Let $f$ be probability density per unit solid angle, with ${\int }_{S^2}fd\Omega =1$ and isotropic initial condition $f(\mathbf{p},0)=1/(4\pi )$. In still fluid, bottom-heavy spherical-cell orientation dynamics gives $\dot{\mathbf{p}}=B^{-1}[\mathbf{k}-(\mathbf{k}\cdot \mathbf{p})\mathbf{p}]$. Adding isotropic rotational diffusion gives the gyrotactic orientation Fokker-Planck equationThe derivatives are tangential to the unit sphere. Rotational symmetry about $\mathbf{k}$ is preserved by both the equation and the initial condition, so $f$ depends only on $\theta ,t$. Since $\dot{\theta }=-\sin \theta /B$, the polar probability flux is $j_{\theta }=-\sin \theta f/B-D_r{f}_{\theta }$. Expanding ${\partial }_{t}f+{\sin }^{-1}\theta {\partial }_{\theta }(\sin \theta j_{\theta })=0$ yieldsRegular solutions have no probability flux through the poles. The mean swimming velocity is ${\mathbf{V}}_c=V_s⟨\mathbf{p}⟩$, with only a vertical component by symmetry.

The linear damped harmonic oscillator has characteristic exponents $-ϵ\pm i\sqrt{1-{ϵ}^2}$. Imposing both initial conditions gives, for $|ϵ|<1$,For completeness, at $|ϵ|=1$ its continuous limiting form is $t{e}^{-ϵt}$; for $|ϵ|>1$ replace $\sin (\omega t)/\omega$ by $\sinh (\sqrt{{ϵ}^2-1}t)/\sqrt{{ϵ}^2-1}$. The initial derivative is one, as the original PDF specifies; the converted TeX loses this derivative and corrupts the time derivatives in both oscillator equations.

The separate nonlinear oscillator has quadratic drag. Introduce the slow time $T=ϵt$ and use the method of multiple scales,The order-$ϵ$ equation isFor positive amplitude, write $\vartheta =t+\varphi$. Its right side is $-2A_T\cos \vartheta +2A{\varphi }_T\sin \vartheta -A^2|\cos \vartheta |\cos \vartheta$. The cosine coefficient of the last angular function iswhile its sine coefficient is zero. The solvability condition in the method of multiple scales removes the resonant harmonics, givingThe leading initial conditions select $A(0)=1$, $\varphi (0)=0$. Therefore the quadratically damped oscillator hasThe same sign follows from the exact energy balance $d[({\dot{x}}^2+x^2)/2]/dt=-ϵ|\dot{x}{|}^3$. Averaging $|\cos \vartheta {|}^3$ gives $4/(3\pi )$, consistent with the amplitude equation.

For $ϵ>0$, the amplitude decays algebraically and stays at most one. On each fixed interval $0\le ϵt\le T_{\ast }$, the omitted displacement correction is $O(ϵ)$, so this approximation is valid through times of order ${ϵ}^{-1}$. The effective drag strength $ϵA$ decreases rather than increases, and the envelope predicts $A\sim 3\pi /(4ϵt)$ at later times. There is no finite-time pole in the positive-drag approximation. A uniform error assertion for arbitrarily long intervals is stronger than the finite-slow-time expansion just derived; the stated $O(ϵ)$ estimate applies to the specified $ϵt=O(1)$ range.

For $ϵ<0$, write $\delta =1-4|ϵ|t/(3\pi )$. The amplitude grows as $1/\delta$, with a formal pole atOn fixed slow-time intervals strictly before that pole, the displacement error remains $O(|ϵ|)$. This estimate is not uniform as $\delta \to 0$: nonresonant corrections have size comparable to $|ϵ|A^2$, and the required weak-drag condition is $|ϵ|A=|ϵ|/\delta \ll 1$. Thus the approximation necessarily fails when $\delta =O(|ϵ|)$, corresponding to a distance of order one in fast time from the formal pole, and may require additional error control when approaching it. The plot therefore stops short of the pole. Its position is a prediction of the leading amplitude equation, not an exact blow-up time for the original nonlinear equation; continuing the formula past it is invalid.

For a slender viscous bubble, approximate the interface normal by the radial direction and its curvature by $1/a$. Axial curvature and axial derivatives of the leading external motion are smaller by the slenderness factors. The exterior leading Stokes flow is locally radial. Incompressibility gives ${\partial }_r(r{u}_r)=0$, while the kinematic boundary condition gives $u_r(a)=a_t$ to leading order, henceThe radial viscous Laplacian of this $1/r$ velocity vanishes, so exterior pressure is constant to leading order; set its reference value to zero. Its radial normal stress at the interface is $2\mu {\partial }_r{u}_r{|}_a=-2\mu a_t/a$. Inner viscous normal stress is smaller by the viscosity ratio $\lambda$, and the bubble's leading normal stress is $-P$. The interfacial stress balance with variable surface tension, here with constant $\gamma$, consequently givesThese are local leading-order relations: the outer axial flow and tangential traction enter at higher slenderness order. The sign also has a useful check: a uniform cylinder with no imposed excess pressure contracts, $a_t=-\gamma /(2\mu )$, under surface tension.

