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A compactly supported top-degree differential form on an oriented smooth manifold is integrated using orientation-preserving coordinates and a partition of unity. The coordinate change of variables formula makes the result independent of the choices. More generally, a form is integrated over a lower-dimensional oriented submanifold by its pullback. The Generalized Stokes theorem states ${\int }_{M}d\beta ={\int }_{\partial M}\beta$ with the induced boundary orientation. Integrating a closed differential form over a closed cycle depends only on its homology class, since its integral over a boundary vanishes.

For an open clamped spline knot sequence and degree $d\ge 1$, the Greville abscissae ${\xi }_i=d^{-1}{\sum }_{r=1}^d{t}_{i+r}$ reproduce the parameter. The B-spline differentiation formula gives derivative coefficients $d({\xi }_i-{\xi }_{i-1})/(t_{i+d}-t_i)=1$. Their lower-degree partition of unity makes the derivative of the weighted sum equal to one. The left endpoint value is the left parameter endpoint, fixing the integration constant to zero. This proves linear precision even with unequal knot intervals.

A locally integrable function has weak derivatives on an open set exactly when it has them near every point. Local derivatives agree almost everywhere on overlaps because they define the same distributional derivative. A countable local cover therefore gives globally defined, locally integrable derivative functions. A smooth partition of unity subordinate to the cover splits each compactly supported test function; summing the local integration identities yields the global weak-derivative identity.

A Riemannian metric gives inverse vector bundle isomorphismsIn coordinates they have matrices $g_{ij}$ and $g^{ij}$. Every smooth manifold admits a Riemannian metric by a partition of unity, so its tangent and cotangent bundles are isomorphic as real smooth vector bundles. This isomorphism depends on the metric, not merely on the manifold.

For a finite open cover $U_0,\dots ,U_d$ of $S^d$, take continuous nonnegative functions ${\varphi }_i$ summing to one and positive only in $U_i$. Such a partition of unity comes from normalizing the distance to a set functions $d(x,S^d\setminus U_i)$. The Borsuk-Ulam theorem applied to $({\varphi }_1,\dots ,{\varphi }_d)$ gives equal values at an antipodal pair. The sum condition makes ${\varphi }_0$ equal too, and some common positive value places both points in one $U_i$.

An oriented atlas is a smooth atlas whose smooth transition maps have positive Jacobian determinants on every overlap. The coordinate bases then give compatible choices of orientation of a vector space on all tangent spaces. A nowhere-vanishing top-degree differential form gives such an atlas: restrict each chart until its coefficient has constant sign, reverse one coordinate if necessary to make that sign positive, and compare the positive coefficients on overlaps. Conversely, a partition of unity glues positive local coordinate top forms to a global volume form.

An almost complex structure is a smooth endomorphism $J:TM\to TM$ with $J^2=-I$; in particular the real dimension is even. Complexifying the tangent bundle splits it into the $i$ and $-i$ eigensubbundlesinterchanged by conjugation. On a complex manifold, these are locally spanned by ${\partial }_{z^j}$ and ${\partial }_{{\bar{z}}^j}$. An integrable almost complex structure is one arising from a holomorphic coordinate atlas. Its $T^{0,1}$ sections are closed under the Lie bracket. The obstruction is the Nijenhuis tensorExpanding this expression in the two eigentypes shows that $N_J=0$ is equivalent to involutivity of $T^{0,1}$. Holomorphic coordinates make this condition necessary; the Newlander-Nirenberg theorem gives its sufficiency for smooth $J$. Integrability also gives $d=\partial +\bar{\partial }$, with ${\partial }^2={\bar{\partial }}^2=0$ and $\partial \bar{\partial }+\bar{\partial }\partial =0$. The Dolbeault operator on $T^{1,0}M=T_M^{\prime }$ defines its holomorphic vector bundle structure.

A real connection on a vector bundle $\nabla$ on $TM$ preserves $J$ when $\nabla J=0$. Its complexification then preserves both eigensubbundles. Restricting to $T_M^{\prime }$ gives a complex connection $D$. Conversely any complex connection on $T_M^{\prime }$ gives such a real connection by adjoining its conjugate on $T^{0,1}$ and restricting to the conjugation-fixed real tangent bundle. Equivalently, transfer $D$ through the real isomorphism $v\mapsto (v-iJv)/2$. These operations are inverse. Preserving $J$ alone does not impose the holomorphic compatibility condition $D^{0,1}={\bar{\partial }}_{T^{\prime }}$.

For integrable $J$, the latter condition has a precise real-connection interpretation: the mixed-type torsion tensor vanishes. In holomorphic coordinates it says $D_{{\partial }_{{\bar{z}}^i}}{\partial }_{z^j}=0$. Reality also gives ${\nabla }_{{\partial }_{z^j}}{\partial }_{{\bar{z}}^i}=0$. Since these coordinate fields commute, their torsion is zero. Conversely a $J$-preserving real connection with zero mixed torsion has these two derivatives equal; they belong to opposite eigentypes, so both vanish, proving $D^{0,1}={\bar{\partial }}_{T^{\prime }}$. This is the holomorphic tangent connection and mixed torsion criterion.

