Can generators of a Coxeter group be redundant?












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I’m just beginning to learn about Coxeter groups and at one point the textbook’s author seems to take it for grant that each generator cannot be “reduced” into a product of other generators. Specifically, it seems the following claim is so obvious that it need not be stated:



If ${s_1, s_2,..., s_n}$ is the set of generators of a coxeter group $W$, then one cannot have $s_i=s_1s_2...s_{j-1}s_js_{j-1}...s_1$ for some $j$ without $i, j$ being $1$.



So I suspected something like $s_1=s_2s_4s_8$ cannot happen too. I know the question looks really stupid but it seems I’m missing something fundamental.










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  • $begingroup$
    This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
    $endgroup$
    – darij grinberg
    Jan 2 at 11:45


















1












$begingroup$


I’m just beginning to learn about Coxeter groups and at one point the textbook’s author seems to take it for grant that each generator cannot be “reduced” into a product of other generators. Specifically, it seems the following claim is so obvious that it need not be stated:



If ${s_1, s_2,..., s_n}$ is the set of generators of a coxeter group $W$, then one cannot have $s_i=s_1s_2...s_{j-1}s_js_{j-1}...s_1$ for some $j$ without $i, j$ being $1$.



So I suspected something like $s_1=s_2s_4s_8$ cannot happen too. I know the question looks really stupid but it seems I’m missing something fundamental.










share|cite|improve this question









$endgroup$












  • $begingroup$
    This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
    $endgroup$
    – darij grinberg
    Jan 2 at 11:45
















1












1








1





$begingroup$


I’m just beginning to learn about Coxeter groups and at one point the textbook’s author seems to take it for grant that each generator cannot be “reduced” into a product of other generators. Specifically, it seems the following claim is so obvious that it need not be stated:



If ${s_1, s_2,..., s_n}$ is the set of generators of a coxeter group $W$, then one cannot have $s_i=s_1s_2...s_{j-1}s_js_{j-1}...s_1$ for some $j$ without $i, j$ being $1$.



So I suspected something like $s_1=s_2s_4s_8$ cannot happen too. I know the question looks really stupid but it seems I’m missing something fundamental.










share|cite|improve this question









$endgroup$




I’m just beginning to learn about Coxeter groups and at one point the textbook’s author seems to take it for grant that each generator cannot be “reduced” into a product of other generators. Specifically, it seems the following claim is so obvious that it need not be stated:



If ${s_1, s_2,..., s_n}$ is the set of generators of a coxeter group $W$, then one cannot have $s_i=s_1s_2...s_{j-1}s_js_{j-1}...s_1$ for some $j$ without $i, j$ being $1$.



So I suspected something like $s_1=s_2s_4s_8$ cannot happen too. I know the question looks really stupid but it seems I’m missing something fundamental.







abstract-algebra coxeter-groups






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share|cite|improve this question











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share|cite|improve this question










asked Jan 2 at 11:18









AhmbakAhmbak

390110




390110












  • $begingroup$
    This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
    $endgroup$
    – darij grinberg
    Jan 2 at 11:45




















  • $begingroup$
    This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
    $endgroup$
    – darij grinberg
    Jan 2 at 11:45


















$begingroup$
This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
$endgroup$
– darij grinberg
Jan 2 at 11:45






$begingroup$
This is not obvious -- I suspect it only becomes clear somewhere in Chapter 4 of Björner/Brenti, not earlier. The easiest way to see the claim is if you know that the length function of a parabolic subgroup $W_I$ of $W$ is the restriction of the length function of $W$ (see, e.g., §9.6 of arXiv:math/0208154v2); thus, if $s_i$ was a product of other $s_j$'s, then $s_i$ would have to be a single other $s_j$, which would contradict the (nontrivial) fact that the elements of $S$ are distinct in $W$.
$endgroup$
– darij grinberg
Jan 2 at 11:45












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