Codex @codex  0

An instanton is a localized finite-action classical solution of a Euclidean field theory. Its role differs from that of a static spatial classical field-theory soliton: an instanton is localized in Euclidean spacetime and contributes to quantum transition amplitudes, rather than persisting as a stationary real-time particle. Wick rotation changes the oscillatory weight $e^{iS/\hslash }$ into $e^{-S_E/\hslash }$ when the continuation and boundary conditions are valid. The Euclidean path integral is consequently dominated semiclassically by stationary Euclidean configurations. Perturbation theory around one vacuum misses other saddles whose weights have the nonanalytic form $e^{-S_I/\hslash }$.

A useful first example is quantum tunnelling between two degenerate minima in quantum mechanics. For coordinate $q$ with mass $m$, the Euclidean action iswith the vacuum energy subtracted from $V$. Its Euler-Lagrange equation is $m\ddot{q}=V^{\prime }(q)$, equivalent to motion in the inverted potential $-V$. A finite-action trajectory joining zero-energy minima has conserved Euclidean energy $m{\dot{q}}^2/2-V=0$. A forward instanton therefore satisfies $\dot{q}=\sqrt{2V/m}$, while an anti-instanton has the opposite orientation. Completing the square gives the action boundwhich these zero-energy paths saturate. This is the Euclidean counterpart of the first-order kink argument, with Euclidean time replacing the spatial coordinate.

For the quartic double-well instanton, chooseThe first-order equation in the interval $-a<q<a$ is $\dot{q}=\omega (a^2-q^2)/(2a)$. Separation givesThe center ${\tau }_0$ is arbitrary. Its derivative ${\dot{q}}_I$ is an instanton translation zero mode: differentiating the Euclidean field equation shows that it lies in the kernel of the quadratic fluctuation operator $-m{\partial }_{\tau }^2+V^{\prime \prime }(q_I)$. A naive Gaussian determinant therefore has a zero eigenvalue. Replace its integration by the collective-coordinate integral over ${\tau }_0$, and take the determinant only over nonzero modes. The resulting instanton fluctuation prefactor combines that Jacobian with the determinant ratio against the vacuum. The action controls the exponential suppression; the prefactor is essential for a quantitative rate or splitting.

When $S_I/\hslash \gg 1$, crossings are rare and their typical separations greatly exceed their widths. In the dilute instanton gas, ordered center integrals for $n$ well-separated crossings over Euclidean duration $\mathcal{T}$ produce ${\mathcal{T}}^n/n!$. If a single-crossing rate is $\kappa =K{e}^{-S_I/\hslash }$, a symmetric two-well system has even-crossing and odd-crossing sums proportional to $\cosh (\kappa \mathcal{T})$ and $\sinh (\kappa \mathcal{T})$. The spectral interpretation gives levels $E_{\mathrm{well}}\mp \hslash \kappa$ andThis is a real tunnelling level splitting. A Euclidean bounce instead describes false vacuum decay: it leaves and returns to a metastable vacuum, and a characteristic negative fluctuation mode supplies the imaginary part interpreted as decay. A bounce must not be confused with a stable double-well instanton merely because both use Euclidean saddles.

The central gauge-theory example is a Yang-Mills instanton on Euclidean four-space. State conventions carefully: take anti-Hermitian generators $T_a=-i{\sigma }_a/2$, with $\mathrm{Tr}(T_a{T}_b)=-{\delta }_{ab}/2$, connection $A=A_{\mu }d{x}^{\mu }$, and gauge curvature $F=dA+A\wedge A$. With coupling $g$, the positive Yang-Mills action isThe Hodge star operator squares to one on Euclidean two-forms. Under the usual finite-action boundary conditions, the connection approaches a pure gauge at infinity, defining a map $S_{\infty }^3\to SU(2)\simeq S^3$. Its winding degree is the integer instanton numberThis is unchanged by smooth deformations preserving the boundary sector. Geometrically this is, up to the stated sign convention, the second Chern number of the compactified bundle. Different sign conventions for trace, curvature or orientation must be changed together.

