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For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

The divergence of a Riemannian vector field $X$ is the trace of the covariant derivative map $Y\mapsto {\nabla }_{Y}X$:It is characterized by ${\mathcal{L}}_X{\omega }_g=(\mathrm{div}X){\omega }_g$ in the oriented case. The metric-dual vector field $X_{\theta }={\theta }^{♯}$ is defined by $g(X_{\theta },Y)=\theta (Y)$ for every $Y$, using the musical isomorphism.

On a compact oriented manifold without boundary, the permitted exactness assertion and Stokes theorem imply ${\int }_M\mathrm{div}Y{\omega }_g=0$ for every smooth vector field $Y$. For a smooth function $a$, the product rule givesSince $⟨da,\theta ⟩=X_{\theta }(a)$, the formal-adjoint identity yieldsThis holds for all $a$; taking $a=\delta \theta +\mathrm{div}{X}_{\theta }$ proves the pointwise formulaIt agrees with $\delta \theta =-\ast d\ast \theta$, since $\ast \theta ={\iota }_{X_{\theta }}{\omega }_g$ and $d{\iota }_{X_{\theta }}{\omega }_g={\mathcal{L}}_{X_{\theta }}{\omega }_g$.

The Bochner formula for one-forms, with the positive Hodge Laplacian, isHere $\Delta =d\delta +\delta d$ acts on one-forms. The covariant derivative $\nabla \theta$ is a section of $T^{\ast }M\otimes T^{\ast }M$, and its formal-adjoint composition, the rough Laplacian, isThis expression is independent of the local orthonormal frame. The Ricci term is the zero-order bundle endomorphism specified byThus the identity separates the differential energy of the form from the curvature contribution; the sign of the Ricci term matches positive sectional curvature on the sphere.

Let $\theta$ be a harmonic one-form on a compact connected manifold with nonnegative Ricci curvature. Pair the formula with $\theta$ and integrate:Both integrands are nonnegative and continuous, so each vanishes everywhere. In particular $\nabla \theta =0$, proving that harmonic one-forms are parallel under nonnegative Ricci curvature. This integration uses the Riemannian volume density and formal adjoints, so orientation is not necessary for the final conclusion.

Fix $p\in M$. The evaluation mapis injective: a parallel form with zero value at $p$ stays zero under parallel transport along every path, and connectedness makes every point reachable from $p$. ThereforeThis is the dimension bound for harmonic one-forms under nonnegative Ricci curvature. The bound is sharp: on a flat $n$-torus the $n$ constant coordinate one-forms are parallel and harmonic. If Ricci curvature is positive definite at even one point, the same integral and parallelism argument force every harmonic one-form to vanish.

For a Killing vector field $\xi$, set $K^{\mu \nu }={\nabla }^{\mu }{\xi }^{\nu }=-K^{\nu \mu }$. Choose surface orientation conventions so that the Komar charge, up to overall normalization, isThis is the Komar current from trace-reversed stress-energy construction. The equality is the spacetime Stokes theorem. With $[{\nabla }_{\mu },{\nabla }_{\nu }]V^{\rho }=R^{\rho }{}_{\sigma \mu \nu }{V}^{\sigma }$, the Killing equation and ${\nabla }_{\nu }{\xi }^{\nu }=0$ give $J_{\xi }^{\mu }=R^{\mu }{}_{\nu }{\xi }^{\nu }$. In four dimensions, at zero cosmological constant, the Einstein field equations give the currentReversing orientation changes the common overall sign, not conservation. The contracted Bianchi identity givesAn isometry preserves scalar curvature, and the second term contracts symmetric and antisymmetric tensors. Equivalently stress-energy conservation, the Killing equation, and ${\mathcal{L}}_{\xi }T=0$ cancel the matter expression; its trace derivative needs this last observation. Hence $\boxed{{\nabla }_{\mu }{J}_{\xi }^{\mu }=0}$. Nonzero cosmological constant adds $\Lambda {\xi }^{\mu }$ to the geometric current, also divergence-free.

