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
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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 Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
$endgroup$
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 Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
$endgroup$
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 Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
$endgroup$
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 Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
edited Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
edited Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
edited Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
edited Mar 14 at 14:27
Carlo Beenakker
78.7k9186289
78.7k9186289
asked Mar 14 at 11:44
James CheungJames Cheung
42816
asked Mar 14 at 11:44
James CheungJames Cheung
42816
asked Mar 14 at 11:44
James CheungJames Cheung
42816
42816
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 Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
$endgroup$
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
$endgroup$
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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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 Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
$endgroup$
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
$endgroup$
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
$endgroup$
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 Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
answered Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
answered Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
answered Mar 14 at 15:12
Johannes HahnJohannes Hahn
6,14222446
6,14222446
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
add a comment |
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
$begingroup$
Thank you for your detailed explanation.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
$endgroup$
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
$endgroup$
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
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 Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
$endgroup$
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 Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
answered Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
answered Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
answered Mar 14 at 14:25
Carlo BeenakkerCarlo Beenakker
78.7k9186289
78.7k9186289
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
add a comment |
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
$begingroup$
Thank you for your notes.
$endgroup$
– James Cheung
Mar 16 at 13:57
add a comment |
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