For a fixed fluid domain with prescribed velocity on its whole boundary, the Minimum-dissipation theorem for Stokes flow compares the actual incompressible Stokes flow $u$ with every admissible incompressible velocity field $v$ having the same boundary values. The trial fields need not themselves satisfy the momentum equation. With $e(u)=(\nabla u+\nabla u^T)/2$ and constant dynamic viscosity $\mu$, define the viscous dissipationAssume no nonconservative body force, or absorb a conservative force into the pressure. Set $w=v-u$, so $\nabla \cdot w=0$ and $w=0$ on the boundary. The Stokes stress tensor $\sigma =-pI+2\mu e(u)$ obeys $\nabla \cdot \sigma =0$. Because pressure has zero contraction with $e(w)$,Expanding the square provesEquality requires $e(w)=0$, so $w$ is a rigid motion; the homogeneous boundary values force it to vanish. The theorem concerns fixed boundary velocities, not fixed applied forces or arbitrary comparisons between different fluid domains.

For the concentric-sphere motion, seek $u=F(r)\mathbf{\Omega }\times x$. Incompressibility is automatic. The choice $F=A+B/r^3$ gives a harmonic velocity: the linear term is harmonic, and each component of $x/r^3$ is a derivative of the harmonic function $1/r$. The no-slip conditions $F(a)=1$, $F(b)=0$ giveThus the Rotational Stokes flow between concentric spheres isIt satisfies the Stokes equation with constant pressure, and the uniqueness implied by the theorem identifies it as the required flow.

In the Unscaled Papkovich–Neuber representation, take $\Phi =-u/2$ and $\chi =0$. Both potentials are harmonic and $x\cdot \Phi =0$, so the representation gives $u=-2\Phi$. Alsoand therefore $p=0$, after choosing the arbitrary pressure constant.

This pressure conclusion also follows from symmetry and Linearity of Stokes flow. Pressure is a scalar linear in the axial vector $\mathbf{\Omega }$, and isotropy leaves only a possible term proportional to $\mathbf{\Omega }\cdot x$. That is a pseudoscalar, forbidden by reflection symmetry of the concentric spherical domain. Hence the pressure perturbation vanishes. Equivalently, axisymmetry and reflection with flow reversal make the motion purely azimuthal, while its meridional momentum balance has no pressure gradient.

Let $v=\mathbf{\Omega }\times x$ in the following stress calculation. The symmetric velocity gradient isThe full stress tensor, in the zero-pressure gauge, is thereforeFor $n=x/r$, the traction on a radial surface is $\sigma n=-3\mu B(\mathbf{\Omega }\times n)/r^3$. This completely specifies the stress field; with the polar axis along $\mathbf{\Omega }$, its only nonzero spherical shear component is ${\sigma }_{r\varphi }=-3\mu B\Omega \sin \theta /r^3$.

The fluid couple on the inner sphere is the integral of $x\times \sigma n$ with the normal pointing from the solid into the fluid. Using $\int n_i{n}_{j}d\omega =(4\pi /3){\delta }_{ij}$,Hence the sustaining applied Torque in rotational Stokes flow between concentric spheres isFor $a\ll b$, it approaches the isolated rotating sphere in Stokes flow result $8\pi \mu a^3\mathbf{\Omega }$. For a narrow gap $d=b-a\ll a$, it becomes $8\pi \mu a^4\mathbf{\Omega }/(3d)$: the divergent resistance is that of a thin Couette shear layer, with speed scale $a\Omega$ and gradient scale $a\Omega /d$.

Now add the force-free, couple-free rigid particles and denote the resulting velocity by $u_p$. Extend $u_p$ through each particle as its rigid translational and rotational velocity. No slip makes this extension continuous across every particle surface; it is incompressible and has zero rate-of-strain tensor inside each particle. It therefore gives an admissible trial velocity in the original particle-free annulus, with exactly the same sphere boundary velocities. Its full-annulus dissipation is the actual fluid dissipation ${\mathcal{D}}_p$, since the added interior integrals are zero. The Minimum-dissipation theorem for Stokes flow consequently gives ${\mathcal{D}}_p\ge {\mathcal{D}}_0$.

