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 $c \sqrt{N}$.

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    $\begingroup$ The limit certainly doesn’t exist, because the terms all have absolute value 1. So whatever orthogonality you want, you have to make sure that the sum actually converges. Maybe throw a $\frac{1}{N}$ in front, and conjugate $\lambda_j(n)$. $\endgroup$ – Bob Jones Dec 26 '17 at 19:57
  • $\begingroup$ Well I think its at least unclear, for the simplest case all $\lambda_j(n)=1$ (would be something like the totally symmetric representation) and $\lambda_i = \lambda$ one gets the summation Liouville function and that seems to have at least infinitely many zeros. Or something like clever summation (Ramanujan, Abel?) suffices? $\endgroup$ – Rudi_Birnbaum Dec 26 '17 at 20:01
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    $\begingroup$ You need the conjugation in there somewhere because if $i=j$ you want the sum to be just a bunch of 1’s but as it’s written now, it’s the square of $\lambda_i$ which has no reason to go to 1. $\endgroup$ – Bob Jones Dec 26 '17 at 20:09
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    $\begingroup$ And now the question that begs to be asked: what if we just take some random root of unity $\zeta\neq1$, and we look at the sequence $a_n=\sum_{i=1}^n \zeta^{\Omega_i}$; is this sequence $o(n)$? A simplistic probability heuristic says its absolute value should be about $\sqrt{n}$, so probably yes. Just a few early thoughts having nothing to do with the function $\Omega_i$. $\endgroup$ – Bob Jones Dec 26 '17 at 20:29
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    $\begingroup$ And I think this book should help you: math.mcgill.ca/darmon/courses/sato-tate/serre-mcgill.pdf in the appendix starting on page I-18. Try applying the stuff there to the Dirichlet series $\sum_{i=1}^{\infty}\frac{\zeta^{\Omega_i}}{i^s}=\prod_{p}\frac{1}{1-\zeta p^{-s}}$. The key theorem is the Wiener Ikehara theorem. $\endgroup$ – Bob Jones Dec 26 '17 at 20:51

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