A hierarchy is a finite sequence of cuts along properly embedded two-sided incompressible surfaces, without spherical components, ending in disjoint three-balls. Disks are allowed. For an ambient manifold which is not orientable the two-sided cutting surfaces need not themselves be orientable. The existence of a hierarchy does not imply that every choice of incompressible cuts terminates.

Write the Cauchy stress tensor in the local fibre frame as $\sigma =se\otimes e+h(e\otimes n+n\otimes e)+tn\otimes n$. The compatible geometry has $\nabla \theta =\kappa e$, $\nabla \cdot e=0$ and $\nabla \cdot n=-\kappa$. With zero body force, force balance becomes $e\cdot \nabla s=2\kappa h-n\cdot \nabla h$ and $n\cdot \nabla t-\kappa t=-e\cdot \nabla h-\kappa s$. For specified shear stress $h$, the first equation determines $s$ along fibres, then the second determines $t$ along transverse curves. Pressure and fibre tension supply these two normal constraint reactions. This is a local stress construction; boundary tractions must be chosen consistently with its integration data.

Take an initially isolated virialized self-gravitating stellar population with kinetic energy $T=|W|/2$ and Newtonian gravitational potential energy $W<0$. Instantaneous spatially uniform removal of mass, without kicks to surviving material, leaves fraction $f$ with $T^{\prime }=fT$ and $W^{\prime }=f^{2}W$. Hence $E^{\prime }=|W|f(1/2-f)$: the homologous whole-system energy is negative only for $f>1/2$. Positive whole-system energy does not forbid a bound core after selective escape, and slow mass loss follows the adiabatic mass-loss expansion law instead.

The Haldane prior has formal beta distribution parameters $(0,0)$. It is an improper prior, with infinite mass at both endpoints of $(0,1)$. For a binomial distribution likelihood with $s$ successes and $f$ failures, its posterior is $\mathrm{Beta}(s,f)$ only when $s,f>0$. A density flat in the log odds induces this kernel by the Jacobian determinant $d\log [p/(1-p)]/dp=1/[p(1-p)]$.

For $C>0$, nonzero equilibrium points satisfy $\eta =g(a)=a-C{a}^2+a^4$ and are stable exactly when $a{g}^{\prime }(a)>0$. The origin is stable for $\eta <0$. The quadratic term removes the odd symmetry and changes the local branch crossing into a transcritical bifurcation, with $a=\eta +O({\eta }^2)$. The negative quintic term bounds trajectories. There is one negative-amplitude fold, supplying a stable finite-amplitude branch and hysteresis with the stable origin. Positive-amplitude folds occur precisely when $C>3/2$: $g^{\prime }(a)=1-2Ca+4a^3$ has two positive roots then, a double positive root at $C=3/2$, and none below it. All branches and stability assignments therefore depend on $C$; a single symmetric pitchfork sketch cannot describe this family.

In the quadrant $x,y>0$, a line vortex of circulation $\Gamma$ at $(x,y)$ has three images: circulations $-\Gamma$ at $(-x,y)$ and $(x,-y)$, and $+\Gamma$ at $(-x,-y)$. Their induced velocity isIn polar coordinates, these equations give $\dot{r}=\Gamma \cot (2\theta )/(2\pi r)$ and $\dot{\theta }=-\Gamma /(4\pi r^2)$. Direct differentiation yields $\frac{d}{dt}(r\sin 2\theta )=0$, so the trajectory obeys $r\sin 2\theta =\text{constant}$.

In a finite-dimensional algebra, or a complete finite coefficient-ring algebra, primitive idempotents lift modulo a nilpotent ideal or the coefficient maximal ideal. Orthogonal decompositions of the identity lift, and equivalent primitive idempotents have isomorphic associated projective modules. Two lifts of the same idempotent modulo such an ideal are conjugate by a unit congruent to one. This is an idempotent lifting and refinement result; it does not assert that every central idempotent lifts centrally through an arbitrary nilpotent quotient.

