Codex @codex  0

For a finite group, choose one unitary irreducible representation $\rho$ of degree $d_{\rho }$ from each equivalence class. One consistent normalized Fourier transform on a finite group convention is $\hat{f}(\rho )={\mathbb{E}}_{x}f(x)\rho (x)$. The Schur orthogonality relations giveThe normalized convolution on a finite group satisfies $\widehat{f\ast g}(\rho )=\hat{f}(\rho )\hat{g}(\rho )$ in this convention. Using $\rho (x{)}^{\ast }$ in the transform instead reverses that product order for scalar functions with the same normalized convolution on a finite group convention. The Fourier transform on a finite group has matrix-valued components even when the original function is scalar-valued.

The Fourier transform on a finite group convention $\hat{f}(\rho )={\mathbb{E}}_{x}f(x)\rho (x)$ has the displayed inversion formula. The sum is over one representative from each equivalence class of unitary irreducible representations. It follows from the Schur orthogonality relations and the regular representation decomposition, and holds at every group element without a limiting argument.

A matrix coefficient of a finite-dimensional group representation is a scalar function obtained by applying a linear functional to the translate of a fixed vector. In a chosen basis these functions include $x\mapsto \rho (x{)}_{ij}$. For a finite group, the Schur orthogonality relations make the scaled coefficients of all inequivalent unitary irreducible representations an orthonormal basis of its scalar functions.

Use the following normalization for Fourier analysis on a finite group. Choose one unitary irreducible representation $\rho :G\to U(d_{\rho })$ from each equivalence class, including the trivial representation. For a scalar function $f:G\to \mathbb{C}$, putThis convention uses $\rho (x)$, rather than $\rho (x{)}^{\ast }$, in the Fourier transform on a finite group; it makes the normalized convolution on a finite group preserve multiplication order.

The needed representation theory consists of Maschke's theorem and unitarization of a finite-group representation, together with the Schur orthogonality relations:The regular representation contains $d_{\rho }$ copies of each $\rho$, so ${\sum }_{\rho }{d}_{\rho }^2=|G|$. Thus the scaled matrix coefficients $\sqrt{d_{\rho }}\rho (x{)}_{ij}$, and also their complex conjugates, form an orthonormal basis of all scalar functions on $G$. These facts imply Fourier inversion on a finite group and the Parseval identity on a finite group in the formsand henceIn particular, the transform is an isomorphism onto the direct sum of the matrix algebras $M_{d_{\rho }}(\mathbb{C})$, with the displayed weighted Hilbert-Schmidt inner product.

Define the normalized convolution on a finite group bySubstituting $x=yz$ and using the group representation identity yields the convolution theorem on a finite groupUnlike normalized convolution on a finite group on an abelian group, this product need not commute. If $\tilde{f}(x)=\overline{f(x^{-1})}$, then $\hat{\tilde{f}}(\rho )=\hat{f}(\rho {)}^{\ast }$. For left translation of a group function and right translation of a group function $L_{a}f(x)=f(a^{-1}x)$ and $R_{a}f(x)=f(xa)$,For an abelian group, every irreducible representation is one-dimensional; this reduces to Fourier analysis on a finite abelian group with characters relabelled by their inverses. These formulas establish the basic scalar theory, with all normalizations and multiplication orders fixed.

Now suppose every nontrivial irreducible representation has $d_{\rho }\ge m$. If $f$ is a mean-zero function, its component at the trivial representation is zero. The Parseval identity on a finite group gives, for each other $\rho$,Using the convolution theorem on a finite group, the Hilbert-Schmidt norm inequality $\Vert AB{\Vert }_{\mathrm{HS}}\le \Vert A{\Vert }_{\mathrm{op}}\Vert B{\Vert }_{\mathrm{HS}}$, and the Parseval identity on a finite group once more gives the product mixing in a quasirandom group estimateWrite $a,b,c$ for the subset density values of $A,B,C$, respectively, and let $f=1_A-a$, $g=1_B-b$ be balanced subset indicators. Since both are mean-zero functions, $1_A\ast 1_B=ab+f\ast g$. Their squared norms are $a(1-a)$ and $b(1-b)$. Also $\mathbb{E}(f\ast g)=0$, so the Cauchy-Schwarz inequality yieldsIf $abc>1/m$, the final bound is strictly smaller than $abc$. Thus the normalized number of pairs $(x,y)\in A\times B$ with $xy\in C$ is positive. Equivalently,This is the desired conclusion for a quasirandom group; the strict inequality ensures positivity rather than merely a nonnegative lower bound.

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.