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For $f:G\to M_n(\mathbb{C})$ on a finite group, use uniform expectation over the solutions of $x{y}^{-1}z{w}^{-1}=e$ and the ordinary, unnormalized trace to define the displayed fourth-order quantity. If $T_{\rho }$ is its matrix Fourier block, this quantity equals ${\sum }_{\rho }{d}_{\rho }\mathrm{Tr}((T_{\rho }{T}_{\rho }^{\ast }{)}^2)$, so it is real and nonnegative. For a scalar function on an abelian group, it reduces to the usual fourth power of the Gowers uniformity norm of order two. The unnormalized trace means a unitary matrix-valued function can have fourth-order quantity as large as $n$.

Use $c>0$, the necessary parameter range for the bounds involving $1/c$. The construction comes from the singular value decomposition of matrix Fourier blocks. For each inequivalent unitary irreducible representation $\rho$ of dimension $d_{\rho }$, consider the Hilbert space ${\mathcal{H}}_{\rho }=M_{n\times d_{\rho }}(\mathbb{C})$ with the Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{tr}(A^{\ast }B)$ and the operatorIts operator norm is at most one, since $\rho (x)$ is a unitary matrix and $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$. Thus all its singular values ${\lambda }_{\rho ,j}$ lie in $[0,1]$.

Two identities control their weighted moments:Here the expectation in $S_4$ is uniform on the $|G{|}^3$ solutions of the constraint, and the trace on ${\mathcal{H}}_{\rho }$ used to compute each moment is the ordinary operator trace. We justify the identities explicitly to fix their normalizations and the noncommutative order.

The adjoint operator is $T_{\rho }^{\ast }(A)={\mathbb{E}}_{y}f(y{)}^{\ast }A\rho (y)$, soFor rectangular matrices, the operator $A\mapsto LAR$ has trace $\mathrm{tr}(L)\mathrm{tr}(R)$, as is seen on the matrix-unit basis. Squaring the previous operator therefore givesThe required representation theory consists of unitarization of a finite-group representation, the Schur orthogonality relations, and the regular representation decomposition. The last gives the character expansion of the identity deltaSumming the two operator traces with weights $d_{\rho }$ selects $x=y$ in the first, and $w=x{y}^{-1}z$ in the second. Multiplication by $|G|$ converts each independent-variable expectation to its conditional expectation. This proves $S_2,S_4$. In particular $S_4$ is real and nonnegative, despite the apparently complex summands in its original formula. Since $S_4\le S_2\le n$, the nonvacuous hypothesis has $0<c\le 1$.

Select every singular value satisfying ${\lambda }_{\rho ,j}\ge \sqrt{c/2}$. Index the selected pairs by $p=1,\dots ,r$, set ${\rho }_p=\rho$, $n_p=d_{\rho }$, and choose unit right singular vectors $A_p$ and unit left singular vectors $B_p$ with $T_{{\rho }_p}{A}_p={\lambda }_p{B}_p$. DefineBoth are $n\times n_p$ matrices, proving (i). Repeated group representations in this list are literally the same chosen representative, rather than different equivalent realizations.

If $\Vert f(x){\Vert }_{\mathrm{op}}\le 1$ and $\Vert f{\Vert }_{U^2}^4\ge cn>0$, select the singular values ${\lambda }_p\ge \sqrt{c/2}$ of the matrix Fourier blocks. Their weighted count $m={\sum }_p{d}_{{\rho }_p}$ lies between $cn/2$ and $2n/c$. Scaling unit right singular vectors and their corresponding unit left singular vectors by $\sqrt{d_{{\rho }_p}}$ gives matrices $U(p),V(p)$ satisfying ${\mathbb{E}}_{x}f(x)U(p){\rho }_p(x{)}^{\ast }={\lambda }_{p}V(p)$ and Hilbert-Schmidt inner product orthogonality within each chosen irreducible representation. The Schur averaging of rectangular matrices then gives ${\mathbb{E}}_x\Vert UP(x{)}^{\ast }b{\Vert }_2^2=\Vert b{\Vert }_2^2$ for the concatenated $U$ and the corresponding block diagonal matrix $P(x)$.