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A variation of the connection one-form has $\dot{F}=D_A\dot{A}$. The Bianchi identity and graded cyclic matrix trace imply ${\partial }_t\mathrm{Tr}(F\wedge F)=2d\mathrm{Tr}(\dot{A}\wedge F)$. Thus the Chern-Weil theory four-form changes by an exact differential form, and its integral on a closed four-cycle is connection-independent. For a general unitary bundle, include the corresponding trace-product term in the Second Chern form.

A coframe dual to a local frame of the tangent bundle is the corresponding local frame of the cotangent bundle. Its elements are differential one-forms. Applying the coframe to covariant derivatives of the frame extracts the connection one-forms; applying it to the torsion tensor extracts the torsion forms.

For a smooth map $\varphi :M^{\prime }\to M$ and a connection on a vector bundle $E\to M$, the curvature form of a connection on the pullback vector bundle satisfiesas an endomorphism of $E_{\varphi (p)}$. In a frame of a vector bundle written as a row, with section components written as columns, the connection one-form is $\Omega$ and its curvature form of a connection is $d\Omega +\Omega \wedge \Omega$. The pullback connection has matrix ${\varphi }^{\ast }\Omega$; commutation of the pullback of a differential form with the exterior derivative and wedge product of differential forms proves the identity. No pushforward to an ambient vector field is required: $d\varphi (X)$ is a vector field along a map.

A connection on a vector bundle induces a connection on its top exterior power by differentiating each factor in turn. Its local connection one-form is $\mathrm{Tr}A$ and its curvature form of a connection is $\mathrm{Tr}{\Theta }_D$.

For an anti-Hermitian $SU(N)$ connection one-form, normalize the Chern-Simons three-form by $1/(8{\pi }^2)$ and use $A^g=g^{-1}Ag+g^{-1}dg$. Its change is $-d\mathrm{Tr}(dg{g}^{-1}\wedge A)/(8{\pi }^2)-\mathrm{Tr}(g^{-1}dg{)}^3/(24{\pi }^2)$. On a closed three-manifold the first term integrates to zero, while the latter is an integer with the fundamental matrix trace. Hence the integral is defined modulo integers; integer-level exponentials are invariant under large Yang-Mills gauge transformations.

In a frame of a vector bundle, write a vector bundle covariant derivative as $\nabla (ev)=e(dv+\Gamma v)$. The associated horizontal subspace at $(x,v)$ consists of tangent pairs $(\xi ,-{\Gamma }_x(\xi )v)$. Under a frame change $e^{\prime }=eg$, the connection one-form changes by ${\Gamma }^{\prime }=g^{-1}\Gamma g+g^{-1}dg$, and this horizontal equation changes to the corresponding one in the new frame. Thus the subspaces glue to a unique smooth linear connection on a vector bundle. Its vertical projection gives back the original derivative. The Leibniz rule ensures locality, allowing this construction from an operator initially defined on global sections.

For a smooth gauge field on ${\mathbb{R}}^3$ with finite positive magnetic energy, dilate its connection one-form by $A_i^{(s)}(x)=s{A}_i(sx)$. Its gauge curvature scales as $F_{ij}^{(s)}(x)=s^2{F}_{ij}(sx)$, so the energy scales as $s^4{s}^{-3}E=sE$. A static stationary configuration must have zero derivative at $s=1$, forcing $E=0$ and hence $F=0$. This Derrick scaling argument assumes the dilation preserves the boundary conditions and that boundary terms vanish. It does not exclude four-dimensional Yang-Mills instantons, whose Euclidean action is scale invariant.

Use an SU(2) connection one-form $A=A_{\mu }d{x}^{\mu }$ with Skew-Hermitian generators $T_a=-i{\sigma }_a/2$, so $\mathrm{tr}(T_a{T}_b)=-{\delta }_{ab}/2$. Absorb the gauge coupling into $A$, and write its gauge curvature as $F=dA+A\wedge A$. The Yang-Mills gauge transformation convention is $A^h=h^{-1}Ah+h^{-1}dh$, giving $F^h=h^{-1}Fh$.