Take the normal vector outward from the cylinder-and-cone solid. The surface consists of the cylindrical side and conical roof, with no bottom disk; this convention fixes the otherwise unspecified orientation of a surface. The boundary is the radius-two circle at $z=-2$. Reversing every normal vector reverses the final sign.

For $\mathbf{F}=(y{z}^2,0,0)$, its curl is $(0,2yz,-z^2)$. For the cylindrical parametrized surface, use ${\mathbf{r}}_c=(2\cos \theta ,2\sin \theta ,z)$, with $-2\le z\le 2$, and use ${\mathbf{r}}_{c,\theta }\times {\mathbf{r}}_{c,z}=(2\cos \theta ,2\sin \theta ,0)$ as the outward area vector. Its flux isFor the cone, let $q=4-z$, $2\le z\le 4$ and ${\mathbf{r}}_k=(q\cos \theta ,q\sin \theta ,z)$. Its outward area vector is ${\mathbf{r}}_{k,\theta }\times {\mathbf{r}}_{k,z}=(q\cos \theta ,q\sin \theta ,q)$. ThusThe total surface integral is therefore $-16\pi$ with the outward convention, or $+16\pi$ for the opposite convention.

For Stokes theorem, the induced bottom-circle orientation is counterclockwise as viewed from above, opposite to the orientation of an outward bottom cap. Use $\mathbf{r}=(2\cos \theta ,2\sin \theta ,-2)$ with increasing $\theta$. Thenagreeing with the two parametrized fluxes. The seam at $z=2$ is internal and cancels between the two pieces; the cone tip does not supply another boundary curve.

The geometric object underlying an $SU(2)$ gauge field is a principal connection on a principal bundle $P\to M$ with structure group SU(2). The base $M$ is spacetime, or an oriented Riemannian manifold in the Euclidean formulation. A choice of local section identifies each fibre with SU(2) and produces a local gauge potential. Such a choice is a choice of internal frame; changing it produces a gauge equivalence of principal connections. The bundle specifies the global gluing of these frames, while the principal connection specifies how frames at neighbouring points are compared.

The SU(2) group consists of $2\times 2$ unitary matrices of determinant one. Its Lie algebra consists of traceless Skew-Hermitian matrices. For definiteness usewhere the ${\sigma }_a$ are Pauli matrices. The commutator fixes all signs below. The positive invariant inner product on this Lie algebra is $-\mathrm{tr}(XY)$. Absorb the coupling into the gauge potential, so the gauge covariant derivative in the defining two-dimensional representation is $D=d+A$ with $A=A^a{t}_a$. A convention using Hermitian Pauli matrices instead moves factors of $i$ and the coupling into $D$; the underlying principal connection is the same geometric data.

A principal connection can be given by a Lie algebra-valued differential one-form $\mathcal{A}$ on $P$ satisfyingHere ${\xi }_P$ is the vertical vector generated by the right group action. The first identity recovers vertical motion and the second makes the construction independent of a fibre frame. Its kernel is the horizontal distribution of a principal connection, complementary to the vertical tangent spaces. A curve is horizontal when $\mathcal{A}$ annihilates its velocity, so a principal connection defines parallel transport along curves in $M$.

For a local section $s$, the local principal connection form is $A=s^{\ast }\mathcal{A}$. Change the section to $s^{\prime }=sh$, with $h:U\to SU(2)$. Differentiating $sh$ has a translated horizontal part and a vertical part from $dh$. The two defining identities for the principal connection therefore giveOn overlapping trivializations this is exactly the compatibility law for the local principal connection forms. In particular, the gauge potentials need not glue as ordinary globally defined differential one-forms on $M$. On a nontrivial principal bundle there need not be any global section with respect to which one could write a single matrix-valued $A$.

An active gauge equivalence of principal connections is generated by a bundle automorphism covering the identity on $M$ and commuting with the right SU(2) action. Its local description has the same transformation formula, but its local group-valued functions must satisfy transition compatibility on overlaps. This distinguishes the structure group SU(2) from the full gauge group of a fixed bundle. Two descriptions related by a passive change of section encode the same principal connection; two connections related by such an automorphism lie in the same physical gauge orbit. The space of principal connections on a fixed bundle is affine: the difference of two connection forms is a differential one-form with values in the adjoint bundle, since their inhomogeneous transformation terms cancel.

