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The complexified cotangent bundle decomposes into $(1,0)$ and $(0,1)$ parts. The Dolbeault operator is the type-$(0,1)$ component of the exterior derivative, $\bar{\partial }:{\Omega }^{p,q}\to {\Omega }^{p,q+1}$, and integrability gives ${\bar{\partial }}^2=0$.

Together with its $(1,0)$ counterpart, the Dolbeault operator decomposes the exterior derivative on complex differential forms.

Because the electromagnetic field tensor is the exterior derivative $F=dA$ of a potential, $dF=d^{2}A=0$. For a torsion-free connection this iswhich is the covariant form of the two homogeneous Maxwell equations.

For a smooth one-form $\alpha$ and smooth vector fields $X,Y$,In coordinates the derivatives of the components of $X,Y$ cancel against the Lie bracket of vector fields, leaving $({\partial }_i{\alpha }_j-{\partial }_j{\alpha }_i)X^i{Y}^j$. This formula extends the coordinate definition of the exterior derivative to arbitrary frames.

At a point in a normal orthonormal frame, torsion-freeness makes the frame brackets vanish there, hence the dual one-forms have zero exterior derivative there. Differentiating the interior product formula then leaves only ${\sum }_i{E}_i(f_i){\omega }_g$, the divergence of a Riemannian vector field. Since the point is arbitrary, the identity holds globally.

If $D={\bigcap }_{i=1}^q\ker {\alpha }^i$ has constant rank, the Frobenius theorem says that $D$ is integrable exactly when its annihilator is closed under the exterior derivative: equivalently,This is also equivalent to involutivity of $D$.

For a nowhere-zero 1-form $\alpha$, the hyperplane distribution $\ker \alpha$ is tangent to hypersurfaces exactly when $\alpha \wedge d\alpha =0$. In general relativity this characterizes hypersurface-orthogonal vector fields and implies vanishing twist.

On an oriented boundaryless Riemannian manifold, apply Stokes theorem to the compactly supported form $f\ast \alpha$. The graded Leibniz rule gives $\int df\wedge \ast \alpha =-\int fd\ast \alpha$. The Hodge star operator identifies these top forms with the scalar integrands displayed above. This proves that the codifferential on one-forms is $\delta =-\ast d\ast$, the formal adjoint of the exterior derivative. Compact support of $f$ suffices.

On a closed oriented Riemannian manifold, let $\alpha$ have degree $r-1$ and $\beta$ degree $r$. The graded Leibniz rule and Stokes theorem give $\int d\alpha \wedge \ast \beta =(-1{)}^r\int \alpha \wedge d\ast \beta$. The codifferential convention $\delta \beta =(-1{)}^{n(r+1)+1}\ast d\ast \beta$, combined with the square of the Hodge star, gives $\ast \delta \beta =(-1{)}^{r}d\ast \beta$. This proves that the codifferential is the formal adjoint of the exterior derivative. A boundary requires the corresponding boundary term or specified boundary conditions.

A nowhere-zero differential one-form $n$ is hypersurface orthogonal if locally $n=Fdt$ with $F\ne 0$. Its kernel consists of tangent vectors to the level hypersurfaces of $t$. Frobenius theorem makes this equivalent to $n\wedge dn=0$, where $d$ is the exterior derivative. With a torsion-free Levi-Civita connection, this is $n_{[\alpha }{\nabla }_{\beta }{n}_{\gamma ]}=0$. For a timelike unit normal, projecting the derivative on both indices gives zero antisymmetric part, which explains why hypersurface orthogonality implies symmetric extrinsic curvature.

On the punctured plane, this one-form is closed and locally the differential of the polar angle. Its line integral around a closed curve is $2\pi$ times the winding number. A once-winding circle therefore has nonzero integral, proving that no global single-valued angle potential exists and the form is not an exact differential form.

