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How to represent SU(N) as polynomials of the irreducible representations of SU(2) in complex vector space of...
0
Suppose $S^x$,$S^y$ and $S^z$ are the irreducible representations of $SU(2)$ in a complex vector space, V, of dimension N. By Burnside's theorem, $I$, $S^x$,$S^y$ and $S^z$ for a basis in the ring of endomorphims of V.
Now take $T_k$ as the generators of $SU(N)$. By the definition of $SU(N)$, $T_k$ is a subset of endomorphims in V.
Thus, I expect I should be able to write:
$
T_k=f_k(S^x,S^y,S^z,I)
$
with $f_k$ being a polynomial.
Is this expecation correct. Did I understand Burnside's theorem and the definition of SU(N) correctly?
- If so, what can I say about $f_k$?
- What is the highest degree polynomial required?
- Do I need products like $S^xS^y$ or do monomials in $(S^i)^n$ suffice
group-theory representation-theory lie-algebras
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
add a comment |
0
Suppose $S^x$,$S^y$ and $S^z$ are the irreducible representations of $SU(2)$ in a complex vector space, V, of dimension N. By Burnside's theorem, $I$, $S^x$,$S^y$ and $S^z$ for a basis in the ring of endomorphims of V.
Now take $T_k$ as the generators of $SU(N)$. By the definition of $SU(N)$, $T_k$ is a subset of endomorphims in V.
Thus, I expect I should be able to write:
$
T_k=f_k(S^x,S^y,S^z,I)
$
with $f_k$ being a polynomial.
Is this expecation correct. Did I understand Burnside's theorem and the definition of SU(N) correctly?
- If so, what can I say about $f_k$?
- What is the highest degree polynomial required?
- Do I need products like $S^xS^y$ or do monomials in $(S^i)^n$ suffice
group-theory representation-theory lie-algebras
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
2
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
2
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16
add a comment |
0
0
0
Suppose $S^x$,$S^y$ and $S^z$ are the irreducible representations of $SU(2)$ in a complex vector space, V, of dimension N. By Burnside's theorem, $I$, $S^x$,$S^y$ and $S^z$ for a basis in the ring of endomorphims of V.
Now take $T_k$ as the generators of $SU(N)$. By the definition of $SU(N)$, $T_k$ is a subset of endomorphims in V.
Thus, I expect I should be able to write:
$
T_k=f_k(S^x,S^y,S^z,I)
$
with $f_k$ being a polynomial.
Is this expecation correct. Did I understand Burnside's theorem and the definition of SU(N) correctly?
- If so, what can I say about $f_k$?
- What is the highest degree polynomial required?
- Do I need products like $S^xS^y$ or do monomials in $(S^i)^n$ suffice
group-theory representation-theory lie-algebras
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
Suppose $S^x$,$S^y$ and $S^z$ are the irreducible representations of $SU(2)$ in a complex vector space, V, of dimension N. By Burnside's theorem, $I$, $S^x$,$S^y$ and $S^z$ for a basis in the ring of endomorphims of V.
Now take $T_k$ as the generators of $SU(N)$. By the definition of $SU(N)$, $T_k$ is a subset of endomorphims in V.
Thus, I expect I should be able to write:
$
T_k=f_k(S^x,S^y,S^z,I)
$
with $f_k$ being a polynomial.
Is this expecation correct. Did I understand Burnside's theorem and the definition of SU(N) correctly?
- If so, what can I say about $f_k$?
- What is the highest degree polynomial required?
- Do I need products like $S^xS^y$ or do monomials in $(S^i)^n$ suffice
group-theory representation-theory lie-algebras
group-theory representation-theory lie-algebras
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
asked Nov 21 '18 at 0:48
Shane P Kelly
1106
1106
What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
2
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
2
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16
add a comment |
What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
2
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
2
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16
What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
2
2
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
2
2
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16
add a comment |
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What "Burnside's theorem" are you referring to?
– Qiaochu Yuan
Nov 21 '18 at 1:01
I didn't know there were multiple: it says that if a moniod of linear transformations acts irreducibly then it is a basis for the ring of endomorphisms, I found it in this paper: arxiv.org/abs/cond-mat/0207106
– Shane P Kelly
Nov 21 '18 at 1:07
2
I don't know this result by that name; for me this is a special case of the Jacobson density theorem (en.wikipedia.org/wiki/Jacobson_density_theorem). In mathematics "Burnside's theorem" refers to a theorem about finite groups: en.wikipedia.org/wiki/Burnside_theorem
– Qiaochu Yuan
Nov 21 '18 at 1:18
2
Your language is somewhat imprecise but you are basically right, if by "polynomial" you mean "noncommutative polynomial" (and "basis" in your second sentence should be "set of generators"). I don't know that there's anything nice to say about the $f_k$ and I would not enjoy attempting to calculate them. You need to make many noncanonical choices to define $f_k$, for example a choice of basis of $mathfrak{su}(2)$, a choice of inner product on $V$, and a choice of basis of $mathfrak{su}(V)$.
– Qiaochu Yuan
Nov 21 '18 at 1:20
@QiaochuYuan Regarding the two Burnside theorems: the lesser known one does really go by that name too, a great proof is prestented here: core.ac.uk/download/pdf/82680953.pdf
– Vincent
Nov 21 '18 at 14:16