A group presentation with a cyclic list of generators $a_i$ and relations $a_i{a}_{i+1}{a}_i^{-1}=a_{i+1}^2$ has no nontrivial finite quotient of a group. In a finite image, all generator orders are odd. Choose the least prime $p$ dividing any nontrivial generator order and let $x$ be the predecessor of a generator $y$ whose order is divisible by $p$. Conjugation by $x$ acts as squaring on $⟨y⟩$, so the multiplicative order of $2$ modulo $p$ divides the order of $x$. That multiplicative order is greater than one and divides $p-1$, giving a prime divisor smaller than $p$ in the order of $x$, a contradiction. This argument applies to any cycle length; it asserts absence of finite quotients, not infinitude of the presented group.

The A2 root system representation $R(2,0)={\mathrm{Sym}}^2{\mathbb{C}}^3$ has dimension six and weights $(2,0),(0,1),(1,-1),(-2,2),(-1,0),(0,-2)$ in Dynkin label coordinates, each with multiplicity one.

The cube of $1+3t$ is $1+9(t+3t^2+3t^3)$. Conversely the polynomial $T+3T^2+3T^3-a$ has a unique simple residue root modulo three for each $a\in {\mathbb{Z}}_3$, so Hensel lemma produces a cube root of every $1+9a$. Thus the principal units modulo their cubes have order three. The change of variable removes the derivative obstruction that prevents direct simple-root lifting for $X^3-u$.

For $r\ge 1$, the map $1+{\pi }^{r}a\mapsto \bar{a}$ is a surjective group homomorphism to the additive residue field with kernel $U_{r+1}$. The extra product term is divisible by ${\pi }^{r+1}$. Thus each quotient has cardinality $|k|$, a useful count when splitting the unit group by roots of unity.

At an active maximum-raising boundary in high-water mark investment taxation, the finite-variation part of the Itô formula is $(V_{\bar{w}}-\tau V_w)d\bar{w}$, so smooth optimality sets this coefficient to zero. If the maximizing policy avoids increasing the maximum, the corresponding inequality $V_{\bar{w}}-\tau V_w\le 0$ is compatible with a wealth-cap investment boundary instead.

A wealth-cap investment boundary prevents wealth from exceeding a fixed historical maximum. In the wealth-variable Legendre dual for wealth normalized by that maximum, $J^{\prime }(z_{\ast })=-1$ and vanishing optimal portfolio volatility gives $J^{\prime \prime }(z_{\ast })=0$. For power utility and a high-water tax, the smooth tax-paying solution has boundary dual curvature $(1-\alpha \tau )/(R{z}_{\ast })$, so it cannot be admissible when $\alpha \tau >1$.

For a complex $n\times n$ matrix $B$, the real linear map $A\mapsto BA{B}^{\ast }$ on the $n^2$-dimensional real vector space of Hermitian matrices has determinant $|\det B{|}^{2n}$. A real basis of Hermitian matrices is also a complex basis of all matrices, so the matrix of this restriction equals the matrix of the complex-linear congruence map. Left and right multiplication separately have determinants $(\det B{)}^n$ and $(\overline{\det B}{)}^n$.

For $\mathbf{k}=k\hat{z}$, these complex vectors form an orthonormal basis of the transverse plane using the Hermitian inner product. They have opposite phases under rotations about $\hat{z}$, so they describe the two helicity components. Since $\mathbf{e}\cdot m^{(\pm )}=\sin \theta e^{\pm i\phi }/\sqrt{2}$, transverse-vector sources carry only the angular sectors $m=\pm 1$ of spherical harmonics. Streaming by $ik\cos \theta$ preserves $m$ and couples neighboring angular degrees.

For the binary phase representation of a Pauli string, these local rules implement $P\mapsto U^{†}PU$ for the Hadamard gate and the phase gate $S=\mathrm{diag}(1,i)$. In particular $S^{†}XS=-iXZ$, so the phase update differs in sign from forward conjugation. Use the old input bits, and reduce $s$ modulo four.

The modular Heisenberg group consists of upper-unitriangular $3\times 3$ matrices over the finite field ${\mathbb{F}}_p$. In coordinates $M(x,y,z)$, multiplication is $M(x,y,z)M(x\prime ,y\prime ,z\prime )=M(x+x\prime ,y+y\prime +xz\prime ,z+z\prime )$, and its order is $p^3$. Its center of a group is $\{M(0,y,0)\}\cong C_p$; the quotient group by the centre is $C_p\times C_p$, represented by the $x,z$ coordinates. The elementary generators $a=M(1,0,0)$, $b=M(0,0,1)$, $c=M(0,1,0)$ satisfy $a^p=b^p=c^p=e$, $ab{a}^{-1}{b}^{-1}=c$, and $c$ central. For $p=2$, $(ab{)}^2=c$, so the whole group does not have exponent two.

For the Heisenberg group law $(x,y,z)(a,b,c)=(x+a,y+b,z+c+xb)$, the right Maurer-Cartan form has coefficients $dx,dy,dz-ydx$. They satisfy $d{\sigma }^1=d{\sigma }^2=0$ and $d{\sigma }^3={\sigma }^1\wedge {\sigma }^2$. A general right-invariant Riemannian metric is a constant positive-definite symmetric matrix in this coframe.