For a counterclockwise circle of radius $R>0$, substitute $x=R\cos t$, $y=R\sin t$. The pullback of a differential form isThe radius cancels. In coordinates, $P=-y/(x^2+y^2)$ and $Q=x/(x^2+y^2)$ satisfy ${\partial }_{x}Q={\partial }_{y}P$ away from the origin, so $d\omega =0$ there. The nonzero integral is a global obstruction to being an exact differential form, not a nonzero local exterior derivative.

Compare this with the smooth one-form $\eta =(-ydx+xdy)/2$ on the whole plane. Here $d\eta =dx\wedge dy$, and its integral around a radius-$R$ circle is $\pi R^2$, the enclosed area. The Generalized Stokes theorem applies to the full disc for $\eta$. For $\omega$, that disc crosses the excluded origin, where the one-form is undefined. On an annulus the theorem applies, but its induced inner boundary is clockwise: the two integrals $2\pi$ and $-2\pi$ cancel. This explains exactly why closedness and nonzero circulation can coexist.

On a compact oriented Riemannian manifold without boundary, the codifferential is the $L^2$ adjoint operator of the exterior derivative. ConsequentlyEvery eigenvalue of the Hodge Laplacian is therefore nonnegative.

For a Killing vector field $K$, its metric-dual differential one-form produces the electromagnetic field $F_{ab}=2{\nabla }_{[a}{K}_{b]}=2{\nabla }_a{K}_b$. The exterior derivative gives $dF=0$, while the second covariant derivative of a Killing vector gives ${\nabla }^a{F}_{ab}=-2R_{bd}{K}^d$. Thus this construction solves the source-free Maxwell equations in a vacuum spacetime with zero cosmological constant. The field is a test field: adding its stress-energy tensor would generally change the original metric tensor.

For a smooth manifold $M$ of dimension $n$, the tangent space $T_{p}M$ can be defined as the space of derivations at $p$: linear maps $v$ on germs of smooth real functions satisfying $v(fg)=f(p)v(g)+g(p)v(f)$. Equivalently, $v$ is the velocity of a smooth curve through $p$, with curves identified when their coordinate velocities agree. A chart $(x^1,\dots ,x^n)$ supplies the basis ${\partial }_i{|}_p$, so $v=v^i{\partial }_i{|}_p$.

The tangent bundle is the disjoint union $TM={\coprod }_{p\in M}{T}_{p}M$, with projection $\pi (v)=p$. Its smooth structure is defined by the local local trivializationsOn overlapping charts, the change of trivialization isThese are smooth changes of coordinates on a $2n$-dimensional smooth manifold, linear and invertible in each fibre. Thus $TM$ is a rank-$n$ vector bundle, and its projection is a submersion. Fibrewise addition, scalar multiplication and the zero section are smooth. A smooth section of a vector bundle $X:M\to TM$, satisfying $\pi \circ X={\mathrm{id}}_M$, is exactly a vector field; locally $X=X^i{\partial }_i$, with components transforming by the displayed Jacobian matrix. Its action on functions is a derivation, and the commutator of two such derivations is their Lie bracket.

The dual bundle of $TM$ is the cotangent bundle $T^{\ast }M$. Its smooth sections of a vector bundle are differential 1-forms, locally $\alpha ={\alpha }_{i}d{x}^i$. The coefficient transformation is the inverse transpose of that for tangent vectors, ensuring that $\alpha (X)$ is a well-defined smooth function. Tensoring these two bundles gives the tensor bundlesTheir sections of a vector bundle are tensor fields; their transition functions of a vector bundle are the corresponding tensor products of the tangent and cotangent transformations. For example, the bundle $\mathrm{End}(TM)=TM\otimes T^{\ast }M$ contains fields of linear endomorphisms, while $S^2{T}^{\ast }M$ contains symmetric bilinear fields. Taking alternating covariant tensors gives ${\Lambda }^k{T}^{\ast }M$, whose sections are differential forms. The wedge product, exterior derivative and pullback of a differential form make these bundles central to integration, Generalized Stokes theorem and de Rham cohomology.

