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The connection is torsion-free, so on a scalar ${\nabla }_a{\nabla }_b\varphi ={\partial }_a{\partial }_b\varphi -{\Gamma }_{ab}^c{\partial }_c\varphi$ is symmetric in $a,b$. A covector also carries a connection acting on its own index, and commuting derivatives gives curvature, generally nonzero.

Use the curvature convention $[{\nabla }_a,{\nabla }_b]v_c=R^d{}_{cab}{v}_d$, consistent with the identity requested in the PDF. Some texts define the opposite Riemann sign, reversing both formulas. Expanding the prescribed derivative combination of the Killing equation givesThe three commutator terms are $[-R^d{}_{cab}+R^d{}_{bac}+R^d{}_{abc}]v_d=2R^d{}_{abc}{v}_d$, by antisymmetry in the last two indices and the first Bianchi identity. ThusIn flat Cartesian Minkowski coordinates this says ${\partial }_a{\partial }_b{v}_c=0$. Hence all Killing fields are affine:The constant term gives four translations and the antisymmetric term six Lorentz generators, rotations and boosts. Conversely every such field satisfies the Killing equation.

For the static block metric, time derivatives vanish and $g_{0i}=0$, $g^{00}=-1/f$. The Christoffel formula gives ${\Gamma }_{00}^0={\Gamma }_{ij}^0=0$ and ${\Gamma }_{0i}^0={\Gamma }_{i0}^0=({\partial }_{i}f)/(2f)$. The vector $v^a=(1,0,0,0)$ has covector components $v_0=-f$, $v_i=0$. Its only potentially nonzero derivatives are ${\nabla }_i{v}_0=-{\partial }_{i}f/2$ and ${\nabla }_0{v}_i={\partial }_{i}f/2$; spatial-spatial and time-time derivatives vanish. Their symmetric sums are zero, proving ${\partial }_t$ is a Killing vector field.

Fix the coefficient-column convention: write the local frame as a row $e=(e_1,\dots ,e_r)$ and a section as $s=eu$ for a column $u$ of functions. Define the connection matrix $\theta =({\theta }_{ij})$ byEach entry is a smooth one-form. Define the curvature form of a connection byThe same Leibniz cancellations as for tangent-bundle curvature show that this is linear over smooth functions in $X,Y,s$. Thus it is a two-form with values in the endomorphisms of $E$. Its curvature matrix $\Theta =({\Theta }_{ij})$ is specified by $R^E(X,Y)e_j={\sum }_i{e}_i{\Theta }_{ij}(X,Y)$.

The covariant exterior derivative on coefficient forms is $d+\theta \wedge$. Squaring it on $u$ givesTherefore the Cartan curvature matrix equation isMatrix-valued wedge products here combine matrix multiplication with the wedge product of differential forms, in the stated order.

Using $d^2=0$ and the graded Leibniz rule,On the other hand, associativity of the endomorphism-valued exterior product givesThe cubic terms cancel, proving the requested Bianchi identity

Let another local frame be $e^{\prime }=eA$ with $A:U\to G{L}_r(\mathbb{R})$ smooth. Since the coefficient columns satisfy $u=A{u}^{\prime }$, the Leibniz rule gives the change of frame of a vector-bundle connection:The curvature form of a connection is linear over smooth functions in its section argument, so its matrix transforms homogeneously:ConsequentlyThe entries of $A$ are functions, of degree zero, so ordinary cyclic invariance of trace applies without a graded sign. The local forms therefore glue to a globally defined two-form $\alpha =\mathrm{tr}\Theta$, the trace of vector-bundle curvature.

Apply the Bianchi identity proved in part (b):For the first term,The middle interchange has sign $(-1{)}^{2\cdot 1}=1$, and the next equality simply renames $i,j$. ThusOne can also check closedness locally: the off-diagonal terms in $\mathrm{tr}(\theta \wedge \theta )={\sum }_{i,j}{\theta }_{ij}\wedge {\theta }_{ji}$ cancel in pairs, while the diagonal terms vanish. Hence $\alpha =d(\mathrm{tr}\theta )$ in each frame, and $d^2=0$ gives the same conclusion.