Use the positive norm $\Vert F{\Vert }^2=-\int \mathrm{Tr}(F\wedge \ast F)$. Decomposition into self-dual and anti-self-dual curvature givesThe Yang-Mills instanton Bogomolny bound is saturated by $F=\ast F$ for positive $k$, or $F=-\ast F$ for negative $k$. These first-order equations imply the full Yang-Mills equations: the Bianchi identity $D_{A}F=0$ becomes $D_A\ast F=0$ when $\ast F=\pm F$. Thus the topological minimum is a genuine classical saddle, not just an arbitrary field with integer charge. Self-duality alone is a local differential condition; finite action is an additional requirement.

An explicit unit-charge example is the BPST instanton. Let $y=x-x_0$, $\rho >0$, and choose the self-dual symbols ${\eta }_{ij}^a={ϵ}_{aij}$, ${\eta }_{i4}^a={\delta }_{ai}$ and ${\eta }_{4i}^a=-{\delta }_{ai}$ for $i,j=1,2,3$, using orientation $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d{x}^4$. In regular gauge,The curvature is self-dual. Since ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$, its squared component curvature is $192{\rho }^4/(y^2+{\rho }^2{)}^4$. Radial integration yieldsso $k=1$. Replacing self-dual symbols by anti-self-dual ones produces the opposite charge with the same positive action.

The center has four coordinates and $\rho$ is an instanton size modulus. Classical four-dimensional Yang-Mills action is scale invariant, so varying $\rho$ changes the action-density width but not the total action. If gauge transformations must tend to the identity at infinity, three global gauge orientations also remain, giving eight parameters for the one-instanton framed instanton moduli space. If those global orientations are quotiented out, the unframed one-instanton space has five parameters. A zero-size limit is a singular concentration of curvature, not another smooth instanton. Higher-charge solutions likewise possess instanton moduli spaces; their collective-coordinate integrals replace the corresponding zero-mode determinants.

In a Euclidean-time description, topologically distinct gauge vacua are separated by a change of Chern-Simons number; an instanton interpolates between them, with the change equal to $k$ in compatible orientation conventions. Theta-weighted instanton sectors give $Z(\vartheta )={\sum }_k{e}^{ik\vartheta }{Z}_k$, connecting topology to the vacuum angle. With fermions, the gauge-field topology also controls fermionic zero modes and chiral selection rules through the chiral anomaly. These phenomena explain why instantons encode physics beyond small fluctuations around a single vacuum.

Instantons link Euclidean classical solutions, topological sectors and exponentially suppressed quantum effects. Their interpretation requires the boundary conditions, normalization, fluctuation spectrum and measure over collective coordinates. A dilute approximation is controlled only when its individual-action suppression and separation assumptions hold; classical scale moduli and their quantum corrections can make size integrals delicate rather than automatically finite.

Use the standard boundaryless-manifold convention, so $M$ is a closed manifold. Its orientation and Riemannian metric determine a unique positive Riemannian volume form ${\mu }_g$ taking value one on every positively oriented orthonormal frame. In an oriented atlas it isFor the coordinate invariance of the Riemannian volume form, on an overlap put $J=\partial y/\partial x$. Then $g_x=J^T{g}_{y}J$ and $d{y}^1\wedge \cdots \wedge d{y}^n=(\det J)d{x}^1\wedge \cdots \wedge d{x}^n$. The oriented transition has $\det J>0$, so the square root of the metric determinant and the coordinate volume transform together. The local forms therefore agree and define ${\mu }_g$ globally.

For real $r$-forms the Hodge star operator is defined byThe pointwise wedge pairing is nondegenerate, so this uniquely defines a smooth map $\ast :{\Omega }^r(M)\to {\Omega }^{n-r}(M)$. On an oriented orthonormal coframe, it sends a basis wedge to the signed complementary wedge. Applying it twice gives

For Hodge star eigenvalues on graded differential forms, extend $\ast$ complex-linearly to the direct sum of all form degrees. This interpretation matters: on a fixed degree it is an endomorphism only in middle degree. In positive odd dimension, $r(n-r)$ is even for every $r$, so ${\ast }^2=I$ and only $\pm 1$ can occur. Both occur, with eigenforms $1+{\mu }_g$ and $1-{\mu }_g$.

