Weight diagram for $(2 , 1)$ representation of $SU(3)$
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Decided to practice my knowledge of representation theory by constructing the weight diagram for the representation $(2 , 1)$ of $SU(3)$. This is apparently the $mathbf{15}$, but when I use what I recall of the method to construct the weights at lower levels using the simple roots, I arrive at dimension 11.
Here's the diagram, where the simple roots are $alpha_1 = (2 ~ bar{1})$ and $alpha_2 = (bar{1} ~ 2)$, and an overbar denotes a minus sign:
$
largeqquad {(2~1)} \
quad {alpha_1} nearrow quad nwarrow alpha_2\
large(0 ~ 2)quadquad(3~bar{1})\
~alpha_1uparrow qquad quad uparrowalpha_1 \
large(bar{2} ~ 3) quadquad (1~0)\
quad alpha_2 nwarrow quadnearrow alpha_1 \
large~~~qquad(bar{1} ~ 1)\
quadalpha_1nearrow quad nwarrow alpha_2\
large(bar{3} ~~ 2) quad quad (0 ~ bar{1})\
~alpha_2uparrow qquad uparrow alpha_2\
large(bar{2} ~ 0)quad quad (1 ~ bar{3})\
quad alpha_2nwarrow quadnearrowalpha_1 \
large~~~~quad(bar1 ~ bar{2})
$
I'm able to correctly get the $mathbf{10}$ and $mathbf{8}$ using what I recall.
group-theory representation-theory lie-groups lie-algebras
asked Nov 15 at 18:50
nonreligious
133
add a comment |
up vote
0
down vote
favorite
Decided to practice my knowledge of representation theory by constructing the weight diagram for the representation $(2 , 1)$ of $SU(3)$. This is apparently the $mathbf{15}$, but when I use what I recall of the method to construct the weights at lower levels using the simple roots, I arrive at dimension 11.
Here's the diagram, where the simple roots are $alpha_1 = (2 ~ bar{1})$ and $alpha_2 = (bar{1} ~ 2)$, and an overbar denotes a minus sign:
$
largeqquad {(2~1)} \
quad {alpha_1} nearrow quad nwarrow alpha_2\
large(0 ~ 2)quadquad(3~bar{1})\
~alpha_1uparrow qquad quad uparrowalpha_1 \
large(bar{2} ~ 3) quadquad (1~0)\
quad alpha_2 nwarrow quadnearrow alpha_1 \
large~~~qquad(bar{1} ~ 1)\
quadalpha_1nearrow quad nwarrow alpha_2\
large(bar{3} ~~ 2) quad quad (0 ~ bar{1})\
~alpha_2uparrow qquad uparrow alpha_2\
large(bar{2} ~ 0)quad quad (1 ~ bar{3})\
quad alpha_2nwarrow quadnearrowalpha_1 \
large~~~~quad(bar1 ~ bar{2})
$
I'm able to correctly get the $mathbf{10}$ and $mathbf{8}$ using what I recall.
group-theory representation-theory lie-groups lie-algebras
asked Nov 15 at 18:50
nonreligious
133
Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Decided to practice my knowledge of representation theory by constructing the weight diagram for the representation $(2 , 1)$ of $SU(3)$. This is apparently the $mathbf{15}$, but when I use what I recall of the method to construct the weights at lower levels using the simple roots, I arrive at dimension 11.
Here's the diagram, where the simple roots are $alpha_1 = (2 ~ bar{1})$ and $alpha_2 = (bar{1} ~ 2)$, and an overbar denotes a minus sign:
$
largeqquad {(2~1)} \
quad {alpha_1} nearrow quad nwarrow alpha_2\
large(0 ~ 2)quadquad(3~bar{1})\
~alpha_1uparrow qquad quad uparrowalpha_1 \
large(bar{2} ~ 3) quadquad (1~0)\
quad alpha_2 nwarrow quadnearrow alpha_1 \
large~~~qquad(bar{1} ~ 1)\
quadalpha_1nearrow quad nwarrow alpha_2\
large(bar{3} ~~ 2) quad quad (0 ~ bar{1})\
~alpha_2uparrow qquad uparrow alpha_2\
large(bar{2} ~ 0)quad quad (1 ~ bar{3})\
quad alpha_2nwarrow quadnearrowalpha_1 \
large~~~~quad(bar1 ~ bar{2})
$
I'm able to correctly get the $mathbf{10}$ and $mathbf{8}$ using what I recall.
group-theory representation-theory lie-groups lie-algebras
asked Nov 15 at 18:50
nonreligious
133
Decided to practice my knowledge of representation theory by constructing the weight diagram for the representation $(2 , 1)$ of $SU(3)$. This is apparently the $mathbf{15}$, but when I use what I recall of the method to construct the weights at lower levels using the simple roots, I arrive at dimension 11.
Here's the diagram, where the simple roots are $alpha_1 = (2 ~ bar{1})$ and $alpha_2 = (bar{1} ~ 2)$, and an overbar denotes a minus sign:
$
largeqquad {(2~1)} \
quad {alpha_1} nearrow quad nwarrow alpha_2\
large(0 ~ 2)quadquad(3~bar{1})\
~alpha_1uparrow qquad quad uparrowalpha_1 \
large(bar{2} ~ 3) quadquad (1~0)\
quad alpha_2 nwarrow quadnearrow alpha_1 \
large~~~qquad(bar{1} ~ 1)\
quadalpha_1nearrow quad nwarrow alpha_2\
large(bar{3} ~~ 2) quad quad (0 ~ bar{1})\
~alpha_2uparrow qquad uparrow alpha_2\
large(bar{2} ~ 0)quad quad (1 ~ bar{3})\
quad alpha_2nwarrow quadnearrowalpha_1 \
large~~~~quad(bar1 ~ bar{2})
$
I'm able to correctly get the $mathbf{10}$ and $mathbf{8}$ using what I recall.
group-theory representation-theory lie-groups lie-algebras
group-theory representation-theory lie-groups lie-algebras
asked Nov 15 at 18:50
nonreligious
133
asked Nov 15 at 18:50
nonreligious
133
asked Nov 15 at 18:50
nonreligious
133
asked Nov 15 at 18:50
nonreligious
133
asked Nov 15 at 18:50
nonreligious
133
133
Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42
add a comment |
Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42
Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42
add a comment |
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Are you sure those weights all occur with multiplicity $1$? (I did not do any calclations myself, but that would be an easy place to get an error). Certainly the irrep with highest weight $(2,1)$ has dimension $15$ by Weyl's dimension formula (assuming you are writing these in terms of the fundamental weights).
– Tobias Kildetoft
Nov 15 at 20:15
I did think that, and indeed some of these weights can be reached by two different arrows - (1 0), (-1 1), (0 -1), (-2 0), (-1 -2) - but they aren't degenerate in the way that the weight (0 0) is in the adjoint of SU(3).
– nonreligious
Nov 15 at 20:42