Use the Levi-Civita connection interpretation of the curvature convention. Its Riemann curvature tensor satisfiesThe first antisymmetry comes from the derivative commutator; the second comes from compatibility with the metric tensor. The cyclic identity is the first Bianchi identity, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric Ricci tensor, $\boxed{R_{ab}=R_{ba}}$. The differential Bianchi identity is ${\nabla }_{[a}{R}_{bc]de}=0$.

There is a hypothesis qualification. A metric connection need not be a torsion-free connection. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}{\nabla }_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with torsion tensor.

For example, with a constant Minkowski metric a connection having only ${\Gamma }^1{}_{02}=1$ and ${\Gamma }^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[{\nabla }_0,{\nabla }_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.

Choose the Riemann curvature tensor conventionwith positive scalar curvature for a round sphere. The connection preserves the gamma matrices. If $D={\gamma }^{\mu }{\nabla }_{\mu }$, its square isHere ${\nabla }^2$ is the rough Laplacian, including the connection on the derivative index. In the product of two antisymmetric gamma pairs, the fully antisymmetric four-gamma term vanishes by the first Bianchi identity. The two-gamma contractions vanish because the Ricci tensor is symmetric, while the scalar contractions give $-2R$. Thus the Lichnerowicz spinor-square formula isMultiplying $(D+m)\psi =0$ by $-D+m$, with constant $m$, provesIn Riemannian signature $D$ in this positive-Clifford convention is formally anti-self-adjoint; the more usual self-adjoint Dirac operator $iD$ has square $-{\nabla }^2+R/4$. Stating this convention avoids an apparent sign conflict between the two forms of the formula.

Now let $D\psi =0$ on the compact Riemannian spin manifold. Use the positive-definite spinor inner product and integrate the squared equation. Integration by parts has no boundary contribution, givingBoth terms are nonnegative. Smoothness then forces $\nabla \psi =0$ pointwise. Consequently every harmonic spinor is parallel, and $R|\psi {|}^2=0$. A nonzero parallel spinor has constant positive norm on its connected component, so its scalar curvature vanishes there. This is compact harmonic-spinor rigidity under nonnegative scalar curvature.

The full Ricci conclusion follows from integrability, not merely from vanishing scalar curvature. Since a parallel spinor has zero curvature for its spin connection, contract its Ricci identity with ${\gamma }^{\nu }$:The three-gamma term vanishes by the Bianchi identity; the remaining contractions give the displayed Ricci term. For each fixed $\mu$, put $v_{\nu }=R_{\mu \nu }$. Clifford multiplication twice givesPositive definiteness and $\psi \ne 0$ imply $v=0$. Thus the Riemannian parallel-spinor Ricci-flatness result isOn a connected manifold a nontrivial parallel solution is nowhere zero, so this holds everywhere. On a disconnected manifold it holds on every component supporting a nonzero spinor; the stated nowhere-vanishing hypothesis ensures all components are covered. The positive-definite metric tensor is essential to the last inference.

The gravitational variable and the fields held fixed in a variation must be specified. For bosonic matter a second-order metric tensor formulation is sufficient. For spinorial matter a vierbein and a spin connection are needed to define the local Lorentz representation and its covariant derivative. “Second order” refers to a connection already expressed through derivatives of the vierbein; “first order” treats that connection as an independent variable.

