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The Riemannian volume form of an oriented $n$-dimensional Riemannian manifold is the unique positive $n$-form taking value one on every positively oriented orthonormal frame. In positive coordinates,The metric induces an inner product on exterior powers of the cotangent bundle, with wedges of distinct orthonormal covectors forming an orthonormal basis. The Hodge star is the pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ specified byIt is an isometry and satisfies $\ast \ast \alpha =(-1{)}^{p(n-p)}\alpha$. With the positive-sign convention, the Hodge Laplacian on differential forms isHere $d$ is the exterior derivative and $\delta$ its formal adjoint; $\delta$ on functions is zero. On functions this is the Laplace-Beltrami operatorThe convention $\mathrm{div}\mathrm{grad}$ gives the negative of this operator. All subsequent Laplacians use the nonnegative convention above.

On a compact oriented manifold without boundary, the Hodge decomposition theorem gives an orthogonal decomposition of smooth real forms,The three component forms are uniquely determined, although the potentials in the last two summands need not be unique. The space ${\mathcal{H}}^p$ is finite-dimensional, and each de Rham cohomology class has a unique harmonic representative. Integration by parts gives $⟨\Delta \alpha ,\alpha {⟩}_{L^2}=\Vert d\alpha {\Vert }_{L^2}^2+\Vert \delta \alpha {\Vert }_{L^2}^2$, so harmonic forms are exactly the closed and coclosed forms. Set ${\Omega }^k=0$ outside $0\le k\le n$. Compactness without boundary is a hypothesis of this version of the theorem; orientation alone does not supply it.

A connection on $TM$ induces a dual connection on $T^{\ast }M$ and then connections on its tensor and exterior powers by the Leibniz rule. Explicitly, for a $p$-form $\alpha$,To prove the parallelism of the Riemannian volume form, use a local positively oriented orthonormal frame $E_i$, its dual coframe ${\theta }^i$, and write ${\nabla }_X{E}_j={\sum }_i{a}_{ij}(X)E_i$. Metric compatibility makes the matrix $a(X)$ skew-symmetric. Since ${\omega }_g={\theta }^1\wedge \cdots \wedge {\theta }^n$, differentiating the wedge yieldsOff-diagonal terms vanish because they repeat a coframe factor. Thus $\nabla {\omega }_g=0$; equivalently parallel transport preserves both the metric and the continuously chosen orientation.

Now let $M$ be compact, oriented and without boundary. For representing functionals on harmonic forms by wedge pairing, choose an $L^2$-orthonormal basis $h_1,\dots ,h_b$ of ${\mathcal{H}}^p$ and defineIf $\phi ={\sum }_a{c}_a{h}_a$, linearity givesThis also works when $b=0$, using the empty sum. Since the Hodge star commutes with the Hodge Laplacian, ${\psi }_0$ is harmonic of degree $n-p$.

For the full ambiguity of harmonic wedge-pairing representatives, write another representative as ${\psi }_0+\chi$. Because the star is an isometry,The forms $\ast \phi$ range over all of ${\mathcal{H}}^{n-p}$. Hence $\chi$ gives zero functional exactly when its harmonic projection is zero. Hodge decomposition proves that the complete answer isBoth summands in this ambiguity are permitted; it is not only an exact-form ambiguity. The exact and coexact component forms are orthogonal to every harmonic form, so every displayed choice works. Conversely a nonzero harmonic component would pair nontrivially with some $\ast \phi$, and therefore cannot occur. The harmonic representative $\ast h_f$ is unique; the arbitrary smooth representative generally is not. The formula also covers the endpoint degrees and any exceptional case where the ambiguity space is zero.

For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

For a prime $p$, the divisor sum defining the real weight is $\varphi (0)-\varphi (\log p/\log X)$. If $p>X^{1/3}$, the second term vanishes and the first is one. Thus $F(p)=1$, while $F(n)\ge 0$ for every integer because $\varphi$ is real and the sum is squared.

Put $h(x)=e^x\varphi (x)$. With the Fourier transform convention in the question, Fourier inversion givesThe function $h$ is smooth and has compact support. Using integration by parts $k$ times, with zero endpoint terms, provesChoose an integer $k\ge A$ to obtain $|\psi (t)|{\ll }_A|t{|}^{-A}$ for $|t|\ge 1$. In particular every polynomially weighted absolute integral of $\psi$ is finite.