Full torsion-freeness imposes the additional symmetry ${\Gamma }_{ij}^k={\Gamma }_{ji}^k$ in $D_{{\partial }_{z^i}}{\partial }_{z^j}={\Gamma }_{ij}^k{\partial }_{z^k}$, and its conjugate. Such a $J$-preserving torsion-free connection exists on any complex manifold: local holomorphic-coordinate flat connections have these properties, and a real partition of unity combines them globally. Preservation of $J$ and zero torsion persist because both conditions are affine in the connection. Conversely if $J$ is merely almost complex, a torsion-free connection preserving it forces $N_J=0$: replace brackets by ${\nabla }_{X}Y-{\nabla }_{Y}X$ in the displayed tensor and use $\nabla J=0$; all terms cancel. Thus this connection condition detects integrability, without imposing any metric condition.

A Hermitian metric is a real Riemannian metric $g$ with $g(JX,JY)=g(X,Y)$. It induces a Hermitian metric $h$ on $T_M^{\prime }$ and the fundamental two-form $\omega (X,Y)=g(JX,Y)$. The unique Chern connection satisfies both $D^{0,1}={\bar{\partial }}_{T^{\prime }}$ and metric compatibility. In a holomorphic frame with metric matrix $H$ and coefficient-column convention, its matrix is $H^{-1}\partial H$. It need not be torsion-free when transferred to $TM$.

In holomorphic coordinates its coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Their symmetry in $i,j$ is equivalent to ${\partial }_i{h}_{j\bar{l}}={\partial }_j{h}_{i\bar{l}}$, which says $\partial \omega =0$. The conjugate gives $\bar{\partial }\omega =0$. The result that a torsion-free Chern tangent connection characterizes a Kähler metric follows: the transferred Chern connection is torsion-free exactly when $\boxed{d\omega =0}$, the Kähler metric condition. In that case it equals the Levi-Civita connection, by uniqueness of the metric-compatible torsion-free connection. Equivalently a Hermitian metric is Kähler exactly when its Levi-Civita connection preserves $J$. In the reverse direction $\nabla J=0$ implies $\nabla \omega =0$, and zero torsion gives $d\omega =0$.

This explains the role of Kähler metrics: they make the real Riemannian and holomorphic connection theories coincide. A general complex manifold has torsion-free $J$-preserving connections and has Hermitian metrics, but need not have a connection satisfying both requirements simultaneously. On a Kähler manifold, the Kähler identities then link the Dolbeault Laplacian and Hodge Laplacian, giving ${\Delta }_d=2{\Delta }_{\bar{\partial }}=2{\Delta }_{\partial }$ and the resulting harmonic Hodge decomposition theorem for compact Kähler manifolds.

For a smooth complex vector bundle $E$, a connection is a complex-linear map $D:\Gamma (E)\to A^1(M;E)$ satisfying $D(fs)=df\otimes s+fDs$ for smooth complex functions $f$. Local trivializations supply flat local connections. A locally finite smooth partition of unity $({\rho }_i)$ subordinate to them gives $D={\sum }_i{\rho }_i{D}_i$; local finiteness makes the sum smooth and ${\sum }_i{\rho }_i=1$ gives the Leibniz rule. Thus connections on a vector bundle always exist on the usual paracompact smooth manifold.

Extend $D$ to bundle-valued forms by $D(\alpha \otimes s)=d\alpha \otimes s+(-1{)}^{\mathrm{deg}\alpha }\alpha \wedge Ds$. Squaring shows that $D^2(fs)=f{D}^{2}s$, so the curvature form is the endomorphism-valued two-form $\Theta =D^2$. With a row frame $e=(e_1,\dots ,e_r)$, write $De=eA$; for coefficient columns $s=ev$, $Ds=e(dv+Av)$. Applying $D$ again and using the graded Leibniz rule givesThe cross terms $-A\wedge dv$ and $A\wedge dv$ cancel. Hence the Cartan curvature matrix equation in this convention isFor a frame change $e^{\prime }=eg$, the connection matrix and curvature transform as $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Trace is invariant under conjugation, so the local forms $\mathrm{Tr}\Theta$ patch into the global trace of vector-bundle curvature.

The induced determinant connection on $\det E={\Lambda }^{r}E$ iswhere the one-form factor of each $D{s}_j$ is placed in front of the section factors. In the local determinant frame $e_1\wedge \cdots \wedge e_r$, only the diagonal terms survive, so its connection form is $\mathrm{Tr}A$. Its curvature is $d\mathrm{Tr}A$. Sinceby pairing off-diagonal terms and using $A_{ii}\wedge A_{ii}=0$, we obtainIt is closed locally by $d^2=0$, and thus closed globally.