The boundary-work identity follows by integrating $\nabla \cdot (\sigma u)$: dissipation equals the mechanical power supplied at all rigid boundaries. The outer sphere is stationary; the inner sphere has no translation, so any force needed to hold its centre does no work; and each added particle contributes zero power because its net force and couple are zero. Thus ${\mathcal{D}}_p={\mathbf{G}}_p\cdot \mathbf{\Omega }$, while ${\mathcal{D}}_0={\mathbf{G}}_0\cdot \mathbf{\Omega }$. ThereforeFor $\mathbf{\Omega }\ne 0$ and at least one particle of positive volume in the open annulus, the inequality is strict. Equality would make the extended velocity identical to the particle-free solution, yet that solution has nonzero strain on every open region away from the rotation axis and cannot be rigid throughout any particle interior. Hence the applied couple component along the imposed angular velocity increases. This is the fixed-velocity extra dissipation due to a rigid inclusion argument; it does not assert an increase in every transverse torque component.

In the Boussinesq approximation, density departures from a constant reference density ${\rho }_0$ are small enough to neglect in inertia and mass continuity, but their gravitational effect is retained. Write the total buoyancy acceleration as $b=-g(\rho -{\rho }_0)/{\rho }_0$, choose a hydrostatic background $\bar{b}(z)$, and set $\sigma =b-\bar{b}$. Thus $\sigma$ is the buoyancy perturbation, positive for a parcel lighter than the background at its present height. The squared buoyancy frequency isStable stratification has $N^2>0$. A small vertical displacement $\xi$ of a parcel conserving its buoyancy gives $\sigma =-N^2\xi$, explaining the restoring acceleration and the name of $N$.

Let $P=p^{\prime }/{\rho }_0$ be pressure with the background hydrostatic pressure removed, $\mathbf{u}=(u,v,w)$, and $D/Dt={\partial }_t+\mathbf{u}\cdot \nabla$ the material derivative. The ideal, nonrotating Boussinesq equations areThe last equation is equivalent to $Db/Dt=0$. There is neither viscosity nor buoyancy diffusion. The reference density occurs in the pressure normalization, while the small density anomaly occurs only through buoyancy.

For the two-dimensional disturbance, linearization about rest givesA plane internal gravity wave proportional to $e^{i(kx+mz-\omega t)}$ therefore satisfiesThese internal-wave polarization relations and the dispersion relation determine the individual responses in parts i and ii. A radiating response has no incoming internal gravity wave from $x=+\infty$; an evanescent response is bounded there. The harmonic solutions describe the response after excluding arbitrary initial transients.

For the localized oscillatory boundary, define $\alpha =\omega /\sqrt{N^2-{\omega }^2}$ and $I(z)=ϵ{\int }_0^{\infty }{\hat{u}}_0(m)e^{imz}dm=f(z)+ig(z)$. Both signs of vertical wavenumber must radiate into $x>0$. Their horizontal wavenumber is $k=\alpha |m|$, so the positive-$m$ contribution involves $Z_1=z+\alpha x$ and the negative-$m$ contribution involves $Z_2=z-\alpha x$. Consequently the complex horizontal velocity isTaking its real part provesThe positive-$m$ internal-wave polarization has $w=-\alpha u$ and the negative-$m$ one has $w=+\alpha u$. HenceIn particular, $2f(z)=ϵu_0(z)$, so the boundary condition is recovered exactly. This is two-beam radiation from a localized oscillating boundary.

Take a smooth localized profile, for example $u_0(z)=e^{-z^2/H^2}$. Its effective support is a neighbourhood of width a few $H$; increasing that neighbourhood makes the tails arbitrarily small. The boundary displacement is $ϵu_0(z)\sin \omega t/\omega$ to first order, so at $t=0$ it is flat and moving at maximum speed. At that instant,There are two narrow oblique beams centered on $z=-\alpha x$ and $z=+\alpha x$. Away from their overlap, the velocity is parallel to the relevant beam, in the directions $(1,-\alpha )$ and $(1,+\alpha )$. The figure uses $\alpha =3/4$ and measures velocities in units of $ϵ$.

At other phases the $g$ terms contribute. For the positive-frequency convention above, $g$ is the Hilbert transform of $f$; it generally has longer tails than $f$. Thus even when the boundary is nearly stationary, the interior need not be stationary or confined to the bright cores shown here. Half a period later all velocities reverse. The beam directions remain the same, but the instantaneous velocity profile and its quadrature component change.

A short pulse has a broad frequency spectrum. Frequencies below $N$ produce outgoing internal gravity wave packets with frequency-dependent ray slopes $\pm \omega /\sqrt{N^2-{\omega }^2}$; frequencies above $N$ contribute an evanescent near-boundary response. Moreover the group velocity depends on wavelength as well as direction. The radiated packet therefore spreads into a dispersive fan or wake, with different frequencies traveling at different angles and different scales separating along those directions. After the forcing ends the propagating packet travels away, rather than maintaining the two permanently driven monochromatic beams.

