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For strictly decreasing integer tuples $\kappa ,\eta$, let $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ on the diagonal maximal torus of the unitary group. Expanding both determinants, Fourier orthogonality makes the integral zero unless the exponent tuples have the same permutation orbit. Strict decreasing order makes that equivalent to $\kappa =\eta$; the $n!$ matching permutation terms then each contribute one. An alternating Laurent polynomial is a finite linear combination of these alternants. For an irreducible character ${\chi }_{\lambda }$, its highest monomial gives coefficient one for $A_{\lambda +\delta }$ in ${\chi }_{\lambda }{A}_{\delta }$, with $\delta =(n-1,\dots ,0)$. The Weyl integration formula and Schur orthogonality relations force the sum of squared coefficients to be one, so all other coefficients vanish. This proves the unitary Weyl character formula directly.

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.

Write the diagonal maximal torus of $U(n)$ as $t=\mathrm{diag}(z_1,\dots ,z_n)$, $|z_i|=1$, and let $\delta =(n-1,n-2,\dots ,0)$. Irreducible complex representations are indexed by dominant integer tuples ${\lambda }_1\ge \cdots \ge {\lambda }_n$, their highest weights. The Weyl character formula in integer-exponent form isThe ratio initially uses distinct $z_i$, and extends to repeated eigenvalues by continuity. Every unitary matrix is conjugate to such a diagonal matrix, so this determines the character on the entire group. The shift $\delta$ differs from the usual Weyl vector $\rho =((n-1)/2,(n-3)/2,\dots ,(1-n)/2)$ by a multiple of $(1,\dots ,1)$; the corresponding determinant factors cancel between numerator and denominator.

We first justify the highest-weight information needed in the proof. Restriction to the torus splits a representation into integer weight spaces. In an irreducible representation choose a lexicographically largest weight $\lambda$ and a nonzero vector $v$ of that weight. A matrix unit $E_{ij}$ with $i<j$ raises the weight by $e_i-e_j$, so it annihilates $v$. The character is invariant under permutations of diagonal coordinates, because permutation matrices conjugate the torus. Hence permutations of $\lambda$ are also weights, and lexical maximality forces ${\lambda }_1\ge \cdots \ge {\lambda }_n$.

Irreducibility implies that $v$ generates the representation under the complexified Lie algebra. Indeed, a Lie-algebra invariant subspace is preserved by exponentials of skew-Hermitian matrices, which generate $U(n)$. Reorder any product of matrix units into lower-triangular units, diagonal units and upper-triangular units, usingInduction on length and on the number of out-of-order pairs gives the required spanning version of the Poincaré-Birkhoff-Witt theorem. Upper-triangular units kill $v$, and diagonal units act by scalars. Every nonempty product of lower-triangular units lowers its weight by a nonzero sum of positive roots. Therefore the top weight space is exactly $\mathbb{C}v$, and its multiplicity is one. This supplies the leading coefficient without assuming a character formula.

For completeness, every dominant integer tuple occurs. Put $d_j={\lambda }_j-{\lambda }_{j+1}\ge 0$ and form the representationIts lexicographically highest weight is $\lambda$, with one-dimensional highest space, represented by the tensor products of $e_1\wedge \cdots \wedge e_j$. A negative determinant exponent is a legitimate one-dimensional unitary-group character. Averaging an inner product over normalized Haar measure gives invariant orthogonal complements, so the tensor representation splits into irreducible summands. One of them has highest weight $\lambda$. Uniqueness will also follow from the formula and Schur orthogonality relations.

Define the alternant $A_{\kappa }(z)=\det (z_i^{{\kappa }_j})$ for a strictly decreasing integer tuple $\kappa$. The denominatoris the Vandermonde determinant, and its modulus is the denominator modulus in the Weyl integration formula. The Laurent polynomial ${\chi }_{\lambda }{A}_{\delta }$ is alternating. Any alternating Laurent polynomial decomposes as a finite summonomials with repeated exponents have zero coefficient, and the remaining permutation orbits are indexed uniquely by strictly decreasing tuples.

The lexicographically highest monomial in the product is $z^{\lambda +\delta }$ with coefficient one. Indeed, the highest monomial of the character is $z^{\lambda }$ with coefficient one, and the unique lexical maximum among permutations of $\delta$ is $\delta$ itself. It follows that $c_{\lambda +\delta }=1$.

Fourier orthogonality on the torus, whose probability measure is ${\prod }_{i}d{\theta }_i/(2\pi )$, gives alternant orthogonality on a unitary torus:Expand both determinants: an integral survives exactly when the two exponent tuples are identical up to permutation. Strict decreasing order then forces $\kappa =\eta$, and its $n!$ matching terms each contribute one.

Finally, Schur orthogonality relations give ${\int }_{U(n)}|{\chi }_{\lambda }{|}^2=1$. Applying the allowed Weyl integration formula to this class function givesSince the coefficient $c_{\lambda +\delta }$ already equals one, every other coefficient vanishes. Thus ${\chi }_{\lambda }{A}_{\delta }=A_{\lambda +\delta }$, proving the formula. If two irreducible representations had the same highest weight, their characters would now coincide; the Schur orthogonality relations imply that they are isomorphic. Apparent singularities at repeated diagonal eigenvalues are removable because the quotient equals the actual finite Laurent-polynomial character. No dimension formula has been assumed.

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.