For two connections, $B=D_1-D_0$ is a global endomorphism-valued one-form: the derivative terms cancel in their Leibniz rules. In a local frame $A_1=A_0+B$, soTaking traces cancels the mixed terms, because both factors are one-forms, and cancels $\mathrm{Tr}(B\wedge B)$. Therefore the trace curvature transgression formula isThe right-hand side is globally exact, proving independence of the de Rham cohomology class without needing a sheaf-cohomology identification.

Finally let $e$ be a holomorphic frame for the holomorphic line bundle, with $h=\Vert e{\Vert }^2>0$. Define $De=e\partial \log h$, extending by the Leibniz rule. If $e^{\prime }=eg$ for a nowhere-zero holomorphic function $g$, then $h^{\prime }=|g{|}^{2}h$ andThis is exactly the line-bundle frame-change law for a connection, so the local definitions glue globally. Its $(0,1)$ part is the Dolbeault operator. Metric compatibility follows from $d\log h=\partial \log h+\bar{\partial }\log h$, or $dh=h(A+\overline{A})$ for $A=\partial \log h$. These conditions uniquely force this $A$, identifying the Chern connection. Since a scalar one-form wedges with itself to zero,This agrees with the local formula for the Chern connection on a line bundle; keeping the order of the two Dolbeault differentials fixes the sign.

Write $N_{i,d}$ for the partition-of-unity-normalized degree-$d$ B-spline, supported on $[t_i,t_{i+d+1}]$. Assume $d\ge 1$. First take distinct knots; endpoint repetitions will be interpreted as confluent limits. For fixed parameter $t$, let $f_t(\tau )=(\tau -t{)}_{+}^d$, and formwhere the brackets denote a divided difference in the knot variable $\tau$. This representation follows from the defining properties, rather than assuming a derivative recurrence. If $t<t_i$, $f_t$ is a degree-$d$ polynomial in all its knot arguments and its order-$(d+1)$ divided difference is zero. If $t>t_{i+d+1}$, every truncated-power term is zero. Between knots it is a degree-$d$ polynomial with $d-1$ continuous derivatives. Those support and smoothness conditions determine a one-dimensional spline space: there are $d+1$ spans, the continuity conditions leave $2d+1$ coefficients, and vanishing through order $d-1$ at both support ends imposes $2d$ conditions.

The remaining scalar is fixed by the prescribed integral. Put $a=t_i$, $b=t_{i+d+1}$. Integrating the truncated power first givesThe order-$(d+1)$ divided difference of this polynomial is its leading coefficient $1/(d+1)$, so the proposed $N_{i,d}$ has integral $(b-a)/(d+1)$, exactly the specified normalization.

Now differentiate with respect to $t$ and apply the final recursion for a divided difference:In terms of normalized lower-degree B-splines, this is the B-spline differentiation formulaA zero denominator corresponds to a collapsed-support basis term and contributes zero. Repeated knots follow by coalescing distinct knots; at knots where the classical derivative is discontinuous, read the formula on each open span or one-sided. The stated maximal interior continuity corresponds to simple interior knots, while the clamped endpoint repetitions do not affect the interior argument.

Let the control-point abscissae be the Greville abscissaeCollect the coefficient of $N_{i,d-1}$ after differentiating:The difference of the two consecutive knot averages telescopes toso every noncollapsed coefficient is one. The lower-degree basis on the endpoint-trimmed knot vector has partition of unity; equivalently the outer lower-degree basis functions on the original vector have collapsed support. Therefore $X^{\prime }(t)=1$ on the parameter interval. With $d+1$ repeated left endpoint knots at $a$, the clamped basis has $N_{0,d}(a)=1$, all other values zero, and ${\xi }_0=a$. Thus $X(a)=a$ fixes the constant of integration. Continuity and endpoint limits now giveThis is linear precision at Greville abscissae. It requires neither equal knot spacing nor a restriction on the other control-point coordinates; those determine the remaining coordinates of the parametric curve.

A differential form of degree $k$ is integrated over an oriented $k$-dimensional smooth manifold by integrating its coefficient in orientation-preserving coordinates. For an oriented parametrization $F:U\to M$, this means integrating the pullback $F^{\ast }\alpha$ over $U$; a partition of unity combines charts. The change of variables formula makes the result independent of the charts. Reversing the orientation of a smooth manifold reverses the integral. The Generalized Stokes theorem says, for a compact oriented $k$-dimensional smooth manifold with boundary and a smooth $(k-1)$-form $\beta$,where the right-hand side includes the boundary pullback, and the boundary has the outward-normal-first boundary orientation. Compact support suffices on a noncompact manifold. This unifies the fundamental theorem of calculus, circulation and flux identities.

