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Test vector for local zeta integral with ramified character

A natural way of thinking of the definition of an Artin $L$-function?A $p$-adic “Jacobi sum” with an unramified characterDefinition of tamely ramifiedWhat is the correct definition of the cuspidal subspace of $L^2$?A critierion for tamely ramified extension for local fieldIntegral formula for the local L factor of a base changed automorphic representationCharacters of a quadratic extension and convergenceConvergence of local zeta functionExample of fully ramified extension of local fieldEisenstein series with character

2

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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?

number-theory modular-forms automorphic-forms l-functions

share|cite|improve this question

edited Mar 18 at 9:33

Sandpiper

asked Mar 15 at 9:30

SandpiperSandpiper

216

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  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 $pi$.
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  – David Loeffler
  Mar 16 at 0:08
  

  
  
  
  


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  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."
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  – Sandpiper
  Mar 17 at 16:43
  

  
  
  
  


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  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?
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  – Sandpiper
  Mar 17 at 17:24
  

  
  
  
  


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  @Sandpiper, I think Gelbart is actually mistaken here; what he has written is only correct if $chi$ is unramified.
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  – Peter Humphries
  Mar 17 at 19:28
  

  
  
  
  


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  @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 $pi otimes chi$.
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  – David Loeffler
  Mar 17 at 23:27
  

  
  
  
  

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show 1 more comment

2

$begingroup$

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?

number-theory modular-forms automorphic-forms l-functions

share|cite|improve this question

edited Mar 18 at 9:33

Sandpiper

asked Mar 15 at 9:30

SandpiperSandpiper

216

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• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
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  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
  

  
  
  
  


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  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?
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  – Sandpiper
  Mar 17 at 17:24
  

  
  
  
  


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  @Sandpiper, I think Gelbart is actually mistaken here; what he has written is only correct if $chi$ is unramified.
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  – Peter Humphries
  Mar 17 at 19:28
  

  
  
  
  


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  @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
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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?

number-theory modular-forms automorphic-forms l-functions

share|cite|improve this question

edited Mar 18 at 9:33

Sandpiper

asked Mar 15 at 9:30

SandpiperSandpiper

216

$endgroup$

share|cite|improve this question

edited Mar 18 at 9:33

Sandpiper

asked Mar 15 at 9:30

SandpiperSandpiper

216

share|cite|improve this question

edited Mar 18 at 9:33

Sandpiper

asked Mar 15 at 9:30

SandpiperSandpiper

216

asked Mar 15 at 9:30

SandpiperSandpiper

216

asked Mar 15 at 9:30

SandpiperSandpiper

216

2 Answers
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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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answered Mar 18 at 13:37

David LoefflerDavid Loeffler

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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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answered Mar 16 at 0:09

KimballKimball

2,0251029

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  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.
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  – Sandpiper
  Mar 18 at 9:35
  
  
  

  
  
  
  


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  @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.
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  – Kimball
  Mar 18 at 13:19
  

  
  
  
  


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  @Kimball It is the correct one.
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  – David Loeffler
  Mar 18 at 13:40
  

  
  
  
  


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  @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.
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  – Kimball
  Mar 18 at 15:28
  

  
  
  
  


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  @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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2

$begingroup$

[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 $pi otimes chi$.)

share|cite|improve this answer

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

$endgroup$

add a comment | 

2

$begingroup$

[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 $pi otimes chi$.)

share|cite|improve this answer

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

$endgroup$

add a comment | 

$begingroup$

[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 $pi otimes chi$.)

share|cite|improve this answer

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

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share|cite|improve this answer

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

answered Mar 18 at 13:37

David LoefflerDavid Loeffler

7,032923

2

$begingroup$

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.

share|cite|improve this answer

answered Mar 16 at 0:09

KimballKimball

2,0251029

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  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
  
  
  

  
  
  
  


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  @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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  @Kimball It is the correct one.
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  – David Loeffler
  Mar 18 at 13:40
  

  
  
  
  


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  $begingroup$
  @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.
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  – Kimball
  Mar 18 at 15:28
  

  
  
  
  


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  @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
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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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answered Mar 16 at 0:09

KimballKimball

2,0251029

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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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  @Kimball It is the correct one.
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  – David Loeffler
  Mar 18 at 13:40
  

  
  
  
  


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  @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.
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  – Kimball
  Mar 18 at 15:28
  

  
  
  
  


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  @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.
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  – David Loeffler
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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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answered Mar 16 at 0:09

KimballKimball

2,0251029

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share|cite|improve this answer

answered Mar 16 at 0:09

KimballKimball

2,0251029

answered Mar 16 at 0:09

KimballKimball

2,0251029

answered Mar 16 at 0:09

KimballKimball

2,0251029