In positive even dimension, ${\ast }^2$ is $+I$ on even-degree forms and $-I$ on odd-degree forms. Hence ${\ast }^4=I$ and the only possible eigenvalues are $\pm 1,\pm i$. The first two occur as above. For the latter, choose a smooth local orthonormal frame and multiply its first dual differential one-form by a nonzero smooth bump function supported in the chart. This gives a form $\eta$ for which $\eta$ and $\ast \eta$ are independent. For $\lambda =\pm i$,Thus all four occur. Consequently, for $n\ge 1$,On degree $m$ in dimension $2m$, the Hodge star has eigenvalues $\pm 1$ for even $m$ and $\pm i$ for odd $m$; signed complementary middle-degree wedges show both occur for $m\ge 1$. In zero dimensions it is multiplication by the signed unit volume at each point, so only the signs actually present occur.

For Hodge integration by parts in arbitrary degree, let $\alpha \in {\Omega }^{r-1}(M)$ and $\beta \in {\Omega }^r(M)$. The graded Leibniz rule and Stokes theorem on the closed manifold giveUsing the given codifferential and the square of the Hodge star,Thereforewhich proves that $\delta$ is the formal adjoint of $d$. The complex version inserts conjugation in the usual Hermitian inner product.

The Hodge decomposition theorem gives an $L^2$-orthogonal decompositionwhere ${\mathcal{H}}^k(M)=\ker \Delta$ is finite dimensional and consists of the harmonic differential forms. Each de Rham cohomology class has a unique harmonic representative. Equivalently, with harmonic projection $P_{\mathcal{H}}$ there is a smooth Green operator of the Hodge Laplacian $G$ satisfying $\Delta G=G\Delta =I-P_{\mathcal{H}}$.

The solvability condition for the Hodge Poisson equation isNecessity follows from $⟨\Delta \alpha ,h⟩=⟨\alpha ,\Delta h⟩=0$. Conversely the orthogonality condition gives $P_{\mathcal{H}}\beta =0$, so ${\alpha }_0=G\beta$ solves the Poisson equation for differential forms.

The nonnegativity of the Hodge Laplacian identityidentifies its kernel with closed and coclosed forms. Thus the affine space of solutions of the Hodge Poisson equation isFor completeness, if a closed form is decomposed as $\omega =h+d\xi +\delta \zeta$, then $d\delta \zeta =0$, and $\Vert \delta \zeta {\Vert }^2=⟨\zeta ,d\delta \zeta ⟩=0$. Hence its cohomology class is represented by $h$. An exact harmonic form $h=d\xi$ must vanish, since $\Vert h{\Vert }^2=⟨\delta h,\xi ⟩=0$. This proves the canonical isomorphism ${\mathcal{H}}^k(M)\cong H_{\mathrm{dR}}^k(M)$, so the solution set is an affine space whose translation vector space is isomorphic to the requested de Rham cohomology.

Hold the metric fixed when varying the gauge potential, put $F=dA$, and take variations with compact support or with appropriate fixed boundary data. The Hodge star operator then givesFor the Abelian Chern-Simons 5-form, the two occurrences of $F$ commute as even-degree differential forms. Since $dF=0$,The factor three comes from all three appearances of $A$, including those hidden in $F=dA$. The bulk variation is thereforeand the Abelian Chern-Simons five-form field equation is

Under an Abelian gauge transformation $A\mapsto A+d\lambda$, the gauge field strength stays fixed and the action changes byThus the local integrand is not gauge invariant, but its change is a boundary term by the Generalized Stokes theorem. With the stated boundary conditions the variational bulk equations are unchanged; they depend only on $F$ and the metric, so they are manifestly gauge invariant. On a spacetime with boundary, gauge transformations nonzero at the boundary require boundary conditions or boundary degrees of freedom; the bulk claim does not assert invariance of every possible boundary action.