In four dimensions the Second Chern form and Second Chern number of the associated fundamental rank-two bundle areHere $M$ is a closed oriented four-manifold. This sign follows by expanding the total Chern class $\det (1+iF/(2\pi ))$ and using $\mathrm{tr}F=0$. The first Chern class vanishes for SU(2); the second is the degree-four characteristic class relevant here. With the same curvature convention the commonly positive self-dual instanton number isBoth sign conventions occur; fixing them at the outset avoids identifying opposite-oriented integers.

The form is gauge-invariant by conjugation and cyclicity of the matrix trace. The gauge-theory Bianchi identity $D_{A}F=0$ makes it closed. To see that its integral is independent of the connection, vary $A$:On a closed $M$ the integral of this exact form vanishes. This is Chern-Weil connection transgression, showing that $k_C$ depends on the bundle's topology rather than on a particular gauge field.

Locally the same four-form is an exterior derivative. The Chern-Simons 3-formsatisfiesFor verification, differentiating gives $\mathrm{tr}(dA\wedge dA+2dA\wedge A\wedge A)$; expansion of $F\wedge F$ gives these terms and $\mathrm{tr}{A}^4$. Graded cyclicity moves the first one-form through the other three, changing the sign, so $\mathrm{tr}{A}^4=0$. This local exactness does not make the Second Chern number zero on every closed manifold: on a nontrivial bundle the gauge potentials and ${\omega }_3$ are only patchwise defined.

For the usual finite-action sector on ${\mathbb{R}}^4$, impose pure-gauge behavior $A\to h^{-1}dh$ at infinity. Compactification gives a bundle over $S^4$, with transition map $h:S^3\to SU(2)$. The Maurer-Cartan equation gives $d(h^{-1}dh)=-(h^{-1}dh{)}^2$, henceWith the outward boundary orientation, Stokes theorem therefore givesThese are opposite choices of the winding number, and either becomes the topological degree of $h$ after specifying the orientation of the target SU(2) sphere. Thus integrality also follows from ${\pi }_3(SU(2))=\mathbb{Z}$. Smooth deformations preserving the asymptotic sector cannot change it.

The positive Euclidean Yang-Mills action isDefine $\Vert X{\Vert }^2=-\int \mathrm{tr}(X\wedge \ast X)$. Since the Hodge star has square one on two-forms in Euclidean dimension four,Choosing $s=\mathrm{sgn}Q$ gives the Yang-Mills instanton Bogomolny boundThe self-dual Yang-Mills equations and their anti-self-dual version imply the Yang-Mills equations $D_A\ast F=0$ by the Bianchi identity. Their finite-action solutions are Yang-Mills instantons.

For a concrete self-dual example, the BPST instanton with center $a$ and size $\rho >0$ haswhere ${\eta }_{\mu \nu }^a$ are self-dual 't Hooft symbols. Their contraction is ${\sum }_{a,\mu ,\nu }({\eta }_{\mu \nu }^a{)}^2=12$. Thus $F_{\mu \nu }^a{F}_{\mu \nu }^a=192{\rho }^4/(|x-a{|}^2+{\rho }^2{)}^4$, andThis gives $S=8{\pi }^2/g^2$ and $Q=1$. Reversing duality gives $Q=-1$. The free center, size and framed gauge orientation illustrate the instanton moduli space.

On a three-dimensional slice $\Sigma$, there is no four-form Chern integral intrinsic to the slice. Its transgression is the Chern-Simons number of a gauge field, in the instanton-charge conventionIt is a real number for a general connection, not necessarily an integer, and is not strictly gauge invariant. Set $\theta =h^{-1}dh$ and $\bar{\theta }=dh{h}^{-1}$. Substituting $A^h=h^{-1}Ah+\theta$, using $d\theta =-{\theta }^2$ and graded trace cyclicity, gives the gauge change of the Chern-Simons three-formThe mixed terms collect into the displayed exact form; the last term is also fixed by the pure-gauge case $A=0$. Therefore on a closed $\Sigma$, or with boundary conditions making the exact-form integral vanish,Thus $N_{\mathrm{CS}}\mathrm{mod}\mathbb{Z}$, or $\exp (2\pi i{N}_{\mathrm{CS}})$, is gauge invariant. Gauge maps homotopic to the identity have $w=0$; maps of nonzero winding are large transformations even when they equal the identity at infinity.

