A velocity increment is the difference of a velocity field between two positions. Its longitudinal component is the projection onto their separation direction. Moments of increments define longitudinal structure functions; the Kolmogorov 1941 theory estimates a typical inertial-range magnitude as $(ϵr{)}^{1/3}$.

Invert the two-dimensional Fourier transform in the long-path projection of a magnetic correlation, then perform the angular integration in polar coordinates. The angular integral is $2\pi J_0(kr)$, a Bessel function of the first kind, giving the displayed Hankel transform with the stated Fourier-Hankel normalization for an isotropic spectrum. The result recovers the scalar tensor coefficient $H$, whose shell-integrated magnetic energy spectrum has an additional factor $k^2$.

Orient a degree-$(d+1)$ regular tree towards one end, and let height $g$ decrease along the unique parent edge and increase along the $d$ child edges. For $w_{\rho }(A)={\sum }_{x\in A}{\rho }^{g(x)}$, suppressing the occupied-target constraint only increases births, so its infinitesimal generator satisfies $L{w}_{\rho }\le [-1+\lambda ({\rho }^{-1}+d\rho )]w_{\rho }$. For $d>1$, choose $\rho =d^{-1/2}$. If $\lambda <1/(2\sqrt{d})$, the expected weight and the infection probability at each fixed site decay exponentially. The expected number of recoveries there equals its expected total occupation time, which is finite. Each infection episode ends almost surely at a recovery, so infinitely late infection is impossible. For $d=1$, take $\rho <1$ sufficiently close to one for each $\lambda <1/2$.

The bipartite inclusion graph between consecutive nonempty ranks of the Boolean lattice is connected. Two rank-$r$ sets differing by one exchanged element share their rank-$(r+1)$ union as a neighbour; successive exchanges connect any two rank-$r$ sets. Every upper vertex has a lower neighbour. For the two middle ranks of an odd-dimensional Boolean lattice the graph is regular, so equality in the Local LYM inequality for a nonempty upper family forces the whole upper rank: equality admits no edges from its lower shadow to the remaining upper vertices.

For a finite Galois extension $L/K$ of non-Archimedean local fields, its relative Brauer group is the kernel of restriction $\mathrm{Br}(K)\to \mathrm{Br}(L)$ and identifies with $H^2(\mathrm{Gal}(L/K),L^{\times })$. Restriction multiplies the local Brauer invariant by $[L:K]$. This kernel is therefore cyclic of order $[L:K]$, and the indicated class is its distinguished generator. In an unramified extension with arithmetic Frobenius generator it is represented by the cyclic algebra with parameter a uniformizer.

With the arithmetic Frobenius normalization, a chosen uniformizer $\pi$ and Lubin–Tate formal group give these actions on the Lubin–Tate tower and the maximal unramified extension. The two fields are linearly disjoint and their compositum is the maximal abelian extension. The inverse on units is essential: using geometric Frobenius would invert the entire Local Artin map. The finite kernels on a compositum of residue degree $d$ and torsion level $r$ are ${\pi }^{d\mathbb{Z}}(1+{\pi }^r{\mathcal{O}}_K)$. They are a basis of finite-index open subgroups, yielding the profinite completion of $K^{\times }$ as the abelian Galois group.

For $G=\mathrm{Gal}(L/K)$, Tate cohomology of a finite group identifies the source as ${\hat{H}}^{-2}(G,\mathbb{Z})$ and the target as ${\hat{H}}^0(G,L^{\times })$. Cup product with the local fundamental class is an isomorphism; its inverse is the finite Local Artin map. The algebraic cup-product isomorphism applies because every subgroup has first cohomology zero by Hilbert theorem 90, and second cohomology cyclic of order equal to its size by the local Brauer invariant. Compatibility with restriction and corestriction yields the tower and norm compatibility of Local Artin reciprocity.

For any initial density operator, average its time-independent Lindblad equation trajectory as ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$. These averages stay in the compact convex set of density operators. Along a convergent subsequence, $\mathcal{L}({\bar{\rho }}_T)=(\rho (T)-\rho (0))/T\to 0$, so the limit is stationary. Thus a finite-dimensional physical Affine Bloch equation has at least one physical equilibrium, even though an arbitrary affine real equation can have none.

With the displacement convention for a Lévy kernel of a Markov jump process, suppose its total rate is bounded, its first moment is zero and its second moment $v$ is bounded by $V$. The bounded rate prevents explosion. The compensated displacement integral is a square-integrable martingale with $\mathbb{E}{X}_t^2\le Vt$. The displayed process is the sum of $2\int X_{s-}d{X}_s$ and the compensated sum of squared jumps. The first term is square integrable since $4\mathbb{E}{\int }_0^t{X}_{s-}^{2}v(X_{s-})ds\le 2V^2{t}^2$; the second is integrable because its expected total jump-square sum is at most $Vt$. Consequently the square compensation is a true martingale, without a fourth-moment assumption.

The knot signature is the signature of the symmetrized Seifert matrix. Equivalently it is the signature of the double branched covering of $B^4$ over a pushed-in Seifert surface. At $\omega =-1$ it agrees with the Levine-Tristram signature. It is additive under connected sum of knots, changes sign under reflection, and obeys $|\sigma (K)|\le 2g_4(K)$. State the ambient orientation and push-off convention when assigning a sign.

