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Under the standard smooth manifold hypotheses, the first answer is yes, although the isomorphism requires a choice. A partition of unity combines Euclidean inner products in manifold charts to produce a Riemannian metric $g$. Nonnegative local weights summing to one preserve positivity, and local finiteness preserves smoothness. The musical isomorphism isIt is smooth and fiberwise invertible, with inverse having local matrix $(g_{ij}{)}^{-1}$. It is therefore a vector bundle isomorphism. There is no preferred such map before a metric or equivalent additional structure is chosen.

The second answer is no. On $S^1=\mathbb{R}/2\pi \mathbb{Z}$, compare the trivial vector bundle $S^1\times \mathbb{R}$ with the Möbius line bundleBoth have rank one. A continuous section of a vector bundle of $L$ is represented on $[0,2\pi ]$ by a continuous function $s$ satisfying $s(2\pi )=-s(0)$. If it were nowhere zero, the intermediate value theorem would prevent its endpoint signs from being opposite. Thus $L$ has no nowhere-zero section. The trivial bundle has the section $1$, and a bundle isomorphism would carry that section to a nowhere-zero section of $L$, a contradiction. Hence

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.

Conversely, suppose $F$ is an entire function with the displayed bound. Its real restriction has polynomial growth, so its inverse Fourier transform defines a tempered distribution byRapid decay of the Schwartz function $\hat{\phi }$ makes this integral absolutely convergent and continuous in the Schwartz space topology.

To determine the support of a distribution, take a test function supported in a half-space $x\cdot \omega \ge H_K(\omega )+\delta$, where $|\omega |=1$ and $\delta >0$. The contour-shift proof of the Paley–Wiener–Schwartz theorem gives, for each $t\ge 0$,Here is the estimate that justifies both the shift and its limiting use. Since $\hat{\phi }(-\xi -it\omega )=\int e^{ix\cdot \xi }{e}^{-tx\cdot \omega }\phi (x)dx$, repeated integration by parts with $1-{\Delta }_x$ yieldsFor fixed $t$, the same estimate is uniform over the imaginary strip from $0$ to $t\omega$. Rotate coordinates so $\omega$ is the first coordinate direction, apply the Cauchy integral theorem on truncated rectangles, and let their real edges tend to infinity. If $2L>N+n$, the boundary integrals vanish and the horizontal integrals converge absolutely; thus the claimed shift is valid. Using $H_K(t\omega )=t{H}_K(\omega )$ now bounds the pairing bywhich tends to zero. A separating hyperplane exists at every point outside the compact convex set $K$. A finite partition of unity on the support of any exterior test function reduces it to such half-spaces, so $\mathrm{supp}u\subseteq K$. Fourier inversion gives uniqueness; the entire transform of this $u$ agrees with $F$ on ${\mathbb{R}}^n$ and hence everywhere, by applying the one-variable identity theorem successively in each coordinate. This completes both directions.

Now apply this theorem to compact support solvability for a constant-coefficient ordinary differential equation. If $P$ is a nonzero constant, its kernel is zero and $u=v/P$ is the unique solution. Otherwise writeThe solution space of $P(-D)\phi =0$ isIndeed these exponentials solve the equation, their initial derivative vectors have a nonzero Vandermonde determinant, and uniqueness for the order-$m$ ordinary differential equation makes them a basis. Pairing $P(D)u=v$ with any such solution proves necessity:

For sufficiency, choose $R$ with $\mathrm{supp}v\subseteq [-R,R]$ and let $V(\zeta )=⟨v,e^{-ix\zeta }⟩$. The assumed annihilation is precisely $V({\alpha }_j)=0$ for every $j$. Since all roots of a polynomial are simple, $U=V/P$ has removable singularities at every root and is an entire function. Outside one disk, $|P(\zeta )|\ge c(1+|\zeta |{)}^m$. Inside that disk, the extended $U$ is bounded. The polynomial division preservation of exponential type therefore givesThe Paley–Wiener–Schwartz theorem produces a compactly supported distribution $u$ with transform $U$, supported in the same interval. Then $\widehat{P(D)u}=PU=V$, so Fourier inversion gives $P(D)u=v$. ThusThe compactly supported solution is unique: $P\hat{u}=0$ forces the entire transform to vanish off the finitely many roots and hence everywhere. With repeated roots, the corresponding condition would also require derivatives $V^{(k)}(\alpha )=0$, or pairing with $x^k{e}^{-i\alpha x}$, up to one less than each multiplicity.

