g) = \int_{F^\times} W((\begin{smallmatrix} y & 0 \\ 0 & 1 \end{smallmatrix})g) \chi(y) |y|^{s-1/2}\, d^\times y,$$ see e.g. (6.28) of Gelbart's book. Then for $W= W_0$ the unique $K = {\rm GL}_2(\mathfrak{o})$-invariant Whittaker function with $W_0(1)=1$, we have $$Z(W_0,s,\chi,1) = L(s, \pi \otimes \chi),$$ see e.g. prop 6.17b of Gelbart. Note that if $\chi$ is ramified, then we have simply $$L(s,\pi \otimes \chi)=1.$$ Let's write $y \in F^\times$ as $p^n x,$ where $x \in \mathfrak{o}^\times$, and $p$ is a uniformizer, and $\chi$ as $\chi(p^nx) = |p^n|^{s'} \chi^*(x),$ where $ \chi^*$ is a character of $\mathfrak{o}^\times$. Then we have
$$Z(W,s, \chi, 1) = \sum_{n \in \mathbb{Z}} W((\begin{smallmatrix} p^n & 0 \\ 0 & 1 \end{smallmatrix}))|p^n|^{s-1/2+s'} \int_{x \in \mathfrak{o}^\times} \chi^*(x)\,dx,$$ which is $0$ if $\chi$ is ramified. Contradiction.

Am I correct to interpret the above to mean that the $W_0$ that I have chosen above is not the correct choice of test vector for the zeta integral when $\chi$ is ramified? If so, what is the correct test vector?

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    $\begingroup$ Are you sure you have quoted Gelbart's Prop 6.17 correctly? It's clear that the local zeta integral with the spherical test vector will be 0 if $\chi$ is ramified. The correct test vector is the new vector of $\pi \otimes \chi$, which is not the same as the image in $\pi \otimes \chi$ of the new vector of $\pi$. $\endgroup$ – David Loeffler Mar 16 at 0:08
  • $\begingroup$ Gelbart Says: "If $\pi_v$ is class one then $W(\pi_v)$ contains exactly one $K_v$ invariant function $W^0_v(g)$ such that $W^0_v(e)=1$ and for this $W^0_v(g)$ [second display of op] obtains." $\endgroup$ – Sandpiper Mar 17 at 16:43
  • $\begingroup$ If one needs to take the new vector for $\pi \otimes \chi$, then the $\chi$ in the definition of $Z(W, \chi, s, 1)$ is pretty redundant, no? Also, if I'm not mistaken, the new vector for $\pi \otimes \chi$ in the Kirillov model is $1_{\mathfrak{o}^\times}$, for which $Z(W, \chi, s, 1)$ is still 0. If I take the new vector for $\pi \otimes \chi$ and use that in a zeta integral of the form $Z(W,s,g)$, then this zeta integral equals $L(s, \pi \otimes \chi)$, of course. Is that what Gelbart means here? $\endgroup$ – Sandpiper Mar 17 at 17:24
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    $\begingroup$ @Sandpiper, I think Gelbart is actually mistaken here; what he has written is only correct if $\chi$ is unramified. $\endgroup$ – Peter Humphries Mar 17 at 19:28
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    $\begingroup$ @Sandpiper What I meant, more precisely, is that the correct test vector is the vector in $\pi$ whose image in $\pi \otimes \chi$ is the new vector of $\pi \otimes \chi$. $\endgroup$ – David Loeffler Mar 17 at 23:27
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[Expanding my comment to an answer]

The new vector of $\pi$ is not the correct test vector to use when $\chi$ is ramified; and if Gelbart really claims this holds for all $\chi$, then he is wrong.

You can decompose the Kirillov model of $\pi$ as a (countably infinite) direct sum of eigenspaces for the action of $\mathfrak{o}^\times$. It's easy to check that any function that is in an eigenspace other than the $\chi^{-1}$-eigenspace will be sent to 0 by $Z(-, s, \chi, 1)$. Conversely, a Schwartz function on $F$ which lies in the $\chi^{-1}$-eigenspace for $\mathfrak{o}^\times$ had better vanish at 0 and hence have compact support in $F^\times$; so this eigenspace is precisely the linear combinations of the functions $\chi^{-1} \cdot 1_{\varpi^n \mathfrak{o}^\times}$, for $n \in \mathbf{Z}$ (exercise). So the space of functions $\{ Z(W, s, \chi, 1) \}$ for varying $W$ is exactly the space of polynomials in $q^{-s}$, and the function with Kirillov model $\chi^{-1} 1_{\varpi^n \mathfrak{o}^\times}$ will be a test vector for any $n$, although $n = 0$ is surely the most obvious choice. (Note that the vector with Kirillov function $\chi^{-1} 1_{\mathfrak{o}^\times}$ maps to the new vector of $\pi \otimes \chi$.)

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Yes. The issue is that $Z(-, s, \chi, 1)$ is a linear functional which transforms by $\chi$ on $\mathfrak o^\times$ (say in the Kirillov model). So if you plug in a function which is $\mathfrak o^\times$-invariant and $\chi$ is ramified, you must get 0.

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  • $\begingroup$ Ok, fair enough, but I shouldn't have written the question so that it could admit a yes/no answer. I've edited the original question to ask for what the right test vector should be. $\endgroup$ – Sandpiper Mar 18 at 9:35
  • $\begingroup$ @Sandpiper You can of course do what David Loeffler suggests in the comments to get a test vector, though I'm not sure if it's the "correct" one. E.g., maybe you want a certain translate of the new vector---see my paper with File and Pitale or Vatsal's recent preprint on the arXiv for the case of ramified test vectors for toric periods. $\endgroup$ – Kimball Mar 18 at 13:19
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    $\begingroup$ @Kimball It is the correct one. $\endgroup$ – David Loeffler Mar 18 at 13:40
  • $\begingroup$ @DavidLoeffler Are you saying that vector is of the form $\pi(diag(a,1))W_0$ where $W_0$ is the newvector in $\pi$? I could be misremembering, but I think this is not true. $\endgroup$ – Kimball Mar 18 at 15:28
  • $\begingroup$ @Kimball No, I am not saying that, it is obviously false; I am saying something else. (The new vector of $\pi \times \chi$ is not the twist of the new vector of $\pi$.) See my detailed answer. $\endgroup$ – David Loeffler Mar 18 at 23:39

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