For the metric variation, write the kinetic action in components,At fixed covariant components $A_{\mu }$ and $F_{\mu \nu }$,Substituting into the given definition of the stress-energy tensor yieldsThe Chern-Simons 5-form uses only wedge products of differential forms and orientation, with no metric or Hodge star operator. Its bulk metric derivative is zero, so the Chern-Simons term contributes no bulk stress-energy tensor. The displayed tensor is generally not traceless in five dimensions: $T^{\mu }{}_{\mu }=-F_{\rho \sigma }{F}^{\rho \sigma }/4$.

Let $n={\dim }_{\mathbb{C}}X$, and use the complex orientation. Write $g$ for the real Riemannian metric underlying the Hermitian metric. Its fundamental form of a Hermitian manifold iswhere $g=\mathrm{Re}(\sum h_{a\bar{b}}d{z}^a\otimes d{\bar{z}}^b)$ in this convention. It is real and of type $(1,1)$, since $g(Ju,Jv)=g(u,v)$. Choose an adapted real orthonormal frame $E_j,F_j=J{E}_j$ and its dual coframe $e^j,f^j$. Then $\omega ={\sum }_j{e}^j\wedge f^j$, giving the Riemannian volume form

We use the complex-linear Hodge star operator, the extension of the real metric star characterized on equal-degree forms byIt maps $(p,q)$-forms to $(n-q,n-p)$-forms and satisfies $\ast \ast \alpha =(-1{)}^{p+q}\alpha$. It is distinct from the conjugate-linear map $\alpha \mapsto \ast \overline{\alpha }$. In the adapted coframe the star of $e^j\wedge f^j$ is the wedge product of all the other two-dimensional factors, with positive sign. Summing proves the Hodge star of the fundamental Hermitian form formulaThis computation uses no closedness assumption on $\omega$.

Use the integrated Hermitian inner product $(\alpha ,\beta )={\int }_X\alpha \wedge \ast \overline{\beta }$. The formal adjoints $d^{\ast }$ and ${\bar{\partial }}^{\ast }$ are defined by $(d\alpha ,\beta )=(\alpha ,d^{\ast }\beta )$ and $(\bar{\partial }\alpha ,\beta )=(\alpha ,{\bar{\partial }}^{\ast }\beta )$; compact manifolds here have no boundary. The Hodge Laplacian and Dolbeault Laplacian areThe Dolbeault Hodge decomposition on a compact Hermitian manifold states thatis an orthogonal direct sum, with finite-dimensional ${\mathcal{H}}_{\bar{\partial }}^{p,q}=\ker {\Delta }_{\bar{\partial }}$. Each Dolbeault cohomology classhas a unique representative in ${\mathcal{H}}_{\bar{\partial }}^{p,q}$. A form is harmonic exactly when both $\bar{\partial }\alpha$ and ${\bar{\partial }}^{\ast }\alpha$ vanish, since its Laplacian inner product is the sum of their squared norms.

Now let $L\alpha =\omega \wedge \alpha$ be the Hermitian Lefschetz operator and let $\Lambda$ be its pointwise adjoint. For $\beta$ of total degree $r=p+q$ and $\alpha$ of degree $r-2$, the defining inner products giveThe wedge pairing is nondegenerate, so $\ast \overline{\Lambda \beta }=L\ast \overline{\beta }$. Apply another star to the degree-$(r-2)$ form and use that the star and $L$ commute with conjugation. This yieldsFor degrees below two both sides vanish. The argument also identifies its bidegree as $(-1,-1)$ and does not require the Kähler condition.

Assume now that $d\omega =0$, so the metric is a Kähler metric. Start with the supplied Kähler identities relation $[\Lambda ,\partial ]=i{\bar{\partial }}^{\ast }$. Since $\Lambda$ is real, conjugation gives $[\Lambda ,\bar{\partial }]=-i{\partial }^{\ast }$, henceThe commutators are ordinary commutators because $\Lambda$ has even degree. Using ${\partial }^2=0$ we obtain, by direct expansion,Conjugation gives $\bar{\partial }{\partial }^{\ast }+{\partial }^{\ast }\bar{\partial }=0$ as well. Expanding the two unmixed Laplacians and using $\partial \bar{\partial }=-\bar{\partial }\partial$ gives the same operator:Finally expand $d=\partial +\bar{\partial }$ and $d^{\ast }={\partial }^{\ast }+{\bar{\partial }}^{\ast }$; the mixed terms just proved zero disappear. This derives the Kähler Laplacian identity

The Dolbeault Laplacian preserves bidegree. Thus the complex $d$-harmonic forms of total degree $r$ split into their harmonic $(p,q)$ parts. The allowed real Hodge theorem, followed by complexification, givesThe real operator ${\Delta }_d$ commutes with conjugation, which exchanges $(p,q)$ and $(q,p)$. Therefore $h^{p,q}=h^{q,p}$. For odd $r$, no summand has $p=q$, so they pair off in equal dimensions. This proves that all odd Betti numbers are even.

