Curvature form ($F(A)$)

The curvature form of a connection on a vector bundle is $F=dA+A\wedge A$ in a local frame. A principal connection has the corresponding Lie algebra-valued form $F=dA+\frac{1}{2}[A\wedge A]$. It measures the failure of horizontal directions to be integrable.

The curvature is $F(A)=d_A^2$ and in a local frame satisfies $F(A)=dA+A\wedge A$.

This is the principal-bundle version of the curvature form. The curvature of a principal connection is the horizontal equivariant two-formFor horizontal vector fields $X,Y$, it satisfies $\mathcal{F}(X,Y)=-\mathcal{A}([X,Y])$.

A principal connection is flat when its curvature vanishes. The identity $\mathcal{F}(X,Y)=-\mathcal{A}([X,Y])$ shows that this is equivalent to integrability of its horizontal distribution.

The trace of the curvature form of a connection is a global closed differential form of degree two. Under a frame change $e^{\prime }=eg$, the connection matrix and curvature change by $A^{\prime }=g^{-1}Ag+g^{-1}dg$ and ${\Theta }^{\prime }=g^{-1}\Theta g$. Thus trace is independent of frame. The Bianchi identity gives $d\Theta =\Theta \wedge A-A\wedge \Theta$. Taking the trace cancels the two terms: one-forms commute with two-forms in the graded sense, and renaming the matrix indices gives equality of their traces. Locally $\mathrm{tr}\Theta =d\mathrm{tr}A$, since the trace of $A\wedge A$ vanishes by pairing its off-diagonal terms. It is the curvature of the determinant connection.

For a connection on a vector bundle written $D=d+A\wedge$ in a local frame, its curvature form of a connection is represented by $\Theta =dA+A\wedge A$. The products combine matrix multiplication with the exterior product. The formula follows by squaring $D$ and using the graded Leibniz rule.

Write $D{e}_i={\sum }_j{\theta }_{ij}\otimes e_j$, with the differentiated input index first. The graded Leibniz rule gives $D^2{e}_i={\sum }_j(d{\theta }_{ij}-{\sum }_k{\theta }_{ik}\wedge {\theta }_{kj})\otimes e_j$. For frames $e^{\prime }=ge$, the connection matrix transforms as ${\theta }^{\prime }=dg{g}^{-1}+g\theta g^{-1}$ and curvature by ${\Theta }^{\prime }=g\Theta g^{-1}$. Transposing to the usual coefficient-column convention gives $A={\theta }^t$ and $F=dA+A\wedge A={\Theta }^t$; the opposite sign is an index convention, not a different curvature.

For the Levi-Civita connection, the curvature form of a connection is the endomorphism-valued two-formIt has tensoriality in all three arguments. Its first Bianchi identity follows from torsion-freeness: the cyclic sum is ${\sum }_{\mathrm{cyc}}({\nabla }_X[Y,Z]-{\nabla }_{[Y,Z]}X)={\sum }_{\mathrm{cyc}}[X,[Y,Z]]=0$ by the Jacobi identity.

If two objects ${\nabla }_1$ and ${\nabla }_2$ are each a connection on a vector bundle and satisfy ${\nabla }_1={\nabla }_2+a$ for an $\mathrm{End}(E)$-valued one-form $a$, thenThis follows by expanding $({\nabla }_2+a{)}^2$ with the covariant exterior derivative on the endomorphism bundle connection.

The difference of two connections on a vector bundle is a global endomorphism-valued one-form. Taking the trace of the curvature difference formula cancels both the commutator terms and the trace of the square of that one-form. Consequently the trace curvatures represent the same de Rham cohomology class.

For a real or complex line bundle, its endomorphism bundle is canonically trivial by scalar multiplication. The Bianchi identity therefore reduces to $d{F}_A=0$. The difference of two connections on a vector bundle is a global endomorphism-valued one-form, which here is an ordinary scalar one-form $a$. Since scalar one-forms have vanishing wedge square, the curvature difference formula reduces to $F_{A+a}-F_A=da$. Thus the de Rham cohomology class is independent of the connection on a vector bundle.

The curvature of a connection satisfies $d_{A}F(A)=0$. In a local frame this follows directly by expanding $d(dA+A\wedge A)+A\wedge F-F\wedge A$.