At a prime not dividing the level, the root-of-unity average contributes $a_{pn}$ and the scaling overlattice contributes the second term. At a prime dividing the level that overlattice loses the marked point's exact order and is excluded. Extending the Dirichlet character by zero gives a uniform formula, with $a_{n/p}=0$ unless $p\mid n$. The bad-prime operator is $U_p$.

A repulsive condensate next to an impenetrable planar wall has Dirichlet boundary condition $\psi (0)=0$. Writing $\psi =\sqrt{n_0}f$ reduces its stationary Gross–Pitaevskii equation to ${\ell }_h^2{f}^{\prime \prime }+(1-f^2)f=0$. The bulk conditions $f\to 1,f^{\prime }\to 0$ give ${\ell }_h^2(f^{\prime }{)}^2=(1-f^2{)}^2/2$ and hence the displayed number density. Omitting $n_0$ describes normalized number density, not dimensional number density.

For the Hawking exponential ray map, regulated integrals ${\int }_0^{\infty }{x}^{ia}{e}^{-(ϵ\pm i{\omega }^{\prime })x}dx=\Gamma (1+ia)(ϵ\pm i{\omega }^{\prime }{)}^{-1-ia}$ have squared-modulus ratio $e^{-2\pi a}$ as $ϵ\downarrow 0$, where $a=\omega /\kappa$. Combining this with the canonical identities for a bosonic Bogoliubov transformation gives the bosonic occupation $[e^{2\pi \omega /\kappa }-1{]}^{-1}$. Continuum modes require wave-packet normalization to interpret finite particle counts.

At parameter $1/2$, primal and dual bond percolation on the square lattice have the same probability distribution. Suppose an infinite percolation cluster exists. Large boxes meet it with probability tending to one. The square-root trick for positively associated events and quarter-turn symmetry imply that each side has an infinite exterior open arm with probability tending to one. The same holds for dual arms. A union bound then gives positive probability of primal top and bottom arms together with dual left and right arms.

Use a primal box $[-r,r{]}^2$ and a dual box with sides at $\pm (r+1/2)$. Open only primal edges whose endpoints both lie in the primal box, preserving all outward and exterior arm edges. The two primal arms are joined inside. Planarity prevents the two alternating dual arms from joining: a proposed dual connection together with the inner boundary arc separates one primal infinite arm from infinity. Thus the modified configuration has two infinite dual clusters. Its positive probability by finite modification of Bernoulli percolation contradicts uniqueness of the infinite percolation cluster. Consequently the percolation probability at $1/2$ is zero. This is often called Zhang's argument.

For planar Brownian motion started at $x+iy$ in $0<y<\pi$, this is its unconditional exit density on the bottom boundary. Its mass is $1-y/\pi$; the remaining mass $y/\pi$ lies on the top boundary. The conformal bijection $z\mapsto e^z$, the Poisson kernel for the upper half-plane and the boundary Jacobian determinant derive the formula. Conditional bottom-boundary density requires division by its mass.

For planar Brownian motion starting on the positive real axis inside a disk, let $T$ be circular exit and $S$ first contact with the imaginary axis. Reflecting the path after $S$ fixes the disk and swaps positive and negative circular exits. The Strong Markov property makes their probabilities equal on $S<T$. Therefore $\mathbb{P}(T<S)=\mathbb{P}(\mathrm{Re}{B}_T>0)-\mathbb{P}(\mathrm{Re}{B}_T<0)$.

A planar Brownian motion beginning at $a\in (0,1)$ in the unit disc has circular exit law obtained by pulling back uniform angular measure through ${\psi }_a(w)=(w-a)/(1-aw)$. This Möbius transformation sends the starting point to zero. The right semicircle maps to an arc with endpoints at angles $\pm (\pi /2+2\arctan a)$, giving positive-half-plane exit probability $1/2+(2/\pi )\arctan a$.

In the intermediate-index range, if an attaching sphere $A_1$ meets a belt sphere $B_1$ at exactly one transverse point, handle slides can remove the intersections of every other attaching sphere with $B_1$. The dual slides then remove the intersections of $A_1$ with the other belt spheres. Each move uses a band to the unique pivot sheet and a local isotopy removing the old and new intersections. If $m_{ji}=A_j\cdot B_i$ and $\delta =m_{11}=\pm 1$, the remaining incidence block is $m_{ji}-m_{j1}{m}_{1i}/\delta$, the Schur complement of the pivot. Extreme-index cases need separate hypotheses.

A simple turning point for a stratified wavespeed occurs at $c(Y_s)=\omega /k$, where $Q={\omega }^2/c^2-k^2$ has a nonzero derivative. If $c_Y(Y_s)>0$, write $q=-Q^{\prime }(Y_s)>0$; then $\ell \sim [q(Y_s-Y){]}^{1/2}$ below it. The vertical group velocity vanishes and the stationary wave amplitude in a stratified wavespeed diverges as $(Y_s-Y{)}^{-1/4}$.