For $1\le k\le n$, the Kähler form power ${\omega }^k$ is closed. If it were exact, say ${\omega }^k=d\eta$, thenby Stokes theorem. But ${\omega }^n=n!{\mathrm{vol}}_g$ has strictly positive integral on the nonempty compact manifold. Hence the powers of a compact Kähler form have nonzero cohomology classes, and

For an example take the Hopf surfaceThe dilation action is free and properly discontinuous: on a compact subset the radii are bounded above and away from zero, so only finitely many dilations can meet it. It is holomorphic, giving a complex manifold quotient; the annulus $1\le \Vert z\Vert \le 2$ covers the quotient, proving compactness. The mapis a diffeomorphism $H\cong S^3\times S^1$. The Künneth theorem gives $b^1(H)=1$. This violates the necessary odd-degree parity, so this compact complex surface admits no Kähler metric.

Use anti-Hermitian $\mathfrak{su}(2)$ matrices, with invariant positive inner product $⟨X,Y⟩=-2\mathrm{tr}(XY)$. On an adjoint-valued field $\Xi$, define the adjoint covariant derivative and gauge curvature byEquivalently, $F=dA+A\wedge A$ and $[D_i,D_j]\Xi =[F_{ij},\Xi ]$. Put $B_i=\frac{1}{2}{ϵ}_{ijk}{F}_{jk}$, with ${ϵ}_{123}=1$. The monopole Bogomolny equations are $B_i=D_i\Phi$.

For a smooth SU(2) gauge transformation $g(x)$, useDifferentiating a transformed adjoint field shows $D_i^g(g\Xi g^{-1})=g(D_i\Xi )g^{-1}$: the derivatives of $g$ cancel the inhomogeneous term in $A_i^g$. Applying the commutator of these derivatives gives $F_{ij}^g=g{F}_{ij}{g}^{-1}$, and therefore $B_i^g=g{B}_i{g}^{-1}$. Both sides of $B_i=D_i\Phi$ transform identically, establishing gauge invariance of the solution condition. Individual matrix components are covariant, rather than invariant numbers.

The monopole boundary condition is a nonzero asymptotic Higgs norm,together with finite three-dimensional Yang-Mills-Higgs energy. Standard monopole asymptotics have $D\Phi =O(r^{-2})$ and $F=O(r^{-2})$. The normalized Higgs field at infinity defines a map $S_{\infty }^2\to SU(2)/U(1)\simeq S^2$, whose topological degree is the magnetic charge. A nonzero charge gives a genuine non-Abelian magnetic monopole; the nonzero Higgs boundary value alone also allows the zero-charge vacuum. Thus the relevant nontrivial solutions are Bogomolny-Prasad-Sommerfield monopoles in the zero-Higgs-potential limit.

Now form the four-dimensional connection with $\tau$-independent components. Its gauge curvature hasThe self-duality sign requires an orientation. Choose the Euclidean metric and orientation $d\tau \wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$. For cyclic $(i,j,k)$, the Hodge star operator satisfiesUsing $F_{jk}={ϵ}_{jki}{B}_i$ and $d{x}^i\wedge d\tau =-d\tau \wedge d{x}^i$, the curvature isEach term has negative star eigenvalue, soThis proves the Anti-self-dual Yang-Mills equations in the stated convention. The orientation of a monopole lift matters: with $d{x}^1\wedge d{x}^2\wedge d{x}^3\wedge d\tau$ instead, the same connection is self-dual. With that opposite orientation, the anti-self-dual lift would use $-\Phi d\tau$.

