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For a nilpotent endomorphism $N$ of a vector space, the commutator action $\mathrm{ad}N:T\mapsto NT-TN$ on its endomorphisms is nilpotent. The commuting left and right multiplication maps give $(\mathrm{ad}N{)}^r(T)={\sum }_{j=0}^r(-1{)}^j\left(\genfrac{}{}{0ex}{}{r}{j}\right)N^{r-j}T{N}^j$. If $N^q=0$ and $r\ge 2q-1$, every summand vanishes. This works in every characteristic of a field and is useful when applying Engel theorem to an Adjoint representation.

In cylindrical coordinates, the Newtonian gravitational stress tensor gives radial angular-momentum flux $r{T}_{r\varphi }=r({\partial }_r\Phi )(r^{-1}{\partial }_{\varphi }\Phi )/(4\pi G)$. Nonaxisymmetric structure can correlate the field components and transport angular momentum. The isotropic negative-pressure part has zero off-diagonal component, so it gives no direct radial transport. This separates a shear flux from a contribution to normal force balance.

For positive masses, let $r$ be the actual minimum pairwise separation and use positive binding energy $U={\sum }_{j<k}G{m}_j{m}_k/r_{jk}$. If $m_{(1)},m_{(2)}$ are the smallest masses, then $G{m}_{(1)}{m}_{(2)}\le rU\le G{\sum }_{j<k}{m}_j{m}_k$. Thus $U^{-1}$ is comparable to $r$ with constants depending only on the masses.

For a binary star viewed far from its centre of mass, let $\mathbf{r}$ be the separation and $\mu$ the reduced mass, $M=M_1+M_2$ the total mass, $r=|\mathbf{r}|$ and $x=|\mathbf{x}|$. The Taylor expansion of the point-mass Newtonian gravitational potential has zero dipole because ${\sum }_A{M}_A{\mathbf{x}}_A=0$. Its second moment is ${\sum }_A{M}_A{x}_{Ai}{x}_{Aj}=\mu r_i{r}_j$, soThis holds for an instantaneous separation in any orbit, not only a circular orbit. The mass quadrupole moment convention $Q_{ij}=\mu (r_i{r}_j-r^2{\delta }_{ij}/3)$ gives ${\Phi }_2=-3G{Q}_{ij}{x}_i{x}_j/(2x^5)$.

A radial Newtonian gravitational potential $\Phi =-A{r}^{-\beta }$ produces inward acceleration $A\beta r^{-\beta -1}$. The point-mass case has $\beta =1$ and $A=GM$. Treating the power-law gravitational potential as prescribed does not require it to be sourced by the flowing gas.

For a flat background dominated by one constant-$w$ barotropic equation of state, assume $\delta P=w\delta \rho$ and no scalar anisotropic stress. The Einstein field equations then reduce the common Newtonian-gauge potential to the stated Fourier transform equation, with $\mathcal{H}=2/[(1+3w)\tau ]$. For radiation the regular solution is $3{\Phi }_0(\sin x-x\cos x)/x^3$, $x=k\tau /\sqrt{3}$, and oscillates with a decaying envelope after sound-horizon entry. For pressureless matter the two solutions are a constant and ${\tau }^{-5}$. A non-adiabatic pressure perturbation would supply an additional source, so constant background $w$ alone does not justify the homogeneous equation.

For a polynomial with nonzero constant coefficient over a complete discretely valued field, plot coefficient exponents against their valuations. A lower Newton polygon segment of slope $s$ and horizontal length $r$ accounts for exactly $r$ roots of valuation $-s$, counted with multiplicity in an algebraic closure. Zero coefficients are omitted. Reflecting the horizontal axis changes the sign of every slope, so slope conventions must be fixed before reading root valuations.

For a degree-$d$ polynomial, reflecting the usual coefficient-index axis replaces $(j,v(a_j))$ by $(d-j,v(a_j))$. The reflected lower polygon has slopes equal to root valuations, rather than their negatives. Horizontal lengths, and therefore multiplicities, are unchanged. This convention makes positive lower ramification groups breaks appear as positive slopes of a reduced ramification polygon.

