Before the cut locus, define $A(X)={\nabla }_X{\partial }_r$ on the tangent space of a distance sphere. This is the negative of the shape operator with outward normal under the convention $S=-\nabla \nu$. A nonsingular matrix $Y$ of radial Jacobi fields satisfies $Y^{\prime \prime }+{\mathcal{R}}_{{\partial }_r}Y=0$ and $A=Y^{\prime }{Y}^{-1}$, yielding the displayed matrix Riccati equation. Taking its trace gives $h^{\prime }+\mathrm{tr}(A^2)+\mathrm{Ric}({\partial }_r,{\partial }_r)=0$, where $h=\mathrm{tr}A={\partial }_r\log \det Y$. This is the local differential identity behind the Bishop-Gromov inequality.

A unit-speed curve with $D_t\dot{\gamma }=f(\gamma )i\dot{\gamma }$ defines the magnetic flow. The vertical term rotates the velocity while the horizontal part transports it along the base. Since $V(f\circ \pi )=0$, the divergence of $X+fV$ in Liouville volume vanishes. The associated two-form on the unit tangent bundle is $d\alpha -f{\pi }^{\ast }{\Omega }_a=-{\iota }_F\mu$, whose kernel is the magnetic flow direction.

A smooth real function $\lambda$ on the unit tangent bundle determines this unit-speed smooth flow. Its base velocity is $v$, so its generator never vanishes. In the canonical frame its divergence in Liouville volume of a surface geodesic flow is $V\lambda$. The special case where $\lambda$ depends only on the base is a magnetic flow; when $\lambda (x,v)=⟨E(x),iv⟩$ it is a Gaussian thermostat.

A perpendicular Jacobi field on a Riemannian surface has scalar equation $\ddot{y}+Ky=0$. For strictly negative curvature on a closed manifold, the displayed derivative is uniformly positive definite, so the quadratic-form criterion for an Anosov flow makes the geodesic flow Anosov. When $K=-1$, the stable and unstable lines in the canonical frame are respectively $\mathbb{R}(H-V)$ and $\mathbb{R}(H+V)$, with factors $e^{-t}$ and $e^t$.

For $D=\{d\in R:d^2=0\}$ in the generic local ring, every internal map $D\to R$ has a unique affine expression $d\mapsto a+bd$. At the stage $A$, such maps are elements of $A[\epsilon ]/({\epsilon }^2)$, which is freely $A\oplus A\epsilon$ as an $A$-module. This decomposition is natural under ring maps and proves the isomorphism. Testing only on ordinary fields loses the generic infinitesimal and would give an incorrect uniqueness argument.

Let $I$ be the set of infinite binary sequences and $G={\mathbb{F}}_p^I$ with its product topology. Index the finite binary strings by the positive integers. For each $n$, take the element whose $r$ coordinate is one exactly when the $n$th string is a prefix of $r$. These countably many elements generate a dense subgroup: given finitely many distinct sequences, sufficiently long prefixes distinguish them individually. No finite family generates $G$, so its topological generator number is countable. Every continuous linear functional depends on finitely many coordinates, however, so $H^1(G,{\mathbb{F}}_p)$ has a basis of size $|I|=2^{{\aleph }_0}$. Thus the arbitrary-rank identity is false for unrestricted topological generating sets.

For a commutative complex unital Banach algebra, a character of an algebra annihilates $a-\varphi (a)1$, so this element cannot be invertible. Conversely a noninvertible $a-\lambda 1$ generates a proper ideal, which lies in a maximal ideal. The corresponding character then has value $\lambda$ at $a$. Thus the Gelfand transform determines every element's spectrum even when the algebra is not semisimple.

