2D representation of 3D algebra
I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, quad [t_3, t_1] = it_1, quad [t_1, t_2] = it_3 tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?
abstract-algebra group-theory lie-groups lie-algebras
asked Nov 18 at 16:09
Vicky
1437
add a comment |
I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, quad [t_3, t_1] = it_1, quad [t_1, t_2] = it_3 tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?
abstract-algebra group-theory lie-groups lie-algebras
asked Nov 18 at 16:09
Vicky
1437
add a comment |
I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, quad [t_3, t_1] = it_1, quad [t_1, t_2] = it_3 tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?
abstract-algebra group-theory lie-groups lie-algebras
asked Nov 18 at 16:09
Vicky
1437
I'm working with a Lie algebra that has 3 generator, $t_i$, that satisfy the following rules:
$$[t_2, t_3] = it_2, quad [t_3, t_1] = it_1, quad [t_1, t_2] = it_3 tag1$$
A 3D representation is easy if we choose the adjoint representation, but I want a 2D representation. I'm trying computing the commutators (with general $2times 2$ matrices) but I got non-sense results like parameter equal its negative or a generator with a zero in all entries. Does anyone know a more efficient method to compute this or know a group with this algebra?
abstract-algebra group-theory lie-groups lie-algebras
abstract-algebra group-theory lie-groups lie-algebras
asked Nov 18 at 16:09
Vicky
1437
asked Nov 18 at 16:09
Vicky
1437
asked Nov 18 at 16:09
Vicky
1437
asked Nov 18 at 16:09
Vicky
1437
asked Nov 18 at 16:09
Vicky
1437
1437
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $epsilon_{ijk}$ is the classical alternate tensor ($epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.
answered Nov 18 at 16:30
Jean Marie
28.8k41949
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $epsilon_{ijk}$ is the classical alternate tensor ($epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.
answered Nov 18 at 16:30
Jean Marie
28.8k41949
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
add a comment |
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $epsilon_{ijk}$ is the classical alternate tensor ($epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.
answered Nov 18 at 16:30
Jean Marie
28.8k41949
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
add a comment |
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $epsilon_{ijk}$ is the classical alternate tensor ($epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.
answered Nov 18 at 16:30
Jean Marie
28.8k41949
It deals with SU(2). Take for the $t_k$ a representation by Pauli matrices $sigma_k$ as well explained in the following document :{http://hepwww.rl.ac.uk/Haywood/Group_Theory_Lectures/Lecture_3.pdf} where $epsilon_{ijk}$ is the classical alternate tensor ($epsilon_{ijk}=1$ if $(i,j,k)$ is an even permutation of $(1,2,3)$, $epsilon_{ijk}=-1$ otherwise). The presence of this tensor explains why the three identities you have given do not look connected by a circular permutation of indices.
answered Nov 18 at 16:30
Jean Marie
28.8k41949
answered Nov 18 at 16:30
Jean Marie
28.8k41949
answered Nov 18 at 16:30
Jean Marie
28.8k41949
answered Nov 18 at 16:30
Jean Marie
28.8k41949
28.8k41949
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
add a comment |
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
Yes, you are absolutely right. How didn't I see it? Thank you very much!
– Vicky
Nov 18 at 16:38
add a comment |
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