In a finite flow network, a maximum flow exists because the feasible flow set contains the zero flow and is a compact set: the capacity constraints bound every coordinate, and they and flow conservation define a closed set. The strength of a flow is a continuous function, so the extreme value theorem gives an attained maximum. An augmenting path in the residual network would permit positive augmentation, contradicting maximality. The source-reachable vertices in the residual network therefore define a cut of a flow network whose outgoing original edges are saturated and incoming original edges carry zero flow. Flow conservation makes the strength of a flow equal to this cut capacity, proving the max-flow min-cut theorem.

Existence and strong duality for finite real flow network edge capacities do not imply finite termination for every choice of augmenting paths in the Ford-Fulkerson algorithm. With irrational capacities, some choices can continue forever. With integer capacities, Integrality of the Ford-Fulkerson algorithm instead forces finite termination; rational capacities reduce to that case by a common denominator. Choosing shortest augmenting paths as in the Edmonds–Karp algorithm avoids dependence on capacity magnitudes.^[ref]

Matter parity assigns $-1$ to every matter chiral superfield, including a singlet $N$ describing a conjugate right-handed neutrino, and $+1$ to Higgs chiral superfields. Thus the quadratic superpotential term $\frac{1}{2}{M}_{ij}{N}_i{N}_j$ is even, although it changes lepton number by two. The higher-degree operators $QQQL$ and $(L{H}_u{)}^2$ are also even and violate baryon number or lepton number. Exact R-parity therefore forbids the renormalizable trilinear violations $LLE$, $LQD$, $UDD$, $N{H}_d{H}_u$, $NNN$, but does not imply conservation of these continuous charges in every allowed interaction.

For $W=C^3+A+b$, $A=a{e}^{-\alpha S}$, the potential is $V=|C^3+b+(1+\alpha s)A{|}^2/(s{t}^3)+3|C{|}^4/(s{t}^2)$. A finite zero-energy vacuum requires $C=0$ and $b+(1+\alpha s)A=0$. With $a,b\ne 0$, such a solution exists exactly when $0<|b/a|\le 2e^{-1/2}$: the magnitude equation is $|b/a|=(1+x)e^{-x/2}$, $x=\alpha s>0$, whose maximum occurs at $x=1$. At that vacuum $F^S=F^C=0$ but $F^T=e^{K/2}t\bar{W}\ne 0$. Both components of $T$ are flat, while $C$ is quartically lifted. If no root exists, the positive potential falls toward zero as $t\to \infty$ and has no finite minimum. If $a=b=0$, the $C=0$ family is supersymmetric instead.

If a no-scale supergravity scalar potential has the displayed form with $y>0$ and nonnegative $A,B$ independent of the volume coordinate $y$, every point with $A+B>0$ has a direction of strictly decreasing potential. Thus a finite stationary vacuum requires $A=B=0$. If that common zero set is empty, the potential has infimum zero along $y\to \infty$ but no finite minimum. If the common zero set is nonempty, varying $y$ on it is a flat direction of a scalar potential. This distinguishes an attained zero-energy vacuum from a decompactification runaway.

For constant positive Hubble parameter $\mu$ and a four-thirds polytropic equation of state, a linear cosmological density perturbation obeys $\ddot{\delta }+2\mu \dot{\delta }+{\eta }^2{e}^{-3\mu t}\delta =0$. Put $\delta =e^{-\mu t}y$ and $x=2|\eta |e^{-3\mu t/2}/(3\mu )$. The resulting Bessel differential equation has order $2/3$; use ordinary Bessel functions for ${\eta }^2>0$ and modified Bessel functions for ${\eta }^2<0$. Their small-$x$ expansions give one constant limiting mode and one $e^{-2\mu t}$ mode. The constant limit need not exceed the initial amplitude without an initial-velocity condition.

The group of invertible linear maps preserving a specified quadratic form over a finite field. In characteristic two one must preserve the quadratic form itself, not merely its polar alternating bilinear form. For a nonsingular form, any two nonzero isotropic vectors can be incorporated into hyperbolic pairs and standard adapted bases, producing an isometry between them.

A classical group is a standard family of linear groups, principally the general/special linear groups and groups preserving nondegenerate alternating, symmetric, Hermitian or quadratic forms. Finite-field versions yield many finite simple groups after imposing the appropriate determinant or derived-subgroup condition and quotienting the scalar center of a group. Small dimensions and small fields have exceptional isomorphisms and nonsimple cases.

This formula allows a matrix and its derivative to fail to commute. Differentiate the absolutely convergent power series of the matrix exponential to obtain ${\sum }_{k\ge 1}k{!}^{-1}{\sum }_{j=0}^{k-1}{f}^j{f}^{\prime }{f}^{k-1-j}$. Expanding the integral gives the same coefficients, since ${\int }_0^1(1-u{)}^a{u}^{b}du=a!b!/(a+b+1)!$. The formula reduces to $e^f{f}^{\prime }$ when $[f,f^{\prime }]=0$.

For a nonzero real three-dimensional skew-symmetric matrix, choose real orthogonal vectors $u,v$ of equal length with $Au=\theta v$, $Av=-\theta u$ and $\theta >0$, and a real unit vector $e_3$ in its kernel. The matrix exponential acts by $e^{A}u=u\cos \theta +v\sin \theta$, $e^{A}v=v\cos \theta -u\sin \theta$ and $e^A{e}_3=e_3$. Its matrix in this orthonormal basis is a planar rotation matrix with an unchanged axis. A purely imaginary nonzero eigenvalue and its complex eigenvector give $u,v$ by taking real and imaginary parts; antisymmetry proves their equal norms and orthogonality.

