The communality is the variance of the shared-factor contribution to variable $i$. In an orthogonal factor model, $\mathrm{Var}(X_i)=h_i^2+{\psi }_i$, where ${\psi }_i$ is its specific variance. For standardized variables, $h_i^2\le 1$. It measures variance shared through the modeled factors, not variance in a chosen principal component.

An orthogonal rotation of factors leaves $L{L}^T$ unchanged and hence preserves the model covariance. It proves rotational nonidentifiability of the unrestrained factor loadings. Choose an orientation for interpretability, for example varimax rotation, or impose identifying restrictions in a confirmatory model. An oblique transformation permits correlated factors: if $L$ becomes $LH$, their covariance becomes $H^{-1}{H}^{-T}$, preserving the same shared covariance.

This collects the gradient, divergence, curl and Laplacian formulas in cylindrical and spherical coordinates. Coordinate unit vectors vary with position, so taking their derivatives is essential when translating vector calculus operators from Cartesian coordinates. Curl in spherical coordinates is one of these formulas.

The weighted strain density in Ornstein-Uhlenbeck multiplicative amplification evolves under ${\mathcal{L}}_{n}p=(\kappa /2)p^{\prime \prime }+{\tau }^{-1}(\sigma p{)}^{\prime }+n\sigma p$. Setting $p=e^{-{\sigma }^2/(2\kappa \tau )}\psi$ eliminates its first derivative. Shifting and scaling to $x=(\sigma -\kappa n{\tau }^2)/\sqrt{\kappa \tau }$ gives ${\psi }_{xx}+[1+\kappa n^2{\tau }^3-2\gamma \tau -x^2]\psi =0$. The quantum harmonic oscillator levels yield the displayed eigenvalues. Its ground state is the dominant positive mode; higher sign-changing modes may occur in a transient expansion.

In global coordinates the de Sitter spacetime metric is $-d{\tau }^2+{\cosh }^2\tau g_{S^3}$. A global frame of a vector bundle orthonormal for the round metric $Y_1,Y_2,Y_3$ on $S^3$ gives the displayed global pseudo-orthonormal frame. For example, identify $S^3$ with the unit quaternions and use $qi,qj,qk$. Choosing the spatial orientation and the future direction consistently gives a section of a fiber bundle for the oriented, time-oriented frame bundle. The principal bundle trivialization by a global section then proves that bundle is trivial.

On the orbit of a face $\tau ⪯\sigma$, the monomial ${\chi }^{\mu }$, with $\mu \in {\sigma }^{\vee }\cap M$, vanishes exactly when $\mu \notin {\tau }^{\perp }$. Its reduced zero locus is consequently the union of those orbit closures. The principal subscheme need not be reduced: $V(x^2)$ and $V(x)$ have the same support but different ideals.

// Target: geometry-and-topology.bigb

Let $Z_j$ be the sum of $j$ independent identically distributed real increments. For a finite convex function $f$, integrability of $f(Z_j)$ and $f(Z_{j+1})$ implies integrability of $f(Z_j+x)$ for every fixed $x$. Its negative part is controlled by negative part of a finite convex function is Lipschitz. For $x>0$, if the increment law is unbounded above, take an independent increment $\eta$ with $p=\mathbb{P}(\eta \ge x)>0$. On this event convexity gives $f(Z_j+x{)}^{+}\le f(Z_j{)}^{+}+f(Z_j+\eta {)}^{+}$, so the expectation is bounded by $\mathbb{E}f(Z_j{)}^{+}+p^{-1}\mathbb{E}f(Z_{j+1}{)}^{+}$. If increments are bounded above by $M$, use $Z_j\le jM$ and bound the positive part by $f(Z_j{)}^{+}+f(jM+x{)}^{+}$. For $x<0$, use the symmetric lower-tail argument. This supplies translated integrability without assuming integrable increments or a growth bound on $f$.

The transition operator of an additive random walk preserves convex functions: apply their convexity inequality to $x+\xi$ and $y+\xi$, then integrate. The pointwise maximum of convex functions preserves convexity too, so induction proves convexity of the Snell envelope Bellman functions. Expectations are allowed to be extended-valued until finiteness is justified. Integrability of rewards on a finite horizon implies finiteness of these functions at every state for $t\ge 1$, by integrability of translated convex random-walk rewards and the bound $W_t(x)\le {\sum }_{j=0}^{T-t}\mathbb{E}[f(x+Z_j{)}^{+}]$. At time zero only $X_0=0$ is observed; if $W_0$ is infinite elsewhere, its finite value at zero has a constant convex extension representing the same Snell envelope. Full statewise Bellman recursion at time zero requires an additional translated-integrability hypothesis.