The curvature of a principal connection isIt is horizontal and equivariant, hence descends to a differential two-form on $M$ valued in the adjoint bundle. In a local section the gauge curvature becomesThe product $A\wedge A$ combines the exterior product with matrix multiplication. The commutator term is the specifically non-Abelian interaction. Substitution of the transformation of $A$, using $d(h^{-1})=-h^{-1}(dh)h^{-1}$, cancels all derivatives of $h$ and givesThus the gauge curvature transforms tensorially even though the gauge potential does not. Geometrically, the curvature of a principal connection measures the failure of the horizontal distribution of a principal connection to be integrable; physically it is the non-Abelian gauge field strength.

Matter fields are sections of associated vector bundles. A field in the defining representation of SU(2) is a section of $E=P{\times }_{SU(2)}{\mathbb{C}}^2$. Its two local components satisfy ${\psi }^{\prime }=h^{-1}\psi$ under the section convention above. The induced gauge covariant derivative obeysThis covariance is why replacing ordinary derivatives by gauge covariant derivatives makes the kinetic terms compatible with changes of internal frame. Applying the gauge covariant derivative twice yields $D^2\psi =F\psi$, so the gauge curvature also measures the noncommutativity of covariant differentiation. Other group representations give other associated vector bundles and replace $A$ by its image under the differential of the representation.

For an adjoint-valued differential form $\Phi$ of degree $r$, setExpansion of $F=dA+A\wedge A$, using $d^2=0$, proves the gauge-theory Bianchi identityThis is an identity for every principal connection, independent of any field equation. It should be distinguished from the dynamical Yang-Mills equations.

To obtain dynamics, equip $M$ with a Riemannian metric and an orientation, and let $\ast$ be its Hodge star operator. With the Skew-Hermitian normalization above, the positive Euclidean Yang-Mills action isThe matrix trace and the Hodge star operator make this independent of the choice of local section. The minus sign compensates for the negative trace form on Skew-Hermitian matrices. Varying the principal connection gives $\delta F=D_A\delta A$, and thereforeFor compactly supported variations, or on a closed base, integration by parts gives the Yang-Mills equationsIn components these are $D_{\mu }{F}^{\mu \nu }=0$. The gauge-theory Bianchi identity provides the complementary geometric identity $D_{A}F=0$. Coupling matter adds the associated gauge current to the dynamical equation; the transformation law of the gauge covariant derivative ensures covariance of that current as well.

The principal connection also supplies nonlocal observables. Along a path $\gamma$, parallel transport is determined byThe ordering is necessary because the Lie algebra matrices at different points need not commute. The resulting Wilson line transforms as $U_{\gamma }^{\prime }=h(\gamma (1){)}^{-1}{U}_{\gamma }h(\gamma (0))$. For a closed path, its matrix trace is a Wilson loop, independent of the frame at the basepoint. These observables record the holonomy of a connection. A flat principal connection has trivial holonomy of a connection around sufficiently small contractible loops but may have nontrivial holonomy of a connection around noncontractible loops. Thus vanishing local gauge curvature need not eliminate global gauge information.

The global topology becomes particularly visible in four dimensions. For a closed oriented four-dimensional base, Chern-Weil theory gives the Second Chern number of the defining associated vector bundle:The plus sign here uses Skew-Hermitian curvature and the mathematical second-Chern convention: expanding $\det (I+iF/(2\pi ))$ with $\mathrm{tr}F=0$ gives precisely the displayed Second Chern form. A convention defining a physics topological charge with a leading minus sign instead calls that charge $-k$. Neither choice changes the absolute-value action bound.