For example, take the unit disk $D$ oriented by $dx\wedge dy$ and the differential one-form $\beta =\frac{1}{2}(xdy-ydx)$. Its exterior derivative is $d\beta =dx\wedge dy$, so the area integral is $\pi$. Parametrizing the positively oriented circle by $(x,y)=(\cos t,\sin t)$ gives the pullback $\beta =\frac{1}{2}dt$ and the boundary integral ${\int }_0^{2\pi }\frac{1}{2}dt=\pi$, directly illustrating Generalized Stokes theorem.

For the distance calculation, write $u=|{\mathbf{r}}_1|$, $v=|{\mathbf{r}}_2|$, $x=u+v$ and $y=u-v$; these are scalar radii, not position vectors. The triangle inequality and its reverse give $r\le u+v=x$ and $|y|=|u-v|\le r$. Also,Thus $r\le x\le 2R-|y|$ and $|y|\le r$. These conditions, with $u,v\ge 0$, are also sufficient for a triangle with side lengths $u,v,r$. Endpoints correspond to collinear configurations.

Assume the two positions are independent random variables, each having constant volume probability density inside the radius-$R$ ball. Independence is needed: uniform marginal distributions alone do not determine the distance distribution. Put $C=3/(4\pi R^3)$ and $\mathbf{s}={\mathbf{r}}_2-{\mathbf{r}}_1$. The change of variables $({\mathbf{r}}_1,{\mathbf{r}}_2)\mapsto ({\mathbf{r}}_1,\mathbf{s})$ has unit Jacobian determinant, so the joint probability volume differential form is the product of the two normalized volume forms. In spherical coordinates for ${\mathbf{r}}_1$ and for $\mathbf{s}$ relative to its axis, it isThe polar coordinates fail on axes and at zero radii, which are sets of zero volume and do not affect the probability. Here $\theta$ is the angle from ${\mathbf{r}}_1$ to $\mathbf{s}$. The law of cosines becomesDifferentiating and taking the wedge product with $du\wedge dr$ eliminates the terms involving $du$ and $dr$:Consequently the positive integration density transforms by $|\sin \theta d\theta |=vdv/(ur)$ at fixed $u,r$. The minus sign is accounted for by reversal of limits: $v$ decreases as $\theta$ increases. The radial variable retained is $r$, while the polar angle is replaced by $v$; equivalently one can start with the angle between the two position vectors and replace that angle by $r$. The wording of the printed hint conflates these two coordinate choices, but its volume form gives the stated transformation directly.

Integrating $\sin {\theta }_{1}d{\theta }_{1}d{\varphi }_1$ over the first orientation gives $4\pi$, and integrating $d\chi$ gives $2\pi$. The probability density function of $r$ is thereforewhere $D_r$ has $0\le u,v\le R$ and $|u-v|\le r\le u+v$. There are no such configurations for $r>2R$.

To evaluate this integral explicitly, the Jacobian determinant of $(u,v)=((x+y)/2,(x-y)/2)$ has absolute value $1/2$. Hence $uvdudv=(x^2-y^2)dxdy/8$ as a positive density. For $0\le r\le 2R$, set $a=\min (r,2R-r)$. The domain is $-a\le y\le a$, $r\le x\le 2R-|y|$, and evenness in $y$ givesAn antiderivative, zero at $y=0$, yieldsSubstituting $a=r$ for $r\le R$ and $a=2R-r$ for $r\ge R$ gives the same polynomial in both intervals:Thus the distance between two uniform points in a three-dimensional ball has the final densityIt is zero outside this interval. Its nonnegativity also follows from $f_R(r)=3r^2(4R+r)(2R-r{)}^2/(16R^6)$, and direct integration gives ${\int }_0^{2R}{f}_R(r)dr=1$. The derivation uses the full six-dimensional probability volume differential form, rather than treating the three scalar distances as independent.

For a positive-dimensional smooth connected manifold, three equivalent formulations of orientability of a smooth manifold are an oriented atlas, a continuously varying orientation of each tangent space, and a nowhere-vanishing smooth top-degree differential form. An orientation of a tangent space means a choice of one class of ordered bases, where two bases agree when their change-of-basis determinant is positive. The second formulation chooses these classes locally continuously; it does not require a global frame.

An oriented atlas has positive Jacobian determinant on every overlap. Its coordinate bases then give a consistent orientation of all tangent spaces. Conversely, a continuous choice of tangent-space orientations makes the sign of a coordinate basis locally constant. Shrink to connected charts and reverse one coordinate when necessary; the resulting positively positively oriented coordinate charts have positive transition determinants. An orientation in atlas language is the maximal collection of charts compatible with that positive sign convention.

