About parabolic Kazhdan Lusztig polynomialsBGG category everywhere implies generalized Kazhdan-Lusztig formula?When are parabolic Kazhdan-Lusztig polynomials nonzero?Implications of non-negativity of coefficients of arbitrary Kazhdan-Lusztig polynomials?Kazhdan-Lusztig Polynomials and Intersection CohomologyPapers/Programs for computing periodic KL polynomials?Recursive formula for inverse Kazhdan-Lusztig polynomialsCombinatorics of $p$-Kazhdan--lusztig polynomialsParabolic Kazhdan-Lusztig polynomial coincide?Examples of non-trivial Kazhdan-Lusztig polynomialsRelationship bewteen Kazhdan-Lusztig Vogan polynomial and Kazhdan-Lusztig polynomial
About parabolic Kazhdan Lusztig polynomials
BGG category everywhere implies generalized Kazhdan-Lusztig formula?When are parabolic Kazhdan-Lusztig polynomials nonzero?Implications of non-negativity of coefficients of arbitrary Kazhdan-Lusztig polynomials?Kazhdan-Lusztig Polynomials and Intersection CohomologyPapers/Programs for computing periodic KL polynomials?Recursive formula for inverse Kazhdan-Lusztig polynomialsCombinatorics of $p$-Kazhdan--lusztig polynomialsParabolic Kazhdan-Lusztig polynomial coincide?Examples of non-trivial Kazhdan-Lusztig polynomialsRelationship bewteen Kazhdan-Lusztig Vogan polynomial and Kazhdan-Lusztig polynomial
$begingroup$
There are two types of parabolic Kazhdan Lusztig polynomials, namely, of type -1: $P_x,w^I,-1$ and of type $q$: $P_x,w^I,q$. See Kazhdan–Lusztig and R-Polynomials,
Young’s Lattice, and Dyck Partitions
My question: What is the meaning of $-1$ and $q$?
rt.representation-theory algebraic-combinatorics kazhdan-lusztig
edited 6 hours ago
Carlo Beenakker
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
$endgroup$
add a comment |
$begingroup$
There are two types of parabolic Kazhdan Lusztig polynomials, namely, of type -1: $P_x,w^I,-1$ and of type $q$: $P_x,w^I,q$. See Kazhdan–Lusztig and R-Polynomials,
Young’s Lattice, and Dyck Partitions
My question: What is the meaning of $-1$ and $q$?
rt.representation-theory algebraic-combinatorics kazhdan-lusztig
edited 6 hours ago
Carlo Beenakker
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
$endgroup$
add a comment |
$begingroup$
There are two types of parabolic Kazhdan Lusztig polynomials, namely, of type -1: $P_x,w^I,-1$ and of type $q$: $P_x,w^I,q$. See Kazhdan–Lusztig and R-Polynomials,
Young’s Lattice, and Dyck Partitions
My question: What is the meaning of $-1$ and $q$?
rt.representation-theory algebraic-combinatorics kazhdan-lusztig
edited 6 hours ago
Carlo Beenakker
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
$endgroup$
There are two types of parabolic Kazhdan Lusztig polynomials, namely, of type -1: $P_x,w^I,-1$ and of type $q$: $P_x,w^I,q$. See Kazhdan–Lusztig and R-Polynomials,
Young’s Lattice, and Dyck Partitions
My question: What is the meaning of $-1$ and $q$?
rt.representation-theory algebraic-combinatorics kazhdan-lusztig
rt.representation-theory algebraic-combinatorics kazhdan-lusztig
edited 6 hours ago
Carlo Beenakker
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
edited 6 hours ago
Carlo Beenakker
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
edited 6 hours ago
Carlo Beenakker
78.5k9185288
edited 6 hours ago
Carlo Beenakker
78.5k9185288
edited 6 hours ago
Carlo Beenakker
78.5k9185288
78.5k9185288
asked 9 hours ago
James CheungJames Cheung
42316
asked 9 hours ago
James CheungJames Cheung
42316
asked 9 hours ago
James CheungJames Cheung
42316
42316
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
These polynomials are connected to the canonical basis of the induction from a parabolic subalgebra $H_I$ up to the whole Hecke algebra $H$. The difference is which module is being induced: The $q$-variant induces the trivial module (on which the generators $T_s$ of $H_I$ act as multiplication by $q$) while the $(-1)$ variant induces the sign module (on which the generators act as multiplication by $-1$).
