These are only going to be a soft questions. And I thought this question is also a case for MO, so I have posted a duplicate there (Does that comply with the etiquette here? In case not I am sorry.)
When looking at the Liouville function, defined as $$ \lambda(n) = (-1)^{\Omega_n},$$ where $\Omega_n$ is the total count of prime factors of $n$ (including multiplicity), it occurred to me, that this in a sense parallels an irreducible representation for $\Omega_n$ and hence also for the multiplicative semigroup of integers. The map $n\rightarrow \Omega_n$ is a multiplicative homomorphism. So one could generalise the Liouville function to $$ \lambda_m(n) = (e^{i\frac{2\pi}{m}})^{\Omega_n \mathrm{mod}\,m} ,$$ which recovers for $m=2$ the normal Liouville function $$ \lambda_2(n) = (e^{i\frac{2\pi}{2}})^{\Omega_n \mathrm{mod}\,2} = (-1)^{\Omega_{n} {\mathrm{mod}}\,2} = \lambda(n).$$ In this way one would get other such functions "mimicking" irreducible representations.
The questions are (i) if this analogy has been exploited already, (ii) if this functions are used already and (iii) if one could show the orthogonality $$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^{N}\overline{\lambda_i(n)}\lambda_j(n)\overset{?}=\delta_{ij}.$$
For $\lambda_j(n)=1\;\forall n$ (the "totally symmetric representation") and $\lambda_i(n)=\lambda(n)$, that reduces to $$\lim_{N\rightarrow\infty}\frac{L_N}{N},$$ with the Liouville sum function $L_N$ wich is known to be bound by $c \sqrt{N}$.