For a metric tensor, a scalar field and an Abelian gauge field, consider the Einstein-Hilbert action with matter,Here $e=\sqrt{-g}$ and the connection in $R$ is Levi-Civita. The metric tensor variation uses $\delta e=-e{g}_{\mu \nu }\delta g^{\mu \nu }/2$ andMetric compatibility makes the last two terms a divergence. ThusStationarity for arbitrary compactly supported variations givesExplicitly,Scalar field and gauge-potential variations give ${\nabla }^2\varphi -V^{\prime }=0$ and ${\nabla }_{\mu }{F}^{\mu \nu }=0$. Their equations imply stress-energy conservation, consistently with the contracted Bianchi identity. If the boundary metric tensor is fixed instead of using compactly supported variations, add the Gibbons–Hawking–York boundary term, ${\kappa }^{-2}{\int }_{\partial M}{\epsilon }_{\partial M}\sqrt{|h|}K$, with the outward-normal orientation. Fixing the metric tensor alone does not otherwise remove the normal derivatives of its variation.

For spinorial matter, write $g_{\mu \nu }=e_{\mu }{}^a{e}_{\nu }{}^b{\eta }_{ab}$ and ${\nabla }_{\mu }\chi ={\partial }_{\mu }\chi +{\omega }_{\mu ab}{\gamma }^{ab}\chi /4$. In the second-order theory set $\omega =\omega (e)$ before varying. A symmetrized Dirac action, in the same no-$i$, mostly-plus convention as the earlier solutions, isVarying $\bar{\chi }$ and integrating by parts gives $(/\nabla +m)\chi =0$; varying $\chi$ gives the adjoint equation. The spinor components held fixed in a gravitational variation are components in the Lorentz frame. Both ${\gamma }^{\mu }=e_a{}^{\mu }{\gamma }^a$ and $\omega (e)$ vary with the vierbein. The former gives the canonical stress, and integrating the spin-connection variation supplies its spin-current improvement. On the Dirac equations the resulting symmetric stress tensor isThe off-shell expression also contains $g_{\mu \nu }{\mathcal{L}}_{\chi }/e$; this vanishes on the matter equations for the displayed symmetrized action. Treating the coordinate gamma matrices as fixed while varying the metric tensor would miss the displayed stress. Local Lorentz invariance removes the independent antisymmetric vierbein equation, leaving the symmetric Einstein field equations with this matter source.

In a first-order vierbein formulation, $e^a$ and an antisymmetric Lorentz connection ${\omega }^{ab}$ are independent. Defineand use $R(e,\omega )=e_a{}^{\mu }{e}_b{}^{\nu }{R}_{\mu \nu }{}^{ab}$. The curvature form of a connection variation is $\delta R^{ab}=D_{\omega }\delta {\omega }^{ab}$. For the Einstein-Hilbert term, integration by parts in form notation gives a connection equation proportional towhen matter has no connection dependence. For an invertible vierbein this forces $T^a=0$, so $\omega =\omega (e)$. The vierbein equation then becomes the second-order Einstein field equations. This applies to the scalar and Maxwell examples: use $F=dA$, which is independent of $\omega$ and remains gauge invariant even if torsion is present.

A Dirac field changes the connection equation. Directly varying its two covariant derivatives givesThus its spin density sources totally antisymmetric torsion. This is Einstein-Cartan theory, and the connection equation is algebraic rather than a propagation equation for an extra field. To see both its solution and its effect, writeFor the totally antisymmetric component sourced by this field, the Ricci scalar is $R(e,\omega )=R(e)-K_{abc}{K}^{abc}$ up to a divergence. The other irreducible contorsion components have no spinor source and their connection equations set them to zero. The auxiliary part of the Lagrangian is thereforeIts algebraic variation gives the algebraic torsion of a Dirac field,Here the first index of $K_{abc}$ labels the derivative, and the torsion convention is $T_{abc}=K_{bac}-K_{cab}$; specifying this convention fixes the sign. Eliminating $K$ leaves the second-order effective action with contact interactionIt is a local axial-current four-fermion interaction. Hence first-order gravity with Dirac spin density is equivalent to torsion-free second-order gravity with this additional interaction. It is not equivalent to simply declaring the spin connection torsion-free and dropping the contact term. In the effective formulation the Einstein field equations have the stress tensor of the full effective matter action. Before eliminating the connection, the antisymmetric stress and its spin current are related by the local Lorentz Noether identity.

