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 $(Y_s-Y{)}^{-1/4}$.

For a spatially flat Friedmann-Lemaître-Robertson-Walker metric, negative extrinsic curvature, and $K=-3H+\kappa$, the Hamiltonian constraint gives the displayed scalar perturbation equation. Indeed $K^2=9H^2-6H\kappa$ and $K^i{}_j{K}^j{}_i=3H^2-2H\kappa$ to first order. The spatial Ricci scalar is $4\Delta \Phi$; subtract the background equation $6H^2=16\pi G\bar{\rho }$ and divide by four.

For a spatial initial slice with vanishing extrinsic curvature, the vacuum momentum constraint is automatic and the Hamiltonian constraint requires zero Ricci scalar. In three spatial dimensions, the indicated conformally flat metric has $\mathcal{R}=-8{\psi }^{-5}\Delta \psi$, with a nonzero conformal factor. Thus a harmonic function supplies the data on its domain. A positive example is $\psi =1+a/r$ on a punctured or exterior region. There is no nonconstant globally smooth example on all of Euclidean three-space tending to one at infinity, by the maximum principle for harmonic functions. Time symmetry of an initial slice does not imply that its entire evolved spacetime is static.

For integer $k\ge 1$, the entrywise power has entries $X_{ij}^k$ and equals the repeated Hadamard product. The zero power is the all-ones matrix, even at zero entries, because it represents the constant function one. Every nonnegative integer Hadamard power of a positive semidefinite matrix is positive semidefinite by the Schur product theorem. It is not an ordinary matrix power.

In the Gaussian white noise model, estimate a Haar scaling function coefficient by $\int {\varphi }_{J,k}dY$. The resulting Haar projection estimator is an unbiased estimator with independent coefficient errors of variance $1/n$. On each cell it is the observed path increment divided by the cell length. This gives a finite-dimensional estimator without imposing smoothness on the drift.

For a condensate with uniform bulk complex field one, its convergent localized momentum isThe subtraction removes the uniform-background phase-gradient contribution. If $\psi -1=O(r^{-1})$ and $\nabla \psi =O(r^{-2})$, its integrand is $O(r^{-3})$, integrable in two dimensions. Under decaying variations its functional derivative is $\delta p_j/\delta {\psi }^{\ast }=-i{\partial }_j\psi$; the subtraction is a boundary correction and does not change the bulk equation.

With $E=\frac{1}{2}\int [|\nabla \psi {|}^2+\frac{1}{2}(1-|\psi {|}^2{)}^2]$ and the renormalized momentum of a condensate, a Gross–Pitaevskii solitary wave is a critical point of $E-v{p}_x$. Along a differentiable solution family, fixed bulk normalization and vanishing variation boundary terms give $dE/d\lambda =vd{p}_x/d\lambda$. Therefore $dE/d{p}_x=v$ wherever $d{p}_x/d\lambda \ne 0$. At a cusp or turning point the parametrized identity is the appropriate statement; one must not assume a globally single-valued dispersion branch.

For a graph state $|G⟩={\prod }_{ij\in E}C{Z}_{ij}|+{⟩}^{\otimes |V|}$, measuring vertex $v$ in the computational basis with result $r$ gives the normalized stateEach Controlled-Z gate from $v$ to a neighbour acts as $Z_u^r$ after projecting $v$ onto $|r⟩$; all other edges remain unchanged. The projection contributes $1/\sqrt{2}$, so either result has probability $1/2$. For a four-cycle, deleting one vertex leaves a path and adds byproducts on its two endpoints.

Let $1$ be a leaf with neighbour $2$ and let the graph on the other vertices be generated by the Controlled-Z gate product $E_{\mathrm{rest}}$. If the leaf has Pauli Z gate frame $Z_1^r$, measuring it at angle $\alpha$ with result $s$ applies one-bit teleportation and leavesAny known Pauli Z gate frames on other vertices remain as additional factors before those vertices' input states. The identity $J(\alpha )Z^r=X^{r}J(\alpha )$ combines the leaf byproduct with the measurement result.

If global dimension is zero, every module is a projective module. Each short exact sequence splits because its quotient is projective. Conversely, splitting every short exact sequence makes each module a direct summand of a free module and therefore a projective module. Splitting also makes every module an injective module: for $A\hookrightarrow B$ choose a retraction $B\to A$ and compose it with any prescribed map out of $A$.

For reference Brownian motion $R=x+W$, stopped at $T=t\wedge {\tau }_{(a,b)}$ with $0<a<x<b$, this expression is the Radon-Nikodym derivative of the stopped Bessel process law of dimension $\delta$. The bounded stopped drift verifies the Novikov condition. The comparison laws must have the same initial value and the same stopping rule.

For $a>0$, the positive amplitude ordinary differential equation $R^{\prime }=aR-R^3$ has solution $R(t)=\sqrt{a}[1+(a/R_0^2-1)e^{-2at}{]}^{-1/2}$. This follows from the logistic differential equation for $R^2$. Every $R_0>0$ approaches $\sqrt{a}$ as a monotone function, and the positive distance to this limit decreases; an oscillating complex wave carrying this amplitude has no ordinary monotone ordering.