Label the six strict orderings of three alternatives \(a,b,c\) by \(k\in \{ 0,\dots ,5\} \). For each ordering let \(x_k\in \{ -1,+1\} \) record the \(a\)-versus-\(b\) comparison, taking the value \(+1\) when \(a\) is preferred to \(b\) and \(-1\) when \(b\) is preferred to \(a\), and let \(y_k\in \{ -1,+1\} \) record the \(b\)-versus-\(c\) comparison in the same way, with \(+1\) when \(b\) is preferred to \(c\) and \(-1\) otherwise. Then

For each of the six strict orderings \(k \in \{ 0,\dots ,5\} \) of three alternatives \(a,b,c\), let \(\sigma \) be the sign encoding, sending \(\mathrm{false}\) to \(+1\) and \(\mathrm{true}\) to \(-1\). Write \(y_k = \sigma (p_k)\), where \(p_k\) is the Boolean preference of ordering \(k\) in the \(b\)-vs-\(c\) comparison, and \(z_k = \sigma (q_k)\), where \(q_k\) is the preference in the \(c\)-vs-\(a\) comparison. Then

For each of the six strict orderings \(k \in \{ 0,\dots ,5\} \) of three alternatives \(a,b,c\), write \(p_{ab}(k)\) and \(p_{ca}(k)\) for the Boolean preferences of ordering \(k\) in the \(a\)-vs-\(b\) and \(c\)-vs-\(a\) comparisons, and let \(\sigma \) be the sign encoding sending \(\mathrm{false}\) to \(1\) and \(\mathrm{true}\) to \(-1\). Then the sum over all six orderings of the products of the encoded signs is

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function. Suppose each of the \(n\) voters chooses, independently and uniformly, one of the six strict orderings of three candidates \(a,b,c\); for such a profile \(p\) write \(x_p\in \{ 0,1\} ^n\) for the vector whose \(i\)-th entry records voter \(i\)’s preference in the \(a\)-versus-\(b\) comparison, and \(y_p\in \{ 0,1\} ^n\) for the vector recording each voter’s preference in the \(b\)-versus-\(c\) comparison. Then, averaging the product of the values of \(f\) on these two vectors over all \(6^n\) profiles,

where \(\hat f(S)\) denotes the Fourier–Walsh coefficient of \(f\) at frequency \(S\).

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function that is odd, so that \(f(\lnot x)=-f(x)\) for all \(x\), and \(\pm 1\)-valued, so that \(f(x)\in \{ -1,1\} \) for all \(x\). Then its Fourier correlation function satisfies

where \(\hat f(S)\) denotes the Fourier–Walsh coefficient of \(f\) at frequency \(S\).

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a \(\pm 1\)-valued Boolean function that is odd, meaning \(f(\lnot x)=-f(x)\) for every \(x\), and suppose its Fourier correlation function attains the value

Then \(\hat f(S)=0\) for every subset \(S\subseteq [n]\) with \(|S|\neq 1\); that is, all Fourier–Walsh weight of \(f\) is carried by the singleton frequencies.

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function that is odd, \(\pm 1\)-valued, and acyclic. Then its Fourier correlation function satisfies

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function that is odd, is \(\pm 1\)-valued, and is unanimous. Suppose further that \(f\) has degree one, in the sense that its Fourier–Walsh coefficient \(\hat f(S)\) vanishes for every set \(S\subseteq [n]\) with \(\left\lvert S\right\rvert \neq 1\). Then \(f\) is a dictatorship: there is a coordinate \(i\) with \(f=\mathrm{dict}_i\).

There are six strict orderings \(k \in \{ 0,1,\dots ,5\} \) of the three alternatives \(a,b,c\). For each such ordering, let \(p_{ab}(k)\) be its \(a\)-vs-\(b\) preference, the Boolean that records whether \(b\) is ranked above \(a\), and apply the sign encoding \(\sigma \), so that \(\sigma (p_{ab}(k))\) equals \(+1\) when ordering \(k\) ranks \(a\) above \(b\) and \(-1\) when it ranks \(b\) above \(a\). Then these six signed preferences sum to zero:

Consider the six strict orderings of three alternatives \(a\), \(b\), and \(c\), indexed by \(k \in \{ 0,\dots ,5\} \), and let \(p(k)\) denote the \(b\)-versus-\(c\) pairwise preference of ordering \(k\). Under the sign encoding \(\sigma \), these preferences sum to zero across all six orderings:

For each of the six strict orderings \(k\in \{ 0,1,\dots ,5\} \) of three alternatives, let \(b_k\in \mathrm{Bool}\) record the \(c\)-vs-\(a\) preference of ordering \(k\), and encode it as a sign by \(\sigma (b_k)=1\) when \(b_k\) is false and \(\sigma (b_k)=-1\) when \(b_k\) is true. Then these six signed preferences sum to zero:

Let \(A \subseteq \mathrm{Fin}\, n\) be a finite set of indices and let \(g : \mathrm{Fin}\, n \to \mathbb {R}\). Then

so that a product taken only over the members of \(A\) may equivalently be written as a product over all of \(\mathrm{Fin}\, n\) in which each factor outside \(A\) is replaced by \(1\).

