Vanishing of certain coefficients coming from Coxeter groupsDouble quotients of Coxeter groups have the chain property?Centralizers of reflections in special subgroups of Coxeter groupsCross between the nil-Hecke ring and the group ring of a Coxeter groupReference request: Reduced reflection length in Coxeter groupsWhat is the relation between Coxeter transformations of generalized Cartan matrices and Coxeter transformations of finite-dimensional algebras?What is the relation between Coxeter transformations of Coxeter systems and Coxeter transformations of generalized Cartan matrices?A spin extension of a Coxeter group?What are the $2 times 2$ matrix generators of $textSL_2big(mathbbZ[i]big)(2+i)$?Coxeter groups generated by one finite conjugacy classRelation between groups $A_n$, $B_n$, $D_n$ and $S_n$ or inversions of random elements in Coxeter groups
Vanishing of certain coefficients coming from Coxeter groups
Double quotients of Coxeter groups have the chain property?Centralizers of reflections in special subgroups of Coxeter groupsCross between the nil-Hecke ring and the group ring of a Coxeter groupReference request: Reduced reflection length in Coxeter groupsWhat is the relation between Coxeter transformations of generalized Cartan matrices and Coxeter transformations of finite-dimensional algebras?What is the relation between Coxeter transformations of Coxeter systems and Coxeter transformations of generalized Cartan matrices?A spin extension of a Coxeter group?What are the $2 times 2$ matrix generators of $textSL_2big(mathbbZ[i]big)(2+i)$?Coxeter groups generated by one finite conjugacy classRelation between groups $A_n$, $B_n$, $D_n$ and $S_n$ or inversions of random elements in Coxeter groups
$begingroup$
Let $left(Wtext, Sright)$ be a Coxeter system. For every $win W$ let us write $left|wright|$ for the length of $w$. Set $lambdaleft(eright)=1$ where $ein W$ denotes the neutral element of the group and define coefficients $lambdaleft(wright)inmathbbZ$ recursively via begineqnarray sum_-leftlambdaleft(vright)=0 endeqnarray for any $win W$. For example this gives us $lambdaleft(eright)=1$, $lambdaleft(sright)=left(-1right)$ for all $sin S$, $lambdaleft(stright)=1$ if $m_st=2$ and $lambdaleft(stright)=0$ if $m_stneq2$ (for the notation see Wikipedia) for all $stext, tin S$ with $sneq t$, and so on.
My question is: Does there always exist some $linmathbbN$ such that $lambdaleft(wright)=0$ for all $win W$ with $left|wright|geq l$?
In the case that $left(Wtext, Sright)$ is right-angled (i.e. $m_stinleft 2text, inftyright$ for all $stext, tin S$ with $sneq t$) this is true and we can take $l=left|Sright|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way?
gr.group-theory coxeter-groups cayley-graphs
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $left(Wtext, Sright)$ be a Coxeter system. For every $win W$ let us write $left|wright|$ for the length of $w$. Set $lambdaleft(eright)=1$ where $ein W$ denotes the neutral element of the group and define coefficients $lambdaleft(wright)inmathbbZ$ recursively via begineqnarray sum_-leftlambdaleft(vright)=0 endeqnarray for any $win W$. For example this gives us $lambdaleft(eright)=1$, $lambdaleft(sright)=left(-1right)$ for all $sin S$, $lambdaleft(stright)=1$ if $m_st=2$ and $lambdaleft(stright)=0$ if $m_stneq2$ (for the notation see Wikipedia) for all $stext, tin S$ with $sneq t$, and so on.
My question is: Does there always exist some $linmathbbN$ such that $lambdaleft(wright)=0$ for all $win W$ with $left|wright|geq l$?
In the case that $left(Wtext, Sright)$ is right-angled (i.e. $m_stinleft 2text, inftyright$ for all $stext, tin S$ with $sneq t$) this is true and we can take $l=left|Sright|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way?
gr.group-theory coxeter-groups cayley-graphs
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
$endgroup$
add a comment |
$begingroup$
Let $left(Wtext, Sright)$ be a Coxeter system. For every $win W$ let us write $left|wright|$ for the length of $w$. Set $lambdaleft(eright)=1$ where $ein W$ denotes the neutral element of the group and define coefficients $lambdaleft(wright)inmathbbZ$ recursively via begineqnarray sum_-leftlambdaleft(vright)=0 endeqnarray for any $win W$. For example this gives us $lambdaleft(eright)=1$, $lambdaleft(sright)=left(-1right)$ for all $sin S$, $lambdaleft(stright)=1$ if $m_st=2$ and $lambdaleft(stright)=0$ if $m_stneq2$ (for the notation see Wikipedia) for all $stext, tin S$ with $sneq t$, and so on.
My question is: Does there always exist some $linmathbbN$ such that $lambdaleft(wright)=0$ for all $win W$ with $left|wright|geq l$?
In the case that $left(Wtext, Sright)$ is right-angled (i.e. $m_stinleft 2text, inftyright$ for all $stext, tin S$ with $sneq t$) this is true and we can take $l=left|Sright|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way?
gr.group-theory coxeter-groups cayley-graphs
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
$endgroup$
Let $left(Wtext, Sright)$ be a Coxeter system. For every $win W$ let us write $left|wright|$ for the length of $w$. Set $lambdaleft(eright)=1$ where $ein W$ denotes the neutral element of the group and define coefficients $lambdaleft(wright)inmathbbZ$ recursively via begineqnarray sum_-leftlambdaleft(vright)=0 endeqnarray for any $win W$. For example this gives us $lambdaleft(eright)=1$, $lambdaleft(sright)=left(-1right)$ for all $sin S$, $lambdaleft(stright)=1$ if $m_st=2$ and $lambdaleft(stright)=0$ if $m_stneq2$ (for the notation see Wikipedia) for all $stext, tin S$ with $sneq t$, and so on.
