For a group acting linearly on $W$, act on the coordinate ring by $(g\cdot f)(w)=f(g^{-1}w)$. The polynomial invariant ring is the fixed subalgebra $\mathbb{C}[W{]}^G$. It inherits the degree grading. Even over the complex numbers it need not be a unique factorization domain, as the quadratic cone invariant ring demonstrates.

For a map $f:M\to N$ of connected closed oriented four-manifolds, naturality of the cup product givesWhen $\mathrm{deg}f\ne 0$, nondegeneracy makes $f^{\ast }:H^2(N;\mathbb{Q})\to H^2(M;\mathbb{Q})$ injective. Consequently the target intersection form, multiplied by the mapping degree, must embed into the source form. For equal second Betti numbers, a matrix $P$ for pullback satisfies $P^{\mathsf{T}}{Q}_{M}P=(\mathrm{deg}f)Q_N$; taking determinants often rules out nonzero degrees. In particular, a two-dimensional indefinite source form cannot realize a nonzero multiple of a two-dimensional definite target form.

An intersection pairing counts signed intersections of transverse representatives of complementary homology classes, using one relative representative when the manifold has boundary. With ${\mathbb{F}}_2$ coefficients it is defined without orientation choices. Poincare-Lefschetz duality makes this a perfect pairing for compact manifolds.

When compact oriented four-manifolds are glued along an entire closed boundary three-manifold, the signature of the result is the sum of their intersection form signatures. It applies to $X_0{\cup }_{\partial X_0}(-X_1)$ with the difference of the two signatures. Gluing along only part of a boundary can require a correction; that is a different situation.

An integral intersection form on a finite-rank free abelian group $L$ is unimodular if its adjoint $x\mapsto \lambda (x,-)$ is an isomorphism to the integral dual. In an integral basis, this is equivalent to the pairing matrix having determinant $1$ or $-1$, not merely nonzero determinant. For example, projective lines generate a unimodular intersection form on the Complex projective plane, with matrix $(1)$ in its complex orientation.

An agreement to exchange interest cash flows according to fixed and floating schedules. For unit accrual periods and unit notional, floating-minus-fixed payments are $r_t-s$ at the specified dates. Zero-coupon bonds value the fixed leg and a rolling floating leg supplies the telescoping floating-leg value.

A treatment consequence is not interchangeable with a baseline covariate in an intention-to-treat analysis. Let randomized $Z$ be independent of $U,\epsilon$, and let actual intervention use be $D=Z+U$, with $Y=D+\epsilon$ and independent centered errors. The assignment changes the mean outcome by one. But the conditional expectation $\mathbb{E}[Y\mid Z,D]=D$ has coefficient zero on $Z$. Thus adjustment for the post-randomization variable $D$ changes the estimand even in a simple identified linear regression.

An orthonormal set ${\varphi }_0,\dots ,{\varphi }_{m-1}$ in an $L^2$ space defines the integral kernel of the orthogonal projection onto its span by this sum. It satisfies $\int K(x,y)K(y,z)d\mu (y)=K(x,z)$ and $\int K(x,x)d\mu (x)=m$. Its evaluation matrices are positive semidefinite Gram matrices.

Suppose an integral kernel reproduces under convolution and has integrable diagonal with integral $r$, and the displayed single-variable integrals exist. In the determinant $D_n=\det [K(x_i,x_j)]$, permutations fixing $n$ contribute $r{D}_{n-1}$ after integration. For every permutation on $n-1$ letters there are $n-1$ ways to insert $n$ into a cycle; convolution contracts the two adjacent factors, and insertion reverses the permutation sign. These terms contribute $-(n-1)D_{n-1}$. This proves the formula without a symmetry assumption on the kernel.

If a field $L$ is integral over a subdomain $A$, every nonzero $a\in A$ has an inverse in $L$ satisfying a monic equation over $A$. Multiplying that equation by $a^{r-1}$ expresses $a^{-1}$ as a polynomial in $a$ with coefficients in $A$. Hence $a^{-1}\in A$, and $A$ is itself a field. The embedding and integrality assumptions are essential; a fraction field is not generally integral over its source domain.

For $V(q)=m{\omega }^2(q^2-a^2{)}^2/(8a^2)$, the zero Euclidean-energy condition is $m{\dot{q}}^2/2=V(q)$. Integrating gives the displayed instanton from $-a$ to $a$. Its action is ${\int }_{-a}^a\sqrt{2mV(q)}dq=2m\omega a^2/3$. The arbitrary center $t_0$ supplies a translation zero mode, treated as a collective coordinate in the instanton fluctuation prefactor.

For a one-dimensional quantum tunnelling instanton, the prefactor combines nonzero Gaussian fluctuation eigenvalues with the translation zero-mode Jacobian. Here $L_I=-{\partial }_t^2+V^{\prime \prime }(q_I)/m$, $L_0=-{\partial }_t^2+{\omega }^2$, and the prime excludes the zero mode. Common endpoint regularization and reference normalization define the determinant ratio. The resulting $K$ has frequency units, while $K{e}^{-S_I/\hslash }$ is the dilute crossing rate. This determinant/collective-coordinate structure is also discussed in Dunne, Sulejmanpasic and Ünsal's treatment; unequal neighbouring well frequencies require care and need not share the symmetric-double-well result.