Let ${\mathcal{C}}_{\mathbb{C}}^{\infty }$ be the additive sheaf of complex-valued smooth functions, and ${\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast }$ its multiplicative sheaf of nowhere-zero functions. The smooth exponential sequence isIts kernel is the integer-valued locally constant functions, namely the integer constant sheaf. It is surjective on stalks because a nowhere-zero smooth function has a smooth logarithm on a sufficiently small neighbourhood.

The additive smooth-function sheaf is a fine sheaf, using a partition of unity, and the smooth manifold is a paracompact space. Its positive-degree sheaf cohomology vanishes. The long exact sequence in sheaf cohomology consequently gives an isomorphism

To connect this with bundles, trivialize a smooth complex line bundle over an open cover. Its transition functions $g_{ij}$ form a multiplicative Čech cocycle, satisfying $g_{ij}{g}_{jk}=g_{ik}$. Changing trivializations changes it by a Čech coboundary. Conversely, any such cocycle glues trivial line bundles, and cohomologous cocycles give isomorphic bundles. Thus the isomorphism classes are $H^1(X,{\mathcal{C}}_{\mathbb{C}}^{\infty ,\ast })$.

The connecting isomorphism is the First Chern class. Locally choose logarithms $g_{ij}=e^{2\pi i{f}_{ij}}$; on triple intersectionsrepresents that class. Combining the gluing classification with $\delta$ gives the requested bijection:

The continuous coefficients are uniformly continuous on a compact neighborhood of $\overline{W}$. For every $x_0\in \overline{W}$, choose a ball $B_r(x_0)⋐U$ small enough thatfor all $i,j$. Part 4(b), with the frozen symmetric matrix $A=(a^{ij}(x_0))$, then applies on this ball.

Choose a finite collection of these balls and a smooth partition of unity $({\eta }_k)$ that sums to one near $\overline{W}$, with each ${\eta }_k$ supported in its corresponding ball. Applying part 4(b) to ${\eta }_{k}u$ givesSince $(a^{ij})$ is symmetric, the Leibniz rule gives the commutator formulaThe coefficients and the finitely many derivatives of the cutoff functions are bounded, soFinally $u={\sum }_k{\eta }_{k}u$ near $\overline{W}$. Summing the finite set of local estimates provesThis is the coefficient-freezing interior second-derivative estimate.

The Paley–Wiener–Schwartz theorem in its closed-ball form says that a tempered distribution $u$ is supported in ${\overline{B}}_R(0)$ if and only if its Fourier transform is the real restriction of an entire function $F$ on ${\mathbb{C}}^n$ satisfying, for some $C>0$ and integer $N\ge 0$,The distribution is unique by Fourier inversion. More generally a compact convex support set $K$ replaces $R|\mathrm{Im}\zeta |$ by its support function $H_K(\mathrm{Im}\zeta )$.

For the forward implication, a compactly supported distribution has finite order of a distribution, say $m$, on a fixed neighborhood of its support. Defineusing a cutoff equal to one near the support. Differentiation with respect to each complex coordinate is allowed in this pairing and gives ${\partial }_{{\zeta }_j}F=⟨u,-i{x}_j{e}^{-ix\cdot \zeta }⟩$, so $F$ is entire.

To obtain the exact radius $R$, rather than an enlarged radius, choose a cutoff function ${\chi }_{\epsilon }$ equal to one near ${\overline{B}}_R$ and supported in $B_{R+\epsilon }$, with derivatives through order $m$ bounded by $C_{\gamma }{\epsilon }^{-|\gamma |}$. Set $\epsilon =(1+|\zeta |{)}^{-1}$. The finite-order bound and the Leibniz rule giveThis proves the required growth estimate.

For the converse, $F$ has at most polynomial growth on real frequencies, so define its inverse tempered distribution byLet a compactly supported test function lie in a half-space $x\cdot \omega \ge R+\eta$, with $|\omega |=1$ and $\eta >0$. By contour shifting,Here is the decay needed to justify the shift and then let $t$ grow. Repeated integration by parts, applying $1-{\Delta }_x$ to $e^{-tx\cdot \omega }\phi (x)$, gives for every integer $M$,Choose $2M>N+n$. For each fixed $t$, this decay makes the vertical sides of a large rectangle vanish in the one complex coordinate parallel to $\omega$; Cauchy integral theorem then supplies the contour identity. Combining the two bounds yieldsThus $u$ vanishes on such test functions. Every point outside ${\overline{B}}_R$ has a neighborhood of this form. A partition of unity therefore proves $\mathrm{supp}u\subseteq {\overline{B}}_R$, completing the converse. The same argument with separating half-spaces proves the compact-convex-set form. This is the contour-shift proof of the Paley–Wiener–Schwartz theorem.

