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The normalized convolution on a finite group and the Fourier transform on a finite group convention with $\rho (x)$ satisfy the displayed identity. Substitute $x=yz$ in the defining expectation and use $\rho (yz)=\rho (y)\rho (z)$. With the convention using $\rho (x{)}^{\ast }$ instead, the scalar convolution has transform $\hat{g}(\rho )\hat{f}(\rho )$; multiplication order matters for noncommutative groups.

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.

For $f:G\to M_n(\mathbb{C})$ and a degree-$d_{\rho }$ unitary irreducible representation, the matrix Fourier block is the operator $T_{\rho }$ on $M_{n\times d_{\rho }}(\mathbb{C})$ defined in the title. Its singular values ${\lambda }_{\rho ,j}$ satisfyPointwise operator norm at most one implies $\Vert T_{\rho }{\Vert }_{\mathrm{op}}\le 1$. These blocks are operators on a matrix space, rather than simply the entrywise scalar Fourier transform on a finite group without any reshaping.

For the Fourier transform on a finite group, the displayed identity uses uniform expectation in the original function space and a weighted Hilbert-Schmidt inner product in the matrix components. In particular $\Vert f{\Vert }_2^2={\sum }_{\rho }{d}_{\rho }\Vert \hat{f}(\rho ){\Vert }_{\mathrm{HS}}^2$. For an abelian group, all $d_{\rho }=1$, and the identity becomes a sum of squared scalar coefficients.

For odd $p$, write $e_p(t)=e^{2\pi it/p}$ and use the discrete paraboloidHere surface integration means $p^{-2}{\sum }_P$, whereas ambient norms below use unnormalized counting measure on ${\mathbb{F}}_p^3$. With the negative-phase transform convention,When $x_3=0$, additive-orthogonality of characters makes this one at $x=0$ and zero at all other points of that plane. When $x_3\ne 0$, completing the two squares givesThe Quadratic Gauss sum identity is $G(t{)}^2=\chi (-1)p$, where $\chi$ is the Legendre symbol. For completeness, counting the preimages of a square gives $G(t)=\chi (t)G(1)$ for nonzero $t$. Also $|G(1){|}^2={\sum }_{u,v}{e}_p(u^2-v^2)=p$, by the invertible change $(u,v)\mapsto (u-v,u+v)$ and orthogonality of characters. Since $\overline{G(1)}=G(-1)=\chi (-1)G(1)$, the asserted square identity follows. Thus for $p\equiv 3(\mathrm{mod}4)$,Using positive phase instead changes the sign inside the character, not the magnitude or any norm estimate.

Define $Eg(x)=p^{-2}{\sum }_{\xi \in P}g(\xi )e_p(x\cdot \xi )$, and define $R^{\ast }(2\to q)$ as the smallest constant forThe adjoint relative to these measures is $E^{\ast }h(\xi )={\sum }_{x}h(x)e_p(-x\cdot \xi )$. Consequently $E{E}^{\ast }h=h\ast \hat{\sigma }(-\cdot )$, where convolution is the unnormalized ambient sum. Put $K=\hat{\sigma }(-\cdot )-1_{\{0\}}$. Its supremum is at most $p^{-1}$, so convolution by $K$ has ${\ell }^1\to {\ell }^{\infty }$ norm at most $p^{-1}$.

Under the unnormalized ambient Fourier transform on a finite group, the convolution multiplier of $\hat{\sigma }(-\cdot )$ is $p^3\sigma =p{1}_P$, up to the harmless reflection convention. Thus the multiplier of $K$ is $p{1}_P-1$, with maximum absolute value at most $p$. The Plancherel theorem gives ${\ell }^2\to {\ell }^2$ norm at most $p$. The Riesz-Thorin interpolation theorem between these two bounds gives ${\ell }^{4/3}\to {\ell }^4$ norm at most one for convolution by $K$. Convolution by $1_{\{0\}}$ is the identity, whose ${\ell }^{4/3}\to {\ell }^4$ norm is at most one for counting measure. Hence $\Vert E{E}^{\ast }h{\Vert }_4\le 2\Vert h{\Vert }_{4/3}$.

By Lp duality and Hölder's inequality, $\Vert E^{\ast }h{\Vert }_{L^2(\sigma )}^2=⟨E{E}^{\ast }h,h⟩\le 2\Vert h{\Vert }_{4/3}^2$. Taking adjoints proves the finite-field paraboloid fourth-moment extension estimateIn fact this argument works for both residue classes of odd primes.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The complete line $\ell =\{(t,it,0):t\in {\mathbb{F}}_p\}$ lies in $P$. For $g=1_{\ell }$ the normalized surface $L^2$ norm is $p^{-1/2}$, and orthogonality of characters gives $Eg(x)=p^{-1}{1}_{\{x_1+i{x}_2=0\}}$. That annihilator plane has $p^2$ points, so the isotropic-line obstruction to finite-field restriction givesThis tends to infinity along these primes for every $q<4$, as required.

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.