A nowhere-vanishing section of ${\Lambda }^n{T}^{\ast }M$ chooses an orientation of a smooth manifold; such a section exists exactly when $M$ is orientable. Without an orientation of a smooth manifold, one can still integrate sections of the density bundle $|{\Lambda }^n{T}^{\ast }M|$, whose coordinate changes use the absolute Jacobian determinant. A Riemannian metric is a positive-definite section $g$ of $S^2{T}^{\ast }M$. It identifies $TM$ with $T^{\ast }M$ by the musical isomorphisms $X\mapsto g(X,\cdot )$ and gives the Riemannian volume densityOn an oriented smooth manifold, this density corresponds to a volume form.

Another organizing construction is the frame bundle $FM$. A frame over $p$ is a linear isomorphism $u:{\mathbb{R}}^n\to T_{p}M$, and the right action $u\cdot A=u\circ A$ makes $FM$ a principal bundle with general linear group $GL(n,\mathbb{R})$ as its structure group. Given a group representation $\rho :GL(n,\mathbb{R})\to GL(V)$, the associated bundle $FM{\times }_{\rho }V$ is obtained from pairs $(u,v)$ by the relation $(uA,v)\sim (u,\rho (A)v)$. The standard representation reconstructs $TM$; dual, tensor and exterior representations reconstruct the associated bundles described above. A Riemannian metric reduces the frame bundle to the orthonormal frame bundle, with orthogonal group $O(n)$ as structure group; an orientation of a smooth manifold reduces it further to the special orthogonal group $SO(n)$. A global section of $FM$ is a global frame, so it exists exactly when $M$ is a parallelizable manifold. Local triviality does not imply a global trivialization: the Hairy ball theorem prevents even one nowhere-zero vector field on $S^2$.

A connection on a vector bundle differentiates its sections in tangent directions and defines parallel transport. For $TM$, the Levi-Civita connection is distinguished by compatibility with a Riemannian metric and vanishing torsion tensor. Its coefficients are not tensor components, because a coordinate change introduces second derivatives; the difference of two connections is a section of $T^{\ast }M\otimes \mathrm{End}(TM)$. The Riemann curvature tensor is a genuine tensor measuring the failure of covariant derivatives to commute.

Finally, a smooth map $f:M\to N$ gives the pullback tangent bundle $f^{\ast }TN$. Its sections are vector fields along a map, including the velocity of a curve and Jacobi fields along a geodesic. For an embedded submanifold $i:M\hookrightarrow N$, the derivative identifies $TM$ with a subbundle of $i^{\ast }TN$, and the quotient is the normal bundle. With an ambient Riemannian metric, the normal bundle is identified with the orthogonal complement of $TM$ in $i^{\ast }TN$; the second fundamental form records the normal component of the ambient derivative of tangent fields. These constructions connect the bundle description to both intrinsic and extrinsic differential geometry.

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.

Let $\pi :T^{\ast }N\to N$ be the projection and $s_{\alpha }(q)=(q,{\alpha }_q)$ the section with image ${\Gamma }_{\alpha }$. For the canonical one-form on a cotangent bundle,Since $\pi \circ s_{\alpha }=\mathrm{id}$, evaluation on a tangent vector $v\in T_{q}N$ givesUse the same position-first standard symplectic form as before, $\omega =-d\theta$. Pullback commutes with the exterior derivative, soThe graph of a differential one-form is an embedded submanifold, because its projection is the inverse of the section on the graph. Its dimension is $\dim N$, half that of $T^{\ast }N$. Therefore the vanishing of the pulled-back symplectic form is exactly the Lagrangian submanifold condition, andThis proves that the graph of a closed one-form is Lagrangian using the intrinsic one-form rather than a coordinate guess.