For a Yang-Mills vacuum on ${\mathbb{R}}^3$, zero magnetic energy requires $F_{ij}=0$. On simply connected space this is a pure gauge $A=h^{-1}dh$, and $N_{\mathrm{CS}}=w(h)$ is an integer after fixing the asymptotic trivialization. Quotienting only by transformations homotopic to the identity leaves integer-labelled vacuum representatives. Allowing all winding transformations identifies them classically; quantum states can instead transform by a phase, producing the theta vacuum. A three-dimensional pure-gauge representative carries a winding number but is not a localized positive-energy monopole. In fact no static finite-energy lump in three-dimensional pure Yang-Mills theory exists with the usual decay conditions: $A_i^{(s)}(x)=s{A}_i(sx)$ gives $E(s)=sE(1)$, and stationarity forces $E=0$.

Finally, on an oriented spacetime slab $[t_{-},t_{+}]\times \Sigma$, with no side-boundary contribution, the transgression identity givesA Yang-Mills instanton therefore interpolates between vacuum representatives whose Chern-Simons numbers differ by its integer charge. In a four-dimensional quantum theory a Yang-Mills theta term weights such sectors by $e^{i\theta Q}$, making $\theta$ periodic modulo $2\pi$. In an intrinsically three-dimensional theory an action $2\pi \ell N_{\mathrm{CS}}$ has a gauge-invariant phase $e^{iS}$ only for integer level $\ell$ under all winding gauge maps. These statements distinguish the gauge-invariant integer four-dimensional Chern charge from the three-dimensional connection-dependent quantity defined modulo integers.

A connection on a vector bundle $E\to M$ is a linear map $D:\Gamma (E)\to {\Omega }^1(M;E)$ with $D(fs)=df\otimes s+fDs$. Write a local frame as a row $e=(e_1,\dots ,e_r)$, and a section as $s=eu$ with a coefficient column $u$. If the connection one-form is the matrix $A={\sum }_i{A}_{i}d{x}^i$, thenThis fixes the matrix convention for all subsequent signs.

For an $E$-valued differential form of degree $k$, represented locally by a column $\alpha$ of ordinary forms, its covariant exterior derivative isThe defining graded Leibniz rule with a scalar form $\omega$ of degree $j$ isIt follows by moving the one-form $A$ past $\omega$, which introduces $(-1{)}^j$. It uniquely extends the connection on sections.

The induced endomorphism bundle connection is characterized on an endomorphism section $T$ by $(D^{\mathrm{End}}T)s=D(Ts)-T(Ds)$. For a matrix-valued form $\beta$ of degree $k$, the extension isIndeed it is exactly the formula that makesThe endomorphism-valued exterior product uses matrix composition as well as the ordinary exterior product, so factors cannot in general be interchanged.

Define the curvature form of a connection by $D^2$ acting on sections, or equivalently byIt is tensorial in both vector arguments and in $s$: the derivatives of multiplying functions cancel by the connection rule. For example $D^2(fs)=D(dfs+fDs)=f{D}^{2}s$, since the two $df\wedge Ds$ terms cancel and $d^{2}f=0$. Thus curvature is a two-form with values in the endomorphism bundle. Locally, applying the graded rule givessoFor a change of frame of a vector-bundle connection $e^{\prime }=eg$, the coefficients satisfy $u^{\prime }=g^{-1}u$, and differentiation givesSquaring yields $F^{\prime }=g^{-1}Fg$, precisely the transformation rule of an endomorphism-valued two-form. This independently proves that the local expressions define one global curvature form of a connection.

For the Bianchi identity, use the endomorphism formula with degree two:Every term cancels with its indicated sign. Thereforewhere $d_A$ here means the induced derivative on the endomorphism bundle.

For a line bundle, the endomorphism bundle is canonically the trivial scalar bundle: every fibre endomorphism is a scalar times the identity. Scalar multiplication commutes, so $A\wedge A=0$ and the endomorphism covariant derivative on scalar-valued forms reduces to $d$. The Bianchi identity consequently says $dF=0$.

If $D^{\prime }$ is another connection, their difference is $C^{\infty }$-linear in sections because the Leibniz terms cancel. Hence $D^{\prime }-D=\eta$ is a global endomorphism-valued one-form; for a line bundle it is an ordinary global scalar one-form. Locally $A^{\prime }=A+\eta$, soThus the two closed curvatures differ by an exact form, proving the connection-independent curvature class of a line bundle:No global trivialization of the line bundle itself was assumed; it is its endomorphism bundle that has the canonical scalar identification.