The Second Chern number is independent of the principal connection on a fixed bundle. Indeed, the gauge-theory Bianchi identity implieswhose integral vanishes on a closed base by Stokes theorem. A nonzero Second Chern number obstructs a global trivialization: for a globally defined $A$, $\mathrm{tr}(F\wedge F)=d\mathrm{tr}(A\wedge dA+\frac{2}{3}A\wedge A\wedge A)$, so its integral on a closed base would be zero. On $S^4$, gluing two trivial bundles over four-balls uses a transition map $S^3\to SU(2)$. Since SU(2) as the three-sphere identifies the target with $S^3$, such gluing maps have integer degree of a map between oriented manifolds, illustrating how distinct topological sectors arise.

In four Euclidean dimensions, the Hodge star operator squares to one on differential two-forms. Decompose $F=F_{+}+F_{-}$ with $\ast F_{\pm }=\pm F_{\pm }$. Define the nonnegative squared norms $a_{\pm }=-\int \mathrm{tr}(F_{\pm }\wedge \ast F_{\pm })$. Orthogonality givesEquality means that one component vanishes, giving the self-dual Yang-Mills equations or the Anti-self-dual Yang-Mills equations. Such finite-action solutions are Yang-Mills instantons. They solve the full Yang-Mills equations because $D_A\ast F=\pm D_{A}F=0$. With the second-Chern convention just specified, anti-self-dual curvature has $k\ge 0$ and self-dual curvature has $k\le 0$. On noncompact spacetime one must impose suitable decay or boundary conditions before treating the topological integral as an integer and using the same bound without boundary terms.

Local gauge potentials, matter gauge covariant derivatives, gauge field strength, parallel transport and Second Chern number are different manifestations of one principal connection. The principal bundle preserves the global information that a single local gauge potential cannot capture; the Yang-Mills action then selects its physical dynamics.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field on a neighborhood of it, Stokes theorem statesThe oriented surface element and boundary direction must agree by the right-hand rule. Every boundary component is included; for an upward-oriented graph over an annulus the outer boundary is anticlockwise and the inner boundary clockwise when viewed from above.

Direct use of partial derivatives givesWith the upward oriented surface element found in part (b), the curl flux integrand over the annulus $A$ becomesEvery term in the numerator is odd in $x$ or $y$, so its integral over the symmetric annulus is zero. Thus the surface integral isEach boundary circle has constant $z$, hence $dz=0$. Its line integral is therefore $\int (-ydx+xdy)$, with no contribution from the third field component. An anticlockwise radius-$r$ circle gives $2\pi r^2$. The outward boundary orientation is anticlockwise at $r=2$ and clockwise at $r=1$, soThe two computed values agree, verifying Stokes theorem with both boundary components. A downward choice would give $-6\pi$ on both sides.

A sufficient condition is that $G$ be an open simply connected domain, with $\mathbf{F}$ continuously differentiable. Under this condition a curl-free vector field has a global potential of a conservative vector field and hence path independence by the fundamental theorem for line integrals. One way to see the global step is to deform the closed loop formed by two paths into a point inside $G$: the integral is unchanged during the deformation because the curl is zero, as expressed by Stokes theorem. Therefore simple connectivity is a sufficient domain hypothesis. It is not a claim that every particular curl-free field needs such a domain; a given field can have a global potential on a domain with holes.

Use $\gamma (t)=(\cos t,\sin t,0)$ for $0\le t\le 2\pi$. On this path $\mathbf{F}=(-\sin t,\cos t,0)={\gamma }^{\prime }(t)$, so its line integral isThe first two components of the curl vanish because the field is independent of $z$ and has zero third component. Writing $q=x^2+y^2$, its third curl component isThus this azimuthal inverse-radius vector field is a curl-free vector field on $G={\mathbb{R}}^3\setminus \{(0,0,z):z\in \mathbb{R}\}$, but its line integral is not path independent. This domain is not simply connected: the circle links the removed axis. A spanning disk through the axis is inadmissible for Stokes theorem, since the vector field is undefined there. Zero curl alone does not force global path independence.