The cotangent fiber translation $f(q,p)=(q,p+{\alpha }_q)$ is a smooth diffeomorphism with inverse $(q,p)\mapsto (q,p-{\alpha }_q)$, and $\pi \circ f=\pi$. For a tangent vector $v$ at $(q,p)$, the intrinsic definition of the canonical one-form on a cotangent bundle givesThus $f^{\ast }\theta =\theta +{\pi }^{\ast }\alpha$. Taking its exterior derivative with the fixed sign convention yieldsIf $\alpha$ is a closed differential one-form, the last term vanishes, proving that $f$ is a symplectomorphism. In fact the converse holds too: the pullback by the zero section shows that ${\pi }^{\ast }(d\alpha )=0$ implies $d\alpha =0$.

The exterior derivative is a linear map $d:{\Omega }^k(M)\to {\Omega }^{k+1}(M)$, agrees with the ordinary differential on functions, satisfies $d^2=0$, and obeys the graded Leibniz ruleIn local coordinates it differentiates the coefficient functions:Antisymmetry cancels the second partial derivatives, proving $d^2=0$ locally. These properties determine the coordinate-independent derivative.

For any smooth $\varphi :N\to M$, the pullback of a differential form is defined byThere is no requirement that $\varphi$ be invertible or an immersion. Pullback preserves wedge products and sends $df$ to $d(f\circ \varphi )$ by the chain rule. Locally every form is a sum of $fd{x}^{i_1}\wedge \cdots \wedge d{x}^{i_k}$; applying the preceding identities and $d^2{\varphi }^i=0$ proves

Describe the brane by an embedding $X:\Sigma \to M$, with nondegenerate induced worldvolume metric $h=X^{\ast }g$. Locally the Poincaré lemma gives $F=dA$ because $F$ is a closed differential form. For a timelike worldvolume, the geometric area and Wess-Zumino brane coupling give the action$T$ is the brane tension and $q$ the charge in this normalization. The geometric term is the higher-dimensional version of the Nambu–Goto action. Both integrals are invariant under changes of worldvolume coordinates.

The additive gauge transformation is $A\mapsto A+d\Lambda$. Its change of action isThe last step is the Generalized Stokes theorem. It vanishes for a closed worldvolume, or with suitable boundary restrictions. An open worldvolume can instead have a compensating boundary coupling, as in gauge invariance of an open-brane coupling.

More directly, for a variation vector $V$ of the embedding, Cartan's magic formula givesThus the interior force depends on $F$, not the gauge potential. Define the trace of the second fundamental form byThe area variation is $-\int \sqrt{|h|}g(V,K)$ plus a boundary term. For compactly supported interior variations the resulting equations areThey exhibit the field-strength force on a charged brane and are gauge invariant.

Two global qualifications are necessary. First, replacing $A$ by $d\Lambda$ alone, as printed, gives $d{A}^{\prime }=0$ and would erase a nonzero $F$; the intended transformation must be the additive shift above. Second, closedness of $F$ only guarantees a potential locally. If a global potential does not exist and the worldvolume bounds a filling $B$, its charged term can be defined as $q{\int }_{B}F$. Different fillings differ by a period of $F$; independence of the quantum phase requires $q{\int }_{C}F\in 2\pi \hslash \mathbb{Z}$ for the corresponding closed cycles. Without the required global potential or patching/filling data, the given closed form still defines the local variational equations, but not automatically a single globally defined real-valued action.

Use the index convention $[e_a,e_c]=C_a{}^b{}_c{e}_b$. For the left- and right-invariant coframes, the Maurer-Cartan equations areTheir dual invariant vector fields satisfyFor example, evaluating $d{\lambda }^b$ on two dual fields gives $-{\lambda }^b([L_a,L_c])$, since its invariant coefficient functions are constant. The opposite right-invariant bracket gives the other sign. The mixed bracket vanishes because left and right translations commute.