Assume global four-dimensional N=1 supersymmetry with canonical kinetic terms and no explicit gaugino mass. Write $W_i={\partial }_{i}W$ and $W_{ij}={\partial }_i{\partial }_{j}W$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated auxiliary field in the component convention of Solution 2. The real D-terms are $D^a=-g^a({\bar{\varphi }}_i(T^a{)}^i{}_j{\varphi }^j+{\xi }^a)$, with Hermitian gauge generators. A constant Fayet–Iliopoulos term is permitted only for an appropriate Abelian factor. The scalar potential isDifferentiate with respect to ${\varphi }^i$, holding $\bar{\varphi }$ independent:Thus stationary point conditions gives the first requested identity. Gauge invariance of the superpotential gives $W_i(T^a{)}^i{}_j{\varphi }^j=0$. Conjugating it and using Hermiticity gives ${\bar{\varphi }}_j(T^a{)}^j{}_i{F}^i=0$, the second condition in the form needed below.

At the vacuum, define $H_{ij}=⟨W_{ij}⟩$ and $G_{ia}=g^a⟨{\bar{\varphi }}_j⟩(T^a{)}^j{}_i$. The two conditions combine asHere the upper equation is stationarity and the lower equation is conjugated gauge invariance. The chiral-superfield fermion mass matrix contributes $-\frac{1}{2}{H}_{ij}{\psi }^i{\psi }^j$, and the gauge Yukawa interaction contributes the mixed chiral-fermion/gaugino bilinear. With canonical Weyl fermion and gaugino kinetic terms, and choosing the gaugino phase to put the mixed bilinear in real-sign convention, these terms areThe factor $\sqrt{2}$ is important: the matrix directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled gauginos, rather than literally the canonical matrix. In detail, for $S=\mathrm{diag}(I,\sqrt{2}I)$,Since supersymmetry breaking makes this vector nonzero, there is at least one zero mass eigenvalue. A normalized massless fermion isWith the consistent gaugino phase, the inhomogeneous vacuum supersymmetry transformation of $(\psi ,\lambda )$ is proportional to the same vector $(F,D/\sqrt{2})$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the Goldstino, the Goldstone fermion of broken global supersymmetry. The argument guarantees at least one Goldstino zero mode, not the absence of other massless fermions.

For the second calculation, first take only canonical chiral superfields and F-term breaking. At a stationary vacuum $v$, let $h=\varphi -v$, $m_{ij}=W_{ij}(v)$ andThese follow by differentiating $V={\sum }_k|W_k{|}^2$ twice. The quadratic scalar potential isThe doubled scalar mass matrix has blocksTransforming to the $2N$ canonically normalized real components of $h$ preserves its trace, so the sum of real scalar mass squares is $\mathrm{tr}{M}_s^2=2\mathrm{tr}A$. The Weyl fermion mass squares are the eigenvalues of $m^{†}m=A$ and their sum is $\mathrm{tr}A$. Each Weyl mass eigenstate has two physical spin states. The printed supertrace has scalar sign $-1$ and fermion weight $+2$, henceThe frequently used boson-minus-fermion definition has the opposite overall sign and gives the same zero. The mixing block $B$ can split scalar masses but cannot change this trace. This proves the tree-level supertrace mass sum rule; it uses canonical kinetic terms and a renormalizable global theory, rather than a general supergravity or noncanonical effective theory.

In the MSSM, trying to make all squarks and sleptons heavy through direct canonical tree-level F-term breaking encounters this constraint. In an unbroken electric-charge and color sector, the scalar mass-squared trace is twice the trace for its ordinary fermionic partners. More explicitly, take a unit vector $u$ minimizing $u^{†}Au$. For $h=e^{i\alpha }u$, choose $\alpha$ so that $\mathrm{Re}(e^{2i\alpha }{u}^{T}Bu)\le 0$. The curvature along this real scalar direction is then at most $u^{†}Au$, so at least one scalar mass square is no larger than the smallest fermion mass square in that sector. For one isolated pair the scalar masses are $m_f^2\pm |B|$. Thus light quarks can force an unacceptably light squark, or a tachyon, under these assumptions. A hidden supersymmetry-breaking sector with mediation produces effective soft supersymmetry breaking; radiative effects, noncanonical kinetic terms, or supergravity can change the assumptions and allow a viable spectrum. The trace over an arbitrary enlarged theory alone would not locate the light state in the visible sector; the MSSM conclusion uses the conserved-charge block structure.

Write $k={\mathbb{F}}_p$. The Nottingham group $\mathcal{N}$ is the set of formal series $f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i$ over $k$, with ordinary composition $f\circ g$ as multiplication and the $t$-adic topology. The identity is $t$. A compositional inverse exists uniquely by solving its coefficients successively, since the linear coefficient is $1$.

