Here $p$ is odd. For a finite p-group, write $N^p=⟨n^p:n\in N⟩$. A powerfully embedded subgroup satisfies $[N,G]\le N^p$; this also makes $N$ normal, since every conjugate of an element of $N$ differs from it by an element of $N$. A powerful p-group is one with $[G,G]\le G^p$, namely one powerfully embedded in itself. Use $[a,b]=a^{-1}{b}^{-1}ab$ and $a^b=b^{-1}ab$ throughout.

The lower p-series is defined byEach quotient $G_i/G_{i+1}$ is an elementary abelian p-group, and $[G_i,G]\le G_{i+1}$. These are generated subgroups, without closure bars, because the group is finite.

A topological group is a group equipped with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous maps. The product uses the product topology. Translation and inversion are therefore homeomorphisms. Hausdorffness is sometimes included in the definition; the two open-subgroup arguments below do not require it.

For the inverse system, take a directed set $I$, topological spaces $X_i$, and continuous maps $f_{ij}:X_j\to X_i$ for $i\le j$, with $f_{ii}=\mathrm{id}$ and $f_{ik}=f_{ij}{f}_{jk}$. An inverse limit is a space $X$ with compatible continuous maps ${\pi }_i:X\to X_i$ such that every compatible family $q_i:Y\to X_i$ factors uniquely as $q_i={\pi }_{i}q$ through a continuous map $q:Y\to X$. Existence follows from the concrete constructionwith the subspace topology of the product topology. The only possible map $q$ is $q(y)=(q_i(y){)}_i$. Compatibility puts its image in $X$, and coordinatewise continuity makes it continuous. This proves the universal property of an inverse limit, even when $X$ is empty. Given two inverse limits, their universal properties supply maps each way; the composites have the same coordinates as the identity, so uniqueness makes them identities. The inverse limit exists and is unique up to the unique homeomorphism respecting its projections.

A profinite group can be defined as an inverse limit of finite discrete groups, or as a compact Hausdorff space that is a totally disconnected space, equipped with continuous group operations. For the first implication, use the Tychonoff theorem. The compatibility equalities define a closed subgroup of the compact Hausdorff product. Two distinct points are distinguished by a finite discrete coordinate, so no connected subset contains both. Consequently the inverse limit is a compact Hausdorff totally disconnected topological group.

For the converse, use the topological result that a compact Hausdorff totally disconnected space has a basis of clopen sets. Choose a clopen set $U$ containing $1$ inside any specified identity neighbourhood. For each $u\in U$, continuity gives an identity neighbourhood $V_u$ and an open neighbourhood $W_u$ of $u$ with $V_u{W}_u\subseteq U$. Finitely many $W_u$ cover the compact set $U$. Intersect their $V_u$ and its inverse to obtain a symmetric identity neighbourhood $V$ with $VU\subseteq U$. Symmetry also gives $V^{-1}U\subseteq U$, and hence $vU=U$ for every $v\in V$. The setis a subgroup containing $V$, so it is open. It is contained in $U$, since $1\in U$ and $h=h\cdot 1\in hU=U$. By part (ii), $H$ has finite index. Its normal core of a subgroup is a finite intersection of conjugates of $H$, hence an open normal subgroup $N$ contained in $U$. This proves the open normal subgroup basis of a profinite group.

The natural group homomorphismis injective because the intersection of these $N$ is $\{1\}$. For a compatible family of cosets, any finitely many have nonempty intersection: their intersection subgroup is another indexing subgroup, and its prescribed coset lies in each. The finite intersection property and compactness give a point belonging to all the cosets, proving surjectivity. Finally a continuous bijection from a compact space to a Hausdorff space is a homeomorphism. The two definitions of profinite group are equivalent.

An experimentally observed example of CP violation is the long-lived neutral kaon decay $K_L\to {\pi }^{+}{\pi }^{-}$. The two-pion spin-zero final state is CP even, whereas $K_L$ would be the CP-odd neutral-kaon state in the CP-conserving limit. Its two-pion decay reveals CP violation. The historical experimental account is cern-courier.web.cern.ch/a/cp-violations-early-days/ .