For a Lenard-Magri recursion, antisymmetry shifts an index: $\{H_i,H_j{\}}_1=\{H_{i-1},H_{j+1}{\}}_1$ when $i\ge 1$. Repeating reaches $\{H_0,H_{i+j}{\}}_1=0$, because $H_0$ is a Casimir function. The recursion then gives $\{H_i,H_j{\}}_2=-\{H_{i+1},H_j{\}}_1=0$. This proof applies to every finite initial portion for the indices that can be shifted inside it, and to all pairs in an infinite sequence.

The next step of a Lenard-Magri recursion requires $-{\Pi }_{2}d{H}_{n-1}$ to be a Hamiltonian vector field for ${\Pi }_1$. It must be tangent to the symplectic leaves, and its contraction with the leaf symplectic form must have a global potential, with a compatible smooth extension between leaves. Even the first step may fail: on ${\mathbb{R}}^2$, take ${\Pi }_1=0$, ${\Pi }_2={\partial }_x\wedge {\partial }_y$ and $H_0=x$. Both Poisson structures are compatible, but ${\Pi }_{1}d{H}_1=-{\Pi }_{2}dx$ cannot hold, because its left side is zero and its right side is nonzero.

With $P_n(1)=1$, the generating function satisfies ${\partial }_x(x^2{\partial }_{x}G)+{\partial }_t((1-t^2){\partial }_{t}G)=0$. Expanding in Legendre polynomials gives $x^2{a}_n^{\prime \prime }+2x{a}_n^{\prime }-n(n+1)a_n=0$. Boundedness at zero removes $x^{-n-1}$, and the value at $t=1$ makes $a_n=x^n$. Orthogonality then gives ${\int }_{-1}^1{G}^{2}dt={\sum }_n{k}_n{x}^{2n}$. The integral is $x^{-1}\log ((1+x)/(1-x))$, so $k_n=2/(2n+1)$. Uniform analyticity on compact subdiscs $|x|<1$ justifies the generating expansion and termwise calculations.

Differentiating the Legendre differential equation gives $((1-x^2{)}^2{R}_n^{\prime }{)}^{\prime }+(n-1)(n+2)(1-x^2)R_n=0$ for $R_n=P_n^{\prime }$. Multiply the equations for two distinct indices by the opposite polynomial and subtract. The endpoint term vanishes because of $(1-x^2{)}^2$, giving the displayed weighted orthogonality. The zero polynomial $R_0$ is not a nonzero eigenfunction.

For the centered two-level advection recurrence, the Fourier roots at Courant number one coalesce at $i$ for phase $\pi /2$. The resulting nontrivial Jordan block admits linearly growing solutions despite both eigenvalues having modulus one. Uniform two-level stability for arbitrary starting perturbations therefore requires the strict inequality $0<\mu <1$. A specially consistent startup can select the exact translating branch at the boundary, which is a restriction on the data rather than a uniform bound on the full update.

For $u_t=u_x+u_y$ with equal spatial mesh sizes, the centered leapfrog recurrence has amplification polynomial of a multilevel finite difference scheme $G^2-2i\mu (\sin \xi +\sin \eta )G-1$. Uniform stability for arbitrary two-level starting perturbations holds exactly for $0<\mu <1/2$. At $\mu =1/2$ the phase pair $(\pi /2,\pi /2)$ gives $(G-i{)}^2$, and its companion matrix has a nontrivial Jordan block, producing growth proportional to the number of steps. For larger $\mu$ one polynomial root has modulus greater than one. Bounds on eigenvalue moduli alone therefore give a misleading non-strict endpoint.

A proper lattice ideal $I$ is prime if $x\wedge y\in I$ implies $x\in I$ or $y\in I$. Its complement is a prime filter of a distributive lattice when the lattice is distributive: downward closure of the ideal makes the complement upward closed, primeness makes it meet-closed, and join closure makes that filter prime. Equivalently, its complement indicator is a homomorphism of lattices to the two-element lattice. Given $b\nleqq a$ in a distributive lattice, maximize an ideal containing $a$ and avoiding $b$ using Zorn lemma. If $x,y$ lie outside it, the two enlarged ideals contain $b$, so $b\le i\vee x$ and $b\le j\vee y$ for some $i,j\in I$. Distributivity shows that $x\wedge y\in I$ would force $b\in I$, a contradiction. Thus this maximal avoiding ideal is prime and separates the two specified elements.

The Laplace radius is the satellite distance where the small-tilt nodal precession contributions from a planet's axisymmetric quadrupole and the stellar tide have equal magnitudes. In the circular hierarchical limit those rates are $3G{M}_p{J}_2{R}_p^2/(2n{a}^5)$ and $3G{M}_{\star }/(4n{a}_p^3)$. Equating them proves the displayed fifth-root expression. Inside this radius the planetary quadrupole coefficient controls the preferred plane; outside it the double-averaged circular tidal potential dominates. When equator and planetary orbit are inclined to one another, their balance determines the local Laplace plane.

For the Lang map, ${\mathcal{L}}_F(gh)=h^{-1}{\mathcal{L}}_F(g)F(h)$. This twisted right-translation identity determines which subgroups act on its inverse images. Fibres are left $G^F$-cosets, by the Lang theorem for algebraic groups.

The signed third-order longitudinal structure function has this leading inertial range form under homogeneity, isotropy and the usual local-equilibrium scale separation. Its coefficient follows from the Kármán-Howarth equation and the mean kinetic energy budget, without assuming the Kolmogorov two-thirds law. Its negative sign encodes a forward energy cascade; it does not establish every higher-order similarity exponent.