In a neutral counterion-only cylindrical cell, send the outer radius to infinity at fixed cylinder charge. Below the Manning parameter threshold, number density tends to zero at each fixed radius. Above threshold, the cylindrical counterion-only Poisson-Boltzmann profile retains line charge $\lambda -q/{\ell }_B$ locally, while the rest recedes to infinity. Therefore the limit of the full finite-cell charge integral is not the charge integral of the pointwise limiting density.

For equal phase thermal properties, a roof undercooling $\Delta T$ and liquid superheat $\Theta$, the planar Stefan problem has interface $a=2\lambda \sqrt{\kappa t}$. The solid error function profile and liquid complementary-error-function profile give $S\lambda =e^{-{\lambda }^2}[1/\mathrm{erf}\lambda -(\Theta /\Delta T)/\mathrm{erfc}\lambda ]/\sqrt{\pi }$. Here $S=L/(c_p\Delta T)$ is the latent-to-sensible heat ratio. Liquid heat supplied to the interface reduces crust growth.

For an object of the nested partial unary operation category ${\mathcal{C}}_n$, adjoining the next operation freely gives one new value at each defined fixed point of the previous operation. Each new value starts an infinite free chain under the first unary operation; the other old operations are undefined on those new points. The new top operation consequently has no fixed points, so subsequent free extensions add no points. This gives a left adjoint to forgetting the last operation; extensions of maps along the chains are forced by the first target operation.

For a linear spin-wave approximation, the ground state is empty of the diagonal quasiparticles but contains the original spin bosons. If $a_k=u_k{\alpha }_k-v_k{\alpha }_{-k}^{†}$, their occupation number is $v_k^2$. The staggered magnetization is reduced from $S$ to $S-N^{-1}{\sum }_k{v}_k^2$. For the one-dimensional nearest-neighbor Heisenberg antiferromagnet, $v_k^2=(|\sin k{|}^{-1}-1)/2$ and the integral diverges logarithmically. This invalidates finite ordered magnetization within the approximation; it is not itself a determination of the exact spectral gap.

The implication introduction rule discharges a temporary assumption $A$ from a derivation of $B$, producing $A\to B$. Under the Curry-Howard correspondence, a body $t:B$ under $a:A$ gives the lambda abstraction $\lambda a^A.t:A\to B$. Unused assumptions may also be discharged.

From $A\to B$ and $A$, implication elimination derives $B$. Under the Curry-Howard correspondence, this is application: from lambda terms $f:A\to B$ and $a:A$ form $fa:B$.

The rule with premises $C,D$ and conclusion $C$ retains the first derivation while allowing an unused second assumption. Under the Curry-Howard correspondence, the output term is the first premise term, with the enlarged context. This allows a lambda abstraction whose bound variable does not occur in its body.

The Lemke-Howson algorithm follows an almost completely labeled path in the product of two best-response polytopes of a bimatrix game. Drop one label at the artificial zero pair, then alternately pivot away the duplicated label until the dropped label returns. A nonzero completely labeled endpoint, after normalization, gives a Nash equilibrium. The two players' vectors must be kept separate even in a symmetric game.

For a symmetric bimatrix game with payoff matrices $P,P^{\mathsf{T}}$, any nonzero $x$ satisfying these equations gives the symmetric Nash equilibrium $(x/{\sum }_i{x}_i,x/{\sum }_i{x}_i)$. Every positive coordinate attains the same maximal payoff. This one-vector construction differs from a two-vector Lemke-Howson algorithm path, which can return asymmetric equilibria.

A symmetric equilibrium of a symmetric game has every player use the same strategy or the same distribution of strategies. Symmetry between players does not require identical behavior across contests with different prizes.

A stagnant mushy-layer model has no bulk advection, equal phase mass densities, pure solid, and local equilibrium on the liquidus. With liquid fraction $\varphi$, specific enthalpy $E=c_{p}T+L\varphi$ and bulk solute concentration $B=\varphi C$, neglecting solute diffusion inside the mushy layer gives $T_t+(L/c_p){\varphi }_t=\kappa T_{zz}$ and $(\varphi C{)}_t=0$. The latter fixes composition at each stationary material level; it does not automatically set every level to the initial liquid concentration when solute can diffuse ahead of the growing mush.

A multiplicative subset is saturated if every factor of a product in the subset also lies in it. We include $1$ in a multiplicative subset. A saturated subset containing $0$ is the whole ring. The units form a saturated subset in a commutative ring.