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9
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edited Nov 26 2009 at 0:56 |
As is well known, the set
$\{a^ib^jc^k | i,j,k \in \mathbb{Z}_{\geq 0}\0},k>0\} \cup \{b^lc^md^n | l,m,n \in \mathbb{N}_{\geq 0}, k > 0\}$mathbb{Z}_{\geq 0}\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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8
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edited Nov 25 2009 at 20:34 |
As is well known, the set
${a^ib^jc^k \{a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}mathbb{Z}_{\geq 0}\} \cup {b^lc^md^n \{b^lc^md^n | l,m,n \in \mathbb{N}{\geq mathbb{N}_{\geq 0}, k > 0}$0\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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7
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edited Nov 25 2009 at 20:19 |
As is well known, the set
$\{a^ib^jc^k {a^ib^jc^k | i,j,k \in \mathbb{Z}_{\geq 0}\mathbb{Z}{\geq 0}} \cup \{b^lc^md^n {b^lc^md^n | l,m,n \in \mathbb{N}_{\geq mathbb{N}{\geq 0}, k > 0\}$0}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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6
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edited Nov 25 2009 at 20:08 |
As is well known, the set
$\{a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq mathbb{Z}_{\geq 0}\} \cup \{b^lc^md^n | l,m,n \in \mathbb{N}{\geq mathbb{N}_{\geq 0}, k > 0\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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5
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edited Nov 25 2009 at 20:07 |
As is well known, the set
${a^ib^jc^k \{a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}0}\} \cup $\{b^lc^md^n | l,m,n \in \mathbb{N}{\geq 0}, k > 0}$0\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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4
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edited Nov 25 2009 at 20:07 |
As is well known, the set
$\{a^ib^jc^k {a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}\0}} \cup \${b^lc^md^n | l,m,n \in \mathbb{N}{\geq 0}, k > 0\}$0}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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3
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edited Nov 25 2009 at 20:07 |
As is well known, the set
${a^ib^jc^k \{a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}0}\} \cup $\{b^lc^md^n | l,m,n \in \mathbb{N}{\geq 0}, k > 0}$0\}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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2
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edited Nov 25 2009 at 18:45 |
As is well known, the set
${a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}} \cup ${b^lc^md^n | l,m,n \in \mathbb{N}{\geq 0}, k > 0}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
My guess would be that a similar result holds. Namely that the set made up of all products of powers of the matrix entries ordered with respect to the canonical ordering such that the first entry in the q-det(n) does not appear would forms a basis. How to prove this, however, I do not know.
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1
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asked Nov 25 2009 at 18:30 |
Basis of quantum SU(n)
As is well known, the set
${a^ib^jc^k | i,j,k \in \mathbb{Z}{\geq 0}} \cup ${b^lc^md^n | l,m,n \in \mathbb{N}{\geq 0}, k > 0}$
forms a basis for quantum $SU(2)$. Does anyone know of a basis for quantum $SU(n)$?
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