Whittaker model equation
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This is from Cogdell and Piateski-Shapiro's paper Derivative and L-functions for $GL_n$.
Let $lambda$ be a non-trivial $psi$-Whittaker functional. The Whittaker model is defined as$W_v(g)=lambda(pi(g)v), vin V_{pi},gin GL_n$. We know that $W_vin W(tau,psi)$ iff there is a compact open subgroup $Ysubset U_n$ such that $$int_Y W_v(py)psi^{-1}(y)dy=0, pin P_n.$$
I am confused why this can lead to: if writing $p=gu, gin GL_{n-1}, uin U_n$, we have
$$
int_YW_v(guy) psi^{-1}(y)dy=W_v(gu)int_Ypsi(gyg^{-1})psi^{-1}(y)dy.
$$
I understand it must be using the fact of $lambda$ being a Whittaker functional, but I am still unable to show $W_v(guy)=W_v(gu)psi(gyg^{-1}),$ frustrated..
functional-analysis number-theory representation-theory automorphic-forms
asked Jan 9 at 4:38
AlahoiAlahoi
215
$endgroup$
add a comment |
$begingroup$
This is from Cogdell and Piateski-Shapiro's paper Derivative and L-functions for $GL_n$.
Let $lambda$ be a non-trivial $psi$-Whittaker functional. The Whittaker model is defined as$W_v(g)=lambda(pi(g)v), vin V_{pi},gin GL_n$. We know that $W_vin W(tau,psi)$ iff there is a compact open subgroup $Ysubset U_n$ such that $$int_Y W_v(py)psi^{-1}(y)dy=0, pin P_n.$$
I am confused why this can lead to: if writing $p=gu, gin GL_{n-1}, uin U_n$, we have
$$
int_YW_v(guy) psi^{-1}(y)dy=W_v(gu)int_Ypsi(gyg^{-1})psi^{-1}(y)dy.
$$
I understand it must be using the fact of $lambda$ being a Whittaker functional, but I am still unable to show $W_v(guy)=W_v(gu)psi(gyg^{-1}),$ frustrated..
functional-analysis number-theory representation-theory automorphic-forms
asked Jan 9 at 4:38
AlahoiAlahoi
215
$endgroup$
$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23
add a comment |
$begingroup$
This is from Cogdell and Piateski-Shapiro's paper Derivative and L-functions for $GL_n$.
Let $lambda$ be a non-trivial $psi$-Whittaker functional. The Whittaker model is defined as$W_v(g)=lambda(pi(g)v), vin V_{pi},gin GL_n$. We know that $W_vin W(tau,psi)$ iff there is a compact open subgroup $Ysubset U_n$ such that $$int_Y W_v(py)psi^{-1}(y)dy=0, pin P_n.$$
I am confused why this can lead to: if writing $p=gu, gin GL_{n-1}, uin U_n$, we have
$$
int_YW_v(guy) psi^{-1}(y)dy=W_v(gu)int_Ypsi(gyg^{-1})psi^{-1}(y)dy.
$$
I understand it must be using the fact of $lambda$ being a Whittaker functional, but I am still unable to show $W_v(guy)=W_v(gu)psi(gyg^{-1}),$ frustrated..
functional-analysis number-theory representation-theory automorphic-forms
asked Jan 9 at 4:38
AlahoiAlahoi
215
$endgroup$
This is from Cogdell and Piateski-Shapiro's paper Derivative and L-functions for $GL_n$.
Let $lambda$ be a non-trivial $psi$-Whittaker functional. The Whittaker model is defined as$W_v(g)=lambda(pi(g)v), vin V_{pi},gin GL_n$. We know that $W_vin W(tau,psi)$ iff there is a compact open subgroup $Ysubset U_n$ such that $$int_Y W_v(py)psi^{-1}(y)dy=0, pin P_n.$$
I am confused why this can lead to: if writing $p=gu, gin GL_{n-1}, uin U_n$, we have
$$
int_YW_v(guy) psi^{-1}(y)dy=W_v(gu)int_Ypsi(gyg^{-1})psi^{-1}(y)dy.
$$
I understand it must be using the fact of $lambda$ being a Whittaker functional, but I am still unable to show $W_v(guy)=W_v(gu)psi(gyg^{-1}),$ frustrated..
functional-analysis number-theory representation-theory automorphic-forms
functional-analysis number-theory representation-theory automorphic-forms
asked Jan 9 at 4:38
AlahoiAlahoi
215
asked Jan 9 at 4:38
AlahoiAlahoi
215
asked Jan 9 at 4:38
AlahoiAlahoi
215
asked Jan 9 at 4:38
AlahoiAlahoi
215
asked Jan 9 at 4:38
AlahoiAlahoi
215
215
$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23
add a comment |
$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23
$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23
$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We just need to show $W_v(guy)=W_v(gu)psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}in U_n$, and it is easy to verify $psi(guyu^{-1}g^{-1})=psi(gyg^{-1})$, hence
$$
W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)psi(gyg^{-1}).
$$
answered Jan 28 at 17:06
AlahoiAlahoi
215
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
We just need to show $W_v(guy)=W_v(gu)psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}in U_n$, and it is easy to verify $psi(guyu^{-1}g^{-1})=psi(gyg^{-1})$, hence
$$
W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)psi(gyg^{-1}).
$$
answered Jan 28 at 17:06
AlahoiAlahoi
215
$endgroup$
add a comment |
$begingroup$
We just need to show $W_v(guy)=W_v(gu)psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}in U_n$, and it is easy to verify $psi(guyu^{-1}g^{-1})=psi(gyg^{-1})$, hence
$$
W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)psi(gyg^{-1}).
$$
answered Jan 28 at 17:06
AlahoiAlahoi
215
$endgroup$
add a comment |
$begingroup$
We just need to show $W_v(guy)=W_v(gu)psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}in U_n$, and it is easy to verify $psi(guyu^{-1}g^{-1})=psi(gyg^{-1})$, hence
$$
W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)psi(gyg^{-1}).
$$
answered Jan 28 at 17:06
AlahoiAlahoi
215
$endgroup$
We just need to show $W_v(guy)=W_v(gu)psi(gyg^{-1})$.
In fact, since $guyu^{-1}g^{-1}in U_n$, and it is easy to verify $psi(guyu^{-1}g^{-1})=psi(gyg^{-1})$, hence
$$
W_v(guy)=W_v(guyu^{-1}g^{-1}gu)=psi(guyu^{-1}g^{-1})W_v(gu)=W_v(gu)psi(gyg^{-1}).
$$
answered Jan 28 at 17:06
AlahoiAlahoi
215
answered Jan 28 at 17:06
AlahoiAlahoi
215
answered Jan 28 at 17:06
AlahoiAlahoi
215
answered Jan 28 at 17:06
AlahoiAlahoi
215
215
add a comment |
add a comment |
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$begingroup$
Perhaps try emailing Cogdell? I had a look at that paper but was also unable to derive this equality.
$endgroup$
– Peter Humphries
Jan 15 at 14:23