For the double integral, useDifferentiating the second formula gives $\int \psi (t)(1+it)e^{-(1+it)u}dt=-{\varphi }^{\prime }(u)$. absolute convergence from the rapid decay permits Fubini's theorem. The derivative energy constant for a smooth sieve cutoff is thereforeThere is no complex conjugate in this integral: the two factors both become $-{\varphi }^{\prime }(u)$, which is real. The constant is positive; in fact the Cauchy-Schwarz inequality applied to $\varphi (1/3)-\varphi (0)=-1$ gives $c_{\varphi }\ge 3$.

Here is the basic structure of the smooth divisor-square sieve asymptotic, with enough uniform estimates to specify the error. Write $L=\log X$ and $R=X^{1/3}$. Expand the square and count multiples of the least common multiple $[d,e]$ in an arbitrary interval $I$ of length $X$:The sum is effectively restricted to $d,e<R$. Each count is $X/[d,e]+O(1)$, independent of the interval's position, so the total error is $O_{\varphi }(X^{2/3})$.

The Fourier representation of a smooth Selberg weight readsHence, with $z^{\prime }=(1+i{t}^{\prime })/L$,The four local terms correspond to the prime dividing neither divisor, only the first, only the second, or both. This is the Euler product for a smoothed divisor-square correlation. Factor it aswhereThe local factors of $H$ are $1+O(p^{-2+4\eta })$ when $|\mathrm{Re}z|,|\mathrm{Re}{z}^{\prime }|\le \eta$ for small fixed $\eta <1/4$. Thus $H$ is a holomorphic function near $(0,0)$, and evaluating each local factor at zero gives exactly one. Consequently $H(0,0)=1$ and $H(z,z^{\prime })=1+O(|z|+|z^{\prime }|)$ there.

The pole expansion $\zeta (1+w)=w^{-1}+O(1)$ now yieldswhen $|t|,|t^{\prime }|\le L^{1/4}$. Integrating the error against the rapidly decreasing transforms gives $O_{\varphi }(L^{-2})$. The complementary tails are harmless uniformly: absolute values in the original divisor series giveAny prescribed power decay of ${\int }_{|t|>L^{1/4}}|\psi (t)|dt$ is available, so these tails and the tails of the limiting kernel are $O_{\varphi }(L^{-2})$ after choosing sufficiently many applications of integration by parts. The same absolute bound justifies the original exchanges of infinite sums and integrals. Using the evaluated double integral,We conclude, uniformly over all interval locations,This proof does not assume that the interval starts near the origin.

Finally choose this one fixed smooth cutoff. primes above $R$ in $(X_0,X_0+X]$ contribute one each to the nonnegative weight, and at most $R$ primes lie below $R$. Thereforefor all sufficiently large $X$, uniformly in $X_0$. The bounded range $2\le X\le X_1$ follows by increasing the fixed constant and using the trivial interval bound $X+1$. Thus the short-interval prime upper bound from a smooth divisor weight is

First use the additional admitted form of the efficient score. Write $\tilde{\ell }=e\zeta (X)$ and let ${\tau }^2=E_f{\epsilon }^2>0$. The difference $\dot{\ell }-\tilde{\ell }$ is the orthogonal projection onto the nuisance tangent space. Part (c) makes every $e\psi (X)$ orthogonal to that space. Therefore, for all $\psi \in L^2(v)$,The function in braces belongs to L2 space; choosing it as $\psi$ shows it vanishes $v$-almost everywhere. Hence the required conditional deduction isUnder the regular tail condition $sf(s)\to 0$ at both infinities, integration by parts gives $\int s{f}^{\prime }(s)ds=-1$, and the formula becomes $e{h}_{\theta }(x)/{\tau }^2$. This also confirms the sign.

The admitted product form is an extra restriction; it does not follow for every independent-error regression model. To locate the restriction precisely, suppose the error-density nuisance statistical paths preserve both normalization and mean to first order. Their closed mean-preserving error tangent space is $\{\gamma :E_f\gamma =E_f(\epsilon \gamma )=0\}$. The full nuisance tangent space is the orthogonal sum of this space and the centered functions of $X$.