For the CKM matrix, diagonalize the up- and down-quark mass matrices independently by biunitary transformations. Writewith $L_u^{†}{M}_u^{\prime }{R}_u$ and $L_d^{†}{M}_d^{\prime }{R}_d$ positive diagonal matrices. Only the left transformations enter the weak charged current, because its chiral projector removes the right-handed fields. Substitution givesThe matrix is unitary but need not be diagonal: the two left mass diagonalizations generally differ.

Using the field transformations in the subparts and reordering the anticommuting Dirac fields, the charged bilinear transforms aswhere the fields on the right are both evaluated at $x_P$. The W boson contributes its second minus sign and Lorentz parity factor, so the product maps into its Hermitian-conjugate product. Crucially CP symmetry acts unitarily: it does not separately complex-conjugate the numerical coefficient $V_{ij}$. Thus the transformed interaction isComparing with the original interaction requires $V_{ij}=V_{ij}^{\ast }$ in this canonical CP convention. The physical condition is quark rephasing and CP conservation:A merely complex-looking matrix in an arbitrary mass-eigenstate phase convention is not itself proof of CP violation.

For CKM parameter counting, a unitary $N\times N$ matrix has $N^2$ real parameters: $N(N-1)/2$ mixing angles and $N(N+1)/2$ phases. Rephasing the $N$ up and $N$ down mass eigenfields removes $2N-1$ phases, since their common phase cancels out of $V$. For generic nondegenerate masses this leavesThere is one physical phase for three families and none for one or two. Degenerate masses would permit additional transformations, so the generic count need not apply unchanged.

For the QCD part of the gauge covariant derivative, define the Hermitian colour matrix ${\mathcal{A}}_{\mu }=A_{\mu }^a{T}^a$ and chooseThis sign convention agrees with the field-strength definition ${\mathcal{F}}_{\mu \nu }=-i[D_{\mu },D_{\nu }]/g_s$ in the source. The kinetic term contains $-g_s\overline{u}{\gamma }^{\mu }{\mathcal{A}}_{\mu }u$. A vector bilinear under CP changes sign, gains the Lorentz parity matrix, and transposes its colour matrix upon exchanging quark and antiquark. Invariance of this interaction therefore requires the CP transformation of a non-Abelian gauge connection:Both the time/spatial distinction and the colour transpose are essential; individual colour components need not all share one charge-conjugation sign.

The same gauge covariant derivative givesApply the transformation above and the chain rule ${\partial }_{\mu }f(x_P)={\Pi }_{\mu }{}^{\alpha }({\partial }_{\alpha }f)(x_P)$. The transpose reverses commutator order, $[X^T,Y^T]=-[X,Y{]}^T$, so the nonlinear term transforms with the same overall minus sign as the derivatives. This proves the CP transformation of non-Abelian field strength:Finally $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$ rewrites the theta density as $2\theta {ϵ}^{\mu \nu \rho \sigma }\mathrm{tr}({\mathcal{F}}_{\mu \nu }{\mathcal{F}}_{\rho \sigma })$. The two minus signs cancel, and transposition followed by cyclicity leaves the colour trace unchanged. But the four parity matrices acting on the Levi-Civita symbol contribute $\det \Pi =-1$. HenceThe density is C even and P odd, therefore CP odd. A generic fixed nonzero theta coefficient violates CP symmetry. This is the source of the Strong CP problem, separate from the charged-current phase in the CKM matrix.

A phase diagram partitions a space of control parameters into regions with distinct equilibrium phases. Its boundaries mark nonanalyticities of the thermodynamic-limit free energy; a first-order phase transition has a jump in a first derivative, whereas a continuous phase transition has a continuously vanishing order parameter and singular fluctuations.