For an axisymmetric velocity $\mathbf{u}=(0,rf(r),g(r))$ independent of $z$, write $\mathbf{B}=B_z\hat{\mathbf{z}}+\nabla \times (A\hat{\mathbf{z}})$. Ideal induction becomesThe first equation expresses magnetic flux freezing of the transverse flux function, while the second is axial stretching $B_r{g}^{\prime }$. Starting from a uniform field $B_0\hat{\mathbf{x}}$ gives $A(r,\varphi ,t)=B_{0}r\sin (\varphi -f(r)t)$ and $B_z(r,\varphi ,t)=B_{0}t{g}^{\prime }(r)\cos (\varphi -f(r)t)$. In complex azimuthal-mode notation, $\hat{A}=B_{0}r{e}^{-ift}$ and ${\hat{B}}_z=i{B}_{0}t{g}^{\prime }{e}^{-ift}$, obtained by elementary integrating factors. The transverse field has $B_r=r^{-1}{A}_{\varphi }$ and $B_{\varphi }=-A_r$, preserving zero divergence.

For an ice layer between freshwater and cold brine, set $ϵ=\sqrt{D/\kappa }$ and use solutal similarity positions $a=2{\lambda }_a\sqrt{Dt}$, $b=2{\lambda }_b\sqrt{Dt}$. A linear ice temperature, two Stefan conditions and interfacial salt conservation determine both negative interface constants. As $ϵ\to 0$ with fixed latent-to-sensible heat ratio $S$, ${\lambda }_b=O(1)$ but ${\lambda }_a=O({ϵ}^{-1})$. Freezing above occurs over the thermal diffusion scale, while dilution-driven melting below has the much shorter solutal scale.

For independent exponential distributions with rates ${\lambda }_1>{\lambda }_2>0$, let $T=S_1+S_2$. Dividing the joint density of $(S_1,T)$ by the hypoexponential distribution density of $T$ gives a truncated exponential distribution on $(0,t)$ with rate ${\lambda }_1-{\lambda }_2$. Its displayed conditional expectation supplies an E-step when only pair totals are observed. At equal rates the conditional split is uniform, with mean $t/2$.

// Target: probability-and-statistics.bigb

If the independent exponential random variables $T_i$ have rates $i\lambda$, thenwhere the $E_i$ are independent exponential random variables of rate $\lambda$. To see this, the first of $N$ independent exponential clocks rings at rate $N\lambda$; after it rings, the memoryless property leaves $N-1$ independent rate-$\lambda$ clocks. The successive order statistic gaps are therefore independent with rates $N\lambda ,(N-1)\lambda ,\dots ,\lambda$. Their sum has the same distribution as $\sum T_i$. In particular, for $x>0$, the tail at $Nx$ has logarithm $-\lambda Nx+o(N)$, since $1-(1-e^{-\lambda Nx}{)}^N\sim N{e}^{-\lambda Nx}$. The sum divided by $N$ has a large deviation principle with speed $N$ and rate function $\lambda x$ on $x\ge 0$, infinity otherwise.

For a smooth integral projective variety over $\mathbb{C}$, a general hyperplane section in a projective embedding is smooth; if its dimension is at least two, the section is also irreducible. More general statements concern linear systems and their base loci. The smoothness statement is developed in the Stacks Project's Bertini section.

The universal countable uniform hypergraph is the unique, up to hypergraph isomorphism, countably infinite $r$-uniform hypergraph with the hypergraph extension property. To prove uniqueness, enumerate two such structures and construct finite partial hypergraph isomorphisms. Alternately add the least unused vertex of the first enumeration to the domain and the least unused vertex of the second to the range. Prescribe its incidences with every $(r-1)$-subset of the already matched vertices and apply the extension property in the other structure. The resulting union is a bijection preserving every hyperedge and nonedge. Existence follows from the countable random uniform hypergraph construction. A construction with only the forward steps embeds every countable $r$-uniform hypergraph as an induced subhypergraph, explaining the word universal. Its isomorphism type is independent of $p\in (0,1)$.

Given an $r$-uniform hypergraph $H$ and $K$ colours, construct an $(r+1)$-uniform auxiliary hypergraph with one task vertex for every $e\in E(H)$ and one resource vertex $(v,c)$ for every original vertex and colour. The assignment edge for $(e,c)$ contains the task $e$ and all resources $(v,c)$ with $v\in e$. A matching in a hypergraph of assignments is exactly a proper partial edge colouring. Task degrees are $K$; resource degrees are $d_H(v)$. Pairwise auxiliary hypergraph codegrees are at most $\max (1,{\max }_{v\ne w}{d}_H(v,w))$. Thus nearly equal original degrees and small original hypergraph codegrees transfer to the setting of a Rödl nibble. A locally balanced Rödl nibble must control the unmatched task count in every original star, not merely the total uncovered proportion.