Equivalently $\mathcal{N}$ is the group of continuous $k$-automorphisms of $k((t))$ whose action on $(t)/(t^2)$ is the identity. Such an automorphism is determined by the image of $t$, which is $t+O(t^2)$. To identify this with ordinary series composition without reversing multiplication, send $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$. Its composite with the automorphism for $g$ is the one for $f\circ g$.

The depth filtration isEach ${\mathcal{N}}_m$ is open and normal, and the coefficient of $t^{m+1}$ identifies ${\mathcal{N}}_m/{\mathcal{N}}_{m+1}$ with $(k,+)$. Thus $[\mathcal{N}:{\mathcal{N}}_m]=p^{m-1}$, and $\mathcal{N}$ is the inverse limit of its finite p-group truncations.

Here is a local-field proof that every finite p-group embeds in $\mathcal{N}$. We first establish the realization step rather than assume the Nottingham embedding itself. Put $F=k((t))$ and $\Gamma =\mathrm{Gal}(F^{\mathrm{sep}}/F)$, and let $I⊴\Gamma$ be the inertia subgroup, the group homomorphism kernel of the action on the algebraic closure of the residue field. We use the standard additive normal basis theorem consequence $H^j(\Gamma ,F^{\mathrm{sep}})=0$ for $j>0$. It follows by taking direct limits of the finite-Galois normal-basis modules, whose additive cohomology vanishes. The Artin-Schreier sequencethen givesThe classes $t^{-m}$ for positive $m$ prime to $p$ are linearly independent: a nonzero finite linear combination has a most negative valuation prime to $p$, whereas a negative valuation of $z^p-z$ is divisible by $p$. Thus there are infinitely many independent continuous characters $\Gamma \to C_p$.

Induct on the order of a finite p-group $E$. Choose a central subgroup $C\cong C_p$ and put $Q=E/C$. By induction there is an onto map $\varphi :\Gamma \to Q$ with $\varphi (I)=Q$, corresponding to a totally ramified $Q$-extension. The central extension $1\to C\to E\to Q\to 1$ is specified by a two-cocycle. Its pullback along $\varphi$ is a coboundary because $H^2(\Gamma ,{\mathbb{F}}_p)=0$. Correcting a set-theoretic section by that coboundary gives a continuous homomorphism ${\rho }_0:\Gamma \to E$ lifting $\varphi$.

We must make the lift onto on inertia, not merely onto on the whole Galois group. Let $A=I\cap \ker \varphi$. On $A$, the lift ${\rho }_0$ takes values in $C$. Characters vanishing on $A$ form a finite-dimensional space: $\Gamma /A$ has a finite subgroup $I/A\cong Q$ and procyclic quotient $\Gamma /I\cong \hat{\mathbb{Z}}$, so a character is determined by finitely many values on generators of these two pieces. The infinite-dimensional character space therefore has infinitely many distinct restrictions to $A$. Choose a character $\chi :\Gamma \to C$ whose restriction to $A$ does not cancel ${\rho }_0{|}_A$, and set $\rho (\gamma )=\chi (\gamma ){\rho }_0(\gamma )$. Centrality of $C$ makes this another homomorphism lifting $\varphi$. Its image on $A$ is nontrivial and hence all of $C$, while its image on $I$ maps onto $Q$. Therefore $\rho (I)=E$. The fixed field of $\ker \rho$ is the required totally ramified finite Galois extension $K/F$ with group $E$. This proves the realization step for every finite p-group.

Since the residue field of $K$ is still $k$, choosing a uniformizer $u$ identifies $K$ with $k((u))$. Indeed successive subtraction of its residue coefficient and division by $u$ gives each element of its valuation ring a unique convergent expansion ${\sum }_{i\ge 0}{b}_i{u}^i$ with $b_i\in k$, and fractions give the Laurent expansions. Each element of $E$ preserves the valuation and fixes $k$, so sends $u$ to $a_{1}u+a_2{u}^2+\cdots$. The linear coefficient is a homomorphism $E\to k^{\times }$. Since $E$ is a $p$-group and $|k^{\times }|=p-1$, that homomorphism is trivial. Thus the faithful Galois action lies in $\mathcal{N}(k)$, completing the embedding proof.

An infinite profinite group is hereditarily just infinite when every open subgroup is infinite and every nontrivial closed normal subgroup of each such open subgroup is open. We sketch the normal-subgroup argument, giving the depth calculation in odd characteristic explicitly. For series of depths $r,s$,where $[f,g]=f^{-1}\circ g^{-1}\circ f\circ g$. The coefficient follows by comparing $f\circ g$ and $g\circ f$ in degree $r+s+1$: their cross terms are $ab(r+1)$ and $ab(s+1)$.