For a semisimple Lie algebra, the Killing form $B(X,Y)=\mathrm{Tr}(\mathrm{ad}X\mathrm{ad}Y)$ is nondegenerate. It is adjoint invariant: using $\mathrm{ad}[Z,X]=[\mathrm{ad}Z,\mathrm{ad}X]$ and cyclicity of trace givesGroup conjugation preserves the trace as well, so this is invariant under the full group adjoint action. Translation supplies a bi-invariant pseudo-Riemannian metric.

Lower the indicated structure index by $C_{abc}=B_{be}{C}_a{}^e{}_c$. ThenThis changes sign under exchange of $a,c$ by the bracket and under exchange of $a,b$ by invariance. Hence $C_{abc}=C_{[abc]}$.

Let $\eta (X,Y,Z)=B(X,[Y,Z])$ at the identity and extend it as a left-invariant differential form. Its value is adjoint invariant, so the extended form is also right-invariant. The three-form in the question is $-\eta$ with the index-lowering convention just specified, an irrelevant overall sign for these properties. To prove closedness rather than only invariance, apply the exterior derivative on four left-invariant fields. Its six bracket terms combine by invariance intoThe final equality is the Jacobi identity. Therefore the Cartan three-formis both bi-invariant and closed.

The exterior derivative is the real-linear degree-one operator $d:{\Omega }^k\to {\Omega }^{k+1}$ characterized by its usual derivative on functions, the graded Leibniz ruleand $d^2=0$. In local coordinates,These rules determine it uniquely, and the coordinate formula shows its existence and invariance under coordinate changes. It also commutes with pullback of a differential form.

On the unit ball let $V=x^i{\partial }_i$ be the radial vector field, and let $F_t(x)=tx$. For $k\ge 1$ define the radial homotopy operator byHere the vectors are identified using the ambient Euclidean coordinates. Equivalently $h_k\omega ={\int }_0^1{t}^{-1}{F}_t^{\ast }({\iota }_V\omega )dt$, where ${\iota }_V$ is the interior product. The apparent factor $t^{-1}$ is harmless: the coordinate coefficients of its integrand contain $t^{k-1}$, so the integral is smooth at $t=0$ and depends smoothly on $x$. This is a linear map to ${\Omega }^{k-1}$; set $h_0=0$.

For $t>0$, differentiation of the coordinate pullback and Cartan's magic formula giveFor completeness, Cartan’s identity follows by checking it on functions and coordinate one-forms: both sides give $V(f)$ on $f$ and $d(V(x^i))$ on $d{x}^i$. Both are degree-zero derivations of the exterior algebra, so the checks imply the identity on every local form. Integrating from zero to one provesFor positive $k$, the pullback by the constant map is zero, so this is the required identity. If $d\omega =0$ and $k\ge 1$, it follows that $\omega =d(h_k\omega )$. A chart about any point contains such a ball, proving the Poincaré lemma. In degree zero the formula is instead $h_{1}df=f-f(0)$: a constant function cannot be the image of the uncorrected homotopy identity. This degree-zero correction to radial homotopy specifies the necessary positive-degree convention in the printed request.

To treat the sphere, cover $S^2$ by the complements of its north and south poles. Each is diffeomorphic to ${\mathbb{R}}^2$, where the same star-shaped homotopy applies. A closed differential one-form $\alpha$ therefore has local primitives $f_N$ and $f_S$. On their overlap $d(f_N-f_S)=0$. This overlap is connected, so the difference is a constant $c$. Replacing $f_N$ by $f_N-c$ makes the primitives agree, and they glue smoothly to a global function. ThusThis proves directly that closed one-forms on the two-sphere are exact.