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
$endgroup$
add a comment |
$begingroup$
these are polynomials in $q$ of two types, which satisfy either of the two recursions:
$$P_v,w^I,q=-P_v,ws^I,q;;textor;;P_v,w^I,-1=qP_v,ws^I,-1,$$
see for example these lecture notes.
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
$endgroup$
add a comment |
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: "504"
;
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%2fmathoverflow.net%2fquestions%2f325436%2fabout-parabolic-kazhdan-lusztig-polynomials%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
These polynomials are connected to the canonical basis of the induction from a parabolic subalgebra $H_I$ up to the whole Hecke algebra $H$. The difference is which module is being induced: The $q$-variant induces the trivial module (on which the generators $T_s$ of $H_I$ act as multiplication by $q$) while the $(-1)$ variant induces the sign module (on which the generators act as multiplication by $-1$).
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
$endgroup$
add a comment |
$begingroup$
These polynomials are connected to the canonical basis of the induction from a parabolic subalgebra $H_I$ up to the whole Hecke algebra $H$. The difference is which module is being induced: The $q$-variant induces the trivial module (on which the generators $T_s$ of $H_I$ act as multiplication by $q$) while the $(-1)$ variant induces the sign module (on which the generators act as multiplication by $-1$).
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
$endgroup$
add a comment |
$begingroup$
These polynomials are connected to the canonical basis of the induction from a parabolic subalgebra $H_I$ up to the whole Hecke algebra $H$. The difference is which module is being induced: The $q$-variant induces the trivial module (on which the generators $T_s$ of $H_I$ act as multiplication by $q$) while the $(-1)$ variant induces the sign module (on which the generators act as multiplication by $-1$).
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
$endgroup$
These polynomials are connected to the canonical basis of the induction from a parabolic subalgebra $H_I$ up to the whole Hecke algebra $H$. The difference is which module is being induced: The $q$-variant induces the trivial module (on which the generators $T_s$ of $H_I$ act as multiplication by $q$) while the $(-1)$ variant induces the sign module (on which the generators act as multiplication by $-1$).
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
answered 5 hours ago
Johannes HahnJohannes Hahn
6,06722445
6,06722445
add a comment |
add a comment |
$begingroup$
these are polynomials in $q$ of two types, which satisfy either of the two recursions:
$$P_v,w^I,q=-P_v,ws^I,q;;textor;;P_v,w^I,-1=qP_v,ws^I,-1,$$
see for example these lecture notes.
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
$endgroup$
add a comment |
$begingroup$
these are polynomials in $q$ of two types, which satisfy either of the two recursions:
$$P_v,w^I,q=-P_v,ws^I,q;;textor;;P_v,w^I,-1=qP_v,ws^I,-1,$$
see for example these lecture notes.
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
$endgroup$
add a comment |
$begingroup$
these are polynomials in $q$ of two types, which satisfy either of the two recursions:
$$P_v,w^I,q=-P_v,ws^I,q;;textor;;P_v,w^I,-1=qP_v,ws^I,-1,$$
see for example these lecture notes.
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
$endgroup$
these are polynomials in $q$ of two types, which satisfy either of the two recursions:
$$P_v,w^I,q=-P_v,ws^I,q;;textor;;P_v,w^I,-1=qP_v,ws^I,-1,$$
see for example these lecture notes.
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
answered 6 hours ago
Carlo BeenakkerCarlo Beenakker
78.5k9185288
78.5k9185288
add a comment |
add a comment |
Thanks for contributing an answer to MathOverflow!
- 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%2fmathoverflow.net%2fquestions%2f325436%2fabout-parabolic-kazhdan-lusztig-polynomials%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
