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On a connected closed oriented surface, a magnetic Anosov flow with $\int f{\Omega }_a=0$ preserves a smooth contact form only when $f=0$. To see the obstruction, take $d\eta =f{\Omega }_a$ and $\sigma =\alpha -{\pi }^{\ast }\eta$. An invariant contact form can be normalized to evaluate to one on $F$; its differential is a constant multiple of ${\iota }_F\mu$, since smooth invariant functions are constant. Averaging closed one-forms forces that multiple to be $-1$, so the contact form differs from $\sigma$ by a closed form. Its fibre-average decomposition gives $Fh=(\eta -\xi )(v)-af$. The degree-zero/one kernel theorem gives $af=0$ and $d\eta =d\xi =aK{\Omega }_a$, forcing $f=0$ whether $a$ vanishes or not. For $f=0$, the canonical form $\alpha$ is invariant and contact.

In a stationary M/M/1 queue with independent Poisson arrivals of rate $\lambda$ and exponential service times of rate $\mu >\lambda$, the departure counting process is a Poisson process of rate $\lambda$. Moreover, the departures before a fixed time $t$ are independent of the queue population $N_t$. This independence concerns the past departures: future departures depend on whether the queue is currently empty.

The proof uses time reversal. The stationary Geometric distribution satisfies Detailed balance, ${\pi }_n\lambda ={\pi }_{n+1}\mu$, so the stationary queue population is a Reversible Markov chain. Departures in forward time are arrivals in reversed time. In the reversed chain the upward rate is again the constant $\lambda$, so those arrivals are an independent Poisson process, including independence of their future increments from the present queue population. Reversing time gives both assertions. Consequently, feeding the output of a stationary M/M/1 queue into another queue supplies Poisson input to that second queue. The claim depends on equilibrium; a queue started empty has no immediate departures even though arrivals have positive rate.

With packet arrival rate lambda, independent exponential payloads of mean one and byte-service rate C, let $\rho =\lambda /C<1$. The number N of unfinished packets has law $(1-\rho ){\rho }^n$. Conditional on N=n, the remaining payloads are n independent unit-mean exponentials, including the in-service residual by memorylessness. Summing their Laplace transforms gives $\mathbb{E}{e}^{-sW}=(1-\rho )(1+s)/(1+s-\rho )$. Thus W has mass $1-\rho$ at zero and, conditional on positivity, is exponential with rate $1-\rho$. The probability of at least r headers is instead ${\rho }^r$ for integer r.

For $L=\dot{F}{F}^{-1}$, the product rule proves the displayed lower-convected derivative identity. Under a superposed rotation $Q(t)$, ${\tau }^{\ast }=Q\tau Q^T$ and $L^{\ast }=QL{Q}^T+\dot{Q}{Q}^T$. The two rotation terms in ${\dot{\tau }}^{\ast }$ cancel those in $(L^{\ast }{)}^T{\tau }^{\ast }+{\tau }^{\ast }{L}^{\ast }$, proving that the lower-convected rate is an objective time derivative.

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.