An explicit three-parameter example is the scalar Landau free energywith parameter axes $(r,u,h)$. These can be realized locally by the temperature, single-ion anisotropy and magnetic field of the Blume–Capel model. The three dimensions here refer to control-parameter space, not necessarily to three spatial dimensions. At $h=0$, minimizing this potential gives a continuous phase transition at $r=0$ for $u>0$. For $u<0$, coexistence occurs at $r=3u^2/(16v)$, whereThere are three global minima, $M=0$ and $M=\pm \sqrt{-3u/(4v)}$. Their order parameter jump decreases to zero as $u\uparrow 0$. Thus $(r,u,h)=(0,0,0)$ is a tricritical point. The two transition curves in the $h=0$ plane meet there.

In the three-dimensional phase diagram, crossing $h=0$ inside the ordered region switches between opposite signs of $M$, so that region is itself a first-order phase transition sheet. For $u<0$, two tricritical wings at opposite nonzero fields separate weakly and strongly magnetized phases. Their wing critical edges can be obtained rather than merely sketched: require $V^{\prime }=V^{\prime \prime }=V^{\prime \prime \prime }=0$. Since $V^{\prime \prime \prime }=6uM+20v{M}^3$, the nonzero solutions satisfyThe fourth derivative there is $-12u>0$, giving ordinary continuous critical endpoints. The wings meet the zero-field coexistence sheet along the tricritical three-phase line and terminate at these edges. Their exact tricritical wing coexistence factorization supplies the surfaces in the figure: for two nonnegative minima $a\le b$, set $s=a+b$, $p=ab$ andDirect expansion gives $V(M)-V(a)=(v/6)(M-a{)}^2(M-b{)}^2[(M+s{)}^2+p]\ge 0$, proving that these surfaces describe coexistence of global, not merely local, minima.

For the unheaded Ising model calculation, assume ferromagnetic exchange $J>0$ and let $M$ denote the mean Ising spin. On the hypercubic lattice the coordination number of a lattice is $z=2D$, even though the original positive-direction bond sum counts each bond only once. The mean-field approximation neglects products of deviations from $M$:There are $ND$ bonds, so the approximate energy is $H_{\mathrm{MF}}=-2DJM{\sum }_i{\sigma }_i+NDJ{M}^2$. Each independent Ising spin therefore sees the effective field $2DJM$. With ${\beta }_{\mathrm{th}}=1/(k_{B}T)$, averaging its two Boltzmann weights gives the self-consistency equationThe slope of its right side at zero is $T_c/T$, whereAbove $T_c$, $\tanh x<x$ for $x>0$ excludes a nonzero solution. Below $T_c$ the initial slope exceeds one, and concavity of the hyperbolic tangent on the positive axis gives one positive solution and its negative partner. The Ising auxiliary mean-field free energy $DJ{M}^2-k_{B}T\log [2\cosh (2{\beta }_{\mathrm{th}}DJM)]$ has positive quartic coefficient near $T_c$, while its quadratic coefficient changes sign there. Its stable minima therefore move continuously away from zero; this is a continuous phase transition in the mean-field theory of the Ising model.

Put $\tau =(T_c-T)/T_c>0$ and $c=T_c/T=(1-\tau {)}^{-1}$. The Taylor expansion $\tanh x=x-x^3/3+O(x^5)$ gives, on the nonzero branch,Consequently $M\sim \sqrt{3}{\tau }^{1/2}$ and the mean-field order-parameter exponent is $\beta =1/2$. The inverse-temperature symbol ${\beta }_{\mathrm{th}}$ used above is distinct from this order-parameter critical exponent.

For integer partitions $\lambda ,\mu$ of $n$, padded by zero parts, the dominance order on partitions isA $\lambda$-tableau fills the Young diagram of $\lambda$ with $1,\dots ,n$, each once. A tabloid $[t]$ remembers the sets of entries in each row, ignoring their order within that row. The Young permutation module $M^{\lambda }=\mathbb{C}{X}_{\lambda }$ has these tabloids as basis, with $S_n$ acting by permuting the entries.

