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0 votes
0 answers
178 views

Dimension of irreducible totally-symmetric tensor product of two su(n) representations

Consider the irreducible representation of $\mathfrak{su}$(n) given by the Young tableu The dimension of the representation is easily obtained by the usual rules, and it is $d=n(n-1)/2$. One can also ...
francesco's user avatar
  • 133
3 votes
1 answer
140 views

Schur functors for $\mathfrak{S}_3$

I have been trying to calculate the explicit images of the Schur functors for the action of $\mathfrak{S}_3$ on $V^{\otimes 3}$ where $V$ is some vector space, for the sake of concreteness of ...
Arnau Mas's user avatar
  • 311
4 votes
1 answer
314 views

Operators commuting with tensor product representations of SU(2)

I am currently investigating $SU(2)$ symmetric qubit systems. In the course of this work I proved the following theorem: Let $S_n$ denote the permutation group of $n$ elements. For $\sigma\in S_n$ ...
oliverkn's user avatar
  • 125
1 vote
1 answer
1k views

$8 \otimes 8$ in $SU(3)$, dimension of the Young-tableau corresponding to the $\bar{10}$

In Georgi's Lie Algebras in Particle Physics, he calculates the decomposition of $8\otimes 8$ in $SU(3)$, and obtains $$8\otimes 8 = 27 \oplus 10 \oplus \bar{10} \oplus 8 \oplus 8 \oplus 1,$$ ...
Ali's user avatar
  • 157
0 votes
0 answers
92 views

Does $\mathbf{n}\otimes \mathbf{m}=\mathbf{m}\otimes \mathbf{n}$ in the Clebsch-Gordan decomposition?

Consider the two irreducible representations, $\mathbf{n}$ and $\mathbf{m}$, of $su_\Bbb{C}(N)$. The tensor product $\mathbf{n}\otimes \mathbf{m}$ can be found using Young tableaux as follows: Write $...
Quantum spaghettification's user avatar
6 votes
2 answers
992 views

Littlewood Richardson rules for the orthogonal group SO(d)

I have a question related to tensor products of Young diagrams. More precisely, I know the Littlewood Richardson rules for the general linear group GL(d) and would like to know the restriction of ...
user264317's user avatar