We use the partition of unity theorem: every open cover of a Hausdorff second-countable smooth manifold has a smooth locally finite subordinate partition. From the oriented atlas, choose such a partition of unity ${\rho }_i$. Extend each ${\rho }_{i}d{x}_i^1\wedge \cdots \wedge d{x}_i^n$ by zero outside its chart and sum the locally finite family. At every point, the nonzero summands are positive multiples of one another and at least one coefficient is positive. Their sum is a smooth nowhere-zero form $\omega$. Conversely, a nowhere-zero form declares a basis positive exactly when $\omega (v_1,\dots ,v_n)>0$, giving a continuously varying orientation and hence an oriented atlas. Two top forms define the same orientation precisely when ${\omega }^{\prime }=h\omega$ for a smooth positive function $h$. Since a nonvanishing coefficient cannot change sign on a connected manifold, these definitions give exactly two opposite orientations.

For the unit sphere in ${\mathbb{R}}^{n+1}$, defineThe radial vector $x$ is a unit normal, so appending any tangent basis gives an ambient basis and the determinant never vanishes. The form is smooth and provides the outward-normal-first orientation. Therefore $\boxed{S^n\text{ is orientable for every }n}$. For $n=0$, this is the nonzero zero-form taking opposite signs on the two sphere points; zero-dimensional manifolds admit orientations by assigning generator signs at their points.

For $n\ge 1$, the quotient map $\pi :S^n\to {\mathbb{RP}}^n$ is a twofold smooth covering with deck transformation $A(x)=-x$. Its action on the sphere form isIf $n$ is odd, the form is invariant and descends using the local covering inverses to a nowhere-vanishing form on the quotient. If $n$ is even, suppose the quotient had a nonvanishing top form $\eta$. Its pullback would be $h\omega$ for a smooth nonzero function $h$ on the connected sphere and would be invariant under $A$. Invariance would require $h(-x)=-h(x)$, impossible because $h$ has constant sign. This proves the orientability of real projective space criterionThe separate endpoint ${\mathbb{RP}}^0$ is one point and is orientable, so the full nonnegative-dimensional answer is $n=0$ or $n$ odd.

A solid body transform is a rigid transformation: it preserves distances within the object. In three-dimensional coordinates it has the form $T(x)=Rx+b$, with translation $b\in {\mathbb{R}}^3$ and an orthogonal matrix $R$. For a proper physical rotation, the coefficient constraints areEquivalently the columns of $R$ are an orthonormal basis with positive orientation. A reflection also preserves distance but has $\det R=-1$ and is excluded when “solid-body motion” means a rotation and translation. There are three rotational and three translational degrees of freedom, not twelve independent matrix coefficients.

Computationally use the homogeneous coordinates representationThe last row is $(0,0,0,1)$; matrix multiplication composes transforms. Alternatively store a unit quaternion together with $b$, maintaining its unit norm. A transpose gives the rotational inverse: $T^{-1}(y)=R^T(y-b)$.

Write the cubic Bézier curve as $P(t)={\sum }_{i=0}^3{B}_i^3(t)p_i$, where $B_i^3(t)=\left(\genfrac{}{}{0ex}{}{3}{i}\right)(1-t{)}^{3-i}{t}^i$. The binomial theorem gives ${\sum }_i{B}_i^3(t)=1$. Transforming the controls producesThis proves pointwise equivariance, including the original parameter values; it is stronger than a statement that the two curves merely have the same shape.

Orthogonality was not used in the calculation. The identical proof applies to every affine map $T(x)=Ax+b$, including scales, shears and even singular affine maps. It applies to every representation $P(u)={\sum }_i{\varphi }_i(u)p_i$ with scalar, control-independent partition of unity basis functions: Bézier curves of every degree, B-splines, surfaces built on a tensor-product surface basis, triangular Bézier patches, and more general linear control-point representations. Positivity is not needed for affine equivariance of a geometric basis. For linear stationary subdivision matrices with row sums one, each refinement step commutes with the affine transformation; taking a convergent limit proves the corresponding assertion for subdivision curves and subdivision surfaces. Rules with geometry-dependent nonlinear weights require separate equivariance tests and are not covered automatically.

The affine class is the largest class that preserves every such Euclidean-control construction with unchanged basis weights, under the usual continuity assumption. In fact the degree-one case would forcefor all $x,y$ and $0\le t\le 1$. This segment identity, followed by continuity, forces $T$ to be affine. Nonlinear transforms such as perspective division generally fail it.

There is a further extension when the representation itself is enlarged: apply a projective transformation linearly to homogeneous controls $(w_i{p}_i,w_i)$, then dehomogenize the weighted sum. The transformed objects are rational Bézier curves or non-uniform rational B-splines, with transformed weights as well as transformed Euclidean controls. This works wherever the new denominator is nonzero. It does not assert projective invariance of ordinary polynomial Bézier curves with their old weights.