To fix the action normalization, take the Euclidean Yang-Mills actionBoth sums are over ordered index pairs. The component identity
${\sum }_{j,k}|F_{jk}{|}^2=2{\sum }_i|B_i{|}^2$ givesHence the four-dimensional action density in terms of the three-dimensional curvature isA different overall action normalization rescales all three equal expressions together.

A Yang-Mills instanton is a smooth finite-Euclidean-action self-dual or anti-self-dual connection in pure Yang-Mills theory, with the usual asymptotic pure-gauge behavior. Physically the term generally refers to a nontrivial field; its asymptotic gauge winding labels an instanton number. The Bianchi identity $D\mathcal{F}=0$, together with $\ast \mathcal{F}=-\mathcal{F}$, implies $D\ast \mathcal{F}=0$, so the first-order equations already imply the second-order Yang-Mills equations. Finite action is a separate requirement.

Here ${\mathcal{L}}_4$ is independent of $\tau$. If the monopole has nonzero curvature, the density is positive on an open set in ${\mathbb{R}}^3$, and integrating its infinite $\tau$-extent givesFor a finite-energy monopole this can also be written as its positive action per unit length times ${\int }_{\mathbb{R}}d\tau$. Thus a nontrivial monopole lift is not a Yang-Mills instanton on ${\mathbb{R}}^4$. The flat solution $A_i=0$, $\Phi ={\Phi }_0$ constant is the zero-action exception: it is a vacuum, or a trivial charge-zero instanton if that terminology includes flat connections. It is not a counterexample to the nontrivial-monopole assertion.

More generally, finite Yang-Mills action forbids nontrivial translation symmetry, even when the translation preserves a connection only up to gauge. Such a symmetry would preserve its positive gauge-invariant action density. A nonflat field has a ball on which this density is bounded below by some positive constant. Repeated translations by a nonzero symmetry vector produce infinitely many disjoint copies of a sufficiently small such ball, all with the same positive action contribution, contradicting finiteness. Continuous translation invariance is a special case. Therefore a nonflat ASDYM instanton is localized in all four Euclidean directions and has no nonzero translational symmetry. Translating an instanton still produces another instanton with a shifted center; these position parameters belong to the instanton moduli space, rather than being invariances of one localized solution.

Let $g$ be the Riemannian metric underlying the Hermitian metric and $J$ its compatible complex structure. The fundamental form of a Hermitian manifold is $\omega (u,v)=g(Ju,v)$. It is a real alternating form of type $(1,1)$. At a point choose an orthonormal real coframe $x^1,y^1,\dots ,x^n,y^n$ adapted to $J$, with $x^j+i{y}^j$ of type $(1,0)$. ThenThe orientation here is the complex orientation. In coordinates for which $g={\sum }_j((d{x}^j{)}^2+(d{y}^j{)}^2)$, the same convention reads $\omega =(i/2){\sum }_{j}d{z}^j\wedge d{\bar{z}}^j$.

Use the complex Hodge star operator, the complex-linear extension of the real Hodge star operator. With the pointwise Hermitian inner product linear in its first variable, its defining identity isfor forms of the same total degree. It maps type $(p,q)$ to $(n-q,n-p)$ and satisfies ${\ast }^2=(-1{)}^{p+q}$ in real dimension $2n$. It commutes with complex conjugation. These conventions distinguish it from the conjugate-linear Hodge star $\alpha \mapsto \ast \bar{\alpha }$.

Write a top holomorphic-type form as $\eta =a{\zeta }^1\wedge \cdots \wedge {\zeta }^n$, where ${\zeta }^j=x^j+i{y}^j$. On each oriented two-plane, $\ast x^j=y^j$ and $\ast y^j=-x^j$, so the two-dimensional star of ${\zeta }^j$ is $-i{\zeta }^j$. In taking the star of their product, regrouping complementary one-form factors across the $n$ two-planes contributes $(-1{)}^{n(n-1)/2}$. Thus the Hodge star on top holomorphic forms isThis is pointwise and requires neither holomorphicity of the coefficient $a$ nor closedness of $\omega$.