The row and column stabilizers $R_t,C_t$ independently permute entries within the rows and columns of $t$. The column antisymmetrizer is ${\kappa }_t={\sum }_{c\in C_t}\mathrm{sgn}(c)c$, and $e_t={\kappa }_t[t]$ is the associated polytabloid. Since $C_{gt}=g{C}_t{g}^{-1}$ and conjugation preserves permutation sign,Thus the Specht module is the submoduleEvery tableau is $gt$ for a fixed $t$, so any single $e_t$ generates it under $S_n$. Also $C_t\cap R_t=\{1\}$, since a permutation preserving both rows and columns must fix every cell. The tabloids $[ct]$ for $c\in C_t$ are consequently distinct, and $e_t$ has coefficient $1$ at $[t]$; in particular $e_t\ne 0$.

We use the following precise column antisymmetrizer facts. If some row of a $\mu$-tableau $s$ contains two entries from one column of $t$, the transposition interchanging them pairs equal tabloids with opposite signs in ${\kappa }_t[s]$, so ${\kappa }_t[s]=0$. Otherwise each row of $s$ meets each column of $t$ at most once. The first $r$ rows of $s$ therefore contain at most $\min (r,{\lambda }_j^{\prime })$ entries from column $j$ of $t$, whenceThis states and explains dominance from a nonzero column antisymmetrizer: ${\kappa }_t[s]\ne 0$ implies $\lambda ⊵\mu$.

When $\mu =\lambda$, the nonzero column antisymmetrizer criterion says more: either ${\kappa }_t[s]=0$, or a column permutation takes $[s]$ to $[t]$, and ${\kappa }_t[s]=\pm e_t$. Thus ${\kappa }_t{M}^{\lambda }=\mathbb{C}{e}_t$. This follows by matching the row positions within each column when all intersections have size at most one; the equal row and column sizes force the Ferrers incidence pattern.

For the inner product, use the positive Hermitian extension of the orthonormal tabloid basis, linear in its first argument. It has the supplied self-adjointness of ${\kappa }_t$. Its orthogonal complement of $S^{\lambda }$ agrees with that obtained from the complex tabloid bilinear form, since the spanning polytabloids have real coefficients. Because the coefficient of $[t]$ in $e_t$ is $1$, self-adjointness and the one-dimensional image give the useful identityIndeed, the scalar multiplying $e_t$ is the $[t]$-coefficient of ${\kappa }_{t}u$, which is $⟨{\kappa }_{t}u,[t]⟩=⟨u,{\kappa }_t[t]⟩=⟨u,e_t⟩$.

An M-group is a finite group for which every irreducible character is induced from a linear character of some subgroup. Let $L◃K$ have both $L$ and $K/L$ abelian, and fix an irreducible $K$-module $V$ with character $\chi$. Since $L$ is abelian, $V_L$ has a linear constituent. Choose, among all pairs $(A,\phi )$ with $L\subseteq A\subseteq K$ and a linear character $\phi$ occurring in ${\chi }_A$, one for which $|A|$ is maximal.

Every subgroup containing $L$ is normal in $K$: its image in the abelian quotient $K/L$ is normal, and it is the full preimage of that image. Thus $A◃K$, so the inertia group of a character $I_K(\phi )$ is defined and contains $A$.

Suppose $x\in I_K(\phi )\setminus A$. The nonzero $\phi$-isotypic component $V_{\phi }$ is preserved by $x$, and every $a\in A$ acts there as the scalar $\phi (a)$ because $\phi (1)=1$. The finite-order operator representing $x$ is diagonalizable over $\mathbb{C}$; choose an eigenvector $0\ne v\in V_{\phi }$. Then $\mathbb{C}v$ is invariant under both $A$ and $x$, hence under $D=⟨A,x⟩$. This line affords a linear character $\psi$ of $D$ with ${\psi }_A=\phi$. Since it is a subrepresentation of $V_D$, it is a constituent of ${\chi }_D$. But $D$ strictly contains $A$ and still contains $L$, contradicting maximality.