Finally, the 1.5-order formalism is a convenient way to vary this effective action. Let $\hat{\omega }(e,\chi )$ solve the algebraic connection equation and let $S_2(e,\chi )=S_1(e,\hat{\omega }(e,\chi ),\chi )$. The chain rule givesThus vary as in first order, then insert the solved connection. One does not need to calculate the complicated induced connection variation. This procedure is justified by its connection equation, not by discarding that variation arbitrarily. In supergravity the gravitino sources a contorsion quadratic in fermions; substituting it generates the four-fermion terms, while the 1.5-order method simplifies the local supersymmetry cancellation through the lower orders considered above.

Let $M$ be a closed manifold of even dimension $n=2m$, with a Riemannian metric and a spin structure. Let $S=S^{+}\oplus S^{-}$ be its spinor bundle, and let $W$ be a Hermitian vector bundle with a unitary connection of vector-bundle curvature $F^W=({\nabla }^W{)}^2$. We use Clifford multiplication satisfying $c(v)c(w)+c(w)c(v)=-2⟨v,w⟩$, so the Twisted Dirac operatoris self-adjoint and $D_W^{-}=(D_W^{+}{)}^{\ast }$. Here $(e_i)$ is an oriented local orthonormal basis and $(e^i)$ its dual. The Atiyah-Singer index theorem in this setting isThe second factor is the Chern character; the first is the A-hat formThe square root has constant term one. All functions of the vector-bundle curvature matrices mean power series truncated above the dimension of $M$. The brackets select the top-degree differential form. The Bianchi identity shows that these forms are closed, and Chern-Weil connection transgression shows that their de Rham cohomology classes do not depend on the chosen connections on a vector bundle.

The global step is supersymmetric quantum mechanics. On the Hilbert space $L^2(S\otimes W)$, take the grading operator $\Gamma$, the odd supercharge $D_W$, and the nonnegative Hamiltonian $H=D_W^2$. They obey $\Gamma D_W=-D_W\Gamma$ and $D_W^2=H$. The principal symbol of $D_W$ is invertible away from the zero covector, so elliptic regularity and Rellich-Kondrachov compactness theorem give a smooth discrete orthonormal eigenbasis for $H$, finite-dimensional eigenspaces, and a trace-class operator $e^{-tH}$ for $t>0$. Define its supertrace by $\mathrm{Str}A=\mathrm{Tr}(\Gamma A)$.

For every positive eigenvalue $\lambda$ of $H$, the map ${\lambda }^{-1/2}{D}_W$ is an isomorphism between its positive and negative grading eigenspaces, with inverse itself. Their contributions to the supertrace cancel. The zero eigenspaces are $\ker D_W^{+}$ and $\ker D_W^{-}$. ConsequentlyThis also proves independence of $t$ without differentiating an unbounded operator inside a trace. If $K_t$ is the heat kernel, the heat kernel trace formula givesThe remaining task is to compute this local integrand as $t\downarrow 0$.

Fix $p\in M$, use geodesic normal coordinates $x$, and trivialize $S\otimes W$ by radial parallel transport. Expanding $D_W^2$ at $p$ gives the Lichnerowicz formula for a twisted Dirac operator:Indeed the terms with identical indices give the rough Laplacian. The other terms are ${\sum }_{i<j}c(e^i)c(e^j)[{\nabla }_i,{\nabla }_j]$. Decompose this vector-bundle curvature into spin curvature and $F^W$; substituting the spin curvature into the Clifford algebra relations and the algebraic Bianchi identity leaves $\mathrm{Scal}/4$. Thus the calculation retains the curvature of the twisting bundle explicitly.