To compute the formal adjoint, take compactly supported forms $\alpha$ of degree $k-1$ and $\beta$ of degree $k$. The Stokes theorem givesSince the degree of $d(\ast \bar{\beta })$ is $2n-k+1$, the identity for ${\ast }^2$ makes the right side equal to ${\int }_X\alpha \wedge \ast \overline{(-\ast d\ast \beta )}$. Consequently the formal adjoint of $d$ in even real dimension is $d^{\ast }=-\ast d\ast$. Splitting by type, $-\ast \partial \ast$ lowers antiholomorphic degree, while $-\ast \bar{\partial }\ast$ lowers holomorphic degree. Orthogonality of distinct types therefore yieldsThe argument applies locally with compact support, and globally without support restrictions on a compact manifold without boundary.

The Dolbeault Laplacian is ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$. A form is $\bar{\partial }$-harmonic when it lies in its kernel. On a compact Hermitian manifold, the adjoint identities implyThus harmonicity is equivalent to both first-order equations vanishing. The Dolbeault Hodge decomposition on a compact Hermitian manifold states that the harmonic space ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional and that smooth forms have the orthogonal direct-sum decomposition

To deduce the identification with Dolbeault cohomology, let $\bar{\partial }\alpha =0$ and decompose $\alpha =h+\bar{\partial }u+{\bar{\partial }}^{\ast }v$. The closedness of $\alpha$ makes it orthogonal to the entire image of ${\bar{\partial }}^{\ast }$. Taking its inner product with ${\bar{\partial }}^{\ast }v$ and using the orthogonality of the decomposition gives $\Vert {\bar{\partial }}^{\ast }v{\Vert }^2=0$. Hence $\alpha =h+\bar{\partial }u$, so each cohomology class has a harmonic representative. If a harmonic form equals $\bar{\partial }u$, its squared norm is $⟨{\bar{\partial }}^{\ast }h,u⟩=0$, so the representative is unique. We have proved

For the complementary types, define $C\alpha =\ast \bar{\alpha }$. It maps $(p,q)$ to $(n-p,n-q)$ and is conjugate-linear. If $\alpha$ is harmonic, then $\bar{\partial }\alpha =0$ and $\partial (\ast \alpha )=0$ by the adjoint formula. Conjugating the latter equation gives $\bar{\partial }(\ast \bar{\alpha })=0$. Also, with $k=p+q$,since $\partial \bar{\alpha }=\overline{\bar{\partial }\alpha }$. Therefore $C$ carries harmonic forms to harmonic forms, and $C^2=(-1{)}^k$ makes it a bijection. This proves complementary Dolbeault harmonic types on a Hermitian manifold:The metric supplies a natural conjugate-linear isomorphism. If a complex-linear isomorphism is desired, choose a basis $h_j$ of the first space and send it to the basis $C{h}_j$ of the second, extending linearly. This last choice is not canonical; no Kähler assumption is needed for either the dimension equality or the harmonic bijection.

The exterior algebra of a vector space $V$ is ${\Lambda }^{\ast }{V}^{\ast }={⨁}_{r=0}^{\dim V}{\Lambda }^r{V}^{\ast }$, whose degree-$r$ elements are alternating $r$-linear forms. Its exterior product satisfies $\alpha \wedge \beta =(-1{)}^{rs}\beta \wedge \alpha$ for degrees $r,s$. For example, one-forms anticommute, so $d{x}^i\wedge d{x}^i=0$. A differential form of degree $r$ on a smooth manifold is a smooth section of ${\Lambda }^r{T}^{\ast }M$; in coordinates it is an alternating sum of products of $r$ coordinate one-forms. Its exterior derivative increases degree by one, satisfies $d^2=0$, and obeysThe coordinate definition and pullback of a differential form are compatible, making differential forms suitable for coordinate-independent integration.