Therefore $I_K(\phi )=A$. Apply the Clifford correspondence with normal subgroup $A$. Its inertia group is $A$ itself, and $\mathrm{Irr}(A\mid \phi )=\{\phi \}$, so the unique irreducible $K$-character lying over $\phi$ is ${\phi }^K$. In particular,Since $\chi$ was arbitrary, every irreducible character is a monomial character, proving that finite metabelian groups are monomial:

Write $m_j=\mathbb{E}{Y}^j$ and $M=M_Y$. The derivatives of the cumulant-generating function areSince $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three cumulants areThe last equality follows by expanding the cube and taking expectations. We use the paper's unstandardized third central moment throughout. The usual skewness coefficient divides this quantity by $\mathrm{Var}(Y{)}^{3/2}$; its sign is the same whenever the variance is positive.

There is a minor domain qualification: finite ordinary moments do not guarantee a finite moment-generating function for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite moments, all formulas also follow from left derivatives at zero, because $Y^j{e}^{tY}\le Y^j$ for $t\le 0$ and dominated convergence applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.

For the random sum of independent claims, condition on $N=k$. Independence of the claims and independence from the count giveTaking expectations and then logarithms yields the random-sum transform identity and the cumulant composition for a random sum:Here $G_N$ is the probability generating function. These identities hold wherever the transforms are finite, in particular for $t\le 0$ with positive claims.

For reuse in the three cases, write $m_j=\mathbb{E}{X}_1^j$ and $a_j={\kappa }_{N,j}$. Applying the chain rule three times givesor equivalentlyThis expresses the requested third central moment in raw claim moments, while keeping the contribution of claim-count variability explicit.

On a fixed finite horizon, a self-financing strategy consists of predictable stock holdings $h^i$, integrable against the price processes, and a bank account holding $h^0$. Its wealth and gains satisfyAn admissible trading strategy has discounted wealth $X_t/B_t$ bounded below by a deterministic constant throughout the horizon. With the bounded short rate in this part, an undiscounted deterministic lower bound is equivalent after adjusting the constant. An arbitrage is an admissible zero-initial-cost strategy with $X_T\ge 0$ almost surely and $\mathbb{P}(X_T>0)>0$.

Put $1=(1,\dots ,1{)}^{\top }$ and define the market price of riskThe bounded coefficients and bounded inverse make $\lambda$ bounded. Define the stochastic exponentialThe Novikov condition states that $\mathbb{E}\exp (\frac{1}{2}{\int }_0^T\Vert {\lambda }_s{\Vert }^{2}ds)<\infty$ makes this exponential a true martingale with expectation one. Boundedness verifies the condition. Therefore $dQ=L_{T}dP$ defines an equivalent probability measure. The Girsanov theorem states thatis a $d$-dimensional Brownian motion under $Q$. Substituting into the asset equations givesThus $Q$ is an equivalent local martingale measure. The sufficient direction of the fundamental theorem of asset pricing says that such a measure excludes arbitrage among admissible self-financing strategies. Here the reason can be seen directly: $X/B$ is a local martingale under $Q$, and its lower bound makes it a supermartingale. Consequently $X_0=0$ implies ${\mathbb{E}}_Q[X_T/B_T]\le 0$, incompatible with the nonnegative terminal payoff of an arbitrage. Therefore the bounded-inverse market has no arbitrage.

For the requested law of one price counterexample, take a broader continuous-time local-martingale model on $[0,T]$. Let $X$ be a Brownian motion, let $\tau =\inf \{u\ge 0:X_u=-1\}$, and set $q(t)=t/(T-t)$ for $t<T$. DefineThe given hitting-time fact gives $\tau <\infty$ almost surely. The corresponding physical hitting time is $T\tau /(1+\tau )<T$, so $S$ reaches one and then stays one. In particular it is continuous at maturity and $S_t\ge 1$.