Let \(x, y : \mathrm{Fin}\, 6 \to \mathrm{Bool}\) be two preference assignments whose sign encodings have balanced marginals, \(\sum _{k=1}^{6}\sigma (x_k) = \sum _{k=1}^{6}\sigma (y_k) = 0\), and whose signed correlation is \(\sum _{k=1}^{6}\sigma (x_k)\, \sigma (y_k) = -2\). For a profile \(p : \mathrm{Fin}\, n \to \mathrm{Fin}\, 6\), write \(x\circ p\) and \(y\circ p\) for the points of the Boolean hypercube whose \(i\)-th coordinates are \(x_{p(i)}\) and \(y_{p(i)}\). Then for all subsets \(S, T \subseteq [n]\), the average of the product of Walsh characters over all \(6^n\) profiles satisfies

Fix \(n\), and for each profile \(p \in \{ 0,\dots ,5\} ^n\) let \(v_{ab}(p), v_{bc}(p) \in \{ 0,1\} ^n\) be the vote vectors recording, voter by voter, each voter’s pairwise preference in the \(a\)-vs-\(b\) and \(b\)-vs-\(c\) comparisons, and for \(S \subseteq [n]\) let \(\chi _S\) be the associated Walsh–Fourier character. Then for all subsets \(S, T \subseteq [n]\), the uniform average of the product \(\chi _S(v_{ab}(p))\, \chi _T(v_{bc}(p))\) over the \(6^n\) profiles satisfies

Let \(S, T \subseteq [n]\), and for a profile \(p \colon \mathrm{Fin}\, n \to \mathrm{Fin}\, 6\) write \(x(p)\) for its \(bc\)-vote vector and \(y(p)\) for its \(ca\)-vote vector, each lying in \(\{ 0,1\} ^n\). Averaging the product of Walsh characters over all \(6^n\) profiles gives

Fix a natural number \(n\) and consider profiles \(p\) of \(n\) voters, each of whom reports one of the six strict orderings of the three alternatives \(a, b, c\); there are \(6^n\) such profiles. For a profile \(p\), let \(x^{ab}(p) \in \{ 0,1\} ^n\) be the vector recording each voter’s preference in the \(a\)-vs-\(b\) comparison, and \(x^{ca}(p) \in \{ 0,1\} ^n\) the vector recording each voter’s preference in the \(c\)-vs-\(a\) comparison. Then for any two subsets \(S, T \subseteq \{ 1,\dots ,n\} \), the averaged product of the corresponding Walsh–Fourier characters satisfies

where the sum ranges over all \(6^n\) profiles.

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function, and let \(\mathrm{votes}_1,\mathrm{votes}_2\) each assign to every profile \(p\in (\mathrm{Fin}\, 6)^{\, n}\) a point of the Boolean hypercube \(\{ 0,1\} ^n\). Suppose the pair satisfies the kernel identity

for all \(S,T\subseteq [n]\), where the sum runs over all \(6^n\) profiles. Then

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function, and for each profile \(p:\mathrm{Fin}\, n\to \{ 0,\dots ,5\} \) write \(x_{bc}(p),\, x_{ca}(p)\in \{ 0,1\} ^n\) for the \(b\)-vs-\(c\) and \(c\)-vs-\(a\) vote vectors it induces. Then the average of the product \(f\bigl(x_{bc}(p)\bigr)\, f\bigl(x_{ca}(p)\bigr)\) over all \(6^n\) profiles equals the Fourier correlation function of \(f\):

Let \(f:\{ 0,1\} ^n\to \mathbb {R}\) be a Boolean function. For a profile \(p\) that assigns each of the \(n\) voters one of the six strict orderings of the three alternatives \(a,b,c\), let \(x^{ab}(p)\in \{ 0,1\} ^n\) record each voter’s preference in the \(a\)-vs-\(b\) comparison and let \(x^{ca}(p)\in \{ 0,1\} ^n\) record each voter’s preference in the \(c\)-vs-\(a\) comparison. Averaging uniformly over all \(6^n\) profiles, the expected product of \(f\) at these two vote vectors equals the Fourier correlation function of \(f\):