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

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

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

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

Use the convention $\nabla =d+A$ with anti-Hermitian connection one-forms. For a unitary complex line bundle, $A$ is locally an $i\mathbb{R}$-valued one-form and $F_{\nabla }=dA$, since the Lie algebra is abelian. Under a change of unitary frame, the vector-bundle curvature transforms by conjugation, which is trivial for $U(1)$. Hence $F_{\nabla }$ is a globally defined imaginary-valued two-form, and $i{F}_{\nabla }$ is an ordinary real two-form. Equivalently, $\mathrm{End}(L)$ has its canonical scalar trivialization, so no nontrivial coefficient bundle remains.

Locally $d{F}_{\nabla }=d^{2}A=0$, and these equalities patch globally. If another unitary connection is ${\nabla }^{\prime }=\nabla +i\beta$ for a global real one-form $\beta$, thenThus the real vector-bundle curvature is closed and its de Rham cohomology class is independent of the unitary connection.

With the First Chern class normalization $c_1(L{)}_{\mathbb{R}}=[i{F}_{\nabla }/(2\pi )]$, the transverse section also identifies this class as $\mathrm{PD}[\Sigma ]$. Indeed, the complex orientation of each normal disk makes the section wind positively once around a transverse zero. The Euler class of the underlying oriented plane bundle counts these zeros with their induced orientation, and equals $c_1(L)$. This records the role of the section and orientation hypotheses; connection independence itself did not require choosing a section.

Write $\Omega =e^{2\rho }$, $F={\partial }_x{A}_2-{\partial }_y{A}_1$, and $s=|\Phi {|}^2$, so the physical magnetic field is $B=F/\Omega$. Use finite-energy Abelian Higgs vortex boundary data: the field tends to unit magnitude at the ideal boundary, its phase winds $N$ times, and the boundary limits in the decomposition exist. This vacuum asymptotic behavior is the setting in which the stated degree gives an energy bound. We retain the current boundary term; degree alone must not be silently equated with magnetic flux on an open disk.

Define $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The gauge-covariant derivative of a charged scalar field satisfies $[D_x,D_y]\Phi =-iF\Phi$. Differentiating the current givesOn the other hand,Combining the two identities and completing the square in the magnetic and potential terms yields the conformal-surface vortex square completionAll integrations on the noncompact metric disk can first be made over $|z|<R<1$ and then passed to the limit. By the Stokes theorem, the two boundary contributions combine to $\frac{1}{2}\oint (A+j)$. At the vacuum boundary write $\Phi =e^{i\chi }$. Then $j=d\chi -A$, so this combined integral is $\frac{1}{2}\oint d\chi =\pi N$, even if the current term does not vanish separately. The usual additional condition of tangential covariant decay makes that current term vanish and yields magnetic flux quantization of an Abelian Higgs vortex, but is not needed if the combined boundary term has this phase limit. Thus for $N\ge 0$, $V\ge \pi N$. Reversing both signs in the square completion gives $V\ge -\pi N$ for $N\le 0$, henceFor positive degree, equality is attained precisely when the Bogomolny vortex equations areDistinguishing the physical field $B$ from the coordinate curvature coefficient $F$ prevents an erroneous conformal factor.

For the proposed field, set $q=x^2+y^2$ and choose the smooth connection one-formIndeed, with ${\partial }_{\bar{z}}=\frac{1}{2}({\partial }_x+i{\partial }_y)$ and $A_{\bar{z}}=\frac{1}{2}(A_1+i{A}_2)$,The second Bogomolny vortex equation follows fromThus both squares vanish, including at the simple zero $z=0$ by smoothness. At $r\to 1$, $\Phi \to e^{i\vartheta }$ and $A=2r^{2}d\vartheta /(1+r^2)\to d\vartheta$, proving degree one. Moreover, $j_{\vartheta }=4r^2(1-r^2)/(1+r^2{)}^3\to 0$, so the current boundary term really vanishes for this solution. The magnetic flux is explicitlyConsequentlyThis is the hyperbolic one-vortex at curvature minus one half; its finite energy follows directly from the vanishing squares and the finite flux.