When individual instantons are much narrower than their typical separation, multi-crossing paths factor at leading semiclassical order. Integrating $n$ ordered centers over time $\tau$ contributes ${\tau }^n/n!$. In a double well, crossings alternate direction and even/odd counts yield hyperbolic cosine/sine amplitudes. This approximation requires large action relative to $\hslash$, matched local well structure and control of corrections from interactions and local anharmonicity.

The orientation-reversed Euclidean trajectory of an instanton. In a symmetric double well it goes from the right vacuum to the left vacuum and has the same positive action as the forward crossing. Its topological orientation changes sign, not its Euclidean action.

Given an integral epimorphism $\alpha :{\pi }_1(X)\to \mathbb{Z}$, the Alexander module is the first integral homology of the corresponding infinite cyclic cover associated to an epimorphism, with $t$ acting by its chosen deck transformation. Its order is the greatest common divisor of maximal presentation matrix minors when it is finitely presented and torsion. For the meridional epimorphism of a knot exterior, this is the Alexander module of a knot.

If two nonnegative finite-valued convex functions are defined on all of a finite-dimensional space, their infimal convolution is finite, convex and continuous everywhere. Convex perturbation duality then gives strong duality and dual attainment at every argument. It does not require or imply attainment of the primal decomposition infimum.

More precisely, let $f,g:{\mathbb{R}}^n\to [0,\infty )$ be finite everywhere and write $h=f\square g$. Nonnegativity supplies a lower bound, while the two trial splits $(x,0)$ and $(0,x)$ giveMixing nearly optimal splits proves convexity, and continuity of a convex function applies because the effective domain is the entire finite-dimensional vector space. These statements concern the optimized value, rather than existence of an optimal split. For the continuity theorem and dual attainment under the equality constraint $u+v=x$, see Bertsekas, Convex Optimization Theory: A Summary, Propositions 1.3.11 and 5.3.3(a). The equality has a feasible point in the whole space for every $x$, and the displayed bound makes its optimal value finite.

A finite value without any optimal split. On $\mathbb{R}$, take the convex functions $f(u)=e^u$ and $g(v)=0$. Both are nonnegative and finite everywhere, butfor every $x$. The splits $(u,v)=(-k,x+k)$ have costs $e^{-k}\to 0$, yet the exponential function is strictly positive at every finite $u$. No split attains the infimum. The value function is even constant and differentiable; regularity of the value function does not prevent optimizing splits from escaping to infinity. For example, $k=1,3,6$ gives costs approximately $0.367879$, $0.049787$ and $0.002479$.

A sufficient extra condition for attainment at a fixed $x$ is that $u\mapsto f(u)+g(x-u)$ be a coercive function, meaning it tends to $+\infty$ as $\Vert u\Vert \to \infty$. Under the finite-valued assumptions it is a continuous function, so a sufficiently large closed ball contains a global minimum by the extreme value theorem. If this cost is also a strictly convex function, the optimal split is unique. The quadratic example above satisfies both conditions; the exponential function example does not satisfy coercivity.

Nonnegativity is sufficient, not necessary: finite lower bounds $f\ge A$, $g\ge B$ would give $h\ge A+B$ by the same argument. Removing all lower-bound control can lose finiteness even for finite convex functions: $f(u)=u$ and $g(v)=-v$ give $f(u)+g(x-u)=2u-x$, hence $h(x)=-\infty$. Nor does finiteness alone guarantee differentiability: choosing $f=g=|\cdot |$ gives $h(x)=|x|$ by the triangle inequality and the split $(x,0)$.

With downward buoyancy $b=g\rho \prime /{\rho }_0$ in a uniform rotating Boussinesq approximation, $q=\zeta -f{b}_z/N^2$ is time-independent. The linear pressure obeys ${\nabla }^2{p}_{tt}+f^2{p}_{zz}+N^2{\nabla }_H^{2}p={\rho }_{0}f{N}^{2}q$. The initial potential vorticity supplies the steady pressure source; nonzero-frequency inertia-gravity waves have zero perturbation PV.

If two versions of a stochastic process both have càdlàg paths on an event of probability one, equality at every rational time holds simultaneously outside one null event. Approaching an arbitrary time by rationals from the right then gives equality at every time on the same event. Thus the stochastic processes are indistinguishable. Right continuity suffices; left limits are unnecessary for this conclusion.

This smooth utility has $U^{\prime }(x)=x^{-1}-x^{-2}+x^{-3}/3>0$ and $U^{\prime \prime }(x)=-(x-1{)}^2/x^4$. Its derivative is strictly decreasing, diverges at zero and tends to zero at infinity, so the utility is strictly concave and satisfies the Inada conditions. Yet its curvature vanishes at one. The inverse marginal utility cannot have a finite derivative at $y=1/3$, since differentiating $U^{\prime }(I(y))=y$ there would give zero equal to one. Its dual is differentiable but not twice differentiable at that price.

Integrate the transverse force along a complete straight path: ${\int }_{-\infty }^{\infty }(1+s^2{)}^{-3/2}ds=2$. This gives the transverse kick. Conservation of relative speed gives longitudinal speed reduction $\Delta v_{\perp }^2/(2V)$ at second order. As an independent check, exact weak hyperbolic scattering has angle $2\arctan \eta$, where $\eta =GM/(b{V}^2)$, and reduction $V(1-\cos (2\arctan \eta ))=2V{\eta }^2/(1+{\eta }^2)$. The small-$\eta$ coefficient agrees. Keeping only one half of the encounter in the transverse kick makes this quadratic reduction four times too small.