Mixing
Matrix coefficients of Supercuspidal representations
$begingroup$
Let $F$ be a finite extension of the $p$-adic field $mathbb{Q}_p$, and let $(pi,V)$ be a supercuspidal representation of $G=GL_n(F)$. In particular, all its matrix coefficients have compact support modulo the center $Z$ of $G$. I want to prove that for every matrix coefficient $phi: G to mathbb{C}$ and every parabolic subgroup $P=MN$ (where $N$ is the unipotent radical of $P$) and every $g in G$, the integral
$$int_N phi(gn)dn = 0$$
Here is my attempt so far: Since $(pi,V)$ is supercuspidal, its Jacquet module $V_{N}$ if trivial for $N$ that arises from every possible parabolic subgroup. Thus, $V=V(N)$, or in other words, any vector in $V$ is expressible as a finite linear combination of vectors of the form $pi(n) v - v $ for some $n in N$ and $v in V$.
Now suppose the matrix coefficient $phi$ corresponds to the vector $w in V$ and $lambda in V^{*}$ so that
$$phi(gn)=lambda ( pi(gn)w)$$
In this case
$$int_N phi(gn)dn = int_N lambda(pi(g)pi(n) w) dn$$
Since $w$ is a finite linear combination of elements of the form $pi(n')v-v$, the above reduces to a sum of integrals of the form
$$int_N lambda( pi(g) pi(n)left(pi(n')v-vright)) dn= int_N lambda(pi(gnn')v)dn - int_N lambda(pi(gn)v)dn$$
But is the above expression $0$?
representation-theory lie-groups p-adic-number-theory
asked Jan 26 at 16:45
BharatRamBharatRam
925619
$endgroup$
add a comment |
$begingroup$
Let $F$ be a finite extension of the $p$-adic field $mathbb{Q}_p$, and let $(pi,V)$ be a supercuspidal representation of $G=GL_n(F)$. In particular, all its matrix coefficients have compact support modulo the center $Z$ of $G$. I want to prove that for every matrix coefficient $phi: G to mathbb{C}$ and every parabolic subgroup $P=MN$ (where $N$ is the unipotent radical of $P$) and every $g in G$, the integral
$$int_N phi(gn)dn = 0$$
Here is my attempt so far: Since $(pi,V)$ is supercuspidal, its Jacquet module $V_{N}$ if trivial for $N$ that arises from every possible parabolic subgroup. Thus, $V=V(N)$, or in other words, any vector in $V$ is expressible as a finite linear combination of vectors of the form $pi(n) v - v $ for some $n in N$ and $v in V$.
Now suppose the matrix coefficient $phi$ corresponds to the vector $w in V$ and $lambda in V^{*}$ so that
$$phi(gn)=lambda ( pi(gn)w)$$
In this case
$$int_N phi(gn)dn = int_N lambda(pi(g)pi(n) w) dn$$
Since $w$ is a finite linear combination of elements of the form $pi(n')v-v$, the above reduces to a sum of integrals of the form
$$int_N lambda( pi(g) pi(n)left(pi(n')v-vright)) dn= int_N lambda(pi(gnn')v)dn - int_N lambda(pi(gn)v)dn$$
But is the above expression $0$?
representation-theory lie-groups p-adic-number-theory
asked Jan 26 at 16:45
BharatRamBharatRam
925619
$endgroup$
$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02
add a comment |
$begingroup$
Let $F$ be a finite extension of the $p$-adic field $mathbb{Q}_p$, and let $(pi,V)$ be a supercuspidal representation of $G=GL_n(F)$. In particular, all its matrix coefficients have compact support modulo the center $Z$ of $G$. I want to prove that for every matrix coefficient $phi: G to mathbb{C}$ and every parabolic subgroup $P=MN$ (where $N$ is the unipotent radical of $P$) and every $g in G$, the integral
$$int_N phi(gn)dn = 0$$
Here is my attempt so far: Since $(pi,V)$ is supercuspidal, its Jacquet module $V_{N}$ if trivial for $N$ that arises from every possible parabolic subgroup. Thus, $V=V(N)$, or in other words, any vector in $V$ is expressible as a finite linear combination of vectors of the form $pi(n) v - v $ for some $n in N$ and $v in V$.
Now suppose the matrix coefficient $phi$ corresponds to the vector $w in V$ and $lambda in V^{*}$ so that
$$phi(gn)=lambda ( pi(gn)w)$$
In this case
$$int_N phi(gn)dn = int_N lambda(pi(g)pi(n) w) dn$$
Since $w$ is a finite linear combination of elements of the form $pi(n')v-v$, the above reduces to a sum of integrals of the form
$$int_N lambda( pi(g) pi(n)left(pi(n')v-vright)) dn= int_N lambda(pi(gnn')v)dn - int_N lambda(pi(gn)v)dn$$
But is the above expression $0$?
representation-theory lie-groups p-adic-number-theory
asked Jan 26 at 16:45
BharatRamBharatRam
925619
$endgroup$
Let $F$ be a finite extension of the $p$-adic field $mathbb{Q}_p$, and let $(pi,V)$ be a supercuspidal representation of $G=GL_n(F)$. In particular, all its matrix coefficients have compact support modulo the center $Z$ of $G$. I want to prove that for every matrix coefficient $phi: G to mathbb{C}$ and every parabolic subgroup $P=MN$ (where $N$ is the unipotent radical of $P$) and every $g in G$, the integral
$$int_N phi(gn)dn = 0$$
Here is my attempt so far: Since $(pi,V)$ is supercuspidal, its Jacquet module $V_{N}$ if trivial for $N$ that arises from every possible parabolic subgroup. Thus, $V=V(N)$, or in other words, any vector in $V$ is expressible as a finite linear combination of vectors of the form $pi(n) v - v $ for some $n in N$ and $v in V$.
Now suppose the matrix coefficient $phi$ corresponds to the vector $w in V$ and $lambda in V^{*}$ so that
$$phi(gn)=lambda ( pi(gn)w)$$
In this case
$$int_N phi(gn)dn = int_N lambda(pi(g)pi(n) w) dn$$
Since $w$ is a finite linear combination of elements of the form $pi(n')v-v$, the above reduces to a sum of integrals of the form
$$int_N lambda( pi(g) pi(n)left(pi(n')v-vright)) dn= int_N lambda(pi(gnn')v)dn - int_N lambda(pi(gn)v)dn$$
But is the above expression $0$?
representation-theory lie-groups p-adic-number-theory
representation-theory lie-groups p-adic-number-theory
asked Jan 26 at 16:45
BharatRamBharatRam
925619
asked Jan 26 at 16:45
BharatRamBharatRam
925619
asked Jan 26 at 16:45
BharatRamBharatRam
925619
asked Jan 26 at 16:45
BharatRamBharatRam
925619
asked Jan 26 at 16:45
BharatRamBharatRam
925619
925619
$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02
add a comment |
$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02
$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02
$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02
add a comment |
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$begingroup$
Yes (use translation invariance of $dn$ to justify deleting $n'$ from the first term of the RHS).
$endgroup$
– Oven
Jan 27 at 15:02