For a regular function $u$, the change of variables $y=tx$ givesThe map $\phi \mapsto \phi (\cdot /t)$ is continuous on the Schwartz space, so duality extends the dilation of a distribution toIts Fourier transform obeys $\widehat{{\delta }_{t}u}(\zeta )=t^{-n}F(\zeta /t)$. If the original support is in the unit ball, the theorem bounds $F$ by $C(1+|\zeta |{)}^N{e}^{|\mathrm{Im}\zeta |}$. HenceThe converse theorem gives the support law for distributional dilation

Let $({\rho }_0,{\rho }_1)$ be a smooth partition of unity subordinate to $(U_0,U_1)$. On $U_0$, extend ${\rho }_1\phi$ by zero away from the overlap, and on $U_1$ extend ${\rho }_0\phi$ similarly. DefineOn the overlap,and therefore

An almost complex structure $J$ on $(M,{\omega }_M)$ is a compatible almost complex structure when $J^2=-I$, ${\omega }_M(Ju,Jv)={\omega }_M(u,v)$, andis a positive-definite inner product. To prove existence, choose any Riemannian metric $h$ and define $A$ by ${\omega }_M(u,v)=h(Au,v)$. The metric construction of a compatible almost complex structureis smooth and compatible, so the space is nonempty.

Identify the compact symplectic manifold $N$ with its image under the symplectic embedding. Along $N$ there is a symplectic splittingChoose compatible almost complex structures on both summands and take their direct sum. Its associated metric makes the two summands orthogonal. Extend this metric from the closed submanifold $N$ to all of $M$ using a partition of unity, and apply the metric construction again. Along $N$ it recovers the prescribed direct sum, so the resulting global compatible $J$ satisfies $J(TN)=TN$. This is the relative extension of a compatible almost complex structure.

For the two compact complex curves, use the supplied holomorphic coordinates at their transverse intersection. There $C_1\cup C_2$ is $\{xy=0\}$. Replace it in a small ball by the complex annulus $\{xy=ϵ\}$ and use a cutoff in a surrounding annulus to rejoin the unchanged curves. For sufficiently small nonzero $ϵ$, the result is an embedded symplectic surface $\Sigma$. This local replacement is a bordism between the old and new cycles, soThis is the symplectic smoothing of a positive transverse node.

Give $\partial X$ the outward-normal-first boundary orientation. The Generalized Stokes theorem states that, for every compactly supported $(n-1)$-form $\omega$,

Choose an oriented coordinate cover by charts into ${\mathbb{R}}^n$ or the half-space ${\mathbb{H}}^n=\{x^1\ge 0\}$, and choose a partition of unity $({\rho }_i)$ subordinate to it. Since the family is locally finite and $\omega$ has compact support, only finitely many ${\rho }_i\omega$ are nonzero. It is therefore legitimate to write both integrals as finite sums and prove the identity for a form supported in one chart.

In an interior chart the integral of an exact compactly supported top form is zero by the fundamental theorem of calculus. In a boundary chart writeIntegrating $d\omega =({\sum }_j{\partial }_j{a}_j)d{x}^1\wedge \cdots \wedge d{x}^n$ coordinate by coordinate kills every tangential derivative. The normal derivative leaves precisely the restriction to $x^1=0$, with the sign selected by the outward-normal-first convention. This is ${\int }_{\partial X}{F}^{\ast }\omega$, proving the theorem.

Choose a Riemannian metric on $X$, using a partition of unity if necessary. For each $p$, letSince $F$ is a submersion, $D_{p}F{|}_{H_p}:H_p\to T_{F(p)}Y$ is an isomorphism. Its inverse depends smoothly on $p$, sodefines a smooth vector field on $X$ satisfying $D_{p}F(v(p))=w(F(p))$. The formula also gives $v(p)=0$ whenever $w(F(p))=0$.

The equivalent formulations of orientability of a smooth manifold are:

A positive chart orients its coordinate frame. Conversely, smoothly oriented frames determine local positive coordinate volume forms. A partition of unity subordinate to an oriented atlas glues these positive forms: at each point they are positive multiples of one another, so their weighted sum cannot vanish. Finally, a nowhere-zero top form $\omega$ declares $(v_1,\dots ,v_n)$ positive exactly when $\omega (v_1,\dots ,v_n)>0$. These constructions are inverse at the level of orientations.

