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Is there an efficient way for constructing gauge invariants given the number of operators one can use is fixed. For example, if I am given some boson in $\mathbf{3}$ of $SU(2)$, and I want to find out the number of possible invariants constructed when I have 20 such objects. One can construct an object like \begin{equation} \mathcal{O}_{a_1,b_1}...\mathcal{O}_{a_{20},b_{20}}\epsilon^{a1,a3}....\epsilon^{a6,b3} \end{equation} I want to find out how many such objects are possible. Is there an efficient way of doing such things for other representations like $\mathbf{5}$, $\mathbf{6}$,..etc, for fermions or maybe mixed objects like $\mathbf{4}$ and $\mathbf{7}$. I was trying to use mathematica for contractions but the number of partitions grows very fast and doing it brute force doesn't seem to be an option. Any suggestion ? Thanks

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    $\begingroup$ Isn't it the multiplicity of the identity rep in the tensor product of all the reps? Maybe need to do some symmetrization/anti-symmetrization if they are all identical particles. $\endgroup$
    – Meng Cheng
    Jun 7 at 14:09

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I am using this reference, Spin Multiplicities, T Curtright, T van Kortryk, and C Zachos, Phys Lett A381 (2017) 422-427.

The character of a spin j, dimension 2j+1 irrep of SU(2) is $$ \chi_j (\theta)= \frac{\sin((2j+1)\theta/2)}{\sin (\theta/2)}, \tag{4} $$ and the multiplicity of spin s in their n-fold composition is $$ M(s;n;j)= \frac{1}{\pi}\int_0^{2\pi} \!\! d\vartheta ~\sin^2{\vartheta} ~~ (\chi_j(2\vartheta))^n ~\chi_s(2\vartheta), \tag{5} $$ whence, in your case, $$ M(0;20;1)= \frac{1}{\pi}\int_0^{2\pi} \!\! d\vartheta ~\sin^2{\vartheta} ~~ (\chi_1(2\vartheta))^{20} , $$ the hypergeometric function of (17). Since this n is large, this is quite close to $$ { 3^{20.5} \over 8\sqrt{\pi} ~ 20^{3/2} }, $$ by (31). Millions...

The method addresses your further questions. Try a few simple examples, to ensure you understand the language.

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  • $\begingroup$ Thanks, it would have been useful if I could get the index structure as well. I need all the singlets, this method gives me the multiplicity only. But, thanks a lot !! $\endgroup$ Jun 7 at 20:20
  • $\begingroup$ It is a non-random walk... steps to the right, standing in place, and steps to the left, starting at the origin. Might look at small n s ... 2, 3, 4, 5, 6... $\endgroup$ Jun 7 at 21:21

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