The complex dual numbers $\mathbb{C}[\epsilon ]/({\epsilon }^2)$, with norm $\Vert a+b\epsilon \Vert =|a|+|b|$ and involution $a+b\epsilon \mapsto \bar{a}+\bar{b}\epsilon$, is a commutative unital Banach algebra with an isometric involution. Its sole character of an algebra kills $\epsilon$, so the Gelfand transform is neither injective nor isometric. The C-star identity excludes this example and gives the Commutative Gelfand--Naimark theorem. Even preservation of the involution on characters can fail without that identity: on ${\mathbb{C}}^2$, the involution $(a,b{)}^{\ast }=(\bar{b},\bar{a})$ exchanges the two evaluation characters.

To match a prescribed initial instantaneous forward rate curve $F_0(T)$ in the Gaussian short-rate model with a deterministic shift, chooseSince $W_0^Q=0$, its model forward rate is then $f_0(T)=g(T)-{\sigma }^2{T}^2/2=F_0(T)$. Moreover,so the initial zero-coupon bond curve matches as well. A continuous initial forward curve gives pointwise calibration; a locally integrable curve gives the corresponding almost-everywhere interpretation of the instantaneous forward rate.

For independent standard Gaussian random variables, the partial sums have characteristic functions $\exp (-u^2{v}_n/2)$, where $v_n={\sum }_{k\le n}{a}_k^2$. A finite almost-sure limit implies convergence of these characteristic functions to one continuous at zero; $v_n\to \infty$ would instead give zero at every nonzero $u$. Conversely, bounded $v_n$ makes the partial sums an martingale bounded in L2 space, which converges almost surely and in L2 space.

For standard one-dimensional Gaussian measure, $\int e^{sx}d\gamma =e^{s^2/2}$. Thus ${\mathrm{Ent}}_{\gamma }(e^{sx})=(s^2/2)e^{s^2/2}$, while $\int |(e^{sx/2}{)}^{\prime }{|}^{2}d\gamma =(s^2/4)e^{s^2/2}$. Every nonzero $s$ therefore attains ratio two, proving optimality of the Gaussian constant.

An odd Gaussian integer is primary when it satisfies the displayed congruence. The four units have distinct images in $(\mathbb{Z}[i]/(1+i{)}^3{)}^{\times }$, a group of order four; hence every odd ideal has a unique primary generator. Indeed $v_{1+i}(-2)=2$ and $v_{1+i}(i-1)=v_{1+i}(-i-1)=1$.

For $d\ge 3$, $A_d=\Gamma (d/2-1)/(2{\pi }^{d/2})$. Substitute $u=|x{|}^2/(2s)$ to obtain the displayed identity for $x\ne 0$. The semigroup property and Tonelli theorem then give $P_t(|\cdot {|}^{2-d})(x)=A_d^{-1}{\int }_t^{\infty }{p}_s(x)ds$. The coefficient multiplying the time integral is the reciprocal of the coefficient in the full Green integral, a distinction important in Brownian motion hitting asymptotics.

For a maturity-$T$ call on a unit maturity-$U$ zero-coupon bond, where $t<T<U$, deterministic forward-rate volatilities make the bond-price ratio lognormal under the T-forward measure. Its integrated variance is $V={\int }_t^T\Vert \Sigma (s,U)-\Sigma (s,T){\Vert }^{2}ds$. For $V>0$, put $d_1=[\log (P(t,U)/(KP(t,T)))+V/2]/\sqrt{V}$ and $d_2=d_1-\sqrt{V}$. Integrating the lognormal distribution above the strike gives the displayed price. At $V=0$, the price is $(P(t,U)-KP(t,T){)}^{+}$. The forward measure is essential: the raw bond price has a stochastic discount rate under the money-market measure.

For time-to-maturity coordinates $g_t(x)=f(t,t+x)$ and time-homogeneous deterministic Heath-Jarrow-Morton model volatility, $A(x)=⟨\sigma (x),{\int }_0^x\sigma (u)du⟩$. The displayed evolution contains the maturity shift. Its mild solution is $g_t(x)=g_0(x+t)+{\int }_0^{t}A(x+s)ds+{\int }_0^t⟨\sigma (x+t-s),d{W}_s⟩$. The whole curve is a time-homogeneous Markov process under suitable function-space well-posedness conditions; the current short rate alone need not be a sufficient state.