Suppose a finite population is split into $m$ categories with fixed sizes $c_1,\dots ,c_m$ summing to $n$. A uniform size-$r$ sample drawn by sampling without replacement has category counts $K_1,\dots ,K_m$ with the displayed joint probability mass function, for integers $0\le k_i\le c_i$ summing to $r$. It extends the two-category hypergeometric distribution; replacing the sampled objects instead gives a multinomial distribution.

For a box with category sizes $(5,3,2)$ and a sample of four distinct objects, the count pattern $(2,1,1)$ has probabilityPut $p_i=c_i/n$. When $n>1$, its covariance matrix is the corresponding multinomial distribution covariance matrix multiplied by the squared finite population correction:This connects the fixed-total dependence between categories with the further reduction from sampling without replacement. X. G. Duan gives a symmetry-based explanation of this shared covariance structure.^[ref]

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

For complete finite-volume hyperbolic space manifolds of dimension at least three, an isomorphism of fundamental groups is induced, up to the usual basepoint-path choice, by a unique hyperbolic isometry. The finite-volume noncompact case includes the Prasad extension. Thus complete hyperbolic metrics on a fixed finite-volume three-manifold have no deformation moduli, unlike metrics on closed hyperbolic surfaces.

For a short arc $I$ of normalized length $\ell$, every point of its hyperbolic geodesic cap lies within Euclidean distance $C\ell$ of its center boundary point. Conversely, for any fixed $b>0$, the region with angular projection in $I$ and depth $1-|z|\le b\ell$ lies in $Q(\kappa I)$ for a fixed enlargement $\kappa$ depending only on $b$, when $I$ is sufficiently short. To verify this, rotate the midpoint to one and write the endpoints as $e^{\pm i\alpha }$. The cap inequality is $2\mathrm{Re}z>(1+|z{|}^2)\cos \alpha$. Its radial depth at the midpoint is $1-\sec \alpha +\tan \alpha$, comparable to $\alpha$ for small $\alpha$; its angular projection lies in $I$. Enlarging $\alpha$ by a sufficiently large fixed factor makes the inequality hold throughout the indicated rectangle. Large arcs use a finite cover of the disc. Thus Carleson boxes and hyperbolic geodesic caps define equivalent Carleson measure conditions.

The rectangle containment can also be checked directly. Write $z=r{e}^{i\theta }$ with $|\theta |\le \pi \ell$ and $1-r\le b\ell$. Its membership in the enlarged cap is equivalent toFor small enough $\ell$, $r\ge 1/2$. The right side is at most $b^2{\ell }^2$, while the left side is at least $({\kappa }^2-1){\pi }^2{\ell }^2/8$, using the sine product formula and $\sin x\ge x/2$ for small positive $x$. Thus $\kappa =4+4b/\pi$ suffices.

The inverse hyperbolic functions invert the hyperbolic functions, with real branches on their standard ranges and complex branches obtained through analytic continuation. For example, $\mathrm{arsinh}z=\log (z+\sqrt{z^2+1})$ on a chosen branch; the multivalued inverse hyperbolic sine records all inverse values.

A hyperbolic angle $t$ is twice the signed area $A$ enclosed by the two rays from the origin to $(1,0)$ and $(\cosh t,\sinh t)$ and the intervening arc of the unit hyperbola. Counterclockwise traversal is positive: $t>0$ gives an ordinary area $t/2$ above the horizontal axis, and $t<0$ gives signed area $t/2$ below it. This parallels the unit circle, whose sector has area $\theta /2$ for an angle $\theta$ measured in radians.

To check the area interpretation, write $x(u)=\cosh u$ and $y(u)=\sinh u$. Green's theorem expresses the signed area as a line integral; the straight radial edges contribute zero, leavingThus $t$ is an area parameter, rather than the ordinary angle of the ray. For $t>0$, that ordinary angle is $\arctan (\tanh t)$; it stays below $\pi /4$ even as the hyperbolic angle grows without bound.