How close can we come to a parity function when there isnt one?

Consider a group G such that there is no homomorphism f:G to {+1,-1}. In that case, how close can we come to such a homomorphism? We show that if f has zero expectation, then the probability that f(xy) = f(x) f(y), where x, y are chosen uniformly and independently from G, is at most 1/2(1+1/sqrt{d}), where d is the dimension of G’s smallest nontrivial irreducible representation. For the alternating group A_n, for instance, d=n-1. On the other hand, A_n contains a subgroup isomorphic to S_{n-2}, whose parity function we can extend to obtain an f for which this probability is 1/2(1+1/{n \choose 2}). Thus the extent to which f can be “more homomorphic” than a random function from A_n to {+1,-1} lies between O(n^{-1/2}) and Omega(n^{-2}).

💡 Research Summary

The paper investigates how closely a ±1‑valued function on a finite group can mimic a group homomorphism when no non‑trivial homomorphism to {+1, −1} exists. The authors focus on functions f : G → {±1} with zero expectation (i.e., the average value of f over G is 0) and ask for the maximal probability that the homomorphism condition f(xy)=f(x)f(y) holds when x and y are drawn independently and uniformly from G.

The central technical tool is the Fourier analysis on finite groups. Any complex‑valued function on G can be expanded in the orthogonal basis provided by the matrix entries of the irreducible representations of G. By examining the Fourier coefficients of f and applying Parseval’s identity together with Cauchy–Schwarz, the authors derive an upper bound that depends only on the smallest dimension d of a non‑trivial irreducible representation of G:

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