For the Hodge decomposition theorem, assume a compact oriented Riemannian manifold without boundary. The Hodge star operator is characterized by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ and obeys $\ast \ast \alpha =(-1{)}^{k(n-k)}\alpha$. The formal $L^2$ adjoint of $d$ on $k$-forms is $\delta =(-1{)}^{n(k+1)+1}\ast d\ast$. Define the Hodge Laplacian $\Delta =d\delta +\delta d$ and the space ${\mathcal{H}}^k=\ker \Delta$ of harmonic differential forms. Integration by parts, with no boundary term, givesHence a form is harmonic exactly when it is both closed and coclosed. The Hodge decomposition theorem states that ${\mathcal{H}}^k$ is finite-dimensional and that there is the orthogonal decompositionIf a closed form has decomposition $\alpha =h+d\beta +\delta \gamma$, then $⟨\alpha ,\delta \gamma ⟩=⟨d\alpha ,\gamma ⟩=0$. Orthogonality consequently forces $\delta \gamma =0$, so its de Rham cohomology class is represented by $h$. An exact harmonic form $h=d\beta$ has $\Vert h{\Vert }^2=⟨\delta h,\beta ⟩=0$, proving uniqueness. Thus

Every top-degree form is $\omega =f{\mathrm{vol}}_g$, and $d\omega =0$ for dimensional reasons. Its Hodge star is $f$, so the formula for $\delta$ shows that $\delta \omega =0$ exactly when $df=0$. On a connected manifold this says that $f$ is constant. Therefore the harmonic top-degree forms on a closed oriented manifold form the spaceAny such smooth manifold admits a Riemannian metric; the assertion is valid for every choice of that metric.

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.

Let ${\eta }_{ab}=g(e_a,e_b)$ be the constant diagonal metric matrix in a local pseudo-orthonormal frame, and define ${\theta }_{ab}={\eta }_{ac}{\theta }^c{}_b$. The metric compatibility of the Levi-Civita connection givesThus the lowered connection matrix is antisymmetric, including in indefinite signature. Ordinary matrix antisymmetry of ${\theta }^a{}_b$ itself is not the corresponding assertion in indefinite signature.

Let ${\omega }^a$ be the dual coframe, and write a vector field as $Y=e_b{\omega }^b(Y)$. ThenSubtract the same expression with $X,Y$ exchanged and subtract ${\omega }^a([X,Y])$. By the definition of the exterior derivative, this gives the components of the torsion tensor:The Levi-Civita connection is torsion-free, so Cartan's first structure equation is

Use $R(X,Y)={\nabla }_X{\nabla }_Y-{\nabla }_Y{\nabla }_X-{\nabla }_{[X,Y]}$ and define its matrix of curvature two-forms by $R(X,Y)e_b=e_a{R}^a{}_b(X,Y)$. Applying two covariant derivatives to $e_b$ yieldsThe first three terms are $d{\theta }^a{}_b(X,Y)$ and the last two are $({\theta }^a{}_c\wedge {\theta }^c{}_b)(X,Y)$. Therefore Cartan's second structure equation isThe associated Riemann curvature tensor components satisfy $R^a{}_b=\frac{1}{2}{R}^a{}_{bcd}{\omega }^c\wedge {\omega }^d$ with this convention.

For a position-dependent change of pseudo-orthonormal frame, write the frame row as $E^{\prime }=E\Lambda$, with ${\Lambda }^T\eta \Lambda =\eta$. The dual coframe obeys ${\omega }^{\prime }={\Lambda }^{-1}\omega$. Applying the Leibniz rule to $\nabla (E\Lambda )$ givesThe connection one-form has an inhomogeneous term because differentiating the moving basis produces $d\Lambda$. The curvature two-form instead transforms homogeneously. Indeed, curvature is tensorial in the vector to which it is applied, so $R(X,Y)(E\Lambda )=(R(X,Y)E)\Lambda =E^{\prime }{\Lambda }^{-1}R(X,Y)\Lambda$. ConsequentlyEquivalently, substituting the transformed connection into $d{\theta }^{\prime }+{\theta }^{\prime }\wedge {\theta }^{\prime }$ and using $d{\Lambda }^{-1}=-{\Lambda }^{-1}(d\Lambda ){\Lambda }^{-1}$ cancels all derivative terms and gives the same result. This transformation law holds for a general invertible frame change as well; the orthogonality condition ensures both frames remain pseudo-orthonormal.