Let $H$ be open and $1\ne K⊴H$ closed. Choose $m$ with ${\mathcal{N}}_m\le H$ and choose $f\in K$ of depth $r$. For every $j\ge m$ with $j\not\equiv r(\mathrm{mod}p)$, commutation with $t+b{t}^{j+1}$ supplies an element of $K$ of depth $r+j$, with any prescribed nonzero leading coefficient by varying $b$. Thus every sufficiently high layer except possibly depths congruent to $2r$ modulo $p$ occurs in $K$.

When $p$ is odd, choose one available depth $u\ge r+m$ with $u\not\equiv 2r,r(\mathrm{mod}p)$. Such a residue exists even for $p=3$. An element $h\in K$ of that depth can be commuted with $t+b{t}^{k-u+1}$ for every sufficiently large missing depth $k\equiv 2r(\mathrm{mod}p)$. The new leading coefficient contains $2u-k\equiv 2(u-r)\ne 0$, so those missing layers occur too. Hence $K$ contains an element with any chosen coefficient in every depth $k\ge M$, for some $M$.

Successive coefficient correction now proves ${\mathcal{N}}_M\le K$: match the coefficient of an arbitrary target in depth $M$ by an element of $K$, correct the error in depth $M+1$, and continue. The partial products converge $t$-adically to the target and remain in $K$, which is closed. Thus $K$ is open. This proves hereditary just infiniteness for odd $p$.

For characteristic two, the leading coefficient alone misses even depths. The standard characteristic-two normal-closure collection lemma supplies the needed extra input: for every $m\ge 1$ and every nonidentity $f\in \mathcal{N}({\mathbb{F}}_2)$, the closed subgroup generated by the conjugates $f^g$ with $g\in {\mathcal{N}}_m$ contains ${\mathcal{N}}_M$ for some $M$. Its proof retains the next coefficients in pairs of group commutators to fill the parity gaps, then uses the same successive coefficient correction. Apply this lemma to $f\in K$ and ${\mathcal{N}}_m\le H$: every such conjugate lies in $K$, so $K$ contains the tail and is open. This states the additional collection result used in the characteristic-two sketch; the odd-prime leading-term argument is not being applied there. Since every open subgroup contains a tail and is infinite, the Nottingham group is hereditarily just infinite for every prime $p$.

For a pro-p group, the lower p-series is the descending sequence of closed characteristic subgroupsThe bar means closure of the subgroup generated by the displayed powers and group commutators. Throughout this question $p$ is odd. For a finitely generated powerful pro-p group, group commutator collection gives $P_i(G)=G^{p^{i-1}}$, each $P_i$ is powerful, and $\Phi (P_i)=P_{i+1}$. All these subgroups are open. A uniform pro-p group is such a group with all subgroup indices $[P_i:P_{i+1}]$ equal to $[G:P_2]$. We use the standard powerful-group torsion criterion: a finitely generated powerful pro-p group is uniform exactly when it is a torsion-free group. For a uniform group of generator number $d$, the indices are $p^d$, and $x\mapsto x^{p^a}$ is a homeomorphism from $G$ onto $G^{p^a}$.

We record explicitly the collection facts used for the operations below. With $U_b=G^{p^b}=P_{b+1}$, one hasThe second statement is the power-congruence property: the $p$th-power map is an isomorphism on adjacent uniform layers, and collection gives $(yu{)}^p\equiv y^p{u}^p(\mathrm{mod}{U}_{b+2})$ for $u\in U_b$. Thus the first nonzero layer of a difference moves down exactly one step under a $p$th power. Iterating gives the asserted equivalence and also justifies taking roots of congruences. These facts follow from powerful-group group commutator collection, using divisibility by $p$ of $\left(\genfrac{}{}{0ex}{}{p}{2}\right)$ and the successive-layer power isomorphisms; they are the uniform-group structure results being used here.

The transported addition on a uniform pro-p group isThe product is in $U_n$ and has a unique $p^n$th root in $G$. For the comparison at $n=1$ define $x{+}_{0}y=xy$. The numbered proofs below establish associativity, approximate commutativity and compatibility of these operations.

The intrinsic Lie algebra of a uniform pro-p group has underlying set $G$, zero equal to the group identity, additive inverse $x^{-1}$, and operationsHere the group commutator is $[x,y]=x^{-1}{y}^{-1}xy$. For $\lambda \in {\mathbb{Z}}_p$, $x^{\lambda }$ is the limit of $x^{m_j}$ for integers $m_j\to \lambda$; the pro-$p$ topology makes it well defined. The addition sequence is Cauchy by the compatibility proved below. The group commutator estimate puts $[x^{p^n},y^{p^n}]$ in $U_{2n+1}$, so its displayed root exists and lies in $U_1$. Collection of group commutators shows that successive bracket approximants differ in layers tending to the identity, giving the other limit. The group commutator identities give additivity of the limiting bracket, and the Hall-Witt identity gives its Jacobi identity. Approximate commutativity makes limiting addition commutative. The resulting lattice $L_G$ is a free module over ${\mathbb{Z}}_p$ of rank $d(G)$, and $[L_G,L_G]\le p{L}_G$.