Write the interpolating spline asThe interpolation equations are $A_{\mathbf{x}}c=f_{\mathbf{x}}$, where $(f_{\mathbf{x}}{)}_i=f(x_i)$, soThe B-splines are nonnegative and form a partition of unity on the spline interval. ConsequentlyTaking the supremum over $t$ and then over $\Vert f{\Vert }_{\infty }\le 1$ proves the operator norm bound

The wedge product of differential forms is the alternating tensor productFor nonzero $\eta \in {\Lambda }^{n-1}(({\mathbb{R}}^n{)}^{\ast })$, choose a volume form $\mu$. There is a nonzero vector $v$ with $\eta ={\iota }_v\mu$. Extend $v$ to a basis and use its dual coframe; then $\eta$ is a scalar multiple of $e^2\wedge \cdots \wedge e^n$, hence equals $\xi \wedge {\eta }_0$. The zero form is immediate.

To integrate a top form on a compact oriented $n$-manifold, choose a finite oriented atlas and a subordinate partition of unity; integrate each compactly supported coordinate expression and sum. A smooth map pulls forms back byIf $F^{\ast }\omega =\omega$ for a nowhere-zero top form, $F$ preserves its orientation. The change-of-variables theorem gives

A continuous map from a smooth manifold to ${\mathbb{R}}^d$ can be approximated by a smooth map with any prescribed positive continuous pointwise error bound. In coordinate charts, smoothing by local convolution and then combining with a smooth partition of unity gives the approximation. On a compact subset of an open chart domain, a smooth cutoff allows the resulting straight-line homotopy to be supported in that domain. This permits the Sard theorem to be used after approximation; the original continuous map can be space-filling and need not omit a point.

Every smooth complex line bundle on ${\mathbb{CP}}^1$ is isomorphic to exactly one twisting bundle $\mathcal{O}(k)$. Trivializing on the two standard affine charts leaves a transition function on ${\mathbb{C}}^{\ast }$; its winding number determines $k$, and a partition of unity removes the zero-winding factor.

A smooth cutoff function is a smooth function used to localize an argument, typically equal to one on one region and zero outside a slightly larger region. Such functions can be built from bump functions and smooth partitions of unity.

For $a<b$ and $0\le \theta \le 1$, the displayed estimate holds on Euclidean space and a torus with the Fourier definition of the Sobolev norm. In the squared norm, write the weighted squared Fourier coefficient as the product of its weight-$a$ value to the power $1-\theta$ and its weight-$b$ value to the power $\theta$. Holder inequality then proves the estimate. On a compact manifold, a finite coordinate cover and a partition of unity give the same estimate up to a constant. Young inequality implies the useful absorption form $\Vert u{\Vert }_{H^{s+1}}\le \epsilon \Vert u{\Vert }_{H^{s+2}}+C_{s,\epsilon }\Vert u{\Vert }_2$ for nonnegative $s$.

The support of a distribution is the complement of the largest open set on which its pairing vanishes for every supported test function. A partition of unity shows that the union of such open sets still has this vanishing property. For example, the Dirac delta distribution and all its derivatives have support $\{0\}$.

Nonnegative tensor-product surface basis factors yield nonnegative product weights. Their two partition of unity sums multiply to one. If the factors are $C^r$ and $C^s$, mixed derivatives through separate orders $r,s$ are continuous, so the resulting finite or locally finite surface representation is jointly $C^{\min (r,s)}$. Control coefficients can cancel derivative jumps and yield more smoothness; inherited continuity is a guaranteed lower bound, not a claim that every surface has exactly the basis's worst regularity.

Products of two univariate bases form a surface basis. Nonnegativity, partition of unity and coordinate-wise linear precision pass to the products by multiplication of the respective univariate identities. If the factors are $C^r$ and $C^s$, each mixed derivative through orders $r,s$ is the product of the corresponding derivatives, giving continuous mixed derivatives and joint $C^{\min (r,s)}$ regularity.

For nonnegative barycentric parameters, these total-degree Bernstein basis functions define a surface over a triangle. The triangular control net has $(n+1)(n+2)/2$ points. Its index constraint couples the three exponents, so it is not a tensor product of two independent univariate bases. The multinomial theorem gives nonnegativity and partition of unity.