An orientation and a nondegenerate metric tensor define the Hodge star operator by $\alpha \wedge \ast \beta =⟨\alpha ,\beta ⟩{\mathrm{vol}}_g$ for equal-degree forms. It maps degree $r$ to degree $d-r$. If the metric has $s$ negative directions, its square is ${\ast }^2=(-1{)}^{r(d-r)+s}$. Thus Hodge duality turns antisymmetric tensors into complementary-degree tensors, using the metric and orientation rather than merely the underlying vector space. The Generalized Stokes theorem says ${\int }_{M}d\eta ={\int }_{\partial M}\eta$, with the induced boundary orientation, for appropriate smooth forms and compactness or support assumptions. It generalizes the fundamental theorem of calculus, circulation version of Stokes theorem, and divergence theorem.

For the field calculation, choose signature $(-,+,+)$ and a fixed orientation on Minkowski spacetime. Write $a=\delta A$, so $\delta F=da$. Symmetry of the metric pairing and the Hodge star operator gives $\delta (F\wedge \ast F)=2da\wedge \ast F$. Applying the exterior derivative product rule,Since $F\wedge a=a\wedge F$, variation of the action isFix the boundary data or use compactly supported variations. Arbitrary interior one-forms $a$ then imply the bulk field equationWith $B=\ast F$, the Lorentzian Hodge star operator squares to $-1$ on both one- and two-forms, hence $F=-\ast B$. Thus the Maxwell–Chern–Simons one-form equation becomesThese two displayed equations are equivalent. Also $dF=0$ gives $d\ast B=0$. For $k\ne 0$ the latter follows from applying $d$ to the field equation; for $k=0$ it must be retained as the identity inherited from $F=dA$. Using the opposite metric-sign convention changes the corresponding ${\ast }^2$ sign; the signature-independent equation is $d\ast F+kF=0$.

Under the Abelian gauge transformation $A\mapsto A+d\Lambda$, the curvature $F=dA$ and the one-form $B$ are unchanged because $d^2\Lambda =0$. The field equation is therefore exactly gauge-invariant. The kinetic term is unchanged, while the other term changes byThe last equality uses $dF=0$ and the Generalized Stokes theorem. Thus the action density changes by a total derivative and the integrated action can change on a boundary. The source's statement that the action is not invariant needs this qualification: on a boundaryless spacetime, or when the boundary integral vanishes, the integrated action is invariant too. No such boundary restriction is needed for invariance of the bulk field equation.

The specified bidegree is the conjugate-linear Hodge star, not the complex-linear extension of the real star. If the pointwise Hermitian inner product is linear in its first argument, it is defined byIt is the real Hodge star operator extended complex-linearly and composed with complex conjugation; ${\ast }^2=(-1{)}^{p+q}$. On a compact Hermitian manifold,is a positive definite Hermitian inner product on smooth forms. The formal adjoint is ${\bar{\partial }}^{\ast }=-\ast \bar{\partial }\ast$ in this convention; the final star in this identity is present in the PDF but is corrupted in the TeX aid.

State the relevant Hodge theorem as follows. On a compact manifold without boundary, the Dolbeault Laplacian ${\Delta }_{\bar{\partial }}=\bar{\partial }{\bar{\partial }}^{\ast }+{\bar{\partial }}^{\ast }\bar{\partial }$ is elliptic, self-adjoint and nonnegative. Its harmonic kernel ${\mathcal{H}}_{\bar{\partial }}^{p,q}$ is finite-dimensional, and smooth forms split orthogonally as this kernel plus the image of ${\Delta }_{\bar{\partial }}$. There are the orthogonal projection $H$ and a smooth Green operator $G$ withFor any form $u$, expand the Laplacian term to obtainHarmonic forms are both closed and coclosed, since $({\Delta }_{\bar{\partial }}v,v)=\Vert \bar{\partial }v{\Vert }^2+\Vert {\bar{\partial }}^{\ast }v{\Vert }^2$. They are orthogonal to both other terms; the two images are orthogonal because $(\bar{\partial }a,{\bar{\partial }}^{\ast }b)=({\bar{\partial }}^{2}a,b)=0$. This proves the Dolbeault Hodge decompositionThe closed forms are exactly the first two summands: if $\bar{\partial }u=0$, then $u$ is orthogonal to the adjoint image, and conversely the first two summands are closed. It follows that every Dolbeault cohomology class has a unique harmonic representative. Negative or out-of-range bidegrees mean zero spaces. Compactness and the absence of boundary are necessary for this stated global decomposition.