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For $\alpha >0$, the Lévy process with characteristic function $\mathbb{E}{e}^{iu{X}_t}=(1+u^2/{\alpha }^2{)}^{-t}$ has zero Gaussian coefficient, zero uncompensated drift, and the displayed Lévy measure. Indeed, the symmetric jump integral is $2{\int }_0^{\infty }(\cos (uy)-1)e^{-\alpha y}dy/y$. Differentiating in $u$ gives $-2u/({\alpha }^2+u^2)$, and integrating back from zero gives $-\log (1+u^2/{\alpha }^2)$. The time-one law is the Laplace distribution of rate $\alpha$. The jump activity is infinite, since $K$ has infinite mass near zero, but the jump variation is finite because $\int (1\wedge |y|)K(dy)<\infty$.

If a Lévy measure satisfies ${\int }_{|y|\le 1}|y|K(dy)<\infty$, the jump integral in the Lévy–Khintchine formula can be written without compensation: $\mathbb{E}{e}^{iu{X}_t}=e^{t\psi (u)}$, with the displayed exponent. Its integral is absolutely convergent because $|e^{iuy}-1|\le |u||y|$ near zero and is at most $2$ elsewhere. If the compensated formula uses drift $\gamma$, the uncompensated drift is $b=\gamma -{\int }_{|y|\le 1}yK(dy)$. The Gaussian coefficient is still ${\sigma }^2\ge 0$.

In the convention $⟨0|J^{\mu }|P(p)⟩=-i\sqrt{2}{F}_P{p}^{\mu }$, the Fermi interaction gives $\mathcal{M}=G_F{F}_P{p}_{\mu }{\bar{u}}_{\ell }{\gamma }^{\mu }(1-{\gamma }_5)v_{\nu }$. With a massless antineutrino, the external Dirac equations reduce this to $G_F{F}_P{m}_{\ell }{\bar{u}}_{\ell }(1-{\gamma }_5)v_{\nu }$. The spin sum is $8G_F^2{F}_P^2{m}_{\ell }^{2}k\cdot q=4G_F^2{F}_P^2{m}_{\ell }^2(M_P^2-m_{\ell }^2)$. The two-body Lorentz-invariant phase space has integrated value $(1-m_{\ell }^2/M_P^2)/(8\pi )$, and the initial flux factor is $1/(2M_P)$, giving the displayed width. If the decay constant is instead defined without the square root of two, $f_P=\sqrt{2}{F}_P$ and the denominator is $8\pi$. A weak mixing coefficient multiplies the amplitude and its modulus squared multiplies the width.

At tree level with universal weak charged current coupling and massless neutrinos, the leptonic pseudoscalar decay width gives the displayed ratio. The pion decay constant, Fermi constant and common quark mixing coefficient cancel. The charged-lepton mass factor is helicity suppression: a spin-zero parent cannot emit the two massless helicities selected by the chiral weak interaction. The ratio tests this suppression and lepton universality; scalar or pseudoscalar four-fermion interactions can modify it. Radiative corrections must be included for a precision comparison.

Lorentz covariance allows only a multiple of $p^{\mu }$ for a vacuum-to-spin-zero one-particle current. In its rest frame only the temporal component can remain. A parity-even temporal vector current between a parity-even vacuum and parity-odd particle changes sign, forcing that component to vanish. A temporal axial current is parity-odd and is allowed. This explains why the pion decay constant enters the axial, rather than vector, part of the weak charged current.

For small categories $\mathcal{C},\mathcal{D}$, a functor $F:\mathcal{C}\to \mathcal{D}$ and $Q:\mathcal{C}\to \mathbf{Set}$, choose an infinite cardinal number $\kappa$ bounding the number of morphisms of $\mathcal{D}$ and the total number of elements of $Q$. Given a natural transformation $\eta :Q\to GF$, generate a subfunctor $G^{\prime }\subseteq G$ from all ${\eta }_c(x)$ by applying every outgoing $\mathcal{D}$-morphism. Each $G^{\prime }(d)$ has at most $\kappa$ elements. Label bounded values by subsets of $\kappa$; only a set of such functors and transformations exists. These factorizations give the solution-set condition for precomposition, whose left adjoint is the left Kan extension.

For small categories and a set-valued functor $Q$, each displayed colimit exists in the Category of sets. A representative $(c,u:Fc\to d,x\in Qc)$ maps under a natural transformation $\eta :Q\to GF$ to $G(u){\eta }_c(x)$. The colimit identifications make this well-defined, proving the adjunction between left Kan extension and precomposition.

For a left-exact reflective subcategory of a Cartesian closed category, an ambient exponential $B^A$ is already fixed by the reflector when $A,B$ are fixed. Indeed maps into $B^A$ transpose to maps from $X\times A$ into $B$, and reflection gives $L(X\times A)\cong LX\times A$. Precomposition with the reflection unit is therefore bijective for maps into $B^A$, characterizing a fixed object.

Declare a global left-invariant, respectively right-invariant, frame parallel to obtain two flat affine connections on a Lie group. For left-invariant vector fields, their derivatives are as displayed and their torsion tensors are $-[X,Y]$ and $+[X,Y]$. The tensors vanish on Abelian groups; two genuinely torsionful connections are not asserted there.

The standard four-torus has trivial complex tangent bundle, so its symplectic canonical class is zero. The symplectic adjunction formula gives $F^2=2g-2$ for a smooth genus-$g$ fiber of a symplectic Lefschetz pencil. Two generic fibers meet once positively at each base point and nowhere else, giving the displayed count.