A powerful Lie algebra over p-adic integers, in the convention relevant here, is a Lie algebra $L$ whose underlying module is finite free over ${\mathbb{Z}}_p$ and whose bracket satisfies $[L,L]\le pL$. For $p=2$ the condition is instead $[L,L]\le 4L$. For odd $p$, define multiplication on its underlying set by the convergent Baker--Campbell--Hausdorff formulaAn iterated bracket of degree $m$ lies in $p^{m-1}L$. The BCH coefficient bound loses at most $\lfloor (m-1)/(p-1)\rfloor$ powers of $p$, so all higher-degree terms are integral, and their valuations tend to infinity. This explains both convergence and why multiplying a lattice with small bracket stays inside the lattice. Formal associativity of BCH passes to the convergent series, the identity is $0$, the inverse is $-x$, and every group power satisfies $x^{\ast m}=mx$. The quotients by $p^{a}L$ are finite p-groups. Hence the resulting compact group is a finitely generated torsion-free group and a pro-p group, and its group commutator lies in $pL$. Thus $(L,\ast )$ is a uniform pro-p group, with $P_i(L,\ast )=p^{i-1}L$ and successive index $p^{{\mathrm{rank}}_{{\mathbb{Z}}_p}L}$.

The standard uniform-group/Lie-lattice correspondence identifies these constructions as mutually inverse: $L_{(L,\ast )}\cong L$ and $(L_G,\ast )\cong G$. To see why the inverse limits recover the lattice operations, the addition approximant for a BCH group is $p^{-n}\mathrm{BCH}(p^{n}x,p^{n}y)$: its linear term is $x+y$ and every higher term tends to zero. Similarly its group commutator approximant has limiting term $[x,y]$. In the other direction, a local logarithm in the complete normed group algebra identifies the intrinsic operations with algebra addition and commutator and identifies group multiplication with BCH; the uniform power maps extend the identification from a sufficiently small power subgroup to all of $G$. Continuous group homomorphisms respect powers, unique roots and limits, while Lie algebra homomorphisms respect every BCH term, so the equivalence also preserves morphisms.

A topological group is a group with a topology for which multiplication $G\times G\to G$ and inversion $G\to G$ are continuous. A profinite group has either of the following equivalent descriptions: a compact Hausdorff totally disconnected topological group, or an inverse limit of finite discrete groups. Totally disconnected means that every connected component is a singleton.

For clarity, an inverse system over a directed set $I$ consists of groups $G_i$ and maps ${\varphi }_{ij}:G_j\to G_i$ for $i\le j$, with ${\varphi }_{ii}=1$ and ${\varphi }_{ik}={\varphi }_{ij}{\varphi }_{jk}$. Its inverse limit iswith coordinatewise multiplication and the subspace topology. Equivalently, it is a group with compatible continuous projections such that every other compatible family of maps into the $G_i$ factors uniquely through it.

The displayed limit is a closed subgroup of the compact Hausdorff product of the finite groups, and is totally disconnected, since two different tuples are separated by a clopen coordinate condition. Conversely a compact Hausdorff totally disconnected group has a neighbourhood basis of open normal subgroups. One way to see the group part of this standard clopen-basis fact is to choose a clopen identity neighbourhood $U$ inside a prescribed neighbourhood. Compactness of $U$ and continuity give a symmetric identity neighbourhood $V$ with $VU\subseteq U$. The stabilizer $H=\{g:gU=U\}$ contains $V$, is open, and is contained in $U$ because $1\in U$. Its index is finite by part (ii); its normal core is the intersection of finitely many conjugates and is open and normal. The canonical mapis injective because these $N$ separate points, and surjective because compatible cosets have the finite intersection property and $G$ is compact. It is a homeomorphism by compactness and the Hausdorff property. This explains the equivalence of the two definitions.

The profinite Frattini subgroup is $\Phi (G)=\bigcap M$, where $M$ ranges over all maximal proper open subgroups; for the trivial group the empty intersection is $G$. For a pro-p group,Here powers and group commutators mean generated subgroups and the bar supplies closure. Every maximal open subgroup in a pro-p group is normal of index $p$, as follows by passing to the finite p-group quotient by its normal core. Such a subgroup therefore contains all powers and group commutators. Conversely the quotient by $D=\overline{G^p[G,G]}$ is an abelian pro-p group of exponent $p$. If an element of this quotient is nontrivial, some finite elementary abelian quotient detects it; a linear functional to ${\mathbb{F}}_p$ then detects it as well. Its group homomorphism kernel pulls back to a maximal open subgroup of $G$. Thus the intersection of these kernels is exactly $D$.