For completeness, bounded functions satisfying the two constraints are dense in the error space. Truncate an arbitrary element, subtract its expected value, and then subtract a multiple of a fixed bounded centered function $q$ with $E_f(\epsilon q)\ne 0$. Such a $q$ exists by truncating $\epsilon$, since ${\tau }^2>0$. The correction coefficients tend to zero by the Cauchy-Schwarz inequality, so the corrected truncations converge in L2 space.

Let $m=E_v{h}_{\theta }(X)$. With $E_f\rho =0$ and $E_f(\epsilon \rho )=1$, the orthogonal projection of $h_{\theta }(X)\rho (\epsilon )$ onto the error nuisance space is $m\{\rho (\epsilon )-\epsilon /{\tau }^2\}$; its projection onto the covariate nuisance space is zero. Thus the general efficient score in independent-error regression isThe orthogonality of the two terms gives the displayed efficient information. For a normal distribution of the error, $\rho (e)=e/{\tau }^2$, so this reduces to the admitted formula. It also does so when $h_{\theta }$ is constant.

A concrete counterexample to the generality of the admission is $g_{\theta }(x)=\theta x$, with $X$ uniform on $[-1,1]$ and independent standard logistic distribution error. Here $m=0$ and $\rho (e)=\tanh (e/2)$, so the actual efficient score is $x\tanh (e/2)$. It cannot equal $e\zeta (x)$ because $\tanh (e/2)/e$ is not constant. This score function is already orthogonal to every nuisance score function: its factor $X$ is centered against error-only directions, and its factor $\rho (\epsilon )$ is centered against covariate-only directions. The requested formula is valid under its stated additional admission, with the general independent-error formula above explaining its limits.

A map $\Phi :S_D\to \mathbb{R}$ is Hadamard differentiable at $s_0$ if there is a continuous linear functional ${\dot{\Phi }}_{s_0}:S\to \mathbb{R}$ such that, whenever $t_j\to 0$ with $t_j\ne 0$, $a_j\to a$ in $S$, and $s_0+t_j{a}_j\in S_D$,For tangential Hadamard differentiability, the limit directions $a$ are restricted to a specified tangent subspace and the derivative is continuous linear on that subspace. The domain condition is indispensable.

Assume $X_n\in S_D$ almost surely and use the usual measurable, separably supported setting for norm-valued convergence in distribution, with the limit in the tangent space. The functional delta method givesTo justify this by Skorokhod representation theorem, a sufficient precise version is: for weakly convergent random elements in a separable complete metric space, one can construct copies of the whole sequence and its limit on one probability space which converge almost surely in the metric. More generally the separable-limit-support version of the representation theorem can be used. Apply it to $Z_n=r_n(X_n-s_0)$, obtaining ${\tilde{Z}}_n\to \tilde{X}$ almost surely. The variables $s_0+r_n^{-1}{\tilde{Z}}_n$ have the same laws as $X_n$ and lie in $S_D$ on a common probability-one event. The definition of Hadamard differentiability, with $t_n=1/r_n$, gives convergence almost surely of the transformed quotients to ${\dot{\Phi }}_{s_0}(\tilde{X})$, and hence the claimed convergence in distribution. If $X$ is a Gaussian measure, the limit is Gaussian because the derivative is linear. Without membership of $X_n$ in $S_D$, the expression $\Phi (X_n)$ in the question is not defined.

The final function space is $C_b(\mathbb{R})$ with supremum norm; it is not separable as a whole. This does not affect the deterministic differentiability proof below. Any statistical application of the preceding representation argument must use its separability or tightness hypotheses, rather than assuming that every normed space automatically satisfies them.

Fix admissible $F,G$ and set $T(F,G)={\int }_0^{u}G(x)F^{\prime }(x)dx$. The candidate derivative, defined even for continuous directions $a,b$ without derivatives, isThis follows formally from the product rule and integration by parts. We now establish the uniform-direction limit required by Hadamard differentiability.

