Mixing
The lattice generated by ${w(rho) - rho,vert,win W}$
Consider an irreducible root system associated to a complex simple Lie algebra $mathfrak{g}$. Let $rho$ be the half sum of positive roots and let $W$ be the Weyl group. Then what is the lattice $L$ generated by ${w(rho) - rho,vert,win W}$?
It is easy to see that $L$ is a sublattice of the root lattice $Q$, and I have checked that $L$ coincides with $Q$ for $A_1, A_2, A_3$ and $G_2$ root systems. Do $L$ and $Q$ always coincide?
lie-algebras root-systems lattices-in-lie-groups
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
add a comment |
Consider an irreducible root system associated to a complex simple Lie algebra $mathfrak{g}$. Let $rho$ be the half sum of positive roots and let $W$ be the Weyl group. Then what is the lattice $L$ generated by ${w(rho) - rho,vert,win W}$?
It is easy to see that $L$ is a sublattice of the root lattice $Q$, and I have checked that $L$ coincides with $Q$ for $A_1, A_2, A_3$ and $G_2$ root systems. Do $L$ and $Q$ always coincide?
lie-algebras root-systems lattices-in-lie-groups
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59
add a comment |
Consider an irreducible root system associated to a complex simple Lie algebra $mathfrak{g}$. Let $rho$ be the half sum of positive roots and let $W$ be the Weyl group. Then what is the lattice $L$ generated by ${w(rho) - rho,vert,win W}$?
It is easy to see that $L$ is a sublattice of the root lattice $Q$, and I have checked that $L$ coincides with $Q$ for $A_1, A_2, A_3$ and $G_2$ root systems. Do $L$ and $Q$ always coincide?
lie-algebras root-systems lattices-in-lie-groups
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
Consider an irreducible root system associated to a complex simple Lie algebra $mathfrak{g}$. Let $rho$ be the half sum of positive roots and let $W$ be the Weyl group. Then what is the lattice $L$ generated by ${w(rho) - rho,vert,win W}$?
It is easy to see that $L$ is a sublattice of the root lattice $Q$, and I have checked that $L$ coincides with $Q$ for $A_1, A_2, A_3$ and $G_2$ root systems. Do $L$ and $Q$ always coincide?
lie-algebras root-systems lattices-in-lie-groups
lie-algebras root-systems lattices-in-lie-groups
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
edited Jan 1 at 22:38
Henry Park
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
asked Jan 1 at 0:47
Henry ParkHenry Park
1,700724
1,700724
Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59
add a comment |
Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59
Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59
add a comment |
1 Answer
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I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $alphain Delta$, there are always a pair of choices of positive roots, say $Delta_+$ and $Delta_+'$ such that $Delta_+ setminus Delta_+' = {alpha}$ and $Delta_+' setminus Delta_+ = {-alpha}$. Let $win W$ be the element of Weyl group that transforms $Delta_+$ to $Delta_+'$. Let $rho$ be the half sum of elements of $Delta_+$. Then, $rho - w(rho) = alpha$. Hence $Lsupseteq Q supseteq L$, and therefore $L = Q$.
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
add a comment |
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1 Answer
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1 Answer
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I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $alphain Delta$, there are always a pair of choices of positive roots, say $Delta_+$ and $Delta_+'$ such that $Delta_+ setminus Delta_+' = {alpha}$ and $Delta_+' setminus Delta_+ = {-alpha}$. Let $win W$ be the element of Weyl group that transforms $Delta_+$ to $Delta_+'$. Let $rho$ be the half sum of elements of $Delta_+$. Then, $rho - w(rho) = alpha$. Hence $Lsupseteq Q supseteq L$, and therefore $L = Q$.
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
add a comment |
I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $alphain Delta$, there are always a pair of choices of positive roots, say $Delta_+$ and $Delta_+'$ such that $Delta_+ setminus Delta_+' = {alpha}$ and $Delta_+' setminus Delta_+ = {-alpha}$. Let $win W$ be the element of Weyl group that transforms $Delta_+$ to $Delta_+'$. Let $rho$ be the half sum of elements of $Delta_+$. Then, $rho - w(rho) = alpha$. Hence $Lsupseteq Q supseteq L$, and therefore $L = Q$.
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
add a comment |
I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $alphain Delta$, there are always a pair of choices of positive roots, say $Delta_+$ and $Delta_+'$ such that $Delta_+ setminus Delta_+' = {alpha}$ and $Delta_+' setminus Delta_+ = {-alpha}$. Let $win W$ be the element of Weyl group that transforms $Delta_+$ to $Delta_+'$. Let $rho$ be the half sum of elements of $Delta_+$. Then, $rho - w(rho) = alpha$. Hence $Lsupseteq Q supseteq L$, and therefore $L = Q$.
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
I just figured that this is an easy question. Yes, $L$ is always the same as $Q$. The proof goes as follows :
A choice of positive roots is the same as the choice of a Weyl chamber, and there is Weyl group-worth of such choices. For any root $alphain Delta$, there are always a pair of choices of positive roots, say $Delta_+$ and $Delta_+'$ such that $Delta_+ setminus Delta_+' = {alpha}$ and $Delta_+' setminus Delta_+ = {-alpha}$. Let $win W$ be the element of Weyl group that transforms $Delta_+$ to $Delta_+'$. Let $rho$ be the half sum of elements of $Delta_+$. Then, $rho - w(rho) = alpha$. Hence $Lsupseteq Q supseteq L$, and therefore $L = Q$.
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
answered Jan 1 at 22:49
Henry ParkHenry Park
1,700724
1,700724
add a comment |
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Hmmm I don't understand why some people just downvoted. I think this is a good question that perfectly fits the purpose of this site. Please let me know how this question can be improved.
– Henry Park
Jan 1 at 4:17
I did not downvote, but you could improve the question by replacing the first sentence with "Consider some irreduicible root system", and by adding what are your thoughts and what you have tried (e.g. some simple examples).
– Torsten Schoeneberg
Jan 1 at 5:44
@TorstenSchoeneberg I added some words, following your comment. Thanks!
– Henry Park
Jan 1 at 18:59