We also use the topological generation criterion: a subset generates $G$ topologically if and only if its image generates $G/\Phi (G)$ topologically. To prove the nontrivial direction, a proper closed generated subgroup $K$ lies in a proper open subgroup $KN$ for a suitably small open normal $N$, and hence in a maximal open subgroup obtained from the finite quotient $G/N$. Since that maximal subgroup contains $\Phi (G)$, the image of $K$ cannot generate the Frattini quotient.

Now suppose $G$ is a nontrivial procyclic pro-p group. If it had distinct maximal open subgroups $M_1,M_2$, the homomorphism to $C_p\times C_p$ induced by the two quotient maps would be onto: its projections are onto, and a proper such subgroup would be a one-dimensional graph, forcing the two kernels to coincide. Hence $G/(M_1\cap M_2)\cong C_p^2$, contradicting procyclicity. Thus there is exactly one maximal open subgroup $M$, and $\Phi (G)=M$. Any $g\notin M$ generates $G/\Phi (G)\cong C_p$, so the criterion givesFor the trivial group use its identity as generator. The additive group ${\mathbb{Z}}_p$ is an infinite example, topologically generated by $1$: its finite continuous quotients are cyclic $\mathbb{Z}/p^n\mathbb{Z}$.

For $0<u<1$ define the generalized inverse, or quantile function, byIt is finite because the cumulative distribution function tends to zero and one at the two infinities. Monotonicity and right continuity of $F$ give the key equivalenceFor the forward implication, choose points in the defining set approaching its infimum from above. Right continuity gives $F(Q(u))\ge u$, and then monotonicity gives $F(x)\ge u$ whenever $Q(u)\le x$. The reverse implication follows immediately because $x$ then belongs to the defining set. Consequently, for a continuous uniform distribution draw $U$,This proves inverse transform sampling for discontinuous as well as continuous distribution functions.

There is a minor endpoint defect in the printed codomain. The true quantile endpoint at probability one is $Q(1)=\inf \{x:F(x)=1\}$, which equals $+\infty$ for an unbounded-above normal distribution. Thus it need not be real at one. Use the extended-real value there, or assign an arbitrary finite value at this single probability level. Since $P(U=1)=0$, either choice gives the same sampled distribution; the latter supplies a real-valued version on all of $(0,1]$. The endpoint has no effect on the inverse-transform law.

For rejection sampling, choose a proposal probability density function $g$ that can be sampled and a finite constant $M$ with $f(y)\le Mg(y)$ wherever the target has positive mass. Generate independent pairs $(Y_j,U_j)$, with $Y_j$ of density $g$ and $U_j$ uniform. Accept $Y_j$ if $U_j\le f(Y_j)/(Mg(Y_j))$; otherwise discard the pair and repeat with fresh independent draws. At points where both densities vanish the acceptance rule may be defined arbitrarily. For any measurable set $A$,Thus one trial accepts with probability $1/M$, and the conditional density given acceptance is $f$. More explicitly, summing over the index of the first accepted proposal givesThe probability of never accepting is zero, since $M<\infty$, and the expected number of proposals is $M$. The algorithm terminates almost surely and samples the target density exactly.

A frailty model represents unobserved heterogeneity in susceptibility by a positive latent frailty random variable $U$. In the proportional frailty model, conditional on $U=u$, the individual hazard is $h(t\mid u)=u{h}_0(t)$, where $h_0$ is a baseline hazard function. If $H_0(t)={\int }_0^t{h}_0(s)ds$, the conditional survivor function isLarge frailty gives earlier events on average. Normalizing $\mathbb{E}U=1$ separates the frailty scale from that of the baseline hazard.

The unheaded final request follows for an arbitrary survival mixture, not only this proportional model. Let a randomly selected individual's type be $U$, with individual survivor function $S(t\mid U)$ and hazard $h(t\mid U)$. The population survivor function is $\overline{S}(t)=\mathbb{E}[S(t\mid U)]$. Where differentiation under the expectation is justified, $S^{\prime }(t\mid U)=-h(t\mid U)S(t\mid U)$ givesThis proves the population hazard of a survival mixture formula: the weights are the individual probabilities of still surviving, normalized to sum or integrate to one. For a finite equally weighted population it becomes ${\sum }_i{h}_i(t)S_i(t)/{\sum }_i{S}_i(t)$. Generally it is not the unweighted average of the original individual hazards. In a proportional frailty model, the frailty distribution among survivors is proportional to $e^{-u{H}_0(t)}$ times the original frailty distribution.