# Constructing gauge invariants

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 $$$$\mathcal{O}_{a_1,b_1}...\mathcal{O}_{a_{20},b_{20}}\epsilon^{a1,a3}....\epsilon^{a6,b3}$$$$ I want to find out how many such objects are possible. Is there an efficient way of doing such things for other representations like $$\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

• 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. Jun 7 at 14:09

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...