If $A$ is real symmetric positive definite and $B$ is real symmetric, let $S=A^{-1/2}$. The matrix $SBS$ is symmetric and has an orthonormal eigenbasis, represented by $Q$. Taking $P=SQ$ gives the two displayed congruences. The diagonal entries are determined by the generalized eigenvalue problem $\det (B-\lambda A)=0$, since $P^T(B-\lambda A)P$ is diagonal with entries ${\lambda }_j-\lambda$. Positive definiteness of the first form is essential to this construction.

The polar decomposition in continuum mechanics and material isotropy remove the reference rotation from an elastic Cauchy stress tensor. Half-turns around principal stretch axes leave the diagonal left stretch tensor fixed, so material frame indifference forces the stress's off-diagonal entries to vanish in those axes. Rotations within repeated-stretch eigenspaces also force isotropic stress there. Consequently stress and the left Cauchy-Green deformation tensor are coaxial symmetric tensors.

For $0<a\le b$, expand the expectation under the weighted matching Markov chain stationary weights to obtain the displayed identity. If the largest matching in a graph has $n$ edges and $b/a\le 1+1/n$, the observable lies in $[1,e]$. Its sample mean has controlled error by the Hoeffding inequality. A product of $J$ ratios, each estimated to relative accuracy $O(\epsilon /J)$, approximates the full ratio to relative accuracy $O(\epsilon )$.

For a four-dimensional $\mathcal{N}=1$ finite-helicity massless supermultiplet, one complex oscillator changes helicity by one-half. A lower-helicity labelling convention assigns $\kappa$ to the pair $(\kappa ,\kappa +1/2)$. A symmetric midpoint convention instead uses $\kappa +1/4$; report the helicity pair when comparing conventions. CPT completion of a supermultiplet pairs $\kappa$ with $-\kappa -1/2$.

The displayed action functional contains no mass term and no $R{\varphi }^2$ curvature coupling. Its Euler-Lagrange field equation is ${\partial }_{\mu }(\sqrt{-g}{g}^{\mu \nu }{\partial }_{\nu }\varphi )=0$. In a flat FLRW metric, a Fourier mode obeys ${\varphi }_k^{\prime \prime }+2(a^{\prime }/a){\varphi }_k^{\prime }+k^2{\varphi }_k=0$ in four dimensions. The expansion can therefore freeze these modes; the curvature term of a conformally coupled scalar field changes that conclusion.

The zero-form field strength deforms massless type IIA supergravity. A D8-brane changes this flux across its worldvolume; it is therefore outside the elementary massless M-theory circle duality dictionary.

// Target: physics.bigb

For a massive four-dimensional N=1 supermultiplet, superspin $j$ is the spin of its Clifford vacuum, the state annihilated by the two lowering supercharges. The creation operators produce spins $j,j,j+1/2,j-1/2$, omitting the last when $j=0$. The corresponding superspin Casimir has an eigenvalue proportional to $j(j+1)$; this representation label is distinct from the spin of any one constituent.

Write $x=m/M_{\odot }$, the initial mass function as $dN=A{x}^{-\alpha }dx$, and a main-sequence mass-luminosity relation as $L/L_{\odot }=x^{\beta }$. On $a<x<b$, the mass-to-light ratio in solar units isFor $\alpha =\beta =3$ this is exactly $1/(ab)$. The result concerns the specified unevolved stellar population; stellar remnants, evolved luminous stars and cooling substellar objects require different accounting.

A dynamical mass-to-light ratio must specify its aperture and whether the mass includes dark matter. A growing dark halo can increase ${\rm Y}(<R)$ outside the bright stellar body even when the stellar ratio is constant. Comparing a tidal-radius satellite mass with a large-aperture host mass is different from comparing both within their half-light radii. From $M\propto {\sigma }^{2}R$ and $L\propto {\sigma }^4$, the inferred ratio scales as $R/{\sigma }^2$, so equal ratios additionally require homologous radius scaling.

For asset dynamics $dS=\mathrm{diag}(S)(\mu dt+\sigma dW)$, let $\lambda ={\sigma }^{-1}\mu$ and $Z=\mathcal{E}(-\int {\lambda }^{T}dW)$. The Itô product rule cancels the drift of each $Z{S}^i$, leaving the displayed stochastic integral. When $\mu$ is bounded and $\sigma {\sigma }^T\ge \epsilon I$, $\lambda$ is bounded and the Novikov condition makes $Z$ a true martingale on every finite horizon. If $\sigma$ is also bounded, each $Z{S}^i/S_0^i$ satisfies the Novikov condition as a stochastic exponential with bounded integrand ${\sigma }_i-\lambda$, so it too is a true martingale.

For a finite-expected value nonnegative random variable and $a>0$, equality in the uncapped Markov's inequality holds exactly when $Y$ takes only $0$ and $a$ almost surely. Indeed the nonnegative difference $Y-a{1}_{\{Y\ge a\}}$ has zero expected value exactly when it vanishes almost surely. Giving $a$ probability $p$ yields $\mathbb{E}Y=pa$ and $\mathbb{P}(Y\ge a)=p$. This shows the bound cannot generally be reduced using only the nonnegative random variable's expected value.