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 a 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 $-(1+t^2+s^2{)}^{-2}<0$.

Here $R(X,Y)=D_{[X,Y]}-[D_X,D_Y]$, the negative of the convention in the ordinary Gauss equation in a curved ambient manifold. Insert the Gauss formula and Weingarten formula into this definition and project tangentially to obtain the displayed identity. Positive sectional curvature uses $⟨R(X,Y)X,Y⟩/|X\wedge Y{|}^2$ with this convention. Thus a Euclidean surface still has Gaussian curvature $K=\det A_{\xi }$; reversing the operator sign without changing the slot order would give the wrong sign.

Let the unit normal satisfy $n^a{n}_a=\epsilon =\pm 1$, let $h_{ab}=g_{ab}-\epsilon n_a{n}_b$, and set $K_{ab}=h_a{}^c{h}_b{}^d{\nabla }_c{n}_d$. The normal component of an ambient derivative of tangent fields is $-\epsilon K(X,Y)n$. Substitute that decomposition into the curvature commutator and project tangentially to obtain the displayed equation. Its scalar contraction is ${}^{\Sigma }R=R-2\epsilon R_{ab}{n}^a{n}^b+\epsilon (K^2-K_{ab}{K}^{ab})$. Both metric signature and curvature-slot convention must be specified when assigning a sign.

For $D=\mathrm{diag}(1,-1)$, $A=D{\lambda }^2+\left(\begin{array}{cc}0& U\\ V& 0\end{array}\right)\lambda +\left(\begin{array}{cc}d& P\\ R& -d\end{array}\right)$, the formal diagonal exponent has $h=d+UV/2$ and $\theta =(UR+VP)/2$. Matching the gauge equation gives the off-diagonal first coefficient $(-U/2,V/2)$ and the diagonal one $\rho D$, where $\rho =UV(d+h)/4-PR/2$. The second coefficient has diagonal entries ${\rho }^2/2-UV/8\pm h\theta /2$ and off-diagonal entries $(U\rho -P)/2,(V\rho +R)/2$. With $G_0=I$ and negative-index coefficients zero, the coefficient comparison isIts diagonal equations at $m=2,3,4,5$ give $h,\theta ,\rho$ and the second diagonal coefficient respectively. Its off-diagonal equations determine the next coefficients because the commutator with $D$ multiplies the upper and lower entries by $2$ and $-2$. This proves the expansion and supplies the recursion for all higher orders. The generally nonzero $\theta$ supplies formal monodromy.

For the left graded Leibniz rule and $s\bar{c}=-b$, $sb=0$, the displayed gauge-fixing fermion gives $b\partial \cdot A+\xi b^2/2+\bar{c}\partial \cdot Dc$. The ghost signs follow from odd parity. BRST nilpotence implies invariance of this density without using field equations. Eliminating the Nakanishi-Lautrup field gives $-(\partial \cdot A{)}^2/(2\xi )$ for nonzero $\xi$.

For a left-acting BRST differential with $sA=Dc$, $s\bar{c}=b$, $sb=0$, choose $\Psi ={\bar{c}}^a(\partial \cdot A^a+\xi b^a/2)$. The graded Leibniz rule gives $s\Psi =b\partial \cdot A+\xi b^2/2-\bar{c}\partial \cdot Dc$. Integrating the auxiliary Gaussian yields $-(\partial \cdot A{)}^2/(2\xi )-\bar{c}\partial \cdot Dc$. At $\xi =1$ this is Feynman gauge. Opposite antighost or exact-action sign conventions require a corresponding change of fermion.

// Target: relativistic-quantum-field.bigb

For integer isospin $I$ and neutral-state charge conjugation value $\eta$, the entire isospin multiplet has $G=\eta (-1{)}^I$. Both charge conjugation and the second-axis half-turn reverse the first and third generators, so their product commutes with the full isospin algebra. Evaluate it on the $M=0$ state and use ladder connectivity to obtain the value for every $M$.