There is an especially strong cancellation in the spinor supertrace. With $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$, every ordered product of distinct Clifford generators of degree less than $2m$ has zero supertrace, whileFor an even product of lower degree, conjugate $\Gamma$ times that product by a missing Clifford generator: it changes sign, so its matrix trace is zero. An odd product exchanges the grading summands and also has zero supertrace. For the full product, $(c(e^1)\cdots c(e^{2m}){)}^2=(-1{)}^m$, and $\dim S=2^m$, giving the displayed constant.

To exploit this, identify the Clifford algebra as a filtered vector space with the exterior algebra by sending an ordered Clifford product to the corresponding wedge product of differential forms. Implement Getzler rescaling: put $x=\epsilon y$, $t={\epsilon }^{2}s$, multiply the degree-$k$ exterior component of the kernel by ${\epsilon }^{-k}$, and multiply the whole kernel by ${\epsilon }^{2m}$. A derivative and a Clifford generator have rescaling degree one; a coordinate has degree minus one. Clifford contractions lower the degree by two relative to the wedge product of differential forms, so the leading product becomes exterior multiplication.

In the radially parallel frame, the linear Taylor term of the connection is determined by its vector-bundle curvature: if $A_i$ is its matrix, then $A_i(x)=\frac{1}{2}{x}^j{F}_{ji}(p)+O(|x{|}^2)$. To check the coefficient, radial gauge means $x^i{A}_i=0$, and integration of $x^j{F}_{ji}$ along the radial segment gives $A_i(x)={\int }_0^{1}s{x}^j{F}_{ji}(sx)ds$. Combining this formula with the spin representation, the symmetries of Riemann curvature tensor, and the Lichnerowicz formula for a twisted Dirac operator, the rescaled square has leading modelHere $\mathcal{R}$ is the skew matrix of tangent-bundle curvature two-forms at $p$; changing the curvature-matrix sign convention changes the sign inside the covariant derivatives but not the determinant below. The coefficients act in $\Lambda T_p^{\ast }M\otimes \mathrm{End}(W_p)$. The scalar-curvature term has degree zero and disappears after multiplication by ${\epsilon }^2$; the linear spin-connection term and the Clifford action of $F^W$ have degree two and survive. Higher Taylor terms have smaller degree and disappear.

For clarity, the analytic meaning of this limiting calculation is the following. Construct the local heat kernel expansion to any fixed order and insert the Taylor expansion of the rescaled coefficients into its transport equations. At each exterior degree, the leading equation is the heat equation of ${\mathcal{H}}_p$; the remaining terms acquire positive powers of $\epsilon$. Equivalently, successive substitutions ingive the same expansion. Polynomial spatial factors are integrable against the local Gaussian kernels, and only finitely many exterior degrees exist. A cutoff outside a fixed normal neighbourhood contributes $O(e^{-c/t})$. Choosing the initial expansion order arbitrarily large bounds its differentiated remainder to any required order. This proves convergence of the rescaled diagonal kernel to the model diagonal kernel, uniformly in $p$ on the compact manifold. Its degree-$2m$ component has no net power of $\epsilon$, precisely the component detected by the spinor supertrace.

We can compute the model kernel rather than invoke a local index formula. Set $F=F_p^W$ and seek its kernel from the origin in the formThe scalar even exterior coefficients of $\mathcal{R}$ commute. On this ansatz the rotational first-order term vanishes because $Q$ is a function of ${\mathcal{R}}^2$. Expanding the model operator gives $-\sum {\partial }_i^2+\frac{1}{2}(\mathcal{R}y)\cdot \nabla +\frac{1}{16}{y}^T{\mathcal{R}}^{2}y+F$, so substitution into the heat equation givesThe Euclidean initial condition $Q(s)\sim s^{-1}I$, $a(s)\sim (4\pi s{)}^{-m}$ selectsThese expressions have well-defined power series at $\mathcal{R}=0$. This is the curvature version of the Gaussian calculation underlying the Mehler formula for Hermite functions.

