Positive numbers $a,b,c$ are the areas of the three coordinate projections of a volume-one Euclidean body exactly when $abc\ge 1$. Necessity is the three-dimensional Loomis--Whitney inequality. For sufficiency take box lengths $x=\sqrt{ac/b}$, $y=\sqrt{ab/c}$ and $z=\sqrt{bc/a}$, and retain the union of the three slabs with first, second or third normalized coordinate at most $t$. All three projections are full rectangles, while volume is $\sqrt{abc}[1-(1-t{)}^3]$. Choosing $t=1-(1-1/\sqrt{abc}{)}^{1/3}$ gives volume one. The slabs have positive thickness, so the construction uses a body with interior rather than zero-volume additions.

Each nonnegative integrable $f_j$ depends on all coordinates except the $j$th. In dimension three, Cauchy-Schwarz inequality first in the third coordinate and then in the remaining pair proves the displayed estimate. Induction integrates the last coordinate by Hölder's inequality on the first $d-1$ factors, then applies Hölder's inequality to the remaining factor and the dimension-$(d-1)$ product. Taking indicators of coordinate projections of a Euclidean body gives the volume version of the Loomis--Whitney inequality.

In isotropic incompressible three-dimensional turbulence, $3S_2(r)/4=\int E(k)H(kr)dk$, with $H(x)=1+3\cos x/x^2-3\sin x/x^3$. It is smooth at zero, where $H=x^2/10+O(x^4)$, and tends to one at large argument. A sharp wavenumber cutoff omits the finite large-scale-gradient contribution.

A smooth velocity varying on a scale larger than $r$ contributes $\Delta U_L\simeq r{\partial }_L{U}_L$ to a velocity increment. Higher powers produce mixed moments with smaller-scale increments. longitudinal structure functions therefore attenuate, but do not exactly filter out, large-scale strain and intensity modulation.

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.