A tensor-product surface basis is formed from two independent univariate basis families ${\varphi }_i(u)$ and ${\psi }_j(v)$ on separate parameter intervals. Given a rectangular control net $p_{ij}$, its parametric surface isThe parameters and indices vary independently over a product domain. The representation can be evaluated first in either parameter, giving a convenient computational advantage. Assume a finite basis, or locally finite spline bases, so differentiation and summation below are justified locally.

For positivity, if each factor basis is nonnegative, then ${\varphi }_i(u){\psi }_j(v)\ge 0$. If both factors are strictly positive at a point, their product is strictly positive there. Nonnegative weights are the usual geometric meaning of positivity; zero boundary weights are allowed. With partition of unity, this gives the local convex hull enclosure of the active control points.

For continuity, suppose the first family is $C^r$ and the second $C^s$, including matching derivatives across their respective spline knots. On every polynomial piece, and by matching at the knot lines,Every factor on the right is continuous in its own variable, so its product is jointly continuous. Thus all these mixed derivatives are continuous, in particular the surface is jointly $C^{\min (r,s)}$. The separate coordinate smoothness can be better: $r$ orders in $u$ and $s$ in $v$. For piecewise polynomials of degrees $d_u,d_v$, an interior knot of multiplicity $k$ in the first factor normally gives $C^{d_u-k}$ matching in $u$, inherited along that knot line; similarly for $v$. The statement is an inherited lower bound. Special control points can cancel derivative jumps, so a particular surface may be smoother than a generic member of its basis.

For summation to unity, independent summation factors exactly:This also explains why translations of all control points translate the surface by the same vector, as in affine equivariance of a geometric basis. These three arguments establish tensor-product inheritance of geometric basis properties directly from the factor identities.

A triangular Bézier patch is a non-tensor-product example. For barycentric parameters $u,v,w\ge 0$ with $u+v+w=1$, a quadratic patch isThe six basis functions have total degree two and sum to $(u+v+w{)}^2=1$. Their index constraint and triangular parameter domain do not arise from two independently indexed univariate factors on a rectangle. An individual triangular polynomial could be embedded in another larger representation, but this triangular basis and control net are not a tensor-product surface definition.

Take a locally finite trivializing cover $(U_{\alpha })$ of the real vector bundle $E$, and a subordinate smooth partition of unity $({\rho }_{\alpha })$. Each trivialization identifies its fibers with ${\mathbb{R}}^m$ and hence supplies a smooth Euclidean inner product $g_{\alpha }$ there. DefineEach weighted term extends by zero outside its chart, since the weight has support within that chart. Local finiteness makes the sum smooth in every bundle chart. At each $b$ at least one weight is positive; for $v\ne 0$ every nonzero contribution to $g_b(v,v)$ is positive. Thus $g$ is a smooth positive-definite fiber metric, which is precisely an inner product varying smoothly with the fibers.

A G-structure on a vector bundle is a reduction of its frame bundle from $GL(m,\mathbb{R})$ to $G$: a principal $G$-subbundle $P\subseteq \mathrm{Fr}(E)$ whose extension of structure group is the full frame bundle. Equivalently, it is a choice of smooth local frames with transition matrices in $G$, and the reduction consists of the right $G$-orbits of these frames. The adapted frames are part of the structure, not merely a statement that an individual fiber has a basis.

Given a fiber metric, the Gram-Schmidt process transforms any smooth local frame into Euclidean orthonormal frames. At each step the squared norm to be divided out is positive, so the divisions and positive square roots depend smoothly on the base point. Two such frames differ by an orthogonal matrix. All the orthonormal frames therefore form a principal $O(m)$ reduction, an orthogonal structure on a real vector bundle.

Conversely, let $P$ be an $O(m)$ reduction. A frame $u:{\mathbb{R}}^m\to E_b$ in $P_b$ definesReplacing $u$ by $uq$ for $q\in O(m)$ leaves this rule unchanged, since $q$ preserves the Euclidean inner product. It is consequently independent of the adapted frame and defines a positive-definite fiber metric. Its coefficients are smooth in local adapted frames. These constructions are mutually inverse. An $O(m)$-structure is equivalent to a smooth fiberwise inner product.

For $SO(m)$, the additional condition is an orientation of a vector bundle. Such a reduction supplies consistently oriented orthonormal frames, since its transition determinants are $+1$. Conversely an orientation together with a fiber metric selects the oriented Euclidean orthonormal frames and gives an $SO(m)$ reduction. ThusIt is not true for every real bundle. The Möbius line bundle is obtained from $[0,2\pi ]\times \mathbb{R}$ by $(2\pi ,v)\sim (0,-v)$. If it had an orientation, hence a nowhere-zero continuous section, its coefficient on the interval would satisfy $f(2\pi )=-f(0)$ while staying nonzero. The intermediate value theorem makes that impossible. Since $SO(1)=\{1\}$, it has no $SO(1)$ reduction. For each $m\ge 1$, its direct sum with a trivial rank-$(m-1)$ bundle has transition matrix $\mathrm{diag}(-1,1,\dots ,1)$ and determinant line equal to the Möbius line. It is also nonorientable and has no $SO(m)$-structure. This obstruction concerns the bundle's orientation, independently of the existence of a metric.