At $s=1$, $y=0$, multiply the top exterior coefficient by the spinor supertrace constant. Since $(4\pi {)}^{-m}(-2i{)}^m=(2\pi i{)}^{-m}$, and every curvature factor is a two-form, the result isIntegration and the constant supertrace established above prove the boxed Atiyah-Singer index theorem. The mechanism is that odd symmetry pairs the nonzero spectrum globally, while the Clifford filtration makes the surviving local contribution an explicitly solvable curvature oscillator. Odd-dimensional spin Dirac operators have no canonical chiral index of this type; their self-adjoint Fredholm index is zero. For comparison of normalization conventions, Raphaël Ponge's local index calculation writes the same unnormalized curvature model and spinor supertrace constant.

For fields ${\Phi }_a$ with a first-derivative Lagrangian density, suppose an infinitesimal global transformation at fixed coordinates is $\delta {\Phi }_a=\alpha {\Delta }_a$ and $\delta \mathcal{L}=\alpha {\partial }_{\mu }{K}^{\mu }$, with constant infinitesimal $\alpha$. Noether theorem gives the on shell conserved currentThis follows by varying the Lagrangian, integrating the terms with ${\partial }_{\mu }{\Delta }_a$ by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, $\int d^{3}x{J}^0$ is a conserved Noether charge. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative $K^{\mu }$.

An example of a theory with a Lorentz four-vector is the source-free Maxwell field with electromagnetic four-potential $A_{\mu }$. Under an Abelian gauge transformation,The change in $F_{\mu \nu }$ is ${\partial }_{\mu }{\partial }_{\nu }\chi -{\partial }_{\nu }{\partial }_{\mu }\chi =0$, so the Maxwell Lagrangian is gauge-invariant. For $A_{\mu }=(\Phi ,-\mathbf{A})$ define $E_i=F_{0i}$ and $F_{ij}=-{ϵ}_{ijk}{B}_k$. With signature $+---$,The minus sign in $-F^2/4$ gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the Hamiltonian's sign for the physical modes. The normalization $1/4$ accounts for antisymmetry and yields the conventional equations. Varying $A_{\nu }$ gives ${\partial }_{\mu }{F}^{\mu \nu }=0$.

For time translation at fixed coordinates take ${\Delta }_{\nu }={\partial }_0{A}_{\nu }$. Since $\alpha$ is constant, $\delta F_{\mu \nu }=\alpha {\partial }_0{F}_{\mu \nu }$, and hence $\delta \mathcal{L}=\alpha {\partial }_0\mathcal{L}$. Thus $K^{\mu }={\delta }^{\mu }{}_0\mathcal{L}$. The momentum derivative isNoether theorem therefore gives the canonical stress-energy tensor's time-translation currentHere ${\dot{A}}_i=E_i+{\partial }_i{A}_0$ was used. The canonical momentum of $A_0$ vanishes, while that of the lower spatial component $A_i$ is $E_i$: $A_0$ acts as a multiplier for Gauss law, not a propagating degree of freedom. The density above is not manifestly gauge-invariant, but its charge agrees with the physical energy on the constraint surface and with appropriate boundary conditions.

The gauge-covariant time translation instead has ${\Delta }_{\nu }={\partial }_0{A}_{\nu }-{\partial }_{\nu }{A}_0=F_{0\nu }$. It is time translation combined with a gauge transformation of parameter $\chi =-\alpha A_0$. Directly,using the definition of $F$, equivalently its Bianchi identity. The total derivative in $\delta \mathcal{L}$ is unchanged. The improved current isIt is the gauge-invariant Maxwell stress-energy tensor. In particular,The spatial current is the Poynting vector $\mathbf{E}\times \mathbf{B}$, so conservation gives ${\partial }_{t}u+\nabla \cdot (\mathbf{E}\times \mathbf{B})=0$. This is the familiar positive local electromagnetic field energy density, expressed entirely in measurable fields.