For a rigorous closed-horizon localization, stop additionally when $S$ hits $n+2$, or at $T$ if it never does. The stopped process is Brownian motion stopped on first exiting $(-1,n)$ and subjected to the deterministic clock; it is a bounded martingale, including at $T$. The stopping times increase to $T$, since every Brownian path on the finite random interval $[0,\tau ]$ has a finite maximum. Thus $S$ is a positive local martingale on the whole horizon. With the time-changed Brownian filtration, the original probability measure is an equivalent local martingale measure. The admissible-wealth supermartingale argument above proves no admissible arbitrage.

Nevertheless $S_T=B_T=1$ while $S_0=2\ne B_0=1$, so the law of one price fails. Shorting one share and buying two units of cash has zero initial cost and terminal gain one, but its wealth is $2-S_t$. It has no deterministic lower bound: for every $n$, the probability of Brownian motion hitting $n$ before $-1$ is $1/(n+1)>0$, obtained by optional stopping of the bounded stopped Brownian motion. That apparently profitable strategy is therefore inadmissible. The counterexample deliberately uses a strict local martingale; the bounded-coefficient class in the first part has true discounted asset martingales and does satisfy the law of one price.

A poset becomes a category whose objects are its elements and whose hom-set $X(x,y)$ is a singleton if $x\le y$ and empty otherwise. Reflexivity supplies identity arrows and transitivity supplies composition; all possible composites are unique, so the category axioms hold. Antisymmetry says that isomorphic objects are equal. A functor between such categories is exactly an order-preserving function: an inequality must be sent to an inequality, and there is no further choice of arrow map.

For a diagram shape $I$, a functor $F:\mathcal{C}\to \mathcal{D}$ preserves categorical limits of shape $I$ if every limiting cone $(L,{\ell }_i)$ over every diagram $D:I\to \mathcal{C}$ is carried to a limiting cone $(FL,F{\ell }_i)$ over $FD$. It reflects categorical limits of that shape if a cone in $\mathcal{C}$ is limiting whenever its image cone is limiting. These properties concern cones which are given; reflection is not an assertion that every cone in the target lifts.

Suppose $F$ is a full and faithful functor and the image of $(L,{\ell }_i)$ is limiting. A cone $(X,x_i)$ in $\mathcal{C}$ becomes a cone in $\mathcal{D}$, so there is a unique $h:FX\to FL$ with $(F{\ell }_i)h=F{x}_i$. Fullness writes $h=Fk$ for some $k:X\to L$. Faithfulness gives ${\ell }_{i}k=x_i$, and also ensures uniqueness: two such $k$ would have the same image, by uniqueness of $h$. Hence the original cone is limiting. This proves every full and faithful functor reflects categorical limits, the limits reflected by full and faithful functors criterion, with no essential-surjectivity requirement.

For the triangle quotient, the universal coefficient theorem for homology gives $H_j(X;{\mathbb{F}}_2)=0$ for every $j>0$, because its only positive integral homology is $\mathbb{Z}/3$. A connected positive-dimensional closed manifold has a nonzero top mod-two fundamental class, independently of orientability. Thus this quotient cannot be homotopy equivalent to such a manifold. A connected zero-dimensional closed manifold is a point, which its nonzero $H_1(X;\mathbb{Z})$ also excludes.

For the attached-band space, the same universal coefficient theorem for homology gives nonzero $H_2(X;{\mathbb{F}}_2)$ and zero homology above degree two. A connected closed manifold homotopy equivalent to it would consequently have dimension two. Its Euler characteristic would be one. The classification theorem for surfaces says that a closed connected surface of Euler characteristic one is the real projective plane. But its integral $H_1$ is $\mathbb{Z}/2$, differing from both possibilities computed in part 2. Thus this space is never homotopy equivalent to a closed manifold.