The expression $d\varphi (X)$ is in general a vector field along a map, rather than a vector field on $M$. Its intended meaning in the required formula is pointwise evaluation of the vector-bundle-valued differential form $\nabla s$:using the canonical identification $({\varphi }^{\ast }E{)}_p=E_{\varphi (p)}$. This interpretation works for every smooth map, including a constant map or a map with self-intersections.

Choose a frame of a vector bundle over $U$, written as a row $e=(e_1,\dots ,e_r)$. Define its connection one-form $\Omega$ byfor component columns $u$. The pulled-back frame $e^{\varphi }=({\varphi }^{\ast }{e}_1,\dots ,{\varphi }^{\ast }{e}_r)$ is a frame of a vector bundle over ${\varphi }^{-1}U$. For any component column $v$ of smooth functions on that open set, defineHere the matrix entries of ${\varphi }^{\ast }\Omega$ are ordinary pullbacks of a differential form. The displayed operator is linear and satisfies the Leibniz rule, hence is locally a connection on a vector bundle.

To prove these local operators glue, change the original frame of a vector bundle to $\tilde{e}=eh$. Expanding $\nabla (ehu)$ gives the change of frame of a vector-bundle connectionThe pullback of a differential form and the chain rule giveThis is precisely the required transformation law for the pulled-back frame ${\tilde{e}}^{\varphi }=e^{\varphi }(h\circ \varphi )$. Thus the definitions agree and give a global pullback connection.

For $s=eu$, substitute $v=u\circ \varphi$ in the defining formula and use $d(u\circ \varphi )(X)=du(d\varphi (X))$. This proves the required identity. Conversely that identity fixes ${\nabla }_X^{\prime }({\varphi }^{\ast }{e}_a)$; the Leibniz rule then fixes the derivative of every ${\sum }_a{v}^a{\varphi }^{\ast }{e}_a$, proving uniqueness. If the identity is initially imposed only on global sections of a vector bundle, multiply each local frame section by a smooth bump function equal to one near the point in question and supported in $U$, and extend by zero. These global sections agree with the frame locally, so the same uniqueness argument applies. The pullback connection exists and is unique.

A connection on a vector bundle $E\to M$ is an $\mathbb{R}$-linear mapsatisfying $\nabla (fs)=df\otimes s+f\nabla s$. Equivalently its directional version is real-bilinear and satisfiesFor a complex bundle it is complex-linear in the section variable and has the same Leibniz property.

Every smooth vector bundle over an ordinary Hausdorff, second-countable smooth manifold has such a connection. Choose trivializing open sets $U_a$, local frames $e_{a,i}$ and a subordinate locally finite partition of unity ${\rho }_a$. In each frame define the flat local connectionThen set $\nabla s={\sum }_a{\rho }_a{\nabla }^{(a)}(s{|}_{U_a})$, extending each weighted term by zero outside $U_a$. The supports lie inside the corresponding trivializing sets, making the extensions smooth, and local finiteness makes the sum smooth. Since ${\sum }_a{\rho }_a=1$, its Leibniz rule isThis is the construction of a vector bundle connection by a partition of unity.

The difference $A={\nabla }^{\prime }-\nabla$ of two connections is $C^{\infty }(M)$-linear in both its vector-field argument and its section argument. The derivative terms cancel, so $A_{X}s$ depends only on $X_p,s_p$ at each point. Smooth local coefficients therefore identify it withConversely, any such $A$ added to a connection preserves its axioms. Thus connections form an affine space modeled on ${\Omega }^1(M;\mathrm{End}E)$. Choosing a base connection ${\nabla }^0$ gives the noncanonical bijection $A\mapsto {\nabla }^0+A$; there is no distinguished zero connection and hence no canonical vector-space identification. This is the affine space of vector-bundle connections.

For a connection on $TM$, the torsion tensor is the alternating operation $\tau (X,Y)={\nabla }_{X}Y-{\nabla }_{Y}X-[X,Y]$. The connection and bracket Leibniz rules giveAlternation then gives $\tau (X,fY)=f\tau (X,Y)$ as well. Hence it is tensorial, of type one contravariant and two covariant indices:It is equivalently a $TM$-valued two-form, a torsion form.

