Parabolic Kazhdan-Lusztig polynomial coincide?

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Let $(W,S)$ be a Coxeter system. For any subset $Isubseteq S$, we can have the parabolic Kazhdan-Lusztig polynomial $P_x,w^I(q)$ with respect to $I$.

Now consider $Isubseteq Jsubseteq S$. Both $(W,S)$, $(W_J,J)$ are Coxeter systems.

Since $Isubseteq J$, we get the parabolic Kazhdan-Lusztig polynomial $overlineP_x,w^I(q)$ with respect to $I$ when considering the Coxeter system $(W_J,J)$.

Does $P_x,w^I(q)=overlineP_x,w^I(q)$ for all $x,win W_J$?

polynomials coxeter-groups weyl-group kazhdan-lusztig

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asked Sep 9 at 12:47

James Cheung

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Let $(W,S)$ be a Coxeter system. For any subset $Isubseteq S$, we can have the parabolic Kazhdan-Lusztig polynomial $P_x,w^I(q)$ with respect to $I$.

Now consider $Isubseteq Jsubseteq S$. Both $(W,S)$, $(W_J,J)$ are Coxeter systems.

Since $Isubseteq J$, we get the parabolic Kazhdan-Lusztig polynomial $overlineP_x,w^I(q)$ with respect to $I$ when considering the Coxeter system $(W_J,J)$.

Does $P_x,w^I(q)=overlineP_x,w^I(q)$ for all $x,win W_J$?

polynomials coxeter-groups weyl-group kazhdan-lusztig

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asked Sep 9 at 12:47

James Cheung

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up vote
5
down vote

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up vote
5
down vote

favorite

Let $(W,S)$ be a Coxeter system. For any subset $Isubseteq S$, we can have the parabolic Kazhdan-Lusztig polynomial $P_x,w^I(q)$ with respect to $I$.

Now consider $Isubseteq Jsubseteq S$. Both $(W,S)$, $(W_J,J)$ are Coxeter systems.

Since $Isubseteq J$, we get the parabolic Kazhdan-Lusztig polynomial $overlineP_x,w^I(q)$ with respect to $I$ when considering the Coxeter system $(W_J,J)$.

Does $P_x,w^I(q)=overlineP_x,w^I(q)$ for all $x,win W_J$?

polynomials coxeter-groups weyl-group kazhdan-lusztig

share|cite|improve this question

asked Sep 9 at 12:47

James Cheung

742

New contributor

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Let $(W,S)$ be a Coxeter system. For any subset $Isubseteq S$, we can have the parabolic Kazhdan-Lusztig polynomial $P_x,w^I(q)$ with respect to $I$.

Now consider $Isubseteq Jsubseteq S$. Both $(W,S)$, $(W_J,J)$ are Coxeter systems.

Since $Isubseteq J$, we get the parabolic Kazhdan-Lusztig polynomial $overlineP_x,w^I(q)$ with respect to $I$ when considering the Coxeter system $(W_J,J)$.

Does $P_x,w^I(q)=overlineP_x,w^I(q)$ for all $x,win W_J$?

polynomials coxeter-groups weyl-group kazhdan-lusztig

polynomials coxeter-groups weyl-group kazhdan-lusztig

share|cite|improve this question

asked Sep 9 at 12:47

James Cheung

742

New contributor

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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asked Sep 9 at 12:47

James Cheung

742

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James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|cite|improve this question

asked Sep 9 at 12:47

James Cheung

742

New contributor

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Sep 9 at 12:47

James Cheung

742

asked Sep 9 at 12:47

James Cheung

742

742

New contributor

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

New contributor

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

James Cheung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

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Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_x,w^I$ are those which are $leq w$ w.r.t. the Bruhat order. If $win W_J$, then all those elements are themselves contained in $W_J$.

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answered Sep 9 at 13:54

Johannes Hahn

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1 Answer
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1 Answer
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active

oldest

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up vote
5
down vote

Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_x,w^I$ are those which are $leq w$ w.r.t. the Bruhat order. If $win W_J$, then all those elements are themselves contained in $W_J$.

share|cite|improve this answer

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

add a comment | 

up vote
5
down vote

Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_x,w^I$ are those which are $leq w$ w.r.t. the Bruhat order. If $win W_J$, then all those elements are themselves contained in $W_J$.

share|cite|improve this answer

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

add a comment | 

up vote
5
down vote

up vote
5
down vote

Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_x,w^I$ are those which are $leq w$ w.r.t. the Bruhat order. If $win W_J$, then all those elements are themselves contained in $W_J$.

share|cite|improve this answer

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

Yes, that's true. The standard recursive constructions will give you this fact easily, because the only group elements involved in $P_x,w^I$ are those which are $leq w$ w.r.t. the Bruhat order. If $win W_J$, then all those elements are themselves contained in $W_J$.

share|cite|improve this answer

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

share|cite|improve this answer

share|cite|improve this answer

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

answered Sep 9 at 13:54

Johannes Hahn

5,48422140

5,48422140

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