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.

The ergodic hypothesis states that long-time averages of equilibrium observables along typical trajectories equal their ensemble averages on the accessible phase space with the same conserved quantities. It permits statistical mechanics to predict measured macroscopic quantities from equilibrium distributions rather than tracking a trajectory. In its microcanonical form the measure is on the constant-energy surface. It is an assumption about the relevant dynamics, not a consequence of particles being noninteracting; the calculations below use the equilibrium canonical ensemble at temperature $T$.

Pontryagin maximum principle leaves the initial costate unknown. Terminal conditions determine these unknown constants and, when terminal time is free, the time. In a maximization convention with terminal reward $\Phi$ and running cost $L$, use $H=p\cdot f-L$. A free terminal coordinate gives $p_i(T)={\partial }_i\Phi$; a fixed terminal coordinate instead allows a constraint multiplier. More generally $p(T)-\nabla \Phi$ is normal to the permitted terminal manifold. The free-terminal-time transversality condition is $H(T)+{\Phi }_T=0$. These conditions convert the adjoint and state equations into a boundary-value problem. Reversing the costate sign gives the equivalent minimization convention.

A model of three-dimensional hyperbolic space is ${\mathbb{H}}^3=\{(z,t):z\in \mathbb{C},t>0\}$ with metric $d{s}^2=(|dz{|}^2+d{t}^2)/t^2$, the complete simply connected metric of curvature $-1$. The translation length of an isometry $g$ is $\ell (g)={\inf }_{P}d(P,gP)$. A hyperbolic isometry has positive translation length attained on its invariant complete geodesic, the axis of a hyperbolic-space isometry. It joins the two fixed ideal points and is translated by distance $\ell (g)$; a three-dimensional axial isometry can also rotate around it.

For the boundary map $z\mapsto kz$, the hyperbolic extension of a Möbius transformation is $(z,t)\mapsto (kz,|k|t)$. Every path from $P$ to $gP$ has length at least the absolute change of $\log t$, namely $|\log |k||$, because the metric length dominates $|dt|/t$. On the vertical line $z=0$ a vertical segment attains that bound. HenceIf $|k|\ne 1$ the vertical line is the axis. If $|k|=1$ the map is a rotation with zero translation length, including the identity when $k=1$; it is not a positive-length hyperbolic isometry.

Set ${\theta }_k=k\pi /(M+1)$ and extend the proposed vector by $v_0=v_{M+1}=0$. For $v_m=i^m\sin (m{\theta }_k)$, the angle-addition identity givesHence the Toeplitz antisymmetric tridiagonal matrix has eigenvaluesThe real sine vectors are eigenvectors of the real symmetric tridiagonal matrix with both off-diagonals one, with distinct eigenvalues $2\cos {\theta }_k$; thus they are orthogonal and form a basis. Multiplication by the unitary diagonal matrix $\mathrm{diag}(i,i^2,\dots ,i^M)$ preserves orthogonality. Their squared norms are $(M+1)/2$, so multiplying by $\sqrt{2/(M+1)}$ gives a common orthonormal basis for all $a,b$, even when some eigenvalues coincide.

Let $K$ have superdiagonal one, subdiagonal minus one and diagonal zero. The prescribed endpoint values remove the exterior unknowns, and the scheme isIn the common orthonormal eigenbasis, the two eigenvalues are $1\mp i(\mu /2)\cos {\theta }_k$. Every eigenvalue of $B$ is nonzero for real $\mu$, and the amplification matrix $Q=B^{-1}C$ has eigenvaluesThus $Q$ is a unitary matrix and $\Vert u^{n+1}{\Vert }_2=\Vert u^n{\Vert }_2$ exactly. The same holds for the mesh-weighted norm. This matrix argument proves $\boxed{\text{stability for every real }\mu \text{, hence every physical }\mu \ge 0}$, without a Fourier stability calculation.

The printed mesh ratio has a consistency error: advection requires $\mu =\Delta t/\Delta x$, while the PDF prints $\Delta t/(\Delta x{)}^2$. Indeed the centered spatial difference is $2\Delta x{u}_x+O((\Delta x{)}^3)$, so division of the scheme by $\Delta t$ gives leading right-hand coefficient $\mu \Delta x/\Delta t$. The printed choice makes this $1/\Delta x$ rather than one. The stability conclusion is valid for the algebraic scheme with either definition, but does not fix this inconsistency. The continuous first-order advection problem also ordinarily specifies only the inflow endpoint, here $x=1$; zero data at both endpoints restrict admissible initial data. For example the initially boundary-zero profile $\sin \pi x$ develops nonzero outflow $u(0,t)=\sin \pi t$ before inflow changes reach that boundary. The matrix norm calculation concerns the stated two-endpoint discretization and does not establish well-posedness for arbitrary continuous initial data.

The left group action on the homogeneous polynomials is ${\rho }_n(g)P(z)=P(g^{-1}z)$, since ${\rho }_n(g){\rho }_n(h)={\rho }_n(gh)$. Every SU(2) matrix is conjugate to $\mathrm{diag}(e^{i\theta },e^{-i\theta })$ with $0\le \theta \le \pi$. In the monomial basis $z_1^{n-j}{z}_2^j$, its eigenvalues are $e^{-i(n-2j)\theta }$. Consequently the character of the homogeneous polynomial representation of SU2 isThe quotient is interpreted by continuity at the endpoints: ${\chi }_n(0)=n+1$ and ${\chi }_n(\pi )=(-1{)}^n(n+1)$. The introductory action and character requests are independent of parts (i)–(iii), so this root Solution slot has been restored.

The angles $\psi$ and $\varphi$ are cyclic. Their conjugate momentum components are $p_{\psi }=I_3(\dot{\psi }+\dot{\varphi }\cos \theta )$ and $p_{\varphi }=I_1\dot{\varphi }{\sin }^2\theta +p_{\psi }\cos \theta$, so both are conserved; consequently ${\omega }_3=p_{\psi }/I_3$ is constant. With $q=I_3{\omega }_3=p_{\varphi }>0$,Eliminating the cyclic velocities gives the nutation effective potentialapart from the constant axial spin energy. ThusThe endpoint $\theta =\pi$ is excluded for these nonzero conserved momenta: the effective potential diverges there. The apparent Euler-angle singularity at $\theta =0$ is interpreted by its smooth small-tilt limit.

Lubrication theory exploits a film thickness $b$ much smaller than its horizontal variation scale $\ell$. Incompressibility then gives vertical velocity of order $(b/\ell )U$, vertical viscous derivatives dominate horizontal ones, and inertia is negligible when $\rho U{b}^2/(\mu \ell )\ll 1$. The leading equations reduce to $\mu u_{yy}=p_x$ and vertical hydrostatic balance $p_y=-\rho g$. Boundary conditions determine the local horizontal velocity profile; integrating it across the film and enforcing mass conservation produces an equation for flux or surface shape. This reduction turns the thin-film problem into a one-dimensional pressure problem while retaining the dominant viscous stresses.