Use compatible orientations for the loop and spanning surface. The relevant Maxwell equations areThe magnetic flux transport for a moving circuit separates the change of the field from the motion of its boundary. To see the boundary term, freeze $\mathbf{B}$ at time $t$. Orient the closed swept surface as $S(t+\delta t)-S(t)+S^{\prime }$. Its side element isto first order. Hence its flux is $-\delta t{\oint }_C\mathbf{B}\cdot (\mathbf{v}\times d\mathbf{r})$. By the divergence theorem and $\nabla \cdot \mathbf{B}=0$, the sum of the three fluxes is zero, giving the purely geometric changeAdding the temporal field change and applying Stokes theorem yieldsThusThe boundary contribution is the motional electromotive force; the field contribution is Faraday's law.

Use the given orientation throughout, and the usual convention that the manifold has no boundary. In an oriented manifold chart the metric volume form isIf $y$ is another oriented chart and $A=\partial x/\partial y$, then $g_y=A^t{g}_{x}A$, so $\sqrt{\det g_y}=(\det A)\sqrt{\det g_x}$ because $\det A>0$. This is exactly the transformation of the top exterior product. The formulas therefore agree on overlaps and define a smooth positive volume form. Equivalently it takes value one on any positively oriented orthonormal tangent frame.

The metric induces an inner product on $p$-forms by making the increasing exterior products of an orthonormal coframe orthonormal. The Hodge star operator is the unique pointwise linear map $\ast :{\Lambda }^p{T}^{\ast }M\to {\Lambda }^{n-p}{T}^{\ast }M$ satisfyingFor an increasing multi-index $I$, $\ast e^I$ is the complementary wedge with the sign making $e^I\wedge \ast e^I={\omega }_g$. Swapping the blocks of $p$ and $n-p$ factors introduces $(-1{)}^{p(n-p)}$. ThereforeIn particular $\ast 1={\omega }_g$ and $\ast {\omega }_g=1$.

For compactly supported smooth $f$, the $(n-1)$-form $f\ast \alpha$ has compact support. Stokes theorem and the graded Leibniz rule giveA top form $\tau$ equals $(\ast \tau ){\omega }_g$. Consequently this Hodge integration by parts for one-forms becomes exactlyCompact support of $f$ is enough; $\alpha$ need not itself have compact support.

Define the codifferential on $p$-forms by $\delta =(-1{)}^{n(p+1)+1}\ast d\ast$ and the Hodge Laplacian by $\Delta =d\delta +\delta d$. A harmonic differential form is a smooth form in $\ker \Delta$. On functions this is the positive Laplace-Beltrami operator, $\Delta f=\delta df=-{\mathrm{div}}_g{\mathrm{grad}}_{g}f$. This sign convention is required by the product identity; it is the negative of the $\mathrm{div}\mathrm{grad}$ convention also commonly used for the Laplace-Beltrami operator.

For a $p$-form $\beta$, the square of the Hodge star operator and the codifferential formula giveApplying the second formula to $d\beta$ and the first to $\delta \beta$ givesThus Hodge star commutes with the Hodge Laplacian. Since $\ast$ is invertible, $\beta$ is harmonic if and only if $\ast \beta$ is harmonic. This holds without compactness; we have not used the generally false noncompact implication that a harmonic form must be closed and coclosed.

For a smooth function $h$ and one-form $\alpha$, the same graded Leibniz rule gives $\delta (h\alpha )=h\delta \alpha -\ast (dh\wedge \ast \alpha )=h\delta \alpha -⟨dh,\alpha {⟩}_g$. Apply this to $d(f_1{f}_2)=f_{2}d{f}_1+f_{1}d{f}_2$ to obtain the product rule for the positive Laplace-Beltrami operator

The Hodge decomposition theorem states that on a compact oriented boundaryless Riemannian manifold, smooth forms have the $L^2$-orthogonal decompositionand each de Rham cohomology class has a unique harmonic differential form representative. To spell out the last conclusion, a harmonic form is closed and coclosed because $⟨\Delta \eta ,\eta ⟩=\Vert d\eta {\Vert }^2+\Vert \delta \eta {\Vert }^2$. If a closed form decomposes as $h+du+\delta v$, then $d\delta v=0$; integration by parts gives $\Vert \delta v{\Vert }^2=⟨v,d\delta v⟩=0$. Thus it represents $h$. A harmonic exact form has zero norm by adjointness, proving uniqueness. The analytic existence of the decomposition is the stated Hodge decomposition theorem.