DefineEvery projective point with $z_3\ne 0$ has a unique unit representative whose third coordinate is positive real, so $F$ is a diffeomorphism. More explicitly, in the affine coordinate $u=(z_1/z_3,z_2/z_3)$ its inverse is $w=u/\sqrt{1+|u{|}^2}$. Lift $F$ to the unit sphere by $\tilde{F}(w)=(w,\sqrt{1-|w{|}^2})$. The last coordinate is real, so its contribution $d{x}_3\wedge d{y}_3$ pulls back to zero. The reduction identity therefore yieldsThis proves the symplectic ball chart in complex projective space directly, including the precise unit radius.

For the final unheaded request, embed the two disjoint balls in this chart. Fix arbitrary $0<{\rho }_i<r_i$, and choose intermediate radii ${\rho }_i<{\rho }_i^{\prime }<r_i$. On the images of the balls of radius ${\rho }_i^{\prime }$, transport the standard complex structure by the given symplectic embeddings. It is compatible with $\Omega$. Their smaller closed neighborhoods are compact and disjoint. Choose a global Riemannian metric agreeing there with the associated compatible metrics, using a partition of unity outside these neighborhoods, and apply the metric construction. The resulting global compatible almost complex structure $J$ agrees with the transported standard structure on each ball of radius ${\rho }_i$ and on a slightly larger neighborhood. This extends metrics, rather than averaging almost complex structures.

Let $p_1,p_2$ be the centres. They are distinct. The degree-one J-holomorphic curve supplied in the question has total area $\pi$, because its homology class is the projective line class. We use the sharp Euclidean version of the Monotonicity theorem for a J-holomorphic curve: a nonconstant holomorphic curve through the centre of a standard complex ball, with no boundary in the ball and counted with its parametrization multiplicities, has area in the radius-$\rho$ ball at least $\pi {\rho }^2$. This includes branched points: their density is a positive integer $m$, and the stronger lower bound is $m\pi {\rho }^2$. One way to see the constant is that complex curves are calibrated minimal surfaces; their stationary area ratio $\mathrm{area}(C\cap B(\rho ))/{\rho }^2$ is nondecreasing and its limit at the centre is $m\pi$.

The closed degree-one curve cannot stay entirely inside either chart ball: there the form is exact, so Stokes theorem would give zero total area, contradicting its positive area. Thus the branch through each centre crosses the surrounding spheres, and the stated monotonicity bound applies to each restriction. The two ball images are disjoint, so their area contributions add without overlap. It follows thatLetting both smaller radii increase to their original radii givesThis two-ball packing obstruction in the projective plane is sharper than volume alone: a volume comparison would only give $r_1^4+r_2^4\le 1$.

Let $M$ be an oriented smooth $m$-dimensional manifold with boundary and $\alpha$ a smooth compactly supported $(m-1)$-differential form. Give $\partial M$ the outward-normal-first orientation. A partition of unity subordinate to oriented interior and boundary charts reduces the proof to a compactly supported form on ${\mathbb{R}}^m$ or on a half-space. Indeed $\alpha ={\sum }_j{\chi }_j\alpha$, so $d\alpha ={\sum }_{j}d({\chi }_j\alpha )$; the sum of terms involving $d{\chi }_j$ is zero. Exterior derivative commutes with chart pullbacks, so each summand can be calculated in coordinates.

Write the local form asFor an interior chart, integrate each derivative using the fundamental theorem of calculus: its integral is zero because $f_j$ has compact support. For a boundary chart modeled on $x^m\ge 0$, all tangential derivatives again integrate to zero, while ${\int }_0^{\infty }{\partial }_m{f}_{m}d{x}^m=-f_m(x^{\prime },0)$. The outward normal is $-{\partial }_m$; its contraction with the volume form gives exactly this sign in the induced boundary integral of $\alpha$. Summing the chart identities proves the Generalized Stokes theorem:Compact support can be replaced by compactness of $M$; on a noncompact manifold, sufficient support or convergence conditions are necessary.

In a convenient cardinal parameter normalization, take the unit $m$-cube and project by the sum of its coordinates. Its pushforward density is the unit-direction univariate box splineEquivalently, with $B_1=1_{[0,1)}$ at its discontinuity points,Here there are $m$ convolution factors; some conventions instead index the spline by its degree $m-1$. The half-open convention for $m=1$ assigns knot values consistently and does not change the pushforward measure.

