Mixing
Supercuspidal representation from compact induction
$begingroup$
Note: this is a homework problem. Any hints would be great.
Consider the group $G=GL_n(mathbb{Q}_p)$ and an open subgroup $K$ that is compact modulo center. Suppose we have a smooth representation $rho$ of $K$ which we compactly induce to obtain a representation $pi$ (with space $V$) of $G$, and suppose this $pi$ is supercuspidal. I need to show that $pi$ has no $K$-fixed vectors.
I have some attempt, but I am unable to take it forward. So suppose there is a non-zero $K$-fixed vector $v$. Then for every $k in K$, $pi(k)v-v$ is zero. Now if we could find a subgroup $N$ contained in $K$ where $N$ is the unipotent subgroup of some parabolic subgroup $P=MN$, then I am thinking we could use this to show that the Jacquet module corresponding to this $N$ is not trivial? But I am unclear on how to show that elements of the form $pi(n) v - v$ do not span the whole of $V$, nor do I know if we can actually find a subgroup $N$ as suggested.
representation-theory lie-groups p-adic-number-theory
asked Jan 28 at 17:55
BharatRamBharatRam
925619
$endgroup$
add a comment |
$begingroup$
Note: this is a homework problem. Any hints would be great.
Consider the group $G=GL_n(mathbb{Q}_p)$ and an open subgroup $K$ that is compact modulo center. Suppose we have a smooth representation $rho$ of $K$ which we compactly induce to obtain a representation $pi$ (with space $V$) of $G$, and suppose this $pi$ is supercuspidal. I need to show that $pi$ has no $K$-fixed vectors.
I have some attempt, but I am unable to take it forward. So suppose there is a non-zero $K$-fixed vector $v$. Then for every $k in K$, $pi(k)v-v$ is zero. Now if we could find a subgroup $N$ contained in $K$ where $N$ is the unipotent subgroup of some parabolic subgroup $P=MN$, then I am thinking we could use this to show that the Jacquet module corresponding to this $N$ is not trivial? But I am unclear on how to show that elements of the form $pi(n) v - v$ do not span the whole of $V$, nor do I know if we can actually find a subgroup $N$ as suggested.
representation-theory lie-groups p-adic-number-theory
asked Jan 28 at 17:55
BharatRamBharatRam
925619
$endgroup$
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05
add a comment |
$begingroup$
Note: this is a homework problem. Any hints would be great.
Consider the group $G=GL_n(mathbb{Q}_p)$ and an open subgroup $K$ that is compact modulo center. Suppose we have a smooth representation $rho$ of $K$ which we compactly induce to obtain a representation $pi$ (with space $V$) of $G$, and suppose this $pi$ is supercuspidal. I need to show that $pi$ has no $K$-fixed vectors.
I have some attempt, but I am unable to take it forward. So suppose there is a non-zero $K$-fixed vector $v$. Then for every $k in K$, $pi(k)v-v$ is zero. Now if we could find a subgroup $N$ contained in $K$ where $N$ is the unipotent subgroup of some parabolic subgroup $P=MN$, then I am thinking we could use this to show that the Jacquet module corresponding to this $N$ is not trivial? But I am unclear on how to show that elements of the form $pi(n) v - v$ do not span the whole of $V$, nor do I know if we can actually find a subgroup $N$ as suggested.
representation-theory lie-groups p-adic-number-theory
asked Jan 28 at 17:55
BharatRamBharatRam
925619
$endgroup$
Note: this is a homework problem. Any hints would be great.
Consider the group $G=GL_n(mathbb{Q}_p)$ and an open subgroup $K$ that is compact modulo center. Suppose we have a smooth representation $rho$ of $K$ which we compactly induce to obtain a representation $pi$ (with space $V$) of $G$, and suppose this $pi$ is supercuspidal. I need to show that $pi$ has no $K$-fixed vectors.
I have some attempt, but I am unable to take it forward. So suppose there is a non-zero $K$-fixed vector $v$. Then for every $k in K$, $pi(k)v-v$ is zero. Now if we could find a subgroup $N$ contained in $K$ where $N$ is the unipotent subgroup of some parabolic subgroup $P=MN$, then I am thinking we could use this to show that the Jacquet module corresponding to this $N$ is not trivial? But I am unclear on how to show that elements of the form $pi(n) v - v$ do not span the whole of $V$, nor do I know if we can actually find a subgroup $N$ as suggested.
representation-theory lie-groups p-adic-number-theory
representation-theory lie-groups p-adic-number-theory
asked Jan 28 at 17:55
BharatRamBharatRam
925619
asked Jan 28 at 17:55
BharatRamBharatRam
925619
asked Jan 28 at 17:55
BharatRamBharatRam
925619
asked Jan 28 at 17:55
BharatRamBharatRam
925619
asked Jan 28 at 17:55
BharatRamBharatRam
925619
925619
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05
add a comment |
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091184%2fsupercuspidal-representation-from-compact-induction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3091184%2fsupercuspidal-representation-from-compact-induction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
It seems to be true in general (as long as the space you induce from is irreducible). Are you sure the supercuspidal assumption is needed?
$endgroup$
– Oven
Jan 29 at 23:46
$begingroup$
@Oven Is irreducibility necessary here? If the representation we induce from is irreducible, then it is easy to show that the compactly induced representation is supercuspidal. All we need to do is define one matrix coefficient that is supported on K. But here, I am already telling you that the representation is supercuspidal, and only interested in the absence of K-fixed vectors.
$endgroup$
– BharatRam
Jan 29 at 23:50
$begingroup$
Irreducibility may not be necessary.
$endgroup$
– Oven
Jan 30 at 0:19
$begingroup$
I suppose you want to assume that $rho$ is not the trivial representation of $K$?
$endgroup$
– WSL
Jan 31 at 22:05