Let $X$ be a smooth projective surface over the complex numbers, and let $H$ be an ample divisor. For sufficiently large integers $m$, a general pencil in $|mH|$ is a Lefschetz pencil. Its smooth fiber represents $mH$, so the arithmetic adjunction formula on a smooth surface givesIf $m=4k$, then $m^2{H}^2/2=8k^2{H}^2$ and $m{K}_X\cdot H/2=2k{K}_X\cdot H$ are both even integers. Therefore $g$ is odd. Positivity is needed to obtain a generic Lefschetz pencil; divisibility by four is the separate parity argument. The genericity assertion here uses complex algebraic geometry, rather than an unqualified positive-characteristic version of the complex Morse local model.^[ref]

A point at which a measurable set occupies an asymptotically full fraction of small balls, as expressed above with Lebesgue measure. The Lebesgue density theorem says that almost every point of a measurable set is a density point. On the real line, the same limiting conclusion holds for arbitrary shrinking intervals containing that point, since each interval lies in a centered interval of at most twice its length.

For any circle null set and any positive sequence ${ϵ}_j$, take infinitely many countable covers with summable total lengths. Regard their arcs as a two-index array. Choose increasing finite-square cutoffs so the length sum outside each square is smaller than the next desired bound. The finite square annuli then give finite arc families with the stated small lengths. Every point belongs to some arc in every row, hence to infinitely many annular families. This handles dense and nonclosed null sets without claiming that they have finite small covers.

Let $a_n=2^{-2^n}$ and, in angular coordinates near zero, take a bounded indicator alternating between one and zero on the rings $a_{n+1}<|t|\le a_n$. Since $a_{n+1}/a_n\to 0$, the mean over $[-a_n,a_n]$ approaches the value of the outermost ring. Its even and odd subsequences approach one and zero. Thus an integrable function can fail to have a limiting centered average at a specified point, although Lebesgue differentiation theorem ensures convergence almost everywhere.

For a finite positive measure on $[0,1)$, average its mass over each half-open dyadic interval. The resulting nonnegative martingale has constant expectation $\mu ([0,1))$: a parent interval's mass is the sum of its children's masses. Its almost-sure limit is integrable but the martingale need not be uniformly integrable. The limit recovers the density of the absolutely continuous component in the Lebesgue decomposition theorem.

For bids $1,2$ and $V>2$, write $p_i$ for player $i$'s probability of bidding $1$. The two expected payoffs are $(V-1)(1-p_j)(1-p_k)$ and $(V-2)p_j{p}_k$. Their unique symmetric equilibrium uses the displayed probability. Every permutation of $(1,0,t)$, $0\le t\le 1$, is also a Nash equilibrium: the mixed bidder is indifferent, while the two pure bidders weakly prefer their prescribed bids. There are no further equilibria. Three interior probabilities must all coincide, since subtracting two indifference equations gives a nonzero positive factor times their difference. Two interior probabilities with a pure opponent are impossible, and a single interior probability requires the other two bids to differ. Thus the nonsymmetric equilibria form six line segments with six pure endpoints, and their number is uncountable.

In the leading interaction of two sedimenting spheres, put $c=3D/4$ and let $\theta$ be the angle from the upward vertical. The equations are $\dot{r}=-(\delta +2c/r)\cos \theta$ and $r\dot{\theta }=(\delta +c/r)\sin \theta$. Their ratio separates toso $r(\delta r+2c){\sin }^2\theta$ is invariant. When $\delta ,c$ have the same nonzero sign, choose labels so both are positive. For a noncollinear encounter the invariant is positive, ensuring a strictly positive closest separation. Meanwhile $\dot{z}=-\delta -c(1+{\cos }^2\theta )/r<0$, so the vertical order reverses and eventually $r\ge |z|\to \infty$. Exactly collinear contact is outside this passing argument.

If $\delta D<0$, the leading interaction of two sedimenting spheres has a formal vertical equilibrium at $r_{\ast }=-3D/(2\delta )$. Choose labels with $\delta >0,D<0$ and put sphere one above sphere two. On that invariant vertical line, $\dot{r}=-\delta +\delta r_{\ast }/r$, so both sides approach $r_{\ast }$ and the radial eigenvalue is $-\delta /r_{\ast }$. A small tilt instead grows at rate $\delta /(2r_{\ast })$: the full configuration-space equilibrium is a saddle, not an attracting node. A physically valid far-field bound pair requires $r_{\ast }\gg a_1+a_2$, or $|D|\gg (a_1+a_2)|\delta |$. Reversing both driving forces reverses the vector field, interchanging radial attraction and repulsion, consistent with kinematic reversibility of Stokes flow.

For integrable random variable $X$ and nested sub-sigma-algebras $\mathcal{H}\subseteq \mathcal{G}$, the tower property saysThe finer-information conditional mean can be averaged within each coarser group, recovering the coarser-information conditional mean. Conversely, conditioning an already $\mathcal{H}$-measurable conditional mean on $\mathcal{G}$ leaves it unchanged. Both orders therefore recover $\mathbb{E}[X\mid \mathcal{H}]$ when the information is nested.^[ref]

For the finite example in law of total expectation, exact knowledge of $X$ gives the finer mean $X$ itself. Averaging within the groups $A,B$ recovers two and eight. Forgetting the group then gives $13/2$. The nesting condition is essential; iterated conditional expectation over nonnested sigma-algebras describes why arbitrary successive conditioning cannot simply be replaced by conditioning on an intersection.

A group of order automorphisms of a totally ordered set that is closed under pointwise maximum and minimum. Products act on the right when writing $\alpha f$; then $\alpha (fg)=(\alpha f)g$. The pointwise lattice operations make it a lattice-ordered group.