For the local frame $e_i$ with dual coframe ${\varphi }_j$, defineThe connection is $C^{\infty }$-linear in $X$, so these are smooth differential one-forms. Expansion in the frame gives ${\nabla }_X{e}_i={\sum }_j{\omega }_{ij}(X)e_j$, and uniqueness follows by applying each ${\varphi }_j$. Here the first index is the input frame vector and the second the output coefficient; this fixes the connection one-form convention.

Expand $Y={\sum }_i{\varphi }_i(Y)e_i$ and apply the connection Leibniz rule:Subtract the corresponding expression with $X,Y$ exchanged, and then subtract ${\varphi }_j([X,Y])$. By the exterior derivative of a one-form evaluated on vector fields,Since all pairs of tangent vectors can be tested, equality of the two-forms follows:This is the Cartan first structure equation with input-first indices. Its plus sign is correct with the stated index order and wedge order; the common output-first convention writes the same identity with a minus sign and reversed wedge factors.

A connection on a vector bundle is a complex-linear map from smooth sections to vector-bundle-valued one-forms satisfyingChoose a locally finite trivializing cover, its componentwise flat connections on a vector bundle $D_i$, and a subordinate smooth partition of unity ${\rho }_i$. The formula $D={\sum }_i{\rho }_i{D}_i$ is globally meaningful: each weighted term extends by zero outside its chart. Its Leibniz rule follows from ${\sum }_i{\rho }_i=1$, proving every smooth complex vector bundle admits a connection.

Extend the connection on a vector bundle to bundle-valued differential forms byThe curvature form of a connection is ${\Theta }_D=D^2$. Applying this twice to $fs$ shows that the $df$ terms cancel, so $D^2$ is tensorial and defines an $\mathrm{End}(E)$-valued two-form. With column coordinates for sections and connection one-form $A$, the local expression is $D=d+A\wedge$. Direct expansion on an arbitrary bundle-valued form givesThe terms containing a derivative of the argument cancel by the graded Leibniz rule. Thus Cartan curvature matrix equation isHere multiplication includes matrix multiplication and the exterior product of form entries. For a frame change $e^{\prime }=eg$, the connection one-form and curvature form of a connection transform asThe trace is consequently frame independent. Alsothe diagonal terms vanish and the off-diagonal terms cancel in pairs. Locally $\mathrm{Tr}\Theta =d\mathrm{Tr}A$, and hence $d\mathrm{Tr}\Theta =0$. The local primitives need not agree, but the two-form does. We obtain a global closed two-form $\mathrm{Tr}{\Theta }_D$.

The determinant connection on $\det E={\bigwedge }^{r}E$ is defined intrinsically bywhere the differential-form coefficient of $D{s}_j$ is placed first. In a local frame, only the diagonal components contribute to the derivative of $e_1\wedge \cdots \wedge e_r$, so its connection one-form is $\mathrm{Tr}A$. A line-bundle connection has curvature $d\mathrm{Tr}A$, because a scalar one-form wedges with itself to zero. ThereforeTo see independence of the de Rham cohomology class, write $D_1-D_0=a$, a globally defined endomorphism-valued one-form. The curvature difference formula and the same trace cancellations give the trace curvature transgressionTheir difference is globally an exact differential form, so their de Rham cohomology classes coincide.

For the Hermitian metric on a holomorphic vector bundle, use the displayed ordering of the local matrices and put $B=(\partial h)h^{-1}$. The identity $\bar{\partial }{h}^{-1}=-h^{-1}(\bar{\partial }h)h^{-1}$ givesThe plus sign comes from applying the graded Leibniz rule to the one-form $\partial h$. By the derivative formula for a determinant,The Hermitian positive-definite matrix $h$ has positive real determinant, so this logarithm is an ordinary smooth real function. The trace of Chern curvature in these conventions is thereforeIt is locally an exact differential form and in particular closed.

A holomorphic local frame change multiplies $\det h$ by $|\det g{|}^2$. A nowhere-zero holomorphic function has a local holomorphic logarithm, so $\bar{\partial }\partial \log |\det g{|}^2=0$. Thus the preceding expression for $\alpha$ is independent of the frame and glues globally. For two Hermitian metrics on a holomorphic vector bundle, the functionis globally defined because the frame-change factors cancel. Their two-forms differ by $d(\partial f)$. Hence the class of $\alpha$ is independent of the metric. The two-form is generally imaginary-valued; the metric independence statement is in complex de Rham cohomology, or equivalently for the real form $i\alpha$.

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.