My question is: Does there always exist some $linmathbbN$ such that $lambdaleft(wright)=0$ for all $win W$ with $left|wright|geq l$?
In the case that $left(Wtext, Sright)$ is right-angled (i.e. $m_stinleft 2text, inftyright$ for all $stext, tin S$ with $sneq t$) this is true and we can take $l=left|Sright|$. I'm wondering if in the general case this property also holds. If no, is it possible to characterize the property in an instructive way?
gr.group-theory coxeter-groups cayley-graphs
gr.group-theory coxeter-groups cayley-graphs
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
asked Jun 20 at 19:00
worldreporter14worldreporter14
2351 silver badge12 bronze badges
asked Jun 20 at 19:00
worldreporter14worldreporter14
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1 Answer
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The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $lambda(w) neq 0$. Specifically, let $vee$ be the join in weak order, which is a semi-lattice. If $ s_1, s_2, ldots, s_j subseteq S$ and $s_1 vee s_2 vee cdots vee s_j$ is defined, then I claim that $lambda(s_1 vee s_2 vee cdots vee s_j)=(-1)^j$, and otherwise I claim that $lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 vee s_2 vee cdots vee s_j$, restricting ourselves to cases where this join is defined.
The condition $|v^-1 w| = |w| - |v|$ says that $v leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order.
The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then
$$mu(x) = sum_B subseteq A, bigvee B = x (-1)^.$$
So $mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
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$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $lambda(w) neq 0$. Specifically, let $vee$ be the join in weak order, which is a semi-lattice. If $ s_1, s_2, ldots, s_j subseteq S$ and $s_1 vee s_2 vee cdots vee s_j$ is defined, then I claim that $lambda(s_1 vee s_2 vee cdots vee s_j)=(-1)^j$, and otherwise I claim that $lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 vee s_2 vee cdots vee s_j$, restricting ourselves to cases where this join is defined.
The condition $|v^-1 w| = |w| - |v|$ says that $v leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order.
The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then
$$mu(x) = sum_B subseteq A, bigvee B = x (-1)^.$$
So $mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
$endgroup$
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
add a comment |
$begingroup$
The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $lambda(w) neq 0$. Specifically, let $vee$ be the join in weak order, which is a semi-lattice. If $ s_1, s_2, ldots, s_j subseteq S$ and $s_1 vee s_2 vee cdots vee s_j$ is defined, then I claim that $lambda(s_1 vee s_2 vee cdots vee s_j)=(-1)^j$, and otherwise I claim that $lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 vee s_2 vee cdots vee s_j$, restricting ourselves to cases where this join is defined.
The condition $|v^-1 w| = |w| - |v|$ says that $v leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order.
The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then
$$mu(x) = sum_B subseteq A, bigvee B = x (-1)^.$$
So $mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
$endgroup$
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
add a comment |
$begingroup$
The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $lambda(w) neq 0$. Specifically, let $vee$ be the join in weak order, which is a semi-lattice. If $ s_1, s_2, ldots, s_j subseteq S$ and $s_1 vee s_2 vee cdots vee s_j$ is defined, then I claim that $lambda(s_1 vee s_2 vee cdots vee s_j)=(-1)^j$, and otherwise I claim that $lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 vee s_2 vee cdots vee s_j$, restricting ourselves to cases where this join is defined.
The condition $|v^-1 w| = |w| - |v|$ says that $v leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order.
The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then
$$mu(x) = sum_B subseteq A, bigvee B = x (-1)^.$$
So $mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
$endgroup$
The answer is yes. I claim there are at most $2^n$ elements of $W$ for which $lambda(w) neq 0$. Specifically, let $vee$ be the join in weak order, which is a semi-lattice. If $ s_1, s_2, ldots, s_j subseteq S$ and $s_1 vee s_2 vee cdots vee s_j$ is defined, then I claim that $lambda(s_1 vee s_2 vee cdots vee s_j)=(-1)^j$, and otherwise I claim that $lambda(w)=0$. Thus $N$ can be taken to be the greatest length of $s_1 vee s_2 vee cdots vee s_j$, restricting ourselves to cases where this join is defined.
The condition $|v^-1 w| = |w| - |v|$ says that $v leq w$ in weak order. Therefore, this is the recursion defining the Mobius function of weak order.
The Mobius function of a finite lattice $L$ can be computed by Rota's crosscut theorem. (The first online reference I could find was Theorem 1.3 here.) Namely, let $A$ be the set of minimal elements of $L$ -- in this case, this is the set $S$. Then
$$mu(x) = sum_B subseteq A, bigvee B = x (-1)^.$$
So $mu(w)=0$ if $w$ is not a join of elements of $S$. If $w$ is a join of elements of $S$, then it is so in only one way (this is a way that weak order is simpler than a general lattice) so we get the description from the first paragraph.
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
edited Jun 20 at 23:20
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
answered Jun 20 at 19:32
David E SpeyerDavid E Speyer
109k10 gold badges289 silver badges554 bronze badges
109k10 gold badges289 silver badges554 bronze badges
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
add a comment |
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
$begingroup$
Thanks for your reply. Great answer!
$endgroup$
– worldreporter14
Jun 24 at 9:34
add a comment |
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