Now let $M$ be compact, connected and oriented. A harmonic function satisfies $0=⟨f,\Delta f⟩=\Vert df{\Vert }^2$, so it is constant. The Hodge star operator identifies ${\mathcal{H}}^0(M)$ with ${\mathcal{H}}^n(M)$, hence ${\mathcal{H}}^n(M)=\mathbb{R}{\omega }_g$. A Riemannian metric exists by Question 3 even if none was initially chosen. Using the harmonic representative of each class, we obtain the top de Rham cohomology of a compact connected oriented manifoldThe class is nonzero also directly from Stokes theorem, since ${\int }_M{\omega }_g>0$ while every exact top form has zero integral. Boundarylessness matters: a compact interval has zero first de Rham cohomology.

Stokes theorem equates the circulation around the induced boundary orientation of an oriented surface to the normal flux of curl:The usual version requires a piecewise smooth oriented surface and a vector field continuously differentiable on a neighbourhood of it. We use the upward orientation wherever the surface is a graph; each outer circle is counterclockwise viewed from above, and an inner circle is clockwise.

Write $\rho =(x^2+y^2{)}^{1/2}$. The surface of revolution has $z=\sqrt{{\rho }^2+1-\lambda }$ with $0\le z\le 1$. For $0<\lambda <1$, it is a disk-shaped cap of a two-sheeted hyperboloid, running from $\rho =0,z=\sqrt{1-\lambda }$ to $\rho =\sqrt{\lambda },z=1$. For $\lambda =1$, it is a right circular cone, $z=\rho$, including its apex. For $\lambda >1$, it is an annular strip of a one-sheet hyperboloid, between the circles $\rho =\sqrt{\lambda -1},z=0$ and $\rho =\sqrt{\lambda },z=1$.

For $z>0$, differentiation of the graph gives $z_x=x/z$ and $z_y=y/z$, so the upward vector surface element of a graph and its scalar area areFor $\lambda >1$ this graph expression is singular at the lower circle because the surface has a vertical tangent there, rather than a singular geometric surface. A regular parametrization isIt proves that the graph surface integral is interpreted by its integrable endpoint limit. The same parametrization works away from the pole or apex in the other regimes.

Direct differentiation of the first vector field givesIts surface integral is twice the projected area, namely $2\pi \lambda$ for $0<\lambda \le 1$ and $2\pi [\lambda -(\lambda -1)]=2\pi$ for $\lambda >1$. On an oriented circle of radius $R$, $\mathbf{F}\cdot d\mathbf{x}=R^{2}d\varphi$. The outer circulation is $2\pi \lambda$; an inner circle, when present, contributes $-2\pi (\lambda -1)$. ThusAt $\lambda =1$, excise a small apex circle of radius $\epsilon$, apply Stokes theorem to the smooth truncated cone, and let $\epsilon \to 0$. The removed flux and its inner circulation are both $2\pi {\epsilon }^2$, so the limiting equality holds. This avoids silently applying a smooth-surface theorem at the cone's nonsmooth apex.

The second vector field is the azimuthal inverse-radius vector field, $\mathbf{G}={\mathbf{e}}_{\varphi }/\rho$. Away from the axis,The other two curl components vanish because the field is independent of $z$ and has zero $z$ component. On any positively oriented circle around the axis, $\mathbf{G}\cdot d\mathbf{x}=d\varphi$, so its circulation is $2\pi$, independent of radius. ThereforeFor $\lambda >1$ the surface avoids the axis, so Stokes theorem applies directly and returns zero curl flux. For $0<\lambda <1$ the disk cap meets the axis, and for $\lambda =1$ the apex lies on it: $\mathbf{G}$ is undefined there, so the neighbourhood hypothesis of Stokes theorem fails. Removing a small circle gives cancelling inner and outer circulations; the inner circulation remains $-2\pi$ as its radius tends to zero. Hence it cannot be discarded as it could for $\mathbf{F}$. A nonzero circulation here is compatible with zero curl away from the excluded axis.