To compare the currents,The last term vanishes on the source-free Maxwell equations, leaving a stress-energy tensor improvement generated by a stress-energy superpotential. At the charge level,when the surface term vanishes and Gauss law holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those boundary conditions were not imposed.

Use $A_{\beta }={\partial }_{\mu }{h}^{\mu }{}_{\beta }$ and $\square ={\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }$. The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives ${\partial }^{\alpha }\delta G_{\alpha \beta }=0$, the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.

A Chern number turns local curvature data into a global integer. Let $E\to M$ be a complex vector bundle over a compact oriented manifold without boundary, with a unitary connection represented locally by an anti-Hermitian matrix-valued one-form $A$. Its curvature form of a connection is $F=dA+A\wedge A$. The Chern-Weil theory representative of the total Chern class isThe coefficients are closed differential forms. They represent the images of integral Chern classes in de Rham cohomology. On an oriented $2r$-manifold, a product $c_{i_1}\cdots c_{i_s}$ with $i_1+\cdots +i_s=r$ pairs with the fundamental class to give a Chern number, an integer independent of the unitary connection.

For example,For an $SU(N)$ bundle with $N\ge 2$, $\mathrm{Tr}F=0$. On an oriented four-manifold the Second Chern number is thenusing the fundamental matrix trace and the stated anti-Hermitian convention. The trace-product correction is required for a general $U(N)$ bundle; it cannot simply be omitted. The sign of a physical instanton number also depends on the trace and orientation convention, so these conventions must accompany the formula.

The local origin of closure is the Bianchi identity $D_{A}F=0$ and the cyclic matrix trace: $d\mathrm{Tr}(F\wedge F)=0$. Independence of the connection one-form follows more concretely by varying a family $A_t$. Since ${\dot{F}}_t=D_{A_t}{\dot{A}}_t$, the Chern-Weil connection transgression isIts integral on a closed four-manifold is zero by the Generalized Stokes theorem. This proves connection independence of $q$; the integrality is the global Chern class statement, not merely a consequence of the local formula.

The local primitive of the Second Chern form is the Chern-Simons three-formHere we continue to use $SU(N)$, so $c_2=\mathrm{Tr}(F\wedge F)/(8{\pi }^2)$. The cubic coefficient is forced by the exterior derivative. In the graded cyclic matrix trace, $d\mathrm{Tr}(A^3)=3\mathrm{Tr}(dA\wedge A^2)$ and $\mathrm{Tr}(A^4)=0$, the latter because cycling one degree-one factor past the other three changes its sign. ThereforeMatrix-valued differential forms require both matrix order and the graded signs; treating all factors as commuting scalars would lose this derivation.

The Chern-Simons three-form depends on a local trivialization and is not itself gauge-invariant. For the Yang-Mills gauge transformation convention $A^g=g^{-1}Ag+g^{-1}dg$, set $u=g^{-1}dg$. The gauge change of the Chern-Simons three-form isThe last term is closed by the Maurer-Cartan equation. On a closed three-manifold its integral is an integer with the fundamental $SU(N)$ normalization. Consequently the Chern-Simons integral is naturally defined modulo integers, while its exponential $\exp (2\pi i\ell \int Y)$ is invariant under large Yang-Mills gauge transformations for integer level $\ell$.

This also explains why a nonzero Chern number is compatible with $dY=c_2$: $Y$ need not be a globally defined three-form. On $S^4$, trivialize over two hemispheres and let $A_N=A_S^g$ on their common equator $\Sigma =S^3$, oriented as the boundary of the northern hemisphere. The Generalized Stokes theorem and the gauge-change formula giveFor $SU(2)\cong S^3$, this is the degree of the transition map, with compatible group orientation; for $SU(N)$ it is the corresponding integer in ${\pi }_3(SU(N))$. Thus the Second Chern number measures the obstruction to choosing one trivialization over the whole four-sphere.