In this temporal sense, the optical coherence length is $L_c=c{\tau }_c$, where $c$ is the vacuum speed of light and ${\tau }_c$ is an explicitly chosen optical coherence time. It measures a delay as a vacuum-equivalent path difference; it does not measure transverse spatial optical coherence. ^[ref].

For the stipulated $1\mathrm{GHz}$ profiles, the $1/e$ convention gives $L_c\approx 9.54269\mathrm{cm}$ for the Lorentzian and $15.8896\mathrm{cm}$ for the Gaussian. In a vacuum Michelson interferometer, changing one mirror position by $x$ changes the round-trip delay by $2x/c$, so the mirror travel to either of these delays is $L_c/2$. In a medium, convert physical travel using the actual group delay, which can depend on frequency. For a narrow line centered at vacuum wavelength ${\lambda }_0$, differentiating $\nu =c/\lambda$ gives $\Delta \nu \approx c\Delta \lambda /{\lambda }_0^2$; this small-width approximation must use consistently defined spectral widths.

Allow terminal positions of an infinite game of perfect information to carry a losing-player label. A clopen set of infinite winning plays still gives a determined infinite game. At a position where all infinite continuations have one winner, the other player can win only by forcing a terminal loss for that winner. The usual winning-position attractor in an infinite game decides this reachability game. The tree of remaining undecided positions is well-founded: every infinite branch would eventually have its clopen payoff decided by a prefix. Assign the already decided winners upwards by well-founded recursion. Terminal labels are essential; a terminal node created by a covering need not be lost by the player whose turn would have come next.

For $0<S<1$ and $V=1-S$, the translational branch of one-sided kinetic solidification dispersion has $\sigma =(S/V-\Gamma ){\alpha }^2+O({\alpha }^4)$. Expand $p=V+(\sigma +{\alpha }^2)/V+O({\alpha }^4)$ and substitute to prove this limit. At zero growth, $p_0=[V+\sqrt{V^2+4{\alpha }^2}]/2$, and the destabilizing right side has sign ${\alpha }^2(SV/p_0^2-\Gamma )$. For nonnegative real $\sigma$, the residual has derivative at least one, so there is one growing root precisely when $\Gamma <SV/p_0^2$. Thus instability exists exactly when $\Gamma <S/V$, with unstable band ${\alpha }^2<p_c(p_c-V)$, $p_c=\sqrt{SV/\Gamma }$ for $\Gamma >0$. At threshold the next term is $\sigma =-2S{\alpha }^4/V^3+O({\alpha }^6)$. Surface energy removes short-wave growth even when long waves remain unstable.

The cubic follows by eliminating $\sigma =p^2-Vp-{\alpha }^2$ from one-sided kinetic solidification dispersion, with $V=1-S$. For $\Gamma =0$, the interface branch tends to $\sigma =SV-SV/\alpha +O({\alpha }^{-2})$. For fixed $0<\Gamma <1$, substituting $p=\sqrt{1-\Gamma }\alpha +d$ gives $2d-V=S\Gamma /(1-\Gamma )$, hence $\sigma =-\Gamma {\alpha }^2+[S\Gamma /\sqrt{1-\Gamma }]\alpha +O(1)$. At $\Gamma =1$ the balance is $p^3\sim S{\alpha }^2$, giving $\sigma =-{\alpha }^2+S^{2/3}{\alpha }^{4/3}+o({\alpha }^{4/3})$. For $1<\Gamma <1+2S/V$, $p\to S/(\Gamma -1)>V/2$ and $\sigma =-{\alpha }^2+p^2-Vp+o(1)$. Stronger capillarity can remove the real isolated interface mode from the chosen thermal branch; it does not create growth. These distinctions explain why replacing $p$ by $\alpha$ is not a controlled fixed-capillarity limit.

The initial cost of the one-period quadratic hedge is $\mathbb{E}[{\rho }^{\ast }\xi ]$. This pricing weight satisfies $\mathbb{E}[{\rho }^{\ast }R]=1$ and $\mathbb{E}[{\rho }^{\ast }{S}_1]=S_0$. Among square-integrable weights satisfying these equations it has the smallest squared norm:For another admissible weight $\rho$, its difference from ${\rho }^{\ast }$ is orthogonal to $1$ and all coordinates of $S_1$, hence to ${\rho }^{\ast }$. The Pythagorean theorem in an inner-product space gives $\mathbb{E}[{\rho }^2]=\mathbb{E}[({\rho }^{\ast }{)}^2]+\mathbb{E}[(\rho -{\rho }^{\ast }{)}^2]$. This weight can be signed; positivity, and hence interpretation as a state-price density, requires additional assumptions.