Use real differential forms and the usual convention that a smooth manifold has no boundary unless one is specified. On a noncompact manifold the formal-adjoint identity below uses compactly supported forms. On a compact manifold it has no boundary contribution; if a boundary were allowed, extra boundary conditions would be necessary.

The Riemannian metric gives the dual inner product on $T_x^{\ast }M$ and hence the inner product on exterior powers of a cotangent space:Extend this bilinearly. Equivalently, the increasing wedge products of an orthonormal coframe are an orthonormal basis of ${\Lambda }^p{T}_x^{\ast }M$. The Hodge star operator is the unique map $\ast :{\Lambda }^p{T}_x^{\ast }M\to {\Lambda }^{n-p}{T}_x^{\ast }M$ satisfying $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$. If $e^1,\dots ,e^n$ is a positively oriented orthonormal coframe, $\ast e^I=\epsilon (I,I^c)e^{I^c}$. Moving the $p$ factors of $I$ across the $n-p$ factors of its complement changes the sign by $(-1{)}^{p(n-p)}$, soThe basis formula also proves that $\ast$ is an isometry.

For $\beta \in {\Omega }^{p-1}(M)$ and $\alpha \in {\Omega }^p(M)$ with compactly supported product, the Stokes theorem and the Leibniz rule for the exterior derivative giveThus $(d\beta ,\alpha {)}_{L^2}=(-1{)}^p{\int }_M\beta \wedge d\ast \alpha$. The formal adjoint operator $\delta$ satisfies $\ast \delta \alpha =(-1{)}^{p}d\ast \alpha$. Applying the inverse star and its square on degree $p-1$ gives the codifferentialThe exponent is congruent to $p+(p-1)(n-p+1)$ modulo two, so this convention is consistent in every degree. Define the Hodge Laplacian by $\Delta =d\delta +\delta d$ and a harmonic differential form by $\Delta \alpha =0$. On a compact manifold,Both nonnegative terms vanish for a harmonic form, giving $d\alpha =\delta \alpha =0$. Conversely these two equations imply harmonicity.

Now let $n=4$. The printed eigenvalue relation omits the subscript on its right-hand side; the intended equation is $\ast {\alpha }_{\pm }=\pm {\alpha }_{\pm }$. On two-forms ${\ast }^2=1$, soThese are respectively self-dual two-forms and anti-self-dual two-forms, and sum to $\alpha$. Their uniqueness follows because the $+1$ and $-1$ eigenspaces have zero intersection. Star is an orthogonal involution here, hence is self-adjoint and its opposite eigenspaces are orthogonal.

For closed two-forms define $Q([\alpha ],[\beta ])={\int }_N\alpha \wedge \beta$. Adding $d\xi$ to $\alpha$ changes the integral by ${\int }_{N}d(\xi \wedge \beta )=0$; adding an exact form to $\beta$ has the same effect. This is the real de Rham intersection form in dimension four. The wedge product of differential forms in degrees two and two is symmetric because $(-1{)}^{2\cdot 2}=1$; hence $Q$ is a symmetric bilinear form on $H_{\mathrm{dR}}^2(N;\mathbb{R})$.

In four dimensions $\delta =-\ast d\ast$ in every degree. If $h$ is a harmonic two-form, $\delta h=0$ gives $d\ast h=0$, and $dh=0$ gives $\delta \ast h=-\ast dh=0$. Thus $\ast h$ is harmonic, and ${\mathcal{H}}^2={\mathcal{H}}^{+}\oplus {\mathcal{H}}^{-}$ is the harmonic star-eigenspace decomposition. Use the permitted unique harmonic representative for each de Rham cohomology class. For harmonic $h,k$,The mixed terms vanish by orthogonality. The first restriction is positive definite and the second negative definite. In particular a nonzero harmonic $h$ has $Q(h,\ast h)=\Vert h{\Vert }_{L^2}^2>0$, proving nondegeneracy. Hence the signature of the intersection form from harmonic duality isThe finite dimensions and the harmonic description also follow from the Hodge decomposition theorem.