Let $F_j=F+t_j{a}_j$ and $G_j=G+t_j{b}_j$ belong to the stated domain, with $a_j\to a$, $b_j\to b$ in supremum norm. The domain gives $\int |F_j^{\prime }|\le 1$ and $\int |F^{\prime }|\le 1$, while $F_j\to F$ uniformly. We first claim that, for every continuous $b$ on $[0,u]$,For a continuously differentiable test function $v$, integration by parts gives convergence directly from uniform convergence of $F_j$: both boundary terms and $\int v^{\prime }{F}_j$ converge. By the Weierstrass approximation theorem, approximate $b$ uniformly on $[0,u]$ by a polynomial $v$. The difference between the two integrals introduced by this approximation is at most $2\Vert b-v{\Vert }_{\infty }$, by the uniform bounded variation bounds. Letting the approximation error decrease proves the claim. This is weak convergence of bounded-variation integrators.

Now decompose the quotient in the order that retains the bounded-variation bound:The first term differs from $\int b{F}_j^{\prime }$ by at most $\Vert b_j-b{\Vert }_{\infty }$, so it tends to $\int b{F}^{\prime }$. For the second term, integration by parts yieldswhich tends to the corresponding expression with $a$. Hence the quotient tends to ${\dot{T}}_{F,G}(a,b)$ for every admissible perturbation sequence. Finally,This proves that the derivative is a continuous linear functional on the product space. Therefore $T$ is Hadamard differentiable at every point of the stated domain. The bounded-variation restriction is what controls the apparently dangerous interaction between uniformly small perturbations and possibly large perturbation derivatives; no convergence of $a_j^{\prime }$ or $b_j^{\prime }$ has been assumed.

Set $\Delta =T/M$, $q_0=q$, and $q_M=q^{\prime }$. The Trotter product formula separates the kinetic and potential energy evolution to first order in $\Delta$. Inserting $M-1$ resolutions of the identity in the position operator basis and using the free short-time kernel givesThe exponent converges formally to the action along the path. The prefactors and intermediate integrations define the time-sliced configuration-space path integral measure, soA real-time path integral is an oscillatory limit with the Feynman i-epsilon prescription, rather than an ordinary probability integral. The operator derivation assumes the usual self-adjointness and product-formula hypotheses for the Hamiltonian.

For a quadratic potential energy write $q=q_c+\eta$, with $\eta (0)=\eta (T)=0$ and ${\ddot{q}}_c+V^{\prime }(q_c)=0$. The first variation vanishes by the Euler-Lagrange equation and the fixed endpoints. Because $V$ is quadratic, the remaining expansion is exact:Here integration by parts has no boundary contribution. Translation of the integration variables leaves the time-sliced measure unchanged. Thus the Gaussian path integral over $\eta$ is independent of the endpoint values, and the entire endpoint dependence lies in $e^{iS[q_c]}$.

Expanding $\eta$ in eigenfunctions satisfying Dirichlet boundary conditions, each real mode contributes an inverse square root of its quadratic eigenvalue, with the oscillatory phase fixed by continuation. Normalizing against $A_0={\partial }_t^2$ gives a precise determinant-ratio form:The determinant ratio is meaningful after a common regulator; the phase is part of the prescription. It is equivalent to writing $N=C(T)({\det }_{D}A{)}^{-1/2}$ with an endpoint-independent normalization.

For example, if $V^{\prime \prime }={\omega }^2$, the Dirichlet oscillator determinant ratio isA linear or constant term in $V$ changes the action but not this prefactor. The formula holds away from conjugate times, when the boundary problem and determinant are nonsingular. At $\sin \omega T=0$, it must be continued as a distribution with its Maslov index phase; the assertion of an ordinary finite prefactor cannot be used literally there. For the unshifted oscillator, at $T=r\pi /\omega$ the kernel is $e^{-ir\pi /2}\delta (q^{\prime }-(-1{)}^{r}q)$.

For the full-line source integral, impose vacuum boundary conditions by adiabatic damping. Take the source initially as a test function, so its pairings with the Green function are defined. After integration by parts, writeWith the Fourier transform convention $G(t)=\int d\omega e^{i\omega t}\tilde{G}(\omega )/(2\pi )$, the inverse isSet $q=\eta -GJ$. Completing the square givesand consequentlyThe Gaussian functional integral $Z[0]$ is source independent and contains the regulated determinant. This is most naturally read as a normalized vacuum generating functional. For $m>0$, closing the frequency contour gives $G(t)=-i{e}^{-im|t|}/(2m)$; the time-ordered oscillator two-point function is $iG(t)$. This checks the sign of the source exponent and identifies the required vacuum prescription.