If $S_1$ has a multivariate normal distribution with mean $\mu$ and invertible covariance $V$, and $g$ has bounded gradient, Gaussian integration by parts gives $\mathrm{Cov}(S_1,g(S_1))=V\mathbb{E}[\nabla g(S_1)]$. Consequently the risky holdings of its one-period quadratic hedge equal the expected payoff gradient. The initial cost is $R^{-1}\mathbb{E}[g(S_1)]+(S_0-R^{-1}\mu )\cdot \mathbb{E}[\nabla g(S_1)]$. Bounded gradient ensures linear growth and square integrability, even if the payoff is not bounded.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

Let $m=\mathbb{E}Y$ and $C=\mathrm{Cov}(Y)$ be an invertible covariance matrix of discounted gains. The displayed signed martingale measure leaves unchanged the mean-zero random variables orthogonal to the martingale part $Y-m$: if $\mathbb{E}L=0$ and $\mathbb{E}[(Y-m)L]=0$, then $\mathbb{E}[Z_{\ast }L]=0$. Conversely, preservation of all these orthogonal directions forces the density to lie in the span of $1,Y-m$, and its pricing constraints give the displayed formula. If $Z_{\ast }>0$ it is an equivalent martingale measure; if $Z_{\ast }\ge 0$ it is a dominated martingale measure. Positivity is an extra condition, not a consequence of absence of arbitrage alone. In one period this density also minimizes $\mathbb{E}{Z}^2$ among square-integrable signed pricing densities, because every other such density differs by a vector orthogonal to $1,Y-m$. The initial capital in the unrestricted one-period quadratic hedge of a discounted payoff $H$ is $\mathbb{E}[Z_{\ast }H]$.

For a discounted contingent claim $H$ and gain vector $Y$, fix initial capital $v$ and minimize $\mathbb{E}(H-v-{\theta }^{T}Y{)}^2$. Differentiation gives the least-squares normal equations $G\theta =\mathbb{E}[Y(H-v)]$, where $G=\mathbb{E}[Y{Y}^T]$ is assumed invertible. This gives the displayed hedge. Optimizing capital as well instead centers the gains and uses their covariance matrix: ${\theta }^{\ast }=\mathrm{Cov}(Y{)}^{-1}\mathrm{Cov}(Y,H)$ and $v^{\ast }=\mathbb{E}H-({\theta }^{\ast }{)}^T\mathbb{E}Y$. Fixing the initial capital is therefore a different quadratic problem.

For positive ${\beta }_0$, the one-loop running coupling tends to zero at high energy and grows as energy decreases. Near the strong-coupling scale, omitted perturbative terms cease to be small; the apparent infrared pole is not a controlled prediction of a physical infinity. Continuing the algebraic formula below $\Lambda$ gives a negative coupling and is not a valid physical continuation. Confinement and hadron formation are nonperturbative phenomena, not a theorem proved by this one-loop pole. The integration constant obeys $\Lambda ={\mu }_0\exp [-1/(2{\beta }_0{\alpha }_0)]$, so an order-one coefficient alone cannot determine an absolute mass scale; a reference coupling or empirical input is also necessary.

If each bond has closed probability at most $q$ in one-independent bond percolation on the square lattice, a set of $\ell$ bonds contains a matching in a graph of size at least $\ell /4$: divide horizontal bonds by the parity of their left endpoint and vertical bonds by the parity of their lower endpoint. Each of the four classes is a matching. Mutual independence in the largest class proves the circuit bound. There are at most $4\ell 3^{\ell }$ dual circuits of length $\ell$ surrounding the origin. For sufficiently small $q$, the sum of these circuit probabilities is less than one, so the origin has a positive chance of belonging to an infinite percolation cluster.

Place side-$2n$ square lattice boxes around the points $3nv$, $v\in {\mathbb{Z}}^2$. The width-$2n$ rectangle joining adjacent boxes has length $5n$. Declare a coarse bond open when its long rectangle is crossed and each endpoint box is crossed in the transverse direction. Nonincident coarse bonds use disjoint fine-bond sets, while incident good bonds give connected open paths inside their common endpoint box. When fixed-aspect crossing probabilities tend to one, these coarse bond probabilities tend to one. The Peierls circuit bound for one-independent percolation then yields an infinite fine-lattice open cluster.