For completeness, the Hodge decomposition theorem on a compact oriented Riemannian manifold without boundary states the finite-dimensional harmonic space and the $L^2$-orthogonal direct sumThe three summand components are unique, although their potentials need not be. Given an exact three-form $\beta =d\omega$ on $N$, decompose its smooth two-form potential as $\omega =h+d\xi +\delta \psi$, with $\psi \in {\Omega }^3$. Since $dh=0$ and $d^2=0$, $\beta =d\delta \psi$. Put $a=\delta \psi$. In dimension four, $a=-\ast d\ast \psi$, and ${\ast }^2=1$ on two-forms gives $\ast a=-d\ast \psi$, so $d\ast a=0$. ConsequentlyThis constructs a self-dual primitive of an exact three-form. The sum contains no factor $1/2$: it is twice the self-dual projection of the coexact potential. Using the unscaled projection alone would give $d\sigma =\beta /2$. This proves the final unheaded request.

We prove the Bott-Chern Poincaré lemma on a polydisc $B$. Let $p,q\ge 1$ and $d\alpha =0$, with $\alpha$ of type $(p,q)$. The allowed Dolbeault-Poincaré lemma first gives $\alpha =\bar{\partial }{b}_0$, where $b_0$ has type $(p,q-1)$. Since $\bar{\partial }\partial b_0=-\partial \alpha =0$, repeatedly applying the same lemma constructsZero spaces outside the allowed bidegrees cause no difficulty. The form $\partial b_{q-1}$ has type $(p+q,0)$, is holomorphic because its Dolbeault operator vanishes, and is $\partial$-closed. The holomorphic Poincare lemma on a polydisc gives a holomorphic form $h$ of type $(p+q-1,0)$ with $\partial h=\partial b_{q-1}$. This holomorphic version follows from the radial homotopy formula: on positive-degree holomorphic forms the ordinary radial Poincare lemma homotopy preserves holomorphic coefficients. If the displayed form is zero, take $h=0$.

Now $b_{q-1}-h$ is $\partial$-closed. The conjugate Dolbeault-Poincaré lemma, obtained by conjugating the allowed lemma, gives $b_{q-1}-h=\partial c_{q-1}$, since its holomorphic degree is positive. Moving backwards, if the last modified form is $\partial c_j$, thenApply the conjugate Dolbeault-Poincaré lemma again to obtain $b_{j-1}+\bar{\partial }{c}_j=\partial c_{j-1}$. The modification does not change its Dolbeault operator, so this process continues down to $b_0$. In the case $q=1$, subtract $h$ at this last step instead. Finally,Thus $\boxed{H_{BC}^{p,q}(B)=0(p,q\ge 1)}$.

This vanishing fails on general complex manifolds. On the complex projective line, let $\omega$ be its Fubini-Study form. It is a $d$-closed $(1,1)$-form with ${\int }_{{\mathbb{P}}^1}\omega >0$. If it were $\partial \bar{\partial }f$, it would be the exact differential form $d(\bar{\partial }f)$, whose integral is zero by Stokes theorem. Therefore $\boxed{H_{BC}^{1,1}({\mathbb{P}}^1)\ne 0}$.

The degree of a map between oriented manifolds measures how many times the domain covers the target, with signs recording the local local orientation of a manifold. Let $M,N$ be connected, oriented closed manifolds of the same dimension $n>0$. A continuous map $f:M\to N$ acts on top-dimensional homology bywhere $[M],[N]$ are their fundamental classes. Connectedness and the choices of local orientation of a manifold identify $H_n(N;\mathbb{Z})$ with $\mathbb{Z}$. Reversing the orientation of either manifold changes the sign; reversing both does not.

For a smooth map, Sard theorem supplies a regular value $y$. Its inverse image is discrete and, by compactness, finite. At each $x\in f^{-1}(y)$ the differential is an isomorphism; let its sign be $+1$ or $-1$ according to whether it preserves or reverses the chosen local orientations. The degree as a sum of local degrees isThe sign is computed in oriented manifold charts. The value is independent of the chosen regular value, even when inverse images appear or disappear: the signed count is the coefficient of $[N]$ in $f_{\ast }[M]$.