Which roots are fixed by simple reflections of the Weyl Group?
Let $Phi$ be a root system of a semisimple Lie Algebra, and $W$ it's Weyl group. Let $Delta = { alpha_1, dots, alpha_l }$ be a root basis, and let $w_i in W$ be the simple reflection corresponding to $alpha_i in Delta$
Is there any way to compute the number $m$ of roots $alpha in Phi$ such that $w_i(alpha) = alpha$ for a fixed $i$?
I know that simple roots perpendicular to $alpha_i$ are fixed. By looking at Dynkin Diagrams, we see that there are either $l-2, l-3, $ or $l-4$ such simple roots. For each fixed simple root, it's negative is also fixed, so immediately we have found $2l-4, 2l-6 $ or $2l-8$ fixed roots.
However, is there a way to compute the exact number $m$? Failing that, is there a way to compute $m pmod 4$?
I am asking because I am interested in comparing $n = |Phi|$ and $m pmod 4$. I am aware they must both be even numbers, and cannot be congruent modulo $4$, but I am trying to see why exactly that is.
Any help would be appreciated, thank you!
lie-algebras weyl-group
asked Jan 4 at 21:16
user366818user366818
933410
add a comment |
Let $Phi$ be a root system of a semisimple Lie Algebra, and $W$ it's Weyl group. Let $Delta = { alpha_1, dots, alpha_l }$ be a root basis, and let $w_i in W$ be the simple reflection corresponding to $alpha_i in Delta$
Is there any way to compute the number $m$ of roots $alpha in Phi$ such that $w_i(alpha) = alpha$ for a fixed $i$?
I know that simple roots perpendicular to $alpha_i$ are fixed. By looking at Dynkin Diagrams, we see that there are either $l-2, l-3, $ or $l-4$ such simple roots. For each fixed simple root, it's negative is also fixed, so immediately we have found $2l-4, 2l-6 $ or $2l-8$ fixed roots.
However, is there a way to compute the exact number $m$? Failing that, is there a way to compute $m pmod 4$?
I am asking because I am interested in comparing $n = |Phi|$ and $m pmod 4$. I am aware they must both be even numbers, and cannot be congruent modulo $4$, but I am trying to see why exactly that is.
Any help would be appreciated, thank you!
lie-algebras weyl-group
asked Jan 4 at 21:16
user366818user366818
933410
It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43
add a comment |
Let $Phi$ be a root system of a semisimple Lie Algebra, and $W$ it's Weyl group. Let $Delta = { alpha_1, dots, alpha_l }$ be a root basis, and let $w_i in W$ be the simple reflection corresponding to $alpha_i in Delta$
Is there any way to compute the number $m$ of roots $alpha in Phi$ such that $w_i(alpha) = alpha$ for a fixed $i$?
I know that simple roots perpendicular to $alpha_i$ are fixed. By looking at Dynkin Diagrams, we see that there are either $l-2, l-3, $ or $l-4$ such simple roots. For each fixed simple root, it's negative is also fixed, so immediately we have found $2l-4, 2l-6 $ or $2l-8$ fixed roots.
However, is there a way to compute the exact number $m$? Failing that, is there a way to compute $m pmod 4$?
I am asking because I am interested in comparing $n = |Phi|$ and $m pmod 4$. I am aware they must both be even numbers, and cannot be congruent modulo $4$, but I am trying to see why exactly that is.
Any help would be appreciated, thank you!
lie-algebras weyl-group
asked Jan 4 at 21:16
user366818user366818
933410
Let $Phi$ be a root system of a semisimple Lie Algebra, and $W$ it's Weyl group. Let $Delta = { alpha_1, dots, alpha_l }$ be a root basis, and let $w_i in W$ be the simple reflection corresponding to $alpha_i in Delta$
Is there any way to compute the number $m$ of roots $alpha in Phi$ such that $w_i(alpha) = alpha$ for a fixed $i$?
I know that simple roots perpendicular to $alpha_i$ are fixed. By looking at Dynkin Diagrams, we see that there are either $l-2, l-3, $ or $l-4$ such simple roots. For each fixed simple root, it's negative is also fixed, so immediately we have found $2l-4, 2l-6 $ or $2l-8$ fixed roots.
However, is there a way to compute the exact number $m$? Failing that, is there a way to compute $m pmod 4$?
I am asking because I am interested in comparing $n = |Phi|$ and $m pmod 4$. I am aware they must both be even numbers, and cannot be congruent modulo $4$, but I am trying to see why exactly that is.
Any help would be appreciated, thank you!
lie-algebras weyl-group
lie-algebras weyl-group
asked Jan 4 at 21:16
user366818user366818
933410
asked Jan 4 at 21:16
user366818user366818
933410
asked Jan 4 at 21:16
user366818user366818
933410
asked Jan 4 at 21:16
user366818user366818
933410
asked Jan 4 at 21:16
user366818user366818
933410
933410
It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43
add a comment |
It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43
It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43
add a comment |
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It seems that starting from the third paragraph, you assume your Lie algebra to be simple / $Phi$ to be irreducible. -- In general, the roots whose number you are looking for are exactly those which are perpendicular to $alpha_i$.Now it seems to me that for any root $alpha$, the roots perpendicular to it form a sub-root system; and if $alpha$ happens to be one of your $alpha_i$, I would hope (but am not sure) that that root system can be read of the Dynkin diagram with all vertices neigbouring $alpha_i$ (and corresponding edges) erased.
– Torsten Schoeneberg
Jan 5 at 5:26
Thank you for this response. I suppose I mistakenly thought that the root system is irreducible when the Lie algebra is semi simple as opposed to simple. I will think about this a bit more
– user366818
Jan 5 at 14:26
@TorstenSchoeneberg sorry i forgot to tag you earlier
– user366818
Jan 5 at 16:43