The local Skyrme baryon density is $\mathcal{B}(x)=-{ϵ}_{ijk}\mathrm{tr}(L_i{L}_j{L}_k)/(24{\pi }^2)$, with $L_i=U^{†}{\partial }_{i}U$ and $B=\int \mathcal{B}{d}^{3}x$. It is invariant under global isorotations and transforms as a scalar under proper spatial rotations, so its contours reveal the geometric shape of a Skyrmion. For the rational map approximation for Skyrmions, $\mathcal{B}=-f^{\prime }{\sin }^{2}f{J}_R/(2{\pi }^2{r}^2)$, where $J_R$ is the angular Jacobian of a rational map. Integrating $J_R$ over the sphere gives $4\pi \mathrm{deg}R$, recovering the integer total charge. A decreasing profile and holomorphic angular map give a nonnegative density, but a general field may have regions of negative density; positivity is not a general topological theorem. The density is different from the energy density.

Under SU(2) as the three-sphere, a Skyrme model field with $U\to I$ at infinity defines $S^3\to S^3$. Choose $T_i=-i{\tau }_i$ and positive left volume form ${\theta }^1\wedge {\theta }^2\wedge {\theta }^3$. The normalized form is $-(24{\pi }^2{)}^{-1}\mathrm{tr}(U^{-1}dU{)}^3$, so the topological baryon number in the Skyrme model equals $\mathrm{deg}U$. Generator and orientation conventions fix the sign.

In the positive definite case this is a nonnegative integer vector with $q_Q(\mathbf{n})=1$, so it is nonzero. The positive definite Tits form indecomposable classification gives one indecomposable isomorphism class per positive root.

For bounded linear operators $K:U\to V$ and $B:U\to W$ between real Hilbert spaces, $\alpha >0$ and $\Vert Bu\Vert \ge \beta \Vert u\Vert$ with $\beta >0$, the unique minimizer of $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$ satisfies $(K^{\ast }K+\alpha B^{\ast }B)u_{\alpha }=K^{\ast }f$. Differentiate the objective in every direction and use the adjoint operator to obtain this equation. The coercive quadratic penalty makes its left-hand operator boundedly invertible. Setting $B=I$ gives the Tikhonov normal equation; omitting the penalty instead gives the normal equation for a linear inverse problem.

Let $K$ be injective, $u^{†}=K^{†}f$, $z={\beta }^{-2}{B}^{\ast }B{u}^{†}$, and ${\eta }_r={\inf }_{\Vert w\Vert \le r}\Vert z-K^{\ast }w\Vert$. Since $\overline{\mathcal{R}(K^{\ast })}=\mathcal{N}(K{)}^{\perp }=U$, ${\eta }_r\to 0$. For $e=R_{\alpha }f-u^{†}$ the normal equation for coercive Tikhonov regularization gives $\Vert Ke{\Vert }^2+\alpha \Vert Be{\Vert }^2=-\alpha {\beta }^2⟨z,e⟩$. Insert an approximate $K^{\ast }w$ and complete the square in $\Vert Ke\Vert$ to obtain $\Vert e{\Vert }^2\le \Vert z-K^{\ast }w\Vert \Vert e\Vert +\alpha {\beta }^2{r}^2/4$. This implies the displayed bound, even with coefficient $1/2$ on its last term, and proves consistency without a fixed exact source condition for quadratic regularization.

The same compact set must work simultaneously for every measure in the family. Prokhorov's theorem relates this condition to relative compactness for the weak topology of probability measures on a Polish space.

With $v_b=v_{\gamma }$ at leading order and $c_s^2=1/[3(1+R)]$, the photon continuity equation and total Euler equation give ${\ddot{\delta }}_{\gamma }+\dot{R}{\dot{\delta }}_{\gamma }/(1+R)+k^2{\delta }_{\gamma }/[3(1+R)]=4\ddot{\varphi }+4\dot{R}\dot{\varphi }/(1+R)-4k^2\psi /3$. The baryon loading parameter changes inertia and shifts the gravitational equilibrium. The leading equations omit diffusion damping.

For $\Theta (\mu )={\sum }_{\ell }(-i{)}^{\ell }{\Theta }_{\ell }{P}_{\ell }(\mu )$ and a Thomson collision operator omitting polarization feedback, the quadrupole relaxes with rate $(9/10){\Gamma }_T$, where ${\Gamma }_T=-{\tau }^{\prime }>0$. Balancing its leading streaming source gives ${\Theta }_2=20k{\Theta }_1/(27{\Gamma }_T)$. This differs from photon quadrupole in tight coupling with polarization; the two formulas expand different collision operators.

A Love number measures a body's tidal response to an applied potential. The degree-two potential response $k_2$ is sensitive to radial density structure and central concentration, and can complement mass and radius constraints on an exoplanet interior. Inferring it from orbital precession requires separating other contributions and assumptions about rotation and the tidal response.

Two equal galaxies each receive the tidal impulse heating of a spherical galaxy, while their relative orbital energy at infinity is $m_g{v}^2/4$. Transfer of more internal energy than that orbital energy gives $pv<[32G^2{m}_g^2⟨r^2⟩/3{]}^{1/4}$. Slow encounters require adiabatic shielding of tidal encounters, so the impulsive inequality does not apply there.

