= Unbounded torsion obstructs uniform control of four-term progressions
{title2=$\|f\|_{U^3}\to0\ \not\Rightarrow\ \Lambda_4\to0\text{ uniformly in }G$}
A bound tending to zero with one factor's <Gowers U3 norm>, uniformly over all finite abelian groups, is false. On $G=\mathbb F_2^m$, choose a random independent sign at each point. A cube with eight distinct vertices has sign product of <expectation> zero. The <probability> of a collision among its vertices is at most $\binom82/|G|$, since each equality of distinct vertices imposes a nonzero linear equation in the increments. Thus some sign <function> $f$ has $\|f\|_{U^3}^8\leq28/|G|$. But the four-term <arithmetic progression> is $(x,x+d,x,x+d)$, so choosing factors $(f,1,f,1)$ gives correlation identically one. As $m\to\infty$, the <norm> tends to zero. Bounded torsion, as in cyclic groups, avoids this obstruction.
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