Support and positivity. A sum of $m$ numbers in $[0,1]$ lies in $[0,m]$, so the closed support is finite. For $0<x<m$, the interval of $u\in (0,1)$ with $0<x-u<m-1$ has positive length. Induction in the convolution recurrence therefore gives $B_m(x)>0$ in the interior. For $m\ge 2$, $B_m(0)=B_m(m)=0$: strict positivity includes the open support interval, not its endpoints.

Partition of unity. The translates of $B_1$ cover the line exactly once under the half-open convention. Nonnegativity permits interchange of sum and integral, and induction givesThus $\boxed{\sum _j{B}_m(x-j)=1}$, a locally finite sum. Each $B_m$ also has integral one, although partition of unity does not mean its maximum value is one.

For $m\ge 2$, repeated integration gives the explicit truncated-power formOn each unit knot interval this is a polynomial of degree at most $m-1$, and on $(0,1)$ it is $x^{m-1}/(m-1)!$, so the degree is attained. Each truncated power has $m-2$ continuous derivatives; the derivative of order $m-1$ jumps at the first knot. Therefore the dependence on the projected box dimension isFor $m=1$ the pieces are constants and the function is discontinuous at the knots. If geometric width is measured under orthogonal projection onto the unit diagonal instead, the projected coordinate is $x/\sqrt{m}$, so the physical support width is $\sqrt{m}$; its unit-mass density is $\sqrt{m}{B}_m(\sqrt{m}x)$. These are different scalings of the same cardinal spline, not different continuity or degree claims.

For a bivariate box spline, choose a spanning direction matrix $\Xi =({\xi }_1,\dots ,{\xi }_m)$ in the plane and project the unit cube by $u\mapsto \Xi u$. Define its normalization byThe support is the zonotope ${\sum }_r[0,1]{\xi }_r$, and the polynomial pieces generally have total degree $m-2$. The grid comes from the chosen direction families and their lattice translates. Two independent directions, with repetitions, give tensor-product B-splines on a parallelogram grid. Three families such as $e_1,e_2,e_1+e_2$ give a three-direction triangular grid; an affine change of basis can make it the usual equilateral triangular grid. The four families $e_1,e_2,e_1+e_2,e_1-e_2$ give the criss-cross square grid and the quadratic four-direction box spline. Changing the direction matrix and repeating directions therefore changes the grid, support and smoothness while retaining the box-projection construction.

Over the Grassmannian ${\mathrm{Gr}}_k({\mathbb{C}}^N)$, the complex tautological bundle on a Grassmannian isIts fibre at $W$ is the k-plane $W$ itself. A neighbourhood of a fixed plane consists of graphs of linear maps $A:W\to W^{\perp }$; $(A,w)\mapsto (\mathrm{graph}A,w+Aw)$ gives a local trivialization. The inclusions of coordinate spaces give the infinite Grassmannian and its tautological rank-k bundle ${\gamma }_k^{\infty }$.

Use the customary compact Hausdorff base convention for the classification of complex vector bundles by a Grassmannian; equivalently the argument uses finite subordinate partitions of unity. Give a rank-k complex vector bundle $E\to X$ a continuous fiber metric given by Hermitian inner products. Choose a finite trivializing cover with local orthonormal coordinates $t_i:E{|}_{U_i}\to U_i\times {\mathbb{C}}^k$, and a subordinate partition of unity ${\rho }_i$ with support contained in $U_i$. Defineextending each component by zero off its support. This is continuous and satisfies $\Vert J_x(v){\Vert }^2={\sum }_i{\rho }_i(x)\Vert v{\Vert }^2=\Vert v{\Vert }^2$, so it is a fibrewise embedding. The image planes vary continuously: in local frames, their orthogonal projection matrix is $J(J^{\ast }J{)}^{-1}{J}^{\ast }$. Thus they define a map $c_E:X\to {\mathrm{Gr}}_k({\mathbb{C}}^{kr})$, and $v\mapsto (x,J_{x}v)$ identifies $E$ with $c_E^{\ast }{\gamma }_k^{\infty }$. This proves that every bundle is a tautological pullback vector bundle, by the finite-dimensional embedding of a complex vector bundle.

We next prove the two independence assertions that turn this construction into a bijection. First, homotopic maps have isomorphic pullback vector bundles. Pulling back along a homotopy gives a bundle over compact Hausdorff $X\times I$; embed that bundle in a finite trivial vector bundle by the construction just given. Its image has a continuous family of rank-k orthogonal projection matrices $P(x,t)$. Compactness gives a subdivision $0=t_0<\cdots <t_s=1$ withRestriction of $P(x,t_{j+1})$ to the image at $t_j$ is injective: if it killed $v$, then $v=(P(x,t_j)-P(x,t_{j+1}))v$, impossible by the strict norm bound unless $v=0$. Equal ranks make it an isomorphism, with continuous local inverses. Composing these bundle maps identifies the endpoint vector bundles. The fact that close orthogonal projections identify their image bundles therefore makes the pullback assignment well-defined on homotopy classes.