The dual polytope consists of the real linear functionals $\alpha$ on $H_2(Y,\partial Y;\mathbb{R})$ such that $|\alpha (u)|\le \Vert u{\Vert }_T$ for every $u$. Thus it is the polar of the unit ball of the Thurston norm and annihilates that seminorm's kernel.

In an irreducible three-manifold with incompressible boundary, a minimizing surface can be cut along a separating incompressible torus after removing inessential intersection circles. The essential pieces have no disk or sphere components, so their ${\chi }_{-}$ costs add. Minimizing independently in the pieces and gluing their matching boundary curves gives the corresponding upper bound. This allows Thurston norm calculations for satellite knots.

For $Y=P\times S^1$ with $P$ a pair of pants and $h$ its circle fiber, a horizontal surface in a Seifert fibered space of degree $d$ has ${\chi }_{-}=|d|$; vertical surfaces in a Seifert fibered space have zero cost. Cutting and pasting these representatives, and classifying essential surfaces, proves $\Vert u{\Vert }_T=|u\cdot h|$. The dual Thurston polytope is therefore the segment between the two functionals $\pm h$ under the natural pairing.

Take sources $k,k,k$, sink $k^2$, and arrows sending $1$ to $e_1,e_2,e_1+e_2$. An endomorphism preserving these three lines must be scalar, so the representation has endomorphism ring $k$ and is a brick module. It is the only indecomposable of the three-subspace quiver with a two-dimensional sink.

For the three-layer stratified piecewise-linear shear flow, set $\alpha =kh/2$. The dispersion relation is quadratic in squared nondimensional phase velocity. Its constant term is $[2\alpha -(1+J)(1+e^{-2\alpha })][2\alpha -(1+J)(1-e^{-2\alpha })]/(4{\alpha }^2)$. Throughout the displayed band this is negative, so one squared speed is negative and gives an exponentially growing normal mode.

For the bounded-shear three-layer profile with density contrasts minus one, zero and plus one, the localized-mode matching determinant is $[a(1-c{)}^2-(1-c)-J][a(1+c{)}^2-(1+c)-J]-b^2(1-c^2{)}^2=0$, where $a=\alpha [1+\coth (2\alpha )]$ and $b=\alpha \mathrm{csch}(2\alpha )$. Dividing by $a^2-b^2=2\alpha a$ gives a quartic in $c$ and a quadratic in $c^2$. Its constant term is $\{[2\alpha -(1+J){]}^2-e^{-4\alpha }(1+J{)}^2\}/(4{\alpha }^2)$, whose negative sign proves the unstable band of a three-layer stratified shear flow.

For a travelling wave of large positive speed in the cylindrical film equation, the leading core is proportional to $1-\cos x$ over an axial interval of length $2\pi$. Edge coordinates $X=(3c{)}^{1/3}(x-x_e)$ reduce the transition equation to $H^3{H}_{XXX}=H-1$. Its real positive exponent gives a monotone trailing approach and its complex stable exponents give damped capillary waves ahead, with physical wavelength $4\pi a/(3^{5/6}{c}^{1/3})$. Matching the two quadratic edge expansions selects the uniform film thickness; this is a matched asymptotic expansion, valid while the pulse itself remains thin relative to the cylinder.

A localized amount $M=\int Cdx$ of insoluble surfactant spreads over a planar film of depth scale $h_0$. Neglect gravity, capillarity and diffusion. Then $C\sim M/\ell$ and Marangoni stress drives $u_s\sim AM{h}_0/(\mu {\ell }^2)$, giving $\ell \sim (AM{h}_{0}t/\mu {)}^{1/3}$. The thin-film equations with insoluble surfactant admit a similarity solution in which the film height and concentration depend on $x/\ell$, conserving total surfactant and fluid volume relative to the undisturbed layer.

In the steady nondiffusive gravity–Marangoni model with capillarity neglected, zero surfactant flux and positive concentration make the surface velocity zero. With $Q=-H^2{\Gamma }_X/2-H^3{H}_X/3$ and $J=-H\Gamma {\Gamma }_X-H^2\Gamma H_X/2$, integration gives $H^4=H_0^4-48QX$ and $\Gamma =1+(H_0^2-H^2)/4$. A concentration drop $\delta$ over unit length therefore selects $Q=-\delta (H_0^2+2\delta )/6$. The liquid flows opposite to the imposed Marangoni traction, with velocity proportional to $H_{X}Y(Y-H)/2$ and an immobilized surface. This reduction is singular at a dry endpoint; its small-slope assumption does not resolve that edge.

In the same gravity–Marangoni reduction, zero liquid flux gives $\Gamma =1+(H_0^2-H^2)/3$ and $6JX=(1+H_0^2/3)(H^3-H_0^3)/3-(H^5-H_0^5)/15$. The lower part of the liquid flows backwards while the surface and upper third flow forwards. For zero initial height and concentration drop $\delta$, $J={\delta }^{3/2}(1-3\delta /5)/(2\sqrt{3})$ is largest formally at complete depletion. Since surface velocity is $J/\Gamma$, that endpoint is singular and requires physical regularization by diffusion, capillarity or an edge model. The algebraic maximum is not by itself a regular zero-concentration outflow.