The antipodal mapping torus is already a closed manifold: gluing the two boundary components of $S^n\times [0,1]$ by a diffeomorphism gives a smooth $(n+1)$-manifold without boundary. For $n\ge 1$, it is orientable exactly when $(-1{)}^{n+1}=1$, namely when $n$ is odd. In that case $-I$ on ${\mathbb{R}}^{n+1}$ is isotopic to the identity through rotations, so the mapping torus is diffeomorphic to $S^n\times S^1$. The $n=0$ case is a circle as above. Only the third family can be homotopy equivalent to a closed manifold.

Put $e_N(t)=e^{2\pi it/N}$ and use normalized averages ${\mathbb{E}}_x=N^{-1}{\sum }_{x\in \mathbb{Z}/N\mathbb{Z}}$. Our convention for Fourier coefficients on a finite abelian group isThe underlying orthogonality of complex exponentials is ${\mathbb{E}}_x{e}_N(rx)=1$ for $r=0$ and zero otherwise. For nonzero $r$ this is the finite geometric series with ratio $e_N(r)\ne 1$; its sum vanishes because that ratio has $N$th power one. The identical orthogonality statement holds when averaging over $r$ instead of $x$.

Using this orthogonality to sum the definition over $r$ proves Fourier inversion on a finite group:Indeed $\mathbb{E}f\overline{g}={\sum }_r\overline{\hat{g}(r)}{\mathbb{E}}_{x}f(x)e_N(-rx)={\sum }_r\hat{f}(r)\overline{\hat{g}(r)}$. This proves the Parseval identity on a finite group and its squared-norm case:For normalized convolution on a finite group, write $(f\ast g)(x)={\mathbb{E}}_{y}f(y)g(x-y)$. Changing variables $x=y+z$ in its transform givesthe convolution theorem on a finite group. Linearity of the transform follows directly from its definition. Translation ${\tau }_{t}f(x)=f(x+t)$ gives $\widehat{{\tau }_{t}f}(r)=e_N(rt)\hat{f}(r)$ by changing variables. Complex conjugation gives $\hat{\overline{f}}(r)=\overline{\hat{f}(-r)}$; for $\tilde{f}(x)=\overline{f(-x)}$, the corresponding identity is $\hat{\tilde{f}}(r)=\overline{\hat{f}(r)}$. Finally $\hat{f}(0)=\mathbb{E}f$ and $|\hat{f}(r)|\le \mathbb{E}|f|$ by the triangle inequality. These identities establish the basic normalized Fourier analysis on a finite abelian group needed below, with all normalizations fixed.

For completeness, pointwise multiplication has the dual convolution formula $\widehat{fg}(r)={\sum }_s\hat{f}(s)\hat{g}(r-s)$: insert the inversion formula for $g$ into the defining average and collect its frequencies. If $c\ne 0$ modulo $N$, the change of variables $u=cx$ gives $\widehat{f(c\cdot )}(r)=\hat{f}(r{c}^{-1})$. These also follow directly from the same additive character orthogonality.

A general feedback shift register of length $d$ over an alphabet $A$ stores $(s_n,\dots ,s_{n+d-1})$, outputs $s_n$, and shifts to $(s_{n+1},\dots ,s_{n+d-1},F(s_n,\dots ,s_{n+d-1}))$ for a fixed function $F:A^d\to A$. When the alphabet is a field $K$, it is a linear-feedback shift register if $F$ is linear over $K$, equivalently $s_{n+d}={\sum }_{j=0}^{d-1}{a}_j{s}_{n+j}$ for fixed coefficients in $K$. The binary case uses the finite field $K={\mathbb{F}}_2$; the examples and counterexamples below are binary.

For a genuinely nonlinear example take length three and $F(a,b,c)=a+bc$. Its values at $(0,1,0)$ and $(0,0,1)$ are both zero, but its value at their sum $(0,1,1)$ is one, violating additivity. Thus this general register is not linear. It even has an invertible state update: from $(b,c,d)$ the old leading bit is $a=d+bc$, so the counterexample does not rely on allowing singular registers.