PFR

4 The 100% version of PFR

Definition 4.1 Symmetry group
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If \(X\) is a \(G\)-valued random variable, then the symmetry group \(\mathrm{Sym}[X]\) is the set of all \(h \in G\) such that \(X +h\) has the same distribution as \(X\).

Lemma 4.2 Symmetry group is a group
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If \(X\) is a \(G\)-valued random variable, then \(\mathrm{Sym}[X]\) is a subgroup of \(G\).

Proof

Direct verification of the group axioms.

Lemma 4.3 Zero Ruzsa distance implies large symmetry group
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If \(X\) is a \(G\)-valued random variable such that \(d[X ;X]=0\), and \(x,y \in G\) are such that \(P[X=x], P[X=y]{\gt}0\), then \(x-y \in \mathrm{Sym}[X]\).

Proof

Let \(X_1,X_2\) be independent copies of \(X\) (from Lemma 3.7). Let \(A\) denote the range of \(X\). From Lemma 3.11 and Lemma 3.10 we have

\[ \mathbb {H}[X_1-X_2] = \mathbb {H}[X_1]. \]

Observe from Lemma 2.12 that

\[ \mathbb {H}[X_1-X_2|X_2] = \mathbb {H}[X_1|X_2] = \mathbb {H}[X_1] \]

and hence by Lemma 2.16

\[ \mathbb {I}[X_1-X_2 : X_1] = 0. \]

By Lemma 2.23, \(X_1-X_2\) and \(X_1\) are therefore independent, thus the law of \((X_1-X_2|X_1=x)\) does not depend on \(x \in A\). The claim follows.

Lemma 4.4 Translate is uniform on symmetry group
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If \(X\) is a \(G\)-valued random variable with \(d[X ;X]=0\), and \(x_0\) is a point with \(P[X=x_0] {\gt} 0\), then \(X-x_0\) is uniformly distributed on \(\mathrm{Sym}[X]\).

Proof

The law of \(X-x_0\) is invariant under \(\mathrm{Sym}[X]\), non-zero at the origin, and supported on \(\mathrm{Sym}[X]\), giving the claim.

Lemma 4.5 Symmetric 100% inverse theorem
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Suppose that \(X\) is a \(G\)-valued random variable such that \(d[X ;X]=0\). Then there exists a subgroup \(H \leq G\) such that \(d[X ;U_H] = 0\).

Proof

Take \(H\) to be the symmetry group of \(X\), which is a group by Lemma 4.2. From Lemma 4.4, \(X-x_0\) is uniform on \(H\), and \(d[X ;X-x_0] = d[X ;X] \leq 0\), and the claim follows.

Corollary 4.6 General 100% inverse theorem
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Suppose that \(X_1,X_2\) are \(G\)-valued random variables such that \(d[X_1;X_2]=0\). Then there exists a subgroup \(H \leq G\) such that \(d[X_1;U_H] = d[X_2;U_H] = 0\).

Proof

Using Lemma 3.18 and Lemma 3.15 we have \(d[X_1;X_1]=0\), hence by Lemma 4.5 \(d[X_1;U_H]=0\) for some subgroup \(H\). By Lemma 3.18 and Lemma 3.15 again we also have \(d[X_2;U_H]\) as required.