For a centered Gaussian forward-rate field with covariance $\mathrm{Cov}(X(s,u),X(r,w))=c(s\wedge r,u,w)$, integrate the short-rate past and the current forward curve together using the displayed mean-square integral. Its variance isFor fixed $T$, its increments are independent of the full earlier forward-rate filtration: their covariance with $X(z,w)$ at $z\le r\le s$ is the integral of $c(s\wedge u\wedge z,u,w)-c(r\wedge u\wedge z,u,w)=0$. Use the uncorrelated jointly Gaussian variables are independent principle first for finite collections and then for their generated sigma-algebra. With $c(0,u,w)=0$, this centered process starts at zero.

For a normal distribution with mean $m$, positive variance $v$, and $\gamma >0$, the target $\mathbb{P}(X\ge C)\le e^{-\gamma }$ is equivalent to the displayed constraint, where $\Phi$ is the standard normal distribution function. This follows from $\mathbb{P}(X\ge C)=1-\Phi ((C-m)/\sqrt{v})$ and strict monotonicity of $\Phi$. Unlike the sufficient Chernoff bound margin $\sqrt{2\gamma v}$, this constraint is necessary as well as sufficient. The exact quantile can be negative when $e^{-\gamma }>1/2$.

For a self-consistent Gaussian angular-momentum suppression in a stellar distribution, put $y=[\psi (r)-\psi (0)]/{\sigma }^2$ and $s=\log (r/r_a)$. At large radius the Poisson equation for Newtonian gravity reduces to $y_{ss}+y_s\sim -\lambda e^y$, where $\lambda =4\pi G{\rho }_c{r}_a^2/{\sigma }^2$. The asymptotic balance $y_s\sim -1/s$ gives $e^y\sim 1/(\lambda s)$, the displayed density, and circular speed $v_c^2\sim {\sigma }^2/\log (r/r_{\ast })$. A sufficiently small regular core can have an intermediate isotropic singular isothermal sphere envelope with $\rho \simeq {\sigma }^2/(2\pi G{r}^2)$ and $v_c^2\simeq 2{\sigma }^2$. At the actual regular centre, however, $\rho \to {\rho }_c$ and $v_c\propto r$; the intermediate power law must not be mistaken for a central limit.

Choose $c_1,\dots ,c_s$ as the polynomial roots of $P_s(2c-1)$ in $(0,1)$, and form the Lagrange interpolation polynomials ${\ell }_j(t)={\prod }_{m\ne j}(t-c_m)/(c_j-c_m)$. The Gauss collocation method has $a_{ij}={\int }_0^{c_i}{\ell }_j(t)dt$ and $b_j={\int }_0^1{\ell }_j(t)dt$. This gives explicit coefficient formulas after finding polynomial roots: if ${\ell }_j(t)={\sum }_r{d}_{jr}{t}^r$, these integrals are ${\sum }_r{d}_{jr}{c}_i^{r+1}/(r+1)$ and ${\sum }_r{d}_{jr}/(r+1)$. Gaussian quadrature gives order $2s$.

A surface locally swept out by ambient geodesics has $II(T,T)=0$ in its unit ruling direction $T$. For a perpendicular unit tangent $E$, the Gauss equation in a curved ambient manifold therefore gives the displayed identity. Its Gaussian curvature is at most the ambient sectional curvature in its tangent plane, with equality precisely when $II(T,E)=0$. An independent variation argument applies the Jacobi field equation to the length of a perpendicular connecting field; its ambient derivative has the extra normal component $II(T,E)$. The Euclidean surface $(t,s,ts)$ is ruled by straight lines and has Gaussian curvature $-(1+t^2+s^2{)}^{-2}<0$.